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Donald Trump Claims the Black Vote Helped Him Win
Not only has President-elect Donald Trump encouraged Russian hackers, disseminated fake news, and named Stephen Bannon (an unapologetic racist and anti-Semite who ran Breitbart News — a website that spouts white supremacy, racism, and Nazi-esque rhetoric) as his White House chief strategist, he’s also praising African Americans for not turning out the vote.
On Friday night (Dec. 9) during a rally in Michigan, Trump said, “The African-American community was great to us,” reports The Washington Post. “They came through, big league. Big league. And frankly, if they had any doubt, they didn’t vote, and that was almost as good …”
What’s more disturbing than having a man who’s a racist, homophobic, Islamaphobic, liar about to become the leader of the free world? Having him also be proud to snag a win all because supposedly not enough people (Black people, according to him) showed up to exercise their constitutional right to vote.
But let’s get one thing out of the way: This Trump monster is not Black folk’s fault.
We did not build a country that relied on white supremacy. We did not foster Jim Crow laws. We did not cause the privileged to believe that equality for all means oppression for them.
Don’t let anyone ever tell you that Black people clinched a win for Trump.
Just peep the exit polls, and you’ll straight up see that 88 percent of Black people voted for Clinton, while 58 percent of white people voted for Trump. Fifty-eight percent. More than half.
And that’s just facts.
But white people aren’t collectively being grilled about their voting choices. They aren’t being scapegoated and blamed for Trump’s win. Think on that.
True, Black voter turnout was slightly lower this year than in the previous two election cycles (Obama won 93% of the Black vote in 2012 and 95% in 2008), but to keep things really real, some reasons for the slight dip can be found in voter suppression and intimidation efforts in key states, such as North Carolina, Texas, Ohio, Georgia, and Wisconsin, just to name a few.
(But no one wants to talk about that.)
The year after Obama’s reelection, the Supreme Court effectively struck down part of the Voting Rights Act of 1965, saying that until Congress can agree on new requirements based on current data for certain states and counties, they can no longer enforce the mandate that those places have to have changes in their voting laws cleared by the federal government. The Court’s reasoning was, to paraphrase Chief Justice Roberts, that our country has changed for the better, that practices like literacy tests and poll taxes have been eradicated. “In the covered jurisdictions, ‘[v]oter turnout and registration rates now approach parity. Blatantly discriminatory evasions of federal decrees are rare. And minority candidates hold office at unprecedented levels,'” the Supreme Court’s decision states.
And even though the Court says it realizes that these changes have been because of the Voting Rights Act, the Court seems to be living under the delusion that the U.S. is a post-racial society because we’ve had a Black president, lots of Black people vote, and Southern states and towns have Black governors, mayors, and representatives.
Bye.
In real life, voter suppression has cropped back up, and less than 24 hours after Trump’s win, there was a spike in hate crimes.
Just add Trump’s claim that Black people not voting won him the presidency to his ongoing list of lies. The real culprit is the undercurrent of hate that’s been brewing for decades upon decades upon decades upon decades…
How can you help fight hate and ensure our democracy lives on? Stay woke + well.
Comment Below!
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It’s 2015 and all of the Social Media companies are gone.
by Michael Maoz | April 20, 2010 | 2 Comments
Is it just me, or do the piles of cash clogging the bank vaults of some of the world’s largest and/or most influential software companies make others think: acquisitions! We spent the past ten years watching companies like Oracle (over 60 acquisitions in the past ten years), IBM and Infor buy the content and maintenance streams to smooth access to markets and future revenue. Even much smaller companies such as Salesforce.com have quietly acquired complementary products (examples for Salesforce.com are Informavores, Instranet, Koral Technologies, CrispyNews, Kieden and Sendia).
It is not a stretch to conjecture that Social Media companies will be a ripe next target. With over 50 vendors in the space, and most with revenues between $3 million and $60 million, they can arguably be viewed as outsourced Research and Development for the larger vendors. Why? Because they are not the core operating system or database or infrastructure or system of record. Because the Enterprise Application vendors and the Unified Communications vendors both need what they have. Because IT governance prefers reduced application sets.
Should you worry? Maybe along the lines of Alfred E. Neuman, you can say, “Me Worry?” These applications and services are mostly delivered in the Cloud, are easy to set up, and give you a great opportunity to learn a lot about an important direction that your business needs to engage in: understanding your consumers / customers / citizens / students. So go ahead and get started.
It does mean, however, that you need to think carefully about contingencies, as 85 percent of these vendors likely will be under different ownership within five: crm customer-centric-web innovation-and-customer-experience leadership saas-and-cloud-computing social-crm social-networking social-software twitter
Thoughts on It’s 2015 and all of the Social Media companies are’s 2015 and all of the Social Media companies are gone. […]
[…] Maoz, VP Distinguished Analyst at Gartner, predicts in a recent blog post that by 2015, 85% of all social media companies will be under different […]
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On State Hwy 82 in Northeast corner of Oregon, coming from Wallowa Buddhist Temple and heading for Walla Walla, WA. About 4-5 miles before Elgin, OR, I was riding into the setting Sun beyond the Blue Mountains which I would have to cross to get to Washington.
I looked into my rearview mirror and saw, above the Wallowa Mountains to my rear, that a 3/5th’s -moon had risen and was brightly reflecting the Sun. Both about the same distance above two separate small mountain ranges that sit about 40 miles apart, with me between them scooting along at 65 miles per hour.
Distance and time traveling, and milage being eaten by trusty truck heading me home. Moon and Sun greeting each other, or so it seemed to me and it isn’t ’til later I realize they see each other so often they probably don’t even bother to say hello. Maybe in the morning, but the morning where?
What I did discover though is something very important. Two things actually, one a deep insight into astronomy, the other something I never knew before about math. I know very little of either subject but because of the the unique situation of time and place and angles and speed and me with eyes and a rearview mirror, I found out this; There is more to the moon than meets the eye. We hear a lot about a Full-moon, A Half-moon, a Quarter-moon and No-moon. I discovered the 3/5th’s-moon.
I know it was all situational, but so was Columbus landing in the New World. Like him, I’m taking the credit. If not me who? If not then, when?
Then, there’s the Math Discovery.
One (1) is an even number! Therefore, all fractions are even numbers! Ha! Take that, all you teachers who flunked a certain German immigrant, repeatedly, in math and other subjects. I best be humble though, the Media is sure to get wind of this.
Anyway, it was a really nice moon and I enjoyed it while it lasted. Safely glancing, not staring. (65mph).
The Sun is there and the Moon is over there and
I am here heading there and came from over there
And looking for Nothing everywhere and, not Anywhere.
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\section{Optimal Sequential Tests}
\label{sec:optimality}
With duality between always valid p-values and sequential tests established, we answer the question: what is an optimal sequential test to use?
Since experimentation online is classified by plentiful data, and a dynamic environment, the user typically wants to run experiments as long as she wants, update preferences along the way, and obtain results quickly. We show the mixture Sequential Probability Ratio Test (mSPRT) achieves this through novel results comparing its expected run time to fixed horizon tests, and further optimize the mixture for fast detection.
The mSPRT is a well studied class of sequential tests first introduced in \cite{robbins1970statistical}. Consider the set of mixture likelihood ratios,
\begin{equation} \label{eq:msprt_definition} \Lambda_t^H(x) = \int_{\Theta} \frac{f_\theta(x)}{f_0(x)} dH(\theta)
\end{equation}
where $H$ is a mixing distribution over the parameter space $\Theta$. The mSPRT is the sequential test with $T(\alpha) = \inf\{n : \Lambda_n^H(S_n) \geq \alpha^{-1} \}$ and $\delta_{\alpha} = 1(T(\alpha) < \infty)$, where $S_n = \sum_{i=1}^n X_i$. \footnote{The choice of threshold $\alpha^{-1}$ on the likelihood ratio ensures Type I error is controlled at level $\alpha$, via standard martingale techniques \cite{siegmund1985sequential}.}
A key feature of the mSPRT is it is a \emph{test of power one}, or $\P_\theta( T(\alpha) < \infty) = 1$ for all $0 < \alpha < 1$ and $\theta \neq 0$ \cite{RS74}. This allows the user to wait essentially indefinitely to detect small effect sizes, or concede inconclusiveness through feedback from always valid p-values.
\subsection{Fast Detection}
\label{sec:fast_detection}
We now show the mSPRT achieves \emph{fast detection} over a prior. Most of the technical work to prove these results is contained in Theorem \ref{thm:ess_truncated} in the Appendix. Compared to previous results such as \cite{L97}, ours are novel in that they examine the truncated mSPRT, $T(\alpha) \wedge n$. This allows for analysis of a wider class of distributions, for example normally distributed observations with a normal prior, and direct comparison to traditional, fixed horizon testing.
For ease of exposition, we specialize to normal data, $X_i \sim N(\theta,1)$, and $\theta_0 = 0$.\footnote{It is possible to extend results to tests on the natural parameter of exponential families, as well as general priors that are positive and continuous on $\Theta$. In fact, some misspecification of $f_\theta$ in (\ref{eq:msprt_definition}) is permitted provided the mean as a function of $\theta$ is correct (see \cite{LS1979})} We also derive results as $\alpha \rightarrow 0$, and consequently, to have non-zero chance to reject, $n \rightarrow \infty$ in the fixed horizon. This is technically for tractability of the mSPRT, and practically, identifies the A/B tester seeking nearly certain results and having lots of data.
Recall the fixed horizon procedure is to determine a MDE level of $\theta$, $\alpha$, and $\beta$, to calculate the required $n(\theta,\alpha,\beta)$. A problem occurs when there is not enough information to get a good estimate of the MDE before starting the test: too low and the user is locked into lengthy experiments; too high and the effective power $1 - \beta$ plummets. We model this uncertainty as a normal prior over the effect size, $\theta \sim N(0,\tau)$.
Compare this to truncating a mSPRT at maximum size $n_{S}$, and admitting an inconclusive result if we ever reach it. This is a sequential test with $T'(\alpha) = T(\alpha) \wedge n_S$ and $\delta'(\alpha) = 0$ on the event $T(\alpha) = n_S$. The following proposition establishes the asymptotic Type II error probability, $\beta$, for both the truncated mSPRT and UMP fixed horizon test when $n \rightarrow \infty$ fast enough.
\begin{proposition} \label{prop:fast_detection_asymptotic_power}
If $\alpha \rightarrow 0$, $n \rightarrow \infty$ such that $\log \alpha^{-1} / n \rightarrow 0$,
\[ \beta_{k} = \E_{\theta \sim N(0,\tau)} 1 - \beta_{k}(\theta) \sim C_k(\alpha) \frac{2 \sqrt{2}}{\tau} \left( \frac{ \log \alpha^{-1}}{n} \right)^{1/2} \]
where $k \in \{ \mathit{(f)ixed} , \ \mathit{m(S)PRT}\}$, $0 < C_k(\alpha) < 1$ and
\begin{equation} \nonumber \begin{split}
C_{f}(\alpha) &= \int_0^1 \bar{\Phi}\left( \sqrt{ \log \alpha^{-1} } (x - 1 ) \right) dx \\
C_{S}(\alpha) &= \int_0^1 \bar{\Phi} \left( \sqrt{ \frac{1}{2} \log \alpha^{-1} } ( x^2 - 1) \right) dx,
\end{split}
\end{equation}
where $\bar{\Phi}(x) = 1 - \Phi(x)$, the right tailed standard Normal CDF.
\end{proposition}
It follows that similar asymptotic power can be achieved by $n_{S} = \left( C_{S}(\alpha) / C_{f}(\alpha) \right)^{2} n$.
The UMP fixed horizon test has no choice but to wait for all of its samples, and so its sample size will always be $n$. Conversely, this is not true for the mSPRT; as the following Theorem shows, there is a clear benefit of stopping early.
\begin{theorem} \label{thm:fast_detection_asymptotic_sample_size}
Let $T$ be the random time when the mSPRT for data $x_i \sim N(\theta,1)$, and hypothesis $H_0: \theta = 0$, first rejects at level $\alpha$. Then for $\alpha \rightarrow 0$, $n \rightarrow \infty$ such that $\log \alpha^{-1} / n \rightarrow 0$,
\[ \E_{\theta \sim N(0,\tau)} \E_\theta ( T \wedge n ) = o(n) \]
up to terms which are $o(1)$.
\end{theorem}
(Explicit coefficients for the preceding result are given in the proof in Appendix \ref{app:optimality_proofs}.)
Thus we find that the mSPRT can achieve similar asymptotic power to the fixed UMP test with average sample size,
\begin{equation}
\E_{\theta \sim N(0,\tau)} \E_\theta ( T \wedge n_S ) = o \left( n_{S} \right) = o(n),
\end{equation}
of smaller order. This surprising superiority is precisely due to the ability of the mSPRT to automatically calibrate its sample size to the effect size of the test.
Yet these gains still understate the benefit from sequential testing. We have not incorporated the additional flexibility offered by the mSPRT---allowing the user to adjust her $\alpha$, $\beta$, $\delta'(\alpha)$, and truncating $n$ after the test has started; we intend to pursue analysis of this impact in future work.
\subsection{Optimal choice of mixture}
The following theorem is an immediate consequence of Theorem \ref{thm:ess_truncated}, along with equation (67) of \cite{LS1979}.
\begin{theorem} \label{thm:optimal_mixture} Let $X_i$, $i = 1,\dots$, be drawn from an exponential family with density $f_\theta(x)$. Suppose $H_\gamma$ is a parametric family, $\gamma \in \Gamma$, with density $h_\gamma$ positive and continuous on $\Theta$. Then up to $o(1)$ terms as $\alpha \rightarrow 0$, $\E_G E_\theta ( T \wedge n )$ is minimized by
\begin{equation} \label{eq:prior_matching} \gamma^* \in \argmin_{\gamma \in \Gamma} - \E_{\theta \sim G} \v{1}_A I(\theta)^{-1} \log h_\gamma(\theta) \end{equation}
for $A = \{ \theta \ : \ I(\theta) \geq ( \log \alpha^{-1} ) / n \}$, $I(\theta) = \theta \Psi'(\theta) - \Psi(\theta)$, and $\Psi$ the log-partition function for $f_\theta$. Optimizing for $h_\gamma$ does not impact first order terms of $\E_G E_\theta ( T \wedge n )$.
\end{theorem}
Returning to our example of normal data, prior and now using a normal mixture, $h_\gamma(\theta) = \frac{1}{\gamma} \phi(\frac{\theta}{\gamma})$, the optimal choice of mixing variance becomes:
\begin{equation} \label{eq:prior_matching_normal} \tau^{2*} = \sigma^2 \frac{ \Phi(-b) }{ \frac{1}{b} \phi(b) - \Phi(-b)}.
\end{equation}
This is a remarkably simple expression: it implies a rough {\em matching} of the mixing variance to the prior variance, with a correction for truncating. The correction tends to $\{0,\infty\}$ with $b$, showing that sampling efficiency is gained from focus on large effects when few samples are available, and smaller effects where there is ample data.
Simulation results support the theory: using (\ref{eq:prior_matching_normal}) matches the average runtime minimizing mixture variance to within a factor of 10. There is also considerable robustness in choosing $\tau$. A factor of $10$ misspecification increases average runtime by less than $5\%$, while a factor of $1000$ misspecification increases average runtime by a factor of 2. Table \ref{tab:opt_matching_sim} in Appendix \ref{app:optimality_proofs} provides more details.
Finally, we note that mixture-prior matching is important in practice as well. Our own numerical analysis of over $40,000$ historical A/B tests on a leading industry platform (Section \ref{sec:abtests}) showed that prior matching has a significant beneficial effect on the run length of experiments.
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Rain FTW
Any kind of ambient music involving rain is amazing. you get bonus point. this is incredible!! +++
Thank you :D
Anima
Rated 4.5 / 5 stars
YAAAAY.
Took me a while to find this song again after so long.
Im glad you found it :D
Rated 5 / 5 stars
Nice
Very nice, relaxing song. The tech beat comes up suddenly, though. Kinda surprised me after setting my headphones a bit louder during the opening :D
Thanks for the review :D
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\section{Potential Extensions}
\label{sec:extensions}
This section briefly addresses some possible extensions of the proposed approach to construct linearly energy stable RBF methods.
These extensions include local RBF-FD methods (\S \ref{sub:local-RBF-methods}), (entropy) stability for nonlinear problems (\S \ref{sub:nonlinear-problems}), and numerical integration (\S \ref{sub:num-int}).
\subsection{Local Radial Basis Function Methods}
\label{sub:local-RBF-methods}
Thus far, we have only considered global RBF methods.
Yet, an obvious concern with these are their computational costs.
In fact, finding a global RBF interpolant or calculating a differentiation matrix each cost $\mathcal{O}(N^3)$ operations for $N$ nodes.
While in the discussed methods this can be done a priori once---before time stepping\footnote{Assuming the nodes do not change over time.}---there are additional $\mathcal{O}(N^2)$ operations each time a differentiation matrix is applied---during time stepping.
Local RBF-FD are considered as one of the leading options to remedy this problem.\footnote{The conference presentation \cite{tolstykh2000using} by Tolstykh in 2000 seems to be the earliest reference to RBF-FD methods.}
Conceptually, these methods can be interpreted as an extreme case of overlapping domain decomposition, with a separate domain surrounding each node.
The basic idea is to center a local RBF-FD stencil at each of the $N$ global nodes, and let it include the $n-1$ nearest neighbors, where $n \ll N$.
For every node, and based on its surrounding stencil, a local FD formula that is exact for all RBF interpolants on that stencil---potentially including polynomilas---is derived then from a system of linear equations similar to \eqref{eq:LS-RBF-interpol}.
The main difference is that the right hand side is replaced by the nodal values of a linear differentiation operator.
For more details, see \cite[Chapter 5]{fornberg2015primer} and references therein.
Unfortunately, the framework discussed in the present manuscript is not immediately applicable to local RBF-FD methods.
Yet, we think that it is mediately transferable by replacing exact integrals and differentiation operators by their discrete counterparts as long as these satisfy certain SBP properties.
In this case, many stability properties which are based on integration by parts (the continuous analogue of SBP) would still be satisfied in a discrete norm.
We intend to address these extensions in future works.
\subsection{Variable Coefficients and Nonlinear Problems}
\label{sub:nonlinear-problems}
Up to this points, only linear problems with constant coefficients have been considered.
In what follows, we provide some comments regarding linear problems with variable coefficients as well as nonlinear problems, namely hyperbolic conservation laws.
Let us consider a linear advection equation
\begin{equation}\label{eq:Model_problem2}
\begin{aligned}
\partial_t u(t,x)+\partial_x \left( a(x) u(t,x) \right) & = 0, \quad
&& x \in (x_L, x_R), \ t>0, \\
u(t,x_L) & = g_L(t), \quad
&& t \geq 0, \\
u(0,x) & = u_{\text{init}}(x), \quad
&& x \in [x_L,x_R]
\end{aligned}
\end{equation}
with variable velocity $a(x)>0$ as well as compatible IC $u_{\text{init}}$ and BC $g_L$.
As described \emph{inter alia} in \cite{nordstrom2006conservative,offner2019error},
the energy is bounded for a fixed time interval if $\partial_x a$ is bounded.
In fact, one has
\begin{equation}
\begin{aligned}
\|u(t)\|^2 \leq & \exp \left(t \|\partial_x a\|_{L^\infty} \right) \cdot \\
& \left( \|u_{\text{init}}\|^2 +
\int_0^t \exp \left( -\tau \| \partial_x a \|_{L^\infty} \right)
\left[ a(x_L) g_L(\tau)^2 - a(x_R)u(\tau,x_R)^2 \right] \intd \tau \right);
\end{aligned}
\end{equation}
see \cite[Section 2]{nordstrom2017conservation}.
To obtain a similar bound for the semidiscretization of \eqref{eq:Model_problem2} by a numerical method, often a skew-symmetric formulation has to be used.
This is, for instance, a well-established technique for SBP-based methods discussed in Remark
\ref{SBP_Stability}.
Assuming an SBP property is holding for the RBF method, the transfer of stability results would be straightforward.
In the one-dimensional setting, a suitable skew-symmetric formulation of \eqref{eq:Model_problem2} might be
\begin{equation}
\partial_x(a(x)u(t,x))= \alpha \partial_x(a(x)u(t,x))+ (1-\alpha) \left( u(t,x) \partial_x a(x) +a(x) \partial_x u(t,x) \right)
\end{equation}
with $\alpha=0.5$.
A more detailed investigation of stability for variable coefficient problems and the usage of skew-symmetric formulations in RBF methods will be provided in future works.
There also nonlinear problems of the form
\begin{equation}
\begin{aligned}
\partial_t u(t,x)+\div f(u(t,x))&=0, &t>0,\; x\in \Omega \subset \R^n,\\
u(0,x)&=u_0(x), & x\in \Omega, \\
L(u(t,x))&= G(t,x) & t>0, x\in \partial \Omega
\end{aligned}
\end{equation}
will be considered.
Here, $L$ and $G$ are boundary operators which are assumed to yield a well-posed problem; see \cite{svard2014review, nordstrom2017roadmap} and references therein.
In the context of nonlinear problems it is often argued---especially in the computational fluid dynamics community---that entropy stability, rather than energy stability, is an appropriate stability concept.
See \cite{hesthaven2019entropy,chen2017entropy,abgrall2018general,abgrall2019reinterpretation, fjordholm2012arbitrarily} and references therein.
Unfortunately, entropy stability alone is not necessarily implying convergence of a numerical scheme.
Yet, experience has demonstrated that entropy stable methods are often more reliable and robust than comparable methods which violate entropy conditions
A crucial tool to ensure entropy stability are numerical (two-point) fluxes developed by Tadmor
\cite{tadmor2003entropy} together with a proper splitting of volume terms \cite{chen2017entropy}.
Another possible approach might be entropy correction as proposed in \cite{abgrall2018general,abgrall2019reinterpretation}.
Of course, also the imposed BCs themselves play an essential role \cite{svard2020entropy}.
An investigation of these techniques in the context of RBF methods will be part of future research as well.
\subsection{Numerical Integration}
\label{sub:num-int}
When discretizing the SAT terms discussed in \S \ref{sec:RBF_SAT}, but also for numerically determining suitable correction functions in \S \ref{sec:RBF_FR}, we are in need of computing certain integrals.
Depending on the number of degrees of freedom and the dimension, this can be computationally costly.
Hence, future works might also address the advantage---and potential pitfalls---of replacing continuous integrals in the RBF method by a discrete quadrature (in one dimension) or cubature (in higher dimensions) formula.
In one dimension, reasonable candidates are the trapezoidal or Gauss--Legendre/Lobatto formula.
In more then one dimension, and assuming a rectangular domain, their tensor products of these rules could be used; see some of the excellent monographs \cite{haber1970numerical,stroud1971approximate,engels1980numerical,davis2007methods,trefethen2017cubature}.
Yet, these would not be available in the notoriously difficult case of non-standard, in particular non-rectangular, domains.
Here, an alternative might be given by classical (quasi-)Monte Carlo methods \cite{metropolis1949monte,niederreiter1992random,caflisch1998monte,dick2013high} or the recently developed high-order least squares cubature formulas \cite{glaubitz2020stableCF,glaubitz2020constructing}, based on the one dimensional works \cite{wilson1970necessary,wilson1970discrete,huybrechs2009stable,glaubitz2020stable}.
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TITLE: Equivalence between two definitions of compactifications of an elliptic curve
QUESTION [4 upvotes]: Suppose we have the curve $y^2=h(x)=x(x-1)(x-2)$ in $\mathbb{C}^2$. Then this is clearly not compact and I saw two definitions for compactifying this and I want to prove that they are equivalent.
Method $1$: Consider the homogenization of the curve given by $y^2z=x(x-z)(x-2z)$. Now just take the zero set in $\mathbb{C}P^2$. Now this is compact since it is a closed subset of a compact set.
Method 2: This is the method I saw in the book "Algebraic curves and Riemann surfaces" by Rick Miranda. The definition used here is two take two patches and glue them by an isomorphism.
So the curve we consider is $w^2=z^4h(\frac{1}{z})=z(1-z)(1-2z)$. Now we have an open set $V=\{(z,w): z \neq 0\}$ and an open set $U=\{(x,y): x \neq 0\}$ and they are isomorphic via a map $\phi(z,w)=(\frac{1}{x},\frac{y}{x^2})$. Now wee can glue these curves using this isomorphism.
Now I am really not sure if these two definitions are equivalent. Any help is appreciated. Thanks.
REPLY [2 votes]: Yes, both constructions are equivalent, in the sense that they give isomorphic projective curves containing your affine curve. Maybe you can show directly that the explicit map from one curve to the other, defined as the identity on the original affine curve and sending one point at infinity to the other, is an isomorphism of Riemann surfaces. I will follow another route.
Key idea: it is not surprising that both constructions are equivalent, because compactifications of algebraic curves are unique. This is a simple consequence of the following theorem.
Theorem. Let $k$ be a field, $X$ be a smooth curve over $k$ and $U\subset X$ be an open subset. If $Y$ is a proper variety over $k$ (not necessarily of dimension 1), then any $k$-morphism $f:U \to Y$ extends uniquely to $X$.
The existence follows from the so-called valuative criterion of properness, and the uniqueness form separatedness. See Proposition 4.1.16 in Q. Liu, Algebraic Geometry and Arithmetic Curves, for a more general statement. For your question, the following statement is sufficient (and easy to prove).
Theorem. Let $X$ be a Riemann surface and $U\subset X$ be an open subset such that $X\setminus U$ is finite. If $Y$ is a compact Riemann surface, then any morphism $f:U \to Y$ extends uniquely to $X$.
Now, let $X_1\subset \mathbb{C} P^2$ be the elliptic curve obtained from the method 1 in your question, $X_2$ be the curve obtained from method 2, and $U$ be the original affine curve given by the equation $y^2=x(x-1)(x-2)$.
The first thing you must do is to prove that $X_2$ is smooth; that is, that it really defines a Riemann surface (this is easy, you must only check one point). Then, it follows from the above theorem that the inclusion $U\to X_1$ extends to a unique morphism $f: X_2 \to X_1$.
Now, notice that, if you prove that $X_2$ is compact (i.e. that it indeed defines a compactification of $U$), then by applying the same theorem, you would obtain a unique morphism $g: X_1 \to X_2$ extending the inclusion $U \to X_2$. By uniqueness, $f$ and $g$ are inverses of each other, and you are done.
Now, how to prove that $X_2$ is compact (or proper, in the language of algebraic geometry)? The idea is to define an explicit closed embedding of $X_2$ in some projective space.
Let $i: X_2 \to \mathbb{C}P^3$ be the morphism defined on $U$ by
$$
(x,y)\mapsto (1:x:x^2:y)
$$
and on $V$ by
$$
(z,w)\mapsto (z^2:z:1:w).
$$
Note that these two morphism indeed glue via the transition map $\phi$. It is easy to prove that $i$ is a closed immersion, and thus it identifies $X_2$ with it's image in $\mathbb{C}P^3$, which is by definition a projective curve (thus compact). This finishes the proof.
Let me remark that I didn't guess the morphism $i$. In algebraic geometry jargon, what is happening here is that we are proving that the line bundle $\mathcal{O}(4\cdot [\infty])$ (of meromorphic functions which are holomorphic on $U$ and have a pole of order at most 4 at the point $\infty \in X_2\setminus U$) is very ample. Of course, if you want to prove that $X_2$ is isomorphic to $X_1$, then this must be true because it is for the curve $X_1$, as the usual theory of elliptic curves show (cf. comments below).
One final remark on why this question is interesting. The method 2 is actually how you should compactify hyperelliptic curves in general, i.e. curves of the form $y^2 = f(x)$. The point is that, if $\deg f\ge 4$, then the curve obtained by method 1 will be singular at infinity. It turns out that for $\deg f = 3$ both methods give smooth curves, so in text books on elliptic curves one usually prefers method 1 as it is simpler: it gives a plane curve, instead of a spacial curve.
| 2,300
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\begin{document}
\title{Evaluation of the matrix exponential function using finite elements in time}
\author{D H Gebremedhin, C A Weatherford, X Zhang, A Wynn III, G Tanaka}
\address{Department of Physics, Florida A \& M University, Tallahassee, FL 32307, USA}
\ead{charles.weatherford@famu.edu}
\begin{abstract}
The evaluation of a matrix exponential function is a classic problem of computational linear algebra. Many different methods have been employed for its numerical evaluation [Moler C and van Loan C 1978 {\it SIAM Review}{\bf \ 20} 4], none of which produce a definitive algorithm which is broadly applicable and sufficiently accurate, as well as being reasonably fast. Herein, we employ a method which evaulates a matrix exponential as the solution to a first-order initial value problem in a fictitious time variable. The new aspect of the present implementation of this method is to use finite elements in the fictitious time variable. [Weatherford C A, Red E, and Wynn A 2002 {\it Journal of Molecular Structure}{\bf \ 592} 47] Then using an expansion in a properly chosen time basis, we are able to make accurate calculations of the exponential of any given matrix as the solution to a set of simultaneous equations.
\end{abstract}
\section*{Introduction}
The evaluation of the exponential of a square matrix $\e^{\bf A}$ is a classic problem of computational linear algebra.[1] A large number of methods have
been proposed and used for its evaluation. None of these methods have produced a generally applicable method which is sufficiently accurate as well as being reasonably fast.
Thus this problem might be considered unsolved. It is clearly an important and pervasive problem which arises in a wide variety of contexts.[2,3] A method which is capable of producing
a solution to essentially arbitrary precision would thus be of great importance. The present work uses a variant of a method described in Ref. 1, namely the introduction of an
artificial time-parameter
which produces an initial-value problem. Instead of calling a \lq canned\rq \ solver, the present work uses a method introduced by several of the present authors [4] to solve quantum mechanical
initial-value problems. In this method, a finite element technique is used to propagate from an initial condition at $t=0$, which is the unit matrix, to the desired result at $t=1$. The time axis
is broken up into an arbitrary number of time elements and the solution is propagated from element to element, using a special basis in time introduced here for the first time.
The next section presents the analysis of the problem and describes the solution algorithm. Then the algorithm is applied to evaluate the exponential of a number of test matrices. Finally, the
conclusions are presented.
\section*{Analysis and Solution Algorithm}
The problem at hand is the evaluation of the exponential of a square (generally complex) matrix $e^{\bf A}$. The present method introduces an
artificial time parameter so as to transform the evaluation into the solution of an initial-value problem. For a given $n\times n$ square matrix $\mathbf A$, consider the following parametrized function definition:
\begin{equation}
\mathbf{\Psi}(t) \equiv \rme^{\mathbf{A}t}, \\
\end{equation}
\noindent
where $\bf \Psi$ is also an $n\times n$ square matrix and the desired solution is $\mathbf{\Psi}(1)=\rme^{\mathbf{A}}$ which evolves
from the initial value given by $\mathbf{\Psi}(0)=\mathbf{1}$ ($\mathbf{1}$ is the diagonal unit matrix). This is a solution of the following linear ordinary differential equation, written for each matrix element,
\begin{equation}
\dot{\Psi}_{ij}(t)=\sum_{k=1}^{n}A_{ik}\Psi_{kj}(t),
\end{equation}
\noindent
where the over-dot stands for the time-derivative. The time axis extends over the interval $[0, 1]$. Now break the time axis
up into elements that extend between nodes $t_{\rm{i}}$ and $t_{\rm{i+1}}$, and define a local time $\tau$ that spans $[-1, 1]$.
The local time transformation is defined by the relation,
\begin{equation}
\tau =qt -p,
\end{equation}
\noindent
where, $q = 2/(t_{\rm{i+1}} - t_{\rm{i}}) $ and $p= (t_{\rm{i+1}} + t_{\rm{i}})/(t_{\rm{i+1}} - t_{\rm{i}})$.
Thus, for an arbitrary time element $e$, Eq. $(2)$ can be written in terms of local time $\tau$ as
\begin{equation}
q\dot{\Psi}^{(e)}_{ij}(\tau)=\sum_{k=1}^{n}A^{(e)}_{ik}\Psi^{(e)}_{kj}(\tau).
\end{equation}
At this point, we will use the following \emph{ansatz} for $\bf \Psi^{(e)}$ to enforce continuity between two consecutive finite elements
\begin{equation}
\Psi^{(e)}_{ij}(\tau) = f^{(e)}_{ij}(\tau)+\Psi^{(e-1)}_{ij}(+1), \qquad f^{(e)}_{ij}(-1) = 0
\end{equation}
\noindent
and expand $f^{(e)}_{ij}(\tau)$ as
\begin{equation}
f^{(e)}_{ij}(\tau) = \sum_{\mu=0}^{m-1} B_{\mu}^{ij(e)} s_\mu (\tau)
\end{equation}
\noindent
in a basis we define by
\begin{equation}
s_{\mu}(\tau) = \int_{-1}^\tau T_{\mu}(\tau) \, \rm{d}\tau
\end{equation}
\noindent
where $T_{\mu}(\tau)$ are Chebyshev Polynomials of the first kind.[5] Note that these basis functions enforce the
initial condition on the $f$'s given in Eq. (5) since $s_{\mu}(-1)=0$. The result for the decomposition of $f$ in $m$ basis functions is
\begin{eqnarray}
\Psi^{(e)}_{ij}(\tau) = \sum_{\mu=0}^{m-1} B_{\mu}^{ij(e)} s_\mu (\tau) +\Psi_{ij}^{(e-1)}(+1) \\
\dot{\Psi}^{(e)}_{ij} (\tau) = \sum_{\mu=0}^{m-1} B_{\mu}^{ij(e)} T_\mu (\tau).
\end{eqnarray}
\noindent
Now, insert (8) and (9) into Eq. (4), and multiply from the left by $w(\tau) s_{\mu'}(\tau)$ and integrate from $-1$ to $+1$
(note that $w(\tau) = (1- \tau^2)^{-1/2}$ is the weighting function for Chebyshev polynomials). Rearranging terms we get,
\begin{equation}
\fl \eqalign {q\sum_{\mu} [\int_{-1}^{1} s_{\mu'}(\tau) \omega (\tau) T_{\mu}(\tau) \, \rm{d} \tau] B_{\mu}^{ij(e)} =
\sum_{k\mu} A^{(e)}_{ik} [\int_{-1}^{1} s_{\mu'}(\tau) \omega (\tau) s_{\mu}(\tau) \, \rm{d} \tau] B_{\mu}^{kj(e)} \cr
+ \sum_{k} A^{(e)}_{ik} [\int_{-1}^{1} s_{\mu'}(\tau) \omega (\tau) T_{0}(\tau) \, \rm{d} \tau] \Psi_{kj}^{(e-1)}(+1)}
\end{equation}
\noindent
where, $T_{0}(\tau) = 1$. Defining, the integrals in the above equation as
\begin{eqnarray}
C_{\mu' \mu} &\equiv \int_{-1}^{1} s_{\mu'}(\tau) \omega (\tau) T_{\mu}(\tau) \, \rm{d} \tau \\
D_{\mu' \mu} &\equiv \int_{-1}^{1} s_{\mu'}(\tau) \omega (\tau) s_{\mu}(\tau) \, \rm{d} \tau \\
g_{\mu'} &\equiv \int_{-1}^{1} s_{\mu'}(\tau) \omega (\tau) T_{0}(\tau) \, \rm{d} \tau
\end{eqnarray}.
\noindent
and substituting Eqs. (11-13) into Eq. (10) gives,
\begin{equation}
{q\sum_{\mu} C_{\mu' \mu} B_{\mu}^{ij(e)} =
\sum_{k\mu} A^{(e)}_{ik} D_{\mu' \mu} B_{\mu}^{kj(e)}
+ g_{\mu'} \sum_{k} A^{(e)}_{ik} \Psi_{kj}^{(e-1)}(+1)}
\end{equation}
\noindent
or, rearranging
\begin{equation}
\sum_{\mu k} (q C_{\mu' \mu} \delta_{ik} - A^{(e)}_{ik} D_{\mu' \mu} )B_{\mu}^{kj(e)}
= g_{\mu'} \sum_{k} A^{(e)}_{ik} \Psi_{kj}^{(e-1)}(+1)
\end{equation}
\noindent
where $\delta_{ik}$ is the usual Kronecker delta function. Then rewrite Eq. (15) as
\begin{equation}
\sum_{\mu k} \Omega^{(e)}_{(\mu' i)(\mu k)}B_{\mu}^{kj(e)}
= \Gamma_{\mu'}^{ij(e,e-1)}
\end{equation}
\noindent
where
\begin{eqnarray}
\Omega^{(e)}_{(\mu' i)(\mu k)} &\equiv (q C_{\mu' \mu} \delta_{ik} - A^{(e)}_{ik} D_{\mu' \mu} ) \\
\Gamma_{\mu'}^{ij(e,e-1)} &\equiv g_{\mu'} \sum_{k} A^{(e)}_{ik} \Phi_{kj}^{(e-1)}(+1).
\end{eqnarray}
Equation (16) is a set of simultaneous equations of size $(n \times m)$, which can be written in matrix form as,
\begin{equation}
\mathbf{\Omega^{(e)} B}^{j (e)}= \mathbf{\Gamma}^{j(e,e-1)}\qquad j = 1, 2,..., n.
\end{equation}
\noindent
Here, $\mathbf{\Omega^{(e)}}$ is a (complex) matrix and for each $j$, $\mathbf{\Gamma^{j(e,e-1)}}$ and $\mathbf{B^{j(e)}}$ are vectors.
Eq. (19) applies for each time element $e$. The solution is propagated from element to element, from $t=0$ to $t=1$.
The above equation can be solved numerically in many ways, but we have chosen the method of LU decomposition.[6]
The present method is ideally suited to high-performance computers where the solver of choice would probably
be iterative. In the present case, we
apply LU decomposition to
$\mathbf{\Omega^{(e)}}$ and back substituting all of the $\mathbf{\Gamma}^{j(e,e-1)}$'s, we will have all the elements for the matrix (which can also be viewed as three dimensional) $\mathbf{B^{j(e)}}$. This LU decomposition only needs to be done once since $\mathbf{\Omega^{(e)}}$ is independent of time. Thus, the propagation
just involves a matrix vector multiply.
Then, we employ Eq. $(8)$ to solve for $\mathbf{\Psi^{(e)}}(\tau = 1)$ for the element e, which, in turn, will be used as $\mathbf{\Psi^{(e + 1)}}(\tau = -1)$ for the next element e + 1. Starting off with a unit matrix for $\mathbf{\Psi^{(1)}}(t = 0)$, we continue this process till we calculate $ \mathbf{\Psi(t = 1)}$ at the last node, which is the exponential of the given matrix $\mathbf{A}$.
\section*{Results}
The calculations presented below were done on a Macintosh Intel laptop using Gnu C++ which has machine accuracy limit of $2.22045\times 10^{-16} $.
As an illustration, let's borrow a 'pathological' matrix from [1], which we have modified slightly to make it even worse. Consider a matrix $\mathbf{M1}$ given by,
\begin{eqnarray}
\eqalign{\mathbf{M1} &= \left[\begin{array}{cc}-73 & 36 \\-96 & 47\end{array}\right] \\
&=\left[\begin{array}{cc}1 & 3 \\2 & 4\end{array}\right] \left[\begin{array}{cc}-1 & 0 \\0 & -25\end{array}\right] {\left[\begin{array}{cc}1 & 3 \\2 & 4\end{array}\right]}^{-1}.}
\end{eqnarray}
\noindent
The exponent of $\mathbf{M1}$ can be easily calculated as,
\begin{eqnarray*}
e^\mathbf{M1} &= \left[\begin{array}{cc}1 & 3 \\2 & 4\end{array}\right] \left[\begin{array}{cc}e^{-1} & 0 \\0 & e^{-25}\end{array}\right] \left[\begin{array}{cc}-2 & 3/2 \\1 & -1/2\end{array}\right] \\
&= \left[\begin{array}{cc}{-2e^{-1}+3e^{-25}} & {(3/2)(e^{-1}-e^{-25})} \\{-4e^{-1}+4e^{-25}} & {3e^{-1}-2e^{-25}}\end{array}\right].
\end{eqnarray*}
\noindent
The above matrix, exact to $16$ decimal places, is given by
\begin{equation}
e^\mathbf{M1}\cong \left[\begin{array}{cc}-0.7357588823012208 & 0.5518191617363316 \\-1.4715177646302175 & 1.1036383234865511\end{array}\right].
\end{equation}
\noindent
The result of our program is displayed below and we run it by using just $8$ time steps and $8$ basis functions. The result is accurate to $13$ decimal places already.
\begin{equation}
e^\mathbf{M1}\cong \left[\begin{array}{cc}-0.7357588823012(181) & 0.5518191617363(358) \\-1.4715177646302(120) & 1.1036383234865(592)\end{array}\right].
\end{equation}
As an example of a non - diagonalizable matrix, consider the following matrix $\mathbf{M2}$, with complex eigenvalues
\begin{equation}
\mathbf{M2} = \left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right]
\end{equation}
\noindent
It can be shown that,
\begin{equation}
e^\mathbf{M2} = \left[\begin{array}{cc}cos(1) & -sin(1) \\sin(1) & cos(1)\end{array}\right]
\end{equation}
\noindent
We are able to achieve $14$ decimal digit accuracy with $8$ time steps and $8$ basis functions.
\begin{equation}
e^\mathbf{M2} \cong \left[\begin{array}{cc} 0.54030230586814 & -0.84147098480790 \\ 0.84147098480790 & 0.54030230586814\end{array}\right]
\end{equation}
\noindent
Table 1 shows the minimum number of basis functions, for a given number of time steps, which were required to achieve a precision of $\pm 1\times 10^{-14}$ on matrices whose exponential is known exactly. The matrices chosen are: the simplest possible matrix - a $2 \times 2$ real unit matrix, for which the result is the constant $e$ on the diagonals, and the matrices $\mathbf {M1}$ and $\mathbf {M2}$.
Let's check our program on matrices, which we picked randomly and for which we had no $\it apriori$ knowledge as to the result of their exponentiation. We fixed 8 time steps and/or 8 basis functions, and varied the other corresponding parameter from 5 to 40 and checked how the results of the program varied in accuracy. For the sake of saving space, we only displayed the result of the last element--the other elements of the matrix exponential behaved similarly.
The matrices chosen are a $5 \times 5$ real matrix $\mathbf{M3}$,
\begin{equation}
\mathbf{M3} = \left[\begin{array}{ccccc}-0.1 & -0.2 & -0.3 & -0.4 & -0.5 \\-0.6 & -0.7 & -0.8 & -0.9 & -1 \\0.1 & 0.2 & 0.3 & 0.4 & 0.5 \\0.6 & 0.7 & 0.8 & 0.9 & 1 \\1 & 2 & 3 & 4 & 0\end{array}\right]
\end{equation}
\noindent
and a $3 \times 3$ complex matrix $\mathbf{M4}$
\begin{equation}
\mathbf{M4} = \left[\begin{array}{ccc}1+i & 1-i & i \\1 & 2i & 0 \\1+2i & -1+i & -1-i\end{array}\right].
\end{equation}
From Table 2, one can see that the numbers up to $12$ decimal digits have saturated after $5$ time steps and/or basis functions. Similarly, Table 3 shows $13$ digits of accuracy as we switch the two parameters from $5$ to $40$, except for the case of $5$ basis functions, which only shows $8$ accurate significant digits. This shows that for complex matrices, there is inherently more work for the program to handle because of the imaginary part of the matrix elements and there is apparently more sensitivity to the number of basis functions used than to the number of time steps.
\Table{\label{tone}Minimum number of basis functions and time steps required for a precision of $\pm 1\times 10^{-14}$ for a $2 \times 2$ unit matrix, $\mathbf{M1}$ and $\mathbf{M2}$.}
\br
\centre{2}{$2\times 2$ unit matrix}&\centre{2}{$\mathbf{M1}$}&\centre{2}{$\mathbf{M2}$}\\
\crule{2}&\crule{2}&\crule{2}\\
Time steps & Basis functions& Time steps & Basis functions& Time steps & Basis functions\\
\mr
\01& 11& \0\005& -& \01& 11\\
\02& \09& \0\008& 7& \02& \09\\
\04& \08& \016& 6& \04& \08\\
\08& \07& \050& 5& \08& \07\\
16& \06& 256& 4& 15& \06\\
58& \05& \0\0-& -& 40& \05\\
\br
\end{tabular}
\end{indented}
\end{table}
\Table{\label{ttwo}Results of matrix $e^{M3}_{5 5}$ for typical runs of 8 time steps and 8 basis functions.}
\br
Result& Time steps & Basis functions\\
\mr
\underline{3.210309305973}118 & \05& \08\\
\underline{3.210309305973}288 & 40& \08\\
\underline{3.210309315373}377& \08& \05\\
\underline{3.210309305973}281& \08& 40\\
\br
\end{tabular}
\end{indented}
\end{table}
\Table{\label{tthree}Results of matrix $e^{M4}_{3 3}$ for typical runs of 8 time steps and 8 basis functions.}
\br
Result& Time steps & Basis functions\\
\mr
\underline{-0.5119771222980}63 - i \underline{0.0897728113135}12 & \05& \08\\
\underline{-0.5119771222980}81 - i \underline{0.0897728113135}26 & 40& \08\\
\underline{-0.51197712}1264660 - i \underline{0.08977281}0979965& \08& \05\\
\underline{-0.5119771222980}82 - i \underline{0.0897728113135}26& \08& 40\\
\br
\end{tabular}
\end{indented}
\end{table}
\section*{Conclusion}
We have presented a robust, easily used, and accurate algorithm for the evaluation of the exponential of a matrix. We did this by introducing an
artificial time parameter and evaluating the matrix exponential as the solution of an initial-value problem in this artificial time. We solved the initial-value problem
by using finite elements in time with a new time basis which we defined here so as to enforce the initial conditions on the solution at the beginning of each time finite element. This resulted in set of simultaneous equations for the expansion coefficients.
The actual algorithm employed here was an LU decomposition which was very fast and efficient. The relative efficiency of the method should be most
apparent when implemented on high-performance computers since the algorithm is highly parallel.
The method was applied to several matrices as a proof of the validity of the algorithm.
The results of our calculations show that we only need about $8$ basis functions and $8$ time steps for the matrices considered for accuracies as great as $13$ significant digits. We trust that this method of numerically calculating the exponential of a matrix will be recognized to be a \emph{nondubious} one! \\
\section*{Acknowlegements}
This work was supported by the NSF CREST Center for Astrophysical Science and Technology under Cooperative Agreement
HRD-0630370.
\section*{References}
| 211,194
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Stone Academy’s Practical Nursing program combines classroom course work with clinical experiences in a variety of area facilities with diverse populations to prepare our future LPNs for entry-level positions in both acute and long-term care settings. Stone Academy’s PN classes offer the flexibility of a full-time day or part-time evening/Saturday program. Upon graduation, our students are eligible to sit for the Connecticut NCLEX-PN licensing examination. Prospective students interested in becoming a Licensed Practical Nurse must first attend an information session. Here you will learn about the application and acceptance process along with specific information regarding the schedules for upcoming PN classes.
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We’re ranging a bit afield from our usual SEO-related topic today to ask an important question: has Microsoft positioned themselves right out of relevancy in today’s internet? We ask this because, having just spent 2 freaking hours trying to recover from an epic “automatic update” fail, it occurs to us that perhaps MS’s business model is in trouble.
MS built their Byzantine Empire on a bedrock philosophy of intimidation. They intimidate their suppliers, their competitors, their potential competitors, their employees, and their customers. They always have and it’s always worked. Being the biggest bully in the room is how Papa Gates and Co. sewed up the operating system market, the office software market, the email client market, the server software market, and the browser market. The resulting tsunami of cash did nothing to persuade them to try any other path.
But here’s the thing: having the power and weight of numerous lucrative near-monopolies has calcified them. They are slow to market, slow to respond to missteps, and most importantly, slow to react to changing market dynamics. In a world where open source alternatives abound, Microsoft’s captive markets have begun to evaporate. First, Firefox (joined later by Chrome) chipped away at ponderous MS Explorer’s 70% market share, now at 50% (and falling). Their office suite dominance suffers threats from Open Office and now, more dangerous perhaps, Google Docs.
MS no longer can claim to be the only dog in the fight. They’re still the biggest dog, but the others are growing and MS is withering away.
And now, introducing the Cloud. All of the applications you will ever need are—or will soon be—available at no cost to anyone with decent internet access, which is fast becoming a basic human right. Word processing? No problem. Spreadsheets? Gotcha covered. Slide shows, drawing, calendars, email, messaging? Already to go. The last thing—and the giant killer—will be the operating system. Up to now, MS has used their Windows systems to beat back every challenge. But this one’s gonna be a bit tougher.
Because the Cloud’s operating systems are agile, flexible, multi-platform, and open source.
Here are five reasons MS will find it hard to migrate its dominance to the cloud:
- Tablets and smart phones and netbooks are browser based and don’t need Windows;
- Open sourced applications are inherently more adaptable and quicker to react to changes, technological, cultural, or legislative;
- MS software is bloated, ungainly, and unnecessarily paternalistic—how many “critical updates” per day are too many?
- Freeware and open source development is typically decentralized to the extent that buying the intellectual property (a standard MS tactic) is useless;
- Google has already built a solid foundation for the future: a gigantic, friendly, all-seeing, all-knowing, and above all useful temple in the clouds that MS will find virtually impossible to keep up with.
Unless Bill thinks of something dramatically different and desperately soon, Cloud computing will break MS’s back, unless they find a way to compete with free, easy, simple, and convenient. And find it fast. Because this cloud is getting darker and pretty soon it’s gonna rain.
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I'd love to see the Mint come to life.Steven Pham/Roadshow
We've got a lot of love for Genesis recently, with cars like the GV70 and G80 in the pipeline, and next year, the Korean luxury brand is ready to go electric. In among a wide-ranging flood of announcements Thursday from Genesis' parent company Hyundai, the automaker said it plans to roll out two new electric cars next year.
Hyundai plans to go big on EVs in the future with more than 12 of them coming by 2025, mostly via Hyundai's new sub-brand Ioniq. Kia will get its own models as well and all three brands, Genesis included, will utilize the new E-GMP EV platform Hyundai detailed earlier this month.
We can only speculate on what form the two Genesis EVs will take, but the brand showed two electric concept cars in the past. The first was the Essentia, which took the form of an electric grand tourer. It stunned crowds at the 2018 New York Auto Show, but the brand never gave a final word on if it was meant for production. That same year, one report said it would enter production as a limited-run model with a lofty price tag. Perhaps Genesis wants to make a big statement as it dives into electrification, should a production Essentia come next year.
The other concept is the Genesis Mint, a darling electric city car that also found a lot of love at the 2019 New York Auto Show. Again, Genesis was tight-lipped on the possibility for production, but the company believes populations will continue to move to city centers. Theory is that will spur a desire for smaller, premium cars. As much as the Mint would be awesome, it seems like an electric SUV is more in line with today's car buyers, especially in the US. It's tough enough to move EVs, and building a super compact one may not reach as many buyers as Hyundai wants.
But we'll see. Genesis has taken some big steps in its short time as an actual Hyundai division. Maybe it will surprise us again.
| 251,192
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\begin{document}
\title[The weak competition system with a free boundary]
{asymptotic spreading speed for the weak competition system with a free boundary}
\author[Z.G. Wang, H. Nie and Y. Du]{Zhiguo Wang$^{\dag}$, Hua Nie$^{\dag}$, Yihong Du$^{\ddag,\ast}$}
\thanks{$^\dag$ School of Mathematics and Information Science,
Shaanxi Normal University, Xi'an, Shaanxi {\rm710119,} China.}
\thanks{$^\ddag$ School of Science and Technology,
University of New England, Armidale, NSW {\rm2351,} Australia.}
\thanks{$^\ast$The corresponding author. E-mail address: ydu@turing.edu.au (Y. Du).}
\keywords{Competition model; Free boundary problem; semi-wave; Asymptotic spreading speed.}
\subjclass{35B40, 35K51, 35R51, 92B05}
\date{\today}
\begin{abstract} \small This paper is concerned with a diffusive Lotka-Volterra type competition system with a free boundary in one space dimension. Such a system may be used to describe the invasion of a new species into the habitat of a native competitor.
We show that the long-time dynamical behavior of the system is determined by a spreading-vanishing dichotomy, and provide sharp criteria for spreading and vanishing of the invasive species. Moreover, we determine the asymptotic spreading speed of the invasive species when its spreading is successful, which involves two systems of traveling wave type equations, and is highly nontrivial to establish.
\end{abstract}
\maketitle
\section{Introduction}
{\setlength{\baselineskip}{16pt}{\setlength\arraycolsep{2pt}
The classical Lotka-Volterra reaction-diffusion system
\begin{linenomath*}\bes\left\{
\begin{aligned}
&u_t=d_1u_{xx}+u(a_1-b_1u-c_1v),&x\in\mathbb{R},&\quad t>0,\\
&v_t=d_2v_{xx}+v(a_2-b_2v-c_2u),&x\in\mathbb{R},&\quad t>0
\end{aligned}
\right.\lbl{section1-a1}\ees\end{linenomath*}
is a model frequently used to describe competitive behavior between two distinct species.
Here $u(x, t)$ and $v(x, t)$ denote the population densities of two competing species at the position
$x$ and time $t$; the constants $d_i, a_i,b_i$ and $c_i\; (i=1,2)$ are the diffusion rates, intrinsic growth rates, intra-specific competition rates, and inter-specific competition rates, respectively, all of which are assumed to be positive. By setting
$$\begin{aligned}
\hat{u}(x,t):=\frac{b_1}{a_1}u\left(\sqrt{\frac{d_1}{a_1}}x, \frac{t}{a_1}\right),
&\quad\hat{v}(x,t):=\frac{b_2}{a_2}v\left(\sqrt{\frac{d_1}{a_1}}x, \frac{t}{a_1}\right),\\
d:=\frac{d_2}{d_1},\quad r:=\frac{a_2}{a_1},&\quad k:=\frac{a_2c_1}{a_1b_2},\quad h:=\frac{a_1c_2}{a_2b_1},
\end{aligned}$$
and dropping the hat signs, system \eqref{section1-a1} becomes the following nondimensional system:
\bes\left\{
\begin{aligned}
&u_t=u_{xx}+u(1-u-kv),&x\in\mathbb{R},\quad t>0,\\
&v_t=dv_{xx}+rv(1-v-hu),&x\in\mathbb{R},\quad t>0.
\end{aligned}
\right.\lbl{section1-a2}\ees
It is easy to see that \eqref{section1-a2} has four
equilibria: $(0,0),(1,0),(0,1)$ and $(u^\ast,v^\ast)=\left(\frac{1-k}{1-hk},\frac{1-h}{1-hk}\right)$, with $(u^*, v^*)$ meaningful only when $(1-k)(1-h)>0$. When the entire real line $\mathbb{R}$ is replaced
by a bounded open interval in $\mathbb{R}$, under the zero Neumann boundary conditions, the asymptotic behavior of the solution $(u(x, t), v(x, t))$
for \eqref{section1-a2} with initial functions $u(x, 0), v(x, 0) > 0$ can be summarized below (see, for example \cite{Mo}):
\begin{itemize}
\item [(\textrm{I})] if $k<1<h$, then $\lim_{t\rightarrow\infty}(u(x, t), v(x, t))=(1,0)$;
\item [(\textrm{II})] if $h<1<k$, then $\lim_{t\rightarrow\infty}(u(x, t), v(x, t))=(0,1)$;
\item [(\textrm{III})] if $h,k<1$, then $\lim_{t\rightarrow\infty}(u(x, t), v(x, t))=(u^\ast,v^\ast)$;
\item [(\textrm{IV})]if $h,k>1$, then $\lim_{t\rightarrow\infty}(u(x, t), v(x, t))=(1,0)$ or $(0,1)$ or $(u^*,v^*)$ (depending on the initial condition).
\end{itemize}
The cases (I) and (II) are usually called the weak-strong competition case, while (\textrm{III}) and (\textrm{IV}) are known as the weak and strong competition cases, respectively.
A number of variations of \eqref{section1-a2} (or \eqref{section1-a1}) have been used to model the spreading of a new or invasive species.
For example, to describe the invasion of a new species into the habitat of a native competitor, Du and Lin \cite{DL14} considered the following
free boundary problem
\bes\left\{
\begin{aligned}
&u_t=u_{xx}+u(1-u-kv),&&0<x<g(t),&\quad t>0,&\\
&v_t=dv_{xx}+rv(1-v-hu),&&0<x<\infty,&\quad t>0,&\\
&u_x(0,t)=v_x(0,t)=0,\quad u(x,t)=0,&&g(t)\leq x<\infty, &\quad t>0,&\\
&g'(t)=-\gamma u_x(g(t),t),&&&t>0,&\\
&g(0)=g_0,\quad u(x,0)=u_{0}(x),&&0\leq x\leq g_0,& &\\
&v(x,0)=v_{0}(x),&& 0\leq x<\infty,& &
\end{aligned}
\right.\lbl{section1-1}\ees
where $x=g(t)$ is usually called a free boundary, which is to be determined together with $u$ and $v$. The initial functions satisfy
\bes
\left\{
\begin{aligned}
&u_0\in C^2([0,g_0]),\; u_0'(0)=u_0(g_0)=0\mbox{ and }u_0(x)>0\mbox{ in }[0,g_0),\\
&v_0\in C^2([0,\infty))\cap L^\infty(0,\infty),\; v_0'(0)=0,\; \liminf_{x\rightarrow\infty}v_0(x)>0\mbox{ and }v_0(x)> 0\mbox{ in }[0,\infty).
\end{aligned}
\right.\lbl{section1-1a}\ees
This model describes how a new species with population density $u$ invades into the habitat of a native competitor $v$. It is assumed that the species $u$ exists initially in the range $0<x<g_0$, invades into new territory through its invading front $x=g(t)$. The native species $v$ undergoes diffusion and growth in the available habitat $0<x<\infty$. Both $u$ and $v$ obey a no-flux
boundary condition at $x=0$. The equation $g'(t)=-\gamma u_x(g(t),t)$ means that the invading speed is proportional to the gradient of the population density of $u$ at the invading front, which coincides with the well-known Stefan free bounary condition. All parameters $d,k,h,r,g_0$ and $\gamma$ are assumed to be positive. For more biological background, we refer to \cite{BDK, DG,DL10,DL14}.
The work \cite{DL14} considers the {\bf weak-strong competition} case only. It is shown in \cite{DL14} that when the invasive species $u$ is the inferior competitor $(k>1>h)$, if the resident species $v$ is already well established initially (i.e., $v_0$ satisfies the conditions in \eqref{section1-1a}), then $u$ can never invade deep into the underlying habitat, and it dies out before its invading front reaches a certain finite limiting position, whereas if the invasive species $u$ is superior $(h>1>k)$, a spreading-vanishing dichotomy holds for $u$ (see Theorem 4.4 in \cite{DL14}). Moreover, when spreading of $u$ happens, the precise asymptotic spreading speed has been given by Du, Wang and Zhou \cite{DWZ}; it concludes that the spreading speed of $u$ has an asymptotic limit as time goes to infinity, which is determined by a certain traveling wave type system.
In this paper, we examine the {\bf weak competition} case of \eqref{section1-1}, namely the case
$$0<k<1,\; 0<h<1.$$
We will show that a similar spreading-vanishing dichotomy holds for the invasive species $u$, but in sharp contrast to the weak-strong competition case $(h>1>k)$ in \cite{DL14}, where when $u$ spreads successfully, $v$ vanishes eventually (namely $(u,v)\to (1,0)$ as $t\to\infty$), here in the weak competition case, when $u$ spreads successfully, the two populations converge to the co-existence steady state $(u^*, v^*)$ as time goes to infinity. In fact, our results here indicate that the native competitor $v$ always
survives the invasion of $u$. Moreover, we also determine the precise spreading speed of $u$ when the invasion is successful,
which turns out to be the most difficult part of this work and consititutes the main body of the paper. We would like to stress that
while the main steps in the approach here are similar in spirit to those in \cite{DL14} and \cite{DWZ}, highly nontrivial changes are needed in the detailed techniques,
due to the different nature of the dynamical behavior of the system under the current weak competition assumption.
\medskip
We now state our main results more precisely. From \cite{DL14} we know that \eqref{section1-1} has a unique solution, which is defined for all $t>0$. Our aim here is to determine its long-time behavior.
\begin{theo}\lbl{f2}Suppose that $h,k\in (0,1)$ and $(u,v,g)$ is the solution of \eqref{section1-1} with $u_0$ and $v_0$ satisying \eqref{section1-1a}. Then, as $t\to\infty$, the following dichotomy holds.
Either {\bf (i) the species $u$ spreads successfully:}
\[\mbox{
$\lim_{t\to\infty}g(t)=\infty$ and \;\; $
\lim_{t\to\infty}(u(\cdot,t),v(\cdot,t))= (u^\ast,v^\ast)$ in $C^2_{loc}([0,\infty))$;}
\]
or
{\bf (ii) the species $u$ vanishes eventually:}
\[\mbox{ $\lim_{t\to\infty}g(t)<\infty$ and\;\; $
\lim_{t\to\infty}(u(\cdot,t),v(\cdot,t))= (0,1)$ in $C^2_{loc}([0,\infty))$.}
\]
\end{theo}
The next theorem provides a sharp criterion for the above spreading-vanishing dichotomy.
\begin{theo}\lbl{f3} Under the assumptions of Theorem \ref{f2}, there exists $\gamma^*\in [0,\infty)$ depending on $(u_0, v_0)$ such that
alternative {\rm (i)} in Theorem \ref{f2} happens if and only if $\gamma>\gamma^*$. Moreover,
$\gamma^*=0$ $($and hence $u$ always spreads successfully$)$ if $g_0\geq \frac{\pi}{2\sqrt{1-k}}$, and $\gamma^*>0$ if
$g_0<\frac{\pi}{2}$.
\end{theo}
When case (i) happens in Theorem \ref{f2}, the spreading speed of $u$ is asymptotically linear, as indicated by the following theorem.
\begin{theo}\lbl{F5} If {\rm (i)} happens in Theorem \ref{f2}, then there exists $c^0>0$ such that
\bess\lim_{t\rightarrow\infty}\frac{g(t)}{t}=c^0.\eess
\end{theo}
As usual, the positive constant $c^0$ in Theorem \ref{F5} is called the {\bf asymptotic spreading speed} of $u$.
The key in the proof of this theorem is to find a way to determine $c^0$. It turns out that two systems of traveling
wave type equations are needed in order to determine $c^0$. The first one is obtained by looking for traveling wave solutions of \eqref{section1-a2},
namely
\bes
\begin{cases}
\Phi''-c\Phi'+\Phi(1-\Phi-k\Psi)=0,\quad\Phi'>0,\quad -\infty<s<\infty,\\
d\Psi''-c\Psi'+r\Psi(1-\Psi-h\Phi)=0,\quad\Psi'<0,\quad -\infty<s<\infty,\\
(\Phi,\Psi)(-\infty)=(0,1),\quad (\Phi(\infty),\Psi(\infty))=(u^\ast,v^\ast).
\end{cases}\lbl{entirewave0}\ees
(The second system is \eqref{section12} below.)
Clearly, if $(\Phi(s),\Psi(s))$ solves \eqref{entirewave0}, then
$$(u(x,t),v(x,t)):=(\Phi(ct-x),\Psi(ct-x))$$
is a solution of \eqref{section1-a2}, which is often called a traveling wave solution with speed $c$.
By Theorem 4.2 and Example 4.2 in \cite{LWL}, we have the following result on \eqref{entirewave0}:
\begin{prop}\lbl{F6}Assume $h, k\in (0,1)$. Then there exists a critical speed $c_{\ast}\geq 2\sqrt{1-k}$ such that
\eqref{entirewave0} has a solution when $c\geq c_{\ast}$ and it has no solution when $c<c_{\ast}$.
\end{prop}
We can further show that $c_{\ast}\leq 2\sqrt{u^\ast}$. Making use of $c_*$, we have the following result
on the system below which gives traveling wave type solutions to \eqref{section1-1}:
\bes\left\{
\begin{aligned}
&\phi''-c\phi'+\phi(1-\phi-k\psi)=0,\quad \phi'>0\quad\mbox{for } 0<s<\infty,\\
&d\psi''-c\psi'+r\psi(1-\psi-h\phi)=0,\quad \psi'<0\quad\mbox{for } -\infty<s<\infty,\\
&\phi(s)\equiv0\mbox{ for }s\leq0,\quad\psi(-\infty)=1,\quad(\phi(\infty),\psi(\infty))=(u^\ast,v^\ast).&\\
\end{aligned}
\right.\lbl{section12}\ees
\begin{theo}\lbl{F4}
Assume $h, k\in (0,1)$. If $c\in[0,c_{\ast})$, then the system \eqref{section12} admits a unique solution $(\phi_c,\psi_c)\in [C(\mathbb{R})\cap C^2(\mathbb{R}^+)]\times C^2(\mathbb{R})$; if $c\geq c_{\ast}$, then \eqref{section12} does not have a solution.
Moreover, the following conclusions hold:
\begin{itemize}
\item[\rm{(i)}] $\lim_{c\nearrow c_{\ast}}(\phi_c,\psi_c)=(0,1)\mbox{ in }C_{loc}^2([0,\infty))\times C_{loc}^2(\mathbb{R}),$
\item[\rm{(ii)}] for any $\gamma>0$, there exists a unique $c_\gamma\in(0, c_\ast)$ such that $\gamma \phi_{c_\gamma}'(0)=c_\gamma$,
\item[\rm{(iii)}]the function $\gamma\longmapsto c_\gamma$ is strictly increasing and $\lim_{\gamma\rightarrow\infty}c_\gamma= c_\ast.$
\end{itemize}
\end{theo}
We will show that the asymptotic spreading speed of $u$ in Theorem \ref{F5} is given by
\[
c^0:=c_\gamma.
\]
Let us note that if $(\phi,\psi, c)$ solves \eqref{section12}, then
\bess\tilde{u}(x,t):=\phi(ct-x),\quad\tilde{v}(x,t):=\psi(ct-x)\eess
satisfy
\bess\left\{
\begin{aligned}
&\tilde{u}_t=\tilde{u}_{xx}+\tilde{u}(1-\tilde{u}-k\tilde{v}),&-\infty<x<ct,\quad t>0,\\
&\tilde{v}_t=d\tilde{v}_{xx}+r\tilde{v}(1-\tilde{v}-h\tilde{u}),&-\infty<x<\infty,\quad t>0,\\
&\tilde{u}(x,t)=0,&\quad ct\leq x<\infty, \quad t>0.
\end{aligned}
\right.\lbl{section1-01}\eess
If $c=c_\gamma$, then we have additionally
$$(ct)'=c=-\gamma\tilde{u}_{x}(ct,t).$$
We call $(\phi_c,\psi_c)$ with $c=c_\gamma$ the \emph{\textbf{semi-wave}} associated with \eqref{section1-1}. This pair of functions (and its suitable variations) will play a crucial role in the proof of Theorem \ref{F5}, and they also provide upper and lower bounds for the solution pair $(u,v)$ (see the proof of Lemmas \ref{T15} and \ref{T14} for details).
Related free boundary problems of two-species competition models have been investigated in many recent works. Apart from \cite{DL14, DWZ} mentioned earlier, one may find various interesting results in
\cite{CLW,GW12,GW15,WJ,WJZ,WZ16,WZ14,Wu13,Wu15,ZW16}, where \cite{CLW,WZ16} considers time-periodic environment, \cite{WJ,WJZ,ZW16} considers space heterogeneous environment, \cite{GW12,Wu13,Wu15} considers the weak competition case and \cite{GW15,WZ14} covers more general situations.
However, except \cite{DWZ}, in all these works the question of whether there is a precise asymptotic spreading speed has been left open. This is in sharp contrast to the corresponding one species models, where the precise spreading speed is obtained in many situations; see, for example, \cite{DDL, DG, DGP, DLiang, DL10, DLe, DMZ, DMZ2, LLS}.
The rest of this paper is organized as follows.
In section 2, we present some basic results including the existence of solutions to \eqref{section1-1}, and the existence of solutions to a more general system than \eqref{section12}.
In section 3, we prove Theorem \ref{F4} based on an upper and lower solution result (Proposition \ref{s7}) established in section 2, and many other techniques. In section 4, we investigate the long time behavior of the solution to \eqref{section1-1}
and prove the spreading-vanishing dichotomy, and obtain sharp criteria for spreading and vanishing, which finish the proof of Theorems \ref{f2} and \ref{f3}. In section 5, we complete the proof of Theorem \ref{F5} by making use of Theorem \ref{F4}.
Section 6 consists of the proof of Proposition \ref{s7} stated in section 2.
{\setlength{\baselineskip}{16pt}{\setlength\arraycolsep{2pt} \indent
\section{Preliminary results }
\setcounter{equation}{0}
In this section, we collect some basic facts which will be needed in our proof of the main results.
We first note that \eqref{section1-1} always has a unique solution. Indeed, by Theorems 2.4 and 2.5 in \cite{DL14}, we have the following results on the solution of system \eqref{section1-1}.
\begin{prop}\label{t1} For any initial function $(u_0, v_0)$ satisfying \eqref{section1-1a}, the free boundary problem \eqref{section1-1} admits a unique solution
$$(u,v,g)\in C^{1+\alpha,(1+\alpha)/2}(D_1)\times C^{1+\alpha,(1+\alpha)/2}(D_2)\times C^{1+\alpha/2}([0,\infty)),$$
where $D_1=\{(x,t):x\in[0,g(t)],t\in[0,\infty)\}$ and $D_2=\{(x,t):x\in[0,\infty),t\in[0,\infty)\}$. Furthermore, there exists a positive constant $M$ depending on $d,r,\gamma,\|u_0\|_\infty$ and $\|v_0\|_\infty$, such that
\bess\begin{aligned}
&0<u(x,t)\leq M,\; 0< g'(t)\leq M\mbox{ for } 0<x<g(t),\quad t>0,\\
&0<v(x,t)\leq M\mbox{ for } 0<x<\infty,\quad t>0.
\end{aligned}\eess
\end{prop}
Next we recall a comparison result for the free boundary problem, which is a special case of Lemma 2.6 in \cite{DL14}.
\begin{prop}\lbl{the comparison principle}\,{\rm (Comparison principle)}
Assume that $T\in(0,\infty)$, $\underline{g},\overline{g}\in C^1([0,T])$, $\underline{u}\in C(\overline{D_T^\ast})\cap C^{2,1}(D_T^\ast)$
with $D_T^\ast=\{(x,t)\in\mathbb{R}^2:x\in(0,\underline{g}(t)),t\in(0,T]\}$, $\overline{u}\in C(\overline{D_T^{\ast\ast}})\cap C^{2,1}(D_T^{\ast\ast})$
with $D_T^{\ast\ast}=\{(x,t)\in\mathbb{R}^2:x\in(0,\overline{g}(t)),t\in(0,T]\}$, $\underline{v},\overline{v}\in (L^\infty\cap C)([0,\infty)\times[0,T])\cap C^{2,1}([0,\infty)\times[0,T])$
and
\bess
\left\{
\begin{aligned}
&\displaystyle\overline{u}_t-\overline{u}_{xx}\geq \overline{u}(1-\overline{u}-k\underline{v}),&0<x<\overline{g}(t),&\quad0<t<T,\lbl{section4-1}\\
&\displaystyle \underline{u}_t-\underline{u}_{xx}\leq \underline{u}(1-\underline{u}-k\overline{v}),&0<x<\underline{g}(t),&\quad0<t<T,\lbl{section4-2}\\
&\displaystyle \overline{v}_t-d\overline{v}_{xx}\geq r\overline{v}(1-\overline{v}-h\underline{u}),&0<x<\infty,&\quad0<t<T,\lbl{section4-3}\\
&\displaystyle \underline{v}_t-d\underline{v}_{xx}\leq r\underline{v}(1-\underline{v}-h\overline{u}),&0<x<\infty,&\quad0<t<T,\lbl{section4-4}\\
&\overline{u}_x(0,t)\leq0,\quad\underline{v}_x(0,t)\geq0,\quad\overline{u}(x,t)=0,&
x=\overline{g}(t),&\quad0<t<T,\lbl{section4-5}\\
&\underline{u}_x(0,t)\geq0,\quad\overline{v}_x(0,t)\leq0,\quad\underline{u}(x,t)=0,&
x=\underline{g}(t),&\quad0<t<T,\lbl{section4-6}\\
&\underline{g}'(t)\leq-\gamma\underline{u}_x(\underline{g}(t),t),
\quad\overline{g}'(t)\geq-\gamma\overline{u}_x(\overline{g}(t),t),&&\quad0<t<T,\\
&\underline{u}(x,0)\leq u_{0}(x)\leq\overline{u}(x,0),&0\leq x\leq g_0,&\lbl{section4-9}\\
&\underline{v}(x,0)\leq v_{0}(x)\leq\overline{v}(x,0),&0\leq x<\infty.&
\end{aligned}
\right.\lbl{section4-10}
\eess
Then the solution $(u,v,g)$ of \eqref{section1-1} satisfies
$$\begin{aligned}
&\underline{g}(t)\leq g(t)\leq\overline{g}(t),\quad\underline{v}(x,t)\leq v(x,t)\leq\overline{v}(x,t)\; \mbox{ for }\; 0\leq x<\infty,\quad t\in[0,T],\\
&\underline{u}(x,t)\leq u(x,t)\mbox{ for }0\leq x<\underline{g}(t),\quad u(x,t) \leq \overline{u}(x,t) \;\mbox{ for }\; 0\leq x<g(t),\quad t\in[0,T].
\end{aligned}$$
\end{prop}
\begin{remark}\lbl{remarkp} In system \eqref{section1-1}, if the boundary conditions at $x=0$ are replaced by
\bess
u(0,t)=m_1(t),\quad v(0,t)=m_2(t),&t>0,
\eess
then Proposition \ref{the comparison principle} also holds if we replace
$$\overline{u}_x(0,t)\leq0,\quad \underline{v}_x(0,t)\geq0,\quad \underline{u}_x(0,t)\geq0,\quad \overline{v}_x(0,t)\leq0$$
by
$$\overline{u}(0,t)\geq m_1(t),\quad \underline{v}(0,t)\leq m_2(t),\quad \underline{u}(0,t)\leq m_1(t),\quad \overline{v}(0,t)\geq m_2(t)$$
for $0<t<T$.
\end{remark}
Lastly in this section, we modify some well known upper and lower solution technique to show the existence of a solution for a
general coorperative system of the form
\bes\begin{cases}
d_1\varphi_1''-c\varphi_1'+f_1(\varphi)=0,&s\in\mathbb{R}^+,\\
d_2\varphi_2''-c\varphi_2'+f_2(\varphi)=0,&s\in\mathbb{R},\\
\varphi_1(s)\equiv0,&s\leq0,
\end{cases}\lbl{function1}
\ees
where $\varphi=(\varphi_1,\varphi_2), c\geq0, d_i>0$, and $f_i:\mathbb{R}^2\rightarrow\mathbb{R}$ ($i=1,2$) satisfy the following conditions:
\begin{itemize}
\item[$(\textbf{A}_1)$] there is a strictly positive vector $\textbf{K}=(k_1,k_2)\in \mathbb{R}^2$ such that $f_i(\textbf{0})=f_i(\textbf{K})=0$ for $i\in\{1,2\}$, and $f_i(u_1,u_2)\not=0$ for $(u_1, u_2)\in \big((0,k_1]\times [0,k_2]\big)\setminus\{{\bf K}\}, i\in\{1,2\}$;
\item[$(\textbf{A}_2)$] the system \eqref{function1} is a cooperative system, that is, $f_i(u_1, u_2)$ is nondecreasing in $u_j$ for $i,j\in\{1,2\},i\neq j$, $(u_1, u_2)\in [0, k_1]\times [0, k_2]$, and there exists a constant $\beta_0\geq0$ such that $\beta_0 u_i+f_i(u_1, u_2)$ is nondecreasing in $u_i$ for $ i=1,2$ and $(u_1, u_2)\in [0, k_1]\times [0, k_2]$;
\item[$(\textbf{A}_3)$] $f_1$ and $f_2$ are locally Lipschitz continuous.
\end{itemize}
We are particularly interested in solutions $\varphi$ of \eqref{function1} that satisfy the asymptotic boundary conditions
\bes
\varphi_2(-\infty)=0,\quad \varphi(\infty)=\textbf{K}.
\lbl{function2a2}
\ees
Indeed, solving \eqref{function1} and \eqref{function2a2} will supply the main step for solving \eqref{section12}. On the other hand,
our method here to solve the more general problems \eqref{function1} and \eqref{function2a2} may have other applications.
We will write $(\varphi_1,\varphi_2)\leq(\tilde{\varphi}_1,\tilde{\varphi}_2)$ if $\varphi_i\leq\tilde{\varphi}_i,i=1,2.$
Let $\mathcal{R}\subset \mathbb{R}^2$ denote the rectangle
$$\mathcal{R}=[0, k_1]\times [0, k_2].$$
It is convenient to introduce the following notations:
\bess\begin{aligned}
&C_{\mathcal{R}}(\mathbb{R}^+,\mathbb{R}^2)=\{\varphi\in C(\mathbb{R}^+,\mathbb{R}^2)
:\varphi(s) \in \mathcal{R} \mbox{ for } s>0\},\\
&C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)=\{\varphi\in C(\mathbb{R},\mathbb{R}^2)
: \varphi(s) \in \mathcal{R} \mbox{ for }s\in\mathbb{R}\}.
\end{aligned}\eess
\begin{definition}\lbl{s2} Suppose that $\overline{\varphi}=(\overline{\varphi}_1,\overline{\varphi}_2)\in C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)$, $\underline{\varphi}=(\underline{\varphi}_1,\underline{\varphi}_2)\in C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)$,
and $\overline{\varphi}_j,\underline{\varphi}_j$ are twice continuously differentiable in $\mathbb{R}\setminus \Omega_j$ for $j=1,2$, where $\Omega_1\subset [0,\infty)$ and $\Omega_2\subset \mathbb{R}$ are two finite sets, say $\Omega_1=\{\xi_i\geq 0: i=1,2,\cdots,m_1\}$ and $\Omega_2=\{\eta_i\in\mathbb{R}: i=1,2,\cdots,m_2\}$. Moreover,
{\rm \bf (i)} the functions $\overline{\varphi}$ and $\underline{\varphi}$ satisfy the inequalities
\bes\left\{\begin{aligned}
&d_1\overline{\varphi}_1''-c\overline{\varphi}_1'-\beta\overline{\varphi}_1+H_1(\overline{\varphi})\leq0\quad\mbox{for }
s\in\mathbb{R}^+\setminus \Omega_1,\\
&d_2\overline{\varphi}_2''-c\overline{\varphi}_2'-\beta\overline{\varphi}_2+H_2(\overline{\varphi})\leq0
\quad\mbox{for } s\in\mathbb{R}\setminus \Omega_2,\\
&\overline{\varphi}_1(s)=0\mbox{ for }s\leq0,\quad\overline{\varphi}_2(-\infty)=0
\end{aligned} \right.\lbl{upper}
\ees
and
\bes\left\{\begin{aligned}
&d_1\underline{\varphi}_1''-c\underline{\varphi}_1'-\beta\underline{\varphi}_1+H_1(\underline{\varphi})\geq0
\quad\mbox{for } s\in\mathbb{R}^+\setminus \Omega_1,\\
&d_2\underline{\varphi}_2''-c\underline{\varphi}_2'-\beta\underline{\varphi}_2+H_2(\underline{\varphi})\geq0
\quad\mbox{for } s\in\mathbb{R}\setminus \Omega_2,\\
&\underline{\varphi}_1(s)=0\mbox{ for }s\leq0,\quad\underline{\varphi}_2(-\infty)=0;
\end{aligned}\right.\lbl{lower}
\ees
{\rm \bf (ii)} the derivatives of $\overline{\varphi}$ and $\underline{\varphi}$ satisfy
\bes\begin{aligned}
&\underline{\varphi}_1'(\xi_i-)\leq\underline{\varphi}_1'(\xi_i+),\quad \overline{\varphi}_1'(\xi_i+)\leq\overline{\varphi}_1'(\xi_i-)\mbox{ for }\xi_i\in\Omega_1\setminus\{0\},\\
&\underline{\varphi}_2'(\eta_i-)\leq\underline{\varphi}_2'(\eta_i+),\quad \overline{\varphi}_2'(\eta_i+)\leq\overline{\varphi}_2'(\eta_i-)\mbox{ for }\eta_i\in\Omega_2.\lbl{lower1}
\end{aligned}\ees
Then $\overline{\varphi}$ and $\underline{\varphi}$ are called a weak \textbf{upper solution} and a weak \textbf{lower solution} of \eqref{function1}-\eqref{function2a2} associated with $\mathcal{R}$, respectively.
\end{definition}
\begin{prop}\lbl{s7} Assume that $(\textbf{A}_1)-(\textbf{A}_3)$ hold. Suppose \eqref{function1}-\eqref{function2a2} has a pair of upper and lower solutions associated with $\mathcal{R}$ satisfying
\[\mbox{
$\underline \varphi_1(s)\not\equiv 0$ \; and\;\;
$\sup_{t\leq s}\underline{\varphi}(t)\leq\overline{\varphi}(s)$ for $s\in\mathbb{R}$.}
\]
Then \eqref{function1} has a monotone non-decreasing solution $\varphi$ satisfying \eqref{function2a2}.
\end{prop}
Proposition \ref{s7} will play an important role in the proof of Theorem \ref{F4} in the next section. The proof of Proposition \ref{s7} is based on some upper and lower solution arguments and involves the Schauder fixed point theorem. Our proof is similar in spirit to that in several works on various different traveling wave problems (see, for example,
\cite{LLR,Ma,WNW,WZ}), but the detailed techniques are rather different. Since the proof is long, it is postponed to Section 6
at the end of the paper.
{\setlength{\baselineskip}{16pt}{\setlength\arraycolsep{2pt} \indent}}
\section{Semi-wave solutions}
\setcounter{equation}{0}
The aim of this section is to prove Theorem \ref{F4}. Firstly, we recall some known results for the Fisher-KPP equation
\bes\left\{\begin{aligned}
&d\chi''-c\chi'+a\chi(b-\chi)=0,\quad0<s<\infty,\\
&\chi(0)=0.
\end{aligned}\right.\lbl{KPP}
\ees
\begin{lem}\lbl{S22}
Let $a>0,b>0$ and $ d>0$ be fixed constants.
\begin{itemize}
\item[(i)] If $c\in[0,2\sqrt{abd})$, then \eqref{KPP} has a unique solution $\chi(s)$.
\item[(ii)] For each $c\in [0, 2\sqrt{abd})$, the solution $\chi(s)$ of \eqref{KPP} is strictly increasing and has the following asymptotic behavior
$$\chi(s)=b-[b_\chi +o(1)]e^{\frac{c-\sqrt{c^2+4ab}}{2d}s} \mbox{ as }s\rightarrow\infty,$$
where $b_\chi$ is a positive constant.
\end{itemize}
\end{lem}
The conclusion (i) can be found in \cite{DL10, BDK}, and the proof of (ii) is standard (see, for example, \cite{S}).
\begin{lem}\lbl{S21} {\rm (\cite{DWZ})}
Let $\tilde{f},\tilde{g}\in C([0,\infty))$ with $\tilde{g}$ nonnegative and not identically 0, and $c,d$ be given constants with $d>0$. Assume that $u_i(s)>0$ in $(0,\infty)$, $u_1(0)\leq u_2(0)$ and
$$cu_1'-du_1''-u_1[\tilde{f}(s)-\tilde{g}(s)u_1]\leq0\leq cu_2'-du_2''-u_2[\tilde{f}(s)-\tilde{g}(s)u_2],\quad 0<s<\infty.$$
If $\limsup_{s\rightarrow\infty}\frac{u_1(s)}{u_2(s)}\leq 1$, then
$$u_1(s)\leq u_2(s)\quad\mbox{ for }\; 0\leq s<\infty.$$
\end{lem}
Next, we consider the problem
\bes\left\{\begin{aligned}
&d\omega''+c\omega'+r(hu^\ast-\omega)(1-hu^\ast+\omega)=0,\quad0<s<\infty,\\
&\omega(0)=0.
\end{aligned}\right.\lbl{KP10}
\ees
\begin{lem}\lbl{S23}
For any constant $d>0$, $c\geq0$, the problem \eqref{KP10} admits a unique positive solution $\omega(s,c)$ satisfying $\omega(\infty, c)=h u^*$. Moreover, $\omega(s,c)$ is strictly increasing in $s$ for $s>0$, and there exists $b_\omega>0$ such that
\bes
\omega(s)=hu^\ast-[b_\omega+o(1)]e^{\frac{-c-\sqrt{c^2+4dr}}{2d}s}\mbox{ as }s\rightarrow\infty.
\lbl{expk}
\ees
\end{lem}
\begin{proof}
It is easily seen that
\bes\underline{\omega}(s):=0,
\quad\overline{\omega}(s):=h u^*
\lbl{KP1}
\ees
are a pair of lower and upper solutions for \eqref{KP10}. Thus, \eqref{KP10} has at least one solution satisfying
$0\leq \omega(s)\leq hu^*$. The strong maximum principle infers that $\omega(s)<hu^*$ for $s>0$.
We claim $\omega(s)$ is increasing and $\omega(\infty)=hu^\ast$. Rewrite \eqref{KP10} as $$-\left(de^{\frac{c}{d}s}\omega'\right)'=re^{\frac{c}{d}s}(hu^\ast-\omega)(1-hu^\ast+\omega).$$
Noting that $0\leq \omega<hu^\ast$, we have
$$-\left(de^{\frac{c}{d}s}\omega'\right)'>0\mbox{ for }s>0.$$
Hence, $e^{\frac{c}{d}s}\omega'$ is a decreasing function. We claim that $\omega(s)$ is monotone in $(R,\infty)$ for some
large $R>0$. Otherwise $\omega(s)$ is oscillating near $s=\infty$ and hence we can find a sequence $s_n\rightarrow\infty$ as $n\rightarrow\infty$ such that $\omega'(s_n)=0$. It follows that for any fixed $s>0$,
$$e^{\frac{c}{d}s}\omega'(s)>\lim_{n\rightarrow\infty}e^{\frac{c}{d}s_n}\omega'(s_n)=0.$$
Thus, we have $\omega'>0$ in $(0,\infty)$, a contradiction to the assumption. Hence, for large $R>0$, $\omega$ is monotone in $(R,\infty)$.
By \eqref{KP10} and $0\leq \omega<hu^*$, we easily obtain $\omega(\infty)=hu^\ast$.
Furthermore, a simple calculation indicates that the ODE
system satisfied by $(\omega,\omega')$ has $(hu^\ast,0)$ as a saddle point. It follows from standard ODE theory that, there exists a constant $b_\omega>0$ such that \eqref{expk} holds, and
\[
\omega'(s)=\frac{c+\sqrt{c^2+4dr}}{2d}[b_\omega+o(1)]e^{\frac{-c-\sqrt{c^2+4dr}}{2d}s}=o(1) e^{-\frac{c}{d}s} \mbox{ as }s\rightarrow\infty.
\]
We thus obtain, for any fixed $s>0$,
$$e^{\frac{c}{d}s}\omega'(s)>\lim_{s\rightarrow\infty}e^{\frac{c}{d}s}\omega'(s)=0.$$
Hence $\omega'>0$ in $(0,\infty)$ and $\omega$ is strictly increasing in $(0,\infty)$.
It remains to show the uniqueness of positive solution of \eqref{KP10} satisfying $\omega(\infty)=hu^*$.
Let $\omega_1$ and $\omega_2$ be two positive solutions of \eqref{KP10} satisfying $\omega_i(\infty)=hu^*$, $i=1,2$.
We easily see that $\omega_i\leq hu^*$ for otherwise there exists $s_i>0$ such that
\[
\omega_i(s_i)=\max_{s\geq 0}\omega_i(s)>hu^*, \; \omega_i'(s_i)=0\geq \omega_i''(s_i),
\]
which gives a contradiction to \eqref{KP10} when evaluated at $s=s_i$. The strong maximum principle then yields
$\omega_i(s)<h u^*$ for $s>0$. We may then argue as for $\omega$ above to obtain $\omega_i(\infty)=h u^*$.
Set $\chi_i(s):=1-hu^*+\omega_i(s)$ and we find that $\chi_i$ satisfies
\[
d\chi''-c\chi'+r(1-\chi)\chi=0 \mbox{ for } s>0, \; \chi(0)=1-hu^*,\; \chi(\infty)=1.
\]
By Lemma \ref{S21} we immediately obtain $\chi_1\equiv \chi_2$ and hence $\omega_1\equiv \omega_1$.
The uniqueness is thus proved.
\end{proof}
Now, we turn to consider system \eqref{section12}.
For convenience, we change it to a coorperative system by setting
\bess
\tilde{\phi}(s):=\phi(s),\quad \tilde{\psi}(s):=1-\psi(s).
\eess
Clearly $(\phi,\psi)$ solves \eqref{section12} if and only if $(\tilde{\phi},\tilde{\psi})$ satisfies
\bes\left\{
\begin{aligned}
&\tilde{\phi}''-c\tilde{\phi}'+\tilde{\phi}(1-k-\tilde{\phi}+k\tilde{\psi})=0,\quad\tilde{\phi}'>0\quad\mbox{for } s\in\mathbb{R}^+,\\
&d\tilde{\psi}''-c\tilde{\psi}'+r(1-\tilde{\psi})(h\tilde{\phi}-\tilde{\psi})=0,\quad\tilde{\psi}'>0\quad\mbox{for } s\in\mathbb{R},\\
&\tilde{\phi}(s)=0\mbox{ for }s\leq0,\quad \tilde{\psi}(-\infty)=0,\quad(\tilde{\phi}(\infty),\tilde{\psi}(\infty))=(u^\ast,hu^\ast).
\end{aligned}
\right.\lbl{sect1}\ees
\medskip
Let $\Sigma=\Sigma_1\times\Sigma_2$, where
\bess\begin{aligned}
&\Sigma_1=\{\tilde{\phi}\in C(\mathbb{R})\cap C^2([0,\infty)):\tilde{\phi}(s)\equiv0\, (s\leq0),\quad\tilde{\phi}'(s)>0\, (s>0),
\quad\tilde{\phi}(\infty)=u^\ast\},\\
&\Sigma_2=\{\tilde{\psi}\in C^2(\mathbb{R}):\tilde{\psi}(-\infty)=0,\quad\tilde{\psi}'(s)> 0\, (s\in\mathbb{R}),
\quad\tilde{\psi}(\infty)=hu^\ast\}.
\end{aligned}\eess
We define a functional $\theta$ on $\Sigma$ by
$$\theta(\tilde{\phi},\tilde{\psi})=\min\left\{\inf_{s\in\mathbb{R}^+}\Theta_1(\tilde{\phi},\tilde{\psi})(s),
\quad\inf_{s\in\mathbb{R}}\Theta_2(\tilde{\phi},\tilde{\psi})(s)\right\},$$
where
$$\begin{aligned}
&\Theta_1(\tilde{\phi},\tilde{\psi})(s)=\frac{\tilde{\phi}''(s)
+\tilde{\phi}(s)(1-k-\tilde{\phi}(s)+k\tilde{\psi}(s))}{\tilde{\phi}'(s)},\quad s\in\mathbb{R}^+,\\
&\Theta_2(\tilde{\phi},\tilde{\psi})(s)=\frac{d\tilde{\psi}''(s)
+r(1-\tilde{\psi}(s))(h\tilde{\phi}(s)-\tilde{\psi}(s))}{\tilde{\psi}'(s)},\quad s\in\mathbb{R}.
\end{aligned}$$
Clearly, if $(\tilde{\phi},\tilde{\psi})$ is a solution of \eqref{sect1} with $c\geq0$, then $(\tilde{\phi},\tilde{\psi})\in\Sigma$ and $\theta(\tilde{\phi},\tilde{\psi})=c$. Therefore,
$$c\leq c_0^\ast:=\sup_{(\tilde{\phi},\tilde{\psi})\in\Sigma}\theta(\tilde{\phi},\tilde{\psi}).$$
We will show $c_0^*=c_*>0$, where $c_*$ is given in Proposition \ref{F6}. For the moment we assume $c_0^*>0$ and
prove that \eqref{sect1} has a solution for every $c\in[0,c_0^\ast)$ by using an upper and lower solution argument.
From Lemma \ref{S22}, the following equation
\bess\left\{\begin{array}{ll}\medskip
\displaystyle
\hat{\chi}''+\hat{\chi}(u^\ast-\hat{\chi})=0,\quad 0<s<\infty,\\
\hat{\chi}(0)=0
\end{array}\right.\lbl{}
\eess
has a unique strictly increasing solution $\hat{\chi}$ satisfying $\hat\chi(\infty)=u^\ast$.
We define
\bes\lbl{upfl}
\overline{\phi}(s)=\left\{\begin{array}{ll}\medskip
\displaystyle
0,&-\infty<s<0,\\
\hat{\chi}(s),&\quad 0\leq s<\infty,
\end{array}\right.
\quad\overline{\psi}(s)=\left\{\begin{array}{ll}\medskip
\displaystyle
hu^\ast-\omega(-s,0),&-\infty<s<0,\\
hu^\ast,&\quad 0\leq s<\infty.
\end{array}\right.
\ees
\begin{lem}\lbl{S24}
For $c\geq0$, the pair of functions $(\overline{\phi}(s),\overline{\psi}(s))$ is an upper solution of \eqref{sect1}
associated with $\mathcal{R}:=[0, u^*]\times [0, hu^*]$.
\end{lem}
\begin{proof} For $s\geq0$, we have
\bess\begin{array}{ll}\medskip
&\overline{\phi}''-c\overline{\phi}'+\overline{\phi}(1-k-\overline{\phi}+k\overline{\psi})\\
=&\hat{\chi}''-c\hat{\chi}'+\hat{\chi}(u^\ast-\hat{\chi})=-c\hat{\chi}'\leq0
\end{array}\eess
and
\bess\begin{array}{ll}\medskip
&d\overline{\psi}''-c\overline{\psi}'+r(1-\overline{\psi})(h\overline{\phi}-\overline{\psi})
=r h(1-hu^\ast)(\hat{\chi}-u^\ast)\leq0.
\end{array}\eess
For $s<0$, we have
\bess\begin{array}{ll}\medskip
&\overline{\phi}''-c\overline{\phi}'+\overline{\phi}(1-k-\overline{\phi}+k\overline{\psi})=0
\end{array}\eess
and
\bes\lbl{lsol}
\begin{array}{ll}\medskip
&d\overline{\psi}''-c\overline{\psi}'+r(1-\overline{\psi})(h\overline{\phi}-\overline{\psi})
=-c\overline{\psi}'\leq0.
\end{array}\ees
Moreover, it is easily seen that
\[
\overline{\psi}'(0-)\geq \overline{\psi}'(0+).
\]
Finally, by definition, $\overline \phi(s)\equiv 0$ for $s\leq 0$, $ \overline\phi(\infty)=u^*$,
$\overline \psi(-\infty)=0$ and $\overline\psi(\infty)=h u^*$. Hence $(\overline\phi,\overline\psi)$ meets all the requirements for an upper solution associated with $\mathcal{R}$ in Definition \ref{s2}.
This completes the proof.\end{proof}
\begin{lem}\lbl{S25} Assume $h, k\in (0,1)$, $c_0^*>0$ and $c\in[0,c_0^\ast)$. Then \eqref{sect1} has a solution $(\tilde{\phi},\tilde{\psi})$.
\end{lem}
\begin{proof}
For $c\in[0,c_0^\ast)$, by the definition of $c_0^\ast$, there exists $(\underline{\phi}(s),\underline{\psi}(s))\in\Sigma$ such that $\theta(\underline{\phi},\underline{\psi})>c$. Thus we have
\bess\left\{\begin{aligned}
&\underline{\phi}''-c\underline{\phi}'+\underline{\phi}(1-k-\underline{\phi}+k\underline{\psi})\geq0,
\quad\underline{\phi}'>0,\quad s\in\mathbb{R}^+,\\
&d\underline{\psi}''-c\underline{\psi}'+r(1-\underline{\psi})(h\underline{\phi}-\underline{\psi})\geq0,
\quad\underline{\psi}'>0,\quad s\in\mathbb{R},\\
&\underline{\phi}(s)=0\mbox{ for }s\leq0,\quad \underline{\psi}(-\infty)=0,\quad(\underline{\phi}(\infty),\underline{\psi}(\infty))=(u^\ast,hu^\ast).
\end{aligned}\right.\eess
Hence, $(\underline{\phi},\underline{\psi})$ is a lower solution of \eqref{sect1} associated with $\mathcal{R}:=[0, u^*]\times [0, hu^*]$.
Next, we show that $(\underline{\phi}(s),\underline{\psi}(s))\leq(\overline{\phi}(s),\overline{\psi}(s))$ for $s\in\mathbb{R}$, where
$(\overline{\phi},\overline{\psi})$ is the upper solution obtained in Lemma \ref{S24}.
Clearly, $\underline{\psi}(s)\leq\overline{\psi}(s)$ for $s>0$ and $\underline{\phi}(s)=\overline{\phi}(s)=0$ for $s<0$.
We only need to show that $\underline{\psi}(s)\leq\overline{\psi}(s)$ for $s\leq0$, and $\underline{\phi}(s)\leq\overline{\phi}(s)$ for $s\geq0$.
Let $\underline{\psi}_1(s)=1-\underline{\psi}(-s)$ and $\overline{\psi}_1(s)=1-\overline{\psi}(-s)$. In view of
$\underline{\phi}(s)=\overline \phi(s)=0$ for $s<0$ and \eqref{lsol},
we have
\bess\begin{aligned}&-d\underline{\psi}_1''-c\underline{\psi}_1'-r\underline{\psi}_1(1
-\underline{\psi}_1)\geq 0\geq
-d\overline{\psi}_1''-c\overline{\psi}_1'-r\overline{\psi}_1(1
-\overline{\psi}_1) \mbox{ for } s>0,\\
&\underline{\psi}_1(\infty)=\overline{\psi}_1(\infty)=1,\ \ \underline{\psi}_1(0)>\overline{\psi}_1(0).
\end{aligned}\eess
By Lemma \ref{S21}, $\underline{\psi}_1(s)\geq\overline{\psi}_1(s)$ for $s\geq0$.
Hence, $\underline{\psi}(s)\leq\overline{\psi}(s)$ for $s\leq0$. Similarly, we can prove $\underline{\phi}(s)\leq\overline{\phi}(s)$ for $s\geq0$. Thus
\[
\mbox{$(\underline{\phi}(s),\underline{\psi}(s))\leq(\overline{\phi}(s),\overline{\psi}(s))$ for $s\in\mathbb{R}$.}
\]
The monotonicity of $\underline{\phi}(s)$ and $\underline{\psi}(s)$ then infers that
\[
\sup_{t\leq s}(\underline{\phi}(t),\underline{\psi}(t))\leq (\overline{\phi}(s),\overline{\psi}(s)) \mbox{ for } s\in\mathbb{R}.
\]
Therefore we can apply Proposition \ref{s7} to conclude that \eqref{sect1} has a positive solution $(\tilde{\phi},\tilde{\psi})$ for each $c\in[0,c_0^\ast)$, except that we only have $\tilde \phi'(s)\geq 0$ for $s>0$ and $\tilde\psi'(s)\geq 0$ for $s\in\mathbb R$.
It remains to prove $\tilde{\phi}'(s)>0$ for $s>0$ and $\tilde{\psi}'(s)>0$ for $s\in\mathbb{R}$.
Since $\tilde{\phi}'(s)\geq0$ for $s\in\mathbb{R}\setminus\{0\}$ and $\tilde{\psi}'(s)\geq0$ for $s\in\mathbb{R}$, and none of them is identically 0, applying the strong maximum principle to the cooperative system satisfied by $(\phi',\psi')$, we have $\tilde{\phi}'(s)>0$ for $s>0$ and $\tilde{\psi}'(s)>0$ for $s\in\mathbb{R}$.
\end{proof}
To prove uniqueness for solutions of \eqref{sect1}, we need to investigate the asymptotic behavior of solutions to \eqref{sect1} as $s\rightarrow\infty$. To this end we consider the linearized equation of \eqref{sect1} at $(u^\ast,hu^\ast)$:
\bes\left\{\begin{array}{ll}\medskip
\check{\phi}''-c\check{\phi}'-u^\ast\check{\phi}+ku^\ast\check{\psi}=0,\\
d \check{\psi}''-c\check{\psi}'-rv^\ast\check{\psi}+rhv^\ast\check{\phi}=0.
\end{array}\right.\lbl{f7}
\ees
If $(\check{\phi},\check{\psi})=(me^{\mu s},ne^{\mu s})$ solves \eqref{f7},
then $(m,n)$ and $\mu$ must satisfy
\bes
A(\mu)(m,n)^T=(0,0)^T,
\lbl{eigenvector}\ees
where
\bess
A(\mu)=\left(
\begin{array}{cc}\medskip
\mu^2-c\mu-u^\ast&ku^\ast\\
hrv^\ast&d\mu^2-c\mu-rv^\ast
\end{array}\right).
\eess
Let
\bess\lbl{poly}
P_1(\mu):={\rm det} (A(\mu))=(\mu^2-c\mu-u^\ast)(d\mu^2-c\mu-rv^\ast)-khru^\ast v^\ast.
\eess
Then \eqref{eigenvector} has a nonzero solution $(m,n)^T$ if and only if $P_1(\mu)=0$.
Let
$$\mu_1^\pm:=\frac{c\pm\sqrt{c^2+4u^\ast}}{2}\mbox{ and } \mu_2^\pm:=\frac{c\pm\sqrt{c^2+4drv^\ast}}{2d}$$
be the two roots of
$$\mu^2-c\mu-u^\ast=0\mbox{ and }d\mu^2-c\mu-rv^\ast=0,$$
respectively.
Clearly
$$
P_1(0)=(1-kh)ru^\ast v^\ast>0,\; P_1(\pm\infty)=\infty\mbox{ and } P_1(\mu_i^\pm)=-khru^*v^*<0\mbox{ for } i=1,2.
$$
Hence, for any $c\geq0$, $P_1(\mu)=0$ has four different real roots $\hat{\mu}_i\,(i=1,2,3,4)$ satisfying
\bes\begin{aligned}
&\hat{\mu}_1<\min\{\mu_1^-,\mu_2^-\}\leq\max\{\mu_1^-,\mu_2^-\}<\hat{\mu}_2<0,\\
&0<\hat{\mu}_3<\min\{\mu_1^+,\mu_2^+\}\leq\max\{\mu_1^+,\mu_2^+\}<\hat{\mu}_4.\end{aligned}\lbl{exp1}\ees
\begin{lem}\lbl{S26}Let $(\tilde{\phi}(s),\tilde{\psi}(s))$ be a solution of \eqref{sect1}. Then there exist positive constants $m$ and $n$ independent of $(\tilde{\phi},\tilde{\psi})$, and a positive constant $\beta$ depending on $(\tilde{\phi},\tilde{\psi})$, such that
\bes
(\tilde{\phi}(s),\tilde{\psi}(s))=(u^\ast,hu^\ast)- \beta e^{\hat{\mu}_2s}(m,n)[1+o(1)]\mbox{ as }s\rightarrow\infty.\lbl{expansion1}\ees
\end{lem}
\begin{proof}\,
Let $(\tilde{\phi}(s),\tilde{\psi}(s))$ be an arbitrary solution of \eqref{sect1}. The inequalities \eqref{exp1} imply that the first
order ODE system satisfied by $(\tilde{\phi}(s),\tilde{\phi}'(s),\tilde{\psi}(s),\tilde{\psi}'(s))$ has a critical point at $(u^\ast,0,hu^\ast,0)$, which is a saddle point. By standard stable manifold theory (see, e.g., Theorem 4.1 and its proof in Chapter 13 of \cite{Co}), we can conclude that
$$(u^\ast,hu^\ast)-(\tilde{\phi}(s),\tilde{\psi}(s))\rightarrow(0,0)\mbox{ exponentially as }s\rightarrow\infty.$$
Let $(\hat{\phi},\hat{\psi})=(u^\ast,hu^\ast)-(\tilde{\phi}(s),\tilde{\psi}(s))$. Then $(\hat{\phi},\hat{\psi})$ satisfies
\bes\left\{\begin{aligned}
& \hat{\phi}''-c\hat{\phi}'-u^\ast\hat{\phi}+ku^\ast\hat{\psi}
+\delta_1(s)\hat{\phi}+\delta_2(s)\hat{\psi}=0,\\
&d \hat{\psi}''-c\tilde{\psi}'-rv^\ast\hat{\psi}+hrv^\ast\hat{\phi}
+\delta_3(s)\hat{\phi}+\delta_4(s)\hat{\psi}=0,
\end{aligned}\right.\lbl{f6}
\ees
where
$$\delta_1(s):=\hat{\phi}(s),\delta_2(s):=-k\hat{\phi}(s),\delta_3(s):=-rh\hat{\psi}(s),\delta_4(s):=r\hat{\psi}(s).$$
Clearly,
\bess
\delta_i(s)\rightarrow0\mbox{ exponentially as }s\rightarrow\infty\mbox{ for }i=1,2,3,4.
\eess
Now we turn to consider the linear system \eqref{f7}.
Recall that $P_1(\mu)=0$ has four different real roots satisfying $\hat{\mu}_1<\hat{\mu}_2<0<\hat{\mu}_3<\hat{\mu}_4$.
Let $(m_i,n_i)$ be an eigenvector corresponding to $\mu=\hat{\mu}_i$ in \eqref{eigenvector}, i.e.,
\[
(m_i, n_i)\not=(0,0) \mbox{ and } A(\hat \mu_i)(m_i,n_i)^T=(0,0)^T.
\]
Then \eqref{f7} has four linearly independent solutions
\[
\Upsilon_i=(m_i,n_i)e^{\hat{\mu}_is},\; i=1,2,3,4,
\]
which form a fundamental system for \eqref{f7}.
Applying Theorem 8.1 in Chapter 3 of \cite{Co} to the system \eqref{f6}, viewed as a perturbed linear system of \eqref{f7}, we conclude that \eqref{f6} has four linearly independent solutions $\tilde{\Upsilon}_i,\; i=1,2,3,4,$ satisfying
$$
\tilde{\Upsilon}_i(s)=(1+o(1))\Upsilon_i(s)\mbox{ as }s\rightarrow\infty,\; i=1,2,3,4,
$$
which form a fundamental system for \eqref{f6} (viewed as a linear system).
So the solution $(\hat{\phi},\hat{\psi})$ of \eqref{f6} can be represented as
$$(\hat{\phi}(s),\hat{\psi}(s))=\sum_{i=1}^4\beta_i\tilde{\Upsilon}_i(s),
$$
where $\beta_i\; (i=1, 2, 3, 4)$ are constants.
Since $(\hat{\phi}(\infty),\hat{\psi}(\infty))=(0,0)$, and $0<\hat \mu_3<\hat\mu_4$, we necessarily have $\beta_3=\beta_4=0$.
We claim that $\beta_2\not=0$. Otherwise we necessarily have $\beta_1\not=0$ and
\[
(\hat{\phi}(s),\hat{\psi}(s))=\beta_1\tilde{\Upsilon}_1(s)=(1+o(1))\beta_1(m_1,n_1)e^{\hat{\mu}_1s}\mbox{ as } s\rightarrow\infty.
\]
However, it is easily checked that all the four elements of the matrix $A(\hat\mu_1)$ are positive, which implies that
$m_1\cdot n_1<0$ and so the two components of the vector $\beta_1(m_1, n_1)$ have opposite signs. It follows that
for all large $s$, $(\hat{\phi}(s),\hat{\psi}(s))$ has a component which is negative, contradicting the fact that $(\hat{\phi}(s),\hat{\psi}(s))>(0,0)$ for all $s>0$. Therefore we must have $\beta_2\not=0$. It is also easily checked that the two rows of the matrix
$A(\hat \mu_2)$ have opposite signs and so $m_2\cdot n_2>0$. For definiteness, we may assume that $m_2$ and $n_2$ are positive. Moreover, due to $\hat \mu_1<\hat \mu_2<0$, we have
\[
(\hat{\phi}(s),\hat{\psi}(s))=\sum_{i=1}^2\beta_i\tilde{\Upsilon}_i(s)=(1+o(1))\beta_2(m_2,n_2)e^{\hat{\mu}_2s}\mbox{ as } s\rightarrow\infty.
\]
Using $(\hat{\phi}(s),\hat{\psi}(s))>(0,0)$ for all $s>0$ we further obtain that $\beta_2>0$, and hence
\eqref{expansion1} holds with $(m,n):=(m_2,n_2)$ and $\beta:=\beta_2$.
\end{proof}
\begin{lem}\lbl{S27}The solution of \eqref{sect1} is unique.
\end{lem}
\begin{proof}\,
Let $(\phi,\psi)$ and $(\phi_1,\psi_1)$ be two arbitrary solutions of \eqref{sect1}. We are going to show that
\bes\lbl{geq}
(\phi(s),\psi(s))\geq (\phi_1(s),\psi_1(s)) \mbox{ for } s\in\mathbb R.
\ees
Note that if we are able to prove \eqref{geq}, then the same argument can also be used to show $(\phi_1,\psi_1)\geq (\phi,\psi)$. Hence uniqueness will follow if we can show
\eqref{geq}.
For $s\in\mathbb R$ and $\xi\geq 0$, define
\bess
\underline{\phi}(s)=\underline \phi^\xi(s):=\phi_1(s-\xi),
\quad\underline{\psi}(s)=\underline{\psi}^\xi(s):=\psi_1(s-\xi).
\eess
We claim that there exists a constant $\xi_0>0$ such that, for every $\xi\geq\xi_0$
\bes\lbl{xi_0}
(\underline{\phi}^\xi(s),\underline{\psi}^\xi(s))\leq(\phi(s),\psi(s))\mbox{ for all }s\in\mathbb{R}.
\ees
Since $\psi_1(-\infty)=0<\psi(0)$, there exists $\xi_1>0$ large enough such that
$\psi_1(-\xi_1)\leq\psi(0)$. Then
$\underline{\psi}^\xi(0)=\psi_1(-\xi)\leq\psi(0)$ for all $\xi\geq\xi_1$, and $\psi(s)$, $\underline{\psi}^\xi(s)$ satisfy
\bess
\begin{aligned}
&d\psi''-c\psi'+r\psi(\psi-1)=0= d\underline{\psi}''-c\underline{\psi}'+r\underline{\psi}(\underline{\psi}-1),\quad s\leq0,\\
&\psi(-\infty)=\underline{\psi}(-\infty)=0,\quad \psi(0)\geq\underline{\psi}(0).
\end{aligned}
\eess
Let $u_1(s):=1-\underline{\psi}(-s)$ and $u_2(s):=1-\psi(-s)$. Then $u_1$ and $u_2$ satisfy
\bess\left\{
\begin{aligned}
&du_2''+cu_2'+ru_2(1-u_2)=0=du_1''+cu_1'+ru_1(1-u_1),\quad s\geq0,\\
&u_2(\infty)=u_1(\infty)=1,\quad u_2(0)\leq u_1(0).
\end{aligned}\right.
\eess
By Lemma \ref{S21}, we deduce that $u_2(s)\leq u_1(s)$ for all $s\geq0$. We thus obtain
\bes\lbl{s<0}
\mbox{ $\underline{\psi}^\xi(s)\leq\psi(s)$ for all $s\leq0$ and $\xi\geq \xi_1$.}
\ees
Applying Lemma \ref{S26}, we can find $\xi_2>\xi_1$ and $s_0\gg1$ such that
$$
\underline{\phi}^{\xi_2}(s) \leq \phi(s) \mbox{ for all } s\geq s_0.
$$
Denote $\xi_0=\xi_2+s_0$. Since $\phi_1$ is nondecreasing in $\mathbb{R}$ and is identically 0 in $(-\infty, 0]$, it follows that
\bes\lbl{R}
\underline{\phi}^\xi(s)\leq\phi(s) \mbox{ for } s\in\mathbb{R},\; \xi\geq\xi_0.
\ees
Similarly, for $s>0$, $\psi(s)$ and $\underline{\psi}^\xi(s)$ satisfy
\bess\left\{
\begin{aligned}
&-d\psi''+c\psi'=r(1-\psi)
(h\phi-\psi),\quad s>0.\\
&-d\underline{\psi}''+c\underline{\psi}'=r(1-\underline{\psi})
(h\underline{\phi}^\xi-\underline{\psi})\leq r(1-\underline{\psi})
(h\phi-\underline{\psi}),\quad s>0,\\
&\psi(\infty)=\underline{\psi}(\infty)=hu^\ast,\quad \psi(0)\geq\underline{\psi}(0).
\end{aligned}
\right.
\eess
Let $1-\underline{\psi}=v_1$ and $1-\psi_1=v_2$.
Then $v_1$ and $v_2$ satisfy
\bess\left\{
\begin{aligned}
&dv_2''-cv_2'=-rv_2
(1-h\phi-v_2),\quad s>0.\\
&dv_1''-cv_1'\leq r(1-\underline{\psi})
(h\phi-\underline{\psi})= -rv_1
(1-h\phi-v_1),\quad s>0,\\
&v_2(0)\leq v_1(0),\quad v_2(\infty)=v_1(\infty)=1-hu^\ast.
\end{aligned}\right.
\eess
Using Lemma \ref{S21} again, we have $v_2(s)\leq v_1(s)$ for all $s\geq0$ and $\xi\geq\xi_0$, and hence
\bes\lbl{s>0}
\mbox{ $\psi(s)\geq\underline{\psi}^\xi(s)$ for all $s>0$ and $\xi\geq\xi_0$.}
\ees
Combining \eqref{s>0}, \eqref{s<0} and \eqref{R}, we immediately obtain \eqref{xi_0}.
Define
\bess
\bar{\zeta}:=\inf\{\zeta_0>0:(\phi(s), \psi(s))\geq(\phi_1(s-\zeta), \psi_1(s-\zeta) \mbox{ for } s\in\mathbb{R},\quad\forall \zeta\geq \zeta_0\}.
\eess
By \eqref{xi_0}, $\bar\zeta$ is well defined. Since $\phi(0)=0<\phi_1(-\zeta)$ for $\zeta>0$, we have $\bar\zeta\geq 0$.
Clearly,
$$
(\phi(s), \psi(s))\geq(\phi_1(s-\bar{\zeta}),\psi_1(s-\bar{\zeta})) \mbox{ for } s\in\mathbb{R}.
$$
If $\bar{\zeta}=0$, then the above inequality already yields \eqref{geq}, and the proof is finished.
Suppose $\bar{\zeta}>0$. We are going to derive a contradiction. To simplify notations we write
\[
(\phi_{\bar\zeta}(s),\psi_{\bar\zeta}(s))=(\phi_1(s-\bar \zeta), \psi_1(s-\bar\zeta)),
\]
and set
\[
P(s):=\phi(s)-\phi_{\bar{\zeta}}(s),\;\; Q(s):=\psi(s)-\psi_{\bar{\zeta}}(s).
\]
Then the nonnegative functions $ P$ and $ Q$ satisfy
\bes\left\{\begin{aligned}
& P''-c P'+(1-k-\phi-\phi_{\bar{\zeta}}+k\psi_{\bar{\zeta}}) P+k\phi Q=0,&s>\bar{\zeta},\\
&d Q''-c Q'+r(\psi+\psi_{\bar{\zeta}}-1-h\psi) Q+rh(1-\psi) P=0,&s\in\mathbb{R},\\
& P(0)= P(\infty)= Q(-\infty)= Q(\infty)=0.
\end{aligned}\right.\lbl{ff10}
\ees
The strong maximum principle implies that $ P(s)>0$ for $s\geq\bar\zeta$ and $ Q(s)>0$ for $s\in\mathbb{R}$. Rewrite \eqref{ff10} as
\bess\left\{\begin{aligned}
& P''-c P'-u^\ast P+ku^\ast Q+\epsilon_1(s) P+\epsilon_2(s) Q=0,&s>\bar\zeta,\\
&d Q''-c Q'-rv^\ast Q+rhv^\ast P+\epsilon_3(s) P+\epsilon_4(s) Q=0,&s\in\mathbb{R},
\end{aligned}\right.\lbl{f10}
\eess
where
\bess\begin{aligned}
&\epsilon_1=1-k-\phi-\phi_{\bar{\zeta}}+k\psi_{\bar{\zeta}}+u^\ast,\quad\epsilon_2=k\phi-ku^\ast,\\
&\epsilon_3=rh(1-\psi)-rhv^\ast,\quad\epsilon_4=r(\psi+\psi_{\bar{\zeta}}-1-h\psi)+rv^\ast.
\end{aligned}\eess
By Lemma \ref{S26}, $\epsilon_i(s)\rightarrow0$ exponentially as $s\rightarrow\infty$ for $i=1,2,3,4$. We may now
repeat the proof process of Lemma \ref{S26} to obtain
$$( P(s), Q(s))=(\bar{C}_1+o(1),\bar{C}_2+o(1))e^{\hat{\mu}_2s}\mbox{ as }s\rightarrow\infty,$$
where $\bar{C}_1,\; \bar C_2$ are positive constants. By Lemma \ref{S26}, there are positive constants $C_*$ and $C$ such that
\bess\begin{aligned}
&(\phi(s),\psi(s))=(u^\ast,hu^\ast)-C_\ast(m+o(1),n+o(1))e^{\hat{\mu}_2s}\mbox{ as }s\rightarrow\infty,\\
&(\phi_{\bar\zeta}(s),\psi_{\bar\zeta}(s))=(u^\ast,hu^\ast)-C(m+o(1),n+o(1))e^{\hat{\mu}_2s}\mbox{ as }s\rightarrow\infty,
\end{aligned}\eess
which lead to
$$\left(m(Ce^{-\hat{\mu}_2\bar{\zeta}}-C_\ast),
n(Ce^{-\hat{\mu}_2\bar{\zeta}}-C_{\ast})\right)=(\bar{C}_1,\bar{C}_2)>(0,0).$$
Therefore, there exists $\epsilon_0>0$ sufficiently small so that for any $\epsilon\in(0,2\epsilon_0]$,
$$\left(m(Ce^{-\hat{\mu}_2(\bar{\zeta}-\epsilon)}-C_{\ast}),
n(Ce^{-\hat{\mu}_2(\bar{\zeta}-\epsilon)}-C_{\ast})\right)>(0,0).$$
It follows that, for all large $s$, say $s\geq M>\bar\zeta$, we have
$$(\phi(s),\psi(s))\geq (\phi_1(s-\bar{\zeta}+\epsilon), \psi_1(s-\bar{\zeta}+\epsilon)) \; (\forall\epsilon\in(0,\epsilon_0]).$$
Since $( P(s), Q(s))>(0,0)$ for $s\in[\bar\zeta,M],$
by continuity, we can find ${\epsilon_1}\in(0,\epsilon_0]$ such that, for every $\epsilon\in (0,\epsilon_1]$,
\[
(\phi(s),\psi(s))\geq (\phi_1(s-\bar{\zeta}+{\epsilon}), \psi_1(s-\bar{\zeta}+{\epsilon}))\mbox{ for } s\in[ \bar\zeta, M].
\]
Hence
\bes\lbl{phi-psi-half}
(\phi(s),\psi(s))\geq (\phi_1(s-\bar{\zeta}+{\epsilon}), \psi_1(s-\bar{\zeta}+{\epsilon}))\mbox{ for } s\geq \bar\zeta,\; \epsilon\in (0,\epsilon_1].
\ees
Since $\phi(\bar\zeta)>0=\phi_1(0)$, by continuity, there exists $\epsilon_2\in (0,\epsilon_1]$ such that
$\phi(\bar\zeta-\epsilon)\geq \phi_1(\epsilon)$ for $\epsilon\in (0,\epsilon_2]$. It follows that
\[
\phi(s)\geq \phi_1(s-\bar\zeta+\epsilon) \mbox{ for } s\in [\bar\zeta-\epsilon,\bar\zeta].
\]
Since $\phi_1(s-\bar\zeta+\epsilon)\equiv 0 $ for $s\leq \bar\zeta-\epsilon$, the above inequality holds for all $s\in(-\infty,\bar\zeta]$.
This and \eqref{phi-psi-half} imply
\bes\lbl{phi-R}
\phi(s)\geq \phi_1(s-\bar\zeta+\epsilon) \mbox{ for } s\in \mathbb R,\;\epsilon\in (0,\epsilon_2].
\ees
Denote
\[
(\phi_\epsilon(s),\psi_\epsilon(s)):=(\phi_1(s-\bar\zeta+\epsilon),\psi_1(s-\bar\zeta+\epsilon)).
\]
We obtain, for any fixed $\epsilon\in (0,\epsilon_2]$,
\[
-d\psi_\epsilon''+c\psi'_\epsilon=r(1-\psi_\epsilon)(h\phi_\epsilon-\psi_\epsilon)\leq r(1-\psi_\epsilon)(h\phi-\psi_\epsilon)
\mbox{ for } s\in \mathbb R.
\]
Moreover, $\psi_\epsilon(-\infty)=0,\; \psi_\epsilon(\infty)=hu^*$.
Hence $u_2(s):=1-\psi_\epsilon(-s)$ satisfies
\[
d u_2''+cu_2'+r u_2(1-h\phi(-s)-u_2)\leq 0 \mbox{ for } s\in\mathbb R,\; u_2(\infty)=1.
\]
Since $u_1(s):=1-\psi(-s)$ satisfies
\[
d u_1''+cu_1'+r u_1(1-h\phi(-s)-u_1)= 0 \mbox{ for } s\in\mathbb R,\; u_1(\infty)=1
\]
and $u_1(-\bar \zeta)\leq u_2(-\bar\zeta)$, we may apply Lemma \ref{S21} to conclude that
$u_1(s)\leq u_2(s)$ for $s\geq -\bar\zeta$, i.e., $\psi(s)\geq \psi_\epsilon(s)$ for $s\leq\bar\zeta$.
Combining this with \eqref{phi-psi-half}, we obtain
\[
\psi(s)\geq \psi_\epsilon(s)=\psi_1(s-\bar\zeta+\epsilon) \mbox{ for } s\in\mathbb R,\;\epsilon\in (0,\epsilon_2].
\]
This and \eqref{phi-R} clearly contradict the definition of $\bar\zeta$.
Hence the case $\bar{\zeta}>0$ can not happen, and the proof is complete.
\end{proof}
Next we will make use of problem \eqref{entirewave0}. Setting $\tilde{\Phi}(s):=\Phi(s),\tilde{\Psi}(s):=1-\Psi(s)$, we may change
\eqref{entirewave0} to the following cooperative system
\bes\left\{
\begin{aligned}
&\tilde{\Phi}''-c\tilde{\Phi}'+\tilde{\Phi}(1-k-\tilde{\Phi}+k\tilde{\Psi})=0,\; \tilde \Phi'>0, \quad s\in\mathbb{R},\\
&d\tilde{\Psi}''-c\tilde{\Psi}'+r(1-\tilde{\Psi})(h\tilde{\Phi}-\tilde{\Psi})=0,\; \tilde \Psi'>0, \quad s\in\mathbb{R},\\
&(\tilde{\Phi},\tilde{\Psi})(-\infty)=(0,0),\quad(\tilde{\Phi},\tilde{\Psi})(\infty)=(u^\ast,hu^\ast ).
\end{aligned}
\right.\lbl{sect1entire}\ees
From Proposition \ref{F6}, we know that there exists $c_*\geq2\sqrt{1-k}$ such that \eqref{sect1entire} possesses a solution if and only if $c\geq c_*$.
In what follows, we shall show $c_0^\ast=c_{\ast}$ and \eqref{sect1} has no solution for $c\geq c_{\ast}$.
\begin{lem}\lbl{aa1} $c_0^\ast\geq c_{\ast}$.
\end{lem}
\begin{proof}\,
Let $(\Phi_0,\Psi_0)$ be a solution of \eqref{sect1entire} with $c=c_{\ast}$. It is easily checked that $(0,0,0,0)$ is a saddle equilibrium point of the ODE system satisfied by $(\Phi_0,\Phi_0',\Psi_0,\Psi'_0)$. It follows that
\[
\Phi_0,\Phi_0',\Psi_0,\Psi'_0\to 0 \mbox{ exponentially as $s\to-\infty$}.
\]
Following the idea in the proof of Lemma \ref{S26}, we rewrite the equation satisfied by $\Phi_0$ as
\[
\Phi_0''-c_*\Phi_0'+(1-k)\Phi_0+\epsilon(s)\Phi_0=0
\]
with
\[\epsilon(s):=k\Psi_0(s)-\Phi_0(s)\to 0 \mbox{ exponentially as $s\to-\infty$},
\]
and view it as a perturbed linear equation to
\[
\Phi''-c_*\Phi'+(1-k)\Phi=0.
\]
Using the fundamental solutions of this latter equation we see that, as
$s\rightarrow-\infty$, the asymptotic behaviour of $(\Phi_0, \Phi_0')$ is given by
\bes\lbl{asymp1}
(\Phi_0(s), \Phi_0'(s))=\left\{\begin{array}{ll}\medskip
\displaystyle
(1, \alpha_1)k_0 e^{\alpha_1 s}(1+o(1)), & \mbox{ when } c_*>2\sqrt{1-k},\\
(1,\alpha_1)k_0|s|e^{\alpha_1 s}(1+o(1))&\\
\hspace{.2cm} \mbox{ or } (1,\alpha_1)k_0e^{\alpha_1 s}(1+o(1)),& \mbox{ when } c_*=2\sqrt{1-k}
\end{array}\right.
\ees
for some $k_0>0$, $\alpha_1\in\left\{\frac 12\left(c_{\ast}+\sqrt{c_{\ast}^2-4(1-k)}\right), \frac 12\left(c_{\ast}-\sqrt{c_{\ast}^2-4(1-k)}\right)\right\}$.
Fix $\epsilon\in(0, k/\alpha_1)$ small. In view of \eqref{asymp1}, there exists a constant $M_0<0$ such that
\bes\lbl{qq1}\frac{\Phi_0'(s)}{\Phi_0(s)}\geq\frac{\alpha_1}{2},\quad\max\{\Phi_0(s),
\Psi_0(s)\}<\min\left\{\frac{\epsilon\alpha_1}{4k},1-k\right\}\mbox{ for }s<M_0.\ees
Next, we prove that system \eqref{sect1} has a solution for $c=c_{\ast}-\epsilon$. To this end, we will treat the cases $c_{\ast}>2\sqrt{1-k}$ and $c_{\ast}=2\sqrt{1-k}$ separately.
{\bf Case 1}: $c_{\ast}>2\sqrt{1-k}$.
Introduce an auxiliary function
$$p_1(s)=0\mbox{ for }s\geq M_0,\quad p_1(s)=e^{\beta_1 s}\mbox{ for }s\leq M_0-1$$
and for $s\in(M_0-1,M_0)$, $p_1(s)>0$, $p_1'(s)\leq0$, where $\beta_1=\left(c_{\ast}-\sqrt{c_{\ast}^2+2dr}\right)/(2d)<0$
and $M_0$ is given by \eqref{qq1}. Moreover, $p_1(s)$ is $C^2$ everywhere. Define
\[
\psi_1(s):=\Psi_0(s)-\epsilon_1p_1(s),
\]
where the positive constant $\epsilon_1$ will be determined later.
We now calculate
\bes\begin{aligned}\lbl{up1}
&d\psi_1''-(c_{\ast}-\epsilon)\psi_1'+r(1-\psi_1)(h\Phi_0-\psi_1)\\
=&r(1-\psi_1)(h\Phi_0-\psi_1)-r(1-\Psi_0)(h\Phi_0-\Psi_0)-d\epsilon_1p_1''-\epsilon\epsilon_1 p_1'+\epsilon\Psi_0'+c_{\ast}\epsilon_1p_1'\\
=&\epsilon\Psi_0'+\epsilon_1[rp_1(1+h\Phi_0-2\Psi_0+\epsilon_1p_1)-dp_1''-\epsilon p_1'+c_{\ast}p_1'].
\end{aligned}\ees
Hence we can fix $\epsilon_1>0$ sufficiently small so that, for $s\in[M_0-1,M_0]$,
\bess\begin{aligned}\lbl{up2}d\psi_1''-(c_{\ast}-\epsilon)\psi_1'
+r(1-\psi_1)(h\Phi_0-\psi_1)>0\end{aligned}\eess
and
$$\psi_1'(s)>0,\quad\psi_1(s)>0.$$
By the definition of $p_1(s)$ for $s\leq M_0-1$, clearly $\psi_1'(s)>0$ for $s\leq M_0-1$, and $\psi_1(s)\rightarrow-\infty$ as $s\rightarrow-\infty$.
Hence there exists a unique constant $M_1<M_0-1$ such that $\psi_1(M_1)=0$. We define
\bess
\underline{\Phi}(s)=\Phi_0(s)\mbox{ for }s\geq M_1,
\quad\underline{\Psi}(s)=\left\{\begin{array}{ll}\medskip
\displaystyle
\Psi_0(s)-\epsilon_1p_1(s),&s>M_1,\\
0,&s\leq M_1.
\end{array}\right.
\eess
For $s\in(M_1,M_0-1)$, by the choice of $\Psi_0\leq1/4$ in \eqref{qq1} for $s\leq M_0$, we have
\bess\begin{aligned}
&rp_1(1+h\Phi_0-2\Psi_0+\epsilon_1p_1)-dp_1''-\epsilon p_1'+c_{\ast}p_1'\\
>&rp_1(1+h\Phi_0-2\Psi_0)-dp_1''+c_{\ast}p_1'\\
>&-dp_1''+c_{\ast}p_1'+\frac{r}{2}p_1=0.
\end{aligned}\eess
Therefore, it follows from \eqref{up1} that
$$d\underline{\Psi}''-(c_{\ast}-\epsilon)\underline{\Psi}'+r(1-\underline{\Psi})(h\underline{\Phi}-\underline{\Psi})>0,\quad s\in(M_1,M_0-1).$$
Since $(\underline{\Phi},\underline{\Psi})=(\Phi_0,\Psi_0)$ for $s>M_0$, and $\underline{\Psi}(s)=0$ for $s\leq M_1$, it is easy to verify that
for any smooth extension of $\underline\Phi(s)$ to $s\leq M_1$ satisfying $\underline{\Phi}(s)\geq0$ in $(-\infty, M_1)$, we have
$$d\underline{\Psi}''-(c_{\ast}-\epsilon)\underline{\Psi}'
+r(1-\underline{\Psi})(h\underline{\Phi}-\underline{\Psi})\geq0 $$
for all $s\in\mathbb{R}$. Moreover, $\underline{\Psi}'(M_1-)=0\leq \underline{\Psi}'(M_1+)$.
In view of \eqref{qq1} and $\psi_1(M_1)=0$, we have $\epsilon_1p_1(M_1)=\Psi_0(M_1)<\frac{\epsilon\alpha_1}{4k}.$
For $s\in(M_1,M_0)$, by \eqref{qq1}, we can fix $\epsilon_1>0$ sufficiently small so that
\bess
\begin{aligned}
\Phi_0''-(c_{\ast}-\epsilon)\Phi_0'+\Phi_0(1-k-\Phi_0+k\underline{\Psi})
=\epsilon\Phi_0'-k\epsilon_1p_1\Phi_0\geq(\frac{\alpha_1}{2}-k\epsilon_1p_1(M_1))\Phi_0
\geq0.
\end{aligned}
\eess
For $s\leq M_1$, we choose $\epsilon>0$ sufficiently small so that the quadratic equation
$\alpha^2-(c_*-\epsilon)\alpha+(1-k)=0$ has a root $\alpha_\epsilon \in (\alpha_1/2, \alpha_1)$.
Define
$$p_2(s)=0\mbox{ for }s\geq M_1,\quad p_2(s)=e^{\alpha_\epsilon s}\mbox{ for }s\leq M_1-1$$
and for $s\in[M_1-1,M_1]$, we define $p_2(s)$ so that $p_2(s)>0$, and $p_2(s)$ is $C^2$ everywhere.
We define
$$\underline{\Phi}(s)=\Phi_0(s)-\epsilon_2p_2(s),$$
with $\epsilon_2>0$ to be determined.
Since $\alpha_\epsilon<\alpha_1$, by \eqref{asymp1} we can find $M_1^\epsilon<M_1-1$ such that
$$\Phi_0'(s)>\alpha_\epsilon\Phi_0(s)\mbox{ for }s\leq M_1^\epsilon.$$
It follows that, for $s\leq M_1^\epsilon$,
\bess
\underline{\Phi}'(s)=\Phi_0'(s)-\epsilon_2p_2'(s)>\alpha_\epsilon\Phi_0(s)
-\epsilon_2\alpha_\epsilon p_2(s)=\alpha_\epsilon\underline{\Phi}(s).
\eess
Recall that $\alpha_\epsilon>\alpha_1/2$. We now choose $\epsilon_2$ sufficiently small such that, for $x\in[M_1^\epsilon,M_1]$,
$$\underline{\Phi}(s)>0,\quad\underline{\Phi}'(s)>0$$
and
\bess
\begin{aligned}
&\underline{\Phi}''-(c_{\ast}-\epsilon)\underline{\Phi}'+\underline{\Phi}(1-k
-\underline{\Phi}+k\underline{\Psi})\\
=&-k\Phi_0\Psi_0+\epsilon\Phi_0'+\epsilon_2[c_{\ast}p_2'-p_2''-\epsilon p_2'
-p_2(1-k-2\Phi_0+\epsilon_2p_2)]\\
>&-k\Phi_0\Psi_0+\epsilon\alpha_\epsilon\Phi_0+\epsilon_2[c_{\ast}p_2'-p_2''-\epsilon p_2'
-p_2(1-k-2\Phi_0+\epsilon_2p_2)]\\
>&\frac{\epsilon}{4}\alpha_1\Phi_0+\epsilon_2[c_{\ast}p_2'-p_2''-\epsilon p_2'
-p_2(1-k-2\Phi_0+\epsilon_2p_2)]>0.
\end{aligned}
\eess
Due to $\alpha_\epsilon<\alpha_1$ and \eqref{asymp1}, we easily deduce $\lim\limits_{s\rightarrow-\infty}\frac{\Phi_0(s)}{e^{\alpha_\epsilon s}}=0.$
It follows that $$\underline{\Phi}(x)=e^{\alpha_\epsilon s}(\frac{\Phi_0(s)}{e^{\alpha_\epsilon s}}-\epsilon_2)<0$$ for all large negative $s.$
Since $\underline{\Phi}(M_1^\epsilon)>0$, by continuity, there exists $M_2<M_1^\epsilon$ such that
$$\underline{\Phi}(M_2)=0,\quad\underline{\Phi}(s)>0\mbox{ for }s\in(M_2,M_1^\epsilon].$$
Thus for $s\in(M_2,M_1^\epsilon)$, we have $-2\Phi_0+\epsilon_2p_2<-\Phi_0<0$ and
\bess
\begin{aligned}
&\underline{\Phi}''-(c_{\ast}-\epsilon)\underline{\Phi}'+\underline{\Phi}(1-k
-\underline{\Phi}+k\underline{\Psi})\\
>&\frac{\epsilon}{4}\alpha_1\Phi_0+\epsilon_2[(c_{\ast}-\epsilon)p_2'-p_2''
-p_2(1-k-2\Phi_0+\epsilon_2p_2)]\\
\geq&\frac{\epsilon}{4}\alpha_1\Phi_0+\epsilon_2[(c_{\ast}-\epsilon)p_2'-p_2''
-(1-k)p_2]=\frac{\epsilon}{4}\alpha_1\Phi_0>0.
\end{aligned}
\eess
Define
\bess
\quad\underline{\Phi}(s)=\left\{\begin{array}{ll}\medskip
\displaystyle
\Phi_0-\epsilon_2p_2(s),&s\geq M_2,\\
0,&s\leq M_2.
\end{array}\right.
\eess
It is easy to see that
$$\underline{\Phi}''-(c_{\ast}-\epsilon)\underline{\Phi}'+\underline{\Phi}(1-k
-\underline{\Phi}+k\underline{\Psi})\geq0$$
for all $s\in\mathbb{R}$. Moreover, $\underline{\Phi}'(M_2-)=0\leq \underline{\Phi}'(M_2+)$.
Finally, we accomplish the proof by the upper and lower solution argument. Define
\[
(\underline{\phi}(s),\underline{\psi}(s)):=(\underline{\Phi}(s+M_2),\underline{\Psi}(s+M_2)).
\]
Then $(\underline{\phi}(s),\underline{\psi}(s))$ is a lower solution for \eqref{sect1} with $c=c_{\ast}-\epsilon$ associated with $\mathcal R:=[0,u^*]\times [0, hu^*]$. Moreover, it is easy to see that $(\overline{\phi}(s),\overline{\psi}(s))$ is a upper solution for \eqref{sect1} with $c=c_{\ast}-\epsilon$
associated with $\mathcal R$, where $(\overline{\phi}(s),\overline{\psi}(s))$ is given by \eqref{upfl}.
We next check that
\[
\mbox{$\sup_{t\leq s}\big(\underline\phi(t), \underline\psi(t)\big)\leq \big(\overline\phi(s),\overline\psi(s)\big)$ holds for all $s\in\mathbb R$.}
\]
By the monotonicity of $\underline \phi$ and $\underline \psi$, it suffices to show
\bes\lbl{lo<up}
\mbox{$\big(\underline\phi(s), \underline\psi(s)\big)\leq \big(\overline\phi(s),\overline\psi(s)\big)$ for all $s\in\mathbb R$.}
\ees
For $s>0$,
\[
\underline \psi(s)\leq \Psi_0(s+M_2)<hu^*=\overline \psi(s).
\]
Since $M_1-M_2>0$ and $\underline \psi(s)=0$ for $s<M_1-M_2$, we thus see that
\[
\underline\psi(s)\leq \overline\psi(s) \mbox{ for all } s\in\mathbb R.
\]
Clearly $\underline \phi(s)=0=\overline\phi(s)$ for $s\leq 0$. For $s>0$, due to $\underline\psi(s)\leq \underline\psi(\infty)=\Psi_0(\infty)=hu^*<1$,
we have
\[
0\leq \underline \phi''-(c_*-\epsilon)\underline \phi'+\underline \phi(1-k-\underline \phi+k\underline \psi)
\leq \underline \phi''+\underline \phi(1-\underline \phi).
\]
Moreover, $\underline\phi(\infty)=\Phi_0(\infty)=u^*=\overline \phi(\infty)$. Hence we can apply Lemma \ref{S21} to conclude that
$\underline \phi(s)\leq \overline \phi(s)$ for $s>0$.
We have thus proved \eqref{lo<up}. Clearly $\underline \varphi(s)\not\equiv 0$ and the nonlinearity functions in \eqref{sect1} satisfy
${\bf (A_1), \; (A_2), \; (A_3)}$.
We may now apply Proposition \ref{s7} to conclude that \eqref{sect1} (with $\tilde\phi'>0$ and $\tilde\psi'>0$ relaxed to
$\tilde\phi'\geq 0$ and $\tilde\psi'\geq 0$) has a solution $(\phi,\psi)$ with $c=c_{\ast}-\epsilon$, which would imply $c_0^{\ast}\geq c_{\ast}-\epsilon$ if we can further prove $\phi'> 0$ and $\psi'> 0$. But these strict inequalities follow easily from the strong maximum principle applied to the coorperate system satisfied by $(\phi', \psi')$. By the arbitrariness of $\epsilon$, it follows $c_0^\ast\geq c_{\ast}$.
{\bf Case 2}: $c_{\ast}=2\sqrt{1-k}$.
It follows from Lemma \ref{S22} that the following problem
\bess\left\{\begin{array}{ll}\medskip
\displaystyle
\tilde{\chi}''-(c_{\ast}-\epsilon)\tilde{\chi}'+\tilde{\chi}(1-k-\tilde{\chi})=0,\quad0<s<\infty,\\
\tilde{\chi}(0)=0
\end{array}\right.\lbl{}
\eess
has a unique strictly increasing solution $\tilde{\chi}$ satisfying $\tilde \chi(\infty)=1-k$.
Define
\bess
\underline{\Phi}(s):=\left\{\begin{array}{ll}\medskip
\displaystyle
0,&-\infty<s<0,\\
\delta\tilde{\chi}(s),&0\leq s<\infty,
\end{array}\right.
\quad\underline{\Psi}(s):=0,
\eess
with $\delta>0$ small such that $\underline{\Phi}(s)\leq\overline{\phi}(s)$ for $s\in\mathbb{R}$, where $\overline{\phi}(s)$ is given by \eqref{upfl}.
It is easy to verify that $(\underline{\Phi}(s),\underline{\Psi}(s))$ is a lower solution of \eqref{sect1} associated with $\mathcal R$.
Moreover, it is easy to see that $(\overline{\phi}(s),\overline{\psi}(s))$ is an upper solution for \eqref{sect1} with $c=c_{\ast}-\epsilon$
associated with $\mathcal R$, and $(\underline{\Phi}(s),\underline{\Psi}(s))\leq (\overline{\phi}(s),\overline{\psi}(s))$ for $s\in\mathbb{R}$. It follows from Proposition \ref{s7} and the strong maximum principle (applied to $(\phi',\psi')$ as in Case (i) above) that \eqref{sect1} has a solution $(\phi,\psi)\in\Sigma$ with $c=c_{\ast}-\epsilon$,
which implies $c_0^{\ast}\geq c_{\ast}-\epsilon$. By the arbitrariness of $\epsilon$, it follows $c_0^\ast\geq c_{\ast}$.
\end{proof}
\begin{lem}\lbl{S28}For $c\geq c_{\ast}$, problem \eqref{sect1} has no solution.
\end{lem}
\begin{proof}\, Suppose on the contrary that for some $c\geq c_{\ast}$, \eqref{sect1} has a solution $(\tilde{\phi},\tilde{\psi})$. By Proposition \ref{F6}, the system \eqref{sect1entire} has a solution $(\tilde{\Phi},\tilde{\Psi})$ for such $c$. We are going to derive a contradiction by making use of $(\tilde{\Phi},\tilde{\Psi})$.
We note that by repeating the arguments in the proof of Lemma \ref{S26}, the monotone increasing functions $\tilde{\phi},\tilde{\psi},\tilde{\Phi}$ and $\tilde{\Psi}$ can be expanded near $\infty$ in the form \eqref{expansion1}. In view of $\tilde{\Phi}'(s)>0$ and $\tilde{\Psi}'(s)>0$, there exists some $\eta_0>0$ such that
$$(\tilde{\Phi}(s+\eta),\tilde{\Psi}(s+\eta))\geq(\tilde{\phi}(s),\tilde{\psi}(s)),\quad\forall s\geq0,\quad\eta\geq\eta_0.$$
Clearly
\bes\lbl{ew0}\tilde{\Phi}(s+\eta)>0=\tilde{\phi}(s)\mbox{ for }s<0.\ees
Now we prove that
$$\tilde{\Psi}(s+\eta)\geq\tilde{\psi}(s)\mbox{ for }s\in\mathbb{R}\mbox{ and }\eta\geq\eta_0.$$
We only need to show this for $s<0$. Denote, for $\eta\geq\eta_0$,
$$\tilde{\Phi}_\eta(s):=\tilde{\Phi}(s+\eta),\quad\tilde{\Psi}_\eta(s):=\tilde{\Psi}(s+\eta),
$$
and let
$$\hat{\Phi}_\eta(s)=1-\tilde{\Phi}_\eta(-s),\quad\hat{\Psi}_\eta(s)=1-\tilde{\Psi}_\eta(-s),\quad\psi(s)=1-\tilde{\psi}(-s).$$
Then
\bes\left\{\begin{aligned}\lbl{wave11}
&-c\hat{\Psi}_\eta'-d\hat{\Psi}_\eta''=r\hat{\Psi}_\eta(1-\hat{\Psi}_\eta-h\tilde{\Phi}_\eta)\leq r\hat{\Psi}_\eta(1-\hat{\Psi}_\eta-h\tilde{\phi}),\quad s\in\mathbb{R},\\
&-c\psi'-d\psi''=r\psi(1-\psi-h\tilde{\phi}),\quad s\in\mathbb{R},\\
&\hat{\Psi}_\eta(\infty)=1=\psi(\infty),\quad\hat{\Psi}_\eta(0)\leq\psi(0).
\end{aligned}
\right.\ees
Using Lemma \ref{S21} we deduce $\hat{\Psi}_\eta(s)\leq\psi(s)$ in $[0, \infty)$, and hence
$\tilde{\Psi}_\eta(s)\geq\tilde{\psi}(s)$ in $(-\infty,0]$.
We are now able to define
\bess\eta^\ast=\inf\left\{\eta_0\in\mathbb{R}:\quad\tilde{\Phi}_\eta(s)\geq\tilde{\phi}(s)\mbox{ in }[0,\infty), \quad\tilde{\Psi}_\eta(s)\geq\tilde{\psi}(s)\mbox{ in }\mathbb{R},\quad\forall \eta\geq\eta_0\right\}.\eess
We claim that $\eta^\ast=-\infty$. Otherwise, $\eta^\ast$ is a finite real number, and by continuity,
\bess\tilde{\Phi}_{\eta^\ast}(s)\geq\tilde{\phi}(s)\mbox{ in }[0,\infty), \quad\tilde{\Psi}_{\eta^\ast}(s)\geq\tilde{\psi}(s)\mbox{ in }\mathbb{R}.\eess
The first inequality of \eqref{wave11} still holds for $\eta=\eta^\ast$, and this inequality is strict for $s<0$ due to \eqref{ew0}. Hence $\tilde{\Psi}_{\eta^\ast}(s)\not\equiv\tilde{\psi}(s)$, and by the strong maximum principle we obtain
$$\tilde{\Psi}_{\eta^\ast}(s)>\tilde{\psi}(s)\mbox{ for }s\in\mathbb{R}.$$
We now have
\bess\begin{aligned}\lbl{wave1}
&c\tilde{\Phi}_{\eta^\ast}'-\tilde{\Phi}_{\eta^\ast}''=\tilde{\Phi}_{\eta^\ast}(1-k-\tilde{\Phi}_{\eta^\ast}
+k\tilde{\Psi}_{\eta^\ast})\geq \tilde{\Phi}_{\eta^\ast}(1-k-\tilde{\Phi}_{\eta^\ast}+k\tilde{\psi}),\quad s\in\mathbb{R}^+,\\
&c\tilde{\phi}'-\tilde{\phi}''=\tilde{\phi}(1-k-\tilde{\phi}+k\tilde{\psi}),\quad s\in\mathbb{R}^+,\\
&\tilde{\Phi}_{\eta^\ast}(0)\geq\psi(0),\quad\tilde{\Phi}_{\eta^\ast}(\infty)=\tilde{\phi}(\infty)=u^\ast.
\end{aligned}\eess
Using Lemma \ref{S21} we deduce
$$\tilde{\Phi}_{\eta^\ast}(s)\geq\tilde{\phi}(s)\mbox{ for }s\in[0,\infty).$$
We may now use the expansion of $(\tilde{\Phi}_{\eta^\ast}-\tilde{\phi},\tilde{\Psi}_{\eta^\ast}-\tilde{\psi})$ near $s=\infty$ as the proof of Lemma \ref{S27} to derive that
\bess
\tilde{\Phi}_{\eta^\ast-\epsilon}(s)\geq\tilde{\phi}(s),\quad\tilde{\Psi}_{\eta^\ast-\epsilon}(s)>\tilde{\psi}(s) \mbox{ for }s\in[0,\infty)\mbox{ and some small }\epsilon>0.
\eess
It then follows from the monotonicity of $\tilde \Phi$ and $\tilde \Psi$ that for all $\eta\geq\eta^\ast-\epsilon$,
$$\tilde{\Phi}_\eta(s)\geq\tilde{\phi}(s)\mbox{ in }[0,\infty), \quad\tilde{\Psi}_\eta(s)\geq\tilde{\psi}(s)\mbox{ in }\mathbb{R},$$
which contradicts the definition of $\eta^\ast$. Hence, $\eta^\ast=-\infty$. The fact $\eta^\ast=-\infty$ implies $\tilde{\Phi}(s+\eta)\geq\tilde{\phi}(s)\mbox{ in }[0,\infty)$ for all $\eta\in\mathbb{R}$. For any fixed $s>0$, letting $\eta\rightarrow-\infty$ and using $\tilde{\Phi}(-\infty)=0$ we obtain $0\geq \tilde{\phi}(s)$. This is a contradiction to the fact that $(\tilde{\phi},\tilde{\psi})$ is a solution of \eqref{sect1}.
\end{proof}
\begin{lem}\lbl{c_*}$c_0^*=c_*$.
\end{lem}
\begin{proof}
Lemmas \ref{aa1} and \ref{S25} imply that \eqref{sect1} has a solution for every $c\in [0, c_0^*)$. Therefore Lemma \ref{S28}
implies $c_*\geq c_0^*$. In view of Lemma \ref{aa1}, we must have $c_0^*=c_*$.
\end{proof}
\begin{lem}\lbl{aa0}$c_0^\ast\leq2\sqrt{u^\ast}$.
\end{lem}
\begin{proof}\, Suppose on the contrary that $c_0^*>2\sqrt{u^*}$. Then system \eqref{sect1} has a solution for some $c>2\sqrt{u^\ast}$. The monotonicity of $(\tilde{\phi}(s),\tilde{\psi}(s))$ implies that $(\tilde{\phi}(s),\tilde{\psi}(s))\leq(u^\ast,hu^\ast)$ on $\mathbb{R}^+$.
We claim that
$\tilde{\phi}'(s)$ and $\tilde{\phi}''(s)$ are uniformly bounded on $\mathbb{R}^+$. Let $\beta=\max\{1+k,r(1+h)\}$; it follows from Lemma 6.1 and \eqref{s002} that
\bess\begin{aligned}
|\tilde{\phi}'(s)|&=\left|\frac{1}{\lambda_{2}-\lambda_{1}}\left[\int_{0}^sK_{1s}(\xi,s)\tilde{\phi}(\xi)(\beta+1
-k-\tilde{\phi}(\xi)+k\tilde{\psi}(\xi))d\xi\right.\right.\\
&\quad\;\;\;\;\;\;\;\;\;\;+\left.\left.\int_s^{\infty}K_{2s}(\xi,s)\tilde{\phi}(\xi)(\beta+1-k
-\tilde{\phi}(\xi)+k\tilde{\psi}(\xi))d\xi\right]\right|\\
&\leq\frac{2(\beta+1+k+u^\ast+khu^\ast)u^\ast}{\lambda_{2}-\lambda_{1}}:=\hat{C}_1.
\end{aligned}\eess
Using the boundedness of $\tilde{\phi},\tilde{\psi},\tilde{\phi}'$ and \eqref{sect1}, we obtain that
\bess\begin{aligned}
|\tilde{\phi}''|&=|c\tilde{\phi}'-\tilde{\phi}(1-k-\tilde{\phi}+k\tilde{\psi})|\\
&\leq c|\tilde{\phi}'|+|\tilde{\phi}(1-k-\tilde{\phi}+k\tilde{\psi})|\\
&\leq c\hat{C}_1+u^\ast(1+k+u^\ast+khu^\ast):=\bar{C}_1.
\end{aligned}\eess
Thus $|\tilde{\phi}'|,|\tilde{\phi}''|<C$ with $C:=\max\{\hat{C}_1,\bar{C}_1\}$.
Thanks to the uniform boundedness of $\tilde{\phi},\tilde{\psi},\tilde{\phi}'$ and $\tilde{\phi}''$, the integrals $\int_0^\infty\tilde{\phi}(s)\tilde{\psi}(s)e^{-\mu s}ds$ and
$\int_0^\infty\tilde{\phi}^{(l)}e^{-\mu s}ds(l=0,1,2)$ are well defined for any $\mu>0$. In view of $c>2\sqrt{u^\ast}$, we know that
$$\mu^2-c\mu+u^\ast=0$$
has two positive roots, say $\tilde{\mu}_1,\tilde{\mu}_2$ with $0<\tilde{\mu}_1<\tilde{\mu}_2$.
Now, choosing $\mu\in(\tilde \mu_1,\tilde\mu_2)$, multiplying the first equation in
\eqref{sect1} by $e^{-\mu s}$ and integrating from $0$ to $\infty$, we obtain
\bes\begin{aligned}
&\tilde{\phi}'(0)+\int_0^\infty\tilde{\phi}^2e^{-\mu s}ds\\
=&(\mu^2-c\mu)\int_0^\infty\tilde{\phi}e^{-\mu s}ds+\int_0^\infty(1-k+k\tilde{\psi})\tilde{\phi} e^{-\mu s}ds\\
\leq&(\mu^2-c\mu+u^\ast)\int_0^\infty\tilde{\phi}e^{-\mu s}ds<0.
\end{aligned}
\lbl{nonexistence}
\ees
Since $\tilde{\phi}'(0)>0$ by the Hopf boundary Lemma, we have $\tilde{\phi}'(0)+\int_0^\infty\tilde{\phi}^2e^{-\mu s}ds>0$, which contradicts \eqref{nonexistence}.
\end{proof}
\begin{lem}\lbl{T9}Let $(\tilde{\phi}_c,\tilde{\psi}_c)$ denote the unique solution of \eqref{sect1}. Then $0\leq c_1<c_2< c_0^\ast $ implies
$$\tilde{\phi}_{c_1}'(0)>\tilde{\phi}_{c_2}'(0),\quad \tilde{\phi}_{c_1}(s)>\tilde{\phi}_{c_2}(s)\mbox{ in }\mathbb{R}^+,
\quad \tilde{\psi}_{c_1}(s)>\tilde{\psi}_{c_2}(s)\mbox{ in }\mathbb{R}.$$
\end{lem}
\begin{proof} From the proof of Lemma \ref{S25} and the uniqueness of solutions to \eqref{sect1}, we have $(\tilde{\phi}_{c_2}(s),\tilde{\psi}_{c_2}(s))\leq(\overline{\phi}(s),\overline{\psi}(s))$, where $(\overline{\phi}(s),\overline{\psi}(s))$ is given by
\eqref{upfl}.
Moreover, due to $0\leq c_1<c_2$ and $\tilde\phi'_{c_2}>0$ in $\mathbb R^+$ and $\tilde\psi'_{c_2}>0$ in $\mathbb R$, we have
\bess\left\{
\begin{aligned}
&-\tilde{\phi}_{c_2}''+c_1\tilde{\phi}_{c_2}'<\tilde{\phi}_{c_2}(1-k-\tilde{\phi}_{c_2}
+k\tilde{\psi}_{c_2}),&0<s<\infty,\\
&-d\tilde{\psi}_{c_2}''+c_1\tilde{\psi}_{c_2}'< r(1-\tilde{\psi}_{c_2})(h\tilde{\phi}_{c_2}-\tilde{\psi}_{c_2}),&-\infty<s<\infty.
\end{aligned}
\right.\eess
Hence, it follows from Proposition \ref{s7} and Lemma \ref{S27} that $\tilde{\phi}_{c_1}(s)\geq\tilde{\phi}_{c_2}(s)$ in $\mathbb{R}^+$ and
$\tilde{\psi}_{c_1}(s)\geq\tilde{\psi}_{c_2}(s)$ in $\mathbb{R}$. Furthermore, the strong maximum principle yields
$\tilde{\phi}_{c_1}(s)>\tilde{\phi}_{c_2}(s)$ for $s>0$,
$\tilde{\psi}_{c_1}(s)>\tilde{\psi}_{c_2}(s)$ for $s\in\mathbb{R}$.
Let $\widetilde{\phi}=\tilde{\phi}_{c_1}-\tilde{\phi}_{c_2}$. Then
$$-\widetilde{\phi}''+c_2\widetilde{\phi}'>\widetilde{\phi}(1-k-\tilde{\phi}_{c_1}-\tilde{\phi}_{c_2}
+k\tilde{\phi}_{c_1}),\quad\widetilde{\phi}(s)>0\mbox{ for }s>0,\quad\widetilde{\phi}(0)=0.$$
By the Hopf boundary lemma, we deduce $\widetilde{\phi}'(0)>0$, that is, $\tilde{\phi}_{c_1}'(0)>\tilde{\phi}_{c_2}'(0)$.
\end{proof}
\begin{lem}\lbl{T10}Let $(\tilde{\phi}_c,\tilde{\psi}_c)$ be the unique monotone solution of \eqref{sect1}. Then the mapping $c\longmapsto(\tilde{\phi}_c,\tilde{\psi}_c)$ is continuous from $[0,c_0^\ast)$ to $C_{loc}^2([0,\infty))\times C_{loc}^2(\mathbb{R})$. Moreover,
\bess\lim_{c\rightarrow c_0^\ast-}(\tilde{\phi}_c,\tilde{\psi}_c)=(0,0)\mbox{ in }C_{loc}^2([0,\infty))\times C_{loc}^2(\mathbb{R}).\lbl{critical}\eess
\end{lem}
\begin{proof}
Suppose $\{c_i\}$ is a sequence in $[0,c_0^\ast)$ such that $c_i\rightarrow \hat{c}\in[0,c_0^\ast]$ as $i\rightarrow\infty$.
Let $(\tilde{\phi}_{c_i},\tilde{\psi}_{c_i})$ be the solution of \eqref{sect1} with $c=c_i$. We claim that $(\tilde{\phi}_{c_i},\tilde{\psi}_{c_i})$ has a subsequence that converges to $(\tilde{\phi}_{\hat{c}},\tilde{\psi}_{\hat{c}})$ in $C_{loc}^2([0,\infty))\times C_{loc}^2(\mathbb{R})$, which clearly implies the continuity of the mapping $c\longmapsto(\tilde{\phi}_c,\tilde{\psi}_c)$.
Firstly, we consider the case $\hat{c}<c_0^\ast$. Let $\bar{c}\in(\hat{c},c_0^\ast)$. Then $c_i\in[0,\bar{c})$ for all large $i$,
and without loss of generality we assume that this is the case for all $i\geq 1$. For simplicity, we denote $(\tilde{\phi}_{c_i},\tilde{\psi}_{c_i})$ by $(\tilde{\phi}_{i},\tilde{\psi}_{i})$. Rewrite equation \eqref{sect1} in the integral form of \eqref{l01} and \eqref{l02}.
Noting that $\tilde{\phi}_{i}$ and $\tilde{\psi}_{i}$ are uniformly bounded, similar arguments as in Lemma \ref{aa0} indicate that $|\tilde{\phi}_{i}'|$ and $|\tilde{\phi}_{i}''|$ are bounded for all $i$ and $s\in\mathbb{R}^+$. Moreover, by similar arguments as in the proof of Lemma \ref{aa0} again, we can prove $|\tilde{\psi}_{i}'|$ and $|\tilde{\psi}_{i}''|$ are bounded for all $i$ and $s\in\mathbb{R}$. Differentiating both sides of \eqref{sect1} with respect to $s$, applying the uniform boundedness of $\tilde{\phi}_{i}^{(j)}$ and $\tilde{\psi}_{i}^{(j)}(j=0,1,2)$, we have $|\tilde{\phi}_{i}'''|$ and $|\tilde{\psi}_{i}'''|$ are bounded for $s\in\mathbb{R}^+$ and $s\in\mathbb{R}$, respectively. Hence, $\{\tilde{\phi}_{i}^{(j)}\}$ and $\{\tilde{\psi}_{i}^{(j)}\}(j=0,1,2)$ are uniformly bounded and equi-continuous for $s\in\mathbb{R}^+$ and $s\in\mathbb{R}$, respectively. Using Arzela-Ascoli's theorem, the nested subsequence argument and Lebesgue's dominated convergence theorem, there is a subsequence $c_{i_k}$ of $\{c_i\}$ such that
$(c_{i_k},\tilde{\phi}_{i_k}^{(j)},\tilde{\psi}_{i_k}^{(j)})\rightarrow(\hat{c},\hat{\phi}^{(j)},\hat{\psi}^{(j)})$
uniformly as $k\rightarrow\infty$ in $C_{loc}^2([0,\infty))\times C_{loc}^2(\mathbb{R})$. Moreover, $(\hat{\phi},\hat{\psi})$ solves \eqref{sect1} with $c=\hat{c}$, except that we only have $\hat{\phi}'\geq0$ and $\hat{\psi}'\geq0$.
Using Lemma \ref{T9}, the required asymptotic behavior of $(\hat{\phi},\hat{\psi})$ at $\pm\infty$ follows from $$(\tilde{\phi}_{\bar{c}},\tilde{\psi}_{\bar{c}})\leq(\hat{\phi},\hat{\psi})\leq(\overline{\phi},\overline{\psi}).$$
Applying the strong maximum principle to the system satisfied by $(\hat{\phi}',\hat{\psi}')$, we deduce $\hat{\phi}'>0$ in $[0,\infty)$ and $\hat{\psi}'>0$ in $\mathbb{R}$. Thus $(\hat{\phi},\hat{\psi})$ is a solution of \eqref{sect1} with $c=\hat{c}$.
By uniqueness, we have $(\hat{\phi},\hat{\psi})=(\tilde{\phi}_{\hat{c}},\tilde{\psi}_{\hat{c}})$.
It remains to consider the case $\hat{c}=c_0^\ast$. Repeating the above arguments, we conclude that, passing to a subsequence,
$$(c_{i_k},\tilde{\phi}_{i_k},\tilde{\psi}_{i_k})\rightarrow( c_0^\ast ,\hat{\phi}_\ast,\hat{\psi}_\ast)\mbox{ in }C_{loc}^2([0,\infty))\times C_{loc}^2(\mathbb{R})\mbox{ as }k\rightarrow\infty$$
and $(\hat{\phi}_\ast,\hat{\psi}_\ast)$ solves \eqref{sect1} with $c= c_0^\ast $, except that we only have $\hat{\phi}_\ast'\geq0$ in $[0,\infty)$ and $\hat{\psi}_\ast'\geq0$ in $\mathbb{R}$. If $\hat{\phi}_\ast\equiv0$, then $\hat{\psi}_\ast$ satisfies
$$d\hat{\psi}_\ast''- c_0^\ast \hat{\psi}_\ast'=r\hat{\psi}_\ast(1-\hat{\psi}_\ast).$$
Let $\tilde{\psi}_\ast=1-\hat{\psi}_\ast$. Then $\tilde{\psi}_\ast$ satisfies
$$-d\tilde{\psi}_\ast''\geq -d\hat{\psi}_\ast''+ c_0^\ast \hat{\psi}_\ast'= r\tilde{\psi}_\ast(1-\tilde{\psi}_\ast).$$
For large $L>1$, assume $u_L$ is the unique positive solution of
$$-du''=ru(1-u),\quad u(\pm L)=0.$$
It is well known that $u_L\rightarrow1$ in $C_{loc}^2(\mathbb{R})$ as $L\rightarrow\infty$. By Lemma 2.1 of \cite{DM}, we have $u_L\leq\tilde{\psi}_\ast\leq1$ in $[-L,L]$. Letting $L\rightarrow\infty$ we obtain $\tilde{\psi}_\ast=1$, as we wanted.
Next, assume that $\hat{\phi}_\ast\not\equiv0$. Let $(\hat{\phi}_\ast,\hat{\psi}_\ast)$ be a solution of \eqref{sect1} with $c=c_0^\ast$. Then we may repeat the proof of Lemma \ref{S28} to conclude $\hat{\phi}_\ast\leq0$, a contradiction.
\end{proof}
\begin{lem}\lbl{tlem13}Let $(\tilde{\phi}_c,\tilde{\psi}_c)$ be the unique monotone solution of \eqref{sect1}. For any $\gamma>0$, there exists a unique $c=c(\gamma)\in(0, c_0^\ast )$ such that $\gamma\tilde{\phi}_c'(0)=c$. Moreover, $\gamma\longmapsto c(\gamma)$ is strictly increasing and $\lim_{\gamma\rightarrow\infty}c(\gamma)= c_0^\ast $.
\end{lem}
\begin{proof}\,By Lemma \ref{T10} and Lemma \ref{T9}, for fixed $\gamma>0$, the function $p(c,\gamma)=\gamma\tilde{\phi}_c'(0)-c$ is continuous and strictly decreasing for $c\in[0, c_0^\ast )$. Note that $\lim_{c\rightarrow c_0^\ast }p(c,\gamma)=- c_0^\ast <0$ and $p(0,\gamma)=\gamma\tilde{\phi}_{0}'(0)>0$. Therefore, there exists a unique $c=c(\gamma)\in(0, c_0^\ast )$ such that $p(c,\gamma)=0$, i.e. $\gamma\phi_c'(0)=c$. Moreover, note that $p(c,\gamma)$ is strictly increasing in $\gamma$ for any given $c\in(0, c_0^\ast )$. Hence, $c(\gamma)$ is strictly increasing in $\gamma$. For any $\epsilon>0$ and $c\in[0, c_0^\ast -\epsilon]$, we have $p(c,\gamma)\geq p( c_0^\ast -\epsilon,\gamma)\rightarrow\infty$ as $\gamma\rightarrow\infty$. It follows that
$ c_0^\ast -\epsilon<c(\gamma)< c_0^\ast $ for all large $\gamma$, which means that $\lim_{\gamma\rightarrow\infty}c= c_0^\ast $.
\end{proof}
Theorem \ref{F4} now follows directly from Lemmas \ref{S25}--\ref{tlem13}.
{\setlength{\baselineskip}{16pt}{\setlength\arraycolsep{2pt} \indent
\section{The spreading-vanishing dichotomy}
\setcounter{equation}{0}
We prove Theorems 1.1 and 1.2 in this section.
Let us recall that for the problem
\bess\left\{
\begin{aligned}
&u_t=du_{xx}+u(a-bu),&0<x<h(t),&\quad t>0,\\
&u_x(0,t)=0,u(x,t)\equiv0,&h(t)\leq x,&\quad t>0,\\
&h'(t)=-\nu u_x(h(t),t),&\nu>0,&\quad t>0,\\
&h(0)=h_0>0,\quad u(x,0)=u_0(x),&0\leq x\leq h_0,&
\end{aligned}
\right.\lbl{(se1)}
\eess
the following result holds.
\begin{lem}\label{t3} {\rm (\cite{DL10})} If $h_0\geq \frac{\pi}{2}\sqrt{\frac{d}{a}}$, then spreading always happens. If $h_0<\frac{\pi}{2}\sqrt{\frac{d}{a}}$, then there exists $\nu^\ast>0$ depending on $u_0$ such that vanishing happens when $\nu\leq\nu^\ast$, and spreading happens when $\nu>\nu^\ast$.
\end{lem}
\begin{lem}\lbl{t4}Let $(u,v,g)$ be the solution of \eqref{section1-1}. If $\lim_{t\rightarrow\infty}g(t)=g_\infty<\infty$, then the solution of equation \eqref{section1-1} satisfies
\bess
\lim_{t\rightarrow\infty}\|u(\cdot, t)\|_{C([0,g(t)])}=0,\;\lim_{t\to\infty}v(\cdot,t)=1 \mbox{ in } C_{loc}([0,\infty)).
\eess
\end{lem}
\begin{proof} The proof is similar to the proof of Lemma 4.6 in \cite{DL14}. For readers' convenience, we give the details here. Define
$$s:=\frac{g_0x}{g(t)},\quad \hat{u}(s,t):=u(x,t),\quad \hat{v}(s,t):=v(x,t).$$
By direct calculation,
$$u_t=\hat{u}_t-\frac{g'(t)}{g(t)}s\hat{u}_s,\quad u_x=\frac{g_0}{g(t)}\hat{u}_s,
\quad u_{xx}=\frac{g_0^2}{g^2(t)}\hat{u}_{ss}.$$
Hence $\hat u$ satisfies
$$\left\{
\begin{aligned}
&\hat{u}_{t}-\frac{g_0^2}{g^2(t)}\hat{u}_{ss}-\frac{g'(t)}{g(t)}s\hat{u}_s
=\hat{u}(1-\hat{u}-k\hat{v}),&0< s<g_0,&\quad t>0,\\
&\hat{u}_{s}(0,t)=\hat{u}(g_0, t)=0,&&\quad t>0,\\
&\hat{u}(s, 0)=u_{0}(s),&0\leq s\leq g_0.&
\end{aligned}
\right.\eqno{}$$
By Proposition 2.1, there exists $M>0$ such that
$$\|1-\hat{u}-k\hat{v}\|_{L^\infty}\leq 1+(1+k)M,\quad \left\|\frac{g'(t)}{g(t)}\right\|_{L^\infty}\leq \frac{M}{g_0}.$$
Since $g_0\leq g(t)<g_\infty<\infty$, the differential operator is uniformly parabolic. Therefore we can apply standard $L^p$ theory to obtain, for any $p>1$,
$$\|\hat{u}\|_{W_p^{2,1}([0,g_0]\times[0,2])}\leq C_1,$$
where $C_1$ is a constant depending on $p,g_0,M$ and $\|u_{0}\|_{C^{1+\alpha}[0,g_0]}$.
For each $T\geq1$, we can apply the partial interior-boundary estimate over $[0,g_0]\times[T, T+2]$ to
obtain $\|\hat{u}\|_{W_p^{2,1}([0,g_0]\times[T+1/2, T+2])}\leq C_2$ for some constant $C_2$ depending on $\alpha, g_0,M$ and $\|u_{0}\|_{C^{1+\alpha}[0,g_0]}$, but
independent of $T$. Therefore, we can use the Sobolev imbedding theorem to obtain, for any $\alpha\in(0,1),$
\bes\|\hat{u}\|_{C^{1+\alpha, (1+\alpha)/2}([0,g_0]\times[0,\infty))}\leq C_3,\lbl{eshu}\ees
where $C_3$ is a constant depending on $\alpha, g_0,M$ and $\|u_{0}\|_{C^{1+\alpha}[0,g_0]}$.
Similarly we may use interior estimates to the equation of $\hat{v}$ to obtain
\bes\|\hat{v}\|_{C^{1+\alpha,(1+\alpha)/2}([0,g_0]\times[0,+\infty))}\leq C_4,\lbl{eshv}\ees
where $C_4$ is a constant depending on $\alpha,g_0,M$ and $\|v_{0}\|_{C^{1+\alpha}[0,g_\infty+1]}$.
Since
\[
g'(t)=-\gamma u_x(g(t),t)=-\gamma \frac{g_0}{g(t)}\hat u_s(g_0,t),
\]
it follows that there exists a constant $\tilde{C}$ depending on
$\alpha,\gamma,g_0,\|(u_{0},v_0)\|_{C^{1+\alpha}[0,g_0]}$ and $g_\infty$ such that
\bes\|g\|_{C^{1+\alpha/2}([0,+\infty))}\leq \tilde{C}.\lbl{se3l1}\ees
For contradiction, we assume that
$$\limsup_{t\rightarrow+\infty}\|u(\cdot ,t)\|_{C([0,g(t)])}=\delta>0.$$
Then there exists a sequence $(x_k,t_k)$ with $0\leq x_k<g(t_k),\ 1<t_k<\infty$ such that $u(x_k,t_k)\geq\frac{\delta}{2}>0$ for all $k\in\mathbb{N}$,
and $t_k\rightarrow+\infty$ as $k\rightarrow+\infty$. By \eqref{se3l1}, we know $|u_x(g(t),t)|$ is uniformly bounded for $t\in[0,+\infty)$, and there exists $\sigma>0$ such that $x_k\leq g(t_k)-\sigma$ for all $k\geq1$. Therefore there exists a subsequence of $\{x_k\}$ that converges to
some $x_0\in[0,g_\infty-\sigma]$. Without loss of generality, we may assume $x_k\rightarrow x_0$ as $k\rightarrow+\infty$, which leads to
$s_k=\frac{g_0x_k}{g(t_k)}\rightarrow s_0=\frac{g_0x_0}{g_\infty}<g_0.$
Set
$$\hat{u}_k(s,t)=\hat{u}(s,t_k+t),\quad \hat{v}_k(s,t)=\hat{v}(s,t_k+t)$$
for $(s,t)\in [0,g_0]\times[-1, 1].$ It follows from \eqref{eshu} and \eqref{eshv} that
$\{(\hat{u}_k,\hat{v}_k)\}$ has a subsequence $\{(\hat{u}_{k_i},\hat{v}_{k_i})\}$ such that
$$\|(\hat{u}_{k_i},\hat{v}_{k_i})-(\hat{u}^*,\hat{v}^*)\|_{C^{1+\alpha',(1+\alpha')/2}([0,g_0]\times[-1, 1])}\rightarrow0 \mbox{ as }i\rightarrow+\infty,$$
where $\alpha'\in(0, \alpha)$.
Since $\|g\|_{C^{1+\alpha/2}([0,+\infty))}\leq \tilde{C}$, $g'(t)>0$ and $g(t)\leq g_\infty$, we necessarily have $g'(t)\rightarrow0$ as $t\rightarrow\infty.$
Hence, $(\hat{u}^*, \hat{v}^*)$ satisfies
$$\begin{cases}
\hat{u}^*_t-(\frac{g_0}{g_\infty})^2\hat{u}^*_{ss}=\hat{u}^*(1-\hat{u}^*-k\hat{v}^*),& 0\leq s<g_0,\quad t\in(-1, 1),\\
\hat{u}^*_s(0, t)=\hat{u}^*(g_0, t)=0,&\quad\quad \quad \quad \quad \quad t\in[-1, 1].
\end{cases}$$
Clearly, $\hat{u}_k(s_k,0)=u(x_k, t_k)\geq\frac{\delta}{2}$, Hence, we have $\hat{u}^*(s_0, 0)\geq\frac{\delta}{2}.$
By the maximum principle, $\hat{u}^*>0$ in $[0,g_0)\times(-1, 1)$. Thus we can apply the Hopf boundary lemma to conclude that $\theta_0:=\hat{u}^*_s(g_0, 0)<0$. It follows that $u_{x}(g(t_{k_i}),t_{k_i})=\partial_s\hat{u}_{k_i}(g_0,0)\frac{g_0}{g(t_{k_i})}\leq\frac{\theta_0g_0}{2g_\infty}<0$ for all large $i$, and hence
$$g'(t_{k_i})=-\gamma u_{x}(g(t_{k_i}), t_{k_i})\geq-\gamma\theta_0/2>0$$
for all large $i$. On the other hand, recalling that $g'(t)\rightarrow0$ as $t\rightarrow+\infty$, we obtain a contradiction.
Hence we must have
\[
\lim_{t\rightarrow+\infty}\|u(\cdot, t)\|_{C([0,g(t)])}=0.
\]
Using this fact and a simple comparison argument we easily deduce $\lim_{t\to\infty} v(\cdot, t)=1$ uniformly in any compact subset of $[0,\infty)$.
\end{proof}
\begin{lem}\lbl{t6}Let $(u,v,g)$ be the solution of \eqref{section1-1} and suppose $g_\infty=\infty$. Then
\bess
(u(x,t),v(x,t))\rightarrow\left(u^\ast,v^\ast\right) \mbox{ as }t\rightarrow\infty
\eess
uniformly for $x$ in any compact subset of $[0,\infty)$.
\end{lem}
\begin{proof} We define
\[
\overline{u}_1= \overline{v}_1=1,\; \underline u_1=1-h,\; \underline v_1=1-k.
\]
Then define inductively for $n\geq 1$,
\[
\overline{u}_{n+1}=1-k\underline{v}_{n},\ \overline{v}_{n+1}=1-h\underline{u}_{n},\ \underline{u}_{n+1}=1-k\overline{v}_{n},\ \underline{v}_{n+1}=1-h\overline{u}_{n}.
\]
It is easily checked that $\{\overline u_n\}$ and $\{\overline v_n\}$ are decreasing, $\{\underline u_n\}$ and $\{\underline v_n\}$ are increasing, and
\bes\lbl{u_n-v_n}
\lim_{n\to\infty} (\overline u_n, \overline v_n)=\lim_{n\to\infty} (\underline u_n, \underline v_n)=(u^*, v^*).
\ees
We claim that, for every $n\geq 1$,
\bes\begin{aligned}
&\liminf_{t\rightarrow\infty}u(x,t)\geq \underline{u}_{n},\; \liminf_{t\rightarrow\infty}v(x,t)\geq \underline{v}_{n},
\\
&\limsup_{t\rightarrow\infty}u(x,t)\leq\overline{u}_{n},\;\limsup_{t\rightarrow\infty}v(x,t)\leq\overline{v}_{n},
\end{aligned}
\lbl{it}\ees
uniformly in any compact subset of $[0,\infty)$.
The conclusion of the Lemma clearly follows directly from \eqref{it} and \eqref{u_n-v_n}. So it suffices to prove \eqref{it}. We do that by an induction argument.
{\bf Step 1}. \eqref{it} holds for $n=1$.
It follows from the comparison principle that $u(x,t)\leq\hat{u}_1(t)$ for $t>0$ and $x\in[0,g(t)]$, where $\hat{u}_1(t)$ satisfies
\bess
\left\{
\begin{aligned}
&\frac{d\hat{u}_{1}}{dt}=\hat{u}_1(1-\hat{u}_1),&t>0,\\
&\hat{u}_1(0)=\|u_0\|_\infty.
\end{aligned}
\right.
\eess
Clearly, $\lim_{t\rightarrow\infty}\hat{u}_1(t)=1$. Hence,
\bes
\limsup_{t\rightarrow\infty}u(x,t)\leq1=\overline{u}_1\mbox{ uniformly for }x\in[0,\infty),\lbl{it1}
\ees
By the same argument as above, one gets
\bes
\limsup_{t\rightarrow\infty}v(x,t)\leq1=\overline{v}_1\mbox{ uniformly for }x\in[0,\infty).\lbl{it2}
\ees
For any given $l>\max\left\{g_0,\frac{\pi}{2}\frac{1}{\sqrt{1-k}},\frac{\pi}{2}\sqrt{\frac{d}{r(1-h)}}\right\}$.
In view of \eqref{it1}, \eqref{it2} and $g_\infty=\infty$, for any small $\epsilon>0$, there exists $t_1>0$ such that
$g(t)>l$ for $t\geq t_1$ and $u(x,t)<\overline{u}_1+\epsilon$, $v(x,t)<\overline{v}_1+\epsilon$
for $x\in [0, l]$, $t>t_1$.
It follows that
\bess\left\{
\begin{aligned}
&v_t\geq dv_{xx}+rv(1-v-h(1+\epsilon)),&0<x<l,\quad t>t_1,\\
&v_x(0,t)=0,\quad v(l,t)>0,&0\leq x<l,\quad t>t_1,\\
\end{aligned}
\right.\lbl{}\eess
which implies that $v$ is an upper solution to the problem
\bess\left\{
\begin{aligned}
&\hat{v}_t=d\hat{v}_{xx}+r\hat{v}(1-\hat{v}-h(1+\epsilon)),&&0<x<l, & t>t_1,\\
&\hat{v}_x(0,t)=0,\quad \hat{v}(l,t)=0,&&& t>t_1,\\
&\hat v(x, t_1)=v(x, t_1), && 0\leq x\leq l.&
\end{aligned}
\right.\lbl{}\eess
Hence
\[
v(x,t)\geq \hat v(x,t) \mbox{ for $x\in[0,l]$ and $t>t_1$.}
\]
In view of $l>\frac{\pi}{2}\sqrt{\frac{d}{r(1-h)}}$, it is well known that $\lim_{t\rightarrow\infty}\hat{v}(x,t)=\hat{v}^\ast(x)$, where $\hat{v}^\ast(x)$ is the unique positive solution of
\bess\left\{
\begin{aligned}
&d\hat{v}^\ast_{xx}+r\hat{v}^\ast(1-\hat{v}^\ast-h(1+\epsilon))=0,&0<x<l,\\
&\hat{v}^\ast_x(0)=0,\quad \hat{v}^\ast(l)=0.
\end{aligned}
\right.\lbl{}\eess
On the other hand, $\hat{v}^\ast\rightarrow1-h(1+\epsilon)$ uniformly in any compact subset of $[0,\infty)$ as $l\rightarrow\infty$
(see, for example, Lemma 2.2 in \cite{DM}). Thanks to the arbitrariness of $l$ and $\epsilon$, we thus obtain from $v(x,t)\geq \hat v(x,t)$ in $[0,l]\times (t_1,\infty)$ that
\bess\liminf_{t\rightarrow\infty}v(x,t)\geq1-h=\underline{v}_1\mbox{ uniformly in any compact subset of }[0,\infty).\lbl{it3}\eess
Similarly, we have
\bess\left\{
\begin{aligned}
&u_t-u_{xx}\geq u(1-u-k(1+\epsilon)),&0<x<l,&\quad t>t_1,\\
&u_x(0,t)=0,\quad u(l,t)>0,&&\quad t>t_1,
\end{aligned}
\right.\eess
which leads to
\bess\liminf_{t\rightarrow\infty}u(x,t)\geq1-k=\underline{u}_1\mbox{ uniformly in any compact subset of }[0,\infty).\lbl{it4}\eess
This completes the proof of Step 1.
{\bf Step 2}. If \eqref{it} holds for $n=j\geq 1$, then it holds for $n=j+1$.
Since \eqref{it} holds for $n=j$, for any small $\epsilon>0$ and large $l>\max\left\{g_0,\frac{\pi}{2}\frac{1}{\sqrt{1-k}},\frac{\pi}{2}\sqrt{\frac{d}{r(1-h)}}\right\}$, there is $t_2>0$ such that
\[
g(t)>l,\; u(x,t)\in[\underline{u}_j-\epsilon, \overline u_j+\epsilon], \;v(x,t)\in [\underline{v}_j-\epsilon, \overline v_j+\epsilon]
\;\mbox{ for } x\in [0, l],\; t>t_2.
\]
It follows from the comparison principle that $u(x,t)\leq\overline{u}(x, t)$ for $x\in[0, l]$ and $t>t_2$, where $\overline{u}(x,t)$ satisfies
\bess
\left\{
\begin{aligned}
&\overline{u}_t-\overline u_{xx}=\overline{u}(1-\overline{u}-k(\underline{v}_j-\epsilon)),&&x\in (0,l),&t>t_2,\\
&\overline{u}_x(0,t)=0,\; \overline u(l, t)=\overline u_j+\epsilon, &&& t>t_2,\\
&\overline u(x,t_2)=u(x, t_2), && x\in [0, l].
\end{aligned}
\right.
\eess
It is well known that this problem has a unique positive steady-state solution $\hat u^*(x)$ and
$\lim_{t\rightarrow\infty}\hat{u}(x,t)=\hat u^*(x)$ uniformly for $x\in [0,l]$. Moreover,
\[
\lim_{l\to\infty} \hat u^*(x)=1-k(\underline v_j-\epsilon) \mbox{ locally uniformly in } [0,\infty).
\]
It follows, since $\epsilon>0$ can be arbitrarily small, that
\[
\limsup_{t\to\infty} u(x,t)\leq 1-k\underline v_j=\overline u_{j+1} \mbox{ locally uniformly in } [0,\infty).
\]
Analogously, from the comparison principle we obtain $u(x,t)\geq\underline{u}(x, t)$ for $x\in[0, l]$ and $t>t_2$, where $\underline{u}(x,t)$ satisfies
\bess
\left\{
\begin{aligned}
&\underline{u}_t-\underline u_{xx}=\underline{u}(1-\underline{u}-k(\overline{v}_j+\epsilon)),&&x\in (0,l),&t>t_2,\\
&\overline{u}_x(0,t)=0,\; \overline u(l, t)=\underline u_j-\epsilon, &&& t>t_2,\\
&\overline u(x,t_2)=u(x, t_2), && x\in [0, l],
\end{aligned}
\right.
\eess
from which we can deduce
\[
\liminf_{t\to\infty} u(x,t)\geq 1-k\overline v_j=\underline u_{j+1} \mbox{ locally uniformly in } [0,\infty).
\]
The proof for
\[
\limsup_{t\to\infty} v(x,t)\leq \overline v_{j+1}, \; \liminf_{t\to\infty} v(x,t)\geq \underline v_{j+1} \mbox{ locally uniformly in } [0,\infty)
\]
is similar, and we omit the details.
\end{proof}
Theorem \ref{f2} now follows directly from Lemmas \ref{t4} and \ref{t6}.
\begin{lem}\lbl{t5} If $g_\infty<+\infty$, then $g_\infty\leq \frac{\pi}{2\sqrt{1-k}}$. Hence $g_0\geq \frac{\pi}{2\sqrt{1-k}}$ implies $g_\infty=+\infty$.
\end{lem}
\begin{proof}\ \ Assume for contradiction that $\frac{\pi}{2\sqrt{1-k}}<g_\infty<\infty$. Then there exists $T_1>0$ such that $g(T_1)>\frac{\pi}{2\sqrt{1-k(1+\epsilon)}}$ for $\epsilon$ sufficiently small. By a simple comparison consideration,
there exists $T>T_1>0$ such that
$$v(x,t)\leq1+\epsilon\mbox{ for }x\in[0,\infty), \quad t>T.$$
Hence $(u,g)$ satisfies
\bess
\left\{
\begin{aligned}
&u_t\geq u_{xx}+u(1-u-k(1+\epsilon)),&0<x<g(t),&\quad t>T,\\
&u_x(0,t)=u(g(t),t)=0,&&\quad t>T,\\
&g'(t)=-\gamma u_x(g(t),t),&&\quad t>T,\\
&u(x,T)>0,&0<x<g(T),&
\end{aligned}
\right.
\eess
which implies that $(u,g)$ is an upper solution to the problem
\bess
\left\{
\begin{aligned}
&w_t=w_{xx}+w(1-w-k(1+\epsilon)),&0<x<\underline{g}(t),&\quad t>T,\\
&w_x(0,t)=w(\underline{g}(t),t)=0,&&\quad t>T,\\
&\underline{g}'(t)=-\gamma w_x(\underline{g}(t),t),&&\quad t>T,\\
&w(x,T)=u(x,T),\quad \underline{g}(T)=g(T),&0<x<\underline{g}(T).&
\end{aligned}
\right.
\eess
Thus, $g(t)\geq\underline{g}(t)$ for $t>T$. Since $\underline{g}(T)=g(T)>g(T_1)>\frac{\pi}{2\sqrt{1-k(1+\epsilon)}}$, it follows from Lemma \ref{t3} that $\underline{g}(t)\rightarrow\infty$ and hence $g_\infty=\infty$. This contradiction leads to $g_\infty<\frac{\pi}{2\sqrt{1-k}}$.\end{proof}
\begin{lem}\label{t7} If $g_0<\frac{\pi}{2\sqrt{1-k}}$, then there exists $\underline{\gamma}\geq0$ depending on $u_0$ and $v_0$ such that spreading happens when $\gamma>\underline{\gamma}$.
\end{lem}
\begin{proof}Since $\limsup_{t\rightarrow\infty}v(x,t)\leq1$ uniformly for $x\in[0,\infty)$, there exists $t_3>0$, which is independent of $\gamma$, such that $v(x,t)\leq 1+\epsilon$ for $x\in[0,\infty),\; t\geq t_3$. Thus $(u,g)$ satisfies
\bess\left\{
\begin{aligned}
&u_t-u_{xx}\geq u(1-u-k(1+\epsilon)),&0<x<g(t),&\quad t>t_3,\\
&u_x(0,t)=0,\quad u(g(t),t)=0,&&\quad t>t_3,\\
&g'(t)=-\gamma u_x(g(t),t),&&\quad t>t_3,\\
&u(x,t_3)>0,&0\leq x<g(t_3).&\quad
\end{aligned}
\right.\lbl{}
\eess
Hence
$(u,g)$ is an upper solution to the problem
\bes\left\{
\begin{aligned}
&\hat{u}_t-\hat{u}_{xx}=\hat{u}(1-\hat{u}-k(1+\epsilon)),&0<x<\hat{g}(t),&\quad t>t_3,\\
&\hat{u}_x(0,t)=0,\quad \hat{u}(\hat{g}(t),t)=0,&&\quad t>t_3\\
&\hat{g}'(t)=-\gamma \hat{u}_x(\hat{g}(t),t),&&\quad t>t_3,\\
&\hat{u}(x,t_3)=u(x,t_3),\quad \hat g(t_3)=g(t_3),&0\leq x<g(t_3).&
\end{aligned}
\right.\lbl{s3-50}
\ees
The comparison principle infers $g(t)\geq \hat g(t)$ for $t>t_3$.
Applying Lemma \ref{t3} to \eqref{s3-50} we see that there exists $\underline{\gamma}\geq0$ depending on $g(t_3)$ and $u(x, t_3)$ (which are uniquely determined by $u_0$ and $v_0$) such that spreading happens for \eqref{s3-50} when $\gamma>\underline{\gamma}$.
Thus $\lim_{t\to\infty} g(t)=\infty$ when $\gamma>\underline \gamma$, and by Lemma \ref{t6}, spreading happens to \eqref{section1-1} for such $\gamma$.
\end{proof}
\begin{lem}\label{t8} There exists $\gamma^\ast\geq0$, depending on $u_0$ and $v_0$, such that $g_\infty<\infty$ if $\gamma\leq\gamma^\ast$, and $g_\infty=\infty$ if $\gamma>\gamma^\ast$.
\end{lem}
\begin{proof}
Set $\Lambda=\{\gamma>0:g_\infty>\frac{\pi}{2\sqrt{1-k}}\}.$
It follows from Lemmas \ref{t5} and \ref{t7} that $\gamma^\ast:=\inf\Lambda\in[0,\infty)$.
The comparison principle infers that
$g_\infty=\infty$ if $\gamma>\gamma^\ast$ and $g_\infty<\infty$ if $0<\gamma<\gamma^\ast$.
It remains to show that $\gamma^\ast\not\in\Lambda$. Otherwise, $\gamma^*>0$ and $g_{\infty}>\frac{\pi}{2\sqrt{1-k}}$ for $\gamma=\gamma^\ast$. Hence we can find $T > 0$ such that $g(T)>\frac{\pi}{2\sqrt{1-k}}$. To emphasize the dependence of the solution of \eqref{section1-1} on $\gamma$, we denote it by $(u_\gamma,v_\gamma,g_\gamma)$ instead of $(u,v,g)$, and so $g_{\gamma^\ast}(T)>\frac{\pi}{2\sqrt{1-k}}$. By the
continuous dependence of $(u_\gamma,v_\gamma,g_\gamma)$ on $\gamma$, we can find $\epsilon>0$ small so that $g_\gamma(T)>\frac{\pi}{2\sqrt{1-k}}$ for $\gamma\in[\gamma^\ast-\epsilon,\gamma^\ast+\epsilon]$. It then follows
from Lemma \ref{t5} that for all such $\gamma$, $\lim_{t\rightarrow\infty}g_\gamma(t)=\infty$. This implies that $[\gamma^\ast-\epsilon,\gamma^\ast+\epsilon]\subset\Lambda$, and $\inf\Lambda\leq\gamma^\ast-\epsilon$, a contradiction to the definition of $\gamma^\ast$.
\end{proof}
\begin{lem}\label{t9}If $g_0<\frac{\pi}{2}$, then there exists $\overline{\gamma}>0$ depending on $u_0$ such that $g_\infty<+\infty$ if $\gamma\leq\overline{\gamma}$.
\end{lem}
\begin{proof}Clearly, $(u,g)$ satisfies
\bess\left\{
\begin{aligned}
&u_t-u_{xx}\leq u(1-u),&0<x<g(t),&\quad t>0,\\
&u_x(0,t)=0,\quad u(g(t),t)=0,&&\quad t>0,\\
&g'(t)=-\gamma u_x(g(t),t),&&\quad t>0,\\
&u(x,0)=u_0(x),&0\leq x\leq g_0.&
\end{aligned}
\right.\lbl{(se111)}
\eess
That is, $(u,g)$ is a lower solution to the problem
\bess\left\{\begin{aligned}
&\bar{u}_t-\bar{u}_{xx}= \bar{u}(1-\bar{u}),&0<x<\bar{g}(t),&\quad t>0,\\
&\bar{u}_x(0,t)=0,\quad\bar{u}(\bar g(t),t)=0,&&\quad t>0,\\
&\bar{g}'(t)=-\gamma \bar{u}_x(\bar{g}(t),t),&&\quad t>0,\\
&\bar{u}(x,0)=u_0(x),&0\leq x\leq g_0.&
\end{aligned}
\right.\lbl{(se111)}
\eess
It follows from the comparison principle that $g(t)\leq\bar{g}(t)$. Since $g_0<\pi/2$, by Lemma \ref{t3} there exists $\bar{\gamma}>0$ depending on $u_0$ such that $\bar{g}(\infty)<\infty$ if $\gamma\leq\bar{\gamma}$. Hence, $g_\infty<+\infty$ if $\gamma\leq\overline{\gamma}$.
\end{proof}
Theorem \ref{f3} now follows directly from Lemmas \ref{t5}-\ref{t9}.
{\setlength{\baselineskip}{16pt}{\setlength\arraycolsep{2pt} \indent
\section{Asymptotic spreading speed}
\setcounter{equation}{0}
We prove Theorem 1.3 in this section.
\begin{lem}\lbl{T15}Suppose spreading occurs, i.e., alternative {\rm (i)} happens in Theorem 1.1. Then
\bess\liminf_{t\rightarrow\infty}\frac{g(t)}{t}\geq c_\gamma.\lbl{section4e}\eess
\end{lem}
\begin{proof}\, Let $V(t)$ be the unique solution of $$V'=rV(1-V),\quad V(0)=\|v_0\|_\infty.$$
Then a simple comparison consideration yields $v(x,t)\leq V(t)$ for $x\geq0$ and $t>0$. Since $\lim_{t\rightarrow\infty}V(t)=1$, we can find $T_0'>0$ such that
\bes\lbl{infe1}
v(x,t)<1+\delta\quad\mbox{for } x\geq0,\quad t\geq T_0'.
\ees
We now consider the auxiliary problem
\bes\left\{
\begin{aligned}
&\phi_\delta''-c\phi_\delta'+\phi_\delta(1-2\delta-\phi_\delta-k\psi_\delta)=0,\quad \phi_\delta'>0\quad\mbox{for } s>0,\\
&d\psi_\delta''-c\psi_\delta'+r\psi_\delta(1+2\delta-\psi_\delta-h\phi_\delta)=0,\quad \psi_\delta'<0\quad\mbox{for } -\infty<s<\infty,\\
&\phi_\delta(s)=0\quad \mbox{for } s\leq 0,\; \gamma \phi_\delta'(0)=c,\quad \psi_\delta(-\infty)=1+2\delta,\\
&(\phi_\delta(\infty),\psi_\delta(\infty))=(u^*_\delta, v^*_\delta):=
\left(\frac{1-2\delta-k(1+2\delta)}{1-hk},\frac{1+2\delta-h(1-2\delta)}{1-hk}\right),
\end{aligned}
\right.\lbl{section401}\ees
where $\delta>0$ is small. We claim that
there exists a unique $c^\delta_{\gamma}>0$ such that \eqref{section401} has a unique solution $(\phi_\delta,\psi_\delta)$ when $c=c_\gamma^\delta$; moreover,
$$
\lim_{\delta\rightarrow0}c^\delta_{\gamma}=c_\gamma.
$$
Indeed, if we define
\[
\sigma_{\pm\delta}=\sqrt{1\pm 2\delta},\; \phi_\delta(s)=\sigma_{-\delta}^2\tilde \phi_\delta(\sigma_{-\delta}s),
\;
\psi_\delta(s)=\sigma_{+\delta}^2\tilde \psi_\delta(\sigma_{-\delta}s),
\]
and
\[
\tilde c=c/\sigma_{-\delta},\; \tilde\gamma=\gamma \sigma_{-\delta}^4,
\;
\tilde r_\delta=r \left(\frac{\sigma_{+\delta}}{\sigma_{-\delta}}\right)^2,\; \tilde h_\delta=h \left(\frac{\sigma_{-\delta}}{\sigma_{+\delta}}\right)^2,\; \tilde k_\delta=k \left(\frac{\sigma_{+\delta}}{\sigma_{-\delta}}\right)^2,
\]
then a direct calculation shows that $(c, \phi_\delta, \psi_\delta)$ solves \eqref{section401} if and only if $(\tilde c, \tilde \phi_\delta,\tilde \psi_\delta)$ satisfies \eqref{section12} and $\tilde\gamma \tilde\phi'(0)=\tilde c$ when
$(r,h,k, u^*, v^*)$ in \eqref{section12} is replaced by $(\tilde r_\delta,\tilde h_\delta,\tilde k_\delta, u^*_\delta, v^*_\delta)$. So the claim follows directly from
Theorem \ref{F4} and Lemma \ref{tlem13}, and the continuous dependence of the unique solution on the parameters.
Since $\psi_\delta(-\infty)=1+2\delta$, $\psi_\delta(+\infty)=v^*_\delta>v^*$ and $\psi_\delta'<0$, there exists $L>1$ such that
\bes\psi_\delta(x)>1+\delta \mbox{ for } x\leq - L,\;
\psi_\delta(x)>v^*_\delta>v^* \mbox{ for }x\geq -L.
\lbl{infe5}\ees
Similarly it follows from $\phi_\delta(+\infty)=u^*_\delta<u^*$ and $\phi_\delta'>0$ that
\[
\phi_\delta(x)<u^*_\delta<u^* \mbox{ for } x\geq 0.
\]
By the spreading assumption, we have
\bess
\lim_{t\to\infty} g(t)=\infty,\; \lim_{t\to\infty}(u(x,t),v(x,t))= (u^*, v^*) \mbox{ in any compact subset of } [0,\infty).
\eess
Hence, in view of $u^*>u^*_\delta$ and $v^*<v^*_\delta$, there exists $T_0>T'_0$ large such that for $x\in [0, L+1]$ and $t\geq T_0$,
\[
g(t)>1,\; u(x,t)>u^*_\delta,\; v(x,t)<v^*_\delta.
\]
We now define
\bess\begin{aligned}
&\underline{g}(t)=c^\delta_{\gamma} (t-T_0)+1,\ \
&\underline{u}(x,t)=\phi_\delta(\underline{g}(t)-x),\ \
&\overline{v}(x,t)=\psi_\delta(\underline{g}(t)-x).
\end{aligned}\eess
Then $\underline g(T_0)=1<g(T_0)$,
\[
\underline u(x, T_0)=\phi_\delta(1-x)<u^*_\delta<u(x, T_0) \mbox{ for } x\in [0,1]=[0,\underline g(T_0)],
\]
and in view of \eqref{infe5} and \eqref{infe1}, we also have
\[\begin{aligned}
&\overline v(x, T_0)=\psi_\delta(1-x)>v^*_\delta>v(x, T_0) \mbox{ for } x\in [0, L+1],
\\
&\overline v(x, T_0)=\psi_\delta(1-x)>1+\delta>v(x, T_0) \mbox{ for } x>L+1.
\end{aligned}
\]
Let us also note that
\[
\underline u(0, t)=\phi_\delta(\underline g(t))<u^*_\delta<u(0, t) \mbox{ for } t\geq T_0,
\]
\[
\overline v(0, t)=\psi_\delta(\underline g(t))>v^*_\delta>u(0, t) \mbox{ for } t\geq T_0,
\]
and moreover,
\bess
\underline{g}'(t)=c^\delta_{\gamma}=\gamma\phi_\delta'(0)=-\gamma\underline{u}_x(\underline{g}(t),t) \mbox{ for } t\geq T_0.
\eess
Furthermore,
\bess\begin{aligned}
\underline{u}_t-\underline{u}_{xx}&=c^\delta_{\gamma}\phi_\delta'-\phi_\delta''\\
&=\phi_\delta(1-2\delta-\phi_\delta-k\psi_\delta)\\
&\leq\phi_\delta(1-\phi_\delta-k\psi_\delta)\\
&=\underline{u}(1-\underline{u}-k\overline{v}),
\end{aligned}\eess
\bess\begin{aligned}
\overline{v}_t-d\overline{v}_{xx}&=c^\delta_{\gamma}\psi_\delta'-d\psi_\delta''\\
&=r\psi_\delta(1+2\delta-\psi_\delta-h\phi_\delta)\\
&\geq r\psi_\delta(1-\psi_\delta-h\phi_\delta)\\
&= r\overline{v}(1-\overline{v}-h\underline{u}).
\end{aligned}\eess
Hence, we can use Proposition \ref{the comparison principle} and Remark \ref{remarkp} to conclude that
$$\begin{array}{l}
g(t)\geq\underline{g}(t) \mbox{ for } t\geq T_0.
\end{array}$$
It follows that
$\liminf_{t\rightarrow\infty}\frac{g(t)}{t}\geq c^\delta_\gamma$, which yields the required inequality by letting $\delta\rightarrow0$.
\end{proof}
\begin{lem}\lbl{T14} Under the assumptions of Lemma \ref{T15} we have
\bess\limsup_{t\rightarrow\infty}\frac{g(t)}{t}\leq c_\gamma.\lbl{section4e}\eess
\end{lem}
\begin{proof}\, For small $\tau>0$ we consider the auxiliary problem
\bes\left\{
\begin{aligned}
&\phi_\tau''-c\phi_\tau'+\phi_\tau(1+2\tau-\phi_\tau-k\psi_\tau)=0,\quad \phi_\tau'>0\quad \mbox{for } 0<s<\infty,\\
&d\psi_\tau''-c\psi_\tau'+r\psi_\tau(1-2\tau-\psi_\tau-h\phi_\tau)=0,\quad \psi_\tau'<0 \quad \mbox{for }-\infty<s<\infty,\\
&\phi_\tau(s)\equiv0\quad\mbox{for } s\leq0,\; \gamma \phi'_\tau(0)=c,\quad \psi_\tau(-\infty)=1-2\tau,\\
&(\phi_\tau,\psi_\tau)(\infty)=(u^*_\tau, v^*_\tau):=\left(\frac{1+2\tau-k(1-2\tau)}{1-hk},\frac{1-2\tau-h(1+2\tau)}{1-hk}\right).
\end{aligned}
\right.\lbl{teif}\ees
As in the proof of Lemma \ref{T15} we can use a change of variable trick to reduce \eqref{teif} to \eqref{section12}, and
then apply Theorem \ref{F4} and Lemma \ref{tlem13} to conclude that there exists a unique $c^\tau_{\gamma}>0$ such that \eqref{teif} has a unique solution $(\phi_\tau,\psi_\tau)$ when $c=c^\tau_\gamma$, and
moreover,
$$ \lim_{\tau\rightarrow0}c^\tau_{\gamma}=c_\gamma.$$
Let us also observe that
\[
u^*<u^*_\tau,\; v^*>v^*_\tau,\; 0<\phi_\tau(x)<u^*_\tau \mbox{ for } x>0,\; v^*_\tau<\psi_\tau(x)<1-2\tau \mbox{ for } x\in (-\infty, \infty).
\]
For clarity we divide the analysis below into three steps.
{\bf Step 1.}
We prove that for any small $\tau>0$, we can find $T'_0>0$ such that for each $T\geq T'_0$, there exists $L(T)>0$
having the following property:
$$
v(x,T)\geq1-\tau\mbox{ for } x\geq L(T).
$$
By \eqref{section1-1a} we have
$$\tilde v_0:=\inf_{x\geq 0}v_0(x)>0.$$
Consider the auxiliary problem
\bes\lbl{tif}
\left\{
\begin{aligned}
&w_t-dw_{xx}=rw(1-w),&x>0,&\quad t>0,\\
&w(0,t)=0,&&\quad t>0,\\
&w(x,0)=\tilde v_0,&x>0.&
\end{aligned}
\right.
\ees
It is well known that the solution of \eqref{tif} satisfies
$$
\lim_{t\rightarrow\infty}w(x,t)=w_\ast(x)\mbox{ locally unformly for }x\in[0,\infty),
$$
where $w_\ast$ is the unique solution of
$$
-w_\ast''=rw_\ast(1-w_\ast)\mbox{ for }x\in[0,\infty),w_\ast(0)=0.
$$
Moreover, $w_*$ has the property that $w_\ast'>0$ and $w_\ast(\infty)=1$. Therefore, there exist positive constants $L_1, T'_0$ large enough such that
$$
w(L_1,t)\geq w_\ast(L_1)-\tau/2\geq1-\tau\mbox{ for }t\geq T'_0.
$$
Applying the maximum principle to the equation satisfied by $w_x(x,t)$, we deduce $w_x(x,t)\geq0$ for $x>0$ and $t>0$. It follows that
$$
w(x,t)\geq1-\tau\mbox{ for }x\geq L_1\mbox{ and }t\geq T'_0.
$$
Fix $T\geq T_0'$ and note that $v$ satisfies
$$\begin{cases}
v_t-dv_{xx}=rv(1-v), & x>g(T),\; 0<t\leq T, \\
v(x,0)\geq \tilde v_0, & x\geq g(T).\end{cases}$$
Set $\tilde{w}(x,t):=w(x-g(T),t).$
Then $\tilde{w}(x,t)$ satisfies
$$
\tilde{w}_t-d\tilde{w}_{xx}=r\tilde{w}(1-\tilde{w})\mbox{ for }x>g(T) \mbox{ and }0<t\leq T.
$$
Since
$$
\tilde{w}(g(T), t)=0<v(g(T), t)\mbox{ for } t\in (0, T],\quad \tilde{w}(x, 0)=\tilde v_0\leq v(x,0)\mbox{ for }x>g(T),
$$
we can use the comparison principle to deduce
$$
v(x,t)\geq\tilde{w}(x,t)=w(x-g(T),t)\mbox{ for }x>g(T)\mbox{ and }0<t\leq T .
$$
Thus we obtain
$$v(x,T)\geq w(x-g(T),T)\geq w(L_1,T)\geq1-\tau\mbox{ for }x\geq L(T):=L_1+g(T).$$
This completes the proof of Step 1.
{\bf Step 2.} We prove that for any small $\tau>0$, there exists $T_1'>0$ such that
\bes\lbl{comp-ode}
u(x,t)\leq u^*_{\tau/2},\; v(x, t)\geq v^*_{\tau/2} \mbox{ for } x\geq 0,\; t\geq T_1'.
\ees
We prove the claimed inequalities in \eqref{comp-ode} by a comparison argument involving the following ODE system
\[\begin{cases}
\check u'(t)=\check u(1-\check u-k\check v),& t>0,\\
\check v'(t)=r\check v(1-\check v-h\check u), & t>0,\\
(\check u(0), \check v(0))=(\|u_0\|_\infty, \tilde v_0).&
\end{cases}
\]
Indeed, by the comparison principle for coorporative system we easily obtain
\[
u(x, t)\leq \check u(t),\; v(x,t)\geq \check v(t) \mbox{ for } x\geq 0,\; t>0.
\]
But it is well known (for example, see \cite{Goh}) that
\[
\lim_{t\to\infty} (\check u(t), \check v(t))=(u^*, v^*).
\]
The inequalties in \eqref{comp-ode} thus follow directly once we recall $u^*<u^*_{\tau/2}<u^*_\tau$ and $v^*>v^*_{\tau/2}>v^*_\tau$.
{\bf Step 3.} We complete the proof of the lemma by
constructing a suitable comparison function triple $(\overline{u}(x,t),
\underline{v}(x,t),\overline{g}(t))$, and applying the comparison principle.
We fix $T_0:=\max\{T_0', T_1'\}$. Then by the conclusions in Steps 1 and 2 we obtain
\[
u(x,T_0)\leq u^*_{\tau/2}<u^*_\tau \mbox{ for } x\geq 0,\; v(x, T_0)\geq v^*_{\tau/2}>v^*_\tau \mbox{ for } x\geq 0,
\]
and
\[
v(x, T_0)\geq 1-\tau \mbox{ for } x\geq L(T_0).
\]
Choose $S>L(T_0)>g(T_0)$ large so that
\[
\phi_\tau(x)>u^*_{\tau/2},\; \psi_\tau(x)<v^*_{\tau/2} \mbox{ for } x\geq S-L(T_0),
\]
and then define
\bess\begin{aligned}
&\overline{g}(t)=c^\tau_{\gamma} (t-T_0)+S,\ \
&\overline{u}(x,t)=\phi_\tau(\overline{g}(t)-x),\ \
&\underline{v}(x,t)=\psi_\tau(\overline{g}(t)-x).
\end{aligned}\eess
Clearly $\overline{g}(T_0)=S> g(T_0)$ and
\bess
\overline{g}'(t)=c^\tau_{\gamma}=\gamma\phi_\tau'(0)=-\gamma\overline{u}_x(\overline{g}(t),t)\quad \mbox{ for }
t\geq T_0.
\eess
Moreover,
\bess\begin{aligned}
&\overline{u}(x,T_0)=\phi_\tau(S-x)\geq\phi_\tau(S-g(T_0))>u^*_{\tau/2}\geq u(x,T_0)\quad\mbox{ for } x\in[0,g(T_0)],\\
&\underline{v}(x,T_0)=\psi_\tau(S-x)\leq\psi_\tau(S-L(T_0))<v^*_{\tau/2}\leq v(x,T_0)\quad\mbox{for } x\in[0,L(T_0)],
\end{aligned}\eess
and for $x>L(T_0)$,
$$\underline{v}(x,T_0)=\psi_\tau(S-x)<\psi_\tau(-\infty)=1-2\tau<v(x,T_0).$$
Furthermore,
\bess\begin{aligned}
&\overline{u}_x(0,t)=-\phi_\tau'(\overline{g}(t))<0,\ \
\overline{u}(\overline{g}(t),t)=\phi_\tau(0)=0,\ \
\underline{v}_x(0,t)=-\psi_\tau'(\overline{g}(t))>0 \quad \mbox{ for } t\geq T_0.
\end{aligned}\eess
Finally, direct calculations show that
\bess\begin{aligned}
\overline{u}_t-\overline{u}_{xx}&=c^\tau_{\gamma}\phi_\tau'-\phi_\tau''\\
&=\phi_\tau(1+2\tau-\phi_\tau-k\psi_\tau)\\
&\geq\phi_\tau(1-\phi_\tau-k\psi_\tau)\\
&=\overline{u}(1-\overline{u}-k\underline{v})
\end{aligned}\eess
and
\bess\begin{aligned}
\underline{v}_t-d\underline{v}_{xx}&=c^\tau_{\gamma}\psi_\tau'-d\psi_\tau''\\
&=r\psi_\tau(1-2\tau-\psi_\tau-h\phi_\tau)\\
&\leq r\psi_\tau(1-\psi_\tau-h\phi_\tau)\\
&=r\underline{v}(1-\underline{v}-h\overline{u}).
\end{aligned}\eess
Hence, we can use Proposition \ref{the comparison principle} to conclude that
$$
g(t)\leq\overline{g}(t)\quad \mbox{ for } t\geq T_0.
$$
It follows that
$\limsup_{t\rightarrow\infty}\frac{g(t)}{t}\leq c^\tau_\gamma$, which gives the required inequality by letting $\tau\rightarrow0$.
\end{proof}
Theorem \ref{F5} now follows directly from Lemmas \ref{T15} and \ref{T14}. So the asymptotic spreading speed of $u$ in Theorem \ref{F5} is given by $c^0=c_\gamma$.
\section{Proof of Proposition \ref{s7}}
\setcounter{equation}{0}
Although we follow some standard steps in the proof of Proposition \ref{s7}, since the first equation of \eqref{function1}
is only satisfied for $s>0$, nontrivial changes are needed. We break the rather long proof into several lemmas.
We start with a second order ODE of the following form
\bes\left\{\begin{aligned}
&d_1y''-cy'-\beta y+f(s)=0,\quad s>0,\\
& y(0)=0,
\end{aligned}\right.\lbl{2.3}
\ees
where the constants $c$ and $\beta$ are positive, and the nonlinear function $f$ is specified below.
Let $$
\lambda_{1}=\frac{c-\sqrt{c^2+4\beta d_1}}{2d_1},\quad\lambda_{2}=\frac{c+\sqrt{c^2+4\beta d_1}}{2d_1}
$$
be the two roots of equation $d_1\lambda^2-c\lambda-\beta=0.$
Then we have the following result.
\begin{lem}\lbl{s1}Assume $f:[0,\infty)\rightarrow\mathbb{R}$ is piecewise continuous and $|f(s)|\leq Ae^{\alpha s}$ for all $s\geq 0$ and some constants $A>0$, $\alpha\in(0,\min\{-\lambda_1,\lambda_2\})$.
Then \eqref{2.3} has a unique solution satisfying $y(s)=O(e^{\alpha s})$ as $s\rightarrow\infty$, and it is given by
\bes
y(s)=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_{0}^sK_1(\xi,s)f(\xi)d\xi
+\int_s^{\infty}K_2(\xi,s)f(\xi)d\xi\right],
\lbl{lem1f0}
\ees
where
\bess
K_1(\xi,s)=e^{\lambda_1s}\big(e^{-\lambda_1\xi}-e^{-\lambda_{2}\xi}\big),\quad
K_2(\xi,s)=\big(e^{\lambda_{2}s}-e^{\lambda_{1}s}\big)e^{-\lambda_2\xi}.
\lbl{lem1f01}
\eess
\end{lem}
\begin{proof}\, By the variation of constants formula, the solutions of \eqref{2.3} are given by
\bes\lbl{lem1f1}
\;\;\;\; y(s)=\gamma\big (e^{\lambda_{1}s}-e^{\lambda_{2}s}\big)+\frac{1}{d_1(\lambda_{2}-\lambda_{1})}
\left[\int_{0}^se^{\lambda_{1}(s-\xi)}f(\xi)d\xi
-\int_{0}^se^{\lambda_{2}(s-\xi)}f(\xi)d\xi\right],\;\gamma\in\mathbb R.
\ees
Multiplying both sides of \eqref{lem1f1} by $e^{-\lambda_{2}s}$, we get
\bess\displaystyle
y(s)e^{-\lambda_{2}s}=\gamma\big(e^{(\lambda_{1}-\lambda_{2})s}-1\big)+
\frac{e^{-\lambda_{2}s}}{d_1(\lambda_{2}-\lambda_{1})}
\left[\int_{0}^se^{\lambda_{1}(s-\xi)}f(\xi)d\xi
-\int_{0}^se^{\lambda_{2}(s-\xi)}f(\xi)d\xi\right].\lbl{lem1f11}
\eess
If $ y(s)=O(e^{\alpha s})$ as $s\rightarrow\infty$, then due to $\lambda_{1}<0<\lambda_{2}$ and $|\lambda_{1}|<\lambda_{2}$, we obtain
\bess
y(s)e^{-\lambda_{2}s}\rightarrow0,\quad e^{(\lambda_{1}-\lambda_{2})s}\rightarrow0
\mbox{ and }\frac{e^{-\lambda_{2}s}}{d_1(\lambda_{2}-\lambda_{1})}
\int_{0}^se^{\lambda_{1}(s-\xi)}f(\xi)d\xi\rightarrow0
\eess
as $s\rightarrow\infty$. Therefore,
\bes
\gamma=\frac{-1}{d_1(\lambda_{2}-\lambda_{1})}
\int_{0}^\infty e^{-\lambda_{2}\xi}f(\xi)d\xi.
\lbl{lem1fu2}
\ees
Substituting \eqref{lem1fu2} into \eqref{lem1f1}, we obtain \eqref{lem1f0}.
If $ y(s)$ is given by \eqref{lem1f0}, then it is easy to check that $ y(s)$ satisfies \eqref{2.3} and $ y(s)=O(e^{\alpha s})$ as $s\rightarrow\infty$.
\end{proof}
Define the operators $H_1:C_{\mathcal{R}}(\mathbb{R}^+,\mathbb{R}^2)\rightarrow C(\mathbb{R^+},\mathbb{R})$ and $H_2:C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)\rightarrow C(\mathbb{R},\mathbb{R})$ by
\bess
\begin{aligned}
&H_1(\varphi)(s):=\beta \varphi_1(s)+f_1(\varphi(s)),\\
&H_2(\varphi)(s):=\beta \varphi_2(s)+f_2(\varphi(s)),
\end{aligned}
\eess
where the positive constant $\beta$ is large enough such that $H_i(\varphi)$ is nondecreasing with respect to $\varphi_1$ and $\varphi_2$, for $(\varphi_1(s),\varphi_2(s))\in \mathcal R=[0, k_1]\times [0,k_2]$.
Let $F_1:C_{\mathcal{R}}(\mathbb{R},\mathbb{R})\rightarrow C(\mathbb{R},\mathbb{R})$ be given by
\bes
F_1(\varphi)(s):=\left\{\begin{aligned}
&\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_{0}^sK_1(\xi,s)H_1(\varphi)(\xi)d\xi\right. &\\
&\quad\quad\hspace{1.5cm}+\left.\int_s^{\infty}K_2(\xi,s)H_1(\varphi)(\xi)d\xi\right],&s>0,\\
&0,&s\leq0,
\end{aligned}\right.\lbl{l01}
\ees
where $K_i(\xi,s)$ is given by \eqref{lem1f01}. By Lemma \ref{s1},
it is easy to see that the operator $F_1$ is well defined and
\bess
\left\{\begin{aligned}
&d_1(F_1(\varphi))''(s)-c(F_1(\varphi))'(s)-\beta F_1(\varphi)(s)+H_1(\varphi)(s)=0,&s>0,\\
&F_1(\varphi)(s)=0,&s\leq0.
\end{aligned}\right.\lbl{slt1}
\eess
Let $$\mu_{1}=\frac{c-\sqrt{c^2+4\beta d_2}}{2d_2},\quad\mu_{2}=\frac{c+\sqrt{c^2+4\beta d_2}}{2d_2}$$
be the two roots of
$$d_2\mu^2-c\mu-\beta=0.$$
Define $F_2:C_{\mathcal{R}}(\mathbb{R},\mathbb{R})\rightarrow C(\mathbb{R},\mathbb{R})$ by
\bes\lbl{l02}\begin{aligned}
F_2(\varphi)(s):&=\frac{1}{d_2(\mu_2-\mu_1)}\left[\int_{-\infty}^se^{\mu_1(s-\xi)}H_2(\varphi)(\xi)d\xi +\int_s^{\infty}e^{\mu_2(s-\xi)}H_2(\varphi)(\xi)d\xi\right].
\end{aligned}\ees
It is easy to show that the operator $F_2$ is well defined and satisfies
\bess
d_2(F_2(\varphi))''(s)-c(F_2(\varphi))'(s)-\beta F_2(\varphi)(s)+H_2(\varphi)(s)=0.\lbl{slt2}
\eess
We now define
$F: C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)\rightarrow C(\mathbb{R},\mathbb{R}^2)$ by
\[
F(\varphi):=(F_1(\varphi), F_2(\varphi)).
\]
Clearly, $\varphi$ is a fixed point of the operator $F$ in $C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)$ if and only if it is a solution of \eqref{function1} in $C_{\mathcal{R}}(\mathbb{R},\mathbb{R}^2)$.
Next, we introduce a Banach space with exponential decay norm. Fix $\sigma\in(0,\min\{|\lambda_{1}|,\lambda_{2},|\mu_{1}|,\mu_{2}\})$. It is easy to see that
$$B_\sigma(\mathbb{R},\mathbb{R}^2):=\left\{\varphi\in C(\mathbb{R},\mathbb{R}^2):\sup_{s\in\mathbb{R}}|\varphi(s)|e^{-\sigma |s|}<\infty\right\}$$
equipped with the norm
$$|\varphi|_\sigma:=\sup_{s\in\mathbb{R}}|\varphi(s)|e^{-\sigma |s|}$$
is a Banach space.
Let $\overline{\varphi}(s)$ and $\underline{\varphi}(s)$ be the upper and lower solutions given in the statement of Proposition \ref{s7}.
Consider the set
$$\Gamma:=\Big\{\varphi=(\varphi_1,\varphi_2)\in B_\sigma(\mathbb{R},\mathbb{R}^2): \underline{\varphi}\leq\varphi\leq\overline{\varphi},\;
\varphi_i\mbox{ is nondecreasing for } s\in\mathbb{R},i=1,2\Big\}.$$
Clearly $\Gamma$ is a nonempty, bounded, closed, convex subset of
the Banach space $B_\sigma(\mathbb{R},\mathbb{R}^2)$.
We are going to show that $F$ maps $\Gamma$ into itself, and is completely comtinuous. Then the Schauder fixed point theorem will yield a fixed point of $F$ in $\Gamma$, and we will then show that
it satisfies \eqref{function1} and \eqref{function2a2}.
\begin{lem}\lbl{s3} {\rm (i)} $F(\hat{\varphi})(s)\leq F(\tilde{\varphi})(s)$ for $s\in\mathbb{R}$ if $\hat{\varphi}\leq \tilde{\varphi}$
and $\hat{\varphi}, \tilde{\varphi}\in\Gamma$;
{\rm (ii)} $F_1(\varphi)(s)$ and $F_2(\varphi)(s)$ are nondecreasing in $s\in\mathbb{R}$ for any $\varphi\in\Gamma$.
\end{lem}
\begin{proof}\, We show that $F_1$ satisfies (i) and (ii) stated in the lemma.
Since $F_1(\varphi)(s)=0$ for $s\leq0$, we only need to consider the case of $s>0$. In view of $\lambda_{1}<0<\lambda_{2}$, it is easy to see that
$K_{1}(\xi,s)>0$ for $0<\xi<s$ and $K_{2}(\xi,s)>0$ for $s<\xi$. Thus, by \eqref{l01} and the hypothesis $(\textbf{A}_2)$ we conclude that $F_1(\hat{\varphi})(s)\leq F_1(\tilde{\varphi})(s)$ for $s\in\mathbb{R}$ if $\hat{\varphi}\leq \tilde{\varphi}$
and $\hat{\varphi}, \tilde{\varphi}\in\Gamma$. This proves (i) for $F_1$.
We next consider (ii).
For $\varphi=(\varphi_1,\varphi_2)\in\Gamma$, the hypothesis $(\textbf{A}_2)$ implies that $H_1(\varphi)$ is nondecreasing in $\varphi_i$.
Since $\varphi_1$ and $\varphi_2$ are nondecreasing in $\mathbb{R}$, we have $\varphi_i(s+\theta)\geq\varphi_i(s)$ for
$\theta>0$ and $i=1, 2$.
This leads to $H_1(\varphi)(s+\theta)-H_1(\varphi)(s)\geq0$. A direct computation gives
$$\begin{aligned}
& F_1(\varphi)(s+\theta)-F_1(\varphi)(s)\\
&=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s+\theta}K_1(\xi,s+\theta)
H_1(\varphi)(\xi)d\xi+\int_{s+\theta}^{\infty}K_2(\xi,s+\theta)H_1(\varphi)(\xi)d\xi\right]\\
& \hspace{3cm} -\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s}K_1(\xi,s)H_1(\varphi)(\xi)d\xi
+\int_{s}^{\infty}K_2(\xi,s)H_1(\varphi)(\xi)d\xi\right]\\
&=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s+\theta}e^{\lambda_{1}(s+\theta-\xi)}H_1(\varphi)(\xi)d\xi
+\int_{s+\theta}^{\infty}e^{\lambda_{2}(s+\theta-\xi)})H_1(\varphi)(\xi)d\xi\right.\\
&\hspace{4cm} -\int_0^{s}e^{\lambda_{1}(s-\xi)}H_1(\varphi)(\xi)d\xi
-\left.\int_{s}^{\infty}e^{\lambda_{2}(s-\xi)}H_1(\varphi)(\xi)d\xi\right]\\
&\hspace{0.5cm} +\frac{1}{d_1(\lambda_{2}
-\lambda_{1})}\left[\int_0^{s}e^{\lambda_{1}s}e^{-\lambda_{2}\xi}H_1(\varphi)(\xi)d\xi
+\int_{s}^{\infty}e^{\lambda_{1}s}e^{-\lambda_{2}\xi}H_1(\varphi)(\xi)d\xi\right.\\
&\hspace{4cm} -\left.\int_0^{s+\theta}e^{\lambda_{1}(s+\theta)}e^{-\lambda_{2}\xi}H_1(\varphi)(\xi)d\xi
-\int_{s+\theta}^{\infty}e^{\lambda_{1}(s+\theta)}e^{-\lambda_{2}\xi}H_1(\varphi)(\xi)d\xi
\right]\\
&=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s+\theta}e^{\lambda_{1}(s+\theta-\xi)}H_1(\varphi)(\xi)d\xi
+\int_{s+\theta}^{\infty}e^{\lambda_{2}(s+\theta-\xi)})H_1(\varphi)(\xi)d\xi\right.\\
&\hspace{5cm} -\int_0^{s}e^{\lambda_{1}(s-\xi)}H_1(\varphi)(\xi)d\xi
-\left. \int_{s}^{\infty}e^{\lambda_{2}(s-\xi)}H_1(\varphi)(\xi)d\xi\right]\\
&\hspace{0.5cm}+\frac{1}{d_1(\lambda_{2}-\lambda_{1})}e^{\lambda_{1}s}\left(1-e^{\lambda_{1}\theta}\right)
\int_0^{\infty}e^{-\lambda_{2}\xi}H_1(\varphi)(\xi)d\xi\\
&\geq\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s+\theta}e^{\lambda_{1}(s+\theta-\xi)}H_1(\varphi)(\xi)d\xi
+\int_{s+\theta}^{\infty}e^{\lambda_{2}(s+\theta-\xi)})H_1(\varphi)(\xi)d\xi\right.\\
&\hspace{5cm} -\int_0^{s}e^{\lambda_{1}(s-\xi)}H_1(\varphi)(\xi)d\xi
-\left.\int_{s}^{\infty}e^{\lambda_{2}(s-\xi)}H_1(\varphi)(\xi)d\xi\right]\\
&=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left\{\int_0^{\theta}e^{\lambda_{1}(s+\theta-\xi)}
H_1(\varphi)(\xi)d\xi+\left[\int_0^{s}e^{\lambda_{1}(s-\xi)}
\Big(H_1(\varphi)(\xi+\theta)-H_1(\varphi)(\xi)\Big)d\xi\right.\right.\\
&\quad\hspace{7.5cm}+\left.\left.\int_{s}^{\infty}e^{\lambda_{2}(s-\xi)}\Big(H_1(\varphi)(\xi+\theta)-H_1(\varphi)(\xi)\Big)d\xi\right]\right\}\\
&\geq 0.
\end{aligned}$$
So $F_2$ satisfies (ii).
Similarly, we can prove $F_2$ satisfies (i) and (ii).
\end{proof}
\begin{lem}\lbl{s4}
$F(\Gamma)\subset\Gamma$.
\end{lem}
\begin{proof}\,Due to Lemma \ref{s3}, it suffices to show that, for all $s\in\mathbb R$,
$$
\underline{\varphi}(s)\leq F(\underline{\varphi})(s),\; F(\overline{\varphi})(s)\leq\overline{\varphi}(s).
$$
We firstly show
\[
\underline{\varphi}_1(s)\leq F_1(\underline{\varphi})(s),
\;\forall s\in\mathbb R.
\]
Since $\underline\varphi_1(s)=F_1(\varphi)(s)=0$ for $s\leq0$, we only need to consider the case of $s>0$. Without loss of generality, we denote $\xi_0=0,\xi_{m_1+1}=\infty$ and assume $\xi_i<\xi_{i+1}$ for $i=0,1,2,\cdots,m_1$. Here $\xi_i$, $i\in \{0,..., m_1\}$, are points in $\Omega_1$ so that $\underline \varphi_1$ satisfies the first inequality \eqref{lower} in $\mathbb R^+\setminus\Omega_1$. According to the definition of $F_1(\varphi)$ and Definition \ref{s2}, for any $s\in(\xi_i,\xi_{i+1})$, we have,
$$\begin{aligned}
F_1(\underline{\varphi})(s)&=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s}K_1(\xi,s)H_1(\underline{\varphi})(\xi)d\xi
+\int_{s}^{\infty}K_2(\xi,s)H_1(\underline{\varphi})(\xi)d\xi\right]\\
&\geq\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s}K_1(\xi,s)\left(\beta\underline{\varphi}_1(\xi)-d_1\underline{\varphi}_1''(\xi)
+c\underline{\varphi}_1(\xi)'\right)d\xi\right.\\
&\quad\hspace{3cm}+\left.\int_{s}^{\infty}K_2(\xi,s)\left(\beta\underline{\varphi}_1(\xi)-d_1\underline{\varphi}_1''(\xi)
+c\underline{\varphi}_1'(\xi)\right)d\xi\right]\\
&=\underline{\varphi}_1(s)+\frac{1}{\lambda_{2}
-\lambda_{1}}\left[\sum_{j=1}^{i}K_1(\xi_j,s)\left(\underline{\varphi}_1'(\xi_j+)
-\underline{\varphi}_1'(\xi_j-)\right)\right.\\
&\quad\hspace{3.5cm}+\left.\sum_{j=i+1}^{m_1}K_2(\xi_j,s)\left(\underline{\varphi}_1'(\xi_j+)
-\underline{\varphi}_1'(\xi_j-)\right)\right]\\
&\geq\underline{\varphi}_1(s).
\end{aligned}$$
The continuity of $\underline{\varphi}(s)$ and $F_1(\underline{\varphi})(s)$ implies that $F_1(\underline{\varphi})(s)\geq\underline{\varphi}_1(s)$ for any $s\in\mathbb{R}^+$.
The proofs of
$$
F_1(\overline{\varphi})(s)
\leq\overline{\varphi}_1(s),
\;
\underline{\varphi}_2(s)\leq F_2(\underline{\varphi})(s),\; F_2(\overline{\varphi})(s)\leq \overline{\varphi}_2(s)
$$
for $s\in\mathbb{R}$ are similar, and we omit the details.
\end{proof}
\begin{lem}\lbl{s5}
$F: \Gamma\rightarrow \Gamma$ is continuous.
\end{lem}
\begin{proof}\, From the hypothesis $(\textbf{A}_3)$, it is easy to see that, for some $L>0$ and all $\hat\varphi,\;\tilde\varphi\in \Gamma$,
$$|H_1(\hat{\varphi})-H_1(\tilde{\varphi})|_\sigma\leq (L+\beta)|\hat{\varphi}-\tilde{\varphi}|_\sigma.$$
By a direct calculation, we have
$$\begin{aligned}
&|F_1(\hat{\varphi})-F_1(\tilde{\varphi})|_\sigma\\
=& \left|\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s}K_1(\xi,s)(H_1(\hat{\varphi})(\xi)-H_1(\tilde{\varphi})(\xi))d\xi
+\int_{s}^{\infty}K_2(\xi,s)(H_1(\hat{\varphi})(\xi)-H_1(\tilde{\varphi})(\xi))d\xi\right]\right|_\sigma\\
\leq& \frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left|\int_0^{s}K_1(\xi,s)|H_1(\hat{\varphi})(\xi)-H_1(\tilde{\varphi})(\xi)|d\xi
+\int_{s}^{\infty}K_2(\xi,s)|H_1(\hat{\varphi})(\xi)-H_1(\tilde{\varphi})(\xi)|d\xi\right|_\sigma\\
\leq&\frac{L+\beta}{d_1(\lambda_{2}-\lambda_{1})}|\hat{\varphi}-\tilde{\varphi}|_\sigma\sup_{s\in\mathbb{R}^+}\left[\int_0^{s}K_1(\xi,s)
e^{\sigma(\xi-s)}d\xi+\int_{s}^{\infty}K_2(\xi,s)e^{\sigma(\xi-s)}d\xi\right]\\
=&\frac{L+\beta}{d_1(\lambda_{2}-\lambda_{1})}|\hat{\varphi}-\tilde{\varphi}|_\sigma\sup_{s\in\mathbb{R}^+}\left(1
-e^{(\lambda_{1}-\sigma) s}\right)\frac{\lambda_{2}-\lambda_{1}}{(\lambda_{2}-\sigma) (\sigma-\lambda_{1})}\\
\leq&\frac{L+\beta}{d_1(\lambda_{2}-\sigma)(\sigma-\lambda_{1})}|\hat{\varphi}-\tilde{\varphi}|_\sigma,
\end{aligned}$$
which clearly implies $F_1: \Gamma\rightarrow B_\sigma(\mathbb{R},\mathbb{R}^2)$ is continuous.
Similarly we can show $F_2: \Gamma\rightarrow B_\sigma(\mathbb{R},\mathbb{R}^2)$ is continuous. Hence $F$ is continuous on $\Gamma$.
\end{proof}
\begin{lem}\lbl{s6}
$F:\Gamma\rightarrow\Gamma$ is compact.
\end{lem}
\begin{proof}\,Since $F$ is continuous on $\Gamma$ by Lemma \ref{s5}, and $\Gamma$ is a bounded set in $B_\sigma(\mathbb R, \mathbb R^2)$,
it suffices to show that $F(\Gamma)$ is a relatively compact set. To this end, let
$$
\rho:=\sup\{|H_i(\varphi)|:\varphi\in\Gamma,i=1,2\}.
$$
In view of $F_1(\varphi)(s)=0$ for $s<0$, we get $(F_1(\varphi))'(s)=0$ when $s<0$. Moreover, for any $\varphi\in\Gamma$ and $s>0$,
\bess(F_1(\varphi))'(s)=\frac{1}{d_1(\lambda_{2}-\lambda_{1})}\left[\int_0^{s} K_{1s}(\xi,s)H_1(\varphi)(\xi)d\xi
+\int_s^{\infty}K_{2s}(\xi,s)H_1(\varphi)(\xi)d\xi\right].\lbl{s001}\eess
Hence,
\bes\begin{aligned}
|(F_1(\varphi))'(s)|&\leq\frac{\rho}{d_1(\lambda_{2}-\lambda_{1})}
\left[\int_{0}^s|K_{1s}(\xi,s)|d\xi
+\int_s^{\infty}|K_{2s}(\xi,s)|d\xi\right]\\
&=\frac{\rho}{d_1(\lambda_{2}-\lambda_{1})}e^{\lambda_{1}s}\left(1-\frac{\lambda_1}{\lambda_2}\right)\leq\frac{\rho}{d_1\lambda_{2}},\;\forall s>0.
\end{aligned}\lbl{s002}\ees
We thus see that $s\to F_1(\varphi)(s)$ is Lipschitz continuous with Lipschitz constant $L_1:=\frac{\rho}{d_1\lambda_{2}}$ independent
of $\varphi\in\Gamma$.
Similarly, from \eqref{l02} we have
\bess
(F_2(\varphi))'(s)&=\frac{1}{d_2(\mu_2-\mu_1)}\left[\mu_1\int_{-\infty}^se^{\mu_1(s-\xi)}H_2(\varphi)(\xi)d\xi +\mu_2\int_s^{\infty}e^{\mu_2(s-\xi)}H_2(\varphi)(\xi)d\xi\right],
\eess
and
\bess\begin{aligned}
|(F_2(\varphi))'(s)|&\leq\frac{\rho}{d_2(\mu_2-\mu_1)}\left[|\mu_1|\int_{-\infty}^se^{\mu_1(s-\xi)}d\xi +\mu_2\int_s^{\infty}e^{\mu_2(s-\xi)}d\xi\right]\\
&\leq\frac{2\rho}{d_2(\mu_2-\mu_1)}.
\end{aligned}\lbl{s005}\eess
Thus $\{F(\varphi)(s): \varphi\in\Gamma\}$ is a family of equi-continuous functions of $s\in\mathbb R$.
Let $\Phi_j$ be a sequence of $\Gamma$ and $\upsilon_j=F(\Phi_j)$. Then the sequence $\upsilon_j$ is equi-continuous.
It follows from Lemma \ref{s3}(ii) that $\upsilon_j(s)$ is nondecreasing in $s\in\mathbb{R}$.
Noting that $\Gamma$ is bounded in $L^\infty(\mathbb{R},\mathbb{R}^2)$,
by the Arzela-Ascoli theorem, we conclude that for any $R>0$, there exists a convergent subsequence of $\upsilon_j|_{[-R,R]}$ in $C([-R,R],\mathbb{R}^2)$. Using a standard diagonal selection scheme, we can extract a subsequence $\upsilon_{j_k}$ that converges in $C([-R,R],\mathbb{R}^2)$ for every $R>0$. Without loss of generality, we assume that the sequence $\upsilon_j$ itself converges in each $C([-R,R],\mathbb{R}^2)$. From this, it follows easily that
$\upsilon_j$ is Cauchy in $B_\sigma(\mathbb{R},\mathbb{R}^2)$, and hence it is convergent. This proves the precompactness of $F(\Gamma)$.
\end{proof}
Since $\Gamma$ is a bounded closed convex set of $B_\sigma(\mathbb R, \mathbb R^2)$, by Lemmas \ref{s4}, \ref{s5} and \ref{s6}, we
can apply Schauder's fixed point theorem to conclude that $F$ has a fixed point $\varphi$ in $\Gamma$, which is a non-decreasing
solution of \eqref{function1}. To complete the proof of Proposition \ref{s7}, it remains to prove the following result.
\begin{lem}\lbl{s8} The fixed point
$\varphi$ obtained above satisfies \eqref{function2a2}.
\end{lem}
\begin{proof}
From $\underline \varphi(s)\leq \varphi(s)\leq \overline\varphi(s)$ and $\underline{\varphi}_1(s)=\overline{\varphi}_1(s)=0$ for $s\leq0$,
$\underline{\varphi}_2(-\infty)=\overline{\varphi}_2(-\infty)=0$, we obtain $\varphi(-\infty)=(0,0)$. Moreover, due to $0\leq \underline \varphi_1(s)\not\equiv 0$ for $s\in\mathbb R$, we have $0\leq \varphi_1(s)\not\equiv 0$ for $s\in\mathbb R$. It then follows
from the monotonicity of $\varphi_1(s)$ that
$\varphi_1(\infty)\in (0, k_1]$. Using \eqref{function1}, it is well known that (cf. lemma 2.2 in \cite{WZ}) $f_1(\varphi(\infty))=f_2(\varphi(\infty))=0$. Thus we may use
$(\textbf{A}_1)$ to conclude that $\varphi(\infty)={\bf K}$. Hence \eqref{function2a2} holds.\end{proof}
\section*{Acknowledgments}
The research in this work was supported by the Natural Science Foundation of China(11671243, 61672021), the Shaanxi New-star Plan of Science and Technology(2015KJXX-21),
the Natural Science Foundation of Shaanxi Province(2014JM1003), the Fundamental Research Funds for the Central Universities(GK201701001, GK201302005), and
the Australian Research Council.
| 46,919
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\begin{document}
\title{A Toeplitz type operator on Hardy spaces in the unit ball}
\author[Jordi Pau]{Jordi Pau}
\address{Jordi Pau \\Departament de Matem\`{a}tiques i Inform\`{a}tica \\
Universitat de Barcelona\\
08007 Barcelona\\
Catalonia, Spain} \email{jordi.pau@ub.edu}
\author[Antti Per\"{a}l\"{a}]{Antti Per\"{a}l\"{a}}
\address{Antti Per\"{a}l\"{a} \\Departament de Matem\`{a}tiques i Inform\`{a}tica \\
Universitat de Barcelona\\
08007 Barcelona\\
Catalonia, Spain\\
Barcelona Graduate School of Mathematics (BGSMath).} \email{perala@ub.edu}
\subjclass[2010]{47B35, 30H10}
\keywords{Hardy spaces, Toeplitz operators, tent spaces, Schatten classes}
\thanks{The first author was partially
supported by DGICYT grant MTM2014-51834-P
(MCyT/MEC) and the grant 2017SGR358 (Generalitat de Catalunya). The second author acknowledges financial support from the Spanish Ministry of Economy and Competitiveness, through the Mar\'ia de Maeztu Programme for Units of Excellence in R\&D (MDM-2014-0445). Both authors are also supported by the grant MTM2017-83499-P (Ministerio de Educaci\'{o}n y Ciencia).}
\begin{abstract}
We study a Toeplitz type operator $Q_\mu$ between the holomorphic Hardy spaces $H^p$ and $H^q$ of the unit ball. Here the generating symbol $\mu$ is assumed to a positive Borel measure. This kind of operator is related to many classical mappings acting on Hardy spaces, such as composition operators, the Volterra type integration operators and Carleson embeddings. We completely characterize the boundedness and compactness of $Q_\mu:H^p\to H^q$ for the full range $1<p,q<\infty$; and also describe the membership in the Schatten classes of $H^2$. In the last section of the paper, we demonstrate the usefulness of $Q_\mu$ through applications.
\end{abstract}
\maketitle
\section{Introduction and main results}
Let $\Bn=\{z\in \mathbb{C}^n:|z|<1\}$ be the open unit ball in $\Cn$, the Euclidian space of complex dimension $n$. For any two points $z=(z_ 1,\dots,z_ n)$ and $w=(w_ 1,\dots,w_ n)$ in $\Cn$ we write
\[
\langle z,w\rangle =z_ 1\bar{w}_ 1+\dots +z_ n \bar{w}_ n,
\]
and
\[
|z|=\sqrt{\langle z,z\rangle}=\sqrt{|z_ 1|^2+\dots +|z_ n|^2}.
\]
For a positive Borel measure $\mu$ on $\Bn$,
the Toeplitz type operator $Q_ {\mu}$ is defined as
\begin{displaymath}
Q_{\mu} f(z)=\int_{\Bn} \!\!\frac{f(w)}{(1-\langle z,w \rangle )^n}\,d\mu(w),\qquad z\in \Bn.
\end{displaymath}
In the one variable setting, the operator $Q_{\mu}$ appeared in \cite{Lu}, where a description of the membership of $Q_{\mu}$ in the Schatten ideals $S_ p$ of the Hardy space $H^2$ was obtained. As mentioned in that paper, this operator is closely related with the study of composition operators, and later on in \cite{AS1} a connection with a Volterra type integration operator was given.
As far as we know, it seems that the operator $Q_{\mu}$ has not been studied in the setting of Hardy spaces in the unit ball.
For $0<p<\infty$, the Hardy space $H^p:=H^p(\Bn)$ consists of those holomorphic functions $f$ in $\Bn$ with
\[ \|f\|_{H^p}^p=\sup_{0<r<1}\int_{\Sn} \!\! |f(r\zeta)|^p \,d\sigma(\zeta)<\infty,\]
where $d\sigma$ is the surface measure on the unit sphere $\Sn:=\partial \Bn$ normalized so that $\sigma(\Sn)=1$. Moreover, any function in $H^p$ has radial limits $f(\zeta)=\lim_{r\to 1^{-}} f(r\zeta)$ for a.e. $\zeta \in \Sn$; and $H^2$ becomes a Hilbert space when endowed with the inner product
$$\langle f,g\rangle_{H^2}= \int_{\Sn} f(\zeta)\overline{g(\zeta)}d\sigma(\zeta).$$
We refer the reader to the books \cite{Rud} and \cite{ZhuBn} for the theory of Hardy spaces in the unit ball.
We completely describe the boundedness of $Q_{\mu}:H^p\rightarrow H^q$ for $1<p,q<\infty$ (the case $p\neq q$ seems to be new even in one dimension), as well as characterizing its membership in $S_ p(H^2)$, thus generalizing Luecking's results to higher dimensions.
Before stating the main results of the paper, we recall the concept of a Carleson measure.
For $\zeta \in \Sn$ and $\delta>0$ consider the non-isotropic metric balls
\[B_{\delta}(\zeta)=\big \{ z\in \Bn: |1-\langle z, \zeta \rangle |<\delta \big \}.\]
A positive Borel measure $\mu$ on $\Bn$ is said to be a Carleson measure if there exists a constant $C>0$ such that
\[\mu \big (B_{\delta}(\zeta)\big )\le C \delta \, ^ n\]
for all $\zeta \in \Sn$ and $\delta>0$. Obviously every Carleson measure is finite. H\"{o}rmander \cite{H} extended to several complex variables the famous Carleson measure embedding theorem \cite{Car0, Car1} asserting that, for $0<p<\infty$, the embedding $I_ d:H^p\rightarrow L^p(\mu):=L^p(\Bn,d\mu)$ is bounded if and only if $\mu$ is a Carleson measure.
More generally, for $s>0$, a finite positive Borel measure on $\Bn$ is called an $s$-Carleson measure if there exists a constant $C>0$ such that $\mu(B_{\delta}(\zeta))\le C \delta \,^{ns}$ for all $\zeta\in \Sn$ and $\delta>0$. We denote by $\|\mu\|_{CM_s}$ the infimum of all possible $C$ above.
It is well known (see \cite[Theorem 45]{ZZ}) that $\mu$ is an $s$-Carleson measure if and only if for each (some) $t>0$ one has
\begin{equation}\label{sCM}
\sup_{a\in \Bn}\int_{\Bn} \!\!\frac{(1-|a|^2)^t}{|1-\langle a,z \rangle |^{ns+t}} \,d\mu(z)<\infty.
\end{equation}
Moreover, with constants depending on $t$, the supremum of the above integral is comparable to $\|\mu\|_{CM_s}$.
In \cite{Du}, Duren gave an extension of Carleson's theorem by showing that, for
$0<p<q<\infty$, one has that $I_ d:H^p\rightarrow L^q(\mu)$ is bounded if and only if $\mu$ is a $q/p$-Carleson measure. Moreover, one has the estimate $\|I_ d\|_{H^p\rightarrow L^q(\mu)}\asymp \|\mu\|^{1/q}_{CM_{q/p}}$. A simple proof of this result, in the setting of the unit ball, can be found in \cite{P1} for example.\\
Our first result characterizes the boundedness of the Toeplitz type operator $Q_{\mu}:H^p\rightarrow H^q$ when $1<p\le q$ in terms of $s$-Carleson measures.
\begin{theorem}\label{T2}
Let $1<p\le q<\infty$ and $\mu$ be a positive Borel measure on $\Bn$. Then $Q_{\mu}:H^p\rightarrow H^q$ is bounded if and only if $\mu$ is an $(1+\frac{1}{p}-\frac{1}{q})$- Carleson measure. Moreover,
\[
\big \| Q_{\mu} \big \|_{H^p\rightarrow H^q} \asymp \big \|\mu \big \|_{CM_ s}.
\]
\end{theorem}
\mbox{}
\\
The notation $A\asymp B$ means that the two quantities are comparable, and $\|T\|_{X\rightarrow Y}$ denotes the norm of the operator $T:X\rightarrow Y$.\\
For $\zeta \in \Sn$ and $\gamma>1$ the admissible approach region $\Gamma_{\gamma}(\zeta)$ is defined as
\begin{displaymath}
\Gamma(\zeta)=\Gamma_{\gamma}(\zeta)=\left \{z\in \Bn: |1-\langle z,\zeta\rangle |<\frac{\gamma}{2} (1-|z|^2) \right \}.
\end{displaymath}
As we will see later, for most of the properties we will use, the choice of $\gamma$ is not important.\\
For a positive Borel measure $\mu$ on $\Bn$, we set
\[
\widetilde{\mu}(\zeta)=\int_{\Gamma(\zeta)} (1-|z|^2)^{-n}d\mu(z),\qquad \zeta\in \Sn.
\]
The characterization of the boundedness of $Q_{\mu}$ from $H^p$ to $H^q$ in the case $1<q<p<\infty$ will be given in terms of the function $\widetilde{\mu}$.
\begin{theorem}\label{T3}
Let $1<q<p<\infty$ and $\mu$ be a positive Borel measure on $\Bn$. Then $Q_{\mu}:H^p\rightarrow H^q$ is bounded if and only if $\widetilde{\mu}$ belongs to $L^{r}(\Sn)$ with $r=pq/(p-q)$. Moreover, one has
\[
\big \| Q_{\mu} \big \|_{H^p\rightarrow H^q} \asymp \big \|\widetilde{\mu} \big \|_{L^r(\Sn)}.
\]
\end{theorem}
For $0<p<\infty$, a compact operator $T$ acting on a separable Hilbert space $H$ belongs to the Schatten class $S_ p:=S_ p(H)$ if its sequence of singular numbers belongs to the sequence space $\ell^p$ (the singular numbers are the square roots of the eigenvalues of the positive operator $T^*T$, where $T^*$ is the Hilbert adjoint of $T$). We refer to \cite[Chapter 1]{Zhu} for a brief account on Schatten classes.
Our next main result (see Theorem \ref{t-SC} in section \ref{SSC}) is a complete characterization of the membership in the Schatten class $S_ p(H^2)$ of the Toeplitz type operator $Q_{\mu}$. One description is similar to the one obtained by Luecking \cite{Lue2} in the one dimensional case, and we also obtain another description in terms of a Berezin type transform.
The paper is organized as follows: first some background and preliminary results are given in section \ref{s2}. In section \ref{S3} we prove Theorem \ref{T2}, and section \ref{s4} is devoted to the proof of Theorem \ref{T3}. A description of the compactness of $Q_{\mu}:H^p\rightarrow H^q$ for $1<p,q<\infty$ is obtained in section \ref{sComp}, and in section \ref{SSC} a characterization of the membership of $Q_{\mu}$ in the Schatten ideal $S_ p(H^2)$ is provided. Finally, the last section contains some applications to weighted composition operators, Volterra type integration operators and Carleson embeddings.
Finally some words on the notation. For $1<p<\infty$, we let $p'$ to denote the conjugate exponent of $p$. We use $dv$ for the normalized volume measure on $\Bn$, and for $\alpha>-1$, we set $dv_{\alpha}(z)=c_{\alpha}(1-|z|^2)^{\alpha} dv(z)$, where $c_{\alpha}$ is a constant taken so that $v_{\alpha}(\Bn)=1$. The notation $a\lesssim b$ means that there is a finite positive constant $C$ with $a\leq C b$. Also, we use the notation $a\gtrsim b$ to indicate that $b\lesssim a$.
\section{Preliminaries}\label{s2}
In this section we collect some facts needed for the proofs of the main results.
\subsection{Admissible maximal and area functions}
For $\zeta \in \Sn$ and $\gamma>1$, recall that the admissible approach region $\Gamma_{\gamma}(\zeta)$ is defined as
\begin{displaymath}
\Gamma(\zeta)=\Gamma_{\gamma}(\zeta)=\left \{z\in \Bn: |1-\langle z,\zeta\rangle |<\frac{\gamma}{2} (1-|z|^2) \right \}.
\end{displaymath}
If $I(z)=\{\zeta \in \Sn: z\in \Gamma(\zeta)\}$, then $\sigma(I(z))\asymp (1-|z|^2)^{n}$, and it follows from Fubini's theorem that, for a positive function $\varphi$, and a finite positive measure $\nu$, one has
\begin{equation}\label{EqG}
\int_{\Bn} \varphi(z)\,d\nu(z)\asymp \int_{\Sn} \left (\int_{\Gamma(\zeta)} \varphi(z) \frac{d\nu(z)}{(1-|z|^2)^{n}} \right )d\sigma(\zeta).
\end{equation}
This fact will be used repeatedly throughout the paper. \\
For $\gamma>1$ and $f$ continuous on $\Bn$, the admissible maximal function $f^*_{\gamma}$ is defined on $\Sn$ by
\[ f^*(\zeta)=f_{\gamma}^*(\zeta)=\sup_{z\in \Gamma_{\gamma}(\zeta)} |f(z)|.\]
We need the following well known result on the $L^p$-boundedness of the admissible maximal function that can be found in \cite[Theorem 5.6.5]{Rud} or \cite[Theorem 4.24]{ZhuBn}.
\begin{otherth}\label{NTMT}
Let $0<p<\infty$ and $f\in H(\Bn)$. Then
\[ \|f^*\|_{L^p(\Sn)}\le C \|f\|_{H^p}.\]
\end{otherth}
Another function we need is the admissible area function $A_{\gamma}f$ defined on $\Sn$ by
\[Af(\zeta)=A_{\gamma}f(\zeta)=\left ( \int_{\Gamma_{\gamma}(\zeta)} |Rf(z)|^2 \,(1-|z|^2)^{1-n}dv(z)\right )^{1/2}.\]
For a function $f$ holomorphic in $\Bn$, here $Rf$ denotes the radial derivative of $f$, that is,
\[Rf(z)= \sum_{k=1}^{n} z_ k \frac{\partial f}{\partial z_ k} (z),\qquad z=(z_ 1,\dots,z_ n)\in \Bn.\]
The following result \cite{AB, FS, P1, Pav1} characterizes the membership in the Hardy space in terms of the admissible area function.
\begin{otherth}\label{AreaT}
Let $0<p<\infty$ and $g\in H(\Bn)$. Then $g\in H^p$ if and only if $Ag\in L^p(\Sn)$. Moreover, if $g(0)=0$ then
\[ \|g\|_{H^p}\asymp \|Ag\|_{L^p(\Sn)}.\]
\end{otherth}
As said before, all the results here are independent of the aperture $\gamma>1$ and, for that reason, from now on we omit it from the notation.\\
\mbox{}
\\
\textbf{Luecking's theorem:}
We will also need the following result essentially due to Luecking \cite{Lue1} (see also \cite{P1}) describing those positive Borel measures for which the embedding from $H^p$ into $L^s(\mu)$ is bounded when $s<p$.
\begin{otherth}\label{LT}
Let $0<s<p<\infty$ and let $\mu$ be a positive Borel measure on $\Bn$. Then the identity $I_ d:H^p\rightarrow L^s(\mu)$ is bounded, if and only if, the function defined on $\Sn$ by
\[ \widetilde{\mu}(\zeta)=\int_{\Gamma(\zeta)} (1-|z|^2)^{-n}d\mu(z) \]
belongs to $L^{p/(p-s)}(\Sn)$. Moreover, one has $\|I_d\|_{H^p\rightarrow L^s(\mu)}\asymp \|\widetilde{\mu}\|_{L^{p/(p-s)}(\Sn)}^{1/s}.$
\end{otherth}
\mbox{}
\\
Finally, we will use the following integral estimate. It can be found in \cite{Ars} and \cite{Jev}.
\begin{otherl}\label{Gamma}
Let $0<s<\infty$ and $\lambda>n\max(1,1/s)$. If $\mu$ is a positive measure, then
$$\int_{\Sn}\left[\int_{\Bn}\left(\frac{1-|z|^2}{|1-\langle z,\zeta\rangle|}\right)d\mu(z)\right]^s d\sigma(\zeta)\leq C \int_{\Sn}\mu(\Gamma(\zeta))^s d\sigma(\zeta).$$
\end{otherl}
\mbox{}
\\
\subsection{Separated sequences and lattices}
A sequence of points $\{z_ j\}\subset \Bn$ is said to be separated if there exists $\delta>0$ such that $\beta(z_ i,z_ j)\ge \delta$ for all $i$ and $j$ with $i\neq j$, where $\beta(z,w)$ denotes the Bergman metric on $\Bn$. This implies that there is $r>0$ such that the Bergman metric balls $D_ j=\{z\in \Bn :\beta(z,z_ j)<r\}$ are pairwise disjoint. \\
Let $D(a,r)=\{z\in \Bn : \beta(a,z)<r\}$ be the Bergman metric ball of radius $r>0$ centered at a point $a\in \Bn$. We need a well-known
result on decomposition of the unit ball $\Bn$.
By Theorem 2.23 in \cite{ZhuBn},
there exists a positive integer $N$ such that for any $0<r<1$ we can
find a sequence $\{a_k\}$ in $\Bn$ with the following properties:
\begin{itemize}
\item[(i)] $\Bn=\cup_{k}D(a_k,r)$.
\item[(ii)] The sets $D(a_k,r/4)$ are mutually disjoint.
\item[(iii)] Each point $z\in\Bn$ belongs to at most $N$ of the sets $D(a_k,4r)$.
\end{itemize}
Any sequence $\{a_k\}$ satisfying the above conditions is called
an $r$-\emph{lattice}
(in the Bergman metric). Obviously any $r$-lattice is a separated sequence.\\
\subsection{Tent spaces of sequences}
For $0<p,q<\infty$ and a fixed separated sequence $Z=\{z_ j\}\subset \Bn$, let the tent space $T^p_ q=T^p_ q(Z)$ consist of those sequences $\lambda=\{\lambda_ j\}$ of complex numbers with
\[ \|\lambda \|_{T^p_ q}^p =\int_{\Sn} \!\!\Big (\!\!\sum_{z_ j\in \Gamma(\zeta)} \!|\lambda_ j |^q \Big )^{p/q} d\sigma(\zeta) <\infty.\]
\begin{otherp}\label{TKL}
Let $Z=\{a_ j\}$ be a separated sequence in $\Bn$ and let $0<p<\infty$. If $b>n\max(1,2/p)$, then the operator $T_{Z}: T^p_ 2(Z)\rightarrow H^p$ defined by
$$T_{Z}(\{\lambda_ j\})=\sum_ j \lambda_ j \,\frac{(1-|a_ j|^2)^{b}}{(1-\langle z, a_ j \rangle )^b}$$ is bounded.
\end{otherp}
\begin{proof}
See for example \cite{Ars, Jev,Lue1} or \cite{P1}.
\end{proof}
The sequence space $T^p_{\infty}=T^p_{\infty}(Z)$ consist of those sequences $\lambda=\{\lambda_ j\}$ of complex numbers with $\sup_{a_ k\in \Gamma(\zeta)} |\lambda_ k|\in L^p(\Sn)$. Set
\[\|\lambda \|_{T^p_ {\infty}}^p =\int_{\Sn}\Big (\sup_{a_ k\in \Gamma(\zeta)} |\lambda_ k| \Big )^p d\sigma(\zeta).\]
\begin{otherth}\label{TTD1}
Let $1<p<\infty$. The dual of $T^p_ 1$ can be identified with $T^{p'}_{\infty}$ under the pairing
\[\langle \lambda ,\mu \rangle =\sum_ k \lambda _ k \,\overline{\mu_ k} (1-|a_ k|^2)^n.\]
Under the same pairing, for $1<q<\infty$, the dual of $T^p_ q$ can be identified with $T^{p'}_{q'}$.
\end{otherth}
\begin{proof}
Again see \cite{Ars, Jev,Lue1}.
\end{proof}
\subsection{Forelli-Rudin type estimates}
We need the following well known integral estimate that has become very useful in this area of analysis (see \cite[Theorem 1.12]{ZhuBn} for example).
\begin{otherl}\label{IctBn}
Let $t>-1$ and $s>0$. There is a positive constant $C$ such that
\[ \int_{\Bn} \frac{(1-|w|^2)^t\,dv(w)}{|1-\langle z,w\rangle |^{n+1+t+s}}\le C\,(1-|z|^2)^{-s}\]
for all $z\in \Bn$.
\end{otherl}
We also need the following well known discrete version of the previous lemma.
\begin{otherl}\label{l2}
Let $\{z_ k\}$ be a separated sequence in $\Bn$, and let $n<t<s$.
Then
$$
\sum_{k}\frac{(1-|z_ k|^2)^t}{|1-\langle z,z_ k \rangle |^s}\le C\,
(1-|z|^2)^{t-s},\qquad z\in \Bn.
$$
\end{otherl}
The following more general version of Lemma \ref{IctBn} will also be needed. The proof can be found in \cite{OF}.
\begin{otherl}\label{FRgeneral}
Let $s>-1$, $s+n+1>r,t>0$, and $r+t-s>n+1$. For $a \in \Bn$ and $z\in \overline{\mathbb{B}}_ n$, one has
$$\int_{\Bn} \frac{(1-|w|^2)^s dv(w)}{|1-\langle z,w\rangle|^r |1-\langle a,w\rangle|^t} \leq C \frac{1}{|1-\langle z,a\rangle|^{r+t-s-n-1}}.$$
\end{otherl}
\mbox{}
\\
\subsection{Differential type operators}
We need to use the differential and integral type operators $R^{\alpha,t}$ and $R_{\alpha,t}$ for $\alpha\ge -1$ and $t\ge 0$ (see \cite[Section 1.4]{ZhuBn}). Recall that $R^{\alpha,t}$ is the unique continuous linear operator on $H(\Bn)$ satisfying
\begin{displaymath}
R^{\alpha,t}\left (\frac{1}{(1-\langle z,w \rangle )^{n+1+\alpha}}\right )=\frac{1}{(1-\langle z,w \rangle )^{n+1+\alpha+t}}
\end{displaymath}
for all $w\in \Bn$. Similarly, $R_{\alpha,t}$ is the unique continuous linear operator on $H(\Bn)$ satisfying
\begin{displaymath}
R_{\alpha,t}\left (\frac{1}{(1-\langle z,w \rangle)^{n+1+\alpha+t}}\right )=\frac{1}{(1-\langle z,w \rangle )^{n+1+\alpha}}
\end{displaymath}
for all $w\in \Bn$. It is well-known that $$R_{\alpha,t}R^{\alpha,t}=R^{\alpha,t}R_{\alpha,t}=I_d.$$
Most of the time we use these operators as follows. If a holomorphic function $f$ in $\Bn$ has an integral representation
\[
f(z)=\int_{\Bn} \frac{d\nu(w)}{(1-\langle z,w \rangle)^{n+1+\alpha}},
\]
then
\[
R^{\alpha,t} f(z)=\int_{\Bn} \frac{d\nu(w)}{(1-\langle z,w \rangle)^{n+1+\alpha+t}}.
\]
\subsection{Khinchine and Kahane inequalities} Consider a sequence of Rademacher functions $r_ k(t)$ (see \cite[Appendix A]{duren1}). For almost every $t\in (0,1)$ the sequence $\{r_ k(t)\}$ consists of signs $\pm 1$. We state first the classical Khinchine's inequality (see \cite[Appendix A]{duren1} for example).\\
\mbox{}
\\
\textbf{Khinchines's inequality:} Let $0<p<\infty$. Then for any sequence $\{c_k\}$ of complex numbers, we have
\begin{displaymath}
\left(\sum_k |c_k|^2\right)^{p/2}\asymp \int_0^1 \left|\sum_k c_kr_k(t)\right|^pdt.
\end{displaymath}
\mbox{}
\\
The next result is known as Kahane's inequality; see for instance Lemma 5 of Luecking \cite{Lue2} or the paper of Kalton \cite{Kal}.\\
\mbox{}
\\
\textbf{Kahane's inequality:} Let $X$ be a Banach space, and $0<p,q<\infty$. For any sequence $\{x_ k\}\subset X$, one has
\begin{displaymath}
\left ( \int_{0}^1 \Big \|\sum_ k r_ k(t)\, x_ k \Big \|_{X}^q dt\right )^{1/q} \asymp \left ( \int_{0}^1 \Big \|\sum_ k r_ k(t)\, x_ k \Big \|_{X}^p dt\right )^{1/p}.
\end{displaymath}
\section{Proof of Theorem \ref{T2}}\label{S3}
\subsection{\textbf{Necessity}}
If $Q_{\mu}:H^p\rightarrow H^q$ is bounded, by the pointwise estimate for $H^p$-functions (see \cite[Theorem 4.17]{ZhuBn}) we get
\[
\big |Q_{\mu} f(z) \big | \le \frac{1}{(1-|z|^2)^{n/q}} \,\big \| Q_{\mu} f \big \|_{H^q}.
\]
Taking the function $f$ to be the reproducing kernel $K_ z$ of $H^2$, that is,
\[
f (w)= K_ z (w)=(1-\langle w,z \rangle )^{-n}
\]
and taking into account that (an immediate application of \cite[Theorem 1.12]{ZhuBn})
\[
\|K_ z\|_{H^p}\lesssim (1-|z|^2)^{-n(p-1)/p},
\]
we obtain
\[
\begin{split}
\int_{\Bn} \frac{d\mu(w)}{|1-\langle z,w \rangle |^{2n}}&=\big |Q_{\mu} K_ z(z) \big |
\\
&\le \frac{1}{(1-|z|^2)^{n/q}} \,\,\big \| Q_{\mu}\|_{H^p\rightarrow H^q} \cdot \|K_ z\|_{H^p}
\\
&\le \frac{C}{(1-|z|^2)^{n \big(1/q+(p-1)/p \big)}} \,\,\big \| Q_{\mu}\|_{H^p\rightarrow H^q}.
\end{split}
\]
Hence, from \eqref{sCM} with $t=n (\frac{1}{q}+\frac{(p-1)}{p} )$ we see that $\mu$ is an $s$-Carleson measure with $s=1+\frac{1}{p}-\frac{1}{q}$, and moreover
\[
\|\mu\|_{CM_ s}\le C \big \| Q_{\mu}\|_{H^p\rightarrow H^q}.
\]
\subsection{\textbf{Sufficiency}}
Let $1<p\le q<\infty$ and let $\mu$ be an $s$-Carleson measure with $s=1+\frac{1}{p}-\frac{1}{q}$. Take $t> 0$ satisfying
$n+t>ns$. By the density of the holomorphic polynomials and duality, it suffices to show that
\begin{equation}\label{E-B1}
I_ t(Q_{\mu}f,g):=\left |\int_{\Bn} R^{-1,t}(Q_{\mu}f)(z)\,\overline{R^{t-1,t}g(z)}\,dv_ {2t-1}(z) \right | \le C \|f\|_{H^p}\cdot \|g\|_{H^{q'}}
\end{equation}
for holomorphic polynomials $f$ and $g$ (it is easy to see that, for holomorphic polynomials $P$ and $Q$, one has $\langle P,Q \rangle _{H^2}=\int_{\Bn} R^{-1,t} P(z)\overline{R^{t-1,t} Q(z)}\, dv_{2t-1}(z)$). Observe that $R^{t-1,t}g$ is also a holomorphic polynomial (this follows from the expression of $R^{t-1,t}g$ in terms of the homogeneous expansion of $g$. See \cite[Chapter 1] {ZhuBn}). This together with \eqref{sCM}, gives
\[
\begin{split}
\int _{\Bn}\left ( \int_{\Bn} \frac{|f(w)|\,d\mu(w)}{|1-\langle w,z\rangle |^{n+t}}\right ) &|R^{t-1,t}g(z)|\,dv_{2t-1}(z)\\
&\le \|f\|_{\infty}\cdot \|R^{t-1,t} g\|_{\infty} \int_{\Bn} \left (\int_{\Bn}\frac{d\mu(w)}{|1-\langle w,z \rangle |^{n+t}}\right ) dv_{2t-1}(z)\\
\\
& \lesssim \|f\|_{\infty}\cdot \|R^{t-1,t}g\|_{\infty} \cdot\|\mu\|_{CM_ s} \int_{\Bn} dv_{t-1+ns-n}(z)<\infty,
\end{split}
\]
and we can use Fubini's theorem and the properties of the operators $R^{\beta,t}$ to get
\[
\begin{split}
\int_{\Bn} R^{-1,t}(Q_{\mu}f)(z)\,\overline{R^{t-1,t}g(z)}\,dv_{2t-1}(z)&=\int_{\Bn} \left ( \int_{\Bn} \frac{f(w)\,d\mu(w)}{(1-\langle z,w \rangle )^{n+t}}\right ) \overline{R^{t-1,t}g(z)}\,dv_{2t-1}(z)
\\
&=\int_{\Bn} f(w) \left (\int_{\Bn} \frac{\overline{R^{t-1,t}g(z)}\,dv_{2t-1}(z)}{(1-\langle z,w \rangle )^{n+t}} \right ) \,d\mu(w)
\\
& =\int_{\Bn} f(w)\,\overline{R_{t-1,t}R^{t-1,t}g(w)}d\mu(w)
\\
& =\int_{\Bn} f(w)\,\overline{g(w)} \, d\mu(w).
\end{split}
\]
This and H\"{o}lder's inequality give
\begin{equation}\label{E-B2}
I_ t(Q_{\mu}f,g) \le \int_{\Bn} |f(w)\,g(w)|\,d\mu(w)\le \|f\|_{L^{\sigma}(\mu)}\cdot \|g\|_{L^{\sigma'}(\mu)}
\end{equation}
with $\sigma=ps\ge p$.
As $\mu$ is a $(\sigma/p)$-Carleson measure, by Carleson-Duren's theorem we have
\[
\|f\|_{L^{\sigma}(\mu)} \le C_ 1 \big \|\mu \big \|^{1/\sigma}_{CM_ s}\cdot \|f\|_{H^p}.
\] Also,
\[
\frac{\sigma '}{q'}=\frac{ps(q-1)}{q(ps-1)}=\frac{ps(q-1)}{q(p-p/q)}=s.
\]
Therefore, we also have
\[
\|g\|_{L^{\sigma'}(\mu)}\le C_ 2 \big \|\mu \big \|^{1/\sigma'}_{CM_ s}\cdot \|g\|_{H^{q'}}.
\]
Bearing in mind \eqref{E-B2}, we see that \eqref{E-B1} holds, proving that $Q_{\mu}:H^{p}\rightarrow H^q$ is bounded, and moreover
\[
\big \|Q_{\mu}\big \|_{H^p\rightarrow H^q} \lesssim \|\mu\|_{CM_ s}.
\]
\section{Proof of Theorem \ref{T3}}\label{s4}
\subsection{\textbf{Sufficiency}}
Suppose first that $\widetilde{\mu}$ belongs to $L^{r}(\Sn)$. Observe that $r>1$ so that by Luecking's theorem it follows that
$$\int_{\Bn} |f(z)|^s \,d\mu(z)\le C \|f\|_{H^p}^s$$
whenever $f$ is in $H^p$, with $s=p+1-\frac{p}{q}$. Testing this inequality on the functions
\[
f_ a(z)=\frac{1}{(1-\langle z,a \rangle )^{\sigma}},\qquad a\in \Bn,
\]
with $\sigma$ big enough, we see that $\mu$ is an $s/p$-Carleson measure. Note that, as $r>1$ and $q<p$, one has $0<s/p<1$. Let $t>0$ with $t+ns/p>n$ (observe that this implies $t+n>ns/p$ because $s/p<1$). As in the previous proof, and with the same notation as in \eqref{E-B1}, we need to show that $$|I_ t(Q_{\mu} f,g) | \le C\, \|\widetilde{\mu}\|_{L^r(\Sn)}\cdot\|f\|_{H^p}\cdot \|g\|_{H^{q'}}$$
for holomorphic polynomials $f$ and $g$. Because $\mu$ is an $(s/p)$-Carleson measure, proceeding as before we can justify the use of Fubini's theorem that gives
\[
\begin{split}
\big | I_ t(Q_{\mu} f,g) \big | & \le \int_{\Bn} |f(z)| \,|g(z)|\,d\mu(z)\asymp \int_{\Sn} \int_{\Gamma(\zeta)}\frac{|fg(z)|}{(1-|z|^2)^{n}} d\mu(z)\,d\sigma(\zeta)\\
\\
&\le \int_{\Sn} |(fg)^*(\zeta)|\,\widetilde{\mu} (\zeta)\,d\sigma(\zeta) \le C \,\|\widetilde{\mu}\|_{L^r(\Sn)} \cdot \|fg \|_{H^{r'}}
\end{split}
\]
with $r'$ being the conjugate exponent of $r$ (the last inequality follows from H\"{o}lder's inequality and Theorem \ref{NTMT}). Finally, H\"{o}lder's inequality gives $\|fg\|_{H^{r'}}\le \|f\|_{H^p}\cdot \|g\|_{H^{q'}}$ finishing the proof of the sufficiency.\\
\subsection{\textbf{Preliminaries for the necessity}}
Set $Q(0)=\Bn$, and given $w \in \Bn\setminus \{0\}$, we write $\zeta_w =w/|w|$ and denote
$$Q(w)=\big \{ z\in \Bn: |1-\langle z, \zeta_w \rangle |<1-|w| \big \}.$$
\begin{lemma}\label{LN1}
Let $w\in \Bn$. Then $1-|w|\asymp |1-\langle z,w \rangle |$ for $z\in Q(w)$.
\end{lemma}
\begin{proof}
The result is trivial for $w=0$. Hence, assume that $w\neq 0$ and $z\in Q(w)$. Then
\begin{displaymath}
\begin{split}
|1-\langle z,w \rangle | &\lesssim |1-\langle z,\zeta_ w \rangle |+|1-\langle \zeta_ w,w \rangle |\\
\\
& \le (1-|w|)+|1-\langle \zeta_ w,w \rangle |=2(1-|w|).
\end{split}
\end{displaymath}
Since the other inequality is trivial, we are done.
\end{proof}
\begin{lemma}\label{discrete}
Let $1<p<\infty$, $\nu$ be a positive Borel measure, finite on compact sets, and consider the general area operator
$$A_\nu g(\zeta)=\left(\int_{\Gamma(\zeta)}|g(z)|^2 d\nu(z)\right)^{1/2}$$
and the discrete maximal operator
$$C_\nu^*(g)(\zeta)=\left(\sup_{a_k \in \Gamma(\zeta)}\frac{1}{(1-|a_k|)^n}\int_{Q(a_k)}(1-|z|^2)^n |g(z)|^2 d\nu(z)\right)^{1/2}
,$$
where $\{a_k\}$ is an $r$-lattice. Suppose that $g\in L^2(\Bn,d\nu_ n)$ with $d\nu_ n(z)=(1-|z|^2)^n d\nu(z)$. If $r$ is small enough and $C_\nu^*g \in L^p(\Sn)$, then $A_\nu g \in L^p(\Sn)$. Moreover, $r>0$ can be chosen to guarantee that
$$\|A_\nu g\|_{L^p(\Sn)}\lesssim \|C_\nu^*g\|_{L^p(\Sn)}+\|g\|_{L^2(\nu_ n)}.$$
\end{lemma}
\begin{proof}
The proof uses some terminology and concepts related to tent spaces, which are only covered in this paper for discrete measures. Moreover, a substantial part of the proof is remarkably similar to a known standard proof, see for instance \cite{Pel}. For completeness, we will prove this lemma by using notation analogous to that of the mentioned reference, and the reader should have no difficulties comparing the two proofs.
We want to show
\[
\big |\langle f,g \rangle_{T^2_2(\nu)}\big |\lesssim \int_{\Sn} A_\nu(f)(\zeta)C^*_\nu(g)(\zeta)d\sigma(\zeta)+ \|A_{\nu} f\|_{L^1(\Sn)}\cdot\|g\|_{L^2(\nu_ n)}.
\]
By duality of tent spaces, it then follows that if $C^*_\nu g \in L^p(\Sn)$, then $g$ belongs to the dual of $T^{p'}_2(\nu)$, which is $T^p_2(\nu)$. This implies that $A_\nu g \in L^p(\Sn)$ with the desired estimate.\\
Fix $K>0$ large enough to be specified later, and set $M_K=\{|z|^2 >1-1/K^2\}$. Split the measure $\nu$ as $\nu=\nu_{|_{M_ K}} +\nu'$ with $\nu'=\nu_{|_{\Bn\setminus M_ K}}$. Then
\[
\langle f,g \rangle_{T^2_2(\nu)}=\langle f,g \rangle_{T^2_2(\nu_{|_{M_ K}})}+\langle f,g \rangle_{T^2_2(\nu')}.
\]
Moreover, we have
\[
\begin{split}
\big |\langle f,g \rangle_{T^2_2(\nu')}\big | & \le \int_{\Bn} |f(z)|\,|g(z)|\,(1-|z|^2)^n\,d\nu'(z)
\\
&\asymp \int_{\Sn} \left (\int_{\Gamma(\zeta)} |f(z)|\,|g(z)|\,d\nu'(z)\right )\,d\sigma(\zeta)
\\
&\le \|g\|_{L^2(\nu_ n)}\int_{\Sn} \left (\int_{\Gamma(\zeta)} |f(z)|^2\,d\nu'(z)\right )^{1/2}\,d\sigma(\zeta)
\\
&\le K^n\,\|A_{\nu} f\|_{L^1(\Sn)}\cdot\|g\|_{L^2(\nu_ n)}.
\end{split}
\]
Therefore, it is enough to show that, for $\nu$ supported on $M_ K$ one has
\[
\big |\langle f,g \rangle_{T^2_2(\nu)}\big |\lesssim \int_{\Sn} A_\nu(f)(\zeta)\,C^*_\nu(g)(\zeta)\,d\sigma(\zeta).
\]
\mbox{}
\\
We note first that if $\gamma>1$, $\zeta \in \Sn$, and $z \in \Gamma(\zeta)$, we have for $R \in (0,1)$ the estimate
\begin{equation}\label{R}
|1-\langle Rz,\zeta\rangle|<\left(1-\frac{\gamma}{2}\right)(1-R)+\frac{\gamma}{2}(1-R|z|^2).
\end{equation}
If $1>|z|^2 > \alpha$ with $\alpha>(2/\gamma-1)$, then for $R$ close enough to $1$, we have $Rz \in \Gamma(\zeta)$. (Note that if $\gamma\geq 2$, any $z\neq 0$ and $R\in (0,1)$ will work.)\\
We will use a variation of a well-known argument, which can be found with details in \cite{Pel}, for instance. Define
$$\Gamma^h(\zeta)=\Gamma(\zeta)\setminus \overline{D(0,1/(h+1))},$$
and
$$A_\nu(g|h)(\zeta)=\left(\int_{\Gamma^h(\zeta)}|f(z)|^2d\nu(z)\right)^{1/2}.$$
Now set
$$h(\zeta)=\sup \big \{h:A_\nu(g|h)(\zeta) \leq C_1 C^*_\nu(g)(\zeta)\big \}.$$
The claim is proven, once we show
\[
\int_{\Bn} k(z)(1-|z|^2)^nd\nu(z)\leq 2\, \int_{\Sn}\left(\int_{\Gamma^{h(\zeta)}(\zeta)}k(z)d\nu(z)\right)d\sigma(\zeta)
\]
for every positive $\nu$-measurable function $k$.
By Fubini's theorem, the integral on the right-hand-side equals
\[
\int_{\Bn}\sigma(I(z)\cap H(z))\,k(z)\,d\nu(z),
\]
with $H(z)=\{\zeta \in \Sn: 1/(1+h(\zeta))\leq |z|\}$. We therefore want to show that
\[
\frac{\sigma(I(z)\cap H(z))}{\sigma(I(z))}\geq \frac{1}{2}
\]
holds for $z\in M_K$. To this end, take $z \in M_K$ and set $z^*$ be the unique point in $\Bn$ satisfying $\zeta_{z^*}=\zeta_z$ and $(1-|z^*|^2)=K(1-|z|^2)$. Suppose, for now, that $K$ is chosen to be large enough so that the conditions on \eqref{R} are met. Then, there exists $r_K>0$ with the following properties.
\begin{itemize}
\item[(i)] If $0<r<r_K$, then $D(z^*,r)\subset \bigcap_{v \in I(z)} \Gamma(v)$;
\item[(ii)] If $z' \in D(z^*,r)$, then $2(1-|z'|^2)\geq (1-|z^*|^2)$;
\item[(iii)] If $z'\in D(z^*,r)$, and $z'_*$ is the unique point in $\Bn$ satisfying $\zeta_{z'}=\zeta_{z'_*}$ and $K(1-|z'_*|^2)=(1-|z'|^2)$, then $|1-\langle z,z'_*\rangle|\leq 2(1-|z|^2)$.
\end{itemize}
Now, suppose that $r>0$ above is the density of the lattice and set $z'$ to be a point of the lattice contained in $D(z^*,r)$. Notice that if $u$ satisfying $|u|\geq |z|$ (so that $1-|u|^2\leq 1-|z|^2$) does not belong to $Q(z')$, then by the property (ii) above (recall that $d(a,b)=|1-\langle a,b\rangle|^{1/2}$ satisfies the triangle inequality on $\overline{\Bn}$)
\begin{align*}
|1-\langle u,\zeta_z\rangle|^{1/2}&\geq |1-\langle u,\zeta_{z'}\rangle|^{1/2}-|1-\langle \zeta_z,\zeta_{z'}\rangle|^{1/2}\\
&\geq (K/2)^{1/2}(1-|z|^2)^{1/2}-|1-\langle \zeta_{z^*},\zeta_{z'}\rangle|^{1/2}.
\end{align*}
An application of Exercise 1.25 from \cite{ZhuBn} together with property (iii) above shows us that
$$|1-\langle \zeta_{z^*},\zeta_{z'}\rangle|\leq 8(1-|z|^2).$$
Let us now set $K$ to be a number big enough so that $((K/2)^{1/2}-8^{1/2})^2>K/3$. Then
$u \notin Q(z')$ implies that $$|1-\langle u,\zeta_z\rangle|\geq (K/3)(1-|z|^2).$$
If now $\zeta \in I(z)$, then
$$|1-\langle u,\zeta\rangle|^{1/2}\geq |1-\langle u,\zeta_z\rangle|^{1/2}-|1-\langle z,\zeta_z\rangle|^{1/2}-|1-\langle z,\zeta\rangle|^{1/2},$$
so
$$|1-\langle u,\zeta\rangle|\geq \left((K/3)^{1/2}-1-(\gamma/2)^{1/2}\right)^2 (1-|z|^2)\geq (\gamma/2)(1-|u|^2),$$
when $K$ is large enough compared to $\gamma$. We now fix $K$ big enough to carry us through all the calculations above.
From this point onwards, we could follow the argument of \cite{Pel}, as the only real difference was that we had to choose the point $z'$ from the lattice. We will present the remaining details for the convenience of the reader.
By our choice of $K$ and $r$, if $|u|\geq |z|$ does not belong to $Q(z')$, then $I(u)\cap I(z)=\emptyset$. Thus, if $x=1/|z|-1$ (so that $\Gamma^x(\zeta)=\Gamma(\zeta)\setminus \overline{D(0,|z|)}$), then
\[
\begin{split}
\frac{1}{\sigma(I(z))}\int_{I(z)} &\left(\int_{\Gamma^x(\zeta)}|g(u)|^{2}d\nu(u)\right)d\sigma(\zeta)\\
&=\frac{1}{\sigma(I(z))}\int_{\{|z|<|u|<1\}}\sigma(I(z)\cap I(u))|g(u)|^2d\nu(u) \\
&\leq\frac{1}{\sigma(I(z))}\int_{Q(z')}\sigma(I(z)\cap I(u))|g(u)|^2d\nu(u)\\
&\leq\frac{C_3}{\sigma(I(z'))}\int_{Q(z')}(1-|u|^2)^n |g(u)|^2 d\nu(u)\\
&\leq C_3 \inf_{v \in I(z)}C^*_\nu(g)(v)^2.
\end{split}
\]
The last inequality holds, because by property (i), we have $z'\in D(z^*,r)\subset \Gamma(v)$ for all $v \in I(z)$.
Now, let us chooce $C_1$ so that $C_1^2>2C_3$. If $E(z)=\Sn\setminus H(z)$, then
\begin{align*}
\sigma(E(z)\cap I(z))&\leq \int_{I(z)}\frac{A_\nu(g|x)(\zeta)^2}{C_1^2 C_\nu^*(g)(\zeta)^2}d\sigma(\zeta)\\
&\leq \frac{1}{C_1^2 \inf_{v \in I(z)}C^*_\nu(g)(v)^2}\int_{I(z)}A_\nu(g|x)(\zeta)^2d\sigma(\zeta)\\
&<\sigma(I(z))/2.
\end{align*}
It follows that $\sigma(I(z)\cap H(z)) \geq \sigma(I(z))/2$ for $z \in M_K$. Note also that the implicit constants in the estimate can be chosen to be independent of the measure $\nu$. This finishes the proof.
\end{proof}
\mbox{}
\\
We also need the following more general area function description of the Hardy spaces. The result is probably known to experts, but we were unable to find a proof in the literature.
\begin{proposition}\label{FracArea}
Let $f$ be holomorphic on $\Bn$, $0<p<\infty$ and $d\lambda_n(z)=dv_{-1-n}(z)$. If $s\geq -1$ and $t>0$, then the following are equivalent.
\begin{enumerate}
\item[(a)] $f \in H^p(\Bn)$;
\item[(b)] $\left(\int_{\Gamma(\zeta)}|Rf(z)|^2 (1-|z|^2)^2 d\lambda_n(z)\right)^{1/2} \in L^p(\Sn);$
\item[(c)] $\left(\int_{\Gamma(\zeta)}|R^{s,t}f(z)|^2 (1-|z|^2)^{2t} d\lambda_n(z)\right)^{1/2} \in L^p(\Sn).$
\end{enumerate}
Moreover, if $f(0)=0$, then the $L^p$ norms involved in all the items above are equivalent.
\end{proposition}
\begin{proof}
The equivalence of (a) and (b) can be found in \cite{P1}. So, we will prove that (b) and (c) are equivalent. To this end, we may clearly assume that $f(0)=0$.
Let us assume (b). Since $f(0)=0$, we have the estimate
$$f(w)=\int_{\Bn}Rf(u)L_\beta(w,u) dv_\beta(u),$$
where
$$L_\beta(w,u)=\int_0^1 \left(\frac{1}{(1-\langle w,\rho u\rangle)^{n+1+\beta}}-1\right)\frac{d\rho}{\rho}$$
valid for large enough $\beta$, see page 51 of \cite{ZhuBn}. Moreover, if $\beta=s+N$ for some positive integer $N$, we have by Proposition 5 of \cite{ZZ},
$$R^{s,t}\frac{1}{(1-\langle w,\rho u\rangle)^{n+1+\beta}}=\frac{\phi(\langle w,\rho u\rangle)}{(1-\langle w,\rho u\rangle)^{n+1+\beta+t}},$$
where $\phi$ is a one variable polynomial of degree $N$. Note that $R^{s,t}1=1$, so putting $\rho=0$ gives $1$ in the above identity. Now, it is straightforward to obtain
$$|R^{s,t}L_\beta (w,u)|\lesssim \frac{1}{|1-\langle w,u\rangle|^{n+\beta+t}}.$$
This allows us to obtain the bound
$$|R^{s,t}f(w)|\lesssim \int_{\Bn} \frac{|Rf(u)|}{|1-\langle w,u\rangle|^{n+\beta+t}}dv_\beta(u).$$
We may assume that $\beta=s+N>n\max(1,2/p)-n+1$. By standard estimates, this leads to
$$(1-|w|^2)^{2t}|R^{s,t}f(w)|^2 \lesssim \int_{\Bn} \frac{|Rf(u)|^2dv_{\beta}(u)}{|1-\langle w,u\rangle|^{n+\beta+t-1}}(1-|w|^2)^{t}.$$
Now, by Lemma \ref{FRgeneral}, if $\theta>n+\beta-1$, we have
\begin{align*}
&\int_{\Gamma(\zeta)}\left[(1-|w|^2)^t |R^{s,t}f(w)|\right]^2d\lambda_n(w)\\
\lesssim&\int_{\Bn}\left[(1-|w|^2)^t |R^{s,t}f(w)|\right]^2\left(\frac{1-|w|^2}{|1-\langle \zeta,w\rangle|}\right)^\theta d\lambda_n(w)\\
\lesssim&\int_{\Bn} \left(\frac{1-|w|^2}{|1-\langle \zeta,w\rangle|}\right)^\theta \left[ \int_{\Bn} \frac{|Rf(u)|^2dv_{\beta}(u)}{|1-\langle w,u\rangle|^{n+\beta+t-1}}\right](1-|w|^2)^{t} d\lambda_n(w)\\
\lesssim& \int_{\Bn} |Rf(u)|^2\left(\int_{\Bn}\frac{(1-|w|^2)^{\theta+t-n-1} dv(w)}{|1-\langle \zeta,w\rangle|^\theta |1-\langle w,u\rangle|^{n+\beta+t-1}}\right) dv_{\beta}(u)\\
\lesssim& \int_{\Bn} |Rf(u)|^2 (1-|u|^2)^2 \left(\frac{1-|u|^2}{|1-\langle \zeta,u\rangle|}\right)^{n+\beta-1} d\lambda_n(u)
\end{align*}
Now, since $n+\beta-1>n\max(1,2/p)$, we can use Lemma \ref{Gamma} to get
\begin{align*}
&\int_{\Sn} \left(\int_{\Gamma(\zeta)}\left[(1-|w|^2)^t |R^{s,t}f(w)|\right]^2d\lambda_n(w)\right)^{p/2} d\sigma(\zeta)\\
\lesssim &\int_{\Sn} \left(\int_{\Bn} |Rf(u)|^2 (1-|u|^2)^2 \left(\frac{1-|u|^2}{|1-\langle \zeta,u\rangle|}\right)^{n+\beta-1} d\lambda_n(u)\right)^{p/2}d\sigma(\zeta)\\
\lesssim &\int_{\Sn} \left(\int_{\Gamma(\zeta)} |Rf(u)|^2 (1-|u|^2)^2 d\lambda_n(u)\right)^{p/2}d\sigma(\zeta),
\end{align*}
so (c) is obtained.\\
Suppose now that (c) holds. By an estimate from \cite[page 20]{ZZ}, for large enough $\beta$ we have
$$(1-|z|^2)|Rf(z)|\lesssim (1-|z|^2)\int_{\Bn}\frac{(1-|w|^2)^t|R^{s,t}f(w)|dv_\beta(w)}{|1-\langle z,w\rangle|^{n+2+\beta}}.$$
We may assume that $\beta>n\max(1,2/p)-n-1$ and let $0<\varepsilon<2$ so that $\beta-\varepsilon>-1$. Then use Cauchy-Schwarz and standard integral estimates to deduce
$$\left[(1-|z|^2)|Rf(z)|\right]^2 \lesssim (1-|z|^2)^2 \int_{\Bn}\frac{(1-|w|^2)^{2t}|R^{s,t}f(w)|^2 dv_{\beta+\varepsilon}(w)}{|1-\langle z,w\rangle|^{n+3+\beta}} (1-|z|^2)^{-\varepsilon}.$$
Now, take $\theta>n\max(1,2/p)+\varepsilon+1+\beta$, and estimate as before by using Lemma \ref{FRgeneral} to obtain
\begin{align*}
&\int_{\Gamma(\zeta)}\left[(1-|z|^2)|Rf(z)|\right]^2 d\lambda_n(z)\\
&\lesssim \int_{\Bn} \left[(1-|z|^2)|Rf(z)|\right]^2 \left(\frac{1-|z|^2}{|1-\langle z,\zeta\rangle|}\right)^{\theta} d\lambda_n(z)\\
&\lesssim \int_{\Bn} (1-|w|^2)^{2t}|R^{s,t}f(w)|^2 \left(\int_{\Bn} \frac{(1-|z|^2)^{2+\theta-\varepsilon-n-1}dv(z)}{|1-\langle z,\zeta\rangle|^\theta \,|1-\langle z,w\rangle|^{n+3+\beta}}\right)dv_{\beta+\varepsilon}(w)\\
&\lesssim \int_{\Bn} (1-|w|^2)^{2t}|R^{s,t}f(w)|^2 \left(\frac{1-|w|^2}{|1-\langle w,\zeta\rangle|}\right)^{n+1+\beta+\varepsilon} d\lambda_n(w).
\end{align*}
Since $n+1+\beta+\varepsilon>n\max(1,2/p)+\varepsilon$, from Lemma \ref{Gamma} we conclude that (c) implies (b). This finishes the proof.
\end{proof}
\subsection{\textbf{Necessity}}
Throughout this proof, we set $\|Q_{\mu}\|=\|Q_{\mu}\|_{H^p\rightarrow H^q}$. Since $\mu(\Bn)=(Q_{\mu}1) (0)$, the measure $\mu$ is finite with $\mu(\Bn)\lesssim \|Q_{\mu}\|$. We split the proof in several cases.
\subsubsection{\textbf{The case $q=2$}}
It is well known that the boundedness is equivalent to
\[
\int_{\Bn} | R^{-1,1}(Q_{\mu} f)(z)|^2 \,dv_ 1(z) \le C \,\|Q_{\mu}\|^2 \cdot \|f\|^2_{H^p}.
\]
For example, this can be deduced from Proposition \ref{FracArea} and the estimate \eqref{EqG}.\\
Let $\{a_ k\}\subset \Bn$ be a separated sequence and define
\[
f_ k(z)=\left (\frac{1-|a_ k|^2}{1-\langle z,a_ k \rangle}\right )^{n+1}, \qquad z\in \Bn.
\]
Apply the previous inequality with the function $$F_ t(z)=\sum_ k \lambda_ k\,r_ k(t)\, f_ k(z)$$ with $\lambda=\{\lambda_ k\}\in T^p_ 2$, and use Proposition \ref{TKL} to get
\[
\int_{\Bn} \Big |\sum_ k \lambda_ k \,r_ k(t)\, R^{-1,1}(Q_{\mu} f_ k)(z)\Big |^2 \,dv_ 1(z) \le C \,\|Q_{\mu}\|^2 \cdot \|\lambda\|^2_{T^p_ 2}.
\]
Integrate respect to $t$ between $0$ and $1$, interchange the order of integration and use Khinchine's inequality to obtain
\[
\sum_ k |\lambda_ k|^2\, \int_{\Bn} \big | R^{-1,1}(Q_{\mu} f_ k)(z)\big |^2 \,dv_ 1(z) \le C \,\|Q_{\mu}\|^2\cdot \|\lambda\|^2_{T^p_ 2}.
\]
By subharmonicity (see \cite[Lemma 2.24]{ZhuBn} for example) this implies
\[
\sum_ k |\lambda_ k|^2\, \big | R^{-1,1}(Q_{\mu} f_ k)(a_ k)\big |^2 \,(1-|a_ k|)^{n+2} \le C \|Q_{\mu}\|^2 \cdot \|\lambda\|^2_{T^p_ 2}.
\]
and therefore
\[
\sum_ k |\lambda_ k|^2\, \big | R^{-1,1}(Q_{\mu} f_ k)(a_ k)\big |^2 \,(1-|a_ k|)^{n+2} \le C \,\|Q_{\mu}\|^2\cdot \|\lambda^2\|_{T^{p/2}_ 1}.
\]
By the duality of tent spaces in Theorem \ref{TTD1}, this is equivalent to
\[
\left \{ \big | R^{-1,1}(Q_{\mu} f_ k)(a_ k)\big |^2 \,(1-|a_ k|)^2\right \} \in T_{\infty}^{p/(p-2)},
\]
with the corresponding $T_{\infty}^{p/(p-2)}$-norm dominated by $\|Q_{\mu}\|^2$. That is,
\[
\sup_{a_ k\in \Gamma (\zeta)} \big | R^{-1,1}(Q_{\mu} f_ k)(a_ k)\,(1-|a_ k|)\big |^2 \in L^{p/(p-2)}(\Sn)
\]
or
\begin{equation}\label{Eqq21}
\sup_{a_ k\in \Gamma (\zeta)} \big | R^{-1,1}(Q_{\mu} f_ k)(a_ k)\big |\,(1-|a_ k|) \in L^{2p/(p-2)}(\Sn)
\end{equation}
with the corresponding $L^{2p/(p-2)}$-norm dominated by $\|Q_{\mu}\|$.
However, since $(1-|a_ k|)\asymp |1-\langle a_ k,w \rangle |$ for $w\in Q(a_ k)$,
\[
\begin{split}
\big | R^{-1,1}(Q_{\mu} f_ k)(a_ k)| \,(1-|a_ k|)&= (1-|a_ k|^2)^{n+2} \left |\int_{\Bn} \!\!\frac{d\mu(w)}{|1-\langle a_ k,w \rangle |^{2(n+1)}} \right | \\
\\
& \ge (1-|a_ k|^2)^{n+2} \int_{Q(a_ k)} \frac{d\mu(w)}{|1-\langle a_ k,w \rangle |^{2(n+1)}} \\
\\
& \gtrsim \frac{1}{(1-|a_ k|)^n} \int_{Q(a_ k)} (1-|w|)^n\frac{d\mu(w)}{(1-|w|)^n}.
\end{split}
\]
Let us restate this estimate for later reference.
\begin{equation}\label{estim}
\frac{\mu(Q(a_k))}{(1-|a_k|)^n} \le C\,\big |R^{-1,1}(Q_{\mu} f_ k)(a_ k)\big |\,(1-|a_ k|) .
\end{equation}
Finally, the result follows from \eqref{Eqq21}, \eqref{estim} and Lemma \ref{discrete} applied to the measure $d\nu(z)=d\mu(z)(1-|z|)^{-n}$ and $g=1$.
\subsubsection{\textbf{The case $q>2$}}
Let $\{a_k\}$ be a separated sequence in $\Bn$. Using the general area function description of $H^q$ given in Proposition \ref{FracArea}, and the same argument (applying Khinchine's inequality) and test functions as in the previous case, we arrive at
\begin{align*}
&\int_{\Sn} \left ( \sum_ k |\lambda_ k|^2 \int_{\Gamma (\zeta)} |R^{-1,1}(Q_{\mu} f_ k)(z)|^ 2\,dv_{1-n}(z)\right )^{q/2} d\sigma(\zeta)\\
\leq & C'\int_{\Sn}\left(\int_0^1 \int_{\Gamma(\zeta)} \left|\sum_k\lambda_k r_k(t)R^{-1,1}(Q_{\mu}f_k)(z)\right|^2dv_{1-n}(z)dt\right)^{q/2}d\sigma(\zeta)\\
\leq& C'\int_{0}^{1} \,\int_{\Sn}\left(\int_{\Gamma(\zeta)}\left|R^{-1,1}(Q_{\mu}F_ t)(z)\right|^2dv_{1-n}(z)\right)^{q/2}d\sigma(\zeta)\,dt \le C \,\|Q_{\mu}\|^q\cdot \|\lambda \|^q_{T^p_ 2}.
\end{align*}
Applying Lemma \ref{Gamma} with $\theta$ big enough we have
$$
\int_{\Sn} \left[\sum_{k}|\lambda_k|^2 \left(\frac{1-|a_k|^2}{|1-\langle a_k,\zeta\rangle|}\right)^\theta\int_{D_k} |R^{-1,1}(Q_\mu f_k)(z)|^2 dv_{1-n}(z)\right]^{q/2}d\sigma(\zeta)\leq C\|Q_\mu\|^q\cdot \|\lambda \|^q_{T^p_ 2},
$$
where $D_ k$ denotes the Bergman metric ball $D(a_ k,r)$. Now, because $|1-\langle a_ k,\zeta \rangle | \asymp 1-|a_ k|$ for $a_ k\in \Gamma(\zeta)$, summing only over indices $k$ so that $a_k \in \Gamma(\zeta)$, we obtain
\[
\int_{\Sn} \left[\sum_{k:a_ k\in \Gamma (\zeta)}|\lambda_k|^2 \int_{D_k} |R^{-1,1}(Q_\mu f_k)(z)|^2 dv_{1-n}(z)\right]^{q/2}d\sigma(\zeta)\leq C\|Q_\mu\|^q\cdot \|\lambda \|^q_{T^p_ 2}.
\]
By subharmonicity and the estimate \eqref{estim}, this gives
\begin{equation}\label{EqQ2b}
\int_{\Sn} \left ( \sum_ {a_ k \in \Gamma (\zeta)} |\lambda_ k|^2 \,\left (\frac{\mu(Q(a_ k))}{(1-|a_ k|)^n}\right )^2 \right )^{q/2} d\sigma(\zeta) \lesssim \|Q_\mu\|^q\|\cdot \|\lambda \|^q_{T^p_ 2}.
\end{equation}
Now, let $\beta=(pq-p-q)/(p-q)$ so that $q(\beta-1)/(q-2)=r$. Since $p>q>2$ we see that $\beta>1$. By \eqref{EqG} we have
\begin{equation}\label{Eq-Q>2}
\begin{split}
\sum_ k |\lambda_ k|^2 &\left ( \frac{\mu(Q(a_ k))}{(1-|a_ k|)^n}\right )^{\beta+1} (1-|a_ k|)^n \asymp \int_{\Sn} \left (\sum_ { a_ k\in \Gamma(\zeta)} |\lambda_ k|^2 \left ( \frac{\mu(Q(a_ k))}{(1-|a_ k|)^n}\right )^{\beta+1}\right )d\sigma(\zeta)\\
\\
& \le \int_{\Sn} \left (\sup_{a_ k\in \Gamma(\zeta)} \frac{\mu(Q(a_ k))}{(1-|a_ k|)^n}\right )^{\beta-1}\left (\sum_ {k: a_ k\in \Gamma(\zeta)} |\lambda_ k|^2 \left (\frac{\mu(Q(a_ k))}{(1-|a_ k|)^n}\right )^2\right )d\sigma(\zeta).
\end{split}
\end{equation}
Hence, applying H\"{o}lder's inequality and \eqref{EqQ2b} we get
\[
\begin{split}
\sum_ k |\lambda_ k|^2 &\left ( \frac{\mu(Q(a_ k))}{(1-|a_ k|)^n}\right )^{\beta+1} (1-|a_ k|)^n \lesssim \|\nu \|_{T^r_{\infty}}^{\beta-1} \cdot \big \|Q_{\mu}\big \|^2 \cdot \|\lambda \|^2_{T^p_ 2},
\end{split}
\]
with $\nu=\{\nu_ k\}$ and $\nu_ k =\mu(Q(a_ k))/(1-|a_ k|)^n$. That is, we have
\[
\sum_ k |\alpha_ k|\,\nu_ k ^{\beta+1} \le C \|\nu \|_{T^r_{\infty}}^{\beta-1} \cdot \big \|Q_{\mu}\big \|^2 \cdot \|\alpha \|^2_{T^{p/2}_ 1}.
\]
By the duality between the tent spaces $T^{p/2}_ 1$ and $T_{\infty}^{(p/2)'}$ (see Theorem \ref{TTD1}), we obtain
\[
\big \|\nu^{\beta+1}\big \|_{T^{p/(p-2)}_{\infty}}\lesssim \|\nu \|_{T^r_{\infty}}^{\beta-1} \cdot \big \|Q_{\mu}\big \|^2 .
\]
It is straightforward to check that
\[
(\beta+1)p/(p-2)=r
\]
so that
\[
\big \|\nu \big \|^{\beta+1}_{T^r_{\infty}}\le \big \|\nu^{\beta+1}\big \|_{T^{p/(p-2)}_{\infty}}\lesssim \big \|\nu \big \|_{T^r_{\infty}}^{\beta-1} \cdot \big \|Q_{\mu}\big \|^2 .
\]
Now, if $\mu$ is compactly supported, then $\|\nu \big \|_{T^r_{\infty}}<\infty$, and therefore we deduce that
\[
\|\nu \big \|_{T^r_{\infty}} \lesssim \big \|Q_{\mu}\big \|
\]
An application of Lemma \ref{discrete} gives $\|\widetilde{\mu}\|_{L^r}\lesssim \|Q_{\mu}\|$ when $\mu$ is compactly supported. The general case follows by an approximation argument. Indeed, let $r_ k \in (0,1)$ with $r_ k\rightarrow 1$, and consider the measures $\mu_ k$ and $\mu_ k^*$ defined on Borel sets $E$ by $\mu_ k(E)= \mu(E\cap D(0,r_ k))$ and $\mu_ k^* (E)=\mu \big (r_ k^{-1}E \cap D(0,r_ k)\big )$. Note that $\widetilde{\mu_ k}\le \widetilde{\mu_ k^*}$ because $$D(0,r_ k)\cap \Gamma (\zeta) \subset D(0,r_ k)\cap r_ k^{-1}\Gamma (\zeta).$$ To be honest, if the aperture of $\Gamma(\zeta)=\Gamma_\gamma(\zeta)$ satisfies $\gamma\geq 2$, then the above inclusion holds as stated, whereas if $1<\gamma<2$, then it will hold modulo a compact set, see the beginning of the proof of Lemma \ref{discrete}. The argument can be completed either way, and we can always just change the aperture. We arrive at
\[
\big \|\widetilde{\mu}\big \|_{L^r} \le \liminf _ k\big \|\widetilde{\mu_ k}\big \|_{L^r} \le \liminf_ k \big \|\widetilde{\mu^*_ k}\big \|_{L^r}
\lesssim \liminf _ k \big \| Q_{\mu^*_ k}\big \|.
\]
Now, if $f$ is a unit vector in $H^p$, we have (by a change of variables) $|Q_{\mu^*_ k}f(z)| \asymp |(Q_{\mu} f_ k)_ k (z)|$, where $g_ k(z)=g(r_ k z)$, which easily gives $\|Q_{\mu_ k^*}\|\lesssim \|Q_{\mu}\|$ because dilatation by $r_ k$ can only decrease the norm. This gives
\[
\big \|\widetilde{\mu}\big \|_{L^r} \lesssim \big \| Q_{\mu}\big \|
\]
finishing the proof in this case.
\subsubsection{\textbf{The case $q<2$}} We will start with the following simple lemma.
\begin{lemma}\label{L-Qmu1}
Let $1<p<\infty$. If $Q_{\mu} f$ is in $H^p$ and $g\in H^{p'}$, then
\begin{equation*}
\int_{\Sn} Q_\mu f(\zeta)\overline{g(\zeta)}d\sigma(\zeta)=\int_{\Bn} \!\!f(w)\,\overline{g(w)}\,d\mu(w).
\end{equation*}
\end{lemma}
\begin{proof}
If $g$ is a holomorphic polynomial, then
$$\Lambda_ g(h)=\int_{\Sn}h(\zeta)\overline{g(\zeta)}d\sigma(\zeta)$$ defines a bounded linear functional on $H^p$. Let $K_ w$ be the reproducing kernel of $H^2$ at the point $w$. Since $Q_{\mu} f\in H^p$, we have
\[
\begin{split}
\int_{\Sn} Q_\mu f(\zeta)\overline{g(\zeta)}d\sigma(\zeta)&=\Lambda_ g(Q_{\mu} f)=\Lambda_ g \left ( \int_{\Bn}\!\frac{f(w)}{(1-\langle \cdot ,w \rangle )^n}\,d\mu(w) \right )\\
\\
&= \int_{\Bn} f(w) \,\Lambda_ g \left ( \frac{1}{(1-\langle \cdot ,w \rangle )^n}\right )\,d\mu(w)=\int_{\Bn} f(w) \,\Lambda_ g (K_ w)\,d\mu(w)\\
\\
&=\int_{\Bn} f(w) \,\langle K_ w,g \rangle _{H^2}\,d\mu(w)=\int_{\Bn} \!\!f(w)\,\overline{g(w)}\,d\mu(w).
\end{split}
\]
As the holomorphic polynomials are dense on $H^{p'}$, the result follows.
\end{proof}
It follows from Lemma \ref{L-Qmu1} that if $Q_\mu:H^p\to H^q$ is bounded, then its adjoint is $Q_\mu:H^{q'}\to H^{p'}$. If $p\le 2$, then $p'\ge 2$ and from the previous cases (as $\frac{p'q'}{q'-p'}=\frac{pq}{p-q}=r$) we obtain that $\widetilde{\mu}$ is in $L^{r}(\Sn)$.
Hence we may assume that $1<q<2<p$. Let $\{a_k\}$ be a separated sequence in $\Bn$. Using the area function description of Hardy spaces obtained in Proposition \ref{FracArea}, Fubini's theorem, and the same test functions as before, we obtain
$$\int_{\Sn}\int_0^1 \left(\int_{\Gamma(\zeta)}\left|\sum_{k} \lambda_k r_k(t)R^{-1,1}(Q_\mu f_k)(z)\right|^2dv_{1-n}(z)\right)^{q/2}dt d\sigma(\zeta) \le C \|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2},$$
where $r_k$ are the Rademacher functions. Proceeding with Kahane's inequality, we get
$$\int_{\Sn}\left(\int_0^1 \int_{\Gamma(\zeta)}\left|\sum_{k} \lambda_k r_k(t)R^{-1,1}(Q_\mu f_k)(z)\right|^2dv_{1-n}(z)dt\right)^{q/2} d\sigma(\zeta) \le C \|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2}.$$
Next, Fubini's theorem together with Khinchine's inequality leads to
$$\int_{\Sn}\left( \int_{\Gamma(\zeta)}\sum_{k} |\lambda_k|^2 |R^{-1,1}(Q_\mu f_k)(z)|^2 dv_{1-n}(z)\right)^{q/2} d\sigma(\zeta) \le C \|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2}.$$
Now, applying Lemma \ref{Gamma}, this gives for $\beta>2n/q$:
$$\int_{\Sn} \left[\sum_{k}|\lambda_k|^2 \int_{\Bn} \left(\frac{1-|z|^2}{|1-\langle z,\zeta\rangle|}\right)^\beta |R^{-1,1}(Q_\mu f_k)(z)|^2 dv_{1-n}(z)\right]^{q/2}d\sigma(\zeta)\leq C\|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2},$$
which implies
$$\int_{\Sn} \left[\sum_{k}|\lambda_k|^2 \left(\frac{1-|a_k|^2}{|1-\langle a_k,\zeta\rangle|}\right)^\beta\int_{D_k} |R^{-1,1}(Q_\mu f_k)(z)|^2 dv_{1-n}(z)\right]^{q/2}d\sigma(\zeta)\leq C\|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2},$$
where $D_ k$ denotes the Bergman metric ball $D(a_ k,r)$. Now, because $|1-\langle a_ k,\zeta \rangle | \asymp 1-|a_ k|$ for $a_ k\in \Gamma(\zeta)$, summing only over indices $k$ so that $a_k \in \Gamma(\zeta)$, we arrive at
\[
\int_{\Sn} \left ( \sum_ {a_ k \in \Gamma (\zeta)} |\lambda_ k|^2 \int_{D_k } |R^{-1,1}(Q_{\mu} f_ k)(z)|^ 2\,dv_{1-n}(z)\right )^{q/2} d\sigma(\zeta) \le C \|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2}.
\]
By subharmonicity, we get
\begin{equation}\label{compare}
\int_{\Sn} \left ( \sum_ {a_ k \in \Gamma (\zeta)} |\lambda_ k|^2 |R^{-1,1}(Q_{\mu} f_ k)(a_ k)|^ 2\,(1-|a_ k|^2)^2\right )^{q/2} d\sigma(\zeta) \le C \|Q_\mu\|^q\|\lambda \|^q_{T^p_ 2}.
\end{equation}
Let us now denote $\nu=(\nu_k)$, where $\nu_k=\mu(Q(a_k))(1-|a_k|^2)^{-n}$ and set
$$s=\frac{p-q}{q(p-2)}.$$
By our choices $p>2>q$, it follows that $2s>qs>1$. Using \eqref{EqG} and two H\"olders (first with $2s$ and $(2s)'$, and then with $qs$ and $(qs)'$), we obtain
\begin{align*}
\sum_k |\lambda_k|^2 \nu_k^{1/s}(1-|a_k|)^n &\asymp \int_{\Sn} \left[ \sum_{a_k \in \Gamma(\zeta)} |\lambda_k|^2 \nu_k^{1/s}\right]d\sigma(\zeta)\\
&\leq \int_{\Sn} \left[ \sum_{a_k \in \Gamma(\zeta)} |\lambda_k|^2\right]^{(2s-1)/2s}\left[ \sum_{a_k \in \Gamma(\zeta)} |\lambda_k|^2 \nu_k^2 \right]^{1/2s}d\sigma(\zeta)\\
&\leq \|\lambda \|^{2-\frac{1}{s}}_{T^p_ 2}\cdot \left\lbrace\int_{\Sn}\left[ \sum_{a_k \in \Gamma(\zeta)}|\lambda_k|^2 \nu_k^2\right]^{q/2}d\sigma(\zeta)\right\rbrace^{1/qs}.
\end{align*}
Here we have used that
$$\frac{(2s-1)}{2s}\frac{qs}{(qs-1)}=\frac{p}{2}.$$
Since, by \eqref{estim}, we have $\nu_k \lesssim |R^{-1,1}(Q_{\mu} f_ k)(a_ k)|\,(1-|a_ k|)$, we can combine this estimate with \eqref{compare}, and arrive at
$$\sum_k |\lambda_k|^2 \nu_k^{1/s}(1-|a_k|)^n \leq C\|Q_\mu\|^{1/s}\|\lambda^2 \|_{T^{p/2}_1}.$$
So, by the duality of tent spaces, we obtain $\|\nu^{1/s}\|_{T^{p/(p-2)}_\infty} \leq C\|Q_\mu\|^{1/s}$. Note that $(1/s)p/(p-2)=r$, so we get
$$\|\nu\|_{T^r_\infty}\leq C \|Q_\mu\|.$$
An application of Lemma \ref{discrete} finally shows that $\widetilde{\mu} \in L^r(\Sn)$ with $\|\widetilde{\mu}\|_{L^r}\lesssim \|Q_{\mu}\|$ as claimed.
\section{Compactness}\label{sComp}
For $1<p,q<\infty$, a linear operator $T:H^p\rightarrow H^q$ is compact if $\|Tf_ n\|_{H^q}\rightarrow 0$ whenever $\{f_ n\}$ is a bounded sequence in $H^p$ converging to zero uniformly on compact subsets of $\Bn$.\\
Recall also that a finite Borel measure on $\Bn$ is called a vanishing $s$-Carleson measure if for every $\zeta \in \Sn$
$$\mu(B_{\delta}(\zeta))\delta^{-ns}\to 0$$ as $\delta \to 0$. Equivalently, one may require
that for each (some) $t>0$ one has
\begin{equation}\label{sCM-1}
\lim_{|a|\rightarrow 1^-}\int_{\Bn} \!\!\frac{(1-|a|^2)^t}{|1-\langle a,w \rangle |^{ns+t}} \,d\mu(w)=0.
\end{equation}
Now we are ready for the description of the compactness of the Toeplitz type operator $Q_{\mu}$ acting between Hardy spaces.
\begin{theorem}\label{mtC-1}
Let $1<p\le q<\infty$ and $\mu$ be a positive Borel measure on $\Bn$. Then $Q_{\mu}: H^p\rightarrow H^q$ is compact if and only if $\mu$ is a vanishing $(1+\frac{1}{p}-\frac{1}{q})$-Carleson measure.
\end{theorem}
\begin{proof}
Assume first that $Q_{\mu}$ is compact. In the proof of the boundedness, we have seen that
\[
\int_{\Bn}\frac{(1-|z|^2)^{n/p'}d\mu(w)}{|1-\langle w,z \rangle |^{2n}} =|Q_{\mu}k_ z (z)| \le (1-|z|^2)^{-n/q}\big \|Q_{\mu}k_ z \big \|_{H^q}.
\]
Assuming that $Q_ {\mu}$ compact, we know that $\|Q_{\mu}k_ z\|_{H^q}\rightarrow 0$ as $|z|\rightarrow 1^{-}$, and the result follows by \eqref{sCM-1}.
Conversely, suppose that $\mu$ is a vanishing $s$-Carleson measure with $s=1+1/p-1/q$. To prove the compactness of $Q_{\mu}$ we must show that $\|Q_{\mu} f_ n\|_{H^q}\rightarrow 0$ if $\{f_ n\}$ is a bounded sequence in $H^p$ that converges to zero uniformly on compact subsets of $\Bn$. As $\mu$ is an $s$-Carleson measure, by Theorem \ref{T2}, the Toeplitz operator $Q_{\mu}:H^p\rightarrow H^q$ is bounded. Hence, by duality, Lemma \ref{L-Qmu1} and Carleson-Duren's theorem (argue as in the proof of Theorem \ref{T2}), we get
\[
\|Q_{\mu} f_ n\|_{H^q}=\sup_{\|g\|_{H^{q'}}=1} |\langle Q_{\mu} f_ n,g \rangle |\le \sup_{\|g\|_{H^{q'}}=1} \int_{\Bn} |f_ n(z)|\,|g(z)| \,d\mu(z)
\lesssim \|f_ n \|_{L^{ps}(\mu)}
\]
Since $\mu$ is a vanishing $s$-Carleson measure, then the embedding $i_{\mu}: H^p\rightarrow L^{ps} (\mu)$ is compact, and therefore $\|f_ n\|_{L^{ps}(\mu)} \rightarrow 0$ which proves that $\|Q_{\mu} f_ n\|_{H^q} \rightarrow 0$ finishing the proof.
\end{proof}
We also present the following, seemingly stronger version of Theorem \ref{T3}.
\begin{theorem}
Let $1<q<p<\infty$ and $\mu$ be a positive Borel measure on $\Bn$. Then $Q_{\mu}: H^p\rightarrow H^q$ is compact if and only if it is bounded, if and only if $\widetilde{\mu}\in L^r(\Sn)$, where $r=pq/(p-q)$.
\end{theorem}
\begin{proof} In view of Theorem \ref{T3}, it suffices to show that the condition $\widetilde{\mu}\in L^r(\Sn)$ implies the compactness of $Q_{\mu}$.
For any compact $K\subset \Bn$, we set $\mu_K=\mu\chi_K$. Suppose that $\{f_n\}$ is a bounded sequence in $H^p$ converging to zero uniformly on compact subsets of $\Bn$. If $g$ is an arbitrary unit vector in $H^{q'}$, then by the standard pointwise estimate we have $|g(z)|\leq C_K$ on $K$ (uniformly on $g$). Hence, by duality and Lemma \ref{L-Qmu1},
\begin{align*}
\big \|Q_{\mu_ K} f_ n \big \|_{H^q}&=\sup_{\|g\|_{H^{q'}}=1}|\langle Q_{\mu_K}f_n,g\rangle|
\leq \sup_{\|g\|_{H^{q'}}=1} \int_K |f_n(z)g(z)|d\mu(z)\\
&\leq \,C_K\int_K|f_n(z)|d\mu(z)\to 0.
\end{align*}
So $Q_{\mu_K}$ is compact. Next, define $\mu_s=\mu\chi_{\overline{D(0,s)}}$, so that $Q_{\mu_s}$ is compact as just shown. Now,
$$\|\widetilde{\mu}-\widetilde{\mu_s}\|_{L^r}^r=\int_{\mathbb{S}^n} \left|\int_{\Gamma(\zeta)}\frac{(1-\chi_{\overline{D(0,s)}})(z)d\mu(z)}{(1-|z|^2)^n}\right|^rd\sigma(\zeta).$$
Write
$$\Phi_s(\zeta)=\left|\int_{\Gamma(\zeta)}\frac{\phi_s(z)d\mu(z)}{(1-|z|^2)^n}\right|^r,$$
where $\phi_s(z)=(1-\chi_{\overline{D(0,s)}})(z)$.
Because $\widetilde{\mu}\in L^r(\Sn)$, the function $\Phi_0$ is defined almost everywhere. Therefore, for almost every $\zeta \in \Sn$, the function
$$\frac{\phi_s(z)d\mu(z)}{(1-|z|^2)^n}$$ has an integrable majorant. Therefore, $\Phi_s(\zeta)\to 0$ as $s\to 1$ for almost every $\zeta \in \Sn$, by the Lebesgue dominated convergence theorem. Remembering that $\widetilde{\mu}\in L^r(\Sn)$, we see that $\Phi_s$ has also an integrable majorant, namely $\Phi_0$. Thus $\widetilde{\mu_s}\to \widetilde{\mu}$ in $L^r(\Sn)$, by another application of the dominated convergence theorem. By the norm estimate for $Q_\mu$ we have
\[
\big \|Q_{\mu}-Q_{\mu_ s} \big \| \lesssim \big \| \widetilde{\mu}-\widetilde{\mu_ s} \big \|_{L^r} \rightarrow 0
\]
as $s\rightarrow 1$, proving that $Q_{\mu}$ is compact.
\end{proof}
\section{Schatten class membership}\label{SSC}
In this section we are going to describe the membership of $Q_{\mu}$ in the Schatten ideals $S_ p(H^2)$. \\
Let $t>0$ and define
\[
S_ t \mu(w)=(1-|w|^2)^{n+t}\int_{\Bn} \frac{d\mu(z)}{|1-\langle z,w \rangle |^{2n+t}}.
\]
We will denote by $d\lambda_n(z):=dv_{-1-n}(z)$ the invariant measure. For technical reasons, it will be convenient to denote
$$t_p=\max(n/p-n,0).
$$
\begin{proposition}\label{S-P1}
Let $\mu$ be a positive Borel measure on $\Bn$, and let $0<p<\infty$. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $S_ t\mu\in L^p(\Bn,d\lambda_ n)$ for all (some) $t>t_p$.
\item[(b)] The sequence $\Big\{\frac{\mu(D(a_ k,r))}{(1-|a_ k|^2)^n}\Big\}$ is in $\ell^p$, for any $r$-lattice $\{a_ k\}$.
\end{enumerate}
\end{proposition}
\begin{proof}
Denote by $D_k=D(a_k,r)$. By the properties of the lattice we have
\[
\begin{split}
\|S_ t \mu \|^p_{L^p(\lambda_ n)}&=\int_{\Bn} \left ( (1-|z|^2)^{n+t}\int_{\Bn} \frac{d\mu(w)}{|1-\langle z,w \rangle |^{2n+t}}\right )^p d\lambda_ n(z)\\
& \gtrsim \sum_ k \int_{D_ k} \left ( (1-|z|^2)^{n+t}\int_{D_ k} \frac{d\mu(w)}{|1-\langle z,w \rangle |^{2n+t}}\right )^p d\lambda_ n(z)\\
& \asymp \sum_ k \left (\frac{\mu(D_ k)}{(1-|a_ k|^2 )^{n}}\right )^p,
\end{split}
\]
because $|1-\langle z,w \rangle | \asymp (1-|z|^2)\asymp (1-|a_ k|^2)$ for $z,w\in D_ k$ (see \cite[Lemma 2.20]{ZhuBn}). Thus (a) implies (b) when $t>0$.
Assume that (b) holds. As the sets $D_ k$ cover $\Bn$ and $|1-\langle z,w \rangle |\asymp |1-\langle a_ k,w \rangle |$ for $z\in D_ k$ (by the estimate (2.20) in p.63 of \cite{ZhuBn}), we have
\[
S_ t \mu (w) \le (1-|w|^2)^{n+t} \sum_ k \int_{D_ k}\frac{d\mu(z)}{|1-\langle z,w \rangle |^{2n+t}}\lesssim (1-|w|^2)^{n+t}\sum_ k \frac{\mu(D_ k)}{|1-\langle a_ k,w \rangle|^{2n+t}}.
\]
This gives
\begin{equation}\label{EqP1}
S_ t \mu(w)^p \lesssim (1-|w|^2)^{np+tp}\left (\sum_ k \frac{\mu(D_ k)}{|1-\langle a_ k,w \rangle|^{2n+t}}\right )^{p}.
\end{equation}
For $p>1$, we take $0<\varepsilon<n\min (\frac{1}{p},\frac{1}{p'})$, and use H\"older's inequality and Lemma \ref{l2} to get
\begin{displaymath}
\begin{split}
S_ t \mu(w)^p
&\lesssim (1-|w|^2)^{(n+t)p} \left (\sum_ k \frac{\mu(D_ k)^p\,(1-|a_ k|^2)^{-n(p-1)-\varepsilon p}}{|1-\langle a_ k,w \rangle|^{2n+tp}}\right )
\left (\sum_ k \frac{(1-|a_ k|^2)^{n+\varepsilon p'}}{|1-\langle a_ k,w \rangle|^{2n}}\right )^{p/p'}
\\
&\lesssim (1-|w|^2)^{\varepsilon p+tp+n} \left (\sum_ k \frac{\mu(D_ k)^p\,(1-|a_ k|^2)^{-n(p-1)-\varepsilon p}}{|1-\langle a_ k,w \rangle|^{2n+tp}}\right ).
\end{split}
\end{displaymath}
Therefore, by the integral type estimate in Lemma \ref{IctBn},
\begin{displaymath}
\begin{split}
\int_{\Bn} S_ t\mu(w)^p \,d\lambda_ n (w)&\lesssim \sum_ k \left (\frac{\mu(D_ k)}{(1-|a_ k|^2)^n} \right )^p (1-|a_ k|^2)^{n-\varepsilon p}
\int_{\Bn} \frac{(1-|w|^2)^{(\varepsilon+t)p-1} dv(w)}{|1-\langle a_ k,w \rangle|^{2n+tp}}
\\
&\lesssim \sum_ k \left (\frac{\mu(D_ k)}{(1-|a_ k|^2)^n} \right )^p,
\end{split}
\end{displaymath}
and we get the result in this case.\\
\mbox{}
\\
If $0<p\le 1$, starting from \eqref{EqP1} we get
\[
S_ t \mu(w)^p \lesssim (1-|w|^2)^{np+tp}\sum_ k \frac{\mu(D_ k)^p}{|1-\langle a_ k,w \rangle|^{2np+tp}}.
\]
Then
\begin{displaymath}
\begin{split}
\int_{\Bn} S_ t \mu(w)^p\,d\lambda_ n(w)
\lesssim \sum_ k \mu(D_ k)^p \int_{\Bn} \frac{(1-|w|^2)^{np+tp}}{|1-\langle w,a_ k \rangle |^{2np+tp}}\,d\lambda_ n(w)
\end{split}
\end{displaymath}
and the result follows from Lemma \ref{IctBn} because $t>t_ p$.
\end{proof}
Now we state the main result of this section, that characterizes the membership in $S_ p(H^2)$ of the Toeplitz type operator $Q_{\mu}$.
\begin{theorem}\label{t-SC}
Let $0<p<\infty$ and $\mu$ be a positive Borel measure on $\Bn$. The following are equivalent:
\begin{enumerate}
\item[(a)] $Q_{\mu}$ belongs to $S_ p(H^2)$;
\item[(b)] $S_ t\mu \in L^p(\Bn,d\lambda_ n)$ for all (some) $t>t_p$;
\item[(c)] for any $r$-lattice $\{a_ k\}$, we have
$$\sum _ k \left ( \frac{\mu(D(a_ k,r))}{(1-|a_ k|^2)^n}\right )^p <\infty.$$
\end{enumerate}
\end{theorem}
From Proposition \ref{S-P1} we already know that (b) and (c) are equivalent. In order to obtain the equivalence with condition (a) we need to introduce some concepts and notation.\\
Recall that $H^2$ is a reproducing kernel Hilbert space with the reproducing kernel function given by
\[K_ z(w)=\frac{1}{(1-\langle w,z\rangle )^{n}},\quad z,w\in \Bn\]
with norm $\|K_ z\|_{H^2}=\sqrt{K_ z(z)}=(1-|z|^2)^{-n/2}$. The normalized kernel functions are denoted by $k_ z=K_ z/\|K_ z\|_{H^2}$. We also need to introduce some ``derivatives" of the kernel functions. For $z,w\in \Bn$ and $t>0$, define
\[K_ z^{t}(w)=\overline{R^{-1,t}K_ w (z)}=\frac{1}{(1-\langle w,z\rangle )^{n+t}}\]
and let $k^{t}_ z$ denote its normalization, that is, $k^{t}_ z=K^{t}_ z/\|K^{t}_ z\|_{H^2}$. In particular, since $f
(z)=\langle f,K_ z\rangle_{H^2}$ whenever $f\in H^2$, one has
\begin{equation}\label{eq1}
R^{-1,t}f (z)=\langle f,K^{t}_ z\rangle_{H^2},\qquad f\in H^2(\Bn).
\end{equation}
\mbox{}
\\
For $\alpha>-1$, the Bergman space $A^2_{\alpha}$ consists of those holomorphic functions $f$ in $\Bn$ with $\|f\|_{A^2_{\alpha}}^2 =\int_{\Bn} |f(z)|^2 dv_{\alpha}(z)<\infty$. It is well known that, for $f\in H^2$ with $f(0)=0$ and $s\geq -1$, one has
\[
\|f\|_{H^2}\asymp \|Rf\|_{A^2_ 1}\asymp \|R^{s,1}f\|_{A^2_ 1}.
\]
\mbox{}
\\
We need first the following lemma that can be found in \cite{P1}.
\begin{lemma}\label{KLS}
Let $T:H^2(\Bn)\rightarrow H^2(\Bn)$ be a positive operator. For $t> 0$ set
\[\widetilde{T^t}(z)=\langle Tk^{t}_ z,k^{t}_ z\rangle_{H^2},\quad z\in \Bn.\]
\begin{enumerate}
\item[(a)] Let $0<p\le 1$. If $\widetilde{T^t} \in L^p(\Bn,d\lambda_ n)$ then $T$ is in $S_ p(H^2)$.
\item[(b)] Let $p\ge 1$. If $T$ is in $S_ p(H^2)$ then $\widetilde{T^t} \in L^p(\Bn,d\lambda_ n)$.
\end{enumerate}
\end{lemma}
\mbox{}
\\
\begin{lemma}\label{p-bound}
Let $0<p<\infty$, $\mu$ be a positive Borel measure on $\Bn$, and suppose that $S_ t\mu \in L^p(\Bn,d\lambda_ n)$ for $t>t_ p$. Then $Q_{\mu}$ is bounded on $H^2$.
\end{lemma}
\begin{proof}
By Theorem \ref{T2} it is enough to show that $\mu$ is a Carleson measure. Let $\{a_ k\}$ be an $r$-lattice on $\Bn$ and set $D_ k=D(a_ k,r)$. By \eqref{sCM} we need to prove that $\sup_{a\in \Bn} I_{\mu}(a)<\infty$, where
\[
I_{\mu}(a):=\int_{\Bn} \left (\frac{1-|a|^2}{|1-\langle a,z \rangle |^2}\right )^n \,d\mu(z).
\]
Since $|1-\langle a,z \rangle |\asymp |1-\langle a,a_ k\rangle |$ for $z\in D_ k$, we get
\[
I_{\mu}(a) \le (1-|a|^2)^n \sum_ k \int_{D_ k} \frac{d\mu(z)}{|1-\langle a,z \rangle |^{2n}}\lesssim (1-|a|^2)^n \sum _ k \frac{\mu(D_ k)}{|1-\langle a_ k, a\rangle |^{2n}}.
\]
If $0<p\le 1$, taking into account Proposition \ref{S-P1}, the result follows directly from our assumption and the fact that $|1-\langle a_k,a\rangle|^2\geq (1-|a_k|)(1-|a|)$. If $p>1$, we use H\"{o}lder's inequality to get
\[
I_{\mu}(a) \lesssim (1-|a|^2)^n \left (\sum _ k \frac{(1-|a_ k|^2)^{np'}}{|1-\langle a_ k,a \rangle |^{2np'}}\right )^{1/p'}
\left (\sum_ k \Big (\frac{\mu(D_ k) }{(1-|a_ k|^2)^n}\Big )^p \right )^{1/p},
\]
and now the result is a consequence of the assumption, Proposition \ref{S-P1} and Lemma \ref{l2}.
\end{proof}
As a consequence of the previous lemmas, we get the following.
\begin{proposition}\label{PS1}
Let $0<p<\infty$, $t>t_ p$, and $\mu$ be a positive Borel measure on $\Bn$.
If $0 < p \leq 1$ and $S_{t}\mu$ is in $L^p(\Bn,d\lambda_{n}),$ then $Q_\mu$ belongs to $S_p(H^2).$ Conversely, if $p\ge 1$ and $Q_\mu$ is in $S_p(H^2)$, then $S_{t}\mu \in L^p(\Bn,d\lambda_{n}).$
\end{proposition}
\begin{proof}
By Lemma \ref{p-bound}, the condition $S_ t\mu \in L^p(\Bn,d\lambda_ n)$ implies the boundedness of $Q_{\mu}$ on $H^2$. Hence, in both cases, we have
\[
\langle Q_{\mu} k_ z^{t/2},k_ z^{t/2}\rangle=\int_{\Bn} |k^{t/2}_ z(w)|^2\,d\mu(w).
\]
As
\[
\|K_ z^{t/2} \|^2_{H^2}\asymp (1-|z|^2)^{-n-t},
\]
we obtain
\[
\langle Q_{\mu} k_ z^{t/2},k_ z^{t/2}\rangle \asymp S_{t} \mu(z).
\]
Therefore, the result is a direct consequence of Lemma \ref{KLS}.
\end{proof}
The next result, together with Proposition \ref{PS1}, gives that condition (b) implies (a) in Theorem \ref{t-SC}.
\begin{proposition}\label{P-SC2}
Let $p>1$, $\mu$ be a positive Borel measure on $\Bn$, and suppose that $S_ t \mu \in L^p(\Bn,d\lambda_ n)$ for $t>t_ p$. Then $Q_{\mu}$ belongs to $S_ p(H^2)$.
\end{proposition}
\begin{proof}
It is easy to modify the proof of Lemma \ref{p-bound} in order to see that the condition implies that $\mu$ is a vanishing Carleson measure. Hence, by Theorem \ref{mtC-1}, the operator $Q_{\mu}$ is compact. Thus, in view of \cite[Theorem 1.27]{Zhu}, it is enough to show that
$$\sum_ k |\langle Q_{\mu} e_ k,e_ k \rangle |^p \le C<\infty$$
for all orthonormal sets $\{e_ k\}$ of $H^2$. In order to prove that, we
follow closely the argument used in \cite{PP}. By Lemma \ref{L-Qmu1} we have
\begin{displaymath}
|\langle Q_{\mu} e_ k,e_ k \rangle |^p =\left (\int_{\Bn} |e_ k(z)|^2 d\mu(z) \right )^p.
\end{displaymath}
By perturbing $Q_\mu$ by a rank-one operator, we may assume that $e_k(0)=0$ for all $k$. Since $e_ k^2 \in H^1 \subset A^2_{n-1}\subset A^1_{n-1}$, we have (see \cite[page 51]{ZhuBn})
\[
|e_ k(z)|^2 \lesssim \int_{\Bn} \frac{|R(e_ k^2)(w)|}{|1-\langle z,w \rangle |^{2n+t}}\,dv_{n+ t}(w)\lesssim \int_{\Bn} \frac{|e_ k(w)|\,|Re_ k(w)|}{|1-\langle z,w \rangle |^{2n+t}}\,dv_{n+ t}(w).
\]
This gives
\begin{equation}\label{Eq-Sc1}
|\langle Q_{\mu} e_ k,e_ k \rangle | \lesssim \int_{\Bn} |e_ k(w)|\,|Re_ k(w)| \,S_ t \mu(w)\,dv(w).
\end{equation}
If $1<p\le 2$, we use H\"{o}lder's inequality in \eqref{Eq-Sc1} and the fact that
\[
\|Re_ k\|_{A^2_ 1}\asymp \|e_ k\|_{H^2} =1
\]
to get
\[
\begin{split}
|\langle Q_{\mu} e_ k,e_ k \rangle |^p & \lesssim \ \int_{\Bn} |e_ k(w)|^p\,|Re_ k(w)|^{2-p} \,S_ t \mu(w)^p \,dv_{1-p}(w)
\end{split}
\]
Since
\[
\begin{split}
\sum_ k |e_ k(w)|^p\,|Re_ k(w)|^{2-p} & \le \Big (\sum_ k |e_ k(w)|^2 \Big )^{p/2}\Big (\sum_ k |Re_ k(w)|^2 \Big )^{(2-p)/2}
\\
& \le \|K_ w\|_ {H^2}^p\cdot \|RK_ w\|_{H^2}^{2-p}
\\
&\lesssim (1-|w|^2)^{-np/2} \,(1-|w|^2)^{-(n+2)(2-p)/2} =(1-|w|^2)^{p-(n+2)}
\end{split}
\]
we get
\[
\begin{split}
\sum_ k |\langle Q_{\mu} e_ k,e_ k \rangle | ^p &\lesssim \int_{\Bn} \left (\sum_ k |e_ k(w)|^p\,|Re_ k(w)|^{2-p} \right ) \,S_ t \mu(w)^p \,dv_{1-p}(w)
\\
&
\lesssim \int_{\Bn} S_ t \mu(w)^p \,d\lambda_ n(w).
\end{split}
\]
This finishes the proof of the case $1<p\leq 2$.\\
If $p>2$, we use Cauchy-Schwarz inequality in \eqref{Eq-Sc1} to obtain
\[
\begin{split}
|\langle Q_{\mu} e_ k,e_ k \rangle |^2
&\lesssim \int_{\Bn} |e_ k(w)|^2\, S_ t\mu(w)^2\, (1-|w|^2)^n\,d\lambda_ n(w).
\end{split}
\]
By the reproducing formula for Bergman spaces, for any $\beta>0$, we have
\[
R^{\beta-1,1} e_ k(w)=\int_{\Bn} \frac{R^{\beta-1,1}e_ k(\zeta)\,dv_{\beta}(\zeta)}{(1-\langle w,\zeta\rangle)^{n+1+\beta}}.
\]
Hence,
\[
e_ k(w)=R_{\beta-1,1} R^{\beta-1,1} e_ k(w)=\int_{\Bn} \frac{R^{\beta-1,1}e_ k(\zeta)\,dv_{\beta}(\zeta)}{(1-\langle w,\zeta\rangle)^{n+\beta}}.
\]
Now, fix $\beta>n$ and take $\varepsilon>0$ with $\varepsilon <\min \big (\beta,\frac{n}{p-1}\big)$. By Cauchy-Schwarz and standard integral estimates we have
\[
\begin{split}
|e_ k(w)|^2 &\lesssim \left ( \int_{\Bn} \frac{|R^{\beta-1,1}e_ k(\zeta)|\,dv_{\beta}(\zeta)}{|1-\langle w,\zeta\rangle|^{n+\beta}} \right )^2
\\
&\lesssim \left ( \int_{\Bn} \frac{|R^{\beta-1,1}e_ k (\zeta)|^2\,dv_{1+\beta+\varepsilon}}{|1-\langle w,\zeta\rangle|^{n+\beta}}\right ) \,(1-|w|^2)^{-\varepsilon}.
\end{split}
\]
This gives
\[
|\langle Q_{\mu} e_ k,e_ k \rangle |^2 \lesssim \int_{\Bn} |R^{\beta-1,1}e_k(\zeta)| ^2\left ( \int_{\Bn} \frac{S_ t\mu(w)^2\, (1-|w|^2)^{n-\varepsilon}\,d\lambda_ n(w)}{|1-\langle w,\zeta\rangle|^{n+\beta}} \right ) \,dv_{1+\beta+\varepsilon}(\zeta).
\]
Because $\|R^{\beta-1,1}e_ k\|_{A^2_ 1}\asymp \|e_ k\|_{H^2}=1$, using H\"{o}lder's inequality with exponent $p/2>1$ we obtain
\[
|\langle Q_{\mu} e_ k,e_ k \rangle |^p \lesssim \int_{\Bn} |R^{\beta-1,1}e_ k(\zeta)|^2 \,C_{\mu}(\zeta)^{p/2} \,dv_{1+(\beta+\varepsilon)p/2} (\zeta),
\]
with
\[
C_{\mu}(\zeta):=\int_{\Bn} \frac{S_ t \mu(w)^2\,dv_{-1-\varepsilon}(w)}{|1-\langle w,\zeta \rangle|^{n+\beta}}.
\]
As
\[
\sum_ k |R^{\beta-1,1}e_ k(\zeta)|^2 \lesssim \big \| R^{\beta-1,1}K_{\zeta}\big \|_{H^2}\lesssim (1-|\zeta|^2)^{-(n+2)},
\]
we get
\[
\sum_ k |\langle Q_{\mu} e_ k,e_ k \rangle |^p \lesssim \int_{\Bn} C_{\mu}(\zeta)^{p/2} \,(1-|\zeta|^2)^{(\beta+\varepsilon)p/2}\,d\lambda_ n (\zeta).
\]
Finally, by H\"{o}lder's inequality and the typical integral estimate,
\[
\begin{split}
C_{\mu}(\zeta)^{p/2}&\lesssim
\left (\int_{\Bn} \frac{S_ t \mu(w)^p}{|1-\langle w,\zeta \rangle |^{n+\beta}} \,dv_{-1-\varepsilon p+\varepsilon}(w) \right )
\left( \int_{\Bn} \frac{dv_{-1+\varepsilon}(w)}{|1-\langle w,\zeta \rangle |^{n+\beta}}
\right )^{\frac{p}{2}-1}\\
\\
&\lesssim \left (\int_{\Bn} \frac{S_ t \mu(w)^p}{|1-\langle w,\zeta \rangle |^{n+\beta}} \,dv_{-1-\varepsilon p+\varepsilon}(w) \right )
(1-|\zeta|^2)^{(\varepsilon-\beta) (\frac{p}{2}-1)}.
\end{split}
\]
Inserting this into the previous estimate, using Fubini's theorem and Lemma \ref{IctBn} we conclude that
\[
\begin{split}
\sum_ k |\langle Q_{\mu} e_ k,e_ k \rangle |^p & \lesssim \int_{\Bn} S_ t \mu(w)^p \left (\int_{\Bn} \frac{(1-|\zeta|^2)^{\beta+\varepsilon (p-1)}}{|1-\langle w,\zeta \rangle|^{n+\beta}} \,d\lambda_ n(\zeta)\right )\, dv_{-1-\varepsilon p+\varepsilon}(w)
\\
& \lesssim \int_{\Bn} S_ t \mu(w)^p\,d\lambda_ n(w).
\end{split}
\]
This finishes the proof.
\end{proof}
\mbox{}
\\
\begin{proposition}\label{FPSC}
Let $0<p\le 1$, $\mu$ be a positive Borel measure on $\Bn$, and suppose that $Q_{\mu}\in S_ p(H^2)$. Then, for any $r$-lattice $\{a_ k\}$ on $\Bn$, we have
\[
\sum_ k \left ( \frac{\mu(D(a_ k,r))}{(1-|a_ k|^2)^n}\right )^p<\infty.
\]
\end{proposition}
\begin{proof}
Fix $R>0$ big enough to be chosen later, and partition the lattice $\{a_ k\}$ into $M$ subsequences such that any two distinct points $b_ j$ and $b_ {\ell}$ in the same subsequence satisfy $\beta(b_ j,b_ {\ell})\ge R$. Let $\{b_ j \}$ be such a subsequence and consider the measure
\[
\nu=\sum_ j \mu \chi_ j,
\]
where $\chi_ j$ denotes the characteristic function of the Bergman metric ball $D_ j:=D(b_ j,r)$. Fix an orthonormal basis $\{e_ j\}$ of $H^2$, and consider the linear operator $A:H^2\rightarrow H^2$ defined by $Ae_ j=h_ j^t$ for a sufficiently large $t$, where
\[
h_ j^t(z)=\frac{(1-|b_ j|^2)^{(n+2t)/2}}{(1-\langle z,b_ j\rangle )^{n+t}}.
\]
For a large $t$, the operator $A$ is bounded on $H^2$. Then the operator $S=A^*Q_{\nu}A$ is in $S_ p(H^2)$ with $\|S\|_{S_ p}\lesssim \|Q_{\nu}\|_{S_ p}$. Split the operator as $S=D+E$, where $D$ is a diagonal operator. We have
\[
\|S\|^p_{S_ p}\ge \|D\|^p_{S_ p}-\|E\|^p_{S_ p}.
\]
We can estimate the diagonal term as follows.
\[
\begin{split}
\|D\|^p_{S_ p}&=\sum_ j \big | \langle De_ j,e_ j\rangle \big |^p=\sum_ j \big | \langle Q_{\nu}h^t_ j,h^t_ j\rangle \big |^p\\
\\
& =\sum_ j \left ( \int_{\Bn} |h_ j^t (z)|^2\,d\nu(z) \right )^p \ge \sum_ j \left ( \int_{D_ j} |h_ j^t (z)|^2\,d\nu(z) \right )^p\\
\\
& \ge C \sum_ j \left ( \frac{\mu(D_ j)}{(1-|b_ j|^2)^n} \right )^p.
\end{split}
\]
For the non-diagonal term, we can proceed almost exactly as in the proof of Lemma 12 in \cite{ZhuNYJM}. We only sketch some details. First, one has the estimate
$$\|E\|_p^p \le C\sum_{i=1}^\infty \mu(D_i)I_i,$$
where
$$I_i=\sum_{j\neq m} |h^t_j(b_i)h^t_m(b_i)|^p=\sum_{j\neq m} \frac{[(1-|b_j|^2)(1-|b_m|^2)]^{pt+pn/2}}{[|1-\langle b_i,b_j\rangle||1-\langle b_i,b_m\rangle|]^{pn+pt}}.$$
Denote
$$G_R=\{(z,w) \in \Bn\times \Bn : \beta(z,w)\geq R-2r\}.$$
Then,
$$I_i\leq C (1-|b_i|^2)^{-pn}\int\int_{G_R} \frac{[(1-|z|^2)(1-|w|^2)]^{pt+pn/2-n-1}dv(z)dv(w)}{[|1-\langle z,b_i\rangle||1-\langle w,b_i\rangle|]^{pt}}.$$
Since $(1-|z|^2)(1-|w|^2)\leq 4|1-\langle z,b_i\rangle||1-\langle w,b_i\rangle|$, we obtain
$$I_i\leq C (1-|b_i|^2)^{-pn}\int\int_{G_R} \frac{dv(z)dv(w)}{[|1-\langle z,b_i\rangle||1-\langle w,b_i\rangle|]^{n+1-pn/2}}.$$
Now, choose $x\in (1,\infty)$ so that
$$A=x(n+1-pn/2)<n+1.$$
It follows that
$$I_i\leq C(1-|b_i|^2)^{-pn} v^2(G_R)^{1/x'},$$
where
$$v^2(G_R)=\int\int_{G_R}dv(z)dv(w)$$ tends to zero as $R\to \infty$.
Therefore, there exists a constant $C'$ so that
$$\|S\|_p^p \geq (C-C'v^2(G_R)^{1/x'})\sum_{i=1}^\infty \left(\frac{\mu(D_i)}{(1-|b_i|^2)^n}\right)^p.$$
Recall that the original lattice was partitioned into $M$ subsequences that all satisfy this bound. Therefore the claim follows.
\end{proof}
\subsection{Proof of Theorem \ref{t-SC}}
The equivalence between (b) and (c) is Proposition \ref{S-P1}. The implication (b) implies (a) is proved in Proposition \ref{PS1} (case $0<p\le1$) and Proposition \ref{P-SC2} (case $p\ge 1$). Finally, Proposition \ref{PS1} gives the implication (a) implies (b) for $p\ge 1$, and the implication (a) implies (c) for $0<p\le 1$ is Proposition \ref{FPSC}.
\section{Applications to weighted composition operators, Volterra type integration operators and Carleson embeddings}
In this section we provide applications of our results on the Toeplitz type operator $Q_{\mu}$ in order to study the membership in the Schatten ideal $S_ p$ of weighted composition and Volterra type integral operators acting on $H^2$. We also obtain a description of when the Carleson embedding $R_{\mu}(f)=f$ is in $S_ p(H^2,L^2(\mu))$ for positive Borel measures $\mu$ in $\Bn$.
\\
For a holomorphic function $\varphi:\Bn \rightarrow \Bn$, the composition operator $C_{\varphi}$ is given by $C_{\varphi} f=f\circ \varphi$ for $f\in H(\Bn)$. Observe that, as $\varphi$ is a bounded holomorphic function on $\Bn$, it has radial limits $\varphi^*$ almost everywhere. Given a function $u\in H(\Bn)$, the weighted composition operator $W_{u,\varphi}$ is given by $W_{u,\varphi} f=uC_{\varphi} f$ for $f\in H(\Bn)$. We refer to \cite{CD1,CD2,Sty,ZC} for studies of weighted composition operators acting on Hardy spaces. If $W_{u,\varphi}$ is acting on $H^2$, then clearly $u\in H^2$, and therefore it has radial limits $u^*$ almost everywhere. Also it is well known and easy to see that $(W_{u,\varphi})^* W_{u,\varphi}=Q_{\mu_{u,\varphi}}$, where $\mu_{u,\varphi}$ is the measure defined on Borel sets $E\subset \Bn$ by
\[
\mu_{u,\varphi}(E)=\int_{ \varphi^{-1}(E)\cap \,\Sn} |u^*(\zeta)|^2\,d\sigma ( \zeta ).
\]
Since $W_{u,\varphi}$ belongs to $S_ p(H^2)$ if and only if $Q_{\mu_{u,\varphi}}\in S_{p/2}(H^2)$, and
\[
\int_{\Bn} \frac{d\mu_{u,\varphi}(z)}{|1-\langle z,w \rangle |^{2n+t}}=\int_{\Sn} \frac{|u^*(\zeta)|^2\,d\sigma(\zeta)}{|1-\langle \varphi^*(\zeta),w \rangle |^{2n+t}},
\]
in view of Theorem \ref{t-SC}, we obtain the following description of when the weighted composition operator $W_{u,\varphi}$ belongs to the Schatten ideal $S_ p(H^2)$.
\begin{theorem}\label{FinalT}
Let $0<p<\infty$, $u\in H(\Bn)$ and $\varphi:\Bn \rightarrow \Bn$ be holomorphic. Then $W_{u,\varphi}\in S_ p(H^2)$ if and only if, for $t>t_ {p/2}$ one has
\begin{equation}\label{FEq2}
\int_{\Bn}\left ((1-|w|^2)^{n+t}\int_{\Sn} \frac{|u^*(\zeta)|^2\,d\sigma(\zeta)}{|1-\langle \varphi^*(\zeta),w \rangle |^{2n+t}}\right )^{p/2}\,d\lambda_ n(w)<\infty.
\end{equation}
\end{theorem}
As a consequence we can recover the result of Luecking and Zhu \cite{LZ} on the membership in the Schatten classes of composition operators acting on the Hardy space of the unit disk $\mathbb{D}$. We use the Nevanlinna counting function $N^*_{\varphi}$ defined as
\[
N^*_{\varphi}(z)=\sum _{w: \varphi(w)=z} (1-|w|^2),\qquad z\in \mathbb{D}\setminus \{\varphi(0)\},
\]
where the last sum is interpreted as being zero if $z\notin \varphi(\mathbb{D})$.
\begin{coro}
Let $\varphi:\mathbb{D}\rightarrow \mathbb{D}$ analytic and $0<p<\infty$. Then $C_{\varphi}\in S_ p(H^2)$ if and only if
\[
\frac{N^*_{\varphi}(z)}{(1-|z|)} \in L^{p/2}(\mathbb{D},d\lambda_ 1).
\]
\end{coro}
\begin{proof}
For $w\in \mathbb{D}$ and $t>t_ {p/2}$, consider the function
$
f_ w(z)=(1-z\overline{w} )^{-1-t/2}
$, and observe that
\[
\int_{\mathbb{S}_1} \frac{d\sigma(\zeta)}{|1- \varphi^*(\zeta)\overline{w} |^{2+t}}=\|C_{\varphi}f_ w \|_{H^2}^2\asymp |C_{\varphi} f_ w(0)|^2 +\|(C_{\varphi} f_ w)'\|^2_{A^2_ 1}.
\]
By the change of variables formula (see \cite{CMc}),
\[
\|(C_{\varphi} f_ w)'\|^2_{A^2_ 1}=2 \int_{\D} |f'_ w(\varphi(z))|^2\,|\varphi'(z)|^2\,(1-|z|^2) \,dA(z)=2\int_{\D} |f'_ w(\zeta)|^2 N^*_{\varphi}(\zeta) dA(\zeta),
\]
where $dA$ is the normalized area measure on $\mathbb{D}$. In view of Theorem \ref{FinalT}, we have $C_{\varphi}\in S_ p(H^2)$ if and only if
\begin{equation}\label{FE4}
I(\varphi):=\int_{\D}\left ((1-|w|^2)^{1+t}\int_{\D} \frac{N^*_{\varphi}(z)\,dA(z)}{|1- z\overline{w}|^{4+t}} \right )^{p/2}\,d\lambda_ 1(w)<\infty.
\end{equation}
From here, the necessity follows from the fact that $N_{\varphi}^*$ satisfies the inequality (a consequence of \cite[Lemma 3.18]{CMc})
\[
\frac{N_{\varphi}^*(w)}{(1-|w|^2)^{2+t}}\lesssim \int_{\D} \frac{N^*_{\varphi}(z)\,dA(z)}{|1-z\overline{w} |^{4+t}}.
\]
When $p\ge 2$, from \eqref{FE4} we obtain the sufficiency after an application of H\"{o}lder's inequality with exponent $p/2$, and the typical integral estimate in Lemma \ref{IctBn}. The case $0<p<2$ requires more work. Take an $r$-lattice $\{a_ k\}$, and let $D_ k=D(a_ k,r)$. Then
\[
\int_{\D} \frac{N^*_{\varphi}(z)\,dA(z)}{|1-z\overline{w} |^{4+t}}\lesssim \sum_ k \frac{1}{|1-a_ k\overline{w} |^{4+t}}\int_{D_ k} N^*_{\varphi}(z)\,dA(z).
\]
By \cite[Proposition 11.4]{Zhu}, for $z\in D_ k$ we have
\[
N^*_{\varphi}(z)^{p/2} \lesssim \frac{1}{(1-|z|^2)^2}\int_{D(z,r)} N^*_{\varphi}(\xi)^{p/2}dA(\xi)\lesssim \int_{\widetilde{D}_ k} N^*_{\varphi}(\xi)^{p/2}d\lambda_ 1(\xi),
\]
with $\widetilde{D}_ k=D(a_ k,2r)$. This gives
\[
\int_{\D} \frac{N^*_{\varphi}(z)\,dA(z)}{|1- z\overline{w} |^{4+t}}\lesssim \sum_ k \frac{(1-|a_ k|^2)^2}{|1- a_ k\overline{w} |^{4+t}}\left (\int_{\widetilde{D}_ k} N^*_{\varphi}(\xi)^{p/2}d\lambda_ 1(\xi)\right )^{2/p}.
\]
Since $0<p/2<1$, we get
\[
\left (\int_{\D} \frac{N^*_{\varphi}(z)\,dA(z)}{|1-z\overline{w} |^{4+t}} \right )^{p/2} \lesssim \sum_ k \frac{(1-|a_ k|^2)^p}{|1- a_ k\overline{w} |^{(4+t)p/2}}\int_{\widetilde{D}_ k} N^*_{\varphi}(\xi)^{p/2}d\lambda_ 1(\xi).
\]
Inserting this into \eqref{FE4} and applying the typical integral estimate, we obtain
\[
\begin{split}
I(\varphi) &\lesssim \sum_ k (1-|a_ k|^2)^p \left (\int_{\mathbb{D}}\frac{(1-|w|^2)^{(1+t)p/2} d\lambda_ 1(w)}{|1-a_ k\overline{w} |^{(4+t)p/2}}\right)\int_{\widetilde{D}_ k} N^*_{\varphi}(\xi)^{p/2}d\lambda_ 1(\xi)
\\
& \lesssim \sum_ k (1-|a_ k|^2)^{-p/2} \int_{\widetilde{D}_ k} N^*_{\varphi}(\xi)^{p/2}d\lambda_ 1(\xi).
\end{split}
\]
As $(1-|a_ k|)\asymp (1-|\xi|)$ for $\xi \in \widetilde{D}_ k$, and because any point $\xi \in \mathbb{D}$ belongs to at most $N$ of the sets $\widetilde{D}_ k$, we finally get
\[
I(\varphi) \lesssim \sum_ k \int_{\widetilde{D}_ k} \left(\frac{N^*_{\varphi}(\xi)}{1-|\xi|}\right )^{p/2}d\lambda_ 1(\xi) \lesssim \int_{\mathbb{D}} \left(\frac{N^*_{\varphi}(\xi)}{1-|\xi|}\right )^{p/2}d\lambda_ 1(\xi).
\]
This finishes the proof.
\end{proof}
\mbox{}
\\
For a function $g\in H(\Bn)$, the Volterra type integration operator $J_ g$ is defined as
\[
J_ g f(z)=\int_{0}^{1} f(tz) Rg(tz) \frac{dt}{t},\qquad z\in \Bn
\]
for $f$ holomorphic in $\Bn$. An easy calculation provides the basic formula $R(J_ g f)=fRg$ involving the radial derivative $R$ and the operator $J_ g$.
If $J_ g:H^2\rightarrow H^2$, then it is well known (and easy to see) that $J_ g^* J_ g =Q_{\mu_ g} $ with $d\mu_ g=|Rg|^2dv_ 1$, where now $J_ g^*$ denotes the Hilbert adjoint respect to the inner product $\langle f,g \rangle_ *=f(0)\overline{g(0)}+\int_{\Bn} Rf\,\overline{Rg}\,dv_ 1$. Therefore, $J_ g$ is in $S_ p(H^2)$ if and only if $Q_{\mu_ g}$ belongs to $S_{p/2}(H^2)$. By Theorem \ref{t-SC}, this is equivalent to
\begin{equation}\label{FEq1}
\int_{\Bn}\left ((1-|w|^2)^{n+t}\int_{\Bn} \frac{|Rg(z)|^2\,dv_ 1(z)}{|1-\langle z,w \rangle |^{2n+t}}\right )^{p/2}\,d\lambda_ n(w)<\infty,
\end{equation}
for $t>t_ {p/2}$. From this condition, it is easy to see that $J_ g \in S_ p(H^2)$ if and only if $g$ belongs to the analytic Besov space $B_ p$ when $p>n$, and $g$ being constant when $0<p\le n$. This recovers the results from \cite{AS1} and \cite{P1}. Indeed, if \eqref{FEq1} holds, by ``subharmonicity" we get
\[
\int_{\Bn} \left ( (1-|w|^2)\,|Rg(w)|\right )^{p}\,d\lambda_ n(w)<\infty.
\]
Hence \eqref{FEq1} implies that $g\in B_ p$ for $p>n$ and $g$ constant when $0<p\le n$. Finally, if $p>n$ and $g\in B_ p$ we see that condition \eqref{FEq1} is satisfied when $p\ge 2$ after an application of H\"{o}lder's inequality and the typical integral estimate. When $n=1$ and $1<p<2$, we apply the inequality
\[
|Rg(z)|^p \lesssim \frac{1}{(1-|z|^2)^{n+1}} \int_{D(z,r)} |Rg(\xi)|^p \,dv(\xi)
\]
and argue in a similar manner as in the case of composition operators.
The result obtained for the integration operator $J_ g$ can be generalized. By the Littlewood-Paley inequalities we have
\[
\|J_ g f \|_{H^2}^2 \asymp \|R(J_ g f)\|^2_{A^2_ 1} =\int_{\Bn} |f(z)|^2 \,d\mu_ g(z).
\]
Hence we can view the operator $J_ g$ on $H^2$ as the Carleson embedding from $H^2$ to $L^2(\Bn,d\mu_ g)$. Next, we are going to characterize, for a positive Borel measure $\mu$, when the Carleson embedding $R_{\mu}: H^2 \rightarrow L^2(\mu):=L^2(\Bn,d\mu)$ is in $S_ p$. An easy computation yields $R^*_{\mu}R_{\mu}=Q_{\mu}$, and from Theorem \ref{t-SC} we obtain the following description.
\begin{theorem}
Let $\mu$ be a positive Borel measure on $\Bn$ and $0<p<\infty$. Then the Carleson embedding $R_{\mu}$ is in $S_ p(H^2,L^2(\mu))$ if and only if $S_ t\mu \in L^{p/2}(\Bn,d\lambda_ n)$ for $t>t_{p/2}$.
\end{theorem}
This characterization can be compared with the result obtained by Smith \cite{Sm} in the case of the unit disk.
| 2,702
|
TITLE: Is Bell nonlocality compatible with, or even predicted by, quantum field theory?
QUESTION [3 upvotes]: Bell's theorem, together with experiments confirming the ordinary quantum mechanical distant correlation predictions which violate its conclusion for local theories, show that any entirely local theory is incorrect. QFT's microcausality postulate forbids an experiment at one space-time point from influencing the probabilities of an experimental outcome at a distant spacelike-related point, but doesn't forbid those distant correlations.
Does the current QFT of the Standard Model actually predict those distant correlations, or if not, does it at least allow them? It is difficult for me to determine this, since I have never taken a QFT course, and QFT is usually stated in the Heisenberg picture rather than the Schrödinger picture. If QFT is entirely local, and forbids those distant correlations, is this viewed as a serious flaw in it, perhaps as serious as the failure, so far, to make any quantum theory fully compatible with general relativity?
REPLY [4 votes]: The pair of entangled photons violating Bell's inequality are perfectly described by a state of the EM field containing two particles. Therefore QFT gives rise to non-local correlations, just in view of the said violation.
There is no problem with micro causality which refers to observables: localized field operators (if they represent observables as is the case for boson fields). They must commute when they are associated to spatially separated regions of the spacetime. Bell's inequality instead concerns outocomes of measurements in peculiar non-local (entanlged) states.
Micro causality does not forbid non-local correlations.
DETAILS
The meaning of locality aka micro causality in QFT can be illustrated as follows.
Consider two regions $A$, $B$ in Minkowski spacetime which are spacelike related and a pair of corresponding (quantum-field) observables $O_A$ and $O_B$ localized at those regions respectively.
Let $P_A$ and $P_B$ be the orthogonal projectors of the spectral measures of $O_A$ and $O_B$ respectively, corrsponding to the oucomes $E_A$ and $E_B$, respectively.
Let $\Psi$ be a normalized vector representing the state of the system.
The locality axiom requires that (*)
$$P_AP_B \Psi = P_BP_A \Psi\tag{1}$$
which, in particular, implies that
$$||P_AP_B \Psi||^2 = ||P_BP_A \Psi||^2\:.$$
In other words,
the probability to measure the outcome $E_A$ first and the outcome $E_B$ next on the state $\Psi$, -- when measuring first $O_A$ and next $O_B$ -- equals the probability to measure the outcome $E_B$ first and the outcome $E_A$ next on the state $\Psi$ -- when measuring first $O_B$ and next $O_A$.
This requirment is very natural since the temporal order of $A$ and $B$ cannot be fixed because these regions are spacelike related.
This natural requirement is in perfect agreement with the setup of Bell experiment. There the quantum field is the EM one and we measure $4$ couples of photon-polarization observables localized in the regions $A$ and $B$ as above
$O_A,O_B\quad$, $O_A,O'_B\quad$, $O'_A,O_B\quad$, $O'_A,O'_B\quad$.
The state $\Psi$ is very peculiar:
(a) it includes two particles only,
(b) it is a special case of an entangled state.
The Bell (actually CHSH) inequality concerns a relation among the expectation values
$\langle \Psi|O_AO_B\Psi\rangle\quad$, $\langle \Psi|O'_AO_B\Psi\rangle\quad$, $\langle \Psi|O_AO'_B\Psi\rangle\quad$, $\langle \Psi|O'_AO'_B\Psi\rangle\quad$.
If there were a (realistic-local) hidden-variable theory capable to explain the results of the above measurment procedures of quantum theory, the expectation values above should satisfy a certain inequality (the CHSH inequality).
Instead, quantum (field) theory predicts -- through an elementary computation -- that this inequality is violated.
DISCUSSION
The crucial fact that, together with the choice of the state (see below), produces the violation is the choice of the pairs $O_A, O'_A$ and $O_B, O'_B$: these pairs are made of certain incompatible observables. Though the observables of each such pair are never simultaneously measured, the total statistics feels this quantum incompatibility and produces the violation.
I finally stress that locality in the hypotheses of Bell inequality is used in a different way with respect to its use in QFT: it means that the result of the mesurement of, e.g., $O_A$ does not depend on the choice of the observable $O_B$ or $O'_B$ simultaneously measured, since this choice is made in a causally separated region.
This requirment holds true in classical physics, but not in quantum physics and the (entangled) quantum states are responsible for that, not the observables.
(*) The postulate is more usually stated in terms of bosonic field operators as
$$[\phi(f),\phi(g)]=0\tag{2}$$ if $\mathrm{supp}(f) \subset A$ and $\mathrm{supp}(g) \subset B$ with $A$ and $B$ spacelike related and $f$, $g$ real.
Assuming the field operators selfadjoint (or taking the closures of the standard operators) and considering the von Neumann algebras generated by them when $f$ and $g$ are taken as above, the commutation relation above extends to all elements of these algebras. These algebras includes the spectral projectors of every observable constructed as said. Therefore the standard requirement imples (1). Conversely (1) immediately produces the standard requirement (2) (on a suitable dense invariant domain) when the considerd observables are the field operators themselves.
| 132,048
|
TITLE: Induction - request for help in proving lemma
QUESTION [1 upvotes]: Could anyone help with proving the following lemma, please?
Let: $n\in \mathbb{N}$, $Z_{n}^{*}:=\{k\in\mathbb{N}: k\in\{1,\dots,n\} \wedge \space GCD(k,n)=1\}$. Then: $\forall n\in \mathbb{N} \space \forall p \in \mathbb{P}: |Z_{p^{n}}^{*}|=p^{n}-p^{n-1}$
I tried to prove this by induction with respect to $n$, but I stuck at general case. I know how induction works, but I can't see how to do main point...
REPLY [1 votes]: You may be overthinking it. ${\rm gcd \ }(k,p^{n}) = 1$ holds if and only if $k$ is not divisible by $p$.
So you need to count the number of elements in $\mathbb Z_{p^n}$ that are not divisible by $p$.
Ask yourself:
(1) How many elements are there in $\mathbb Z_{p^n}$ in total?
(2) What fraction of these elements are divisible by $p$ and what fraction are not?
| 185,261
|
Description. developpers Conctact preimarly_filter_cf7_delay_select2_launch’
This allows you manually launch the select2 jquery dropdonw fields. This is required if you need to customise the select dropdown on windows load event with your own jquery scripts before the select2 transformation is applied. Please read the FAQ on custom scripts to make sure you trigger your script after the form is mapped.
add_filter( 'cf7_2_post_filter_cf7_delay_select2_launch', '__return_true');
‘cf7_2_post_filter_cf7_taxonomy_select_optgroup’
This filter expects a boolean, by default it is
false and disables
optgroup on select dropdown options.
To enable grouped options for hierarchical taxonomy top level term, you can use this filter. Note however, that child term options will be grouped by their top-level parent only as nested
optgroup are not allowed. Furthermore the parent term will not be selectable. Therefore this option only makes sense if you have a hierarchical taxonomy with only single level child terms which can be selected. To enable grouped options,
add_filter('cf7_2_post_filter_cf7_taxonomy_select_optgroup','enable_grouped_options',10,4); function enable_grouped_options($enable, $cf7_post_id, $form_field, $parent_term){ if(20 == $cf7_post_id && 'your-option' == $form_field){ //we assume here that cf7 form 20 has a dropdown field called 'your-option' which was mapped to a hierarchical taxonomy //you can even filter it based on the parent term, so as to group some and not others. //the attribute $parent_term is a WP_Term object switch($parent_term->name){ case 'Others': $enable=false; break; default: $enable=true; break; } } return $enable; }
`cf7_2_post_filter_user_draft_form_query`
This filter is useful to change the behaviour in which previously submitted values can be edited by a user. By default the plugin loads forms values that have been saved using the ‘save’ button. However, you can modify this by changing the default post query such as the example below,
add_filter('cf7_2_post_filter_user_draft_form_query','load_published_submissions',10,2); function load_published_submissions($query_args, $post_type){ if('submitted-reports' == post_type){ //we assume a cf7 form allowing users to submit online reports has been mapped to a post_type submitted-reports' if(is_user_logged_in()){ $user = wp_get_current_user(); //load this user's previously submitted values $query_args = array( 'posts_per_page' => 1, 'post_type' => post_type, 'author' => $user->ID, 'post_status' => 'published' ); } } return $query_args; }
`cf7_2_post_form_append_output`
this filter is fired when the cf7 shortcode output is printed, it allows you to add a custom script at the end of your form should you need; }
`cf7_2_post_form_values`
This filter allows you to load default values into your mapped forms. If the current user has a saved form, this filter will override any values you set.
add_filter('cf7_2_post_form_values', 'set_default_location', 10, 3); function set_default_location($values, $$cf7_id, $mapped_post_type, $cf7_key){ if('travel_post' != mapped_post_type){ return values; } if('contact-us' == cf7_key){ //check this is your form $field_name = 'location-rental'; $values[$field_name] .= 'Paris'; } return $values; },
Screenshots.
Installation
- save it as a draft or publish it. Once published you cannot edit the mapping anymore, so be warned. As of version 1.2.0 you will have to delete the whole form and start again to remap it. Subsequent versions may introduce a ‘delete’ button.
3. I made a mistake in my form mapping, how do I correct it once it is created?
as of v2.0.0 you can now quick-edit (inline edit) your form in the forms table listing and reset your form mapping to
draftmode which will allow you?
You can quick-edit a form in the Contact Form table and delete the mapping. herlper code is availabe have a form which your users can save before submitting and you need to customise some additional functionality on your form on
$(document).ready(), then you need to make sure it fires after the plugin’s scripts have finished. In order to achieve this, the script fires a custom event on the form,
cf7Mapped, which you can use to ensure you script fires in the right order, here is how you would enable this,
(function( $ ) { 'use strict'; $( function() { //the jQuery equivalent of document.ready() var cf7Form = $('div.cf7_2_post form.wpcf7-form'); //this ensures you target the mapped forms cf7Form.on('formMapped', function(){ //fire your script }); }); })( jQuery );.']');
15. I have enabled a save button on my form, but draft submissions are not being validated!
This is the default functionality for saving draft submissions. This is especially useful for averyable on the front-end.
If you have created a custom taxonomy for your post, you can include these into your main menu by enabling them in the Appearance->Menu screen opttions yoru dashboard using meta-boxes. You can custom create these meta-boxes in your functions.php file or you can also use a plugin. However, without these meta-boxees you won’t be able to see your field values when you edit your posts.
19. Why does my form page have no-cache metas ?
As of v3.0.0 the no-cache metas have been added by default to pages with embeded attribtues.
Reviews
Contributors & Developers
“Post My CF7 Form” is open source software. The following people have contributed to this plugin.Contributors
Translate “Post My CF7 Form” into your language.
Interested in development?
Browse the code, check out the SVN repository, or subscribe to the development log by RSS.
Changelog
4.1.9
- make mapped post types default to _cf7_2_post_form_submitted = yes for posts created in the dashboard.
4.1.8
- fix single select values.
- improved form admin tables.
- fixed js issue in admin mapping page.
4.1.7
- fix cfadmin table class clash with Smart-grid.
4.1.6
- fix cf7 table clash with cf7sg.
4.1.5
- fix hidden field bugs introduced by CF7 v5.2.
4.1.4
- fix file upload bug, use $_FILES instead of CF7 submission.
4.1.3
- fix add_filter helper code.
4.1.2
- added submitted files to cf7_2_post_form_submitted_to_$post_type hook.
- fixed helper functions.
4.1.
| 281,987
|
Aug
Books I've read:
Blood Memory by Margaret Coel
Sex Lies and Snickerdoodles by Wendy Delany
Coming this week:
Aug 18th- Guest Post -Molly MacRae
Aug 19th- Book Review- Murder in the Air by Marilyn levinson
Aug 20th- Author Spotlight- Carola Dunn
Aug 21st- First in a New Series- Death of a Crabby Cook by Penny Pike
Aug 22nd- Friday 56
Aug 23rd- Book Trivia- Mary Anne Edwards
Hope to see you!!
| 21,216
|
\begin{document}
\begin{abstract}
We sharpen the ellipticity criteria for random walks in i.i.d. random environments introduced by Campos and Ram\'\i rez which ensure ballistic behavior. Furthermore, we construct new examples of random environments for which the walk satisfies the polynomial ballisticity criteria of Berger, Drewitz and Ram\'\i rez. As a corollary we can exhibit a new range of values for the parameters of Dirichlet random environments in dimension $d=2$ under which the corresponding random walk is ballistic.
\end{abstract}
\maketitle
\section{Introduction}
We continue the study initiated in \cite{CR} sharpening the ellipticity
criteria which ensure ballistic behavior of random walks in random
environment. Furthermore, we apply our results to exhibit a new
class of ballistic random walks in Dirichlet random environments
in dimensions $d=2$.
For $x \in \R^d$, denote by $|x|_1$ and $|x|_2$ its $L_1$ and $L_2$ norm respectively. Call $U:=\{ e \in \Z^d : |e|_1 = 1 \} = \{ e_1, \ldots ,e_{2d} \}$ the canonical vectors with the convention that $ e_{d+i} = -e_i $ for $1\leq i\leq d$. We set ${\mathcal P}:=\{p(e): p(e)\geq 0,\sum_{e\in U}p(e)=1\}$.
An environment is an element $\omega := \{ \omega(x): x \in \Z^d \}$ of the environment space $ \Omega := \mathcal{P} ^{\Z^d}$. We denote the components of $\omega(x)$ by $\omega(x,e)$.
The random walk in the environment $\omega$ starting from $x$ is the Markov chain $ \{ X_n: n \geq 0 \} $ in $\Z^d$ with law $P_{x,\omega}$ defined by the condition $ P_{x,\omega}( X_0 = x )=1 $ and the
transition probabilities
\[ P_{x,\omega}(X_{n+1} = x+e | X_n = x) = \omega (x,e) \]
for each $x \in \Z^d$ and $e\in U$.
Let $\P$ be a probability measure defined on the environment space $\Omega$ endowed with its Borel $\sigma$-algebra. We choose $ \{ \omega(x): x \in \Z^d \} $ i.i.d. under $\P$. We call $P_{x,\omega}$ the quenched law of the random walk in random environment (RWRE) starting from $x$, and $ P_x: = \int P_{x,\omega} d\P $ the averaged or annealed law of the RWRE starting from $x$.
The law $\P$ is said to be elliptic if for every $ x \in \Z^d $ and $ e \in U $, $ \P( \omega(x,e) > 0 ) = 1 $. We say that $\P$ is uniformly elliptic if there exists a constant $\gamma>0$ such that for every $ x \in \Z^d $ and $ e \in U $, $ \P ( \omega(x,e) \geq \gamma ) = 1 $.
Given $ l \in \mathbb{S} ^{d-1}$ we say that the RWRE is transient in direction $l$ if
\[ P_0 (A_l) = 1 ,\]
with
\[ A_l := \{ \lim _{n \to \infty} X_n \cdot l = \infty \} .\]
Furthermore, it is ballistic in direction $l$ if $P_0$-a.s.
\[ \liminf _{ n \to \infty } \frac{ X_n \cdot l }{ n } > 0 .\]
Given $ \Lambda \subset \Z^d$, we denote its outer boundary by
\[ \partial \Lambda := \{ x \notin \Lambda : |x-y|_1 = 1 \text{ for some } y \in \Lambda \} .\]
We denote any nearest neighbour path with $n$ steps joining two points $ x,y \in \Z^d $ by $ ( x_1, x_2,\ldots, x_n ) $, where $x_1=x$ and $x_n=y$.
\subsection{Polynomial condition, ellipticity condition}
In \cite{BDR}, Berger, Drewitz and Ramírez introduced a polynomial ballisticity condition within the uniformly elliptic context, which was later extended to the elliptic case by Campos and Ram\'\i rez in \cite{CR}. This condition will be of interest for our results. It is effective, in the sense that it can a priori be verified explicitly for a given environment.
To define it, we need for each $ L,\tilde L > 0 $ and $ l \in \mathbb S ^{d-1} $ to consider the box
\[ B _{l,L,\tilde L} := R \( ( -L,L ) \times ( -\tilde L,\tilde L ) ^{d-1} \) \cap \Z^d ,\]
where $R$ is a rotation of $\R^d$ that verifies $ R(e_1) = l $.
For each subset $A\subset\mathbb Z^d$ we note the first exit time from the set $A$ as
\[ T_A := \inf \{ n \geq 0: X_n \notin A \} .\]
\begin{de}
Given $ M \geq 1 $ and $ L \geq 2 $, we say that condition $ (P) _M $ in direction $l$ is satisfied on a box of size $L$ (also written as $ (P) _M | l $) if there exists an $\tilde L \leq 70 L^3 $ such that one has the following upper bound for the probability that the walk does not exit the box $B_{l,L,\tilde L}$ through its front side :
\[ P _0 ( X_{ T_{ B_{l,L,\tilde L} } } \cdot l < L ) \leq \frac{1}{L^M}.\]
\end{de}
This condition has proven useful in the uniformly elliptic case. Indeed, $(P)_M$ for $ M \geq 15 d+5 $ implies ballisticity (see \cite{BDR}).
For non uniformly elliptic environments in dimensions $d\geq 2$, there exist elliptic random walks which are transient in a given direction but not ballistic in that direction (see for example Sabot-Tournier \cite{ST}, Bouchet \cite{Bouchet}).
In \cite{CR}, Campos and Ramírez introduced ellipticity criteria on the law of the environment
which ensure ballisticity if condition $(P)_M$ is satisfied for $M\geq 15d+5$. In this article we will sharpen this ellipticity criteria.
Define $ \eta _\alpha := \max _{e \in U} \E \( \frac{1}{ \omega(0,e) ^\alpha
} \) $, $ \overline{\alpha} := \sup \{ \alpha \geq 0 : \eta _\alpha < \infty
\} $. We need the constant $ c_0 := \frac{2}{3} 3 ^{ 120 d^4 + 3000 d ( \log
\eta _{ \overline{\alpha}/2 } ) ^2 } $ : throughout the rest of this paper,
whenever we assume that the polynomial condition $ (P)_M $ is satisfied, it will be understood that this happens on a box of size $ L \geq c_0 $.
Let us first recall the ellipticity condition of \cite{CR}. The polynomial condition $(P)_M$ implies the existence of an asymptotic direction (see for example Simenhaus \cite{Si}): there exists $ \hat v \in \mathbb{S} ^{d-1} $ such that $P_0$-a.s.,
\[ \lim _{n\to\infty} \frac{X_n}{|X_n|_2} = \hat v .\]
We call $ \hat v $ the asymptotic direction. Under $(P)_M$ we can then define the half space
\[ H_{\hat v} := \{ l \in \R^{d} : l \cdot \hat v \geq 0 \} .\]
\begin{de}
Let $ \beta > 0 $. We say that the law of the environment satisfies the ellipticity condition $ (E') _\beta $ if there exists an $ \{ \alpha(e) : e \in U \} \in (0,\infty)^{2d} $ such that
\begin{equation}
\label{kappa}
\kappa \( \{ \alpha(e) : e \in U \} \) := 2 \sum_{e'} \alpha(e') - \sup _{e\in U} \( \alpha(e) + \alpha(-e) \) > \beta
\end{equation}
and for every $ e \in U $
\begin{equation}
\label{finiteness}
\E \( e ^{ \sum _{e'\neq e} \alpha(e') \log \frac{1}{\omega(0,e')} } \) < \infty .
\end{equation}
Furthermore, when $ \hat v $ exists, we say that the ellipticity condition $ (E') _\beta $ is satisfied towards the asymptotic direction if there exists an $\{ \alpha(e) : e \in U \}$ satisfying (\ref{kappa}) and (\ref{finiteness}) and such that there exists $ \alpha_1 > 0 $ that satisfies $ \alpha(e) = \alpha_1 $ for $ e \in H _{ \hat v} \cap U $ while $ \alpha(e) \leq \alpha_1 $ for $ e \in U \setminus H _{ \hat v} $.
\end{de}
\begin{rmq}
Replacing (\ref{finiteness}) by $ \E \( e ^{ \sum _{e'} \alpha(e') \log \frac{1}{\omega(0,e')} } \) < \infty $ gives an equivalent condition. The direct implication is straightforward. And since $ 1 \leq \sum _{e \in U} \1 _{ \{ \omega(0,e) \geq \frac{1}{2d} \} } $, we get
\[ \E \( e ^{ \sum _{e'} \alpha(e') \log \frac{1}{\omega(0,e')} } \)
\leq
\sum _{e \in U} e ^{ \alpha(e) \log (2d) } \E \( e ^{ \sum _{e'\neq e} \alpha(e') \log \frac{1}{\omega(0,e')} } \) .\]
This gives the reverse implication.
\end{rmq}
\begin{rmq}
\label{eta-alpha}
The results of \cite{CR} still hold (with no changes in the proofs) if we replace the previous definition of $\eta _\alpha$ by
\[ \eta _\alpha := \max _{e \in H _{ \hat v}} \E \( \frac{1}{ \omega(0,e) ^\alpha
} \) .\]
It means that the constant $c_0$ in the definition of condition $ (P)_M $ depends only of the integrability towards the asymptotic direction.
But knowing the existence of $ \hat v $ does not mean that we know its value. In most cases, $ \hat v $ is found to be inaccessible. A notable exception is the result of Tournier (\cite{T2}) that gives the value of $ \hat v $ in the case of random walks in Dirichlet environments.
\end{rmq}
\subsection{Ballisticity results}
Our main results are a generalization of theorems 1.2 and 1.3 of \cite{CR} where we remove the
"towards the asymptotic direction" condition of Theorems 1.2 and 1.3 of \cite{CR}.
Let $\tau^{\hat v}_1$ be the first renewal time in the direction $\hat v$, its precise definition is recalled in the next section. We prove the following tail estimate on renewal times, which improves proposition 5.1 of \cite{CR}.
\begin{theo}
\label{tails-improved}
Let $l \in \mathbb S^{d-1}$, $\beta > 0$ and $M \geq 15d+5$. Assume that $(P)_M|l$ is satisfied and that $(E')_\beta$ holds (cf. (\ref{kappa}), (\ref{finiteness})).
Then
\[ \limsup _{u \to \infty} (\log u)^{-1} \log P_0 ( \tau_1 ^{\hat v} > u ) \leq -\beta .\]
\end{theo}
The condition $(E'_\beta)$ is sharp in a sense that is made precise in remark \ref{sharpness} below. Together with previous results of Sznitman, Zerner, Sepp\"al\"ainen and Rassoul-Agha, cf \cite{SZ,Z,Sz00,RAS}, it implies the following:
\begin{theo}[Law of large numbers]
\label{LLN-improved}
Consider a random walk in an i.i.d. environment in dimensions $d \geq 2$. Let $l \in \mathbb S^{d-1}$ and $M \geq 15 d+5$. Assume that the random walk satisfies condition $(P)_M|l$ and the ellipticity condition $(E')_1$.
Then the random walk is ballistic in direction $l$ and there is a $v \in \R^d$, $v \neq 0$ such that
\[ \lim _{n \to \infty} \frac{X_n}{n} = v, \quad P_0-a.s. \]
\end{theo}
\begin{theo}[Central limit theorems]
\label{CLT-improved}
Consider a random walk in an i.i.d. environment in dimensions $d \geq 2$. Let $ l \in \mathbb S^{d-1} $ and $ M \geq 15 d+5 $. Assume that the random walk satisfies condition $(P)_M|l$.
\begin{itemize}
\item[a)] (Annealed central limit theorem) If $(E')_2$ is satisfied then
\[ \epsilon ^{1/2} (X _{ [\epsilon^{-1} n] } - [\epsilon^{-1} n] v) \]
converges in law under $P_0$ as $\epsilon \to 0$ to a Brownian motion with non-degenerate covariance matrix.
\item[b)] (Quenched central limit theorem) If $(E')_{176d}$ is satisfied, then $\P$-a.s. we have that
\[ \epsilon ^{1/2} (X _{ [\epsilon^{-1} n] } - [\epsilon^{-1} n] v) \]
converges in law under $P_{0,\omega}$ as $\epsilon\to 0$ to a Brownian motion with non-degenerate covariance matrix.
\end{itemize}
\end{theo}
Removing the "towards the asymptotic direction" is a real improvement: in section~\ref{example-dirichlet}, we will give some examples of environments (in the class of Dirichlet environments) that satisfy $(E'_\beta)$ but not towards the asymptotic direction. For those environments, our new theorems allows to prove a LLN or CLT. Furthermore, our final goal would be to get a ballisticity condition that depends only locally on the environment (i.e. a condition that depends only on the law of the environment at one point). Condition $(E')_\beta$ is local, whereas $(E')_\beta$ towards the asymptotic dimension is not: removing the "towards the asymptotic direction" is then a first step in this direction. Ideally, we would also need to get rid of condition $(P)_M|l$, that is not local either. This is a much more difficult problem, not solved even in the uniformly elliptic case.
\begin{rmq}\label{sharpness}
The condition of theorem \ref{tails-improved} is sharp under the following assumption on the tail behavior of the environment at one site: there
exists some $(\beta_{e})_{e\in U}$, $\beta_e\geq 0$, and a positive constant $C>1$ such that for all $e\in U$
$$
C^{-1} \left( \prod_{e'\in U, e'\neq e} t_{e'}^{\beta_{e'}}\right) \leq \P \( \omega(0,e')\leq t_{e'}, \; \forall e'\in U, e'\neq e \) \leq C \left( \prod_{e'\in U, e'\neq e} t_{e'}^{\beta_{e'}}\right)
$$
for all $(t_{e'})_{e'\in U\setminus\{e\}}$, $0 \leq t_{e'}\le1$. Dirichlet environment (cf the next section)
is a typical example of environment that satisfies this condition.
Indeed, in this case we easily see that $(E')_\beta$ is satisfied if and only if $\beta < 2 \sum_{e'} \beta(e') - \sup _{e\in U} \( \beta(e) + \beta(-e) \)$. On the other hand
if $\beta \geq 2 \sum_{e'} \beta(e') - \sup _{e\in U} \( \beta(e) + \beta(-e) \)$ then $\E\left((\tau_1^{\hat v})^\beta\right)=\infty$. Indeed, consider a direction $e_0$ which realizes
the maximum in $\sup _{e\in U} \( \beta(e) + \beta(-e) \)$ and set $K=\{0,e_0\}$. We denote by $\partial_+ K$ the set of edges that exit the set $K$, which is composed of
the edges $\{(0,e)\}_{e\neq e_0}$ and $\{(e_0, e)\}_{e\neq -e_0}$.
For small $t>0$, under the condition that $\omega(x,y)\leq t$ for all $(x,y)\in \partial_+ K$ we have $P_{0,\omega} (T_K\geq n)\geq (1-(2d-1)t)^n$. Hence,
\begin{eqnarray*}
P_0(T_K \geq n)
&\geq &
(1-(2d-1)/n)^n \P \( \omega(x,y) \leq {1/ n}, \; \forall (x,y)\in \partial_+K \)
\\
&\geq &(1-(2d-1)/n)^n C^{-1} n^{- \( \(\sum_{e'} \beta(e')\) - \( \beta(e_0) + \beta(-e_0) \) \) }
\end{eqnarray*}
which implies that $E_0(T_K^\beta)=\infty$. Since $T_K$ is clearly a lower bound for the first renewal time it gives the result.
\end{rmq}
\begin{rmq}
\label{P_M}
Theorem 1.1 of \cite{CR} states that for i.i.d. environments in dimensions $d\geq 2$ satisfying the ellipticity condition $(E')_0$, the polynomial condition $(P)_M|l$ (for $ l \in \mathbb S^{d-1} $ and $M \geq 15 d+5$) is equivalent to Sznitman's condition $(T')|l$ (see for example \cite{Sz01} for the definition).
We can therefore replace $(P)_M|l$ by $(T')|l$ in the statements of Theorems \ref{LLN-improved} and \ref{CLT-improved}.
\end{rmq}
\subsection{New examples of random walks satisfying the polynomial condition}
In this article we also introduce new examples of RWRE in environments which are not uniformly elliptic and which satisfy the polynomial condition $(P)_M$ for $M\geq 15d+5$. In subsection \ref{example1} we prove the polynomial condition for a subset of marginal nestling random walks, including a particular environment introduced by Campos and Ram\'\i rez in \cite{CR}. In subsection \ref{example-dirichlet}, we prove the polynomial condition for a class of random walks in Dirichlet random environments which do not necessarily satisfy Kalikow's condition.
\subsubsection{Example within the class of marginal nestling random walks}
\label{example1}
Following Sznitman \cite{Sz00}, we say that a law $\P$ on $\Omega$ is {\it marginal nestling} if the convex hull $K_o$ of the support of the law of
$$
d(0,\omega):=\sum_{e\in U}\omega(0,e)e
$$
is such that $0\in\partial K_o$. We will prove in section \ref{last-section1} that a certain subset of the marginal nestling laws satisfies the polynomial condition.
\begin{theo}
\label{marginal}
Consider an elliptic law $\P$ under which $\{\omega(x):x\in\mathbb Z^d\}$ are i.i.d. Assume that there exists an $r>1$ such that $\omega(0,e_1)=r\omega(0,e_{1+d})$. Then the polynomial condition $(P)_M|_{e_1}$ is satisfied for some $M\geq 15d+5$.
\end{theo}
\begin{rmq}
This theorem is valid for all i.i.d. elliptic environments satisfying $ \omega(0,e_1) = r \omega(0,e_{1+d}) $, including uniformly elliptic environments. However, the environments are marginal nestling only in the non-uniformly elliptic case.
\end{rmq}
The above result includes an example suggested in \cite{CR}, by Campos and Ramírez, of an environment which satisfies the polynomial condition and for which the random walk is directionally transient but not ballistic. They showed that on this environment, $ (E') _{\alpha} $ is satisfied for $\alpha$ smaller but arbitrarily close to $1$, and that the walk is transient but not ballistic in a given direction. The proof that this environment satisfies the polynomial condition was left for a future work.
Let us define the environment introduced in \cite{CR}. Let $ \phi $ be any random variable taking values on the interval $ (0 , 1/4) $ and such that the expected value of $ \phi ^{-1/2} $ is infinite, while for every $ \epsilon > 0 $, the expected value of $ \phi ^{ - (1/2 - \epsilon) } $ is finite. Let $X$ be a Bernoulli random variable of parameter $1/2$. We now define $ \omega (0,e_1) = 2 \phi $, $ \omega (0,-e_1) = \phi $, $ \omega (0,e_2)= X \phi + (1-X) (1 - 4 \phi) $ and $ \omega (0,-e_2) = X (1 - 4 \phi) + (1-X) \phi $.
For every $\epsilon>0$ this environment satisfies $(E')_{1-\epsilon}$: traps can appear because the random walk can get caught on two edges of the type $ ( x , e_2 ) , ( x + e_2 , -e_2 ) $. Furthermore, it is transient in direction $e_1$ but not ballistic in that direction.
\subsubsection{Examples within the class of Dirichlet random environments}
\label{example-dirichlet}
Random Walks in Dirichlet Environment (RWDE) are interesting because of the analytical simplifications they offer, and because of their link with reinforced random walks. Indeed, the annealed law of a RWDE corresponds to the law of a linearly directed-edge reinforced random walk (\cite{ES}, \cite{P}).
Given a family of positive weights $ (\beta _1, \dots , \beta _{2d}) $, a random i.i.d. Dirichlet environment is a law on $\Omega$ constructed by choosing independently at each site $x \in \Z^d$ the values of $ \( \omega ( x, e_i ) \) _{i \in [\![1,2d]\!]} $ according to a Dirichlet law with parameters $ (\beta _1, \dots , \beta _{2d})$. That is, at each site we choose independently a law with density
\[ \frac{ \Gamma \( \sum _{i=1} ^{2d} \beta _i \) }{ \prod_{i=1} ^{2d} \Gamma \( \beta _i \) } \( \prod _{i=1} ^{2d} x_i ^{\beta _i - 1} \) dx_1 \dots dx_{2d-1} \]
on the simplex $\{(x_1,\dots, x_{2d}) \in ]0,1]^{2d}, \sum _{i=1} ^{2d} x_i = 1 \}$. Here $\Gamma$ denotes the Gamma function $ \Gamma (\beta) = \int _0 ^\infty t ^{\beta - 1} e ^{-t} dt$ , and $dx_1 \dots dx_{2d-1}$ represents the image of the Lebesgue measure on $ \R^ {2d-1} $ by the application $( x_1, \dots ,x_{2d-1} ) \to ( x_1, \dots , x_{2d-1}, 1- x_1 - \dots - x_{2d-1} ) $. Obviously, the law does not depend on the specific role of $ x_{2d}$.
\begin{rmq}
Given a Dirichlet law of parameters $ ( \beta_1, \dots, \beta_{2d} ) $, the ellipticity condition $ (E') _\beta $ is satisfied if and only if
\[ \kappa \( ( \beta_1, \dots, \beta_{2d} ) \) = 2 \( \sum_{i=1} ^{2d} \beta _i \) - \max _{i=1,\dots, d} (\beta _i + \beta _{i+d}) > \beta .\]
As stated in remark~\ref{sharpness}, this ellipticity condition is optimal to get theorem~\ref{tails-improved} in the case of Dirichlet environments. Remark that for Dirichlet environments, $ (E') _\beta $ is much sharper that $ (E') _\beta $ towards the asymptotic direction. Indeed, $ (E') _\beta $ is satisfied towards the asymptotic direction $ \hat{v} $ if and only if we also have the existence of a constant $ c > 0 $ such that for all $i$, $ \beta_i \leq c $, and for all $i$ such that $ e_i \cdot \hat{v} \geq 0 $, $ \beta_i = c $. The result of Tournier (\cite{T2}) gives us the value of $\hat{v}$ in terms of the $ ( \beta_1, \dots, \beta_{2d} ) $. But there are many obvious cases when we do not need to find $\hat{v}$ to notice that $ (E') _\beta $ is not satisfied toward the asymptotic direction, for example when we cannot find $d$ of the $\beta_i$ with the same value.
\end{rmq}
In the case of RWDE, it has been proved that Kalikow's condition, and thus the $ (T') $ condition, is satisfied whenever
\begin{equation}
\label{kalikow}
\max _{1 \leq i \leq d} | \beta_i - \beta_{i+d} | > 1
\end{equation}
(see Enriquez and Sabot in \cite{ES2} and Tournier in \cite{T}). The characterization of Kalikow's condition in terms of the parameters of a RWDE remains an open question.
On the other hand, we believe that for RWDE condition $(T')$ is satisfied if and only if
$ \max _{1 \leq i \leq d} | \beta_i - \beta_{i+d} | > 0 $. Nevertheless, in this article we are able to prove the following result.
\begin{theo}
\label{theorem-dirichlet}
Let $\beta_1, \beta_2,\ldots,\beta_d,\beta_{d+2},\ldots,\beta_{2d}$ be fixed positive numbers. Then, there exists an $\epsilon\in (0,1)$ depending on these numbers such that if $\beta_{1+d}$ is chosen so that $ \beta_{1+d} \leq \epsilon $, the Random Walk in Dirichlet Environment with parameters $ ( \beta_1, \dots, \beta_{2d} ) $ satisfies condition $(P)_M|e_1$ for $ M \geq 15d + 5 $.
\end{theo}
\medskip
Theorem \ref{theorem-dirichlet} gives as a corollary new examples of RWDE which are ballistic in dimension $d=2$ since they do not correspond to ranges of the parameters satisfying condition \ref{kalikow} of Tournier \cite{T} and Sabot and Enriquez \cite{ES2}.
Indeed, by Theorem \ref{LLN-improved}, if
$$
\sum_{i=1}^{2d}\beta_i-\sup_{1\leq i\leq d}(\beta_i+\beta_{i+d})>1
$$
and one of the parameters $\{\beta_i:1\leq i\leq d\}$ is small enough, the walk is ballistic.
\begin{rmq}
In dimension $d\geq 3$, in \cite{S, Bouchet}, precise conditions on the existence of an invariant measure viewed from the particle absolutely continuous with respect to the law have been given ; this allows to characterize parameters for which there is ballisticity, but it fails to give information on the (T') condition and on the tails of renewal times. It also fails to give a CLT.
Theorem~\ref{CLT-improved} then gives us annealed CLTs for Dirichlet laws when the parameters $ ( \beta_1, \dots, \beta_{2d} ) $ satisfy $\sum_{i=1}^{2d}\beta_i-\sup_{1\leq i\leq d}(\beta_i+\beta_{i+d})>2$ along with condition (\ref{kalikow}) or the hypothesis of theorem~\ref{theorem-dirichlet}.
\end{rmq}
\begin{rmq}
For the Dirichlet laws in dimension $d=2$ with parameters $ ( \beta_1, \dots, \beta_{4} ) $ satisfying $\sum_{i=1}^{4}\beta_i-\sup_{1\leq i\leq 2}(\beta_i+\beta_{i+d})>1$, with one of the parameters $\{\beta_i:1\leq i\leq 4\}$ small enough, but for which there are no vector $v$ and constant $c$ such that for all $i$, $ \beta_i \leq c $, and for all $i$ such that $ e_i \cdot v \geq 0 $, $ \beta_i = c $, our theorem~\ref{LLN-improved} gives the ballisticity when the results of \cite{CR} would not have been enough.
For the Dirichlet laws in dimension $d\geq2$ with parameters $ ( \beta_1, \dots, \beta_{2d} ) $ satisfying $ \sum _{i=1} ^{2d} \beta_i - \sup_{1\leq i\leq d}( \beta_i + \beta_{i+d} ) > 2 $, with condition (\ref{kalikow}) or the hypothesis of theorem~\ref{theorem-dirichlet}, but for which there are no vector $v$ and constant $c$ such that for all $i$, $ \beta_i \leq c $, and for all $i$ such that $ e_i \cdot v \geq 0 $, $ \beta_i = c $, our theorem~\ref{CLT-improved} gives the annealed CLT when the results of \cite{CR} would not have been enough.
This illustrates the relevance of having removed the "toward the asymptotic direction" hypothesis in theorem~\ref{tails-improved}.
\end{rmq}
\section{First tools for the proofs}
In this section we will introduce some tools that will prove necessary for the proof of theorem \ref{tails-improved}.
\subsection{Regeneration times}
The proofs in \cite{CR} are based on finding bounds on the regeneration times. We thus begin by giving the definition and some results about the regeneration times with respect to a direction $l$.
We define $\{ \theta_n : n \geq 1 \}$ as the canonical time shift on ${\Z^d}^\N$. For $ l \in \mathbb S^{d-1} $ and $u \geq 0$, we define the times
\[ T ^l _u := \inf \{ n \geq 0: X_n \cdot l \geq u \} \]
and
\[ \tilde T ^l _u := \inf \{ n \geq 0: X_n \cdot l \leq u \} .\]
Set
\begin{equation}
\label{defa}
a > 2 \sqrt{d}
\end{equation}
and
\[ D^l := \min \{ n \geq 0 : X_n \cdot l < X_0 \cdot l \} .\]
We define
\[ S_0 := 0 , \quad M_0 := X_0 \cdot l ,\]
\[ S_1 := T^l _{M_0+a}, \quad R_1 := D^l \circ \theta_{S_1} + S_1 ,\]
\[ M_1 := \sup \{ X_n \cdot l : 0 \leq n \leq R_1 \} ,\]
and recursively for $k \geq 1$,
\[ S_{k+1} := T ^l _{M_k+a} , \quad R_{k+1} := D^l \circ \theta _{S_{k+1}} + S_{k+1} ,\]
\[ M_{k+1} := \sup \{ X_n \cdot l : 0 \leq n \leq R_{k+1} \} .\]
The first regeneration time is then defined as
\[ \tau_1 := \min \{ k \geq 1 : S_k <\infty , R_k = \infty \} .\]
We can now define recursively in $n$ the $(n + 1)$-th regeneration time $ \tau _{n+1} $ as $ \tau_1 (X_\cdot) + \tau_n \( X_{\tau_1+\cdot} - X _{\tau_1} \) $. We will occasionally write $ \tau_1^l, \tau_2^l, \ldots $ to emphasize the dependence on the chosen direction.
\begin{rmq}
The condition (\ref{defa}) on $a$ is only necessary to prove the non-degeneracy of the covariance matrix of part $(a)$ of theorem \ref{CLT-improved}.
\end{rmq}
It is a standard fact (see for example Sznitman and Zerner \cite{SZ}) to show that the sequence $ ( ( \tau_1, X _{(\tau_1+\cdot) \land \tau_2} - X _{\tau_1} ), ( \tau_2 - \tau_1, X _{(\tau_2+\cdot) \land \tau_3} - X _{\tau_2}), \ldots ) $ is independent and (except for its first term) i.i.d.. Its law is the same as the law of $\tau_1$ with respect to the conditional probability measure $ P_0 ( \cdot |D^l = \infty )$.
Those regeneration times are particularly useful to us because of the two following theorems :
\begin{theo}[Sznitman and Zerner \cite{SZ}, Zerner \cite{Z}, Sznitman \cite{Sz00}]
\label{LLN-CLT-ren}
Consider a RWRE in an elliptic i.i.d. environment. Let $l \in \mathbb S^{d-1}$ and assume that there is
a neighbourhood $V$ of $l$ such that for every $l' \in V$ the random walk is transient in the direction $l'$. Then there is a deterministic $v$ such that $P_0$-a.s.
\[ \lim _{n \to \infty} \frac{X_n}{n} = v .\]
Furthermore, the following are satisfied.
\begin{itemize}
\item[a)] If $ E_0 ( \tau_1 ) < \infty $, the walk is ballistic and $v \neq 0$.
\item[b)] If $ E_0 ( \tau_1^2 ) < \infty $,
\[ \epsilon ^{1/2} \( X_{ [\epsilon^{-1} n] } - [\epsilon^{-1} n] v \) \]
converges in law under $P_0$ to a Brownian motion with non-degenerate covariance matrix.
\end{itemize}
\end{theo}
\begin{theo}[Rassoul-Agha and Seppäläinen \cite{RAS}]
\label{CLTq-ren}
Consider a RWRE in an elliptic i.i.d. environment. Take $ l \in \mathbb S^{d-1} $ and let $\tau_1$ be the corresponding regeneration time. Assume that
\[ E_0 ( \tau_1 ^p )< \infty ,\]
for some $ p > 176 d $. Then $\P$-a.s. we have that
\[ \epsilon ^{1/2} \( X_{ [\epsilon^{-1} n] } - [\epsilon^{-1} n] v \) \]
converges in law under $P_{0,\omega}$ to a Brownian motion with non-degenerate covariance matrix.
\end{theo}
\subsection{Atypical Quenched Exit Estimate}
The proof of theorem \ref{tails-improved} is based on an atypical quenched exit estimate proved in \cite{CR}. We will also need this result, and thus recall it in this section. Let us first introduce some notations.
Without loss of generality, we can assume that $e_1$ is contained in the open half-space defined by the asymptotic direction so that
\[ \hat v\cdot e_1>0. \]
We define the hyperplane :
\[ H:=\{ x \in \R^d: x \cdot e_1 = 0 \}. \]
Let $ P := P_{\hat v} $ be the projection on the asymptotic direction along the hyperplane $H$ defined for $ z \in \Z^d $ by
\[ P(z) := \( \frac{ z \cdot e_1 }{ \hat{v} \cdot e_1 } \) \hat{v} ,\]
and $ Q := Q_l $ be the projection of $z$ on $H$ along $\hat v$ so that
\[ Q(z) : = z - P(z) .\]
Now, for $ x \in \Z^d$ , $ \beta > 0 $, $\rho>0$ and $ L > 0 $, we define the tilted boxes with
respect to the asymptotic direction $\hat v$ by :
\begin{equation}
\label{tilted-box}
B_{\beta,L}(x) :=
\left\{ y \in \Z^d \text{ s.t. } - L^{\beta} < (y-x) \cdot e_1 < L \text{ and } \| Q(y-x) \| _\infty < \rho L^{\beta} \right\}.
\end{equation}
and their front boundary by
\[ \partial ^{+} B_{\beta,L} (x) := \{ y \in \partial B_{\beta,L}(x) \text{ s.t. } (y -x) \cdot e_1 = L \} .\]
We have :
\begin{prop}[Atypical Quenched Exit Estimate, proposition 4.1 of \cite{CR}]
\label{aqee}
Set $ \alpha>0 $ such that $ \eta_\alpha := \sup_{e\in U} \E \( \( \frac{1}{\omega(0,e)} \)^\alpha \) < \infty $. Take $ M \geq 15d+5 $ such that $(P)_M|l$ is satisfied. Let $\beta_0 \in (1/2,1)$, $\beta \in \( \frac{\beta_0+1}{2} , 1 \)$ and $ \zeta \in (0,\beta_0) $.
Then, for each $\gamma > 0$ we have that
\[ \limsup_{L\to\infty} L ^{-g(\beta_0,\beta,\zeta)} \log \P \( P_{0,\omega} \( X _{ T_{B_{\beta,L}(0)} } \in \partial^{+} B_{\beta,L}(0) \) \leq e^{-\gamma L^\beta} \) < 0 ,\]
where
\[ g (\beta_0,\beta,\zeta) := \min \{ \beta + \zeta, 3\beta - 2 + (d-1) (\beta-\beta_0) \} .\]
\end{prop}
\subsection{Some results on flows}
The main tools that enables us to improve the results of \cite{CR} is the use of flows and max-flow-min-cut theorems. We need some definitions and properties that we will detail in this section. In the following we consider a finite directed graph $ G = ( V , E ) $, where $V$ is the set of vertices and $E$ is the set of edges. For all $ e \in E $, we denote by $\underline{e}$ and $\overline{e}$ the vertices that are the head and tail of the edge $e$ (the edge $e$ goes from $\underline{e}$ to $\overline{e}$).
\begin{de}
We consider a finite directed graph $ G = ( V , E ) $. A flow from a set $ A \subset V $ to a set $ Z \subset V $ is a non-negative function $ \theta : E \to \R _+ $ such that :
\begin{itemize}
\item $ \forall x \in (A \cup Z) ^c $, $ \text{ div } \theta (x) = 0 $.
\item $ \forall x \in A $, $ \text{ div } \theta (x) \geq 0 $.
\item $ \forall x \in Z $, $ \text{ div } \theta (x) \leq 0 $.
\end{itemize}
where the divergence operator is $ \text{ div } : \R ^E \to \R ^V $ such that for all $ x \in V $,
\[ \text{ div } \theta (x) = \sum _{e \in E ,\underline{e}=x} \theta(e) - \sum _{e \in E ,\overline{e}=x} \theta(e) .\]
A unit flow from $A$ to $Z$ is a flow such that $ \sum _{x \in A} \text{ div } \theta (x) = 1 $. (Then we have also $ \sum _{x \in Z} \text{ div } \theta (x) = -1 $ ).
\end{de}
We will need the following generalized version of the max-flow-min-cut theorem :
\begin{prop}[proposition 1 of \cite{S}]
\label{max-flow-min-cut}
Let $ G = (V,E) $ be a finite directed graph. Let $ ( c(e) )_{e \in E} $ be a set of non-negative reals (called capacities). Let $x_0$ be a vertex and $ ( p_x ) _{x \in V} $ be a set of non-negative reals. There exists a non-negative function $ \theta : E \to \R _+ $ such that
\begin{equation}
\label{divergence-condition}
\text{ div } \theta = \sum _{ x \in V } p_x ( \delta _{x_0} - \delta _x ) ,
\end{equation}
\begin{equation}
\label{capacities-condition}
\forall e \in E, \; \theta(e) \leq c(e) ,
\end{equation}
if and only if for all subset $ K \subset V $ containing $x_0$ we have
\begin{equation}
\label{cutset-big-enough}
c ( \partial _+ K ) \geq \sum _{ x \in K ^c } p_x ,
\end{equation}
where $ \partial _+ K = \{ e \in E, \underline{e} \in K, \overline{e} \in K^c \} $ and $ c ( \partial _+ K ) = \sum _{ e \in \partial _+ K } c(e) $. The same is true if we restrict the condition (\ref{cutset-big-enough}) to the subsets $K$ such that any $ y \in K $ can be reached from $0$ following a directed path in $K$.
\end{prop}
We will give here an idea of the proof, that explains why we call this result a generalized version of the classical max-flow-min-cut theorem. The complete proof can be found in \cite{S}.
\begin{proof}[Idea of the proof]
If $\theta$ satisfies (\ref{divergence-condition}) and (\ref{capacities-condition}) then
\[ \sum _{e, \; \underline{e} \in K, \; \overline{e} \in K^c} \theta(e) - \sum _{e, \; \overline{e} \in K, \; \underline{e} \in K^c} \theta(e)
=
\sum _{x \in K} \text{ div } \theta (x)
=
\sum_{x \in K^c} p_x .\]
It implies (\ref{cutset-big-enough}) by (\ref{capacities-condition}) and positivity of $\theta$.
The reversed implication is an easy consequence of the classical max-flow min-cut theorem on finite directed graphs (see for example \cite{LP} section 3.1). If $ ( c(e) )_{e \in E} $ satisfies (\ref{cutset-big-enough}), we consider the new graph $\tilde{G} = ( V \cup \{ \delta \}, \tilde{E} )$, where
\[ \tilde{E} = E \cup \{ (x,\delta), \; x \in V \} .\]
We define a new set of capacities $( \tilde{c} (e) ) _{e \in \tilde{E}}$ where $ c(e) = \tilde{c} (e) $ for $ e \in E $ and $ \tilde{c} ( ( x , \delta ) ) = p_x $. The strategy is to apply the max-flow min-cut theorem with capacities $\tilde{c}$ and with source $x_0$ and sink $\delta$.
It gives a flow $ \tilde{\theta} $ on $ \tilde{G} $ between $ x_0 $ and $ \delta $ with strength $ \sum _{x \in V} p_x $ and such that $ \tilde{\theta} \leq \tilde{c} $. The function $\theta$ obtained by restriction of $ \tilde{\theta} $ to $ E $ satisfies (\ref{capacities-condition}) and (\ref{divergence-condition}).
\end{proof}
For the proof of theorem \ref{tails-improved}, we will consider the oriented graph $ ( \Z ^d , E _{\Z ^d} ) $ where $ E _{\Z ^d} := \{ (x,y) \in (\Z ^d)^2 \text{ s.t. } | x - y |_1 = 1 \} $. This graph is not finite, but we will only consider flows with compact support ($ \theta (e) = 0 $ for all $e$ except in a finite subset of $E _{\Z ^d}$). We can then proceed as if the graph were finite, and use the previous definition and proposition.
\section{Proof of theorem \ref{tails-improved}}
Let $l \in \mathbb S^{d-1}$, $\beta > 0$ and $M \geq 15d+5$. Assume that $(P)_M|l$ is satisfied and that $(E')_\beta$ holds.
Let us take a rotation $\hat R$ such that $ \hat R(e_1) = \hat v $. We fix $ \beta ' \in \( \frac{5}{6}, 1 \)$, $M>0$ and for simplicity we will write $\tau _1$ instead of $\tau _1 ^{\hat{v}}$.
For $ u > 0 $, take
\[ L = L(u) := \( \frac{1}{4M\sqrt{d}} \) ^{\frac{1}{\beta '}} ( \log u) ^{\frac{1}{\beta '}} ,\]
\[ C_L := \left\{ x \in \Z ^d : \frac{-L}{2(\hat{v}\cdot e_1)} \leq x \cdot \hat{R}(e_i) \leq \frac{L}{2(\hat{v}\cdot e_1)} ,\text{ for } 0 \leq i \leq 2d \right\} \]
Following the proof of proposition 5.1 in \cite{CR}, we write
\[ P_0 (\tau_1 >u) \leq P_0 \( \tau_1 > u, T_{C_{L(u)}} \leq \tau_1 \) + \E \( F_1 ^c, P_{0,\omega} \(T_{C_{L(u)}}>u \) \) + \P(F_1) ,\]
with
\[ F_1 := \left\{ \omega \in \Omega: t_{\omega} \( C_{L(u)} \) > \frac{u}{(\log u)^{\frac{1}{\beta '}}} \right\} \]
and
\[ t_{\omega}(A) := \inf \left\{ n \geq 0: \sup_{x} P_{x,\omega} \( T_A > n \) \leq \frac{1}{2} \right\} .\]
As in \cite{CR}, the term $P_0 \( \tau_1 > u, T_{C_{L(u)}} \leq \tau_1 \)$ is bounded thanks to condition $ (P) _M | l $, and the term $\E \( F_1 ^c, P_{0,\omega} \(T_{C_{L(u)}}>u \) \)$ is bounded thanks to the strong Markov property. This part of the original proof is not modified, so we will not give more details here. It gives the existence for every $ \gamma \in ( \beta ' , 1 ) $ of a constant $c > 0$ such that :
\[ P_0 (\tau_1 >u) \leq \frac{ e ^{- c L ^\gamma (u) } }{c} + \( \frac{1}{2} \) ^{ \floor{ ( \log u) ^{\frac{1}{\beta '}} } } + \P (F_1) .\]
It only remains to show that we can find a constant $ C > 0$ such that $ \P (F_1) \leq C u ^{-\beta} $ for $u$ big enough.
For each $\omega \in \Omega$, still as in \cite{CR}, there exists $x_0 \in C_{L(u)}$ such that
\[ P _{x_0,\omega} ( \tilde{H}_{x_0} > T_{C_{L(u)}} ) \leq \frac{ 2 |C_{L(u)}| }{ t_{\omega}(C_{L(u)}) } \]
where for $y \in \Z^d$, $\tilde{H}_y = \inf \{ n \geq 1: X_n =y \}$. It gives
\[ \P (F_1) \leq \P \( \omega \in \Omega \text{ s.t. } \exists x_0 \in C_{L(u)} \text{ s.t. } P_{x_0,\omega} ( \tilde{H}_{x_0} > T _{C_{L(u)}} ) \leq \frac{2 (\log u) ^{\frac{1}{\beta '}}}{u} \abs{C_{L(u)}} \) .\]
We define for each point $x \in C_{L(u)}$ a point $y_x$, closest from $ x + 2 \frac{L^{\beta '}}{\hat{v}\cdot e_1} \hat{v}$. To bound $ \P (F_1)$, we will need paths that go from $x$ to $y_x$ with probability big enough and the atypical quenched exit estimate (proposition \ref{aqee}).
Define :
\[ N:= \frac{ \abs{\hat{v}} \log u }{ 2 M \sqrt{d} (\hat{v}\cdot e_1) } .\]
It is straightforward that
\[ N - 1 \leq | y_x - x | _1 \leq N+1 .\]
The following of the proof will be developed in three parts : first we will construct unit flows $ \theta _{i,x} $ going from $ \{ x , x + e_i \} $ to $ \{ y_x , y_x + e_i \} $, for all $x \in C_{L(u)}$. Then we will construct paths with those flows, and use the atypical quenched exit estimate to bound $\P (F_1)$ in the case that those paths are big enough. We will conclude by bounding the probability that the paths are not big enough.
\subsection{Construction of the flows $\theta _{i,x}$}
We consider the oriented graph $ ( \Z ^d , E _{\Z ^d} ) $ where $ E _{\Z ^d} := \{ (x,y) \in (\Z ^d)^2 \text{ s.t. } | x - y |_1 = 1 \} $. We want to construct unit flows $ \theta _{i,x} $ going from $ \{ x , x + e_i \} $ to $ \{ y_x , y_x + e_i \} $, for all $x \in C_{L(u)}$. But there are additional constraints, as we will need them to construct paths that have a probability big enough. The aim of this section is to prove the following proposition :
\begin{prop}
\label{existence-of-flows}
For all $x \in C_{L(u)}$, for all $ \alpha_1, \dots, \alpha_{2d} $ positive constants, there exists $2d$ unit flows $ \theta _{i,x} : E _{\Z ^d} \to \R _+ $, respectively going from $ \{ x , x + e_i \} $ to $ \{ y_x , y_x + e_i \} $, such that :
\begin{equation}
\label{bound-on-theta}
\forall e \in E _{\Z ^d} , \; \theta_{i,x}(e) \leq \frac{\alpha(e)}{\kappa_i} ,
\end{equation}
where $ \kappa _i := 2 \sum _{j=1} ^{2d} \alpha_j - (\alpha _i + \alpha _{i+d})$, and $ \alpha(e) := \alpha_i $ for $e$ of the type $(z,e_i)$.
Furthermore, we can construct $\theta_{i,x}$ with a compact support, and in a way that allows to find $\gamma$ and $ S \subset E _{\Z ^d} $, $\abs{S}$ independent of $u$, such that $\theta_{i,x}(e) \kappa_i \leq \gamma < \alpha(e) $ for all $ e \in S ^c $.
\end{prop}
We will construct the $ \theta _{i,x} $ to prove their existences. For this we need three steps. Let $ B(x,R) $ be the box of $ \Z ^d $ of center $x$ and radius $R$, and $ B_i(x,R) $ be the same box, where the vertices $x$ and $x+e_i$ are merged (and we suppress the edge between them). We note $ E _{B(x,R)} := \{ (x,y) \in E _{\Z ^d} \cap ( B(x,R) )^2 \} $ and $ E _{B_i(x,R)} := \{ (x,y) \in E _{\Z ^d} \cap ( B_i(x,R) )^2 \} $ the corresponding sets of edges. We will construct a unit flow in the graph $ ( B_i(x,R) , E _{B_i(x,R)} ) $ from $ \{ x , x + e_i \} $ to $ B_i(x,R) ^c $, a unit flow in the graph $ ( B_i(y_x,R) , E _{B_i(y_x,R)} ) $ from $ B_i(y_x,R) ^c $ to $ \{ y_x , x + e_i \} $, and then connect them. At each step, we will ensure that condition \ref{bound-on-theta} is fulfilled.
\textbf{First step} : construction of a unit flow from $ \{ x , x + e_i \} $ to $ B_i(x,R) ^c $ :
\begin{lemma}
\label{flow-in-the-ball}
Set $x \in C_{L(u)}$, and $ \alpha_1, \dots, \alpha_{2d} $ positive constants. If $ R \geq \frac{ \max _{i} \kappa_i }{ \min_j \alpha_j }$, there exists $2d$ unit flows $ \theta _{i,x} : E _{B_i(x,R)} \to \R _+ $ such that :
\[ \text{ div } \theta_{i,x} = \sum _{z \in \partial B_i(x,R)} \frac{1}{\abs{\partial B_i(x,R)}} ( \delta_x - \delta_z) \]
and \[ \forall e \in E _{B_i(x,R)}, \; \theta_{i,x}(e) \leq \frac{\alpha(e)}{\kappa_i} \]
where $ \partial B_i(x,R) = \{ z \in B_i(x,R) \text{ that has a neighbour in } B_i(x,R) ^c \} $.
\end{lemma}
The divergence condition ensures that the flow will be a unit flow, that it goes from $x$, and that it leaves $B_i(x,R)$ uniformly on the boundary of the box.
\begin{proof}
The result is a simple application of proposition \ref{max-flow-min-cut}. We fix $ x \in C_{L(u)} $ and $i$ between $1$ and $2d$. Define $ p_z = \frac{1}{\abs{\partial B_i(x,R)}} $ if $ z \in \partial B_i (x,R) $, $ p_z = 0 $ if $ z \notin \partial B_i (x,R) $.
To prove the result we only have to check that $ \forall K \subset B_i(x,R) $ containing $x$, $ \sum _{e \in \partial_+ K} \frac{\alpha(e)}{\kappa_i} \geq \sum _{z \notin K} p_z $, where $ \partial_+ K = \{ e \in E _{B_i(x,R)} \text{ s.t. } \underline{e} \in K \text{ and } \overline{e} \notin K \} $.
We have two cases to examine:
\begin{itemize}
\item If $ K \bigcap \partial B_i (x,R) = \emptyset $, $\sum _{z \notin K} p_z = 1$. We then need $ \sum _{e \in \partial_+ K} \alpha(e) \geq \kappa _i $. For $ K = \{ x \} $, $ \sum _{e \in \partial_+ K} \alpha(e) = \kappa _i $ as we merged $x$ and $x+e_i$. For bigger $K$, we consider for all $j \neq i $ the paths $ (x + n e_j) _{n \in \N} $ and for all $j \neq i + d $ the paths $ (x + e_i + n e_j) _{n \in \N} $. They intersect the boundary of $K$ in $2d+1$ different points, and the exit directions give us the corresponding $\alpha_j$, that sum to $ \kappa_i $. It gives that $ \sum _{e \in \partial_+ K} \alpha(e) \geq \kappa _i $.
\item If $ K \bigcap \partial B_i (x,R) \neq \emptyset $, $\sum _{z \notin K} p_z < 1$. As $K$ contains a path from $x$ to $ \partial B_i (x,R) $, $\sum _{e \in \partial_+ K} \frac{\alpha(e)}{\kappa_i} \geq \frac{R \min_j (\alpha_j + \alpha_{j+d})}{\kappa_i}$. It is bigger than $1$ thanks to the hypothesis on $R$. It gives the result.
\end{itemize}
\end{proof}
\textbf{Second step} : by the same way, we construct a flow $ \theta _{i,x} : E _{B_i(y_x,R)} \to \R _+ $ such that
\[ \text{ div } \theta_{i,x} = \sum _{z \in \partial B_i(y_x,R)} \frac{1}{\abs{\partial B_i(y_x,R)}} ( \delta_z - \delta_{y_x}) .\]
and \[ \forall e \in E _{B_i(y_x,R)}, \; \theta_{i,x}(e) \leq \frac{\alpha(e)}{\kappa_i} .\]
\textbf{Third step} : we will join the flows on $ E _{B_i(x,R)} $ and $ E _{B_i(y_x,R)} $ with simple paths, to get a flow on $ E _{\Z^d} $. Take $ R \geq \frac{ \max _{i} \kappa_i }{ \min_j \alpha_j }$, and make sure that $ \frac{1}{\abs{\partial B(x,R)}} < \frac{\alpha(e)}{\kappa_i} $ for all $ e \in E _{\Z^d} $ (always possible by taking $R$ big enough, $R$ depends only on the $\alpha_i$ and the dimension).
We can find $\partial B(x,R)$ simple paths $ \pi_j \subset E _{\Z^d} $ satisfying :
\begin{itemize}
\item $\forall j$, $ \pi_j $ connects a point of $\partial B(x,R)$ to a point of $\partial B(y_x,R)$.
\item $\forall j$, $ \pi_j $ stays outside of $B(x,R)$ and $B(y_x,R)$, except from the departure and arrival points.
\item If two paths intersect, they perform jumps in different direction after the intersection (no edge is used by two paths). If $ (x, e_i) $ is in a path, then $(x+e_i, -e_i )$ is not in any path.
\item The number of steps of each path is close to $N$ : there exists constants $K_1$ and $K_2$ independent of $u$ such that the length of $ \pi_j $ is smaller than $K_1 N + K_2$.
\end{itemize}
(For example we can use the paths $ \pi ^{(i,j)} $ p45 of \cite{CR}, and make them exit the ball $B(x,R)$ instead of $ \{ x , x + e_i \} $).
For all $i$, $\partial B_i(x,R) = \partial B(x,R)$ and $\partial B_i(y_x,R) = \partial B(y_x,R)$ as soon as $ R > 1 $. By construction, $ - \text{ div } \theta _{i,x} ( z_1 ) = \text{ div } \theta _{i,x} ( z_2 ) = \frac{1}{\abs{\partial B(x,R)}} $ for any $ z_1 \in \partial B_i(x,R)$ and $ z_2 \in \partial B_i(y_x,R) $. We can then join the flows of the first two steps by defining a flow $ \theta _{i,x} (e) = \frac{1}{\abs{\partial B(y_x,R)}} $ for all $e \in \pi_j$ (and $0$ on all the other edges of $ E _{\Z^d} $).
We have thus constructed a unit flow $ \theta _{i,x} $ on $ E _{\Z^d} $, from $ \{ x , x + e_i \} $ to $ \{ y_x , y_x + e_i \} $, satisfying (\ref{bound-on-theta}) ( (\ref{bound-on-theta}) is satisfied on $ E _{B_i(x,R)} $ and $ E _{B_i(y_x,R)} $ as $ R \geq \frac{ \max _{i} \kappa_i }{ \min_j \alpha_j } $ thanks to lemma \ref{flow-in-the-ball}, and outside those balls as $ \frac{1}{\abs{\partial B(x,R)}} < \frac{\alpha(e)}{\kappa_i} $ for all $ e \in E _{\Z^d} $). It concludes the proof of the first part of proposition \ref{existence-of-flows}.
As $ \theta _{i,x} (e) = 0 $ out of the finite set $ E _{B_i(x,R)} \cup E _{B_i(y_x,R)} \cup \{ e \in \pi_j , 1 \leq j \leq \partial B(x,R) \} $, the flow has a compact support. And as we made sure that $ \frac{1}{\abs{\partial B(x,R)}} < \frac{\alpha(e)}{\kappa_i} $, we can take $S = B(x,R) \cup B(y_x,R)$ and $ \gamma = \frac{\kappa_i}{\abs{\partial B(x,R)}} $ to conclude the proof.
\subsection{Bounds for $\P(F_1)$}
We apply proposition \ref{existence-of-flows} for the $ \alpha_1, \dots, \alpha_{2d} $ of the definition of $(E')_\beta$ (see (\ref{kappa}) and (\ref{finiteness})). It gives flows $ \theta _{i,x} $ on $ E _{\Z^d} $, constructed as in the previous section.
We can decompose a given $ \theta _{i,x} $ (for $i$ and $x$ fixed) in a finite set of weighted paths, each path starting from $x$ or $x+e_i$ and arriving to $y_x$ or $y_x + e_i$. It suffices to choose a path $\sigma$ where the flow is always positive, to give it a weight $ p_\sigma := \min _{e \in \sigma} \theta _{i,x} (e) > 0 $ and to iterate with the new flow $ \theta (e) := \theta _{i,x} (e) - p _\sigma \1 _{ e \in \sigma} $.
The weigh $p_\sigma$ of a path $\sigma$ then satisfies : for all $e \in E _{\Z^d} $, $ \theta _{i,x} (e) = \sum _{\sigma \text{ containing } e} p_\sigma $. As $\theta _{i,x}$ is a unit flow we get $ \sum _{\sigma \text{ path of } \theta_{i,x}} p_\sigma = 1 $. We will use those weights in the next section, to prove that those paths are "big enough" with high probability.
We now introduce :
\[ F_{2,i} = \left\{ \omega \in \Omega \text{ s.t. } \forall x \in C_{L(u)}, \forall \sigma \text{ path of } \theta_{i,x}, \omega _{\sigma} := \prod _{e \in \sigma} \omega _e \geq u ^{\frac{1}{M} - 1} \right\} \]
and
\[ F_{2} = \bigcap _{i=1} ^{2d} F_{2,i} .\]
Define
\[ F_3 := \left\{ \omega \in \Omega \text{ s.t. } \exists x_0 \in C_{L(u)} \text{ s.t. } P_{x_0,\omega} ( \tilde{H}_{x_0} > T _{C_{L(u)}} ) \leq \frac{2 (\log u) ^{\frac{1}{\beta '}}}{u} \abs{C_{L(u)}} , F_2 \right\} .\]
We get immediately :
\[ \P (F_1) \leq \P(F_3) + \P(F_2 ^c) .\]
It gives two new terms to bound. We start by bounding $\P(F_3)$. For this we will use the same method as in \cite{CR} : on the event $F_3$, for all $ 1 \leq i \leq 2d $ we can use a path $\sigma$ of $ \theta _{i,x} $ to join $x$ or $x+e_i$ to $y_x$ or $y_x + e_i$. It gives :
\[ \omega ( x_0, e_i ) u ^{ \frac{1}{M} - 1 } \inf _{z \in \{ y_{x_0} , y_{x_0} + e_i \} } P_{z,\omega} ( T _{C_{L(u)}} < H_{x_0} ) \leq P_{x_0,\omega} ( T _{C_{L(u)}} < \tilde{H}_{x_0} ) \leq \frac{2 (\log u) ^{\frac{1}{\beta '}}}{u} \abs{C_{L(u)}} ,\]
where the factor $\omega ( x_0, e_i )$ corresponds to the probability of jumping from $x$ to $x+e_i$, in the case where the path $\sigma$ starts from $x+e_i$.
As $ \sum _{i=1} ^{2d} \omega ( x_0, e_i ) = 1 $, it gives
\[ u ^{ \frac{1}{M} - 1 } \inf _{z \in V( y_{x_0} ) } P_{z,\omega} ( T _{C_{L(u)}} < H_{x_0} ) \leq \frac{4d (\log u) ^{\frac{1}{\beta '}}}{u} \abs{C_{L(u)}} ,\]
where $ V( y_{x_0} ) := \{ y_{x_0} , (y_{x_0} + e_i) _{i=1, \dots, 2d} \} $.
In particular, on $F_3$, we can see that for $u$ large enough $ V( y_{x_0} ) \subset C_{L(u)}$. As a result, on $F_3$, we have for $u$ large enough
\[ \inf _{z \in V( y_{x_0} ) } P_{z,\omega} ( X _{T_{z} + U _{\beta ',L}} \cdot e_1 > z \cdot e_1 )
\leq \inf _{z \in V( y_{x_0} ) } P_{z,\omega} ( T _{C_{L(u)}} < H_{x_0} )
\leq \frac{1}{ u ^\frac{1}{2M} } = e ^{ -2 \sqrt d L(u) ^{\beta '} } ,\]
where
\[ U _{\beta ',L} := \{ x \in \Z^d : -L ^{\beta '} < x \cdot e_1 < L \} .\]
From this and using the translation invariance of the measure $\P$, we conclude that :
\begin{align*}
& \P \( \exists x_0 \in C_{L(u)} \text{ s.t. } P_{x_0,\omega} ( \tilde{H}_{x_0} > T _{C_{L(u)}} ) \leq \frac{4d (\log u) ^{\frac{1}{\beta '}}}{u} \abs{C_{L(u)}} , F_2 \) &\\
& \leq \P \( \exists x_0 \in C_{L(u)} \text{ s.t. } \inf _{z \in V( y_{x_0} ) } P_{z,\omega} ( X _{T_{z} + U _{\beta ',L}} \cdot e_1 > z \cdot e_1 ) \leq e ^{ -2 \sqrt d L(u) ^{\beta '} } \) &\\
& \leq (2d+1) \abs{C_{L(u)}} \P \( P_{0,\omega} ( X _{ T _{U _{\beta ' , L(u)}} } \cdot e_1 > 0 ) \leq e ^{ -2 \sqrt d L(u) ^{\beta '} } \) &\\
& \leq (2d+1) \abs{C_{L(u)}} \P \( P_{0,\omega} ( X _{ T _{B _{\beta ' , L(u)}} } \cdot e_1 > 0 ) \leq e ^{ -2 \sqrt d L(u) ^{\beta '} } \) ,&
\end{align*}
where the titled box $ B _{\beta ' , L(u)} $ is defined as in (\ref{tilted-box}).
We conclude with the atypical quenched exit estimate (proposition \ref{aqee}) : there exists a constant $c>0$ such that for each $ \beta_0 \in (\frac{1}{2},1)$ one has :
\[ \P(F_3) \leq \frac{1}{c} e ^{-c L(u) ^{g(\beta_0, \beta ', \zeta)}} ,\]
where $g(\beta_0,\beta ',\zeta)$ is defined as in proposition \ref{aqee}.
Note that for each $ \beta ' \in \( \frac{5}{6},1 \)$ there exists a $ \beta_0 \in \( \frac{1}{2},\beta \) $ such that for every $ \zeta \in \( 0,\frac{1}{2} \)$ one has $ g ( \beta_0, \beta ', \zeta ) > \beta ' $. Therefore, replacing $L$ by its value, we proved that there exists $c > 0$ such that :
\[ \P(F_3) \leq c u^{-\beta} .\]
\subsection{Bound for $\P(F_2 ^c)$}
To conclude the bound for $ \P (F_1) $ and the proof of theorem \ref{tails-improved}, it only remains to control $\P(F_2 ^c)$. It is in this section that we will use the conditions that were imposed on $ \theta _{i,x} $ during the construction of the flows.
\begin{align*}
\P(F_2 ^c)
& \leq \sum _{i=1} ^{2d} \P(F_{2,i}^c) &\\
& \leq \sum _{i=1} ^{2d} \sum_{x \in C_{L(u)}} \P( \forall \sigma \text{ path of } \theta_{i,x}, \omega _{\sigma} \leq u ^{\frac{1}{M} - 1} ) &
\end{align*}
As $\theta_{i,x}$ is a unit flow, if $\forall \sigma \text{ path of } \theta_{i,x}$, $\omega _{\sigma} \leq u ^{\frac{1}{M} - 1}$ then :
\[ \sum _{\sigma \text{ path of } \theta_{i,x}} p_\sigma \omega _{\sigma} \leq u ^{\frac{1}{M} - 1} \sum_{\sigma \text{ path of } \theta_{i,x}} p_\sigma = u ^{\frac{1}{M} - 1} .\]
Jensen's inequality then gives :
\[ \prod _{\sigma \text{ path of } \theta_{i,x}} \omega _{\sigma} ^{p_\sigma} = \prod _{e \in E _{\Z^d} } \omega_e ^{\theta_{i,x}(e)} \leq u ^{\frac{1}{M} - 1} .\]
It allows to write :
\begin{align*}
\P(F_2 ^c)
& \leq \sum _{i=1} ^{2d} \sum_{x \in C_{L(u)}} \P \( \prod _{e \in E _{\Z^d} } \omega_e ^{\theta_{i,x}(e)} \leq u ^{\frac{1}{M} - 1} \) &\\
&\leq \sum _{i=1} ^{2d} \sum_{x \in C_{L(u)}} \frac{ \E \( \prod _{e \in E _{\Z^d} } \omega_e ^{-\kappa_i \theta_{i,x}(e)} \) }{ u ^{-\kappa_i (\frac{1}{M} - 1)} } &
\end{align*}
We will use the integrability given by the flows to bound the expectations. The independence of the environment gives (for $i$ and $x$ fixed) :
\begin{align*}
\E \( \prod _e \omega_e ^{-\kappa_i \theta_{i,x}(e)} \)
& = \prod_{z \in \Z^d} \E \( \prod_{e \text{ s.t. } \underline{e} = z} \omega_e ^{-\kappa_i \theta_{i,x}(e)} \) &\\
& = \prod_{z \in S} \E \( \prod_{e \text{ s.t. } \underline{e} = z} \omega_e ^{-\kappa_i \theta_{i,x}(e)} \) \prod_{z \notin S} \E \( \prod_{e \text{ s.t. } \underline{e} = z} \omega_e ^{-\kappa_i \theta_{i,x}(e)} \) &
\end{align*}
where we recall that $S = B(x,R) \cup B(y_x,R)$.
As $ \theta_{i,x} $ satisfies (\ref{bound-on-theta}), the ellipticity condition $ (E') _\beta $ gives that each of the expectations $\E \( \prod_{e \text{ s.t. } \underline{e} = z} \omega_e ^{-\kappa_i \theta_{i,x}(e)} \)$ are finite.
By construction $ \abs{S} $ is finite and does not depend on $u$ : $\prod_{z \in S} \E \( \prod_{e \text{ s.t. } \underline{e} = z} \omega_e ^{-\kappa_i \theta_{i,x}(e)} \)$ is a finite constant independent on $u$.
It remains to deal with the case of $z \notin S$. As we chose $R$ to get $ \theta_{i,x}(e) \kappa_i < \gamma $ for the edges outside $S$, and thanks to the bounds on the number of edges with positive flow (there is a finite number of paths, and each path has a bounded length), we have :
\[ \prod_{z \notin S} \E \( \prod_{e \text{ s.t. } \underline{e} = z} \omega_e ^{-\kappa_i \theta_{i,x}(e)} \) \leq \E \( \prod_{e \text{ s.t. } \underline{e} = 0} \omega_e ^{- \gamma} \) ^{c_1 N + c_2} ,\]
where $c_1$ and $c_2$ are positive constants, independent of $u$. Then, putting all of those bounds together,
\begin{align*}
\P(F_2 ^c)
& \leq \sum _{i=1} ^{2d} \sum_{x \in C_{L(u)}} C_1 C_2 ^{C_3 N} u ^{\kappa_i (\frac{1}{M} - 1)} &\\
&\leq \sum _{i=1} ^{2d} \sum_{x \in C_{L(u)}} C_4 u ^{ \frac{C_5 + \kappa_i}{M} - \kappa_i} &\\
&\leq C_6 (\log u)^{C_7} u ^{\frac{C_8}{M} - \min_i \kappa_i} &
\end{align*}
where all the constants $C_i$ are positive and do not depend on $u$. As remark \ref{P_M} tells us that we can choose $M$ as large as we want, we can get $\frac{C_8}{M}$ as small as we want.
Then we can find a constant $C > 0$ such that $\P(F_2 ^c) \leq C u ^{-\beta} $ for $u$ big enough. It concludes the proof.
\section{New examples of random walks satisfying the polynomial condition}
\label{last-section1}
\subsection{Proof of Theorem \ref{marginal}}
Consider the box $B_{e_1,L,\tilde L}$ for
$ \tilde{L} = 70 L^3 $. We want to find some
$L > c_0 $ such that
$$
P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \)
\leq \frac{1}{L^M},
$$
for some $M\geq 15d+5$.
We first decompose this probability according to whether the exit point of the random walk from the box $B_{e_1,L,\tilde L}$ is on the
bottom or on one of the sides of the box, so that,
\begin{align*}
& P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \) &\\
& = P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 = - L \) + P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \) + P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = - \tilde{L} \) .&\\
\end{align*}
We will first bound the probability to exit through the sides. We do the computations for $ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \) $ but the other term can be dealt with in the same way. Suppose that $ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} $, and define $ n_0, \dots, n_{ \tilde{L} - 1 } $ the finite hitting times of new levels in direction $e_2$ as follows :
\[ n _k := \inf \{ n \geq 0 \text{ s.t. } X_n \cdot e_2 \geq k \} .\]
To simplify notation define $\phi(x):=\omega(x,e_{1+d})$.
We now choose a constant $ 1 > \delta > 0 $, and we will call "good point"
any $ x \in \Z ^2 $ such that $ \phi ( x ) > \delta $. We define $p := \P ( \phi ( x ) > \delta )$. Note that
$p$ does not depend on $x$ since the
environment is i.i.d., so that it depends only on $ \delta $ and the law of $ \phi $.
We now introduce the event that a great number of the $X _{n _k}$ are good points :
\[ C_1 := \left\{ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \text{ and at least } \frac{p}{2} \tilde{L} \text{ of the } X _{n _k} , 1 \leq k \leq \tilde{L} - 1, \text{ are good } \right\} .\]
We get immediately
\[ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \) = P _0 ( C _1 ) + P _0 \( \left\{ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \right\} \cap (C_1 ^c) \) .\]
By construction of the $ X _{n _k} $ and independence of the environment, and with $Z$ an independent random variable following a binomial law of parameters $ p $ and $ \tilde{L} $, we can bound the second term of the sum :
\begin{align*}
P _0 \( \left\{ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \right\} \cap (C_1 ^c) \)
& \leq P \( Z \leq \frac{p}{2} \tilde{L} \) & \\
& \leq \exp \( - 2 \frac{ ( p \tilde{L} - p \tilde{L} / 2 ) ^2 }{ \tilde{L} } \) & \\
& = \exp \( - \frac{ p ^2 \tilde{L} }{ 2 } \) &
\end{align*}
where the last inequality is Hoeffding's inequality.
It only remains to bound $ P _0 ( C_1 ) $. For that, we introduce the following new event
\[ C_2 := \left\{ C_1 \text{ and } X _{n _k + 1} - X _{n _k} = e_1 \text{ for at least } \frac{\delta p}{4} \tilde{L} \text{ of the good } X _{n _k} \right\} ,\]
that states that the walk goes often in direction $e_1$ just after reaching a $X _{n _k}$ that is a good point.
We can then write
\[ P _0 ( C_1 ) = P _0 ( C_2 ) + P _0 ( C_1 \cap (C_2 ^c) ) .\]
To bound $ P _0 ( C_1 \cap (C_2 ^c) ) $, we use the uniform bound
"$ \phi ( x ) > \delta$" for good points that gives us that
$ \omega ( x , e_1 ) > r \delta$ on those points. And we get $Z'$ an
independent random variable following a binomial law of parameters $ r \delta $ and $ \frac{p}{2} \tilde{L} $ :
\begin{align*}
P _0 ( C_1 \cap (C_2 ^c) )
& \leq P \( Z' \leq \frac{\delta p}{4} \tilde{L} \) & \\
& \leq \exp \( - p \delta ^2 \tilde{L} \( r - \frac{1}{2} \)^2 \) . &
\end{align*}
It only remains to bound $ P _0 ( C_2 ) $. Set $n^+$ (respectively $n^-$) the total number of jumps in direction $e_1$ (respectively $-e_1$) before exiting the box $B_{e_1,L,\tilde L}$. We will need a third new event
\[ C_3 := \left\{ n^+ \geq \frac{1+r}{r} n^- \right\} ,\]
that allows us to write
\[ P _0 ( C_2 ) = P _0 ( C_2 \cap C_3 ) + P _0 ( C_2 \cap (C_3 ^c)) .\]
First notice that for $L$ big enough, $ C_2 \cap C_3 = \emptyset $. Indeed,
$C_1$ implies that we exit the box $B_{e_1,L,\tilde L}$ by the side "$ x \cdot
e_2 = \tilde{L} $". Now, since the vertical displacement
of the walk before exiting the box $B_{e_1,L,\tilde L}$ is $n^+-n^-$,
on the event $C_3$ we know that this displacement is at
least equal to $\frac{1}{1+r}n^+$. Therefore, since
on $C_2$ the walk makes at least
$ \frac{\delta
p}{4} \tilde{L} = \frac{35 \delta p}{2} L ^3 $ moves in the direction $ e_1$,
on $C_2\cap C_3$ its vertical displacement before exiting the box
is at least $\frac{35\delta p}{2(1+r)} L^3$. Since on $C_2\cap C_3$ the
walk exits the box by the "$ x \cdot e_2 = \tilde{L} $" side
we see that for $L$ larger than $L_1 := \sqrt{ \frac{2(1+r)}{35 \delta p} }$ the event $C_2\cap C_3$ is empty.
We now want to bound $ P _0 ( C_2 \cap (C_3 ^c)) $.
\begin{align*}
P _0 ( C_2 \cap (C_3 ^c))
&\leq P _0 \( n^+ \geq \frac{\delta p}{4} \tilde{L} \text{ and } n^+ <
\frac{1+r}{r} n^- \) &\\
&\leq P _0 \( n^+ + n^- \geq \frac{\delta p}{4} \tilde{L} \text{ and } (n^+ +
n^-) \frac{r}{1+2r} < n^- \) .&
\end{align*}
Now note that whenever we go through a vertical
edge from a point $x$, the law of the environment tells us that it is an edge
$ (x,e_1) $ with probability $\frac{r}{1+r} $, and $ (x,-e_1) $ with probability $\frac{1}{1+r} $. Then, defining $Z''$ as a random variable following
a binomial law of parameters $ \frac{1}{1+r} $ and $ \frac{p \delta}{4} \tilde{L} $, we have the bound :
\begin{align*}
P _0 ( C_2 \cap (C_3 ^c))
&\leq P \( Z'' \geq \frac{r}{1+2r} \frac{p \delta}{4} \tilde{L} \) &\\
&\leq \exp \( - 4 p \delta \tilde{L} \( \frac{r}{r+1} + \frac{r}{1+2r} - \frac{1}{4 p \delta \tilde{L}} \) ^2 \) ,&
\end{align*}
where we need $ 1 \leq \frac{ r p \delta \tilde{L} }{4} \( \frac{1}{1+r} + \frac{1}{1+2r} \) $ to apply Hoeffding's inequality in the last inequality. We can find $ L_2 $ such that this is true for $ L \geq L_2 $.
Choose $ M \geq 15d+5$. By putting all of our previous bounds together, we finally get, for all $ L \geq L_2 $,
\begin{align*}
& P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \) &\\
& \leq \exp \( - \frac{ p ^2 \tilde{L} }{ 2 } \) + \exp \( - p \delta ^2 \tilde{L} \( r - \frac{1}{2} \)^2 \) + \exp \( - 4 p \delta \tilde{L} \( \frac{r}{r+1} + \frac{r}{1+2r} - \frac{1}{4 p \delta \tilde{L}} \) ^2 \) ,&
\end{align*}
where we recall that $ \tilde{L} = 70 L^3 $, $ \delta > 0 $ and $ p = \P (
\phi(x) > \delta ) $. Then, for any choice of $ \delta $, we can find $ L_3 \geq \max ( c_0, L_1 , L_2 ) $ such that for all $ L \geq L_3 $,
\[ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = \tilde{L} \) + P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_2 = -\tilde{L} \) \leq \frac{1}{2 L ^M} .\]
We now only need to bound $ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 =
-L \)$ to prove $ (P)_M | e_1 $. We will use again the notations $n^+$
(respectively $n^-$) for the total number of jumps in direction $e_1$
(respectively $-e_1$) before exiting the box $B_{e_1,L,\tilde L}$. Suppose
that $ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 = -L $. Then necessarily $ n^+
< n^- $, which gives $ n^+ < \frac{ n^+ + n^- }{2} $. As $n^+$ conditioned to
$n^++n^-$ follows a binomial law of parameters $ \frac{r}{1+r} $ and $ n^+ + n^- $, Hoeffding's inequality gives the bound :
\[ P_0 \( n^+ < \frac{ n^+ + n^- }{2}
\left| n^++n^- \right. \) \leq \exp \( - 2 ( n^+ + n^- ) \( \frac{r}{1+r} - \frac{1}{2} \) ^2 \) .\]
But $ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 = -L $ also gives that necessarily, $ n^- \geq L $. Then
\begin{align*}
P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 = -L \)
& \leq P_0 \( n^+ < \frac{ n^+ + n^- }{2} \text { and } n^- \geq L \) &\\
& \leq \sum_{m=L}^\infty \exp \( - 2 m \( \frac{r}{1+r} - \frac{1}{2} \) ^2 \) .&
\end{align*}
Therefore we can find $ L_4 \geq L_3 $ such that for all $ L \geq L_4 $,
\[ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 = -L \) \leq \frac{1}{2 L ^M} ,\]
from where we conclude that for all $ L \geq L_4 $,
\[ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \) \leq \frac{1}{L ^M}. \]
\subsection{Proof of Theorem \ref{theorem-dirichlet}}
It is classical to represent Dirichlet distributions with independent gamma
random variables : if $ \gamma_1, \dots, \gamma _N $ are independent gamma
random variables with parameters $ \beta_1, \dots, \beta_N $, then $
\frac{\gamma_1}{\sum \gamma_i}, \dots, \frac{\gamma_N}{\sum \gamma_i} $ is a
Dirichlet random variable with parameters $ ( \beta_1, \dots, \beta_N )
$. We get a restriction property as an easy consequence of this representation (see \cite{W}, pages 179-182) : for $J$ a non-empty subset of $ \{ 1, \dots, N \} $, the random variable $ \( \frac{x_j}{\sum _{i \in J} x_i} \) _{j \in J} $ follows a Dirichlet law with parameters $ ( \beta _j ) _{j \in J} $ and is independent of $ \sum _{i \in J} x_i $. This property will be useful in the following.
We consider the box $B_{e_1,L,\tilde L}$ for $ \tilde{L} = 70 L^3 $, and want to find some $ L > c_0 $ such that $ P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \) \leq \frac{1}{L^M} $ to prove $(P)_M | e_1$.
Let $ l_i := \{ x \in \Z ^d \text{ s.t. } x \cdot e_1 = i \} $ and $t_i := \inf \{ n \geq 0 : X_n \in l_i, X _{n+1} \notin l_i \}$. We first consider the events that, when the walk arrives on $l_i$ for the first time, it gets out of it by an edge in direction $e_1$ (the alternative being getting out by an edge in direction $-e_1$) :
\[ G _{1,i} := \{ X _{ t_i +1 } - X _{ t_i } = e_1 \} .\]
At the point $X _{ t_i }$, we know that the walk will go either to $X _{ t_i } + e_1$ or to $X _{ t_i } - e_1$. Thanks to the restriction property of the Dirichlet laws, we know that $ \frac { \omega ( X _{ t_i } , e_1) }{ \omega ( X _{ t_i } , e_1) + \omega ( X _{ t_i } , - e_1) } $ follows a beta law of parameters $ ( \beta _1, \beta _1 + \beta _{1+d} ) $ and is independent of the previous trajectory of the walk on $l_i$. Then
\[ P _0 ( G _{1,i} ) = \frac{ \beta _1 }{ \beta _1 + \beta _{1+d} } .\]
Now define
\[ G _1 := \bigcap _{i=0} ^{ L-1 } G _{1,i} ,\]
and note that
\[ P _0 ( G _1^c ) \leq L\frac{\beta_{1+d}}{\beta_1+\beta_{1+d}}.
\]
We can now write
\begin{align*}
P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \)
& \leq P _0 \( \left\{ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \right\} \cap G _1 \) + P _0 ( G _1 ^c ) & \\
& \leq P _0 \( \left\{ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \right\}
\cap G _1 \) + L\frac{\beta_{1+d}}{\beta_1+\beta_{1+d}},&
\end{align*}
and we only need to bound the first term of this sum. If $G _1$ is satisfied, the walk cannot get out of the box $B_{e_1,L,\tilde L}$ by the "lower boundary" $ \{ x \in \Z^d \text{ s.t. } x \cdot e_1 = -L \} $. Then the walk has to get out by one of the $2d-2$ "side boundaries" :
\begin{equation*}
P _0 \( \left\{ X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \right\} \cap G _1 \)
= P _0 \left( \cup_{j=2}^d\left\{ X_{ T_{ B_{e_1,L,\tilde L} } }
\cdot e_j = \pm\tilde{L} \right\} \cap G _1 \right).
\end{equation*}
On the event $ \cup_{j=1}^d\{X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_j =
\pm\tilde{L}\}$ define $ n_0, \dots, n_{ \tilde{L} - 1 }$ as the finite
hitting times of new levels in any direction perpendicular to $e_1$ as follows :
\[ n _k := \inf \{ n \geq 0 \text{ s.t. } \sup_{2\leq j\leq d}|
X_n \cdot e_j| \geq k \} .\]
Let now $p = \frac{ \beta_1 }{1 + \sum_{i \neq 1+d} \beta_i}$ and
consider the event
\[ G _{3} := \left\{ G_1 \text{ and } X _{n _k + 1} - X _{n _k } = e_1 \text{ for at least } \frac{ p}{2} \tilde{L} \text{ of the points } X _{n _k} \right\} .\]
Suppose $ \beta _{1+d} \leq 1 $, then $ p \leq \E ( \omega (0,e_1) ) $.
Consider now a random variable $Z$ with a binomial law of
parameters $p$ and $\tilde L$. Using Hoeffding's inequality, we see that
$$
P(G_3^c)\leq P\left(Z\le\frac{p\tilde L}{2}\right) \le\exp{\left(
-\frac{p^2}{2}\tilde L\right)}.
$$
But clearly $G_1\cap G_3=\emptyset$ for $ L \geq L_0 := \sqrt{\frac{1}{35p}}$. Therefore we have in this case
$$
P _0 \left( \cup_{j=2}^d\left\{ X_{ T_{ B_{e_1,L,\tilde L} } }
\cdot e_j = \pm\tilde{L} \right\} \cap G _1 \right)
\leq \exp{\left(
-\frac{p^2}{2}\tilde L\right)}.
$$
Putting the previous bounds together, we finally get for all $ L \geq L_0 $ :
\begin{equation*}
P _0 \( X_{ T_{ B_{e_1,L,\tilde L} } } \cdot e_1 < L \)\leq L\frac{\beta_{1+d}}{\beta_1+\beta_{1+d}}+\exp{\left(
-\frac{p^2}{2}\tilde L\right)}.
\end{equation*}
Let now $L_1$ be such that for all $L \geq L_1$
$$
\exp \left(
-\frac{p^2}{2}\tilde L\right) \leq \frac{1}{2 L^M}.
$$
Take now $L_2:=\max\{c_0,L_0,L_1\}$
and then choose $\beta_{1+d}$ ($\leq 1$) so that
$$
L_2\frac{\beta_{1+d}}{\beta_1 + \beta_{1+d}}\le\frac{1}{2 L_2^M}.
$$
We then conclude that for this choice of $\beta_{1+d}$
there exists an $L\geq c_0$ such that
$$
P _0 \( X_{ T_{ B_{e_1,L_2,\tilde L} } } \cdot e_1 < L \) \leq
\frac{1}{L ^M}.
$$
{\bf Acknowledgements.} The authors would like
to thank Alexander Drewitz for pointing out the ideas
of the proof of Theorem \ref{marginal}.
| 172,809
|
The flip, "Touch the Toes," is a bit more surprising. It's still loop-based, but has enough to keep you interested in its first few seconds for the nigh-on ten minutes of its length. More congas: but more eccentrically accented, bobbling and syncopated, and a slinky "come with me" vocal swinging smoothly around them. A very funky mix indeed, and the way they're toned together to give a lush Amazonian vibe conjures images of firelit ceremonial dances in jungle clearings.
- Published /
Thu / 23 Dec 2010
- Words /
Daniel Petry
-
- Tracklist /
A Almost There
B Touch the Toes
| 43,419
|
Mulpuna
1995
- Artist/s name
- Pinta Pinta Tjapanangka
Pintupi
- Medium
- synthetic polymer paint on canvas
- Measurements
- 28.2 x 107.0 cm
- Place/s of Execution
- Kintore, Northern Territory
- Accession Number
- 2005.266
- Credit Line
- National Gallery of Victoria, Melbourne
Presented through the NGV Foundation by anonymous donors, 2005
- Gallery Location
- Not on display
| 353,168
|
Our Sponsors
156 / 267
Vet Day Lafayette Park Ele
Students and veterans participate in the Veterans Day celebration at Lafayette Park Elementary on November 10, 2017. After the school program each classroom prepared a program for their invited veterans. These kids sing a song thanking veterans.
Tim Bath | Kokomo Tribune
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| 201,935
|
TITLE: $\mathbb{P}^1_{k'}$ is not smooth over $k$ if $k'/k$ finit inseparable
QUESTION [1 upvotes]: Let $k'$ a finite inseparable extension of a field $k$, with $k$ not perfect. I want to prove that $\mathbb{P}^1_{k'}$ is not smooth over $k$.
Because all is local I guess this is equivalent with $\mathbb{A}^1_{k'}$ is not smooth over $k$ that is $\operatorname{Spec}(k'[x]\otimes_k\overline{k})$ is not regular, but I don't see how to simplify this tensor product.
REPLY [3 votes]: Say $k = \Bbb F_p(t^p)$ and $k' = \Bbb F_p(t)$.
The first thing to do is to find out what exactly is $k' \otimes_k k'$. Since $k'$ is a $k$-vector space of dimension $p$, this is going to be a $k$-vector space of dimension $p^2$, or also a $k'$-vector space of dimension $p$.
If you consider $k' \otimes_k k'$ as a $k'$-vector space with multiplication on the right, a basis is the family $1 \otimes 1, t \otimes 1, t^2 \otimes 1, \ldots, t^{p-1} \otimes 1$.
Let $\varepsilon = 1 \otimes 1 - t \otimes t^{-1}$. It should be easy to check that with respect to this basis the matrix of the family $1 \otimes 1, \varepsilon, \varepsilon^2 \ldots, \varepsilon^{p-1}$ is a triangular matrix with nonzero entries on the diagonal, so that it also is a $k'$-basis for $k' \otimes_k k'$.
Finally, $\varepsilon^p = 1 \otimes 1 - (t^p \otimes t^{-p}) = 1 \otimes 1 - 1 \otimes 1 = 0$. We conclude that $k' \otimes_k k'$ is isomorphic to $k'[\varepsilon]/(\varepsilon^p)$. And yes, everyone has quite a lot of $p$-th roots in this ring.
Similarly, our ring of interest $A = k'[x] \otimes_k \bar k \simeq \bar k[x,\varepsilon]/(\varepsilon^p)$.
The spectrum of this is the same as the regular $\operatorname{Spec}(\bar{k}[x])$, but where we add $\varepsilon$ to every prime ideal : Since $\varepsilon^p = 0$, $\varepsilon$ is in the radical of $A$, so is included in any prime ideal. Prime ideals of $A$ then correspond to prime ideals of $A/(\varepsilon) = \bar k[x]$.
The localization at the prime $(x,\varepsilon)$ is a local ring whose maximal ideal $\frak m$ is generated by $x$ and $\varepsilon$. It is not regular because it is generated by $2$ elements while the dimension of the ring is $1$.
if $k \subset k'$ is Galois, then $k' \otimes_k k' = k'^{[k':k]}$ (this is actually a characterisation of Galois-ness), and the spectrum of $k' \otimes_k \bar k$ is $[k':k]$ separated geometric points. From there you can hopefully check the smoothness without problem.
| 212,704
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:: On a Dividing Function of the Simple Closed Curve into Segments
:: by Yatsuka Nakamura
environ
vocabularies NUMBERS, PRE_TOPC, EUCLID, ARYTM_1, SUBSET_1, XXREAL_0, ARYTM_3,
RCOMP_1, XBOOLE_0, TOPREAL2, PSCOMP_1, JORDAN6, TOPREAL1, JORDAN3,
TARSKI, FUNCT_1, BORSUK_1, RELAT_1, TOPS_2, ORDINAL2, STRUCT_0, FINSEQ_1,
CARD_1, XXREAL_1, REAL_1, PARTFUN1, MEASURE5, GOBRD10, ORDINAL4, NAT_1;
notations TARSKI, XBOOLE_0, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0, XREAL_0,
REAL_1, NAT_1, RCOMP_1, NAT_D, FINSEQ_1, FUNCT_1, RELSET_1, PARTFUN1,
VALUED_0, TOPREAL1, TOPREAL2, TBSP_1, GOBRD10, PRE_TOPC, TOPS_2,
FINSEQ_6, STRUCT_0, COMPTS_1, EUCLID, PSCOMP_1, JORDAN5C, JORDAN6,
TOPMETR, XXREAL_0;
constructors REAL_1, RCOMP_1, NAT_D, TOPS_2, COMPTS_1, TBSP_1, TOPMETR,
TOPREAL1, PSCOMP_1, GOBRD10, JORDAN5C, JORDAN6, XXREAL_2, FINSEQ_6;
registrations RELSET_1, XXREAL_0, XREAL_0, MEMBERED, STRUCT_0, PRE_TOPC,
EUCLID, TOPREAL1, BORSUK_2, VALUED_0, COMPTS_1, XXREAL_2, FINSEQ_1,
FUNCT_1, ORDINAL1;
requirements REAL, SUBSET, BOOLE, NUMERALS, ARITHM;
definitions TARSKI, XBOOLE_0, FUNCT_1;
equalities XBOOLE_0, STRUCT_0;
expansions TARSKI, XBOOLE_0, FUNCT_1;
theorems TARSKI, JORDAN6, TOPREAL1, TOPREAL5, JORDAN5C, UNIFORM1, PRE_TOPC,
FUNCT_2, TOPS_2, RELAT_1, FUNCT_1, FINSEQ_1, FINSEQ_2, NAT_1, FINSEQ_4,
GOBRD10, SPRECT_1, XBOOLE_0, XBOOLE_1, FINSEQ_3, BORSUK_1, COMPTS_1,
RELSET_1, JORDAN5B, PARTFUN2, XCMPLX_1, XREAL_1, XXREAL_0, PARTFUN1,
XXREAL_1, XXREAL_2, SEQM_3, NAT_D, XREAL_0, FINSEQ_6;
schemes DOMAIN_1, NAT_1;
begin :: Definition of the Segment and its property
reserve p,p1,p2,p3,q for Point of TOP-REAL 2;
Lm1: 2-'1 = 2-1 by XREAL_1:233
.=1;
Lm2: for i, j, k be Nat holds i-'k <= j implies i <= j + k
proof
let i, j, k be Nat;
assume
A1: i-'k <= j;
per cases;
suppose
A2: i >= k;
i-'k +k <= j + k by A1,XREAL_1:6;
hence thesis by A2,XREAL_1:235;
end;
suppose
A3: i <= k;
k <= j + k by NAT_1:11;
hence thesis by A3,XXREAL_0:2;
end;
end;
Lm3: for i, j, k be Nat holds j + k <= i implies k <= i -' j
proof
let i, j, k be Nat;
assume
A1: j + k <= i;
per cases by A1,XXREAL_0:1;
suppose
j + k = i;
hence thesis by NAT_D:34;
end;
suppose
j + k < i;
hence thesis by Lm2;
end;
end;
theorem Th1:
for P being compact non empty Subset of TOP-REAL 2 st P is
being_simple_closed_curve holds W-min(P) in Lower_Arc(P) & E-max(P) in
Lower_Arc(P) & W-min(P) in Upper_Arc(P) & E-max(P) in Upper_Arc(P)
proof
let P be compact non empty Subset of TOP-REAL 2;
assume P is being_simple_closed_curve;
then
Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) & Lower_Arc(P) is_an_arc_of
E-max(P),W-min(P) by JORDAN6:50;
hence thesis by TOPREAL1:1;
end;
theorem Th2:
for P being compact non empty Subset of TOP-REAL 2,q st P is
being_simple_closed_curve & LE q,W-min(P),P holds q=W-min(P)
proof
let P be compact non empty Subset of TOP-REAL 2,q;
assume P is being_simple_closed_curve & LE q,W-min(P),P;
then
LE q,W-min(P),Upper_Arc(P),W-min(P),E-max(P) & Upper_Arc(P) is_an_arc_of
W-min(P),E-max(P) by JORDAN6:def 8,def 10;
hence thesis by JORDAN6:54;
end;
theorem Th3:
for P being compact non empty Subset of TOP-REAL 2,q st P is
being_simple_closed_curve & q in P holds LE W-min(P),q,P
proof
let P be compact non empty Subset of TOP-REAL 2,q;
assume that
A1: P is being_simple_closed_curve and
A2: q in P;
A3: q in Upper_Arc(P) \/ Lower_Arc(P) by A1,A2,JORDAN6:50;
A4: Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by A1,JORDAN6:50;
A5: W-min(P) in Upper_Arc(P) by A1,Th1;
per cases by A3,XBOOLE_0:def 3;
suppose
A6: q in Upper_Arc(P);
then LE W-min(P),q,Upper_Arc(P),W-min(P),E-max(P) by A4,JORDAN5C:10;
hence thesis by A5,A6,JORDAN6:def 10;
end;
suppose
A7: q in Lower_Arc(P);
per cases;
suppose
not q=W-min(P);
hence thesis by A5,A7,JORDAN6:def 10;
end;
suppose
A8: q=W-min(P);
then LE W-min(P),q,Upper_Arc(P),W-min(P),E-max(P) by A5,JORDAN5C:9;
hence thesis by A5,A8,JORDAN6:def 10;
end;
end;
end;
definition
let P be compact non empty Subset of TOP-REAL 2, q1,q2 be Point of TOP-REAL
2;
func Segment(q1,q2,P) -> Subset of TOP-REAL 2 equals
:Def1:
{p: LE q1,p,P &
LE p,q2,P} if q2<>W-min(P) otherwise {p1: LE q1,p1,P or q1 in P & p1=W-min(P)};
correctness
proof
ex B being Subset of TOP-REAL 2 st (q2<>W-min(P) implies B={p: LE q1,p
,P & LE p,q2,P})& (not q2<>W-min(P) implies B={p1: LE q1,p1,P or q1 in P & p1=
W-min(P)})
proof
per cases;
suppose
A1: q2<>W-min(P);
defpred P[Point of TOP-REAL 2] means LE q1,$1,P & LE $1,q2,P;
{p: P[p]} is Subset of TOP-REAL 2 from DOMAIN_1:sch 7;
then reconsider
C = {p: LE q1,p,P & LE p,q2,P} as Subset of TOP-REAL 2;
q2<>W-min(P) implies C={p: LE q1,p,P & LE p,q2,P};
hence thesis by A1;
end;
suppose
A2: not q2<>W-min(P);
defpred P[Point of TOP-REAL 2] means LE q1,$1,P or q1 in P & $1=W-min(
P);
{p1: P[p1]} is Subset of TOP-REAL 2 from DOMAIN_1:sch 7;
then reconsider
C = {p1: LE q1,p1,P or q1 in P & p1=W-min(P)} as Subset of
TOP-REAL 2;
not q2<>W-min(P) implies C={p1: LE q1,p1,P or q1 in P & p1=W-min(P )};
hence thesis by A2;
end;
end;
hence thesis;
end;
end;
theorem
for P being compact non empty Subset of TOP-REAL 2 st P is
being_simple_closed_curve holds Segment(W-min(P),E-max(P),P)=Upper_Arc(P) &
Segment(E-max(P),W-min(P),P)=Lower_Arc(P)
proof
let P be compact non empty Subset of TOP-REAL 2;
assume
A1: P is being_simple_closed_curve;
then
A2: Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by JORDAN6:def 8;
A3: {p1: LE E-max(P),p1,P or E-max(P) in P & p1=W-min(P)} = Lower_Arc(P)
proof
A4: {p1: LE E-max(P),p1,P or E-max(P) in P & p1=W-min(P)} c= Lower_Arc(P)
proof
let x be object;
assume x in {p1: LE E-max(P),p1,P or E-max(P) in P & p1=W-min(P)};
then consider p1 such that
A5: p1=x and
A6: LE E-max(P),p1,P or E-max(P) in P & p1=W-min(P);
per cases by A6;
suppose
A7: LE E-max(P),p1,P;
per cases by A5,A7,JORDAN6:def 10;
suppose
x in Lower_Arc(P);
hence thesis;
end;
suppose
A8: E-max(P) in Upper_Arc(P) & p1 in Upper_Arc(P) & LE E-max(P)
,p1,Upper_Arc(P),W-min(P),E-max(P);
A9: Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by A1,JORDAN6:50;
then LE p1,E-max(P),Upper_Arc(P),W-min(P),E-max(P) by A8,JORDAN5C:10;
then
A10: p1=E-max(P) by A8,A9,JORDAN5C:12;
Lower_Arc(P) is_an_arc_of E-max(P),W-min(P) by A1,JORDAN6:def 9;
hence thesis by A5,A10,TOPREAL1:1;
end;
end;
suppose
E-max(P) in P & p1=W-min(P);
then x in {W-min(P),E-max(P)} by A5,TARSKI:def 2;
then x in Upper_Arc(P) /\ Lower_Arc(P) by A1,JORDAN6:def 9;
hence thesis by XBOOLE_0:def 4;
end;
end;
Lower_Arc(P) c= {p1: LE E-max(P),p1,P or E-max(P) in P & p1=W-min(P) }
proof
let x be object;
assume
A11: x in Lower_Arc(P);
then reconsider p2=x as Point of TOP-REAL 2;
Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by A1,JORDAN6:50;
then
not (E-max P in P & p2=W-min P) implies E-max P in Upper_Arc P & p2
in Lower_Arc P & not p2=W-min P by A11,SPRECT_1:14,TOPREAL1:1;
then LE E-max(P),p2,P or E-max P in P & p2=W-min P by JORDAN6:def 10;
hence thesis;
end;
hence thesis by A4;
end;
A12: E-max(P)<>W-min(P) by A1,TOPREAL5:19;
{p: LE W-min(P),p,P & LE p,E-max(P),P} = Upper_Arc(P)
proof
A13: {p: LE W-min(P),p,P & LE p,E-max(P),P} c= Upper_Arc(P)
proof
let x be object;
assume x in {p: LE W-min(P),p,P & LE p,E-max(P),P};
then consider p such that
A14: p=x and
LE W-min(P),p,P and
A15: LE p,E-max(P),P;
per cases by A15,JORDAN6:def 10;
suppose
p in Upper_Arc(P) & E-max(P) in Lower_Arc(P) & not E-max(P)= W-min(P);
hence thesis by A14;
end;
suppose
p in Upper_Arc(P) & E-max(P) in Upper_Arc(P) & LE p,E-max(P),
Upper_Arc(P),W-min(P),E-max(P);
hence thesis by A14;
end;
suppose
A16: p in Lower_Arc(P) & E-max(P) in Lower_Arc(P) & not E-max(P)=
W-min(P) & LE p,E-max(P),Lower_Arc(P),E-max(P),W-min(P);
Lower_Arc(P) is_an_arc_of E-max(P),W-min(P) by A1,JORDAN6:def 9;
then p=E-max(P) by A16,JORDAN6:54;
hence thesis by A2,A14,TOPREAL1:1;
end;
end;
Upper_Arc(P) c= {p: LE W-min(P),p,P & LE p,E-max(P),P}
proof
let x be object;
assume
A17: x in Upper_Arc(P);
then reconsider p2=x as Point of TOP-REAL 2;
E-max(P) in Lower_Arc(P) by A1,Th1;
then
A18: LE p2,E-max(P),P by A12,A17,JORDAN6:def 10;
A19: W-min(P) in Upper_Arc(P) by A1,Th1;
LE W-min(P),p2,Upper_Arc(P),W-min(P),E-max(P) by A2,A17,JORDAN5C:10;
then LE W-min(P),p2,P by A17,A19,JORDAN6:def 10;
hence thesis by A18;
end;
hence thesis by A13;
end;
hence thesis by A12,A3,Def1;
end;
theorem Th5:
for P being compact non empty Subset of TOP-REAL 2, q1,q2 being
Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P holds q1 in
P & q2 in P
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2 be Point of TOP-REAL
2;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P;
A3: Upper_Arc(P) \/ Lower_Arc(P)=P by A1,JORDAN6:50;
per cases by A2,JORDAN6:def 10;
suppose
q1 in Upper_Arc(P) & q2 in Lower_Arc(P);
hence thesis by A3,XBOOLE_0:def 3;
end;
suppose
q1 in Upper_Arc(P) & q2 in Upper_Arc(P);
hence thesis by A3,XBOOLE_0:def 3;
end;
suppose
q1 in Lower_Arc(P) & q2 in Lower_Arc(P);
hence thesis by A3,XBOOLE_0:def 3;
end;
end;
theorem Th6:
for P being compact non empty Subset of TOP-REAL 2, q1,q2 being
Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P holds q1 in
Segment(q1,q2,P) & q2 in Segment(q1,q2,P)
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2 be Point of TOP-REAL
2;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P;
hereby
per cases;
suppose
A3: q2<>W-min(P);
q1 in P by A1,A2,Th5;
then LE q1,q1,P by A1,JORDAN6:56;
then q1 in {p: LE q1,p,P & LE p,q2,P} by A2;
hence q1 in Segment(q1,q2,P) by A3,Def1;
end;
suppose
A4: q2=W-min(P);
q1 in P by A1,A2,Th5;
then LE q1,q1,P by A1,JORDAN6:56;
then q1 in {p1: LE q1,p1,P or q1 in P & p1=W-min(P)};
hence q1 in Segment(q1,q2,P) by A4,Def1;
end;
end;
per cases;
suppose
A5: q2<>W-min(P);
q2 in P by A1,A2,Th5;
then LE q2,q2,P by A1,JORDAN6:56;
then q2 in {p: LE q1,p,P & LE p,q2,P} by A2;
hence thesis by A5,Def1;
end;
suppose
A6: q2=W-min(P);
q2 in {p1: LE q1,p1,P or q1 in P & p1=W-min(P)} by A2;
hence thesis by A6,Def1;
end;
end;
theorem Th7:
for P being compact non empty Subset of TOP-REAL 2, q1 being
Point of TOP-REAL 2 st q1 in P & P is being_simple_closed_curve holds q1 in
Segment(q1,W-min P,P)
proof
let P be compact non empty Subset of TOP-REAL 2, q1 be Point of TOP-REAL 2
such that
A1: q1 in P;
assume P is being_simple_closed_curve;
then LE q1,q1,P by A1,JORDAN6:56;
then q1 in {p1: LE q1,p1,P or q1 in P & p1=W-min(P)};
hence thesis by Def1;
end;
theorem
for P being compact non empty Subset of TOP-REAL 2, q being Point of
TOP-REAL 2 st P is being_simple_closed_curve & q in P & q<>W-min(P) holds
Segment(q,q,P)={q}
proof
let P be compact non empty Subset of TOP-REAL 2, q be Point of TOP-REAL 2;
assume that
A1: P is being_simple_closed_curve and
A2: q in P and
A3: q <> W-min P;
for x being object holds x in Segment(q,q,P) iff x=q
proof
let x be object;
hereby
assume x in Segment(q,q,P);
then x in {p: LE q,p,P & LE p,q,P} by A3,Def1;
then ex p st p=x & LE q,p,P & LE p,q,P;
hence x=q by A1,JORDAN6:57;
end;
assume
A4: x=q;
LE q,q,P by A1,A2,JORDAN6:56;
then x in {p: LE q,p,P & LE p,q,P} by A4;
hence thesis by A3,Def1;
end;
hence thesis by TARSKI:def 1;
end;
theorem
for P being compact non empty Subset of TOP-REAL 2, q1,q2 being Point
of TOP-REAL 2 st P is being_simple_closed_curve & q1<>W-min(P) & q2<>W-min(P)
holds not W-min(P) in Segment(q1,q2,P)
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2 be Point of TOP-REAL
2;
assume that
A1: P is being_simple_closed_curve and
A2: q1<>W-min(P) and
A3: q2<>W-min(P);
A4: Segment(q1,q2,P)={p: LE q1,p,P & LE p,q2,P} by A3,Def1;
now
A5: Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by A1,JORDAN6:def 8;
assume W-min(P) in Segment(q1,q2,P);
then consider p such that
A6: p=W-min(P) and
A7: LE q1,p,P and
LE p,q2,P by A4;
LE q1,p,Upper_Arc(P),W-min(P),E-max(P) by A6,A7,JORDAN6:def 10;
hence contradiction by A2,A6,A5,JORDAN6:54;
end;
hence thesis;
end;
theorem Th10:
for P being compact non empty Subset of TOP-REAL 2, q1,q2,q3
being Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P & LE
q2,q3,P & not(q1=q2 & q1=W-min(P)) & not(q2=q3 & q2=W-min(P)) holds Segment(q1,
q2,P)/\ Segment(q2,q3,P)={q2}
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2,q3 be Point of
TOP-REAL 2;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P and
A3: LE q2,q3,P and
A4: not(q1=q2 & q1=W-min(P)) and
A5: not(q2=q3 & q2=W-min(P));
A6: Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by A1,JORDAN6:def 8;
thus Segment(q1,q2,P)/\ Segment(q2,q3,P) c= {q2}
proof
let x be object;
assume
A7: x in Segment(q1,q2,P)/\ Segment(q2,q3,P);
then
A8: x in Segment(q2,q3,P) by XBOOLE_0:def 4;
A9: x in Segment(q1,q2,P) by A7,XBOOLE_0:def 4;
now
per cases;
case
q3<>W-min(P);
then x in {p: LE q2,p,P & LE p,q3,P} by A8,Def1;
then
A10: ex p st p=x & LE q2,p,P & LE p,q3,P;
per cases;
suppose
q2<>W-min(P);
then x in {p2: LE q1,p2,P & LE p2,q2,P} by A9,Def1;
then ex p2 st p2=x & LE q1,p2,P & LE p2,q2,P;
hence x=q2 by A1,A10,JORDAN6:57;
end;
suppose
A11: q2=W-min(P);
then LE q1,q2,Upper_Arc(P),W-min(P),E-max(P) by A2,JORDAN6:def 10;
hence x=q2 by A4,A6,A11,JORDAN6:54;
end;
end;
case
A12: q3=W-min(P);
then x in {p1: LE q2,p1,P or q2 in P & p1=W-min(P)} by A8,Def1;
then consider p1 such that
A13: p1=x and
A14: LE q2,p1,P or q2 in P & p1=W-min(P);
p1 in {p: LE q1,p,P & LE p,q2,P} by A5,A9,A12,A13,Def1;
then ex p st p=p1 & LE q1,p,P & LE p,q2,P;
hence x=q2 by A1,A3,A12,A13,A14,JORDAN6:57;
end;
end;
hence thesis by TARSKI:def 1;
end;
let x be object;
assume x in {q2};
then x=q2 by TARSKI:def 1;
then x in Segment(q1,q2,P) & x in Segment(q2,q3,P) by A1,A2,A3,Th6;
hence thesis by XBOOLE_0:def 4;
end;
theorem Th11:
for P being compact non empty Subset of TOP-REAL 2, q1,q2 being
Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P & q1 <>
W-min P & q2 <> W-min P holds Segment(q1,q2,P)/\ Segment(q2,W-min P,P)={q2}
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2 be Point of TOP-REAL
2;
set q3 = W-min P;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P and
A3: q1<>q3 and
A4: not q2=W-min(P);
thus Segment(q1,q2,P)/\ Segment(q2,W-min P,P) c= {q2}
proof
let x be object;
assume
A5: x in Segment(q1,q2,P)/\ Segment(q2,q3,P);
then x in Segment(q2,q3,P) by XBOOLE_0:def 4;
then x in {p1: LE q2,p1,P or q2 in P & p1=W-min P} by Def1;
then consider p1 such that
A6: p1=x and
A7: LE q2,p1,P or q2 in P & p1=W-min P;
x in Segment(q1,q2,P) by A5,XBOOLE_0:def 4;
then p1 in {p: LE q1,p,P & LE p,q2,P} by A4,A6,Def1;
then
A8: ex p st p=p1 & LE q1,p,P & LE p,q2,P;
per cases by A7;
suppose
LE q2,p1,P;
then x=q2 by A1,A6,A8,JORDAN6:57;
hence thesis by TARSKI:def 1;
end;
suppose
q2 in P & p1=W-min(P);
hence thesis by A1,A3,A8,Th2;
end;
end;
let x be object;
assume x in {q2};
then
A9: x=q2 by TARSKI:def 1;
q2 in P by A1,A2,Th5;
then
A10: x in Segment(q2,q3,P) by A1,A9,Th7;
x in Segment(q1,q2,P) by A1,A2,A9,Th6;
hence thesis by A10,XBOOLE_0:def 4;
end;
theorem Th12:
for P being compact non empty Subset of TOP-REAL 2, q1,q2 being
Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P & q1<>q2 &
q1<>W-min(P) holds Segment(q2,W-min(P),P)/\ Segment(W-min(P),q1,P)={W-min(P)}
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2 be Point of TOP-REAL
2;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P and
A3: q1<>q2 and
A4: q1<>W-min(P);
thus Segment(q2,W-min(P),P)/\ Segment(W-min(P),q1,P) c= {W-min(P)}
proof
let x be object;
assume
A5: x in Segment(q2,W-min(P),P)/\ Segment(W-min(P),q1,P);
then x in Segment(q2,W-min(P),P) by XBOOLE_0:def 4;
then x in {p1: LE q2,p1,P or q2 in P & p1=W-min(P)} by Def1;
then consider p1 such that
A6: p1=x and
A7: LE q2,p1,P or q2 in P & p1=W-min(P);
A8: x in Segment(W-min(P),q1,P) by A5,XBOOLE_0:def 4;
now
per cases by A7;
case
A9: LE q2,p1,P;
x in {p: LE W-min(P),p,P & LE p,q1,P} by A4,A8,Def1;
then ex p2 st p2=x & LE W-min(P),p2,P & LE p2,q1,P;
then LE q2,q1,P by A1,A6,A9,JORDAN6:58;
hence contradiction by A1,A2,A3,JORDAN6:57;
end;
case
q2 in P & p1=W-min(P);
hence x=W-min(P) by A6;
end;
end;
hence thesis by TARSKI:def 1;
end;
let x be object;
assume x in {W-min(P)};
then
A10: x=W-min(P) by TARSKI:def 1;
q2 in P by A1,A2,Th5;
then x in {p1: LE q2,p1,P or q2 in P & p1=W-min(P)} by A10;
then
A11: x in Segment(q2,W-min(P),P) by Def1;
q1 in P by A1,A2,Th5;
then LE W-min(P),q1,P by A1,Th3;
then x in Segment(W-min(P),q1,P) by A1,A10,Th6;
hence thesis by A11,XBOOLE_0:def 4;
end;
theorem Th13:
for P being compact non empty Subset of TOP-REAL 2, q1,q2,q3,q4
being Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P & LE
q2,q3,P & LE q3,q4,P & q1<>q2 & q2<>q3 holds Segment(q1,q2,P) misses Segment(q3
,q4,P)
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2,q3,q4 be Point of
TOP-REAL 2;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P and
A3: LE q2,q3,P and
A4: LE q3,q4,P and
A5: q1<>q2 and
A6: q2<>q3;
set x = the Element of Segment(q1,q2,P)/\ Segment(q3,q4,P);
assume
A7: Segment(q1,q2,P)/\ Segment(q3,q4,P)<>{};
then
A8: x in Segment(q1,q2,P) by XBOOLE_0:def 4;
A9: x in Segment(q3,q4,P) by A7,XBOOLE_0:def 4;
per cases;
suppose
q4=W-min(P);
then q3=W-min(P) by A1,A4,Th2;
hence contradiction by A1,A3,A6,Th2;
end;
suppose
A10: q4<>W-min(P);
q2<>W-min(P) by A1,A2,A5,Th2;
then x in {p2: LE q1,p2,P & LE p2,q2,P} by A8,Def1;
then
A11: ex p2 st p2=x & LE q1,p2,P & LE p2,q2,P;
x in {p1: LE q3,p1,P & LE p1,q4,P} by A9,A10,Def1;
then ex p1 st p1=x & LE q3,p1,P & LE p1,q4,P;
then LE q3,q2,P by A1,A11,JORDAN6:58;
hence contradiction by A1,A3,A6,JORDAN6:57;
end;
end;
theorem Th14:
for P being compact non empty Subset of TOP-REAL 2, q1,q2,q3
being Point of TOP-REAL 2 st P is being_simple_closed_curve & LE q1,q2,P & LE
q2,q3,P & q1<>W-min P & q2<>q3 holds Segment(q1,q2,P) misses Segment(q3,W-min P
,P)
proof
let P be compact non empty Subset of TOP-REAL 2, q1,q2,q3 be Point of
TOP-REAL 2;
assume that
A1: P is being_simple_closed_curve and
A2: LE q1,q2,P and
A3: LE q2,q3,P and
A4: q1<>W-min P and
A5: q2<>q3;
set x = the Element of Segment(q1,q2,P)/\ Segment(q3,W-min P,P);
assume
A6: Segment(q1,q2,P)/\ Segment(q3,W-min P,P)<>{};
then
A7: x in Segment(q1,q2,P) by XBOOLE_0:def 4;
x in Segment(q3,W-min P,P) by A6,XBOOLE_0:def 4;
then x in {p1: LE q3,p1,P or q3 in P & p1=W-min P} by Def1;
then
A8: ex p1 st p1=x &( LE q3,p1,P or q3 in P & p1=W-min P);
q2 <> W-min P by A1,A2,A4,Th2;
then x in {p: LE q1,p,P & LE p,q2,P} by A7,Def1;
then ex p3 st p3 = x & LE q1,p3,P & LE p3,q2,P;
then LE q3,q2,P by A1,A4,A8,Th2,JORDAN6:58;
hence contradiction by A1,A3,A5,JORDAN6:57;
end;
begin :: A function to divide the simple closed curve
reserve n for Nat;
theorem Th15:
for P being non empty Subset of TOP-REAL n, f being Function of
I[01], (TOP-REAL n)|P st f is being_homeomorphism ex g being Function of I[01],
TOP-REAL n st f=g & g is continuous & g is one-to-one
proof
let P be non empty Subset of TOP-REAL n, f be Function of I[01], (TOP-REAL n
)|P;
A1: [#]((TOP-REAL n)|P)= P by PRE_TOPC:def 5;
the carrier of (TOP-REAL n)|P = [#]((TOP-REAL n)|P) .=P by PRE_TOPC:def 5;
then reconsider g1=f as Function of I[01],TOP-REAL n by FUNCT_2:7;
assume
A2: f is being_homeomorphism;
then
A3: f is one-to-one by TOPS_2:def 5;
A4: [#]((TOP-REAL n)|P) <> {} & f is continuous by A2,TOPS_2:def 5;
A5: for P2 being Subset of TOP-REAL n st P2 is open holds g1"P2 is open
proof
let P2 be Subset of TOP-REAL n;
reconsider B1=P2 /\ P as Subset of (TOP-REAL n)|P by A1,XBOOLE_1:17;
f"(rng f) c= f"P by A1,RELAT_1:143;
then
A6: dom f c= f"P by RELAT_1:134;
assume P2 is open;
then B1 is open by A1,TOPS_2:24;
then
A7: f"B1 is open by A4,TOPS_2:43;
f"P c= dom f by RELAT_1:132;
then f"B1 = f"P2 /\ f"P & f"P=dom f by A6,FUNCT_1:68;
hence thesis by A7,RELAT_1:132,XBOOLE_1:28;
end;
[#]TOP-REAL n <> {};
then g1 is continuous by A5,TOPS_2:43;
hence thesis by A3;
end;
theorem Th16:
for P being non empty Subset of TOP-REAL n, g being Function of
I[01], (TOP-REAL n) st g is continuous one-to-one & rng g = P ex f being
Function of I[01],(TOP-REAL n)|P st f=g & f is being_homeomorphism
proof
let P be non empty Subset of TOP-REAL n, g be Function of I[01], TOP-REAL n;
assume that
A1: g is continuous one-to-one and
A2: rng g = P;
the carrier of (TOP-REAL n)|P = [#]((TOP-REAL n)|P);
then
A3: the carrier of (TOP-REAL n)|P = P by PRE_TOPC:def 5;
then reconsider f=g as Function of I[01],(TOP-REAL n)|P by A2,FUNCT_2:6;
take f;
thus f = g;
A4: [#]((TOP-REAL n)|P)= P by PRE_TOPC:def 5;
A5: dom f = the carrier of I[01] by FUNCT_2:def 1
.= [#]I[01];
A6: [#]TOP-REAL n<>{};
for P2 being Subset of (TOP-REAL n)|P st P2 is open holds f"P2 is open
proof
let P2 be Subset of (TOP-REAL n)|P;
assume P2 is open;
then consider C being Subset of TOP-REAL n such that
A7: C is open and
A8: C /\ [#]((TOP-REAL n)|P) = P2 by TOPS_2:24;
g"P = [#]I[01] by A3,A5,RELSET_1:22;
then f"P2 = f"C /\ [#]I[01] by A4,A8,FUNCT_1:68
.= f"C by XBOOLE_1:28;
hence thesis by A1,A6,A7,TOPS_2:43;
end;
then
A9: f is continuous by A4,TOPS_2:43;
rng f = [#]((TOP-REAL n)|P) by A2,PRE_TOPC:def 5;
hence thesis by A1,A9,COMPTS_1:17;
end;
Lm4: now
let h2 be Nat;
h2<h2+1 by NAT_1:13;
then
A1: h2-1<h2+1-1 by XREAL_1:9;
then h2-1-1<h2-1 by XREAL_1:9;
hence h2-1-1<h2 by A1,XXREAL_0:2;
end;
Lm5: 0 in [.0,1.] by XXREAL_1:1;
Lm6: 1 in [.0,1.] by XXREAL_1:1;
theorem Th17:
for A being Subset of TOP-REAL 2, p1,p2 being Point of TOP-REAL
2 st A is_an_arc_of p1,p2 ex g being Function of I[01], TOP-REAL 2 st g is
continuous one-to-one & rng g = A & g.0 = p1 & g.1 = p2
proof
let A be Subset of TOP-REAL 2, p1,p2 being Point of TOP-REAL 2;
assume
A1: A is_an_arc_of p1,p2;
then reconsider A9 = A as non empty Subset of TOP-REAL 2 by TOPREAL1:1;
consider f being Function of I[01], (TOP-REAL 2)|A9 such that
A2: f is being_homeomorphism and
A3: f.0 = p1 & f.1 = p2 by A1,TOPREAL1:def 1;
consider g being Function of I[01], TOP-REAL 2 such that
A4: f=g and
A5: g is continuous one-to-one by A2,Th15;
take g;
thus g is continuous one-to-one by A5;
rng f=[#]((TOP-REAL 2)|A9) by A2,TOPS_2:def 5;
hence rng g = A by A4,PRE_TOPC:def 5;
thus thesis by A3,A4;
end;
theorem Th18:
for P being non empty Subset of TOP-REAL 2, p1, p2, q1, q2 being
Point of TOP-REAL 2, g being Function of I[01], TOP-REAL 2,
s1, s2 being Real
st P is_an_arc_of p1,p2 & g is continuous one-to-one & rng g = P & g.0 = p1 & g
.1 = p2 & g.s1 = q1 & 0 <= s1 & s1 <= 1 & g.s2 = q2 & 0 <= s2 & s2 <= 1 & s1 <=
s2 holds LE q1,q2,P,p1,p2
proof
let P be non empty Subset of TOP-REAL 2, p1, p2, q1, q2 be Point of TOP-REAL
2, g be Function of I[01], TOP-REAL 2, s1, s2 be Real such that
A1: P is_an_arc_of p1,p2 and
A2: g is continuous one-to-one & rng g = P;
ex f being Function of I[01],(TOP-REAL 2)|P st f=g & f is
being_homeomorphism by A2,Th16;
hence thesis by A1,JORDAN5C:8;
end;
theorem Th19:
for P being non empty Subset of TOP-REAL 2, p1, p2, q1, q2 being
Point of TOP-REAL 2, g being Function of I[01], TOP-REAL 2, s1, s2 being Real
st g is continuous one-to-one & rng g = P & g.0 = p1 & g.1 = p2 & g.s1 = q1 & 0
<= s1 & s1 <= 1 & g.s2 = q2 & 0 <= s2 & s2 <= 1 & LE q1,q2,P,p1,p2 holds s1 <=
s2
proof
let P be non empty Subset of TOP-REAL 2, p1, p2, q1, q2 be Point of TOP-REAL
2, g be Function of I[01], TOP-REAL 2, s1, s2 be Real;
assume g is continuous one-to-one & rng g = P;
then ex f being Function of I[01],(TOP-REAL 2)|P st f=g & f is
being_homeomorphism by Th16;
hence thesis by JORDAN5C:def 3;
end;
theorem :: Dividing simple closed curve into segments.
for P being compact non empty Subset of TOP-REAL 2, e being Real
st P is being_simple_closed_curve & e > 0
ex h being FinSequence of the carrier of TOP-REAL 2
st h.1=W-min(P) & h is one-to-one & 8<=len h & rng h c= P &(for i
being Nat st 1<=i & i<len h holds LE h/.i,h/.(i+1),P) &(for i being
Nat,W being Subset of Euclid 2 st 1<=i & i<len h & W=Segment(h/.i,h
/.(i+1),P) holds diameter(W)<e) & (for W being Subset of Euclid 2 st W=Segment(
h/.len h,h/.1,P) holds diameter(W)<e) & (for i being Nat st 1<=i & i
+1<len h holds Segment(h/.i,h/.(i+1),P)/\ Segment(h/.(i+1),h/.(i+2),P)={h/.(i+1
)}) & Segment(h/.len h,h/.1,P)/\ Segment(h/.1,h/.2,P)={h/.1} & Segment(h/.(len
h-' 1),h/.len h,P)/\ Segment(h/.len h,h/.1,P)={h/.len h} & Segment(h/.(len h-'1
),h/.len h,P) misses Segment(h/.1,h/.2,P) &(for i,j being Nat st 1<=
i & i < j & j < len h & i,j aren't_adjacent holds Segment(h/.i,h/.(i+1),P)
misses Segment(h/.j,h/.(j+1),P)) & for i being Nat st 1 < i & i+1 <
len h holds Segment(h/.len h,h/.1,P) misses Segment(h/.i,h/.(i+1),P)
proof
let P be compact non empty Subset of TOP-REAL 2, e be Real;
assume that
A1: P is being_simple_closed_curve and
A2: e > 0;
A3: Upper_Arc P is_an_arc_of W-min P, E-max P by A1,JORDAN6:def 8;
then consider g1 being Function of I[01], TOP-REAL 2 such that
A4: g1 is continuous one-to-one and
A5: rng g1 = Upper_Arc P and
A6: g1.0 = W-min P and
A7: g1.1 = E-max P by Th17;
consider h1 being FinSequence of REAL such that
A8: h1.1=0 and
A9: h1.len h1=1 and
A10: 5<=len h1 and
A11: rng h1 c= the carrier of I[01] and
A12: h1 is increasing and
A13: for i being Nat,Q being Subset of I[01], W being Subset
of Euclid 2 st 1<=i & i<len h1 & Q=[.h1/.i,h1/.(i+1).] & W=g1.:Q holds diameter
(W)<e by A2,A4,UNIFORM1:13;
h1 is FinSequence of the carrier of I[01] by A11,FINSEQ_1:def 4;
then reconsider h11=g1*h1 as FinSequence of the carrier of TOP-REAL 2 by
FINSEQ_2:32;
A14: 2<len h1 by A10,XXREAL_0:2;
then
A15: 2 in dom h1 by FINSEQ_3:25;
A16: 1<=len h1 by A10,XXREAL_0:2;
then
A17: 1 in dom h1 by FINSEQ_3:25;
A18: 1+1 in dom h1 by A14,FINSEQ_3:25;
then
A19: h1.(1+1) in rng h1 by FUNCT_1:def 3;
A20: h11 is one-to-one by A4,A12;
A21: Lower_Arc P is_an_arc_of E-max P, W-min P by A1,JORDAN6:def 9;
then consider g2 being Function of I[01], TOP-REAL 2 such that
A22: g2 is continuous one-to-one and
A23: rng g2 = Lower_Arc P and
A24: g2.0 = E-max P and
A25: g2.1 = W-min P by Th17;
consider h2 being FinSequence of REAL such that
A26: h2.1=0 and
A27: h2.len h2=1 and
A28: 5<=len h2 and
A29: rng h2 c= the carrier of I[01] and
A30: h2 is increasing and
A31: for i being Nat,Q being Subset of I[01], W being Subset
of Euclid 2 st 1<=i & i<len h2 & Q=[.h2/.i,h2/.(i+1).] & W=g2.:Q holds diameter
(W)<e by A2,A22,UNIFORM1:13;
h2 is FinSequence of the carrier of I[01] by A29,FINSEQ_1:def 4;
then reconsider h21=g2*h2 as FinSequence of the carrier of TOP-REAL 2 by
FINSEQ_2:32;
A32: h21 is one-to-one by A22,A30;
A33: 1<=len h2 by A28,XXREAL_0:2;
then
A34: len h2 in dom h2 by FINSEQ_3:25;
then
A35: h21.len h2=W-min P by A25,A27,FUNCT_1:13;
reconsider h0=h11^(mid(h21,2,len h21-' 1)) as FinSequence of the carrier of
TOP-REAL 2;
A36: len h0=len h11+len (mid(h21,2,len h21 -'1)) by FINSEQ_1:22;
set i = len h0-'1;
take h0;
A37: rng h1 c= dom g1 by A11,FUNCT_2:def 1;
then
A38: dom h1=dom h11 by RELAT_1:27;
then
A39: len h1=len h11 by FINSEQ_3:29;
then
A40: h0.2=h11.2 by A14,FINSEQ_1:64;
A41: h0.(1+1)=h11.(1+1) by A39,A14,FINSEQ_1:64;
then
A42: h0.(1+1)=g1.(h1.(1+1)) by A18,FUNCT_1:13;
set k=len h0-'len h11+2-'1;
0+2<=len h0-'len h11 +2 by XREAL_1:6;
then
A43: 2-'1<=len h0-'len h11 +2-'1 by NAT_D:42;
A44: 0 in dom g1 by Lm5,BORSUK_1:40,FUNCT_2:def 1;
A45: len h1 in dom h1 by A16,FINSEQ_3:25;
dom g2=the carrier of I[01] by FUNCT_2:def 1;
then
A46: dom h2=dom h21 by A29,RELAT_1:27;
then
A47: len h2=len h21 by FINSEQ_3:29;
then
A48: 2<=len h21 by A28,XXREAL_0:2;
len h21<=len h21 +1 by NAT_1:12;
then
A49: len h21 -1<=len h21 +1-1 by XREAL_1:9;
then
A50: len h21-'1<=len h21 by A28,A47,XREAL_0:def 2;
2<=len h21 by A28,A47,XXREAL_0:2;
then
A51: 1+1-1<=len h21-1 by XREAL_1:9;
then
A52: len h21-'1=len h21 -1 by XREAL_0:def 2;
3<len h21 by A28,A47,XXREAL_0:2;
then
A53: 2+1-1<len h21 -1 by XREAL_1:9;
then
A54: 2<len h21 -'1 by A51,NAT_D:39;
then
A55: len h21 -'1-'2=len h21 -'1-2 by XREAL_1:233;
A56: 1<=len h21-'1 by A51,XREAL_0:def 2;
then
A57: len (mid(h21,2,len h21 -'1)) =(len h21-'1)-'2+1 by A48,A50,A54,
FINSEQ_6:118;
3+2-2<=len h2 -2 by A28,XREAL_1:9;
then
A58: 5+3<=len h1+(len h2 -2) by A10,XREAL_1:7;
then
A59: len h0 > 1+1+1 by A39,A47,A36,A52,A55,A57,XXREAL_0:2;
then
A60: len h0-'1 > 1+1 by Lm2;
then
A61: 1<i by XXREAL_0:2;
A62: 3+2-2<=len h2 -2 by A28,XREAL_1:9;
then
A63: 1<=len h2 -2 by XXREAL_0:2;
then
A64: len h1+1<=len h0 by A39,A47,A36,A52,A55,A57,XREAL_1:7;
then
A65: len h0>len h1 by NAT_1:13;
then
A66: 1<len h0 by A16,XXREAL_0:2;
then
A67: 1 in dom h0 by FINSEQ_3:25;
then
A68: h0/.1 = h0.1 by PARTFUN1:def 6;
A69: dom g1=[.0,1.] by BORSUK_1:40,FUNCT_2:def 1;
then
A70: 1 in dom (g1*h1) by A8,A17,Lm5,FUNCT_1:11;
then
A71: h11.1=W-min P by A6,A8,FUNCT_1:12;
hence
A72: h0.1=W-min P by A70,FINSEQ_1:def 7;
then
A73: h0/.1=W-min(P) by A67,PARTFUN1:def 6;
A74: len h0 in dom h0 by A66,FINSEQ_3:25;
then
A75: h0/.len h0=h0.len h0 by PARTFUN1:def 6;
A76: 1 in dom h2 by A33,FINSEQ_3:25;
then
A77: h21.1=E-max P by A24,A26,FUNCT_1:13;
thus
A78: h0 is one-to-one
proof
let x,y be object;
assume that
A79: x in dom h0 and
A80: y in dom h0 and
A81: h0.x=h0.y;
reconsider nx=x,ny=y as Nat by A79,A80;
A82: 1<=nx by A79,FINSEQ_3:25;
A83: nx<=len h0 by A79,FINSEQ_3:25;
A84: 1<=ny by A80,FINSEQ_3:25;
A85: ny<=len h0 by A80,FINSEQ_3:25;
per cases;
suppose
nx<=len h1;
then
A86: nx in dom h1 by A82,FINSEQ_3:25;
then
A87: h1.nx in rng h1 by FUNCT_1:def 3;
A88: h0.nx=h11.nx by A38,A86,FINSEQ_1:def 7
.=g1.(h1.nx) by A38,A86,FUNCT_1:12;
then
A89: h0.nx in Upper_Arc(P) by A5,A37,A87,FUNCT_1:def 3;
per cases;
suppose
ny<=len h1;
then
A90: ny in dom h1 by A84,FINSEQ_3:25;
then
A91: h1.ny in rng h1 by FUNCT_1:def 3;
h0.ny=h11.ny by A38,A90,FINSEQ_1:def 7
.=g1.(h1.ny) by A90,FUNCT_1:13;
then h1.nx=h1.ny by A4,A37,A81,A87,A88,A91;
hence thesis by A12,A86,A90,FUNCT_1:def 4;
end;
suppose
A92: ny>len h1;
A93: 0+2<=ny-'len h11 +2 by XREAL_1:6;
then
A94: 1<=(ny-'len h11 +2-'1) by Lm1,NAT_D:42;
len h1 +1<=ny by A92,NAT_1:13;
then
A95: len h1 +1-len h1<=ny-len h1 by XREAL_1:9;
then 1<=ny-'len h11 by A39,A92,XREAL_1:233;
then 1+1<=ny-'len h11+1+1-1 by XREAL_1:6;
then
A96: 2<=ny-'len h11+2-'1 by A93,Lm1,NAT_D:39,42;
A97: ny-len h11=ny-'len h11 by A39,A92,XREAL_1:233;
ny-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A85,
XREAL_1:9;
then
A98: ny-'len h11+2<=len h2-2+2 by A39,A97,XREAL_1:6;
then (ny-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
then
A99: (ny-'len h11 +2-'1) in dom h21 by A94,FINSEQ_3:25;
ny-'len h11+2-1<=len h2-1 by A98,XREAL_1:9;
then
A100: ny-'len h11+2-'1<=len h2-1 by A93,Lm1,NAT_D:39,42;
A101: ny<=len h11 + len (mid(h21,2,len h21 -'1)) by A85,FINSEQ_1:22;
then
A102: ny-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by XREAL_1:9
;
len h11+1<=ny by A39,A92,NAT_1:13;
then
A103: h0.ny=(mid(h21,2,len h21 -'1)).(ny -len h11) by A101,FINSEQ_1:23;
then
A104: h0.ny=h21.(ny-'len h11 +2-'1) by A39,A48,A56,A50,A54,A97,A102,A95,
FINSEQ_6:118;
then h0.ny in rng h21 by A99,FUNCT_1:def 3;
then h0.ny in rng g2 by FUNCT_1:14;
then h0.nx in Upper_Arc(P)/\ Lower_Arc(P) by A23,A81,A89,XBOOLE_0:def 4
;
then
A105: h0.nx in {W-min(P),E-max(P)} by A1,JORDAN6:50;
per cases by A105,TARSKI:def 2;
suppose
h0.nx=W-min(P);
then h21.(ny-'len h11 +2-'1)=W-min(P) by A39,A48,A56,A50,A54,A81,A103
,A97,A102,A95,FINSEQ_6:118;
then len h21=(ny-'len h11 +2-'1) by A46,A47,A34,A35,A32,A99;
then len h21+1<=len h21-1+1 by A47,A100,XREAL_1:6;
then len h21+1-len h21<=len h21 -len h21 by XREAL_1:9;
then len h21+1-len h21<=0;
hence thesis;
end;
suppose
h0.nx=E-max(P);
then 1=(ny-'len h11 +2-'1) by A46,A76,A77,A32,A81,A104,A99;
hence thesis by A96;
end;
end;
end;
suppose
A106: nx>len h1;
then len h1 +1<=nx by NAT_1:13;
then
A107: len h1 +1-len h1<=nx-len h1 by XREAL_1:9;
then 1<=nx-'len h11 by A39,A106,XREAL_1:233;
then
A108: 1+1<=nx-'len h11+1+1-1 by XREAL_1:6;
A109: nx<=len h11 + len (mid(h21,2,len h21 -'1)) by A83,FINSEQ_1:22;
then
A110: nx-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by XREAL_1:9;
A111: nx-len h11=nx-'len h11 by A39,A106,XREAL_1:233;
nx-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A83,
XREAL_1:9;
then
A112: nx-'len h11+2<=len h2-2+2 by A39,A111,XREAL_1:6;
then
A113: nx-'len h11+2-1<=len h2-1 by XREAL_1:9;
A114: (nx-'len h11 +2-'1)<=len h21 by A47,A112,NAT_D:44;
A115: 0+2<=nx-'len h11 +2 by XREAL_1:6;
then 1<=(nx-'len h11 +2-'1) by Lm1,NAT_D:42;
then
A116: (nx-'len h11 +2-'1) in dom h21 by A114,FINSEQ_3:25;
len h11+1<=nx by A39,A106,NAT_1:13;
then
A117: h0.nx=(mid(h21,2,len h21 -'1)).(nx -len h11) by A109,FINSEQ_1:23;
then
A118: h0.nx=h21.(nx-'len h11 +2-'1) by A39,A48,A56,A50,A54,A111,A110,A107,
FINSEQ_6:118;
then h0.nx in rng h21 by A116,FUNCT_1:def 3;
then
A119: h0.nx in Lower_Arc(P) by A23,FUNCT_1:14;
per cases;
suppose
ny<=len h1;
then
A120: ny in dom h1 by A84,FINSEQ_3:25;
then
A121: h1.ny in rng h1 by FUNCT_1:def 3;
h0.ny=h11.ny by A38,A120,FINSEQ_1:def 7
.=g1.(h1.ny) by A38,A120,FUNCT_1:12;
then h0.ny in rng g1 by A37,A121,FUNCT_1:def 3;
then h0.ny in Upper_Arc(P)/\ Lower_Arc(P) by A5,A81,A119,XBOOLE_0:def 4
;
then
A122: h0.ny in {W-min(P),E-max(P)} by A1,JORDAN6:50;
A123: nx-'len h11+2-'1<=len h2-1 by A115,A113,Lm1,NAT_D:39,42;
A124: 2<=nx-'len h11+2-'1 by A108,A115,Lm1,NAT_D:39,42;
per cases by A122,TARSKI:def 2;
suppose
h0.ny=W-min(P);
then len h21=(nx-'len h11 +2-'1) by A46,A47,A34,A35,A32,A81,A118,A116
;
then len h21+1<=len h21-1+1 by A47,A123,XREAL_1:6;
then len h21+1-len h21<=len h21 -len h21 by XREAL_1:9;
then len h21+1-len h21<=0;
hence thesis;
end;
suppose
h0.ny=E-max(P);
then h21.(nx-'len h11 +2-'1)=E-max(P) by A39,A48,A56,A50,A54,A81,A117
,A111,A110,A107,FINSEQ_6:118;
then 1=(nx-'len h11 +2-'1) by A46,A76,A77,A32,A116;
hence thesis by A124;
end;
end;
suppose
A125: ny>len h1;
then
A126: ny-len h11=ny-'len h11 by A39,XREAL_1:233;
len h1 +1<=ny by A125,NAT_1:13;
then
A127: h0.ny=(mid(h21,2,len h21 -'1)).(ny -len h11) & len h1 +1-len h1
<=ny-len h1 by A39,A36,A85,FINSEQ_1:23,XREAL_1:9;
0+2<=ny-'len h11 +2 by XREAL_1:6;
then
A128: 1<=(ny-'len h11 +2-'1) by Lm1,NAT_D:42;
ny-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A85,
XREAL_1:9;
then
A129: h0.ny=h21.(ny-'len h11 +2-'1) by A39,A48,A56,A50,A54,A126,A127,
FINSEQ_6:118;
ny-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A85,
XREAL_1:9;
then ny-'len h11+2<=len h2-2+2 by A39,A126,XREAL_1:6;
then (ny-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
then (ny-'len h11 +2-'1) in dom h21 by A128,FINSEQ_3:25;
then nx-'len h1+2-'1=ny-'len h1+2-'1 by A39,A32,A81,A118,A116,A129;
then nx-'len h1+2-1=ny-'len h1+2-'1 by A39,A115,Lm1,NAT_D:39,42;
then nx-'len h1+(2-1)=ny-'len h1+2-1 by A39,A128,NAT_D:39;
then len h1+nx-len h1=len h1+(ny-len h1) by A39,A111,A126,XCMPLX_1:29;
hence thesis;
end;
end;
end;
then
A130: h0/.len h0 <> W-min P by A16,A72,A65,A74,A75,A67;
A131: dom g2 = [.0,1.] by BORSUK_1:40,FUNCT_2:def 1;
thus 8<=len h0 by A38,A47,A36,A52,A55,A57,A58,FINSEQ_3:29;
rng (mid(h21,2,len h21 -'1)) c= rng h21 & rng (g2*h2) c= rng g2 by
FINSEQ_6:119,RELAT_1:26;
then rng (g1*h1) c= rng g1 & rng (mid(h21,2,len h21 -'1)) c= rng g2 by
RELAT_1:26;
then rng h11 \/ rng (mid(h21,2,len h21 -'1)) c= Upper_Arc(P) \/ Lower_Arc(P
) by A5,A23,XBOOLE_1:13;
then rng h0 c=Upper_Arc(P) \/ Lower_Arc(P) by FINSEQ_1:31;
hence rng h0 c= P by A1,JORDAN6:def 9;
A132: dom g1 =[.0,1.] by BORSUK_1:40,FUNCT_2:def 1;
thus for i being Nat st 1<=i & i<len h0 holds LE h0/.i,h0/.(i+1),
P
proof
let i be Nat;
assume that
A133: 1<=i and
A134: i<len h0;
A135: i+1<=len h0 by A134,NAT_1:13;
A136: i<i+1 by NAT_1:13;
A137: 1<i+1 by A133,NAT_1:13;
per cases;
suppose
A138: i<len h1;
then
A139: i+1<=len h1 by NAT_1:13;
then
A140: i+1 in dom h1 by A137,FINSEQ_3:25;
then
A141: h1.(i+1) in rng h1 by FUNCT_1:def 3;
then
A142: h1.(i+1)<=1 by A11,BORSUK_1:40,XXREAL_1:1;
h0.(i+1)=h11.(i+1) by A39,A137,A139,FINSEQ_1:64;
then
A143: h0.(i+1)=g1.(h1.(i+1)) by A140,FUNCT_1:13;
then
A144: h0.(i+1) in Upper_Arc(P) by A5,A132,A11,A141,BORSUK_1:40,FUNCT_1:def 3;
i in dom h0 by A133,A134,FINSEQ_3:25;
then
A145: h0/.i=h0.i by PARTFUN1:def 6;
A146: i in dom h1 by A133,A138,FINSEQ_3:25;
then
A147: h1.i in rng h1 by FUNCT_1:def 3;
then
A148: 0<=h1.i & h1.i<=1 by A11,BORSUK_1:40,XXREAL_1:1;
A149: g1.(h1.i) in rng g1 by A132,A11,A147,BORSUK_1:40,FUNCT_1:def 3;
h0.i=h11.i by A39,A133,A138,FINSEQ_1:64;
then
A150: h0.i=g1.(h1.i) by A146,FUNCT_1:13;
i+1 in dom h0 by A135,A137,FINSEQ_3:25;
then
A151: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
h1.i<h1.(i+1) by A12,A136,A146,A140,SEQM_3:def 1;
then LE h0/.i,h0/.(i+1),Upper_Arc(P),W-min(P),E-max(P) by A3,A4,A5,A6,A7
,A150,A148,A143,A142,A145,A151,Th18;
hence thesis by A5,A150,A145,A151,A149,A144,JORDAN6:def 10;
end;
suppose
A152: i>=len h1;
per cases by A152,XXREAL_0:1;
suppose
A153: i>len h1;
then len h11+1<=i by A39,NAT_1:13;
then
A154: h0.i=(mid(h21,2,len h21 -'1)).(i -len h11) by A36,A134,FINSEQ_1:23;
A155: i+1-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A135
,XREAL_1:9;
i+1>len h11 by A39,A153,NAT_1:13;
then
A156: i+1-len h11=i+1-'len h11 by XREAL_1:233;
A157: len h1 +1<=i by A153,NAT_1:13;
then
A158: len h1 +1-len h1<=i-len h1 by XREAL_1:9;
A159: i-len h11=i-'len h11 by A39,A153,XREAL_1:233;
A160: len h1 +1<=i+1 by A157,NAT_1:13;
then
A161: len h1 +1-len h1<=i+1-len h1 by XREAL_1:9;
then
A162: 1<i+1-'len h11+(2-1) by A39,A156,NAT_1:13;
then
A163: 0<i+1-'len h11+2-1;
h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) by A39,A36,A135,A160,
FINSEQ_1:23;
then
A164: h0.(i+1)=h21.(i+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A156,A155,A161
,FINSEQ_6:118;
set j=i-'len h11+2-'1;
len h2<len h2+1 by NAT_1:13;
then
A165: len h2-1<len h2+1-1 by XREAL_1:9;
A166: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A167: 1<=(i-'len h11 +2-'1) by Lm1,NAT_D:42;
then
A168: 1<j+1 by NAT_1:13;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A134,
XREAL_1:9;
then
A169: i-'len h11+2<=len h2-2+2 by A39,A159,XREAL_1:6;
then (i-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
then
A170: j in dom h2 by A46,A167,FINSEQ_3:25;
i-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A134,
XREAL_1:9;
then h0.i=h21.(i-'len h11 +2-'1) by A39,A48,A56,A50,A54,A154,A159,A158,
FINSEQ_6:118;
then
A171: h0.i=g2.(h2.j) by A170,FUNCT_1:13;
A172: h2.j in rng h2 by A170,FUNCT_1:def 3;
then
A173: h0.i in Lower_Arc(P) by A23,A131,A29,A171,BORSUK_1:40,FUNCT_1:def 3;
i+1 in dom h0 by A135,A137,FINSEQ_3:25;
then
A174: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
j+1=i-'len h11+1+1-1+1 by A166,Lm1,NAT_D:39,42
.=i-'len h11+2;
then
A175: j+1 in dom h2 by A169,A168,FINSEQ_3:25;
then
A176: h2.(j+1) in rng h2 by FUNCT_1:def 3;
then
A177: h2.(j+1)<=1 by A29,BORSUK_1:40,XXREAL_1:1;
A178: j+1=i-len h11+2-1+1 by A159,A166,Lm1,NAT_D:39,42
.=i+1-'len h11+2-'1 by A156,A163,XREAL_0:def 2;
then
A179: h0.(i+1)=g2.(h2.(j+1)) by A164,A175,FUNCT_1:13;
then
A180: h0.(i+1) in Lower_Arc(P) by A23,A131,A29,A176,BORSUK_1:40,FUNCT_1:def 3
;
A181: i+1-'len h11+2-1=i+1-'len h11+2-'1 by A162,XREAL_0:def 2;
i+1-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A135,
XREAL_1:9;
then
A182: i+1-'len h11+2<=len h2-2+2 by A39,A156,XREAL_1:6;
then i+1-'len h11+2-1<=len h2-1 by XREAL_1:9;
then i+1-'len h11+2-'1<len h2 by A181,A165,XXREAL_0:2;
then
A183: i+1-'len h11+2-'1 in dom h2 by A178,A168,FINSEQ_3:25;
A184: now
assume h0/.(i+1)=W-min(P);
then len h21=(i+1-'len h11 +2-'1) by A46,A47,A34,A35,A32,A164,A174
,A183;
then len h21+1-len h21<=len h21 -len h21 by A47,A181,A182,XREAL_1:9;
then len h21+1-len h21<=0;
hence contradiction;
end;
j<j+1 by NAT_1:13;
then
A185: h2.j<h2.(j+1) by A30,A170,A175,SEQM_3:def 1;
i in dom h0 by A133,A134,FINSEQ_3:25;
then
A186: h0/.i=h0.i by PARTFUN1:def 6;
0<=h2.j & h2.j<=1 by A29,A172,BORSUK_1:40,XXREAL_1:1;
then LE h0/.i,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23
,A24,A25,A171,A179,A177,A185,A186,A174,Th18;
hence thesis by A186,A174,A173,A180,A184,JORDAN6:def 10;
end;
suppose
A187: i=len h1;
then h0.i=h11.i & i in dom h1 by A39,A133,FINSEQ_1:64,FINSEQ_3:25;
then
A188: h0.i=E-max(P) by A7,A9,A187,FUNCT_1:13;
i in dom h0 by A133,A134,FINSEQ_3:25;
then h0/.i=E-max(P) by A188,PARTFUN1:def 6;
then
A189: h0/.i in Upper_Arc(P) by A1,Th1;
i+1 in dom h0 by A135,A137,FINSEQ_3:25;
then
A190: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
set j=i-'len h11+2-'1;
len h11-'len h11= len h11-len h11 by XREAL_1:233
.=0;
then
A191: j=2-1 by A39,A187,XREAL_1:233;
then j+1<=len h2 by A28,XXREAL_0:2;
then
A192: j+1 in dom h2 by A191,FINSEQ_3:25;
then
A193: h2.(j+1) in rng h2 by FUNCT_1:def 3;
2<=len h21 by A28,A47,XXREAL_0:2;
then
A194: 2 in dom h21 by FINSEQ_3:25;
A195: i+1-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A135
,XREAL_1:9;
A196: i+1-'len h11+2-'1=1+2-'1 by A39,A187,NAT_D:34
.=2 by NAT_D:34;
h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) & i+1-len h11=i
+1-'len h11 by A39,A36,A135,A136,A187,FINSEQ_1:23,XREAL_1:233;
then
A197: h0.(i+1)=h21.(i+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A187,A195,
FINSEQ_6:118;
then h0.(i+1)=g2.(h2.(j+1)) by A191,A196,A192,FUNCT_1:13;
then
A198: h0.(i+1) in Lower_Arc(P) by A23,A131,A29,A193,BORSUK_1:40,FUNCT_1:def 3
;
len h21 <> i+1-'len h11 +2-'1 by A28,A47,A196;
then h0/.(i+1) <> W-min P by A46,A47,A34,A35,A32,A197,A196,A190,A194;
hence thesis by A189,A190,A198,JORDAN6:def 10;
end;
end;
end;
A199: i<len h0 by A66,JORDAN5B:1;
then
A200: i+1<=len h0 by NAT_1:13;
A201: 1+1<=len h0 by A65,A14,XXREAL_0:2;
then
A202: 1<=i by Lm3;
then
A203: i in dom h0 by A199,FINSEQ_3:25;
then
A204: h0/.i=h0.i by PARTFUN1:def 6;
A205: 1+1<=len h0 by A66,NAT_1:13;
then 1+1 in dom h0 by FINSEQ_3:25;
then
A206: h0/.(1+1)=h0.(1+1) by PARTFUN1:def 6;
A207: now
A208: 1+1 in dom h1 by A14,FINSEQ_3:25;
then
A209: h1.(1+1) in rng h1 by FUNCT_1:def 3;
A210: h0.(1+1)=h11.(1+1) by A39,A14,FINSEQ_1:64;
then h0.(1+1)=g1.(h1.(1+1)) by A208,FUNCT_1:13;
then
A211: h0.(1+1) in Upper_Arc(P) by A5,A132,A11,A209,BORSUK_1:40,FUNCT_1:def 3;
assume
A212: h0/.(1+1)=h0/.i;
per cases;
suppose
i<=len h1;
then h0.i=h11.i & i in dom h1 by A39,A202,FINSEQ_1:64,FINSEQ_3:25;
hence contradiction by A38,A20,A60,A204,A206,A212,A208,A210;
end;
suppose
A213: i>len h1;
i in dom h0 by A202,A199,FINSEQ_3:25;
then
A214: h0/.i=h0.i by PARTFUN1:def 6;
A215: i-len h11=i-'len h11 by A39,A213,XREAL_1:233;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A199,
XREAL_1:9;
then i-'len h11+2<=len h2-2+2 by A39,A215,XREAL_1:6;
then
A216: (i-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A199,
XREAL_1:9;
then
A217: i-'len h11+2<=len h2-2+2 by A39,A215,XREAL_1:6;
set k=i-'len h11+2-'1;
A218: i-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A199,
XREAL_1:9;
A219: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A220: i-'len h11+2-'1=i-'len h11+2-1 by Lm1,NAT_D:39,42;
1<=(i-'len h11 +2-'1) by A219,Lm1,NAT_D:42;
then
A221: k in dom h2 by A46,A216,FINSEQ_3:25;
then h2.k in rng h2 by FUNCT_1:def 3;
then
A222: g2.(h2.k) in rng g2 by A131,A29,BORSUK_1:40,FUNCT_1:def 3;
A223: len h1 +1<=i by A213,NAT_1:13;
then h0.i=(mid(h21,2,len h21 -'1)).(i -len h11) & len h1 +1-len h1<=i-
len h1 by A39,A36,A199,FINSEQ_1:23,XREAL_1:9;
then
A224: h0.i=h21.(i-'len h11 +2-'1) by A39,A48,A56,A50,A54,A215,A218,FINSEQ_6:118
;
then h0.i=g2.(h2.k) by A221,FUNCT_1:13;
then h0.i in Upper_Arc(P) /\ Lower_Arc(P) by A23,A206,A212,A211,A214,A222
,XBOOLE_0:def 4;
then h0.i in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
then
A225: h0.i=W-min(P) or h0.i=E-max(P)by TARSKI:def 2;
len h11+1-len h11<=i-len h11 by A39,A223,XREAL_1:9;
then 1<=i-'len h11 by NAT_D:39;
then 1+2<=i-'len h11+2 by XREAL_1:6;
then 1+2-1<=i-'len h11+2-1 by XREAL_1:9;
then
A226: 1<k by A220,XXREAL_0:2;
i-'len h11+2-'1<i-'len h11+2-1+1 by A220,NAT_1:13;
hence contradiction by A46,A76,A34,A77,A35,A32,A224,A217,A226,A221,A225;
end;
end;
A227: 1 in dom g2 by Lm6,BORSUK_1:40,FUNCT_2:def 1;
A228: len h2-1-1<len h2 by Lm4;
A229: now
per cases;
case
A230: i<=len h1;
A231: h0/.(1+1) in Upper_Arc(P) by A5,A132,A11,A206,A42,A19,BORSUK_1:40
,FUNCT_1:def 3;
A232: 0 <= h1.(1+1) & h1.(1+1) <= 1 by A11,A19,BORSUK_1:40,XXREAL_1:1;
A233: i in dom h1 by A61,A230,FINSEQ_3:25;
then
A234: h1.i in rng h1 by FUNCT_1:def 3;
then
A235: h1.i<=1 by A11,BORSUK_1:40,XXREAL_1:1;
h0.i=h11.i by A39,A61,A230,FINSEQ_1:64;
then
A236: h0.i=g1.(h1.i) by A233,FUNCT_1:13;
then
A237: h0/.i in Upper_Arc(P) by A5,A132,A11,A204,A234,BORSUK_1:40,FUNCT_1:def 3;
h1.(1+1)<h1.i by A12,A60,A18,A233,SEQM_3:def 1;
then LE h0/.(1+1),h0/.i,Upper_Arc(P),W-min(P),E-max(P) by A3,A4,A5,A6,A7
,A204,A206,A42,A236,A235,A232,Th18;
hence LE h0/.(1+1),h0/.i,P by A231,A237,JORDAN6:def 10;
end;
case
A238: i>len h1;
1+1 in dom h1 by A14,FINSEQ_3:25;
then
A239: h11.(1+1)=g1.(h1.(1+1)) by FUNCT_1:13;
A240: i-len h11=i-'len h11 by A39,A238,XREAL_1:233;
i+1-1<=len h1+(len h2-2)-1 by A39,A47,A36,A52,A55,A57,A200,XREAL_1:9;
then i-len h11<=len h1+((len h2-2)-1)-len h11 by XREAL_1:9;
then i-'len h11+2<=len h2-2-1+2 by A39,A240,XREAL_1:6;
then
A241: i-'len h11+2-1<=len h2-1-1 by XREAL_1:9;
A242: len h1 +1<=i by A238,NAT_1:13;
then
A243: len h1 +1-len h1<=i-len h1 by XREAL_1:9;
h1.(1+1) in rng h1 by A15,FUNCT_1:def 3;
then
A244: g1.(h1.(1+1)) in rng g1 by A132,A11,BORSUK_1:40,FUNCT_1:def 3;
0+2<=i-'len h11 +2 by XREAL_1:6;
then
A245: 2-'1<=(i-'len h11 +2-'1) by NAT_D:42;
set k=i-'len h11+2-'1;
0+1<=i-'len h11+1+1-1 by XREAL_1:6;
then
A246: i-'len h11+2-'1=i-'len h11+2-1 by NAT_D:39;
A247: i-len h11 <=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A199,
XREAL_1:9;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A199,
XREAL_1:9;
then i-'len h11+2<=len h2-2+2 by A39,A240,XREAL_1:6;
then i-'len h11 +2-'1<=len h21 by A47,NAT_D:44;
then
A248: (i-'len h11 +2-'1) in dom h21 by A245,Lm1,FINSEQ_3:25;
A249: h0.i=(mid(h21,2,len h21 -'1)).(i -len h11) by A39,A36,A199,A242,
FINSEQ_1:23;
then h0.(i)=h21.(i-'len h11 +2-'1) by A39,A48,A56,A50,A54,A240,A247,A243,
FINSEQ_6:118;
then
A250: h0.i=g2.(h2.k) by A46,A248,FUNCT_1:13;
h2.k in rng h2 by A46,A248,FUNCT_1:def 3;
then
A251: h0.i in Lower_Arc(P) by A23,A131,A29,A250,BORSUK_1:40,FUNCT_1:def 3;
1<=i-len h11 by A38,A243,FINSEQ_3:29;
then h0.i=h21.k by A48,A56,A50,A54,A240,A247,A249,FINSEQ_6:118;
then h0/.i <> W-min P by A228,A46,A34,A35,A32,A204,A246,A248,A241;
hence LE h0/.(1+1),h0/.i,P by A5,A204,A206,A41,A239,A244,A251,
JORDAN6:def 10;
end;
end;
A252: len h0-len h11=len h0-'len h11 by A39,A65,XREAL_1:233;
then (len h0-'len h11 +2-'1)<=len h21 by A36,A52,A55,A57,NAT_D:44;
then (len h0-'len h11 +2-'1) in dom h21 by A43,Lm1,FINSEQ_3:25;
then
A253: h21.k=g2.(h2.k) & h2.k in rng h2 by A46,FUNCT_1:13,def 3;
h1.len h1 in dom g1 by A9,A69,XXREAL_1:1;
then
A254: len h1 in dom (g1*h1) by A45,FUNCT_1:11;
then
A255: h11.len h1=E-max(P) by A7,A9,FUNCT_1:12;
A256: for i being Nat st 1<=i & i+1<=len h0 holds LE h0/.i,h0/.(i
+1),P & h0/.(i+1)<>W-min(P) & h0/.i<>h0/.(i+1)
proof
let i be Nat such that
A257: 1<=i and
A258: i+1<=len h0;
A259: i<i+1 by NAT_1:13;
A260: 1<i+1 by A257,NAT_1:13;
then i+1 in dom h0 by A258,FINSEQ_3:25;
then
A261: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A262: i<len h0 by A258,NAT_1:13;
then i in dom h0 by A257,FINSEQ_3:25;
then
A263: h0/.i=h0.i by PARTFUN1:def 6;
per cases;
suppose
A264: i<len h1;
then
A265: i+1<=len h1 by NAT_1:13;
then
A266: i+1 in dom h1 by A260,FINSEQ_3:25;
then
A267: h1.(i+1) in rng h1 by FUNCT_1:def 3;
then
A268: (h1.(i+1)) <= 1 by A11,BORSUK_1:40,XXREAL_1:1;
A269: i+1 <> 1 & i+1 <> i by A257,NAT_1:13;
A270: i in dom h1 by A257,A264,FINSEQ_3:25;
then
A271: h1.i in rng h1 by FUNCT_1:def 3;
then
A272: 0 <= (h1.i) & (h1.i) <= 1 by A11,BORSUK_1:40,XXREAL_1:1;
A273: h0.(i+1)=h11.(i+1) by A39,A260,A265,FINSEQ_1:64;
then
A274: h0.(i+1)=g1.(h1.(i+1)) by A266,FUNCT_1:13;
then
A275: h0.(i+1) in Upper_Arc(P) by A5,A132,A11,A267,BORSUK_1:40,FUNCT_1:def 3;
A276: h0.i=h11.i by A39,A257,A264,FINSEQ_1:64;
then
A277: g1.(h1.i) = h0/.i by A263,A270,FUNCT_1:13;
g1.(h1.i) in rng g1 by A132,A11,A271,BORSUK_1:40,FUNCT_1:def 3;
then
A278: h0.i in Upper_Arc(P) by A5,A276,A270,FUNCT_1:13;
h1.i<h1.(i+1) by A12,A259,A270,A266,SEQM_3:def 1;
then LE h0/.i,h0/.(i+1),Upper_Arc(P),W-min(P),E-max(P) by A3,A4,A5,A6,A7
,A261,A274,A277,A272,A268,Th18;
hence
thesis by A38,A17,A71,A20,A263,A261,A276,A270,A273,A266,A278,A275,A269,
JORDAN6:def 10;
end;
suppose
A279: i>=len h1;
per cases by A279,XXREAL_0:1;
suppose
A280: i>len h1;
i-len h11<len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A262,
XREAL_1:9;
then
A281: i-len h11+2<len h2-2+2 by XREAL_1:6;
i+1>len h11 by A39,A280,NAT_1:13;
then
A282: i+1-len h11=i+1-'len h11 by XREAL_1:233;
set j=i-'len h11+2-'1;
A283: i+1-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A258
,XREAL_1:9;
A284: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A285: j+1=i-'len h11+(1+1)-1+1 by Lm1,NAT_D:39,42
.=i-'len h11+(1+1);
A286: len h1 +1<=i by A280,NAT_1:13;
then
A287: len h1 +1-len h1<=i-len h1 by XREAL_1:9;
i+1 in dom h0 by A258,A260,FINSEQ_3:25;
then
A288: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A289: len h1 +1<=i+1 by A286,NAT_1:13;
then
A290: len h1 +1-len h1<=i+1-len h1 by XREAL_1:9;
then 1<i+1-'len h11+(2-1) by A39,A282,NAT_1:13;
then
A291: 0<i+1-'len h11+2-1;
h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) by A39,A36,A258,A289,
FINSEQ_1:23;
then
A292: h0.(i+1)=h21.(i+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A282,A283,A290
,FINSEQ_6:118;
A293: i<=len h11 + len (mid(h21,2,len h21 -'1)) by A262,FINSEQ_1:22;
len h11+1<=i by A39,A280,NAT_1:13;
then
A294: h0.i=(mid(h21, 2,len h21 -'1)).(i -len h11) by A293,FINSEQ_1:23;
A295: i-len h11=i-'len h11 by A39,A280,XREAL_1:233;
A296: 1<=i-'len h11 +2-'1 by A284,Lm1,NAT_D:42;
then 1<j+1 by NAT_1:13;
then
A297: j+1 in dom h2 by A295,A281,A285,FINSEQ_3:25;
then
A298: h2.(j+1) in rng h2 by FUNCT_1:def 3;
then
A299: (h2.(j+1)) <= 1 by A29,BORSUK_1:40,XXREAL_1:1;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A262,
XREAL_1:9;
then i-'len h11+2<=len h2-2+2 by A39,A295,XREAL_1:6;
then i-'len h11 +2-'1<=len h21 by A47,NAT_D:44;
then
A300: j in dom h2 by A46,A296,FINSEQ_3:25;
j<j+1 by NAT_1:13;
then
A301: h2.j<h2.(j+1) by A30,A300,A297,SEQM_3:def 1;
i in dom h0 by A257,A262,FINSEQ_3:25;
then
A302: h0/.i=h0.i by PARTFUN1:def 6;
i-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A293,
XREAL_1:9;
then
A303: h0.i=h21.(i-'len h11 +2-'1) by A39,A48,A56,A50,A54,A294,A295,A287,
FINSEQ_6:118;
then
A304: h0.i=g2.(h2.j) by A300,FUNCT_1:13;
A305: j+1=i-len h11+2-1+1 by A295,A284,Lm1,NAT_D:39,42
.=i+1-'len h11+2-'1 by A282,A291,XREAL_0:def 2;
then
A306: h0/.(i+1) <> W-min P by A46,A34,A35,A32,A295,A292,A281,A285,A297,A288;
A307: h0.(i+1)=g2.(h2.(j+1)) by A292,A305,A297,FUNCT_1:13;
then
A308: h0/.(i+1) in Lower_Arc(P) by A23,A131,A29,A298,A288,BORSUK_1:40
,FUNCT_1:def 3;
A309: h2.j in rng h2 by A300,FUNCT_1:def 3;
then
A310: j < j+1 & h0/.i in Lower_Arc(P) by A23,A131,A29,A304,A302,BORSUK_1:40
,FUNCT_1:def 3,NAT_1:13;
0 <= (h2.j) & (h2.j) <= 1 by A29,A309,BORSUK_1:40,XXREAL_1:1;
then LE h0/.i,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23
,A24,A25,A304,A307,A301,A302,A288,A299,Th18;
hence thesis by A46,A32,A303,A292,A305,A300,A297,A302,A288,A306,A310
,A308,JORDAN6:def 10;
end;
suppose
A311: i=len h1;
then
A312: i-len h11=i-'len h11 by A39,XREAL_1:233;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A262,
XREAL_1:9;
then
A313: i-'len h11+2<=len h2-2+2 by A39,A312,XREAL_1:6;
then
A314: (i-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
set j=i-'len h11+2-'1;
A315: j+1 <> j;
A316: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A317: j+1=i-'len h11+(1+1)-1+1 by Lm1,NAT_D:39,42
.=i-'len h11+2;
2-'1<=(i-'len h11 +2-'1) by A316,NAT_D:42;
then 1<j+1 by Lm1,NAT_1:13;
then
A318: j+1 in dom h2 by A313,A317,FINSEQ_3:25;
then
A319: h2.(j+1) in rng h2 by FUNCT_1:def 3;
i+1 <=len h11 + len (mid(h21,2,len h21 -'1)) by A258,FINSEQ_1:22;
then
A320: i+1-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by
XREAL_1:9;
A321: i+1-len h11=i+1-'len h11 by A39,A259,A311,XREAL_1:233;
then
A322: 0<i+1-'len h11+2-1 by A39,A311;
h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) by A39,A36,A258,A311,
FINSEQ_1:23;
then
A323: h0.(i+1)=h21.(i+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A311,A321,A320
,FINSEQ_6:118;
A324: h0.i=E-max(P) by A39,A255,A257,A311,FINSEQ_1:64;
then
A325: h0.i in Upper_Arc(P) by A1,Th1;
len h1-'len h11=len h11-len h11 by A39,XREAL_0:def 2;
then 0+2-1=len h1-'len h11+2-1;
then
A326: h0.i=g2.(h2.j) by A24,A26,A311,A324,NAT_D:39;
1<=(i-'len h11 +2-'1) by A316,Lm1,NAT_D:42;
then
A327: j in dom h2 by A46,A314,FINSEQ_3:25;
then
A328: h21.j=g2.(h2.j) by FUNCT_1:13;
i in dom h0 by A257,A262,FINSEQ_3:25;
then
A329: h0/.i=h0.i by PARTFUN1:def 6;
i+1 in dom h0 by A258,A260,FINSEQ_3:25;
then
A330: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A331: j+1=i-len h11+2-1+1 by A39,A311,Lm1,XREAL_0:def 2
.=i+1-'len h11+2-'1 by A321,A322,XREAL_0:def 2;
then h0.(i+1)=g2.(h2.(j+1)) by A323,A318,FUNCT_1:13;
then
A332: h0.(i+1) in Lower_Arc(P) by A23,A131,A29,A319,BORSUK_1:40,FUNCT_1:def 3
;
i-len h11<len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A262,
XREAL_1:9;
then i-len h11+2<len h2-2+2 by XREAL_1:6;
then j+1<len h2 by A39,A311,A317,XREAL_0:def 2;
then h0/.(i+1) <> W-min P by A46,A34,A35,A32,A323,A331,A318,A330;
hence thesis by A46,A32,A323,A331,A327,A328,A326,A318,A329,A330,A332
,A325,A315,JORDAN6:def 10;
end;
end;
end;
then
A333: LE h0/.1,h0/.(1+1),P & h0/.1<>h0/.(1+1) by A205;
A334: E-max P in Upper_Arc P by A1,Th1;
thus for i being Nat,W being Subset of Euclid 2 st 1<=i & i<len
h0 & W=Segment(h0/.i,h0/.(i+1),P) holds diameter(W)<e
proof
let i be Nat,W be Subset of Euclid 2;
assume that
A335: 1<=i and
A336: i<len h0 and
A337: W=Segment(h0/.i,h0/.(i+1),P);
A338: i+1<=len h0 by A336,NAT_1:13;
A339: i<i+1 by NAT_1:13;
A340: 1<i+1 by A335,NAT_1:13;
per cases by XXREAL_0:1;
suppose
A341: i<len h1;
then
A342: i in dom h1 by A335,FINSEQ_3:25;
then
A343: h1.i in rng h1 by FUNCT_1:def 3;
then
A344: h1.i<=1 by A11,BORSUK_1:40,XXREAL_1:1;
A345: 0<=h1.i by A11,A343,BORSUK_1:40,XXREAL_1:1;
A346: h1/.i=h1.i by A335,A341,FINSEQ_4:15;
A347: h11.i=g1.(h1.i) by A342,FUNCT_1:13;
then
A348: h0.i=g1.(h1.i) by A39,A335,A341,FINSEQ_1:64;
then
A349: h0.i in Upper_Arc(P) by A5,A132,A11,A343,BORSUK_1:40,FUNCT_1:def 3;
i in dom h0 by A335,A336,FINSEQ_3:25;
then
A350: h0/.i=h0.i by PARTFUN1:def 6;
i+1 in dom h0 by A338,A340,FINSEQ_3:25;
then
A351: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A352: i+1<=len h1 by A341,NAT_1:13;
then
A353: i+1 in dom h1 by A340,FINSEQ_3:25;
then
A354: h1.i<h1.(i+1) by A12,A339,A342,SEQM_3:def 1;
A355: h1/.(i+1)=h1.(i+1) by A340,A352,FINSEQ_4:15;
A356: h1.(i+1) in rng h1 by A353,FUNCT_1:def 3;
then reconsider Q1=[.h1/.i,h1/.(i+1).] as Subset of I[01] by A11,A343
,A346,A355,BORSUK_1:40,XXREAL_2:def 12;
A357: h0.i=h11.i by A39,A335,A341,FINSEQ_1:64;
A358: h0.(i+1)=h11.(i+1) by A39,A340,A352,FINSEQ_1:64;
then
A359: h0.(i+1)=g1.(h1.(i+1)) by A353,FUNCT_1:13;
then
A360: h0.(i+1) in Upper_Arc(P) by A5,A132,A11,A356,BORSUK_1:40,FUNCT_1:def 3;
A361: Segment(h0/.i,h0/.(i+1),P) c= g1.:([.h1/.i,h1/.(i+1).])
proof
let x be object;
A362: h0/.(i+1) <> W-min P by A38,A17,A71,A20,A340,A358,A353,A351;
assume x in Segment(h0/.i,h0/.(i+1),P);
then x in {p: LE h0/.i,p,P & LE p,h0/.(i+1),P} by A362,Def1;
then consider p being Point of TOP-REAL 2 such that
A363: p=x and
A364: LE h0/.i,p,P and
A365: LE p,h0/.(i+1),P;
A366: h0/.i in Upper_Arc(P) & p in Lower_Arc(P)& not p=W-min(P) or h0
/.i in Upper_Arc(P) & p in Upper_Arc(P) & LE h0/.i,p,Upper_Arc(P),W-min(P),
E-max(P) or h0/.i in Lower_Arc(P) & p in Lower_Arc(P)& not p=W-min(P) & LE h0/.
i,p,Lower_Arc(P),E-max(P),W-min(P) by A364,JORDAN6:def 10;
A367: p in Upper_Arc(P) & h0/.(i+1) in Lower_Arc(P) or p in Upper_Arc(
P) & h0/.(i+1) in Upper_Arc(P) & LE p,h0/.(i+1),Upper_Arc(P),W-min(P),E-max(P)
or p in Lower_Arc(P) & h0/.(i+1) in Lower_Arc(P)& LE p,h0/.(i+1),Lower_Arc(P),
E-max(P),W-min(P) by A365,JORDAN6:def 10;
now
per cases by A352,XXREAL_0:1;
suppose
i+1<len h1;
then
A368: h0/.(i+1) <> E-max P by A38,A45,A255,A20,A358,A353,A351;
A369: now
assume h0/.(i+1) in Lower_Arc(P);
then
h0/.(i+1) in Upper_Arc(P)/\ Lower_Arc(P) by A351,A360,XBOOLE_0:def 4;
then h0/.(i+1) in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A362,A368,TARSKI:def 2;
end;
then
A370: LE p,h0/.(i+1),Upper_Arc(P),W-min(P),E-max(P) by A365,
JORDAN6:def 10;
then
A371: p<>E-max(P) by A3,A368,JORDAN6:55;
A372: p in Upper_Arc(P) by A365,A369,JORDAN6:def 10;
per cases by A335,XXREAL_0:1;
suppose
i>1;
then
A373: h0/.i <> W-min P by A38,A17,A71,A20,A342,A347,A348,A350;
A374: h11.i <> E-max(P) by A38,A45,A255,A20,A341,A342;
now
assume h0/.i in Lower_Arc(P);
then h0/.i in Upper_Arc(P)/\ Lower_Arc(P) by A350,A349,
XBOOLE_0:def 4;
then h0/.i in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A357,A350,A373,A374,TARSKI:def 2;
end;
then
A375: h0/.i in Upper_Arc(P) & p in Lower_Arc(P) & not p=W-min(P)
or h0/.i in Upper_Arc(P) & p in Upper_Arc(P) & LE h0/.i,p,Upper_Arc(P),W-min(P)
,E-max(P) by A364,JORDAN6:def 10;
then
A376: p <> W-min P by A3,A373,JORDAN6:54;
A377: now
assume p in Lower_Arc(P);
then p in Upper_Arc(P)/\ Lower_Arc(P) by A372,XBOOLE_0:def 4;
then p in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A371,A376,TARSKI:def 2;
end;
then consider z being object such that
A378: z in dom g1 and
A379: p=g1.z by A5,A375,FUNCT_1:def 3;
reconsider rz=z as Real by A132,A378;
A380: rz<=1 by A378,BORSUK_1:40,XXREAL_1:1;
h1.(i+1) in rng h1 by A353,FUNCT_1:def 3;
then
A381: 0<=h1/.(i+1) & h1/.(i+1)<=1 by A11,A355,BORSUK_1:40,XXREAL_1:1;
reconsider z as set by TARSKI:1;
take z;
thus z in dom g1 by A378;
A382: g1.(h1/.i)=h0/.i & h1/.i<=1 by A335,A341,A357,A347,A344,A350,
FINSEQ_4:15;
g1.(h1/.(i+1))=h0/.(i+1) by A340,A352,A359,A351,FINSEQ_4:15;
then
A383: rz<=h1/.(i+1) by A4,A5,A6,A7,A370,A379,A381,A380,Th19;
0<=rz by A378,BORSUK_1:40,XXREAL_1:1;
then h1/.i<=rz by A4,A5,A6,A7,A375,A377,A379,A382,A380,Th19;
hence z in [.h1/.i,h1/.(i+1).] by A383,XXREAL_1:1;
thus x = g1.z by A363,A379;
end;
suppose
A384: i=1;
now
per cases;
case
A385: p<>W-min(P);
now
assume p in Lower_Arc(P);
then p in Upper_Arc(P)/\ Lower_Arc(P) by A372,
XBOOLE_0:def 4;
then p in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A371,A385,TARSKI:def 2;
end;
then consider z being object such that
A386: z in dom g1 and
A387: p=g1.z by A5,A366,FUNCT_1:def 3;
reconsider rz=z as Real by A132,A386;
A388: h1/.1<=rz by A8,A346,A384,A386,BORSUK_1:40,XXREAL_1:1;
h1.(1+1) in rng h1 by A353,A384,FUNCT_1:def 3;
then
A389: 0<=h1/.(1+1) & h1/.(1+1)<=1 by A11,A355,A384,BORSUK_1:40
,XXREAL_1:1;
A390: g1.(h1/.(1+1))=h0/.(1+1) by A352,A359,A351,A384,FINSEQ_4:15;
take rz;
rz<=1 by A386,BORSUK_1:40,XXREAL_1:1;
then rz<=h1/.(1+1) by A4,A5,A6,A7,A370,A384,A387,A390,A389
,Th19;
hence
rz in dom g1 & rz in [.h1/.1,h1/.(1+1).] & x = g1.rz by A363
,A386,A387,A388,XXREAL_1:1;
end;
case
A391: p=W-min(P);
thus 0 in [.h1/.1,h1/.(1+1).] by A8,A354,A346,A355,A384,
XXREAL_1:1;
thus x = g1.0 by A6,A363,A391;
end;
end;
hence ex y being set
st y in dom g1 & y in [.h1/.i,h1/.(i+1).] &
x = g1.y by A44,A384;
end;
end;
suppose
A392: i+1=len h1;
then
A393: h0/.(i+1)=E-max(P) by A7,A9,A45,A358,A351,FUNCT_1:13;
A394: now
assume that
A395: p in Lower_Arc(P) and
A396: not p in Upper_Arc(P);
LE h0/.(i+1),p,Lower_Arc(P),E-max(P),W-min(P) by A21,A393,A395,
JORDAN5C:10;
hence contradiction by A334,A21,A367,A393,A396,JORDAN5C:12;
end;
p in Upper_Arc(P) or p in Lower_Arc(P) by A364,JORDAN6:def 10;
then
A397: LE p,h0/.(i+1),Upper_Arc(P),W-min(P),E-max(P) by A3,A393,A394,
JORDAN5C:10;
per cases;
suppose
A398: p<>E-max(P);
now
per cases;
case
A399: p<>W-min(P);
A400: now
assume p in Lower_Arc(P);
then p in Upper_Arc(P)/\ Lower_Arc(P) by A394,
XBOOLE_0:def 4;
then p in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A398,A399,TARSKI:def 2;
end;
then consider z being object such that
A401: z in dom g1 and
A402: p=g1.z by A5,A366,FUNCT_1:def 3;
reconsider rz=z as Real by A132,A401;
A403: rz<=1 by A401,BORSUK_1:40,XXREAL_1:1;
h1.(i+1) in rng h1 by A353,FUNCT_1:def 3;
then
A404: 0<=h1/.(i+1) & h1/.(i+1)<=1 by A11,A355,BORSUK_1:40
,XXREAL_1:1;
g1.(h1/.(i+1))=h0/.(i+1) by A340,A352,A359,A351,FINSEQ_4:15;
then
A405: rz<=h1/.(i+1) by A4,A5,A6,A7,A397,A402,A404,A403,Th19;
take rz;
0<=rz by A401,BORSUK_1:40,XXREAL_1:1;
then h1/.i<=rz by A4,A5,A6,A7,A357,A347,A344,A350,A346,A366
,A400,A402,A403,Th19;
hence
rz in dom g1 & rz in [.h1/.i,h1/.(i+1).] & x = g1.rz by A363
,A401,A402,A405,XXREAL_1:1;
end;
case
A406: p=W-min(P);
then h11.i=W-min(P) by A3,A357,A350,A366,JORDAN6:54;
then i=1 by A38,A17,A71,A20,A342;
then 0 in [.h1/.i,h1/.(i+1).] by A8,A354,A346,A355,XXREAL_1:1
;
hence ex y being set st y in dom g1 & y in [.h1/.i,h1/.(i+1)
.] & x = g1.y by A6,A44,A363,A406;
end;
end;
hence ex y being set st y in dom g1 & y in [.h1/.i,h1/.(i+1).] &
x = g1.y;
end;
suppose
A407: p=E-max(P);
1 in [.h1/.i,h1/.(i+1).] by A9,A354,A346,A355,A392,XXREAL_1:1;
hence ex y being set
st y in dom g1 & y in [.h1/.i,h1/.(i+1).] &
x = g1.y by A7,A69,A363,A407,Lm6;
end;
end;
end;
hence thesis by FUNCT_1:def 6;
end;
A408: h1.(i+1)<=1 by A11,A356,BORSUK_1:40,XXREAL_1:1;
g1.:([.h1/.i,h1/.(i+1).]) c= Segment(h0/.i,h0/.(i+1),P)
proof
A409: g1.(h1/.i)=h0/.i & 0<=h1/.i by A335,A341,A348,A345,A350,FINSEQ_4:15;
let x be object;
assume x in g1.:([.h1/.i,h1/.(i+1).]);
then consider y being object such that
A410: y in dom g1 and
A411: y in [.h1/.i,h1/.(i+1).] and
A412: x=g1.y by FUNCT_1:def 6;
reconsider sy=y as Real by A411;
A413: sy<=1 by A410,BORSUK_1:40,XXREAL_1:1;
A414: x in Upper_Arc(P) by A5,A410,A412,FUNCT_1:def 3;
then reconsider p1=x as Point of TOP-REAL 2;
A415: h1/.i<=1 by A335,A341,A344,FINSEQ_4:15;
h1/.i<=sy by A411,XXREAL_1:1;
then LE h0/.i,p1,Upper_Arc(P),W-min(P),E-max(P) by A3,A4,A5,A6,A7,A412
,A409,A415,A413,Th18;
then
A416: LE h0/.i,p1,P by A350,A349,A414,JORDAN6:def 10;
sy<=h1/.(i+1) & 0<=sy by A410,A411,BORSUK_1:40,XXREAL_1:1;
then LE p1,h0/.(i+1),Upper_Arc(P),W-min(P),E-max(P) by A3,A4,A5,A6,A7
,A359,A408,A351,A355,A412,A413,Th18;
then LE p1,h0/.(i+1),P by A351,A360,A414,JORDAN6:def 10;
then
A417: x in {p: LE h0/.i,p,P & LE p,h0/.(i+1),P} by A416;
not h0/.(i+1)=W-min P by A38,A17,A71,A20,A340,A358,A353,A351;
hence thesis by A417,Def1;
end;
then W=g1.:Q1 by A337,A361;
hence thesis by A13,A335,A341;
end;
suppose
A418: i>len h1;
i-len h11<len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A336,
XREAL_1:9;
then
A419: i-len h11+2<len h2-2+2 by XREAL_1:6;
A420: len h1 +1<=i by A418,NAT_1:13;
then
A421: len h1 +1-len h1<=i-len h1 by XREAL_1:9;
A422: i-len h11=i-'len h11 by A39,A418,XREAL_1:233;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A336,
XREAL_1:9;
then
A423: i-'len h11+2<=len h2-2+2 by A39,A422,XREAL_1:6;
i+1>len h11 by A39,A418,NAT_1:13;
then
A424: i+1-len h11=i+1-'len h11 by XREAL_1:233;
i+1 in dom h0 by A338,A340,FINSEQ_3:25;
then
A425: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A426: i<=len h11 + len (mid(h21,2,len h21 -'1)) by A336,FINSEQ_1:22;
len h11+1<=i by A39,A418,NAT_1:13;
then
A427: h0.i=(mid(h21,2,len h21 -'1)).(i -len h11) by A426,FINSEQ_1:23;
A428: i+1-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A338,
XREAL_1:9;
i in dom h0 by A335,A336,FINSEQ_3:25;
then
A429: h0/.i=h0.i by PARTFUN1:def 6;
set j=i-'len h11+2-'1;
len h2<len h2+1 by NAT_1:13;
then
A430: len h2-1<len h2+1-1 by XREAL_1:9;
A431: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A432: j+1=i-'len h11+1+1-1+1 by Lm1,NAT_D:39,42
.=i-'len h11+(1+1);
A433: 1<=(i-'len h11 +2-'1) by A431,Lm1,NAT_D:42;
then
A434: 1<j+1 by NAT_1:13;
then
A435: j+1 in dom h2 by A423,A432,FINSEQ_3:25;
then
A436: h2.(j+1) in rng h2 by FUNCT_1:def 3;
then
A437: h2.(j+1) <= 1 by A29,BORSUK_1:40,XXREAL_1:1;
A438: h2/.(j+1)=h2.(j+1) by A423,A432,A434,FINSEQ_4:15;
(i-'len h11 +2-'1)<=len h21 by A47,A423,NAT_D:44;
then
A439: j in dom h2 by A46,A433,FINSEQ_3:25;
i-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A336,
XREAL_1:9;
then
A440: h0.i=h21.(i-'len h11 +2-'1) by A39,A48,A56,A50,A54,A427,A422,A421,
FINSEQ_6:118;
then
A441: h0.i=g2.(h2.j) by A439,FUNCT_1:13;
A442: h2.j in rng h2 by A439,FUNCT_1:def 3;
then g2.(h2.j) in rng g2 by A131,A29,BORSUK_1:40,FUNCT_1:def 3;
then
A443: h0.i in Lower_Arc(P) by A23,A440,A439,FUNCT_1:13;
i-'len h11+2-1<=len h2-1 by A423,XREAL_1:9;
then i-'len h11+2-1<len h2 by A430,XXREAL_0:2;
then
A444: j< len h2 by A431,Lm1,NAT_D:39,42;
then
A445: h2/.j=h2.j by A431,Lm1,FINSEQ_4:15,NAT_D:42;
then reconsider
Q1=[.h2/.j,h2/.(j+1).] as Subset of I[01] by A29,A442,A436,A438,
BORSUK_1:40,XXREAL_2:def 12;
A446: 0 <= h2.j & h2.j <= 1 by A29,A442,BORSUK_1:40,XXREAL_1:1;
A447: i-'len h11+2-'1=i-'len h11+2-1 by A431,Lm1,NAT_D:39,42;
A448: len h1 +1<=i+1 by A420,NAT_1:13;
then
A449: len h1 +1-len h1<=i+1-len h1 by XREAL_1:9;
then 1<i+1-'len h11+(2-1) by A39,A424,NAT_1:13;
then 0<i+1-'len h11+2-1;
then
A450: j+1 =i+1-'len h11+2-'1 by A422,A424,A447,XREAL_0:def 2;
h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) by A39,A36,A338,A448,
FINSEQ_1:23;
then
A451: h0.(i+1)=h21.(i+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A424,A428,A449,
FINSEQ_6:118;
then h0.(i+1)=g2.(h2.(j+1)) by A450,A435,FUNCT_1:13;
then
A452: h0.(i+1) in Lower_Arc(P) by A23,A131,A29,A436,BORSUK_1:40,FUNCT_1:def 3;
len h1 +1<=i by A418,NAT_1:13;
then len h11+1-len h11<=i-len h11 by A39,XREAL_1:9;
then 1<=i-'len h11 by NAT_D:39;
then 1+2<=i-'len h11+2 by XREAL_1:6;
then 1+2-1<=i-'len h11+2-1 by XREAL_1:9;
then
A453: 1<j by A447,XXREAL_0:2;
A454: Segment(h0/.i,h0/.(i+1),P) c= g2.:([.h2/.j,h2/.(j+1).])
proof
h2.(j+1) in rng h2 by A435,FUNCT_1:def 3;
then
A455: 0<=h2/.(j+1) & h2/.(j+1)<=1 by A29,A438,BORSUK_1:40,XXREAL_1:1;
A456: g2.(h2/.(j+1))=h0/.(i+1) by A451,A450,A435,A425,A438,FUNCT_1:13;
j<len h2 by A423,A432,NAT_1:13;
then
A457: h0/.i <> W-min(P) by A46,A34,A35,A32,A440,A439,A429;
A458: h2/.j<=1 by A29,A442,A445,BORSUK_1:40,XXREAL_1:1;
let x be object;
assume
A459: x in Segment(h0/.i,h0/.(i+1),P);
h0/.(i+1) <> W-min P by A46,A34,A35,A32,A422,A451,A419,A432,A450,A435
,A425;
then x in {p: LE h0/.i,p,P & LE p,h0/.(i+1),P} by A459,Def1;
then consider p being Point of TOP-REAL 2 such that
A460: p=x and
A461: LE h0/.i,p,P and
A462: LE p,h0/.(i+1),P;
A463: h0/.i in Upper_Arc(P) & p in Lower_Arc(P)& not p=W-min(P) or h0
/.i in Upper_Arc(P) & p in Upper_Arc(P) & LE h0/.i,p,Upper_Arc(P),W-min(P),
E-max(P) or h0/.i in Lower_Arc(P) & p in Lower_Arc(P)& not p=W-min(P) & LE h0/.
i,p,Lower_Arc(P),E-max(P),W-min(P) by A461,JORDAN6:def 10;
A464: h21.j <> E-max P by A46,A76,A77,A32,A453,A439;
A465: now
assume h0/.i in Upper_Arc(P);
then h0/.i in Upper_Arc(P)/\ Lower_Arc(P) by A429,A443,XBOOLE_0:def 4
;
then h0/.i in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A440,A429,A457,A464,TARSKI:def 2;
end;
then
A466: LE h0/.i,p,Lower_Arc P, E-max P, W-min P by A461,JORDAN6:def 10;
A467: h0/.i <> E-max(P) by A46,A76,A77,A32,A440,A453,A439,A429;
A468: now
assume p in Upper_Arc(P);
then p in Upper_Arc P /\ Lower_Arc P by A463,A465,XBOOLE_0:def 4;
then p in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
then p=W-min(P) or p=E-max(P) by TARSKI:def 2;
hence contradiction by A21,A463,A465,A467,JORDAN6:54;
end;
A469: h0/.(i+1) <> E-max P & h21.(j+1) <> W-min P by A46,A76,A34,A77,A35,A32
,A422,A451,A419,A432,A450,A434,A435,A425;
now
assume h0/.(i+1) in Upper_Arc(P);
then h0/.(i+1) in Upper_Arc(P)/\ Lower_Arc(P) by A425,A452,
XBOOLE_0:def 4;
then h0/.(i+1) in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
hence contradiction by A451,A450,A425,A469,TARSKI:def 2;
end;
then p in Lower_Arc(P) & h0/.(i+1) in Lower_Arc(P)& not h0/.(i+ 1)=
W-min(P) & LE p,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) or p in Upper_Arc(P) &
h0/.(i+1) in Lower_Arc(P) & not h0/.(i+1)=W-min(P) by A462,JORDAN6:def 10;
then consider z being object such that
A470: z in dom g2 and
A471: p=g2.z by A23,A468,FUNCT_1:def 3;
reconsider rz=z as Real by A131,A470;
A472: rz<=1 by A470,BORSUK_1:40,XXREAL_1:1;
LE p,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) by A462,A468,
JORDAN6:def 10;
then
A473: rz<=h2/.(j+1) by A22,A23,A24,A25,A471,A456,A472,A455,Th19;
0<=rz by A470,BORSUK_1:40,XXREAL_1:1;
then h2/.j<=rz by A22,A23,A24,A25,A441,A429,A445,A466,A471,A458,A472
,Th19;
then rz in [.h2/.j,h2/.(j+1).] by A473,XXREAL_1:1;
hence thesis by A460,A470,A471,FUNCT_1:def 6;
end;
A474: g2.(h2.(j+1)) = h0/.(i+1) by A451,A450,A435,A425,FUNCT_1:13;
g2.:([.h2/.j,h2/.(j+1).]) c= Segment(h0/.i,h0/.(i+1),P)
proof
let x be object;
assume x in g2.:([.h2/.j,h2/.(j+1).]);
then consider y being object such that
A475: y in dom g2 and
A476: y in [.h2/.j,h2/.(j+1).] and
A477: x=g2.y by FUNCT_1:def 6;
reconsider sy=y as Real by A476;
A478: sy<=h2/.(j+1) by A476,XXREAL_1:1;
A479: x in Lower_Arc(P) by A23,A475,A477,FUNCT_1:def 3;
then reconsider p1=x as Point of TOP-REAL 2;
A480: h2.(j+1) <> 1 by A27,A30,A34,A422,A419,A432,A435,FUNCT_1:def 4;
A481: now
assume p1=W-min(P);
then 1=sy by A22,A25,A227,A475,A477;
hence contradiction by A437,A438,A478,A480,XXREAL_0:1;
end;
A482: sy<=1 by A475,BORSUK_1:40,XXREAL_1:1;
h2/.j<=sy by A476,XXREAL_1:1;
then LE h0/.i,p1,Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23,A24,A25
,A441,A429,A446,A445,A477,A482,Th18;
then
A483: LE h0/.i,p1,P by A429,A443,A479,A481,JORDAN6:def 10;
A484: h0/.(i+1) <> W-min P by A46,A34,A35,A32,A422,A451,A419,A432,A450,A435
,A425;
0<=sy by A475,BORSUK_1:40,XXREAL_1:1;
then LE p1,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23,A24
,A25,A474,A437,A438,A477,A478,A482,Th18;
then LE p1,h0/.(i+1),P by A425,A452,A479,A484,JORDAN6:def 10;
then x in {p: LE h0/.i,p,P & LE p,h0/.(i+1),P} by A483;
hence thesis by A484,Def1;
end;
then W=g2.:(Q1) by A337,A454;
hence thesis by A31,A433,A444;
end;
suppose
A485: i=len h1;
A486: i+1-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A338,
XREAL_1:9;
then 1<=i+1-'len h11 by A39,A339,A485,XREAL_1:233;
then 1<i+1-'len h11+(2-1) by NAT_1:13;
then
A487: 0<i+1-'len h11+2-1;
i in dom h0 by A335,A336,FINSEQ_3:25;
then
A488: h0/.i=h0.i by PARTFUN1:def 6;
A489: h0.i=E-max(P) by A39,A255,A335,A485,FINSEQ_1:64;
set j=i-'len h11+2-'1;
A490: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A491: j+1=i-'len h11+1+1-1+1 by Lm1,NAT_D:39,42
.=i-'len h11+(1+1);
A492: len h1-'len h11=len h11-len h11 by A39,XREAL_0:def 2;
then
A493: 0+2-1=len h1-'len h11+2-1;
then
A494: h0.i=g2.(h2.j) by A24,A26,A485,A489,NAT_D:39;
A495: i-len h11=i-'len h11 by A39,A485,XREAL_1:233;
i-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A336,
XREAL_1:9;
then
A496: i-'len h11+2<=len h2-2+2 by A39,A495,XREAL_1:6;
i-len h11<len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A336,
XREAL_1:9;
then
A497: i-len h11+2<len h2-2+2 by XREAL_1:6;
then
A498: j+1<len h2 by A39,A485,A491,XREAL_0:def 2;
A499: h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) by A39,A36,A338,A485,
FINSEQ_1:23;
len h2<len h2+1 by NAT_1:13;
then
A500: len h2-1<len h2+1-1 by XREAL_1:9;
i+1 in dom h0 by A338,A340,FINSEQ_3:25;
then
A501: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A502: 1<=(i-'len h11 +2-'1) by A490,Lm1,NAT_D:42;
then
A503: 1<j+1 by NAT_1:13;
then
A504: j+1 in dom h2 by A496,A491,FINSEQ_3:25;
then
A505: h2.(j+1) in rng h2 by FUNCT_1:def 3;
then
A506: h2.(j+1) <= 1 by A29,BORSUK_1:40,XXREAL_1:1;
(i-'len h11 +2-'1)<=len h21 by A47,A496,NAT_D:44;
then
A507: j in dom h2 by A46,A502,FINSEQ_3:25;
then
A508: h2.j in rng h2 by FUNCT_1:def 3;
then g2.(h2.j) in rng g2 by A131,A29,BORSUK_1:40,FUNCT_1:def 3;
then
A509: h0.i in Lower_Arc P by A23,A24,A26,A485,A489,A493,NAT_D:39;
A510: h2/.(j+1)=h2.(j+1) by A496,A491,A503,FINSEQ_4:15;
i-'len h11+2-1<=len h2-1 by A496,XREAL_1:9;
then
A511: i-'len h11+2-1<len h2 by A500,XXREAL_0:2;
then
A512: j<len h2 by A490,Lm1,NAT_D:39,42;
then h2/.j=h2.j by A490,Lm1,FINSEQ_4:15,NAT_D:42;
then reconsider
Q1=[.h2/.j,h2/.(j+1).] as Subset of I[01] by A29,A508,A505,A510,
BORSUK_1:40,XXREAL_2:def 12;
A513: i+1-len h11=i+1-'len h11 by A39,A339,A485,XREAL_1:233;
j+1=i-len h11+2-1+1 by A39,A485,Lm1,XREAL_0:def 2
.=i+1-'len h11+2-'1 by A513,A487,XREAL_0:def 2;
then
A514: h0
.(i+1)=h21.(j+1) by A39,A48,A56,A50,A54,A485,A499,A513,A486,FINSEQ_6:118;
then
A515: h0.(i+1)=g2.(h2.(j+1)) by A504,FUNCT_1:13;
then
A516: h0.(i+1) in Lower_Arc P by A23,A131,A29,A505,BORSUK_1:40,FUNCT_1:def 3;
A517: h21.j=g2.(h2.j) by A507,FUNCT_1:13;
A518: Segment(h0/.i,h0/.(i+1),P) c= g2.:([.h2/.j,h2/.(j+1).])
proof
j+1<len h2 by A39,A485,A497,A491,XREAL_0:def 2;
then j<len h2 by NAT_1:13;
then
A519: h0/.i<>W-min(P) by A46,A34,A35,A32,A507,A517,A494,A488;
A520: g2.(h2/.(j+1))=h0/.(i+1) by A496,A491,A503,A515,A501,FINSEQ_4:15;
h2.(j+1) in rng h2 by A504,FUNCT_1:def 3;
then
A521: 0<=h2/.(j+1) & h2/.(j+1)<=1 by A29,A510,BORSUK_1:40,XXREAL_1:1;
A522: h0/.(i+1) in Lower_Arc(P) by A23,A131,A29,A515,A505,A501,BORSUK_1:40
,FUNCT_1:def 3;
let x be object;
assume
A523: x in Segment(h0/.i,h0/.(i+1),P);
h0/.(i+1) <> W-min P by A46,A34,A35,A32,A498,A514,A504,A501;
then x in {p: LE h0/.i,p,P & LE p,h0/.(i+1),P} by A523,Def1;
then consider p being Point of TOP-REAL 2 such that
A524: p=x and
A525: LE h0/.i,p,P and
A526: LE p,h0/.(i+1),P;
A527: h0/.i in Upper_Arc(P) & p in Lower_Arc(P)& not p=W-min(P) or h0
/.i in Upper_Arc(P) & p in Upper_Arc(P) & LE h0/.i,p,Upper_Arc(P),W-min(P),
E-max(P) or h0/.i in Lower_Arc(P) & p in Lower_Arc(P)& not p=W-min(P) & LE h0/.
i,p,Lower_Arc(P),E-max(P),W-min(P) by A525,JORDAN6:def 10;
dom (g1*h1) c= dom h0 by FINSEQ_1:26;
then
A528: h0/.i=E-max(P) by A254,A485,A489,PARTFUN1:def 6;
A529: now
assume
A530: not p in Lower_Arc(P);
then p=E-max(P) by A3,A527,A528,JORDAN6:55;
hence contradiction by A1,A530,Th1;
end;
A531: now
assume p in Upper_Arc(P);
then p in Upper_Arc(P)/\ Lower_Arc(P) by A529,XBOOLE_0:def 4;
then
A532: p in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
p <> W-min P by A3,A527,A519,JORDAN6:54;
hence p = E-max P by A532,TARSKI:def 2;
end;
then p in rng g2 by A1,A23,A525,Th1,JORDAN6:def 10;
then consider z being object such that
A533: z in dom g2 and
A534: p=g2.z by FUNCT_1:def 3;
reconsider rz=z as Real by A131,A533;
0<=rz by A533,BORSUK_1:40,XXREAL_1:1;
then
A535: h2/.j<=rz by A26,A485,A511,A492,Lm1,FINSEQ_4:15;
A536: not h0/.(i+1)=E-max(P) by A46,A76,A77,A32,A503,A514,A504,A501;
now
assume h0/.(i+1) in Upper_Arc(P);
then h0/.(i+1) in Upper_Arc(P)/\ Lower_Arc(P) by A522,XBOOLE_0:def 4;
then h0/.(i+1) in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
then h21.(j+1)=W-min(P) by A514,A501,A536,TARSKI:def 2;
hence contradiction by A46,A34,A35,A32,A498,A504;
end;
then p in Lower_Arc(P) & h0/.(i+1) in Lower_Arc(P) & not h0/.(i +1)=
W-min(P) & LE p,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) or p in Upper_Arc(P) &
h0/.(i+1) in Lower_Arc(P) & not h0/.(i+1)=W-min(P) by A526,JORDAN6:def 10;
then
A537: LE p,h0/.(i+1),Lower_Arc P, E-max P, W-min P by A21,A531,JORDAN5C:10;
rz<=1 by A533,BORSUK_1:40,XXREAL_1:1;
then rz<=h2/.(j+1) by A22,A23,A24,A25,A537,A534,A520,A521,Th19;
then rz in [.h2/.j,h2/.(j+1).] by A535,XXREAL_1:1;
hence thesis by A524,A533,A534,FUNCT_1:def 6;
end;
A538: g2.(h2.(j+1)) = h0/.(i+1) by A514,A504,A501,FUNCT_1:13;
g2.:([.h2/.j,h2/.(j+1).]) c= Segment(h0/.i,h0/.(i+1),P)
proof
let x be object;
assume x in g2.:([.h2/.j,h2/.(j+1).]);
then consider y being object such that
A539: y in dom g2 and
A540: y in [.h2/.j,h2/.(j+1).] and
A541: x=g2.y by FUNCT_1:def 6;
reconsider sy=y as Real by A540;
A542: sy<=h2/.(j+1) by A540,XXREAL_1:1;
A543: x in Lower_Arc(P) by A23,A539,A541,FUNCT_1:def 3;
then reconsider p1=x as Point of TOP-REAL 2;
A544: h2.(j+1) <> 1 by A27,A30,A34,A498,A504,FUNCT_1:def 4;
A545: now
assume p1=W-min(P);
then 1=sy by A22,A25,A227,A539,A541;
hence contradiction by A506,A510,A542,A544,XXREAL_0:1;
end;
A546: 0<=sy & sy<=1 by A539,BORSUK_1:40,XXREAL_1:1;
then LE h0/.i,p1,Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23,A24,A25
,A489,A488,A541,Th18;
then
A547: LE h0/.i,p1,P by A488,A509,A543,A545,JORDAN6:def 10;
A548: h0/.(i+1) <> W-min(P) by A46,A34,A35,A32,A498,A514,A504,A501;
LE p1,h0/.(i+1),Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23,A24,A25
,A538,A506,A510,A541,A542,A546,Th18;
then LE p1,h0/.(i+1),P by A501,A516,A543,A548,JORDAN6:def 10;
then x in {p: LE h0/.i,p,P & LE p,h0/.(i+1),P} by A547;
hence thesis by A548,Def1;
end;
then W=g2.:(Q1) by A337,A518;
hence thesis by A31,A502,A512;
end;
end;
A549: len h0=len h1+(len h2-2) by A38,A47,A36,A52,A55,A57,FINSEQ_3:29;
thus for W being Subset of Euclid 2 st W=Segment(h0/.len h0,h0/.1,P) holds
diameter W < e
proof
set i=len h0;
let W be Subset of Euclid 2;
set j=i-'len h11+2-'1;
A550: 0+2<=i-'len h11 +2 by XREAL_1:6;
then
A551: 1<=(i-'len h11 +2-'1) by Lm1,NAT_D:42;
len h11+1-len h11<=i-len h11 by A47,A36,A52,A55,A57,A62,XXREAL_0:2;
then 1<=i-'len h11 by NAT_D:39;
then 1+2<=i-'len h11+2 by XREAL_1:6;
then
A552: 1+2-1<=i-'len h11+2-1 by XREAL_1:9;
len h0<=len h11 + len mid(h21,2,len h21 -'1) by FINSEQ_1:22;
then
A553: h0.i=(mid(h21,2,len h21 -'1)).(i -len h11) by A39,A64,FINSEQ_1:23;
A554: i-len h11=i-'len h11 by A39,A65,XREAL_1:233;
then (i-'len h11 +2-'1)<=len h21 by A36,A52,A55,A57,NAT_D:44;
then
A555: j in dom h2 by A46,A551,FINSEQ_3:25;
then
A556: h2.j in rng h2 by FUNCT_1:def 3;
i-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 & len h1 +
1-len h1 <=i-len h1 by A549,A62,FINSEQ_1:22,XXREAL_0:2;
then
A557: h0.i=h21.(i-'len h11 +2-'1) by A39,A48,A56,A50,A54,A553,A554,FINSEQ_6:118
;
then
A558: h0.i=g2.(h2.j) by A555,FUNCT_1:13;
then
A559: h0.i in Lower_Arc P by A23,A131,A29,A556,BORSUK_1:40,FUNCT_1:def 3;
A560: 2-'1<=(i-'len h11 +2-'1) by A550,NAT_D:42;
then
A561: 1<j+1 by Lm1,NAT_1:13;
then
A562: h2/.(j+1)=h2.(j+1) by A47,A36,A52,A55,A57,A554,FINSEQ_4:15;
len h2<len h2+1 by NAT_1:13;
then
A563: len h2-1<len h2+1-1 by XREAL_1:9;
then
A564: h2/.j=h2.j by A47,A36,A52,A55,A57,A554,A560,Lm1,FINSEQ_4:15;
j+1 in dom h2 by A46,A36,A52,A55,A57,A554,A561,FINSEQ_3:25;
then h2.(j+1) in rng h2 by FUNCT_1:def 3;
then reconsider Q1=[.h2/.j,h2/.(j+1).] as Subset of I[01] by A29,A556,A564
,A562,BORSUK_1:40,XXREAL_2:def 12;
i in dom h0 by A66,FINSEQ_3:25;
then
A565: h0/.i=h0.i by PARTFUN1:def 6;
i-'len h11+2-'1=i-'len h11+2-1 by A550,Lm1,NAT_D:39,42;
then
A566: 1<j by A552,XXREAL_0:2;
A567: now
assume h0.i in Upper_Arc(P);
then h0.i in Upper_Arc(P)/\ Lower_Arc(P) by A559,XBOOLE_0:def 4;
then h0.i in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
then h0.i=W-min(P) or h0.i=E-max(P) by TARSKI:def 2;
hence contradiction by A46,A47,A76,A34,A77,A35,A36,A52,A55,A57,A32,A554
,A557,A563,A566,A555;
end;
A568: h2.j<=1 by A29,A556,BORSUK_1:40,XXREAL_1:1;
A569: Segment(h0/.i,h0/.1,P) c= g2.:([.h2/.j,h2/.(j+1).])
proof
let x be object;
assume
A570: x in Segment(h0/.i,h0/.1,P);
h0/.1=W-min(P) by A72,A67,PARTFUN1:def 6;
then
A571: x in {p: LE h0/.i,p,P or h0/.i in P & p=W-min(P)} by A570,Def1;
A572: j+1 in dom h2 by A46,A36,A52,A55,A57,A554,A561,FINSEQ_3:25;
j<j+1 & j in dom h2 by A47,A36,A52,A55,A57,A554,A560,A563,Lm1,FINSEQ_3:25
;
then h2.j<h2.(j+1) by A30,A572,SEQM_3:def 1;
then
A573: h2/.(j+1) in [.h2/.j, h2/.(j+1).] by A564,A562,XXREAL_1:1;
consider p being Point of TOP-REAL 2 such that
A574: p=x and
A575: LE h0/.i,p,P or h0/.i in P & p=W-min(P) by A571;
A576: h2/.(j+1)=1 by A27,A47,A34,A36,A52,A55,A57,A554,PARTFUN1:def 6;
now
per cases by A575;
suppose
A577: LE h0/.i,p,P & (p<>W-min(P) or not h0/.i in P);
then p in Lower_Arc(P) by A567,A565,JORDAN6:def 10;
then consider z being object such that
A578: z in dom g2 and
A579: p=g2.z by A23,FUNCT_1:def 3;
take z;
thus z in dom g2 by A578;
reconsider rz=z as Real by A131,A578;
A580: rz<=1 by A578,BORSUK_1:40,XXREAL_1:1;
then
A581: rz<=h2/.(j+1) by A27,A47,A36,A52,A55,A57,A554,A561,FINSEQ_4:15;
A582: LE h0/.i,p,Lower_Arc(P),E-max(P),W-min(P) by A567,A565,A577,
JORDAN6:def 10;
0<=rz by A578,BORSUK_1:40,XXREAL_1:1;
then h2/.j<=rz by A22,A23,A24,A25,A558,A568,A565,A564,A582,A579,A580
,Th19;
hence z in [.h2/.j,h2/.(j+1).] & x = g2.z by A574,A579,A581,
XXREAL_1:1;
end;
suppose
h0/.i in P & p=W-min(P);
hence ex y being object
st y in dom g2 & y in [.h2/.j,h2/.(j+1).] & x =
g2.y by A25,A227,A574,A573,A576;
end;
end;
hence thesis by FUNCT_1:def 6;
end;
A583: 0 <= h2.j & h2.j <= 1 by A29,A556,BORSUK_1:40,XXREAL_1:1;
A584: g2.:([.h2/.j,h2/.(j+1).]) c= Segment(h0/.i,h0/.1,P)
proof
A585: Upper_Arc(P) \/ Lower_Arc(P)=P by A1,JORDAN6:def 9;
let x be object;
assume x in g2.:([.h2/.j,h2/.(j+1).]);
then consider y being object such that
A586: y in dom g2 and
A587: y in [.h2/.j,h2/.(j+1).] and
A588: x=g2.y by FUNCT_1:def 6;
reconsider sy=y as Real by A587;
A589: x in Lower_Arc(P) by A23,A586,A588,FUNCT_1:def 3;
then reconsider p1=x as Point of TOP-REAL 2;
h2/.j<=sy & sy<=1 by A586,A587,BORSUK_1:40,XXREAL_1:1;
then
A590: LE h0/.i,p1,Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23,A24,A25,A558
,A565,A583,A564,A588,Th18;
now
per cases;
case
p1=W-min(P);
hence LE h0/.i,p1,P or h0/.i in P & p1=W-min(P) by A559,A565,A585,
XBOOLE_0:def 3;
end;
case
p1<>W-min(P);
hence LE h0/.i,p1,P or h0/.i in P & p1=W-min(P) by A559,A565,A589
,A590,JORDAN6:def 10;
end;
end;
then
A591: x in {p: LE h0/.i,p,P or h0/.i in P & p=W-min(P)};
h0/.1=W-min(P) by A72,A67,PARTFUN1:def 6;
hence thesis by A591,Def1;
end;
assume W=Segment(h0/.len h0,h0/.1,P);
then W=g2.:Q1 by A569,A584;
hence thesis by A31,A47,A36,A52,A55,A57,A554,A560,A563,Lm1;
end;
A592: for i being Nat st 1<=i & i+1<len h0 holds LE h0/.(i+1),h0
/.(i+2),P
proof
let i be Nat;
assume that
A593: 1<=i and
A594: i+1<len h0;
A595: i+1+1<=len h0 by A594,NAT_1:13;
A596: i+1<i+1+1 by NAT_1:13;
A597: 1<i+1 by A593,NAT_1:13;
then
A598: 1<i+1+1 by NAT_1:13;
per cases;
suppose
A599: i+1<len h1;
then
A600: i+1 in dom h1 by A597,FINSEQ_3:25;
then
A601: h1.(i+1) in rng h1 by FUNCT_1:def 3;
then
A602: 0<=h1.(i+1) & h1.(i+1)<=1 by A11,BORSUK_1:40,XXREAL_1:1;
A603: 1<i+1+1 by A597,NAT_1:13;
then (i+1)+1 in dom h0 by A595,FINSEQ_3:25;
then
A604: h0/.(i+1+1)=h0.(i+1+1) by PARTFUN1:def 6;
A605: i+1+1<=len h1 by A599,NAT_1:13;
then
A606: i+1+1 in dom h1 by A603,FINSEQ_3:25;
then
A607: h1.((i+1)+1) in rng h1 by FUNCT_1:def 3;
then
A608: h1.(i+1+1)<=1 by A11,BORSUK_1:40,XXREAL_1:1;
h0.(i+1+1)=h11.(i+1+1) by A39,A605,A603,FINSEQ_1:64;
then
A609: h0.(i+1+1)=g1.(h1.(i+1+1)) by A606,FUNCT_1:13;
then
A610: h0/.(i+1+1) in Upper_Arc P by A5,A132,A11,A607,A604,BORSUK_1:40
,FUNCT_1:def 3;
i+1 in dom h0 by A594,A597,FINSEQ_3:25;
then
A611: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A612: h0.(i+1)=h11.(i+1) by A39,A597,A599,FINSEQ_1:64;
then
A613: h0.(i+1)=g1.(h1.(i+1)) by A600,FUNCT_1:13;
g1.(h1.(i+1)) in rng g1 by A132,A11,A601,BORSUK_1:40,FUNCT_1:def 3;
then
A614: h0/.(i+1) in Upper_Arc P by A5,A612,A600,A611,FUNCT_1:13;
h1.(i+1)<h1.(i+1+1) by A12,A596,A600,A606,SEQM_3:def 1;
then LE h0/.(i+1),h0/.((i+1)+1),Upper_Arc(P),W-min(P),E-max(P) by A3,A4
,A5,A6,A7,A613,A602,A609,A608,A611,A604,Th18;
hence thesis by A614,A610,JORDAN6:def 10;
end;
suppose
A615: i+1>=len h1;
per cases by A615,XXREAL_0:1;
suppose
A616: i+1>len h1;
set j=(i+1)-'len h11+2-'1;
A617: (i+1)+1-len h11 <=len h11 + len (mid(h21,2,len h21 -'1))-len h11
by A36,A595,XREAL_1:9;
A618: 0+2<=(i+1)-'len h11 +2 by XREAL_1:6;
then
A619: 1<=((i+1)-'len h11 +2-'1) by Lm1,NAT_D:42;
A620: j+1=(i+1)-'len h11+(1+1)-1+1 by A618,Lm1,NAT_D:39,42
.=(i+1)-'len h11+2;
A621: len h1 +1<=i+1 by A616,NAT_1:13;
then
A622: len h1 +1-len h1<=(i+1)-len h1 by XREAL_1:9;
A623: (i+1)-len h11=(i+1)-'len h11 by A39,A616,XREAL_1:233;
(i+1)+1>len h11 by A39,A616,NAT_1:13;
then
A624: (i+1)+1-len h11=(i+1)+1-'len h11 by XREAL_1:233;
A625: len h1 +1<=(i+1)+1 by A621,NAT_1:13;
then
A626: len h1 +1-len h1<=(i+1)+1-len h1 by XREAL_1:9;
then 1<(i+1)+1-'len h11+(2-1) by A39,A624,NAT_1:13;
then
A627: 0<(i+1)+1-'len h11+2-1;
(i+1)-'len h11+2-'1=(i+1)-'len h11+2-1 by A618,Lm1,NAT_D:39,42;
then
A628: j+1=(i+1)+1-'len h11+2-'1 by A623,A624,A627,XREAL_0:def 2;
(i+1)-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57
,A594,XREAL_1:9;
then
A629: (i+1)-'len h11+2<=len h2-2+2 by A39,A623,XREAL_1:6;
then ((i+1)-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
then
A630: j in dom h2 by A46,A619,FINSEQ_3:25;
2-'1<=((i+1)-'len h11 +2-'1) by A618,NAT_D:42;
then 1<j+1 by Lm1,NAT_1:13;
then
A631: j+1 in dom h2 by A629,A620,FINSEQ_3:25;
then
A632: h2.(j+1) in rng h2 by FUNCT_1:def 3;
then
A633: (h2.(j+1)) <= 1 by A29,BORSUK_1:40,XXREAL_1:1;
j<j+1 by NAT_1:13;
then
A634: h2.j<h2.(j+1) by A30,A630,A631,SEQM_3:def 1;
A635: i+1<=len h11 + len (mid(h21,2,len h21 -'1)) by A594,FINSEQ_1:22;
len h11+1<=i+1 by A39,A616,NAT_1:13;
then
A636: h0.(i+1)=(mid(h21,2,len h21 -'1)).((i+1) -len h11) by A635,FINSEQ_1:23;
(i+1)-len h11<=len h11 + len (mid(h21,2,len h21 -'1)) -len h11
by A635,XREAL_1:9;
then h0.(i+1)=h21.((i+1)-'len h11 +2-'1) by A39,A48,A56,A50,A54,A636
,A623,A622,FINSEQ_6:118;
then
A637: h0.(i+1)=g2.(h2.j) by A630,FUNCT_1:13;
A638: h2.j in rng h2 by A630,FUNCT_1:def 3;
then
A639: h0.(i+1) in Lower_Arc(P) by A23,A131,A29,A637,BORSUK_1:40,FUNCT_1:def 3
;
(i+1)+1 in dom h0 by A595,A598,FINSEQ_3:25;
then
A640: h0/.((i+1)+1)=h0.((i+1)+1) by PARTFUN1:def 6;
h0.((i+1)+1)=(mid(h21,2,len h21 -'1)).((i+1)+1 -len h11) by A39,A36
,A595,A625,FINSEQ_1:23;
then
A641: h0.((i+1)+1)=h21.((i+1)+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A624
,A617,A626,FINSEQ_6:118;
then
A642: h0.((i+1)+1)=g2.(h2.(j+1)) by A628,A631,FUNCT_1:13;
then
A643: h0.((i+1)+1) in Lower_Arc(P) by A23,A131,A29,A632,BORSUK_1:40
,FUNCT_1:def 3;
(i+1)-len h11<len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A594,
XREAL_1:9;
then (i+1)-len h11+2<len h2-2+2 by XREAL_1:6;
then
A644: h0/.(i+1+1) <> W-min P by A46,A34,A35,A32,A623,A641,A620,A628,A631,A640
;
i+1 in dom h0 by A594,A597,FINSEQ_3:25;
then
A645: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
0<=h2.j & h2.j<=1 by A29,A638,BORSUK_1:40,XXREAL_1:1;
then LE h0/.(i+1),h0/.((i+1)+1),Lower_Arc(P),E-max(P),W-min(P) by A21
,A22,A23,A24,A25,A637,A642,A634,A645,A640,A633,Th18;
hence thesis by A645,A640,A639,A643,A644,JORDAN6:def 10;
end;
suppose
A646: i+1=len h1;
then len h1+1 <=len h11 + len (mid(h21,2,len h21 -'1)) by A595,
FINSEQ_1:22;
then
A647: (i+1)+1-len h11 <=len h11 + len (mid(h21,2,len h21 -'1))-len h11
by A646,XREAL_1:9;
then 1<=(i+1)+1-'len h11 by A39,A596,A646,XREAL_1:233;
then 1<(i+1)+1-'len h11+(2-1) by NAT_1:13;
then
A648: 0<(i+1)+1-'len h11+2-1;
A649: (i+1)+1-len h11=(i+1)+1-'len h11 by A39,A596,A646,XREAL_1:233;
len h1 in dom h0 by A594,A597,A646,FINSEQ_3:25;
then
A650: h0/.len h1=h0.len h1 by PARTFUN1:def 6;
set j=(i+1)-'len h11+2-'1;
A651: 0+2<=(i+1)-'len h11 +2 by XREAL_1:6;
then
A652: j+1=(i+1)-'len h11+(1+1)-1+1 by Lm1,NAT_D:39,42
.=(i+1)-'len h11+(1+1);
2-'1<=((i+1)-'len h11 +2-'1) by A651,NAT_D:42;
then
A653: 1<j+1 by Lm1,NAT_1:13;
len h1-len h11=len h1-'len h11 & (i+1)-len h11<=len h1+(len h2-2
)-len h11 by A39,A47,A36,A52,A55,A57,A594,XREAL_1:9,233;
then (i+1)-'len h11+2<=len h2-2+2 by A39,A646,XREAL_1:6;
then
A654: j+1 in dom h2 by A652,A653,FINSEQ_3:25;
then
A655: h2.(j+1) in rng h2 by FUNCT_1:def 3;
h0.len h1 =E-max(P) by A39,A255,A597,A646,FINSEQ_1:64;
then
A656: h0.(i+1) in Upper_Arc(P) by A1,A646,Th1;
(i+1)+1 in dom h0 by A595,A598,FINSEQ_3:25;
then
A657: h0/.((i+1)+1)=h0.((i+1)+1) by PARTFUN1:def 6;
h0.((i+1)+1)=(mid(h21,2,len h21 -'1)).((i+1)+1 -len h11) by A39,A36
,A595,A646,FINSEQ_1:23;
then
A658: h0.((i+1)+1)=h21.((i+1)+1-'len h11 +2-'1) by A39,A48,A56,A50,A54,A646
,A649,A647,FINSEQ_6:118;
A659: j+1=(i+1)-len h11+2-1+1 by A39,A646,Lm1,XREAL_0:def 2
.=(i+1)+1-'len h11+2-'1 by A649,A648,XREAL_0:def 2;
then h0.((i+1)+1)=g2.(h2.(j+1)) by A658,A654,FUNCT_1:13;
then
A660: h0.(i+1+1) in Lower_Arc P by A23,A131,A29,A655,BORSUK_1:40
,FUNCT_1:def 3;
(i+1)-len h11<len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A594,
XREAL_1:9;
then (i+1)-len h11+2<len h2-2+2 by XREAL_1:6;
then j+1<len h2 by A39,A646,A652,XREAL_0:def 2;
then h0/.(i+1+1) <> W-min P by A46,A34,A35,A32,A658,A659,A654,A657;
hence thesis by A646,A650,A657,A660,A656,JORDAN6:def 10;
end;
end;
end;
thus for i being Nat st 1<=i & i+1<len h0 holds Segment(h0/.i,h0
/.(i+1),P)/\ Segment(h0/.(i+1),h0/.(i+2),P) ={h0/.(i+1)}
proof
let i be Nat;
assume
A661: 1<=i & i+1<len h0;
then
A662: LE h0/.i,h0/.(i+1),P & h0/.(i+1)<>W-min(P) by A256;
h0/.i<>h0/.(i+1) & LE h0/.(i+1),h0/.(i+2),P by A592,A256,A661;
hence thesis by A1,A662,Th10;
end;
A663: 2 in dom h0 by A201,FINSEQ_3:25;
i <> 1 by A59,Lm2;
then
A664: h0/.i <> h0/.1 by A67,A78,A203,PARTFUN2:10;
A665: len h1 in dom h1 by A16,FINSEQ_3:25;
thus Segment(h0/.len h0,h0/.1,P)/\ Segment(h0/.1,h0/.2,P)={h0/.1}
proof
defpred P[Nat] means $1+2<=len h0 implies LE h0/.2,h0/.($1+2),P;
set j=len h0-'len h11+2-'1;
A666: len h0 -len h11 <=len h11 + len mid(h21,2,len h21 -'1) -len h11 by
FINSEQ_1:22;
A667: h0/.2=h0.2 by A663,PARTFUN1:def 6;
A668: for k being Nat st P[k] holds P[k+1]
proof
let k be Nat;
assume
A669: k+2<=len h0 implies LE h0/.2,h0/.(k+2),P;
now
A670: k+1+1=k+2;
A671: k+1+2=k+2+1;
assume
A672: k+1+2<=len h0;
then k+2<len h0 by A671,NAT_1:13;
then LE h0/.(k+2),h0/.(k+2+1),P by A592,A671,A670,NAT_1:11;
hence LE h0/.2,h0/.(k+1+2),P by A1,A669,A672,JORDAN6:58,NAT_1:13;
end;
hence thesis;
end;
len h0 -'2=len h0-2 by A65,A14,XREAL_1:233,XXREAL_0:2;
then
A673: len h0-'2+2=len h0;
0+2<=len h0-'len h11 +2 by XREAL_1:6;
then
A674: 1<=(len h0-'len h11 +2-'1) by Lm1,NAT_D:42;
(len h0-'len h11 +2-'1)<=len h21 by A36,A52,A55,A57,A252,NAT_D:44;
then
A675: j in dom h2 by A46,A674,FINSEQ_3:25;
h0.2=g1.(h1.2) & h1.2 in rng h1 by A15,A40,FUNCT_1:13,def 3;
then
A676: h0/.2 in Upper_Arc P by A5,A132,A11,A667,BORSUK_1:40,FUNCT_1:def 3;
Upper_Arc(P) \/ Lower_Arc(P)= P by A1,JORDAN6:50;
then h0/.2 in P by A676,XBOOLE_0:def 3;
then
A677: P[0] by A1,JORDAN6:56;
A678: for i being Nat holds P[i] from NAT_1:sch 2(A677,A668);
A679: h11.2 <> W-min(P) by A38,A17,A71,A15,A20;
len h1 +1-len h1<=len h0 -len h1 & h0.len h0=(mid(h21,2,len h21 -'1)
).(len h0 -len h11) by A39,A36,A549,A62,A64,FINSEQ_1:23,XXREAL_0:2;
then h0.len h0=h21.(len h0-'len h11 +2-'1) by A39,A48,A56,A50,A54,A252,A666
,FINSEQ_6:118;
then
A680: h0.len h0=g2.(h2.j) by A675,FUNCT_1:13;
A681: now
h2.j in rng h2 by A675,FUNCT_1:def 3;
then
A682: g2.(h2.j) in rng g2 by A131,A29,BORSUK_1:40,FUNCT_1:def 3;
assume h0/.2=h0/.len h0;
then h0/.2 in Upper_Arc(P)/\ Lower_Arc(P) by A23,A75,A676,A680,A682,
XBOOLE_0:def 4;
then h0/.2 in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
then h11.2=E-max(P) by A40,A679,A667,TARSKI:def 2;
hence contradiction by A38,A665,A255,A14,A15,A20;
end;
h0/.2<>W-min(P) by A663,A40,A679,PARTFUN1:def 6;
hence thesis by A1,A73,A681,A678,A673,Th12;
end;
A683: i+1 = len h0 by A16,A65,XREAL_1:235,XXREAL_0:2;
then LE h0/.i,h0/.(i+1),P & h0/.(i+1)<>W-min(P) by A256,A202;
hence Segment(h0/.i,h0/.len h0,P)/\ Segment(h0/.len h0,h0/.1,P)={h0/.len h0}
by A1,A73,A683,A664,Th11;
LE h0/.i,h0/.(i+1),P by A256,A202,A200;
hence Segment(h0/.i,h0/.len h0,P) misses Segment(h0/.1,h0/.2,P) by A1,A683
,A333,A229,A207,Th13;
thus for i,j being Nat st 1<=i & i < j & j < len h0 &
i,j aren't_adjacent
holds Segment(h0/.i,h0/.(i+1),P) misses Segment(h0/.j,h0/.(j+1),P)
proof
let i,j be Nat;
assume that
A684: 1<=i and
A685: i < j and
A686: j < len h0 and
A687: i,j aren't_adjacent;
A688: 1<j by A684,A685,XXREAL_0:2;
i<len h0 by A685,A686,XXREAL_0:2;
then
A689: i+1<=len h0 by NAT_1:13;
then
A690: LE h0/.i,h0/.(i+1),P & h0/.i<>h0/.(i+1) by A256,A684;
A691: i+1<=j by A685,NAT_1:13;
then
A692: (i+1)<len h0 by A686,XXREAL_0:2;
A693: not j=i+1 by A687,GOBRD10:def 1;
then
A694: i+1<j by A691,XXREAL_0:1;
A695: now
assume
A696: h0/.(i+1)=h0/.j;
per cases;
suppose
A697: i+1<=len h1;
A698: 1<i+1 by A684,NAT_1:13;
then
A699: i+1 in dom h1 by A697,FINSEQ_3:25;
then
A700: h1.(i+1) in rng h1 by FUNCT_1:def 3;
i+1 in dom h0 by A689,A698,FINSEQ_3:25;
then
A701: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A702: h0.(i+1)=h11.(i+1) by A39,A697,A698,FINSEQ_1:64;
then h0.(i+1)=g1.(h1.(i+1)) by A699,FUNCT_1:13;
then
A703: h0.(i+1) in Upper_Arc(P) by A5,A132,A11,A700,BORSUK_1:40,FUNCT_1:def 3;
per cases;
suppose
A704: j<=len h1;
j in dom h0 by A686,A688,FINSEQ_3:25;
then
A705: h0/.j=h0.j by PARTFUN1:def 6;
h0.(j)=h11.(j) & j in dom h1 by A39,A688,A704,FINSEQ_1:64,FINSEQ_3:25
;
hence contradiction by A38,A20,A693,A696,A699,A701,A702,A705;
end;
suppose
A706: j>len h1;
j in dom h0 by A686,A688,FINSEQ_3:25;
then
A707: h0/.j=h0.j by PARTFUN1:def 6;
A708: j-len h11=j-'len h11 by A39,A706,XREAL_1:233;
j-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A686,
XREAL_1:9;
then j-'len h11+2<=len h2-2+2 by A39,A708,XREAL_1:6;
then
A709: (j-'len h11 +2-'1)<= len h21 by A47,NAT_D:44;
j-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A686,
XREAL_1:9;
then
A710: j-'len h11+2<=len h2-2+2 by A39,A708,XREAL_1:6;
set k=j-'len h11+2-'1;
j<= len h11 + len (mid(h21,2,len h21 -'1)) by A686,FINSEQ_1:22;
then
A711: j-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by
XREAL_1:9;
A712: 0+2<=j-'len h11 +2 by XREAL_1:6;
then
A713: j-'len h11+2-'1=j-'len h11+2-1 by Lm1,NAT_D:39,42;
1<=(j-'len h11 +2-'1) by A712,Lm1,NAT_D:42;
then
A714: k in dom h2 by A46,A709,FINSEQ_3:25;
then h2.k in rng h2 by FUNCT_1:def 3;
then
A715: g2.(h2.k) in rng g2 by A131,A29,BORSUK_1:40,FUNCT_1:def 3;
A716: len h1 +1<=j by A706,NAT_1:13;
then h0.j=(mid(h21,2,len h21 -'1)).(j -len h11) & len h1 +1-len h1
<=j-len h1 by A39,A36,A686,FINSEQ_1:23,XREAL_1:9;
then
A717: h0.j=h21.(j-'len h11 +2-'1) by A39,A48,A56,A50,A54,A708,A711,
FINSEQ_6:118;
then h0.j=g2.(h2.k) by A714,FUNCT_1:13;
then h0.j in Upper_Arc(P) /\ Lower_Arc(P) by A23,A696,A701,A703,A707
,A715,XBOOLE_0:def 4;
then
A718: h0.j in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
len h11+1-len h11<=j-len h11 by A39,A716,XREAL_1:9;
then 1<=j-'len h11 by NAT_D:39;
then 1+2<=j-'len h11+2 by XREAL_1:6;
then 1+2-1<=j-'len h11+2-1 by XREAL_1:9;
then 1<k by A713,XXREAL_0:2;
then
A719: h0.j <> E-max P by A46,A76,A77,A32,A717,A714;
j-'len h11+2-'1<j-'len h11+2-'1+1 by NAT_1:13;
then h0.j <> W-min P by A46,A34,A35,A32,A717,A713,A710,A714;
hence contradiction by A718,A719,TARSKI:def 2;
end;
end;
suppose
A720: i+1>len h1;
then
A721: j>len h1 by A691,XXREAL_0:2;
then
A722: len h1 +1<=j by NAT_1:13;
then
A723: len h1 +1-len h1<=j-len h1 by XREAL_1:9;
len h11+1-len h11<=j-len h11 by A39,A722,XREAL_1:9;
then
A724: j-'len h11=j-len h11 by NAT_D:39;
A725: len h1 +1<=i+1 by A720,NAT_1:13;
then len h11+1-len h11<=(i+1)-len h11 by A39,XREAL_1:9;
then (i+1)-'len h11=(i+1)-len h11 by NAT_D:39;
then i+1-'len h11<j-'len h11 by A694,A724,XREAL_1:9;
then
A726: i+1-'len h11+2<j-'len h11+2 by XREAL_1:6;
set k=j-'len h11+2-'1;
set j0=(i+1)-'len h11+2-'1;
A727: j<= len h11 + len (mid(h21,2,len h21 -'1)) by A686,FINSEQ_1:22;
A728: (i+1)-len h11 <=len h11 + len (mid(h21,2,len h21 -'1))-len h11
by A36,A689,XREAL_1:9;
A729: j-len h11=j-'len h11 by A39,A691,A720,XREAL_1:233,XXREAL_0:2;
j-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A686,
XREAL_1:9;
then j-'len h11+2<=len h2-2+2 by A39,A729,XREAL_1:6;
then
A730: (j-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
A731: 0+2<=j-'len h11 +2 by XREAL_1:6;
then
A732: j-'len h11+2-'1=j-'len h11+2-1 by Lm1,NAT_D:39,42;
1<=(j-'len h11 +2-'1) by A731,Lm1,NAT_D:42;
then
A733: k in dom h2 by A46,A730,FINSEQ_3:25;
A734: (i+1)-len h11=(i+1)-'len h11 by A39,A720,XREAL_1:233;
len h11+1<=j by A39,A721,NAT_1:13;
then
A735: h0.j=(mid(h21,2,len h21 -'1)).(j -len h11) by A727,FINSEQ_1:23;
j-len h11<=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A686,
XREAL_1:9;
then
A736: h0.j=h21.(j-'len h11 +2-'1) by A39,A48,A56,A50,A54,A735,A729,A723,
FINSEQ_6:118;
1<=i+1 by A684,NAT_1:13;
then i+1 in dom h0 by A689,FINSEQ_3:25;
then
A737: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
j in dom h0 by A686,A688,FINSEQ_3:25;
then
A738: h0/.j=h0.j by PARTFUN1:def 6;
(i+1)-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57
,A689,XREAL_1:9;
then (i+1)-'len h11+2<=len h2-2+2 by A39,A734,XREAL_1:6;
then
A739: ((i+1)-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
A740: 0+2<=(i+1)-'len h11 +2 by XREAL_1:6;
then 1<=((i+1)-'len h11 +2-'1) by Lm1,NAT_D:42;
then
A741: j0 in dom h2 by A46,A739,FINSEQ_3:25;
(i+1)-'len h11+2-'1=(i+1)-'len h11+2-1 by A740,Lm1,NAT_D:39,42;
then
A742: (i+1)-'len h11+2-'1<j-'len h11+2-'1 by A732,A726,XREAL_1:9;
h0.(i+1)=(mid(h21,2,len h21 -'1)).((i+1) -len h11) & len h1 +1-
len h1<=(i+1) -len h1 by A39,A36,A689,A725,FINSEQ_1:23,XREAL_1:9;
then h0.(i+1)=h21.((i+1)-'len h11 +2-'1) by A39,A48,A56,A50,A54,A734
,A728,FINSEQ_6:118;
hence contradiction by A46,A32,A696,A741,A737,A736,A742,A733,A738;
end;
end;
A743: j+1<=len h0 by A686,NAT_1:13;
A744: 1<i+1 by A684,NAT_1:13;
A745: 1<=(i+1) by A684,NAT_1:13;
A746: i+1<len h0 by A686,A691,XXREAL_0:2;
A747: LE h0/.(i+1),h0/.j,P
proof
per cases;
suppose
A748: i+1<=len h1;
per cases;
suppose
A749: j<=len h1;
A750: 1<j by A694,A745,XXREAL_0:2;
then
A751: j in dom h1 by A749,FINSEQ_3:25;
then
A752: h1.j in rng h1 by FUNCT_1:def 3;
then
A753: (h1.j) <= 1 by A11,BORSUK_1:40,XXREAL_1:1;
j in dom h0 by A686,A750,FINSEQ_3:25;
then
A754: h0/.j=h0.j by PARTFUN1:def 6;
h0.j=h11.j by A39,A749,A750,FINSEQ_1:64;
then
A755: g1.(h1.j) = h0/.j by A751,A754,FUNCT_1:13;
then
A756: h0/.j in Upper_Arc(P) by A5,A132,A11,A752,BORSUK_1:40,FUNCT_1:def 3;
i+1 in dom h0 by A745,A692,FINSEQ_3:25;
then
A757: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A758: Upper_Arc(P) is_an_arc_of W-min(P),E-max(P) by A1,JORDAN6:def 8;
A759: (i+1) in dom h1 by A745,A748,FINSEQ_3:25;
then
A760: h1.(i+1) in rng h1 by FUNCT_1:def 3;
then
A761: 0 <= (h1.(i+1)) & (h1.(i+1)) <= 1 by A11,BORSUK_1:40,XXREAL_1:1;
h0.(i+1)=h11.(i+1) by A39,A745,A748,FINSEQ_1:64;
then
A762: g1.(h1.(i+1)) = h0/.(i+1) by A759,A757,FUNCT_1:13;
then
A763: h0/.(i+1) in Upper_Arc(P) by A5,A132,A11,A760,BORSUK_1:40
,FUNCT_1:def 3;
(h1.(i+1)) <= (h1.j) by A12,A694,A759,A751,SEQM_3:def 1;
then LE h0/.(i+1),h0/.j,Upper_Arc(P),W-min(P),E-max(P) by A4,A5,A6,A7
,A758,A762,A761,A755,A753,Th18;
hence thesis by A763,A756,JORDAN6:def 10;
end;
suppose
A764: j>len h1;
set k=j-'len h11+2-'1;
0+2<=j-'len h11 +2 by XREAL_1:6;
then
A765: 2-'1<=j-'len h11 +2-'1 by NAT_D:42;
A766: j-len h11=j-'len h11 by A39,A764,XREAL_1:233;
j-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A686,
XREAL_1:9;
then j-'len h11+2<=len h2-2+2 by A39,A766,XREAL_1:6;
then (j-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
then
A767: (j-'len h11 +2-'1) in dom h21 by A765,Lm1,FINSEQ_3:25;
j+1-1<=len h1+(len h2-2)-1 by A39,A47,A36,A52,A55,A57,A743,XREAL_1:9;
then j-len h11<=len h1+((len h2-2)-1)-len h11 by XREAL_1:9;
then j-'len h11+2<=len h2-2-1+2 by A39,A766,XREAL_1:6;
then
A768: j-'len h11+2-1<=len h2-1-1 by XREAL_1:9;
A769: h0.(i+1)=h11.(i+1) by A39,A744,A748,FINSEQ_1:64;
i+1 in dom h1 by A744,A748,FINSEQ_3:25;
then h1.(i+1) in rng h1 by FUNCT_1:def 3;
then
A770: g1.(h1.(i+1)) in rng g1 by A132,A11,BORSUK_1:40,FUNCT_1:def 3;
0+1<=j-'len h11+1+1-1 by XREAL_1:6;
then
A771: j-'len h11+2-'1=j-'len h11+2-1 by NAT_D:39;
len h1 +1<=j by A764,NAT_1:13;
then
A772: h0.j=(mid(h21,2,len h21 -'1)).(j -len h11) & len h1 +1-len h1
<=j-len h1 by A39,A36,A686,FINSEQ_1:23,XREAL_1:9;
A773: j-len h11 <=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36
,A686,XREAL_1:9;
then h0.j=h21.(j-'len h11 +2-'1) by A39,A48,A56,A50,A54,A766,A772,
FINSEQ_6:118;
then
A774: h0.j=g2.(h2.k) by A46,A767,FUNCT_1:13;
j-len h11=j-'len h11 by A39,A764,XREAL_1:233;
then
A775: h0.j=h21.k by A39,A48,A56,A50,A54,A773,A772,FINSEQ_6:118;
j in dom h0 by A686,A688,FINSEQ_3:25;
then
A776: h0/.j=h0.j by PARTFUN1:def 6;
h2.k in rng h2 by A46,A767,FUNCT_1:def 3;
then
A777: h0.j in Lower_Arc(P) by A23,A131,A29,A774,BORSUK_1:40,FUNCT_1:def 3;
i+1 in Seg len h1 by A745,A748,FINSEQ_1:1;
then i+1 in dom h1 by FINSEQ_1:def 3;
then
A778: h11.(i+1)=g1.(h1.(i+1)) by FUNCT_1:13;
(i+1) in dom h0 by A745,A692,FINSEQ_3:25;
then
A779: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
len h2-1-1<len h2 by Lm4;
then h0/.j <> W-min P by A46,A34,A35,A32,A771,A767,A768,A775,A776;
hence thesis by A5,A769,A778,A779,A776,A770,A777,JORDAN6:def 10;
end;
end;
suppose
A780: i+1>len h1;
set j0=(i+1)-'len h11+2-'1;
set k=j-'len h11+2-'1;
A781: 0+2<=(i+1)-'len h11 +2 by XREAL_1:6;
then
A782: 1<=((i+1)-'len h11 +2-'1) by Lm1,NAT_D:42;
A783: j-len h11=j-'len h11 by A39,A691,A780,XREAL_1:233,XXREAL_0:2;
len h1<j by A691,A780,XXREAL_0:2;
then
A784: len h11+1<=j by A39,NAT_1:13;
then
A785: len h1 +1-len h1<=j-len h1 by A39,XREAL_1:9;
j<=len h11 + len (mid(h21,2,len h21 -'1)) by A686,FINSEQ_1:22;
then
A786: h0.j=(mid(h21,2,len h21 -'1)).(j -len h11) by A784,FINSEQ_1:23;
A787: i+1-len h11<len h11 + len (mid(h21,2,len h21 -'1)) - len h11 by A36
,A746,XREAL_1:9;
then j-len h11 <= len (mid(h21,2,len h21 -'1)) by A36,A686,XREAL_1:9;
then
A788: h0.j=h21.(k) by A39,A48,A56,A50,A54,A783,A786,A785,FINSEQ_6:118;
A789: (i+1)-len h11=(i+1)-'len h11 by A39,A780,XREAL_1:233;
then (i+1)-'len h11+2<=len h2-2+2 by A47,A52,A55,A57,A787,XREAL_1:6;
then ((i+1)-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
then
A790: j0 in dom h2 by A46,A782,FINSEQ_3:25;
then
A791: h2.j0 in rng h2 by FUNCT_1:def 3;
then
A792: 0<=h2.j0 & h2.j0<=1 by A29,BORSUK_1:40,XXREAL_1:1;
A793: g2.(h2.j0) in rng g2 by A131,A29,A791,BORSUK_1:40,FUNCT_1:def 3;
A794: j-len h11=j-'len h11 by A39,A691,A780,XREAL_1:233,XXREAL_0:2;
j+1-1<=len h1+(len h2-2)-1 by A39,A47,A36,A52,A55,A57,A743,XREAL_1:9;
then j-len h11<=len h1+((len h2-2)-1)-len h11 by XREAL_1:9;
then j-'len h11+2<=len h2-2-1+2 by A39,A794,XREAL_1:6;
then
A795: j-'len h11+2-1<=len h2-1-1 by XREAL_1:9;
0+1<=j-'len h11+1+1-1 by XREAL_1:6;
then
A796: j-'len h11+2-'1=j-'len h11+2-1 by NAT_D:39;
j-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A686,
XREAL_1:9;
then j-'len h11+2<=len h2-2+2 by A39,A794,XREAL_1:6;
then
A797: (j-'len h11 +2-'1)<=len h21 by A47,NAT_D:44;
0+2<=j-'len h11 +2 by XREAL_1:6;
then 2-'1<=j-'len h11 +2-'1 by NAT_D:42;
then
A798: j-'len h11 +2-'1 in dom h21 by A797,Lm1,FINSEQ_3:25;
then
A799: h2.k in rng h2 by A46,FUNCT_1:def 3;
then
A800: h2.k<=1 by A29,BORSUK_1:40,XXREAL_1:1;
j-len h11 <=len h11 + len (mid(h21,2,len h21 -'1))-len h11 by A36,A686,
XREAL_1:9;
then
A801: h0.j=h21.(j-'len h11 +2-'1) by A39,A48,A56,A50,A54,A794,A786,A785,
FINSEQ_6:118;
then h0.j=g2.(h2.k) by A46,A798,FUNCT_1:13;
then
A802: h0.(j) in Lower_Arc(P) by A23,A131,A29,A799,BORSUK_1:40,FUNCT_1:def 3;
A803: (i+1)-'len h11+2-'1=(i+1)-'len h11+2-1 by A781,Lm1,NAT_D:39,42;
A804: i+1<len h11 + len (mid(h21,2,len h21 -'1)) by A746,FINSEQ_1:22;
(i+1)-len h11<j-len h11 by A694,XREAL_1:9;
then (i+1)-'len h11+2<j-'len h11+2 by A783,A789,XREAL_1:6;
then j0<k by A796,A803,XREAL_1:9;
then
A805: h2.(j0)<h2.(k) by A30,A46,A798,A790,SEQM_3:def 1;
(i+1) in dom h0 by A745,A692,FINSEQ_3:25;
then
A806: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
len h1+1<=i+1 by A780,NAT_1:13;
then
A807: len h1+1-len h1<=i+1-len h1 by XREAL_1:9;
then
A808: i+1-'len h11=i+1-len h11 by A39,NAT_D:39;
j in dom h0 by A686,A688,FINSEQ_3:25;
then
A809: h0/.j=h0.(j) by PARTFUN1:def 6;
len h11+1<=i+1 by A39,A780,NAT_1:13;
then h0.(i+1)=(mid(h21,2,len h21 -'1)).(i+1 -len h11) by A804,
FINSEQ_1:23
.=h21.(i+1-'len h11+2-'1) by A39,A48,A56,A50,A54,A787,A807,A808,
FINSEQ_6:118;
then
A810: h0.(i+1)=g2.(h2.j0) by A790,FUNCT_1:13;
len h2-1-1<len h2 by Lm4;
then
A811: h0/.j <> W-min P by A46,A34,A35,A32,A796,A798,A795,A788,A809;
h21.k=g2.(h2.k) by A46,A798,FUNCT_1:13;
then LE h0/.(i+1),h0/.j,Lower_Arc(P),E-max(P),W-min(P) by A21,A22,A23
,A24,A25,A801,A800,A810,A792,A805,A806,A809,Th18;
hence thesis by A23,A810,A806,A809,A793,A802,A811,JORDAN6:def 10;
end;
end;
LE h0/.j,h0/.(j+1),P by A256,A688,A743;
hence thesis by A1,A690,A747,A695,Th13;
end;
let i be Nat such that
A812: 1 < i and
A813: i+1 < len h0;
A814: 1<i+1 by A812,NAT_1:13;
then
A815: i+1 in dom h0 by A813,FINSEQ_3:25;
A816: 1<=len h0-len h1 by A549,A62,XXREAL_0:2;
A817: now
assume
A818: h0/.(i+1)=h0/.len h0;
per cases;
suppose
A819: i+1<=len h1;
then
A820: i+1 in dom h1 by A814,FINSEQ_3:25;
h0.(i+1)=h11.(i+1) by A39,A814,A819,FINSEQ_1:64;
then
A821: h0.(i+1)=g1.(h1.(i+1)) by A820,FUNCT_1:13;
h1.(i+1) in rng h1 by A820,FUNCT_1:def 3;
then
A822: h0.(i+1) in Upper_Arc(P) by A5,A132,A11,A821,BORSUK_1:40,FUNCT_1:def 3;
i+1 in dom h0 by A813,A814,FINSEQ_3:25;
then
A823: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
1+2<=len h0-'len h11+2 by A47,A36,A52,A55,A57,A63,A252,XREAL_1:6;
then
A824: 1+2-1<=len h0-'len h11+2-1 by XREAL_1:9;
set k=len h0-'len h11+2-'1;
A825: 0+2<=len h0-'len h11 +2 by XREAL_1:6;
then
A826: 2-'1<=(len h0-'len h11 +2-'1) by NAT_D:42;
len h0-'len h11 +2-'1 <= len h21 by A36,A52,A55,A57,A252,NAT_D:44;
then
A827: k in dom h2 by A46,A826,Lm1,FINSEQ_3:25;
then h2.k in rng h2 by FUNCT_1:def 3;
then
A828: g2.(h2.k) in rng g2 by A131,A29,BORSUK_1:40,FUNCT_1:def 3;
h0.len h0=(mid(h21,2,len h21 -'1)).(len h0 -len h11) by A39,A36,A64,
FINSEQ_1:23;
then
A829: h0.len h0=h21.(len h0-'len h11 +2-'1) by A39,A36,A48,A56,A50,A54,A816
,A252,FINSEQ_6:118;
then h0.len h0=g2.(h2.k) by A827,FUNCT_1:13;
then h0.len h0 in Upper_Arc(P) /\ Lower_Arc(P) by A23,A75,A818,A823,A822
,A828,XBOOLE_0:def 4;
then
A830: h0.len h0 in {W-min(P),E-max(P)} by A1,JORDAN6:def 9;
len h0-'len h11+2-'1=len h0-'len h11+2-1 by A825,Lm1,NAT_D:39,42;
then 1<k by A824,XXREAL_0:2;
then
A831: h0.len h0 <> E-max P by A46,A76,A77,A32,A829,A827;
len h0-'len h11+2-'1<len h0-'len h11+2-'1+1 by NAT_1:13;
then h0.len h0 <> W-min P by A46,A47,A34,A35,A36,A52,A55,A57,A252,A32
,A829,A827;
hence contradiction by A830,A831,TARSKI:def 2;
end;
suppose
A832: i+1>len h1;
set k=len h0-'len h11+2-'1;
set j0=(i+1)-'len h11+2-'1;
A833: 0+2<=len h0-'len h11 +2 by XREAL_1:6;
then
A834: 2-'1<=(len h0-'len h11 +2-'1) by NAT_D:42;
len h0-'len h11 +2-'1 <= len h21 by A36,A52,A55,A57,A252,NAT_D:44;
then
A835: k in dom h2 by A46,A834,Lm1,FINSEQ_3:25;
i+1 <= len h11 + len mid(h21,2,len h21 -'1) by A813,FINSEQ_1:22;
then
A836: i+1-len h11 <=len h11 + len mid(h21,2,len h21 -'1)-len h11 by XREAL_1:9;
A837: len h1 +1<=i+1 by A832,NAT_1:13;
then len h11+1-len h11<=(i+1)-len h11 by A39,XREAL_1:9;
then
A838: (i+1)-'len h11=(i+1)-len h11 by NAT_D:39;
len h0-'len h11=len h0-len h11 by A36,A57,XREAL_0:def 2;
then i+1-'len h11<len h0-'len h11 by A813,A838,XREAL_1:9;
then
A839: i+1-'len h11+2<len h0-'len h11+2 by XREAL_1:6;
1<=i+1 by A812,NAT_1:13;
then i+1 in dom h0 by A813,FINSEQ_3:25;
then
A840: h0/.(i+1)=h0.(i+1) by PARTFUN1:def 6;
A841: len h0-'len h11+2-'1=len h0-'len h11+2-1 by A833,Lm1,NAT_D:39,42;
A842: (i+1)-len h11=(i+1)-'len h11 by A39,A832,XREAL_1:233;
(i+1)-len h11<=len h1+(len h2-2)-len h11 by A39,A47,A36,A52,A55,A57,A813,
XREAL_1:9;
then (i+1)-'len h11+2<=len h2-2+2 by A39,A842,XREAL_1:6;
then
A843: (i+1)-'len h11 +2-'1 <= len h21 by A47,NAT_D:44;
A844: 0+2<=(i+1)-'len h11 +2 by XREAL_1:6;
then 2-'1<=((i+1)-'len h11 +2-'1) by NAT_D:42;
then
A845: j0 in dom h2 by A46,A843,Lm1,FINSEQ_3:25;
h0.(i+1)=(mid(h21,2,len h21 -'1)).((i+1) -len h11) & len h1 +1-
len h1<=(i+1) -len h1 by A39,A36,A813,A837,FINSEQ_1:23,XREAL_1:9;
then
A846: h0.(i+1)=h21.((i+1)-'len h11 +2-'1) by A39,A48,A56,A50,A54,A842,A836,
FINSEQ_6:118;
(i+1)-'len h11+2-'1=(i+1)-'len h11+2-1 by A844,Lm1,NAT_D:39,42;
then
A847: (i+1)-'len h11+2-'1<k by A841,A839,XREAL_1:9;
h0.len h0=(mid(h21,2,len h21 -'1)).(len h0 -len h11) by A39,A36,A64,
FINSEQ_1:23;
then h0.len h0=h21.(len h0-'len h11 +2-'1) by A47,A36,A51,A52,A49,A48,A53
,A55,A57,A63,A252,FINSEQ_6:118;
hence contradiction by A46,A75,A32,A818,A846,A845,A840,A847,A835;
end;
end;
h0.len h0=mid(h21,2,len h21 -'1).(len h0 -len h11) by A39,A36,A64,FINSEQ_1:23
;
then
A848: h0.len h0=h21.(len h0-'len h11 +2-'1) by A39,A36,A48,A56,A50,A54,A816
,A252,FINSEQ_6:118;
then
A849: h0.len h0 in Lower_Arc(P) by A23,A131,A29,A253,BORSUK_1:40,FUNCT_1:def 3;
A850: LE h0/.(i+1),h0/.len h0,P
proof
per cases;
suppose
A851: i+1<=len h1;
then i+1 in dom h1 by A814,FINSEQ_3:25;
then h1.(i+1) in rng h1 by FUNCT_1:def 3;
then
A852: g1.(h1.(i+1)) in rng g1 by A132,A11,BORSUK_1:40,FUNCT_1:def 3;
A853: h0/.(i+1)=h0.(i+1) by A815,PARTFUN1:def 6;
i+1 in dom h1 by A814,A851,FINSEQ_3:25;
then
A854: h11.(i+1)=g1.(h1.(i+1)) by FUNCT_1:13;
h0.(i+1)=h11.(i+1) by A39,A814,A851,FINSEQ_1:64;
hence thesis by A5,A75,A130,A849,A854,A853,A852,JORDAN6:def 10;
end;
suppose
A855: i+1>len h1;
then len h1+1<=i+1 by NAT_1:13;
then
A856: len h1+1-len h1<=i+1-len h1 by XREAL_1:9;
then
A857: i+1-'len h11=i+1-len h11 by A39,NAT_D:39;
A858: i+1-len h11<len h11 + len (mid(h21,2,len h21 -'1)) - len h11 by A36,A813,
XREAL_1:9;
A859: i+1<len h11 + len (mid(h21,2,len h21 -'1)) by A813,FINSEQ_1:22;
len h11+1<=i+1 by A39,A855,NAT_1:13;
then
A860: h0.(i+1) = mid(h21,2,len h21 -'1).(i+1 -len h11) by A859,FINSEQ_1:23
.= h21.(i+1-'len h11+2-'1) by A39,A48,A56,A50,A54,A858,A856,A857,
FINSEQ_6:118;
set j0=(i+1)-'len h11+2-'1;
set k=len h0-'len h11+2-'1;
0+1<=len h0-'len h11+1+1-1 by XREAL_1:6;
then
A861: len h0-'len h11+2-'1=len h0-'len h11+2-1 by NAT_D:39;
A862: (len h0-'len h11 +2-'1)<=len h21 by A36,A52,A55,A57,A252,NAT_D:44;
then
A863: (len h0-'len h11 +2-'1) in dom h21 by A43,Lm1,FINSEQ_3:25;
then h2.k in rng h2 by A46,FUNCT_1:def 3;
then
A864: h2.k<=1 by A29,BORSUK_1:40,XXREAL_1:1;
len h0-'len h11 +2-'1 in dom h21 by A43,A862,Lm1,FINSEQ_3:25;
then
A865: h21.k=g2.(h2.k) by A46,FUNCT_1:13;
A866: (i+1)-len h11=(i+1)-'len h11 by A39,A855,XREAL_1:233;
(i+1)-len h11<=len h11+(len h2-2)-len h11 by A47,A36,A52,A55,A57,A813,
XREAL_1:9;
then (i+1)-'len h11+2<=len h2-2+2 by A866,XREAL_1:6;
then
A867: (i+1)-'len h11 +2-'1 <= len h21 by A47,NAT_D:44;
h0.(len h0) in Lower_Arc(P) by A23,A131,A29,A848,A253,BORSUK_1:40
,FUNCT_1:def 3;
then
A868: h0/.len h0 in Lower_Arc P by A74,PARTFUN1:def 6;
A869: 0+2<=(i+1)-'len h11 +2 by XREAL_1:6;
then 2-'1<=((i+1)-'len h11 +2-'1) by NAT_D:42;
then
A870: j0 in dom h2 by A46,A867,Lm1,FINSEQ_3:25;
then
A871: h2.j0 in rng h2 by FUNCT_1:def 3;
then
A872: 0<=h2.j0 & h2.j0<=1 by A29,BORSUK_1:40,XXREAL_1:1;
(i+1)-len h11<len h0-len h11 by A813,XREAL_1:9;
then
A873: (i+1)-'len h11+2<len h0-'len h11+2 by A252,A866,XREAL_1:6;
h0.len h0=(mid(h21,2,len h21 -'1)).(len h0 -len h11) by A39,A36,A64,
FINSEQ_1:23;
then
A874: h0.len h0=h21.(len h0-'len h11 +2-'1) by A39,A36,A48,A56,A50,A54,A816
,A252,FINSEQ_6:118;
A875: h0/.(i+1)=h0.(i+1) by A815,PARTFUN1:def 6;
g2.(h2.j0) in rng g2 by A131,A29,A871,BORSUK_1:40,FUNCT_1:def 3;
then
A876: h0.(i+1) in Lower_Arc(P) by A23,A860,A870,FUNCT_1:13;
(i+1)-'len h11+2-'1=(i+1)-'len h11+2-1 by A869,Lm1,NAT_D:39,42;
then j0<k by A861,A873,XREAL_1:9;
then
A877: h2.(j0)<h2.(k) by A30,A46,A863,A870,SEQM_3:def 1;
h0.(i+1)=g2.(h2.j0) by A860,A870,FUNCT_1:13;
then LE h0/.(i+1),h0/.len h0,Lower_Arc(P),E-max(P),W-min(P) by A21,A22
,A23,A24,A25,A75,A874,A865,A864,A872,A877,A875,Th18;
hence thesis by A130,A875,A876,A868,JORDAN6:def 10;
end;
end;
i<len h0 by A813,NAT_1:13;
then
A878: i in dom h0 by A812,FINSEQ_3:25;
then h0/.i = h0.i by PARTFUN1:def 6;
then
A879: h0/.i<>W-min P by A72,A67,A78,A812,A878;
LE h0/.i,h0/.(i+1),P by A256,A812,A813;
hence thesis by A1,A72,A68,A879,A850,A817,Th14;
end;
| 37,455
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Dear Friends…
Rejection. Nobody likes it, but everyone will and does experience it. No matter how great your relationships may be, at some point on the road of relationships you will get your feelings hurt, feel left out, not considered or flat out abandoned by someone you care about. It’s hard to take, and often leaves us with a resolve to protect our heart in attempts to not get hurt like that again. But is this the right solution? Think about it. This would mean never getting close to another and that leads us to isolation.
So what do we do when we experience rejection? While we have no control over how others treat us, we do have control over how we handle things or react when we get hurt.
Here are the three typical psychological responses a person might have:
Fight: You react; you engage or attack the other person not worried about the outcome. (“I hate you. You never pay attention to me.”)
Flight: You “run” or escape internally in order to protect yourself. (“If you don’t care, I just won’t talk to you.”)
Freeze: You act like nothing is wrong and keep your emotions to yourself. The rejection is kept inside which shuts down your emotions. (“Nothing’s wrong. I’m okay”.)
Or there is a fourth option: Focus: We can remember the truths of God and choose to focus on those truths instead of the initial feeling of rejection. Remember: whatever you magnify becomes the biggest influence in your life. Don’t allow others to define your truth and tell you who you are. Use God’s word to rewire your brain.
God’s truth says this…
“He was despised and rejected by mankind, a man of suffering, and familiar with pain. Like one from whom people hide their faces he was despised, and we held him in low esteem.” Isaiah 53:3 (He knows how you feel!)
“He will never leave you or forsake you.” Hebrews 13:5 (Our earthly relationships will always fall short, but not so with God.)
“This is my comfort and consolation in my affliction: that Your word has revived me and given me life”. Psalm 119:50 (God’s reviving comfort is available when experience rejection.)
“For we are His workmanship, created in Christ Jesus for good works, which God prepared beforehand so that we would walk in them.” Ephesians 2:10 (You are chosen and loved by Him.)
“I praise you because I am fearfully and wonderfully made; your works are wonderful, I know that full well.” Psalm 139:14 (God created you for a purpose. It is perfect and good.)
“So don’t be afraid; you are more valuable to God than a whole flock of sparrows.” Matthew 10:31 (God loves you, you are valuable to Him and you are worthy of love.)
"So don't be afraid, little flock. For it gives your Father great happiness to give you the Kingdom”. Luke12: 32 (God has good gifts to give you even when others do not. You are never alone.)
Next time you experience rejection, turn to God and pour your pain out to Him. Allow His truths to soothe your heart and heal your wounds. Practice remembering His truths and watch your feelings be transformed!
Blessings~
Angel
| 275,945
|
\begin{document}
\begin{abstract}
Following Nori's original idea we here provide certain motivic categories with a canonical tensor structure. These motivic categories are associated to a cohomological functor on a suitable base category and the tensor structure is induced by the cartesian tensor structure on the base category via a cohomological K\"unneth formula.
\end{abstract}
\maketitle
\section*{Introduction}
We here make use of Nori's approach to motives by considering the universal abelian category ${\sf ECM}_\cC^H$ corresponding to a cohomology $H$ on a given category $\cC$ which is endowed with some good geometric properties, as reformulated and generalised in \cite{BV} and \cite{BVP}. Making use of \cite{BVHP} we show that a suitable K\"unneth formula for $H$, along with a cellularity condition, provide ${\sf ECM}_\cC^H$ with a canonical tensor structure: this is our main theorem, see Theorem~\ref{thm:Tmot} below. In particular, this framework applies to Nori's original category of motives and to several other geometric situations, e.g., it applies to \cite{Ar}, \cite{I1}, \cite{I2}, \cite{FJ} and \cite{NP}. Recall that Nori's category, i.e., the category of (effective, cohomological) Nori motives over a subfield $k$ of the complex numbers, is obtained in this way as ${\sf ECM}_\cC^H$ for $H = H_{\rm sing}$ singular cohomology and $\cC$ being the category of (affine) schemes over the field $k$ (e.g., see \cite{HMS} for details on Nori's original construction and \cite{BCL} for its reconstruction).
\subsection*{Cohomology theories} Following \cite{BV}, a cohomology theory on a given (small) category $\cC$, regarded as a category of geometric objects, can be increasingly specified by imposing conditions which reflect the geometric properties of $\cC$ and which we want to hold true for those $H$ that we wish to regard as cohomologies. Some of these conditions may be expressed as axioms within the framework of a formal language. As a starting point, a cohomology theory should express the basic idea of associating $(X,Y)\leadsto \{H^n (X,Y)\}_{n\in\Z}$ in such a way that
\begin{itemize}
\item[{\it i)}] $H = \{H^n\}_{n\in\Z}$ is a family of contravariant functors $H^n : \cC^{\square}\to \cA$ on a suitable category of pairs $\cC^\square$, taking values in an abelian category $\cA$,
\item[{\it ii)}] for the triple $(X,Y)$, $(X,Z)$ and $(Y,Z)$ we have an extra connecting morphism $\delta^n : H^n (Y,Z)\to H^{n+1} (X,Y)$, and
\item[{\it iii)}] we have the following long exact sequence
\begin{equation}\label{longexact}
\cdots\to H^n (X,Y)\to H^n (X,Z)\to H^n (Y,Z)\to H^{n+1} (X,Y)\to \cdots
\end{equation}
which is also assumed to be natural in a canonical way (see \cite[\S 3.1]{BV} for more details).
\end{itemize}
Note that we may weaken the additivity assumption on $\cA$, as we may appeal to the notion of exact category (in the sense of Barr \cite{Ba}) where we still can talk of exact sequences of pointed objects. For example, $\cA$ can be the category of sets or that of groups, these being Barr-exact. In fact, to express this basic idea of a cohomology $H: \cC^{\square}\to \cA$, the assumptions on $\cA$ can be weakened further: we just need a regular category (see also \cite{Ba} for this notion), i.e., a category with finite limits where every arrow factorises as a regular epi followed by a mono and these factorisations are stable under pullback. Recall that regular epis are just coequalizers of some parallel pair of morphisms, and, in a regular category, the regular epi-mono factorisation is unique, so we can talk of the image of a morphism.
There is a general key relation between regular categories and logic, which we outline (see, e.g., \cite{Bu} for more detail).
The first important fact is that regular categories are well suited for the interpretation of any regular theory. A regular theory $\R$ is a set of sequents $\top\vdash_{x,y} \varphi \rightarrow \psi$, also called axioms, where $\varphi$ and $\psi$ are regular formulas involving sorts for which $x$ and $y$ are terms and/or variables. The regular formulas (also called positive primitive formulas) are those (first order) formulas built up from the atomic formulas (= equations, in any algebraic theory) using only meet $\wedge$ (``and") and existential quantification $\exists$. For example, to express exactness in the long exact sequence \eqref{longexact} displayed above we may include the axioms $\top\vdash_{y} fg(y) = * $ and $\top\vdash_{x} f (x) = * \rightarrow (\exists y) g (y)= x$ for any composable function symbols $g$ and $f$ and where $*$ is the constant of each sort ($0$ in the additive case). In the case of an abelian category, and using an alternative notation, these axioms would read $\forall y\,\, fg(y)=0$ and $\forall x \, (f(x)=0 \rightarrow \exists y (g(y)=x))$.
Now, given a regular theory $\R$, we can interpret it in any regular category $\cA$; any such interpretation in which every sequent in $\R$ is valid is said to be a model of the theory (e.g., see \cite[\S 3]{Bu}). A functor $F: \cA\to \cB$ between regular categories is called exact (or regular) if it preserves finite limits and regular epis. It follows that models are preserved by exact functors. We can form the category of models $\Rmod (\cA)$ of $\R$ in $\cA$, and this gives a functor $\Rmod ( - )$ on the category of regular categories with exact functors. The second important fact is that this functor is representable. That is, given $\R$, there exists a regular category $\cC_{\R}^{\rm reg}$ and an equivalence between the models in a regular category $\cA$ and the exact functors from $\cC_{\R}^{\rm reg}$ to $\cA$, i.e., we have
\begin{equation}\label{repreg}
\Rmod (\cA) \cong {\sf Ex} (\cC_{\R}^{\rm reg}, \cA )
\end{equation}
which is natural in $\cA$ (e.g., see \cite[\S 6]{Bu} for details). The category $\cC_{\R}^{\rm reg}$ is the so-called regular syntactic category for $\R$ and the model $U$ in $\Rmod (\cC_{\R}^{\rm reg})$ corresponding to the identity functor on $\cC_{\R}^{\rm reg}$ is called the universal (or generic) model. A third fact is that any regular category is the regular syntactic category of some regular theory (see \cite[\S 5]{Bu}).
Finally, let $\cC_{\R}^{\rm reg/ex}$ be the canonical (Barr-)exact completion (e.g., see \cite{La}) of the syntactic category; so
\begin{equation}\label{regex}
{\sf Ex} (\cC_{\R}^{\rm reg}, \cA )\cong {\sf Ex} (\cC_{\R}^{\rm reg/ex}, \cA )
\end{equation}
whenever $\cA$ is exact. Note that the equivalence in \eqref{regex} is natural in $\cA$ (and implicitly defines what we mean by exact completion). Since the exact completion preserves additivity we have that $\cC_{\R}^{\rm reg}$ additive yields $\cC_{\R}^{\rm reg/ex}$ abelian. Indeed, Tierney's theorem is that an exact additive category $\cA$ is abelian and conversely.
These are the main facts arising from the general theory of regular categories that we will apply to the study of cohomology. In \cite[\S 2]{BV}, for a fixed base category $\cC$, a {\em regular cohomology theory} $\T^{op}$ consists of the initial set of regular axioms, which express the properties {\it i) -- iii)}\, on any
$H = \{H^n\}_{n\in \Z}$ as indicated above: any such specific cohomology $H$ is then referred to as a model of the theory $\T^{op}$; that is $H$ is an interpretation of the axioms in some category $\cA$ which should at least be regular. So, in the following, saying that $H$ is a \emph{$\T^{op}$-model} is synonymous with saying that $H$ is a \emph{cohomology}.
\subsection*{Effective cohomological motives} Therefore, following \cite{BV}, for a fixed category $\cC$ a category of {\em effective constructible $\T^{op}$-motives} can be defined as follows
\begin{equation}\label{deftmot}
\cA[\T^{op}] := \cC_{\T^{op}}^{\rm reg/ex}
\end{equation}
via the exact completion of the regular syntactic category associated to the regular cohomology theory $\T^{op}$ mentioned above.
For any cohomology $H: \cC^{\square}\to \cA$ with $\cA$ exact we get an induced realisation exact functor $r_H : \cA[\T^{op}]\to \cA$ as it follows directly from \eqref{repreg} and \eqref{regex} above.
Moreover, for any such cohomology $H$ we then get the regular theory $\T_H^{op}$ of that model, a richer cohomology theory obtained by adding to $\T^{op}$ all regular axioms which are valid in $H$, and, therefore, we obtain another universal category
$\cA[\T_H^{op}]$ (see \cite[\S 3 \& \S 4]{BV}, \cite[\S 2.2]{BVP}, \cite{BCL} and the next \S 1.1 for more details). Note that we always have the larger category of $\T^{op}_H$-motives $\Ind (\cA[\T_H^{op}])$ provided by indization as well as an even larger $\T_H^{op}$-motivic topos $\cE [\T^{op}_H]$ given by the topos of sheaves on $\cA[\T^{op}_H]$ regarded as a site for the regular topology (see \cite{BV}).
Actually, as explained in \cite{BVP}, for any cohomology $H: \cC^{\square}\to \cA$ where $\cA$ is an abelian category, there is a canonical equivalence
\begin{equation}\label{motives}
\cA[\T_H^{op}]\cong \cA (H)
\end{equation}
with the universal abelian category $\cA (H)$ given by $H$ regarded as a representation of Nori's diagram $D^\Nori$ associated to $\cC^\square$ (see \cite[Cor. 2.9]{BVP}). This latter category $\cA (H)$ is obtained by a direct application of Freyd's universal construction (see \cite{Frey}) and Serre localisation. For instance, one has Freyd's universal abelian category ${\rm Ab} (D^\Nori)$ on the preadditive category generated by the diagram $D^\Nori$ and then an exact functor $F_H : {\rm Ab} (D^\Nori)\to \cA$ which is induced by the representation $H$ of $D^\Nori$ in $\cA$; then one obtains $\cA (H)$ as the Serre quotient of ${\rm Ab} (D^\Nori)$ by the kernel of $F_H$. Therefore, following Nori's idea, in the additive case, we may simply refer to both the abelian categories in \eqref{motives} as the \emph{category of (effective, cohomological) motives} associated to $\cC$, a chosen category of geometric objects, and $H$, a paradigmatic cohomology, and we shall denote it by
\begin{equation}\label{Nori}
{\sf ECM}_\cC^H
\end{equation} following Nori's notation. It may well be that a different cohomology $H'$ gives rise to an equivalent category of motives ${\sf ECM}_\cC^H\cong {\sf ECM}_\cC^{H'}$. The ambiguity in the choice of the cohomology is somewhat the motivic analogue of the Tannakian formalism of fibre functors even if we cannot get for free any tensor structure on this category of motives.
\subsection*{Tensor product of motives} In order to get a canonical tensor structure on the category ${\sf ECM}_\cC^H$ we need to appeal to an additional K\"unneth formula for the cohomology $H$, i.e., a K\"unneth formula for the $\T^{op}$-model $H$. First we need to consider the restriction of the cohomology $H$ to {\em good pairs}\, (see Definition~\ref{gp}). We can then express {\em cellularity}\, of $H$ (see Definition~\ref{cell}), a condition which ensures that the cohomology $H$ is determined by its values on good pairs. For the full subcategory $\cC^\good\subseteq \cC^\square$, of good pairs for $H$, we consider $H^\good$, the cohomology $H$ restricted to $\cC^\good$, and we show that there is an equivalence of abelian categories
\begin{equation}\label{good}
\cA (H)\cong\cA (H^\good)
\end{equation}
as a consequence of the cellularity assumption (see Lemma~\ref{lemma:goodpair}).
For any cohomology $H$ which is {\em cellular and satisfies the K\"unneth formula} (see Definition~\ref{kf}), we thus can provide the abelian category of motives ${\sf ECM}_\cC^H$ with a canonical tensor structure, by making use of the results of \cite{BVHP}, see our Theorem~\ref{thm:Tmot} for the precise statement. Actually, all this is just following Nori's original idea but the constructions are completely different and fairly general. The key assumptions on $H$ are made just in order to transfer the cartesian structure of $\cC^\square$ to a canonical tensor structure on $\cA[\T_H^{op}]$ via the cohomology $H$ regarded as a representation of the corresponding subquiver
$D^\good$ of $D^\Nori$, as from \eqref{motives} and \eqref{good} we get
\begin{equation}\label{end}
\cA[\T_H^{op}]\cong \cA (H^\good)
\end{equation}
and $\cA (H^\good)$ has a canonical tensor structure. Note that the cartesian structure of the category $\cC^\good$ of good pairs is obtained by the K\"unneth formula and the tensor structure on $\cA (H^\good)$ is given by $H^\good$ regarded as a $\otimes$-representation of the restricted Nori (graded) $\otimes$-diagram $D^\good$ (see \cite[Thm. 2.18]{BVHP}). Freyd's universal abelian category ${\rm Ab} (D^\good)$ carries an induced right-exact tensor structure which is also universal with respect to tensor functors
in an abelian tensor category $\cA$ with a right exact tensor product. Therefore, such an assumption on the category $\cA$ where our cohomology $H$ is taking values has to be satisfied. This sheds some light on the geometric meaning of K\"unneth formulas in the additive case but also raises the question of getting a canonical tensor structure directly on $\cA[\T_H^{op}]$ without making reference to Freyd's universal construction, i.e., not using the equivalence \eqref{end}.
Finally, all this clearly applies to the construction of effective cohomological Nori motives for several different (new and old) geometric categories.
\subsubsection*{Acknowledgements} We would like to thank A. Huber for helpful discussions on some of the matters treated herein and the referee for his/her suggestions that considerably improved the exposition.
\section{Good pairs and cellularity} We here assume to be given a base category $\cC$ along with a subcategory $\mathcal{M}$ of distinguished monomorphisms, containing all isomorphisms of $\cC$, and saturated, in the sense that if the composition of a distinguished monomorphism with a morphism is a distinguished monomorphism then the morphism is a distinguished monomorphism. We consider $\cC^{\square}$, the category whose objects are the arrows in $\mathcal{M}$ and whose arrows are commutative squares in $\cC$. We adopt the conventions of \cite[\S 2.1]{BV} regarding this category. For example we shall denote by $(X, Y)$ the object of $\cC^{\square}$ which is a monomorphism $Y\to X$ in $\mathcal{M}$. We also assume the hypotheses of \cite[\S 4.4]{BV} on the subcategory $\cM$. In particular, we assume that the distinguished monomorphisms are stable under direct and inverse images (in the sense of \cite[Remark 4.4.1]{BV}). Also assume that we have joins $Y\cup Z$ of $\mathcal{M}$-subobjects $Y,Z$ of any object $X$, a strict initial object $\emptyset$ of $\cC$ and that $\emptyset\to X$ is in $\mathcal{M}$ for each object $X$ of $\cC$. The key examples are given by $\cC$ being a category of nice topological spaces or schemes (over a base) and $\mathcal{M}$ the subcategory of closed subspaces or subschemes.
We consider the regular cohomological theory $\T^{op}$ on the signature $\Sigma^{op}$ as defined in \cite[Def. 2.3.3]{BV}. The signature $\Sigma^{op}$ of the theory $\T^{op}$ contains sorts $h^n(X, Y)$ for $n\in \Z$ and $Y\hookrightarrow X\in \mathcal{M}$, function symbols $\square^n : h^n(X', Y')\to h^n(X, Y)$ for all arrows $\square: (X,Y) \to (X', Y')$ in $\cC^{\square}$ and additional function symbols $\partial^n :h^n (Y,Z)\to h^{n+1} (X,Y)$ associated to the morphism $\partial : (Y, Z)\to (X, Y)$ in $\cC^{\square}$ given by any pair of composable arrows $Z\hookrightarrow Y\hookrightarrow X$ in $\mathcal{M}$. Such a theory $\T^{op}$ gives rise to the universal abelian category $\cA [\T^{op}]$ introduced in \eqref{deftmot} above, i.e., the category of constructible effective $\T^{op}$-motives (see \cite[\S 4]{BV}). Note that there is also a homological regular theory $\T$ which yields a universal abelian category $\cA [\T]$ and we actually have a duality equivalence
$$\cA [\T]^{op}\cong \cA [\T^{op}]$$
as proven in \cite[Prop. 4.1.4]{BV}. This means that the abelian category of theoretical motives associated with the cohomology theory is just given by the opposite of that for the homology theory.
We may further assume the existence of an interval object $I^+$ in $\cC$ and then add the regular axiom of $I^+$-invariance and we still get a regular theory $\T^+$ (see \cite[Def. 2.5.1 \& \S 3.8]{BV}). Adding the axiom of $cd$-exactness, which translates Mayer-Vietoris (see \cite[Def. 2.5.2 \& Lemma 3.8.2]{BV}), also provides a regular theory. Moreover, in order to get that $\cA [\T^{op}]$ is an abelian category we here have to assume that each $h^n(X, Y)$ is an abelian group (see \cite[Lemma 4.1.1 \& Prop. 4.1.3]{BV}). A non additive version of the cohomology theory, which we may also denote $\T^{op}$, is directly obtained by weakening the algebraic structure of $h^n(X, Y)$, e.g., removing the assumption that the group is abelian. In the non additive case the corresponding category $\cA [\T^{op}]$ shall be (Barr) exact only.
Finally, for any such regular theory $\T'$ obtained by adding or removing regular axioms from $\T^{op}$, to give a $\T'$-cohomology $H$ on $\cC$ with values in an abelian (or just exact) category $\cA$, i.e., a $\T'$-model $H$ in $\cA$, is equivalent to giving an exact functor $$r_H: \cA [\T']\to \cA$$ which is the so-called realisation functor (see \cite[Prop. 4.1.3 \& Def. 4.1.5]{BV}). This property says that $\cA [\T']$ is universal with respect to such a $\T'$-models. In fact, this is essentially an instance of the general theorem \cite[Thm. 6.5]{Bu} on regular theories.
\subsection{The theory of a model}
A model $H$ of the theory $\T^{op}$ in an abelian (or exact) category $\cA$ (see \cite[\S 3]{BV}), a cohomology $H$ for short, is given by $$\{H^n (X,Y)\}_{n\in\Z}\in \cA$$ a $\Z$-indexed family of objects which is a family of contravariant functors
$$H: \cC^{\square}\to \cA \hspace*{0.5cm} (X,Y)\leadsto \{H^n (X,Y)\}_{n\in\Z}$$
and, further, $H$ is required to satisfy the exactness axiom involving $\partial$, i.e., for any pair of composable arrows $Z\hookrightarrow Y\hookrightarrow X$ in $\mathcal{M}$, we have the long exact sequence \eqref{longexact} of the triple in $\cA$, which is natural (with respect to $\partial$-cubes, see \cite[\S 3.1 \& (2.1)]{BV}). It follows that $H^n (X,Y)=0$ if $Y\cong X$ is an isomorphism (see \cite[Lemma 3.1.1]{BV}). We shall denote $H^n(X)= H^n(X,\emptyset )$ and note that $H^n(\emptyset) =0$.
Furthermore, we may consider the regular theory $\T_H^{op}$ of the model $H$, i.e., the theory obtained from $\T^{op}$ by adding all regular axioms which are satisfied by the specific cohomology $H$; these additional axioms are all properties of $H$ that can be expressed by making use of regular sequents on the signature $\Sigma^{op}$. For example, if $H$ satisfies $I^+$-invariance (in the sense of \cite[Def. 2.5.1]{BV}) then this axiom is automatically included in the theory $\T_H^{op}$ as it can be formulated by the equality of two function symbols.
Suppose now that excision holds for the cohomology $H$, i.e., for $Y\hookrightarrow X$ and $Z\hookrightarrow X$ such that $X = Y\cup Z$ we have that $\triangle : (Y, Y\cap Z)\to (X,Z)$ induces
$$\triangle^n_H : H^n(X,Z)\stackrel{\simeq}{\longrightarrow} H^n(Y, Y\cap Z)$$
an isomorphism in $\cA$ for all $n\in\Z$. Since we have the function symbol $\triangle^n : h^n(X,Z)\to h^n(Y, Y\cap Z)$ in the signature $\Sigma^{op}$, excision can be expressed in the language of the theory $\T^{op}$ by the following regular sequents: $\top\vdash_{x} \triangle^n (x) = * \rightarrow x = *$ and $\top\vdash_{y} y=y \rightarrow (\exists x) \triangle^n (x)= y$. The validity of these sequents for a model $H$ is equivalent to the fact that $\triangle^n_H$ is an isomorphism of group objects in the regular category $\cA$ (by using \cite[I.5.11]{Ba}). Therefore, adding these axioms to the theory $\T^{op}$ is guaranteeing that the inverse of $\triangle^n$ exists in any model; conversely, if the cohomology $H$ satisfies excision we then have these sequents in the theory $\T_H^{op}$ of the model $H$.
However, this axiom is not equationally expressible in the theory $\T_H^{op}$; for that we would have to add a function symbol $h^n(Y, Y\cap Z)\to h^n(X,Z)$ corresponding to the inverse of $\triangle^n$ (that is not included in $\Sigma^{op}$).
As usual, if excision holds true for $H$, we obtain the following Mayer-Vietoris exact sequence $$\cdots\to H^n(X) \to H^n(Y)\oplus H^n(Z)\to H^n (Y\cap Z) \to H^{n+1} (X) \to \cdots$$
but, similarly, to express exactness of this sequence as axioms of $\T_H^{op}$ we would have to enlarge the signature
$\Sigma^{op}$ to include function symbols $h^n (Y\cap Z) \to h^{n+1} (X)$. We invite the interested reader to compare with $cd$-exactness in \cite[Def. 2.5.2]{BV}.
As already stated, setting $\T'=\T_H^{op}$ on the signature $\Sigma^{op}$ we obtain a universal category $\cA [\T_H^{op}]$, according with the notation adopted above.
Nori's quiver $D^{\rm Nori}$ naturally arises from the sorts and function symbols of the signature $\Sigma^{op}$.
Recall that the quiver $D^{\rm Nori}$ corresponding to $\cC^\square$ is given by the vertices $(X,Y, n)$ for each $n\in\Z$ and edges $(X', Y', n)\to (X, Y, n)$ for each morphism $\square : (X, Y)\to (X', Y')$ in $\cC^\square$ and an additional edge $(Y, Z, n)\to (X, Y, n+1)$ for the morphism $\partial : (Y,Z) \to (X,Y)$. We then get a representation
$$H: D^{\rm Nori}\to \cA \hspace*{0.5cm} (X,Y, n)\leadsto H^n (X,Y)$$
of Nori's quiver. This representation $H$ of $D^{\rm Nori}$ in $\cA$ abelian yields a universal abelian category $\cA (H)$ (as indicated in the introduction, see \cite[\S 1]{BVP}) along with the canonical equivalence \eqref{motives} in the additive case. In fact, for any $\T^{op}$-model $H$ in $\cA$
along with its realisation functor $r_H$ we obtain a universal factorisation through the Serre quotient as follows (see \cite[Prop. 4.3.4]{BV} and \cite[Cor. 2.9]{BVP})
$$\xymatrix{\cA [\T^{op}] \ar@{->>}[r] \ar@/^1.7pc/[rr]^-{r_H} & \cA [\T_H^{op}] \ar@{^{(}->}[r]^-{} & \cA}$$
Note that the realisation $r_H$ is faithful exactly if the model $H$ is conservative, meaning that $\T_H^{op}$ and $\T^{op}$ have equivalent categories of models (compare with \cite[Cor. 2.8]{BVP}). This means that if a regular sequent is valid in the regular theory $\T_H^{op}$ of the model $H$ then it is already valid in $\T^{op}$, and conversely.
In particular, the universal model $U$ in $\cA [\T^{op}]$, whose realisation is the identity functor, is conservative (see also \cite[Prop. 6.4]{Bu}), that is, the regular theory of the universal model is exactly $\T^{op}$ and we have (see \cite[Thm 2.7]{BVP})
$$\catN{U} \cong \cA[\T^{op}].$$
This also shows, as a consequence of the universal representation theorem of \cite{BVP}, that the category $\cA[\T^{op}]$ itself is a Serre quotient of Freyd's free abelian $\catF{D^\Nori}$ on the quiver $D^\Nori$.
\subsection{Good pairs}
Following Nori's original idea, for a cohomology $H: \cC^{\square}\to \cA$ where $(\cA, \tensor )$ we now assume to be a tensor abelian category (in the sense of \cite{DMT}), we consider those pairs $(X, Y)\in \cC^\square$ such that $H^{m}(X, Y)$ is non-zero in a single degree $n$ and zero otherwise. With the same assumptions and notations as in \cite{BVHP}, we shall further assume that $H^{n}(X, Y)$ belongs to a $\flat$-subcategory $\cA^\flat\subset\cA$. Recall that such a $\flat$-subcategory is a full additive subcategory $\cA^\flat$ of $\cA$ such that all objects of $\cA^\flat$ are flat w.r.t.\,$\tensor$ and $\cA^\flat$ is closed under kernels (see \cite[Def. 1.7]{BVHP}). We may additionally assume that the objects of $\cA^\flat$ are projectives in $\cA$ (and this is coherent with the framework explained in \cite[Rmk. 1.10]{BVHP}) but this is not really needed.
\begin{defn}\label{gp}
Let $(\cA, \tensor)$ be an abelian tensor category, with a right exact tensor $\tensor$. Define the \emph{good pairs} for a model $H$ as above to be those $(X, Y)\in \cC^\square$ such that there exists an integer $n$ such that $H^{m}(X, Y)=0$ for $m\neq n$ and $H^{n}(X, Y)\in \cA^\flat\subset\cA$. \end{defn}
\begin{defn}\label{gf} An \emph{$\mathcal{M}$-filtration of dimensional type $d$} of $X\in \cC$ is a finite filtration by $\cM$-subojects $$\Phi: X=X_d \supset \cdots \supset X_p\supset X_{p-1}\supset \cdots \supset X_{-1}=\emptyset$$
where $d\geq -1$ is an integer.
We say that such an $\mathcal{M}$-filtration $\Phi$ is \emph{good} if each $(X_p, X_{p-1})$ is a good pair with $H^{n}(X_p, X_{p-1})=0$ if $n\neq p$.
\end{defn} Clearly we can order (good) $\mathcal{M}$-filtrations by inclusion at each level. In the same way we can also form the join of two $\mathcal{M}$-filtrations of the same dimensional type obtaining an $\mathcal{M}$-filtration but this is not necessarily preserving that the filtration be good. A dual version of \cite[Lemma 4.4.2]{BV} (without passing to ``lim'') yields the following converging \emph{coniveau spectral sequence} in $\cA$
$${ }^\Phi E_1^{p , q}(X) = H^{p+q}(X_{p}, X_{p-1})\Rightarrow H^{p+q}(X)$$
depending on the choosen $\mathcal{M}$-filtration $\Phi$. For a refinement of filtrations $\Phi\subseteq \Psi$ we obtain a morphism of spectral sequences from ${ }^\Psi E_1^{p , q}(X)$ to ${ }^\Phi E_1^{p , q}(X)$. For $\Phi$ a $\mathcal{M}$-filtration on $X$, $\Phi^\prime$ a $\mathcal{M}$-filtration on $X^\prime$ and $f : X\to X'$ a morphism in $\cC$, which is compatible with the filtrations, i.e., it induces $(X_{p}, X_{p-1})\to (X^\prime_{p}, X^\prime_{p-1})$ in $\cC^\square$, we obtain a map of the spectral sequences ${ }^\Phi E_1^{p , q}(X^\prime) \to { }^{\Phi^\prime} E_1^{p , q}(X)$, the induced morphism on the abutments $H^{p+q}(X')\to H^{p+q}(X)$ being the one given the functoriality.
\subsection{Cellularity} Assume that we have a well defined notion of dimension, for any object $X\in \cC$ an integer $\dim (X)=d$, and that we have good $\mathcal{M}$-filtrations of the same dimensional type $d$. In the key geometric examples the dimension exists, e.g., the dimension of a scheme of finite type over a field or the dimension of a CW complex. In general, we here simply assume that $X'\subseteq X$ implies $\dim (X')\leq \dim (X)$ and $\dim (\emptyset) = -1$.
\begin{defn}\label{cell}
Say that the cohomology $H$ is \emph{cellular} with respect to $\cC^\square$ if the following assumptions and conditions are satisfied: {(i)}\, for any object $X\in\cC$ there exists a non-empty family of good $\mathcal{M}$-filtrations of the same dimensional type $d = \dim (X)$, {(ii)}\, for any two $\mathcal{M}$-filtrations of $X$ there is a third good $\mathcal{M}$-filtration containing the join of the given ones and {(iii)}\, if $f : X\to X'$ is a morphism in $\cC$ then there is a good $\mathcal{M}$-filtration on $X'$ containing the direct image under $f$ of any given good $\mathcal{M}$-filtration of $X$.
\end{defn}
Now let $\cC^\good\subseteq \cC^\square$ be the full subcategory of good pairs for $H$. We have that $H$ is a model of $\T^{op}$ for the restricted signature $\Sigma^{op}_\good$ corresponding to $\cC^\good$. We shall denote by $H^\good$ this model.
Let $D^\good\subset D^{\rm Nori}$ be the full subquiver given by $\cC^\good$ so that the restriction of $H$ to $D^\good$ is a representation
$$H^\good: D^\good\to \cA^\flat\subset \cA \hspace*{0.5cm} (X,Y, n)\leadsto H^n (X,Y)$$ in the subcategory $\cA^\flat\subset \cA$ as above. Because of cellularity, we shall see that a dual version of \cite[Lemma 4.4.4]{BV} applies here (without passing to ``lim''). We have that:
\begin{lemma}\label{lemma:goodpair}
If the cohomology $H$ is cellular then $$ \cA[\T_{H^\good}^{op}]\cong \catN{H^\good}\cong \catN{H}\cong \cA[\T_{H}^{op}]$$ are all equivalent abelian categories.
\end{lemma}
\begin{proof} Note that since $D^\good\subset D^{\rm Nori}$ we have that
${\rm Ab} (D^\good) \hookrightarrow {\rm Ab} (D^{\rm Nori})$ is faithful and we therefore obtain faithful exact functors
$$F^\good :\cA[\T_{H^\good}^{op}]\cong\catN{H^\good}\hookrightarrow \catN{H}\hookrightarrow \cA$$
making use of \eqref{motives} for $H^\good$.
We further have that each $X\in \cC$ is provided with a good $\mathcal{M}$-filtration $\Phi$ of dimensional type $d= \dim (X)$ (by Definition \ref{cell} (i)); for any such good $\mathcal{M}$-filtration $\Phi$, $H^{p+q}(X_{p}, X_{p-1})=0$ for $q\neq 0$ and the coniveau spectral sequence ${ }^\Phi E_1^{p , q}(X)$ degenerates in $\cA$. We then have that $H^p(X)\in \cA$ is canonically isomorphic to the $p$-cohomology of the following complex ${ }^\Phi E_1^{* , 0}(X)= { }^\Phi C^{*}_H(X)\in D^b(\cA)$:
$$ \cdots \to H^{p-1}(X_{p-1}, X_{p-2})\to H^{p}(X_{p}, X_{p-1})\to H^{p+1}(X_{p+1}, X_{p})\to \cdots $$ where ${ }^\Phi C^{p}_H(X):= H^{p}(X_{p}, X_{p-1})$. Note that ${ }^\Phi C^{*}_H(X)$ is bounded in $[0, d]$ for $X\neq \emptyset$. Moreover, for two good $\cM$-filtrations $\Phi$ and $\Phi^\prime$ there is a third good filtration $\Psi$ such that $\Phi \subseteq \Psi$ and $\Phi^\prime\subseteq \Psi$ so that the coniveau spectral sequence ${ }^\Psi E_1^{p , q}(X)$ maps to both ${ }^\Phi E_1^{p , q}(X)$ and ${ }^{\Phi^\prime} E_1^{p , q}(X)$ (by Definition \ref{cell} (ii)); therefore, we obtain cochain maps from ${ }^\Psi C^{p}_H(X)$ to both complexes ${ }^\Phi C^{p}_H(X)$ and ${ }^{\Phi^\prime} C^{p}_H(X)$ in such a way that the choice of two good $\mathcal{M}$-filtrations $\Phi$ and $\Phi^\prime$ yields a quasi-isomorphism ${ }^\Phi C^{p}_H(X)\cong { }^{\Phi^\prime} C^{p}_H(X)$ in $D^b(\cA)$.
If $f : Y\to X$ is a morphism and $\Phi$ is a good $\mathcal{M}$-filtration of $Y$ we pick a good $\mathcal{M}$-filtration $\Psi$ on $X$ which is compatible with $f$ (by Definition \ref{cell} (iii)), whence we obtain a cochain map ${ }^\Psi C^{*}_H(X)\to { }^\Phi C^{*}_H(Y)$ showing that the isomorphism $H^n({ }^{\dag}C^{*}_H( - ))\cong H^n( - )$ can be made functorial on $\cC$. In particular, if $(X,Y)$ is an object of $\cC^\square$ we get $H^p({ }^{\Psi, \Phi}C^{*}_H(X,Y))\in \cA$, where ${ }^{\Psi, \Phi}C^{*}_H(X,Y)$ is the fiber (= the shifted mapping cone) of ${ }^{\Psi}C^{*}_H(X)\to { }^{\Phi}C^{*}_H(Y)$ and
$H^n({ }^{\dag}C^{*}_H( - ))\cong H^n( - )$ can be made functorial on $\cC^\square$.
Thus the universal model and/or representation in $\cA[\T_{H^\good}]\cong\catN{H^\good}$ can be extended to a $\T_{H}$-model and/or representation of $D^{\rm Nori}$ compatibly with $H^\good$. By the universality we then easily obtain a faithful exact functor $F_H: \cA[\T_{H}]\cong\catN{H}\hookrightarrow \cA[\T_{H^\good}]\cong\catN{H^\good}$ providing a quasi-inverse of $F^\good$.
\end{proof}
\begin{remark}
Observe that the original argument due to Nori (c.f. \cite[Rem. 9.2.5]{HMS}) in the proof that singular cohomology is cellular with respect to $\cC$ being the category of affine schemes over a field $k$ and $\cM$ the subcategory of closed subschemes follows from the so called ``basic Lemma'' (see \cite[\S 2.5]{HMS}). The previous Lemma \ref{lemma:goodpair} is sufficient to detect Nori motives of all algebraic schemes because the inclusion of affine schemes in all algebraic schemes induces an equivalence on the corresponding categories of motives (e.g., see \cite[Thm. 9.2.22]{HMS}). Similarly, the same holds for Voevodsky motives (e.g., see \cite[Cor. 5.16]{CG}). We leave it to the interested reader to determine general conditions on a cohomology theory $H$ which allow such extension from the affine case. For example, assuming that $I^+$-invariance holds for $H$ (see \cite[Def. 2.5.1 \& \S 3.8]{BV}) and the existence of an affine homotopy replacement one can adapt the arguments in the proof of \cite[Thm. 4.3.2]{FJ}.
\end{remark}
\section{K\"unneth axiom}
Here we express an axiom which corresponds to the usual K\"unneth formula. First of all we need to assume that $\cC$ has finite products $\times$ and a final object $1$ and that the distinguished monomorphisms in $\cM$ are stable under products. Moreover, we shall assume distributivity of joins with respect to products, i.e., for $\cM$-subojects $Y, Z\subset X'$ we assume that $X\times (Y\cup Z) = X\times Y \cup X\times Z$. We have that $\cC^\square$ is provided with a product, as follows
$$(X, Y)\times (X', Y') := (X \times X', X \times Y' \cup Y\times X')$$
Also assume that we have a cohomology satisfying excision in an abelian tensor category with a right exact tensor.
\subsection{External product} Fixing $(\cA, \tensor)$ an abelian tensor category (with a right exact tensor) we are now going to describe a \emph{$\otimes$-model $H$} in $(\cA, \tensor)$ or a cohomology $H$ provided with an \emph{external product $\kappa^H$} by the following data and conditions. We assume given a cohomology $H :\cC\to \cA$ together with a morphism
$$\kappa_{n,n'}^H : H^n(X,Y)\otimes H^{n'}(X',Y')\to H^{n+n'}(X \times X', X \times Y' \cup Y\times X')$$
for all objects $(X, Y)$ and $(X', Y')$ in $\cC^\square$ and $n, n'\in \Z$.
Note that $\kappa^H$ is providing, as usual, by composition with the diagonal $\Delta : (X, Y\cup Z)\to (X \times X, X \times Y \cup Z\times X)$ a cup product $$\cup : H^n(X,Y)\otimes H^{m}(X,Z)\to H^{n+m}(X,Y\cup Z)$$ on cohomology and conversely. We shall assume that $\kappa^H$ satisfies the following axioms:
\begin{itemize}
\item[Ax.0] $\kappa_{}^H$ is compatible with the associativity and commutativity constraints given by the product in $\cC^\square$,
\item[Ax.1] $\kappa_{}^H$ is natural in both variables with respect to morphisms in $\cC^\square$,
\item[Ax.2] for $Z\subseteq Y \subseteq X$ and $Y'\subseteq X'$ the following diagram
\[\begin{xy}\xymatrix{
H^n(Y,Z)\otimes H^{n'}(X',Y')\ar[r]^{\kappa_{}^H\ \ }\ar[d]_{\partial^n\tensor \id}&H^{n+n'}(Y \times X', Y \times Y' \cup Z\times X')\ar[d]^{\delta^{n+n'}}\\
H^{n+1}(X,Y)\otimes H^{n'}(X',Y')\ar[r]^{\kappa_{}^H\ \ }&H^{n+n'+1}(X \times X', X \times Y' \cup Y\times X')
}\end{xy}\] commutes with sign $(-1)^{n'}$, where $\delta^{k}$ for $k= n+n'$ is the composition of the following three morphisms:
\begin{itemize}
\item[{\it i)}] the morphism $H^{k}(Y \times X', Y \times Y' \cup Z\times X') \to H^{k}(Y \times X', Y \times Y')$ induced by functoriality
\item[{\it ii)}] the inverse of $\Delta^k$, the excision isomorphism
$$H^{k}(Y \times X', Y \times Y')\stackrel{\simeq}{\longleftarrow} H^{k}(X \times Y'\cup Y \times X', X \times Y')$$
\item[{\it iii)}] the morphism $\partial^k : H^{k}(X \times Y'\cup Y \times X', X \times Y')\to H^{k+1}(X \times X', X \times Y' \cup Y\times X')$
\end{itemize}
\item[Ax.3] and the following commutes
\[\begin{xy}\xymatrix{
H^{n'}(X',Y')\tensor H^n(Y,Z)\ar[r]^{\kappa_{}^H\ \ }\ar[d]_{\id \tensor \partial^n}&H^{n'+n}(X' \times Y ,
X'\times Z \cup Y' \times Y )\ar[d]^{\delta^{n'+n}}\\
H^{n'}(X',Y')\otimes H^{n+1}(X,Y)\ar[r]^{\kappa_{}^H\ \ }&H^{n'+n+1}(X' \times X, X'\times Y\cup Y' \times X).
}\end{xy}\]
where $\delta^{k}$ for $k= n'+n$ is now induced by the composition of the following three morphisms:
\begin{itemize}
\item[{\it i)}] the morphism $H^{k}(X' \times Y ,
X'\times Z \cup Y' \times Y )\to H^{k}(X' \times Y, Y' \times Y)$ induced by functoriality
\item[{\it ii)}] the inverse of $\Delta^k$, the excision isomorphism
$$H^{k}(X' \times Y, Y' \times Y)\stackrel{\simeq}{\longleftarrow} H^{k}(X' \times Y\cup Y' \times X, Y' \times X)$$
\item[{\it iii)}] the morphism $\partial^k : H^{k}(X' \times Y\cup Y' \times X, Y' \times X)\to H^{k+1}(X' \times X, X'\times Y\cup Y' \times X)$
\end{itemize}
\end{itemize}
We further assume that there is an isomorphism $\varepsilon :H^0(1,\emptyset ) \cong \unit$ with the unit of the tensor structure in such a way that
\begin{itemize}
\item[Ax.4] $H^0(1, \emptyset )\otimes H^n(X, Y)\xrightarrow{\kappa} H^n(1 \times X, 1\times Y\cup \emptyset\times X)\xrightarrow{u^*} H^n(X, Y)$
is the canonical isomorphism $\unit \tensor H^n(X, Y)\cong H^n(X, Y)$ via $\varepsilon$, where $u^*$ is induced by the canonical isomorphism $u: (X, Y) \to (1 \times X, 1\times Y\cup \emptyset\times X)$ (note that here we have that $\emptyset\times - = \emptyset$ as $\emptyset$ is strict initial).
\end{itemize}
\begin{remark} We could consider a theory $\T^\tensor$ which is the extension of the theory $\T^{op}$ on a new signature $\Sigma^\tensor$ containing $\Sigma^{op}$, such that the $h^n(X, Y)\otimes h^{n'}(X', Y')$ are additional sorts, requiring that they are abelian groups and the $\kappa_{n,n'} : h^n(X,Y)\otimes h^{n'}(X',Y')\to h^{n+n'}(X \times X', X \times Y' \cup Y\times X')$
are function symbols which are homomorphisms for each $n, n'\in \Z$.
In that case we would have to assume that $h^n(X, Y)\otimes h^{n'}(X', Y')$ are functorial in each variable and also introduce other sorts and function symbols in order to express the above axioms and the assumption that $h^0(1, \emptyset )$ is a unit. For example, we would need to add function symbols for each variable of $-\tensor +$ corresponding to the function symbols of $\Sigma$, that is, when $\square^n$ is the function symbol associated to a morphism $\square$, or $\partial^n$ for $\partial$, and for the identity function symbol $\boxdot^{m}$ associated to the identity in $\cC^\square$ (refer to \cite{BV} for notation).
Note that we could then also include axioms expressing the strong K\"unneth formula, i.e. that the $``\oplus" \kappa_{i, n-i}$ are isomorphisms, by regular sequents. For example the surjectivity condition can be easily expressed by the regular sequent $$\top\vdash_{y} y =y \rightarrow \exists \overline{x} (\kappa_{0, n}(x_0) + \kappa_{1, n-1}(x_1) + \cdots +\kappa_{n, 0}(x_n)=y).$$
However, few models actually satisfy the strong K\"unneth formula and even in those cases it is not clear how to provide a tensor structure on $\T^{op}$-motives going through $\T^\tensor$-motives. However, it appears to be interesting to understand the differences between the corresponding syntactic categories: the regular syntactic category for $\T^{op}$ on the signature $\Sigma^{op}$ which yields $\cA[\T^{op}]$ and the exact completion of the syntactic category for $\T^\tensor$ on the extended signature $\Sigma^\tensor$.
\end{remark}
\subsection{Nori's graded $\tensor$-quiver}
For $(X, Y, n)$ and $(X', Y', n')$ in $D^\Nori$ we set
$$(X, Y, n)\otimes (X', Y', n') := (X \times X', X \times Y' \cup Y\times X', n+n')$$
which is a vertex of $D^\Nori$. The grading $| \cdot | : D^\Nori\to \Z_2$ given by $|(X, Y,n)| =n$ modulo 2 can be considered here as in \cite[\S 2]{BVHP}. We have, following the notation of that reference, the arrows, which we denote $\alpha$, $\beta$, $\beta'$, for expressing the commutativity and associativity constraints, the unit $\unit = (1, \emptyset , 0)$ and arrows $u$ for expressing its properties all given by the canonical choices in $\cC^\square$.
We shall denote the graded Nori $\tensor$-quiver by $D^{\tensor} = (D^{\rm Nori}, \id,\tensor, \alpha,\beta,\beta',\unit,u)$ with the relations listed in \cite[Def. 2.11 \& Def. 2.12]{BVHP}.
\subsection{K\"unneth formula} Let $H$ be a cohomology in an abelian tensor category $\cA$ with an external product $\kappa^H$. Following Nori's original idea we can actually make use of the K\"unneth formulas in order to get a graded $\tensor$-representation of Nori's $\tensor$-subquiver $D^\good$ which consists of the good pairs for the cohomology $H$. We have a canonical comparison map
$$\oplus \kappa_{i, n-i}^H : ``\bigoplus_{i\in \Z}{}" H^i(X,Y)\otimes H^{n-i}(X',Y')\to H^{n}(X \times X', X \times Y' \cup Y\times X')$$
Clearly, for good pairs, we have that only a single pair $(i, n-i)$ of degrees gives a non-zero component in the sum $``\oplus "\kappa_{i, n-i}^H$.
\begin{defn}\label{kf}
We say that a cohomology $H$ in $\cA$, provided with an external product $\kappa^H$, satisfies the \emph{K\"unneth formula} if, for any $n, n'\in \Z$ and for all pairs $(X, Y)$ and $(X', Y')$ in $\cC^\good$, we have that
$$\kappa_{n, n'}^H : H^n(X,Y)\otimes H^{n'}(X',Y')\to H^{n+n'}(X \times X', X \times Y' \cup Y\times X')$$
is an isomorphism and
\[\begin{xy}\xymatrix{
H^{n}(X,Y)\otimes H^{n'}(X',Y')\ar[r]^{\kappa_{}^H\ \ }\ar[d]_{}&H^{n+n'}(X \times X', X \times Y' \cup Y\times X')\ar[d]^{\alpha^*}\\
H^{n'}(X',Y')\otimes H^{n}(X,Y)&H^{n'+n}(X' \times X, X'\times Y\cup Y' \times X)\ar[l]^{(\kappa_{}^H)^{-1}\ \ }
}\end{xy}\]
commutes with sign $(-1)^{nn'}$.
\end{defn}
\begin{lemma}\label{lemma:kunneth}
If $H :\cC\to \cA$ is a cohomology provided with an external product $\kappa^H$ that satisfies the K\"unneth formula, then $D^\good$ is a graded $\tensor$-subquiver of $D^{\rm Nori}$ and $H^\good$ is a graded $\tensor$-representation of $D^\good$ in $\cA^\flat\subset\cA$ via the external product $\kappa^H$.
\end{lemma}
\begin{proof} In fact, $(X, Y, n)\otimes (X', Y', n')= (X \times X', X \times Y' \cup Y\times X', n+n')$ is a good pair and $H^\good$ is a graded $\tensor$-representation (see \cite[Def. 2.15]{BVHP}) by construction.
\end{proof}
Let us summarise the assumptions on the category $\cC$ in order to state our main result. Making reference to \cite[\S 4.4]{BV} for the explanation of some terminology we assume that
\begin{itemize}
\item[(a)] the category $\cC$ is provided with a final object $1$, products $X \times X'$ for $X, X'\in \cC$ and a strict initial object $\emptyset \in \cC$;
\item[(b)] there is a subcategory $\cM$ of distinguished monomorphisms which are stable under products and such that:
\begin{itemize}
\item[(b.1)] $\cM$ is saturated, contains all isomorphisms $Y\cong X$ and all $\emptyset\to X$ for $X\in \cC$,
\item[(b.2)] for $\cM$-subojects $Y\subset X$ and $Y'\subset X'$ in $\cM$ and any morphism $f : X\to X'$ in $\cC$ there is a direct and inverse image, respectively $f_*(Y)\subset X'$ and $f^*(Y')\subset X$ in $\cM$, and
\item[(b.3)] there are joins $Y\cup Z\subset X$ of $Y\subset X$ and $Z\subset X$ of $\cM$-subojects and they are distributive with respect to products; finally, we assume that
\end{itemize}
\item[(c)] there is a dimension function $\dim : {\rm Ob}\, \cC \to \Z$ such that $\dim (\emptyset ) = -1$ and $X'\subseteq X$ in $\cM$ implies $\dim (X')\leq \dim (X)$.
\end{itemize}
Note that in the concrete geometric categories $\cC$ endowed with a closure operator the subcategory $\cM$ can be given by the subcategory of closed monomorphisms.
\begin{thm}\label{thm:Tmot} Let $\cC$ be a category and $\cM$ a subcategory, satisfying the assumptions and conditions (a), (b) and (c) here above. Let $H :\cC^{\square}\to \cA$ be a cohomology taking values in an abelian tensor category $\cA$ with a right exact tensor and let $\cA^\flat\subset\cA$ be a $\flat$-subcategory. If $H$ is cellular with respect to $\cC^\square$ and $\cA^\flat$ (see Definition \ref{cell}), and it is provided with an external product and satisfies the K\"unneth formula (see Definition \ref{kf}) then the abelian category of motives ${\sf ECM}_\cC^H$ associated to $\cC$ and $H$ is provided with a canonical tensor structure such that the faithful exact functor ${\sf ECM}_\cC^H\to \cA$ is a tensor functor.
\end{thm}
\begin{proof} According with the notation introduced in \eqref{Nori} we here just use $\catN{H^\good} \cong \cA[\T_H]$ by Lemma \ref{lemma:goodpair}. In fact, Lemma \ref{lemma:kunneth} yields that the subdiagram $D^\good\subset D^{\rm Nori}$ given by good pairs along with $H^\good: D^\good\to \cA^\flat$ satisfies the assumptions of the universal graded $\tensor$-representation theorem \cite[Thm 2.18]{BVHP}.
\end{proof}
\subsection{Applications}
All this clearly applies to the well known case of $H$ being cellular cohomology on the category $\cC$ of CW complexes, canonically filtered by $n$-skeletons. Moreover, it applies to $H$ being singular cohomology on the category $\cC$ of affine algebraic $k$-schemes for $k$ a subfield of the complex numbers $\C$; the singular cohomology is cellular because of Nori's basic lemma and we can apply Theorem \ref{thm:Tmot} where $\cA$ is the category of abelian groups and where $\cA^\flat$ is the subcategory of free abelian groups (see also \cite[Thm. 4.5]{BVHP}). Similarly, the interested reader can see that our formalism applies to \cite{Ar}. Note that the representation of singular cohomology in the abelian tensor category $\cA = {\rm MHS}$ of mixed Hodge structures yields back Nori's original category: as soon as we consider the relative case, considering Betti representation in Saito's mixed Hodge modules $\cA = {\rm MHM}$ we get Ivorra's perverse Nori motives \cite{I1}.
Following \cite{FJ}, for pairs $(X, f)$ where $X$ is an algebraic variety defined over a subfield $k$ of $\C$ and $f:X\to \C$ a regular function consider $H^n(X, f)$ the rapid decay cohomology; for $\cC$ the category of pairs $(X, f)$ with $X$ affine this is yielding a $\T^{op}$-model $H$ in the category $\cA$ of finite dimensional $\Q$-vector spaces; one has an exponential basic lemma and a K\"unneth formula (see \cite[Chap. 3-4]{FJ}) in such a way that our assumptions in Theorem \ref{thm:Tmot} are satisfied by rapid decay cohomology, yielding the desired tensor structure on exponential motives. Similarly, it appears that our construction applies in the context of hypergeometric motives \cite{NP} as well.
Finally, as a conjectural application, one aims at constructing a ``sharp" singular cohomology $H$ on a suitable category $\cC$ based on algebraic varieties (e.g., see \cite[\S 0.2]{BB} for more details) and where now $\cA =$ FHM, formal Hodge structures, or EHM, enriched Hodge structures, or, lastly, MHSM, mixed Hodge structures with modulus (see \cite{IY} for work in this direction). One seeks a ``sharp" singular cohomology $H$ that satisfies the hypotheses in Theorem \ref{thm:Tmot} in such a way that we would get an abelian tensor category of ``sharp" Nori motives (or motives with modulus).
| 122,909
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Today, the 24th of Septemeber, the Dominican Republic celebrates a national holiday called “El Día de la Virgen de las Mercedes” or simply “Día de las Mercedes”.
“El Día de la Virgen de las Mercedes” has been celebrated in the Dominican Republic since 1218, when a new religious Order was founded in Spain. Later the practice was adopted in this country – just one of the many examples of the Dominican Republic’s rich cultural heritage that incorporates the celebrations of so many diverse countries.
Although the holiday was originally established in 1218 in Spain, as the Dominican Republic was at that time not yet a Spanish colony, it wasn’t until several hundred years later that Catholicism became the predominant religion in the Dominican Republic. So it was in 1844, the year of the Dominican Independence, that the Virgen de las Mercedes was declared “Patroness of the Dominican Republic”, and the holiday began to be celebrated in this country.
The sanctuary of the Virgen de las Mercedes is located in the ‘Santo Cerro’ (Holy Hill), a small hill on which stood the ancient city of La Concepción de La Vega, this holy place is located in the middle of the Dominican Republic, in between La Vega and Santiago.
The Virgen de las Mercedes is considered a spiritual guide for many cities in the country, and this day is celebrated with great devotion annually on the 24th of September – mostly in the towns of Cabrera, Constanza, Hato Mayor, Imbert, Pimentel and Sabana Grande de Palenque, which are distributed all over the country.
In addition, many hundreds of people travel up to the “Santo Cerro” to pray and offer the Virgen de las Mercedes flowers, as well as taking part in other symbolic rituals.
Happy Mercedes Day, and we hope this can be for all, a beautiful day to enjoy with the family!
| 227,427
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TITLE: What does it mean for a sequence of self-homeomorphism of $\mathbb{R}^n$ to converge to a point?
QUESTION [3 upvotes]: Let $\{f_j\}$ be a family of self-homeomorphisms of $\overline{\mathbb{R}}^n$ and $x,y\in\overline{\mathbb{R}}^n$, where $\overline{\mathbb{R}}^n$ is the one-point compactification of $\mathbb{R}^n$. What does it mean for $\{f_j\}$ to converge c-uniformly to y in $\overline{\mathbb{R}}^n\setminus \{x\}$? That is, what does "$\lim_{j\rightarrow\infty} f_j =y$, c-uniformly in $\overline{\mathbb{R}}^n\setminus {x}$" mean? I'm only confused about what function the point $y$ represents. I would have assumed it meant the function that sends everything to the point $y$, but this is not a bijection on $\overline{\mathbb{R}}^n$.
The source for this question is the paper "Discrete Quasiconformal Groups I" by Gehring and Martin, Proc. London Math. Soc. (3) 55, 1987. Definition 3.3 and Theorem 3.7. My interest here is in understanding convergence groups, in particular their use in defining relatively hyperbolic groups.
REPLY [2 votes]: I'm pretty sure it is meant that for every compact subset $K \subset \overline{\mathbb{R}}^{n} \smallsetminus \{x\}$ and every neighborhood $U$ of $y$ there is $N$ such that $f_{j}(K) \subset U$ for all $j \geq N$. Think of a hyperbolic automorphism of the (closed) Poincaré disk. It has an attractive point $y$ and a repelling point $x$ and precisely this behaviour.
| 175,815
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\begin{document}
\nc{\tred}[1]{\textcolor{red}{#1}}
\nc{\li}[1]{\tred{Li:#1 }}
\title{Nijenhuis algebras, NS algebras and N-dendriform
algebras}
\author{Peng Lei}
\address{
Department of Mathematics, Lanzhou University,
Lanzhou, Gansu 730000, China}
\email{leip@lzu.edu.cn}
\author{Li Guo}
\address{
Department of Mathematics and Computer Science,
Rutgers University,
Newark, NJ 07102, USA}
\email{liguo@newark.rutgers.edu}
\begin{abstract}
In this paper we study (associative) Nijenhuis algebras, with emphasis
on the relationship between the category of Nijenhuis algebras and
the categories of NS algebras. This is in analogy
to the well-known theory of the adjoint functor from the category of
Lie algebras to that of associative algebras, and the more recent
results on the adjoint functor from the categories of dendriform and tridendriform algebras to that of Rota-Baxter algebras. We first give an
explicit construction of free Nijenhuis algebras and
then apply it to obtain the universal enveloping Nijenhuis algebra of an NS algebra. We further apply the construction to determine the binary quadratic nonsymmetric algebra, called the N-dendriform algebra, that is compatible with the Nijenhuis algebra. As it turns out, the N-dendriform algebra has more relations than the NS algebra.
\end{abstract}
\maketitle
\tableofcontents
\setcounter{section}{0}
\section{Introduction}
\mlabel{sec:intro}
Through the antisymmetry bracket $[x,y]:=xy-yx$, an associative algebra $A$ defines a Lie algebra structure on $A$.
The resulting functor from the category of associative algebras to that of Lie algebras and its adjoint functor have played a fundamental role in the study of these algebraic structures.
A similar relationship holds for Rota-Baxter algebras and dendriform algebras.
This paper studies a similar relationship between (associative) Nijenhuis algebras and NS algebras.
\medskip
A {\bf Nijenhuis algebra} is a nonunitary associative algebra $N$ with a linear endomorphism $P$ satisfying the {\bf Nijenhuis equation}:
\begin{equation}
P(x)P(y) = P(P(x)y) + P(xP(y)) - P^2(xy),\quad \forall x,y \in N.
\mlabel{eq:Nij}
\end{equation}
The concept of a Nijenhuis operator on a Lie algebra originated from the important concept of a Nijenhuis tensor that was introduced by Nijenhuis~\mcite{N} in the study of pseudo-complex manifolds in the 1950s and was related to the well-known concepts of Schouten-Nijenhuis bracket, the Fr\"olicher-Nijenhuis bracket~\mcite{FN} and the Nijenhuis-Richardson bracket. Nijenhuis operator operators on a Lie algebra appeared in~\mcite{KM} in a more general study of Poisson-Nijenhuis manifolds and then more recently in~\mcite{GS1,GS2} in the context of the classical Yang-Baxter equation.
The Nijenhuis operator on an associative algebra was introduced by Carinena and coauthors~\mcite{CGM} to study quantum bi-Hamiltonian systems.
In~\mcite{Uc}, Nijenhuis operators are constructed by analogy with Poisson-Nijenhuis geometry, from relative Rota-Baxter operators.
Note the close analogue of the Nijenhuis operator with the more familiar {\bf Rota-Baxter operator} of weight $\lambda$ (where $\lambda$ is a constant) defined to be a linear endomorphism $P$ on an associative algebra $R$ satisfying
$$ P(x)P(y)=P(P(x)y) + P(xP(y)) +\lambda P(xy), \quad \forall x, y\in R.$$
The latter originated from the probability study of G. Baxter~\mcite{Ba}, was studied by Cartier and Rota and is closely related to the operator form of the classical Yang-Baxter equation. Its study has experienced a quite remarkable renascence in the last decade with many applications in mathematics and physics, most notably the work of Connes and Kreimer on renormalization of quantum field theory~\mcite{CK,EGK,EGM}. See~\mcite{Gub} for further details and references.
The recent theoretic developments of Nijenhuis algebras have largely followed those of Rota-Baxter algebras. Commutative Nijenhuis algebras were constructed in~\mcite{EF2,EL} following the construction of free commutative Rota-Baxter algebras~\mcite{GK1}.
Another development followed the relationship between Rota-Baxter algebras and dendriform algebras.
Recall that a {\bf dendriform algebra}, defined by Loday~\mcite{Lo1}, is a vector space $D$ with two binary operations $\prec$ and $\succ$ such that
$$(x\prec y)\prec z=x\prec(y\star z),\; (x\succ y)\prec z=x\succ(y\prec
z),\; (x\star y)\succ z=x\succ(y\succ z), x, y, z\in D,$$
where $\star:=\prec+\succ$. Similarly a {\bf tridendriform algebra}, defined by Loday and Ronco~\mcite{L-R1}, is a vector space $T$ with three binary operations $\prec, \succ$ and $\cdot$ that satisfy seven relations. Aguiar~\mcite{Ag3} showed that for a Rota-Baxter algebra $(R,P)$ of weight 0, the binary operations
$$ x \prec_Py:=xP(y), \quad x\succ_P y:=P(x)y, \quad \forall x, y\in R,$$
define a dendriform algebra on $R$. Similarly, Ebrahimi-Fard~\mcite{EF1} showed that, for a Rota-Baxter algebra $(R,P)$ of non-zero weight, the binary operations
$$ x \prec_Py:=xP(y), \quad x\succ_P y:=P(x)y, \quad x\cdot_P y:= \lambda xy, \quad \forall x, y\in R,$$
define a tridendriform algebra on $R$.
As an analogue of the tridendriform algebra, the concept of an {\bf NS algebra} was introduced by Leroux~\mcite{Le2}, to be a vector space $M$ with three binary
operations $\prec$, $\succ$ and $\bullet$ that satisfy four relations (see Eq.~(\mref{eq:ns})).
As an analogue of the Rota-Baxter algebra case, it was shown~\mcite{Le2} that, for a Nijenhuis algebra $(N,P)$, the binary operations
$$ x \prec_Py:=xP(y), \quad x\succ_P y:=P(x)y, \quad x\bullet_P y:= -P( xy), \quad \forall x, y\in R,$$
defines an NS algebra on $R$.
Considering the adjoint functor of the functor induced by the above mentioned map from Rota-Baxter algebras to (tri-)dendriform algebras, the Rota-Baxter universal enveloping algebra of a (tri-)dendriform algebra was constructed in~\mcite{EG2}. For this purpose, free Rota-Baxter algebras was first constructed.
\smallskip
In this paper we give a similar approach for Nijenhuis algebras, but we go beyond the case of Rota-Baxter algebras. Our first goal is to give an explicit construction of free Nijenhuis algebras in Section~\mref{sec:nonua}.
We consider both the cases when the free Nijenhuis algebra is generated by a set and by another algebra. Other than its role in the theoretical study of Nijenhuis algebras, this construction allows us to construct the universal enveloping algebra of an NS algebra. We achieve this in Section~\mref{sec:adj}.
Knowing that a Nijenhuis algebra gives an NS algebra, it is natural to ask what other dendriform type algebras that Nijenhuis algebras can give in a similar way. As a second application of our construction of free Nijenhuis algebras, we determine all ``quadratic nonsymmetric" relations that can be derived from Nijenhuis algebras and find that one can actually derive more relations than defined by the NS algebra in Eq.~(\mref{eq:ns}). This discussion is presented in Section~\mref{sec:sdn}. \smallskip
\noindent
{\bf Notation:} In this paper $\bfk$ is taken to be a field. A $\bfk$-algebra is taken to be nonunitary associative unless otherwise stated.
\section{Free Nijenhuis algebra on an algebra}
\mlabel{sec:nonua}
We start with the definition of free Nijenhuis algebras.
\begin{defn}
{\rm
Let $A$ be a $\bf k$-algebra. A free Nijenhuis
algebra over $A$ is a Nijenhuis algebra $\FN(A)$ with a
Nijenhuis operator $P_A$ and an algebra homomorphism $j_A:
A\to \FN(A)$ such that, for any Nijenhuis algebra $N$ and
any algebra homomorphism $f:A\to N$, there is a unique
Nijenhuis algebra homomorphism $\free{f}: \FN(A)\to N$
such that $\free{f}\circ j_A=f$:
$$ \xymatrix{ A \ar[rr]^{j_A}\ar[drr]^{f} && \FN(A) \ar[d]_{\free{f}} \\
&& N}
$$
}
\mlabel{de:fna}
\end{defn}
For the construction of free Nijenhuis algebras, we follow the construction of free Rota-Baxter algebras~\mcite{EG2,Gub} by bracketed words. Alternatively, one can follow~\mcite{EG3} to give the construction by rooted trees that is more in the spirit of operads~\mcite{LV}. One can also follow the approach of Gr\"obner-Shirshov bases~\mcite{BCQ}.
Because of the lack of a uniform approach (see~\mcite{GSZ,GSZ2} for some recent attempts in this direction) and to be notationally self contained, we give some details.
We first display a $\bfk$-basis of the free Nijenhuis algebra in
terms of bracketed words in \S~\mref{ss:base}. The product on the free
Nijenhuis algebra is given in \S~\mref{ss:prodao} and the universal
property of the free Nijenhuis algebra is proved in \S~\mref{ss:proof}.
\subsection{A basis of the free Nijenhuis algebra}
\mlabel{ss:base} Let $A$ be a $\bfk$-algebra with a
$\bfk$-basis $X$. We first display a $\bfk$-basis $\frakX_\infty$ of
$\FN(A)$ in terms of bracketed words from the alphabet set $X$.
Let $\lc$ and $\rc$ be symbols, called brackets, and let $X'=X\cup
\{\lc,\rc\}$. Let $M(X')$ denote the free semigroup generated by $X'$.
\begin{defn} (\mcite{EG2,Gub})
{\rm
Let $Y,Z$ be two subsets of $M(X')$. Define the {\bf alternating
product} of $Y$ and $Z$ to be \allowdisplaybreaks{
\begin{eqnarray}
\altx(Y,Z)&=&\Big( \bigcup_{r\geq 1} \big (Y\lc Z\rc \big)^r \Big)
\bigcup
\Big(\bigcup_{r\geq 0} \big (Y\lc Z\rc \big)^r Y\Big)
\bigcup \Big( \bigcup_{r\geq 1} \big( \lc Z\rc Y \big )^r \Big)
\bigcup \Big( \bigcup_{r\geq 0} \big (\lc Z\rc Y\big )^r \lc Z\rc \Big).
\mlabel{eq:wordsao}
\end{eqnarray}}
}
\mlabel{de:alt}
\end{defn}
We construct a sequence $\frakX_n$ of subsets of $M(X')$ by the
following recursion. Let $\frakX_0=X$ and, for $n\geq 0$, define
\allowdisplaybreaks{
\begin{equation}
\frakX_{n+1}=\altx(X,\frakX_n). \notag
\end{equation}
Further, define
\begin{eqnarray}
\frakX_\infty &=& \bigcup_{n\geq 0} \frakX_n = \dirlim \frakX_n.
\mlabel{eq:x3ao}
\end{eqnarray}}
Here the second equation in Eq. (\mref{eq:x3ao}) follows since
$\frakX_1\supseteq \frakX_0$ and, assuming $\frakX_n\supseteq
\frakX_{n-1}$, we have
$$\frakX_{n+1}=\altx(X,\frakX_n) \supseteq \altx(X,\frakX_{n-1})
\supseteq \frakX_n.$$
By~\mcite{EG2,Gub} we have the disjoint union
\allowdisplaybreaks{
\begin{eqnarray}
\frakX_\infty & =&
\Big( \bigsqcup_{r\geq 1} \big (X\lc \frakX_{\infty}\rc\big )^r \Big) \bigsqcup
\Big(\bigsqcup_{r\geq 0} \big (X\lc \frakX_{\infty}\rc\big )^r X\Big) \notag\\
&& \bigsqcup \Big( \bigsqcup_{r\geq 1} \big (\lc \frakX_{\infty}\rc X\big )^r \Big)
\bigsqcup \Big( \bigsqcup_{r\geq 0} \big( \lc \frakX_{\infty}\rc X \big)^r
\lc \frakX_{\infty}\rc \Big).
\mlabel{eq:words3}
\end{eqnarray}}
Further, every $\frakx\in \frakX_\infty$ has a unique decomposition
\begin{equation}
\frakx=\frakx_1 \cdots \frakx_b,
\mlabel{eq:st}
\end{equation}
where $\frakx_i$, $1\leq i\leq b$, is alternatively in $X$ or in
$\lc \frakX_\infty\rc$. This decomposition will be called the {\bf
standard decomposition} of $\frakx$.
For $\frakx$ in ${\frakX}_\infty$ with standard decomposition
$\frakx_1 \cdots \frakx_b$, we define $b$ to be the {\bf breadth}
$b(\frakx)$ of $\frakx$, we define the {\bf head} $h(\frakx)$ of
$\frakx$ to be 0 (resp. 1) if $\frakx_1$ is in $X$ (resp. in $\lc
\frakX_\infty \rc$). Similarly define the {\bf tail} $t(\frakx)$ of
$\frakx$ to be 0 (resp. 1) if $\frakx_b$ is in $X$ (resp. in $\lc
\frakX_\infty \rc$).
\subsection{The product in a free Nijenhuis
algebra} \mlabel{ss:prodao} Let
$$\FN(A)=\bigoplus_{\frakx\in \frakX_\infty} \bfk \frakx.$$
We now define a product $\shpr$ on $\FN(A)$ by defining $\frakx\shpr
\frakx'\in \FN(A)$ for $\frakx,\frakx'\in \frakX_\infty$ and then
extending bilinearly. Roughly speaking, the product of $\frakx$ and
$\frakx'$ is defined to be the concatenation whenever $t(\frakx)\neq
h(\frakx')$. When $t(\frakx)=h(\frakx')$, the product is defined by
the product in $A$ or by the Nijenhuis relation in
Eq.~(\mref{eq:Nij}).
To be precise, we use induction on the sum $n:=d(\frakx)+d(\frakx')$
of the depths of $\frakx$ and $\frakx'$. Then $n\geq 0$. If $n=0$,
then $\frakx,\frakx'$ are in $X$ and so are in $A$ and we define
$\frakx\shpr \frakx'=\frakx \spr \frakx'\in A \subseteq \FN(A)$. Here
$\spr$ is the product in $A$.
Suppose $\frakx\shpr \frakx'$ have been defined for all
$\frakx,\frakx'\in \frakX_\infty$ with $n\geq k\geq 0$ and let
$\frakx, \frakx'\in \frakX_\infty$ with $n=k+1$.
First assume the breadth $b(\frakx)=b(\frakx')=1$. Then $\frakx$ and
$\frakx'$ are in $X$ or $\lc \frakX_\infty\rc$. Since $n=k+1$ is at least one, $\frakx$ and $\frakx'$ cannot be both in $X$.
We accordingly define
\begin{equation}
\frakx\shpr \frakx'=\left \{ \begin{array}{ll}
\frakx \frakx', & {\rm if\ } \frakx\in X, \frakx'\in \lc \frakX_\infty\rc,\\
\frakx \frakx', & {\rm if\ } \frakx\in \lc \frakX_\infty\rc, \frakx'\in X,\\
\lc \lc \ox\rc \shpr \ox'\rc +\lc \ox \shpr \lc \ox'\rc \rc -\lc \lc
\ox \shpr \ox' \rc\rc, & {\rm if\ } \frakx=\lc \ox\rc, \frakx'=\lc
\ox'\rc \in \lc \frakX_\infty \rc.
\end{array} \right .
\mlabel{eq:shprod0}
\end{equation}
Here the product in the first and second case are by concatenation and in the third case is
by the induction hypothesis since for the three products on the
right hand side we have
\begin{eqnarray*}
d(\lc\ox \rc)+ d(\ox') &=& d(\lc \ox \rc)+d(\lc \ox' \rc)-1
= d(\frakx)+d(\frakx')-1,\\
d(\ox)+d(\lc \ox'\rc) &=& d(\lc \ox \rc)+d(\lc \ox'\rc)-1
= d(\frakx)+ d(\frakx')-1,\\
d(\ox)+ d(\ox') &=& d(\lc \ox \rc)-1+ d(\lc \ox' \rc)-1 =
d(\frakx)+d(\frakx')-2
\end{eqnarray*}
which are all less than or equal to $k$.
Now assume $b(\frakx)>1$ or $b(\frakx')>1$. Let
$\frakx=\frakx_1\cdots\frakx_b$ and
$\frakx'=\frakx'_1\cdots\frakx'_{b'}$ be the standard decompositions
from Eq.~(\mref{eq:st}). We then define
\begin{equation}
\frakx \shpr \frakx'= \frakx_1\cdots \frakx_{b-1}(\frakx_b\shpr
\frakx'_1)\,
\frakx'_{2}\cdots \frakx'_{b'}
\end{equation}
where $\frakx_b\shpr \frakx'_1$ is defined by
Eq.~(\mref{eq:shprod0}) and the rest is given by concatenation. The concatenation is well-defined since by Eq.~(\mref{eq:shprod0}), we
have $h(\frakx_b)=h(\frakx_b\shpr \frakx'_1)$ and
$t(\frakx'_1)=t(\frakx_b\shpr \frakx'_1)$. Therefore,
$t(\frakx_{b-1})\neq h(\frakx_b\shpr \frakx'_1)$ and
$h(\frakx'_2)\neq t(\frakx_b\shpr \frakx'_1)$.
\medskip
We have the following simple properties of $\shpr$.
\begin{lemma} Let $\frakx,\frakx'\in \frakX_\infty$. We have the following
statements.
\begin{enumerate}
\item $h(\frakx)=h(\frakx\shpr \frakx')$ and $t(\frakx')=t(\frakx\shpr \frakx')$.
\mlabel{it:mat0}
\item If $t(\frakx)\neq h(\frakx')$, then
$\frakx \shpr \frakx' =\frakx \frakx'$ (concatenation).
\mlabel{it:mat1}
\item If $t(\frakx)\neq h(\frakx')$, then for any $\frakx''\in \frakX_\infty$,
$$(\frakx\frakx')\shpr \frakx'' =\frakx(\frakx' \shpr \frakx''), \quad
\frakx''\shpr (\frakx \frakx') =(\frakx'' \shpr \frakx) \frakx'.$$
\mlabel{it:mat2}
\end{enumerate}
\mlabel{lem:match}
\end{lemma}
Extending $\shpr$ bilinearly, we obtain a binary operation
$$ \FN(A)\otimes \FN(A) \to \FN(A).$$
For $\frakx\in \frakX_\infty$, define
\begin{equation}
N_A(\frakx)=\lc \frakx \rc. \mlabel{eq:RBop}
\end{equation}
Obviously $\lc \frakx \rc$ is again in $\frakX_\infty$. Thus $N_A$
extends to a linear operator $N_A$ on $\FN(A)$. Let
$$j_X:X\to \frakX_\infty \to \FN(A)$$
be the natural injection which extends to an algebra injection
\begin{equation}
j_A: A \to \FN(A). \mlabel{eq:jo}
\end{equation}
The following is our first main result which will be proved in the
next subsection.
\begin{theorem}
Let $A$ be a $\bfk$-algebra with a $\bfk$-basis $X$.
\begin{enumerate}
\item
The pair $(\FN(A),\shpr)$ is an algebra.
\mlabel{it:alg}
\item
The triple $(\FN(A),\shpr,N_A)$ is a Nijenhuis algebra.
\mlabel{it:RB}
\item
The quadruple $(\FN(A),\shpr,N_A,j_A)$ is the free
Nijenhuis algebra on the algebra $A$. \mlabel{it:free}
\end{enumerate}
\mlabel{thm:freeao}
\end{theorem}
The following corollary of the theorem will be used later in the
paper.
\begin{coro}
Let $M$ be a $\bfk$-module and let $T(M)=\bigoplus_{n\geq 1} M^{\ot
n}$ be the reduced tensor algebra over $M$. Then $\FN(T(M))$, together with
the natural injection $i_M: M\to T(M) \xrightarrow{j_{T(M)}}
\FN(T(M))$, is a free Nijenhuis algebra over $M$, in the
sense that, for any Nijenhuis algebra $N$ and
$\bfk$-module map $f: M\to N$ there is a unique Nijenhuis
algebra homomorphism $\freev{f}: \FN(T(M)) \to N$ such that
$\freev{f} \circ k_M = f$. \mlabel{co:vecfree}
\end{coro}
\begin{proof}
This follows immediately from Theorem~\mref{thm:freeao} and the fact that the construction of the free algebra on a module (resp. free Nijenhuis algebra on an algebra, resp. free Nijenhuis on a module) is the left adjoint functor of the forgetful functor from algebras to modules (resp. from Nijenhuis algebras to algebras, resp. from Nijenhuis algebras to modules), and the fact that the composition of two left adjoint functors is the left adjoint functor of the composition.
\end{proof}
\subsection{The proof of Theorem~\mref{thm:freeao}}
\mlabel{ss:proof}
\begin{proof}
\mref{it:alg}. We just need to verify the associativity. For this we
only need to verify
\begin{equation}
(\frakx'\shpr \frakx'')\shpr \frakx''' =\frakx'\shpr(\frakx'' \shpr \frakx''')
\mlabel{eq:assx}
\end{equation}
for $\frakx',\frakx'',\frakx'''\in \frakX_\infty$. We will do this
by induction on the sum of the depths
$n:=d(\frakx')+d(\frakx'')+d(\frakx''')$. If $n=0$, then all of
$\frakx',\frakx'',\frakx'''$ have depth zero and so are in $X$. In
this case the product $\shpr$ is given by the product $\spr$ in $A$
and so is associative.
Assume the associativity holds for $n\leq k$ and assume that
$\frakx',\frakx'',\frakx'''\in \frakX_\infty$ have
$n=d(\frakx')+d(\frakx'')+d(\frakx''')=k+1.$
If $t(\frakx')\neq h(\frakx'')$, then by Lemma~\mref{lem:match},
$$ (\frakx' \shpr \frakx'') \shpr \frakx'''=(\frakx'\frakx'')\shpr \frakx'''
= \frakx' (\frakx'' \shpr \frakx''') =\frakx'\shpr (\frakx''\shpr
\frakx''').$$
A similar argument holds when $t(\frakx'')\neq h(\frakx''')$.
Thus we only need to verify the associativity when
$t(\frakx')=h(\frakx'')$ and $t(\frakx'')=h(\frakx''')$. We next
reduce the breadths of the words.
\begin{lemma}
If the associativity
$$(\frakx' \shpr \frakx'')\shpr \frakx'''=
\frakx'\shpr (\frakx'' \shpr \frakx''') $$
holds for all $\frakx', \frakx''$ and $\frakx'''$ in $\frakX_\infty$
of breadth one, then it holds for all $\frakx', \frakx''$ and
$\frakx'''$ in $\frakX_\infty$. \mlabel{lem:ell}
\end{lemma}
\begin{proof}
We use induction on the sum of breadths
$m:=b(\frakx')+b(\frakx'')+b(\frakx''')$. Then $m\geq 3$. The case
when $m=3$ is the assumption of the lemma. Assume the associativity
holds for $3\leq m \leq j$ and take $\frakx', \frakx'',\frakx'''\in
\frakX_\infty$ with $m = j+1.$ Then $j+1\geq 4$. So at least one of
$\frakx',\frakx'',\frakx'''$ have breadth greater than or equal to
2.
First assume $b(\frakx')\geq 2$. Then $\frakx'=\frakx'_1\frakx'_2$
with $\frakx'_1,\, \frakx'_2\in \frakX_\infty$ and $t(\frakx'_1)\neq
h(\frakx'_2)$. Thus by Lemma~\mref{lem:match}, we obtain
$$
(\frakx'\shpr \frakx'') \shpr \frakx'''=
((\frakx'_1\frakx'_2)\shpr \frakx'')\shpr \frakx'''
= (\frakx'_1 (\frakx'_2 \shpr \frakx''))\shpr \frakx'''
= \frakx'_1 ((\frakx'_2 \shpr \frakx'') \shpr \frakx'''). $$
Similarly,
$$ \frakx'\shpr (\frakx'' \shpr \frakx''')=
(\frakx'_1\frakx'_2)\shpr (\frakx''\shpr \frakx''')
= \frakx'_1 (\frakx'_2 \shpr (\frakx''\shpr \frakx''')).
$$
Thus
$ (\frakx'\shpr \frakx'') \shpr \frakx'''=
\frakx'\shpr (\frakx'' \shpr \frakx''')$
whenever
$ (\frakx'_2 \shpr \frakx'') \shpr \frakx'''=
\frakx'_2 \shpr (\frakx''\shpr \frakx''').$
The latter follows from the induction hypothesis.
A similar proof works if $b(\frakx''')\geq 2.$
Finally if $b(\frakx'')\geq 2$, then $\frakx''=\frakx''_1\frakx''_2$
with $\frakx''_1,\,\frakx''_2\in \frakX_\infty$ and
$t(\frakx''_1)\neq h(\frakx''_2)$. By applying Lemma~\mref{lem:match}
repeatedly, we obtain
$$
(\frakx' \shpr \frakx'')\shpr \frakx'''=
(\frakx' \shpr (\frakx''_1 \frakx''_2)) \shpr \frakx''' \\
= ((\frakx' \shpr \frakx''_1)\frakx''_2)\shpr \frakx'''
= (\frakx'\shpr \frakx''_1)(\frakx''_2 \shpr \frakx''').
$$
In the same way, we have
$$(\frakx'\shpr \frakx''_1)(\frakx''_2 \shpr \frakx''')
= \frakx'\shpr (\frakx'' \shpr \frakx''').$$ This again proves the
associativity.
\end{proof}
To summarize, our proof of the associativity has been reduced to the
special case when $\frakx',\frakx'',\frakx'''\in \frakX_\infty$ are
chosen so that
\begin{enumerate}
\item
$n:= d(\frakx')+d(\frakx'')+d(\frakx''')=k+1\geq 1$ with the
assumption that the associativity holds when $n\leq k$.
\mlabel{it:sp1}
\item
the elements have breadth one and \mlabel{it:sp2}
\item
$t(\frakx')=h(\frakx'')$ and $t(\frakx'')=h(\frakx''')$.
\mlabel{it:sp3}
\end{enumerate}
By item \mref{it:sp2}, the head and tail of each of the elements are
the same. Therefore by item \mref{it:sp3}, either all the three
elements are in $X$ or they are all in $\lc \frakX_\infty \rc$. If
all of $\frakx',\frakx'',\frakx'''$ are in $X$, then as already
shown, the associativity follows from the associativity in $A$.
So it remains to consider the case when $\frakx',\frakx'',\frakx'''$ are all in $\lc
\frakX_\infty \rc$. Then $\frakx'=\lc \ox'\rc, \frakx''=\lc \ox''
\rc, \frakx'''=\lc \ox'''\rc$ with $\ox',\ox'',\ox'''\in
\frakX_\infty$. Using Eq.~(\mref{eq:shprod0}) and bilinearity of the
product $\shpr$, we have
\allowdisplaybreaks{
\begin{eqnarray*}
(\frakx'\shpr \frakx'')\shpr \frakx'''
&=& \big( \lc \lc \ox'\rc \shpr
\ox ''\rc +\lc\ox'\shpr \lc\ox''\rc\rc
-\lc\lc\ox'\shpr \ox''\rc \rc\big ) \shpr \lc \ox'''\rc \\
&=& \lc\lc \ox'\rc \shpr \ox''\rc \shpr \lc\ox'''\rc
+ \lc\ox'\shpr \lc \ox''\rc \rc\shpr \lc \ox'''\rc
-\lc \lc \ox'\shpr \ox''\rc\rc \shpr \lc\ox'''\rc \\
&=& \lc\lc\lc \ox'\rc\shpr \ox''\rc\shpr \ox''' \rc
+ \lc\big(\lc\ox'\rc \shpr \ox''\big) \shpr \lc\ox'''\rc\rc
-\lc \lc\big(\lc\ox'\rc \shpr\ox''\big)\shpr \ox'''\rc\rc\\
&& + \lc\lc\ox'\shpr\lc\ox''\rc\rc \shpr \ox'''\rc
+ \lc\big(\ox'\shpr\lc \ox''\rc\big) \shpr\lc \ox'''\rc\rc
-\lc \lc\big(\ox'\shpr \lc \ox''\rc \big) \shpr \ox'''\rc\rc \\
&& - \lc \lc \lc \ox'\shpr \ox''\rc\rc\shpr \ox'''\rc
-\lc \lc \ox'\shpr \ox''\rc\shpr \lc \ox'''\rc \rc
+\lc\lc \lc \ox'\shpr \ox''\rc \shpr \ox'''\rc\rc.
\end{eqnarray*}}
Applying the induction hypothesis in $n$ to the fifth term $\big
(\ox'\shpr\lc \ox''\rc\big) \shpr\lc \ox'''\rc$ and the eighth term, and then use
Eq.~(\mref{eq:shprod0}) again, we obtain
\allowdisplaybreaks{
\begin{eqnarray*}
(\frakx'\shpr \frakx'')\shpr \frakx'''
&=& \lc\lc\lc \ox'\rc\shpr \ox''\rc\shpr \ox''' \rc
+ \lc\big(\lc\ox'\rc \shpr \ox''\big) \shpr \lc\ox'''\rc\rc
-\lc \lc\big(\lc\ox'\rc \shpr\ox''\big)\shpr \ox'''\rc\rc\\
&& + \lc\lc\ox'\shpr\lc\ox''\rc\rc \shpr \ox'''\rc
+ \lc\ox'\shpr\lc \lc \ox''\rc \shpr\ox'''\rc\rc
+ \lc\ox'\shpr\lc \ox'' \shpr\lc \ox'''\rc\rc\rc\\
&& - \lc\ox'\shpr\lc \lc\ox'' \shpr \ox'''\rc\rc\rc
-\lc \lc\big(\ox'\shpr \lc \ox''\rc \big) \shpr \ox'''\rc\rc
- \lc \lc \lc \ox'\shpr \ox''\rc\rc\shpr \ox'''\rc
\\
&&-\lc \lc \lc\ox'\shpr \ox''\rc\shpr \ox'''\rc \rc
-\lc \lc (\ox'\shpr \ox'')\shpr \lc \ox'''\rc \rc\rc
+\lc \lc \lc(\ox'\shpr \ox'')\shpr \ox'''\rc\rc \rc\\
&& +\lc\lc \lc \ox'\shpr \ox''\rc \shpr \ox'''\rc\rc\\
&=& \lc\lc\lc \ox'\rc\shpr \ox''\rc\shpr \ox''' \rc
+ \lc\big(\lc\ox'\rc \shpr \ox''\big) \shpr \lc\ox'''\rc\rc
-\lc \lc\big(\lc\ox'\rc \shpr\ox''\big)\shpr \ox'''\rc\rc\\
&& + \lc\lc\ox'\shpr\lc\ox''\rc\rc \shpr \ox'''\rc
+ \lc\ox'\shpr\lc \lc \ox''\rc \shpr\ox'''\rc\rc
+ \lc\ox'\shpr\lc \ox'' \shpr\lc \ox'''\rc\rc\rc\\
&& - \lc\ox'\shpr\lc \lc\ox'' \shpr \ox'''\rc\rc\rc
-\lc \lc\big(\ox'\shpr \lc \ox''\rc \big) \shpr \ox'''\rc\rc
- \lc \lc \lc \ox'\shpr \ox''\rc\rc\shpr \ox'''\rc
\\
&& -\lc \lc (\ox'\shpr \ox'')\shpr \lc \ox'''\rc \rc\rc
+\lc \lc \lc(\ox'\shpr \ox'')\shpr \ox'''\rc\rc \rc.
\end{eqnarray*}}
By a similar computation, we obtain
\allowdisplaybreaks{
\begin{eqnarray*}
\frakx' \shpr \big(\frakx''\shpr \frakx'''\big)
&=& \lc\lc\lc\ox'\rc\shpr \ox''\rc\shpr \ox'''\rc
+ \lc \lc \ox'\shpr \lc\ox''\rc\rc \shpr \ox'''\rc
-\lc \lc\lc\ox'\shpr\ox''\rc\rc\shpr\ox'''\rc\\
&& +\lc \ox'\shpr \lc \lc \ox''\rc \shpr \ox'''\rc\rc
- \lc\lc \ox'\shpr \big(\lc\ox''\rc\shpr \ox'''\big)\rc\\
&& + \lc\lc \ox'\rc\shpr \big(\ox''\shpr \lc \ox'''\rc \big) \rc
+ \lc \ox' \shpr \lc \ox'' \shpr \lc \ox'''\rc\rc\rc
-\lc \lc \ox'\shpr \big( \ox''\shpr \lc \ox'''\rc \big)\rc\rc\\
&& - \lc\lc\lc\ox'\rc\shpr (\ox''\shpr \ox''') \rc\rc
+\lc\lc\lc\ox'\shpr (\ox''\shpr \ox''') \rc\rc\rc
- \lc \ox'\shpr \lc\lc \ox''\shpr \ox'''\rc\rc\rc.
\end{eqnarray*}}
Now by induction, the $i$-th term in the expansion of $(\frakx'\shpr
\frakx'')\shpr \frakx'''$ matches with the $\sigma(i)$-th term in
the expansion of $\frakx'\shpr(\frakx'' \shpr \frakx''')$. Here the
permutation $\sigma\in \Sigma_{11}$ is given by
\begin{equation}
\sigma= \left ( \begin{array}{ccccccccccc} 1&2&3&4&5&6&7&8&9&10&11\\
1&6&9&2&4&7&11&5&3&8&10\end{array} \right ).
\mlabel{eq:sigma}
\end{equation}
This completes the proof of Theorem~\mref{thm:freeao}.\mref{it:alg}.
\mref{it:RB}. The proof follows from the definition
$N_A(\frakx)=\lc \frakx\rc$ and Eq. (\mref{eq:shprod0}).
\mref{it:free}. Let $(N,\ast,P)$ be a Nijenhuis algebra with multiplication $\ast$. Let $f:A\to N$ be a $\bfk$-algebra homomorphism. We will construct a
$\bfk$-linear map $\free{f}:\FN(A) \to N$ by defining
$\free{f}(\frakx)$ for $\frakx\in \frakX_\infty$. We achieve this by
defining $\free{f}(\frakx)$ for $\frakx\in \frakX_n,\ n\geq 0$,
inductively on $n$. For $\frakx\in \frakX_0:=X$, define
$\free{f}(\frakx)=f(\frakx).$ Suppose $\free{f}(\frakx)$ has been
defined for $\frakx\in \frakX_n$ and consider $\frakx$ in
$\frakX_{n+1}$ which is, by definition and Eq.~(\mref{eq:words3}),
\allowdisplaybreaks{
\begin{eqnarray*} \altx(X,\frakX_{n})& =&
\Big( \bigsqcup_{r\geq 1} (X\lc \frakX_{n}\rc)^r \Big) \bigsqcup
\Big(\bigsqcup_{r\geq 0} (X\lc \frakX_{n}\rc)^r X\Big) \\
&& \bigsqcup \Big( \bigsqcup_{r\geq 0} \lc \frakX_{n}\rc (X\lc \frakX_{n}\rc)^r \Big)
\bigsqcup \Big( \bigsqcup_{r\geq 0} \lc \frakX_{n}\rc (X\lc \frakX_{n}\rc)^r X\Big).
\end{eqnarray*}}
Let $\frakx$ be in the first union component $\bigsqcup_{r\geq 1}
(X\lc \frakX_{n}\rc)^r$ above. Then
$$\frakx = \prod_{i=1}^r(\frakx_{2i-1} \lc \frakx_{2i} \rc)$$
for $\frakx_{2i-1}\in X$ and $\frakx_{2i}\in \frakX_n$, $1\leq i\leq
r$. By the construction of the multiplication $\shpr$ and the
Nijenhuis operator $N_A$, we have
$$\frakx= \shpr_{i=1}^r(\frakx_{2i-1} \shpr \lc \frakx_{2i}\rc)
= \shpr_{i=1}^r(\frakx_{2i-1} \shpr N_A(\frakx_{2i})).$$
Define
\begin{equation}
\free{f}(\frakx) = \ast_{i=1}^r \big(\free{f}(\frakx_{2i-1})
\ast N\big (\free{f}(\frakx_{2i})) \big).
\mlabel{eq:hom}
\end{equation}
where the right hand side is well-defined by the induction
hypothesis. Similarly define $\free{f}(\frakx)$ if $\frakx$ is in
the other union components. For any $\frakx\in \frakX_\infty$, we
have $P_A(\frakx)=\lc \frakx\rc\in \frakX_\infty$, and by the definition of $\free{f}$ in (Eq. (\mref{eq:hom})), we have
\begin{equation}
\free{f}(\lc \frakx \rc)=P(\free{f}(\frakx)). \mlabel{eq:hom1-2}
\end{equation}
So $\free{f}$ commutes with the Nijenhuis operators. Combining this
equation with Eq.~(\mref{eq:hom}) we see that if
$\frakx=\frakx_1\cdots \frakx_b$ is the standard decomposition of
$\frakx$, then
\begin{equation}
\free{f}(\frakx)=\free{f}(\frakx_1)*\cdots * \free{f}(\frakx_b).
\mlabel{eq:staohom}
\end{equation}
Note that this is the only possible way to define $\free{f}(\frakx)$
in order for $\free{f}$ to be a Nijenhuis algebra homomorphism
extending $f$.
It remains to prove that the map $\free{f}$ defined in
Eq.~(\mref{eq:hom}) is indeed an algebra homomorphism. For this we
only need to check the multiplicity
\begin{equation}
\free{f} (\frakx \shpr \frakx')=\free{f}(\frakx) \ast
\free{f}(\frakx') \mlabel{eq:hom2}
\end{equation}
for all $\frakx,\frakx'\in \frakX_\infty$. For this we use induction
on the sum of depths $n:=d(\frakx)+d(\frakx')$. Then $n\geq 0$. When
$n=0$, we have $\frakx,\frakx'\in X$. Then Eq.~(\mref{eq:hom2})
follows from the multiplicity of $f$. Assume the multiplicity holds
for $\frakx,\frakx' \in \frakX_\infty$ with $n\geq k$ and take
$\frakx,\frakx'\in \frakX_\infty$ with $n=k+1$. Let
$\frakx=\frakx_1\cdots \frakx_b$ and
$\frakx'=\frakx'_1\cdots\frakx'_{b'}$ be the standard
decompositions. Since $n=k+1\geq 1$, at least one of $\frakx_b$ and $\frakx'_{b'}$ is in $\lc \frakX_\infty\rc$. Then by Eq.~(\mref{eq:shprod0}) we have,
\begin{align*}
\free{f}(\frakx_b\shpr \frakx'_1)&= \left \{\begin{array}{ll}
\free{f}(\frakx_b \frakx'_1), & {\rm if\ } \frakx_b\in X, \frakx'_1\in \lc \frakX_\infty\rc,\\
\free{f}(\frakx_b \frakx'_1), & {\rm if\ } \frakx_b\in \lc
\frakX_\infty\rc,
\frakx'_1\in X,\\
\free{f}\big( \lc \lc \ox_b\rc \shpr \ox'_1\rc +\lc \ox_b \shpr \lc
\ox'_1\rc \rc -\lc \lc \ox_b \shpr \ox'_1 \rc\rc\big), & {\rm if\ }
\frakx_b=\lc \ox_b\rc, \frakx'_1=\lc \ox'_1\rc \in \lc \frakX_\infty
\rc.
\end{array} \right .
\end{align*}
In the first two cases, the right hand side is
$\free{f}(\frakx_b)*\free{f}(\frakx'_1)$ by the definition of
$\free{f}$. In the third case, we have, by Eq.~(\mref{eq:hom1-2}),
the induction hypothesis and the Nijenhuis relation of the operator $P$ on $N$,
\begin{align*}
&\free{f}\big( \lc \lc \ox_b\rc \shpr \ox'_1\rc
+\lc \ox_b \shpr \lc \ox'_1\rc \rc
-\lc \lc \ox_b \shpr \ox'_1 \rc\rc\big)\\
=&\free{f}(\lc \lc \ox_b\rc \shpr \ox'_1\rc) + \free{f}(\lc \ox_b
\shpr \lc \ox'_1\rc \rc)
-\free{f}(\lc \lc \ox_b \shpr \ox'_1 \rc\rc)\\
=&P(\free{f}(\lc \ox_b\rc \shpr \ox'_1)) + P(\free{f}(\ox_b \shpr
\lc \ox'_1\rc ))
-P(\free{f}(\lc\ox_b \shpr \ox'_1 \rc))\\
=&P(\free{f}(\lc \ox_b\rc)*\free{f}(\ox'_1)) + P(\free{f}(\ox_b)
*\free{f}( \lc \ox'_1\rc ))
-P(P(\free{f}(\ox_b) * \free{f}(\ox'_1)) )\\
=&P(P(\free{f}(\ox_b))*\free{f}(\ox'_1)) + P(\free{f}(\ox_b)
*P(\free{f}(\ox'_1)))
-P(P((\free{f}(\ox_b) * \free{f}(\ox'_1)) )\\
=& P(\free{f}(\ox_b))*P(\free{f}(\ox'_1))\\
=& \free{f}(\lc \ox_b\rc) * \free{f}(\lc\ox'_1\rc)\\
=& \free{f} (\frakx_b) *\free{f}(\frakx'_1).
\end{align*}
Therefore $\free{f}(\frakx_b\shpr
\frakx'_1)=\free{f}(\frakx_b)*\free{f}(\frakx'_1)$. Then
\begin{align*}
\free{f}(\frakx\shpr \frakx')&=
\free{f}\big(\frakx_1\cdots\frakx_{b-1}(\frakx_b\shpr
\frakx'_1)\frakx'_2\cdots
\frakx'_{b'}\big) \\
&= \free{f}(\frakx_1)*\cdots *\free{f}(\frakx_{b-1})*
\free{f}(\frakx_b\shpr \frakx'_1)*\free{f}(\frakx'_2)\cdots
\free{f}(\frakx'_{b'})\\
&= \free{f}(\frakx_1)*\cdots *\free{f}(\frakx_{b-1})*
\free{f}(\frakx_b)* \free{f} (\frakx'_1)*\free{f}(\frakx'_2)\cdots
\free{f}(\frakx'_{b'})\\
&= \free{f}(\frakx)*\free{f}(\frakx').
\end{align*}
This is what we need.
\end{proof}
\section{NS algebras and their universal enveloping algebras}
\mlabel{sec:adj}
The concept of an NS algebra was introduced by Leroux~\mcite{Le2} as an analogue of the dendriform algebra of Loday~\mcite{Lo1} and the tridendriform algebra of Loday and Ronco~\mcite{L-R1}.
\begin{defn}
{\rm
An {\bf NS algebra} is a module $M$ with three binary
operations $\prec$, $\succ$ and $\bullet$ that satisfy the following
four relations
\begin{eqnarray}
&(x\prec y)\prec z=x\prec (y\star z),\quad
(x\succ y)\prec z=x\succ (y\prec z), & \notag \\
&(x\star y)\succ z=x\succ
(y\succ z),\quad
(x\star y)\bullet z+(x\bullet y)\prec z = x \succ (y\bullet z)
+x\bullet (y\star z).&
\mlabel{eq:ns}
\end{eqnarray}
for $x,y,z\in M$. Here $\star$ denotes $\prec+\succ+\,
\bullet$.
}
\end{defn}
NS algebras share similar properties as dendriform algebras. For example, the operation
$\star$ defines an associative operation. Another similarity is the following theorem which is an analogue of the results of Aguiar~\mcite{Ag3} and Ebrahimi-Fard~\mcite{EF1} that a Rota-Baxter algebra gives a dendriform algebra or a tridendriform algebra.
\begin{theorem} $($\mcite{Le2}$)$ A Nijenhuis algebra $(N,P)$ defines an NS algebra $(N,\prec_P,\succ_P,\bullet_P)$, where
\begin{equation} x\prec_P y=xP(y),\ x\succ_P y=P(x)y, x\bullet_P y=-P(xy).
\mlabel{eq:eqs}
\end{equation}
\mlabel{thm:le}
\end{theorem}
Let $\NA$ denote the category of Nijenhuis algebras and let $\NS$ denote the category of NS algebras.
It is easy to see
that the map from $\NA$ to $\NS$ in Theorem~\mref{thm:le} is compatible with the morphisms in the two categories. Thus we obtain a functor
\begin{equation}
\cale: \NA \to \NS.
\mlabel{eq:nsdn}
\end{equation}
We will study its left adjoint functor.
Motivated by the enveloping algebra of a Lie
algebra and the Rota-Baxter enveloping algebra of a tridendriform algebra~\mcite{EG2}, we are naturally led to the following definition.
\begin{defn}
{\rm
Let $M$ be an NS-algebra. A {\bf universal enveloping Nijenhuis algebra} of
$M$ is a Nijenhuis algebra
$\UN(M)\in \NA$ with a homomorphism $\rho: M\to \UN(M)$ in $\NS$ such
that for any $N\in \NA$ and homomorphism $f:M\to N$ in $\NS$, there is a unique $\den{f}: \UN(M)\to N$ in $\NA$ such that $\den{f} \circ
\rho =f$.
}\mlabel{de:env}
\end{defn}
Let $M:=(M,\prec,\succ,\bullet)\in \NS$. Let $T(M)=\bigoplus_{n\geq 1}
M^{\ot n}$ be the tensor algebra. Then $T(M)$ is the free algebra generated by the $\bfk$-module
$M$.
By Corollary~\mref{co:vecfree}, $\FN(T(M))$, with the natural
injection $i_M: M\to T(M) \to \FN(T(M))$, is the free Nijenhuis
algebra over the vector space $M$.
Let $J_M$ be the Nijenhuis ideal of $\FN(T(M))$ generated by the set
\begin{equation}
\big \{ x\prec y - xP( y),\;
x\succ y - P( x) y,\;x\bullet y=P(x\otimes y) \big|\ x,y\in M \big\}
\mlabel{eq:gendend}
\end{equation}
Let $\pi: \FN(T(M))\to \FN(T(M))/J_M$ be the quotient map.
\begin{theorem}
Let $(M,\prec,\succ,\bullet)$ be an NS algebra. The quotient Nijenhuis algebra $\FN(T(M))/J_M$, together with $\rho:=
\pi \circ i_M$, is the universal enveloping Nijenhuis algebra of $M$. \mlabel{thm:envdend}
\end{theorem}
\begin{proof}
The proof is similar to the case of tridendriform algebras and Rota-Baxter algebras~\mcite{EG2}. So we skip some of the details.
Let $(N,P)$ be a Nijenhuis algebra and let $f:M\to N$ be a homomorphism
in $\NS$. More precisely, we have $f:(M,\prec,\succ,\bullet) \to (N,\prec_P',\succ_P',\bullet_P)$. We will complete the following commutative
diagram, using notations from Corollary~\mref{co:vecfree}.
\begin{equation}
\xymatrix{ & T(M) \ar[rd]^{j_{T(M)}} \ar@{.>}[lddd]^{\freea{f}} & \\
M \ar[rr]^{i_M} \ar[dd]_f \ar[ru]^{k_M} && \FN(T(M)) \ar[dd]^\pi \ar@{.>}[ddll]_{\freev{f}} \\
&& \\
N && \FN(T(M))/J_M \ar@{.>}[ll]_{\den{f}} }
\end{equation}
By the universal property of the free algebra $T(M)$ over $M$, there
is a unique homomorphism $\freea{f}:T(M)\to N$ such that
$\freea{f}\circ k_M =f$. So $\freea{f}(x_1\ot \cdots \ot
x_n)=f(x_1) * \cdots * f(x_n)$. Here $*$ is the product in $N$. Then
by the universal property of the free Nijenhuis algebra $\FN(T(M))$
over $T(M)$, there is a unique morphism $\free{\freea{f}}:\FN(T(M))
\to N$ in $\NA$ such that $\free{\freea{f}}\circ j_{T(M)}
=\freea{f}$. By Corollary~\mref{co:vecfree},
$\free{\freea{f}}=\freev{f}.$ Then
\begin{equation} \freev{f}\circ i_M =\freev{f} \circ j_{T(M)} \circ
k_M = \freea{f} \circ k_M = f. \mlabel{eq:free2}
\end{equation}
So for any $x,y\in M$, we check that
\begin{eqnarray*}
\freev{f}(x\prec y - x P( y))=0, \quad
\freev{f}(x\succ y - P(x)y)=0, \quad
\freev{f}(x\bullet y - P(x\otimes y))=0.
\end{eqnarray*}
Thus $J_M$ is in $\ker(\freev{f})$ and there is a morphism $\den{f}:
\FN(T(M))/J_M\to N$ in $\NA$ such that $\freev{f}=\den{f} \circ \pi$.
Then by the definition of $\rho=\pi \circ i_M$ in the theorem and
Eq. (\mref{eq:free2}), we have
$$ \den{f}\circ \rho = \den{f} \circ \pi \circ i_M=\freev{f}\circ i_M=f.$$
This proves the existence of $\den{f}$.
Suppose $\den{f}':\FN(T(M))/J_M \to N$ is also a homomorphism in $\NA$
such that $\den{f}'\circ \rho=f$. Then
$$ (\den{f}' \circ \pi)\circ i_M = f = (\den{f}\circ \pi)\circ i_M.$$
By Corollary~\mref{co:vecfree}, the free Nijenhuis algebra
$\FN(T(M))$ over the algebra $T(M)$ is also the free Nijenhuis
algebra over the vector space $M$ with respect the natural injection
$i_M$. So we have $\den{f}'\circ \pi = \den{f} \circ \pi$ in $\NA$.
Since $\pi$ is surjective, we have $\den{f}'=\den{f}$. This proves
the uniqueness of $\den{f}$.
\end{proof}
\section{From Nijenhuis algebras to N-dendriform algebras}
\mlabel{sec:sdn}
In this section, we consider an inverse of Theorem~\mref{thm:le} in the following sense. Suppose $(N,P)$ is a Nijenhuis algebra and define binary operations $$
x\prec_P y=xP(y),\ x\succ_P y=P(x)y, x\bullet_P y=-P(xy).$$
By Theorem~\mref{thm:le}, the three operations satisfy the NS relations in Eq.~(\mref{eq:ns}). Our inverse question is, what other quadratic nonsymmetric relations could
$(N,\prec_P,\succ_P,\bullet_P)$ satisfy? We recall some background on binary quadratic nonsymmetric operads in order to make the question precise. We then determine all the quadratic nonsymmetric relations that are consistent with the Nijenhuis operator.
\subsection{Background and the statement of Theorem~\mref{thm:wdn}}
For details on binary quadratic nonsymmetric operads, see~\mcite{Gub,LV}.
\begin{defn}
{\rm
Let $\bfk$ be a field.
\begin{enumerate}
\item
A {\bf graded vector space} is a sequence $\calp:=\{\calp_n\}_{n\geq 0}$ of $\bfk$-vector spaces $\calp_n, n\geq 0$.
\item
A {\bf nonsymmetric (ns) operad} is a graded vector space $\calp=\{\calp_n\}_{n\geq 0}$ equipped with {\bf partial compositions}:
\begin{equation}
\circ_i:=\circ_{m,n,i}: \calp_m\ot \calp_n\longrightarrow \calp_{m+n-1}, \quad 1\leq i\leq m,
\mlabel{eq:opc}
\end{equation}
such that, for $\lambda\in\calp_\ell, \mu\in\calp_m$ and $\nu\in\calp_n$, the following relations hold.
\begin{enumerate}
\item[(i)] $
(\lambda \circ_i \mu)\circ_{i-1+j}\nu = \lambda\circ_i (\mu\circ_j\nu), \quad 1\leq i\leq \ell, 1\leq j\leq m.$
\mlabel{it:esc}
\item[(ii)]$(\lambda\circ_i\mu)\circ_{k-1+m}\nu =(\lambda\circ_k\nu)\circ_i\mu, \quad
1\leq i<k\leq \ell.$
\mlabel{it:epc}
\item[(iii)]
There is an element $\id\in \calp_1$ such that $\id\circ \mu=\mu$ and $\mu\circ\id=\mu$ for $\mu\in \calp_n, n\geq 0$.
\mlabel{it:id}
\end{enumerate}
\end{enumerate}
}
\end{defn}
An ns operad $\calp=\{\calp_n\}$ is called {\bf binary} if $\calp_1=\bfk.\id$ and $\calp_n,
n\geq 3$ are induced from $\calp_2$ by composition. Then in particular, for the free operad, we have
\begin{equation}
\calp_3=(\calp_2 \circ_1 \calp_2) \oplus (\calp_2\circ_2 \calp_2),
\mlabel{eq:bq}
\end{equation}
which can be identified with $\calp_2^{\ot 2}\oplus \calp_2^{\ot 2}$. A binary ns operad $\calp$ is called {\bf quadratic} if
all relations among the binary operations in $\calp_2$ are derived
from $\calp_3$.
Thus a binary, quadratic, ns operad
is determined by a pair $(\dfgen,\dfrel)$ where $\dfgen=\calp_2$, called
the {\bf space of generators},
and $\dfrel$ is a subspace of $\dfgen^{\otimes 2} \oplus \dfgen^{\otimes 2}$,
called the {\bf space of relations.} So we can denote $\calp=\calp(\dfgen)/(\dfrel)$.
Note that a typical element of $\dfgen^{\ot 2}$ is of the form $\sum\limits_{i=1}^k\dfoa_i\otimes \dfob_i$ with $\dfoa_i,\dfob_i\in V, 1\leq i\leq k$. Thus a typical element of
$\dfgen^{\otimes 2} \oplus \dfgen^{\otimes 2}$ is of the form
$$\left (\sum_{i=1}^k\dfoa_i\otimes \dfob_i, \sum_{j=1}^m\dfoc_j\otimes \dfod_j \right), \quad
\dfoa_i,\dfob_i,\dfoc_j,\dfod_j\in \dfgen,
1\leq i\leq k, 1\leq j\leq m, k, m\geq 1.$$
For a given binary quadratic ns operad $\calp=\calp(\dfgen)/(\dfrel)$, a $\bfk$-vector space $A$ is called a {\bf $\calp$-algebra}
\index{$\calp$-algebra}
if $A$ has binary operations (indexed by) $\dfgen$ and if,
for
$$\big (\sum_{i=1}^k\dfoa_i\otimes \dfob_i, \sum_{j=1}^m\dfoc_j\otimes \dfod_j \big)
\in \dfrel \subseteq
\dfgen^{\otimes 2} \oplus \dfgen^{\otimes 2}$$ with
$\dfoa_i,\dfob_i,\dfoc_j,\dfod_j\in \dfgen$, $1\leq i\leq k$, $1\leq j\leq m$,
we have
\begin{equation}
\sum_{i=1}^k(x\dfoa_i y) \dfob_i z = \sum_{j=1}^m x \dfoc_j (y \dfod_j z), \quad \forall\ x,y,z\in A.
\mlabel{eq:rel}
\end{equation}
For example, from Eq.~(\mref{eq:ns}) the NS algebras are precisely the $\calp$-algebras where $\calp=\calp(V)/(R)$ with $R$ being the subspace of $V^{\ot 2}\oplus V^{\ot 2}$ spanned by the four elements
\begin{eqnarray*}
&(\prec\ot\prec,\prec \ot\star),\quad
(\succ \ot\prec,\succ\ot\prec), & \\
&(\star \ot\succ,\succ\ot\succ),\quad
(\star\ot\bullet +\bullet \,\ot\prec, \succ\ot\bullet
+\bullet\ot\star),&
\end{eqnarray*}
where $\star=\prec+\succ+\,\bullet$.
\begin{theorem}
Let $V=\bfk\{\prec,\succ,\bullet\}$ be the vector space with basis $\{\prec,\succ,\bullet\}$ and let $\calp=\calp(\dfgen)/(\dfrel)$ be a binary quadratic ns operad. The following statements are equivalent.
\begin{enumerate}
\item
For every Nijenhuis algebra $(N,P)$, the quadruple $(N,\prec_P,\succ_P,\bullet_P)$ is a $\calp$-algebra.
\mlabel{it:nap}
\item
The relation space $\dfrel$ of $\calp$ is contained in the subspace of $V^{\ot 2}\oplus V^{\ot 2}$ spanned by
\begin{eqnarray}
&&(\prec \ot \prec,\prec\ot \star),\notag\\
&&(\succ \ot \prec,\succ\ot \prec),\notag\\
&&(\succ\ot \star, \succ\ot \succ),
\mlabel{eq:wdn}\\
&&(\prec\ot\bullet, \bullet\,\ot\succ),\notag\\
&&(\succ\ot\bullet+\bullet\ot\prec+\bullet\ot\bullet,
\succ\ot\bullet+\bullet\ot\prec+\bullet\ot\bullet), \notag
\end{eqnarray}
where $\star=\prec+\succ+\,\bullet\,$.
More precisely, any $\calp$-algebra $A$ satisfies the relations
\begin{eqnarray}
(x\prec y)\prec z&=&x \prec(y \prec z)+x \prec(y\prec
z)+x\prec(y\bullet z), \notag\\
(x\succ y)\prec z&=&x\succ(y\prec z), \notag\\
(x\prec y)\succ z+(x\succ y)\succ z+(x\bullet
y)\succ z&=&x\succ(y\succ z), \quad \forall x,y,z\in A \mlabel{eq:wdna}\\
(x\prec y)\bullet z&=&x\bullet (y\succ z), \notag\\
(x\succ y)\bullet z+(x\bullet y)\prec z+(x\bullet y)\bullet
z&=&x\succ(y\bullet z)+x\bullet (y\prec z)+x\bullet (y\bullet z). \notag
\end{eqnarray}
\mlabel{it:pna}
\end{enumerate}
\mlabel{thm:wdn}
\end{theorem}
Note that the relations of the NS algebra in Eq.~(\mref{eq:ns}) is contained in the space spanned by the relations in Eq.~(\mref{eq:wdn}). We call $\calp$ defined by the relations in Eq.~(\mref{eq:wdn}) the {\bf N-dendriform operad} and call a quadruple $(A,\prec,\succ,\bullet)$ satisfying Eq.~(\mref{eq:wdna}) an {\bf N-dendriform algebra}. Let $\ND$ denote the category of N-dendriform algebras. Then we have the following immediate corollary of Theorem~\mref{thm:wdn}.
\begin{coro}
\begin{enumerate}
\item
There is a natural functor
\begin{equation}
\calf: \NA \to \ND, \quad (N,P)\mapsto (N,\prec_P,\succ_P,\bullet_P).
\mlabel{eq:nasdn}
\end{equation}
\mlabel{it:nawdn}
\item
There is a natural (inclusion) functor
\begin{equation}
\calg: \ND\to \NS, \quad (M,\prec,\succ,\bullet) \mapsto (M,\prec,\succ,\bullet).
\mlabel{eq:sdndn}
\end{equation}
\mlabel{it:dnwdn}
\item
The functors $\calf$ and $\calg$ give a refinement of the functor $\cale:\NA\to \NS$ in Eq.~(\mref{eq:ns}) in the sense that the following diagram commutes
\begin{equation}
\xymatrix{ \NA \ar[rr]^{\calf} \ar[rrd]^{\cale} && \ND \ar[d]^{\calg} \\
&& \NS
}
\end{equation}
\mlabel{it:comm}
\end{enumerate}
\end{coro}
\subsection{The proof of Theorem~\mref{thm:wdn}}
With $V=\bfk \{\prec, \succ, \bullet\}$, we have
$$V^{\ot 2} \oplus V^{\ot 2} = \bigoplus\limits_{\dfop_1,\dfop_2,\dfop_3,\dfop_4\in \{\prec,\succ,\bullet\}} \bfk (\dfop_1\ot\dfop_2, \dfop_3\ot\dfop_4).$$
Thus any element $r$ of $\dfgen^{\otimes 2} \oplus \dfgen^{\otimes 2}$ is of the form
\begin{eqnarray*}
r&:=&a_1(\prec\ot \prec,0)+a_2(\prec\ot\succ,0)+a_3(\prec\ot \bullet,0)\\
&&+b_1(\succ\ot \prec,0)+b_2(\succ\ot \succ,0)+b_3(\succ\ot\bullet,0)\\
&&+c_1(\bullet\,\ot \prec,0)+c_2(\bullet\,\ot\succ,0)+c_3(\bullet\ot\bullet,0)\\
&&+d_1(0,\prec\ot \prec)+d_2(0,\prec\ot \succ)+d_3(0,\prec\ot \bullet)\\
&&+e_1(0,\succ\ot \prec)+e_3(0,\succ\ot \succ)+e_3(0,\succ\ot\bullet)\\
&&+f_1(0,\bullet\ot\prec)+f_2(0,\bullet\ot\succ) +f_3(0,\bullet\ot\bullet)
\end{eqnarray*}
where the coefficients are in $\bfk$.
\smallskip
\noindent
(\mref{it:nap} $\Rightarrow$ \mref{it:pna})
Let $\calp=\calp(V)/(R)$ be an operad satisfying the condition in Item~\mref{it:nap}. Let $r$ be in $R$ expressed in the above form. Then for any Nijenhuis algebra $(N,P)$, the quadruple $(N,\prec_P,\succ_P,\bullet_P)$ is a $\calp$-algebra. Thus \begin{eqnarray*}
&&a_1(x\prec_P y)\prec_P z+a_2(x\prec_P y)\succ_P z +a_3(x\prec_P y)\bullet_P z \\
&& +b_1(x\succ_P y)\prec_P z+b_2(x\succ_P y)\succ_P z+b_3(x\succ_P y)\bullet_P
z\\
&&+c_1(x\bullet_P y)\prec_P z+c_2(x\bullet_P y)\succ_P z+c_3(x\bullet_P y)\bullet_P z\\
&&+d_1 x\succ_P(y\succ_P z)+d_2 x\succ_P(y\prec_P z)+d_3 x\succ_P(y\bullet_P z)\\
&&+e_1 x\prec_P(y\succ_P z)+e_2 x\prec_P(y\prec_P z)+e_3x\prec_P(y\bullet_P z) \\
&&+f_1x \bullet_P(y\succ_P z)+f_2x \bullet_P (y\prec_P z)+f_3x \bullet_P (y\bullet_P z)
=0, \forall x,y,z\in N.
\end{eqnarray*}
By the definitions of $\prec_P,\succ_P,\bullet_P$ in Eq.(\mref{eq:eqs}), we have
\begin{eqnarray*}
&&a_1xP(y)P(z)+a_2P(xP(y))z-a_3P(xP(y)z)+b_1P(x)yP(z) \\
&&+b_2P(P(x)y)z-b_3P(P(x)yz)-c_1P(xy)P(z)-c_2P(P(xy))z \\ &&+c_3P(P(xy)z)+d_1P(x)P(y)z+d_2P(x)yP(z) \\
&&-d_3P(x)P(yz)+e_1xP(P(y)z)+e_2xP(yP(z))-e_3xP(P(yz)) \\ &&-f_1P(xP(y)z)-f_2P(xyP(z))+f_3P(xP(yz))
=0.
\end{eqnarray*}
Since $P$ is a Nijenhuis operator, we further have
\begin{eqnarray*}
&&a_1xP(yP(z))+a_1xP(P(y)z)-a_1xP^{2}(yz)+a_2P(xP(y))z -a_3P(xP(y)z) \\
&&+b_1P(x)yP(z)+b_2P(P(x)y)z-b_3P(P(x)yz)\\
&&-c_1P(xyP(z))-c_1P(P(xy)z)+c_1P^{2}(xyz) -c_2P(P(xy))z+c_3P(P(xy)z)\\
&& +d_1P(x)P(y)z+d_2P(x)yP(z)-d_3P(xP(yz))-d_3P(P(x)yz))
+d_3P^{2}(xyz)\\
&&+e_1xP(P(y)z)+e_2xP(yP(z))-e_3xP(P(yz))\\
&&-f_1P(xP(y)z)-f_2P(xyP(z))+f_3P(xP(yz))=0.
\end{eqnarray*}
Collecting similar terms, we obtain
\begin{eqnarray*}
&&(a_1+e_2) xP(yP( z) ) +(a_1+e_1) xP( P( y) z) -(a_1+e_3) xP(P( yz) +(a_2+d_1) P( xP( y) ) z\\
&&-(a_3+f_1) P( xP( y) z) +(b_1+d_2) P( x) yP( z) +(b_2+d_1) P( P( x) y) z-(b_3+d_3) P( P( x) yz) \\
&&-(c_1+f_2) P( xyP( z) ) +(c_3-c_1) P( P( xy) z) +(c_1+d_3) P^2( xyz) \\
&&-(c_2+d_1) P( P( xy) ) z+(f_3-d_3) P( xP( yz) ) =0.
\end{eqnarray*}
Now we take the special case when $(N,P)$ is the free Nijenhuis algebra $(\FN(T(M)),P_{T(M)})$ defined in Corollary~\mref{co:vecfree} for our choice of $M=\bfk\{x,y,z\}$ and $P_{T(M)}(u)=\lc u\rc$. Then the above equation is just
\begin{eqnarray*}
&&(a_1+e_2) x\lc y\lc z\rc \rc +(a_1+e_1) x\lc \lc y\rc z\rc -(a_1+e_3) x\lc\lc yz\rc +(a_2+d_1) \lc x\lc y\rc \rc z\\
&&-(a_3+f_1) \lc x\lc y\rc z\rc +(b_1+d_2) \lc x\rc y\lc z\rc +(b_2+d_1) \lc \lc x\rc y\rc z-(b_3+d_3) \lc \lc x\rc yz\rc \\
&&-(c_1+f_2) \lc xy\lc z\rc \rc +(c_3-c_1) \lc \lc xy\rc z\rc +(c_1+d_3) \lc\lc xyz\rc\rc \\
&&-(c_2+d_1) \lc \lc xy\rc \rc z+(f_3-d_3) \lc x\lc yz\rc \rc =0.
\end{eqnarray*}
Note that the set of elements
$$x\lc y\lc z\rc \rc, x\lc \lc y\rc z\rc, x\lc\lc yz\rc, \lc x\lc y\rc \rc z, \lc x\lc y\rc z\rc, \lc x\rc y\lc z\rc,$$
$$
\lc \lc x\rc y\rc z, \lc \lc x\rc yz\rc, \lc xy\lc z\rc \rc, \lc \lc xy\rc z\rc, \lc\lc xyz\rc\rc, \lc \lc xy\rc \rc z, \lc x\lc yz\rc \rc $$
is a subset of the basis $\frakX_\infty$ of the free Nijenhuis algebra $\FN(T(M))$ and hence is linearly independent. Thus the coefficients must be zero, that is,
\begin{eqnarray*}
&&a_1=-e_1=-e_2=-e_3, \\
&&a_2=b_2=c_2=-d_1, \\
&&a_3=-f_1,b_1=-d_2, \\
&&b_3=c_1=c_3=-f_2=-f_3=-d_3.
\end{eqnarray*}
Substituting these equations into the general relation $r$, we find that the any relation $r$ that can be satisfied by $\prec_P,\succ_P,\bullet_P$ for all Nijenhuis algebras $(N,P)$ is of the form
\begin{eqnarray*}
r&=&a_1\Big((x\prec y)\prec z-x \prec(y \prec z)-x \prec(y\prec
z)-x\prec(y\bullet z)\Big)\\
&&+b_1\Big((x\succ y)\prec z-x\succ(y\prec z)\Big)\\
&&+d_1\Big(x\succ(y\succ z)-(x\prec y)\succ z-(x\succ y)\succ z-(x\bullet
y)\succ z\Big)\\
&&+ a_3\Big((x\prec y)\bullet z-x\bullet (y\succ z)\Big)\\
&&+b_3\Big((x\succ y)\bullet z+(x\bullet y)\prec z+(x\bullet y)\bullet
z-x\succ(y\bullet z)-x\bullet (y\prec z)-x\bullet (y\bullet z)\Big),
\end{eqnarray*}
where $a_1,b_1,d_1,a_3,b_3\in \bfk$ can be arbitrary. Thus $r$ is in the subspace prescribed in Item~\mref{it:pna}, as needed.
\smallskip
\noindent
(\mref{it:pna} $\Rightarrow$ \mref{it:nap})
We check directly that all the relations in Eq.~(\mref{eq:wdna}) are satisfied by $(N,\prec_P,\succ_P,\bullet_P)$ for every
Nijenhuis algebra $(N,P)$. First of all
\begin{eqnarray*}
(x\prec_P y) \prec_P z&=&xP(y)P(z) \\
&=&xP(yP(z))+xP(P(y)z)-xP^2(yz)\\
&=&x\prec_P (y\prec_P z)+x\prec_P (y\succ_P z)
+x\prec_P (y\bullet_P z),
\end{eqnarray*}
proving the first equation in Eq.~(\mref{eq:wdna}). The proofs of the second and third equations are similar. For the fourth equation, we have
$$ (x\prec_P y)\bullet_P z=-P((xP(y))z)=-P(x(P(y)z))=x\bullet_P (y\succ_P z).$$
Finally for the last equation, we verify
\begin{eqnarray*}
&&(x\succ_P y)\bullet_P z+(x\bullet_P y)\prec_P z +(x\bullet_P y)\bullet_P z \\
&=& -P((P(x)y)z)-P(xy)P(z)+P(P(xy)z)\\
&=& -P(P(x)yz)-P(xyP(z))-P(P(xy)z)+P^2(xyz)+P(P(xy)z)\\
&=& -P(P(x)yz)-P(xyP(z))+P^2(xyz),
\end{eqnarray*}
and
\begin{eqnarray*}
&& x\succ_P (y\bullet_P z)+x\bullet_P(y\prec_P z) +x\bullet_P(y\bullet_P z)\\
&=& -P(x)P(yz) -P(x(yP(z))) +P(xP(yz))\\
&=& -P(xP(yz))-P(P(x)yz)+P^2(xyz)-P(xyP(z))+P(xP(yz))\\
&=& -P(P(x)yz)+P^2(xyz)-P(xyP(z)).
\end{eqnarray*}
So the two sides of the last equation agree.
Thus if the relation space $R$ of an operad $\calp=\calp(V)/(R)$ is contained in the subspace spanned by the vectors in Eq.~(\mref{eq:wdn}), then the corresponding relations are linear combinations of the equations in Eq.~(\mref{eq:wdna}) and hence are satisfied by $(N,\prec_P,\succ_P,\bullet_P)$ for each Nijenhuis algebra $(N,P)$. Therefore $(N,\prec_P,\succ_P,\bullet_P)$ is a $\calp$-algebra. This completes the proof of Theorem~\mref{thm:wdn}.
| 216,755
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TITLE: Question about inequalities and sets
QUESTION [0 upvotes]: Theorem: For every $x \in \mathbb{R}$, show there exists a unique $n \in \mathbb{Z}$ such that $n\leq x<n+1$.
Prove for that $x \in \mathbb{R}$, there is exactly one integer $n$ which satisfies $x < n \leq x+1.$
My attempt: From the theorem there is exactly one integer $n$ such that $n-1\le x<n$.This means that $n-1\le x\implies n\le x+1$. Therefore, this unique $n$ satisfies $x<n\le x+1$.
However, I don't think this is correct. Could someone please help? Thanks!
REPLY [1 votes]: It's valid.
There is a unique element $m$ so that $m\le x < m+1$. Let $n=m+1$ and $n-1 \le x < n$ so $x< n$ and $n-1+1 \le x+1$ so $n \le x+1$ so $x< n \le x+1$.
But we have prove this $n$ is unique. Which is just a matter of doing in reverse. If $x < m \le x+1$ then we have $x-1< m-1 \le x$ so $m-1\le x < m$. But there The theorem says there is only one unique integer so that $n \le x < n+1$ so $m$ must equal $n+1$.
We can do this is a single argument.
There exists exact one unique integer $m$ so that $m \le x < m+1$. Thus $m-1 \le x-1 < m \le x < m+1 \le x+1 < m+1$ and $m$ is unique in that regard. So $n =m+1$ is an existing and unique integer so that $x< m \le x+1$
| 32,578
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I am not easily impressed by web sites because normally they are glitzy and shallow, or worse, shallow and ugly. I just hate the ugly ones.
« Pura Belpré Awards 2009 | Main | e-books »
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| 185,148
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The time of the year when Europe’s top soccer clubs play exhibition matches on United States soil is just around the corner, with Yankee Stadium set to host two world powerhouses for the second year running.
It has just been announced that reigning Champions League winners Chelsea will take on English Premier League champions Manchester City on May 25 at 5:30 pm ET, just one week after the Premiership campaign ends. Tickets for this high-profile encounter are already available for sale on TicketCenter.com ( - Receive a 5% discount when you join their Facebook page!), so make sure to reserve your tickets if you don’t want to miss the once-in-a-lifetime opportunity.
Some month and a half after the two sides battle each other at Wembley in the FA Cup semi-final, they will meet again at a stadium that replaced a truly historical ground, giving the fans in New York a chance to see some genuine world-class players at first hand. With neither side too impressive this season, the match at the Yankee Stadium might be a perfect chance for the two teams to prove their worth and still finish the season on a high.
This is the clash of two up-and-coming soccer clubs that have been making waves in both England and Europe of late, with Chelsea enjoying a bit more success over the past few years. The Blues may have failed to challenge for the Premier League title recently, but last season’s Champions League triumph and this year’s solid showings in Europa League prove the Stamford Bridge outfit are slowly turning into true European giants.
As far as their star-studded squad is concerned, seasoned professionals like Fernando Torres, Frank Lampard, John Terry and Ashley Cole still hold important roles, but the US soccer fans are also going to enjoy seeing the likes of Eden Hazard, Oscar and Juan Mata, who are setting the Premier League alight with their scintillating displays.
Chelsea will be up against their familiar foes Manchester City, and the club that recently won their first league title after 44 years in one of the most exciting Premier League campaigns ever. The Citizens have been making steady progress since the club were purchased by Abu Dhabi United Group in the summer of 2008, although their current season hasn’t quite progressed according to plan. This is going to be an additional motive for Roberto Mancini’s side to do well in the exhibition game and no doubt his men possess enough quality to hurt their opposition.
While the likes of Joe Hart, Vincent Kompany and Gareth Barry are all considered important players at Etihad, there is every chance that people will be flocking to see players like Yaya Toure, David Silva, Carlos Tevez and Sergio Aguero strut their stuff at the Yankee Stadium in New York. This is shaping up to be quite a match, so make sure not to miss out.
Tickets for the exhibition game taking place on on May 23 at Busch Stadium, Saint Louis, MO, can be purchased here:
For more soccer tickets, visit our ticket section:
| 317,353
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Special mention goes to Billy Mays, hawking health insurance for iCan with the same glib tones and forced gestures he employs when demonstrating cleaners. Writes Stevenson,
This ad is kind of amazing. Watch Billy Mays' hand gestures—they never stop! Hands go out, palms facing each other. Hands come back in, fists balled. Hands go back out, index fingers extended. It's either OCD or some sort of primitive sign language. I think it's happening because he has no tangible product to demonstrate. They should have let him use a sheaf of insurance documentation to wipe up spills with. Or maybe a wallet card for dabbing on grout?If you're a speech and debate coach, don't-be-like-Billy gives you at least a full week's lesson plan.
| 260,827
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TITLE: Integration of Torque for a Circular Current Loop in Magnetic Field
QUESTION [1 upvotes]: I am trying to derive the formula for Torque on a circular current loop inside a magnetic field.
I know the formula is:
$\tau = IAB\sin{\theta}$
Where I is the current, B is the magnetic field and A is the Area.
My attempt so far:
$d\vec{F} = I\,d\vec{s}\times \vec{B} = IB\,ds\cdot\sin{\alpha}$
Now, if the formula for Torque is: $\tau=bF\sin{\theta}$, and $b = r\sin{\alpha}$, then
$d\tau = r\cdot sin{\alpha}\cdot IB\sin{\theta}ds\cdot \sin{\alpha} = rIBsin{\theta}\cdot\sin^2{\alpha}\,ds$
Ultimately, if I take the integral of this last equation, I cannot exactly understand how to integrate $\sin{\alpha}^2\,ds$.
I guess that my underlying misunderstanding lies here: I can tell what the integral of $d\vec{s}\times \vec{B}$ will be, since I know the diameter of the circle. However, I think there is no way to express $\sin{\alpha}$ with respect to $ds$.
Am I getting this wrong? Thank you
REPLY [2 votes]: You didn't use vector notations so it seems to be quite terrible. Also, you've used $M$ for torque (it should be $\tau$) rather than for magnetic moment (which are generally accepted symbols).
Proof:
A circular loop lies in $x-y$ plane with raduis $r$ and center at origin $O$. It is carrying a constant current in anti-clockwise direction. There is uniform magnetic field $\vec B$ directed along positive $x$-axis.
Consider an element $d\vec s$ on the ring at an angle $\theta$ subtending an angle $d\theta$ at the origin. Torque on this element is given by
$$\begin{align}d\tau&=\vec r\times d\vec F=\vec r\times(Id\vec s\times\vec B)\\
&=I(r\cos\theta\ \hat i+r\sin\theta\ \hat j)\times\bigg((-rd\theta\sin\theta\ \hat i+rd\theta\cos\theta\ \hat j)\times(B_0\ \hat i)\bigg)\\
\tau&=I\bigg(\int_0^{2\pi}B_0r^2\cos^2\theta\ d\theta\ (\hat j)-\int_0^{2\pi}B_0r^2\sin\theta\cos\theta\ d\theta\ (\hat i)\bigg)\\
&=I(\pi r^2)B_0\ \hat j=(I\pi r^2\ \hat k)(B_0\ \hat i)\\
&=\vec M\times\vec B
\end{align}$$
Note: I've skipped the calculation part. Also, you can also take $\vec B=B_x\ \hat i+B_y\ \hat j +B_z\ \hat k$, I've taken only $x$-component for simplicity. The result will reamin the same. Same with the shape of conductor, doesn't matter whether square or circle.
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He was there when I emerged into the world for the very first time.
He stood by my mother in tears and held her hand in desperation and anger when they realised their firstborn had severe birth defects.
He bullied the nurses in the hospital when they were late with my feeding bottle.
He would leave work every day at 4:00pm on the dot to take his little girl to the beach.
He taught me how to pet dogs and love cats.
He showed me how to fold a viable paper plane and construct stable Lego towers.
He reminded me every day that I was supposed to be a good girl, not a tomboy or playground bully.
He insisted that I coloured within the lines and painted as little abstract as possible.
Average or mediocre was not acceptable.
Wobbly or fragile did not exist in our house.
He introduced me to a camera.
We collected stones and music together.
He took the training wheels off my bike.
He made sure I knew how to handle a screwdriver, a hammer, a wrench, and car jacks.
I wasn’t allowed to drive after finishing driving school until I passed his test.
Curiosity for technology and understanding of functionality was essential in tools, machines and management.
Strange and new food was never a problem.
Being there for others was at the core of his being.
He was someone else’s big brother.
He was my mother’s life partner, husband and friend for 50 years.
He had no clue how to navigate my teenage years, but what parent does…
He stood outside the operating room when I gave birth and photographed the first moment I held my baby.
He slept on the floor in the nursery beside the crib for three weeks because he didn’t trust the cat with the baby.
He loved sushi, pizza and fresh corn.
He taught me the meaning of faith and the strength of prayer.
He insisted on good, sturdy luggage.
I was never allowed to chase away birds, stray cats or dogs.
He worried across two continents when I had to install the washing machine and dishwasher on my own.
He would have been 81 today.
Our was far from being the perfect father-daughter relationship, and we had more than our fair share of difficult moments, but he was my first anchor. Showing emotion was never his strength, nor was opening up to talk, but he was always there for me. When he could no longer walk or talk, he made sure he found a way to communicate.
Happy birthday Daddy.
3 Comments
Hi Maritess
Allow me to share your feelings of happiness and sadness in remembering the birthday of your daddy… my brother. I still feel bitter, remorseful, regretful for not being with him when he passed on. The only little consolation i have is the lingering thought that I was able to hold his hand nearing the end, when I know he was still aware that I was there beside him.
Happy Birthday dear brother.. not a day passes without a prayer that may you rest in the arms of our Lord forever.
Take care Maritess.. and a big hug 🤗 ninang
Sent from my iPhone
>
A shining example of how the rest of us fathers should aspire to be. Happy Birthday Sir.
Thank you John! Lovely to hear from you. I do miss Daddy, and the list of questions I will never get an answer to anymore keeps growing
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RECIPES: Recipe Details
Patty O'Rueben's Corned Beef Irish Burger
Well Now........ As my sainted Grandmother, when she had me on her knee, would say to me,"Tommy me boy, take the best from the old sod and marry it with the new goodness of the U S and A" Through spells and incantations, in the full light of the moon, we have wedded Irish corned beef and cabbage with a Yankee burger. The result is a magical gift from the Gods for one and all to enjoy!
(Sauerkraut Dressing)
1 1/2 cups sauerkraut
1 cup mayonnaise
1/4 cup sweet relish
1/4 cup cocktail sauce
juice of 1/2 lemon
3 teaspoons worcestershire sauce
2 tablespoons Grey Poupon Dijon mustard
1/4 cup Irish Mist liqueur
1 teaspoon kosher salt
1/2 teaspoon freshly ground pepper
(Patties)
1 cup Jewish rye bread crumbs
2 Lbs packaged corn beef brisket,point cut (have your market butcher grind it for you)
1/2 teaspoon freshly ground pepper
(Cheese Chips)
1 1/2 cups Dubliner cheese, shredded vegetable oil for brushing grill rack
7 Jewish rye split rolls (available at most supermarket bakeries) or Jewish rye bread sliced 3/4 inch thick
To make the sauerkraut dressing- Drain but do not rinse the sauerkraut. Spread on paper towls and pat dry. Place sauerkraut in a medium bowl adding mayonnaise,relish,cocktail sauce,lemon juice,worcestershire sauce,dijon mustard,Irish Mist liqueur,salt and pepper. Mix well and refrigerate until serving. To make the patties- Take 1 rye burger roll(or 2 rye slices,crusts removed) Cut into 1 inch pieces and gring into crumbs in food processer. In a large bowl, combine and mix well with the ground corned beef brisket and pepper, handling the meat as little as possible to avoid compacting it. Divide into 6 equal portions and form into patties. Press the center of each patty down with finger tips (to about 1/2 inch thick) to prevent puffing when grilling. Refrigerate until ready to grill. To make cheese chips- Heat a 10 inch nonstick skillet over medium-high heat on grill or side burner if available. Sprinkle 1/4 cup shredded Dubliner cheese in a disk shape keeping the cheese light around the edges so it looks lacy. When golden brown, remove with a nonstick spatula and set aside on paper towels. Repeat for all 6 chips. Refrigerate until ready to assemble. Prepare a medium-hot fire in a charcoal grill with a cover or preheat a gas grill to medium-high. When the grill is ready, brush the rack with vegetable oil. Place the burgers on the grill divot side up, cover and cook for 3 minutes. Turn the patties and continue grilling for another 3 minutes for medium rare. Remove burgers and cover loosely with foil letting rest for 5 minutes. Place the burger rolls on the grill and toast the cut sides lightly. If using sliced bread, toast both sides. To assemble the burgers, place 1/4 cup sauerkraut dressing on the bottom roll,followed by the corned beef patty,cheese chip and top roll. Makes 6 burgers.
| 242,136
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Gangrene rots away 65-year-old man’s penis until it DROPS OFF after medics botch a routine procedure during cancer surgery
- The unnamed man, 65, went to hospital in Lucknow in the north of India
- Flesh-eating infection spread along the man’s swollen penis and turned it black
- His case is thought to only be the third of its kind caused by Fournier’s gangrene
A man’s entire penis fell off after it rotted away because of gangrene triggered during surgery to treat cancer in his neck.
The unnamed 65-year-old had cancer surgery just two weeks earlier and, for 10 days, lived with ‘blackish discolouration’ of the shaft of his penis, a case report revealed.
Medics realised the unfortunate man had developed an infection in his penis after attempts to put in a catheter – a tube in the penis to drain urine – during the operation had been botched.
During the process, the inside of his penis had been damaged and he developed Fournier’s gangrene, an infection which eats away at human flesh.
Then, two weeks after surgeons tried to cut away the infection, the man’s penis ‘autoamputated’ – meaning the whole member fell off on its own.
The unnamed man, 65, lost his entire penis because of an infection he caught after medics failed to properly insert a tube for him to urinate through during his cancer surgery (pictured, a scan of the remains of the man’s urethra connected to his bladder)
The case is thought to be only the third of its kind – the man now urinates out of a tube and is said to be ‘doing well’ after recovering from his ordeal.
Admitted to King George’s Medical University Hospital in Lucknow in India’s northern Uttar Pradesh, the man was diagnosed with a condition called Fournier’s gangrene.
Fournier’s gangrene describes necrotising fasciitis, a rare bacterial infection which destroys tissue beyond repair, when it is specifically on the genitals.
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The unnamed man had an operation to treat cancer in his thyroid – in the neck – two weeks before being diagnosed with the penis infection, according to BMJ Case Reports.
WHAT IS GANGRENE?
Gangrene refers to the decay and death of tissue resulting from an interruption of blood flow to a certain area of your body. fatty buildup in the arteries.
Treatments for gangrene include surgery to remove dead tissue, antibiotics and other approaches.
The prognosis for recovery is good if gangrene is identified early and treated quickly.
Source: Mayo Clinic
During that operation, medics had tried to put a catheter into his penis to drain urine from his bladder, but failed and damaged the tissue inside his urethra.
After his procedure the man’s penis swelled up and began to turn black, filling with fluid.
Symptoms worsened for 10 days and the man returned to hospital where medics realised he had Fournier’s gangrene.
The gangrene had spread along the penis shaft and, although surgeons tried to cut away the infected flesh to save the man’s penis, it eventually became so destroyed it dropped off.
The man had surgery to remove what was left of his penis, and has now been left with no appendange, just a tube to urinate from above his scrotum.
Doctor Siddharth Pandey wrote in his report: ‘The patient had a [cancer] which might have contributed to him developing this condition.’
He suggested the man’s weakened immune system because of his recent surgery and cancer could have also worsened the gangrene, as well as his age.
The shocking case is thought to only be the third ever case of someone’s entire penis rotting away because of Fournier’s gangrene.
Despite having lost his penis, the patient was ‘doing well’ three months later, Dr Pandey said.
Source: Read Full Article
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How to play Sizzling Hot 6 Extra GoldClick here for a print friendly version
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RTP: 95.09%
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TITLE: Find all of the elements of a subgroup generated by a $3$-cycle notation
QUESTION [1 upvotes]: Question: Let $H$ be the subgroup of $A_n$ generated by $(123)$. Write down all elements of $H$. Here is my attempt: $H\le \langle 123 \rangle$ $H$=$\langle a \rangle$ where the order of $a$=$n$ where $\langle a \rangle$=$\langle e,a,a^{2},....,a^{n-1} \rangle$ since $\langle 123 \rangle$ has order $3$, $\langle 123 \rangle$=$\langle (e),(123),(23)(31) \rangle$ I computed $(123)^{2}$ and $(123)^{3}$ since I had to order $3$ to come up with my answer for $\langle 123 \rangle$ Did I do this right?
REPLY [0 votes]: Yes, your answer is correct. $H = \langle (123) \rangle = \{e, (123), (132) \}$.
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.
"Love it"
Queen Betsy is great Maryjanice Davidson has done it again This book and series is a good time to be had 😄
.
"Fun Easy Read"
I really enjoyed this book as well as the rest of the series. I hope audible gets the 9th book Undead and Unfinished when it releases.
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By Dr. Mercola
One in four Americans aged 45 and over take a cholesterol-lowering statin drug, despite its and EU Approve New Class of Cholesterol-Lowering Drugs
The drug, alirocumab (sold under the brand name Praluent) received approval from the US Food and Drug Administration (FDA) on July 24.1,2,3
Two days earlier, the European Commission (EC) granted marketing authorization for evolocumab4 (brand name Repatha). (Some of the history behind this drug discovery was recently published by Time Magazine.5)
.
PCSK9 Inhibitors Appear7,8 published in theNew England Journal of Medicine in April this year, shows that alirocumab (Praluent) causes higher incidence of many of the same side effects as stat
PCSK9 Inhibitors — Another Deadly Wealth Generator?..
How Do PCSK9 Inhibitors Work?
PCSK9 is an enzyme that causes degradation of the LDL receptor that binds to LDL. If you don’t have that receptor, the cholesterol stays in your blood. PCSK9 inhibitors employ man-made antibodies to block the PCSK9 enzyme, which means you end up with more LDL receptors on the surface of the cell (typically a cell in the liver), allowing more cholesterol to be brought inside the cell. This leaves less cholesterol to circulate in your blood stream.
PCSK9 is also a so-called protein convertase, a class of high-level metabolic regulators that are still poorly understood, but they are thought to play a role in maintaining homeostasis in your body, according to Dr. Seneff. PCSK9 is unusual in that, unlike other protein convertases, it does not require calcium uptake to be activated. It’s also unusual in that it becomes sulfated, and Dr. Seneff, who has studied sulfate and the pathology associated with calcium uptake, believes these two observations are red flags indicating PCSK9 inhibition may cause significant harm.
Moreover, the earlier research name for PCSK9 was NARC-1, which stands for “neural apoptosis-regulated convertase 1.” “Neural,” of course, means it relates to or involves your nerves and/or nervous system, and “apoptosis” means “programmed cell death,” and this may hint at some of the potential side effects you might expect when you inhibit this protein.
A 2006 study11 looking at zebrafish found that NARC-1/PCSK9 is expressed in neurons in the cerebral cortex and cerebellum in association with neurogenesis (i.e. the brain’s ability to adapt and grow new brain cells).
When PCSK9 was inactivated in zebrafish embryos, the embryos died approximately 96 hours after fertilization, with the mid- and hindbrain knotted together in a disorganized array of neurons due to the loss of hindbrain-midbrain boundaries. NARC-1/ PCSK9 is also expressed in the human liver, small intestine, kidney, the cerebral hemispheres, and the cerebellum.12.
“I think it will be a disaster if this new class of drugs is widely used,” Dr. Seneff warns. “I predict liver disease and perhaps even liver cancer. Not to mention the increase in cholesterol and sulfate deficiency throughout the rest of the body, due to the reduction in LDL levels in the blood.”.
PCSK9-Deficient Mice Exhibit Impaired Glucose Tolerance
As noted earlier, unsulfated PCSK9 restricts your liver’s uptake of LDL by degrading LDL receptors (LDLR). It likely does this to protect the liver from taking in more cholesterol than it can properly handle in the context of sulfate deficiency. Research13shows that “PCSK9 and LDLR are also expressed in insulin-producing pancreatic islet beta cells, possibly affecting the function of these cells.”
The study in question found that PCSK9-deficient mice had less insulin in their pancreas compared to normal mice. The PCSK9-deficient mice were also hypoinsulinemic, hyperglycemic, glucose-intolerant, and their pancreatic islets showed signs of malformation, apoptosis (cellular death), and inflammation. According to the authors:
“Collectively, these observations suggest that PCSK9 may be necessary for the normal function of pancreatic islets.”
Assess Your Actual Need for a Cholesterol-Lowering Drug
As a general rule, cholesterol-lowering drugs are not required or prudent for the majority of people — especially if high cholesterol and longevity run in your family. Also keep in mind that your overall cholesterol level says very little about your risk for heart disease. The following tests can give you a far better assessment of your heart disease risk than your total cholesterol alone:
Strategies to Reduce Your Risk of Heart Disease Without Drugs
Heart disease is predominantly the end result of unhealthy lifestyle choices, and cholesterol-lowering medications are far from being magic bullets to lower your risk of dying from heart disease. It’s really quite tragic that the medical system has reduced it down to a cholesterol problem — especially since:
| 349,449
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\begin{document}
\begin{abstract}
This is the continuation of ~\cite{Li}. We describe the affine canonical basis
elements in the case when the affine quiver has arbitrary orientation.
This generalizes the description in ~\cite{lusztig2}.
\end{abstract}
\maketitle
\section{Introduction}
Let $\mbf U^-$ be the negative part of the quantized enveloping algebra $\mbf U$
of the Kac-Moody Lie algebra
associated to a symmetric generalized Cartan matrix $C$.
One of the milestones in the Lie theory is the discovery of Lusztig's canonical basis
$\mbf B$ (or Kashiwara's global crystal basis ~\cite{Kashiwara}) of $\mbf U^-$.
The canonical basis $\mbf B$ possesses many remarkable properties
such as total positivity and integrality (see ~\cite{lusztig1} and ~\cite{lusztig3}).
There are two different approaches of defining the canonical basis $\mbf B$.
One is algebraic and the other geometric.
For the algebraic approach, see ~\cite{Lus2} and ~\cite{Kashiwara}.
The geometric approach is also done in ~\cite{Lus2} when $C$ is positive definite and
is extended to all cases in ~\cite{lusztig1}.
It is shown in ~\cite{Grojnowski-Lusztig} that the two different approaches
produce the same basis in $\mbf U^-$.
Let $Q$ be a quiver such that the associated Cartan matrix is $C$.
In the geometric approach, Lusztig studies certain perverse sheaves over the representation
space $E_{V,Q}$ of $Q$.
An algebra $\mathcal K_Q$ is constructed and is shown to be isomorphic to $\mbf U^-$.
The simple perverse sheaves in $\mathcal K_Q$ form a basis $\mathcal B_Q$ of
$\mathcal K_Q$.
The canonical basis $\mbf B$ is then defined to be $\mathcal B_Q$ if one identifies
$\mathcal K_Q$ with $\mbf U^-$. In other words, the geometric approach gives a geometric
realization of $\mbf U^-$ and $\mbf B$ for each $Q$ such that the associated Cartan matrix
is $C$.
(Note that for each $C$,
there may be several quivers such that the associated Cartan matrices are $C$.)
It is well-known that Lie theory and the representation theory of quivers
have very deep connections ever since Gabriel's theorem ~\cite{Gabriel} was found.
The interaction between the two theories attracts a lot of attentions since then.
Note that $\mbf U^-$ and $\mbf B$ are objects coming purely from Lie theory.
Although the framework of Lusztig's geometric realizations of $\mbf U^-$ and $\mbf B$
is the representation space $E_{V,Q}$ of the quiver $Q$.
No representation theory of quivers is used explicitly.
Nor are there clear connections between the representation theory of quivers and
Lusztig's geometric realizations of $\mbf U^-$ and $\mbf B$.
One may ask to what extent the representation theory of quivers can help in understanding
the quantum group $\mbf U$ and the canonical basis $\mbf B$ or $\mathcal B_Q$.
Keeping this in mind, one may ask the following natural questions.
\begin{Qst}
\label{q1}
Characterize the elements in $\mathcal B_Q$
by using the representation theory of
quivers. More precisely, describe what the supports and the corresponding
local systems of the elements in $\mathcal B_Q$ are in terms of
representations of quivers.
\end{Qst}
\begin{Qst}
\label{q2}
Recover the canonical basis via bases arising in Ringel-Hall algebras.
\end{Qst}
\begin{Qst}
\label{q3}
Show the positivity or the integrality of the canonical basis $\mbf B$
in an elementary way with the help of the representation theory of quivers,
instead of the theory of perverse sheaves.
\end{Qst}
It seems very difficult at this moment to answer Question ~\ref{q3}.
For an attempt to answer Question ~\ref{q2}, see ~\cite{Lus2} when $Q$
is of finite type and ~\cite{LXZ} when $Q$ is of affine type and the
references therein.
By now there are only partial answers to Question ~\ref{q1}.
The answer to Question ~\ref{q1} is known when $Q$ is of finite type, i.e.,
when the Cartan matrix is positive definite.
The elements in $\mathcal B_Q$ are simple perverse sheaves whose supports are
$G_V$-orbits $O$ in $E_{V,Q}$ and whose restrictions to $O$ are the constant sheaf on $O$.
The answer to Question ~\ref{q1} is known when $Q$ is a cyclic quiver.
Elements in $\mathcal B_Q$ are simple perverse sheaves whose supports are
the aperiodic $G_V$-orbits $O$ in $E_{V,Q}$
and whose restrictions to $O$ are the constant sheaf on $O$.
From now on in the introduction,
we assume that $Q$ is an affine quiver but not a cyclic quiver.
When $Q$ is a McKay quiver, i.e., all vertices are either a sink or a source,
the answer to Question ~\ref{q1} is given in ~\cite[Theorem 6.16]{lusztig2} by using
the representation theory of McKay quivers. The theory of representation of
McKay quivers in \cite{lusztig2} is based on McKay's correspondence.
In ~\cite{lusztig2}, only McKay quivers and cyclic quivers are considered.
Note that the construction of $\mathcal K_Q$ and $\mathcal B_Q$ applies to any quivers and
the language used in ~\cite{DR} for the representation theory of affine quivers
works for all affine quivers, not just McKay quivers.
Given any two affine quivers $Q$ and $Q'$ such that the associated
generalized Cartan matrices coincide. Although the inverse images
$\phi^{-1}(\mathcal B_Q)$ and $(\phi')^{-1}(\mathcal B_{Q'})$ coincide
(see Theorem ~\ref{coincide}), the isomorphism classes of simple
equivariant perverse sheaves in $\mathcal B_Q$ and $\mathcal B_{Q'}$
may be quite different. This becomes obvious when $Q$ and $Q'$ are the Kronecker
quiver and the cyclic quiver with two vertices, respectively.
The goal of this paper is to give a complete answer
to Question ~\ref{q1} for any affine quiver, by
carrying out Lusztig's arguments in ~\cite{lusztig2} to any affine quiver via
the representation theory developed in ~\cite{DR}.
In ~\cite{Li}, such a goal has been accomplished when $Q$ is of type $\tilde{A}$.
Similar to [Li],
the crucial part on the way of generalizing Lusztig's argument
is still to construct a $'$nice$'$ functor
from the category of nilpotent representations of a cyclic quiver $\mrm C_p$
to the full subcategory $\mrm{Rep}(T)$ generated by a tube $T$ of period $p$
which will give an equivariant morphisms between the representation
varieties and transporting the equivariant simple perverse sheaves to
equivariant simple perverse sheaves.
It turns out that the Hall functor in ~\cite{FMV}
will suffice to overcome this difficulty.
This functor gives us all the properties we need in the proof of
Lemma \ref{tube} in Section \ref{noncyclic}.
It should be mentioned that after the preprint of the paper is written, the
second author received a preprint \cite{nakajima} by Nakajima, in which he
outlines an approach in answering Question ~\ref{q1}
when the quivers are of type $ \tilde{D}$ and $ \tilde{E}$
by using the description of $\mathcal B_{Q}$ for McKay quivers by Lusztig
(\cite{lusztig2} Theorem 6.16)
and then apply the reflection functors. Our argument does not depend on
Theorem 6.16 in ~\cite{lusztig2}.
{\bf Acknowledgements.}
We wish to thank Professor Bangming Deng and Professor Jie Xiao
for very helpful discussions. The second author thanks Professor Hiraku
Nakajima for discussing questions in his paper \cite{nakajima}.
The first author thanks the second author for allowing him to use the
results in this paper as part of his Ph.D. thesis.
\section{Representation theory of Affine quivers}
In this section, we give a brief review of representation theory of
affine quivers. See ~\cite{DR} and ~\cite{BGP} for more details.
\subsection{Preliminary}
\label{preliminary}
A $quiver$ is an oriented graph.
It is a quadruple $Q=(I,\Omega, h, t)$,
where $I$ and $\Omega$ are two finite sets
and $h$, $t$ are two maps from $\Omega$ to $I$ such that $h(\omega)\neq t(\omega)$
for any $\omega\in \Omega$.
$I$ and $\Omega$ are called the vertex and arrow sets respectively.
Pictorially, $t(\omega) \overset{\omega}{\to} h(\omega)$ stands
for any arrow $\omega\in \Omega$
and we call $h(\omega)$ and $t(\omega)$ the head and the tail of
the arrow $\omega$ respectively.
An {\em affine quiver} is a quiver whose underlying graph is of type $\tilde{A}_n$,
$\tilde{D}_n$, $\tilde{E}_6$, $\tilde{E}_7$ or $\tilde{E}_8$.
{\em Throughout this paper, all quivers considered will be affine.
We fix an algebraically closed field $K$ once and for all.}
A $representaion$ of a quiver over $K$ is a pair $(V, x)$,
where $V=\oplus_{i\in I}V_i$ is an $I$-graded finite dimensional
$K$-vector space and
$x=\{x_{\omega}:V_{t(\omega)}\to V_{h(\omega)}|\;\omega\in \Omega\}$
is a collection of linear maps.
A morphism $f: (V,x)\rightarrow (W,y)$ is a collection of linear maps
$\{f_i:V_i\to W_i|\;i\in I\}$ such that
$f_{h(\omega)}\;x_{\omega}=y_{\omega}\;f_{t(\omega)}$, for any $\omega\in \Omega$.
This defines a (abelian) category, denoted by $\mrm{Rep}(Q)$.
A {\em nilpotent} representation is a representation $(V,x)$ having the property: there is
an $N$ such that for any $\omega_1,\cdots,\omega_N$ in $\Omega$ satisfying
$t(\omega_k)=h(\omega_{k-1})$ for any $k$, the composition of morphisms
$x_{\omega_N}\circ \cdots\circ x_{\omega_1}$ is zero.
Denote by $\mrm {Nil}(Q)$ the full subcategory of $\mrm{Rep}(Q)$
of all nilpotent representations of $Q$.
Note that when $Q$ has no oriented cycles, every representation is
nilpotent,
i.e., $\mrm{Nil}(Q)=\mrm{Rep}(Q)$.
Denote by $\mrm{Ind}(Q)$ the set of representatives of all pairwise
nonisomorphic
indecomposable representations in $\mrm{Rep}(Q)$.
The Euler form $<\ ,\ >$ on $\mathbb Z[I]$ is defined by
\[<\alpha,\beta> \, =\sum_{i\in I}\alpha_i\beta_i-
\sum_{\omega\in \Omega}\alpha_{t(\omega)}\beta_{h(\omega)},\]
for any $\alpha=\sum_i \alpha_ii$, $\beta=\sum_i \beta_ii\in \mathbb Z[I]$.
The symmetric Euler form $(,)$ on $\mathbb Z[I]$ is defined to be
\[(\alpha,\beta)=\,<\alpha,\beta>+<\beta,\alpha>,\]
for any $\alpha=\sum_i \alpha_ii$, $\beta=\sum_i \beta_ii\in \mathbb Z[I]$.
Given $\mbf V=(V,x)\in \mrm{Rep}(Q)$, denote by $|\mbf V|$ its dimension vector
$\sum_{i\in I}(\mrm{dim}V_i)i\in \mathbb Z[I]$.
For any $M$, $N\in \mrm{Rep}(Q)$, we have from \cite{DR}
\[<|M|,|N|>\,=\mrm{dim} \;\mrm{Hom}_Q(M,N)- \mrm{dim}\; \mrm{Ext}^1_Q(M,N).\]
Given any $I$-graded $K$-vector space $V$, let
\[E_{V,\Omega}=\oplus_{\omega\in \Omega}\, \mrm{Hom}(V_{t(\omega)},V_{h(\omega)})
\quad \text{and}\quad
G_{V}=\prod_{i \in I} \,\mrm{GL}(V_i),\]
where $\mrm{GL}(V_i)$ is the general linear group of $V_i$ for all $i\in I$.
$G_V$ acts on $E_{V,\Omega}$ naturally, i.e.,
$g.x=y$, where $y_{\omega}=g_{h(\omega)}\,x_{\omega}\,g_{t(\omega)}^{-1}$ for all
$\omega\in \Omega$.
Note that for any $x\in E_{V,\Omega}$, $(V,x)$ is a representation of $Q$. We call
that $x$ is nilpotent if $(V,x)$ is a nilpotent representation.
\subsection{BGP reflection functors}
A vertex $i \in I$ is called a sink (resp., a source) if
$i\in \{t(\omega),h(\omega)\}$ implies $h(\omega)=i$ (resp.,
$t(\omega)=i$) for any $\omega\in \Omega$.
For any $i \in I$,
let $\sigma_{i}Q=(I, \Omega,h', t')$ be the quiver with the orientation
$(t', h')$ defined by
\begin{enumerate}
\item[] $t'(\omega)=t(\omega)$ and $ h'(\omega)=h(\omega)$
if $i\notin \{t(\omega), h(\omega)\}$;
\item[] $t'(\omega)=h(\omega)$ and $h'(\omega)=t(\omega)$
if $i\in \{t(\omega), h(\omega)\}$;
\end{enumerate}
for any $\omega\in \Omega$.
In other words, $\sigma_i Q$ is the quiver obtained by reversing
the arrows in $Q$ that start or terminate at $i$.
If $i$ is a sink in $Q$,
define a functor
\[\Phi_{i}^+:\mrm{Rep}(Q) \to \mrm{Rep} (\sigma_{i}Q)\]
in the following way.
For any $(V, x) \in \mrm{Rep}(Q)$,
$\Phi_{i}^+(V,x)=(W,y)\in \mrm{Rep}(\sigma_{i}Q)$,
where $W=\oplus_{j\in I} \, W_j$ such that
\begin{enumerate}
\item[] $W_j=V_j$, if $j\neq i$;
\item[] $W_i$ is the kernel of the linear map
$\sum_{\omega \in \Omega:\;h(\omega)=i} x_{\omega}\; :
\oplus_{\omega\in \Omega:h(\omega)=i} V_{t(\omega)} \to V_i$;
\end{enumerate}
and $y=(y_{\omega}\; |\; \omega \in \Omega)$ such that
\begin{enumerate}
\item[] $y_{\omega}=x_{\omega}$, if $t'(\omega)\neq i$;
\item[] $y_{\omega}:W_i\to W_{h'(\omega)}$ is the composition of the natural maps:
\[W_i \to \oplus_{\omega\in \Omega:\;h(\omega)=i}V_{t(\omega)}
=\oplus_{\omega\in \Omega:t'(\omega)=i}W_{h'(\omega)} \to
W_{h'(\omega)},\] if $t'(\omega)=i$.
\end{enumerate}
Note that the assignments extend to a functor.
$\Phi^+_i$ is called the BGP reflection functor with respect to the sink
$i$.
Since $Q$ has no oriented cycles,
we can order the vertices in $I$,
say $(i_1 ,\cdots,i_n)$ ($|I|=n$),
in such a way that $i_r$ is a sink
in the quiver $\sigma_{i_{r-1}}\cdots \sigma_{i_1}Q$.
Then from ~\cite{BGP},
\[Q=\sigma_{i_n}\cdots \sigma_{i_1} Q.\]
Define the Coxeter functor $\Phi^+: \mrm{Rep} (Q) \to
\mrm{Rep} (Q)$ to be
\[\Phi^+ = \Phi_{i_n}^+\circ \cdots \circ \Phi_{i_1}^+.\]
Similarly, if $i$ is a source in $Q$.
For any $(V, x) \in \mrm{Rep}(Q)$,
let $\Phi_{i}^-(V, x)=(W, y)$ be a representation of
$\sigma_{i}Q$,
where $W_j=V_j$ if $j\neq i$ and
$W_i$ equals the cokernel of the linear map
$\sum_{\omega\in \Omega: t(\omega)=i}x_{\omega}:
V_i \to \oplus_{\omega\in \Omega: t(\omega)=i}V_{h(\omega)}$;
for any $\omega \in \Omega$,
$y_{\omega}=x_{\omega}$ if $h'(\omega)\neq i$,
otherwise if $h'(\omega)=i$, $y_{\omega}$ is the composition of the natural maps
$W_{t'(\omega)} \to \oplus_{\omega\in \Omega:t(\omega)=i}V_{h(\omega)} \to W_i$.
This extends to a functor
\[\Phi_i^-: \mrm{Rep}(Q) \to \mrm{Rep}(\sigma_iQ).\]
The Coxeter functor
$\Phi^- :\mrm{Rep}(Q) \to \mrm{Rep}(Q)$
is defined to be
\[\Phi^-=\Phi_{i_1}^- \circ \cdots \circ \Phi_{i_{n}}^-.\]
\subsection{Classification of indecomposable representations}
Assume that $Q$ has no oriented cycles.
By using the two Coxeter functors $\Phi^+$ and $\Phi^-$,
the representations $M \in \mrm{Ind}(Q)$ are classified into four
classes: preprojective, inhomogeneous regular, homogeneous regular and
preinjective. More precisely, a representation $M \in \mrm{Ind}(Q)$
is called preprojective if $(\Phi^+)^r M=0$, for $r \gg 0$;
inhomogeneous regular if $(\Phi^+)^rM\simeq M$ for some $r\geq2$;
homogeneous regular if $(\Phi^+)^rM\simeq M$ for any positive integer
$r$; and
preinjective if $(\Phi^-)^r M =0$, for $r \gg 0$.
In general, $M\in \mrm{Rep}(Q)$ is preprojective
(resp. inhomogeneous regular, homogeneous regular, preinjective) if all indecomposable direct summands of
$M$ are preprojective (resp. inhomogeneous regular, homogeneous
regular, preinjective).
Let $S_{i}$ be the simple representation corresponding to the vertex $i$.
It's a representation $(V, x)$ such that $V_i=k$, $V_j=0$
if $j\neq i$ and all linear maps $x_{\omega}$ are 0.
Note that given a graph, the definition of the simple representation
works for any orientation of the graph.
By abuse of notation, we always denote by $S_{i}$
the simple representation corresponding to the vertex $i$ regardless of the orientation.
Fix a sequence $(i_1,\cdots,i_n)$ such that $i_r$ is a sink of the quiver
$\sigma_{i_{r-1}}\cdots\sigma_{i_1}(Q)$.
Let
\[ P(i_r)=\Phi_{i_1}^-\circ\cdots\circ\Phi_{i_{r-1}}^-(S_{i_r})
\quad \text{and} \quad
I(i_r)=\Phi_{i_n}^+\circ\cdots\circ\Phi_{i_{r+1}}^+(S_{i_r}).\]
Then we have
\begin{align*}
&M\in \mrm{Ind}(Q)\; \text{ is projective iff}\;
M\simeq P(i_r). \tag{1}\\
&M\in \mrm{Ind}(Q)\; \text{ is injective iff}\;
M\simeq I(i_r). \tag{1'}\\
&M\in \mrm{Ind}(Q)\;\text{ is preprojective iff} \;
M=(\Phi^-)^r\, P(i). \tag{2}\\
&M \in \mrm{Ind}(Q)\; \text{is preinjective iff}\;
M =(\Phi^+)^r\, I(i)\tag{3}
\end{align*}
Let $\mrm{Reg}(Q)$ be the full subcategory of $\mrm{Rep}(Q)$
whose objects are inhomogeneous regular and
homogeneous representations. Then we have
\begin{align*}
&\mrm{Reg}(Q)\;\text{is an extension-closed
full subcategory of}\; \mrm{Rep}(Q). \tag{4}\\
&\Phi^+\;\text{is an autoequivalence on}\; \mrm{Reg}(Q). \;\Phi^-\;\text{ is its inverse.}
\tag{5}
\end{align*}
The simple objects in $\mrm{Reg}(Q)$ are called regular simple representations.
For each regular simple representation $R$, there exists a positive integer $r$ such that
$(\Phi^+)^r\, R=R$.
We call the smallest one, $p$, the period of $R$ under $\Phi^+$.
The set $\{R,\cdots,(\Phi^+)^{p-1} \, R\}$
is called the $\Phi^+$-orbit of $R$.
Given a $\Phi^+$-orbit of a regular simple representation,
the corresponding $tube$, say $T$,
is a set of isoclasses of all indecomposable regular representations whose
regular composition factors belong to this orbit.
Let $\mrm{Rep}(T)$ be the full subcategory of $\mrm{Rep} (Q)$ the
objects of which are direct sums of indecomposable representations in
$T$.
We have the following facts:
\begin{align*}
&\text{Every regular indecomposable representation belongs to a unique
tube;} \tag{6}\\
&\text{Every indecomposable object in a tube has the same period under
$\Phi^+$;}\tag{7}\\
&\text{All but finitely many regular simple objects have period
one;}\tag{8}
\end{align*}
Let $p$ be the cardinality of the set of isomorphism classes of
simple objects in $\mrm{Rep}(T)$ and $p_T$ be the period of $T$.
Then
\begin{align*}
&p=p_T
\tag{9}
\end{align*}
Given any representation $M\in \mrm{Rep}(T)$ with $T$ of period $p$,
$M$ is called aperiodic if for any $N\in T$ not all the indecomposable representations
\[
N, (\Phi^+)\, N, \cdots, (\Phi^+)^{p-1}\, N
\]
are direct summands of $M$.
Given any $x\in E_{V, \Omega}$ such that the representation $(V,x)$ is
aperiodic, we call the $G_V$-orbit $O_x$ of $x$ aperiodic.
The following lemma will be needed in the proof of Proposition ~\ref{generalcases}.
\begin{lem} \label{vanishing}
Let $M$ and $ N\in \mrm{Ind}(Q)$ be one of the following cases.
\begin{enumerate}
\item $M=(\Phi^+)^m I(i_r)$ and $ N=(\Phi^+)^{n} I(i_{r'})$,
for $m> n$ or $m=n$ and $r\leq r'$.
\item $M=(\Phi^-)^{m}P(i_r)$ and
$N=(\Phi^-)^{n}P(i_{r'})$,
for $m<n$ or $m=n$ and $r\leq r'$.
\item $M$ and $N$ are both regular, but they are not in the same tube.
\item $M$ is nonpreinjective and $ N$ is preinjective.
\item $M$ is preprojective and $N$ is nonpreprojective.
\end{enumerate}
Then we have
\begin{align*}
&\mrm{Ext}^1(M, N)=0; \tag{i}\\
&\mrm{Hom}(N, M)=0;\quad \text{ if}\; M\; \text{ and}\; N\;\text{ are not isomorphic}.
\tag{ii}\\
\end{align*}
\end{lem}
The Lemma follows from ~\cite[Chapter 6, Lemma 1]{Crawley-Boevey}.
\subsection{Tubes in a noncyclic quiver}
\label{noncyclicquivers}
Let $Q=(I,\Omega; s,t)$ be an affine quiver other than $\mrm C_p$.
Let $R=(\oplus_{i\in I}R_i, r)$ be an inhomogeneous regular simple in $\mrm{Ind}(Q)$.
Let $T$ be a tube in $\mrm{Rep}(Q)$, consisting of all indecomposable
regular representations such that there exists a filtration of subrepresentations
whose consecutive quotients
are isomorphic to regular simple representations in the $\Phi^+$-orbit of $R$.
Let $\mrm{Rep}(T)$ be the full subcategory of $\mrm{Rep}(Q)$
consisting of direct sums of indecomposable representations in $T$.
The $\Phi^+$-orbit of $R$ forms a complete list of simple objects in the subcategory
$\mrm{Rep}(T)$. Let $p$ be the minimal $ k\geq 1 $ such that $(\Phi^+)^k(R)\cong R$.
For convenience, we write $R_z=(\oplus_{i\in I}R_{z,i},r_z)$ for $(\Phi^+)^k(R)$
if $ [k]=z$ in $\mathbb{Z}/p\mathbb{Z}$, where $[k]$ is the class of $k$ in
$\mathbb{Z}/p\mathbb{Z}$.
Note that $ R_{z,i}$ is a vector space and $r_z=(r_{z, \omega}) $
with $ r_{z, \omega}:R_{z,t(\omega)}\rightarrow R_{z,h(\omega)}$ is a linear transformation
for each $ \omega \in \Omega$.
These regular simples in $ T$ satisfy the following properties.
\begin{enumerate}
\item[(1)] $\mrm{Hom}_Q\;(R_z,R_z)=K$ and $\mrm{Hom}_Q\;(R_z,R_{z'})=0$
if $z\neq z'$;
\item[(2)] $\mrm{Ext}^1_Q(R_z,R_{z-1})=K$ and $\mrm{Ext}^1_Q(R_z,R_{z'})=0$
if $z\neq z'+1$;
\item[(3)] All higher extension groups of the simple objects in $\mrm{Rep}(T)$ vanish.
\end{enumerate}
For each $z\in \mathbb{Z}/p\mathbb{Z}$,
fix an extension $E_z=(\oplus_{i \in I} E_{z,i}, e_z)$ of $R_z$ by $R_{z-1}$
such that the exact sequence
$0\to R_{z-1}\to E_z \to R_z \to 0$ is nonsplit. Notice that
$E_{z,i}=R_{z-1,i}\oplus R_{z,i}$ for any $i\in I$.
(In another word, we fix
a basis of the extension group $\mrm{Ext}^1_Q(R_z,R_{z-1})$.)
For each arrow $\omega\in \Omega$,
the linear map $e_{z,\omega}:E_{z,t(\omega)}\to E_{z, h(\omega)}$
induces a linear map
\[
l_{z,\omega}:R_{z,t(\omega)}\to R_{z-1,h(\omega)}.
\tag{4}
\]
\subsection{Cyclic quivers}
\label{cyclic}
Let $\mrm C_p$ be a cyclic quiver of $p$ vertices. More precisely,
$\mrm C_p$ is defined to be the quadruple $\mrm C_p=(I_p, \Omega_p; s_p,t_p)$, where
\begin{enumerate}
\item[(i)] $I_p=\mathbb{Z}/p\mathbb{Z}$,
\item[(ii)] $\Omega_p=\{\omega_z|\;z\in \mathbb{Z}/p\mathbb{Z}\}$,
\item[(iii)] $t_p(\omega_z)=z$ and $h_p(\omega_z)=z-1$ for any $z\in \mathbb{Z}/p\mathbb{Z}$.
\end{enumerate}
We sometimes write $z\to z-1$ for the arrow $\omega_z$.
For each vertex $z\in \mathbb{Z}/p\mathbb{Z}$, denote by $s_z$ the corresponding
simple representation, i.e., $s_z$ is the representation whose associated vector
space is $K$ at vertex $z$ and $0$ elsewhere and whose associated linear maps are zero.
For each $\lambda \in K^*$, define a representation $t_{\lambda}$
by associating the one dimensional vector space $K$
to each vertex and the identity map to each arrow except the arrow $0 \to p-1$
which is associated with the scalar map $\lambda \mrm{Id}$,
where $\mrm{Id}$ is the identity map.
The union
$\{s_z| z\in \mathbb{Z}/p\mathbb{Z}\} \cup \{t_{\lambda}|\lambda \in K^*\}$
forms a complete list of pairwise nonisomorphic simple representations
in $\mrm{Rep}(\mrm C_p)$.
In particular, the subset $\{s_z|\;z\in \mathbb{Z}/p\mathbb{Z}\}$
is a complete list of pairwise nonisomorphic simple objects
in $\mrm{Nil}(\mrm C_p)$.
Moreover, they satisfy the following properties.
\begin{itemize}
\item[(1)] $\mrm{Hom}_{\mrm C_p}\;(s_z,s_z)=K$ and $\mrm{Hom}_{\mrm
C_p}(s_z,s_{z'})=0$
if $z\neq z'$;
\item[(2)] $\mrm{Ext}^1_{\mrm C_p}(s_z,s_{z-1})=K$ and
$\mrm{Ext}^1_{\mrm C_p}(s_z,s_{z'})=0$ if $z'\neq z-1$;
\item[(3)] $\mrm{Hom}_{\mrm C_p}(t_{\lambda},t_{\lambda})=K$ and
$\mrm{Hom}_{\mrm C_p}(t_{\lambda},t_{\lambda'})=0$ if $\lambda \neq \lambda'$;
\item[(4)] $\mrm{Ext}^1_{\mrm C_p}(t_{\lambda}, t_{\lambda})=K$ and
$\mrm{Ext}^1_{\mrm C_p}(t_{\lambda}, t_{\lambda'})=0$ if $\lambda \neq \lambda'$;
\item[(5)] $\mrm{Hom}_{\mrm C_p}(s_z, t_{\lambda})=0$ and
$\mrm{Hom}_{\mrm C_p}(t_{\lambda}, s_z)=0$
for any $z\in \mathbb Z/p\mathbb{Z}$ and $\lambda \in K^*$;
\item[(6)] $\mrm{Ext}^1_{\mrm C_p}(s_z,t_{\lambda})=0$ and
$\mrm{Ext}^1_{\mrm C_p}(t_{\lambda}, s_z)=0$
for any $z\in \mathbb Z/p\mathbb{Z}$ and $\lambda \in K^*$;
\item[(7)] All higher extension groups of the simple objects vanish.
\end{itemize}
Let $s_{z,l}$ be the indecomposable representation of $\mrm C_p$ such
that its socle is $s_i$ and its length is $l$ for $z\in I_p$ and
$l\in \mathbb N$. Given any representation $M$ in $\mrm{Rep}(\mrm C_p)$,
$M$ is called aperiodic if for each $l\in \mathbb N$ not all of the
indecomposable representations
\[
s_{0, l}, s_{1, l}, \cdots, s_{p-1,l}
\]
are direct summands of $M$.
Given any $x \in E_{\mathbb V, c_p}$,
if the representation $(V,x)$ is aperiodic, then we call the
$G_{\mathbb V}$-orbit $O_x$ of $x$ aperiodic.
\section{Hall functors and Hall morphisms}
\subsection{The Hall functor $F$}
Let $T$ be a tube (Section \ref{noncyclicquivers}) in $\mrm{Rep}(Q)$. Recall
that $R_z=(\oplus_{i\in I}R_{z,i},r_z)$ ($z\in \mathbb Z/p\mathbb Z$)
are the pairwise nonisomorphic simple objects in $\mrm{Rep}(T)$.
$l_{z,\omega}:R_{z,t(\omega)}\to R_{z-1,h(\omega)}$
($\omega\in\Omega$) is a linear map defined Section
~\ref{noncyclicquivers}.
For any representation $(\mathbb V,\theta)\in \mrm{Rep}(\mrm C_p)$
(Section ~\ref{cyclic}),
define a representation
\[F(\mathbb V,\theta)=(F(\mathbb V),F(\theta))\in \mrm{Rep}(Q)\]
by
\begin{enumerate}
\item[(i)] $F(\mathbb V)_i=\oplus_{z\in \mathbb{Z}/p\mathbb{Z}}\mathbb V_z\otimes R_{z,i}$
for any $i\in I$,
\item[(ii)] $F(\theta)_{\omega}=\sum_{z\in \mathbb{Z}/p\mathbb{Z}}
(\I_z \otimes r_{z,\omega}+\theta_{z\to z-1}\otimes l_{z,\omega})$,
for each arrow $\omega\in \Omega$.
\end{enumerate}
(Here $\I_z: \mathbb{V}_z\rightarrow \mathbb{V}_z$ in (ii) is the identity map.)
Note that $F(\theta)_{\omega}$ is a linear map for $F(\mathbb
V)_{t(\omega)}$ to $F(\mathbb V)_{h(\omega)}$.
This map extends to
a functor $F:\mrm{Rep}(\mrm C_p)\to \mrm{Rep}(Q)$.
By construction, we have
\begin{lem}
\begin{itemize}
\item [(a)] $F$ is an exact functor.
\item [(b)] $F(s_z)=R_z$ for all $z \in \mathbb{Z}/p\mathbb{Z}$.
\item [(c)] $F(\mathbb V,\theta)\in \mrm{Rep}(T)$
for any $(\mathbb V,\theta) \in \mrm{Nil}(\mrm C_p)$.
\item [(d)] $F(t_{\lambda})$ is a homogeneous regular simple for any $\lambda \in K^*$.
\end{itemize}
\end{lem}
\begin{proof}
From the construction, $F$ is exact on each vector space. To show (a), one only
needs to check that $F$ is a functor which is straightforward
verified. (b) can be checked directly.
For any $(\mathbb V,\theta)\in \mrm{Nil}(\mrm C_p)$, it is well-known that
there exists a sequence of subrepresentations
\[(\mathbb V,\theta)\supseteq (\mathbb V^1,\theta^1)\supseteq \cdots
\supseteq (\mathbb \mathbb V^n,\theta^n)=0,\]
such that $(\mathbb V^l,\theta^l)/(\mathbb V^{l+1},\theta^{l+1})$ are simple representations
of type $s_1, \cdots s_p$.
From this and the construction of $F$,
there is a sequence of subrepresentations of $F(\mathbb V,\theta)$:
\[F(\mathbb V,\theta)\supseteq F(\mathbb V^1,\theta^1)\supseteq \cdots
\supseteq F(\mathbb V^n,\theta^n)=0,\]
satisfying $F(\mathbb V^l,\theta^l)/F(\mathbb
V^{l+1},\theta^{l+1})\simeq R_z$ by (a)
for some $z\in \mathbb{Z}/p\mathbb{Z}$. Therefore, (c) follows from (b).
To prove (d),
when $Q$ is of type $\tilde{A}_n$,
the functor $F$ coincides with the functor $G$ in [Li].
From the construction of $G$, $F(t_{\lambda})$ is a homogeneous regular simple.
Furthermore, $\{F(t_{\lambda})|\lambda \in K^*\}$ is a complete list of pairwise nonisomorphic
homogeneous regular simples in $\mrm{Rep}(Q)$.
When $Q$ is of type $\tilde{D}_n$ or $\tilde E_m$ ($m=6, 7$ and $8$),
we choose the orientation $\Omega'$ given in [DR, P. 46-49].
We denote by $Q'$ the corresponding quiver.
By direct computation, $F(t_{\lambda})$ is a homogeneous regular simple.
Furthermore, $\{F(t_{\lambda})|\lambda \in K^*\}$ is a complete list of pairwise nonisomorphic
homogeneous regular simples in $\mrm{Rep}(Q')$.
On the other hand,
by [BGP], there exists a sequence $i_1, \cdots, i_k$ of vertices in $I$ such that
it is $(+)$-accessible with respect to $Q$ and
$\sigma_{i_k}\sigma_{i_{k-1}}\cdots \sigma_{i_1} Q=Q'$.
Denote by $\Psi^+:\mrm{Rep}(Q)\to \mrm{Rep}(Q')$ the composition of
the corresponding reflection functors $\Phi_i^+$.
We write $R_z'$ for $\Psi^+(R_z)$, for any $z\in \mathbb Z/p\mathbb Z$.
Denote by $T'$ the tube in $\mrm{Rep}(Q')$ generated by $R_z'$,
for all $z\in \mathbb Z/p\mathbb Z$.
Define the functor $F': \mrm{Rep}(\mrm C_p)\to \mrm{Rep}(Q')$ as the functor $F$ by
replacing $R_z$ by $R_z'$.
We then have
\[F'\simeq \Psi^+ \circ F.\]
In fact,
it suffices to prove the case when $\sigma_i Q=Q'$.
In this case it can be verified directly by the construction of
the functors $F$, $F'$ and $\Phi_i^+$.
Note that $\Psi^+$ sends homogeneous regular simples to homogeneous regular simples.
This proves (d).
\end{proof}
We denote by $\mrm{HT}$ the full subcategory of $\mrm{Rep}(Q)$ generated
by $R_z$ and $F(t_{\lambda})$, for all $z\in \mathbb Z/p\mathbb Z$, $\lambda\in K^*$.
From the above Lemma, $F$ induces a functor from $\mrm{Rep}(\mrm C_p)$ to $\mrm{HT}$,
still denoted by $F$. We have
\begin{lem} \label{lem:2.6}
The induced functor $F:\mrm{Rep}(\mrm C_p) \to \mrm{HT}$ is a categorical equivalence.
\end{lem}
\begin{proof}
By Lemma 2.5 (a), $F$ induces maps:
\begin{enumerate}
\item $F_z: \mrm{Ext}^1_{\mrm C_p}(s_z,s_{z-1}) \to \mrm{Ext}^1_Q(F(s_z),F(s_{z-1}))$
for any $z\in\mathbb{Z}/p\mathbb{Z}$.
\item $F_{\lambda}:\mrm{Ext}^1_{\mrm C_p}(t_{\lambda},t_{\lambda})\to
\mrm{Ext}^1_Q(F(t_{\lambda}),F(t_{\lambda}))$ for any $\lambda \in K^*$.
\end{enumerate}
hey are all injective and $K$-linear. So $F_z$ and $F_{\lambda}$ are bijective,
for any $z\in \mathbb{Z}/p\mathbb{Z}$ and $\lambda \in K^*$.
We then have the equivalence by the following Lemma stated in
\cite[P. 129]{GRK}.
\begin{lem}
Let $E: \mathscr B \to \mathscr C$ be an exact functor between two abelian aggregates
whose objects have finite Jordan-H\"older series. Then $E$ is an equivalence if and
only if the following two conditions are satisfied:
\begin{itemize}
\item[(i)]
$E$ maps simples onto simples and induces a bijection between their sets of isoclasses.
\item[(ii)] For all simples $S, T\in \mathscr B$, the map
$\mrm{Ext}^i_{\mathscr B}(S,T)\to \mrm{Ext}^i_{\mathscr C}(ES,ET)$ induced by $E$ is bijective
for $i=1$ and injective for $i=2$.
\end{itemize}
\end{lem}
\end{proof}
\noindent
{\bf Remark.}
The functor $F$ is a Hall functor in \cite{FMV}.
A prototype of the functor $F$ can be found in [GRK].
The restriction $F:\mrm{Nil}(\mrm C_p) \to \mrm{Rep}(T)$ is equivalent.
\subsection{The Hall morphism $F$}
\label{sec:2.7}
The assignment $(\mathbb V, \theta)\mapsto (F(\mathbb V),F(\theta))$
in the above subsection gives a morphism of varieties:
\[ \tag{a} F: E_{\mathbb V, \Omega_p}\to E_{F(\mathbb V),\Omega}, \;\;\theta \mapsto F(\theta).\]
Note that both $E_{\mathbb V,\Omega_p}$ and $E_{F(\mathbb V), \Omega}$
are $K$-vector spaces and the map is affine linear ( a linear map plus
a constant map).
We set $V=F(\mathbb V)$.
By definition, $V$ is an $I$-graded $K$-vector space.
Denote by $E_1$ the image of $E_{\mathbb V,\Omega_p}$ under $F$.
(Note that $E_1$ is a translate of a vector subspace of $E_{V, \Omega}$.)
Denote by $E_2$ the set of all elements $x \in E_{V,\Omega}$ such that
$x$ is in the same $G_V$-orbit of some element in $E_1$. Clearly,
\[\tag{b} E_1\subseteq E_2\; \text{ and } \;E_2 \; \text{ is a $G_V$-stable subvariety
of $E_{V, \Omega}$}.\]
Moreover, the assignment $(\mathbb V, \theta)\mapsto (F(\mathbb V),F(\theta))$
induces an algebraic group homomorphism defined by
\[\tag{c} F: G_{\mathbb V} \to G_V, \;\; g \mapsto g \otimes 1, \; \text{
for any } \;g \in G_{\mathbb V},\]
where
$(g \otimes 1)_i=\oplus_{z\in \mathbb{Z}/ p \mathbb{Z}}\, g_z \otimes \I_{z,i} :
\oplus_z \mathbb V_z \otimes R_{z,i} \to \oplus_z \mathbb V_z \otimes R_{z,i}$
for all $i\in I$, with $\I_{z,i}: R_{z,i}\rightarrow R_{z,i}$ being the identity map.
For simplicity, we write $H_V$ for $F(G_{\mathbb V})$.
It is an algebraic subgroup of $G_V$.
Note that $G_{\mathbb V} \simeq H_V$.
Also there is an action of $H_V$ on $E_1$ induced by the action of
$G_{\mathbb V}$ on $E_{\mathbb V,\Omega_p}$.
By definitions, the action of $G_{\mathbb V}$ on $E_{\mathbb V,\Omega_p}$ is compatible with
the action of $H_V$ on $ E_1$, i.e.,
\[\tag{d}
F(g\theta)=F(g)F(\theta),\; \text{ for any }\; g\in G_{\mathbb V} \;
\text{ and }\theta\in E_{\mathbb V,\Omega_p}.\]
Furthermore, we have
\[ \tag{e} \;
\text{The map $F: E_{\mathbb V, \Omega_p} \to E_1$ is an isomorphism of varieties.}\]
\[\tag{f} \text{ For $x, x_1\in E_1$, $\{\xi \in G_V\;|\; \xi
x=x_1\} \subseteq H_V$. In particular, $ H_V=\mrm{Stab}_{G_V}(F(0))$.}\]
In fact,
$x=F(\theta)$ and $x_1=F(\theta_1)$,
for some $\theta, \theta_1\in E_{\mathbb V,\Omega_p}$.
So $\xi F(\theta)=F(\theta_1)$.
That is
$\xi \in \mrm{Hom}_Q((V,F(\theta)), (V,F(\theta_1)))$.
Since the functor $F:\mrm{Rep}(\mrm C_p)\to \mrm{HT}$ is a
categorical equivalence,
it is then fully faithful. So
$\mrm{Hom}_Q((V,F(\theta)), (V,F(\theta_1)))=
F(\mrm{Hom_{C_p}}((\mathbb V,\theta), (\mathbb V,\theta_1)))$.
Therefore, $\xi=F(g)$ for some $g\in G_{\mathbb V}$.
Let
\[
\mrm{GL}(V_i)^0=\prod_{z\in \mathbb{Z/Z}p} \mrm{GL}(\mathbb V_z\otimes R_{z,i})
\quad
\text{and}
\quad
G_V^0=\prod_{i\in I} \mrm{GL}(V_i)^0.
\]
Then $G_V^0$ is a Levi subgroup of the reductive group $G_V$.
Let
\begin{equation*}
\begin{split}
E_{V,\Omega}^0=\oplus_{\omega\in \Omega} \oplus_{z\in \mathbb{Z/Z}p}
\mrm{Hom}_K(\mathbb V_z\otimes R_{z,t(\omega)}, \mathbb V_z\otimes R_{z, h(\omega)}),\\
E_{V,\Omega}^1=\oplus_{\omega\in \Omega} \oplus_{z\in \mathbb{Z/Z}p}
\mrm{Hom}_K(\mathbb V_z\otimes R_{z,t(\omega)}, \mathbb V_{z-1}\otimes R_{z-1, h(\omega)}).
\end{split}
\end{equation*}
By definition
\[
G_V^0E_1 \subseteq E_{V,\Omega}^0\oplus E_{V,\Omega}^1.
\]
The map $\phi: G_V^0 E_1 \to G_V^0F(0) \quad (gF(\theta)\mapsto gF(0))$ is then a
restriction of the second projection $E_{V,\Omega}^0\oplus E_{V,\Omega}^1 \to E_{V,\Omega}^0$.
Hence $\phi$ is a morphism of varieties.
By ~\cite[Lemma 4 in 3.7]{slodowy} and the fact that $G_V^0/H_V\simeq G_V^0 F(0)$, we have
\[
\label{k}
G_V^0\times ^{H_V}E_1 \simeq G_V^0 E_1.
\tag{g}
\]
Define an action of $H_V$ on $G_V \times E_1$ by
$(\zeta, x).\xi=(\zeta\xi, \xi^{-1}x)$ for any
$(\zeta,x)\in G_V \times E_1$ and $\xi \in H_V$.
Define a map $\tau: G_V \times E_1 \to E_2$ by $(\zeta, x) \mapsto \zeta.x$
for any $(\zeta, x)\in G_V \times E_1 $.
By (j), the morphism $\tau$ is an $H_V$-orbit map, i.e., all the
fibres of $\tau$ over $x\in E_2$ are $H_V$-orbit.
Thus it induces a bijective morphism of varieties
\[
\bar\tau: G_V\times^{H_V} E_1 \to E_2.
\]
Note that $H_V \subseteq G_V^0$. By ~\cite{slodowy} and (\ref{k}), we have
\[
G_V\times^{H_V}E_1 \simeq G_V\times^{G_V^0}(G_V^0\times ^{H_V} E_1)\simeq
G_V\times ^{G_V^0} (G_V^0E_1).
\]
Thus $\bar\tau$ induces a bijective morphism of varieties
\[
\bar\tau: G_V\times^{G_V^0}(G_V^0E_1) \to E_2.
\tag{h}
\]
Let $U^+$ (resp. $U^-$) be the block upper (resp. lower) triangle matrices
with respect to $G_V^0$ in $G_V$.
$U^+$ and $U^-$ are unipotent radicals of opposite parabolics with
$G_V^0$ as Levi subgroup. Then $U^+\times G^0_V \times U^-$ is an affine open
subvariety in $G_V$. Thus $U:=U^+\times U^- \times G_V^0E_1$ is an affine open subvariety in
$G_V\times^{G_V^0} G_V^0E_1$. By restricting to the affine open subvariety
$U$, the morphism $\bar\tau$ becomes an isomorphism onto its image by
direct matrix computations (the inverse of $\bar \tau$ is algebraic). Now that
$\{gU\,|\, g\in G_V\}$ is an open cover of $G_V\times^{G_V^0}(G_V^0E_1)$, we have
\begin{lem}
\label{geometric-F}
$G_V\times^{H_V} E_1 \simeq E_2$ as $G_V$-varieties.
\end{lem}
In particular,
\[\tag{i} O_x=G_V \times^{H_V} O'_x, \]
where $x\in E_1$, $O_x$ is the $G_V$-orbit of $x$ in
$ E_2$ and $O'_x$ is the $H_V$-orbit of $x$ in $ E_1$.
\section{The canonical basis}
In this section, we recall Lusztig's geometric realization of the
canonical basis of $\mbf U^-$.
\subsection{Notations}
We fix some notations, most of them are consistent with the notations
in ~\cite{lusztig3}.
Fix a prime $l$ that is invertible in $K$.
Given any algebraic variety $X$ over $K$,
denote by $\mathcal{D}_c^b(X)$ the bounded derived category of complexes
of $l$-adic sheaves on $X$ (\cite{BBD}).
Let $\mathcal M(X)$ be the full subcategory of $\mathcal D_c^b(X)$
consisting of all perverse sheaves on $X$ (\cite{BBD}).
Let $G$ be a connected algebraic group. Assume that $G$ acts on $X$
algebraically. Denote by $\mathcal D_G(X)$ the full subcategory of
$\mathcal D_c^b(X)$ consisting of all $G$-equivariant complexes over
$X$.
Similarly, denote by $\mathcal M_G(X)$ the full subcategory of
$\mathcal M(X)$ consisting of all $G$-equivariant perverse sheaves (\cite{lusztig3}).
Let $\bar{\mathbb Q}_l$ be an algebraic closure of the field of $l$-adic numbers.
By abuse of notation, denote by $\bar{\mathbb Q}_l=(\bar{\mathbb Q}_l)_X$
the complex concentrated on degree zero, corresponding to
the constant $l$-adic sheaf over $X$.
For any complex $K \in \mathcal D_c^b(X)$ and $n\in \mathbb Z$,
let $K[n]$ be the complex such that $K[n]^i=K^{n+i}$ and the
differential is multiplied by a factor $(-1)^n$.
Denote by $\mathcal M(X)[n]$ the full subcategory of
$\mathcal D_c^b(X)$ objects of which are of the form $K[n]$ with
$K\in \mathcal M(X)$.
For any $K\in \mathcal D_c^b(X)$ and $L\in \mathcal D_c^b(Y)$, denote
by $K\boxtimes L$ the external tensor product of $K$ and $L$ in
$\mathcal D_c^b(X\times Y)$.
Let $f: X\to Y$ be a morphism of varieties, denote by
$f^*: \mathcal D_c^b(Y) \to \mathcal D_c^b(X)$ and
$f_!: \mathcal D_c^b(X) \to \mathcal D_c^b(Y)$
the inverse image functor and the direct image functor with compact support, respectively.
If $G$ acts on $X$ algebraically and $f$ is a principal $G$-bundle,
then $f^*$ induces a functor (still denote by $f^*$) of equivalence
between $\mathcal M(Y)[\dim G]$ and $\mathcal M_G(X)$.
Its inverse functor is denoted by $f_{\flat}: \mathcal M_G(X) \to
\mathcal M(Y)[\dim G]$ (\cite{lusztig3}).
\subsection{Lusztig's induction functor}
\label{inductionfunctor}
We recall the (geometric) definition of the canonical basis
from ~\cite{lusztig1} (or ~\cite{lusztig3}).
Let $V=\oplus_{i\in I} V_i$ be an $I$-graded $K$-vector space.
We have the following data:
\begin{enumerate}
\item The set $\mathcal{X}_{|V|}$. It consists of all sequences
$\boldsymbol{\nu}=(\nu^1,\cdots, \nu^n)$
such that $\sum_{m=1}^{n}\nu^m=|V|$ and
$\nu^m_{s(\omega)}\cdot \nu^m_{t(\omega)}=0$,
for any $\omega \in \Omega$
and $1\leq m \leq n$.
\item The variety $\mathcal{F}_{\boldsymbol{\nu}}$, for any
$\boldsymbol{\nu}=(\nu^1,\cdots,\nu^n)\in \mathcal{X}_{|V|}$.
It consists of all sequences,
$(V=V^0\supseteq V^1\supseteq \cdots \supseteq V^n=0)$,
of $I$-graded subspaces of $V$ such that $|V^m/V^{m+1}|=\nu^{m+1}$,
for $0\leq m \leq n-1$.
\item The variety $\tilde{\mathcal{F}}_{\boldsymbol{\nu}}$, for any
$\boldsymbol{\nu}=(\nu^1,\cdots,\nu^n)\in \mathcal{X}_{|V|}$.
It consists of all pairs $(x,\mbf f)$,
where $x\in E_{V,\Omega}$ and $\mbf f\in \mathcal{F}_{\boldsymbol{\nu}}$,
such that $\mbf f$ is $x$-stable.
(Here $\mbf f$ is $x$-stable means that for any vector subspace,
$V^m$ in $\mbf f$,
$x_{\omega}(V^m_{s(\omega)})\subseteq V^m_{t(\omega)}$, for all $\omega \in \Omega$.)
\end{enumerate}
The first projection
$\boldsymbol{\pi_{\nu}}: \tilde{\mathcal F}_{\boldsymbol{\nu}} \to E_{V, \Omega}$
then induced a right derived functor
\[(\boldsymbol{\pi_{\nu}})_!:\mathcal{D}_c^b(\tilde{\mathcal{F}}_{\boldsymbol{\nu}})
\to \mathcal{D}_c^b(E_{V,\Omega}).\]
Note that
$\boldsymbol{\pi_{\nu}}$ is proper and $\tilde{\mathcal F}_{\boldsymbol{\nu}}$ is smooth.
By the Decomposition theorem in ~\cite{BBD},
$(\boldsymbol{\pi_{\nu}})_!(\bar{\mathbb Q}_l)$ is a semisimple complex in
$\mathcal{D}_c^b(E_{V,\Omega})$.
Moreover, $(\boldsymbol{\pi_{\nu}})_!(\bar{\mathbb Q}_l)$ is $G_V$-equivariant since
$\boldsymbol{\pi_{\nu}}$ is $G_V$-equivariant.
Let $\mathcal{B}_V$ be the set consisting of all isomorphism classes of
simple $G_V$-equivariant perverse sheaves on $E_{V,\Omega}$
that are in the direct summand of the semisimple complex
$(\boldsymbol{\pi_{\nu}})_!(\bar{\mathbb Q}_l)$
(up to shift) for some $\boldsymbol{\nu}\in \mathcal{X}_{|V|}$.
By abuse of language, we say ``a complex is in $\mathcal B_V$''
instead of ``the isomorphism classes of a complex is in $\mathcal B_V$''.
\begin{example}
When $|V|=i$, $E_{V,\Omega}$ consists of a single point. The isomorphism classes of
the complex $\bar{\mathbb Q}_l$ is the only element in $\mathcal B_V$. We denote it by
$F_i^{[1]}$.
\end{example}
Let $\mathcal{Q}_V$ be the full subcategory of $\mathcal D_c^b(E_{V,\Omega})$
consisting of all complexes on $E_{V,\Omega}$
isomorphic to a direct sum of shifts of finitely many complexes in $\mathcal{B}_V$.
Let $W\subseteq V$ be an $I$-graded $K$-subspace of $V$ and $T=V/W$.
For any $x\in E_{V,\omega}$ such that
$x_{\omega}(W_{t(\omega)})\subseteq W_{h(\omega)}$ for all $\omega \in
\Omega$, we call that $W$ is $x$-stable and
it then induces elements $x_W$ and $x_{T}$ in $E_{W,\Omega}$ and
$E_{T,\Omega}$, respectively.
Define $E''$ to be the variety consisting of all pairs $(x, V')$, where
$x\in E_{V,\Omega}$ and $V'$ is an $I$-graded subspace of $V$, such that
$|V'|=|W|$ and $V'$ is $x$-stable.
Define $E'$ to be the variety consisting of all quadruples
$(X, V'; R', R'')$ where $(x, V')$ is in $E''$, and $R': V'\to W$ and
$R'': V/V' \to T$ are $I$-graded linear isomorphisms.
Consider the following diagram
\[
\label{diagram}
\begin{CD}
E_{T,\Omega}\times E_{W,\Omega} @<p_1<< E' @>p_2>> E'' @>p_3>> E_{V,\Omega},
\end{CD}
\tag{*}
\]
where the maps are defined as follows.
$p_3: (x, V') \mapsto x$, $p_2:(x, V'; R', R'') \mapsto (x,V')$, and
$p_1: (x,V';R',R'') \mapsto (x', x'')$, where
$x'_{\omega}=R'_{h(\omega)} x_{V'} (R'_{t(\omega)})^{-1}$ and
$x''_{\omega}=R''_{h(\omega)} x_{V/V'} (R''_{t(\omega)})^{-1}$
for all $\omega \in \Omega$.
Note that $p_3$ is proper, $p_2$ is a principal $G_T\times G_W$-bundle
and $p_1$ is smooth with connected fibres.
From (\ref{diagram}), we can form a functor
\[
(p_3)_! (p_2)_{\flat} (p_1)^*: \mathcal D_c^b(E_{T,\Omega}\times
E_{W,\Omega}) \to \mathcal D_c^b(E_{V,\Omega}).
\]
We write $K\star L:=(p_3)_! (p_2)_{\flat} (p_1)^*(K\boxtimes L)$ for any
$K\in \mathcal D_c^b(E_{T,\Omega})$ and $L\in \mathcal D_c^b(E_{W,\Omega})$.
\begin{lem}
\label{induction}
\begin{enumerate}
\item $K\star L \in \mathcal Q_V$
for any $K\in \mathcal Q_T$ and $L\in \mathcal Q_W$.
\item $(K\star L) \star M=K \star (L \star M)$ for any $K\in \mathcal
Q_T$, $L\in \mathcal Q_W$ and $M\in \mathcal Q_U$.
\item $(\boldsymbol{\pi_{\nu}})_!(\bar{\mathbb Q}_l)=
(\boldsymbol{\pi}_{(\nu^1)})_! (\bar{\mathbb Q}_l)
\star \cdots \star
(\boldsymbol{\pi}_{(\nu^n)})_! (\bar{\mathbb Q}_l)$
where $(\nu^m)$ are sequences with only one entry $\nu^m$ for $m=1,\cdots,n$.
\end{enumerate}
\end{lem}
See ~\cite{lusztig1} for a proof.
By Lemma ~\ref{induction} (2), the expression $K\star L \star M$ makes no confusion.
Let $\boldsymbol{\nu}=(\nu_1, \cdots, \nu_n)$ be a sequence of elements in $\mathbb N[I]$.
(($\nu_1, \cdots, \nu_n$) need not satisfy the conditions
in Section ~\ref{inductionfunctor} (1).)
Let $V$ be an $I$-graded $K$-space of dimension $\sum_{m=1}^n \nu_m$.
Assume that $V^{(m)}$ ($m=1,\cdots, n$) are $I$-graded subspaces of $V$ such that
$V=\oplus_{m=1}^n V^{(m)}$.
Define $F''$ to be the variety consisting of all pairs $(x, V^{\bullet})$
where $x\in E_{V,\Omega}$ and $V^{\bullet}$ is a flag of type $\boldsymbol{\nu}$ such that
$V^{\bullet}$ is $x$-stable.
Define $F'$ to be the variety consisting of all triples $(x, V^{\bullet}; \mbf g)$
where $(x, V^{\bullet})$ is in $F'$ and $\mbf g$ is a sequence of linear isomorphisms
$(g_m: V^{m-1}/V^m\to V^{(m)}\;|\; m=1, \cdots, n)$.
(Here $V^{\bullet}=(V=V^0\supseteq V^1 \supseteq \cdots \supseteq V^n=0)$.)
Consider the following diagram
\[
\label{n-sequence}
\begin{CD}
E_{V^{(1)}, \Omega}\times \cdots \times E_{V^{(n)},\Omega} @<q_1<< F' @>q_2>> F'' @>q_3>> E_{V,\Omega}
\end{CD}
\tag{**}
\]
where the maps are defined by $q_3: (x, V^{\bullet}) \mapsto x$,
$q_2: (x, V^{\bullet},\mbf g)\mapsto (x, V^{\bullet})$, and
$q_1: (x, V^{\bullet}, \mbf g)\mapsto (x^{(1)}, \cdots, x^{(n)})$ with
$x^{(m)}_{\omega}= (g_m)_{h(\omega)} x_{V^{m-1}/V^m} (g_m)_{t(\omega)}^{-1}$ for $m=1\cdots,n$.
Similar to $p_1, p_2$ and $p_3$,
the morphisms $q_1, q_2$ and $q_3$ are smooth with connected fibres, principal
$G_{V^{(1)}}\times \cdots \times G_{V^{(n)}}$-bundle and proper, respectively.
Note that when $n=2$, diagram (\ref{n-sequence}) coincides with diagram (\ref{diagram}).
\begin{lem}
\label{n-mult}
$K^{(1)}\star \cdots \star K^{(n)}
=(q_3)_! (q_2)_{\flat} (q_1)^* (K^{(1)}\boxtimes \cdots \boxtimes K^{(n)})$
for any $K^{(m)}\in \mathcal Q_{V^{(n)}}$ where $m=1, \cdots,n$.
\end{lem}
\begin{proof}
The statement follows from Lemma ~\ref{induction} when $n=2$.
When $n> 2$, the statement can be proved by induction.
\end{proof}
\subsection{Lusztig's algebras}
\label{algebra}
Let $\mathcal{K}_V=\mathcal K(\mathcal Q_V)$ be the Grothendieck group of the category
$\mathcal Q_V$, i.e., it is
the abelian group with one generator $\langle L\rangle$ for each isomorphism class
of objects in $\mathcal{Q}_V$ with relations:
$\langle L\rangle+\langle L'\rangle=\langle L''\rangle$
if $ L'' $ is isomorphic to $L\oplus L'$.
Let $v$ be an indeterminate. Set $\mathbb A=\mathbb Z[v,v^{-1}]$.
Define an $\mathbb A$-module structure on $\mathcal K_V$ by
$v^n\langle L\rangle =\langle L[n]\rangle $
for any generator $\langle L\rangle\in\mathcal{Q}_V$ and
$n\in \mathbb Z$.
From the construction,
it is a free $\mathbb A$-module with basis $\langle L\rangle$
where $\langle L\rangle$ runs over $\mathcal{B}_V$.
From the construction, we have
$\mathcal{K}_V \cong \mathcal{K}_{V'}$,
for any $V$ and $V'$ such that $|V|=|V'|$.
For each $\nu\in \mathbb N[I]$, fix an $I$-graded vector space $V$ of dimension $\nu$.
Let
\[
\mathcal{K}_{\nu}=\mathcal{K}_V,\quad
\mathcal{K}=\oplus_{\nu\in \mathbb{N}[I]}\mathcal{K}_{\nu}\quad
\text{and} \quad
\mathcal{K}_Q=\mathbb{Q}(v)\otimes_{\mathbb A]}\mathcal{K}.
\]
Also let
\[
\mathcal B_{\nu}=\mathcal B_V
\quad \text{and} \quad
\mathcal B_Q=\cup_{\nu\in \mathbb N[I]} \mathcal B_{\nu}.
\]
For any $\alpha, \beta\in \mathbb N[I]$, the operation $\star$ induces an
$\mathbb A$-linear map
\[
\star: \mathcal K_{\alpha} \otimes_{\mathbb A} \mathcal K_{\beta}
\to \mathcal K_{\alpha +\beta}.
\]
By adding up these linear maps, we have a linear map
\[
\star: \mathcal K \otimes_{\mathbb A} \mathcal K \to \mathcal K.
\]
Similarly, the operation $\star$ induces a $\mathbb Q(v)$-linear map
\[
\star:\mathcal K_Q \otimes_{\mathbb Q(v)} \mathcal K_Q \to \mathcal K_Q.
\]
\begin{prop}
\begin{enumerate}
\item $(\mathcal K, \star)$ (resp. $(\mathcal K_Q, \star)$) is an associative algebra
over $\mathbb A$ (resp. $\mathbb Q(v)$).
\item $\mathcal B_Q$ is an $\mathbb A$-basis of $(\mathcal K, \star)$
and a $\mathbb Q(v)$-basis of $(\mathcal K_Q, \star)$.
\end{enumerate}
\end{prop}
\begin{proof}
The associativity of $\star$ follows from Lemma ~\ref{induction} (2).
\end{proof}
Define a new $\mathbb A$-linear map
$\circ: \mathcal K_{\alpha} \otimes \mathcal K_{\beta} \to \mathcal K_{\alpha+\beta}$
by
\[x\circ y=v^{m(\alpha,\beta)} x\star y\]
where
\[m(\alpha, \beta)=\sum_{i\in I}\alpha_i \beta_i+
\sum_{\omega\in \omega}\alpha_{t(\omega)}\beta_{h(\omega)}.\]
This induces a bilinear map
\[
\circ: \mathcal K\otimes \mathcal K \to \mathcal K.
\]
Then
\begin{cor}
$(\mathcal K, \circ)$ is an associative algebra over $\mathbb A$.
\end{cor}
The linear map $\circ$ satisfies the associativity due to the fact that $m(-,-)$ is a cocycle.
Similarly, we have an associative algebra $(\mathcal K_Q, \circ)$ over $\mathbb Q(v)$.
\subsection{The canonical basis $\mbf B$ of the algebra $\mbf U^-$}
\label{canonical}
Given any quiver $Q$, let $c_{ii}=2$ and
\[c_{ij}=-\#\{ \omega\in \Omega\;|\;
\{t(\omega), h(\omega)\}=\{i,j\}\} \quad \text{for} \quad i\neq j.\]
$C=(c_{ij})_{i,j\in I}$ is then a symmetric generalized Cartan matrix.
Note that the Cartan matrix $C$ is independent of changes of the orientation of $Q$.
For any $m\leq n\in \mathbb N$, let
\[
[n]=\frac{v^n-v^{-n}}{v-v^{-1}}, \quad
[n]^!=\prod_{m=1}^n [m]
\quad
\text{and}
\quad
\bin{n}{m}=\frac{[n]^!}{[m]^![n-m]^!}.
\]
Denote by $\mbf{U}^-$ the negative part of the quantized
enveloping algebra attached to the Cartan matrix $\mrm{C}$.
$\mbf U^-$ is the quotient of the free algebra with generators $F_i$, $i\in I$ by
the two-sided ideal generated by
\begin{equation}
\sum_{p=0}^{1-c_{ij}} (-1)^p \bin{1-c_{ij}}{p} F_i^p F_j F_i^{1-c_{ij}-p},
\end{equation}
for $i\neq j\in I$.
Let $F^{(n)}_i=\frac{F_i^n}{[n]^!}$ for all $i\in I$ and $n\in \mathbb N$.
Let $_{\mathbb A}\mbf U^-$ be the $\mathbb A$-subalgebra of $\mbf U^-$
generated by $F^{(n)}_i$ for $i\in I$ and $n\in \mathbb N$.
We have
\begin{thm}(\cite{lusztig1}, \cite{lusztig3})
\label{iso}
The map $F_i^{(1)} \mapsto F_i^{[1]}$ induces an $\mathbb A$-algebra isomorphism
\[_{\mathbb A}\phi: \,_{\mathbb A} \mbf U^- \to (\mathcal K, \circ)\]
and a $\mathbb Q(v)$-algebra isomorphism
\[
\phi: \mbf U^- \to(\mathcal K_Q, \circ).
\]
\end{thm}
\noindent
{\bf Remark.} See ~\cite{lusztig3} for a more general treatment that
works for any symmetrisable generalized Cartan matrix $C$.
Given another quiver $Q'=(I,\Omega,t',h')$ such that
\[\{t(\omega), h(\omega)\}=\{t'(\omega),h'(\omega)\}\] for all $\omega\in \Omega$.
From Theorem ~\ref{iso}, the map $F_i^{(1)}\mapsto F_i^{[1]}$ induces
an $\mathbb Q(v)$-algebra isomorphism
\[
\phi': \mbf U^- \to (\mathcal K_{Q'}, \circ).
\]
We have
\begin{thm}(\cite{lusztig1}, \cite{lusztig3})
\label{coincide}
$\phi^{-1}(\mathcal B_Q)=(\phi')^{-1} (\mathcal B_{Q'}).$
\end{thm}
\begin{Def}
$\mbf B=\phi^{-1}(\mathcal B_Q)$ is called the Canonical Basis of $\mbf U^-$.
\end{Def}
For each $Q$, $\mathcal B_Q$ gives a presentation of $\mbf B$.
The main goal of this paper is to describe the elements in $\mathcal B_Q$ by specifying
their supports and the corresponding local systems when $Q$ is affine.
\section{The description of the elements in $\mathcal B_Q$ via quiver representations}
\subsection{Simple equivariant perverse sheaves}
Note that elements in $\mathcal B_Q$ are isomorphism classes of simple equivariant perverse
sheaves. We give a brief description of simple equivariant perverse sheaves.
Let $X$ be an algebraic variety over $K$ with a connected algebraic group $G$ acting on it.
Let $Y$ be a smooth, locally closed, irreducible $G$-invariant subvariety of $X$ and
$\mathcal L$ an irreducible, $G$-equivariant, local system on $Y$.
Denote by $j: Y\to X$ the natural embedding.
\begin{thm} (\cite{BBD}, ~\cite{BL})
The complex
\[\mrm{IC}(Y, \mathcal L):=j_{!\star}(\mathcal L)[\dim Y]\]
is a simple $G$-equivariant perverse sheaf on $X$.
Moreover, all simple $G$-equivariant perverse sheaves on $X$ are of this form.
\end{thm}
\subsection{Cyclic quivers}
When the quiver $Q$ is the cyclic quiver $\mrm C_p$ for some $p\in \mathbb N$.
The description of the elements in $\mathcal B$ is given as follows.
Let $\mathbb V$ be a $\mathbb Z/p\mathbb Z$-graded $K$-vector space.
Recall that a $G_{\mathbb V}$-orbit $O$ in $E_{\mathbb V, \Omega_p}$
is aperiodic if for any $x \in O$, the representation $(\mathbb V, x)$ is aperiodic
(see Section ~\ref{cyclic}).
Let $\mathcal O_{\mathbb V}^a$ be the set of all aperiodic $G_{\mathbb V}$-orbits
in $ E_{\mathbb V,\Omega_p}$.
Given $O \in \mathcal O_{\mathbb V}^a$,
let $\mrm{IC}(O, \bar{\mathbb Q}_l)$ be the intersection cohomology complex on
$E_{\mathbb V,\Omega}$ determined by the subvariety $O$ and the constant sheaf
$\bar{\mathbb Q}_l$ on $O$.
The assignment $O\mapsto \mrm{IC}(O,\bar{\mathbb Q}_l)$
defines a map $\mathcal O_{\mathbb V}^a\to \mathcal B_{\mathbb V}$.
Furthermore, we have
\begin{thm}
(\cite[5.9]{lusztig2})
\label{cycliccase}
The map $\mathcal O_{\mathbb V}^a\to \mathcal B_{\mathbb V}$ is bijective.
\end{thm}
\subsection{Noncyclic quivers}
\label{noncyclic}
From now on, we assume that the affine quiver $Q$ is not $\mrm C_p$,
for any $p\in \mathbb N$. We follow Lusztig's argument in
~\cite[Section 6]{lusztig2}.
We study three special cases in this section.
First,
given $M \in \mrm{Ind}(Q)$ of dimension vector $\nu$.
Assume that $M$ is either preprojective or preinjective.
Let $V$ be a $K$-vector space such that $|V|=\nu$.
Let $O_M$ be the $G_V$-orbit in $E_{V,\Omega}$ corresponding to $M$.
Then we have
\begin{lem} (\cite[Lemma 6.8]{lusztig2})
\label{Lemma-1}
$\mrm{IC}(O_M,\bar{\mathbb Q}_l)\in \mathcal B_V$.
\end{lem}
\begin{proof}
Since $M$ is either preprojective or preinjective, its self-extension group
$\mrm{Ext}^1_Q(M,M)=0$, so the corresponding $G_V$-orbit
$O_M$ is open in $E_{V,\Omega}$ (see ~\cite{Crawley-Boevey}).
Since $E_{V,\Omega}$ is smooth, $\mrm{IC}(O_M,\bar{\mathbb Q}_l)$
is the constant sheaf $\bar{\mathbb Q}_l$ on $E_{V,\Omega}$
up to shift (see ~\cite[Lemma 4.3.2]{BBD}).
Now that $Q$ has no oriented cycles,
we can order the vertices in $I$
$i_1,\cdots,i_n$ ($n=|I|$) in a way
such that $i_r$ is a source of the full subquiver $Q_r$
with vertex set $I-\{i_1,\cdots,i_{r-1}\}$.
For any $V$ of dimension vector $\nu$, let
$\boldsymbol{\nu}=(\nu_{i_1}\;i_1,
\cdots,\nu_{i_n}\;i_n)$.
By definition,
$\mathcal{F}_{\boldsymbol{\nu}}$ consists of a single flag.
Also any $x$ in $E_{V,\Omega}$ stabilizes this flag.
So the first projection
$\boldsymbol{\pi_{\nu}}:\tilde{\mathcal{F}}_{\boldsymbol{\nu}} \to E_{V,\Omega}$
is an isomorphism.
Therefore, we have
$(\boldsymbol{\pi_{\nu}})_!(\bar{\mathbb{Q}}_l)[d]=
\bar{\mathbb{Q}}_l[d]\in \mathcal B_V $.
\end{proof}
Second, we assume that $V$ is an $I$-graded $K$-vector space of dimension vector
$q \delta$, where $\delta\in \mathbb N[I]$ is the minimal positive imaginary root
of the symmetric Euler form (see Section ~\ref{preliminary}) associated to Q.
We define two varieties as follows.
\begin{enumerate}
\item The variety $X(0)$.
It is the subvariety of
$E_{V,\Omega}$ consisting of all elements $x$ such that
$(V,x) \simeq R_1 \oplus \cdots \oplus R_q$,
where $R_1,\cdots,R_q$ are pairwise nonisomorphic homogeneous regular simples.
\item The variety $\tilde{X}(0)$.
This variety consists of all pairs $\{x, (R_1,\cdots,R_q)\}$,
where $x\in X(0)$ and $(R_1,\cdots,R_q)$ is a sequence of
representations in $\mrm{Ind}(Q)$,
such that $(V,x)\simeq R_1\oplus\cdots\oplus R_q$.
\end{enumerate}
Note that the dimension vectors of $R_m$ in (1) have to be $\delta$,
for all $m\in \{1, \cdots, q\}$
and once $x$ is fixed, the set of the representations $R_1,\cdots, R_q$ in (2)
is completely determined.
Note also that the closure of $X(0)$ equals $E_{V,\Omega}$.
In fact, by \cite{Ringel2}, we have $\mrm{dim}\; X(0)=\mrm{dim}\; O_x+q$.
Since
\[
\mrm{dim} \; G_V-\mrm{dim} \; E_{V,\Omega}
=\, <q\delta, q\delta>\,=
\mrm{dim}\;\mrm{Hom}_Q((V,x),(V,x))-\mrm{dim}\;\mrm{Ext}^1((V,x),(V,x)),
\]
we have
\[
\mrm{dim}\; E_{V,\Omega}=\mrm{dim} \; G_V-\mrm{dim}\;\mrm{Hom}((V,x),(V,x))+q.
\]
So $\mrm{dim}\; E_{V,\Omega}=\mrm{dim}\; O_x+q=\mrm{dim} \; X(0)$.
Therefore $E_{V, \Omega}$ is the closure of $X(0)$.
The first projection
\[
\pi_1:\tilde{X}(0)\to X(0)
\quad
(x, (R_1,\cdots, R_q)) \mapsto x
\]
is an $S_q$-principal covering
where $S_q$ is the symmetric group of $q$ letters.
$S_q$ acts naturally on $(\pi_1)_{\star}(\bar{\mathbb{Q}}_l)$.
Given any irreducible representation $\chi$ of $S_q$,
denote by $\mathscr L_{\chi}$ the irreducible local system corresponding to
the representation $\chi$ via the monodromy functor (see ~\cite{Iversen}).
Note that $\mathscr L$ is a direct summand of
$(\pi_1)_{\star}(\bar{\mathbb{Q}}_l)$.
Let $\mrm{IC}(X(0),\mathscr L_{\chi})$ be the intersection complex on
$E_{V,\Omega}$ determined by $X(0)$ and $\mathscr L_{\chi}$.
We then have:
\begin{lem} (\cite[6.10 (a)]{lusztig2})
\label{Lemma-2}
$\mrm{IC}(X(0),\mathscr L_{\chi})\in \mathcal B_V.$
\end{lem}
\begin{proof}
Let $\boldsymbol{\delta}=(\delta,\cdots,\delta)$
such that
$|\boldsymbol{\delta}|=q\;\delta$.
Let
$\boldsymbol{\pi_{\delta}}:\tilde{\mathcal{F}}_{\boldsymbol{\delta}}
\rightarrow E_{V,\Omega}$
be the first projection defined as the morphism $\boldsymbol{\pi_{\nu}}$
in Section ~\ref{inductionfunctor}.
Given any $x\in X(0)$,
let
\[\mbf f=(V=V^0\supseteq V^1\supseteq\cdots\supseteq V^n=0)\]
be a flag in
$\mathcal F_{\boldsymbol{\nu}}$ such that $\mbf f$ is $x$-stable. Then
\[
\tag{1}
\text{The subrepresentation $(V^m,x)$ is regular, for any $m\in \{0,\cdots,n-1\}$}.
\]
This is because if $x\in X(0)$, $(V,x)$ is a regular representation.
So $(V^r,x)$ can not have preinjective subrepresentations.
Now that the dimension vector of $(V^r,x)$ is $\delta$,
the defect of $(V^r,x)$ is zero.
Thus, $(V^r,x)$ can not have preprojective subrepresentations.
Therefore, $(V^r,x)$ is regular.
Fix an element $x\in X(0)$,
we decompose $V=\oplus_{r=1}^q V(r)$ such that $V(r)$ is $x$-stable and $|V(r)|=\delta$.
By the definition of $X(0)$ and (1), we have $(V,x) \simeq \oplus_r (V(r),x)$.
\[
\tag{2}
\text{ Moreover, this decomposition is unique up to order.}
\]
In fact, if $V=\oplus_r W(r)$ is another decomposition,
we can reorder the $W(r)$'s such that the subrepresentations
$(W(r),x)$ and $(V(r),x)$ are isomorphic, for any $r$.
Fix an isomorphism $f_r:(W(r),x)\to (V(r),x)$ for each $r$, then
they induce an isomorphism
\[f:=\sum_r f_r:(V,x)\to (V,x)\]
satisfying $f(W(r))\subseteq V(r)$, for any $r$.
On the other hand, the composition
\[
p_{r'} \circ f \circ i_r: V(r) \overset{i_r}{\to}V \overset{f}{\to} V
\overset{p_{r'}}{\to} V(r')
\]
is naturally a homomorphism of representations
in $\mrm{Hom}_Q ((V(r),x),(V(r'),x))$,
where $p_{r'}$ and $i_r$ are natural projection and inclusion, respectively.
Note that
\[
\text{$\mrm{Hom}_Q((V(r),x), (V(r'),x))=0$, if $r\neq r'$.}
\]
We have $p_{r'} \circ f \circ i_r=0$, for $r\neq r'$.
So, $f(V(r))\subseteq V(r)$.
But by definition, $f(V(r))\subseteq W(r)$.
Thus, $V(r)=W(r)$.
Therefore the decomposition is unique up to order.
From (2), we can define an injective map
$\alpha:\tilde{X}(0)\to \tilde{\mathcal F}_{\boldsymbol{\delta}}$
by $\{x,(R_1,\cdots,R_q)\} \mapsto (x,\mbf f)$,
where $\mbf f$ is the flag
($V=V^1\supseteq V^2 \supseteq \cdots \supseteq V^{q+1}=0$)
such that $V^r=\oplus_{k=r}^qV(k)$ and $(V(k), x)\simeq R_k$, for $r=1,\cdots,q$.
We then have the following commutative diagram:
\[
\begin{CD}
\tilde{X}(0) @>\pi_1>> X(0)\\
@V\alpha VV @VVV\\
\tilde{\mathcal F}_{\boldsymbol{\delta}} @>\boldsymbol{\pi_{\delta}}>> E_{V,\Omega}\;.
\end{CD}
\]
Note that this diagram is Cartesian. So the restriction of
$\boldsymbol{\pi_{\delta}}_!(\bar{\mathbb Q}_l)$ to $X(0)$ is
$(\pi_1)_!(\bar{\mathbb Q}_l)$.
Recall that $\mathscr L_{\chi}$ is a direct summand of $(\pi_1)_!(\bar{\mathbb Q}_l)$
and $X(0)$ is open in $E_{V,\Omega}$.
So
\[
\tag{3}
\text{
$\mrm{IC}(X(0),\mathscr L_{\chi})$ is a direct summand of
$(\boldsymbol{\pi_{\delta}})_!(\bar{\mathbb{Q}}_l)$, up to shift.}
\]
By Lemma ~\ref{induction} (3),
\[
\tag{4}
(\boldsymbol{\pi_{\delta}})_!(\bar{\mathbb{Q}}_l)
=(\boldsymbol{\pi}_{\delta})_!(\bar{\mathbb Q}_l)\star \cdots
\star (\boldsymbol{\pi}_{\delta})_!(\bar{\mathbb Q}_l),\]
where $\delta$ is regarded as a sequence with only one entry.
Observe that $(\boldsymbol{\pi}_{\delta})_!(\bar{\mathbb Q}_l)=\bar{\mathbb Q}_l$.
By the proof of Lemma ~\ref{Lemma-1}, they are all in $\mathcal B$ (up to shifts).
From (3) and (4), $\mrm{IC}(X(0),\mathscr L_{\chi}) \in \mathcal B_V$.
Lemma ~\ref{Lemma-2} is proved.
\end{proof}
Finally,
let $T$ be a tube of period $p \neq 1$.
Let $\mathcal O_{V,T}^a$ be the set of all aperiodic
$\mrm G_V$-orbits $O_x$ in $E_{V,\Omega}$ (see ~\ref{noncyclicquivers}).
Given any $O\in \mathcal O_{V,T}^a$,
denote by $\mrm{IC}(O,\bar{\mathbb Q}_l)$ the intersection complex on $E_{V,\Omega}$
determined by $O$ and the constant local system $\bar{\mathbb{Q}}_l$ on $O$.
Then, we have
\begin{lem}
(\cite[6.9 (a)]{lusztig2})
\label{tube}
$\mrm{IC}(O,\bar{\mathbb Q}_l) \in \mathcal B_V$, for any $O \in \mathcal O_{V,T}^a$.
\end{lem}
\begin{proof}
Fix a regular simple $R$ in $T$, following the construction in Section ~\ref{sec:2.7},
we have an categorical equivalence $F:\mrm{Rep}(\mrm C_p)\to \mrm{HT}$,
where $\mrm{HT}$ is the full subcategory of $\mrm{Rep}(Q)$ generated by $T$ and
all the homogeneous regular simples such that $HT$ is closed under
extensions in $\mrm{Rep}(Q)$ and taking kernel and cokernels of
morphisms in $HT$.
Given an element $x$ in $O$, there exists a representation
$(\mathbb V,\theta)\in \mrm{Rep}(\mrm C_p)$ such that $F(\mathbb V,\theta)\simeq (V,x)$.
In particular, $F(\mathbb V)\simeq V$ as $K$-vector spaces.
Consequently, $E_{F(\mathbb V),\Omega}\simeq E_{V,\Omega}$.
So we can identify $F(\mathbb V)$ with $V$ and identify
the $G_{F(\mathbb V)}$-orbit of $F(\theta)$ in $E_{F(\mathbb V),\Omega}$ with $O$
in $E_{V,\Omega}$.
Since $F$ is equivalent and $O$ is aperiodic,
we have $O_{\theta}$ is aperiodic in $\mrm{Rep}(\mrm C_p)$.
By Theorem ~\ref{cycliccase}, we have
\[
\tag{1}
\text{
$\mrm{IC}(O_{\theta},\bar{\mathbb Q}_l)\in \mathcal B_{\mathbb V}$.
}
\]
In other words,
\[
\tag{2}
\text{
$\mrm{IC}(O_{\theta},\bar{\mathbb Q}_l)$ is a direct summand of
$(\pi_{\mbf z})_!(\bar{\mathbb Q}_l)$, (up to shift)
}
\]
for some $\mbf z=(z_1,\cdots,z_n)$, $z_s\in \mathbb Z/p\mathbb Z$.
Define $\mathcal{F}_{\boldsymbol{\nu}}'$ to be the variety consisting of
all flags of the form
\[F(f)=(F(\mathbb V)\supseteq F(\mathbb V^1)\supseteq
\cdots\supseteq F(\mathbb V^n)),\]
where
$f=(\mathbb V\supseteq \mathbb V^1\supseteq \cdots \supseteq \mathbb V^n)$ is a flag of type
$\mbf z$.
Define $\tilde{\mathcal F}_{\boldsymbol{\nu}}'$ to be the variety consisting of
all pair $(x,\mbf f)$, where $x\in E_1$ and $\mbf f \in \mathcal F_{\boldsymbol{\nu}}'$,
such that $\mbf f$ is $x$-stable.
Define $\tilde{\mathcal F}''_{\boldsymbol{\nu}}$ to be the variety consisting of
all pairs $(x,\mbf f)$, where $x\in E_2$ and $\mbf f\in \mathcal F_{\boldsymbol{\nu}}$
such that $\mbf f$ is $x$-stable. Then we have
$\tilde{\mathcal F}''_{\boldsymbol{\nu}}=G_V\times^{H_V}
\tilde{\mathcal F}_{\boldsymbol{\nu}}$
Consider the following commutative diagram
\[
\begin{CD}
\tilde{\mathcal F}_{\mbf z}
@>\tilde F>>
\tilde{\mathcal F}'_{\boldsymbol{\nu}}
@>\tilde i_1>>
\tilde{\mathcal F}''_{\boldsymbol{\nu}}
@>\tilde{i}>>
\tilde{\mathcal F}_{\boldsymbol{\nu}}\\
@V\boldsymbol{\pi}_{\mbf z}VV @V\boldsymbol{\pi_{\nu}}'VV
@V \boldsymbol{\pi_{\nu}}''VV @V \boldsymbol{\pi_{\nu}}VV\\
E_{\mathbb V, \Omega_p} @>F>> E_1 @>i_1>> E_2 @>i>> E_{V,\Omega},
\end{CD}
\]
where the vertical maps are first projections, $i, i_1, \tilde i$ and
$\tilde i_1$ are inclusions, and
$\tilde F: (\theta, f) \mapsto (F(\theta), F(\mbf f))$.
Note that all squares are Cartesians.
Let $O'$ be the $H_V$-orbit of $F(\theta)$ in $E_1$.
Denote by $\mrm{IC}(O',\bar{\mathbb Q}_l)$ the intersection complex on
$E_1$ determined by $O'$ and the trivial local system $\bar{\mathbb Q}_l$.
From Section ~\ref{sec:2.7} (e),
the statement (2) and the Cartesian square on the left in the diagram above, we
have
\[
\tag{3}
\text{
$\mrm{IC}(O',\bar{\mathbb Q}_l)$ is a direct summand of
$(\pi_{\boldsymbol{\nu}}')_!(\bar{\mathbb Q}_l)[d']$, for some $d'$.
}
\]
Let $O''$ be the $G_V$-orbit of $F(\theta)$ in $E_2$.
(In fact, $O''=O$.)
Denote by $\mrm{IC}(O'', \bar{\mathbb Q}_l)$ be the intersection
complex on $E_2$ determined by $O''$ and $\bar{\mathbb Q}_l$.
By Section ~\ref{sec:2.7} (i) and ~\cite[Theorem 2.6.3]{BL}, the derived functor
\[i_1^*: \mathcal D_{G_V}(E_2) \to \mathcal D_{H_V}(E_1)\]
is a categorical equivalence. In particular,
\[
i_1^*(\mrm{IC}(O'', \bar{\mathbb Q}_l))=\mrm{IC}(O',\bar{\mathbb Q}_l).
\tag{4}
\]
Since the middle square in the above diagram is Cartesian and
$\boldsymbol{\pi_{\nu}}''$ is proper, by the base change Theorem for
proper morphism (\cite[Theorem 6.2.5]{BBD}), We have
\[\tag{5} i_1^*(\pi''_{\boldsymbol{\nu}})_!(\bar{\mathbb Q}_l)=
(\pi_{\boldsymbol{\nu}}')_!(\bar{\mathbb Q}_l).\]
Thus by (3), (4) and (5), we have
\[
\tag{6}
\text{
$\mrm{IC}(O'',\bar{\mathbb Q}_l)$ is a direct summand of
$(\pi''_{\boldsymbol{\nu}})_!(\bar{\mathbb Q}_l)[d'']$, for some $d''$.
}
\]
The right square is Cartesian, so we have
\[
i^*(\pi_{\boldsymbol{\nu}})_!(\bar{\mathbb Q}_l)
=(\pi''_{\boldsymbol{\nu}})_!(\bar{\mathbb Q}_l).
\tag{7}\]
Note that the closure of $E_2$ is
$\boldsymbol{\pi_{\nu}}(\tilde{\mathcal F}_{\boldsymbol{\nu}})$
and $E_2$ is open in its closure. By (6) and (7), we have
\[
\text{
$\mrm{IC}(O,\bar{\mathbb Q}_l)$ is a direct summand of
$(\pi_{\boldsymbol{\nu}})_!(\bar{\mathbb Q}_l)[d]$, for some $d$.
}
\tag{8}
\]
But $(\pi_{\boldsymbol{\nu}})_!(\bar{\mathbb Q}_l)$ is a direct sum of
simple perverse sheaves from $\mathcal B$ with shifts. Therefore,
$\mrm{IC}(O,\bar{\mathbb Q}_l)\in \mathcal B_V$. Lemma ~\ref{tube} follows.
\end{proof}
\subsection{General Cases}
\label{general}
In this section, we study general cases.
Recall that $\mrm{Ind}(Q)$ is the set of representatives of pairwise
nonisomorphic indecomposable representations of $Q$.
Given any $\nu \in \mathbb N[I]$, denote by $\Delta_{\nu}$ the set
of all pairs $(\sigma,\lambda)$ where
$\sigma: \mrm{Ind}(Q) \to \mathbb N$ is a function and $\lambda$ is
the sequence $(0)$
or a sequence $(\lambda_1,\cdots,\lambda_n)$ of decreasing positive integers
satisfying the following properties:
\begin{itemize}
\item [(a)] $\prod_{m=0}^{r-1} \sigma((\Phi^+)^m(R))=0$
for any regular representation $R\in
\mrm{Ind}(Q)$ of period $r$;
\item[(b)] $\sum_{M\in \mrm{Ind}(Q)}\sigma(M)|M| +\sum_{m=1}^n \lambda_m \delta=
\nu$ if $\lambda=(\lambda_1,\cdots,\lambda_n)$;
\item[(c)] $\sum_{M\in \mrm{Ind}(Q)}\sigma(M)|M|=\nu$ if $\lambda=(0)$.
\end{itemize}
From (a), if $R$ is homogeneous, $\sigma(R)=0$.
From (b) and (c), the function $\sigma$ has finite support.
Given any $(\sigma,\lambda)\in \Delta_{\nu}$,
fix a $K$-vector space $V$ of dimension vector $\nu$.
Define the varieties $X(\sigma, \lambda)$ and $\tilde
X(\sigma,\lambda)$, the map $\pi_1$ and
the irreducible local system $\mathscr L_{\lambda}$ as follows.
If $\lambda=(\lambda_1,\cdots,\lambda_n)$, $X(\sigma,\lambda)$ is the
subvariety of $E_{V,\Omega}$ consisting of all elements $x$ such that
\[
(V,x)\simeq \oplus_{M\in \mrm{Ind}(Q)}
M^{\sigma(M)} \oplus R_1 \oplus \cdots \oplus R_q,\]
where $ M^{\sigma(M)}$ is the direct sum of $\sigma(M)$ copies of $M$,
$R_1,\cdots,$ and $R_q$ are pairwise nonisomorphic homogeneous regular simples.
The variety
$\tilde{X})(\sigma,\lambda)$ is the variety consisting of all pairs
\[
(x, (R_1,\cdots, R_q))\]
where $x\in X(\sigma, \lambda)$ and
$(R_1,\cdots, R_q)$ is a sequence of homogeneous regular simples in $\mrm{Ind}(Q)$
completely determined by $x$ up to order.
The map $\pi_1: \tilde{X}(\sigma, \lambda) \to X(\sigma, \lambda)$ is the first projection.
Note that the first projection
$\pi_1:\tilde{X}(\sigma, \lambda) \to X(\sigma,\lambda)$ is a
$S_q$-principal covering.
The sequence $\lambda$ determines an irreducible representation
$\chi(\lambda)$
of the symmetric group $S_q$.
Define $\mathscr L_{\lambda}$ to be
the direct summand of $(\pi_1)_{\star}(\bar{\mathbb Q}_l)$
corresponding to the irreducible representation of $S_q$ determined by the partition $\lambda$.
If $\lambda=(0)$, the variety $X(\sigma, \lambda)$ is the subvariety
of $E_{V,\Omega}$ consisting of all elements $s$ such that
$
(V,x)\simeq \oplus_{M\in \mrm{Ind}(Q)} M^{\sigma(M)}.
$
$\tilde{X}(\sigma,\lambda)$ is $X(\sigma,\lambda)$.
$\pi_1$ is the identity map $\tilde{X}(\sigma, \lambda) \to X(\sigma,
\lambda)$.
Denote by $\mathscr L_{\lambda}$ the trivial local system on $X(\sigma,\lambda)$.
Note that when $\lambda=(0)$, the variety $X(\sigma,\lambda)$ is a
$G_V$-orbit in $E_{V,\Omega}$.
Let
$\mrm{IC}(\sigma,\lambda)=\mrm{IC}(X(\sigma,\lambda),\mathscr L_{\lambda})$
be the simple perverse sheaf on $\mrm E_{V,\Omega}$
determined by $X(\sigma, \lambda)$ and $\mathscr L_{\lambda}$.
\begin{prop}
(\cite[Proposition 6.7]{lusztig2})
\label{generalcases}
$\mrm{IC}(\sigma,\lambda)\in \mathcal B_V$.
\end{prop}
The proof will be given in the next section.
By Proposition ~\ref{generalcases}, the assignment
$(\sigma,\lambda)\mapsto \mrm{IC}(\sigma,\lambda)$
defines a map $\Delta_{\nu} \to \mathcal B_V$.
This map is injective due to the fact that different pairs
$(\sigma,\lambda)$ determine different perverse sheaves.
Moreover, the cardinalities
$|\Delta_{\nu}|\overset{(1)}{=}|\mrm{Irr}\;\Lambda_V|\overset{(2)}{=} |\mathcal P_V|$,
where $\mrm{Irr}\;\Lambda_V$ is the set of all irreducible component of the variety
$\Lambda_V$ constructed in \cite{lusztig1}. The equality (1) holds by
\cite[corollary 5.3]{Ringel2} and
the equality (2) holds by \cite[Theorem 4.16 (b)] {lusztig2}.
By definitions, the two sets are of finite order.
Therefore we have
\begin{thm}
(\cite[Theorem 6.16 (b)]{lusztig2})
The map $\Delta_{\nu} \to \mathcal B_V$ is bijective.
\end{thm}
\subsection{Proof of Proposition ~\ref{generalcases}}
We preserve the setting of Section ~\ref{general}.
Given any element $(\sigma,\lambda)\in \Delta_{\nu}$,
recall that the variety $X(\sigma, \lambda)$ contains all elements
$x\in E_{V,\Omega}$ such that
\[
(V,x)\simeq \oplus M^{\sigma(M)} \, \oplus\, (R_1 \oplus \cdots \oplus R_q),
\]
where $R_1,\cdots R_q$ are pairwise nonisomorphic homogeneous regular
simple representations in $\mrm{Ind}(Q)$.
We can write the representation
$\oplus M^{\sigma(M)} \, \oplus\, (R_1 \oplus \cdots \oplus R_q)$ as
\begin{align*}
&O_1 \oplus \cdots \oplus O_n,
\end{align*}
such that
\begin{align*}
&\mrm{Ext}^1(O_m, O_{m'})=0\;\text{and}\; \mrm{Hom}(O_{m'}, O_m)=0,\;\text{
if}\; m < m',
\tag{a}
\end{align*}
and $O_m$ has one of the following forms:
\begin{align*}
&O_m=M^{\sigma(M)}\quad \text{ where}\; M \in \text{Ind}(Q) \;
\text{is preprojective or preinjective} ;\tag{1}\\
&O_m=\oplus_{M\in T} \, M^{\sigma(M)},\quad \text{ where }\; T\;
\text{is a tube};\tag{2}\\
&O_m=R_1\oplus \cdots \oplus R_q.\tag{3}
\end{align*}
The condition (a) can be accomplished by
putting the preprojective (resp. preinjective) $O_m$'s in case (1)
in the first (resp. last) part of the
sequence and putting the $O_m$'s in cases (2) and (3) in the middle part of the sequence,
then adjusting the $O_m$'s in case (1) such that they satisfy the
condition (a). (This can be done due to Lemma ~\ref{vanishing}.)
Note that any order of the $O_m$'s in case (2) and (3)
already satisfies the condition (a).
For each $m$, let $\nu_m=|O_m|$.
By the definition of $O_m$, this is well-defined.
Fix a $K$-vector space $V(m)$ such that $|V(m)|=\nu_m$.
By abuse of notations, denote by $O_m$ the subvariety in
$E_{V(m),\Omega}$ consisting of all elements $x$ such that
$(V,x)\simeq O_m$.
(Note that $O_m$ in case (3) is nothing but $X(0)$ in
$E_{V(m),\Omega}$ in Section ~\ref{noncyclic}.)
Define the irreducible local system $\mathscr L_m$ on $O_m$ by
\begin{itemize}
\item $\mathscr L_m=\bar{\mathbb Q}_l$ when $O_m$ is case (1) or (2);
\item $\mathscr L_m=\mathscr L_{\chi(\lambda)}$ (see Lemma ~\ref{Lemma-2}) when $O_m$ is case (3).
\end{itemize}
Denote by $\mrm{IC}(O_m,\mathscr L_m)$ the intersection complex
on $E_{V(m),\Omega}$ determined by $O_m$ and $\mathscr L_m$.
Then $\mrm{IC}(O_m,\mathscr L_m)$ is in $\mathcal P_{V(m)}$ by Lemma
~\ref{Lemma-1},~\ref{tube} and ~\ref{Lemma-2} for
$O_m$ in the case (1), (2) and (3), respectively.
So the semisimple complex
\[
\mrm{IC}(O_1,\mathscr L_1)\star\cdots\star\mrm{IC}(O_n,\mathscr L_n)
\]
on $E_{V, \Omega}$ is in $\mathcal Q_V$ (see Section ~\ref{inductionfunctor}).
To prove Proposition ~\ref{generalcases}, it suffices to show that
$\mrm{IC}(\sigma,\lambda)$ is a direct summand of the semisimple complex
$\mrm{IC}(O_1,\mathscr L_1)\star\cdots\star\mrm{IC}(O_n,\mathscr L_n)$
up to shift.
For simplicity, denote by $E_m$ the variety $E_{V(m),\Omega}$ for
$m=1,\cdots, n$. Let $\bar O_m$ the closure of $O_m$ in $E_m$ for
$m=1,\cdots,n$.
Recall from Section ~\ref{inductionfunctor},
we have the following diagram
\[
\begin{CD}
E_1\times \cdots \times E_n @<q_1<< F' @>q_2>> F'' @>q_3>> E_{V,\Omega}.
\end{CD}
\tag{**}
\]
By Lemma ~\ref{n-mult}, we have
\[
\mrm{IC}(O_1,\mathscr L_1)\star\cdots\star\mrm{IC}(O_n,\mathscr L_n)
=(q_3)_! (q_2)_{\flat} (q_1)^*(\mrm{IC}(O_1, \mathscr L_1)\boxtimes \cdots \boxtimes
\mrm{IC}(O_n,\mathscr L_n)).
\tag{4}
\]
Let $\tilde{\mathcal F}''$ be the subvariety of $F''$ consisting of
all elements $(x, V^{\bullet})$ such that the induced representations
$(V^{m-1}/V^m, x)$ is in $\bar{O}_m$
for $m=1,\cdots,n$.
Let $\tilde{\mathcal F}'$ be the subvariety of $F''$ consisting of
all elements $(x, V^{\bullet})$ such that the induced representations
$(V^{m-1}/V^m,x)$ is in $O_m$ for any $m=1,\cdots,n$.
Denote by $A''$ the subvariety of $F'$ consisting of all triples $(x,V^{\bullet},\mbf g)$
in $E'$ such that the induced representations of $x$ are in the $\bar{O}_m$'s.
Denote by $A'$ the subvariety of $F'$ consisting of all triples $(x, V^{\bullet},\mbf g)$
such that the induced representations of $x$ are in the $O_m$'s.
Consider the following commutative diagram:
\[
\begin{CD}
O_1\times\cdots\times O_n @<q_1''<< A' @>q_2''>> \tilde{\mathcal F}'\\
@Vj'VV @Vi'VV @ViVV\\
\bar{O}_1\times\cdots\times \bar{O}_n @<q_1'<< A'' @>q_2'>> \tilde{\mathcal F}''\\
@VjVV @V\rho 'VV @V\rho VV\\
E_1\times\cdots\times E_n @<q_1<< F' @>q_2>> F'' @>q_3>> E_{V,\Omega}
\end{CD}
\]
where the bottom row is the diagram (**),
$q_1'$ and $q_1''$ are the restrictions of $q_1$,
$q_2'$ and $q_2''$ are the restrictions of $q_2$, and
the vertical maps are natural embeddings.
From the definitions, the squares in the above diagram are Cartesian.
Since $q_3$ and $\pi:=q_3\rho$ are proper, $\rho$ is proper. Hence $\rho'$ is proper.
By the base change theorem for proper morphisms, we have
\[
q_1^* j_!=\rho_! (q_1')^* \quad \text{and} \quad q_2^* \rho_!=\rho'_!(q_2')^*.
\]
Note that
\[
\boxtimes_{m=1}^n \mrm{IC}(O_m,\mathscr L_m)=j_{!*} j'_{!*} (\boxtimes
\mathscr L_m)[\dim O_1\times\cdots \times O_n].
\]
Set $d=\dim O_1\times \cdots \times O_n$. Note that $j_{!*}=j_!$.
So
\begin{equation*}
\begin{split}
&(q_3)_! (q_2)_{\flat} (q_1)^* (\boxtimes_{m=1}^n\mrm{IC}(O_m,\mathscr L_m))
=(q_3)_! (q_2)_{\flat} (q_1)^* j_! j_{!*}' (\boxtimes_{m=1}^n
\mathscr L_m)[d]\\
&=(q_3)_! (q_2)_{\flat} \rho_! (q_1')^* (j_{!*}' (\boxtimes_{m=1}^n
\mathscr L_m)[d]
=(q_3)_! \rho_! (q_2')_{\flat} (q_1')^* j_{!*}'(\boxtimes_{m=1}^n
\mathscr L_m)[d]\\
&=\pi_! (q_2')_{\flat} (q_1')^* j_{!*}' (\boxtimes_{m=1}^n \mathscr L_m)[d].
\end{split}
\end{equation*}
Denote by $\mathscr L$ the complex
$(q_2')_{\flat} (q_1')^* j_{!*}' (\boxtimes_{m=1}^n \mathscr L_m)[d]$.
So (4) becomes
\[
(q_3)_! (q_2)_{\flat} (q_1)^* (\boxtimes_{m=1}^n\mrm{IC}(O_m,\mathscr
L_m))=\pi_!(\mathscr L).
\tag{5}
\]
Note that $j_{!*}(\boxtimes_{m=1}^n\mathscr L_m)[d]$ is a simple
perverse sheaf. Recall that $q_1'$ is smooth with connected fibres, by
\cite[Proposition 4.2.5]{BBD},
\[
(q_1')^* [d_1](j_{!*}(\boxtimes_{m=1}^n(\mathscr L_m))[d] \quad \text{ is
a simple perverse sheaf on $A''$.}
\]
Since $q_2'$ is a principal $G_{V(1)}\times \cdots \times G_{V(n)}$-bundle,
$\mathscr L$ is a simple perverse sheaf on $\tilde{\mathcal F}''$ up
to shift.
Note that $O_1\times \cdots \times O_n$ is a smooth variety, by
\cite[Lemma 4.3.2]{BBD},
\[
(j')^* j'_{!*}(\boxtimes_{m=1}^n\mathscr L_m)=\boxtimes_{m=1}^n
\mathscr L_m.
\]
Since the top square in the above diagram are Cartesian,
\[
i^*\mathscr L=(q_2'')_{\flat} (q_1'')^* (j')^*
j'_{!*}(\boxtimes_{m=1}^n\mathscr L_m)[d]=
(q_2'')_{\flat} (q_1'')^* (\boxtimes_{m=1}^n \mathscr L_m)[d].
\tag{6}
\]
We set
$X=X(\sigma, \lambda)-X(\sigma, \lambda)\cap
\pi(\tilde{\mathcal F}''-\tilde{\mathcal F}')$ and
$Y=\pi^{-1}(X)$. Then
\[
\text{$X$ is open dense in $X(\sigma,\lambda)$ and the restriction
$\pi^0:Y\to X$ is an isomorphism.}
\]
(See [L2, 6.12 (a) (b)] or [Li, 5.6], this is where the condition (a) is used.)
Thus we have the following diagram:
\[
\begin{CD}
O_1\times\cdots\times O_n @<q_1^0<< F^0 @>q_2^0>>
Y @>\pi^0>> X,
\end{CD}
\]
where $F^0=(q_2')^{-1}(Y)$ and $q_1^0$, $q_2^0$, $\pi^0$ are the natural restrictions of
$q_1'$, $q_2'$ and $\pi$, respectively.
By the definition of $\mathscr L_{\chi(\lambda)}$, we have
\[
(\pi^0 q_2^0)^*(\mathscr L_{\chi(\lambda)}|_{X})
=(q_1^0)^*(\boxtimes_m\mathscr L_m).
\]
Also by (6), $(q_1^0)^*(\boxtimes_m\mathscr L_m)=(q_2^0)^*\mathscr L|_Y$.
So $(\pi^0)^*(\mathscr L_{\chi(\lambda)}|_X)=\mathscr L|_Y$.
Since $\pi^0:Y\to X$ is an isomorphism,
\[
(\pi)_!(\mathscr L)[-d]|_X \simeq \mathscr L_{\chi(\lambda)}|_X.\]
Therefore,
the intersection complex $\mrm{IC}(X,\mathscr L_{\chi(\lambda)}|_X)$ is a direct summand
of $\pi_!(\mathscr L)[d]$.
Since $X$ is open dense in the closure of $X(\sigma,\lambda)$,
$\mrm{IC}(X,\mathscr L_{\chi(\lambda)}|_X)=\mrm{IC}(\sigma,\lambda)$.
Proposition ~\ref{generalcases} follows.
\section{Comments}
Note that we deal with the characterizations of the canonical bases in the symmetric cases.
It may be of interest to characterize the canonical bases in the
nonsymmetric cases.
From the proof of Proposition ~\ref{generalcases}, the set
\[
\{\mrm{IC}(O_1,\mathscr L_1)\star \cdots \star \mrm{IC}(O_n,\mathscr
L_n)\;|\; (\sigma, \lambda) \in \Delta_{\nu}, \nu\in \mathbb N[I]\}
\]
is a $\mathbb Q(v)$-basis of the algebra $(\mathcal K_Q, \star)$ (see ~\ref{algebra}).
Moreover by looking closer to the shifts, one can show that
\begin{cor}
The set
\[
\{
\mrm{IC}(O_1,\mathscr L_1)\circ \cdots \circ \mrm{IC}(O_n,\mathscr
L_n)\;|\; (\sigma,\lambda)\in \Delta_{\nu},\nu\in \mathbb N[I]
\}
\]
is an $\mathbb A$-basis of the algebra
$(\mathcal K,\circ)=\,_{\mathbb A}\mbf U^-$ (see ~\ref{algebra}) and
stable under bar involution. The
transition matrix between this basis and the canonical basis is upper triangular with
entries in the diagonal equal 1 and entries above the diagonal in
$\mathbb A$.
\end{cor}
\begin{proof}
For simplicity, we write $\mathbf C_{\sigma,\lambda}$ for the complex
\[
\mrm{IC}(O_1,\mathscr L_1)\circ \cdots \circ \mrm{IC}(O_n,\mathscr L_n).
\]
Since the bar involution commutes with the multplication $\circ$ (see \cite{lusztig3})
and the intesection cohomology complexes are self-dual,
the complex $\mathbf C_{\sigma, \lambda}$ is stable under the bar involution.
From the proof of Proposition ~\ref{generalcases}, we see that
\[
\mathbf C_{\sigma,\lambda}=\mrm{IC}(\sigma,\lambda)[d] \oplus P,
\]
for some $d$ and
$supp(P)\subseteq \overline{X(\sigma,\lambda)}- X(\sigma, \lambda)$.
But $\mathbf C_{\sigma,\lambda}$ is bar invariant, $d$ has to be zero. Corollary follows.
\end{proof}
The relationship between this
basis and the ``canonical basis'' defined in ~\cite{LXZ} deserves
further investigation.
| 213,935
|
Dr. Grandel Forever 39 Ampoules (1 amps)
For a youthful, firmed facial contour firming
active ingredient ampoule for a firm facial V line
For the customers of Forever 39, a youthful, firmed facial contour is essential. They attach extreme importance to lasting reinforcement and protection of the skinâ?¢s
INGREDIENT HIGHLIGHTS:
Macro-algae extract, Caffeine, Ruby powder
DR. GRANDEL Forever 39 Amp)
| 218,985
|
Thank you for visiting
Thank you for visiting the home of ITmeNOW, where we bring Commercial and Residential IT solutions to you. Feel free to submit a form and we'll contact you as soon as possible, use the online chat or reach us at (561) 406-9993What we offer our clients Contact us!
| 277,842
|
A Local Scooby-Doo Enthusiast Keeps Sending White Powder to Churches and Schools last week were apparently harmless, but the FBI is not laughing. Even if one of the letters referenced Scooby Doo..
Which, come to think of it, does ring a bell.!
| 393,998
|
TITLE: Confusion about definition of ergodicity in Markov Chains
QUESTION [2 upvotes]: On Wikipedia I found the following section on ergodicity
Ergodicity
A state i is said to be ergodic if it is aperiodic and positive
recurrent (*). In other words, a state i is ergodic if it is recurrent,
has a period of 1, and has finite mean recurrence time. If all states
in an irreducible Markov chain are ergodic, then the chain is said to
be ergodic.[dubious – discuss]
It can be shown that a finite state irreducible Markov chain is
ergodic if it has an aperiodic state. More generally, a Markov chain
is ergodic if there is a number N such that any state can be reached
from any other state in any number of steps less or equal to a number
N (**). In case of a fully connected transition matrix, where all
transitions have a non-zero probability, this condition is fulfilled
with N = 1.
A Markov chain with more than one state and just one out-going
transition per state is either not irreducible or not aperiodic, hence
cannot be ergodic.
I added the asterisks myself. What's confusing to me is that (**) seems much less strict than (*) because it doesn't mention periodicity. For example the transition matrix
$$P=\pmatrix{0&1&0\\0&0&1\\1&0&0}$$
seems to fullfil (**) because every state can reach any other state in 3 steps. Is (**) correct? Am I reading this wrong?
REPLY [2 votes]: Yes, the definition given there is confusing.
(**) is more general than (*).
Actually there are two definitions there. This reflects the fact that there are several inequivalent definitions of ergodicity for Markov chains in the literature.
Some authors ask for a single absorbing class, which is sligthly more general than (**), others say that a chain is ergodic iff it is both irreducible, positively recurrent and aperiodic, which is (*).
| 99,385
|
TITLE: Can find the angles of the triangle created by 3 points if I have each points compass bearing?
QUESTION [1 upvotes]: I am currently researching using magnetometers and radio field strength of 3 points for localisation. Is it possible to use the compass heading of 3 points to work out the angles of the triangle they create?
the diagram can be found at this link (unfortunately not enough rep to post images yet) http://i.stack.imgur.com/2HY7o.jpg
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
In the diagram, the solid black arrows are the bearings I will have with the angle relevant to north. I want to find the angles between each of the blue lines. I will also have a rough estimate of the length of the blue lines. I fear though that as the angles give no information about the other robots, there isn't much hope of finding the triangle's angles.
REPLY [0 votes]: If you have the compass heading from each point to each of the other two, you have measured the angles-just take the difference of the headings. Even if it is a large triangle so the compass deviation varies between the points and the angles don't add to $180^\circ$, the measurement of the two angles is local and applies.
REPLY [0 votes]: If you have all three edge lengths of a triangle, all three angles can be found using the Law of Cosines. If the triangle is large enough for the curvature of the Earth to make any difference, there are two versions of a spherical Law of Cosines. Spherical geometry and trigonometry are well documented, as, for many centuries, they have formed the basis for celestial navigation.
EDIT: if I were doing this, with approximate edge lengths, I would check for errors using both sum of the angles being 180 degrees and the Law of Sines.
| 71,046
|
TITLE: If $f$ is monotone increasing and $f$ is differentiable at $x_{0}$, then $f'(x_{0}) \geq 0$.
QUESTION [2 upvotes]: Let $X$ be a subset of $\textbf{R}$, let $x_{0}\in X$ be a limit point of $X$, and let $f:X\rightarrow\textbf{R}$ be a function. If $f$ is monotone increasing and $f$ is differentiable at $x_{0}$, then $f'(x_{0}) \geq 0$. If $f$ is monotone decreasing and $f$ is differentiable at $x_{0}$, then $f'(x_{0})\leq 0$.
MY ATTEMPT
Lemma
Let $X\subseteq\textbf{R}$, $f:X\rightarrow\textbf{R}$, $g:X\rightarrow\textbf{R}$, $x_{0}\in X$ is an adherent point, $f(x) \leq g(x)$ for every $x\in X$ and $\displaystyle\lim_{x\rightarrow x_{0}}f(x) = L$ and $\displaystyle\lim_{x\rightarrow x_{0}}g(x) = M$. Then we have that $L \leq M$.
Proof
According to the definition of limit, for every $\varepsilon > 0$, there are $\delta_{1} > 0$ and $\delta_{2} > 0$ such that
\begin{align*}
\begin{cases}
0 < |x - x_{0}| < \delta_{1}\\\\
0 < |x - x_{0}| < \delta_{2}
\end{cases} \Longrightarrow
\begin{cases}
|f(x) - L| < \varepsilon\\\\
|g(x) - M| < \varepsilon
\end{cases} \Longrightarrow L - \varepsilon < f(x) \leq g(x) < M + \varepsilon
\end{align*}
Let us assume that $L > M$. In this case, we can choose $\displaystyle\varepsilon = \frac{L - M}{3}$, whence we get that
\begin{align*}
M - L + 2\varepsilon > M - L + \frac{2(L - M)}{3} = \frac{M - L}{3} > 0 \Longrightarrow M > L
\end{align*}
which leads to a contradiction. Therefore the original claim is true and $L \leq M$.
Solution
Assuming that $f$ is monotone increasing at $x_{0}$, we have that
\begin{align*}
\frac{f(x) - f(x_{0})}{x - x_{0}} \geq 0
\end{align*}
Taking the limit from both sides to $x_{0}$ we conclude that
\begin{align*}
\lim_{x\rightarrow x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}} = f'(x_{0}) \geq 0 = \lim_{x\rightarrow x_{0}}0
\end{align*}
simliar reasoning applies to the monotone decreasing case, and we are done.
Could someone please verify if I am arguing correctly? Any other solution is welcome.
REPLY [1 votes]: Why did you even write the lemme? It looks like your solution is self-contained and correct.
Perhabs this helps you : $$(D_{x_0}f)(x):=\frac{f(x)-f(x_0)}{x-x_0}$$ is continuous everywhere, by the definition of the $f$. Then by your argument, ($D_{x_0}f\geq 0$ everywhere) $$\lim_{x\rightarrow x_0}(D_{x_0}f)(x)\geq 0.$$
| 126,425
|
The proposed OHV park in Forest County Wisconsin has been in the local paper lately with some opposition. Support from the OHV community is needed to help stop misconceptions of off roading, its not just MUDDING! In fact I would say a large majority of us keep away from it.
Please take a moment and write an email or letter or perhaps make a call.
Below is the Forest County website.
Here are the last 2 copies of the Pioneer Express;
| 308,420
|
Earthquake damage in Bhutan
Excellent post Mark. I'm an engineer in the UK working with a Bhutanese consultancy on the design of buildings in Bhutan. My particular interest is in rammed earth buildings, but we are looking generally at the behaviour of the buildings in Bhutan, and making them more earthquake resistant. The images you show are very interesting, do you have any more, would it be possible to view these images at full size? That way is is easier to understand how the buildings behave in an earthquake and thus design more seismically resilient ones. Many thanks Paul
| 280,583
|
\begin{document}
\title{Some restrictions on weight enumerators of
singly even self-dual codes II}
\author{
Masaaki Harada\thanks{
Corresponding author.
Research Center for Pure and Applied Mathematics,
Graduate School of Information Sciences,
Tohoku University, Sendai 980--8579, Japan.}
and
Akihiro Munemasa\thanks{
Research Center for Pure and Applied Mathematics,
Graduate School of Information Sciences,
Tohoku University,
Sendai 980--8579, Japan.}
}
\maketitle
\begin{abstract}
In this note,
we give some restrictions on the number of vectors
of weight $d/2+1$ in the shadow of a singly even
self-dual $[n,n/2,d]$ code.
This eliminates some of the possible weight enumerators
of singly even self-dual $[n,n/2,d]$ codes
for $(n,d)=(62,12)$, $(72,14)$, $(82,16)$, $(90,16)$ and $(100,18)$.
\end{abstract}
\smallskip
\noindent
{\bf Keywords:} self-dual code, weight enumerator, shadow
\smallskip
\noindent
{\bf Mathematics Subject Classification:} 94B05
\section{Introduction}
Let $C$ be a singly even self-dual code and
let $C_0$ denote the
subcode of codewords having weight $\equiv0\pmod4$.
Then $C_0$ is a subcode of codimension $1$.
The {\em shadow} $S$ of $C$ is defined to be $C_0^\perp \setminus C$.
Shadows for self-dual codes were introduced by Conway and
Sloane~\cite{C-S}
in order to
derive new upper bounds for the minimum weight of
singly even self-dual codes, and to
provide restrictions on the weight enumerators of
singly even self-dual codes.
The largest possible minimum weights of
singly even self-dual codes of lengths $n \le 72$ were
given in~\cite[Table~I]{C-S}.
The work was extended to lengths $74 \le n \le 100$ in~\cite[Table~VI]{DGH}.
We denote by $d(n)$ the largest possible minimum weight given
in~\cite[Table~I]{C-S} and \cite[Table~VI]{DGH}
throughout this note.
The possible weight enumerators of singly
even self-dual codes having minimum weight $d(n)$
were also given in~\cite{C-S}
for lengths $n \le 64$ and $n=72$ (see
also~\cite{DGH} for length $72$), and the work was
extended to lengths up to $100$ in~\cite{DGH}.
It is a fundamental problem to find which weight enumerators
actually occur among the possible weight enumerators
(see~\cite{C-S} and \cite{Huffman}).
Some restrictions on the number of vectors
of weight $d/2$ in the shadow of a singly even
self-dual $[n,n/2,d]$ code were given in~\cite{HM}.
Also, some restrictions on the number of vectors
of weight $d/2+1$ in the shadow of a singly even
self-dual $[n,n/2,d]$ code were given in~\cite{BHM}
for $n \equiv 0 \pmod 4$.
In this note,
we improve the result in~\cite{BHM} about
the restriction on the number of vectors
of weight $d/2+1$ in the shadow of a singly even
self-dual $[n,n/2,d]$ code for $n \equiv 0 \pmod 4$.
We also give a restriction on the number of vectors
of weight $d/2+1$ in the shadow of a singly even
self-dual $[n,n/2,d]$ code for $n \equiv 2 \pmod 4$.
These restrictions
eliminate some of the possible weight enumerators determined
in~\cite{C-S} and \cite{DGH}
for the parameters
$(n,d)=(62,12)$, $(72,14)$, $(82,16)$, $(90,16)$ and $(100,18)$
\section{Preliminaries}
A (binary) $[n,k]$ {\em code} $C$ is a $k$-dimensional vector subspace
of $\FF_2^n$,
where $\FF_2$ denotes the finite field of order $2$.
All codes in this note are binary.
The parameter $n$ is called the {\em length} of $C$.
The {\em weight} $\wt(x)$ of a vector $x \in \FF_2^n$ is
the number of non-zero components of $x$.
A vector of $C$ is a {\em codeword} of $C$.
The minimum non-zero weight of all codewords in $C$ is called
the {\em minimum weight} $d(C)$ of $C$ and an $[n,k]$ code with minimum
weight $d$ is called an $[n,k,d]$ code.
The \textit{dual code} $C^{\perp}$ of a code
$C$ of length $n$ is defined as
$
C^{\perp}=
\{x \in \FF_2^n \mid x \cdot y = 0 \text{ for all } y \in C\},
$
where $x \cdot y$ is the standard inner product.
A code $C$ is called \textit{self-dual} if $C = C^{\perp}$.
A self-dual code $C$ is {\em doubly even} if all
codewords of $C$ have weight divisible by four, and
{\em singly even} if there exists at least one codeword of
weight $\equiv 2 \pmod 4$.
Rains~\cite{Rains} showed that
the minimum weight $d$ of a self-dual code $C$ of length $n$
is bounded by
$d \le 4 \lfloor{\frac {n}{24}} \rfloor + 6$
if $n \equiv 22 \pmod {24}$,
$d \le 4 \lfloor{\frac {n}{24}} \rfloor + 4$
otherwise.
In addition, if $n \equiv 0 \pmod{24}$ and $C$ is singly even,
then $d \le 4 \lfloor{\frac {n}{24}} \rfloor + 2$.
A self-dual code meeting the bound is called {\em extremal}.
Let $A_i$ and $B_i$ be the numbers of vectors of
weight $i$ in $C$ and $S$, respectively.
The weight enumerators of $C$ and $S$ are given by
$\sum_{i=0}^n A_i y^i$
and
$\sum_{i=d(S)}^{n-d(S)} B_i y^i$, respectively,
where $d(S)$ denotes the minimum weight of $S$.
Let $C$ be a singly even
self-dual code of length $n$ and let $S$ be the
shadow of $C$.
Let $C_0$ denote the subcode of codewords having weight
$\equiv0\pmod4$.
There are cosets $C_1,C_2,C_3$ of $C_0$ such that
$C_0^\perp = C_0 \cup C_1 \cup C_2 \cup C_3 $, where
$C = C_0 \cup C_2$ and $S = C_1 \cup C_3$.
\begin{lem}[Conway and Sloane~\cite{C-S}]\label{lem:C-S}
Let $x_1,y_1$ be vectors of $C_1$ and
let $x_3$ be a vector of $C_3$.
Then
$x_1+y_1 \in C_0$,
$x_1+x_3 \in C_2$ and
$\wt(x_1) \equiv \wt(x_3) \equiv \frac{n}{2} \pmod 4$.
\end{lem}
\begin{lem}[Brualdi and Pless~\cite{BP}]\label{lem:BP}
Let $x_1,y_1$ be vectors of $C_1$ and
let $x_3$ be a vector of $C_3$.
\begin{enumerate}
\renewcommand{\labelenumi}{\rm \arabic{enumi})}
\item Suppose that $n \equiv 0 \pmod 4$.
Then $x_1 \cdot y_1=0$ and $x_1 \cdot x_3=1$.
\item Suppose that $n \equiv 2 \pmod 4$.
Then $x_1 \cdot y_1=1$ and $x_1 \cdot x_3=0$.
\end{enumerate}
\end{lem}
\section{$n \equiv 2 \pmod 4$ and $d(S)=\frac{d(C)}{2}+1$}
\label{sec:2mod4}
Recall that the Johnson graph $J(v,d)$ has the collection $X$ of all
$d$-subsets of $\{1,2,\dots,v\}$ as vertices, and two distinct
vertices are adjacent whenever they share $d-1$ elements in common.
Assume $v\geq2d$ and set
\[R_i=\{(x,y)\in X\times X\mid |x\cap y|=d-i\}.\]
Then $\{R_i\}_{i=0}^d$ is a partition of $X\times X$.
The following lemma is known as Delsarte's inequalities since
it is the basis of Delsarte's linear programming bound.
We refer the reader to \cite{D} for an explicit formula
for the second eigenmatrix $Q$ appearing in the lemma.
\begin{lem}[{\cite[Proposition~2.5.2]{bcn}}]\label{lem:D}
Let $Y$ be a subset of vertices of $J(v,d)$, and set
\[a_i=\frac{1}{|Y|}|(Y\times Y)\cap R_i|\quad(0\leq i\leq d).\]
If we denote by $Q=(q^{(v)}_j(i))$ the second eigenmatrix
of $J(v,d)$, then every entry of the vector
$(a_0,\dots,a_d)Q$ is nonnegative.
\end{lem}
Suppose that $Y$ is a subset of vertices of $J(v,d)$
such that two distinct members intersect at exactly
one element. Then by Lemma~\ref{lem:D}, every entry of the vector
\[
(1,0,\ldots,0,0,|Y|-1,0)Q
\]
is nonnegative, i.e.,
\[
q^{(v)}_j(0)+(|Y|-1)q_j^{(v)}(d-1)\geq0
\quad(1\leq j\leq d).
\]
Thus, we obtain
\begin{equation}\label{eq:bound1}
|Y|\leq M_{v,d},
\end{equation}
where
\[
M_{v,d}=\min\{ 1-\frac{q^{(v)}_j(0)}{q^{(v)}_j(d-1)}
\mid 1\leq j\leq d\text{ and }q_j^{(v)}(d-1)<0\}.
\]
If we define
\[
M_{v,d}=\begin{cases}
2&\text{if $v=2d-1$,}\\
1&\text{if $d\leq v\leq 2d-2$,}\\
0&\text{if $0\leq v\leq d-1$,}
\end{cases}
\]
then~\eqref{eq:bound1} also holds for all $v,d$.
Now, let $C$ be a singly even self-dual code of length $n$
and let $S$ be the shadow of $C$.
For the remainder of this section, we assume that
\begin{equation}\label{eq:2mod4}
n \equiv 2 \pmod 4 \text{ and } d(S)=\frac{d(C)}{2}+1.
\end{equation}
By Lemma~\ref{lem:C-S},
$d(C) \equiv n-2 \pmod 8$, and hence $d(S)$ is odd.
For each of $i=1,3$, let $Y_i$ be
the set of supports of vectors of weight $d(S)$ in
$C_i$, and let $S_i$ be the union of the members
of $Y_i$.
From Lemma~\ref{lem:BP} and~\eqref{eq:2mod4}, we have the following:
\begin{equation}\label{eq:s}
| x \cap y| =
\begin{cases}
1 & \text{ if } x,y \in Y_i,\; x\ne y, \\
0 & \text{ if } x \in Y_1,\; y \in Y_3.
\end{cases}
\end{equation}
Then by \eqref{eq:bound1}, we have
\[
|Y_i|\leq M_{|S_i|,d(S)}.
\]
It follows from~\eqref{eq:s} that $S_1\cap S_3=\emptyset$.
Thus, we have
\begin{equation}\label{bd}
B_{d(S)}=|Y_1|+|Y_3|\leq
\max\{ M_{v,d(S)}+M_{n-v,d(S)} \mid 0\leq v \leq n/2 \}.
\end{equation}
For $42 \le n \le 98$ and $d(C) =d(n)$,
the parameters $(n,d(C),d(S))$ satisfying Condition~\eqref{eq:2mod4}
are listed in Table~\ref{Tab:2mod4},
where the values $d(n)$ are also listed in the table.
For some lengths $n$, the existence of a singly even self-dual
code of length $n$ and minimum weight $d(n)$
is currently not known. In this case, we consider
the case $d(C)=d(n)-2$.
We calculated the upper bound~\eqref{bd},
where the results are listed in Table~\ref{Tab:2mod4}.
This calculation was done by
the program written in {\sc Magma}~\cite{Magma},
where the program is listed in~\ref{appendix}.
\begin{table}[thb]
\caption{Parameters satisfying~\eqref{eq:2mod4}}
\label{Tab:2mod4}
\begin{center}
{\small
\begin{tabular}{c|c|c|c|c}
\noalign{\hrule height0.8pt}
$n$ & $d(n)$ & $d(C)$&$d(S)$ & $B_{d(S)}$\\
\hline
42 & 8 & 8 & 5 &$\le 42$ \\
62 & 12 &12 & 7 &$\le 48$ \\
70 & 14 &12 & 7 &$\le 52$ \\
82 & 16 &16 & 9 &$\le 74$ \\
90 & 16 &16 & 9 &$\le 76$ \\
98 & 18 &16 & 9 &$\le 78$ \\
\noalign{\hrule height0.8pt}
\end{tabular}
}
\end{center}
\end{table}
We discuss the possible weight enumerators for the case
$d(n)=d(C)$ in Table~\ref{Tab:2mod4}.
The possible weight enumerators $W_{42}$ and $S_{42}$
of an extremal singly even self-dual $[42,21,8]$ code
with $d(S) \ge 5$
and its shadow are as follows~\cite{C-S}:
\[
\begin{array}{lll}
W_{42} &=& 1 + (84+8\beta)y^8 + (1449-24\beta)y^{10} + \cdots,\\
S_{42} &=&
\beta y^5+ (896-8\beta)y^9 + (48384+28\beta)y^{13}+ \cdots, \\
\end{array}
\]
respectively,
where $\beta$ is an integer.
It was shown in~\cite{BHM42} that $0 \le \beta \le 42$.
Table~\ref{Tab:2mod4} gives an alternative proof.
The possible weight enumerators $W_{62}$ and $S_{62}$
of an extremal singly even self-dual $[62,31,12]$ code
with $d(S)\ge 7$
and its shadow are as follows~\cite{C-S} (see also~\cite{DH}):
\[
\begin{array}{lll}
W_{62} &=& 1 + (1860+32\beta)y^{12} + (28055-160\beta)y^{14}+ \cdots, \\
S_{62} &=& \beta y^7 + (1116-12\beta)y^{11}
+(171368 + 66 \beta) y^{15} + \cdots,
\end{array}
\]
respectively,
where $\beta$ is an integer with $0 \le \beta \le 93$.
Table~\ref{Tab:2mod4} gives the following:
\begin{prop}
If there exists an extremal singly even self-dual $[62,31,12]$ code
with weight enumerator $W_{62}$,
then $0 \le \beta \le 48$.
\end{prop}
It is known that there exists an extremal singly even self-dual
$[62,31,12]$ code with weight enumerator $W_{62}$
for $\beta=0,2,9,10,15,16$ (see~\cite{Y14}).
The possible weight enumerators $W_{82}$ and $S_{82}$
of an extremal singly even self-dual $[82,41,16]$ code
with $d(S)\ge 9$
and its shadow are as follows~\cite{DGH}:
\[
\begin{array}{lll}
W_{82} &=& 1 + (39524 + 128 \alpha) y^{16} + (556985 - 896 \alpha) y^{18}
+ \cdots, \\
S_{82} &=& \alpha y^9 + (1640 - \alpha) y^{13} +
(281424+ 120 \alpha) y^{17} + \cdots, \\
\end{array}
\]
respectively,
where $\alpha$ is an integer with
$0 \leq \alpha \leq \lfloor \frac{556985}{896} \rfloor=621$.
Table~\ref{Tab:2mod4} gives the following:
\begin{prop}
If there exists an extremal singly even self-dual $[82,41,16]$ code
with weight enumerator $W_{82}$,
then $0 \le \alpha \le 74$.
\end{prop}
It is unknown whether there exists an extremal singly even self-dual
code for any of these cases.
The possible weight enumerators $W_{90}$ and $S_{90}$
of an extremal singly even self-dual $[90,45,16]$ code
with $d(S)\ge 9$
and its shadow are as follows~\cite{DGH}:
\[
\begin{array}{lll}
W_{90} &=& 1 + (9180 + 8 \beta)y^{16} + (-512 \alpha - 24 \beta +
224360)y^{18} + \cdots,\\
S_{90} &=& \alpha y^9 + (\beta -18 \alpha) y^{13} +(112320 + 153 \alpha -16
\beta)y^{17} + \cdots,
\end{array}
\]
respectively,
where $\alpha$ and $\beta$ are integers with
$0 \le \alpha \le \frac{1}{18} \beta \le
\lfloor \frac{224360}{24} \rfloor =9348$.
Table~\ref{Tab:2mod4} gives the following:
\begin{prop}
If there exists an extremal singly even self-dual $[90,45,16]$ code
with weight enumerator $W_{90}$,
then $0 \le \alpha \le 76$.
\end{prop}
It is unknown whether there exists an extremal singly even self-dual
code for any of these cases.
\section{$n \equiv 0 \pmod 4$ and $d(S)=\frac{d(C)}{2}+1$}
\label{sec:0mod4}
Let $C$ be a singly even self-dual code of length $n$
and let $S$ be the shadow of $C$.
In this section, we write $d(C)=d$ and $d(S)=s$ for short,
and assume that
\begin{equation}\label{eq:0mod4}
n \equiv 0 \pmod 4 \text{ and } s=\frac{d}{2}+1.
\end{equation}
By Lemma~\ref{lem:C-S},
$d \equiv n-2 \pmod 8$, and hence $s$ is even.
\begin{prop}[\cite{BHM}]\label{prop:BHM}
Suppose that $n \equiv 0 \pmod 4$ and $s=\frac{d}{2}+1$.
Let $B_{s}$ denote the number of vectors of weight $s$ in $S$.
\begin{enumerate}
\renewcommand{\labelenumi}{\rm (\roman{enumi})}
\item If $2n>(d+2)^2$, then
\[B_{s}\leq\frac{2n}{d+2}.\]
\item If $(d+2)^2\leq 4n\leq2(d+2)^2$, then
\[B_{s}\leq d+2,\quad B_{s}\neq d+1.\]
\item If $4n<(d+2)^2$, then
\[B_{s}\leq 2\frac{2n-d-2}{d}.\]
\end{enumerate}
\end{prop}
The above proposition was essentially established by
showing $B_{s}\leq\max\{l_1,l_2\}$, where
\begin{align*}
l_1&=\frac{2n}{d+2},\\
l_2&=\min\left\{d+2,2\frac{2n-d-2}{d}\right\}.
\end{align*}
We recall part of the proof of Proposition~\ref{prop:BHM}
for later use.
Denote the set of all vectors in $C_i$ of weight $s$ by
$\cB_i$ ($i = 1,3$). Denote by $v * w$ the entrywise product
of two vectors $v,w$.
If $v, w \in \cB_i$, then $\wt(v * w) = 0$ and
hence these vectors have disjoint supports. This implies
\begin{equation}\label{l1}
|\cB_i|\leq l_1\quad(i=1,3).
\end{equation}
If $v\in \cB_1$ and $w \in \cB_3$, then $\wt(v * w) = 1$.
Thus, if
$\cB_1$ and $\cB_3$ are both nonempty, then
\begin{equation}\label{l2}
|\cB_i|\leq s.
\end{equation}
Using the following lemmas,
we give an improvement of the upper bound by showing
$B_{s}\leq\max\{l'_1,l'_2\}$, where
\begin{align*}
l'_1&=
\begin{cases}
l_1&\text{if $n$ is divisible by $2s$,}\\
2\left\lceil\frac{n-d+2}{d+2}\right\rceil-1
&\text{otherwise,}
\end{cases}\\
l'_2&=\begin{cases}
d+2-\left\lceil\sqrt{(d+2)^2-4n}\,\right\rceil
&\text{if $4n<(d+2)^2$,}\\
\min\left\{d+2,
4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2\right\}
&\text{otherwise.}
\end{cases}
\end{align*}
Since
\begin{equation}\label{n2s}
\left\lceil\frac{n-d+2}{d+2}\right\rceil
=\left\lceil\frac{n/4-(s/2-1)}{s/2}\right\rceil
\leq
\frac{n}{2s},
\end{equation}
we have
\begin{equation}\label{l1'}
l'_1\leq l_1,
\end{equation}
and
\[4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2
\leq
\frac{2n}{s}-2
<2\frac{2n-d-2}{d}.\]
The latter implies $l'_2\leq l_2$ provided $4n\geq(d+2)^2$.
If $4n<(d+2)^2$, then
\begin{align*}
&2\frac{2n-d-2}{d}-\left(d+2-\sqrt{(d+2)^2-4n}\right)
\\&=
\frac{\sqrt{(d+2)^2-4n}}{d}(d-\sqrt{(d+2)^2-4n})
\\&\geq0.
\end{align*}
Thus $l'_2\leq l_2$ holds in this case as well.
Therefore, the bound $B_s\leq\max\{l'_1,l'_2\}$ which will be
shown in Proposition~\ref{prop:imp} below is an improvement
of the bound given in Proposition~\ref{prop:BHM}.
\begin{lem}\label{lem:1}
Let
\[k=\left\lceil\frac{n-d+2}{2s}\right\rceil.\]
If $n$ is not divisible by $2s$, then
$|\cB_i|\leq2k-1$ for $i=1,3$.
\end{lem}
\begin{proof}
Suppose, to the contrary, $|\cB_i| \ge 2k$.
Then the sum of the all-one vector and the
$2k$ vectors of weight $s$
belongs to $C_0$ and has weight
$n-2ks\leq d-2$.
This forces $n-2ks=0$, contradicting the assumption.
\end{proof}
\begin{lem}\label{lem:maxab}
Let $n$ and $s$ be positive integers with $n<s^2$. Then
\[\max\{a+b\mid a,b\in\ZZ,\;0\leq a,b\leq s,\;s(a+b)-ab\leq n\}
=2s-\left\lceil2\sqrt{s^2-n}\,\right\rceil.\]
\end{lem}
\begin{proof}
Since $n<s^2$, we have
\begin{align*}
&\max\{a+b\mid a,b\in\RR,\;0\leq a,b\leq s,\;s(a+b)-ab\leq n\}
\\&=\max\{a+b\mid 0\leq a,b\leq s,\;(s-a)b\leq n-sa\}
\\&=\max\{a+\min\{(n-sa)/(s-a),s\}\mid 0\leq a< s\}
\\&=\max\{(n-a^2)/(s-a)\mid 0\leq a< s\}.
\end{align*}
The function $f(x)=(n-x^2)/(s-x)$ defined on the interval $[0,s)$
has maximum $f(\alpha)=2\alpha$, where $\alpha=s-\sqrt{s^2-n}$.
Thus, we have
\begin{align*}
&\max\{a+b\mid a,b\in\ZZ,\;0\leq a,b\leq s,\;s(a+b)-ab\leq n\}
\\&\leq
\left\lfloor
\max\{a+b\mid a,b\in\RR,\;0\leq a,b\leq s,\;s(a+b)-ab\leq n\}
\right\rfloor
\\&=
\left\lfloor2\alpha\right\rfloor.
\end{align*}
Define $a,b\in\ZZ$ by
$a=\left\lfloor\alpha\right\rfloor$ and
\[b=\begin{cases}
\left\lfloor\alpha\right\rfloor&\text{if $\alpha-
\left\lfloor\alpha\right\rfloor<\frac12$,}\\
\left\lfloor\alpha\right\rfloor+1&\text{otherwise.}
\end{cases}\]
Then $a+b=\left\lfloor2\alpha\right\rfloor=
2s-\left\lceil2\sqrt{s^2-n}\,\right\rceil$. Since
$\alpha<s$, we have $b\leq s$. It remains to show
$s(a+b)-ab\leq n$, or equivalently,
\begin{equation}\label{eq:sab}
ab-s(a+b)+n\geq0.
\end{equation}
Observe
\[s-\left\lfloor\alpha\right\rfloor=\left\lceil\sqrt{s^2-n}\right\rceil.\]
If $\alpha-\left\lfloor\alpha\right\rfloor<\frac12$, then
\begin{align*}
ab-s(a+b)+n&=
\left\lfloor\alpha\right\rfloor^2-2s\left\lfloor\alpha\right\rfloor+n
\\&=
(s-\left\lfloor\alpha\right\rfloor)^2-(s^2-n)
\\&=
\left\lceil\sqrt{s^2-n}\right\rceil^2-(s^2-n)
\\&\geq0.
\end{align*}
Thus, \eqref{eq:sab} holds.
If $\alpha-\left\lfloor\alpha\right\rfloor\geq\frac12$, then
\[s-\left\lfloor\alpha\right\rfloor\geq\sqrt{s^2-n}+\frac12.\]
Thus
\begin{align*}
ab-s(a+b)+n&=
\left\lfloor\alpha\right\rfloor
(\left\lfloor\alpha\right\rfloor+1)
-s(2\left\lfloor\alpha\right\rfloor+1)+n
\\&=
(\left\lfloor\alpha\right\rfloor-s)
(\left\lfloor\alpha\right\rfloor+1-s)-(s^2-n)
\\&\geq
\left(\sqrt{s^2-n}+\frac12\right)
\left(\sqrt{s^2-n}-\frac12\right)-(s^2-n)
\\&=
-\frac14.
\end{align*}
Since $ab-s(a+b)+n$ is an integer, \eqref{eq:sab} holds.
\end{proof}
\begin{prop}\label{prop:imp}
Suppose that $n \equiv 0 \pmod 4$ and $s=\frac{d}{2}+1$.
Let $B_{s}$ denote the number of vectors of weight $s$
in $S$. Then
\begin{equation}\label{bd12}
B_{s}\leq\max\{l'_1,l'_2\}.
\end{equation}
More precisely,
\begin{enumerate}
\renewcommand{\labelenumi}{\rm (\roman{enumi})}
\item If $2n>d^2+6d$, then
\[B_{s}\leq
\begin{cases}
\frac{2n}{d+2}&\text{if $n$ is divisible by $2s$,}\\
2\left\lceil\frac{n-d+2}{d+2}\right\rceil-1
&\text{otherwise.}
\end{cases}
\]
\item If $(d+2)^2<2n\leq d^2+6d$, then
\[B_{s}\leq
\begin{cases}
\frac{2n}{d+2}&\text{if $n$ is divisible by $2s$,}\\
d+2&\text{otherwise.}
\end{cases}\]
\item If $d^2+8d-4<4n\leq2(d+2)^2$, then
\[B_{s}\leq d+2,\quad B_{s}\neq d+1.\]
\item If $(d+2)^2\leq 4n\leq d^2+8d-4$, then
\[B_{s}\leq 4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2.\]
\item If $4n<(d+2)^2$, then
\[B_{s}\leq d+2-\left\lceil\sqrt{(d+2)^2-4n}\,\right\rceil.\]
\end{enumerate}
\end{prop}
\begin{proof}
If one of $\cB_1$ and $\cB_3$ is empty, then
\eqref{l1} and Lemma~\ref{lem:1} imply $B_{s}\leq l'_1$.
If
$\cB_1$ and $\cB_3$ are both nonempty, then by \eqref{l2},
we have $B_{s}\leq2s=d+2$.
Moreover, suppose $n<s^2$.
Observe
\[\left|\bigcup_{x\in\cB_1\cup\cB_3}\supp(x)\right|=
s(|\cB_1|+|\cB_3|)-|\cB_1||\cB_3|,\]
and this is at most $n$. By \eqref{l2}, we can apply
Lemma~\ref{lem:maxab} to conclude
\[B_{s}\leq2s-\left\lceil2\sqrt{s^2-n}\,\right\rceil.\]
Thus $B_{s}\leq l'_2$.
Therefore, \eqref{bd12} holds.
Next, we determine $\max\{l'_1,l'_2\}$.
If $2n>d^2+6d$, then
\[\frac{n-d+2}{d+2}>\frac12(d+2)\in\ZZ,\]
so
\begin{align*}
l'_1&\geq
2\left\lceil\frac{n-d+2}{d+2}\right\rceil-1
&&\text{(by \eqref{n2s})}
\\&\geq
2\left(\frac12(d+2)+1\right)-1
\\&=d+3
\\&\geq l'_2.
\end{align*}
Thus $\max\{l'_1,l'_2\}=l'_1$, and (i) holds.
Next suppose $(d+2)^2<2n\leq d^2+6d$.
Since
\begin{align*}
4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2-(d+2)
&\geq
4\frac{n-d+2}{d+2}-2-(d+2)
\\&>\frac{d^2-2d+8}{d+2}
\\&>0,
\end{align*}
we have $l'_2=d+2$.
Since
\[\frac{n-d+2}{d+2}\leq \frac12(d+2)\in\ZZ,\]
we have
\[2\left\lceil\frac{n-d+2}{d+2}\right\rceil-1<d+2<l_1.\]
These imply
\[\max\{l'_1,l'_2\}=\begin{cases}
l_1&\text{if $n$ is divisible by $2s$,}\\
l'_2&\text{otherwise,}
\end{cases}\]
and (ii) holds.
Next suppose $(d+2)^2\leq 4n\leq2(d+2)^2$.
We claim
\[l'_2=\begin{cases}
d+2&\text{if $4n\leq d^2+8d-4$,}\\
4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2
&\text{otherwise.}
\end{cases}\]
Indeed, since $(d+4)/4=(s+1)/2\notin\ZZ$, we have
\begin{align*}
d+2>4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2
&\iff
\frac{s}{2}\geq\left\lceil\frac{n-d+2}{d+2}\right\rceil
\\&\iff
\frac{s}{2}\geq\frac{n-d+2}{d+2}
\\&\iff
4n\leq d^2+8d-4.
\end{align*}
Since $4n\geq(d+2)^2$ and $d\neq4$, we have $n\geq 3d-2$. Thus
\[4\left\lceil\frac{n-d+2}{d+2}\right\rceil-2\geq\frac{2n}{d+2}.\]
This, together with $2n\leq(d+2)^2$ implies $l_1\leq l'_2$. Therefore,
$\max\{l'_1,l'_2\}=l'_2$. Now (iii) and (iv) hold
by Proposition~\ref{prop:BHM}~(ii).
Finally, suppose $4n<(d+2)^2$. Then it is easy to verify
\[\frac{2n}{d+2}\leq d+2-\sqrt{(d+2)^2-4n},\]
hence $\max\{l'_1,l'_2\}=l'_2$ by \eqref{l1'}. Thus (v) holds.
\end{proof}
\begin{rem}
In Proposition~\ref{prop:imp}~(v), it is sometimes possible to
draw a stronger conclusion
\[|\cB_i|=\frac12
\left(d+2-\left\lceil\sqrt{(d+2)^2-4n}\,\right\rceil\right)
\quad(i=1,3).\]
This is when a pair $\{a,b\}$ achieving the maximum in
Lemma~\ref{lem:maxab} is unique. For the parameters
$(n,d,s)=(128,22,12)$,
we necessarily have $|\cB_i|=8$ for $i=1,3$.
In general, a pair $\{a,b\}$ achieving the maximum in
Lemma~\ref{lem:maxab} is not unique. For example,
when $(n,d,s)=(120,22,12)$, both
$\{6,8\}$ and $\{7,7\}$ achieve the maximum.
\end{rem}
For only the parameters
$(n,d,s)=
(72, 14, 8)$ and
$(100,18,10)$,
Proposition~\ref{prop:imp} gives an improvement over
Proposition~\ref{prop:BHM},
for $44 \le n \le 100$ and $d=d(n)$.
The bounds on $B_{s}$ obtained by Proposition~\ref{prop:imp}
are listed in Table~\ref{Tab:0mod4} for these parameters,
together with the part of Proposition~\ref{prop:imp} used,
where the bounds by Proposition~\ref{prop:BHM} are listed
in the last column.
The values $d(n)$ are also listed in the table.
\begin{table}[thb]
\caption{Parameters satisfying~\eqref{eq:0mod4}}
\label{Tab:0mod4}
\begin{center}
{\small
\begin{tabular}{c|c|c|c|c|c}
\noalign{\hrule height0.8pt}
$n$ & $d(n)$ & $d$&$s$ & Proposition~\ref{prop:imp} & Proposition~\ref{prop:BHM}\\
\hline
72 & 14 & 14 & 8 & $B_{s}\le 14$ (iv)& $B_{s}\le 16$, $\ne 15$ \\
100 & 18 & 18 & 10 & $B_{s}\le 18$ (iv)& $B_{s}\le 20$, $\ne 19$ \\
\hline
108 & - & 18 &10 &$B_{s}\le 18$ (iv)& $B_{s}\le 20$, $\ne 19$ \\
116 & - & 18 &10 &$B_{s}\le 18$ (iv)& $B_{s}\le 20$, $\ne 19$ \\
128 & - & 22 &12 &$B_{s}\le 16$ (v)& $B_{s}\le 21$ \\
\noalign{\hrule height0.8pt}
\end{tabular}
}
\end{center}
\end{table}
We discuss the possible weight enumerators for the case
$d(n)=d$ in Table~\ref{Tab:0mod4}.
The possible weight enumerators of an extremal singly even
self-dual $[72,36,14]$ code with $s\ge 8$ and the shadow are as follows:
\[
\begin{array}{lll}
W_{72} &=& 1 + (8640 - 64 \alpha)y^{14} + (124281 + 384 \alpha) y^{16}
+ \cdots, \\
S_{72} &=&
\alpha y^8 + (546 - 14 \alpha) y^{12} + (244584+91 \alpha)y^{16}
+ \cdots,
\end{array}
\]
respectively,
where $\alpha$ is an integer with
$0\leq \alpha\leq \lfloor \frac{546}{14} \rfloor=39$~\cite{DGH}.
We remark that Conway and Sloane~\cite{C-S} give only
two weight enumerators as the possible weight enumerators
of an extremal singly even self-dual $[72,36,14]$ code with
$s\ge 8$ without reason, namely $\alpha=0,1$ in
$W_{72}$.
Table~\ref{Tab:0mod4} shows the following:
\begin{prop}
If there exists an extremal singly even
self-dual $[72,36,14]$ code with weight enumerator $W_{72}$,
then $0 \le \alpha \le 14$.
\end{prop}
It is unknown whether there exists an extremal singly even self-dual
code for any of these cases.
The possible weight enumerators of a singly even
self-dual $[100,50,18]$ code with $s\ge 10$ and the shadow are as follows:
\[
\begin{array}{lll}
W_{100} &=& 1 + (16 \beta + 52250)y^{18}
+ (1024 \alpha - 64 \beta +972180)y^{20}
+ \cdots, \\
S_{100} &=&
\alpha y^{10} + (-20 \alpha - \beta) y^{14} + (190 \alpha + 104500 +
18 \beta)y^{18}
+ \cdots,
\end{array}
\]
respectively,
where $\alpha,\beta$ are integers with
$0 \leq \alpha \leq \frac{-1}{20} \beta \leq \frac{5225}{32}$~\cite{DGH}.
Table~\ref{Tab:0mod4} shows the following:
\begin{prop}
If there exists a singly even
self-dual $[100,50,18]$ code with weight enumerator $W_{100}$,
then $0 \le \alpha \le 18$.
\end{prop}
It is unknown whether there exists a singly even self-dual $[100,50,18]$
code for any of these cases.
We give more sets of parameters for which
the bound on $B_{s}$ obtained by
Proposition~\ref{prop:imp}
improves the bound obtained by Proposition~\ref{prop:BHM}:
\[
(n,d,s)=
(108,18,10),
(116,18,10),
(128,22,12).
\]
These bounds are also listed in Table~\ref{Tab:0mod4}.
\bigskip
\noindent
{\bf Acknowledgment.}
This work was supported by JSPS KAKENHI Grant Number 15H03633.
| 62,020
|
TITLE: Show that the sum of the products in pairs of the number 1,2,3...p-1 is divisible by p, where p is prime
QUESTION [1 upvotes]: If $p ≥ 5$ is prime, show that the sum of the products in pairs of the
numbers $1, 2, . . . , p−1$ is divisble by p. We do not count $1×1$, and
$1 × 2$ precludes $2 × 1$.
REPLY [2 votes]: $1,2,\ldots,p-1$ are the $p-1$ roots of the polynomial $x^{p-1}-1$ modulo $p$. By Vieta's formulae, the mentioned sum is congruent (modulo $p$) to the coefficient of $x^{p-3}$ in $x^{p-1}-1$, which is $0$.
Note that this relies on the fact that $\mathbb Z/p\mathbb Z$ is a field.
| 53,810
|
“Robo-investing” – the use of automated digital platforms to create and manage portfolios for investors—has gained substantial traction in recent years. A convenient alternative to consulting human advisors, digital investing has proven to be an efficient wealth management model for millions of investors across America.
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Investing digitally also allows consumers to obtain financial advice much faster than they would by consulting a human advisor. Since these “robo-investing” services are available 24/7, use a well-developed algorithm, and present information to users in an easily digestible format, the consumer saves time while receiving top-notch personalized financial advice.
Consumers receive a higher quality of service through digital investing: the algorithm-based platform gives users more consistent and accurate advice than a human advisor who may be prone to poor judgment, partiality, or simply human error. Digital investing ensures that the advice users receive is accurate and objective every time.
You’re in Control
One of the greatest benefits of digital investing is the user’s ability to personalize their portfolio. The consumer receives efficient and accurate advice on managing their investments while still staying fully informed and in control of their strategy. The “freshness” of the financial advice is also a major draw of going digital. Users get investment advice based on the most current market data, allowing them to receive real-time, personalized feedback. Whereas human advisors may be slower to track constantly-shifting market developments, a digital advisor streamlines this process and brings the user the most up-to-date information.
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So, let’s put all those advisers and handlers out of work. Following the advice of a robot makes one think of a robot wiping one’s hiney. No thanks. Anyway. unless you are knowledgeable in the first place, you wouldn’t know good advice if it bit you where the sun don’t shine.
Why is this advertisement disguised as a stand alone information article? I trust a knowledgeable financial adviser far more than the algorithm of someone I don’t know.
What a tremendous disservice to your subscribers and members to recommend the use of robo advisors instead of the real thing. Agreed, we need more real advisors. Especially ones who know anything about investing and don’t just follow their corporate lead. But it’s an eyeball-to-eyeball kind of activity and pointing them to a robotic solution is demeaning, degrading, and certainly makes one wonder who provides you with this advice.
Cleaver of AMAC! It has to bring in money somehow.
| 396,461
|
Lawrence King Publishing released this new title in fall 2012 that
profiles selected lighting design projects and reveals unique detailing that
makes these projects special. The
Joplin Residence by Derek Porter Studio is included along with the detailing of
the integrated lighting design of the exterior skin that blurs boundaries
between interior and exterior domains.
| 297,182
|
TITLE: Confused by simplification of operators
QUESTION [0 upvotes]: I have two operators defined as:
$$ \hat{A} \psi_n(x) = \sqrt{n} \psi_{n-1}(x), $$
$$ \hat{A}^+ \psi_{n-1}(x) = \sqrt{n} \psi_n(x). $$
And then I have this summation with its simplification:
$$ \hat{A} \hat{A} +\hat{A}^+\hat{A}^+ + \hat{A}^+\hat{A} + \hat{A}\hat{A}^+ $$
simplifies to
$$ \hat{A}^+\hat{A} + \hat{A}\hat{A}^+.$$
I don't understand why the first two terms cancel out, its confusing me. I am still new to operators so my brain can't quite think correctly with them yet.
Hope some one can explain what is going on here and why they cancel out, thanks!
REPLY [0 votes]: These two terms cannot cancel each other, I think, this equation is meant in the sense of action on a state with definite number of particles $\psi_n$.
The operator $\hat{A}^{+} \hat{A}^{+}$ increases the occupation number by 2 - maps $\psi_n$ to $\psi_{n+2}$, whereas $\hat{A} \hat{A}$ maps $\psi_n$ to $\psi_{n-2}$.
Being sandwiched by any state $\psi_n = | n \rangle$ first two terms give zero:
$$
\langle n | \hat{A}^{+} \hat{A}^{+} | n \rangle =
\langle n | \hat{A} \hat{A} | n \rangle = 0
$$
Because $\langle m | n \rangle \sim \delta_{mn}$, states with different occupation numbers are orthogonal to each other.
These combinations vanish in the expectation values, but are not equal as operators.
| 26,900
|
TITLE: 3D rotation matrix and direction cosines of the transformed coordinate system.
QUESTION [0 upvotes]: I have an orthogonal basis coordinate system and a rotated coordinate system of which I know the orthonormal direction vectors $ (\overrightarrow{e}_1, \overrightarrow{e}_2, \overrightarrow{e}_3) $ with respect to the basis system. In a reference I found that the rotation matrix consists then of the components of the direction cosines (written as columns). This seems contra-intuitive to me; is there a simple method to prove / disprove it?
REPLY [1 votes]: The components of a unit vector wrt the basis system are its direction cosines.
The columns of a transformation matrix are the images of the basis vectors.
So this property works for any linear transformation, provided you normalize the column vectors.
| 129,727
|
Operator Standing Applications Console Enclosures Datasheets
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| 85,820
|
\section{Main Results}\label{sec:results}
This section establishes rates of convergence for Algorithm~\ref{alg:QDSG} in the cases where $f_i$ are convex and strongly convex. These results show that under adaptive quantization, the convergence rates of the distributed subgradient algorithm are essentially unaffected by the finite communication bandwidths, except for a constant factor that captures the size of these bandwidths.
The main steps of our analysis are as follows. As we observed in the previous section, the adaptive quantization scheme forces the quantization error $\Delta_i(k)$ to decay to zero at the same rate as the step sizes $\alpha(k)$.
Our first step is to use this fact to show that the distance between the estimates $\xbf_i(k)$ to the average $\xbarbf(k)$ converges to zero, implying the nodes eventually reach consensus.
Next we will show that the update of (descent on) $\xbarbf(k)$ mirrors the update of standard centralized subgradient methods.
This allows us to study the convergence rate of Algorithm \ref{alg:QDSG} using the standard outline for the analysis of centralized subgradient methods.
When the $f_i$ are convex, and we use stepsizes $\alpha(k)= 1/ \,\sqrt{k+1}$, we show that the time-weighted average $\zbf_i(k)$ in \eqref{alg_QDSG:zi} obeys
\[
f(\zbf_i(k)) - f^* ~\lesssim~
\frac{1}{(2^b-1)^2(1-\sigma_2)}\cdot \frac{\ln k}{\sqrt{k}},
\]
where $1-\sigma_2$ is the spectral gap that quantifies the connectivity of the underlying network, and $ 1\,/\,(2^b-1)$ is the resolution of the quantizer using $b$ communication bits. When the objective function is strongly convex, using stepsizes $\alpha(k)= a/ \,(k+1)$ for appropriately chosen constant $a$, we have a refined rate on the convergence of the decision variables themselves,
\[
\|\zbf_i(k)-\xbf^*\|^2 ~\lesssim~
\frac{1}{(2^b-1)^2(1-\sigma_2)}\cdot \frac{\ln k}{k}.
\]
Aside from the constant $ 1\,/\,(2^b-1)$, these results match the standard results for the distributed subgradient method with perfect communication, meaning that the quantization does not qualitatively affect the behavior of the algorithm.
\subsection{Preliminaries}
\label{subsec:prelim}
Given a vector $\vbf\in\Rset^d$ we denote by $\xibf(\vbf)$ the error due to the projection of $\vbf$ on to $\Xcal$,
\[
\xibf(\vbf) = \vbf - \left[\vbf\right]_{\Xcal},
\]
and rewrite Eq.\ \eqref{distributed:QDSG} as
\begin{align}
\begin{aligned}
&\vbf_i(k) = \left(\sum_{j\in\Ncal_i} a_{ij}\xbf_j(k)\right) + \xbf_i(k)-\qbf_i(k) + \sum_{j\in\Ncal_i}a_{ij}(\qbf_j(k)-\xbf_j(k)) - \alpha(k)\gbf_i(\xbf_i(k)), \\
&\xbf_i(k+1) = \left[\vbf_i(k)\right]_{\Xcal} = \vbf_i(k) - \xibf_i(\vbf_i(k)).
\end{aligned}\label{prelim:xi}
\end{align}
Stacking the $\xbf_i^T$ as rows in the $n\times d$ matrix $\Xbf$ (and doing likewise with the $\vbf_i,\qbf_i,\xibf_i$), we can write the above in matrix form as
\begin{align}
\begin{aligned}
\Vbf(k) &= \Abf\Xbf(k) + (\Ibf-\Abf)(\Xbf(k)-\Qbf(k)) - \alpha(k)\Gbf(\Xbf(k)),\\
\Xbf(k+1) &= \Vbf(k) - \Xibf(\Vbf(k)),
\end{aligned}\label{prelim:X}
\end{align}
where $\Abf$ is the adjacency matrix in Assumption \ref{assump:doub_stoch}.
Let $\xbarbf(k)$ and $\bar{\xibf}(k)$ be the averages of $\xbf_i(k)$ and $\xibf_i(\vbf_i(k))$ across all nodes at time $k$:
\begin{align*}
\xbarbf(k) = \frac{1}{n}\sum_{i=1}^n\xbf_i(k) = \frac{1}{n}\Xbf^T\1 \in\Rset^{d}\quad\text{and}\quad \bar{\xibf}(k) = \frac{1}{n}\sum_{i=1}^n\xibf_i(\vbf_i(k)) =\frac{1}{n} \Xibf(\Vbf(k))^T\1\in\Rset^{d}.
\end{align*}
Since $\1^T\Abf = \1^T$, \eqref{prelim:X} gives
\begin{align}
\begin{aligned}
\vbarbf(k) &= \xbarbf(k) - \frac{\alpha(k)}{n}\sum_{i=1}^n\gbf_i(\xbf_i(k)),\\
\xbarbf(k+1) &= \vbarbf(k) - \bar{\xibf}(k).
\end{aligned}\label{prelim:xbar}
\end{align}
Recall that $\Delta_i(k) = \xbf_i(k) - \qbf_i(k)\in\Rset^{d}$ and let $\Delta(k)$ be defined as
\begin{align*}
\Delta(k) = \left[\begin{array}{cc}
-\;\Delta_1(k)^T\;- \\
\cdots\\
-\;\Delta_n(k)^T\;-
\end{array} \right]\in\Rset^{n\times d}\cdot
\end{align*}
We now consider the following sequence of lemmas, which provides fundamental preliminaries for our main results given in the next sections. In the sequel, we will consider two choices of the step sizes $\alpha(k)$, that is, $\alpha(k) = 1/\sqrt{k+1}$ or $\alpha(k) = 1/(k+1)$. These choices of step sizes also are used to establish our main results in the next section. For ease of exposition we delay all the proofs of the results in this section to Appendix \ref{sec:proofs}.
We first provide an upper bound for the projection error $\xibf_i$ in the following lemma.
\begin{lemma}\label{lem:projection}
Suppose that Assumption \ref{assump:doub_stoch} holds. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. Then for all $i\in\Vcal$ we have
\begin{align}
\|\,\xibf_i(\vbf_i(k))\,\| \leq \sum_{j\in\Ncal_i}a_{ij}\left\|\Delta_i(k) - \Delta_j(k)\right\| + L_i\alpha(k).\label{lem_project:Ineq1}
\end{align}
In addition, let $L=\sum_{i\in\Vcal}L_i$. Then, we obtain
\begin{align}
\sum_{i=1}^n\|\,\xibf_i(\vbf_i(k))\,\|^2 \leq 8\sum_{i=1}^n\|\Delta_i(k)\|^2 + 2L^2\alpha^2(k).\label{lem_project:Ineq2}
\end{align}
\end{lemma}
Next we provide an upper bound for the consensus errors $\|\xbf_i(k)-\xbarbf(k)\|$ in the following lemma.
\begin{lemma}\label{lem:consensus}
Suppose that Assumption \ref{assump:doub_stoch} holds. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. In addition, let $\{\alpha(k)\}$ be a nonnegative nonincreasing sequence of stepsizes. Then, we have
\begin{align}
\|\Xbf(k+1)-\1\xbarbf(k+1)^T\| &\leq 6\sum_{t=0}^{k}\sigma_2^{k-t}\|\Delta(t)\| + 3L\sum_{t=0}^{k}\sigma_2^{k-t}\alpha(t).\label{lem_consensus:consensus_bound}
\end{align}
\end{lemma}
As mentioned, the main motivation of the adaptive quantization is to eliminate the impact of quantized errors. In particular, we will show that the quantized errors produced by Algorithm \ref{alg:QDSG} decrease to zero at the same rate with the step size $\alpha(k)$. To do that, we require the following technical condition.
\begin{assump}\label{assump:bits}
Let $\gamma = 48(2+ L)\,/\,(1-\sigma_2)$. Then the number bits $b$ of the communication bandwdith satisfies
\begin{align}
\sqrt{nd}\gamma \leq 2^b - 1.\label{assumption_bits:Ineq}
\end{align}
\end{assump}
The following lemma is to show that the quantized error $\|\Delta_i(k)\|\lesssim~ \alpha(k)$, for all $i\in\Vcal$, for the case $\sigma_{2}^{k}\leq \alpha(k)$ for all $k\geq0$. When $\sigma_{2}^k\geq \alpha(k)$ for some small $k$, e.g., $\sigma_2$ is closed to $1$ but not equal, one can show from our analysis that $\|\Delta_i(k)\|\lesssim~ \sigma_2^{k} $, which is eventually converge to zero faster than $\alpha(k)$. Since this issue has been captured by the rate of consensus in Eq.\ \eqref{lem_consensus:consensus_bound}, we skip it here for simplicity.
\begin{lemma}\label{lem:quantized_error}
Suppose that Assumption \ref{assump:doub_stoch} and \ref{assump:bits} hold. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. Let $\alpha(k)$ be either $\alpha(k) = 1\,/\,\sqrt{k+1}$ or $\alpha(k) = 1\,/\,(k+1)$. Then we have all $k\geq 0$
\begin{align}
\xbf_i(k+1)\in \Rcal_i(k+1) \triangleq \left[\qbf_i(k) - \frac{\gamma}{2}\alpha(k)\1,\,\qbf_i(k) + \frac{\gamma}{2}\alpha(k)\1\right].\label{lem_quantized_error:Ineq}
\end{align}
In addition, we also have
\begin{align}
\|\Delta_i(k)\| \leq \frac{\sqrt{d}\gamma}{2^b-1}\alpha(k).\label{lem_quantized_error:error_bound}
\end{align}
\end{lemma}
The following lemma is a consequence of Lemmas \ref{lem:consensus} and \ref{lem:quantized_error}.
\begin{lemma}\label{lem:consensus_bound}
Suppose that Assumptions \ref{assump:doub_stoch} and \ref{assump:bits} hold. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. Let $\alpha(k)$ be either $\alpha(k) = 1\,/\,\sqrt{k+1}$ or $\alpha(k) = 1\,/\,(k+1)$. Then, we have
\begin{align}
\lim_{k\rightarrow\infty} \xbf_i(k) = \lim_{k\rightarrow\infty} \xbf_j(k),\qquad \forall\, i,j\in\Vcal.\label{lem_consensus:asymp_conv}
\end{align}
In addition, if $\alpha(k)$ is also square-summable, i.e.,
\begin{align}
\sum_{k=0}^{\infty}\alpha^2(k) < \infty,\label{lem_consensus:square_summable}
\end{align}
then for all $k\geq0$ we have
\begin{align}
\sum_{t=0}^{k}\alpha(t) \|\Xbf(t)-\1\xbarbf(t)^T\| \leq \left(\frac{6\sqrt{nd}\gamma+3L2^b}{(1-\sigma_2)(2^b-1)}\right)\sum_{t=0}^{k}\alpha^2(t)<\infty.\label{lem_consensus:finite_sum}
\end{align}
\item If $\alpha(k) = 1\,/\,\sqrt{k+1}$ then we have for all $k\geq0,$
\begin{align}
\sum_{t=0}^{k}\alpha(t)\|\Xbf(t)-\1\xbarbf(t)^T\|\leq \left(\frac{6\sqrt{nd}\gamma+3L2^b}{(1-\sigma_2)(2^b-1)}\right)(\ln(k+1)+1). \label{lem_consensus:rate}
\end{align}
\end{lemma}
Finally, we provide an upper bound for the optimal distance $\|\xbarbf(k)-\xbf^*\|^2$ in the following lemma.
\begin{lemma}\label{lem:opt_dist}
Suppose that Assumptions \ref{assump:doub_stoch} and \ref{assump:bits} hold. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. In addition, let $\xbf^*\in\Xcal^*$ be a solution of problem \eqref{prob:obj}. Let $\alpha(k)$ be either $\alpha(k) = 1\,/\,\sqrt{k+1}$ or $\alpha(k) = 1\,/\,(k+1)$. Then, we have
\begin{align}
\|\xbarbf(k+1)-\xbf^*\|^2&\leq \|\xbarbf(k)-\xbf^*\|^2-\frac{2\alpha(k)}{n}\sum_{i=1}^n\gbf_i(\xbf_i(k))^{T}(\xbf_i(k)-\xbf^*) \notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k)\cdot\label{lem_opt_dist:Ineq}
\end{align}
\end{lemma}
\subsection{Convergence Results of Convex Functions}
We now present the first main result of this paper, which is the rate of convergence of Algorithm \ref{alg:QDSG} to the optimal value of problem \eqref{prob:obj} when the local functions $f_i$ are convex. Since the update of $\xbarbf(k)$ in Eq.\ \eqref{prelim:xbar} can be viewed as a variant of a centralized projected subgradient methods used to solve problem \eqref{prob:obj}, we utilize standard techniques in the analysis of these methods to derive the rate of convergence of Algorithm \ref{alg:QDSG}. Specifically, at any time $k \geq 0$ if each node $i \in \Vcal$ maintains a variable $\zbf_i(k)$ to compute
the time-weighted average of its estimate $\xbf_i(k)$ and if the stepsize $\alpha(k)$ decays as $\alpha(k) = 1/ \,\sqrt{k+1}$, the objective function value $f$ in Eq.\ \eqref{prob:obj} estimated at each $\zbf_i(k)$ converges to the optimal value with a rate $\mathcal{O}\left(\eta \ln(k+1)\, /\,\sqrt{k+1}\right)$, where $\eta$ is some constant depending on the algebraic connectivity $1-\sigma_2$ of the network, the number of quantized bits $b$, and the Lipschitz constants $L_i$ of $f_i$. We also note that this condition on the stepsizes is also used to study the convergence rate of centralized subgradient methods \cite{Nesterov2004}. The following theorem is used to show the convergence rate of Algorithm \ref{alg:QDSG}.
\begin{theorem}\label{thm_convex:rate}
Suppose that Assumptions \ref{assump:doub_stoch} and \ref{assump:bits} hold. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. In addition, let $\alpha(k) = 1\,/\,\sqrt{k+1}$. Moreover, suppose that each node $i$, for all $i\in\Vcal$, stores a variable $\zbf_i\in\Rset^{d}$ initiated arbitrarily in $\Xcal$ and updated as
\begin{align}
\zbf_i(k) = \frac{\sum_{t=0}^k\alpha(t)\xbf_i(t)}{\sum_{t=0}^k\alpha(t)},\quad\forall i\in\Vcal.\label{thm_convex_rate:zi}
\end{align}
Then for all $i\in\Vcal$ and $k\geq 0$ we have
\begin{align}
f(\zbf_{i}(k)) - f^*&\leq\frac{n \|\xbarbf(0)-\xbf^*\|^2}{2\sqrt{k+1}}+ \frac{\sqrt{n}(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2}\frac{(\ln(k+1)+1)}{\sqrt{k+1}}\cdot\label{thm_convex_rate:Ineq}
\end{align}
\end{theorem}
\begin{proof}
For convenience, let $\rbf(k) = \xbarbf(k)-\xbf^*$, where $\xbf^*\in\Xcal^*$ is a solution of problem \eqref{prob:obj}. By Eq.\ \eqref{lem_opt_dist:Ineq} we have
\begin{align*}
\|\rbf(k+1)\|^2 &\leq \|\rbf(k)\|^2-\frac{2\alpha(k)}{n}\sum_{i=1}^n\gbf_i(\xbf_i(k))^{T}(\xbf_i(k)-\xbf^*) + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k) \notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|,
\end{align*}
which by the convexity of $f_i$ yields
\begin{align}
\|\rbf(k+1)\|^2&\leq \|\rbf(k)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k) \notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad - \frac{2\alpha(k)}{n}\left(\sum_{i=1}f_i(\xbf_i(k))-f_i(\xbf^*))\right)
,\label{thm_convex_rate:Eq1}
\end{align}
We now analyze the last term on the right-hand side of Eq.\ \eqref{thm_convex_rate:Eq1}. Indeed, by Eq.\ \eqref{prop:bounded_subg_ineq} and using $f = \sum_{i=1}^n f_i$ and $f^* = f(x^*)$, we have for a fixed $\ell\in\Vcal$
\begin{align}
&- \sum_{i=1}^n f_i(\xbf_i(k)) - f_i(\xbf^*) = - \sum_{i=1}^n\Big( f_i(\xbf_i(k)) -f_i(\xbarbf(k)) + f_i(\xbarbf(k)) - f_i(\xbf^*)\Big)\notag\\
&\qquad\leq \sum_{i=1}^n L_i |\,\xbf_i(k)-\xbarbf(k)\,| - \Big(f(\xbarbf(k)) - f^*\Big)\notag\\
&\qquad \leq L \|\,\Xbf(k)-\1\xbarbf(k)^T\,\| - \Big(f(\xbarbf(k)) -f(\xbf_{\ell}(k)) +f(\xbf_{\ell}(k)) - f^*\Big)\notag\\
&\qquad \leq 2L\|\,\Xbf(k)-\1\xbarbf(k)^T\,\| - \Big(f(\xbf_{\ell}(k)) - f^*\Big),
\label{thm_convex_rate:Eq1a}
\end{align}
which when substituting into Eq.\ \eqref{thm_convex_rate:Eq1} yields
\begin{align}
\|\rbf(k+1)\|^2 &\leq \|\rbf(k)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k) - \frac{2}{n}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big) \notag\\
&\qquad +\left(\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)})+ \frac{4L}{n}\right)\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&= \|\rbf(k)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k) - \frac{2}{n}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big)\notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+5L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\label{thm_convex_rate:Eq1b},
\end{align}
which when iteratively updating over $k = 0,\ldots,K$ for some $K\geq0$ we have
\begin{align*}
\|\rbf(K+1)\|^2 &\leq \|\rbf(0)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\sum_{k=0}^{K}\alpha^2(k) \notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+5L2^{b}\right)}{\sqrt{n}(2^b-1)}\sum_{k=0}^{K}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad - \frac{2}{n}\sum_{k=0}^{K}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big).
\end{align*}
Since $\alpha(k) = 1/\sqrt{k+1}$ we now use Eq.\ \eqref{lem_consensus:rate} into the preceding relation to have
\begin{align*}
\|\rbf(K+1)\|^2 & \leq \|\rbf(0)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}(\ln(K+1)+1) \notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+5L2^{b}\right)}{\sqrt{n}(2^b-1)}\left(\frac{6\sqrt{nd}\gamma+3L2^b}{(1-\sigma_2)(2^b-1)}\right)(\ln(K+1)+1)\notag\\
&\qquad - \frac{2}{n}\sum_{k=0}^{K}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big)\notag\\
&\leq \|\rbf(0)\|^2 + \frac{4(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2\sqrt{n}}(\ln(K+1)+1) \notag\\
&\qquad - \frac{2}{n}\sum_{k=0}^{K}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big).
\end{align*}
Rearranging the preceding relation and dropping the nonnegative $\|\rbf(K+1)\|$ we obtain
\begin{align*}
\sum_{k=0}^{K}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big)&\leq \frac{n \|\rbf(0)\|^2}{2}+ \frac{2\sqrt{n}(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2}(\ln(K+1)+1),
\end{align*}
which by dividing both sides by $\sum_{k=0}^{K}\alpha(k)$ and using the convexity of $f$ gives Eq.\ \eqref{thm_convex_rate:Ineq}, i.e.,
\begin{align*}
f(\zbf_{\ell}(K)) - f^*&\leq \frac{n \|\rbf(0)\|^2}{2\sqrt{K+1}}+ \frac{2\sqrt{n}(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2}\frac{(\ln(K+1)+1)}{\sqrt{K+1}},
\end{align*}
where in the last inequality we use the integral test for $K\geq 0$ to have
\begin{align*}
\sum_{k=0}^{K}\alpha(k) = \sum_{k=0}^{K}\frac{1}{\sqrt{k+1}}\geq \int_{t=0}^{K+1}\frac{1}{\sqrt{t+1}}dt = 2(\sqrt{K+2}-1)\geq \sqrt{K+1}.
\end{align*}
\end{proof}
It is worth to mention that under the choice of $\alpha(k) = 1\,/\,(k+1)$, for all $k\geq 0$, one can show that $\xbf_i(k)$ asymptotically converges to $\xbf^*$ for all $i\in\Vcal$. This is a consequence of Lemmas \ref{lem:consensus} and \ref{lem:opt_dist}, and some standard analysis. The following lemma states this result. The analysis is omitted and can be found in \cite[Theorem 3]{DoanMR2018a}.
\begin{lemma}
Suppose that Assumptions \ref{assump:doub_stoch} and \ref{assump:bits} hold. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. Let $\alpha(k) = 1\,/\,(k+1)$. Then we obtain
\begin{align}
\lim_{k\rightarrow\infty} \xbf_i(k) = \xbf^*,\qquad \text{for all } i\in\Vcal,
\end{align}
for some $\xbf^*$ that is a solution of problem \eqref{prob:obj}.
\end{lemma}
\subsection{Convergence Results of Strongly Convex Case}
In this section, our goal is to study the convergence rate of Algorithm \ref{alg:QDSG} when the local functions $f_i$ are strongly convex, that is, we make the following assumption on $f_i$
\begin{assump}\label{assump:sconvexity}
Each function $f_i$ is strongly convex with some positive constant $\mu_i$, i.e., the condition \eqref{notation:sc} holds.
\end{assump}
Under this assumption, we show that if each node $i \in \Vcal$ maintains a variable $\zbf_i(k)$ to compute the time average of its estimate $\xbf_i(k)$ and if the stepsize $\alpha(k)$ decays as $\alpha(k) = a\,/\, (k+1)$ for some properly chosen constant $a$, the variable $\zbf_i(k)$ converges to the optimal solution $x^*$ of problem \eqref{prob:obj} with a rate $\mathcal{O}\left(\eta \ln(k+1)\, /\,(k+1)\right)$, where $\eta$ is some constant depending on the algebraic connectivity $1-\sigma_2$ of the network, the number of quantized bits $b$, and the constants $L_i$ and $\mu_i$ of $f_i$. The following theorem is used to show the convergence rate of Algorithm 1 under Assumption \ref{assump:sconvexity}.
\begin{theorem}\label{thm_sconvex:rate}
Suppose that Assumptions \ref{assump:doub_stoch} and \ref{assump:sconvexity} hold. Let the sequence $\{\xbf_i(k)\}$, for all $i\in\Vcal$, be generated by Algorithm \ref{alg:QDSG}. We denote by $\mu=\min_{i\in\Vcal}\mu_i$. In addition, let $\{\alpha(k)\} = a\,/\,k+1$ for some $a\geq 1\,/\,\mu$. Moreover, suppose that each node $i$, for all $i\in\Vcal$, stores a variable $\zbf_i\in\Rset$ initiated arbitrarily in $\Xcal$ and updated as
\begin{align}
\zbf_i(k) = \frac{\sum_{t=0}^k \xbf_i(t)}{k+1},\quad\forall i\in\Vcal.\label{thm_sconvex_rate:zi}
\end{align}
Let $\xbf^*\in\Xcal^*$ be a solution of problem \eqref{prob:obj}. Then for all $i\in\Vcal$ and $k\geq 0$ we have
\begin{align}
\|\zbf_{i}(k)-\xbf^*\|^2 \leq \frac{4\sqrt{n}\alpha(0)(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2}\frac{1+\ln(k+1)}{k+1}\cdot\label{thm_sconvex_rate:Ineq}
\end{align}
\end{theorem}
\begin{proof}
Let $\xbf^*$ be a solution of problem \eqref{prob:obj}. For convenience, let $\rbf(k) = \xbarbf(k)-\xbf^*.$ By Eq.\ \eqref{lem_opt_dist:Ineq} we have
\begin{align}
\|\rbf(k+1)\|^2 &\leq \|\rbf(k)\|^2-\frac{2\alpha(k)}{n}\sum_{i=1}^n\gbf_i(\xbf_i(k))^{T}(\xbf_i(k)-\xbf^*) \notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k)\notag\\
&\leq \|\rbf(k)\| + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k)\notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad -\frac{2\alpha(k)}{n}\sum_{i=1}^n \Big(f_i(\xbf_i(k)) - f_i(\xbf^*)+\frac{\mu_i}{2}\|\xbf_i(k)-\xbf^*\|^2\Big),\label{thm_sconvex_rate:Eq1}
\end{align}
where the last inequality is due to the strong convexity of $f_i$, i.e., Eq.\ \eqref{notation:sc}. First, using the Jensen's inequality on quadratic function $(\cdot)^2$ we have
\begin{align*}
-\frac{1}{n}\sum_{i=1}^n\mu_i\|\xbf_i(k)-\xbf^*\|^2\leq -\mu\,\frac{1}{n}\sum_{i=1}^n\|\xbf_i(k)-\xbf^*\|^2\leq -\mu\|\xbarbf(k)-\xbf^*\|^2 = -\mu\,\|\rbf(k)\|^2.
\end{align*}
Fix some $\ell\in\Vcal$. Then, substituting the preceding relation into Eq.\ \eqref{thm_sconvex_rate:Eq1} and using Eq.\ \eqref{thm_convex_rate:Eq1a} yield
\begin{align*}
\|\rbf(k+1)\|^2
&\leq \left(1-\mu\alpha(k)\right)\|\rbf(k)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k)\notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+3L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad -\frac{2\alpha(k)}{n}\sum_{i=1}^n \Big(f_i(\xbf_i(k)) - f_i(\xbf^*)\Big)\notag\\
&\stackrel{\eqref{thm_convex_rate:Eq1a} }{\leq} \left(1-\mu\alpha(k)\right)\|\rbf(k)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k)\notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+5L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad - \frac{2}{n}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big),
\end{align*}
Note that $\alpha(k) = a\,/\,(k+1)$ with $a\geq 1\,/\,\mu$, implying $\mu\alpha(k)\geq 1\,/\,(k+1)$. Thus, the preceding equation gives
\begin{align*}
\|\rbf(k+1)\|^2 &\leq \frac{k}{k+1}\|\rbf(k)\|^2 + \frac{2(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha^2(k)\notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+5L2^{b}\right)}{\sqrt{n}(2^b-1)}\alpha(k)\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad - \frac{2}{n}\alpha(k)\Big(f(\xbf_{\ell}(k)) - f^*\Big).
\end{align*}
Multiplying both sides of the preceding equation by $k+1$, and using $(k+1)\,/\,k\leq 2$ and $\alpha(k) = \alpha(0) / (k+1)$ we have
\begin{align}
(k+1)\|\rbf(k+1)\|^2 &\leq k\|\rbf(k)\|^2 + \frac{4\alpha(0)(4\sqrt{nd}\gamma + 3L2^{b})^2}{(2^b-1)^2\sqrt{n}}\alpha(k)\notag\\
&\qquad +\frac{2\left(4\sqrt{nd}\gamma+5L2^{b}\right)}{\sqrt{n}(2^b-1)}\|\Xbf(k)-\1\xbarbf(k)^T\|\notag\\
&\qquad - \frac{2\alpha(0)}{n}\Big(f(\xbf_{\ell}(k)) - f^*\Big),\label{thm_sconvex_rate:Eq1a}
\end{align}
By Eq.\ \eqref{lem_consensus:consensus_bound} and using $\|\Delta(t)\|\leq \alpha(t)$ we have
\begin{align}
\sum_{k=0}^{K}\|\Xbf(k)-\1\xbarbf(k)^T\| &\leq (3L+6)\sum_{k=0}^{K}\sum_{t=0}^{k-1}\sigma_2^{k-t}\alpha(t)\notag\\
&\leq (3L+6)\sum_{k=0}^{K-1}\alpha(k)\sum_{t=k+1}^{K}\sigma_2^{t}\leq \frac{3L+6}{1-\sigma_2}\sum_{k=0}^{K-1}\alpha(k).
\end{align}
Next, summing up both sides of Eq. \eqref{thm_sconvex_rate:Eq1a} over $k = 0,\ldots,K$ for some $K\geq 0$, using the preceding relation, and rearranging we obtain
\begin{align*}
\frac{2\alpha(0)}{n}\sum_{k=0}^{K}\Big(f(x_{\ell}(k)) - f^*\Big) &\leq \frac{2\alpha(0)(6\sqrt{nd}\gamma + 5L2^{b})^2}{\sqrt{n}(1-\sigma_2)(2^b-1)^2}\sum_{k=0}^{K}\alpha(k)\notag\\
&\leq \frac{2\alpha(0)(6\sqrt{nd}\gamma + 5L2^{b})^2}{\sqrt{n}(1-\sigma_2)(2^b-1)^2}(\ln(K+1)+1),
\end{align*}
which when dividing both sides by $(K+1)\,/\,n$ and using the convexity of $f$ yields
\begin{align*}
&2\alpha(0)\Big[f(\zbf_{\ell}(K)) - f^*\Big] \leq \frac{2\sqrt{n}\alpha(0)(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2}\frac{1+\ln(K+1)}{K+1}.
\end{align*}
Since the functions $f_i$ are strongly convex with constant $\mu_i$, $f$ is strongly convex with constant $\mu$. Thus, using the preceding equation and $\alpha(0) = a\geq 1\,/\,\mu$ gives Eq.\ \eqref{thm_sconvex_rate:Ineq}, i.e.,
\begin{align*}
&\|\zbf_{\ell}(K)-\xbf^*\|^2 \leq \frac{2}{\mu}\left[f(\zbf_{\ell}(K)) - f^* \right]\leq \frac{2\sqrt{n}\alpha(0)(6\sqrt{nd}\gamma + 5L2^{b})^2}{(1-\sigma_2)(2^b-1)^2}\frac{1+\ln(K+1)}{K+1}\cdot
\end{align*}
\end{proof}
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How of Elimination?
When we talk about elimination, there are five major channels your body uses to help remove wastes from the body. These include:
- the skin;
- the bowel;
- the lymphatic system;
- the kidneys
- the lungs;
Even though sometimes we talk about these systems separately, all are equally important and all need to interoperate to effectively keep your body toxin free.
The signs of congestion are most frequently seen in the skin, where skin conditions such as acne, dermatitis, eczema or psoriasis can result from an overburdened bowel or lymphatic systems. However, there are exceptions. The constant production of concentrated urine, ongoing bad breath despite good dental hygiene and heavily blood shot eyes may also be indicators that the bowel or lymph needs attention.
How Can You Keep the Plumbing Clean?
A key principle of natural medicine has long been that good health starts with the correct support of the bodies elimination processes.
How Can Diet and Lifestyle Help?
The best method for ensuring healthy elimination is a healthy diet and lifestyle. Follow these simple tips to make the most of your elimination channels.
- Adequate intake of plant-based fibre and water is essential for the channels of elimination to function properly.
- Enjoy a minimum of three cups of fresh vegetables daily: Choose from a variety of vegetables and aim to have a rainbow of colours on your dinner plate.
- Enjoy a minimum of two pieces or one cup of fresh fruit daily: Berries are an excellent choice as they are rich in antioxidants.
- Limit starchy carbohydrates to two small serves weekly: Aim for only 1 to 2 serves of bread, rice, pasta, cereal or potatoes.
- Drink a minimum of eight glasses of pure water daily: To add flavour, opt for natural flavourings such as fresh lemon, lime and mint in water instead of soft drinks and cordials.
- Be active – exercise helps increase blood flow to the skin, and encourages the removal of toxins through sweat; so make daily exercise a part of your skin health program today!
How Can You Improve Lymphatic Function?
One of the most neglected organs of elimination is the lymphatic system. As a network of fine vessels that help move toxins back to the kidney and bowel for elimination, constricting, or confining clothing, lack of exercise and inadequate water intake can all lead to congestion and a slow down in the lymphatic system.
Lymphatic massage is a powerful way to re-invigorate the flow of the lymphatic system and remove congestion or blockages. Our therapists are qualified in lymphatic drainage, or for something you can do at home try dry skin brushing. Dry skin brushing prior to your morning shower is a simple technique to stimulate lymphatic circulation, helping your body to clear the build up of waste products that can aggravate skin conditions.
A number of herbs have also been traditionally used to improve lymphatic function including:
- cleavers (Galium aparine);
- burdock (Arctium lappa);
- nettle (Urtica dioica);
- red sage (Salvia miltiorrhiza);
- blue flag (Iris versicolour).
We have had great results using these in our clinic and highly recommend anyone concerned about lymphatic function talk to our Naturopaths today.
How Can You Improve Bowel Function?
The diets we eat today do not promote a healthy bowel. We regularly find people are affected by either slow transit times, or irregular bowel movements. Like your garbage bin, a bowel that doesn’t move frequently is smelly, unhygienic and a home for bad bacteria and pathogens.
A properly functioning digestive system and moving your bowels regularly is an integral part of achieving overall health. While diet is the major component for ensuring good bowel health, sometimes herbal support may be necessary. In our clinic we have had great results improving bowel function using herbs such as:
- ginger (Zingiber officinale);
- german chamomile (Matricaria recutita);
- cascara (Rhamnus purshianus);
- cinnamon (Cinnamomum cassia); and
- rhubarb (Rheum officinale);
It is important to use these herbs only to kick start a bowel that Is moving sluggishly. Persistent problems with bowel function may indicate more serious problems that need the assistance of a Naturopath.
Looking for Other Ways to Improve Elimination?
Problems with elimination can often lead to other, more serious health complications. If you are experiencing problems with any part of the elimination process book online or talk to our Naturopathic team today. We can help you implement dietary and lifestyle changes that support healthy elimination and will help to improve your overall health.
| 37,723
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IkiWiki::Plugin::syntax::Simple - Simple engine for syntax highlight
This documentation refers to IkiWiki::Plugin::syntax::Simple version 0.1
use IkiWiki::Plugin::syntax::Simple; my $engine = IkiWiki::Plugin::syntax::Simple->new(); my $htmlized_text = $engine->syntax_highlight( source => q(....), language => q(pod), linenumbers => 1, );
This module provides a simple syntax highlight engine for use with ikiwiki on installations where don't install third party modules.
The code return the source text received without special CSS marks inside with the exception of the
PRE html paragraph.
Returns a hash with information about his capabilities.
This method returns always true because it don't make any real work with the source.
This module don't raise any exceptions.
This module don't need any special configuration nor environment.
| 275,865
|
by Lisa Huff
Last week I wrote about Avago’s new miniaturized transmitters and receivers so today I’d like to introduce you to a similar product from privately-held Luxtera. Well known for its CMOS photonics technology, Luxtera actually introduced its OptoPHY transceivers first – in late 2009.
Luxtera took a different approach to its new high-density, optical interconnect solution. It is a transceiver module and is based on LW (1490nm) optics. Just like Avago’s devices, the transceivers use 12-fiber ribbon cables provided by Luxtera, but that’s really were the similarities end. The entire 10G–per-lane module only uses about 800mW compared to Avago's 3W, and they are true transceivers as opposed to separate transmitters and receivers. Luxtera is shipping its device to customers, but have not announced which ones yet.
In addition to the projected low cost for these devices, what should also be noted is that all of the solutions mentioned in the last three entries – Intel’s Light Peak; Avago’s MicroPOD and Luxtera’s OptoPHY – have moved away from the pluggable module product theme to board-mounted devices. This in and of itself may not seem significant until you think about why there were pluggable products to begin with. The original intent was to give OEMs and end users flexibility in design so they could use an electrical, SW optical or LW optical device in a port depending on what length of cable needed to be supported. You could also grow as you needed to – so only populate those ports required at the time of installation and add others when necessary. The need for this flexibility has seemed to have waned in recent years in favor of density, lower cost and lower power consumption. The majority of pluggable ports are now optical ones, so why not just move back to board-mounted products that can achieve the miniaturization, price points and lower power consumption?
Note: Only a member of this blog may post a comment.
| 71,231
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TITLE: $||x||^2/||A^{-1}||\leq x^TAx$ for positive definite symmetric matrix $A$
QUESTION [0 upvotes]: I am trying to prove the inequality on Page 605 of Nocedal and Wright (Numerical Optimization);
$$\sigma_n(A)||x||^2 = ||x||^2/||A^{-1}||\leq x^TAx \leq ||A|| ||x||^2 = \sigma_1(A) ||x||^2$$
where $A$ is a symmetric positive definite matrix and $\sigma_n(A)$ is the $nth$ singular value of $A$ and $\sigma_1(A)$ is the first singular value of $A$.
I have been able to prove every part of the inequality except for;
$$||x||^2/||A^{-1}||\leq x^TAx$$
If someone could provide a proof or a reference to a book that proves this it would be greatly appreciated.
Below are the details of the proofs of the other parts of the inequality;
First here is the proof of $x^TAx\leq ||x||^2 \cdot ||A||$;
$$\begin{aligned} x^TAx &= ||x|| \cdot ||Ax|| \cos (\theta) \\ &= ||x||^2 \cdot \frac{||Ax||}{||x||} \cos(\theta)\\ &\leq ||x||^2 \max_{x \neq 0} \frac{||Ax||}{||x||} \quad \quad \cos(\theta)\geq 0 \text{ since } x^TAx \geq 0\\ &= ||x||^2 \cdot ||A|| \end{aligned}$$
and also here is the proof of $$||x||^2/||A^{-1}||\leq ||A|| \cdot ||x||^2$$;
$$\begin{aligned}||x||^2/||A||^{-1} &=||x||^2 \min_{y \neq 0} \frac{||y||}{||A^{-1}y||} \\ &= ||x||^2 \min_{y \neq 0} \frac{||AA^{-1}y||}{||A^{-1}y||} \\ & \leq ||x||^2 \min_{y \neq 0} \frac{||A|| \cdot ||A^{-1}y||}{||A^{-1}y||} \\ &= ||x||^2 ||A|| \end{aligned}$$
REPLY [1 votes]: Since $A$ is symmetric positive definite, we can write $A = \sum_k \sigma_k v_k v_k^T$, and we can assume the $\sigma_k$ are ordered.
It is straightforward to check by multiplying that $A^{-1} = \sum_k {1 \over \sigma_k} v_k v_k^T$, and so $\|A^{-1}\| = {1 \over \sigma_n}$.
Then $x^TAx = \sum_k \sigma_k ( v_k^T x)^2$ and so
$\sigma_n \sum_k ( v_k^T x)^2 \le x^T Ax \le \sigma_1 \sum_k ( v_k^T x)^2$.
Since the $v_k$ are orthonormal, $\sum_k ( v_k^T x)^2 = \|x\|^2$.
| 108,057
|
TITLE: Why is the height of persons in a country a random variable of the continuous type?
QUESTION [1 upvotes]: If we define a measurable function on a probability space as a function that maps from the sample space to the real numbers, and treat all the people in a Country as the sample space, shouldn't the image of the random variable be a finite set? I don't understand why we would say that the random variable is countable.
REPLY [2 votes]: Assuming data are normal is a practical convenience. A discrete random variable that has very many possible values requires a long list of possible values and associated probabilities.
In the US, heights of people are usually rounded to the nearest inch. (A person's height can fluctuate by about half an inch during the day--usually taller in the AM shorter in the PM. So rounding to the nearest inch doesn't lose crucial information.)
Almost all US adults are between 50 and 90 inches tall, so in using a discrete random variable there would be about 40 different values to list along with probabilities of each.
Saying that heights are approximately normally distributed with mean 68 and standard deviation 3.5 allows us to get by with 2 parameters.
Theoretically, a normal distribution takes all values on the real line, but an interval $(56,89)$ with $99.9\%$ of
the probability is manageably short--even if it theoretically contains uncountably many values. Also one ignores the fact that this normal distribution allows a
very tiny probability of (practically impossible)
negative heights.
According to this normal distribution, what is the probability that a randomly chosen person is 68in
tall? The theoretical answer is $0.$ (if you have uncountably many points you can't assign positive probabilities to any of them.) If you mean $68.00000$ inches tall, then the practical answer is also essentially $0.$ (Who could measure that?) It is more useful to ask for
the probability that person is between
$67.5$ and $68.5$ (so that the height would be rounded to "68". Then the answer is $0.1136.$
If pilots for
a small military airplane need to be shorter than
65 inches, then almost 20% of the population (including lots of women) are still potential
candidates as pilots. All such computations are
understood to be approximate. Very few, if any,
statisticians believe any natural phenomenon
is exactly normally distributed.
Sometimes, the gulf between measure theoretic probability and applied statistics can be very wide.
Computations using R software, where pbinom is
a continuous normal cumulative distribution function, and qnorm is its inverse (called
a quantile function),
qnorm(c(.0005,.9995), 68, 3.5)
[1] 56.48316 79.51684
pnorm(0, 68, 3.5)
[1] 2.212535e-84
diff( pnorm(c(67.5,68.5), 68, 3.5) )
[1] 0.113597
pnorm(65, 68, 3.5)
[1] 0.195683
| 146,467
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- Slingbox Givaway: First off, Dave Zatz (who is employed by Sling Media) is giving away a Slingbox Pro with nothing required by you except your comment on his blog. Read more on how to win this Slingbox at ZatzNotFunny.com.
- Seems to be a lot of rumbling going on at the various Software PVR makers forums/blogs. Snapstream, Vista MCE's Chris Lanier Blog & Charlie Owen's Blog, and SageTV to name a few. Snapstreamers are unhappy about not seeing a new, single interface for TV/PVR, DVD and Music; Chris Lanier and others are unhappy about the lack of true convergence with Microsoft's Media Center Products (the comments on Charlie Owen's Blog are a must read); and SageTV users are lamenting the difficulty in getting encrypted HD. I'll be posting on this in more detail soon as it's very near and dear to me. There really is no Holy Grail of HTPC's even today.
- Halloween is almost here and judging by the amount of traffic I've seen on this blog, many people are on the net searching for last minute Jack-O-Lantern Patterns/Ideas and Costume Ideas. Once my daughter and I carve our pumpkins Halloween is sort of a non-event for us except for the many trick & treaters and trick & drinkers (it's really a little mini-party :) at our house.
- I'll be reviewing the Firefly Remote Control for Home Theater PC's made by Snapstream soon. Haven't started the review, but it's coming.
- Last week I had the honor of meeting with Rakesh Agrawal, one of the founders and the current CEO of Snapstream Media, Inc. during his visit to Kansas City. I'm glad I had the opportunity for this meetup - it was a great experience. I'll be sharing some of what I learned hopefully later this week or early next week at the latest. As many of my readers know, I use Snapstream's Beyond TV and Beyond Media for my Home Theater PC software so I was particularly interested in meeting Rakesh.
- I'm borrowing my dad's handheld GPS device for the next few months. I plan to take it with me to the Caribbean to "play" with it. Gotta love toys :)
- Speaking of trips, I'm trying to figure out a way to make the "Indiana Jones Map Effect" for a vacation video I'll be making for the trip. You know the way they show a line mark on a map to represent the movement from one destination to the next. Anyone have any thoughts on how to do this?
| 79,369
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LAKELAND, Fla. — Brandon..
Detroit April 2.
Zawadzki, who split 2012 with between three triple-A teams, jump-started the Toronto offence with his grand slam off Ryan Robowski to snap a 3-3 tie. Goins then made Melvin Mercedes pay for a pair of walks one inning later.
Zawadzki, 27, had ‘D’.
Right-handed pitcher Anibal Sanchez retired the Jays 1-2-3 in the first and left after two innings without yielding a run, thanks to the Toronto base-running.
The Jays had men on first and second with two outs in the second inning but Gose’s hit was wasted when Sierra was thrown out at third before Lind crossed home plate.
Toronto went ahead in the third, taking advantage of a pair of Shawn Hill walks when Lawrie — with the bases loaded — singled in a pair of runs. Lind then hit a sacrifice fly to make it 3-2.
Cabrera wasted little time pulling Detroit even, with a solo homer off Delabar in the bottom of the third.
Both teams threw in fresh troops in the sixth.
The game marked the official return of John Gibbons as manager. He went 305-305 from 2004-08 at the Jays’ helm.
And while only a spring training affair, it drew the attention of Prime Minister Stephen Harper.
“Good luck to the game Saturday,.
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Why are astronomers fascinated by Uranus? Some have even gone so far as to claim it is the best darn planet in the Solar System. The gringa says, “Hey! What about Earth and Mars? Aren’t they the ones NASA is making such a fuss over? Trying to save one and explore the other?”
Well, astronomical appreciation for Uranus is because it is just so bizarre. Bizarre, huh? Like, odd rainbow colored creatures with spiny noses and squishy springs for appendages and gumballs for tails? Well, no, not quite that bizarre.
Uranus is bizarre because, apparently, it’s a bit lazy. You see, other planets spin around on their axis, or axi, the gringa’s not quite sure about the plural spelling of axis, but you know what I mean. To get back to the point, yes, Uranus is lazy. It does not spin on its axis like other planets. It lays on its side.
Another oddity is that, even though it is not the farthest planet from the Sun, it is the coldest. Perhaps that’s because it’s so darn lazy. It’s never up, spinning around creating friction and heat and all that good, heat-generating stuff that movement creates.
Also, Uranus is confused and misplaces things, things like its magnetic field. Its magnetic field is NOT where it’s supposed to be. Uranus is kind of like a teenager. It lays about doing a lot of nothing and is messy, laying other things about where they are not supposed to be.
One more thing that makes Uranus like a teenager is its greenish atmosphere. It’s moody. It vacillates between dull boredom and doing absolute nothing to crazy business.
Uranus also has an identity crisis. You know how a parent names a child, like, say, the gringa named her eldest son, Zachary, then he goes off to school and engages in some mild rebellion to assert his independence and comes home with a name like, say, Milkshake? Yes, that’s Uranus, too. While all the other planets were named after Roman gods, Uranus had to go and be different and have a name after a Greek god instead, Ouranos, the sky father, who beget Saturn (aka Cronus) and Jupiter (aka Zeus).
Another aspect of Uranus is, because of its laziness and slow motion movement, for a very long time it was thought to be a star. It wasn’t until 1781 that Sir William Hershel discovered that it was actually a planet. Poor Uranus, so misunderstood and underestimated.
If we chose to colonize Uranus instead of Mars, life would be rather odd living on a sideways planet. Summer would last for twenty years without a single sunset and winter would be just as long, spent entirely in darkness. The gringa would surely go mad.
It takes the planet over eighty years to orbit the sun. Surrounded by its 27 moons (that we know about) and ringed about by 13 circlets of rock and spacedust, Uranus plods along at its own snail’s pace. And those moons and rings are just as odd as the planet they surround.
One ring is made up completely of spacedust astronomers think came from the moon named Mab when it was hit by asteroids. Another ring has simply disappeared since the last image received while another moved about and is now somewhere else. But, perhaps the strangest ring of all is the one that “breathes”. Every few hours it expands and then contracts throughout a five kilometer difference. Now that’s just weird.
The moons don’t just orbit Uranus but seem to engage in a dance. They are not considered stable because they are constantly pushing and pulling one another with their different gravities. Scientists expect a few will eventually crash into one another and then who knows what kind of changes will develop. Maybe the planet will get another ring or two.
And with an atmosphere of hydrogen, helium, methane, ammonia and hydrogen sulfide, the gringa’s pretty sure it would be a very unpleasant place to set up household. Everyone would speak ridiculously, no more opera and musicals to appreciate. The air would also smell like a big fart, everywhere, and your eyes would sting and tear. There are also storms with winds over 100 mph that can last for years.
However, one oddity that the gringa thinks may just make all that nastiness about the stinky, unpleasant air worth the sacrifice is what scientists think about the “oceans” of Uranus. Underneath those thick, smelly gas clouds there could be an “ocean” of liquid diamonds! For heaven’s sake! You don’t say?! The gringa has just GOT to know if this is true! Can you imagine! If it is, every single woman I know who loves sparkly things will be on the first commercial rocket, no matter the cost. Goodbye Earth!
So, what the heck happened to Uranus? What got it knocked off its axis? Some experts theorize a large moon, that is long since extinct, had a powerful gravitational pull that overpowered the planet. Others consider that perhaps it had a cosmic collision with something larger than Earth.
Unfortunately, NASA doesn’t expect to dig in to a deeply involved study of this mysterious planet anytime soon. We just don’t have the technology developed that can effectively get an orbiter that far away (almost 2 billion miles) and successfully cope with the instability of all the oddities of Uranus. But, with NASA, the word is never “impossible”. The word is always, well, the two words are always, “not yet”. So, the gringa hopes somewhere there are some NASA scientists as incredibly curious as herself and are being some Johnnies-on-the-spot getting this technology developed. I just have to know more!
Source & Photo Credit:
One thought on “Uranus – The Teenager Planet”
Fun reading this, teenager seems apt: what a moody unpredictable planet! Lots of moons sounds so romantic, and think what a fistfull of diamonds could do for our lives! But the farty smell is rather off-putting.
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\begin{document}
\title{Infinitesimal Legendre symmetry in the Geometrothermodynamics programme}
\author{D. Garc\'ia-Pel\'aez}
\email{dgarciap@up.edu.mx}
\affiliation{Instituto de Ciencias Nucleares, Universidad Nacional Aut\'onoma de M\'exico,\\
A.P. 70-543, 04510 M\'exico D.F., M\'exico}
\affiliation{Universidad Panamericana,\\ Tecoyotitla 366.
Col. Ex Hacienda Guadalupe Chimalistac, 01050 M\'exico D.F., M\'exico}
\author{C. S. L\'opez-Monsalvo}
\email{cesar.slm@correo.nucleares.unam.mx}
\affiliation{Instituto de Ciencias Nucleares, Universidad Nacional Aut\'onoma de M\'exico,\\
A.P. 70-543, 04510 M\'exico D.F., M\'exico}
\begin{abstract}
The work within the Geometrothermodynamics programme rests upon the metric structure for the thermodynamic phase-space. Such structure exhibits discrete Legendre symmetry. In this work, we study the class of metrics which are invariant along the infinitesimal generators of Legendre transformations. We solve the Legendre-Killing equation for a $K$-contact general metric. We consider the case with two thermodynamic degrees of freedom, i.e. when the dimension of the thermodynamic phase-space is five. For the generic form of contact metrics, the solution of the Legendre-Killing system is unique, with the sole restriction that the only independent metric function -- $\Omega$ -- should be dragged along the orbits of the Legendre generator. We revisit the ideal gas in the light of this class of metrics. Imposing the vanishing of the scalar curvature for this system results in a further differential equation for the metric function $\Omega$ which is not compatible with the Legendre invariance constraint. This result does not allow us to use the regular interpretation of the curvature scalar as a measure of thermodynamic interaction for this particular class.
\end{abstract}
\maketitle
\section{Introduction}
The mathematical nature of thermodynamics has been largely explored over the past few decades. Motivated by the early works on metric formulations for thermodynamics of Rao and Fisher \cite{fisher} together with the later developments in contact geometry applied to thermodynamics of Mrugala \cite{mrugala1, mrugala2}, the central goal of the GTD programme has been to provide a Legendre covariant description of thermal phenomena \cite{quevedo}. However, a consistent way for obtaining Legendre invariant metrics whose associated geometry reveals some physical property has been lacking since the birth of the programme. The difficulty of such a task lies in the fact that \emph{discrete} Legendre transformations do not form a group. A summary of the various challenges for the formalism can be found in \cite{conformal,conformalII}.
The metric based programme for geometric thermodynamics presented here considers Legendre symmetry as a fundamental building block for the formalism. From a physical point of view, this is consistent with the fact that in an experiment certain thermodynamic quantities may not be suitable for direct measurement, e.g. in a system with two degrees of freedom represented by the molar internal energy, where the molar entropy and volume are the relevant variables. In such cases, one can use `conjugate' variables, e.g. temperature and pressure, which are easier to control without changing the physical content of the conclusions one can reach from such experiment. This is an indication that thermodynamics is independent of the potential used to describe a particular phenomena. Therefore, in any metric theory of thermodynamics, if one assigns some physical significance to quantities derived from the metric tensor, one should demand Legendre covariance. In this respect, the GTD programme has been successful.
A natural way to set up a geometric theory of thermodynamics has been to endow the working space with a metric. It is at this point that different geometric programmes for thermodynamics diverge. On the one hand, there is the fluctuation based formalism of Ruppeiner \cite{rupp}, where the metric tensor and its associated line element account for the probability of fluctuations. On the other hand, the GTD programme equips the thermodynamic phase-space (c.f. Section \ref{secII}) with a Legendre invariant metric which is carried over to the equilibrium space by means of an embedding map. This guarantees that the curvature of the equilibrium space is also preserved by a Legendre transformation, that is, independent of the thermodynamic potential. However, its relationship with thermodynamic fluctuation theory is a bit less clear. Finally, let us note that the metrics used within the GTD programme, albeit being manifestly invariant under discrete Legendre transformations, they are not invariant along the orbits of the generating vector field of such transformations. This should be contrasted with the case of Special Relativity, where the flat Minkowski metric is Lorentz invariant and is carried along the orbits of the infinitesimal generators of the Lorentz group. \cite{stewart}
In this work, we explore the infinitesimal generators of the Legendre transformations and construct a family of metrics whose isometry group contains these generators. This turns out to be a very strong constraint on the possibilities for Legendre invariant metrics. In particular, none of the metrics used within the GTD programme satisfies such a constraint.
We solve the system defined by the Killing equation associated with the generators of Legendre transformations for an arbitrary metric in the thermodynamic phase-space with the sole restriction of being a K-contact, i.e. that the Reeb vector field associated with the choice of contact form generates an extra symmetry. We perform the analysis by considering a system with two thermodynamic degrees of freedom which can be easily extended to the general case.
The paper is organised as follows. In section \ref{secII}, we provide a brief introduction to the geometry of thermodynamics and the general framework of the GTD programme. In section III, we introduce the contact Hamiltonians as the generating functions for the infinitesimal generators of Legendre transformations. This approach to Legendre symmetry in thermodynamics has been largely studied by Rajeev \cite{rajeev}. In section IV, we present a class of metrics whose isometries are given by infinitesimal Legendre transformations. We note that these metrics are \emph{conformally} associated with the contact structure of the thermodynamic phase space. Additionally, we show that none of the metrics used in the GTD programme belongs to this class. We explore the geometry of the space of equilibrium in the light of the induced metrics for this class and observe that we can no longer use the traditional interpretation for the curvature scalar. Finally, in section V we provide some closing remarks.
\section{The geometric structure of thermodynamics}
\label{secII}
Let us consider the standard set-up (c.f. reference \cite{conformal}). To geometrise a thermodynamic system with $n$ degrees of freedom, we use a $2n+1$ dimensional contact manifold -- the thermodynamic phase-space $\mathcal{T}$ -- whose coordinates represent the thermodynamic variables. This allows us to take into account the First-Law of thermodynamics in a natural way through the maximal Legendre sub-manifold -- the equilibrium space $\mathcal{E}$ -- which is uniquely determined once the thermodynamic fundamental relation is known, i.e. when the thermodynamic potential has been fully specified.
Let us choose a contact 1-form field -- $\eta \in T^*\mathcal{T}$ -- whose elements belong to the class generating the contact structure of the phase-space at each point. Such a field satisfies the non-integrability condition
\beq
\eta \wedge \left(\d \eta \right)^n \neq 0.
\eeq
Contact manifolds have the property that, in every patch $\mathcal{U}\subset\mathcal{T}$ there is a set of local coordinates --Darboux's coordinates $\{\Phi,q^a,p_a \}$, with $a = 1...n$ -- such that at every point $x \in \mathcal{U}$ the contact 1-form can be written as
\beq
\eta = \d \Phi - p_a \d q^a,
\eeq
where the Einstein's sum convention have been used. We will follow such convention unless otherwise stated.
The space of equilibrium states is the maximal dimensional integral sub-manifold embedded in $\mathcal{T}$ satisfying the isotropic condition
\beq
\varphi^*(\eta) = 0.
\eeq
Here, the map $\varphi:\mathcal{E} \rightarrow \mathcal{T}$ is called a Legendre embedding. In Darboux coordinates, this is equivalent to the First-Law since
\beq
\varphi^*(\eta) = \varphi(\d \Phi - p_a \d q^a) = \left(\frac{\partial \Phi}{\partial q^a} - p_a \right) \d q^a = 0.
\eeq
The last equality serves as the definition of the conjugate variables to the $q^a$'s through
\beq
\label{equil}
\Phi= \Phi(q^a) \quad \text{and} \quad p_a = \frac{\partial \Phi}{\partial q^a}.
\eeq
This is simply the well known statement that the fundamental relation $\Phi(q^a)$ completely specifies the thermodynamic system and, thus, the emebedding $\varphi$.
Additional to the contact structure, the GTD programme endows the thermodynamic phase-space with two families of metrics (c.f. section III in \cite{conformal}),
\beq
\label{quevedoGII}
G_{\rm T} = \eta \otimes \eta + \Omega \left (\xi^a_{\ b}q^b p_a \right) \left(\chi^c_{\ d}\ \d q^d \otimes \d p_c \right)
\eeq
and
\beq
\label{quevedoGIII}
G_{\rm P} =\eta \otimes \eta +\Omega \sum_{i=1}^n \left[\left(q^i p_i \right)^{2k+1} \d q^i \otimes \d p_i \right].
\eeq
Both have the discrete Legendre transformations as isometries, the former is only invariant under \emph{total} Legendre transformations and the latter is also invariant under partial Legendre transformations. We will clarify the role of Legendre symmetry in the GTD programme in the next section. Here, $\Omega$ is a Legendre invariant function of the Darboux coordinates, $\xi^a_{\ b}$ and $\chi^a_{\ b}$ are $n \times n$ diagonal arrays of numbers providing the algebraic structure of each family and $k$ is an integer.
These families have been tested in various situations where the thermodynamic behaviour of a particular system is fully specified, allowing us to interpret the manifestly Legendre invariant curvature scalar associated with the induced metric in $\mathcal{E}$ as a ``measure of thermodynamic interaction'' (c.f. Quevedo \cite{quevedo}).
\section{Legendre Transformations and Contact Hamiltonians}
Legendre transformations are particular examples of contact diffeomorphisms, i.e. symmetries of the contact structure of the thermodynamic phase-space. Notice that a transformation of $\mathcal{T}$ which leaves invariant its contact structure necessarily preserves the equilibrium space, since the tangent bundle of a Legendre sub-manifold is completely contained in the contact structure of $\mathcal{T}$ (c.f. Arnold \cite{arnold}). In this sense, a contact diffeomorphism of the thermodynamic phase-space, is a symmetry compatible with the First-Law.
In thermodynamics, a Legendre transformation exchanges the role of the extensive and intensive variables of a given fundamental function $\Phi(q^a)$. Consider the map $\phi:\mathcal{T} \rightarrow \mathcal{T}$ which leaves the contact structure invariant, that is, such that
\beq
\phi^*\eta = f\eta \quad \text{where} \quad f:\mathcal{T}\rightarrow \mathbb{R}^+.
\eeq
The requirement that the function $f$ to be strictly positive simply means that we are considering only diffeomorphisms which preserve the orientation of the contact structure.
Let us call a differentiable function $h:\mathcal{T} \rightarrow \mathbb{R}$ a contact Hamiltonian. The Hamiltonian vector field generated by $h$ is defined through the relation
\beq
h = \eta\left[X_h\right].
\eeq
In local Darboux coordinates it takes the form \cite{arnold}
\beq
\label{generic.ham}
X_h = \left( h - p_a \frac{\partial h}{\partial p_a}\right)\frac{\partial}{\partial \Phi} + \left(\frac{\partial h}{\partial q^a} + p_a \frac{\partial h}{\partial \Phi}\right)\frac{\partial }{\partial p_a} - \left(\frac{\partial h }{\partial p_a} \right)\frac{\partial }{\partial q^a}.
\eeq
Note that for every contact 1-form in the class defining the contact structure, there is a unique vector field $R_\eta$ such that
\beq
\eta[R_\eta] = 1 \quad \text{and} \quad \d\eta[R_\eta] = 0.
\eeq
The vector field $R_\eta$ is called the Reeb vector field associated with the contact form $\eta$, and it is generated by the Hamiltonian $h=1$. In local Darboux coordinates, the Reeb vector field is simply
\beq
\label{reeb}
R_\eta = \frac{\partial}{\partial \Phi}.
\eeq
Consider the contact Hamiltonian given by
\beq
\label{hamiltonian}
h = \sum_{i=1}^n\frac{1}{2} \left({q^i}^2 + {p_i}^2 \right),
\eeq
then, its associated vector field is
\beq
\label{XL}
X_{L} = \left[\sum_{i=1}^n\frac{1}{2} \left({q^{i}}^2 - {p_{i}}^2 \right) \right]\frac{\partial}{\partial \Phi} + q^{a} \frac{\partial }{\partial p_{a}} - p_{a} \frac{\partial}{\partial q^{a}}.
\eeq
It is straightforward to obtain the flow of $X_L$. To acquire an intuition of the orbits of the contact Hamiltonian \eqref{XL}, let us consider the simpler generating Hamiltonian
\beq
\label{hams}
h_i = \frac{1}{2} \left({q^i}^2 + {p_i}^2\right).
\eeq
Note that the Hamiltonian \eqref{hamiltonian} is simply the sum over $i$ of the individual functions \eqref{hams} and, therefore, the Hamiltonian vector field \eqref{XL} will be simply the sum of each Hamiltonian vector field
\beq
X_{L_i} = \frac{1}{2} \left({q^{i}}^2 - {p_{i}}^2 \right)\frac{\partial}{\partial \Phi} + q^{i} \frac{\partial }{\partial p_{i}} - p_{i} \frac{\partial}{\partial q^{i}} \quad \text{(no sum over $i$)}.
\eeq
To find the contact transformation generated by $X_{L_i}$, we need to integrate the flow
\beq
\frac{\d }{\d t} Z^A_i = X_{L_i}^a,
\eeq
where the $Z^A$ are the Darboux coordinates of $\mathcal{T}$. Writting up explicitly the equations for the coordinate transformation we have
\begin{align}
\label{z1}
\frac{\d}{\d t} \Phi_{(i)} &= \frac{1}{2} \left[ {q^{i}}^2 - {p_{i}}^2\right],\\
\label{z2}
\frac{\d }{\d t} p_{i} &= q^{i},\\
\label{z3}
\frac{\d }{\d t} q^{i} &= -p_{i}.
\end{align}
These are the contact equivalent to Hamilton's equations. The last two, equations \eqref{z2} and \eqref{z3}, can be integrated immediately to obtain
\begin{align}
p_{(i)}(t) & = p_{(i)} \cos(t) + q^{(i)} \sin(t),\\
q^{(i)}(t) & = -p_{(i)} \sin(t) + q^{(i)} \cos(t),
\end{align}
and substituting these into \eqref{z1}, we find
\beq
\Phi_{(i)}(t) = \frac{1}{2} \left[ {q^{i}}^2 - {p_{i}}^2\right] \sin(t)\cos(t) - p_{(i)} q^{(i)} \sin^2(t) + \Phi \quad \text{(no sum over $i$)}.
\eeq
Note that the value of $\Phi_{(i)}(t)$ is not fixed along the orbit of $X_{L_i}$ (see figure \ref{fig1}). Given an initial condition on the $q^i$ axis where the initial value $\Phi(0) = \Phi$ is a constant, there are exactly four points along the orbit of $X_{L_i}$ with the same value of $\Phi$, each corresponding to a $n \pi/2$ rotation in the $\left(q^{i},p_{i}\right)$-plane. Thus, we see that a $\pi/2$ rotation yields the partial Legendre transformation interchanging the $i$th pair of conjugate variables,
\begin{align}
\label{transp}
\tilde \Phi_{(i)} \equiv & \Phi_{(i)}\left(\frac{\pi}{2} \right) = \Phi - p_i q^i \quad \text{(no sum over $i$)},\\
\tilde p_i \equiv & p_i\left(\frac{\pi}{2} \right) = q^i,\\
\tilde q^i \equiv & q^i\left(\frac{\pi}{2} \right) = - p_i.
\end{align}
In the rest of this work, we will consider the generator for the total Legendre transformations, which interchanges every pair of conjugate variables and the expression for the Legendre transformed potential $\tilde \Phi$, equation \eqref{transp} above, will have an implicit sum over all the indices $i$.
\begin{figure}
\includegraphics[width=0.35\columnwidth]{figure1}\hskip1cm\includegraphics[width=0.45\columnwidth]{figure2}
\caption{Orbits of the infinitesimal Legendre symmetry generator $X_{L_i}$. On the left, we show the projected orbits on the $(q^i,p_i)$-plane for the initial conditions $(2,0,0)$, $(1,0,0)$ and $(1/2,0,0)$. On the right, we show the shape of the orbits of $X_{L_i}$ on the space $(q^i,p_i,\Phi)$ for the same initial conditions}
\label{fig1}
\end{figure}
\section{Infinitesimal Legendre Isometries}
The aim of this work is to obtain a class of metrics which remain invariant along the flow of the Hamiltonian vector field $X_L$ generating total Legendre transformations. For simplicity, let us consider the particular case with two degrees of freedom. All the results in this section can be easily extended to the general case and shall be presented elsewhere. Thus, consider the metric for the five-dimensional thermodynamic phase-space
\beq
\label{metric}
G = G_{AB}(Z^C)\ \d Z^A \otimes \d Z^B,
\eeq
where the capital indices are taken over the set $\{1 ... 5\}$. As before, we work in \emph{ordered} Darboux coordinates, therefore, $Z^A \in \{\Phi,q^a,p_a \}$ for each $A$.
From the symmetry of the metric tensor, there are 15 independent components in \eqref{metric} that must be determined by solving the Killing equation
\begin{equation}
\dlie_{X_L} G = 0\,.
\label{legendresymm}
\end{equation}
This system of equations has a freedom along the direction of $\Phi$ which can be removed once a fundamental representation is chosen (c.f. reference \cite{conformal}). Thus, choosing a contact 1-form -- $\eta$ -- in the class generating the contact structure, together with its associated Reeb vector field $R_\eta$ [c.f equation \eqref{reeb}, above], we see that $\eta$ is dragged along the orbits of the Reeb flow. We will demand that the metric \eqref{metric} shares this symmetry. A metric satisfying the equation
\beq
\lie_{R_\eta} G = 0,
\eeq
is called a $K$-contact. This imposes the restriction on the metric components of being independent of the coordinate $\Phi$. Therefore, instead of solving the general Legendre-Killing system, equation \eqref{legendresymm}, we will find the family of $K$-contact metrics with Legendre symmetries obtained from
\beq
\label{kill2}
\lie_{X_L} \mathcal{G} = 0,
\eeq
where the $K$-contact $\mathcal{G}$ has the form
\beq
\label{metric2}
\mathcal{G} = \mathcal{G}_{AB}(q^a,p_a)\ \d Z^A \otimes \d Z^B.
\eeq
The form of the metric $\mathcal{G}$ remains quite general. Following the standard formulation of contact metric manifolds \cite{blair}, a contact metric is typically written in blocks. Thus, let us assume that the structure of $\mathcal{G}$ is of the form
\beq
\label{metric2.5}
\mathcal{G} = \eta \otimes \eta + 2 \Omega_a^{\ b}(q^c,p_c) \ \d q^a \otimes \d p_b,
\eeq
where we have fixed six out of fifteen independent components of \eqref{metric2} and we only need to determine the nine remaining functions $\Omega_a^{\ b}(q^c,p_c)$. This form of the metric is very restrictive with respect to the system \eqref{kill2}, and yields a unique class of solutions of the form
\beq
\label{metric3}
\mathcal{G} = \eta \otimes \eta + 2 \Omega(q^c,p_c)\ \epsilon_a^{\ b}\ \d q^a \otimes \d p_b,
\eeq
where $\Omega(q^a,p_a)$ is a non-vanishing, manifestly Legendre invariant function, i.e. it is a non-trivial solution to the constraint
\beq
\label{constraint}
\{h, \Omega\} \equiv \cp{a}\prtf{\Omega}{\cx{a}}-\cx{a}\prtf{\Omega}{\cp{a}} =0 \,,
\eeq
and $\epsilon_a^{\ b}$ are the components of the symplectic matrix
\beq
\epsilon = \left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array} \right].
\eeq
This result is relevant since it excludes all the metric tensors used in the GTD programme. The similarity of \eqref{metric3} with the GTD metrics, equations \eqref{quevedoGII} and \eqref{quevedoGIII}, is deceiving since the arrays $\xi^a_{\ b}$ and $\chi^a_{\ b}$ are diagonal, whereas $\epsilon_a^{\ b}$ is not. Thus, one can question the invariance of the GTD metrics altogether. One can see here that discrete invariance does not imply infinitesimal invariance, whilst the converse is true. Indeed, the GTD metrics change along the orbits of $X_L$ but recover their initial value every $\pi/2$ rotation.
This structure yields results which are significantly different with those stemming from the standard metric approaches for geometric thermodynamics.
\subsection{Thermodynamic interpretation: Revisiting the ideal gas}
Let us consider the equilibrium space as defined by the relations \eqref{equil}. In this case, the induced metric -- $g = \varphi^*\mathcal{G}$ -- is
\beq
g = 2\ \Omega\ \epsilon_a^{\ b}\ \Phi_{,bc} \ \d q^a \otimes \d q^c,
\eeq
where $\Phi_{,ab}$ stands for the second derivatives of the thermodynamic potential $\Phi$ with respect to the coordinates of $\mathcal{E}$.
In the particular case of the molarised ideal gas in the entropy representation, let us choose $q^1 = u$ the molar energy and $q^2 = v$ the molar volume. The potential is written as
\beq
s(u,v) = c_{v} \ln\left(u \right) + \log(v).
\eeq
Thus, the metric takes the simple, non-diagonal form
\beq
g_{{\rm ig}}= 2\ \Omega \left(\frac{c_v}{u^2} - \frac{1}{v^2} \right)\ \d u \otimes \d v,
\eeq
and its determinant is given by
\beq
\det\left(g_{\rm ig} \right) = -\Omega^2 \left[\frac{\left(c_v v^2 - u^2\right)^2}{\left(u v\right)^4} \right].
\eeq
It is straightforward to compute the curvature scalar in terms of the equilibrium space coordinates $u$ and $v$. However, it is more convenient to write it up in terms of the energy density $\rho = u/v$ so that
\beq
\label{RS}
R_{\rm ig} = \frac{2 \rho^2}{\Omega^3} \left[\frac{v^2 \left(\Omega \Omega_{,uv} - \Omega_{,u} \Omega_{,v} \right)}{\rho^2 - c_v} + \frac{4 \Omega^2 c_v \rho}{\left(\rho^2 - c_v\right)^3} \right].
\eeq
Note that if we demand that such a scalar vanishes, we have an extra equation for $\Omega$, apart from \eqref{constraint}. However, such a system is inconsistent. Moreover, assuming that the Legendre invariant function $\Omega$ (as a function of $\rho$) is regular for values of $\rho>0$, we observe that the curvature scalar \eqref{RS} is singular when the value of the energy density coincides with the square root of the heat capacity. Beyond that point, i.e. for values of $\rho>\sqrt{c_v}$, the curvature scalar decays very rapidly to zero. Thus, for this type of metrics, one can no longer keep the interpretation associating the curvature of the space of equilibrium states as a measure of thermodynamic interaction without any restriction, as it is normally done in every other metric programme for geometric thermodynamics. It is interesting that the curvature of the class of metrics we presented here exhibits a limit of applicability, presumably related with the thermodynamic limit itself (see figure \ref{fig2}).
\begin{figure}
\includegraphics[width=0.45\columnwidth]{figure3}
\caption{Scalar curvature of the space of equilibrium states for the ideal gas. Note that if the metric function $\Omega$ is constant, the scalar curvature depends only on one parameter, the energy density $\rho$. Here we have used the heat capacity for a monatomic gas, $c_v=3/2$, and $\Omega=1$.}
\label{fig2}
\end{figure}
\section{Conclusions}
In this work we have obtained a class of metrics for the thermodynamic phase-space $\mathcal{T}$ that are carried along the orbits of the infinitesimal generator of the total Legendre transformations in the case of two thermodynamic degrees of freedom. We have restricted the analysis to metrics which are fully compatible with the choice of one-form $\eta$ in the class generating the contact structure of $\mathcal{T}$. We do this by demanding that the metric has the Reeb vector field associated with $\eta$ as a symmetry. Such metrics are called $K$-contacts. The Legendre-Killing system resulting from \eqref{kill2} is a set of fifteen independent first order partial differential equations. By imposing the usual block decomposition for the metric of Riemannian contact manifolds [c.f. equation \eqref{metric2.5}], we fix six of the fifteen unknowns and the remaining nine are solved to yield a unique class of solutions of a single manifestly Legendre invariant function $\Omega(q^a,p_a)$.
The solution obtained here, excludes all the metrics used in various programmes of geometric thermodynamics. This should be expected since, in general, discrete invariance does not imply infinitesimal symmetry. However, the class presented here stems from a differential system that probes the geometric structure of the thermodynamic phase-space consistently with the choice of contact 1-form.
We revisited the space of equilibrium states for the ideal gas. We observed that we can no longer interpret the curvature of $\mathcal{E}$ as a measure of thermodynamic interaction since such requirement is inconsistent with the constraint that the metric function $\Omega$ should satisfy. Moreover, rewriting the scalar curvature in terms of the energy density of the system, it becomes singular when the numerical value of the energy density coincides with the square root of the heat capacity at constant volume. This might be an indication of the domain of applicability of the formalism in terms of the thermodynamic limit. This issue remains to be explored.
\section*{Acknowledgments}
DGP was funded by a CONACYT Scholarship. CSLM acknowledges financial support of a DGAPA-UNAM Post-doctoral Fellowship. The authors are thankful to Alessandro Bravetti and Francisco Nettel for helpful comments.
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It is a total honor to be featured on APhotoEditor.com I met with Jonathan Blaustein at PhotoNOLA a couple of weeks ago and got some pretty hard core criticism as I showed him images from my series, “Collect Moments Not Things”. I’m not sure anyone’s ever been so directly honest about my work. With a little time, I realized that as much as it hurt to hear his opinion, it was also some of the best advice I’ve ever gotten. Thankfully, at the end of our conversation, he gave me the opportunity to come home and send him a re-edit of 8 images that might be more tightly connected and I took that challenge to heart. So I’m pretty excited to have my work featured on A Photo Editor today.
You can read the entire post here.
| 141,496
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Thesaurus
Synonyms and Antonyms of odd ducks
- That old man's an odd duck who keeps his extravagant holiday lighting display up all year long.
characters, codgers, cracks, crackbrains, crackpots, cranks, eccentrics, flakes, fruitcakes, head cases, kooks, nuts, nutcases, nutters [British slang], oddballs, oddities, originals, quizzes, screwballs, weirdos, zanies
Seen and Heard
What made you want to look up odd ducks? Please tell us where you read or heard it (including the quote, if possible).
| 153,185
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TITLE: Closed immersions are stable under base change
QUESTION [22 upvotes]: My question can be summarized as:
I want to prove that closed immersions are stable under base change.
This is exercise II.3.11.a in Hartshorne's Algebraic Geometry. I researched this for about half a day. I consulted a number of books and online notes, but I found the proofs to be vague. Vakil's notes (9.2.1) hint that this is an immediate consequence of the canonical isomorphism $ M/IM\cong M\otimes_{A}A/I $, and so does Liu's book (1.23). The proof can't be this simple. I need to show that this can be reduced to the affine case, and to do so, I need to show that a closed immersion into an affine scheme is affine. I haven't been able to do so yet. Edit: I found a proof for this later but I still don't know how to use it to prove the question above.
As for the Stacks project, I had to traverse a tree propositions, until I eventually found a proof that uses concepts like quasi-coherent sheaves, something not introduced in Hartshorne's book at this point.
I also consulted Gortz & Wedhorn. The book cites section 4.11 as a proof for this. However the section is a general introduction to the categorical fibred product. It's unrelated. In fact the official errata mentions this error and cites proposition 4.20 instead. It is unclear to me how the proposition shows the result. I suspect the book uses a different - but equivalent - definition.
At this point, I'm frustrated. I'm self-studying and don't have anybody to ask. Could somebody please be kind enough to show me a self-contained proof?
REPLY [0 votes]: In the below, we consider a criterion for stability of base change that works in this exercise.
Proof of this criterion using abstract non-sense is given at the end.
Criterion for Stability under Base Change
Let $\mathbf{P}$ be a property that one can talk about on morphisms of schemes.
Suppose the following holds :
Given $f:X\to Y$, for an open subset $U$ of $Y$, if $f$ has property $\mathbf{P}$, so does the base change of $f$ along the inclusion of the open subscheme $U\subset Y$.
Given $f:X\to Y$, for an open cover $\{U_i\}_i$ of $Y$, if the base change of each $f$ along the inclusion of the open subscheme $U_i\subset Y$ has property $\mathbf{P}$, so does $f$.
then $\mathbf{P}$ is stable under base change iff $\mathbf{P}$ is stable under base change along morphisms with target and domain between affine schemes.
Solution
Closed immersions clearly satisfy condition 1., 2. : on the level of topological spaces, things are clear; on the level of sheaves, use stalks to check epimorphicity.
In Hartshorne's Exercise 3.11(b), closed immersions with target an affine scheme are characterized as the $\operatorname{Spec}$ of some $A\to A/\mathfrak{a}$). Therefore, after verifying condition 1., 2. in the above, we have reduced the problem to the computation of the pushout of some span $A\leftarrow B\rightarrow B/I$ (which gives $A\to A/IA$).
Proof of Criterion
Consider $Z\rightarrow X\leftarrow Y$ with pullback $W$, where $Y\rightarrow X$ has property $\mathbf{P}$.
The case $X$ is affine : choose an open affine cover $\{Z_i\}_i$ of $Z$, note that
$$(W\times_{Z}Z_i\to Z_i)\simeq (Y\times_XZ_i\to Z_i)$$
As $X,Z_i$ are affine, condition 2. shows $W\to Z$ has property $\mathbf{P}$.
The general case : choose an affine cover $\{X_i\}_i$ of $X$, let $Z_i=X_i\times_XZ$, we have
$$(W\times_{Z_i}Z)\simeq(Y\times_XX_i)\times_{X_i}Z_i$$
By condition 1., $(Y\times_XX_i)\to X_i$ has property $\mathbf{P}$, and as $X_i$ is affine, this concludes the proof of the criterion.
| 193,468
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Have a great weekend, everyone!
1. Had a little date at the aquarium on Sunday morning.
I liked seeing live versions of Finding Nemo and Happy Feet.
I liked seeing live versions of Finding Nemo and Happy Feet.
2. I love taco nights.
3. Pacific Blue is back! This is easily one of my favorite polishes.
I wrote more on it here if you're interested.
I wrote more on it here if you're interested.
4. Christmas cards mailed. :)
5. I enjoy the Glee cast version of Christmas music on repeat.
See more H54F posts on Lauren's blog!
See more H54F posts on Lauren's blog!
Hi there! I'm Hanna!11 I d=found your LOVELY blog on the hippity hop!!! Now your newest follower!!! HAPPY HOLIDAYS!!!! You can find me anytime at
xooxxo Hanna
those tacos look delish!
I LOVE the glee christmas album!
YAAY for getting your cards mailed!
My nails are a similar color right now too. Love the pop of blue this winter! :)
| 142,244
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Oil prices headed toward $125/barrel: Pickens
WASHINGTON (Reuters) - Crude oil prices are still headed upward and could top $125 a barrel in the near-term, legendary oil investor T. Boone Pickens said on Thursday.
"It will go up," said Pickens, who heads the BP Capital hedge fund with over $4 billion under management. "Oil is moving to a substantially higher level -- say above $125 a barrel."
U.S. crude futures hit a record $115.54 on Thursday. Oil prices have more than quintupled since 2002, propelled higher by soaring demand from emerging economies like China alongside slow increases in global production capacity.
Despite new production from the Canadian oil sands and elsewhere, Pickens said global crude oil production is unlikely to rise above its current rate of about 85 million barrels per day, while global demand will likely hit 87 million bpd in the third quarter of 2008.
Pickens also expects U.S. natural gas prices to rise from current levels near $10 per million British thermal units to $12-$14 this upcoming winter.
Pickens, in Washington on Thursday to deliver a speech about energy at Georgetown University, made more than $1 billion in 2006 by betting on rising oil prices.
Pickens' hedge fund lost over 20 percent in the first three months of 2008 on a bet that oil prices would fall.
Pickens said his fund is now looking for oil and natural gas prices to rise.
"The position is long, not short," he told reporters. "I covered the short position - it was a mistake on my part."
(Reporting by Chris Baltimore;
| 253,501
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TITLE: On quantization of relativistic quantum theory on curved space
QUESTION [4 upvotes]: In his book "Lectures on quantum mechanics", at the end of chapter 3, Dirac states that
"it does not seem possible to fulfill the conditions which are necessary for building up a relativistic quantum theory on curved surfaces".
Is that true?
REPLY [5 votes]: No, it not true (from a pure theoretical point of view, the last word is of Nature as usual).
The point is that Dirac relied his statement upon the principles of QM as known more than 90 years ago. There are other viewpoints nowadays where, for instance, ccrs of position and momentun do not play a crucial role and are obtained as a byproduct in Minkowski spacetime. There is a huge literature on the subject. From Birrell-Davies' textbook of the last century to some modern treatise based on general covariance and locality like this Advances in Algebraic Quantum Field Theory.
However everything depends on what one think are the basic requirements of a quantum theory and there are really many possibilities each leading to different answers, equally respectable (see Charles Francis' answer)... before some experimental check.
| 205,059
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Can You Be Denied Credit Despite Having A Good Credit Score?
Do you have a good credit score? Well, that’s great buddy! By the way, are you planning to opt for a credit card? If yes, you can get a credit card with a suitable interest rate! But the irony is, having a good credit score does not necessarily mean your credit card application will […]
| 24,893
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New Hand Knit Products Just Added - Hats, Caps and Cowls Druid Hill Designs Team Oct 10, 2017 0 Comments Hand Knit Hats, Caps and Cowls. These are one of a kind in most cases, so check them out! Tags: caps cowls handknit hats sewn Shares Leave your comment Comments have to be approved before showing up Post Comment
| 252,834
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How do guitar effects pedals work?
You need two guitar cables for each pedal you buy. One cable to plug into the input of the pedal and another cable to plug into the output of the pedal. Most pedals will label the input and output jacks, but the standard is for the input to be on the right side of the pedal and the output on the left side.
What pedals should every guitarist have?
15 Must-Have Guitar Effects Pedals
- Distortion Pedal.
- Overdrive Pedal.
- Fuzz Pedal.
- Delay Pedal.
- Reverb Pedal.
- Wah Pedal.
- Chorus Pedal.
- Phaser Pedal.
What guitar pedals should I get first?
Which Guitar Pedals Should I Start With?
- Tuner Pedal. It may sound boring, but a tuner pedal is undoubtedly the most important thing to have on your pedalboard. …
- Distortion Pedal. …
- Overdrive Pedal. …
- Wah Pedal. …
- Chorus Pedal. …
- Reverb Pedal. …
- Delay Pedal. …
- Pedaltrain Pedalboards.
Why are my guitar pedals not working?
If both jack leads are not fully inserted, the pedal won’t work. This is a power-saving feature. A pedal will not switch on unless your jack leads are inserted correctly and into the guitar end, otherwise known as the Input. … Make sure both the input (guitar end) and output (amp end) are inserted properly.28 мая 2020 г.
What order should I buy guitar pedals?.
Do you need a noise gate pedal?
Guitar processors usually have this feature in which you can control noise by using the noise gate option but players using pedals/amp distortion would need to get a noise gate pedal to solve this issue. Threshold and Decay are two important components that factor into any noise gate pedal.
Did Jimi Hendrix use pedals?
Hendrix used a variety of pedals, guitars and amps over his career but he is most well known for his iconic white Strat strung upside down, his feedback-inducing fuzz, articulate wah and custom-designed octave pedal alongside the warm wash of his Uni-Vibe..
How many guitar pedals is too many?
there are people that have actually told me THAT is too much. yeah, right. I’ve seen guitarists use 4 times that amount of pedals. look, the bottom line is: if all the pedals on your board are honestly and truly that essential to your sound, and they ALL get used at some point throughout your set, then I say go for it.
Should I get a distortion or overdrive pedal?
If you are a straight up rock/metalhead, your amp isn’t cutting it, you need the high gain; go for the distortion pedal. Want even more gain, get both a distortion and overdrive. Stacking overdrive and distortions can yield some pretty good results if done correctly.
Do I need an overdrive pedal?
An overdrive pedal is great for many situations: when you want to be able to play both clean and overdriven within the same show, or even within the same song without futzing with amp knobs; also great for people with high-headroom amps playing smaller shows or bedroom practicing (or people with clean solid state amps …
How long do guitar pedals last?
3 to 4 years.
Can you leave guitar pedals plugged in?
The answer is yes. While neither guitar nor your equipment can actually wear out, it is possible that you, or someone, will trip on the cable and damage the instrument or the equipment. … There are several reasons why leaving everything plugged in can cause harm to either the guitar or your equipment.
| 285,450
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\begin{document}
\title{A singularity theorem for Einstein--Klein--Gordon theory}
\author{Peter J. Brown}
\author{Christopher J. Fewster}
\author{Eleni-Alexandra Kontou}
\affiliation{Department of Mathematics, University of York, Heslington, York YO10 5DD, United Kingdom}
\begin{abstract}
Hawking's singularity theorem concerns matter obeying the strong energy condition (SEC), which means that all observers experience a nonnegative effective energy density (EED), thereby guaranteeing the timelike convergence property. However, there are models that do not satisfy the SEC and therefore lie outside the scope of Hawking's hypotheses, an important example being the massive Klein--Gordon field.
Here we derive lower bounds on local averages of the EED for solutions to the Klein--Gordon equation, allowing nonzero mass and nonminimal coupling to the scalar curvature. The averages are taken along timelike geodesics or over spacetime volumes, and our bounds are valid for a range of coupling constants
including both minimal and conformal coupling. Using methods developed by Fewster and Galloway,
these lower bounds are applied
to prove a Hawking-type singularity theorem for solutions to the Einstein--Klein--Gordon theory, asserting that solutions with sufficient initial contraction at a compact Cauchy surface will be future timelike geodesically incomplete.\\
\noindent\emph{Dedicated to the memory of S.W.~Hawking}
\end{abstract}
\maketitle
\noindent
\section{Introduction}
\label{sec:introduction}
The conditions under which cosmological models either originate or terminate in a singularity provided an active subject of research in the decades prior to the breakthroughs made by Penrose~\cite{Penrose_prl:1965} and Hawking~\cite{Hawking:1966sx}. Results from that era mainly concern solutions with symmetries, as represented by the survey~\cite{HeckmannSchuecking:1962}. Raychaudhuri's work in 1955 represented a decisive step forward, because he was able to analyse inhomogeneous models using (a forerunner of) the equation that now carries his name. In their general
and modern form~\cite{Ehlers:1993} the Raychaudhuri equations present the
evolution of timelike geodesic congruences in a physically transparent fashion. For the special case of
an irrotational congruence with velocity field $U^\mu$, the expansion $\theta=\nabla_\mu U^\mu$ satisfies
\be
\label{eqn:ray0}
\nabla_U\theta =R_{\mu \nu} U^\mu U^\nu-2\sigma^2 -\frac{\theta^2}{n-1} \,,
\ee
where $n$ is the spacetime dimension, $\sigma$ is the shear scalar and $R_{\mu\nu}$ is the Ricci tensor. Assuming that the
geometry is a solution to the Einstein equations
\be \label{eqn:Einstein}
G_{\mu\nu} = - 8\pi T_{\mu\nu},
\ee
the Raychaudhuri equation \eqref{eqn:ray0} becomes
\be
\label{eqn:ray1}
\nabla_U\theta=-8\pi\rho -2\sigma^2 -\frac{\theta^2}{n-1} \,,
\ee
where
\begin{equation}\label{eq:EED_def}
\rho= T_{\mu \nu} U^\mu U^\nu-\frac{T}{n-2}
\end{equation}
and $T=T^{\mu}_{\phantom{\mu}\mu}$. The quantity $\rho$ has appeared in general relativity
since the works of Whittaker~\cite{Whittaker:1935} and Synge~\cite{Synge:1937}, playing
the role of the mass-energy density in general relativistic versions of the Gauss law; Pirani~\cite{Pirani:2009} likewise identifies it as the `effective density of gravitational mass'.
Here, imputing units of energy rather than mass, we will use the term \emph{effective energy density} (EED) for $\rho$.
Evidently, the sign of $\rho$ is crucial. If $\rho\ge 0$, that is, if the strong energy condition (SEC) holds, then the right-hand side of \eqref{eqn:ray1} is negative, driving $\theta\to -\infty$ in finite proper time. This is incompatible with geodesic completeness and implies the existence of a singularity.
Senovilla~\cite{Senovilla:2014} has described the skeleton of the singularity theorems in terms of a `pattern theorem' with three ingredients. An \emph{energy condition} establishes a focussing effect for geodesics, while a \emph{causality condition} removes the possibility of closed timelike curves and a \emph{boundary or initial condition} establishes the existence of some trapped region of spacetime. The goal of the singularity theorems is to show that the spacetime contains at least one incomplete causal geodesic;
we will divide singularity theorems into `Hawking-type' and `Penrose-type',
depending on whether they demonstrate timelike or null geodesic incompleteness respectively. While Hawking-type results are based on the SEC, Penrose-type results assume the null energy condition (NEC), $T_{\mu\nu}k^\mu k^\nu\ge 0$ for all null $k^\mu$.
Hawking wrote that `[the energy conditions] are properties that any normal matter should have'~\cite[\S 5]{Hawking:1966sx} and indeed many models do respect the SEC. However, not all do, and in fact the massive minimally coupled Klein--Gordon field obeying $(\Box+m^2)\phi=0$ has EED
\begin{equation}
\rho = (\nabla_U \phi)^2 -\frac{m^2\phi^2}{n-2},
\end{equation}
which is easily made negative at individual points. Similarly, it is easily seen that the SEC and NEC fail for the nonminimally coupled Klein--Gordon field. This situation is exacerbated in quantum field theory, in which none of the pointwise energy conditions can hold~\cite{Epstein:1965zza}. We refer the reader to recent reviews of energy conditions \cite{Curiel:2017,MartinMorunoVisser:2017}.
For these reasons there has long been interest in establishing singularity theorems under weakened energy assumptions. Examples include \cite{Tipler:1978zz,ChiconeEhrlich:1980,Borde:1987b, Roman:1988vv,Wald:1991xn,Borde:1994}, in which various averages of the energy density or related quantities are
required to be nonnegative if the average is taken over a sufficiently large portion of a (half-)complete causal geodesic, or at least is intermittently nonnegative~\cite{Borde:1987b}.
Our approach in this paper follows~\cite{Fewster:2010gm}, in which (generalising results from~\cite{Galloway:1981}) it was shown among other things that suitable lower bounds on local weighted averages of $\rho$ are sufficient to derive singularity theorems of Hawking and Penrose type, even if $\rho$ is not everywhere positive or has a negative long-term average.
The bounds adopted in~\cite{Fewster:2010gm} were inspired by the Quantum Energy Inequalities (QEIs) that have been established in various models of quantum field theory (see~\cite{Fewster2017QEIs} for a recent review). However, there is a significant gap between the results of~\cite{Fewster:2010gm} and a semiclassical Hawking-type singularity theorem, because there is so far no QEI version of the SEC. The purpose of this paper is to show that the classical nonminimally coupled massive Klein--Gordon field obeys lower bounds on $\rho$ of the type considered in~\cite{Fewster:2010gm}. The general approach is parallel to methods used in~\cite{Fewster:2006ti} to obtain averaged versions of the weak and null energy conditions for the classical nonminimally coupled scalar field. Elsewhere, we will use our results to establish QEI analogues of the SEC (cf.~\cite{Fewster:2007ec}); here, we use them to derive a new Hawking-type singularity theorem for the Einstein--Klein--Gordon system. In a completely different direction, we mention that the methods of~\cite{Fewster:2010gm}, and therefore bounds of the type developed here, could be used in other problems in relativity. See, for example~\cite{Lesourd:2017}, in which a version of Hawking's area theorem is proved under weakened hypotheses.
The paper is structured as follows. In \sec\ref{sec:nonmfield} we recall the energy-momentum tensor for the non-minimally coupled scalar field and the manner in which it can violate the pointwise SEC. Next, in \sec\ref{sec:worldline}, we consider local averages of $\rho$ of the form
\begin{equation}
\int\rho(\gamma(\tau)) f(\tau)^2 \,d\tau,
\end{equation}
where $\gamma$ is a timelike geodesic parameterised by proper time and $f$ is a real-valued smooth and compactly supported function. Here, it not is assumed that the background spacetime and field together solve the Einstein--Klein--Gordon equations.
We derive lower bounds on such averages that depend only the values of $\phi$, but not its derivatives. The bounds also depend on $\gamma$ and $f$ together with its derivatives and are valid for all values of the coupling $\xi$ in the interval $[0,2\xi_c]$ where $\xi_c$ is the conformal coupling constant ($\xi_c=1/6$ for $n=4$). We investigate the behaviour of these lower bounds under scaling
of $f$ and also derive constraints on the time for which $\rho$ can be more negative than some given value. Section~\ref{sec:worldvolume} addresses similar questions for worldvolume averages of $\rho$ obtaining bounds valid on an interval containing $[0,\xi_c]$ for dimensions $n\ge 4$. In the special case of flat spacetime, one
may prove that the average value of $\rho$ over all spacetime is nonnegative.
In \sec\ref{sec:ekg}, we return to worldline bounds, now adapted to the special case of solutions to the Einstein--Klein--Gordon system and obtaining a slightly refined bound, which is used in our discussion of singularity theorems in \sec\ref{sec:singularity}. There, we first establish a Hawking-type singularity theorem using methods taken from~\cite{Fewster:2010gm} and apply it to the Einstein--Klein--Gordon theory using our worldline bounds. This provides an
analogue to the Penrose-type singularity theorem for the nonminimally coupled
scalar field discussed in~\cite{Fewster:2010gm}. Finally, we conclude in \sec~\ref{sec:disc} with
a discussion of the magnitude of the initial contraction needed to ensure
timelike geodesic incompleteness according to our results.
Our sign conventions are the $[-,-,-]$ of Misner, Thorne and Wheeler \cite{MTW}. We write the d'Alembertian with respect to the metric $g$ as $\Box_g = g^{\mu\nu}\nabla_\mu\nabla_\nu$ and work in $n$ spacetime dimensions unless otherwise stated. Except in \sec\ref{sec:disc} we adopt units in which $G=c=1$.
\section{The non-minimally coupled field}
\label{sec:nonmfield}
The field equation for non-minimally coupled scalar fields is
\be \label{eqn:field}
(\Box_g+m^2+\xi R)\phi=0 \,,
\ee
where $\xi$ is the coupling constant and $R$ is the Ricci scalar.
The constant $m$ has dimensions of inverse length, which would be the
inverse Compton wavelength if one regarded~\eqref{eqn:field} as the
starting-point for a quantum field theory with massive particles.
The Lagrangian is
\be
\label{eqn:lagrangian}
L[\phi]=\frac{1}{2} [(\nabla \phi)^2-(m^2+\xi R)\phi^2 ] \,,
\ee
from which the stress energy tensor is obtained by varying the action with respect to the metric, giving
\be
\label{eqn:tmunu}
T_{\mu \nu}=(\nabla_\mu \phi)(\nabla_\nu \phi)+\frac{1}{2} g_{\mu \nu} (m^2 \phi^2-(\nabla \phi)^2)+\xi(g_{\mu \nu} \Box_g-\nabla_\mu \nabla_\nu-G_{\mu \nu}) \phi^2 \,,
\ee
where $G_{\mu \nu}$ is the Einstein tensor. The trace of the stress-energy tensor is given by
\be
\label{eqn:trace}
T = \left(1-\frac{n}{2}\right) (\nabla\phi)^2 + \frac{n}{2}m^2\phi^2 + \xi\left((n-1)\Box_g - \left(1-\frac{n}{2}\right)R \right)\phi^2 \,.
\ee
We should observe here that the field equation and the Lagrangian reduce to those of minimal coupling for flat spacetimes but the stress energy tensor does not. The effective energy density $\rho$ of
Eq.~\eqref{eq:EED_def} obtained from the stress-energy tensor Eq.~(\ref{eqn:tmunu}) for a timelike observer with $4$-velocity $U^\mu$ is
\bea \label{eqn:SED}
\rho &=& (1-2\xi) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi)-\frac{1-2\xi}{n-2} m^2 \phi^2 -\frac{2\xi}{n-2} (\nabla \phi)^2 \nonumber\\
&&\quad -2\xi U^\mu U^\nu \phi \nabla_\mu \nabla_\nu \phi -\xi U^\mu U^\nu R_{\mu \nu} \phi^2+\frac{2\xi^2}{n-2} R \phi^2-\frac{2\xi}{n-2} (\phi P_\xi \phi) \,,
\eea
where $P_\xi=\Box_g+m^2+\xi R$ is the Klein-Gordon operator. The last term can be discarded ``on shell'' i.e. for $\phi$ satisfying Eq.~(\ref{eqn:field}). For $\xi=0$ the EED further reduces to
\be \label{eqn:minSED}
\rho= U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi)-\frac{1}{n-2} m^2 \phi^2 \,.
\ee
From Eq.~\eqref{eqn:minSED} we can see that, even for minimally coupled fields, we find a violation of the SEC at any point in spacetime at which $ m^2\phi^2 \geq (n-2) (\nabla_U\phi)^2$, and the violation can be made arbitrarily large if $m$ or $\phi$ can be made large. We also observe a guaranteed violation whenever the field derivatives vanish, as we are left with a manifestly negative term.
\section{Worldline strong energy inequality}
\label{sec:worldline}
We will study the EED of the stress-energy tensor of Eq.~\eqref{eqn:tmunu} with respect to freely falling observers. Let $\gamma$ be a a timelike geodesic parametrised by proper time $\tau$.
Let $f$ be a real-valued and compactly supported function $f \in C_0^2(\mathbb{R})$. We are interested in expressions of the form
\be \label{eqn:clASED}
\int_\gamma d\tau \,\rho \, f^2(\tau)=\int_\gamma d\tau \left( T_{\mu \nu}\dot{\gamma}^\mu \dot{\gamma}^\nu-\frac{1}{n-2} T \right)f^2(\tau) \,.
\ee
Eq.~(\ref{eqn:clASED}) ``on shell" reduces to
\bea\label{eqn:onshell-clASED}
\int_\gamma d\tau \,\rho\, f^2(\tau)&=&\int_\gamma d\tau \bigg((1-2\xi) (\nabla_{\dot{\gamma}}\phi)^2-\frac{1-2\xi}{n-2}m^2 \phi^2 -\frac{2\xi}{n-2} (\nabla \phi)^2 \nonumber\\
&& \qquad \qquad \qquad \qquad -2\xi \phi (\nabla_{\dot{\gamma}}^2 \phi)-\xi \dot{\gamma}^\mu \dot{\gamma}^\nu R_{\mu \nu} \phi^2+\frac{2\xi^2}{n-2} R \phi^2 \bigg) f^2(\tau) \,.
\eea
From Eq.~(9) of Ref.~\cite{Fewster:2006ti} we have
\be \label{eqn:posdif}
-2\xi \int_\gamma d\tau\, f^2(\tau) \phi \nabla_{\dot{\gamma}}^2 \phi=2 \xi \int_\gamma d\tau [\nabla_{\dot{\gamma}} (f(\tau) \phi)]^2-2\xi \int_\gamma d\tau \phi^2 ( f'(\tau))^2 \,,
\ee
which is a difference of positive terms for positive coupling constant. Additionally we can write
\be
(1-2\xi) (\nabla_{\dot{\gamma}}\phi)^2-\frac{2\xi}{n-2} (\nabla \phi)^2=\left(1-\frac{\xi}{2\xi_c}\right) (\nabla_{\dot{\gamma}}\phi)^2+\frac{2\xi}{n-2} h^{\mu \nu} \nabla_\mu \phi \nabla_{\nu} \phi \,,
\ee
where $h^{\mu \nu}=\dot{\gamma}^\mu \dot{\gamma}^\nu-g^{\mu \nu}$ is a positive definite metric and $\xi_c$ is the conformal coupling constant defined as
\be
\xi_c=\frac{n-2}{4(n-1)} \,.
\ee
Applying the previous two identities to Eq.~\eqref{eqn:onshell-clASED}, we find that all the curvature independent terms are either positive or negative for $\xi \in[0,2 \xi_c]$. As a result, we have proved the following theorem.
\begin{theorem}
\label{the:clline}
Let $\gamma$ be a timelike geodesic parametrized by proper time $\tau$ in $(M,g)$, where $M$ is a manifold with dimension $n\geq 2$. Let $T_{\mu \nu}$ be the stress-energy tensor of a scalar field with coupling constant $\xi \in[0,2\xi_c]$ and $f$ a real valued function of compact support. Then ``on shell''
\bea
\label{eqn:cline}
\int_\gamma d\tau \, \rho\, f^2(\tau)&\geq& - \int_\gamma d\tau \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau)+\xi \bigg(2( f'(\tau))^2+R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu f^2(\tau) \nonumber\\
&& \qquad \qquad - \frac{2\xi}{n-2} Rf^2(\tau) \bigg) \bigg\}\phi^2 \,.
\eea
\end{theorem}
In fact, we have proved a slightly stronger bound that will be useful later on, namely
\bea
\label{eqn:cline2}
\int_\gamma d\tau \, \rho\, f^2(\tau)&\geq& - \int_\gamma d\tau \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau)+\xi \bigg(2( f'(\tau))^2+R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu f^2(\tau) \nonumber\\
&& \qquad - \frac{2\xi}{n-2} Rf^2(\tau) \bigg) \bigg\}\phi^2
+ \int_\gamma d\tau
\left(1-\frac{\xi}{2\xi_c}\right) (\nabla_{\dot{\gamma}}\phi)^2 f^2(\tau)
\,.
\eea
However \eqref{eqn:cline} has the advantage that only the field $\phi$, and not its derivative, appears on the right-hand side.
For flat spacetimes the bound of Theorem \ref{the:clline} becomes
\be \label{eqn:clasflat}
\int_\gamma d\tau \, \rho\, f^2(\tau) \geq - \int_\gamma d\tau \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau) +2 \xi ( f'(\tau))^2 \bigg\}\phi^2 \,,
\ee
while for minimally coupled fields, regardless of the curvature
\be
\int_\gamma d\tau \, \rho\, f^2(\tau) \geq - \frac{1}{n-2}m^2 \int_\gamma d\tau \, f^2(\tau) \phi^2 \,.
\ee
In order to understand the significance of these results, it is useful to discuss
some consequences of the flat spacetime bound Eq.~\eqref{eqn:clasflat}. First,
let us consider its behaviour under rescaling of the smearing function $f$.
Writing $\fmax$ for the maximum field amplitude of the field along the inertial trajectory $\gamma$,
\be
\label{eqn:phimax}
\fmax=\sup_\gamma |\phi | \,,
\ee
Eq.~\eqref{eqn:clasflat} implies
\be
\int_\gamma d\tau \, \rho\, f^2(\tau) \geq - \fmax^2 \int_\gamma d\tau \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau) +2 \xi ( f'(\tau))^2 \bigg\} \,
\ee
for any compactly supported real-valued $f$. Let us now assume that $f$ has unit $L^2$-norm. Introducing the rescaled function
\be
f_{\lambda}(\tau)=\frac{f(\tau/\lambda)}{\sqrt{\lambda}} \,,
\ee
chosen so that its normalization is independent of the choice of $\lambda>0$
\be
\int d\tau f^2_\lambda (\tau)= \int d\tau f^2 (\tau)=1 \,,
\ee
we can write
\be
\int_\gamma d\tau \, \rho \, f^2_\lambda (\tau) \geq - \fmax^2 \int_\gamma d\tau \frac{1}{\lambda} \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau/\lambda) +\frac{2\xi}{\lambda^2} (f'(\tau/\lambda))^2 \bigg\} \,.
\ee
Changing variables to $\tau \to \tau \lambda$ on the right-hand side and taking the limit $\lambda \to \infty$ we get
\be
\liminf_{\lambda \to \infty} \int_\gamma d\tau \, \rho \, f^2_\lambda (\tau) \geq - \frac{1-2\xi}{n-2} m^2 \fmax^2 \,.
\ee
This result may be interpreted as providing a lower bound on the long-term average value of $\rho$, which leaves open the possibility that the long-term average in the case $m>0$ can be negative, even for $\xi=0$. This can be contrasted with analogous results for the null energy condition in~\cite{Fewster:2006ti}, which establish ANEC in an appropriate limit.
A slightly different approach is to estimate the supremum of the EED over an open interval $I$ of proper time with duration $\tau_0$. This gives
\be\label{eq:longterm}
\sup_{\gamma(I)} \rho \geq - \left\{ \frac{1-2\xi}{n-2}m^2+\frac{2\xi \pi^2}{\tau_0^2} \right\} \sup_{\gamma(I)} |\phi|^2\,,
\ee
where use the fact that
\begin{equation}
\inf_f \frac{\|f'\|^2}{\|f\|^2}= \frac{\pi^2}{\tau_0^2}
\end{equation}
where $\|\cdot\|$ denotes the $L^2$-norm and the infimum is
taken over all smooth $f$ with compact support in an interval of length $\tau_0$ (see Ref.~\cite{Fewster:1998xn,Fewster:2006ti} for similar arguments). Thus violations of the
SEC beyond the level of the long term average bound in Eq.~\eqref{eq:longterm} are possible only on timescales $\tau_0\ll \xi^{1/2}m^{-1}$, and not at all if $\xi=0$.
\section{Worldvolume strong energy inequality}
\label{sec:worldvolume}
Instead of averaging the EED over a worldline we can average over spacetime volumes. Let $U^\mu$ be a future-directed timelike unit vector field. Introducing $f(x)$ as a smearing function with compact support, and writing $V^\mu = f(x)U^\mu$, the averaged EED for the nonminimally coupled scalar field is ``on shell"
\bea \label{eqn:ASEDvol}
&& \int dVol \, \rho\, f^2(x)=\int dVol \bigg\{ (1-2\xi) f^2(x) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi) -\frac{1-2\xi}{n-2} m^2 \phi^2 f^2(x) \nonumber\\
&&\qquad \qquad -\frac{2\xi}{n-2} (\nabla \phi)^2 f^2(x)-\xi V^\mu V^\nu (2\phi \nabla_\mu \nabla_\mu \phi+ R_{\mu \nu} \phi^2)+\frac{2\xi^2}{n-2} R\phi^2 f^2(x) \bigg\} \,,
\eea
where $\rho$ is given by Eq.~\eqref{eqn:SED}. From Eq.~(34) of Ref.~\cite{Fewster:2006ti} we have
\bea \label{eqn:posdifvol}
-\xi \int dVol \, V^\mu V^\nu( 2\phi \nabla_\mu \nabla_\nu \phi+R_{\mu \nu} \phi^2)&& =2\xi \int dVol \, [\nabla_\mu (V^\mu \phi)]^2 \nonumber\\
&&-\xi \int dVol \, [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \phi^2 \,,
\eea
which is a generalization of Eq.~(\ref{eqn:posdif}) that was used for the worldline average. We can also write
\bea \label{eqn:posmetr}
(1-2\xi) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi)-\frac{2\xi}{n-2} (\nabla \phi)^2&=&\left(1-2\xi\frac{n-1}{n-2}\right) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi)\nonumber \\
&&\qquad \qquad +\frac{2\xi}{n-2} h^{\mu \nu} (\nabla_\mu \phi) (\nabla_\nu \phi)
\,,
\eea
where $h^{\mu \nu}=U^\mu U^\nu - g^{\mu \nu}$ is a positive definite metric. Now all curvature-independent terms are either positive or negative for $\xi \in[0,2\xi_c]$, and we have the following bound for the averaged EED
\bea \label{eqn:volbound1}
\int dVol \, \rho\, f^2(x) &\geq& -\int dVol \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(x)+\xi [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \nonumber\\
&& \qquad \qquad \qquad -\frac{2\xi^2}{n-2} R f^2(x) \bigg\} \phi^2 \,.
\eea
This bound retains many features of the worldline bound Eq.~\eqref{eqn:cline}. In particular the mass-dependent term $(1-2\xi) m^2 f^2(x)/(n-2)$ appears in both. In the worldline case, this term
prevented us from showing that the long-term worldline average of $\rho$ is positive. For worldvolume averaging, however, we can use the field equation \eqref{eqn:field} along with successive integration-by-parts to derive an alternative bound that has no explicit mass-dependence and remains free from any field derivatives. This is achieved as follows.
The field equation allows us to rewrite the stress-energy tensor as
\be
T_{\mu \nu}=(1-2\xi) (\nabla_\mu \phi)(\nabla_\nu \phi)-\frac{1}{2} (1-4\xi)( \phi \Box_g \phi+(\nabla \phi)^2)-2\xi \phi \nabla_\mu \nabla_\mu \phi-\xi R_{\mu \nu} \phi^2 +\frac{1}{2} g_{\mu \nu} (\phi P_\xi \phi)\,,
\ee
resulting in an alternative expression for the EED,
\bea
\label{eqn:SECbox}
\rho &=& (1-2\xi) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi) +\frac{1-2\xi}{n-2}(\phi \Box_g \phi)-\frac{2\xi}{n-2} (\nabla \phi)^2 \nonumber\\
&& -2 \xi U^\mu U^\nu \phi \nabla_\mu \nabla_\nu \phi-\xi U^\mu U^\nu R_{\mu \nu} \phi^2+\frac{1}{n-2}\xi R \phi^2 -\frac{1}{n-2}(\phi P_\xi \phi) \,.
\eea
Thus, we may write the averaged EED for ``on shell" field configurations, as
\bea \label{eqn:ASED2}
&& \int dVol \, \rho\, f^2(x) =\int dVol \bigg\{ (1-2\xi) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi)f^2(x)+\frac{1-2\xi}{n-2} f^2(x) \phi \Box_g \phi \nonumber\\
&&\qquad \qquad -\frac{2\xi}{n-2} (\nabla\phi)^2 f^2(x)-\xi V^\mu V^\nu( 2\phi \nabla_\mu \nabla_\nu \phi+R_{\mu \nu} \phi^2)+\frac{1}{n-2} \xi R\phi^2 f^2(x) \bigg\} \,.
\eea
Writing
\be
\phi \Box_g \phi=\frac{1}{2} \Box_g \phi^2-g^{\mu \nu} (\nabla_\mu \phi)(\nabla_\nu \phi) \,,
\ee
the EED becomes
\bea \label{eqn:ASED3}
\int dVol \, \rho\ f^2(x) &=&\int dVol \bigg\{ \left( \frac{n-3}{n-2}-2\xi \right) U^\mu U^\nu (\nabla_\mu \phi)(\nabla_\nu \phi)f^2(x) \nonumber\\
&&\qquad \qquad +\frac{h^{\mu \nu}}{n-2}(\nabla_\mu \phi)(\nabla_\nu \phi)f^2(x)+\frac{1-2\xi}{2(n-2)} (\Box_g \phi^2)f^2(x) \nonumber \\
&&\qquad \qquad-\xi V^\mu V^\nu (2\phi \nabla_\mu \nabla_\nu \phi+R_{\mu \nu} \phi^2)+\frac{\xi R}{n-2} \phi^2 f^2(x) \bigg\} \,.
\eea
By integrating by parts we can rewrite the third term of the integral
\be
\int d Vol \, \frac{1-2\xi}{2(n-2)} (\Box_g \phi^2)f^2(x) = \frac{1-2\xi}{2(n-2)} \int d Vol \, (\Box_g f^2(x)) \phi^2 \,,
\ee
where we used the fact that the boundary terms vanish. Using Eq.~(\ref{eqn:posdifvol}) and discarding the positive terms from the bound for $\xi \in [0,\xi_v]$, where
\be
\xi_v= \frac{n-3}{2(n-2)} \,,
\ee
we can write
\bea \label{eqn:volbound2}
\int dVol \, \rho\, f^2(x) &\geq& - \int dVol \bigg\{ -\frac{1-2\xi}{2(n-2)}(\Box_g f^2(x))+\xi [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \nonumber\\
&& \qquad \qquad \qquad \qquad- \frac{1}{n-2} \xi R f^2(x) \bigg\} \phi^2 \,.
\eea
Note that $\xi_v < 2\xi_c$ for any spacetime dimension $n>2$, while $\xi_c<\xi_v$ for $n\ge 4$. Using Eqs.~(\ref{eqn:volbound1},\ref{eqn:volbound2}) we have proved following theorem:
\begin{theorem}
\label{the:clasvol}
If $M$ is a manifold with metric $g$ and dimension $n \geq 3$, $T_{\mu \nu}$ the stress-energy tensor of a scalar field with coupling constant $\xi \in [0,\xi_v]$ and $f$ a real valued function evaluated on a spacetime point $x$ then ``on shell ''
\be
\int dVol \, \rho \, f^2(x) \geq -\min \{ \mathcal{B}_1,\mathcal{B}_2\} \,,
\ee
where
\be
\mathcal{B}_1=\int dVol \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(x) -\frac{2\xi^2 R}{n-2} f^2(x) + \xi \left[(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)\right] \bigg\} \phi^2 \,,
\ee
and
\be
\mathcal{B}_2=\int dVol \bigg\{ -\frac{1-2\xi}{2(n-2)}(\Box_g f^2(x)) - \frac{\xi R}{n-2} f^2(x)+\xi [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \bigg\} \phi^2 \,.
\ee
\end{theorem}
Note that this result is restricted to minimal coupling in $n=3$.
For flat spacetimes the bounds of Theorem \ref{the:clasvol} become
\be \label{eqn:B1flat}
\mathcal{B}_1=\int dVol \left\{ \frac{1-2\xi}{n-2} m^2 f^2(x) +\xi [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \right\} \phi^2 \,,
\ee
and
\be \label{eqn:B2flat}
\mathcal{B}_2=\int dVol \bigg\{ -\frac{1-2\xi}{2(n-2)} (\Box f^2(x)) + \xi [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \bigg\} \phi^2 \,,
\ee
while for minimally coupled fields on any spacetime,
\be
\mathcal{B}_1=\frac{1}{n-2} m^2 \int dVol \, f^2(x) \phi^2 \,, \text{ and } \mathcal{B}_2= -\frac{1}{2(n-2)} \int dVol \, (\Box_g f^2(x)) \phi^2 \,.
\ee
We now investigate the behaviour of Eq.~(\ref{eqn:B2flat}) under rescaling of the smearing function $f$. First let $\fmax$ be the maximum amplitude of the field
\be
\fmax=\sup_M |\phi | \,,
\ee
so we can take it out of the bound, yielding
\be
\label{eqn:fmax}
\int dVol \, \rho\, f^2(x) \geq - \fmax^2 \int dVol \bigg\{ -\frac{1-2\xi}{2(n-2)} (\Box_g f^2(x)) + \xi [(\nabla_\mu V^\mu)^2+(\nabla_\mu V^\nu)(\nabla_\nu V^\mu)] \bigg\} \,.
\ee
(Eq.~\eqref{eqn:fmax} also holds if the supremum in the definition of $\phi_{\max}$ is taken over the support of $f$. However, in order to keep $\phi_{\max}$ constant for all rescaled smearings, we extend its definition to the entire manifold.)
Consider a translationally invariant unit timelike vector field $U^\mu$ and define the rescaled smearing function $f_\lambda$ for $\lambda>0$ to be
\be
f_\lambda(x)=\frac{f(x/\lambda)}{\lambda^{n/2}} \,,
\ee
so that its normalization is independent of the choice of $\lambda$
\be
\int dVol \, f^2_\lambda (x)= \int dVol \, f^2 (x)=1 \,.
\ee
Replacing $f$ by $f_\lambda$, the right-hand side of Eq.~(\ref{eqn:fmax}) becomes
\bea
&&-\frac{1}{4} \int dVol \bigg( -\frac{1-2\xi}{2(n-2)} \Box f^2_\lambda(x)+ \xi [(U^\mu [\nabla_\mu f_\lambda (x)] )^2+(U^\nu [\nabla_\mu f_\lambda(x)])(U^\mu [\nabla_\nu f_\lambda(x)])] \bigg) \fmax^2 \nonumber\\
&&= -\frac{1}{4} \int dVol \frac{1}{\lambda^{n+2}} \bigg( -\frac{1-2\xi}{2(n-2)} \Box f^2(x/\lambda) + \xi [(U^\mu [\nabla_\mu f (x/\lambda)] )^2\\
&&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad+(U^\nu [\nabla_\mu f(x/\lambda)])(U^\mu [\nabla_\nu f(x/\lambda)])] \bigg) \fmax^2 \nonumber \,,
\eea
where we used the fact that $U^\mu$ is translationally invariant and so its derivatives vanish. Changing variables $x \to \lambda x$ gives
\be
-\frac{1}{4} \int dVol \frac{1}{\lambda^2} \bigg( -\frac{1-2\xi}{2(n-2)} \Box f^2(x)+ \xi [(U^\mu [\nabla_\mu f (x)] )^2+(U^\nu [\nabla_\mu f(x)])(U^\mu [\nabla_\nu f(x)])] \bigg) \fmax^2 \,.
\ee
In the limit of large $\lambda$ the bound goes to zero and we have
\be
\liminf_{\lambda \to \infty} \int dVol \, \rho\, f^2_\lambda(x) \geq 0 \,,
\ee
thus establishing an averaged SEC for flat spacetimes. A similar calculation for the $\mathcal{B}_1$ bound gives a weaker, negative, bound in this case.
\section{A worldline inequality for the Einstein--Klein--Gordon system}
\label{sec:ekg}
The inequalities proved in Sections~\ref{sec:worldline} and~\ref{sec:worldvolume}
are valid for solutions to the Klein--Gordon equation on an arbitrary fixed background spacetime.
In this section we discuss how our worldline bound can be adapted to provide
more specific information about solutions to the full Einstein--Klein--Gordon system.
In our discussion it will be important that the field magnitude is constrained below
a critical value. To see why, recall from Eq.~\eqref{eqn:tmunu} that the stress-energy tensor of the nonminimally coupled scalar field contains a term proportional to the Einstein tensor. Therefore the Einstein equations $G_{\mu\nu}=-8\pi T_{\mu\nu}$ can be rearranged into the form
\begin{equation}
G_{\mu\nu} =\frac{[\text{terms in $\phi$, $\nabla\phi$ and $\nabla\nabla\phi$}]_{\mu\nu}}{1-8\pi\xi\phi^2} ,
\end{equation}
where the numerator on the right-hand side no longer contains the Einstein tensor.
For this reason, values of $|\phi|$ larger than the critical value $(8\pi\xi)^{-1/2}$ are considered unphysical since they correspond to a change of sign of the physical Newton's constant. See for example Ref.~\cite{Barcelo:2000zf}.
We now adapt our worldline bounds of \sec~\ref{sec:worldline} to solutions of the Einstein--Klein--Gordon theory. Taking the trace of the Einstein equation $G_{\mu\nu}=-8\pi T_{\mu\nu}$ gives
\begin{equation}\label{eq:trEins}
\left(\frac{n}{2}-1\right)R = 8\pi T \,.
\end{equation}
This can be used, in combination with the Klein--Gordon equation, to rearrange the trace of the stress-energy tensor of the non-minimally coupled scalar field which is given by Eq.~\eqref{eqn:trace}
\begin{align}
(1-8\pi\xi\phi^2)T &= \left(1-\frac{n}{2}\right) (\nabla\phi)^2 + \frac{n}{2}m^2\phi^2 + \xi (n-1)\Box_g\phi^2 \nonumber \\
&=\left(1-\frac{n}{2}+2\xi(n-1)\right) (\nabla\phi)^2 + \frac{n}{2}m^2\phi^2 - 2\xi (n-1)(m^2+\xi R)\phi^2
\end{align}
and using \eqref{eq:trEins} again,
\begin{equation}
\left(1-8\pi\xi(1-\xi/\xi_c)\phi^2\right)T = -2(n-1)(\xi_c-\xi) (\nabla\phi)^2 + \left(1+2(n-1)(\xi_c-\xi) \right)m^2\phi^2 \,.
\end{equation}
Therefore, if $\xi\le \xi_c$,
\begin{equation}
\left(1-8\pi\xi(1-\xi/\xi_c)\phi^2\right)T \ge -2(n-1)(\xi_c-\xi) (\nabla_{\dot{\gamma}}\phi)^2 \,,
\end{equation}
where we used the fact that $h^{\mu \nu}=\dot{\gamma}^\mu \dot{\gamma}^\nu-g^{\mu \nu}$ is a positive definite metric. If we take the maximum value of the field less than the critical value so $8\pi\xi\phi^2\le 1$, we have
\begin{equation}
\xi^2 R\phi^2 \ge - (\xi_c-\xi) (\nabla_{\dot{\gamma}}\phi)^2 \,,
\end{equation}
where we used Eq.~\eqref{eq:trEins}.
Now we can replace the term including the Ricci scalar in the bound of Eq.~\eqref{eqn:cline2} using the inequality
\begin{equation}\label{eqn:Ricci_ineq}
\int \frac{2\xi^2 R\phi^2}{n-2} f^2(\tau) d\tau \ge -\int \frac{2(\xi_c-\xi)}{n-2} (\nabla_{\dot{\gamma}}\phi)^2 f^2(\tau) d\tau \,.
\end{equation}
This gives the following bound for any solution to the Einstein--Klein--Gordon system
\bea
\label{eqn:cline3}
\int_\gamma d\tau \, \rho\, f^2(\tau)&\geq& - \int_\gamma d\tau \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau)+\xi \bigg(2( f'(\tau))^2+R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu f^2(\tau) \bigg) \bigg\}\phi^2 \,,
\eea
valid for $0\le \xi\le\xi_c$. To get the bound of Eq.~\eqref{eqn:cline3} from Eqs.~\eqref{eqn:cline2} and~\eqref{eqn:Ricci_ineq}, we discarded
\begin{equation}
\int_\gamma \left(1-\frac{1}{2(n-1)}-2\xi\right) (\nabla_{\dot{\gamma}}\phi)^2 f^2(\tau) d\tau \geq \frac{1}{2} \int_\gamma (\nabla_{\dot{\gamma}}\phi)^2 f^2(\tau) d\tau \geq 0 \,,
\end{equation}
for $\xi\le\xi_c$. Now noticing that
\be
R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu=-8\pi \rho \,,
\ee
we can move this term to the left side of the inequality of (\ref{eqn:cline3})
\be
\int_\gamma d\tau \, R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu f^2(\tau) (1-8\pi \xi \phi^2) \leq \int_\gamma d\tau \bigg\{ \frac{1-2\xi}{n-2} m^2 f^2(\tau)+2 \xi ( f'(\tau))^2 \bigg\}8\pi \phi^2 \,.
\ee
Since $\phi$ is less than $(8\pi \xi)^{-1/2}$ we can absorb the factor $(1-8\pi \xi \phi^2)$ in $f(\tau)$ and state the following theorem
\begin{theorem}
\label{the:ekgineq}
Suppose $(M,g,\phi)$ is a solution to the Einstein--Klein--Gordon equation in dimension $n>2$ with coupling constant $\xi\in[0,\xi_c]$ and $|\phi| \le (8\pi\xi)^{-1/2}$. Let $\gamma$ be a timelike geodesic parametrized by proper time $\tau$ in $(M,g)$ and $f$ a real valued function. Then
\be
\label{eqn:ffactor}
\int_\gamma d\tau \, R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu f^2(\tau) \leq \int_\gamma d\tau \bigg\{ \left(\frac{1-2\xi}{n-2}\right) \frac{m^2 f^2(\tau)}{1-8\pi \xi \phi^2} +2 \xi \left(\frac{d}{d\tau}\frac{f(\tau)}{\sqrt{1-8\pi \xi \phi^2}}\right)^2 \bigg\}8\pi \phi^2 \,.
\ee
\end{theorem}
This inequality has the advantage that the left-hand side is geometric, while only non-geometric terms appear on the right-hand side. It will enable us to prove
a singularity theorem for this system.
\section{A Hawking-type singularity theorem}
\label{sec:singularity}
In this section we establish a Hawking-type singularity theorem with a weakened energy condition. A similar Penrose-type singularity theorem was discussed in \cite{Fewster:2010gm}.\footnote{Note that \cite{Fewster:2010gm} employs $+++$ sign conventions.}
We then use the result of \sec\ref{sec:ekg} to obtain a Hawking-type singularity theorem for
the non-minimally coupled Einstein--Klein--Gordon theory.
\begin{theorem}\label{thm:sing}
Let $(M,g)$ be a globally hyperbolic spacetime of dimension $n>2$, and let $S$ be a
smooth compact spacelike Cauchy surface for $(M,g)$. Suppose that
\begin{enumerate}\renewcommand{\theenumi}{\alph{enumi}}
\item\label{it:lifetime} there exists $\tau_0>0$ such that the congruence of
future-directed unit-speed geodesics issuing orthogonally from $S$ can be continued to the past of $S$ for a proper time of at least $\tau_0$ with a smooth velocity field $U^\mu$ and expansion $\theta=\nabla_\mu U^\mu$;
\item\label{it:Rbdsing} there are positive constants $Q$ and $\tilde{Q}$ such that, along each future complete unit speed timelike geodesic $\gamma:[-\tau_0,\infty) \to M$ issuing orthogonally from $S$
one has an inequality
\begin{equation}\label{eq:Rbdsing}
\int R_{\mu \nu}\dot{\gamma}^\mu \dot{\gamma}^\nu f(\tau)^2\,d\tau \le Q(\|f'\|^2+ \tilde{Q}^2\|f\|^2),
\end{equation}
where $|| \cdot ||$ is the $L^2$-norm, for all smooth, real-valued $f$ compactly supported in $(-\tau_0,\infty)$;
\item\label{it:singcon} for some $K>0$, (i) the inequality
\begin{equation}\label{eq:singcon}
\nabla_U \theta|_{\gamma(\tau)}+\frac{\theta(\gamma(\tau))^2}{n-1}\ge Q(\tilde{Q}^2-K^2) \qquad \textrm{on $(-\tau_0,0]$}
\end{equation}
holds along every future-directed unit-speed geodesic $\gamma(\tau)$
issuing orthogonally from $S$ at $\tau=0$, and \\
(ii) the expansion $\theta$ on $S$ obeys
\begin{equation}\label{eq:theta0bd}
\theta|_S <-\tilde{Q}\sqrt{Q(n-1)+Q^2/2}-\frac{1}{2}Q K \coth{(K\tau_0)}\,.
\end{equation}
Alternatively,
it is sufficient if (c)(i) holds with
\be
\label{eqn:singconb}
\nabla_U \theta|_{\gamma(\tau)} \geq 0 \qquad \textrm{on $(-\tau_0,0]$,}
\ee
in place of \eqref{eq:singcon}, and (c)(ii) holds either as before (for some $K>0$) or with
\eqref{eq:theta0bd} replaced by
\be
\label{eqn:theta0b}
\theta|_S < -\tilde{Q} \sqrt{Q(n-1)+Q^2/2}-\frac{Q}{2 \tau_0} \,.
\ee
\end{enumerate}
Then the spacetime is future geodesically incomplete.
\end{theorem}
{\noindent\em Remarks:}
1. Note that hypotheses~\eqref{it:lifetime} and~\eqref{it:singcon} refer to the recent past of the Cauchy surface $S$, and therefore would in principle be amenable to observational confirmation.
2. The proof shows that the expansion of the geodesic congruence normal to $S$ must actually diverge to $-\infty$ at finite time. From this perspective it may seem strange that~\eqref{eq:singcon} can be satisfied if $\dot{\theta}$ is large and positive on $(-\tau_0,0]$. However this is just an expression of the averaged bound~\eqref{eq:Rbdsing}: large positive values of $R_{\mu\nu}\dot{\gamma}^\mu\dot{\gamma}^\nu$ in the recent past must be counterbalanced by (even larger) negative values in the future (this follows by exactly the same reasoning used in explorations of `quantum interest'~\cite{Ford:1999qv, Fewster:1999kr}) which drive the expansion to $-\infty$. 3. The constants $Q$ and $\tilde{Q}$ can be global or be allowed to vary between geodesic congruences if that leads to a tighter bound.
4. Clearly $\tau_0$ may be replaced in hypothesis (c) by any $\tilde{\tau}_0\in (0,\tau_0]$, giving useful additional freedom. Reducing $\tilde{\tau}_0$ means that \eqref{eq:singcon} can perhaps be satisfied with a smaller value of $K$, although this needs to be weighed against any consequent increase in $Q K \coth{(K\tilde{\tau}_0)}$. In any case there is an optimum value of $\tilde{\tau}_0$ for any fixed function $\theta$.
\begin{proof}
The beginning of the proof is exactly the same as the singularity theorem 5.1 in Ref.~\cite{Fewster:2010gm}. We suppose that the spacetime is timelike geodesically complete, and aim for a contradiction. General properties of globally hyperbolic spacetimes with compact Cauchy surfaces guarantee the existence of a future-directed
\emph{$S$-ray} $\gamma$ --- that is, $\gamma$ is a unit-speed geodesic, issuing orthogonally from $S$, so that the Lorentzian distance from each $\gamma(\tau)$ to $S$ is precisely $\tau$, for all $\tau\in [0,\infty)$ --- $\gamma$ is complete by assumption. There is a neighbourhood of $\gamma$ in $J^+(S)$ in which the Lorentzian distance $\rho_S(p)$ from $p$ to $S$ is smooth. (We choose conventions so that $\rho_S$ is positive for timelike separation.) In this neighbourhood, the velocity field $U^\mu=\nabla^\mu\rho_S(p)$ is
a smooth future-directed unit timelike vector field which is irrotational and orthogonal to $S$.
We now restrict to the geodesic $\gamma$ and write $\theta(\tau):=\nabla_\mu U^\mu|_{\gamma(\tau)}$ for the expansion along $\gamma$. By the above properties, $\theta(\tau)$
is a smooth solution to Raychaudhuri's equation
\be
\label{eqn:ray}
\dot{\theta}(\tau)=r(\tau) -\frac{1}{n-1} \theta(\tau)^2 \,.
\ee
where $r(\tau)=R_{\mu \nu} \dot{\gamma}^\mu \dot{\gamma}^\nu-2\sigma(\gamma(\tau))^2$ and $\sigma$ is the shear scalar. By our assumption (\ref{it:lifetime}) together with the assumption of future geodesic completeness, this equation is valid on $(-\tau_0,\infty)$. Additionally note that if condition (b) holds then,
\be
\label{eqn:rcondb}
\int r(\tau) f(\tau)^2\,d\tau \le Q(\|f'\|^2+ \tilde{Q}^2\|f\|^2) =: |||f|||^2\,,
\ee
is also true for all $f$. We proceed to prove that, contrary to what has just been shown,
Eq.~\eqref{eqn:ray} can have no smooth solution on $(-\tau_0,\infty)$ under these conditions; indeed, the solution must tend to $-\infty$ at finite positive time.
Suppose first that $r(t)\equiv r$ is constant and note that \eqref{eq:theta0bd} or \eqref{eqn:theta0b} imply that $\theta_0=\theta(0)<0$. If $r<0$ then the unique
solution to~\eqref{eqn:ray} is
\begin{equation}
\theta(\tau)=-\sqrt{-(n-1)r} \cot{\left[\sqrt{\frac{-r}{n-1}}(\tau_*-\tau)\right]} \,, \quad \tau_*=\sqrt{\frac{n-1}{-r}} \cot^{-1}{\left[\frac{-\theta_0}{\sqrt{-(n-1)r}}\right]} \,,
\end{equation}
using the branch of arc-cotangent valued in $(0,\pi)$. As $\theta_0<0$, we have $\tau_*>0$ and the solution blows up as $\tau\to\tau_*$.
Similarly, if $r=0$, the solution is
\begin{equation}
\theta(\tau)=\frac{n-1}{\tau-\tau^*}, \qquad \tau_*=-\frac{n-1}{\theta_0}
\end{equation}
and again there is blow-up as $\tau\to\tau_*>0$. If $r>0$, then \eqref{eqn:rcondb} implies that
$r\leq Q\tilde{Q}^2$. Using Eq.~\eqref{eq:theta0bd}, we have $\theta_0<-\tilde{Q}\sqrt{Q(n-1)}\leq -\sqrt{-(n-1)r}$
and the solution is
\begin{equation}
\theta(\tau)=-\sqrt{(n-1)r} \coth{\left[\sqrt{\frac{r}{n-1}}(\tau_*-\tau)\right]} \,, \qquad \tau_*=\sqrt{\frac{n-1}{r}} \coth^{-1}{\left[\frac{-\theta_0}{\sqrt{(n-1)r}}\right]} \,,
\end{equation}
again blowing up at finite positive time. Therefore, Raychaudhuri's equation has no solution on $(-\tau_0,\infty)$ if $r(\tau)$ is constant, contradicting the existence of the solution shown above. Therefore $r(\tau)$ must be nonconstant.
We may choose $\epsilon>0$ so that $\theta_0+\epsilon$ is also less than the right-hand side of
\eqref{eq:theta0bd}. By Lemma~6.1 of~\cite{Fewster:2010gm}, there exists $c>0$ and a smooth function
$h$ supported on $[-\tau_0,\infty)$ with $h(\tau)=e^{-c\tau/(n-1)}$ on $[0,\infty)$ and
so that the second inequality in
\begin{equation}
\label{eqn:thefour}
-\theta_0 >
\epsilon + \tilde{Q}\sqrt{Q(n-1) + Q^2/2} + \frac{1}{2}QK\coth K\tau_0
\geq \frac{c}{2} + |||h|||^2 - \int_{-\tau_0}^0 h^2(\tau) r(\tau) \,d\tau \,.
\end{equation}
holds (the first one holds by virtue of our choice of $\epsilon$).
Eq.~\eqref{eqn:thefour} implies that \eqref{eqn:ray} has no solution on $[0,\infty)$ by Theorem~4.1 of~\cite{Fewster:2010gm} (applied with $z=-\theta$, $r_0\equiv 0$, $s=n-1$ and with the opposite sign convention for $r$). This contradicts the existence of the solution shown above and therefore demonstrates that the spacetime is future timelike geodesically incomplete.
It remains to show that the alternative conditions stated in hypothesis (c) also suffice.
For \eqref{eqn:singconb}, observe that, together with \eqref{eq:theta0bd},
it implies that $\theta \leq \theta_0<0$ on $(-\tau_0,0]$ and that
\begin{equation}
\dot{\theta} + \frac{\theta^2}{n-1} \geq \frac{\theta_0^2}{n-1} \geq \frac{\tilde{Q}^2(Q(n-1) + Q^2/2)}{n-1} \geq Q\tilde{Q}^2 \qquad \textrm{on $(-\tau_0,0]$}
\end{equation}
as a result of \eqref{eq:theta0bd}. We obtain \eqref{eq:singcon} in consequence and
therefore conditions (c)(i,ii) both hold.
Now if \eqref{eqn:theta0b} holds then Eq.~\eqref{eq:theta0bd} holds
for some $K>0$ because the former is just the $K\to 0+$ limit of the latter. Therefore
we have conditions~\eqref{eqn:singconb} and~\eqref{eq:theta0bd} and the previous paragraph
shows that (c)(i,ii) both hold.
\end{proof}
We now apply this theorem to the more specific setting of Einstein--Klein--Gordon theory, using the worldline bound of Theorem \ref{the:ekgineq}
\begin{corollary}\label{cor:sing}
Let $(M,g,\phi)$ be a solution to the Einstein--Klein--Gordon equation in dimension $n>2$ and with coupling $\xi\in[0,\xi_c]$. Suppose that $(M,g)$ is globally hyperbolic, let $S$ be a smooth
spacelike Cauchy surface for $(M,g)$ and suppose that $\phi$ obeys global bounds
$|\phi|\le\phi_{\text{max}}< (8\pi\xi)^{-1/2}$ and $|\nabla_{\dot{\gamma}}\phi|\le\phi'_{\textrm{max}}$ along all unit speed timelike geodesics $\gamma$ issuing orthogonally from $S$. Assume hypotheses~(\ref{it:lifetime}) and~(\ref{it:singcon}) of Theorem~\ref{thm:sing}, where $Q$ and $\tilde{Q}$ are given by
\be\label{eq:QQtdef}
Q=\frac{32\pi \xi \fmax^2}{1-8\pi \xi \fmax^2} \,, \qquad \tilde{Q}^2=\frac{(1-2\xi)m^2}{4\xi(n-2)}+\left(\frac{8\pi \xi \fmax \fmax'}{1-8\pi \xi \fmax^2}\right)^2 \,.
\ee
Then $(M,g)$ is future geodesically incomplete.
\end{corollary}
\begin{proof}
First, we estimate the right-hand side of Eq.~\eqref{eqn:ffactor}, noting first that $\phi^2/(1-8\pi\xi\phi^2)$ is increasing with $\phi$, and therefore
\begin{equation}
\int_\gamma d\tau \left(\frac{1-2\xi}{n-2}\right) \frac{8\pi m^2 \phi^2 f^2(\tau)}{1-8\pi \xi \phi^2} \le \left(\frac{1-2\xi}{n-2}\right) \frac{8\pi m^2 \fmax^2}{1-8\pi \xi \fmax^2}\|f\|^2 =\frac{Qm^2}{4\xi} \left(\frac{1-2\xi}{n-2}\right)\|f\|^2
\end{equation}
for all smooth compactly supported $f$. Next,
using the inequality
\be
||(fg)'||^2 \leq 2( ||f'||^2 ||g||^2_\infty +||f||^2 ||g'||^2_\infty) \,,
\ee
and also the global bound on $\phi$, we can write
\be
\int_\gamma d\tau \left(\frac{d}{d\tau}\frac{f(\tau)}{\sqrt{1-8\pi \xi \phi^2}}\right)^2 16\pi\xi\phi^2 \leq \frac{32\pi\xi\fmax^2}{1-8\pi \xi \fmax^2} \left( ||f'||^2+||f||^2 \left(\frac{8\pi \xi \fmax \fmax'}{1-8\pi \fmax^2}\right)^2 \right)
\ee
for all smooth compactly supported $f$.
Thus Eq.~(\ref{eqn:ffactor}) implies
that Eq.~\eqref{eq:Rbdsing} holds with $Q$ and $\tilde{Q}$ given by Eq.~\eqref{eq:QQtdef}. This supplies hypothesis~\eqref{it:Rbdsing} of Theorem~\ref{thm:sing} and as we also assume hypotheses~\eqref{it:lifetime} and~\eqref{it:singcon} the result follows.
\end{proof}
As in Theorem~\ref{thm:sing}, one could replace the hypotheses~\eqref{eq:singcon} and~\eqref{eq:theta0bd} by the alternatives Eqs.~\eqref{eqn:singconb} and~\eqref{eqn:theta0b}.
\section{Discussion}
\label{sec:disc}
This paper has accomplished two main goals. First, we have established worldline and worldvolume bounds on the effective energy density of the nonminimally coupled scalar field. Elsewhere, these bounds will be used as the basis for a quantum energy inequality variant of the strong energy condition for the quantized field. Second, we have shown that our bounds can be used to derive
a Hawking-type singularity theorem for the Einstein--Klein--Gordon theory, by applying methods
developed in~\cite{Fewster:2010gm}. The overall message here is that sufficient initial contraction on a compact Cauchy surface is enough to guarantee timelike geodesic incompleteness, even though the non-minimally coupled Klein--Gordon theory does not obey the SEC.
To conclude, we discuss the hypotheses of the singularity theorem in more depth, aiming for an understanding the physical magnitude of the initial contraction required. Reinserting units and thus
restoring $G$ and $c$, the constants $Q$ and $\tilde{Q}$ become
\be\label{eq:QQtdefdim}
Q=\frac{32\pi \xi G\fmax^2/c^4}{1-8\pi \xi G\fmax^2/c^4} \,, \qquad \tilde{Q}^2=\frac{(1-2\xi)(mc)^2}{4\xi(n-2)}+\left(\frac{8\pi \xi G\fmax \fmax'/c^4}{1-8\pi \xi G\fmax^2/c^4}\right)^2 \,,
\ee
where $m$ has units of inverse length. Thus $Q$ is dimensionless, while $\tilde{Q}$ has dimensions of inverse time, as required for dimensional correctness in Eq.~\eqref{eq:theta0bd} with $\tau_0$ having dimensions of time and consequently
$K$ being an inverse time. Evidently both $Q$ and $\tilde{Q}$ become very large if
$\fmax^2$ is allowed to be close to the critical value $c^4/(8\pi G\xi)$. However,
$Q\le 4$ if $\fmax^2$ does not exceed half the critical value, for example, so it is no great restriction to take $Q$ of order $1$. Turning to $\tilde{Q}$, the second term is $Q^2(\fmax'/\fmax)^2$ up to numerical factors. The ratio $\fmax'/(mc\fmax)$
is dimensionless and it would be reasonable to assume bounds so that this quantity does not
greatly exceed unity. Therefore $\tilde{Q}$ would be reasonably expected not greatly to exceed $mc$.
The remaining ingredient in the contraction bound Eq.~\eqref{eq:theta0bd} are the timescale $\tau_0$ and constant $K$, which depend
on the history of the solution prior to $S$; we may assume for the purposes of
discussion that the corresponding terms in Eq.~\eqref{eq:theta0bd} are not large.
Accordingly, the initial contraction required to ensure geodesic incompleteness might be expected to be of the order of $mc$, i.e., the characteristic frequency
of the (minimally coupled) Klein--Gordon operator (recall that $m$ is an inverse length in this discussion).
For the purely classical Einstein--Klein--Gordon theory, this seems very reasonable. But---with a view to semiclassical quantum gravity---what if the scalar field is supposed to describe an elementary particle, with a correspondingly small mass? In this situation $m$ is replaced in Eq.~\eqref{eq:QQtdefdim} by $mc/\hbar$, the inverse Compton length for a particle of mass $m$, so $\tilde{Q}$ is of the order of the inverse Compton time and our previous
reasoning would suggest that a huge initial contraction would be required to guarantee geodesic incompleteness. For example, with the physical values of $G$, $\hbar$ and $c$ in $n=4$ dimensions, and with $m$ on the order of the pion mass, $m=140\textrm{MeV}/c^2$, the initial contraction would be of the order $2\times 10^{23}{\textrm{s}}^{-1}$ (using the value $\hbar=6.6\times 10^{-22}\textrm{MeV.s}$) by such arguments. This would call into question the utility of the singularity result for this situation.
However, the value of $\fmax$ should be reconsidered in this hybrid model.
To indicate the values that would be reasonable,
we consider a quantized scalar field in Minkowski spacetime of dimension $n$, in a thermal state of temperature $T<T_m$, where $T_m=mc^2/k$ sets a natural temperature scale for the model, beyond which its applicability might be doubtful. Here, $k$ is Boltzmann's constant. In this regime,
the expectation value of the Wick square is
\begin{equation}
\label{eqn:estim}
\langle {:}\phi^2{:}\rangle_T \sim C_n \hbar c\left(\frac{kT_m}{\hbar c}\right)^{n-2}
(T/T_m)^{(n-2)/2} K_{(n-2)/2}(T_m/T) ,
\end{equation}
where the numerical constant is
\begin{equation}
C_n=\frac{S_{n-2}2^{(n-2)/2}}{(2\pi)^{n-1}\sqrt{\pi}} \Gamma((n-1)/2)
\end{equation}
and takes the value $C_4=0.05$ in $n=4$ dimensions, for example.
Therefore
\begin{equation}
\frac{G\langle {:}\phi^2{:}\rangle_T}{c^4}\sim C_n\left(\frac{T_m}{T_{\text{Pl}}}\right)^{n-2}(T/T_m)^{(n-2)/2} K_{(n-2)/2}(T_m/T),
\end{equation}
where $T_{\text{Pl}}$ is the Planck temperature (in $n$ dimensions). The derivation of this estimate is given in Appendix \ref{app:temp}.
If we take $\fmax^2\sim\langle {:}\phi^2{:}\rangle_T$ then
the factor $Q$ is reasonably assumed to be given by
\begin{equation}
Q\sim (m/m_{\textrm{Pl}})^{n-2}(T/T_m)^{(n-2)/2} K_{(n-2)/2}(T_m/T)
\end{equation}
where $m_{\textrm{Pl}}$ is the Planck mass. Maintaining the previous expectation that
$\fmax'/\fmax\sim mc$, and using $T$ to parameterise the maximum acceptable field amplitude in this way leads to a contraction estimate
\be
\theta_0 \lesssim -\frac{mc^2}{\hbar}Q^{1/2} \sim -\frac{mc^2}{\hbar} (m/m_{\textrm{Pl}})^{(n-2)/2} (T/T_m)^{(n-2)/4} (K_{(n-2)/2}(T_m/T) )^{1/2} \,,
\ee
to guarantee geodesic incompleteness.
For a pion in $n=4$, the leading factor on the right-hand side $m/m_{\textrm{Pl}}=5.9 \times 10^{-20}$, while $mc^2/ \hbar$, as we mentioned, is $2\times 10^{23} \textrm{s}^{-1}$ and $T_m=1.7 \times 10^{12}\textrm{K}$. However, the remaining factor decays very rapidly as $T/T_m$ is reduced below $1$. For example, if $T=10^{-2}T_m$ then $(T/T_m)^{(n-2)/4} (K_{(n-2)/2}(T_m/T) )^{1/2}=6.8 \times 10^{-24}$, thus bringing the overall contraction needed down to the order of $10^{-19} \textrm{s}^{-1}$. This is indeed very small: If a volume were subject to contraction at this constant fractional rate, it would require approximately 100 times the current age of the universe to halve its size. Our calculation has allowed for a maximum temperature scale of $T=10^{10}\textrm{K}$ -- the temperature of the universe approximately one second after the Big Bang. Repeating the calculation for the Higgs mass $125\textrm{GeV}/c^2$,
a minimum contraction of order $10^{-14}\textrm{s}^{-1}$ is required, but this time allowing
a temperature up to $T=10^{13}\textrm{K}$, the temperature of the Universe at age $0.0001\textrm{s}^{-1}$.
Summarising this discussion: a model in which the field mass is taken equal to
an elementary particle mass would need very little initial contraction to guarantee that either there is geodesic incompleteness
or that, at the least, the solution evolves to a situation where the natural energy scales associated with the field approach those of the early universe. At this stage, a macroscopic observer might be forgiven for believing that a singularity had occurred!
Hawking and Ellis \cite{HawkingEllis:1973} discuss the violation of an average SEC by a pion. Their heuristic analysis led them to argue that the convergence of timelike geodesics would not be influenced by SEC violation on scales greater than $10^{-14} \textrm{m}$. By contrast, our analysis instead rigorously shows that even for extremely small initial contractions a singularity is inevitable.
The obvious extension of this work is the derivation of a quantum strong energy inequality for the non-minimally coupled field. To examine if such an inequality can be used to prove a Hawking-type singularity theorem, it is also necessary to find estimates on the timescales for averaging over which the curved spacetime approximates the Minkowski results.
It should be noted that to fully understand whether the dynamics are driven towards
singularities at the semi-classical level requires a semiclassical analysis that takes backreaction into account in a dynamical way. Positive results include \cite{Drago:2014eoa}, which calculates geometrical fluctuations from (passive) quantum fields, and a result on ANEC with transverse smearing \cite{Flanagan:1996gw}. However, the calculation of backreaction, even in the second order in perturbation theory brings significant technical challenges to the problem. Finally, the question of whether full quantum gravity can resolve singularities via a ``big bounce'' or not remains open.
\section*{Acknowledgments}
We thank Roger Colbeck for comments on the text. This work is part of a project that has received funding from the European Union's Horizon 2020 research and innovation programme under the Marie Sk\l odowska-Curie grant agreement No. 744037 ``QuEST''. P.J.B.\ thanks the WW Smith Fund for their financial support.
\appendix
\section{Temperature dependence of Wick square}
\label{app:temp}
Consider the KMS state of the Klein--Gordon field with mass $m$ in $n$-dimensional Minkowski space for $n> 3$ (or $m>0$ if $n=3$) corresponding to a thermal equilibrium state at temperature $T$. The two-point function of that state is
\be
W^{(2)}_T(t,\mathbf{x},t',\mathbf{x}')=\int d\mu(\mathbf{k}) \left( \frac{e^{-i((t-t')\omega(\mathbf{k})-(\mathbf{x}-\mathbf{x}') \mathbf{k})}}{1-e^{-\beta\omega(\mathbf{k})}}+\frac{e^{i((t-t')\omega(\mathbf{k})-(\mathbf{x}-\mathbf{x}') \mathbf{k})}}{e^{\beta\omega(\mathbf{k})}-1} \right) \,,
\ee
where $\beta=(kT)^{-1}$ and $\mu(\mathbf{k})$ is given by
\be
\label{eqn:measure}
d\mu(\bfk)=\int \frac{d^{n-1}\bfk }{(2\pi)^{n-1}}\frac{1}{2\omega(\bfk)} \,,
\ee
with $\omega(\mathbf{k})=\sqrt{\mathbf{k}^2+m^2}$. Here we use $\hbar=c=1$. After renormalizing with the ground state two-point function
\be
\label{eqn:vac}
W^{(2)}_0(t,\mathbf{x},t',\mathbf{x}')=\int d\mu (\mathbf{k}) e^{-i[(t-t')\omega(\mathbf{k})-(\mathbf{x}-\mathbf{x}')\mathbf{k}]} \,,
\ee
in the coincidence limit we find
\be
\langle {:}\phi^2{:}\rangle_T =[ W^{(2)}_T - W^{(2)}_0]_c
= \frac{2 S_{n-2}}{(2\pi)^{n-1}} \int_m^\infty d\omega (\omega^2-m^2)^{(n-3)/2} \frac{1}{e^{\beta \omega(k)}-1} \,,
\ee
where $S_{n-2}$ is the area of the unit $(n-2)$-sphere. Changing variables to $x= \omega/m$ gives
\be \label{eqn:betaone}
\langle {:}\phi^2{:}\rangle_T=\frac{2 S_{n-2}m^{n-2}}{(2\pi)^{n-1}} \int_1^\infty dx (x^2-1)^{(n-3)/2} \frac{1}{e^{\beta x/m}-1} \,.
\ee
Reinserting units gives
\begin{equation}
\langle {:}\phi^2{:}\rangle_T = \hbar c\frac{S_{n-2}(k T_m/(\hbar c))^{n-2}}{(2\pi)^{n-1}}\int_{1}^\infty dx\frac{(x^2-1)^{(n-3)/2}}{e^{T_m x/T}-1} \,.
\end{equation}
As we are interested in $T<T_m$, we may expand the integrand using a geometric series
\begin{equation}
\langle {:}\phi^2{:}\rangle_T = \hbar c\frac{S_{n-2}(k T_m/(\hbar c))^{n-2}}{(2\pi)^{n-1}}\int_{1}^\infty dx \sum_{r=1}^\infty (x^2-1)^{(n-3)/2}e^{-rT_m x/T}\,,
\end{equation}
and exchanging sum and integral (which is valid) one obtains
\begin{equation}
\langle {:}\phi^2{:}\rangle_T =\hbar c \frac{S_{n-2}(k T_m/(\hbar c))^{n-2}\Gamma((n-1)/2)}{(2\pi)^{n-1}\sqrt{\pi}} \sum_{r=1}^\infty
\left(\frac{2T}{rT_m}\right)^{(n-2)/2} K_{(n-2)/2}(rT_m/T) \,,
\end{equation}
in terms of modified Bessel functions. For order of magnitude purposes, the first term in the sum is perfectly adequate for $T<T_m$. This becomes more obvious if we subtract the first term from the sum
\be
\frac{1}{e^{T_m x/T}-1}-\frac{1}{e^{T_m x/T}} \leq 2 e^{-2 T_m x/T}
\,,
\ee
which is double the second term of the series. This gives the estimate of Eq.~\eqref{eqn:estim}.
\bibliography{class}
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TITLE: Number of ways to pick an equal number of elements from two sets?
QUESTION [3 upvotes]: There are two groups such that one group contains $m$ elements and another group contains $n$ elements. We have to find number of ways to pick same number of elements from both the groups.
My approach is
$1+\dbinom m1.\dbinom n1+\dbinom m2 . \dbinom n2 +.... $upto $\dbinom mm$ or $\dbinom nn$ whatever is smaller.
Is there any faster way to do the same?
Example:
$3$ and $2$
$1+\dbinom 31.\dbinom 21+\dbinom 32 . \dbinom 22=10$ ways
REPLY [1 votes]: Yes, you can do better! The number of objects as you've described can be written as:
$$ \sum_{k=0}^{\min(m,n)} \binom{n}{k}\binom{m}{k}. $$
Notice that we could choose to stop the sum at either $m$ or $n$, regardless of which one is larger. For instance, even if $m>n$, whenever $k>n$ the first factor vanishes. We'll choose to stop the summation at $m$, and then apply the symmetry of binomial coefficients to get
$$ \sum_{k=0}^m \binom{n}{k}\binom{m}{m-k} $$
I claim that this can simplify in the following way:
Proposition: $\qquad\qquad\displaystyle \sum_{k=0}^m \binom{n}{k}\binom{m}{m-k} = \binom{m+n}{m}$.
The right-hand side clearly counts the number of ways to choose $m$ objects from $m+n$. The left-hand side also counts this because you must choose some number of the first $n$; call that $k$. By definition, the other $m-k$ must be chosen from the remaining $m$. Apply the product principle and we're done.
| 53,358
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| 381,974
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TITLE: Mathematical Induction in constructive setting
QUESTION [1 upvotes]: I'm little confused. As we don't have a proof hence we can't say : let the equation holds till(for) fixed n, and then we are going to show (prove) it holds for n + 1.
From this argument mathematical induction is nearly useless in constructive setting ? Normally we prove nearly 3/4 of stuff using this technique.
The confusion started by diving equality as Extensional Equality and Intentional Equality. where in former case you need a proof and in later case you can just enumerate all possible instances.
Example :
Lets say we are inductively defining Natural Numbers. In extensional equality you have to prove anything about it, so u'll need principle of mathematical induction while in case of Intentional equality u can't use it (Or there is no such principle, {I get to know by Bob Harper's lectures , i might not be presenting facts right}).
My argument are not perfect and might not be completely formal, i'm just a beginner. I have doubts and would love to iteratively understand it. So, stop flagging things as "put on hold". It looks like "Bullying" to me. Stop doing that and just ask or help me to refine my question. Stop showing that you are an elite, its disgusting.
REPLY [2 votes]: See Richard Dedekind, Stetigkeit und irrationale Zahlen (1872), Engls.translation into:
Richard Dedekind, Essays on the Theory of Numbers, page 4:
I regard the whole of arithmetic as a necessary, or at least natural, consequence of the simplest arithmetic act, that of counting, and counting itself as nothing else than the successive creation of the infinite series of positive integers in which each individual is defined by the one immediately preceding; the simplest act is the passing from an already-formed individual to the consecutive new one to be formed. The chain of these numbers forms in itself an exceedingly useful instrument for the human mind; it presents an inexhaustible wealth of remarkable laws obtained by the introduction of the four fundamental operations of arithmetic. Addition is the combination of any arbitrary repetitions of the above-mentioned simplest act into a single
act; from it in a similar way arises multiplication.
We can see the Axiom of induction as a rule: a sort of "iterated" modus ponens.
For an arithmetical "property" $P$, we start proving the base case:
(1) "$P(0)$ holds";
then we prove the inductive step:
(2) "for $n$ whatever, if $P(n)$ holds, then aso $P(n+1)$ holds".
Having done this, we can "assembly" the two result into an unlmited chain of deductions: first, we have that $P(0)$ holds by (1); then we can "instantiate" (2) to $P(0)→P(1)$ and by modus ponens with $P(0)$ we derive $P(1)$.
Now we use again (2): $P(1)→P(2)$ and by modus ponens again we derive $P(2)$.
And so on. The "and so on" is "built in" into the axiom, and it is the "inner nature" of the progression of natural numbers.
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| 57,981
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TITLE: Why is this famous proof of the chain rule called "technically incorrect" in this pdf?
QUESTION [13 upvotes]: So I was looking through various proofs of the chain rule...and I came across this paper. The first proof given is complete and quite well-explained. But another simplistic proof is given in the end...which is mentioned as "technically incorrect". Can anyone tell me why?
Here is the incorrect proof in question:
$$\begin{aligned}
(f \circ g)'(x) &= \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{h} \\
\implies (f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) &= \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)\\
&= \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}\right) \\
&= f'(g(x)) \\
\end{aligned}$$
REPLY [28 votes]: If $g'(x) = 0$, the proof fails, but as mentioned in the comments, this can be handled as a separate case.
A more serious issue is the calculation
$$(f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) = \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)$$
If we write
$$a(h) = \frac{f(g(x+h))-f(g(x))}{h}$$
and
$$b(h) = \frac{h}{g(x+h)-g(x)}$$
then the above calculation is the assertion that
$$\lim_{h \to 0}\ a(h) \cdot \lim_{h \to 0}\ b(h) = \lim_{h \to 0}\ a(h)b(h),$$
in other words, that the limit of a product is the product of the limits.
This is true, provided that both limits on the left-hand side exist. But the existence of $\displaystyle \lim_{h \to 0}\ a(h)$ is exactly what we are trying to prove, so the argument is circular.
Edit to add:
Another issue worth noting is the final assertion, namely
$$f'(g(x)) = \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}$$
In fact, the definition of the derivative of $f$, evaluated at $g(x)$, gives us
$$f'(g(x)) = \lim_{k \to 0}\frac{f(g(x)+k) - f(g(x))}{k}$$
The only reason we are able to conclude that these two expressions are equal (even after handling the $g(x+h) = g(x)$ case properly) is because $g$ is continuous at $x$. This of course follows from the differentiability of $g$ at $x$, but a careful proof would point this out.
| 128,362
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TITLE: A diffeomorphism between one-sheet hyperboloid and $\mathbb{R}^2 -\{0\}$
QUESTION [0 upvotes]: As shown in the following image, in the book "Analysis and Algebra on Differentiable Manifolds" on page 54, it is proven that there is a diffeomorphism between one-sheet hyperboloid and $\mathbb{R}^2 -\{0\}$:
I would like to understand that how can obtain The equations of $\phi$ are given by
$$x' = x(1 − \frac{z}{1+z^2})\qquad ,\qquad y' = y(1 − \frac{z}{1+z^2}).$$
My attempt: Projecting point $p(x , y , z)$ onto plane $z=0$, one gets $p'(x , y , 0)$. If $\phi(p) = (x' , y' , 0)$, then by using the Thales theorem one gets:
$$\frac{x'}{x} = \frac{y'}{y} = \frac{\sqrt{x'^2 + y'^2}}{x^2 + y^2}.$$
Since $p$ is on hyperboloid, we have $x^2 + y^2 = 1+ z^2$. Then
$$x' = x \frac{\sqrt{x'^2 + y'^2}}{\sqrt{1 + z^2}} \qquad , \qquad y' = y \frac{\sqrt{x'^2 + y'^2}}{\sqrt{1 + z^2}}.$$
but I don't know how prove that $\sqrt{x'^2 + y'^2} = \sqrt{x^2 + y^2} - z$.
Thank you for any help.
REPLY [1 votes]: Let $\textbf{p} = (X,Y,Z)$ be fixed. (We use capital letters for the coordinates of $p$ to differentiate from the coordinate functions on $\mathbb{R}^3$.) We consider the plane through the origin spanned by the $z$-axis and the vector $\frac{1}{\sqrt{X^2+Y^2}}(X,Y,0)$. If we slice the hyperboloid by this plane, we get a hyperbola whose asymptotes have direction vectors
$\textbf{v} = (X,Y, \sqrt{X^2+Y^2})$ and $\textbf{w} = (X,Y, -\sqrt{X^2+Y^2})$.
According to the description of $\varphi$, to get from $\textbf{p}$ to $\varphi(\textbf{p})$, we should travel along the line through $\textbf{p}$ with direction vector $\textbf{v}$ until we reach the $z = 0$ plane. This line can be parameterized by
$$
\textbf{g}(t) = (X,Y,Z) + t\left(X,Y, \sqrt{X^2+Y^2} \right).
$$
One can compute that this line intersects the $z=0$ plane when we take
$$
t_0 = - \frac{Z}{\sqrt{X^2+Y^2}}.
$$
That is, $X’$ and $Y’$ are the first and second coordinates of $\textbf{g}(t_0)$, which means that
$$
X’ = X \left( 1 - \frac{Z}{\sqrt{X^2+Y^2}} \right) = X \left( 1 - \frac{Z}{\sqrt{Z^2+1}}\right)
$$
and
$$
Y’ = Y \left( 1 - \frac{Z}{\sqrt{X^2+Y^2}}\right) = Y \left( 1 - \frac{Z}{\sqrt{Z^2+1}}\right),
$$
as claimed in your book.
| 168,443
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TITLE: Calculate confidence interval
QUESTION [0 upvotes]: The distribution of days that are required to complete a certain
activity can be approximated by a standard normal distribution with a
mean of 500 days and a standard deviation of 12 days. What is the
approximate 95% confidence interval in days for the activity ?
I have read this example, but found it hard to understand. I hope maybe someone can give me a clearer explanation. I have to admit, I'm not very good with Math!
REPLY [1 votes]: The basic idea for any "Gaussian based" (read: pretty much all of them in a basic stats class) is that you do $\hat{x} \pm \mbox{critical value} \times sd$.
Here, $\hat{x}$ is your estimate, $sd$ is your standard deviation of the estimate and the ``critical value" is what you look up from a distribution.
You know that you want a 95% confidence interval. Using a normal distribution, you know that if $z = 1.96$ you have 2.5% of the area to the right of that and 2.5% of the area to the left of $-z$. Since that equals 5% total, that's your "critical value".
The rest is just plug and play.
Your estimate of the time of the activity is 500. Your standard deviation is 12 days.
Slight addendum: When I said $sd$ is the standard deviation, I meant the standard deviation of your estimate. Sometimes, you want a confidence interval for the mean, $\bar{x}$. In that case, the $sd(\bar{x}) = sd(x)/\sqrt{n}$. In your case, they just give you the final standard deviation so it's easy to use.
Thus the confidence interval is:
$(500 - 1.96 *12, 500 + 1.96*12)$ = (476.48, 523.52).
Notice that the confidence interval is symmetric around 500.
| 123,544
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| 132,355
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TITLE: Probability- conditional and binomial
QUESTION [1 upvotes]: 6) A hospital receives 1/5 of its flu vaccine shipments from Company X and the remainder of its shipments from other companies. Each shipment contains a very large number of vaccine vials.
For Company X’s shipments, 10% of the vials are ineffective. For every other company, 2% of the vials are ineffective. The hospital tests 30 randomly selected vials from a shipment and finds that one vial is ineffective.
What is the probability that this shipment came from Company X?
How on earth am I supposed to come to the conclusion that this is a conditional probability? This may even seem like a even more stupid question but why is the probability that there is one ineffective vial equal to probability that it came from company x plus the probability that it came from another company?
10) A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year period of time. Individual participants in the study drop out before the end of the study with probability 0.2 (independently of the other participants).
What is the probability that at least 9 participants complete the study in one of the two groups, but not in both groups?
Probability of dropping = .20
Probability of staying = .80
Probability that at least 9 complete the study in either group is equal to the probability:Of 9 +Of 8 completing + 7 completing ….+of 0 completing
$P[9]= b(9,10,.80)= {10 \choose 9} \times .80^9 \times .20^1= .2684$
$P[8]= b(8,10,.80= {10 \choose 8} \times .80^8 \times .20^2=$
$P[7]= b(7,10,.80)= {10 \choose 7} \times .80^7 \times .20^3=$
$P[6]= b(6,10,.80)= {10 \choose 6} \times .80^6 \times .20^4=$
$P[5]= b(5,10,.80)= {10 \choose 5} \times .80^5 \times .20^5=$
$P[4]= b(4.,10,.80)= {10 \choose 4} \times .80^4 \times .20^6$
$P[3]= b(3,10,.80)= {10 \choose 3} \times .80^3 \times .20^7=$
$P[2]=b(2,10,.80)= {10 \choose 2} \times .80^2 \times .20^8=$
$P[1]= {10 \choose 1} \times .80 \times .20^9=$
$P[0]= {10 \choose 0} \times 1 \times .20^10=$
${2 \choose 1} \times P{x \leq 9}= $
Since we want either group one or group 2 we have {2 \choose 1} ways either one of the two groups has at least 9 complete the study…This was the reasoning behind my answer. Yet why is this incorrect?
$P[N_i ≥ 9] = {10 \choose 9} \times (0.8)^9 (0.2) + {10 \choose 10} \times (0.8)^{10} = 0.3758 $
This tells me the probability that we have at least 9 complete the study from either group. Correct?
$P[N1 ≥ 9,N2 < 9] + P[N1 < 9,N2 ≥ 9] = 0.3758·(1−0.3758) + (1−0.3758)·0.3758=.469$
Then they take 1 minus this probability which gives the probability that at most 9 complete the study times the probability that at least 9 complete the study why?
REPLY [0 votes]: How on earth am I supposed to come to the conclusion that this is a conditional probability?
You seek a probability under a given condition. Ie: the probability for the origin being company X given the sample was one ineffective among thirty.
This may even seem like a even more stupid question but why is the probability that there is one ineffective vial equal to probability that it came from company x plus the probability that it came from another company?
That is called the Law of Total Probability: $$\begin{align}\mathsf P(B) &= \mathsf P(B\cap(A\cup A^\complement)) \\[1ex] &= \mathsf P(B\cap A)+\mathsf P(B\cap A^\complement) \\[1ex] &= \mathsf P(B\mid A)\mathsf P(A)+\mathsf P(B\mid A^\complement)\mathsf P(A^\complement)\end{align}$$ After which, you apply Bayes' Rule $$\begin{align}\mathsf P(A\mid B) ~&= ~\dfrac{\mathsf P(B\mid A)\mathsf P(A)}{\mathsf P(B\mid A)\mathsf P(A)+\mathsf P(B\mid A^\complement)\mathsf P(A^\complement)} \end{align}$$
This was the reasoning behind my answer. Yet why is this incorrect?
Your reasoning by symmetry is okay, however, you neglected the possibility that both groups might fail to have enough.
Further you are after the probability that one group succeeds while the other fails, to have enough. $$\mathsf P((F_1\cup F_2)\setminus(F_1\cap F_2))~=~\mathsf P((F_1\cap F_2^\complement)\cup (F_1^\complement \cap F_2)) ~=~ 2 \,\mathsf P(F_\star)\,(1-\mathsf P(F_\star))$$
Then they take 1 minus this probability which gives the probability that at most 9 complete the study times the probability that at least 9 complete the study why?
The rule of complements: The probability for an event and the probability for its complement sum to one.
Thus: $\mathsf P(F_\star^\complement)=1-\mathsf P(F_\star)$
In this case $\mathsf P(N_\star< 9)=1-\mathsf P(N_\star\geq 9)$ and it is much easier to find $\mathsf P(N_\star\geq 9)$ . $$\mathsf P(N_\star\geq 9) = \tbinom{10}9 0.8^90.2 + 0.8^{10}$$
| 219,500
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\begin{document}
\begin{abstract}
A manifold with fibered cusp metrics $X$ can be considered as a geometrical generalization of locally symmetric spaces of $\Rational-$rank one at infinity. We prove a Hodge-type theorem for this class of Riemannian manifolds, i.e.\ we find harmonic representatives of the de Rham cohomology $H^p(X)$. Similar to the situation of locally symmetric spaces, these representatives are computed by special values or residues of generalized eigenforms of the Hodge-Laplace-Operator on $\Omega^p(X)$.
\end{abstract}
\maketitle
\section*{Introduction and Statement of Results}
The classical theorems of geometric topology, such as the Hodge theorem, the signature
theorem and the index theorem, reveal a profound relationship between analysis of
classical linear operators over a smooth compact manifold and the topology of the manifold.
Since their proofs in the middle of the last century, there has been much interest in
extending these theorems to more general settings, where the manifolds may either
be noncompact or may have singularities. The analytic approaches have often involved
creation of pseudodifferential operator calculi suited to certain geometric settings,
such as the $b$-calculus of Melrose. Another approach has sprung from techniques
in analytic number theory, and the analysis of the Laplacian over noncompact
locally symmetric spaces, as for example in the work of G.~Harder.
Recall that the classical Hodge theorem states that the natural map from harmonic
forms over a compact smooth manifold to de Rham cohomology classes is an isomorphism.
In the situation when $X$ is not compact or is not smooth, there is no such general statement. One possible extension is to consider square integrable harmonic forms $\H_{(2)}(X)$ and identify them with a topologically defined space.
For several geometric situations, including manifolds with cylindrical ends (\cite{APS}), conical singularities (\cite{chgr}) and locally symmetric spaces (e.g. \cite{zucker}, \cite{saper}), such theorems of ``Hodge type'' have been found.
Manifolds with fibered cusp metrics can be considered as a geometrical generalization of $\Rational$-rank one locally symmetric spaces at ``infinity'' as well as of manifolds with cusps or cylindrical ends.
In \cite{hunsicker1} methods from the $\phi-$calculus developed by Melrose \cite{Mel}, Mazzeo, Vaillant \cite{vaill} and others
have been used
to find an identification of $\H_{(2)}^p(X)$ with a subspace of the intersection cohomology.
We want to take another approach and identify the de Rham cohomology of a manifold with fibered
cusps with a space of harmonic forms.
Generally these forms will not be square integrable.
In the early paper \cite{harder} of G.~Harder such a theorem for locally symmetric spaces of $\Rational$-rank one was proved. In this situation the representatives of the de Rham cohomology classes are either $L^2-$harmonic forms, or they are defined by special harmonic values of Eisenstein series. In \cite{mu-unpub} W.~M\"uller suggested to use analytical arguments to find a similar theorem in the context of manifolds with cusps, by replacing Eisenstein series with generalized eigenforms of the Laplacian. His method relies on
an explicit parametrix construction for the resolvent which allows
an investigation of the scattering- and spectral theory of the underlying manifold.
We take these ideas as the starting point to prove a Hodge-type theorem for a large class of manifolds with fibered cusp metrics.
For a Riemannian submersion $M\to B$ with $f$-dimensional standard fiber $F$ we equip
$Z=\mathbb{R}^+\times M$ with the Riemannian metric
$g^Z=du^2+\pi^*g^B+e^{-2u} g^{F_b}$, where $g^B$ and $g^{F_b}$ are the Riemannian metrics on
$B$ resp.~the vertical tangent bundle $TF$.
A Riemannian manifold $X$ is called manifold with fibered cusp metric, if $X$ is isometric to $(Z, g^Z)$ outside a compact set.
As long as the fibers are not points, $X$ is a complete manifold with finite volume. The splitting
$TM=\pi^*TB\oplus TF$ of the tangent bundle induces an isomorphism $\Omega^*(M)=\Omega^*(B,W)$, where
$W$ is the vector bundle whose fiber over $b\in B$ is $C^\infty(F_b, \Lambda(T^*F)|_{F_b})$. On $W$ one can define a Hermitean metric and thus the fiber-wise ``vertical''
Laplacian. In this way we can
define `fiber-harmonic forms' over $M$, and from there, over $Z$. These turn out to play
an crucial role in the spectral theory of the Laplacian on $X$.
For the analysis of the spectral theory it is necessary to ensure that fiber-harmonic forms are an invariant subspace of $\Delta_Z$. To that end we impose two obstructions on the submersion $M\to B$, namely (A) that the horizontal distribution is integrable and (B) that fiber-harmonic forms are an invariant subspace of the ``horizontal Laplacian'' $\Delta_{1,0}=d^{1,0}\delta^{1,0}+\delta^{1,0}d^{1,0}$.
To find a geometrical interpretation of condition (B) is an open question, but
we can prove the following sufficient criterion
\begin{introenv}{Proposition}{satzproj}
If the mean curvature $H$ of the fibers is projectable, then the decomposition \eqref{faserorth} is invariant under $d^{1,0}$ and $\delta^{1,0}$.
\end{introenv}
Let $r:H^p(X)\to H^p(M)$ be the standard restriction map induced from the inclusion $M \subset X$. Using generalized eigenforms of $\Delta_X$ we will explicitely construct a map $\Xi$ from $\H^p(M)$ into the smooth harmonic forms on $X$, which extends to a map $H^p(M)\to H^p(X)$.
We obtain the following theorem:
\begin{introenv}{Theorem}{thm1}
Let $H_!^p(X)\defgleich\bild (H_c^p(X)\to H^p(X))$ be the image of cohomology with compact support in the de~Rham-cohomology.
Let $H_{\mathrm{inf}}^p(X)$ be a complementary space to $H_!^p(X)$ in $H^p(X)$,\index{$H_{\rm inf}^p(X)$}\index{$H_{\shriek}^p(X)$}
\begin{equation}
H^p(X)=H_!^p(X)\oplus H_{\mathrm{inf}}^p(X).\label{thsplit}
\end{equation}
Let $R^p\defgleich\bild (r: H^p(X)\to H^p(M))$.
Then $\Xi(R^p)$ is isomorphic to $H_{\text{\upshape{inf}}}^p(X)$
and\/ $\Xi(H^p(M))=\Xi(R^p).$
\end{introenv}
Since all classes on the right hand side of \eqref{thsplit} have harmonic representatives, this indeed is a ``Hodge-type'' theorem.
The paper is organized as follows.
First we investigate the spectral theory of the Hodge-Laplace operator on $X$. To that end we first examine the spectral theory of the noncompact end $Z$ using the Friedrichs extension of the Laplacian on compactly supported forms.
As mentioned above, fiber-harmonic forms play a key role. In the classical theory of automorphic forms, eigenforms of the Laplacian which are orthogonal to fiber-harmonic forms are known as ``cusp forms''.
It turns out that fiber-harmonic forms determine the essential spectrum, whereas
the cusp forms are associated with the point spectrum of $\Delta_Z$, i.e. they are $L^2-$eigenforms (\textbf{Propositions \ref{kptr}} and \textbf{\ref{wspek}}).
One important consequence of the two conditions $A$ and $B$ introduced above is \textbf{Proposition \ref{satzspec}}, which states that the de Rham cohomology of $M$ can be identified with $\Delta_{1,0}$-harmonic forms on $B$ with values in harmonic sections in the fibers:
\[
H^p(M)\cong\bigoplus_{r+s=p} \H^r(B,\H^s(F)).
\]
This allows the parametrix construction of the resolvent for the Laplacian from the setting of manifolds with cusps (e.g. \cite{mu-rank1}) to be carried out in our more general fibered setting (section \ref{parakapform}).
With this explicit knowledge of the resolvent kernel, the spectral decomposition of $\Delta_Z$ can be computed.
Furthermore, an argument from mathematical scattering theory shows
\begin{introenv}{Proposition}{acspec1}
The absolutely continuous part $L^2_{ac}\Omega^p(X)$ of $\dom\Delta_X$ is unitarily equi\-valent to the fiber-harmonic forms $\Pi_0L^2\Omega^p(Z)$.
\end{introenv}
In this sense, the spectral theory on $X$ is determined by the spectral theory of $\Delta$ on $Z$.
The spectral resolution of $L^2_{ac}\Omega^p(X)$ is given by generalized eigenforms (GEs). In the theory of locally symmetric spaces, these are given by Eisenstein series. Since GEs are used frequently in this paper, it is useful to briefly recall some facts about them. Their construction in section \ref{keif} is similar to \cite{mu-cusprk1}.
For every fiber-harmonic form on $M$--or every cohomology class in $H^p(M)$--there exists a GE on $X$ which depends on a parameter on an infinite covering of $\Complex$, the spectral surface. Near $0$ this covering is twofold, and the $E(s,\phi)$ are parametrized by harmonic forms $\phi$ on $M$.
Let $\phi\in \H^{p-k}(B,\H^k(F))$ and set $d_k=|f/2-k|$. Then
$s\mapsto E(s,\phi)$ is meromorphic for $s\in U\subset \Complex$ and $E(s,\phi)\in \Omega^p(X)$ satisfies a growth condition on the end $Z$. Furthermore
\[
\Delta E(s,\phi)=s(2d_k-s)E(s,\phi),
\]
and $E(s,\phi)$ is uniquely determined by these conditions.
From the conditions one can also conclude that the asymptotic expansion on the end $Z$ of the fiber-harmonic part of $E$ is given by
\begin{equation}
\Pi_0 E(s,\phi) = e^{(f/2-k-d_k+s) r}\phi+\sum_{l=0}^f e^{(f/2-l+d_l-s) r}T^{[l]}(s)(\phi)
+G(s,\phi),\label{introasym}
\end{equation}
where $G(s,\phi)\in L^2\Omega^p(X)$ for $\Re(s)>d_k$ and $T^{[l]}(s)$ are linear operators $\H^*(M)\to\H^*(B,\H^l(F))$, which are meromorphic in $s$. In the context of mathematical scattering theory, the $T^{[l]}(s)$ are referred to as scattering operators.
In view of the Hodge-theorem the spectral value $s=2d_k$ is of particular interest, because if $E(s,\phi)$ is defined there, it is harmonic. Thus the next task is to identify the poles of $s\mapsto E(s,\phi)$.
In the theory of automorphic forms, information about the poles of generalized eigenforms can be read off from product formulas known as the Maa{\ss}--Selberg relations. We will derive a similar formula for the inner product of GEs which are perpendicular to fiber-harmonic forms outside of a compact set (\textbf{Proposition \ref{maass-selb1}}).
This provides us with detailed information about the location and order of poles of $s\mapsto E(s,\phi)$. In particular, the order of a pole in $s=2d_k$ coincides with the maximal order of a pole of the scattering operator $T(s)$ and is at most one (section \ref{epole}).
Moreover the Maa{\ss}--Selberg relations allow us to derive important properties of the residues of the scattering operator and GEs: Let ${\tC}^{[k]}\defgleich \res_{s=2d_k}T^{[l]}(s) $ and $\tE(\phi)\defgleich\res_{s=2d_k}E(s,\phi)$.
Then $\H^{*,k}(M)\defgleich\H^*(B,\H^k(F))$ splits into the orthogonal direct sum
\[
\H^{*,k}(M)=\ker {\tC}^{[k]}\oplus \bild {\tC}^{[k]},
\]
and the residue $\tE(\phi)$ is in $L^2\Omega^p(X)$.
For the middle fiber degree we have even more information about $\H^{*,f/2}(M)$.
Again from the Maa{\ss}--Selberg relations and the functional equations for $E$ we can conclude that
$T^{[f/2]}(s)$ is regular at $s=2d_k=0$ and that
$T^{[f/2]}(0)$ is selfadjoint and an involution on $\H^{*,f/2}(M)$. Thus
$\H^{*,f/2}(M)=\H_+\oplus \H_-$, where $\H_\pm$ are the $\pm 1$-eigenspaces of $T^{[f/2]}(0)$.
Now we have all the information required for the classification of harmonic representatives of $H^p(X)$. We use
the following idea that goes back to G.~Harder.
Let $\phi\in\H^r(B,\H^k(F))$.
If the generalized eigenform $E(\cdot,\phi)$ does not have a pole at $s=2d_k$, then it is a solution of $\Delta E=0$ which is not square integrable.
Then it remains to determine under which conditions $E(2d_k,\phi)$ is closed.
For that, we employ the functional equations from
\textbf{Corollary \ref{funceq}} that are derived from the asymptotic expansion \eqref{introasym}.
If there is a pole at $2d_k$, we instead consider the residue, $\tE(\phi)$. This is a closed $L^2-$harmonic form due to \textbf{Corollary \ref{zerlres}}, thus a representative in both $H^p(X)$ and $H_{(2)}^p(X)$.
In this way to every $\phi\in \H^p(M)$ we may associate a so-called singular value, that is, a closed harmonic form $\Xi(\phi)\in \Omega^p(X)$. We interpret $\Xi$ as a linear map $H^p(M)\to H^p(X)$.
This is the map that is needed for Theorem \ref{thm1} and the final step is to prove the isomorphism $\Xi(\bild r)\simeq H_{\rm{inf}}$.
The image of the restriction map $r$ is of independent interest, in fact, the proof of Theorem \ref{thm1} relies heavily on its explicit description given in
\begin{introenv}{Theorem}{thm2}
Let
\[
\mathcal{A}^p=\bigoplus_{k=0}^f \mathcal{A}^{(p-k,k)}\qquad\text{with}\qquad \mathcal{A}^{(p-k,k)}\defgleich
\begin{cases}
\bild {\tC}^{[k]}, & k<f/2\\
\H_+^p,&k=f/2\\
\astm\ker {\tC}^{[f-k]}, & f/2<k\\
0,& \text{otherwise}
\end{cases}
\]
Then under the Hodge-isomorphism
\[
\bild (r: H^p(X)\to H^p(M))\simeq \mathcal{A}^p.
\]
\end{introenv}
The detailed knowledge we have obtained about $H_{\mathrm{inf}}^p(X)$ also allows us to compute
the signature of $X$ directly in \textbf{Proposition \ref{gleichesign}}, by showing that there are involutions $\tau$ on $L^2-$harmonic forms, which commute with the construction of $\tE$:
\[
\tau_X \tE(\omega) = \tE(\tau_Z\omega).
\]
This recovers the identity
$
L^2-\sign(X)= \sign(X_0,\partial X_0)
$
proven in \cite{dai}.
\subsection*{Acknowledgements}
This paper includes part of my Ph.D.~thesis. I am indebted to my advisor W.~M\"uller for his mathematical guidance and deep insight. Also I am grateful to Eugenie Hunsicker, Alexander Strohmaier and Gregor Weingart for their interest and helpful discussions.
Finally I would like to thank MSRI for hospitality during the program ``Analysis on Singular Spaces''.
\section{Manifolds with fibered cusp metrics}
\subsection{Manifolds with fibered cusp metrics}
\label{mfgspitz}
Let $(M,g^M)$ and $(B,g^B)$\index{$M$}\index{$B$} be closed, connected orientable Riemannian manifolds and $\pi: M\to B$ a Riemannian submersion.
The fibers $F_b\defgleich\pi^{-1}(\{b\})$\index{$F$} are closed, and we assume that they are connected.
Since $(M,g^M)$ is complete, $\pi: M\to B$ is a fiber bundle and all fibers are diffeomorphic, the diffeomorphism being given by horizontal lift of curves in $B$. In particular $f\defgleich\dim F$ is constant.
Let $TF$\index{$TF$} be the vertical tangent bundle, i.e. $(TF)_y=T_y F_{\pi(y)}$. Let $g^F$ denote the family of Riemannian metrics on $TF$ induced by $g^M$.
Let $T^H M$\index{$T^H M$} be a horizontal distribution for $\pi:M\to B$, i.e.
$
TM=T^H M\oplus TF.
$
By definition of a Riemannian submersion
$
T^H M \simeq \pi^* T B
$
is an isometry.
We define a family of metrics on $M$ by
\[
g_u^M =\pi^* g^B + e^{-2u} g^{F}, \qquad u\in\Reell^+,
\]
and equip $Z\defgleich\Reell^+\times M$ with the metric $g^Z\defgleich du^2+g_u^M$.\index{$Z$}
A manifold
$X$\index{$X$} is called \emph{manifold with fibered cusp metric}, if $X$ is isometric to $Z$ outside of a compact set $X_0$\index{$X_0$}. $X$ is a complete Riemannian manifold.
\subsubsection{Examples}
\begin{enumerate}[a)]
\item
If the base $B$ is a point, the metric on $Z$ is
\[
du^2+e^{-2u} g,
\]
so that $Z$ is a cusp with base $M$ and $X$ is a manifold with cusp end.
Similarily, if $\pi=\text{id}$, then $X$ is a manifold with cylindrical end as considered in \cite{APS}.
\item In \cite{mu-cusprk1}, \cite{mu-rank1} W.~M\"uller considers locally symmetric spaces of $\Rational-$rank $1$.
Here we consider only the case of a single cusp: Let $X=X_0\cup_M Z$, where $X_0$ is a compact manifold with boundary $M$, and $Z=\Reell^+\times M$ isometric
to a cusp of a locally symmetric space with $\Rational-$rank $1$.
Then there is a fiber bundle $M\to B$ where the fiber $F=\Gamma\cap N\backslash N$ is a compact nilmanifold, and $B$ is a locally symmetric space. The metric on $\Reell^+\times M$ then locally takes the form
\begin{equation}
g^Z=du^2+\pi^*g^B+e^{-2 a u} g_1(b)+e^{-4 a u} g_2(b),\label{cprk1}
\end{equation}
where $a>0$ and $g^B$ is the metric on $B$. $g_1(b), g_2(b)$ have support along the fibers $F_b$ over $b\in B$.
The volume form on $Z$ is given by
\[
\vol_Z= e^{-q u} du\, \vol_B \vol_{F_b},
\]
for some $q>0$. If as the symmetric space we choose the $n$-dimensional hyperbolic space
$
SO_e(n,1)/SO(n)
$, then the metric \eqref{cprk1} takes the form
\[
g^Z=du^2+\pi^*g^B+e^{-2 a u} g_1(b),
\]
see Proposition 2.9 in \cite{weber} and e.g. \cite{carped}.
A similar situation is considered in the work \cite{harder2}, there $X=\Gamma\backslash(\mathds{H}\times \ldots \times \mathds{H}\times Y\times\ldots \times Y)$
with the upper complex half plane $\mathds{H}$ and the 3-dimensional hyperbolic space $Y,$ and a torsion free conguence subgroup $\Gamma\subset SL(2,\mathfrak{O})$.
\end{enumerate}
Now we want to recall some facts about connections in the Riemannian fiber bundle $M\to B$, following
\cite{bismut-lott}, pg. 323-329 and \cite{bismut-cheeger}, pg. 53-55.
As a bundle of $\Ganz-$graded algebras over $M$ we have the isomorphism
\begin{equation} \label{bm3.3}
\Lambda(T^*M)\simeq \pi^*(\Lambda(T^*B))\otimes \Lambda(T^* F).
\end{equation}
Let $W$\index{$W$} be the smooth infinite dimensional $\Ganz-$graded vector bundle over $B$, whose fiber $W_b$ over $b\in B$ is
$C^\infty(F_b, \Lambda(T^*F)|_{F_b})$.
This means
\[
C^\infty(B,W^{(k)})\simeq C^\infty(M,\Lambda^k(T^*F))
\]
and we have an isomorphism of $\Ganz-$graded vector spaces
\[
\Omega^\bullet(M)\simeq \Omega^\bullet(B,W)\defgleich C^\infty(B, \Lambda^\bullet(T^*B)\otimes W).
\]
Let $d^M$ be the outer derivative on $\Omega(M)$. Then as usual $(d^M)^2=0$.
We consider the decomposition of $d^M$ in horizontal and vertical components.
Let
\[
\Omega^{a,b}(M)\defgleich \Omega^a(B,W^{(b)})\;,\qquad \Omega^{*,k}\defgleich\bigoplus_{j=0}^{n-k} \Omega^{j,k},\quad n=\dim M
\]
and consider the decomposition $d^M=\sum_{i+j=1}d^{i,j}$
with
$
d^{i,j}:\Omega^{a,b}\to\Omega^{a+i,b+j}.
$
The fiber-degree operator $\kappa$ is the linear operator \index{$\kappa$}$\kappa:\Omega^*(M)\to \Omega^*(M)$, which is defined by
$
\kappa \phi= k \phi
$
for $\phi\in \Omega^{*,k}$.
In \cite[Proposition 3.4]{bismut-lott} and \cite[Proposition 10.1]{BGV} it is proven that
\begin{equation} \label{dmzerl}
d^M=d^{0,1}+d^{1,0}+d^{2,-1}
\end{equation}
where
\begin{itemize}
\item $d^{0,1}=d^F\in C^\infty(B, \text{Hom}(W^\bullet,W^{\bullet+1}))$ is the differential along the fibers.
\index{$d^F$}
\item $d^{1,0}$ is given as follows. Let $X$ be a smooth vector field on $B$ with horizontal lift $X^H\in C^\infty(M, T^H M)$.
For $s\in C^\infty(B,W)$ and a vector field $X$ on $B$, we define a covariant derivative on $W$ that preserves the $\Ganz-$grading by
\[
\nabla^W_X s= \text{Lie}_{X^H}s.
\]
$\nabla^W$ is extended in a unique way to a covariant derivative
\[
d^{1,0}:\Omega^\bullet(B,W)\to \Omega^{\bullet+1}(B,W),
\]
so that Leibniz' rule
\begin{equation}
d^{1,0}(\alpha\wedge\theta)=d^B\alpha\wedge \theta +(-1)^r \alpha\wedge d^{1,0}\theta,\qquad \alpha\in \Omega^r(B),\theta\in \Omega^\bullet(B,W)\label{leibniz}
\end{equation}
holds.
\item Finally
\[
d^{2,-1}\in \Omega^2(B,\text{Hom}(W^\bullet,W^{\bullet-1}))
\]
is given by inner multiplication with curvature of the fibers, i.e. for a pair $(X,Y)$ of vector fields on $B$
\begin{equation}
d^{2,-1}(X,Y)(\omega)=i_{\Vert[Y^H,X^H]} \omega\label{defd21}
\end{equation}
where $\Vert$\index{$\Vert$} is the projection from $TM$ onto $TF$.
\end{itemize}
\subsection{Fibre-harmonic forms}\label{kfash}
Let $\ast$ be the fiberwise Hodge-star-operator with respect to $g^F$. This is an operator on
\[
C^\infty(M, \Lambda(T^*F))\simeq C^\infty(B,W).
\]
In this way $W$ obtains a Hermitean metric $h^W$, so that for $s, s'\in C^\infty(B,W)$ and $b\in B$
\begin{equation*}
(\sce{s}{s'}_{h^W})(b) = \int_{F_b} s(b)\wedge \ast s'(b) = \int_{F_b} \sce{s(b)}{s'(b)}_{F_b} \vol_{F_b}.
\end{equation*}
Here $\sce{\cdot}{\cdot}_{F_b}$ is the scalar product on $\Omega^*(F_b)$ induced by $g^F$.
Using the metric $g^B$ on $B$, the hermitean fiber product $h^W$ extends to $\Omega^*(B,W)$.
\begin{defn}
Let $\delta^{i,j}$ be the fiberwise formal adjoint of $d^{i,j}$, and $\delta^F\defgleich\delta^{0,1}$.
\end{defn}\index{$\delta^F$}\index{$\delta^{i,j}$}
Let $H^*(F)=\bigoplus_{i=0}^f H^i(F)$ be the $\Ganz-$graded vector bundle over $B$, whose fiber over $b\in B$
is the cohomology $H^*(F_b)$ of the complex $(W_b, d^{F_b})$.
Let $V=\delta^F-d^F \in C^\infty(B, \text{End}(W))$. The Hodge-theorem gives an isomorphism
$
\H(F_b)\defgleich \ker(V_b)\cong H(F_b)
$
and this induces an isomorphism
\begin{equation} \label{h234}
\H^*(F)\defgleich \ker(V)\cong H^*(F)
\end{equation}
of $\Ganz-$graded finite dimensional vector bundles.
\begin{defn}
An element in $\omega\in \Omega^*(B,\H^*(F))$ is called \emph{fiberwise harmonic}, or simply \emph{fiber-harmonic}.
\end{defn}
Let $\Delta^F\defgleich d^F\delta^F+\delta^F d^F$ be the Hodge-Laplace-operator along the fibers. Since the fibers of $M\to B$ are closed submanifolds, $\omega\in \Omega^*(B,W)$ is fiber-harmonic if and only if $\Delta^F\omega=0$.
Since $\ker(V)$ is a subbundle of $W$, it inherits a hermitean metric from $h^W$ by projection, and
with respect to $h^W$ there is a direct orthogonal decomposition
\begin{equation}
\Omega^*(B,W)=\Omega^*(B,\H^*(F))\oplus \Omega^*(B,\H^*(F)^\perp)\label{faserorth}.
\end{equation}
Let $\Pi_0$\index{$\Pi_0$} resp. $\Pi_\perp$\index{$\Pi_\perp$} denote the projections onto $\Omega^*(B,\H^*(F))$ resp. $\Omega^*(B,\H^*(F)^\perp)$.
Then $\Pi_0 d^{1,0}$ is a connection on $\Omega(B,\H(F))$, which can be understood as a connection on
$\Omega(B,H(F))$ by \eqref{h234}.
The following statement follows directly from Propositions 2.5 and 2.6 in \cite{bismut-lott}.
\begin{satz}\label{flachzus}
The connection $\nabla^{H(F)}=\Pi_0 d^{1,0}$ on $\Omega(B, H(F))$ is flat, i.e. $(\nabla^{H(F)})^2=0$.
\end{satz}
Generally the decomposition \eqref{faserorth} is not invariant under $d^{1,0}$, i.e. $\Pi_\perp d^{1,0} \Pi_0 \neq 0$. We will give a sufficient criterion for invariance in Proposition \ref{satzproj} below.
\begin{defn}
Let\index{$H$} $\Hor:C^\infty(M, TM) \to C^\infty(M, T^HM)$ be the projection onto the horizontal distribution and $\{U_j\}_{j=1}^r\subset C^\infty(M, TF)$ be a vertical orthonormal frame. With the Levi-Civita connection $\nabla^M$ on $C^\infty(M, TM)$ we define the \emph{mean curvature $H$ of the fibers} of $\pi:M\to B$ to be the horizontal vector field\index{$H$}
\[
H\defgleich\sum_{j=1}^{\dim F}\Hor\nabla^M_{U_j}{U_j}.
\]
\end{defn}
Recall that a vector field on $M$ is called \emph{projectable}, if there is a vector field $\bH$ on $B$ such that $d\pi H=\bH$. A projectable horizontal vector field is called \emph{basic}.
\begin{satz}\label{satzproj}
If the mean curvature $H$ of the fibers is projectable, then the decomposition \eqref{faserorth} is invariant under $d^{1,0}$ and $\delta^{1,0}$.
\end{satz}
\begin{proof}
By definition of $d^{1,0}$ it is sufficient to prove that $\nabla^W$ leaves
$
C^\infty(B,\H^*(F))\oplus C^\infty(B,\H^*(F)^\perp),
$ invariant,
then claim for $\Omega^*(B,W)$ follows from \eqref{leibniz}.
From $d^2=0$ we already know that $(d^2)^{1,1}=d^{1,0}d^F+d^F d^{1,0}=0$. To prove $\nabla^W C^\infty(B,\H^*(F))\subset C^\infty(B,\H^*(F))$ it remains to show
\[
s\in C^\infty(B,\H^*(F)) \Longrightarrow \delta^F \nabla^W_X s=0
\]
for a basic vector field $X=\pi^* \bX$.
Let $\Phi_t^{\bX}$ be the flow of ${\bX}$. The flow $\Phi_t^{X}$ of $X$ projects onto $\Phi_t^{\bX}$ and is a diffeomorphism of the fibers:
\begin{equation}
\Phi_t^{X}: F_b\to F_{\Phi_t^{\bX}(b)}.\label{fluss}
\end{equation}
Let $\omega$ be the volume form of the fibers, i.e. $\omega$ is the volume form on $M$, such that $\omega|_{F_b}=\vol_{F_b}$ is the volume form on $F_b$.
Then because of \eqref{fluss}
\begin{align*}
\bX(h^W(s_1, s_2))&=\text{Lie}_{\bX} \Big( \int_{F_b} \sce{s_1}{s_2}_{F_b} \vol_{F_b} \Big) \\
&= \int_{F_b} \text{Lie}_X (\sce{s_1}{s_2}_{F_b}) \vol_{F_b} + \int_{F_b} \sce{s_1}{s_2}_{F_b} (\text{Lie}_X \omega)|_{F_b}
\end{align*}
Let $f_X= -g^M(H,X)\in C^\infty(M)$ with the mean curvature $H$ of the fibers. It is well known\footnote{Lemma 10.4 in \cite{BGV} or Theorem 6.6 in \cite{lang}} that
\[
(\text{Lie}_X \omega)|_{F_b} = f_X\vol_{F_b}.
\]
Because $\nabla^W$ is compatible with $\sce{}{}_{F_b}$,
\begin{equation}
\bX(h^W(s_1,s_2))
= h^W(\nabla^W_X s_1,s_2)+h^W(s_1,\nabla^W_X s_2) +h^W(f_X s_1, s_2)\label{bed6}
\end{equation}
Now let $s_2$ be fiber-harmonic, i.e. $\delta^F s_2=0,$ $d^F s_2=0$. Using $\nabla^W_X d^F= - d^F\nabla^W_X$, we obtain
\begin{align*}
0&= \bX(h^W(s_1,\delta^F s_2))
= \bX(h^W(d^F s_1,s_2))\\
&= -h^W( \nabla^W_X s_1, \delta^F s_2)+h^W( s_1, \delta^F \nabla^W_X s_2)+ h^W(f_X d^F s_1, s_2)\\
&= h^W( s_1, \delta^F \nabla^W_X s_2)+ h^W(d^F (f_X s_1), s_2)-h^W(d^F(f_X) s_1, s_2),
\end{align*}
so that
\[
h^W( s_1, \delta^F \nabla^W_X s_2) = h^W(s_1, d^F(f_X) s_2).
\]
In particular for $s_1= \delta^F \nabla^W_X s_2$,
\[
h^W(s_1, s_1)
= h^W(\nabla^W_X s_2 ,d^F(d^F(f_X) s_2))=0
\]
which implies $s_1=0$. This shows $\nabla^W C^\infty(B,\H^*(F))\subset C^\infty(B,\H^*(F))$.
Now let $s_1\in C^\infty(B,\H^*(F)^\perp),$ $s_2 \in C^\infty(B,\H^*(F))$.
If $H$ is projectable with $\pi_* H={\bH}$, then $f_X=-\pi^*g^B(\bX,\bH)$ and
from \eqref{bed6}
\begin{align*}
h^W(\nabla^W_X s_1, s_2)&= h^W(\nabla^W_X s_1, s_2)+h^W( s_1, \nabla^W_X s_2)-g^B(\bX,\bH)h^W(s_1,s_2)\\
&=\bX(h^W(s_1, s_2)) = 0
\end{align*}
so that $\nabla^W C^\infty(B,\H^*(F)^\perp)\subset C^\infty(B,\H^*(F)^\perp)$.
The statement about $\delta^{1,0}$ finally follows from $\Pi_0 \delta^{1,0}\Pi_\perp=(\Pi_\perp d^{1,0}\Pi_0)^*$.
\end{proof}
\subsection[Laplace-operator on $\Omega^*(Z)$]{Laplace-operator on $\mathbf{\Omega^*(Z)}$}\label{klap}
Let $\Delta_Z$ be the Hodge-Laplace-operator on $Z$, where we consider $\Delta_Z$ as linear operator on $L^2\Omega^p(Z)$ with domain
$\dom \Delta_Z=\Omega_0^p(Z)$. Let $\bar\Delta_Z$ be the Friedrichs extension of $\Delta_Z$.
More precisely, we define
$q(\phi,\psi)\defgleich(\phi, \Delta_Z\psi)_{L^2\Omega^p(Z)}$ with $\phi, \psi\in \dom \Delta_Z$.
This is quadratic form with domain $Q(q)=\Omega_0^p(Z)$. Let $\tilde q$ the closure of this form. The Friedrichs extension $\bar \Delta_Z$ of $\Delta_Z$ then
is the selfadjoint operator on $L^2\Omega^p(Z)$ that is defined by $(\phi, \bar\Delta_Z\psi)=\tilde q(\phi,\psi)$.
By definition the domain of the quadratic form of $\bar\Delta_Z$ is the closure of $\dom \Delta_Z=\Omega_0^p(Z)$ in the norm
\[
\|\phi\|_1^2=\|\phi\|^2+q(\phi,\phi),
\]
i.e. the Sobolev-space $H_0^1\Omega^p(Z)$. The domain of $\bar\Delta_Z$ is $H^2\Omega^p(X)\cap H_0^1\Omega^p(Z)$. For $\varphi\in \dom \bar\Delta_Z\cap \Omega^p(\bar Z)$ we have
\begin{equation}\label{relrbd}
i^*\varphi=0,\qquad i^*(\ast\varphi)=0
\end{equation}
with the inclusion $i:M\hookrightarrow Z$. From now on we write \index{$\Delta_Z$}$\Delta_Z$ for $\bar \Delta_Z$.
By\index{$\rho_u$}
$
\rho_u(\omega)=e^{\kappa u} \omega
$
an isometry
\[
\rho_u:(\Omega^*(M),g_u^M)\to (\Omega^*(M),g^M)
\]
is defined.
Let $\alpha, \beta$ be elements in $\Omega^{*,k}(M)$, which depend on the real parameter $u$, i.e.
$\alpha, \beta \in \pi_2^*\Omega^{*,k}(M)$ for the projection $\pi_2: Z=\Reell^+\times M \to M$.
We extend $\rho_u$ to an operator on $\Omega^\ast(Z)$ by setting $\rho_u^{-1}(du\wedge\beta)\defgleich du\wedge \rho_u^{-1}\beta$.
Also it will be useful to consider the isometry\index{$\tilde\rho_u$}
\[
\tilde\rho_u: L^2\Omega^*(Z,g_Z)\to L^2\Omega^*(Z,du^2+g),
\]
which is given by
\[
\tilde\rho_u(\alpha+du\wedge\beta)\defgleich e^{(k-f/2)u}(\alpha+du\wedge\beta)
\]
Finally let\index{$a_k$}\index{$\ua$}\index{$d_k$}\index{$\ud$}
\begin{align*}
a_k&=f/2-k, &d_k&=|a_k|\\
\ua&=f/2\cdot\text{id}-\kappa,& \ud&=|\ua|.
\end{align*}
The local form of $\Delta_Z$ is described in
\begin{satz}\label{s3.9}
For $\omega\in \Omega_0^*(Z)$ we write $\omega=(\alpha,\beta)\defgleich\alpha+du\wedge\beta$, where $\alpha, \beta \in \pi_2^*\Omega^{*}(M)$.
Moreover let
\begin{eqnarray*}
d_u^M&=&e^{-u}d^{2,-1}+d^{1,0}+e^{u}d^{F},\quad D_u^M=d_u^M+ \delta_u^M\\\
\tilde{T}_u&=& -{\textstyle\frac{\partial^2}{\partial u^2}}+\big({\textstyle\frac{f}{2}}-k\big)^2+(D_u^M)^2\\
Q_u&=&e^{-u}(d^{2,-1}-\delta^{2,-1})-e^{u}(d^{F}-\delta^{F}).
\end{eqnarray*}
With respect to the above decomposition
\begin{equation}\label{deltazf}
\tilde\rho_u(d+\delta)^2\tilde\rho_u^{-1} \omega =
\begin{pmatrix}
\tilde{T}_u& -Q_u\\
Q_u & \tilde{T}_u
\end{pmatrix}
\begin{pmatrix}
\alpha\\
\beta
\end{pmatrix}.
\end{equation}
\end{satz}
\begin{proof}
Let $\astm$ be the Hodge-star-operator on $\Omega^*(M)$ with respect to $g^M$.
Then \eqref{deltazf} follows with a straightforward calculation from $\Delta_Z=d^Z\delta^Z+\delta^Z d^Z$ and
\begin{align*}
d^Z(\rho_u^{-1} \alpha+du\wedge \beta) &= \rho_u^{-1}(d_u^M\alpha+du\wedge(\partial_u\alpha-\kappa \alpha))-\rho_u^{-1} (du\wedge d_u^M\beta)\\
\delta^Z(\rho_u^{-1}\alpha+du\wedge \beta)&=\rho_u^{-1}\delta_u^M\alpha -\rho_u^{-1} (du\wedge \delta_u^M\beta+\partial_u\beta-(f-\kappa)\beta).\qedhere
\end{align*}
\end{proof}
\subsection{Cohomology and Harmonic Forms}\label{kohom}
In this chapter we introduce some additional notation and recall known results which will be referred to later.
Let $\Omega_0(X)$ denote the space of differential forms on $X$ with compact support.
As usual let $L^2\Omega^p(X)$ be the closure of $\Omega_0^p(X)$ in the norm induced from the scalar product
\[
\Scr{\phi}{\psi}_{L^2\Omega^p(X)}\defgleich\int_X \phi\wedge \ast\psi.
\]
Most statements about cohomology in this article refer to de~Rham-cohomology\index{$H^p$}
\[
H^p(X)=\frac{\{\omega\in \Omega^p(X)\mid d\omega=0 \}}{d \Omega^{p-1}(X)}
\]
and to de~Rham-cohomology with compact support\index{$H_c^p$}
\[
H_c^p(X)=\frac{\{\omega\in \Omega_0^p(X)\mid d\omega=0 \}}{d \Omega_0^{p-1}(X)}.
\]
Because $\Omega_0^p(X)$ is dense in $L^2\Omega^p(X)$, $d$ has a well-defined strong closure (again denoted by $d$), which is usually called the maximal closed extension of $d$. Note that, since $X$ is complete in our case, all closed extensions of $d$ have the same domain due to a classical result of Gaffney.
The domain of the differential $d_p: L^2\Omega^p(X)\to L^2\Omega^{p+1}(X)$ is
\[
\dom d_p=\big\{ \phi\in \Omega^p(X)\cap L^2\Omega^p(X) \mid d\phi\in L^2\Omega^{p+1}(X) \big\},
\]
where $d\phi$ is understood in the distributional sense.
We define the $p-$th $L^2-$cohomology group\index{$H^p_{(2)}$}
\[
H^p_{(2)}=\frac{\ker d_p}{\bild d_{p-1}}
\]
and the $L^2-$harmonic $p-$forms\index{$\H_{(2)}^p$}
\[
\H_{(2)}^p(X)=\big\{ \omega\in L^2\Omega^p(X)\mid \Delta_X \omega\defgleich(d\delta +\delta d)\omega=0 \big\}.
\]
Here $\Delta_X$ is the selfadjoint extension of Laplace-operator on compactly supported forms $\Delta_X:\Omega_0^p(X)\to \Omega_0^p(X)$ to an operator on $L^2\Omega^p(X)$.
The regularity theorem for elliptic operators states that forms in $\H_{(2)}^p(X)$ are smooth.
For $\omega\in \H_{(2)}^p(X)$ we have $\scr{\Delta \omega}{\omega}=\|d\omega\|^2+\|\delta\omega\|^2=0$, so that
$\H_{(2)}^p\subset \ker d_p$, which induces a map
\[
\H_{(2)}^p\to H^p_{(2)},\qquad\omega\mapsto [\omega].
\]
In general this map is neither injective nor surjective. However, the space of $L^2-$harmonic forms is isomorphic to the \emph{reduced} $L^2-$cohomology:
\[
\H_{(2)}^p\cong H^p_{(2),\text{red}}=\frac{\ker d_p}{\;\overline{\bild d_{p-1}}\;}.
\]
Let $H^p_!(X)\defgleich \bild\big(H_c^p(X)\to H^p(X)\big)$.
It is a well known result (e.g. \cite{anders}) that for a \emph{complete} Riemannian manifold $X$ there is a natural injective map
\[
H^p_!(X)\to H^p_{(2),\text{red}}(X).
\]
In particular every class in $H^p_!(X)$ has a unique $L^2-$harmonic representative.
Finally we want to mention the following theorem of Kodaira:
\begin{equation*}
L^2\Omega^p(X)=\H_{(2)}^p(X)\oplus \overline{\delta \Omega_0^{p+1}(X)} \oplus \overline{d \Omega_0^{p-1}(X)}
\end{equation*}
\section{Spectral Theory}
\subsection[Point spectrum of $\Delta_Z$]{Point spectrum of $\mathbf{\Delta_Z}$}
The decomposition \eqref{faserorth} admits an extension to
\begin{equation}
L^2\Omega^p(Z)=\Pi_\perp L^2\Omega^p(Z)\oplus \Pi_0 L^2\Omega^p(Z) \label{decoml2}
\end{equation}
where $\Pi_0 L^2\Omega^p(Z)$ is the closure of compactly supported forms, which are fiberharmonic in each cross-section $\{u\}\times M\subset Z$.
In analogy to the theory of classical automorphic forms, we define
\begin{defn}\label{spifo}
The \emph{cusp forms on $Z$} are elements in
\[
L_{\text{cusp}}^2\Omega^p(Z)\defgleich\{\omega \in L^2\Omega^p(Z)\mid\exists \lambda\ge 0: \Delta_Z\omega=\lambda\omega, \quad\Pi_0(\omega)=0 \}
\]
Here $\Delta_Z\omega$ first has to be understood in the sense of distributions, but from elliptic regularity we get the smoothness of cusp forms.
The significance of cusp forms in the setting of manifolds with fibered cusps comes from
\begin{satz}\label{kptr}
The restriction of $\Delta_Z$ to the orthogonal complement of fiber-harmonic forms
\[
\Pi_\perp \Delta_Z \Pi_\perp: {\Pi_\perp L^2\Omega^*(Z)}\cap\dom\Delta_Z \to {\Pi_\perp L^2\Omega^*(Z)}
\]
has pure point spectrum.
\end{satz}
\begin{proof}
The proof uses the Minimax-principle \cite[Theorem XIII.1]{reedsimon4} for $A\defgleich\Pi_\perp \Delta_Z \Pi_\perp$.
This states that the real numbers
\[
\lambda_n(A) = \sup_{v_1,\ldots,v_{n-1}\in Q(A)} \inf_{\substack{v_n\in H_0^{1}\\ v_n\perp v_k \forall k<n}}
\frac{\norm{\tilde\rho d^Z\tilde\rho^{-1}v_n}^2+\norm{\tilde\rho\delta^Z\tilde\rho^{-1}v_n}^2}{ \|v_n\|^2}
\]
are either eigenvalues of $A$ or accumulate at the beginning of the essential spectrum of $A$. Here $Q(A)\subset H_0^1\Omega^p(Z)$ is the form domain of $A$.
Because of Theorem XIII.64 in \cite{reedsimon4},
the resolvent of $A$ is compact, if
\begin{equation}
\lim_{n\to\infty}\lambda_n=\infty.\label{ziel1}
\end{equation}
Let $\omega=\alpha+du\wedge \beta, \alpha,\beta\in \Omega_0^{*,k}(Z)$ with $\supp \omega \subset [a,\infty)\times M$ and $\|\alpha\|_{H^1}=1, \|\beta\|_{H^1}=1$;
$\alpha, \beta$ depend on the parameter $u\in \Reell^+$.
Let
\[
V_u\defgleich
\begin{pmatrix}
(D_u^M)^2& -Q_u\\
Q_u & (D_u^M)^2
\end{pmatrix}
\]
so that
\[
\tilde\rho\Delta_Z\tilde\rho^{-1}=(-{\textstyle \frac{\partial^2}{\partial u^2}}+(f/2-k)^2)\begin{pmatrix}
1& 0\\
0 & 1
\end{pmatrix}
+ V_u,\qquad
\]
as sum of quadratic forms.
We will show
\begin{equation}\label{gwz}
|\Scr{V_u\omega}{\omega}| \to\infty\quad\text{for}\quad a\to\infty.
\end{equation}
Then since $-{\textstyle \frac{\partial^2}{\partial u^2}}+(f/2-k)^2$ has pure absolutely continuous spectrum $[(f/2-k)^2,\infty)$, a standard argument using the Minimax-principle shows \eqref{ziel1}.
In the following, all norms and scalar products are meant to be those in $L^2\Omega^p(Z,du^2+g_0)$.
\begin{eqnarray*}
\Scr{V_u\omega}{\omega} &= &\Scr{\pfrac{(D_u^M)^2\alpha-Q_u\beta}{Q_u\alpha+(D_u^M)^2\beta}}{\pfrac{\alpha}{\beta}} \\
&=&\Scr{(D_u^M)^2\alpha}{\alpha}+\Scr{(D_u^M)^2\beta}{\beta}-\Scr{Q_u\beta}{\alpha}+\Scr{Q_u\alpha}{\beta}\\
&=& \norm{D_u^M \alpha}^2 + \norm{D_u^M \beta}^2 + 2 \Scr{Q_u \alpha}{\beta}\\[2ex]
|\Scr{V_u\omega}{\omega}| &\ge& \norm{D_u^M \alpha}^2+ \norm{D_u^M \beta}^2 -2\norm{Q_u\alpha}\norm{\beta}
\end{eqnarray*}
Because $\|\alpha\|_{H^1}=1$, all $\|d^{i,j}\alpha\|, \|\delta^{i,j}\alpha\|$ are bounded by a constant independent of $a$, and the same holds for $\beta$:
\begin{equation*}
|\Scr{V_u\omega}{\omega}|\ge \norm{D_u^M \alpha}^2+ \norm{D_u^M \beta}^2 -C_0(\norm{e^u d^F\alpha}+\norm{e^u\delta^F\alpha})+C_1.\label{xyz888}
\end{equation*}
A simple calculation shows
\begin{equation*}
\norm{D_u^M \alpha}^2 \ge \norm{e^u d^F\alpha}^2+\norm{e^u \delta^F\alpha}^2-C_2\norm{e^u d^F\alpha}
-C_3 \norm{e^u \delta^F\alpha}+C_4, \label{xyz777}
\end{equation*}
and an analogous estimate holds for $ \norm{D_u^M \beta}^2$.
But now $\norm{e^u d^F\alpha}\ge e^{a}\norm{d^F\alpha}$. Thus for every real constant $c$
\[
\norm{d^F\alpha}\neq 0\quad\Rightarrow\quad \lim_{a\to\infty} \big(\norm{e^u d^F\alpha}^2-c \norm{e^u d^F\alpha}\big) = +\infty,
\]
and so we conclude from \eqref{xyz888} and \eqref{xyz777} that \eqref{gwz} holds, i.e.
\[
|\Scr{V_u\omega}{\omega}| \to\infty\quad\text{for}\quad a\to\infty,
\]
if $\norm{d^F\omega}\neq 0$ or $\norm{\delta^F\omega}\neq 0$.
This proves the claim.
\end{proof}
\end{defn}
\subsection[Essential spectrum of $\Delta_Z$]{Essential spectrum of $\mathbf{\Delta_Z}$}
With respect to the decomposition \eqref{decoml2}
we can write
\[
\tilde\rho\Delta_Z\tilde\rho^{-1}= \mathfrak{L}+
\begin{pmatrix}
0 & \Pi_\perp \Delta_Z \Pi_0 \\
\Pi_0 \Delta_Z \Pi_\perp & 0\end{pmatrix}
\quad\text{with}\quad
\mathfrak{L}\defgleich\begin{pmatrix}
\Pi_\perp\Delta_Z \Pi_\perp & 0 \\[1ex]
0 & \Pi_0\Delta_Z \Pi_0\\
\end{pmatrix}.
\]
Here the off-diagonal terms
\begin{equation*}
\Pi_\perp \Delta_Z \Pi_0 =
\Pi_\perp \begin{pmatrix}
(D_u^M)^2 & -Q_u \\
Q_u & (D_u^M)^2
\end{pmatrix}
\Pi_0
\quad\text{and} \quad\Pi_\perp \Delta_Z \Pi_0
\end{equation*}
are bounded operators in $L^2\Omega^p(Z)$.
Now we want to examine the contribution of fiberharmonic forms to the spectrum of $\Delta_Z$.
It is easy to see
that $\Pi_0\tilde\rho\Delta_Z\tilde\rho^{-1}\Pi_0$
is a relatively compact perturbation of
$
-{\textstyle \frac{\partial^2}{\partial u^2}}+(f/2-\kappa)^2,
$
and as such has pure absolutely continuous spectrum. Together with Proposition \ref{kptr} we get
\begin{satz}\label{wspek}
$\Delta_Z$ and $\mathfrak{L}$ have the same essential spectrum.
\end{satz}
\begin{proof}
The proof is similar to Theorem 2 in \cite{lott}, see \cite{jmthesis}
\end{proof}
\subsection{Two conditions}\label{Bedingungen}
For the remaining chapters we will make the assumptions
\begin{enumerate}[(A)]
\item The horizontal distribution is integrable, i.e. $d^{2,-1}=0$.
\item $\Pi_\perp \delta^{1,0} \Pi_0 =0$
\end{enumerate}
An immediate consequence of (A) is
\[
(d^{1,0})^2=d^{2,-1}d^{0,1}+(d^{1,0})^2+d^{0,1}d^{2,-1}=({(d^M)}^2)^{2,0}=0
\]
so that
\[
\Delta_{1,0}\defgleich d^{1,0}\delta^{1,0}+\delta^{1,0}d^{1,0}=(d^{1,0}+\delta^{1,0})^2.
\]
From the Hodge-decomposition we have $[\Pi_0, d^F]=[\Pi_0 ,\delta^F]=0$.
Together with condition (B) this implies $[d^Z, \Pi_0]=[\delta^Z, \Pi_0]=0$,
thus $\Delta_Z$ leaves the splitting of $L^2\Omega^*(Z)$ into fiber-harmonic forms and their orthogonal complement invariant. Also
$\Pi_0\Delta_Z\Pi_0$ takes the especially simple form
\[
\tilde\rho_u\Pi_0(d^Z+\delta^Z)^2\Pi_0\tilde\rho_u^{-1}=(-\partial_u^2+(f/2-\kappa)^2 +\Delta_{1,0})\Pi_0
\]
in this case.
\begin{lem}\label{delta10ell}
Under the given conditions (A) and (B),
\[
\Delta_{1,0}=d^{1,0}\delta^{1,0}+\delta^{1,0}d^{1,0}:\Omega^*(B,\H^k(F))\to \Omega^*(B,\H^k(F))
\]
is a non-negative symmetric elliptic operator.
\end{lem}
\begin{proof}
From Proposition \ref{flachzus} we get $\Delta_{1,0}=(d^{1,0}+\delta^{1,0})^2$, which is non-negative.
From the local formulas for $\nabla^W$ and $(\nabla^W)^*$ in \cite{bismut-lott} (Proposition 3.5 and 3.7 there; also see \cite{gilkeybook})
one concludes that $\Delta_{1,0}$ is a generalized Laplace-operator; as such it is elliptic.
\end{proof}
A sufficient criterion for condition (B) is given by Proposition \ref{satzproj}. In particular (B) is fulfilled
when the fibers of $\pi: M\to B$ are minimal, or in the case of a warped product.
A necessary condition for (A) is given by
\begin{lem}[{\cite[Lemma 1.7.2]{gilkeybook}}]
If the horizonal distribution of a Riemannian submersion $\pi:M\to B$ is integrable, there are local coordinates $m=(y,b)$ in $M$, so that $\pi(m)=b$. In these coordinates the metric on $M$ takes the form
\begin{equation}
g^M= h_{ij}(b)db^i\otimes db^j+ f_{\alpha\beta}(y,b)dy^\alpha\otimes dy^\beta \label{prodmed1}
\end{equation}
If in addition $B$ is simply connected, then $\pi:M\to B$ is a global product with metric \eqref{prodmed1}.
\end{lem}
\textbf{Example.\hspace{1em}}
Let $G/K$ be a symmetric space of non-compact type, where $K\subset G$ is a maximal compact subgroup of the non-compact semisimple Lie group $G$.
More precisely $G$ is the group of real points of a semisimple algebraic group $\mathbf{G}\subset GL(n,\Complex)$, which is defined over $\Rational$.
Let $\Gamma\subset G$ be an arithmetic lattice\footnote{see e.g. \cite{weber}, Definition 2.1} of $\Rational-$rank $1$ so that $X=\Gamma\backslash G/K$
is a locally symmetric space of $\Rational-$rank $1$.
This situation is also considered in the articles \cite{harder} and \cite{harder2}.
Let $\mathbf{P}$ be a rational parabolic subgroup of $\mathbf{G}$ and $P=\mathbf{P}(\Reell)$.
Then $P$ is a parabolic subgroup of $G$ and there is a ``rational horocyclic decomposition'' (\cite[S.141]{weber})
\[
G/K\cong A_P\times N_P\times X_P.
\]
Furthermore $\Gamma\cap P$ induces a
discrete group $\Gamma_P$, that operates on $X_P$. The ``cusp'' $Z$ corresponds to
\[
(\Gamma\cap P)\backslash (G/K)\cong A_P\times (\Gamma\cap P)\backslash
(N_P\times X_P)
\]
The base of the cusp is $M=(\Gamma\cap P)\backslash(N_P\times X_P)$ and
the canonical projection $N_P\times X_P\to X_P$ induces a fibration
$
M\to B,
$
with $B=\Gamma_P\backslash X_P$.
Proposition 2.9 in \cite{weber} or Proposition 4.3 in \cite{borel} describe the local form on $M$, and it follows that the horizontal distribution of $\pi:M\to B$
is integrable. Furthermore Borel shows in the proof of \cite[Corollary 4.4]{borel} that horizontal parallel transport preserves the volume form of the fibers of $\pi$.
Under the assumption that the horizontal distribution is integrable, Lemma 10.4 in \cite{BGV} shows
\begin{equation}
d^{1,0}\vol_{F_b}= -H^\vee \wedge \vol_{F_b}, \quad H^\vee \text{ dual to $H$ w.r.t. $g^M$},\label{hint1}
\end{equation}
so the fibers of $\pi$ are minimal and (B) is fulfilled.\qed
\subsubsection{Spectral sequence}\label{kspecseq}
At the end of this chapter we want to prove an important decomposition of the de~Rham-cohomology $H^p(M)$.
\begin{satz}\label{satzspec}
If the mean curvature $H$ of the fibers is projectable and $\pi:M\to B$ is flat, then\index{$\H^a(B,\H^b(F))$}
\[
H^r(B,\H^s(F))\cong \H^r(B,\H^s(F))\defgleich\{\omega\in \Omega^r(B,\H^s(F))\mid \Delta_{1,0}\omega=0\}
\]
and
\[
H^p(M)\cong\bigoplus_{r+s=p} \H^r(B,\H^s(F))
\]
\end{satz}
\begin{proof}
The proof uses the Leray-Serre spectral sequence.
For $0\le n\le\dim M$ let
\[
\FF_i \Omega^n(M)\defgleich \big\{\omega\in \Omega^n(M) \mid \omega(Y_1,\ldots, Y_n)=0,\text{ if }n-i+1\\
\text{ of the $Y_l$ are vertical}\big\}.
\]
and $\FF_{q}\defgleich \FF_0$ for $q<0$, $\FF_q=0$ for $q>n$.
A form $\omega\in \FF_i \Omega^n(M)$ can be expressed as sum of elements $\pi^*\eta^{(k)} \wedge \psi^{(n-k)}$ with
$\eta^{(k)}\in \Omega^k(B), \psi^{(n-k)}\in \Omega^V(M)$ for $k\ge i$. Here $ \psi\in\Omega^V(M)$ by definition means $X\lrcorner \omega=0$ for all horizontal vector fields $X$.
The $\FF_i \Omega^n(M)$ define a filtration
\[
\Omega^n=\FF_0\supset \FF_1 \supset \ldots \supset \FF_n \supset \FF_{n+1}=0,
\]
of $\Omega^n$ which is compatible with $d^M$, i.e. $d^M(\FF_i\Omega^n)\subset \FF_i\Omega^{n+1}$. This filtration gives rise to a spectral sequence $E^{p,q}$ as usual, see e.g. \cite{mccleary}, \cite{bott-tu}. In \cite{dai} the first terms of this spectral sequence were calculated, with the result
\[
E_0^{i,n-i}=\Omega^i(B,W^{(n-i)}),\qquad E_1^{i,n-i}=\Omega^i(B,H^{n-i}(F)).
\]
The Hodge theorem then gives the identification
\begin{equation*}
E_1^{i,n-i}=\Omega^i(B,\H^{n-i}(F))
\end{equation*}
of $E_1$ with fiber-harmonic forms.
Finally for $p+q=n$
\begin{equation*}
E_2^{p,q}=H^p(B,\H^q(F)),
\end{equation*}
and under the further condition $d^{2,-1}=0$, Dai shows that the spectral sequence degenerates at $E_2$, so that
\[
E_\infty^{p,q}=E_2^{p,q}=H^p(B, H^q(F)).
\]
Under the assumption of projectable mean curvature of the fibers, there is a Hodge decomposition of $\Omega^r(B,\H^s(F))$ with respect to $\Delta_{1,0}$,
\[
\Omega^{r,s}=\H^r(B,\H^s(F))\oplus\ker d^{1,0}\oplus \im \delta^{1,0}.
\]
This proves the first statement in Proposition \ref{satzspec}, and the second statement follows from $ H^{p}(M)=E_\infty^{p}=\bigoplus_{r+s=p} E_\infty^{r,s}$.
\end{proof}
\subsection[Spectrum of the Laplacian on $X$]{Spectrum of the Laplacian on $\mathbf{X}$}\label{kstreu}
As before let $\Delta_X:\Omega_0^p(X)\to \Omega_0^p(X)$ be the Laplace-operator on compactly supported forms. $\Delta_X$ admits an extension to a selfadjoint operator on $L^2\Omega^p(X)$, which we will again denote by $\Delta_X$.
Let $\Delta_{X;D}$ be the Friedrichs extension of $\Delta_{X}:\Omega_0^p(X\setminus (\{0\}\times M))\to \Omega_0^p(X\setminus (\{0\}\times M))$ to $L^2\Omega^p(X)$. This decomposes as $\Delta_{X;D}=\Delta_{X_0;D}\oplus \Delta_{Z}$, where $\Delta_Z$ was defined in section \ref{klap} and $\Delta_{X_0;D}$ is the Friedrichs extension of the Laplacian on $\Omega_0^p(X_0)$. $\Delta_{X_0;D}$ is a selfadjoint elliptic operator on a compact manifold with boundary, as such it has pure point spectrum.
We recall some results from mathematical scattering theory.
The \emph{wave operators} $W^\pm$ are defined by
\begin{equation}
W^\pm(\Delta_{X},\Delta_{X;D})=s-\lim_{t\to\mp\infty} e^{i \Delta_X t} J e^{-i \Delta_{X;D} t} P_{ac}(\Delta_{X;D})\label{wof}
\end{equation}
with the inclusion $J:\dom \Delta_{X;D}\hookrightarrow \dom \Delta_{X}$ and the projection $P_{ac}$ onto the absolutely continuous subspace of $\dom\Delta_{X;D}$ in $L^2\Omega^p(X)$.
It is well known that if the wave operators exist, they are partial isometries $ W^\pm: P_{ac}(\Delta_{X;D})\to \bild W^\pm$, i.e. isometries on the complement of $\ker W^\pm$.
In this case the $W^\pm$ are called \emph{complete}, if $\bild W^\pm=\bild P_{ac}(\Delta_X)$. Then $W^\pm: \bild P_{ac}(\Delta_{X;D})\to \bild \Delta_{X}$ are unitary equivalences, in particular $\Delta_{X;D}$ and $ \Delta_{X}$ have the same absolutely continuous spectrum.
There are several methods to show existence and completeness of $W^\pm$. In the present case most information can be obtained from the \emph{Enss method}, see e.g. \cite{BaWo} and \cite{mu-cusprk1}.
If we assume conditions (A) and (B) from section \ref{Bedingungen}, the reasoning in the case of the manifold $X$ with fibered cusp metric is analogous to that in chapter 6 of \cite{mu-cusprk1}. Not only does the Enss method show existence and completeness of the wave operators, but also that $\Delta_X$ has empty singular continuous spectrum, and that the point spectrum of $\Delta_X$ has no points of accumulation outside of $\text{spec}(\Delta_{X;D})$
Altogether this shows
\begin{satz}\label{acspec1}
The absolutely continuous part of $\dom\Delta_X$ is unitarily equivalent to the fiber-harmonic forms $\Pi_0L^2\Omega^p(Z)$.
\end{satz}
\subsection{Analytic continuation of the Resolvent}\label{parakapform}
\subsubsection{Parametrix}
Under the conditions from chapter \ref{Bedingungen} the construction of a parametrix for the resolvent of the Laplace-operator
$\Delta_X$ on $L^2\Omega^*(X)$ is entirely parallel to \cite{mu-cusprk1} or \cite{mu-rank1}. We recall only the essential steps.
Let $X_1=X_0\cup([0,1]\times M)$ and let $\hat X$ be a closed manifold into which $X_1$ is embedded isometrically. For the Hodge-Laplace-operator $\Delta_{\hat X}$ on $\hat X$
the resolvent $(\Delta_{\hat X}-\lambda)^{-1}$ is an operator valued function which is meromorphic in
$\lambda\in \Complex$ with poles in the eigenvalues of $\Delta_{\hat X}$. Let $Q_1(x,x',\lambda)$
be the restriction of the resolvent kernel of $(\Delta_{\hat X}-\lambda)^{-1}$ on $X_1\times X_1$.
This is the ``inner'' part of the parametrix. The ``outer'' part is given by the resolvent of $\Delta_Z$, i.e. $Q_2(\lambda)=(\Delta_Z-\lambda)^{-1}$. The latter will be examined further now.
Let
\[
\Delta_{\text{cusp}}=\Pi_\perp\Delta_Z\Pi_\perp=\Delta_Z\Pi_\perp,\qquad \Delta_{1;Z}=\Pi_0\Delta_Z\Pi_0=\Delta_Z\Pi_0
\]
be the restrictions of $\Delta_Z$ on $L_{\text{cusp}}^2\Omega^p(Z)$ resp. $\Pi_0 L^2\Omega^p(Z)$.
Since the splitting
\[
L^2\Omega^p(Z)=L_{\text{cusp}}^2\Omega^p(Z)\oplus \Pi_0 L^2\Omega^p(Z)
\]
into cusp forms (Definition \ref{spifo}) and their orthogonal complement is invariant under $\Delta_Z$,
we obtain for the resolvent
\[
Q_2(\lambda)=(\Delta_Z-\lambda)^{-1}=(\Delta_{\text{cusp}}-\lambda)^{-1}+(\Delta_{1;Z}-\lambda)^{-1}.
\]
According to Proposition \ref{kptr} the resolvent of $\Delta_{\text{cusp}}$ is compact. It has the kernel
\[
K_{\text{cusp}}(\lambda,x_1,x_2)=\sum_i \frac{1}{\lambda_i-\lambda} \psi_i(x_1)\otimes\overline{\psi_i(x_2)}
\]
for eigenforms $\psi_i$ of $\Delta_{\text{cusp}}$ to the eigenvalue $\lambda_i$.
Let
\[
\tau_0=\inf\big\{d_k^2+\mu\mid 0\le k\le f,
\mu\in\text{spec} \{\Delta_{1,0}:\Omega^{p-k}(B,\H^k(F))\to\Omega^{p-k}(B,\H^k(F))\} \big\}.
\]
The spectrum of $\Delta_{1;Z}$ is $[\tau_0,\infty)$ with branch points at $(f/2-k)^2+\nu, 0\le k\le f $ for each eigenvalue
$\nu$ of $\Delta_{1,0}$.
Certainly $\Complex\smallsetminus \Reell^+$ is contained in the resolvent set of $\Delta_{1;Z}$.
We want to compute the integral kernel of $(\Delta_{1;Z}-\lambda)^{-1}$ for $\lambda\in \Complex\smallsetminus \Reell^+$
explicitly.
With respect to the decomposition $\Omega^p(Z)=\pi_2^*\Omega^p(M)\oplus \pi_2^*\Omega^{p-1}(M)$ we have (chapter \ref{Bedingungen})
\[
\tilde\rho_u\Delta_{1;Z}\tilde\rho_u^{-1}=\begin{pmatrix}
\check{T}_u& 0\\
0& \check{T}_u
\end{pmatrix} ,
\quad \check{T}_u= -\partial_u^2+(f/2-\kappa)^2 +\Delta_{1,0}.
\]
A simple calculation shows that the integral kernel $K_1^{(p)}$ of $(\tilde\rho_u\check{T}_u\tilde\rho_u^{-1}-\lambda)^{-1}$ for $\lambda\in\Complex\setminus\Reell$ is given by
\begin{multline} \label{reskerf}
K_1^{(p)}(\lambda,(u,y),(r,z)) \\ = \sum_k\sum_{\mu^{(k)}}^\infty \frac{i}{2} e^{a_k (u+r)}\frac{ e^{i |u-r| \sqrt{\lambda-d_k^2-\mu^{(k)}} }- e^{i (u+r) \sqrt{\lambda-d_k^2-\mu^{(k)}} }}{\sqrt{\lambda-d_k^2-\mu^{(k)}}}
\:(\phi_\mu^{(k)})(y) \otimes (\overline{\phi}_\mu^{(k)})(z),
\end{multline}
where $\phi_\mu^{(k)}\in \Omega^{p-k}(B, \H^k(F))$ form a local orthonormal basis for every $k$, with $\Delta_{1,0} \phi_\mu^{(k)}=\mu^{(k)} \phi_\mu^{(k)}$.
Finally
$
K_{\text{cusp}}+K_1^{(p)}+du\wedge K_1^{(p-1)}
$
is the integral kernel of $(\Delta_Z-\lambda)^{-1}$.
Let $\xi_1,\xi_2,\chi_1,\chi_2$ be suitable cut-off functions with the properties as in e.g. \cite{APS}.
Then we define an operator $Q(\lambda):L^2\Omega^p(X)\to L^2\Omega^p(X)$ by its integral kernel
\[
Q(\lambda,x_1,x_2)= \chi_1(x_1) Q_1(\lambda,x_1,x_2) \xi_1(x_2)+\chi_2(x_1) Q_2(\lambda,x_1,x_2) \xi_2(x_2).
\]
From this formula we get as in \cite{mu-rank1} that
\[
Q(\lambda)(\Delta_X-\lambda)=\text{Id}+\mathcal{K}(\lambda),\qquad\lambda\in \Complex \setminus\Reell^+
\]
where $\mathcal{K}(\lambda)$ are compact operators in $L^2\Omega^p(X)$, and that
\[
(\Delta_X-\lambda)^{-1}-Q(\lambda)
\]
for $\lambda\in \Complex \setminus\Reell^+$ is a family of compact operators which is meromorphic in $\lambda$.
Altogether this shows that $Q(\lambda)$ is a parametrix for $(\Delta_X-\lambda)^{-1}$.
\subsubsection{Continuation to the Spectral Surface}\label{spektrfl}
The square roots $\lambda\mapsto \sqrt{\lambda-d_k^2-\mu_j}$ in \eqref{reskerf} are holomorphic in the complex plane $\Complex\setminus[\tau_0,\infty)$ and
they cannot be extended holomorphically to the whole complex plane.
Let $\mathfrak{I}=\{\tau_i\}_{i\in\Nat}$ a discrete set of real numbers with $-\infty<\tau_0<\tau_1<\ldots$.
We recall the construction of the \emph{spectral surface} $\Sigma_s$. This is a Riemannian surface on which all square roots $z\mapsto\sqrt{z-\tau_i}$ are holomorphic.
\index{$\Sigma_s$}As in \cite{guil} we define
\[
\Sigma_s=\left\{\Lambda=(\Lambda_\mu)\in \Complex^{\sharp \mathfrak{I}} \mid \forall \mu, \nu \in \mathfrak{I}: \Lambda_\mu^2+\mu=\Lambda_\nu^2+\nu \right\}
\]
and the projection $\pi_s:\Sigma_s\to \Complex$, \index{$\pi_s$}$\pi_s(\Lambda)=\Lambda_\mu^2+\mu$. Then $(\Sigma_s,\pi_s)$ is a covering of $\Complex$ with infinitely many leafs and branch points in $\mathfrak{I}$. For $\mu\in\mathfrak{I}$ the square roots are given by
\begin{equation}\label{wufu}
\sqrt{\Lambda-\mu} \defgleich\Lambda_\mu.
\end{equation}
We define the \emph{physical domain} \index{$\FP$}$\FP$, i.e. the leaf of ``positive'' square roots, as
\[
\FP\defgleich\left\{\Lambda\in \Sigma_s \mid \forall \mu \in \mathfrak{I}: \text{Im } \Lambda_\mu>0\right\}.
\]
This can be identified with $\Complex\setminus [\tau_0,\infty)$ via $\pi_s$.
The boundary of $\FP$ consists of two rays $\partial_\pm \FP \simeq [\tau_0, \infty)$. Here $\Lambda\in \partial_\pm \FP$ means that
$\pi_s^{-1}(\pi_s(\Lambda)\pm i\eps) \to \Lambda$ for $\eps\to 0$.
Let $\gamma_{\tau}:\Sigma_s\to\Sigma_s$ be a deck transform, which is given by a closed curve which encircles
the single branch point $\tau$.
Then square roots \eqref{wufu} fulfill the following identities:
\begin{equation}\label{wurzelverh}
\sqrt{\gamma_\mu\Lambda-\mu}=-\sqrt{\Lambda-\mu},\qquad
\sqrt{\gamma_\mu\Lambda-\nu}=\sqrt{\Lambda-\nu},\quad \mu\neq\nu
\end{equation}
Finally let $\Sigma_s^\mu$ for $\mu\in \mathfrak{I}$ be the connected component of $\FP$ in $\pi_s^{-1}(\Complex\setminus[\mu,\infty))$.
Now in our case
\[
\mathfrak{I}=\{\mu+d_k^2\mid \mu\in\text{spec}(\Delta_{1,0}:\Omega^*(B, \H^{k}(F))\to\Omega^*(B, \H^{k}(F)),\quad 0\le k\le f \}.
\]
On the corresponing spectral surface $\Sigma_s$ all square roots in \eqref{reskerf} are holomorphic functions.
Outside $\FP$ the integral kernel in \eqref{reskerf} does not define a continuous operator on $L^2\Omega^p(X)$.
However we can introduce weighted $L^2-$spaces, on which the analytic continuation of the kernel will be a continuous operator.
For $\delta\in\Reell$ consider the weight operator $\omega_\delta$ on differential forms $\phi\in \Omega^p(Z)$ defined by
\[
\omega_\delta(\phi)(u,y)\defgleich e^{\delta u} \phi(u,y)
\]
The corresponding weighted $L^2-$space is
\[
L_\delta^2\Omega^p(Z)=\left\{\phi: Z\to\Reell\mid \phi\text{ measurable and }
\omega_\delta(\phi) \in L^2\Omega^p(Z)\right\}
\]
For $\delta>0$ we have
\[
L_\delta^2\Omega^p(Z) \subset L^2\Omega^p(Z) \subset L_{-\delta}^2\Omega^p(Z).
\]
For $\tau>0$ let $D_\tau(0)=\{z\in \Complex \mid |z|<\tau \}$ and $\tilde{\mu}(\tau)$
the smallest eigenvalue of $\Delta_{1,0}$ with $\tilde{\mu}(\tau)+d_k^2>\tau$. Let
\[
\Omega_\tau = ( \FP \cup \pi_s^{-1}(D_\tau(0)) ) \cap \Sigma_s^{\tilde{\mu}(\tau)},
\]
and choose $\delta>0$ with $\delta^2>\tau$. Then $|\Im \sqrt{\Lambda-d_k^2-\nu}|<\delta$ for $\Lambda\in \Omega_\tau$ and $\nu\le\tilde{\mu}(\tau)$. This implies
\begin{lem}
For all $\eps>0$ and $\delta>0$ with $\delta^2>\eps$ the parametrix $Q(\lambda)$ has a continuation to a meromorphic family of continuous operators $Q(\Lambda): L_\delta^2\Omega^p(X)\to L_{-\delta}^2\Omega^p(X),
\Lambda \in \Omega_\eps$.
\end{lem}
For our later applications it is sufficent to know the analytic continuation of the resolvent in a neighbourhood of $\lambda=0$, i.e. the continuation to $\Sigma_s^{\tau_1}$, the Riemann surface for $z\mapsto\sqrt{z}$.
Let
\[
\tau_1=\min (\mathfrak{I}\smallsetminus \{0\})
\]
and $0<\eps<\tau_1$.
If
\[
\Lambda\in \Omega_\eps \defgleich\big(\FP\cup \pi_s^{-1}(B_\eps(0))\big)\cap \Sigma_s^{\tau_1},
\]
then all coefficients in the resolvent kernel \eqref{reskerf} lie in $L_{-\alpha}^2(\Reell^+,du)=e^{\alpha u} L^2(\Reell^+,du)$ with $\alpha^2>\tau_1$.
Thus for $\Lambda\in\Omega_\eps$ the resolvent kernel defines a continuous operator $L_{\alpha}^2\to L_{-\alpha}^2$.
As in \cite{mu-cusprk1}
we obtain
\begin{thm}\label{resfortform}
For $\delta^2>\eps>0,$ $\eps<\tau_1,$ the resolvent $(\Delta-\lambda)^{-1}$ can be continued analytically to a family of operators
$R(\Lambda)\in \mathscr{B}(L_\delta^2\Omega^p(X), L_{-\delta}^2\Omega^p(X))$ which is meromorphic in $\Lambda\in \Omega_\eps$.
\end{thm}
\begin{bem}
A similar statement holds for any $\eps>0$, then $\delta$ must be chosen sufficiently large.
\end{bem}
Since $\Omega_0^p(X)\subset L_a^2\Omega^p(X)\subset L^2_{\text{loc}}\Omega^p(X)$ for all $a\in\Reell$ und $\Sigma_s=\bigcup_{\eps>0} \Omega_\eps$, from the above theorem follows
\begin{cor}
The resolvent $(\Delta-\lambda)^{-1}$ has an analytic continuation to a family of operators $R(\Lambda):\Omega_0^p(X)\to L^2_{\text{\upshape loc}}\Omega^p(X)$ which is meromorphic in $\Lambda\in \Sigma_s$.
\end{cor}
\subsection{Generalized Eigenforms}\label{keif}
The generalized eigenforms provide the spectral resolution of the absolutely continuous part of $\dom(\Delta_X)$. Once the analytic continuation of the resolvent of $\Delta_X$ is known, it is possible to construct generalized eigenforms of $\Delta_X$ explicitely as follows.
An element in $\Omega^{l,k}=\Omega^l(B,\H^k(F))$ is a section in the finite dimensional vector bundle
$W_0^{k+l}=\Lambda^l T^* B\tensor \H^k(F)$ over $B$. There is an orthonormal basis $\{\phi_\mu\}$
of $L^2(B, W_0^{k+l})$ consisting of eigenforms of $\Delta_{1,0}$.
Let $z\mapsto\sqrt{z}$ be the branch of the square root, for which $\Im \sqrt{z}>0$ if $z\in\Complex\setminus\Reell^+$.
Let $\psi\in \Omega^{*,k}$ fiber harmonic with degree $k$ in the fibers, so that
$
\Delta_{1,0}\psi=\mu\psi.
$
Let
\[
e_{\mu,k,\pm}(\lambda,\psi,(u,x))\defgleich\tilde\rho_u^{-1} e^{\pm i\sqrt{\lambda-\mu-(f/2-k)^2}\, u} \psi(x), \quad x\in M
\]
be the solutions of
\[
\Delta_Z e_{\mu,k,\pm}(\lambda,\psi,(u,x)) = \lambda e_{\mu,k,\pm}(\lambda,\psi,(u,x)).
\]
For $\lambda\in\Complex\setminus[(f/2-k)^2+\mu,\infty)$ the section $e_{\mu,k,+}$ lies in $L^2\Omega^*(Z,g_Z)$ and $e_{-}$ is not square integrable.
These $e_{\mu,\pm}$ do not satisfy the boundary conditions \eqref{relrbd}. Therefore we consider the linear combination
$
\text{SIN}_\mu\defgleich\frac{1}{2i}(e_{\mu,+}-e_{\mu,-}).
$
A direct computation shows that
\begin{multline}\label{tmp01}
\Big\{ \text{SIN}_{\mu,k,\pm}(\psi),\;du\wedge \text{SIN}_{\mu,k,\pm}(\psi) \mid 0\le k\le f, (\mu,\psi) \text{ so that }\\ \psi \in\Omega^*(B,\H^k(F)),\; \Delta_{1,0}\psi=\mu\psi \Big\}
\end{multline}
span a complete system of generalized eigenforms of $\Delta_Z$.
The square roots appearing in \eqref{tmp01} and with them $e_\pm$ admit an analytic continuation to $\Sigma_s$.
Let $\psi\in \Omega^{p-k}(B,\H^k(F))$ be an eigenform of $\Delta_{1,0}^{p-k}$ for the eigenvalue $\mu$, $\Lambda\in\Sigma_s$ and $\lambda=\pi_s(\Lambda)$.
Let $\chi$ be a smooth cutoff-function on $X$ with $\chi =1$ on $ [1,\infty)\times M\subset Z$ and $\chi=0$ on $X_0$.
Then $(\Delta_X-\lambda)(\chi e_{\mu,k,-}(\Lambda,\psi))\in \Omega_0^p(X)$ lies in the domain of the analytic continuation $R(\Lambda)$ of the resolvent of $\Delta_X$.
Now the generalized eigenforms of $\Delta_X$ are defined as
\index{$E_{\mu}(\Lambda,\psi)$}
\begin{equation}
\begin{split}
E_{\mu}(\Lambda,\psi)(p) &\defgleich \chi(p) e_{\mu,k,-}(\Lambda,\psi,p)-R(\Lambda)(\Delta_X-\lambda)(\chi(p) e_{\mu,k,-}(\Lambda,\psi,p))\label{geneig}
\end{split}
\end{equation}
and an analogous formula for $ E_{\mu}(\Lambda,du\wedge \psi)\in \Omega^{p+1}(X)$.
The generalized eigenform $E_\mu(\Lambda,\psi)$ is uniquely determined by the three properties
\begin{enumerate}[1)]
\item $E_\mu(\Lambda,\psi) \in \Omega^{p}(X)$ and $E_\mu(\Lambda,\psi)$ is meromorphic in $\Lambda\in\Sigma_s$.
\item $\Delta E_\mu(\Lambda,\psi) =\pi_s(\Lambda) E_\mu(\Lambda,\psi)$ for $\Lambda\in\Sigma_s$
\item $E_\mu(\Lambda,\psi) - \chi e_{-,\mu}(\Lambda,\psi) \in L^2\Omega^{p}(X)$ for $\Lambda\in \FP$.\label{egs3}
\end{enumerate}
\begin{proof}
The proof is the same as in \cite{mu-cusprk1}: Property 1) follows from the corresponding properties of the resolvent and elliptic regularity; 2) and 3) from the definition \eqref{geneig} of $E$.
To show uniqueness, we assume there is a second form $G$ with properties 1)-3). For $\Lambda\in\FP$ we have
$(E-G)(\Lambda)\in L^2\Omega^p(X)$ because of 3), and from 2)
\[
(\Delta_\text{loc}-\pi_s(\Lambda))(E-G)(\Lambda)=0.
\]
Since $\pi_s(\Lambda)\notin\Reell_+$, this is a contradiction to $\Delta$ being self-adjoint, so that $E=G$.
\end{proof}
By computing the wave operators \eqref{wof},
it can be shown as in \cite{mu-cusprk1} that $\{E_\mu\}$ are a complete system of generalized eigenforms of $\Delta_X P_{ac}$.
We have seen in Proposition \ref{wspek} that the essential spectrum of $\Delta_X$ is determined by fiber-harmonic forms. The fiber-harmonic part
$\Pi_0(E_\mu|_{Z})$ is called the \emph{constant term} of $E_\mu$\index{constant term}.
Finally we want to associate a generalized eigenform to each cohomology class $[\phi]\in H^*(M), [\phi]\neq 0$.
Because of the horizontal Hodge-decomposition Proposition \ref{satzspec} we only need to consider generalized eigenforms for $\mu=0$ and define
\[
E(\Lambda,[\phi])\defgleich E_0(\Lambda,\phi_0)\quad\text{for the unique}\quad\phi_0\in \H^{p-k}(B,\H^k(F))\quad\text{with}\quad [\phi_0]=[\phi].
\]
\subsection{Asymptotics of the constant term}\label{kasym}
We want to determine the asymptotic expansion on the end $Z$ of the constant term $\Pi_0 E(\Lambda,\psi)$ for
$
\psi\in \H^{p-k}(B,\H^k(F))\subset \H^p(M).
$
As before we define
\[
e_{\pm,\nu}(\Lambda, \phi) = e^{(\ua\pm i\sqrt{\Lambda-\nu-\ud^2})u} \phi,\qquad \phi\in \Omega^*(B,\H^*(F)),\quad
\Delta_{1,0}\phi=\nu\phi,\quad \Lambda\in\Sigma_s.
\]
We expand in a basis of eigenforms of $\Delta_{1,0}:\Omega^{p-l,l}\to\Omega^{p-l,l}$ for each base-degree
$p-l$. Then the identity $(\Delta_{Z}-\pi_s(\Lambda))\Pi_0 E(\Lambda,\psi)=0$ gives a system of ordinary differential equations for the coefficients leading to
\begin{multline}\label{asymt:0}
\Pi_0 E(\Lambda,\psi) = e_{-,0}(\Lambda,\psi)+\sum_{l=0}^f e_{+,0}(\Lambda,\psi_0^{l})
+du\wedge \sum_{l=0}^f e_{+,0}(\Lambda,\hat\psi_0^{l})\\
+\sum_{l=0}^f \sum_{\nu_l>0}e_{+,\nu_l}(\Lambda,\psi_{\nu_l}^{l})
+\sum_{l=0}^f\sum_{\gamma_l>0}du\wedge e_{+,\gamma_l}(\Lambda,\hat\psi_{\gamma_l}^{l}).
\end{multline}
Here $\psi_\nu^l\in \Omega^{p-l,l} ,\;\hat\psi_\gamma^l \in \Omega^{p-l+1,l-1}$ are the eigenforms of $\Delta_{1,0}$,
\[
\Delta_{1,0}\psi_{\nu_l}^l=\nu_l \psi_{\nu_l}^l,\qquad \Delta_{1,0}\hat\psi_{\gamma_l}^l=\gamma_l \hat\psi_{\gamma_l}^l.
\]
For better readability the indices of the eigenvalues of $\Delta_{1,0}$ will be suppressed in the following.
Let $\mathcal{E}_\nu^{a,b} \subset \Omega^a(B,\H^b(F))$ be the eigenspace of $\Delta_{1,0}$ for the eigenvalue $\nu$.
Now we define linear, in $\Lambda\in\Sigma_s$ meromorphic operators
\[
S_{0\nu}^{[l]}(\Lambda), \quad T_{0\nu}^{[l]}(\Lambda):\mathcal{E}_0^{*,l}\to\mathcal{E}_\nu^{*,l}
\]
by \eqref{asymt:0} as\index{$T_{0\nu}^{[l]}$}
\[
T_{0\nu}^{[l]}(\Lambda,\psi)\defgleich\psi_\nu^{l},\qquad S_{0\gamma}^{[l]}(\Lambda,\psi)\defgleich\hat\psi_\gamma^{l}.
\]
We are particularily interested in the eigenvalue $0$ and the corresponding operator\index{$T_{00}^{[l]}$}
\[
T_{00}^{[l]}(\Lambda,\cdot): \H^{p}(M)\to\H^{p-l}(B,\H^l(F)),\qquad 0\le l\le f.
\]
Sometimes we will use the abbreviated notation \index{$T_{00}$}$T_{00}=\sum_{l=0}^f T_{00}^{[l]}.$
\begin{satz}\label{sasym8}
The asymptotic expansion of the constant term of $E(\psi,\Lambda)$ with $\psi\in \H^*(B,\H^k(F))$ is
\begin{multline}\label{aympt1}
\Pi_0 E(\Lambda,\psi) = e_{-,0}(\Lambda,\psi)+e_{+,0}(\Lambda,T_{00}(\Lambda,\psi))\\
+\sum_{\nu>0}e_{+,\nu}(\Lambda,T_{0\nu}(\Lambda,\psi))
+\sum_{\gamma>0}du\wedge e_{+,\gamma}(\Lambda,S_{0\gamma}(\Lambda,\psi))
\end{multline}
with linear maps $T_{0\nu}(\Lambda), S_{0\nu}(\Lambda): \mathcal{E}_0\to \mathcal{E}_\nu$ which are meromorphic in $\Lambda$.
Similarily
\begin{multline}\label{aympt2}
\Pi_0 E(\Lambda,du\wedge \psi) = du\wedge e_{-,0}(\Lambda,\psi)+du\wedge e_{+,0}(\Lambda,\check{T}_{00}(\Lambda,\psi))\\
+\sum_{\nu>0}du\wedge e_{+,\nu}(\Lambda,\check{T}_{0\nu}(\Lambda,\psi))
+\sum_{\gamma>0} e_{+,\gamma}(\Lambda,\check{S}_{0\gamma}(\Lambda,\psi))
\end{multline}
\end{satz}
\begin{proof}
It remains to show that in the asymptotic expansion
\begin{multline}\label{asymt:1}
\Pi_0 E(\Lambda,\psi) = e_{-,0}(\Lambda,\psi)+e_{+,0}(\Lambda,T_{00}(\Lambda,\psi))\\
+du\wedge e_{+,0}(\Lambda,S_{00}(\Lambda,\psi))\\
+\sum_{\nu>0}e_{+,\nu}(\Lambda,T_{0\nu}(\Lambda,\psi))
+\sum_{\gamma>0}du\wedge e_{+,\gamma}(\Lambda,S_{0\gamma}(\Lambda,\psi)).
\end{multline}
the term $du\wedge e_{+,0}(\Lambda,S_{00}(\Lambda,\psi))$ is actually zero.
The two series in the last line in \eqref{asymt:1} are exponentially decreasing in $u$, and the term $e_{-,0}(\Lambda,\psi)$ is the only one in \eqref{asymt:1}, which is not square integrable for
$\Lambda\in \FP$.
Similar to \eqref{asymt:1}, $E(\Lambda,du\wedge \psi)$ has the asymptotic expansion
\begin{multline}\label{asymt:du}
\Pi_0 E(\Lambda,du\wedge \psi) = du\wedge e_{-,0}(\Lambda,\psi)+du\wedge e_{+,0}(\Lambda,\check{T}_{00}(\Lambda,\psi))\\
+ e_{+,0}(\Lambda,\check{S}_{00}(\Lambda,\psi))\\
+\sum_{\nu>0}e_{+,\nu}(\Lambda,\check{S}_{0\nu}(\Lambda,\psi))
+\sum_{\nu>0}du\wedge e_{+,\gamma}(\Lambda,\check{T}_{0\gamma}(\Lambda,\psi))
\end{multline}
with certain linear maps $\check{T}_{0,\nu}(\Lambda), \check{S}_{0,\nu}(\Lambda):\mathcal{E}_0\to \mathcal{E}_\nu$ which are meromorphic in $\Lambda$ as above.
Because $d^{1,0}(T_{00}\psi)=0,$ $d^F(T_{0\nu}\psi)=0,$ and $d^M(S_{00}\psi)=0$, for $\Lambda\in \FP$
\begin{multline}\label{dzasympt}
d^Z\Pi_0 E_0(\Lambda,\psi)
=(\ua-i\sqrt{\Lambda-\ud^2}) du\wedge e_{-,0}(\Lambda,\psi) \\
+(\ua+i\sqrt{\Lambda-\ud^2}) du\wedge e_{+,0}(\Lambda,T_{00}(\Lambda,\psi)) \\
+\sum_{\nu>0}\Big\{(\ua+ i\sqrt{\Lambda-\nu-\ud^2}) du\wedge e_{+,\nu}(\Lambda,T_{0\nu}(\Lambda,\psi ))\\
+ e_{+,\nu}(\Lambda,d^{1,0}(\Lambda,T_{0\nu}(\Lambda,\psi))\Big\}\\
-\sum_{\gamma>0}\ du\wedge d^{1,0}(e_{+,\gamma}(\Lambda,S_{0\gamma}(\Lambda,\psi))).
\end{multline}
Comparing \eqref{dzasympt} and \eqref{asymt:du} and using the uniqueness of generalized eigenforms gives for $\Lambda\in\Sigma_s$
\begin{equation} \label{devse1}
d E(\Lambda, \psi) = (\ua- i\sqrt{\Lambda-\ud^2})(\psi) E(\Lambda, du\wedge \psi).
\end{equation}
But then because there is no corresponding term in \eqref{dzasympt}, $e_{+,0}(\Lambda,\check{S}_{00}(\Lambda,\psi))$ can not appear in
\eqref{asymt:du}.
Furthermore
\begin{multline}
\ast \Pi_0 E(\Lambda,du\wedge\psi) = e_{-,0}(\Lambda,\astm \psi) + e_{+,0}(\Lambda,\astm\check{T}_{00}(\Lambda,\psi)) \\
+\sum_{\nu>0}du\wedge e_{+,\nu}(\Lambda,\astm \check{T}_{0\nu}(\Lambda,\psi))
+\sum_{\nu>0}(-1)^p du\wedge e_{+,\nu}(\Lambda,\astm \check{S}_{0\nu}(\Lambda,\psi)).
\end{multline}
If this is compared with \eqref{asymt:1} for $\astm\psi$, we first conclude
\[
E(\Lambda,\astm \psi)= \ast E(\Lambda, du\wedge \psi),
\]
and then, that the term $du\wedge e_{+,0}(\Lambda,S_{00}(\Lambda,\psi))$ does not appear in \eqref{asymt:1}.
\end{proof}
From this asymptotic expansion we can read off several functional equations for $E$ and the leading coefficient $T_{00}$.
Let $\{\gamma_{\nu k}\}$ be the generators of $\text{Aut}(\Sigma_s)$ as described in section \ref{spektrfl}. Here $\gamma_{\nu k}$ is associated with the ramification point $\nu+d_k^2$.
\begin{cor}\label{funceq}
The generalized eigenforms satisfy the functional equations
\begin{subequations}
\begin{eqnarray}
d E(\Lambda, \psi) &=& (\ua- i\sqrt{\Lambda-\ud^2})(\psi) E(\Lambda, du\wedge \psi) \label{devse}\\
E(\Lambda,\astm \psi)&=& \ast E(\Lambda, du\wedge \psi),\label{emitstern}\\
(\ua- i\sqrt{\Lambda-\ud^2})(\psi)T_{00}(\Lambda,\astm \psi) &=&\astm (\ua+i\sqrt{\Lambda-\ud^2})( T_{00}(\Lambda,\psi)) \label{Tbez}\\
E(\Lambda,\psi) &=& E(\gamma_{0k}\Lambda, T_{00}^{[k]}(\Lambda,\psi)) \label{fktgle}
\end{eqnarray}
\end{subequations}
\end{cor}
\begin{proof}
Equations \eqref{devse} and \eqref{emitstern} have been shown in the proof of Proposition \ref{sasym8}.
Comparing the asymptotic expansion in these gives
\begin{equation} \label{fftmat}
(\ua- i\sqrt{\Lambda-\ud^2})(\psi)\check{T}_{00}(\Lambda,\psi) =(\ua+i\sqrt{\Lambda-\ud^2})( T_{00}(\Lambda,\psi)).
\end{equation}
From \eqref{emitstern} and \eqref{fftmat}
\begin{equation*}
T_{00}(\Lambda,\astm \psi)= \astm \check{T}_{00}(\Lambda,\psi),
\end{equation*}
which leads to the identity \eqref{Tbez}.
Using \eqref{wurzelverh} in the expansion \eqref{aympt1}, we get
\begin{multline*}
\Pi_0 E(\gamma_{0k}\Lambda,\psi) = e_{+,0}(\gamma_{0k}\Lambda,\psi)
+e_{-,0}(\Lambda,T_{00}^{[k]}(\gamma_{0k}\Lambda,\psi))
+e_{+,0}(\Lambda,T_{00}^{[\neq k]}(\gamma_{0k}\Lambda,\psi))\\
+\sum_{\nu>0}e_{+,\nu}(\Lambda,T_{0\nu}(\gamma_{0k}\Lambda,\psi))
+\sum_{\eta>0}du\wedge e_{+,\eta}(\Lambda,S_{\eta}(\gamma_{0k}\Lambda,\psi)).
\end{multline*}
The only term on the right hand side, which is not in $L^2$, is $e_{-,0}(\Lambda,T_{00}^{[k]}(\gamma_{0k}\Lambda,\psi))$.
Uniqueness of the generalized eigenforms gives
\begin{equation*}
E(\gamma_{0k}\Lambda,\psi) = E(\Lambda, T_{00}^{[k]}(\gamma_{0k}\Lambda,\psi))
\end{equation*}
and the claim follows from $\gamma_{0k}^2=\text{id}$.
\end{proof}
\subsection{Maa{\ss}--Selberg Relations}\label{kmsb}
The Maa{\ss}--Selberg relations will provide us with important information about the position of poles and the asymptotic behavior of generalized eigenforms. The proceeding is similar to \cite{roelcke}.
For $\psi\in \H^p(M)$ let $\psi^{[k]}$ be the projection of $\psi$ onto $\H^{p-k}(B,\H^k(F))$.
Write $X=X_r\sqcup_M Z_r$ with $Z_r\defgleich[r,\infty)\times M$.
Let $E=E(\phi,\Lambda)=E_{\mu=0}(\phi,\Lambda)$ and define
\[
E^r = E-\Pi_0(E|_{Z_r}).
\]
Because of property \ref{egs3}) of the generalized eigenforms, we have $E^r(\phi,\Lambda)\in L^2\Omega^*(X)$ for $\Lambda\in\FP$.
\begin{satz}\label{maass-selb1} Let $\phi\in \H^*(B,\H^k(F))$.
Let $\nu_1$\index{$\nu_1$} be the smallest positive eigenvalue of $\Delta_{1,0}$ and
$\tau$ with $0<\tau<\min\{1/4,\nu_1\}$ not a pole of $E(.,\phi)$.
If the fiber-degree of $\phi$ is $k\neq f/2$, the cut-off generalized eigenform $E^r(\tau,\phi)$ has the norm
\begin{multline*}
\|E^r(\tau,\phi)\|_{L^2(X)}^2 =
\frac{e^{2 r \sqrt{d_k^2-\tau}}}{2\sqrt{d_k^2-\tau}} \| \phi \|^2 \\
+r\cdot \sqrt{d_k^2-\tau}\; \Big\{ \sce{T_{00}(\tau,\phi)}{\phi }-
\sce{\phi }{T_{00}(\tau,\phi)}\Big\}
+r\cdot \| T_{00}^{[f/2]}(\tau,\phi)\|^2\\
-i \sqrt{\tau} \Big\{ \Sce{\textstyle{\frac{d}{d\Lambda}|_\tau} T_{00}^{[f/2]}(.,\phi)}{T_{00}^{[f/2]}(\tau,\phi)} -
\Sce{T_{00}^{[f/2]}(\tau,\phi)} {\textstyle{\frac{d}{d\Lambda}|_\tau} T_{00}^{[f/2]}(.,\phi)}\Big\}\\
-\sum_{\substack{\nu\ge 0; l\\ \nu+d_l^2>0}} \frac{e^{-2 r \sqrt{d_l^2+\nu-\tau}}}{2\sqrt{d_l^2+\nu-\tau}} \| T_{0\nu}^{[l]}(\tau,\phi) \|^2
-\sum_{\gamma> 0; l} \frac{e^{-2 r \sqrt{d_l^2+\gamma-\tau}}}{2\sqrt{d_l^2+\gamma-\tau}} \| S_{0\gamma}^{[l]}(\tau,\phi) \|^2.
\end{multline*}
For $\phi\in \H^*(B,\H^{f/2}(F))$,
\begin{multline*}
\|E^r(\tau,\phi)\|_{L^2(X)}^2 =
r (\| \phi\|^2 + \| T_{00}^{[f/2]}(\tau,\phi) \|^2) \\
-i \sqrt{\tau} \Big\{ \Sce{\textstyle{\frac{d}{d\Lambda}|_\tau} T_{00}^{[f/2]}(.,\phi)}{T_{00}^{[f/2]}(\tau,\phi)} -
\Sce{T_{00}^{[f/2]}(\tau,\phi)} {\textstyle{\frac{d}{d\Lambda}|_\tau} T_{00}^{[f/2]}(.,\phi)}\Big\}\\
+\frac{1}{2i\sqrt{\tau}}\Big\{ e^{2 i \sqrt{\tau} r}\sce{T_{00}(\tau,\phi)}{\phi }
-e^{-2 i \sqrt{\tau} r}\sce{\phi }{T_{00}(\tau,\phi)} \Big\} \\
-\sum_{\substack{\nu\ge 0; l\\ \nu+d_l>0}} \frac{e^{-2 r \sqrt{d_l^2+\nu-\tau}}}{2\sqrt{d_l^2+\nu-\tau}} \| T_{0\nu}^{[l]}(\tau,\phi) \|^2
-\sum_{\gamma> 0; l} \frac{e^{-2 r \sqrt{d_l^2+\gamma-\tau}}}{2\sqrt{d_l^2+\gamma-\tau}} \| S_{0\gamma}^{[l]}(\tau,\phi) \|^2.
\end{multline*}
\end{satz}
\begin{proof}
Let $\phi\in \H^*(M)$ and $\lambda\in\Complex\smallsetminus \Reell_+$.
By definition $E^r\upharpoonleft Z_r=\Pi_\perp(E\upharpoonleft Z_r)$, and under the given conditions $\Delta=\Delta_Z$ leaves fiber-harmonic forms on $Z_r$ invariant. Let $M_r=\{r\}\times M$ with the Riemannian metric $\pi^* g^B+e^{-2r}g^F$.
The Green's formula gives
\begin{multline*}
\Scr{\Delta E^r}{E^r}_{L^2(X)}-\Scr{E^r}{\Delta E^r}_{L^2(X)} \\
=\Sce{\Pi_0 E|_{M_r}}{\nabla^Z_{\partial/\partial u} \Pi_0 E|_{M_r}}_{\!L^2(M_r)}
-\Sce{\nabla^Z_{\partial/\partial u} \Pi_0 E|_{M_r}}{\Pi_0 E|_{M_r}}_{\!L^2(M_r)}
\end{multline*}
Here we used that $\frac{\partial}{\partial u}$ is outer unit normal vector field to $X_r$ and inner unit normal vector field to $Z_r$.
But $\nabla^Z_{\partial/\partial u}=\frac{\partial}{\partial u}+\kappa$ and
$\tilde\rho_u=e^{-(f/2-\kappa) r}:L^2\Omega^*(M_r,g_u^M)\to L^2\Omega^*(M,g^M)$ is an isometry. Let $g^M=\pi^* g^B+g^F$ the unscaled Riemannian metric on $M$. With $\Sce{}{}_M$ denote the induced $L^2-$norm on $M$.
Let $\Lambda_1, \Lambda_2\in \FP=\Complex\setminus\Reell^+$, d.h. $\Im \Lambda_1>0, \Im \Lambda_2>0$ and $\lambda_1=\pi_s(\Lambda_1), \lambda_2=\pi_s(\Lambda_2)$. Also let $\lambda_1\neq \overline\lambda_2$,
and $\Lambda_1, \Lambda_2$ should not be poles of $E(\phi)$ and $E(\psi)$ respectively.
\begin{align*}
(\overline\lambda_2-\lambda_1)&\Scr{E^r(\phi,\Lambda_1)}{E^r(\psi,\Lambda_2)}_{L^2(X)} \\
=\;& \Scr{E^r(\phi,\Lambda_1)}{\Delta E^r(\psi,\Lambda_2)}
-\Scr{\Delta E^r(\phi,\Lambda_1)}{E^r(\psi,\Lambda_2)}\\
=\;& \Sce{\textstyle\frac{\partial}{\partial u}\bei{r} \tilde\rho_u\Pi_0 E(\phi,\Lambda_1)}{\tilde\rho_u\Pi_0 E(\psi,\Lambda_2)(r,\cdot)}_{\!M} \\
&-\Sce{\tilde\rho_u\Pi_0 E(\phi,\Lambda_1)(r,\cdot)}{\textstyle\frac{\partial}{\partial u}\bei{r} \tilde\rho_u\Pi_0 E(\psi,\Lambda_2)}_{\!M}
\end{align*}
In the sequel we use the notation
\begin{equation} \label{notwu}
s_\nu^\pm(\Lambda_1,\Lambda_2)=\sqrt{\Lambda_1 -\nu-\ud^2}\pm \sqrt{\overline\Lambda_2 -\nu-\ud^2},\qquad \nu\in\sigma(\Delta_{1,0}).
\end{equation}
Here the branch of the square root, for which $\Im \sqrt{z}>0$ for $z\in \Complex\setminus\Reell$ is chosen;
this implies $\overline{\sqrt{\lambda}}=-\sqrt{\overline{\lambda}}$.
From the asymptotic expansion \eqref{aympt1}
of the constant term $\Pi_0 E(\Lambda,\phi)$ we obtain
\begin{subequations}
\begin{equation} \label{g3}
\begin{aligned}
(\overline\lambda_2-\lambda_1) & \Scr{E^r(\phi,\Lambda_1)}{E^r(\psi,\Lambda_2)}\\
=\;& -i s_0^- e^{-is_0^+ r }\sce{\phi}{\psi}
+\sum_{\nu\ge 0} i s_\nu^- e^{i s_\nu^+ r } \sce{T_{0\nu}(\Lambda_1,\phi)}{T_{0\nu}(\Lambda_2,\psi)}\\
&+ i s_0^+ \Big\{ e^{i s_0^- r} \Sce{T_{0 0}(\Lambda_1,\phi)}{\psi}
-e^{-i s_0^-r} \Sce{\phi}{T_{00}(\Lambda_2,\psi)}\Big\}\\
&+\sum_{\gamma> 0} i s_\gamma^- e^{i s_\gamma^+ r } \sce{S_{0\gamma}(\Lambda_1,\gamma)}{S_{0\gamma}(\Lambda_2,\psi)}\\
\end{aligned}
\end{equation}
Here we used the notation
$\theta(\ua)\sce{\phi}{\psi}\defgleich\sum_l \theta(a_l)\sce{\phi^{[l]}}{\psi^{[l]}}$ for a function $\theta$, where $\psi^{[l]}$ is the projection of $\psi$ onto $\H^{p-l}(B, \H^l(F))$.
Finally because of $s^+ s^-(\Lambda_1,\Lambda_2)=\lambda_1-\overline\lambda_2$
\begin{multline} \label{g4}
\Scr{E^r(\phi,\Lambda_1)}{E^r(\psi,\Lambda_2)}=\\
\frac{i}{s_0^+}e^{-is_0^+ r }\sce{\phi}{\psi}
- \frac{i}{s_0^-} \Big\{ e^{i s_0^- r} \Sce{T_{0 0}(\Lambda_1,\phi)}{\psi}
-e^{-i s_0^-r} \Sce{\phi}{T_{00}(\Lambda_2,\psi)}\Big\}\\
-\sum_{\nu\ge 0} \frac{i}{s_\nu^+} e^{+i s_\nu^+ r } \sce{T_{0\nu}(\Lambda_1,\phi)}{T_{0\nu}(\Lambda_2,\psi)}
-\sum_{\gamma> 0} \frac{i}{s_\gamma^+} e^{+i s_\gamma^+ r } \sce{S_{0\gamma}(\Lambda_1,\phi)}{S_{0\gamma}(\Lambda_2,\psi)}.
\end{multline}
\end{subequations}
Let $\tau \in \partial_+\FP\simeq \Complex\setminus\Reell^+$, so that $0<\tau<\min\{1/4,\nu_1\}$ and $\eps>0$. In addition we assume that $E(\Lambda,\phi)$
is holomorphic for $\Lambda$ in a neighbourhood of
$\pi_s^{-1}(\tau)$.
The remaining steps are the same as in the proof of Proposition 9.17 in \cite{mu-cusprk1}:
We let $\lambda_1, \lambda_2 \to \tau$ in the upper half plane $\Im \lambda_1>0,$ $\Im \lambda_2>0$.
Then by choice of the square root
\begin{align*}
s_\nu^-(\lambda_1,\lambda_2)&\to 2\sqrt{\tau-\nu-\ud^2},&
s_\nu^+(\lambda_1,\lambda_2)&\to 0 &&\text{ for } \tau>\nu+\ud^2\\
s_\nu^+(\lambda_1,\lambda_2)&\to 2 i \sqrt{\nu+\ud^2-\tau},&
s_\nu^-(\lambda_1,\lambda_2)&\to 0 &&\text{ for } \tau<\nu+\ud^2.
\end{align*}
Only for terms in $\H^{p-f/2}(B,\H^{f/2}(F))$, that is for fiber-degree $f/2$ and $\nu=0$, the inequality $\tau > \nu +\ud^2$ holds.
We set $\lambda_1=\lambda_2=\tau+i\eps$ in \eqref{g3}.
For $\eps\to 0$, because $\tau$ is no pole:
\begin{multline}\label{awehncv}
\sqrt{\tau} \| \phi^{[f/2]}\|^2 = \sqrt{\tau} \| T_{00}^{[f/2]}\phi \|^2 \\
+i \sqrt{\ud^2-\tau}\;\cdot \;\Big\{ \sce{T_{00}^{[\neq f/2]}\phi}{\phi^{[\neq f/2]} }- \sce{\phi^{[\neq f/2]} }{T_{00}^{[\neq f/2]}\phi}\Big\}
\end{multline}
Then we divide
\eqref{g3} with the given choice of $\lambda_1, \lambda_2$
by $-2 i \eps$ and employ \eqref{awehncv} and let $\eps\to 0$ under the assumption that $\tau$ is not a pole of $T_{00}^{[f/2]}$.
Choosing $\phi\in\H(B,\H^{f/2}(F))$ or $\phi\in\H(B,\H^{k}(F))$ with $k\neq f/2$ now gives the claim.
\end{proof}
\subsubsection{Reparametrization}
For the remainder we consider only $0<|\lambda|<\tau_1=\min\{1/4,\nu_1\}$, where $\nu_1$ is the smallest positive eigenvalue of $\Delta_{1,0}$.
The preimage in $\Sigma_s$ under $\pi_s$ of this domain lies in a double covering of $\Complex$.
For $\lambda\in\Complex\setminus \Reell^+$\index{$s$} let
$
s=s_{\nu,k}(\lambda)=d_k-i\sqrt{\lambda-\nu-d_k^2},
$
so that $
\lambda =s(2d_k-s).
$
In particular in the asymptotic expansion \eqref{aympt1} for $\psi\in \H^*(B,\H^k(F))$ we choose the parameter
$s=s_{0,k}$, so that
\begin{multline}\label{aymptneu}
\Pi_0 E(s,\psi) = e^{(a_k-d_k+s) r}\psi+e^{(\ua+\ud-s) r}T_{00}(s,\psi)\\
+\sum_{\nu>0}e^{(\ua+i\sqrt{s(2d_k-s)-\ud^2})r}T_{0\nu}(s,\psi)
+\sum_{\gamma>0}du\wedge e^{(\ua+i\sqrt{s(2d_k-s)-\ud^2})r}S_{0\gamma}(s,\psi).
\end{multline}
Note that $\Re s > d_k$ corresponds to $\Lambda\in \FP$.
With the notation from \eqref{notwu}
\begin{align*}
s_{\nu,k}^+(\lambda_1,\lambda_2)=i(s_{\nu,k}(\lambda_1)+\overline{s_{\nu,k}(\lambda_2)}-2d_k),\quad
s_{\nu,k}^-(\lambda_1,\lambda_2)=i(s_{\nu,k}(\lambda_1)-\overline{s_{\nu,k}(\lambda_2)}).
\end{align*}
Now \eqref{g4} for $\phi, \psi\in \H^{*}(B,\H^k(F))$, $\hat s\neq \bar s$ and $ \hat s+\bar s\neq 2d_k$ becomes
\begin{multline} \label{g4s}
(E^r(\hat s,\phi) ,\: E^r(s,\psi))
=\: \frac{1}{\hat s+\overline s - 2d_k}e^{(\hat s+\overline s - 2d_k) r }\sce{\phi}{\psi} \\
+\frac{1}{\overline s-\hat s} \Big\{ e^{(\overline s-\hat s) r} \Sce{T_{0 0}(\hat s,\phi)}{\psi}
-e^{(\hat s-\overline s)r} \Sce{\phi}{T_{00}(s,\psi)}\Big\}\\
-i \sum_{\nu\ge 0; l} \frac{e^{i s_{\nu,l}^+ r }}{ s_{\nu,l}^+} \sce{T_{0\nu}^{[l]}(\hat s,\phi)}{T_{0\nu}^{[l]}(s,\psi)}
-i \sum_{\gamma> 0; l} \frac{e^{i s_{\gamma,l}^+ r }}{ s_{\gamma,l}^+} \sce{S_{0\gamma}^{[l]}(\hat s,\phi)}{S_{0\gamma}^{[l]}(s,\psi)}.
\end{multline}
Equations \eqref{devse}, \eqref{emitstern} become
\begin{eqnarray}
dE(s,\psi)&=&(a_k-d_k+s) E(s, du\wedge \psi) \label{deformel}\\
E(s,\astm\psi)&=&\ast E(s,du\wedge\psi), \label{duformel}
\end{eqnarray}
and \eqref{Tbez} translates to
\begin{multline}
(a_k-d_k+s) T_{00}(s,\astm\psi)= (a_k+d_k-s) \astm T_{00}^{[k]}(s,\psi)\\
+\sum_{\substack{l=0\\ l\neq k}}^f \big(a_l+i\sqrt{s(2d_k-s)-d_l^2}\,\big) \astm T_{00}^{[l]}(s,\psi).\label{sternswitch}
\end{multline}
\subsection[Poles of $E(s, \phi)$]{Poles of $\mathbf{E(s, \phi)}$}\label{epole}
Let $U$ be a neighbourhood of $0$ and $2d_k$ which is open in $\Complex$, so that $s(2d_k-s)\in B_{t}(0)$ for $t<\tau_1=\min\{1/4,\nu_1\}$.
\begin{satz}\label{polposition}
Let $\phi\in \H^{*}(B,\H^k(F))$. Let $\Re(s)\ge d_k$ and $s\in U$, i.e. $s$ lies in the same connected component of $U$ as the point $2d_k$.
For every $\eps>0$ there is a constant $C(\eps)$, so that $\|T_{0\nu}^{[l]}(s,\phi)\|<C(\eps)$ for $s\in U\cap\{\Re s\ge d_k, \Im s\ge \eps\}$.
In particular poles of $E(s,\phi)$ in $U\cap\{\Re s\ge d_k\}$ must lie in the interval $(d_k,2d_k]$. The order of a pole is 1.
In addition, $T_{00}^{[f/2]}$ is holomorphic at $s=2d_k$.
\end{satz}
\begin{proof}
In \eqref{g4s} set $\hat s=s$, $\sigma=\Re s\ge d_k, \Im s\neq 0$ and $s\in U$. The left hand side of \eqref{g4s} is non-negative for $\phi=\psi$. Because $\Re(s)\ge d_k$, i.e.
$\lambda=s(2d_k-s)\in \Complex\setminus \Reell^+$, we have
\[
s_{\nu,l}^+=2 i \:\Im\sqrt{\lambda-\nu-d_l^2}\quad\text{where}\quad\Im\sqrt{\lambda-\nu-d_l^2}\ge 0.
\]
Thus
\begin{align*}
\sum_{\nu\ge 0; l} &\frac{i e^{i s_{\nu,l}^+ r }}{ s_{\nu,l}^+} \|T_{0\nu}^{[l]}(s,\phi)\|^2
+ \sum_{\gamma> 0; l} \frac{i e^{i s_{\gamma,l}^+ r }}{ s_{\gamma,l}^+} \|S_{0\gamma}^{[l]}(s,\phi)\|^2 \\
&\le \Big| \frac{e^{( s+\overline s - 2d_k))r }}{ s+\overline s - 2d_k}\Big| \|\phi\|^2
+2 \Big|\frac{ e^{(\overline s- s) r}}{\overline s- s}\Big| \| T_{0 0}^{[k]}(s,\phi) \| \|\phi\| \nonumber \\
&= \frac{e^{2(\sigma - d_k)r }}{ 2|\sigma - d_k|} \|\phi\|^2
+\frac{ 1}{|\Im s|}\| T_{0 0}^{[k]}(s,\phi) \| \|\phi\|.
\end{align*}
Now let $\|\phi\|_{L^2(M,g^M)}=1$. After multiplication with $e^{-2(\sigma - d_k)r}$ we obtain an estimate of the form
\begin{equation} \label{fab1}
\sum_{\nu\ge 0; l} \frac{i\,C_{\nu,l}}{s_{\nu,l}^+} \|T_{0\nu}^{[l]}(s,\phi)\|^2
+\sum_{\gamma> 0; l} \frac{i\,C_{\gamma,l}}{s_{\gamma,l}^+} \|S_{0\gamma}^{[l]}(s,\phi)\|^2
\le \frac{1}{ 2|\sigma - d_k|}
+\frac{c}{|\Im s|}\| T_{00}^{[k]}(s,\phi) \| \\
\end{equation}
with non-negative numbers $c, C_{\nu,l}$, which depend on $r$, but remain bounded for $r\to\infty$.
Now the asymptotics of both sides of \eqref{fab1} in a neighbourhood of $s_0$ will be compared.
We distinguish between four cases:
\begin{enumerate}[1)]
\item Let $\Re s_0\ge d_k, \Im s_0\neq 0$. Then $(s_{\nu,l}^+)^{-1}$ is bounded for $s\to s_0$ and all $(\nu,l)$ and \eqref{fab1} becomes
\begin{equation} \label{ba247}
C \|T_{00}^{[k]}\|^2+\sum_{\nu>0} C_\nu \|T_{0\nu}\|^2
+\sum_{\gamma>0} C_\gamma \|S_{0\gamma}\|^2\le \frac{c}{|\Im s|} \|T_{00}^{[k]}\|.
\end{equation}
Let $n$ be the order of a pole of $T_{00}^{[k]}(.,\phi)$ at $s_0$. Because of \eqref{ba247} for $s$ sufficiently close to $s_0$,
\begin{equation}
\gamma_1 |s-s_0|^{-2n}\le \|T_{00}^{[k]}(s,\phi)\|^2 \le \frac{c}{|\Im s|} \|T_{00}^{[k]}(s,\phi)\| \le \frac{\gamma_2}{|\Im s|} |s-s_0|^{-n}\label{fab2}
\end{equation}
for certain positive constants $\gamma_1,\gamma_2$.
If $s_0$ is not real, then $\frac{c}{|\Im s|}$ remains bounded for $s\to s_0$. This implies
\[
\gamma |s-s_0|^{-2n} \le |s-s_0|^{-n}\qquad \Longrightarrow\quad n\le 0.
\]
Thus $T_{00}^{[k]}(\cdot,\phi)$ is regular in $s_0$, and due to \eqref{ba247} all $T_{0\nu}(\cdot,\phi), S_{0\gamma}(\cdot,\phi)$ must be regular in $s_0$.
\item\label{itemoben} Now let $ \Im s_0=0$ with $d_k<s_0<2 d_k$, in particular $k\neq f/2$.
In \eqref{fab1} we have to note that possibly
\[
s_{\nu,l}^+(\tau,\tau)=2\sqrt{\tau-\nu-d_l^2}=0
\]
for $\tau\defgleich s_0(2d_k-s_0)$, if $\nu+d_l^2\le d_k^2$.
Therefore we first choose $s_0$ so that $s_{\nu,l}^+(\tau,\tau)\neq 0$. Then \eqref{fab2} shows for $s$ sufficiently close to $s_0$
\[
\gamma |s-s_0|^{-2n}\le |s-s_0|^{-n-1},\;\gamma>0 \qquad\Longrightarrow \quad n\le 1,
\]
which means the order of a pole of $T_{00}^{[k]}(\cdot,\phi)$ at $s_0$ can be at most 1.
If the left hand side of \eqref{ba247} has a pole of order $m$ in $s_0$, as above we conclude $2m\le n+1\le 2$.
So all $T_{0\nu}$ have a singularity of order less or equal to 1 in $s_0$.
For a given $s_0$ in $(d_k,2d_k)$ choose $(\nu,l)$ so that $s_{\nu,l}^+(\tau,\tau)=0$. Then
\begin{eqnarray*}
s_{\nu,l}^+ &=& 2 i \:\Im \sqrt{s(2d_k-s)-\nu-d_l^2}\\
&=& 2 i \:\Im \sqrt{-(s-s_0)^2-2(s_0-d_k)(s-s_0)}
=O(|s-s_0|^{1/2})
\end{eqnarray*}
This means the order $q$ of a pole of $T_{0\nu}^{[l]}(.,\phi)$ in $s_0$ satisfies $2q+1/2\le 1$, i.e. $q=0$.
The corresponding terms in the asymptotic expansion are holomorphic. For terms corresponding to other $(\nu,l)$ again the maximal order of a pole in $s_0$ is $1$.
\item\label{itemdavor} Now we consider $s_0=2 d_k$ and $d_k>0$, i.e. $k\neq f/2$.
\[
s_{0,f/2}^+=\sqrt{s(2d_k-s)}+\sqrt{\bar s(2d_k-\bar s)}=O(|s-2d_k|^{1/2})\quad\text{for}\quad s\to s_0.
\]
The term $(s_{\nu,l}^+)^{-1}$ remains bounded in the limit $s\to s_0$ for all other $(\nu,l)$.
Thus \eqref{fab1} takes the form
\begin{multline*}
\gamma_1 |s-2d_k|^{-1/2}\|T_{00}^{[f/2]}(s,\phi)\|^2+ C \|T_{00}^{[k]}(s,\phi)\|^2\\
+\sum_{\nu>0} C_\nu \|T_{0\nu}(s,\phi)\|^2+\sum_{\gamma>0} C_\gamma \|S_{0\gamma}(s,\phi)\|^2
\le \frac{c}{|\Im s|} \|T_{00}^{[k]}(s,\phi)\|
\end{multline*}
Analoguous to \ref{itemoben}) we conclude that $T_{00}^{[k]}$ and $T_{0\nu}^{[l]}$ can have a pole of order at most one in $s_0$ and that
$T_{00}^{[f/2]}$ is regular in $s_0$.
\item Finally let $s_0=0$ for $k=f/2$.
Then $|s_{0,f/2}^+|<\gamma |s|$ for $s$ near $0$.
Here we have to consider the inequality
\begin{multline*}
\gamma_1 |s|^{-1}\|T_{00}^{[f/2]}(s,\phi)\|^2+\sum_{\nu>0} C_\nu \|T_{0\nu}(s,\phi)\|^2
+\sum_{\gamma>0} C_\gamma \|S_{0\gamma}(s,\phi)\|^2 \\
\le \frac{c}{|\Im s|} \|T_{00}^{[f/2]}(s,\phi)\|+\frac{c'}{|s|}.
\end{multline*}
Again this implies that $T_{0\nu}$ and $S_{0\gamma}$ must be regular in $s_0$.
\end{enumerate}
\end{proof}
\begin{satz}\label{satzpolord}
The order of a pole of $E(\cdot ,\phi)$ in $s_0\in U$ is the maximum of the orders of poles of $T_{0\nu}^{[l]}(\cdot ,\phi)$ in $s_0$.
In particular $E(\cdot ,\phi)$ has a pole of order at most 1 in $2d_k$.
\end{satz}
\begin{proof}
This follows directly from \eqref{g4s} and the proof of Proposition \ref{polposition}.
\end{proof}
\begin{lem}\label{lemma3}
Let $\bar s$ be not a pole of $E^r(.,\phi)$ and $s$ not a pole of $E^r(.,\psi)$.
Then
\[
\Sce{T_{00}(\bar s,\phi)}{\psi}= \Sce{\phi}{T_{00}(s,\psi)}.
\]
\end{lem}
\begin{proof}
Under the assumption that neither $\bar s$ is a pole of $E^r(.,\phi)$ nor $s$ is a pole of $E^r(.,\psi)$, from \eqref{g3} (or \eqref{g4s}) we see for $\hat s=\bar s$
\[
0= 2(\bar s-2d_k) \{ \Sce{T_{00}(\bar s,\phi)}{\psi} - \Sce{\phi}{T_{00}(s,\psi)}\}.
\]
Because of Proposition \ref{satzpolord} the $T_{00}$ are holomorphic in $\bar s$ and $s$, and the claim follows.
\end{proof}
\subsection{Residues}
We have seen that poles in a neighbourhood $U$ of $2d_k$ can lie only in $U\cap(d_k,2d_k]$, and can only have order one. Let $s_0\in U\cap(d_k,2d_k]$ and $\tau= s_0(2d_k-s_0)$.
First we choose $s_0$ such that $s_{\nu,l}^+(\tau,\tau)\neq 0$ for all $l, \nu$.
After multiplication of \eqref{g4s} with $(\hat s-s_0)(\overline s-s_0)$ we first let $s\to s_0$ and then $\hat s\to s_0$.
\begin{align}
&(\res_{s_0} E^r(\cdot,\phi),\res_{s_0} E^r(\cdot,\psi))\nonumber\\
&= \Sce{\phi}{\res_{s_0} T_{00}(\cdot,\psi)}
-\sum_{\nu\ge 0; l} \frac{e^{-2r \sqrt{\nu+d_l^2-\tau} }}{ 2\sqrt{\nu+d_l^2-\tau}}\sce{\res_{s_0} T_{0\nu}^{[l]}(\hat s,\phi)}{\res_{s_0} T_{0\nu}^{[l]}(s,\psi)} \nonumber\\
&\qquad-\sum_{\gamma> 0; l} \frac{e^{-2r \sqrt{\gamma+d_l^2-\tau} }}{ 2\sqrt{\gamma+d_l^2-\tau}}\sce{\res_{s_0} S_{0\gamma}^{[l]}(\hat s,\phi)}{\res_{s_0} S_{0\gamma}^{[l]}(s,\psi)}
\end{align}
Both sums converge to $0$ for $r\to \infty$.
Now let $s_0=2d_k$, that is $\tau=0$. Then we have
\[
s_{\nu,l}^+(\tau,\tau)=0\iff (\nu=0 \land l=f/2).
\]
First we consider $d_k>0$.
\begin{align*}
& \lim_{\hat s\to 2 d_k} \lim_{s\to 2d_k}\frac{e^{i s_{0,f/2}^+(\hat s, s) r }}{ s_{0,f/2}^+(\hat s, s)}\sce{(\hat s-2d_k)T_{00}^{[f/2]}(\hat s,\phi)}{(s-2d_k) T_{00}^{[f/2]}(s,\psi)}\\
=& \lim_{\hat s\to 2 d_k}\frac{e^{i \sqrt{\hat s(2d_k-\hat s)} r }}{ \sqrt{\hat s(2d_k-\hat s)}}\sce{(\hat s-2d_k)T_{00}^{[f/2]}(\hat s,\phi)}{\res_{2d_k} T_{00}^{[f/2]}(\psi)}=0,
\end{align*}
where in the last line we used that $T_{00}^{[f/2]}$ is holomorphic at $2d_k$.
Analogous for $d_k=0$, i.e. $\psi\in \H^{*}(B,\H^{f/2}(F))$:
\begin{align*}
& \lim_{\hat s\to 0} \lim_{s\to0}\frac{e^{i s_{0,f/2}^+(\hat s, s) r }}{ s_{0,f/2}^+(\hat s, s)}\sce{\hat s T_{00}^{[f/2]}(\hat s,\phi)}{s T_{00}^{[f/2]}(s,\psi)}\\
=& \lim_{\hat s\to 0}\frac{e^{\hat s r }}{\hat s}\sce{\hat s T_{00}^{[f/2]}(\hat s,\phi)}{\res_{0} T_{00}^{[f/2]}(\psi)}=0.
\end{align*}
By interchanging the limits we arrive at
\begin{satz}\label{residuen}
Let $\H^{*}(B,\H^{k}(F))$. Let $s_0=2d_k$ or $s_0\in U \cap (d_k, 2 d_k)$, so that $s_{\nu,l}^+(\tau,\tau)\neq 0$ for all $l, \nu$. Then
$\res_{s_0} E(\cdot,\phi) \in L^2\Omega^*(X)$ and
\[
\Scr{\res_{s_0} E(\cdot,\phi)}{\res_{s_0} E(\cdot,\psi)}
= \Sce{\phi}{\res_{s_0} T_{00}(\cdot,\psi)}= \Sce{\res_{s_0} T_{00}(\cdot,\phi)}{\psi}.
\]
\end{satz}
The residues at $2d_k$ are of special interest. From now on we will use the notation\index{$\tC(\phi)$}\index{$\tE(\phi)$}
\[
\tC(\phi)\defgleich\res_{2d_k} T_{00}(.,\phi),\quad
{\tE}(\phi)\defgleich\res_{2d_k} E(.,\phi)
\]
and
$\H^{*,k}(M)\defgleich \H^{*}(B,\H^{k}(F)).$
An immediate consequence of Proposition \ref{residuen} is
\begin{cor}\label{zerlres}
\begin{enumerate}[a)]
\item ${\tC}^{[k]}=({\tC}^{[k]})^*, \quad {\tC}^{[k]}\ge 0$ and $\H^{*,k}(M)$ splits into the orthogonal direct sum
\[
\H^{*,k}(M)=\ker {\tC}^{[k]}\oplus \bild {\tC}^{[k]}.
\]
\item For $\phi\in\H^{*,k}(M)$,
\[
E(s,\phi) \text{ is holomorphic at } s=2d_k \iff \phi \in \ker {\tC}^{[k]} \iff \phi \in \ker {\tC}
\]
\item For $\phi\in \H^{*,k}(M)$ the residues $\tE(\phi)$ and $\widetilde E(du\wedge \phi)$ are in $L^2\Omega^*(X)$.
\end{enumerate}
\end{cor}
\begin{proof}
a) is immediate from Proposition \ref{residuen}. For b), let $\phi \in \ker {\tC}^{[k]}$. Proposition \ref{residuen} shows
\[
\| {\tE}(\phi) \|^2=\sce{\widetilde{C}^{[k]}(\phi)}{\phi}=0 \Rightarrow {\tE}(\phi)=0.
\]
If conversely $ {\tE}(\phi)=0$, from Proposition \ref{satzpolord} we conclude that $T_{00}(.,\phi)$ is holomorphic at $2d_k$.
Finally c) follows from Proposition \ref{residuen} and $E(s,\astm\psi)=\ast E(s,du\wedge\psi)$.
\end{proof}
In particular the residues are closed, because
\[
\|d \tE\|^2+ \|\delta \tE\|^2=\scr{\Delta \tE}{\tE} =0.
\]
This gives a map $\tE\mapsto [\tE]$ from a subspace of $\H_{(2)}^p(X)$ to $H^p(X)$.
\section{A Hodge--type Theorem}
\subsection{Harmonic representatives}
Using generalized Eigenforms, to each class in $H^p(M)$ a harmonic representative of a class in $H^p(X)$ will be associated.
Let $\phi\in \H^{*,k}(M)$ be a representative of a class in $H^p(M)$.
First we consider the case $k\neq \frac{f}{2}$.
\begin{satz}\label{satz47}
Let $k>f/2$ and $\phi\in \H^{*,k}(M)$.
Then $E(\cdot,\phi)$ is holomorphic at $s=2d_k$.
$E(2d_k,\phi)$ is closed if and only if ${\tC}^{[f-k]}(\astm\phi)=0$.
Let $k<f/2$ and $\phi\in \H^{*,k}(M)$. Then $E(\cdot,\phi)$ is holomorphic at $s=2d_k$ if and only if $\phi\in\ker {\tC}^{[k]}$.
\end{satz}
\begin{proof}
Let $\phi\in \H^{*,k}(M)$.
For $k>f/2$, that is $a_k<0$, in \eqref{sternswitch} we consider only the contribution coming from fiber degree $f-k$:
\begin{equation}
(2d_k-s) T_{00}^{[f-k]}(s,\astm\phi) = s \astm T_{00}^{[k]}(s,\phi). \label{reszue}
\end{equation}
So if $T_{00}^{[f-k]}(s,\astm\phi)$ at $2d_k$ has a singularity of order $n$, then
$T_{00}^{[k]}(s,\phi)$ has a singularity of order $n-1$ there. But the order of a pole at $2d_k$ is at most $1$ because of Proposition \ref{satzpolord}, so that
$T_{00}^{[k]}(s,\psi)$ must be holomorphic at $s=2d_k$.
From Corollary \ref{zerlres} we conclude that $E(s,\phi)$ is holomorphic at $2d_k$. Equations \eqref{deformel} and \eqref{duformel} give
\begin{align*}
d E(s,\phi) &= (a_k-d_k+s) E(s,du\wedge \phi)\\
&= (-2d_k+s) E(s,du\wedge \phi) = \pm(-2d_k+s) \ast E(s,\astm\phi) .
\end{align*}
Thus $E(2d_k,\phi)$ is closed if and only if $E(.,\astm\phi) $ is holomorphic at $2d_k$. From Corollary
\ref{zerlres} we know
$
\H^{*,f-k}(M)=\ker {\tC_{2d_k}}^{f-k}\oplus \bild{\tC_{2d_k}}^{f-k}
$
and the remaining statements follow.
\end{proof}
\begin{satz}\label{satz48}
$E(s,du\wedge \phi)$ is holomorphic and exact in $s=2d_k$ for $k<f/2$.
$\widetilde{E}(du\wedge \phi)=dE(2d_k,\phi)$ for $k>f/2$.
In particular both $E(2d_k, du\wedge\phi)$ (if existent) and $ \widetilde{E}(du\wedge \phi)$ are zero in the de~Rham cohomology of $X$ for all $\phi\in \H^{*,k}(M)$ with $k\neq f/2$.
\end{satz}
\begin{proof}
Let $k<f/2$ and $\phi\in \H^{*,k}(M)$. In equation \eqref{duformel}
the left hand side is holomorphic at $s=2d_k$.
For the residues follows
$
\widetilde{E}(du\wedge \phi)=0,
$ so that $E(.,du\wedge \phi)$ is holomorphic in $2d_k$.
For the exactness we use \eqref{deformel},
$
E(2d_k, du\wedge\phi)=\frac{1}{2d_k} d E(2d_k,\phi).
$
If otherwise $\phi\in \H^{*,k}(M)$ for $k>f/2$,
then again from \eqref{deformel}
\[
d E(s,\phi)=(s-2d_k) E(s, du\wedge\phi),
\]
and so the claim about the residue follows in the limit $s\to 2d_k$.
\end{proof}
Next we consider the middle fiber degree, $\phi\in \H^{*,f/2}(M)$.
From Proposition \ref{polposition} we know that $T_{00}^{[f/2]}(s,\phi)$ is holomorphic in $s=0$.
The functional equation \eqref{fktgle}
here implies
\[
T_{00}^{[f/2]}(0, T_{00}^{[f/2]}(0,\phi))=\phi\in \H^{*,f/2}(M).
\]
Additionally, $(T_{00}^{[f/2]})^*=T_{00}^{[f/2]}$ by Lemma \ref{lemma3}.
Thus for $p\ge f/2$ there is a decomposition of $\H^{*,f/2}(M)$ into the direct orthogonal sum
\begin{equation} \label{splitf2}
\H^{p-f/2,f/2}(M)=\H_+^p\oplus \H_-^p
\end{equation}
with\index{$\H_\pm^p$}
\[
\H_\pm^p=\{\phi\in \H^{p-f/2,f/2}(M)\mid T_{00}^{[f/2]}(0,\phi)=\pm\phi\}.
\]
The Hodge-isomorphism $\H^p(M)\to H^p(M)$ gives a corresponding splitting
\[
H^{p-f/2}(B,\H^{f/2})=:H^{p-f/2,f/2}(M)=\H_+^p(M)\oplus \H_-^p(M).
\]
\begin{lem}\label{hodgeiso}
The Hodge-star-operator
\[
\astm: \H_\pm^{p} \to \H_\mp^{n-p}
\]
is an isomorphism.
\end{lem}
\begin{proof}
This follows from the functional equation \eqref{sternswitch},
\[
T_{00}^{[f/2]}(s,\astm\phi)=-\astm T_{00}^{[f/2]}(s,\phi).\qedhere
\]
\end{proof}
\begin{satz}\label{satz49}
Let $\phi\in \H^{*,f/2}(M)$.
The generalized eigenform $E(s,\phi)$ is holomorphic and closed in $s=0$. $E(s, du\wedge\phi)$ is holomorphic in $0$.
\end{satz}
\begin{proof}
That $E$ is holomorphic in $0$ follows from Propositions \ref{polposition} and \ref{residuen}.
Then $E(s,du\wedge \phi)$ is holomorphic in $s=0$ because of \eqref{duformel} and \eqref{deformel}
implies $d E(s,\phi)=0$.
\end{proof}
\begin{satz}\label{satz50}
$E(0,\phi)=0$ for $\phi\in \H_-$, and $E(0,du\wedge\phi)=0$ for $\phi\in \H_+$.
\end{satz}
\begin{proof}
From Proposition \ref{maass-selb1} for real $\tau$ near $0$:
\begin{multline}
\|E^r(\tau,\phi)\|^2 =
r (\| \phi\|^2 + \| T_{00}^{[f/2]}(\tau,\phi) \|^2) \\
-i \sqrt{\tau} \Big\{ \Sce{\textstyle{\frac{d}{d\Lambda}|_\tau} T_{00}^{[f/2]}(.,\phi)}{T_{00}^{[f/2]}(\tau,\phi)} -
\Sce{T_{00}^{[f/2]}(\tau,\phi)} {\textstyle{\frac{d}{d\Lambda}|_\tau} T_{00}^{[f/2]}(.,\phi)}\Big\}\\
+\frac{1}{2i\sqrt{\tau}}\Big\{ e^{2 i \sqrt{\tau} r}\sce{T_{00}(\tau,\phi)}{\phi }
-e^{-2 i \sqrt{\tau} r}\sce{\phi }{T_{00}(\tau,\phi)} \Big\} \\
-\frac{e^{-2 i r \sqrt{\tau}}}{2 i\sqrt{\tau}} \| T_{00}^{[f/2]}(\tau,\phi) \|^2
+G_r
\end{multline}
where the remainder term $G_r\xrightarrow[r\to\infty]{}0$.
Since $T_{00}^{[f/2]}(\tau, \phi)$ is regular in $\tau=0$ and $\|\phi\|^2=\|T_{00}^{[f/2]}(0,\phi)\|^2$, we conclude in the limit $\tau\to 0$
\begin{equation*}
\|E^r(0, \phi)\|^2 =
2 r\cdot \: \| \phi\|^2
+r \cdot\Big\{ \sce{T_{00}^{[f/2]}(0, \phi)}{\phi} +\sce{\phi}{T_{00}^{[f/2]}(0, \phi)} \Big\}
+G_r.
\end{equation*}
Now if $\phi\in \H_-$, this gives the equality
$
\|E^r(0, \phi)\|^2 = G_r,
$
and the first claim follows by taking the limit $r\to\infty$.
Now let $\phi\in \H_+ \Longrightarrow \astm\phi\in \H_-$. The second statement then is a consequence of
\[
E(s,du\wedge \phi)=\pm\ast E(s,\astm\phi)=0.\qedhere
\]
\end{proof}
Now we want to collect all information about harmonic representatives defined by generalized eigenforms in a single map.
Let $\phi\in \H^{p-k}(B,\H^k(F))$.
We define a map $\Xi: \H^p(M)\to \Omega^p(X)$ by
\begin{equation*}
\Xi(\phi)=
\begin{cases}
\tE(\phi),&k<f/2\quad\text{and}\quad\phi\in \bild \tC^{[k]}\\
E(0,\phi),&k=f/2\quad\text{and}\quad\phi\in \H_+\\
E(2d_k,\phi),&k>f/2\quad\text{and}\quad\astm\phi\in \ker \tC^{[f-k]}\\
0,&\text{otherwise}
\end{cases}
\end{equation*}
and linear extension to $\H^p(M)$. The differential forms $\Xi(\phi)$ will be called \emph{singular values}.
From Corollary \ref{zerlres}, Propositions \ref{satz47} and \ref{satz49} it follows that singular values are closed harmonic differential forms; as such they represent classes in $H^p(X)$. Thus $\Xi$ extends to a map $\Xi:H^p(M)\to H^p(X)$ by setting
\[
\Xi([\phi])= [\Xi(\phi)],\qquad \phi\in \H^p(M).
\]
\subsection{Restriction map}
We want to consider the restriction of classes of singular values to the ``boundary'' $M$. More precisely, we identify $Y_s=\{s\}\times M\subset Z$ for $s>0$ with $M$ and
let $r=i_s^*:H^p(X)\to H^p(M)$ with the inclusion $i_s:Y_s\hookrightarrow X$. To see that $r$ is well-defined, let $\gamma$ be a cycle in $M$ and let $\gamma_s, \gamma_t$ denote the corresponding cycles in $Y_s$ and $Y_t$, respectively. From Stokes' theorem for $[\theta]\in H^p(X)$
\begin{equation} \label{zyka}
\int_{\gamma_s} i_s^*\theta- \int_{\gamma_t}i_t^*\theta= \int_{\partial([s,t]\times \gamma)} \theta= \int_{[s,t]\times \gamma} d\theta =0.
\end{equation}
The Theorem of de~Rham states that the map $\Psi^*:H^p(M)\to H^p_{\text{sing}}(M)$ induced by $\Psi(\theta)(\gamma)=\int_\gamma\theta$ is an isomorphism, so that $[i_t^*\theta]=[i_s^*\theta]$.
\begin{lem}\label{randrham}
Let $\theta\in \Omega^*(X)$ be a closed form such that the restriction $\theta|_Z=\theta(u,y)$ is rapidly decreasing for $u\to\infty$.
Then $r([\theta])=0$.
Furthermore if $\theta\in\Omega^p(X)$ is a closed form with $\Pi_0 (\theta|_Z)=0$, then $r[\theta]=0$.
\end{lem}
\begin{proof}
Since $i_t^*\theta$ is rapidly decreasing for $t\to\infty$, \eqref{zyka} shows
\[
\int_{\gamma} i^*\theta=0,
\]
which implies $r[\theta]=0\in H^p(M).$
Now let $\Pi_0 (\theta|_Z)=0$. In particular there is a $u>0$ such that
\[
i_u^*\theta \perp \Omega^*(B,\H^*(F))\quad \Longrightarrow\quad i_u^*\theta \perp \H^*(M).
\]
This shows $[i_u^*\theta]=0\in H^p(M)$, and we have seen that this class is independent of $u$.
\end{proof}
\begin{cor}
Let $\phi\in \H^{p-k}(B,\H^k(F))$. Then the restriction $(r\circ\Xi)[\phi]$ of singular values to $H^p(M)$ is given by
\begin{subequations}
\[
(r\circ\Xi)[\phi]=\left\{
\text{
\begin{minipage}{0.8\textwidth}
\begin{flalign}
&[\sum_{l<f/2}{\tC}^{[l]}(\phi)] ,&&\text{for }k<f/2,\quad \phi\in \bild \tC^{[k]}, \label{einsch1}\\
& 2[\phi]+\sum_{l<f/2}[T_{00}^{[l]}(2d_k,\phi)] ,&& \text{for }k=f/2, \quad\phi\in \H_+\label{einsch2} \\
& [\phi]+\sum_{l\le f/2}[T_{00}^{[l]}(2d_k,\phi)] ,&&\text{for } k>f/2,\quad \astm\phi\in \ker \tC^{[f-k]}\label{einsch3}\\
&0&&\text{otherwise}\nonumber
\end{flalign}\nonumber
\end{minipage}
}
\right.
\]
\end{subequations}
\end{cor}
\begin{proof}
From the second statement of Lemma \ref{randrham} we obtain $r[E]=r[\Pi_0 E]$ if $E$ is closed, and $r[\tE]=r[\Pi_0 \tE]$. Thus it is sufficient to examine the restrictions of the respective constant terms to the ``boundary'' $M$.
From the asymptotic expansion \eqref{aymptneu} of $\Pi_0 E$ it follows for $k>f/2$
\[
\Pi_0 E(2d_k,\phi)=\phi + \sum_{l\le f/2} T_{00}^{[l]}(2d_k,\phi) + \sum_{l> f/2} e^{-2d_l u}T_{00}^{[l]}(2d_k,\phi)+\theta
\]
with exponentially decreasing $\theta$.
Then Lemma \ref{randrham} gives
\[
r[\Xi[\phi]]=r[E(2d_k,\phi)]=r[\Pi_0 E(2d_k,\phi)]= [\phi]+\sum_{l\le f/2}[T_{00}^{[l]}(2d_k,\phi)].
\]
Similarily we derive \eqref{einsch2} for $r[E(0,\phi)]$ in the case $k=f/2$ and $\phi\in \H_+$.
Finally in the case $k<f/2$,
\[
\Pi_0 \tE(2d_k,\phi)=\sum_{l<f/2} {\tC}^{[l]}(\phi) + \sum_{l> f/2} e^{-2d_l u}{\tC}^{[l]}(\phi)+\tilde\theta,
\]
where we used that $T_{00}^{[f/2]}$ is holomorphic (Proposition \ref{polposition}).
This gives \eqref{einsch1}.
\end{proof}
\subsection{A Hodge-type theorem}\label{ktheorem}
Let $X$ be a manifold with fibered cusp metric, such that conditions (A) and (B) hold.
Let
\[
\mathcal{A}^p=\bigoplus_{k=0}^f \mathcal{A}^{(p-k,k)}\qquad\text{with}\qquad \mathcal{A}^{(p-k,k)}\defgleich
\begin{cases}
\bild {\tC}^{[k]}, & k<f/2\\
\H_+^p,&k=f/2\\
\astm\ker {\tC}^{[f-k]}, & f/2<k\le \min\{f,p\}\\
0,& k> \min\{f,p\}
\end{cases}
\]
\begin{xthm}\label{thm2}
Let $\mathfrak{h}:\H^p(M)\to H^p(M), \phi\mapsto [\phi]$ be the Hodge-isomorphism. Then
$\bild (r: H^p(X)\to H^p(M))=\mathfrak{h}(\mathcal{A}^p)$.
\end{xthm}
\begin{proof}
Let $R^p\defgleich\bild (r: H^p(X)\to H^p(M))$ and $R^{(j,k)}\defgleich\bild\big(r^{(j,k)}: H^{j+k}(X)\to H^{(j,k)}(M)\big)$ be the image of the restriction map $r$, projected onto $H^{(j,k)}(M)\defgleich H^j(B,\H^k(F))$.
From Stokes' theorem and the Hodge decomposition on the closed manifold $M$ it follows that
\begin{equation}
q([\phi], [\psi])=\int_M \phi\wedge\psi= (\astm \phi, \psi)_{L^2\Omega^{p}(M)}\label{qformel}
\end{equation}
defines a non-degenerate bilinear form $q: H^{(a,k)}(M)\times H^{(n-a-f,f-k)}(M)$ such that
$R^{(a,k)}$ and $R^{(n-f-a,f-k)}$ are $q-$orthogonal.
We claim
\begin{equation} \label{step3}
R^{(*,k)}=\mathfrak{h}(\mathcal{A}^{(*,k)}).
\end{equation}
To prove \eqref{step3}, first let $k<f/2$.
From \eqref{einsch1} we get $\mathfrak{h}(\bild\tC^{[k]})\subset R^{(j,k)}$, so that
\[
\bild\tC^{[k]}\subset \mathfrak{h}^{-1}(R^{(j,k)}) \subset \H^{j,k}(M)=\bild\tC^{[k]}\oplus \ker\tC^{[k]}
\]
by Corollary \ref{zerlres}. To show
\begin{equation}
\mathfrak{h}(\bild\tC^{[k]})=R^{(j,k)},\qquad k<f/2,\label{pres2}
\end{equation}
it is sufficient to prove $\mathfrak{h}^{-1}(R^{(j,k)})\cap \ker\tC^{[k]}=\{0\}$.
For $v\in \ker\tC^{[k]}$, from \eqref{einsch3}
\[
R^{(n-f-j,f-k)}\ni (r^{n-f-j,f-k}\circ\Xi)([\astm v])=[\astm v],\qquad k<f/2.
\]
This shows
\begin{equation}
\mathfrak{h}(\astm \ker\tC^{[k]})\subset R^{(\cdot,f-k)},\qquad k<f/2.\label{pres1a}
\end{equation}
In particular if $[v]\in R^{(j,k)}$ with $v\in \ker\tC^{[k]}$, then $\|v\|^2=q([\astm v],[v])=0$ and
\eqref{pres2} follows.
Now let $v\in \H^{*,f-k}(M)$ with $[v]\in R^{(\cdot,f-k)}$. From \eqref{pres2} and \eqref{qformel}
\[
[v]\in (R^{(\cdot,k)})^{\perp_q}=\mathfrak{h}((\bild\tC^{[k]})^{\perp_q}) \Longrightarrow \astm v\in (\bild\tC^{[k]})^\perp=\ker\tC^{[k]},
\]
so that
\begin{equation}
R^{(\cdot,f-k)}\subset \mathfrak{h}(\astm \ker{\tC}^{[k]}),\qquad k<f/2.\label{pres1b}
\end{equation}
Now let $k=f/2$. From \eqref{einsch2} and \eqref{splitf2}
\begin{equation*}
H_+^{(j,f/2)}\subset R^{(j,f/2)} \subset H^{(j,f/2)}=H_+^{(j,f/2)}\oplus H_-^{(j,f/2)}
\end{equation*}
But $\astm: \H_-^{(j,f/2)}\to \H_+^{(n-j-f,f/2)}$ is an isomorphism due to Lemma \ref{hodgeiso}.
Then
\[
[v]\in H_-^{(j,f/2)}\cap R^{(j,f/2)} \Rightarrow [\astm v]\in H_+^{(n-f-j,f/2)}\subset R^{(n-f-j,f/2)} \subset (R^{(j,f/2)})^{\perp_q}
\]
Again $[v]=0$ and thus
\begin{equation}
H_+^{(j,f/2)}(M)=R^{(j,f/2)}.\label{pres3}
\end{equation}
Equations \eqref{pres1a}, \eqref{pres1b}, \eqref{pres2} and \eqref{pres3} prove \eqref{step3} and the theorem.
\end{proof}
An example is given by the case of a manifold with cylindrical end, that is the metric on $Z$ is $du^2+g^M$. Then the fibers are points, and Theorem \ref{thm2} shows
\[
\im\big(H^p(X)\to H^p(M)\big)=\H^p_+.
\]
This is Theorem 3.1 in \cite{MuStro}.
The main result of this article is
\begin{xthm}\label{thm1}
Let $H_!^p(X)\defgleich\bild (H_c^p(X)\to H^p(X))$ be the image of cohomology with compact support in the de~Rham-cohomology.
Let $H_{\mathrm{inf}}^p(X)$ be a complementary space to $H_!^p(X)$ in $H^p(X)$,\index{$H_{\rm inf}^p(X)$}\index{$H_{\shriek}^p(X)$}
\[
H^p(X)=H_!^p(X)\oplus H_{\mathrm{inf}}^p(X).
\]
Let $R^p\defgleich\bild (r: H^p(X)\to H^p(M))$.
Then $\Xi(R^p)$ is isomorphic to $H_{\text{\upshape{inf}}}^p(X)$
and\/ $\Xi(H^p(M))=\Xi(R^p).$
\end{xthm}
\begin{proof}
By the definition of relative de~Rham-cohomology we have
\[
\ker (r: H^p(X)\to H^p(M))\cong\bild (H^p(X,M)\to H^p(X))
\]
and it is well known that $H_c^p(X)$ is isomorphic to $H^p(X,M)$.
This implies that
\begin{equation}
r: H_{\text{inf}}^p(X) \to \bild (r: H^p(X)\to H^p(M))
\end{equation}
is an isomorphism.
We show that $r\circ\Xi : R^p \to R^p $ is an isomorphism, then
\[
(r|_{\bild r})^{-1}\circ r :\Xi(R^p)\xrightarrow[]{\simeq} H_{\text{inf}}^p(X)
\]
gives the desired isomorphism.
First, from equations \eqref{einsch1}, \eqref{einsch2} and \eqref{einsch3} we obtain
\[
(r^{(p-k,k)}\circ \Xi)(\mathcal{A}^{(p-k,k)})\subset \mathcal{A}^{(p-k,k)},
\]
and altogether $r\circ \Xi(\mathcal{A})\subset \mathcal{A}$.
Next we show that $r\circ \Xi$ is injective on $\mathcal{A}$.
Let $\phi\in \H^{p-k,k}(M)$ and $r(\Xi([\phi]))=0$. For $k\ge f/2$, from \eqref{einsch2} and \eqref{einsch3} it follows that $[\phi]=0$.
If otherwise $k<f/2$ and $\phi\in \bild{{\tC}^{[k]}}$, then by \eqref{einsch1}
\[
r(\Xi([\phi]))=0 \Longrightarrow {\tC}^{[k]}(\phi)=0.
\]
Then Corollary \ref{zerlres} shows $\phi=0$.
From Theorem \ref{thm2} we get $R^p=\mathcal{A}$ and hence $r\circ\Xi:R^p\to R^p$ is an isomorphism. Furthermore this shows the second claim of the theorem, as
$
\Xi(H^p(M))= \Xi(\mathcal{A})
$
by definition of $\Xi$.
\end{proof}
Singular values are harmonic, so every class in $H_{\text{\upshape{inf}}}^p(X)$ has a harmonic representative.
As noted in section \ref{kohom}, every class in $H^p_!(X)$ has a unique $L^2-$harmonic representative, because $X$ is a complete Riemannian manifold.
This leads to
\begin{cor}\label{co1}
Every class in $H_{\text{\upshape{inf}}}^p(X)$ has a representative in the singular values.
Every class in $H^p(X)$ has a harmonic representative.
\end{cor}
Unlike classes in $H_!^p(X)$, in general the classes in $H_{\text{inf}}^p(X)$ do not belong to $L^2\Omega^p(X)$, unless they are represented by residues.
Because of Lemma \ref{randrham}, closed cusp forms are in $\ker r$.
Therefore we have a further decomposition
\[
H_!^p(X)=H_0^p(X)\oplus H_{\text{Eis}}^p(X)
\]
Here by definition harmonic representatives of $H_0^p(X)$ are rapidly decreasing on $Z$ and thus orthogonal to fiber harmonic forms. In \cite[\S 2]{harder2} it is shown that $H_{\text{Eis}}^p(X)$ generally is not empty.
Finally we emphasize that the decomposition
$
H^p(X)=H_!^p(X)\oplus H_{\mathrm{inf}}^p(X)
$
does not give rise to a corresponding decomposition of the $L^2-$cohomology $H_{(2)}(X)$, since there are classes in $H^p_{(2)}(X)$, which are not in $H^p(X)$.
\subsection{A Signature Formula}
Theorem \ref{thm1} can be applied to compute the $L^2-$signature of $X$ as follows. Let $N=\dim X=4 l$.
Then $L^2-\sign(X)$ is the signature defined by $H^{2l}_{(2),\text{red}}(X)\cong \H_{(2)}^{2l}(X)$, i.e. it equals the signature of the quadratic form
\[
Q(\phi,\psi)=\int_X \phi\wedge\psi,\quad \phi,\psi\in \H_{(2)}^{2 l}(X).
\]
Similarily let $\sign_c(X)=\sign(X_0,\partial X_0)$ be the signature of $Q$ on $H_!(X)$.
Since every class in $H_!(X)$ has a unique $L^2-$harmonic representative, we get an isomorphism between $H_!(X)$ and the space $\H_!(X)$ spanned by all $L^2-$harmonic representatives of $H_!(X)$. This shows that $\sign(X_0,\partial X_0)$ equals the signature of $Q$ on $\H_!^{2l}(X)$.
Now lets recall how $L^2-\sign(X)$ can be computed.
The operator
\[
\tau_X\defgleich i^{p(p-1)+2 l}\ast : \H_{(2)}^p(X)\to \H_{(2)}^{N-p}(X)
\]
is an involution on $\H_{(2)}^{2l}(X)$.
Let $\H_{\pm}^{2l}(X)$ be the $\pm 1$-Eigenspaces of $\tau_X=\ast$ on $\H_{(2)}^{2l}(X)$.
Then
\[
L^2-\sign(X)=\dim \H_{+}^{2l}(X)-\dim \H_{-}^{2l}(X).
\]
\begin{satz}\label{gleichesign}
Under the conditions (A) and (B) for the manifold $X$ with fibered cusp metric the equality
\[
L^2-\sign(X)= \sign(X_0,\partial X_0)
\]
holds.
\end{satz}
\begin{bem}
The results of \cite{dai} and \cite{vaill} show
\[
L^2-\sign(X)=\sign(X_0,\partial X_0)+\tau
\]
where $\tau$ is a topological invariant of the spectral sequence of $M\to B$. Under the conditions (A) and (B) one can conclude as in Dai's paper that $\tau=0$.
Instead of using this result, in the following proof the difference of the signatures under consideration will be calculated directly. In this way also additional information about singular values is obtained.
\end{bem}
\begin{proof}
Let
\[
\hat\H^p(Z)=\H^p(M)\oplus du\wedge \astm \H^p(M).
\]
Define $\tau_Z \psi\defgleich i^{p(p-1)+2 l}\astz \psi$ for the Hodge-Star-Operator $\astz$ on $Z=\Reell^+\times M$ equipped with the metric $du^2+g^M$.
For each $\omega\in\hat\H^p(Z)$ an $L^2-$harmonic form is defined by
\[
\tE(\omega) =\text{res}_{s=2\ud(\omega)} E(s,\omega)\defgleich\sum_k\text{res}_{s=2 d_k} E(s,\omega^{[k]}).
\]
\begin{lem}
$\tau$ commutes with the construction of $\tE$,
\begin{equation}
\tau_X \tE(\omega) = \tE(\tau_Z\omega)\label{tauswap}
\end{equation}
\end{lem}
\begin{proof}
Let $\phi\in \H^{p-k,k}(M)$ with $k<f/2$, $\phi\in \bild \tC^{[k]}$. Equations \eqref{duformel} and \eqref{deformel} show
\[
dE(s,\astm\phi)=(s-2d_k)E(s,du\wedge\astm\phi)=(s-2d_k)\ast^2(\astm)^2\ast E(s,\phi)
\]
For $s\to 2d_k$:
\[
dE(2d_k,\astm\phi)=\tE(du\wedge\astm\phi)=(-1)^p \ast\tE(\phi),
\]
and both $\tE(\phi)$ and are $\ast\tE(\phi)$ square integrable.
But now
\[
\astz(du\wedge \phi)=\astm \phi, \quad \astz\phi=(-1)^p du\wedge \astm\phi,
\]
so that
\[
\ast\tE(\phi)=\tE(\astz\phi).
\]
Finally $\tE(\phi)=0$ for $\phi \in \H^{p-k,k}(M)$ with $k\ge f/2$ or $\phi\in \ker\tC^{[k]}$.
\end{proof}
Let $h=2l$ and $\mathfrak{I}^k(M)\defgleich\bild \tC^{[k]}\subset \H^h(M), k<f/2$.
\[
\mathfrak{I}^k(Z)\defgleich \mathfrak{I}^k(M)+du\wedge \astm \mathfrak{I}^k(M)\subset \hat\H^h(Z),\qquad 0\le k<f/2
\]
is invariant under $\tau_Z$.
Let $Q_{k,\pm}\subset \mathfrak{I}^k(Z)$ be the $\pm 1-$eigenspaces of $\tau_Z$ and $Q_{\pm}=\bigoplus_{k<f/2} Q_{k,\pm}$.
Let $W\defgleich\tE(\hat\H^h(Z))\subset \H_{(2)}^h(X)$ and
$W_\pm\defgleich\tE(Q_\pm)$.
Because of \eqref{tauswap} we have $W_\pm\subset \H_{\pm}^h(X)$.
\begin{lem}\label{gleichdim}
\[
W=W_+\oplus_Q W_-\qquad\text{and}\qquad \dim W_+=\dim W_-
\]
\end{lem}
\begin{proof}
All differential forms under consideration have even total degree
$h$, so that $\tau=\ast$. First we show
\[
Q_{k\pm}=\{\phi\pm\tau_Z\phi\mid \phi\in \mathfrak{I}^k(M)\}, \quad 0\le k<f/2.
\]
Obviously the right-hand side is contained in $Q_{k\pm}$. Let $\psi\in Q_{k\pm}$ and
write $\psi=\phi+du\wedge\astm \phi_1$ with $\phi, \phi_1\in \mathfrak{I}^k(X)$.
Then
\[
\tau_Z \psi=du\wedge \astm \phi+\phi_1 =\pm\psi.
\]
Contraction with $\frac{\partial}{\partial u}$ shows $\phi_1=\pm \phi$, which implies
$\dim Q_{k+}=\dim Q_{k-}$.
The further statements now follow from
$\tE(\hat\H^h(Z))=\tE(Q_+\oplus Q_-)$ (by construction), and from $\tE(\eta)\neq 0$ for $\eta\in Q_+\oplus Q_-$.
\end{proof}
Let $\psi\in\H_{(2)}^h(X)$ with $r[\psi]\neq 0$. We claim that then already
$
r[\psi]\in\bigoplus_{k<f/2}H^{*,k}.
$
Because $\psi$ is square integrable,
$e^{-\ua u}\psi|_Z$ must lie in $L^2\Omega(Z, du^2+g^M)$. For fiber-degrees $k\ge f/2$ by definition $\ua\leq 0$, so that
$\|\psi^{[k]}|_{\{u\}\times M}\|\xrightarrow[u\to 0]{} 0$. Because in addition $d\psi=0$, we conclude $r[\psi^{[k]}]=0$ as in Lemma \ref{randrham}.
Now Theorem \ref{thm1} proves the existence of a unique
\[
\phi_0\in \bild \sum_{0\le j<f/2} \mathfrak{I}^j(M)\subset \H^h(M)
\quad\text{ so that }\quad
r[\tE(\phi_0)]=r[\psi].
\]
If $\psi\in\H_{(2)}^h(X)$ with $r[\psi]= 0$, let $\phi_0=0$.
Let $\psi_!=\psi-\tE(\phi_0)$. Since both $\psi$ and $\tE(\phi_0)$ are $L^2-$harmonic, the same must be true for $\psi_!$. In addition
$
r[\psi_!]=0,
$
so that $\psi_!\in\H_!(X)$.
We have
\[
\psi
=\frac{1}{2}\tE(\phi_0+\tau_Z\phi_0)
+\frac{1}{2}\tE(\phi_0-\tau_Z\phi_0)
+\psi_! .
\]
In this way we have defined an isomorphism of vector spaces
\begin{align*}
\rho:\H_{(2)}^h(X)&\to (W_+\oplus_Q W_-)\oplus \H_!(X)\\
\psi&\mapsto (\tE(\phi_0+\tau_Z\phi_0),\;\tE(\phi_0-\tau_Z\phi_0),\;\:\psi_!).
\end{align*}
To proof injectivity of $\rho$, let $\psi\in\H_{(2)}^h(X)$ with $\rho(\psi)=0$. From $\tE(\phi_0\pm\tau_Z\phi_0)=0$ we conclude $\phi_0=0$ and $\psi=\psi_!=0$.
To see surjectivity, let $(w_+,w_-,\psi_!)\in (W_+\oplus_Q W_-)\oplus \H_!$. Lemma \ref{gleichdim} shows the existence of $\phi_1, \phi_2 \in \bigoplus_{0\le k<f/2}\mathfrak{I}^k(M)$ with
\[
w_+=\tE(\phi_1+\tau_Z\phi_1),\quad w_-= \tE(\phi_2-\tau_Z\phi_2),
\]
and
$
\psi\defgleich\tE(\phi_1+\tau_Z\phi_2)+\psi_!
$
is mapped to $\rho(\psi)=(w_+,w_-,\psi_!)$.
Because $W=W_+\oplus_Q W_-$ and $\H_{(2)}^h(X)$ are $\ast-$invariant and $\H_{(2)}^h(X)=W\oplus \H_!(X)$, also
$\H_!(X)$ must be $\ast-$invariant. Finally
\begin{eqnarray*}
L^2-\sign(X)&=&\dim \H_{+}^{h}(X)-\dim \H_{-}^{h}(X) \\
&=&\dim ( \H_!^h(X)\cap \H_{+}^{h}(X))-\dim ( \H_!^h(X)\cap \H_{-}^{h}(X)).
\end{eqnarray*}
This proves Proposition \ref{gleichesign}.
\end{proof}
\input{literatur}
\end{document}
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The Exchange
Toni Rinde interview by Dr. Carolyn Ellis
TR: In the beginning, when they were first put into the ghetto, they were still given the opportunity to go out during the day and go to work and come back at night for—you know, to sleep. And then, after a while, that changed. But while they were able to go out, my parents were walking in the street one day pushing a carriage, and a lady comes up to my father and says, “What are you going to do with this beautiful child, taking—bringing her to the ghetto?” And my father looked at her and—it’s tough.
CE: I know. This is hard stuff.
TR: And said, “Well, I have nothing to do with her. Will you help me?” And she said, “Yes, I’ll take her.” Arrangements were made to meet the next day on a street corner, certain time. They were going to bring the carriage, put the carriage down on the street corner, turn around and walk away, and she was going to take it. That’s exactly what happened, because my parents watched while this was going on, to make sure that she indeed took me. And that was the beginning of my life.
| 4,014
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Past Honors Projects
- Kevin Donohue '21, "A poet and a scholar is what I was told": Analyses and Poems Inspired by the Two Men in A. E. Housman
- Naomi Meron '21, "What is your sickness?": Madness in Greek Tragedy and Medical Literature
- Clayton Howard '21, The Oxford History of the Archaic Greek World: Naxos and Paros
- Sophia Kocher '21, Mater Patriae: Livia in Nummis after Augustus
- Noah Simmons '20, The Upwind Performance of Roman Period Sailing Vessels
- Grace Caldwell '19, Gorgias' Encomium of Helen and the logos - a jeu de l'amour et du hazard
- Leah Alpern '18, Choosing to be grateful: ethical reasoning in Seneca's On Benefits
- Maeve Lentricchia '17 "The Theme You Left in Death": Between Euripides and Heidegger
- Emma Lape '16, Flowers that Bend with the Rainfall: Time and Identity in Greek Epic and Tragedy
- Thomas Rover '16, The Combat Archaeology of the Fifth-Century BCE Kopis
- Aaron Pellowski '15, De Tempore Discendum Est: Seneca's Philosophy of Time
- Kathleen Wahl '15, Lessons in Rape: Sexual Violence in the Roman Declamations
- Chloe Lee '14, The "Good Men" of 100 BCE: A Study of Cicero's Pro Rabirio Perduellionis
- Margarita S. Montgomery '14, Poetry of Politics in the 30s BC: A Reading of Vergil's Eclogues and Horace's Epodes
- Anna Leah Berstein-Simpson '13, Nostoi: Cultural Repatriation and National Identity
- Joyce Cho '13, Constructions of the Body of the Mistress By the Men of Latin Love Elegy
- Sarah Loucks '13, The Rhetoric of Wealth in Demosthenes' On the Crown
- Joel Malkin '13, Pindar's Delphi: The Delphic Authority behind Epinician Praise
- Kyle McGinty '13, Circles of Framing and Light: Analyzing the Nimbus in the Mediterranean
(image corpus overview)
- Chloe Moon '13, Early Athenian Gendering Through Art: Representations of Mixed-Gender Dancing from the Late Geometric to the Late Archaic Period
- Elizabeth Neill '13, Pillars of Society: Women as Bearers of Burdens in Archaic and Classical Art
- Emily Stronski '13i, A Representation of Immortality and the Afterlife in the Third Century CE: the Capitoline Prometheus Sarcophagus
- Emma Vance '13, Before the Alba Mater: Classics, Civilization, and Race at Moor's Indian Charity School
- Gregory Knight '12, Adeia as Immunity in 5th Century BC Athens
- Scott O'Brien '12, Seneca on the Renaissance University Stage: The Plays of William Gager and the Function of Moral Agency in the Tragic Genre
- Peter Osorio '12, A Classicist Under Constraint
- Chelsea Perfect '12, Fortuna Redux in Early Imperial Coinage
- Charles H. Clark '11, The Gortyn Laws in Architectural Context
- Kathryn T. Mammel '11, Bodies in Bloom: The Association of Flora and Female Figures in Late Bronze Age Aegean Iconography
(catalogue) (appendix)
- Sarah C. Spangenberg '11, Issues of Planning in Diocletian's Palace at Split: Imperial Cult and the Late Antique Palace
(images)
- Margaret Bell '10, The Gadde of Palmyra and Dura-Europos: Images of Fortune from Late 2nd Century Roman Syria
- Ryan Marnell '10, "Betwixt the Joys of Sex:" Scholarship and Marriage in Michael Field's treatment of Sappho
- Christopher Blankenship '09, The Role of the Individual in Ephoros' Histories
- Ray DiCiaccio '09, DAMP: A Digital Archaeological Mapping Program: Problems with its Applicability to Surface Survey Information Due to Issues of Data Comparability
- Catherine Lacey '09, Ovid's Art of Meaning: Interpretation and Authority in Tristia 2
- Radha D. S. Kulkarni '09, Visualizing the Past: Tu Marcellus Eris
- Dominic Machado '09, Marcus Claudius Marcellus: Rebellious Republican or Traditional Triumphator?
- Peter Mathias '09, Capital Punishment and Jurisprudence in Ancient Rome
- Debra M. Aboodi '08, A Log Cabin Out of Stone: Translating Horace's Epodes
- Aindriu C. Colgan '08, Understanding Imperial Rome and the Evolving Meaning of Empire
- Kyle Jazwa '08, Depictions of Boxing in Late Archaic Etruscan Tomb Paintings
- Alisa R. Koonce '08 (Senior Fellow), Philosophical Studies in early monastic schools: Ontology and mereology in marginal glosses of commentaries to Aristotle's Categories and related writings (10-11 c)
- Briar (Teron) Dent '08, The Pompeian Bakeries: An Analysis of the Urban Distribution of the Bakeries and Pastry Shops at Pompeii
(images)
- Craig W. Dent '07, Etruscan Tomb Markers: The Archaic and Classical Production of Clusium and Felsina (600-420 BCE)
- Jacqueline T. Olson '07, At Death's Doors: An Iconographical Study of the Velletri Sarcophagus
- Katherine B. Harrington '06, A Comparative Study of the Pebble Mosaics of Ancient Greece
- Sophia S. Khan '06, Polis under Threat: The Enduring Power of Euripides' Hecuba
- Bradley G. Wolcott '06, Striking a Balance: The numismatic Evidence for the Religious and Political Policy of Constantine the Great
- Matthew R. Jedreski '05, Minoan Religious Architecture in Glyptic Art: A Complete Contextual Analysis
- Tori L. McKee '05, Representations of Wives and Marriageable Women in Pliny the Younger and Ps. - Quintilian
- Matthew I. Kenney '04, The Military Nike
- Sarah C. Murray '04, Man Overboard? A Re-evaluation of the Under-representation of Sailors and Naval Warfare in Classical Athenian Art
- Magdalena M. Panz '04, The Aesthetics of Desire in Sappho: A Comparative Reading with Charles Baudelaire
- George M. Storm '04, Damnatio Memoriae: Case Studies from the Roman Republic through the First Century AD
- Christopher M. Chan '03, Plautus and the Learned Comedy of the Renaissance
- Rose B. MacLean '03, Fashioning a Freed Self: Representations of Civic and Familial Identity in the Ostian Epigraphic Monuments
- Dongngan Truong '02, A Novel Theme for Elegy: Abortion in Ovid, Amores 2.13 and 2.14
- Julia Tzeng '02, Hadrian's Provincial Policies: An Examination of Achaea, Britannia, and Judea under the Reign of Hadrian
- Michael F. O'Donnell '02, Poet as Playwright: Characterization through Speech in Homer's Iliad
- Julie B. Axelrod '01, Lifting the Veil: A Critical Model for the Interpretation of Juvenal's Satires
- Seth L. Button '01, Chariots in War and and Art in the Late Bronze Age Aegean
- Abigail J. Gillard '01, Politics in the Kerameikos and Demosion Sema: Civic Identity and Funerary Custom in Fifth-Century BC Atheos
| 64,141
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Vi’s inspiring story. How one woman reclaimed her voice and found her freedom to speak up after years of suffering from a history of abuse which had made her a target for further abuse from employers, relationships, and members of the medical and legal profession. Here’s her story:
‘Finding a Voice’
getting over abuse
Like the flowing river making its way back to the sea, life had led me to enact the part of the farmer’s wife in the African Folk Tale “The Woman from the Stars” at the point when she discovered that her husband had broken her only requirement of their relationship.
I could not have known that the words I spoke were the manifestation of what I needed to say to many people in whom I had put trust, in my life. I could not have known that this enactment would be preparing me for events that were soon to follow.
An affinity to cattle
Having grown up on a small farm, my affinity to cattle started very young and my mother had often told of how, when I started to crawl, I would make my way out of the farmhouse and crawl down the farmyard and into the cattle pen, where I would often be found amongst their hooves amusing myself. No harm came to me (except for developing ringworm on my face, body and in my hair).
Life in the country was very isolated and solitary, and I had no way to express and deal with the pain of the neglect, abuse, abandonment of parental betrayal of not fulfilling the precious role of protecting a vulnerable, innocent, dependent child (me).
It was not until my 20’s – following a court case where my abusive ex-lover (who was extremely wealthy and a prominent businessman in the community) had been taken to court by the police and then had been bound over by the courts to keep away from me – that I found my way to the big city, away from his clutches and to take up employment.
The pain and damage from childhood traumas and my career had led me to become involved in numerous types of therapies, and these were an excellent foundation for the workshop I was to find myself in. My recollection of drama at school was not positive, and this kind of workshop had in no way appealed to me. But having met Claire through other mutual activities, I had attended the Making Moves “Bringing the Stars Down to Earth” workshop.
Betrayal – Acting for Real
To set the picture, in the first scene I had acted herding the cows, and in the second scene, I became the farmer’s wife who, while tending her crops on the land, had heard her/my husband’s laughing. She/I instantly knew that the mocking laughter had meant that he had done the one thing she/I had asked him not to.
As the scene unfolded, I as the wife stood facing my/her husband. The moment was so real. I said with all my might and every ounce of my being:
“You betrayed me, our love and the only request I had made of our relationship”.
The pain I felt was unbearable, and in that moment I saw the look on the face of the other participant who was acting the part of my husband. It was only a moment, but it felt minutes, and I knew from the expression on his face that the words I had spoken and the pain I felt had at last been heard and understood.
It was a very powerful and healing moment.
The pain I experienced was beyond words. It was the first time I had dared to give voice to the unbearable pain of betrayal I had been living with for so many years. I had talked about the pain with friends, in therapy sessions and group sessions but talking about it did not release the devastation I had experienced.
Speaking-Up
I could not have then known that the words I spoke in the workshop were to have huge repercussions. I would have the opportunity to speak to them again in a variety of situations.
First, it was to my employer, and to numerous members of the medical and legal professions who were involved in the case, I was fighting as the result of a car accident. Medical experts failed to diagnose a head injury and its impact. Then others changed their medical diagnosis when they saw what more senior medical staff had written, and then the limitations and hidden agendas of the legal profession resulted in paltry compensation.
My employers terminated my employment when they realised my condition, saying that they would be equally culpable if my condition were to get worse. I had felt numerous betrayals and injustices from the incompetencies of these professionals in whom I had put my trust. I could no longer trust my own judgment, and it was important for me to state that I felt betrayed by recording my experience in written statements for all involved to read.
A bigger opportunity
But a much bigger opportunity arose when my sister telephoned me to inform me that my father had died. I knew that if I were to go to the funeral, I would have to face the terrible past and a family who had colluded in abuse and who made me a scapegoat rather than face their own shadows.
Initially, I was told that I could not see my father’s body prior to the funeral. And, in a subsequent telephone call, my other sister basically warned me against attending the funeral, saying she knew what would happen if I did. What could possibly be the reason behind this statement?
I was to find out on the day of the funeral.
The man I had fallen in love with and in whom I had put my trust some 25 years earlier had to my utter disbelief and horror been invited to be a pallbearer for my father’s coffin at the funeral.
This man had betrayed me with affairs with other women, emotional abuse and violence and who had driven a wedge between me and my family.
My parents had chosen to support him over me, their vulnerable daughter, during a humiliating court case in which the police had pressed charges against him for his violent behaviour and threats on my life.
Healing words
My longest-standing friend who had lived through the dreadful ordeal with me in my early 20’s arranged to get me to the chapel of rest to see my father’s body. There I was able to say how hurt and betrayed I had felt as a result of his failings – and to make some peace.
I continually prayed that I might get through the ordeal with dignity and grace. Then in the middle of the funeral service, my brother announced that my former lover, abuser and betrayer would be reading my father’s tribute.
All eyes were on me as he got up from the family pew (where he had been invited to sit with his son by the woman he betrayed me for) and proceeded to speak about my father’s life.
Speaking the words of Betrayal
The pain and humiliation were becoming unbearable. At the graveside, I had comforted my mother as she said goodbye to my father, and as I walked backwards, my ex-lover/abuser took me in his arms and kissed me on the lips – saying that he was pleased to see me, knowing that I had no escape.
As I pulled away from him, I saw the flash of a camera going off in the distance and then realised that my family had arranged to photograph the funeral. To say it was surreal was an understatement and I knew whatever I did would be captured on film for posterity.
I then found myself speaking the words of betrayal I had spoken in the enactment of the African Folk Tale – except this time I was saying them calmly, coherently and with total power.
To all those looking on, it was as if I were having an agreeable exchange with a man from the past, but to me and my friend who knew, this was a man who had caused me immense pain and who had destroyed my confidence, self-worth, and any relationship with my family for the past 20 years.
Articulate and in total control
I know it is good to own and express my feelings, but sometimes that can work against us women because we are seen as over-emotional and sadly the meaning of what we have to say can be lost.
This time I was articulate and in total control of what I had needed to say. I was able to fully represent myself in a way I had never been able to do before.
I believe that if we surrender, the universe will lead us to what we need to heal. I give thanks for meeting Claire and for the opportunity to attend the perfect workshop at the perfect time which was to be a perfect healing catalyst for me.
I continue my journey with renewed spirit and affinity to cattle.
Vi
From Making Moves Newsletter No 5 Autumn 2004
Commentary by Claire
Synchronicity was first observed by Jung who saw chance happenings, coincidences, not as meaningless occurrences but as highly meaningful and evidence of our connection with an unseen pool of experience that he called “the collective unconscious”.
In her article, Vi described how her participation in the Bringing the Stars Down to Earth Workshop had precipitated a momentous series of events. Read her article, and you will learn how this courageous woman, who is to my mind a Woman from the Stars (the title of the African tale we worked with), had managed to triumph over the very raw deal that life has dealt her.
Both of us reflected on how her experience in the workshop had in some way paved the way for the events that had followed. Was it by chance that her father died just at the moment that she had found the strength to confront people in her life. Jung would probably say “Yes, absolutely”. When we do healing work or make a breakthrough, then this is registered within the collective consciousness. And then things start to happen in mysterious and unforeseen ways.
I know that enacting myths and folk tales which have been told hundreds and thousands of times, mostly through oral tradition, have powerful effects. Working with an African tale, which is even more likely to have a current, alive oral tradition, has the potential to be even more potent – as these tales often reflect the injustices and pain suffered by its people.
It is a way of invoking the collective unconscious through which healing can happen. See this explanation of how this process works.
The story is very simple. A farmer was upset because he believes that someone is stealing the milk from his cattle at night. He watches out for the thiefs and is amazed to discover that a group of Star Women have climbed down from the stars and have been gaily milking his cows. When they run away, he catches one of them. The Woman agrees to become his bride on the condition that he must never look in her basket.
For a while, the farmer is happy with his wife, but as time passes, his curiosity eats at him. Finally, he peeps into his wife’s basket and discovers that there is nothing in the basket. Triumphantly he faces her with the “truth” about her basket.
This is the scene that Vi played out that had had so much significance for her. The Woman from the Stars spoke the words of betrayal that she had never before been able to speak – and this liberated her to find her own words.
This is the magical power of drama. To give us permission to speak and act in ways that open up doors to freedom
At the end of the story, the Woman from the Stars leaves the farmer and returns to her sisters. The farmer is left to grieve for what he has lost, the magical woman who will never return. After speaking her truth, Vi too was able to turn her back on the man and the family who had betrayed her, and who would never do so again. She truly was a Woman from the Stars returning to her “true home”, the Stars and her “sisters” (good friends who believed in her and supported her)- able to take her place in the world.
The beauty of working with a story like the “Woman from the Stars” is that it raises human experience into the realm of the magical and the otherworldly. It offers a way to elevate the mundanity of this life into something that is beyond us.
Claire Schrader, Making Moves Newsletter
As Vi discovered, the healing and transformational potential of working in this way can never be known in advance. It is a kind of magic that is quite simply beyond words and the capacity of the logical mind.
I would recommend the Breakthrough Plus Course or the Breaking Family Patterns Course for working with this kind of issue.
Useful Links:
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Join the mailing list
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\begin{document}
\author{G\'erard Endimioni}
\address{C.M.I-Universit\'{e} de Provence,
39, rue F. Joliot-Curie, F-13453 Marseille Cedex 13}
\email{endimion@gyptis.univ-mrs.fr}
\title[Normal automorphisms]{Normal automorphisms of a free metabelian nilpotent group}
\subjclass[2000]{20E36, 20F28.}
\keywords{Normal Automorphism; Free Metabelian Nilpotent Group}
\begin{abstract}
An automorphism $\varphi$ of a group $G$ is said to be normal
if $\varphi(H)=H$ for each normal subgroup $H$ of $G$.
These automorphisms form a group containing the group of inner
automorphisms.
When $G$ is a nonabelian free (or free soluble) group, it is known that
these groups of automorphisms coincide,
but this is not always true when
$G$ is a free metabelian nilpotent group. The aim of this paper is to
determine the group of normal automorphisms in this last case.
\end{abstract}
\maketitle
\section{Preliminary results}
\noindent In a group $G$, consider a map $\varphi: G\to G$ of the form
$$\varphi:x\to
x\left[x,u_{1}\right]^{\lambda(1)}\ldots\left[x,u_{m}\right]^{\lambda(m)},$$
where $u_{1},\ldots,u_{m}$ are elements
of $G$, the exponents $\lambda(1),\ldots,\lambda(m)$ being integers (as usual, the
commutator $[a,b]$ is defined by $[a,b]=a^{-1}b^{-1}ab$).
When $G$ is metabelian, using the relation
$[xy,u]=y^{-1}[x,u]y[y,u]$, it is easy to see that $\varphi$ is an
endomorphism of group. These endomorphisms appear in \cite{KU} as a
solution to a problem of extension from $G'$
to $G$ for certain endomorphisms (also see \cite{CS}).
Now consider in an arbitrary group $G$ the map
$\varphi_{u,v}:x\to x[x,u][x,v]$, with $u,v\in G$ and suppose that $\varphi_{u,v}$ is
an endomorphism. It is then easy to show that the equality
$\varphi_{u,v}(xy^{-1})=\varphi_{u,v}(x)\varphi_{u,v}(y^{-1})$ is equivalent to the
relation $[u,y][x,v]=[x,v][u,y]$. Consequently, if $\varphi_{u,v}$ is
an endomorphism for any $u,v\in G$, the group $G$ is metabelian.
One can summarize these preliminary remarks like this.
\begin{prp}
\bigskip
In a group $G$, the following
conditions are equivalent:\\
{\rm (i)} every map $\varphi: G\to G$ of the form
$\displaystyle\varphi:x\to
x\prod_{i=1}^m\left[x,u_{i}\right]^{\lambda(i)}$
$(u_{i}\in G,\: \lambda(i)\in {\mathbb Z})$
is an endomorphism;\\
{\rm (ii)} $G$ is metabelian.
\end{prp}
It is very easy to see that in an metabelian group, such endomorphisms
are not necessarily
automorphisms (consider for example the endomorphism
$x\to x[x,b]^{-1}$ in the nonabelian group of order 6
defined by the presentation
$\langle a,b\, \vert\, a^3=b^2=(ab)^2=1\rangle$).
But in a nilpotent group, each map of
the form
$$x\to w_{0}x^{\lambda(1)}w_{1}x^{\lambda(2)}\ldots
x^{\lambda(n)}w_{n}\;\;\; ({\rm with}\;
\lambda(1)+\lambda(2)+\cdots+\lambda(n)=\pm 1)$$
is bijective \cite[Theorem 1]{EN1}. Hence we have:
\begin{prp}
In a metabelian nilpotent group $G$, every map $\varphi: G\to G$ of the form
$\displaystyle\varphi:x\to
x\prod_{i=1}^m\left[x,u_{i}\right]^{\lambda(i)}$
$(u_{i}\in G,\: \lambda(i)\in {\mathbb Z})$
is an automorphism.
\end{prp}
For convenience sake, in a metabelian nilpotent group, an
automorphism of the form $\displaystyle x\to
x\prod_{i=1}^m\left[x,u_{i}\right]^{\lambda(i)}$
will be called a {\em generalized inner automorphism}.
Obviously, every inner automorphism
is a generalized inner automorphism, since $u^{-1}xu=x[x,u]$.
Notice that in a nilpotent group $G$ of class $\leq 2$,
we may write
$$\varphi(x)=x\prod_{i=1}^m\left[x,u_{i}\right]^{\lambda(i)}
=x\prod_{i=1}^m\left[x,u_{i}^{\lambda(i)}\right]
=x\left[x,\prod_{i=1}^m u_{i}^{\lambda(i)}\right].$$
Hence $\varphi(x)=x[x,u]=u^{-1}xu$, where
$\displaystyle u=\prod_{i=1}^m u_{i}^{\lambda(i)}$.
Consequently, in a nilpotent group of class $\leq 2$, the notions
of inner automorphism and generalized inner automorphism coincide.
On the other hand, in the free nilpotent group of class 3 and rank 2
freely generated by $a$ and $b$, it is easy to see that the map
$x\to x[x,a,a]$
is a generalized inner automorphism but is not an inner automorphism.
As usual, in a group, the left-normed commutator
$[x_{1},\ldots,x_{n}]$ is defined inductively by
\begin{eqnarray*}
[x_{1},\ldots,x_{n}] & = & [x_{1},\ldots,x_{n-1}]^{-1}[x_{1},\ldots,x_{n-1}]^{x_{n}} \\
& = & [x_{1},\ldots,x_{n-1}]^{-1}x_{n}^{-1}[x_{1},\ldots,x_{n-1}]x_{n}.
\end{eqnarray*}
The next technical result will be useful in the following.
\begin{prp}
In a group $G$, consider a map
$\varphi:G\to G$ of the form
$$\varphi(x)=x\prod_{i=1}^{n}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{\eta(i)}
\quad ({\eta(i)}\in {\mathbb Z}),$$
for some function $\sigma :\{ 1,\ldots,n\}\to {\mathbb N}\setminus\{ 0\}$ and
elements $v_{i,j}\in G$ ($1\leq i\leq n$, $1\leq j\leq \sigma(i)$).
Then $\varphi(x)$ can be written in the form
$$\varphi(x)=
x\left[x,u_{1}\right]^{\lambda(1)}\ldots\left[x,u_{m}\right]^{\lambda(m)}
\quad (\lambda(i)\in {\mathbb Z},\: u_{i}\in G).$$
\end{prp}
\begin{proof} By induction, using the relation
$[x,y,z]=[x,y]^{-1}[x,z]^{-1}[x,yz]$
\end{proof}
Frequently in this paper we shall make use of well-known
commutator identities (see for example \cite[5.1.5]{RO}). In particular,
we have the following relations,
valid in a metabelian group $G$, for any $x,y,z\in G$, $t\in G'$ and
$\lambda\in {\mathbb Z}$:
\begin{center}
\begin{tabular}{cccc}
$[xt,y]=[x,y][t,y]$ & {} & {} & $[t^{\lambda},y]=[t,y]^{\lambda}$ \\
$[x,y,z][y,z,x][z,x,y]=1$ & {} & {} & $[t,x,y]=[t,y,x]$ \\
\end{tabular}
\end{center}
\begin{prp}
The set of generalized inner automorphisms of a
metabelian nilpotent group $G$ forms a
(normal) subgroup of the group of automorphisms of $G$.
\end{prp}
\begin{proof} If $\varphi$ and $\psi$ are generalized inner
automorphisms respectively defined by
$$\varphi(x)=x\prod_{i=1}^m [x,u_{i}]^{\lambda(i)},\;
\psi(x)=x\prod_{i=1}^n [x,v_{i}]^{\mu(i)},$$
an easy calculus shows that
$$\psi\circ\varphi(x)=x\prod_{i=1}^m [x,u_{i}]^{\lambda(i)}
\prod_{i=1}^n [x,v_{i}]^{\mu(i)}
\prod_{i=1}^m \prod_{j=1}^n [x,u_{i},v_{j}]^{\lambda(i)\mu(j)}.$$
Thus $\psi\circ\varphi$ is a generalized inner automorphism by
Proposition 1.3.\\
It remains to prove that $\varphi^{-1}$ is a
generalized inner automorphism.
For that, it suffices to construct for each integer
$k\geq 1$ a generalized inner automorphism $\psi_{k}$ such that
$\psi_{k}\circ \varphi$ is of the form
$$\psi_{k}\circ \varphi :x\to
x\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{\eta(i)}$$
for some function $\sigma :\{ 1,\ldots,m\}\to {\mathbb N}\setminus\{ 0\}$ and
elements $v_{i,j}\in G$ ($1\leq i\leq m$, $1\leq j\leq \sigma(i)$),
and where each commutator is of weight $\geq 1+2^{k-1}$
(namely, $\sigma(i)\geq 2^{k-1}$ for $i=1,\ldots,m$).
Indeed, since $G$ is nilpotent, this implies that
$\psi_{k}\circ \varphi(x)=x$ for $k$ large enough, thus
$\varphi^{-1}=\psi_{k}$ is a generalized inner automorphism, as required.
We argue by induction on $k$. The result is clear when $k=1$
by taking for $\psi_{1}$ the identity map. Now
suppose that for some integer $k\geq 1$, there exists
a generalized inner automorphism $\psi_{k}$ such that
$\displaystyle \psi_{k}\circ \varphi(x)=
x\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{\eta(i)}$, with
$\sigma(i)\geq 2^{k-1}$ for $i=1,\ldots,m$. Put $\psi_{k+1}=\psi'\circ\psi_{k}$,
where $\psi'$ is defined by
$\displaystyle \psi'(x)=
x\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{-\eta(i)}$.
We have
$$\psi_{k+1}\circ\varphi(x)=x\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{\eta(i)}$$
$$\times \prod_{j=1}^{m}\left[x\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{\eta(i)},
v_{j,1},\ldots,v_{j,\sigma(j)}\right]^{-\eta(j)}.$$
Since
$$ \prod_{j=1}^{m}\left[x\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)}]^{\eta(i)},
v_{j,1},\ldots,v_{j,\sigma(j)}\right]^{-\eta(j)}$$
$$=\prod_{i=1}^{m}[x,v_{j,1},\ldots,v_{j,\sigma(j)}]^{-\eta(j)}
\prod_{j=1}^{m}\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)},
v_{j,1},\ldots,v_{j,\sigma(j)}]^{-{\eta(i)}\eta(j)}, $$
we obtain
$$\psi_{k+1}\circ\varphi(x)=x\prod_{j=1}^{m}\prod_{i=1}^{m}[x,v_{i,1},\ldots,v_{i,\sigma(i)},
v_{j,1},\ldots,v_{j,\sigma(j)}]^{-{\eta(i)}\eta(j)}$$
and this completes the proof of the proposition.
\end{proof}
\section{Main result}
We recall that a normal automorphism $\varphi$ of a group $G$ is an
automorphism such that $\varphi(H)=H$ for each normal subgroup
$H$ of $G$. These automorphisms form a subgroup of the group of all
automorphisms of $G$. Obviously, this subgroup contains the subgroup
of inner automorphisms of $G$. It happens these subgroups coincide,
for instance when $G$ is a nonabelian free group \cite{LU},
a nonabelian free soluble group \cite{ROM},
or a nonabelian free nilpotent group of class $2$ \cite{EN2}.
On the other hand, the subgroup of inner automorphisms
is of infinite index in the group of normal automorphisms when $G$
is a nonabelian free nilpotent group of class $k\geq 3$ \cite{EN2}.
Also note there are exactly two normal
automorphisms in a (nontrivial) free abelian group: $x\to x$ and
$x\to x^{-1}$.
The determination of all normal automorphisms
in a free nilpotent group of class $k\geq 3$ seems to be an open problem.
Note however that for $k=3$, Theorem 2.1 below provides such a
determination, since the free nilpotent metabelian group of class $3$
(and of given rank) coincide with the corresponding free nilpotent group of class
$3$.
One can find in \cite{FG} information about the structure of the group
of normal automorphisms of a nilpotent group.
Certainly, in a metabelian nilpotent group, each
generalized inner automorphism is a normal automorphism, but
a normal automorphism need not to be a generalized inner automorphism.
However, our main result states that the converse holds in a
nonabelian free metabelian nilpotent group.
\begin{thm}
In a nonabelian free metabelian nilpotent
group, the group of normal automorphisms coincides with the group of
generalized inner automorphisms.
\end{thm}
Before to tackle the proof of this theorem, we give a few consequences.
Let $\varphi$ be an automorphism of a group $G$. According to Schweigert \cite{SC},
one says that $\varphi$ is a polynomial automorphism of $G$
if there exist integers $\epsilon_{1},\ldots,\epsilon_{m}\in{\mathbb Z}$
and elements $u_{1},\ldots,u_{m}\in G$ such that
$$\varphi(x)=(u_{1}^{-1}x^{\epsilon_{1}}u_{1})\ldots
(u_{m}^{-1}x^{\epsilon_{m}}u_{m})$$
for all $x\in G$. For instance, in a metabelian nilpotent group, any
generalized inner automorphism is a polynomial automorphism. It turns out that in a
nonabelian free metabelian nilpotent group, these notions coincide.
\begin{cor} An automorphism $\varphi$ of a nonabelian free
metabelian nilpotent group $G$ is a generalized inner automorphism if and only if
it is polynomial.
\end{cor}
\begin{proof} Only the part 'if' needs a proof. Therefore, suppose that $\varphi:G\to G$ is a polynomial automorphism
defined by
$$\varphi(x)=(u_{1}^{-1}x^{\epsilon_{1}}u_{1})\ldots
(u_{m}^{-1}x^{\epsilon_{m}}u_{m})$$
and put
$\epsilon= \epsilon_{1}+\cdots+\epsilon_{m}$.
If $\gamma_2(G)=[G',G]$ is the second term of the lower central series of $G$,
it is not difficult to see that $\varphi$ induces the automorphism $\overline{\varphi}:x\mapsto x^{\epsilon}[x,u]$
on the free nilpotent group $G/\gamma_2(G)$, where
$u=u_1^{\epsilon_{1}}\ldots u_m^{\epsilon_{m}}\gamma_2(G)$. An easy calculation
shows that the relation $\overline{\varphi}(xy)=\overline{\varphi}(x)\overline{\varphi}(y)$
is equivalent to the relation $(xy)^{\epsilon}=x^{\epsilon}y^{\epsilon}$. This implies that $\epsilon=1$, and so $\varphi^{-1}$ is polynomial by
\cite[Theorem 1]{EN1}. Now it is clear that $\varphi$ is a normal automorphism, hence
$\varphi$ is a generalized inner automorphism by Theorem 2.1.
\end{proof}
In \cite{EN3}, it is proved that the group generated by all
polynomial automorphisms of a metabelian group is itself metabelian.
Therefore, we can state:
\begin{cor} The group of normal automorphisms of a free
metabelian nilpotent group is metabelian.
\end{cor}
Note that this result is valid even if the relatively free group is abelian, since the
group of normal automorphisms is abelian in this case.\\
Recall that an automorphism of a group $G$ is said to be an IA-automorphism if it induces the identity automorphism on $G/G'$. We shall write $\textrm{IA}(G)$ for the group of
IA-automorphisms of $G$. In a nonabelian free metabelian nilpotent group $G$
of class $k$, the group of normal automorphisms is a subgroup of $\textrm{IA}(G)$.
Since $\textrm{IA}(G)$ stabilizes the lower central series of $G$, it is nilpotent of class $k-1$ (see for instance \cite[p. 9]{SE}), and thus so is the group of normal automorphisms. Also this result is a consequence of the fact that in a nilpotent group of class $k\geq 2$, the group generated by all polynomial automorphisms is nilpotent of class $k-1$ \cite{EN3}.\\
If $d>1$ denotes the rank of the free metabelian nilpotent group $G$
of class $k$, note that $\textrm{IA}(G)$ is metabelian if $d=2$ (by a result of C.K. Gupta
\cite{GU}) or if $k\leq 4$ (in this case, $\textrm{IA}(G)$ is nilpotent of class $\leq 3$).
On the other hand, it is worth pointing out that contrary to the group of normal automorphisms, $\textrm{IA}(G)$ is not metabelian if $d>2$ and $k>4$.
Indeed, without loss of generality, we can assume that $G$ is the free metabelian nilpotent group of class $5$ and of rank 3, freely generated by $a,b,c$. In this group, we define three IA-automorphisms $f,g,h$ by
$$f(a)=a[a,b],\: f(b)=b,\: f(c)=c,$$
$$g(a)=a,\: g(b)=b[b,c],\: g(c)=c,$$
$$h(a)=a,\: h(b)=b,\: h(c)=c[c,a].$$
It follows
$$[f,g](a)=a[c^{-1},b,a],\: [f,g](b)=b,\: [f,g](c)=c,$$
$$[f,h](a)=a,\: [f,h](b)=b,\: [f,h](c)=c[a,b^{-1},c],$$
whence
$$[f,h]\circ[f,g](c)=c[a,b^{-1},c],$$
$$[f,g]\circ[f,h](c)=c[a,b^{-1},c][c,b,a,b,c].$$
Consequently, $[f,g]$ and $[f,h]$ do not commute, hence $\textrm{IA}(G)$ is not metabelian.
\section{Proof of Theorem 2.1.}
In all this section, we consider a fixed set $S$ of cardinality $\geq 2$
and we denote by $M_{k}$ the free metabelian nilpotent group
of class $k$ freely
generated by $S$. In other words, $M_{k}=F/F''\gamma_{k+1}(F)$, where
$F$ is the free group freely generated by $S$ and $\gamma_{k+1}(F)$
the $(k+1)$th term of the lower central series of $F$.
The normal closure in a group $G$ of an element $a$ is written
$\langle a^{G}\rangle$.
\begin{lem}
For any distinct elements $a,b\in S$ and
any integer $t$, the subgroup
$\langle (a^{t}b)^{M_{k}} \rangle\cap \gamma_{k}(M_{k})\unlhd M_{k}$ is generated
by the set of elements of the form
$$\left[ a^{t}b,c_{1},\ldots,c_{k-1} \right],\;\; with\;
c_{1},\ldots,c_{k-1}\in S \;\;(we \; suppose \; k>1). $$
Moreover, for any
subset $\{ a=a_{0},a_{1},\ldots,a_{r}=b\}\subseteq S$ containing $a$
and $b$, the subgroup
$$\langle (a^{t}b)^{M_{k}} \rangle\cap \gamma_{k}(M_{k})\cap
\langle a_{0},a_{1},\ldots,a_{r}\rangle$$
is generated
by the set of elements of the form
$\left[ a^{t}b,c_{1},\ldots,c_{k-1} \right]$, with
$c_{1},\ldots,c_{k-1}\in \{ a_{0},a_{1},\ldots,a_{r}\}$.
\end{lem}
\begin{proof} First suppose that $t=0$ and consider an
element $w\in \langle b^{M_{k}} \rangle\cap
\gamma_{k}(M_{k})$. Hence $w$ is a product of elements of the form
$[c_{0},c_{1},\ldots,c_{k-1}]^{\pm 1}$, with $c_{i}\in S$. More
precisely, we can write $w=w_{0}w_{1}$, where $w_{0}$ (resp. $w_{1}$)
is a product of elements of the form $[c_{0},c_{1},\ldots,c_{k-1}]^{\pm 1}$
with $c_{i}\in S\setminus\{ b\}$, (resp. with $c_{i}\in S$, the element $b$
occuring once at least in $[c_{0},c_{1},\ldots,c_{k-1}]$).
In fact, substituting 1 for the indeterminate $b$ in the relation
$w=w_{0}w_{1}$ and using the fact that $w$ lies in
$\langle b^{M_{k}} \rangle$, we obtain $w_{0}=1$.
Thus $w$ is a product of elements
of the form $[c_{0},c_{1},\ldots,c_{k-1}]^{\pm 1}$, with $c_{i}=b$ for
some $i\in \{0,\ldots,k-1\}$. If $i=1$, we can write
$[c_{0},b,\ldots,c_{k-1}]=[b,c_{0},\ldots,c_{k-1}]^{-1}$. If $i>1$,
we have
$[c_{0},c_{1},\ldots,b,\ldots,c_{k-1}]=[c_{0},c_{1},b,\ldots,c_{k-1}]$;
by using the relation $[c_{0},c_{1},b]=[c_{1},b,c_{0}]^{-1}[b,c_{0},c_{1}]^{-1}$,
it follows
$$[c_{0},c_{1},\ldots,b,\ldots,c_{k-1}]=
[b,c_{1},c_{0},\ldots,c_{k-1}]
[b,c_{0},c_{1},\ldots,c_{k-1}]^{-1}.$$
Thus we have shown that any element of $\langle b^{M_{k}} \rangle\cap
\gamma_{k}(M_{k})$ is a product of elements of the form
$\left[ b,c_{1},\ldots,c_{k-1} \right]^{\pm 1}$, with
$c_{i}\in S$. Since $\left[ b,c_{1},\ldots,c_{k-1} \right]\in
\langle b^{M_{k}} \rangle\cap \gamma_{k}(M_{k})$, the first
part of our lemma is proved when $t=0$.\\
Now consider the general case. Actually, since clearly
$S'=\{ a^tb\}\cup S\setminus\{ b\}$ is a free basis of $M_{k}$, we
can use the result obtained in the particular case. It follows that
$\langle (a^{t}b)^{M_{k}} \rangle\cap \gamma_{k}(M_{k})\unlhd M_{k}$ is generated
by the set of elements of the form
$\left[ a^{t}b,c_{1},\ldots,c_{k-1} \right]$, with
$c_{i}\in S'$. But in fact we may take $c_{i}\in S$ and so conclude,
since
$$\left[ a^{t}b,c_{1},\ldots, a^{t}b,\ldots,c_{k-1} \right]=
\left[ a^{t}b,c_{1},\ldots, a,\ldots,c_{k-1} \right]^{t}
\left[ a^{t}b,c_{1},\ldots, b,\ldots,c_{k-1} \right].$$
Finally, consider an element $w\in \langle (a^{t}b)^{M_{k}} \rangle\cap \gamma_{k}(M_{k})\cap
\langle a_{0},a_{1},\ldots,a_{r}\rangle$.
we can express $w$ in the form $w=w'w''$, where $w'$ (resp. $w''$)
is a product of elements of the form $[a^tb,c_{1},\ldots,c_{k-1}]^{\pm 1}$
with $c_{1},\ldots,c_{k-1}\in \{ a_{0},a_{1},\ldots,a_{r}\}$
(resp. with $c_{1},\ldots,c_{k-1}\in S$, an element of
$S\setminus \{ a_{0},a_{1},\ldots,a_{r}\}$ occuring once at least in
the sequence $c_{1},\ldots,c_{k-1}$).
Substituting 1 for all indeterminates in
$S\setminus \{ a_{0},a_{1},\ldots,a_{r}\}$, the equality $w=w'w''$
gives $w=w'$. This completes the proof of the lemma.
\end{proof}
As usual, the expression $[x,_{n}y]$ is defined in a group by
$[x,_{0}y]=x$ and $[x,_{n}y]=[[x,_{n-1}y],y]$ for each positive
integer $n$.
For a fixed subset $\{ a_{0},\ldots,a_{r}\}\subseteq S$
and a function $\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$, we define
in $M_{k}$ the symbol $\left[x,y,\Delta\right]$ ($x,y\in M_{k}$) by
$$\left[x,y,\Delta\right]=
\left[x,y,_{\Delta(0)}a_{0},_{\Delta(1)}a_{1},\ldots,_{\Delta(r)}a_{r}\right].$$
Note that for any sequence $b_{1},\ldots,b_{k}$ of elements of
$\{ a_{0},\ldots,a_{r}\}$, there is a function
$\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$ such that
$\left[x,y,b_{1},\ldots,b_{k}\right]=\left[x,y,\Delta\right]$, with
$\Delta(0)+\cdots+\Delta(r)=k$ (it suffices to apply the relation
$[u,v,w]=[u,w,v]$, valid in any metabelian group whenever $u$ belongs
to the derived subgroup).
If $j,j'$ are distinct given integers in
$\{ 0,\ldots,r\}$ and if $\Delta(j)\not = 0$, we define the function
$\Delta_{(j)}^{(j')}: \{ 0,\ldots,r\}\to {\mathbb N}$
by
$$\Delta_{(j)}^{(j')}(j)=\Delta(j)-1,\;\;
\Delta_{(j)}^{(j')}(j')=\Delta(j')+1\;\; {\rm and} $$
$$\Delta_{(j)}^{(j')}(i)=\Delta(i)\;\;
{\rm for \; all} \; i\in\{ 0,\ldots,r\}\setminus\{ j,j'\}.$$
When $\Delta$ is not the
null-function, we shall denote by $m(\Delta)$
the least integer $j$ such that $\Delta(j)\not = 0$.
If $S$ is ordered, we may define in $M_{k}$ {\em basic commutators}
(see for example \cite[Chapter 3]{NE}. Recall that a basic commutator of
weight $k'$ ($2\leq k'\leq k$) is a commutator of the form
$\left[b_{1},b_{2},\ldots,b_{k'}\right]$ ($b_{i}\in S$), with
$b_{1}>b_{2}$ and $b_{2}\leq b_{3}\leq\cdots \leq b_{k'}$. Any set of
these commutators freely generates a free abelian subgroup of $M_{k}'$.
In the next lemma, we aim to express a product of commutateurs
of the form $[a_{s},a_{i},\Delta]$ as a product where only basic
commutators occur.
\begin{lem}
Let $\{ a_{0},\ldots,a_{r}\}$ be a
finite subset of $S$ ($r>0$). Choose an integer $s\in \{0,\ldots,r\}$
and consider an element $w\in M_{k+2}$ ($k>0$) of the form
$$w=\prod_{i,\Delta}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}\quad
(\epsilon(i,\Delta)\in {\mathbb Z}),$$
where the product is taken over all integers
$i\in \{ 0,\ldots,r\}$ and all functions
$\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$ such that
$\Delta(0)+\cdots+\Delta(r)=k$.
Then:\\
{\rm (i)} We have
$$w=\prod_{i<s,\, i\leq m(\Delta)}[a_{s},a_{i},\Delta]
^{\epsilon(i,\Delta)}
\prod_{s<i,\, i\leq
m(\Delta)}[a_{i},a_{s},\Delta]^{-\epsilon(i,\Delta)}$$
$$\times \prod_{s\leq m(\Delta),\, m(\Delta)<i}[a_{i},a_{s},\Delta]
^{-\epsilon(i,\Delta)}
\prod_{m(\Delta)<s,\, m(\Delta)<i}[a_{i},a_{m(\Delta)},\Delta^{(s)}_{(m(\Delta))}]
^{-\epsilon(i,\Delta)}$$
$$ \times \prod_{m(\Delta)<s,\, m(\Delta)<i}
[a_{s},a_{m(\Delta)},\Delta^{(i)}_{(m(\Delta))}]
^{\epsilon(i,\Delta)}$$
(in all these products, $i$ lies in
$\{0,\ldots,r\}\setminus\{s\}$).\\
{\rm (ii)} We have $w=1$ only if all exponents $\epsilon(i,\Delta)$
with $i\in \{0,\ldots,r\}\setminus\{s\}$ occuring in the expression of $w$ are zero.
\end{lem}
\begin{proof} (i) First we write $w$ as a product of two factors:
$$w=\prod_{i\leq m(\Delta)}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}
\prod_{m(\Delta)<i}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}.$$
The first factor can be expressed in the form
\begin{eqnarray*}
\prod_{i\leq m(\Delta)}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)} & = &
\prod_{i<s,\, i\leq m(\Delta)}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}
\prod_{s<i,\, i\leq m(\Delta)}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}\\
{} & = & \prod_{i<s,\, i\leq m(\Delta)}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}
\prod_{s<i,\, i\leq m(\Delta)}[a_{i},a_{s},\Delta]^{-\epsilon(i,\Delta)}.
\end{eqnarray*}
Now we deal with the second facteur. We have
\begin{eqnarray*}
\prod_{m(\Delta)<i}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)} & = &
\prod_{s\leq m(\Delta),\, m(\Delta)<i}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}
\prod_{m(\Delta)<s,\, m(\Delta)<i}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)} \\
{} & = & \prod_{s\leq m(\Delta),\, m(\Delta)<i}[a_{i},a_{s},\Delta]^{-\epsilon(i,\Delta)}
\prod_{m(\Delta)<s,\, m(\Delta)<i}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}.
\end{eqnarray*}
Therefore Lemma 3.2(i) is proved if we show the relation
\begin{equation}
\prod_{m(\Delta)<s,\,
m(\Delta)<i}[a_{s},a_{i},\Delta]^{\epsilon(i,\Delta)}=
\end{equation}
$$ \prod_{m(\Delta)<s,\, m(\Delta)<i}[a_{i},a_{m(\Delta)},\Delta^{(s)}_{(m(\Delta))}]
^{-\epsilon(i,\Delta)}
\times \prod_{m(\Delta)<s,\, m(\Delta)<i}
[a_{s},a_{m(\Delta)},\Delta^{(i)}_{(m(\Delta))}]
^{\epsilon(i,\Delta)}.
$$
For that, write more explicitly the commutator $[a_{s},a_{i},\Delta]$
(in the following equalities, we write $m$ instead of $m(\Delta)$):
\begin{eqnarray*}
[a_{s},a_{i},\Delta] & = &
\left[a_{s},a_{i},_{\Delta(0)}a_{0},\ldots,_{\Delta(r)}a_{r}\right] \\
& = & \left[a_{s},a_{i},_{\Delta(m)}a_{m},\ldots,_{\Delta(r)}a_{r}\right] \\
& = & \left[a_{s},a_{i},a_{m},_{\Delta(m)-1}a_{m},\ldots,_{\Delta(r)}a_{r}\right].
\end{eqnarray*}
Since
$\left[a_{s},a_{i},a_{m}\right]=
\left[a_{i},a_{m},a_{s}\right]^{-1}\left[a_{m},a_{s},a_{i}\right]^{-1}
=\left[a_{i},a_{m},a_{s}\right]^{-1}\left[a_{s},a_{m},a_{i}\right]$,
we obtain
\begin{equation}
[a_{s},a_{i},\Delta]=[a_{i},a_{m},\Delta_{(m)}^{(s)}]^{-1}
[a_{s},a_{m},\Delta_{(m)}^{(i)}].
\end{equation}
Relation (1) is now an immediate consequence of
(2). \\
(ii) Choose an order in $S$ such that $a_{0}<a_{1}<\cdots <a_{r}$.
Under this condition, all commutators occuring in the expression of
$w$ obtained in the first part of the lemma are basic commutators.
Suppose that $w=1$.
Since basic commutators of the form $[a_{j},a_{i},\ldots]$
($i,j\in\{0,\ldots,r\}\setminus \{ s\}$) occur only in the fourth
factor, we have $\epsilon(i,\Delta)=0$ whenever
$m(\Delta)<s$ and $m(\Delta)<i$.
It follows
$$w=\prod_{i<s,\, i\leq m(\Delta)}[a_{s},a_{i},\Delta]
^{\epsilon(i,\Delta)}
\prod_{s<i,\, i\leq
m(\Delta)}[a_{i},a_{s},\Delta]^{-\epsilon(i,\Delta)}$$
$$\times \prod_{s\leq m(\Delta),\, m(\Delta)<i}[a_{i},a_{s},\Delta]
^{-\epsilon(i,\Delta)}=1.$$
But all basic commutators occuring in this equality are distinct. Thus
$\epsilon(i,\Delta)=0$ for all integers $i\not = s$ and all functions $\Delta$, as required.
\end{proof}
\begin{lem}
Let $\varphi$ be a normal automorphism
of $M_{k+2}$ ($k>0$) acting trivially on
$M_{k+2}/\gamma_{k+2}(M_{k+2})$. Then, for all distinct elements
$a,b\in S$, there exists a generalized inner automorphism
$\psi$ of $M_{k+2}$ such that $\varphi(a)=\psi(a)$ and $\varphi(b)=\psi(b)$.
\end{lem}
\begin{proof} Let $a,b$ be two distinct elements of $S$. Then
$a^{-1}\varphi(a)$ and $b^{-1}\varphi(b)$ belong to
$\langle a^{M_{k+2}}\rangle \cap \gamma_{k+2}(M_{k+2})$ and
$\langle b^{M_{k+2}}\rangle \cap \gamma_{k+2}(M_{k+2})$ respectively.
By Lemma 3.1, there is a finite subset
$\{ a=a_{0},a_{1},\ldots,a_{r}=b\}\subseteq S$ such that
$$
\varphi(a)=\varphi(a_{0})=a_{0}\prod_{i,\Delta}[a_{0},a_{i},\Delta]^{\alpha(i,\Delta)}\quad
(\alpha(i,\Delta)\in {\mathbb Z}),$$
$$
\varphi(b)=\varphi(a_{r})=a_{r}\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\beta(i,\Delta)}\quad
(\beta(i,\Delta)\in {\mathbb Z}),$$
where the two products are taken over all integers
$i\in \{ 0,\ldots,r\}$ and all functions
$\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$ with
$\Delta(0)+\cdots+\Delta(r)=k$ (as in Lemma 3.2, $[a_{s},a_{i},\Delta ]$ is
defined by
$[a_{s},a_{i},\Delta]=[a_{s},a_{i},_{\Delta(0)}a_{0},\ldots,_{\Delta(r)}a_{r}]$).
Note that if $|S|=2$ (and so $r=1$), Lemma 3.3 is easily verified by taking the
generalized inner automorphism $\psi$ defined by
$$\psi(x)=x\prod_{\Delta}[x,a_{1},\Delta]^{\alpha(1,\Delta)}
\prod_{\Delta}[x,a_{0},\Delta]^{\beta(0,\Delta)}.$$
Thus we can assume in the following that $|S|>2$.
For any positive integer $t$, $(a_{0}^ta_{r})^{-1}\varphi(a_{0}^ta_{r})=
(a_{0}^ta_{r})^{-1}\varphi(a_{0})^t\varphi(a_{r})$ belongs to
$\langle a_{0},a_{1},\ldots,a_{r}\rangle$. Therefore, once again by Lemma
2.1, $\varphi(a_{0}^ta_{r})$ can be expressed in the form
$$\varphi(a_{0}^ta_{r})=a_{0}^ta_{r}\prod_{i,\Delta}[a_{0}^ta_{r},a_{i},\Delta]^{\eta_{t}(i,\Delta)}\quad
(\eta_{t}(i,\Delta)\in {\mathbb Z}).$$
Since $[a_{0},a_{i},\Delta]$ is a commutator of weight $k+2$ in a
nilpotent group of class $k+2$, we obtain
$$\varphi(a_{0}^ta_{r})=a_{0}^ta_{r}\prod_{i,\Delta}[a_{0},a_{i},\Delta]^{t\eta_{t}(i,\Delta)}
\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\eta_{t}(i,\Delta)}.$$
Thus the relation $\varphi(a_{0}^ta_{r})=\varphi(a_{0})^t\varphi(a_{r})$ implies that
\begin{equation}
\prod_{i,\Delta}[a_{0},a_{i},\Delta]^{t\eta_{t}(i,\Delta)}
\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\eta_{t}(i,\Delta)}
\end{equation}
$$=\, \prod_{i,\Delta}[a_{0},a_{i},\Delta]^{t\alpha(i,\Delta)}
\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\beta(i,\Delta)}. $$
Choose an order in $S$ such that $a_{0}<a_{1}<\cdots <a_{r}$.
Then we can use Lemma 3.2(i) (with $s=0$ or $s=r$) to express each product in (3) as a
product of basic commutators (or their inverses). The first product gives
$$\prod_{i,\Delta}[a_{0},a_{i},\Delta]^{t\eta_{t}(i,\Delta)}
=\prod_{0<i,\, i\leq m(\Delta)}[a_{i},a_{0},\Delta]^{-t\eta_{t}(i,\Delta)}
\prod_{0\leq m(\Delta),\, m(\Delta)<i}
[a_{i},a_{0},\Delta]^{-t\eta_{t}(i,\Delta)}$$
and so
$\displaystyle \prod_{i,\Delta}[a_{0},a_{i},\Delta]^{t\eta_{t}(i,\Delta)}
=\prod_{i\not = 0,\, \Delta}
[a_{i},a_{0},\Delta]^{-t\eta_{t}(i,\Delta)},$
where this product is taken over all integers
$i\in \{ 1,\ldots,r\}$ and all functions
$\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$ with
$\Delta(0)+\cdots+\Delta(r)=k$.\\
Likewise we have $\displaystyle
\prod_{i,\Delta}[a_{0},a_{i},\Delta]^{t\alpha(i,\Delta)}
=\prod_{i\not = 0,\, \Delta}
[a_{i},a_{0},\Delta]^{-t\alpha(i,\Delta)}$.\\
In the same way, applying Lemma 3.2(i) with $s=r$,
the second product of (3) gives
$$\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\eta_{t}(i,\Delta)}=
\prod_{i\not = r,\, i\leq m(\Delta)}[a_{r},a_{i},\Delta]
^{\eta_{t}(i,\Delta)}$$
$$\times \prod_{i\not = r,\, m(\Delta)<i}[a_{i},a_{m(\Delta)},\Delta^{(r)}_{(m(\Delta))}]
^{-\eta_{t}(i,\Delta)}
\prod_{i\not = r,\, m(\Delta)<i}
[a_{r},a_{m(\Delta)},\Delta^{(i)}_{(m(\Delta))}]
^{\eta_{t}(i,\Delta)}.$$
Finally, the last product of (3) gives
$$\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\beta(i,\Delta)}=
\prod_{i\not = r,\, i\leq m(\Delta)}[a_{r},a_{i},\Delta]
^{\beta(i,\Delta)}$$
$$\times \prod_{i\not = r,\, m(\Delta)<i}[a_{i},a_{m(\Delta)},\Delta^{(r)}_{(m(\Delta))}]
^{-\beta(i,\Delta)}
\prod_{i\not = r,\, m(\Delta)<i}
[a_{r},a_{m(\Delta)},\Delta^{(i)}_{(m(\Delta))}]
^{\beta(i,\Delta)}.$$
Thus relation (3) can be written in the form
\begin{equation}
\prod_{i\not = 0,\, \Delta}
[a_{i},a_{0},\Delta]^{-t\eta_{t}(i,\Delta)}
\prod_{i\not = r,\, i\leq m(\Delta)}[a_{r},a_{i},\Delta]
^{\eta_{t}(i,\Delta)}
\end{equation}
$$\times \prod_{i\not = r,\, m(\Delta)<i}[a_{i},a_{m(\Delta)},\Delta^{(r)}_{(m(\Delta))}]
^{-\eta_{t}(i,\Delta)}
\prod_{i\not = r,\, m(\Delta)<i}
[a_{r},a_{m(\Delta)},\Delta^{(i)}_{(m(\Delta))}]
^{\eta_{t}(i,\Delta)}$$
$$=\quad \prod_{i\not = 0,\, \Delta}
[a_{i},a_{0},\Delta]^{-t\alpha(i,\Delta)}
\prod_{i\not = r,\, i\leq m(\Delta)}[a_{r},a_{i},\Delta]
^{\beta(i,\Delta)}$$
$$
\times \prod_{i\not = r,\, m(\Delta)<i}[a_{i},a_{m(\Delta)},\Delta^{(r)}_{(m(\Delta))}]
^{-\beta(i,\Delta)}
\prod_{i\not = r,\, m(\Delta)<i}
[a_{r},a_{m(\Delta)},\Delta^{(i)}_{(m(\Delta))}]
^{\beta(i,\Delta)}
$$
(so each commutator occuring in this relation is a basic commutator).\\
Now consider an integer $i\in \{ 1,\ldots,r-1\}$ and a function
$\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$, with
$\Delta(0)+\cdots+\Delta(r)=k$ (we can always suppose that $r>1$ since
$|S|>2$).
By identifying the exponents of the basic commutator
$[a_{i},a_{0},\Delta]$ of each side of relation (4), it is easy to see
that
\begin{equation}
t\eta_{t}(i,\Delta)+\eta_{t}\left(i,\Delta_{(r)}^{(0)}\right)=
t\alpha(i,\Delta)+\beta\left(i,\Delta_{(r)}^{(0)}\right)
\end{equation}
if $\Delta(r)>0$, and
$\eta_{t}(i,\Delta)=\alpha(i,\Delta)$ if $\Delta(r)=0$.
We prove by induction on $\Delta(r)$ that actually, we have always
the equality $\eta_{t}(i,\Delta)=\alpha(i,\Delta)$.
At first observe that if
$\Delta(r)>0$, we have $\Delta_{(r)}^{(0)}(r)=\Delta(r)-1$ and so
$\eta_{t}\left(i,\Delta_{(r)}^{(0)}\right)=\alpha\left(i,\Delta_{(r)}^{(0)}\right)$ by
induction. Hence relation (5) implies that
$$\alpha\left(i,\Delta_{(r)}^{(0)}\right)-\beta\left(i,\Delta_{(r)}^{(0)}\right)=
t\left\{\alpha(i,\Delta)-\eta_{t}(i,\Delta)\right\}.$$
Consequently, each positive integer $t$ divides the integer
$\alpha\left(i,\Delta_{(r)}^{(0)}\right)-\beta\left(i,\Delta_{(r)}^{(0)}\right)$,
which is independant of $t$.
It follows that
$\alpha\left(i,\Delta_{(r)}^{(0)}\right)=\beta\left(i,\Delta_{(r)}^{(0)}\right)$ and so
$\eta_{t}(i,\Delta)=\alpha(i,\Delta)$, as required.\\
Using theses relations and taking $t=1$, relation (3) implies
$$\prod_{\Delta}[a_{0},a_{r},\Delta]^{\eta(r,\Delta)}
\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\eta(i,\Delta)}
=\prod_{\Delta}[a_{0},a_{r},\Delta]^{\alpha(r,\Delta)}
\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\beta(i,\Delta)}$$
(we write $\eta$ for $\eta_{1}$) and so
$$\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\beta(i,\Delta)}=
\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\eta(i,\Delta)}
\prod_{\Delta}[a_{r},a_{0},\Delta]^{\alpha(r,\Delta)-\eta(r,\Delta)}.$$
Since
$\displaystyle \varphi(a_{r})=
a_{r}\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\beta(i,\Delta)}$,
we obtain
\begin{equation}
\varphi(a_{r})=a_{r}\prod_{i,\Delta}[a_{r},a_{i},\Delta]^{\eta(i,\Delta)}
\prod_{\Delta}[a_{r},a_{0},\Delta]^{\alpha(r,\Delta)-\eta(r,\Delta)}
\end{equation}
Now consider the generalized inner automorphism $\psi$ defined by
$$\psi(x)=x\prod_{i=1,\ldots,r,\, \Delta}
[x,a_{i},\Delta]^{\alpha(i,\Delta)}
\prod_{\Delta}[x,a_{0},\Delta]^{\alpha(r,\Delta)+\eta(0,\Delta)-\eta(r,\Delta)}.$$
We have
$$\psi(a_{0})=a_{0}\prod_{i=1,\ldots,r,\, \Delta}
[a_{0},a_{i},\Delta]^{\alpha(i,\Delta)}=\varphi(a_{0}).$$
In the same way,
\begin{eqnarray*}
\psi(a_{r}) & = & a_{r}\prod_{i=1,\ldots,r-1,\, \Delta}
[a_{r},a_{i},\Delta]^{\alpha(i,\Delta)}
\prod_{\Delta}[a_{r},a_{0},\Delta]^{\alpha(r,\Delta)+\eta(0,\Delta)-\eta(r,\Delta)} \\
{} & = & a_{r}\prod_{i=1,\ldots,r-1,\, \Delta}
[a_{r},a_{i},\Delta]^{\eta(i,\Delta)}
\prod_{\Delta}[a_{r},a_{0},\Delta]^{\alpha(r,\Delta)+\eta(0,\Delta)-\eta(r,\Delta)} \\
{} & = & a_{r}\prod_{i,\, \Delta}
[a_{r},a_{i},\Delta]^{\eta(i,\Delta)}
\prod_{\Delta}[a_{r},a_{0},\Delta]^{\alpha(r,\Delta)-\eta(r,\Delta)}
\end{eqnarray*}
and so $\psi(a_{r})=\varphi(a_{r})$ by (6). This completes the proof of
Lemma 3.3.
\end{proof}
Lemma 3.3 is actually a weak form of the next result.
\begin{lem}
Let $\varphi$ be a normal automorphism
of $M_{k+2}$ ($k>0$) acting trivially on
$M_{k+2}/\gamma_{k+2}(M_{k+2})$. Then $\varphi$ is
a generalized inner automorphism of $M_{k+2}$.
\end{lem}
\begin{proof} We can assume that $|S|>2$ (otherwise Lemma 3.4
is a consequence of Lemma 3.3). Consider two distinct elements
$a,b\in S$. According to Lemma 3.3, there exists a generalized inner
automorphism $\psi$ such that $\varphi(a)=\psi(a)$ and $\varphi(b)=\psi(b)$.
It suffices to prove that for any $c\in S\setminus\{ a,b\}$, we have $\varphi(c)=\psi(c)$.
For that, apply again Lemma 3.3: there are generalized inner
automorphisms $\psi',\psi''$ such that
$\varphi(a)=\psi'(a)$, $\varphi(c)=\psi'(c)$ and $\varphi(b)=\psi''(b)$,
$\varphi(c)=\psi''(c)$. There exists a finite subset
$\{ a_{0},\ldots,a_{r}\}\subseteq S$, containing $a,b,c$, such that
$\psi,\psi',\psi''$ can be defined by the equations
\begin{eqnarray*}
\psi(x) & = &
x\prod_{i,\,\Delta}[x,a_{i},\Delta]^{\epsilon(i,\Delta)}\\
\psi'(x) & = &
x\prod_{i,\,\Delta}[x,a_{i},\Delta]^{\epsilon'(i,\Delta)} \\
\psi''(x) & = & x\prod_{i,\,\Delta}[x,a_{i},\Delta]^{\epsilon''(i,\Delta)}
\end{eqnarray*}
(the products are taken over all integers
$i\in \{ 0,\ldots,r\}$ and all functions
$\Delta : \{ 0,\ldots,r\}\to {\mathbb N}$ with
$\Delta(0)+\cdots+\Delta(r)=k$). Since $\psi(a)=\psi'(a)$, we have
$$a\prod_{i,\,\Delta}[a,a_{i},\Delta]^{\epsilon(i,\Delta)}=
a\prod_{i,\,\Delta}[a,a_{i},\Delta]^{\epsilon'(i,\Delta)}$$ and so
$$\prod_{i,\,\Delta}[a,a_{i},\Delta]^{\epsilon(i,\Delta)-\epsilon'(i,\Delta)}=1.$$
Applying Lemma 3.2(ii), we obtain
$\epsilon(i,\Delta)=\epsilon'(i,\Delta)$ for all functions $\Delta$ and
all integers $i\in \{0,\ldots,r\}$ such that $a_{i}\not = a$.
Similarly, we have $\epsilon(i,\Delta)=\epsilon''(i,\Delta)$ if
$a_{i}\not = b$ and $\epsilon'(i,\Delta)=\epsilon''(i,\Delta)$ if
$a_{i}\not = c$. It follows that
$\epsilon(i,\Delta)=\epsilon'(i,\Delta)$ for all function $\Delta$ and
all integer $i\in \{0,\ldots,r\}$, hence $\psi=\psi'$.
Thus $\varphi(c)=\psi'(c)=\psi(c)$, as required.
\end{proof}
\begin{proof}[Proof of Theorem 2.1] We argue by induction on the
nilpotency class $k$ of $M_{k}$. If $k=2$, the result follows from
\cite[Theorem 2(ii)]{EN2} (in this case, each normal automorphism is
inner). Now consider a normal automorphism $\varphi$ of $M_{k}$, with
$k>2$. Then $\varphi$ induces a normal automorphism on the quotient
group $M_{k}/\gamma_{k}(M_{k})$. By induction, since this quotient is isomorphic to
$M_{k-1}$, there exists a generalized inner automorphism $\psi:M_{k}\to M_{k}$
such that $\varphi(x)=\psi(x)\theta(x)$, where $\theta(x)$ is an
element of $\gamma_{k}(M_{k})$. It follows that
$\psi^{-1}(\varphi(x))=x\psi^{-1}(\theta(x))$. Thus
$\psi':=\psi^{-1}\circ\varphi$ is a normal automorphism
of $M_{k}$ acting trivially on
$M_{k}/\gamma_{k}(M_{k})$. By Lemma 3.4, $\psi'$ is a
generalized inner automorphism, and so is $\varphi=\psi\circ\psi'$.
This completes the proof of Theorem 2.1.
\end{proof}
We end with a question: assume $\varphi$ is a normal automorphism of
a free nilpotent group $G$ of class $k\geq 4$ (and of rank $>1$). Is $\varphi$ of
the form
$$\varphi:x\to
x\left[x,u_{1}\right]^{\lambda(1)}\ldots\left[x,u_{m}\right]^{\lambda(m)}\;\;\;
(u_{i}\in G,\; \lambda(i)\in{\mathbb Z})\, ?$$
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\begin{document}
\maketitle
\begin{abstract}
We present, in dimension $n \geq 3$, a survey of examples to: the Jacobian conjecture, the weak Markus--Yamabe conjecture. Furthermore, we show
and construct new examples of vector fields where the origin is almost globally asymptotically stable by using the novel concept of density functions
introduced by Rantzer.
\end{abstract}
\section{Introduction}
One of the central problems on dynamical systems is to determine
conditions under which certain points or sets are attractors for
some dynamics, that is, the orbits through
points in a neighborhood of the attractor converge to them.
In the case of continuous-time, that is, flows
associated to vector fields, an analytic condition ensuring
that an equilibrium point $x^{*}$ is a local attractor is given by the negativeness of
the real part of the eigenvalues of the Jacobian matrix at $x^{*}$. Motivated by this simple observation, in \cite{MY}, L. Markus and
H. Yamabe establish their well known global stability conjecture
\medskip
\noindent {\bf{Markus--Yamabe Conjecture (MYC):}} Let $ F: \R^n
\to \R^n$ be a $C^1-$ vector field with $ F(0) = 0 $. If for any $ x
\in \R^n$ all the eigenvalues of $JF(x)$, the Jacobian matrix of $ F $ at $ x $,
have negative real part, then the origin is a global attractor of
the system $\dot{x} = F(x)$.
\medskip
Let us recall that the vector fields satisfying the hypothesis of {\textbf{MYC}} are called Hurwitz vector fields. It is known that the MYC is true
when $ n \leq 2 $ and false when $ n \geq 3$ (see \cite{CEGMH} for a counterexample). The proofs in the planar context, both the polynomial case (G. Meisters and C. Olech in \cite{MO}) as the $ C^1-$ case (R. Fe{\ss}ler in \cite{F}, A.A. Glutsyuk in \cite{Glu} and C. Guti\'errez in \cite{Gu}) are based on a remarkable result of C. Olech \cite{O}, where the author showed that MYC (in dimension two) is equivalent to the injectivity of the map $F.$ In $\R^n $, the problem of knowing if a Hurwitz vector field is injective is known as the Weak Markus--Yamabe conjecture.
\medskip
\noindent {\bf{Weak Markus--Yamabe Conjecture (WMYC):}} If $F:
\R^n \to \R^n$ is a $C^1-$ Hurwitz map, then $F$ is injective.
\medskip
The {\textbf{WMYC}} is true when $n \leq 2$ and, to the best of our knowledge, it has been proved in dimension $n \geq 3$ for $C^1$ Lipschitz Hurwitz maps by A. Fernandes, C. Guti\'errez and R. Rabanal in \cite[Corollary 4]{Fernandez}.
For $ n = 2 $, the strong injectivity theorem of Guti\'errez \cite{Gu} is the following:
``A $ C^1-$map $ f : \R^2 \to \R^2 $ is injective if $ [0,\infty) \cap \textrm{Spec}(Jf(x)) = \emptyset $, for all $ x \in \R^2 $". However, this result fails in high dimensions as shown B. Smyth and F. Xavier \cite[Theorem 4]{Smyth}:
``There exist integers $ n > 2 $ and non--injective polynomial maps $ f : \R^n \to \R^n $ with $ [0,\infty) \cap \textrm{Spec}(Jf(x)) = \emptyset $, for all $ x \in \R^n $".
In addition, a new tool (density functions) introduced by A. Rantzer in \cite{R}, gives sufficient conditions ensuring almost global stability of an equilibrium point for a $ C^1-$vector field in $ \mathbb{R}^n $ (\emph{i.e.}, all trajectories, except for a set of initial states with zero Lebesgue measure, converge to the equilibrium point). Recently, this tool was used by R. Potrie and P. Monz\'on \cite{PM} to construct a vector field $ X $ in $ \R^3 $ where the origin is almost globally stable but is not a local attractor for the differential system generated for $X$. We point out that $X$ is an almost Hurwitz field (Hurwitz vector field except in a zero Lebesgue measure set).
This article is focused on two tasks: Firstly, we will construct (a formal description will be given later) polynomial maps $ F = \lambda I + H\colon \mathbb{R}^{3}\to \mathbb{R}^{3}$ with $JH$ nilpotent, such that the{\textbf{ WMYC}} and the {\bf{Jacobian Conjecture}} are true, giving the inverse of $ F $ explicitly. The results obtained are strongly related with the works \cite{C} of L.A. Campbell and \cite{ChE} of M. Chamberland and A. van den Essen. Secondly, we construct two families of three dimensional vector fields having the Rantzer's density functions stated above. The vector fields of the first family are a generalization of the
Potrie--Monz\'on's example \cite{PM} in the sense that are almost Hurwitz and the vector field restricted to the invariant plane $ z = 0 $ is a centre. Moreover, perturbing these vector fields by $ \lambda I $, we obtain a new family a Hurwitz vector fields with the origin as global attractor which are not included in the examples of \cite[Theorem 2.7]{GC} and \cite[Theorem 2.5]{CG}.
The vector fields of the second family are not almost Hurwitz and the the existence of a density function seems to be the only way to demonstrate the almost global stability of the origin.
The paper is organized as follows. The Section $2$ is devoted to generalize the procedure of L.A. Campbell in order to obtain new examples to {\bf{WMYC}} and the {\bf{Jacobian Conjecture}} on $\R^n.$ The Section $3,$ using density functions, shows two family of vector fields such that the origin is almost globally stable. Moreover, we construct new examples to {\bf{MYC}}.
\section{Examples to Jacobian conjecture and WMYC}
In dimension three, all existing examples and counterexamples to MYC are maps of the form $\lambda I + H$ with $\lambda < 0$ and $JH$ nilpotent (see \cite{CEGMH},\cite{CGM}, \cite{CG}). This kind of maps, in any dimension, also are important in the study of the Jacobian conjecture since H. Bass, E. Connell and D. Wright showed
in \cite{BCW} that it suffices to solve this conjecture for such maps. Indeed, the Jacobian conjecture follows from the injectivity of these maps for all dimensions.
It is worth to emphasize that the result above triggered new questions and problems as the following one. Let $ \kappa $ be a field of characteristic zero.
\medskip
\noindent {\bf{(Homogeneous) dependence problem.}} Let $H=(H_1, \ldots, H_n) \in \kappa[x_1, \ldots, x_n]^n$ (homogeneous of degree $d \geq 1$) such that $JH$ is nilpotent and $H(0) = 0.$
Does it follow that $H_1, \ldots, H_n$ are linearly dependent over $ \kappa $?
\medskip
The homogeneous problem is true in dimension three (\cite{Bondt1}) and false in dimensions bigger than five (\cite{Bondt2}).
A counterexample for the inhomogeneous problem in dimension three was given by E. Hubbers in \cite{vE}. The map
$$H =(y-x^2, z + 2x(y-x^2), -(y-x^2)^2)$$
verifies that $JH$ is nilpotent, $\rank(JH) = 2$ and $H_1, H_2, H_3$ are linearly independent over $\kappa.$
Moreover, L.A. Campbell in \cite{C} generalizes this counterexample obtaining
$$ H = (\phi(y - x^2),z + 2 x \phi(y - x^2), -(\phi(y - x^2))^2) $$
with the same properties. Here $ \phi $ is a $ C^1-$ function of a single variable. Notice that the inverse of $ F = I + H $ can be computed explicitly as follows
$$ F^{-1} = (x - \phi(y - x^2 - z),y - z - 2 x \phi(y - x^2 - z) + (\phi(y - x^2 - z))^2,z + (\phi(y - x^2 - z))^2) \, . $$
Therefore $ F = I + H $ with $ \phi \in \R[t] $ is an example in dimension three to
\medskip
\noindent{\bf{Jacobian Conjecture on $\R^n.$}} Every polynomial map $F: \R^n \to \R^n$ such that $\det JF \equiv 1$ is a bijective map with a polynomial inverse.
\bigskip
Let us emphasize that the previous
examples have the special form
$$H=(u(x,y,z), v(x,y,z), h(u,v)), $$
which is completely studied in \cite{ChE}, which deals with polynomial maps satisfying $ H(0) = 0 $, $ h $ has no linear part and the components of $ H $ are linearly independent over $ \kappa $.
Given $A = v_x u_z - u_x v_z$ and $B = v_y u_z - u_y v_z,$ they show that if $JH$ is nilpotent
and $\deg_z u A \neq \deg_z v B$, then there exists $T \in GL_3(k)$ such that $THT^{-1}$ takes the form
\begin{equation} \label{ache}
(g(t),v_1 z
- (b_1 + 2 v_1 \alpha x) g(t),\alpha (g(t))^2) \, ,
\end{equation}
with $ t = y + b_1 x + v_1 \alpha x^2 $, $v_1 \alpha \neq 0 $ and $ g(t) \in k[t], g(0) = 0$ and
$\deg_t g(t) \geq 1.$
The expressions $A$ and $B$ are very useful for determine if a map
$H$ is nilpotent \cite[Proposition 3.1]{ChE}. Also they are useful for decide if the rows of $H$ are linearly independent over $ \kappa $. In fact, we obtain the following result with $ \kappa = \R.$
\begin{prop}\label{independencia}
Let $H=(u(x,y,z),v(x,y,z),h(u(x,y,z), v(x,y,z)))$ be a polynomial map such that the components of $H$ are linearly dependent over $\R$ and $h$ has no linear part in $u$ and $v.$ Then $A = B = 0.$
\end{prop}
\begin{proof}
Let $\alpha, \beta \in \mathbb{R}^*$ such that
\begin{displaymath}
h_u \cdot (u_x, u_y, u_z) + h_v \cdot (v_x, v_y, v_z) \equiv \alpha \cdot (u_x, u_y, u_z) + \beta \cdot (v_x, v_y, v_z)
\end{displaymath}
which is equivalent to
\begin{equation}
\label{10}
(h_u - \alpha) \cdot (u_x,u_y,u_z) + (h_v - \beta) \cdot (v_x, v_y, v_z) \equiv 0.
\end{equation}
Thus, we have the following systems of equations
\begin{equation}\label{sistemas}
\left \{
\begin{array}{rcl}
(h_u - \alpha) u_x + (h_v - \beta) v_x &= &0\\
(h_u - \alpha) u_y + (h_v - \beta) v_y &= &0\\
(h_u - \alpha) u_z + (h_v - \beta) v_z &= &0.
\end{array}
\right.
\end{equation}
By using (\ref{sistemas}), we see that
$$
\begin{array}{rcl}
u_z A - v_z B & = &u_z^2 v_x - u_z v_z (u_x + v_y) + v_z^2 u_y\\\\
& = & \frac{v_z^2}{(h_u - \alpha)^2} \{ (h_v - \beta)^2 v_x + (h_v - \beta)(h_u - \alpha)(u_x + v_y) +(h_u - \alpha)^2 u_y\}\\\\
& = & \frac{v_z^2}{(h_u - \alpha)^2} \{(h_v - \beta)^2 v_x\\\\ & & \hspace{1.5 cm} -(h_v - \beta)(h_u - \alpha)(\frac{(h_v - \beta)}{(h_u- \alpha)} v_x - v_y) - \frac{(h_u - \alpha)^2(h_v - \beta)}{(h_u -\alpha)} v_y\}\\\\
& = & 0
\end{array}
$$
over the set of points where $ h_u \neq \alpha $. Otherwise, if $h_u = \alpha$ then in (\ref{10}) we have that $(v_x, v_y, v_z) =0$ or $h_v = \beta.$ The first case immediately implies that $A = B = 0,$ the second case is not
possible due to is a contradiction with no linearity of $h$ with respect to $u$ and $v.$
In similar way is proved that $u_z A + v_z B =0$ and the result follows.
\end{proof}
\begin{teo}
\label{11}
Consider a polynomial map of the form $ H=(u,v,h(u, v))$ such that $ H(0) = 0 $, $ h $ has no linear part and the components of $ H $ are linearly independent over $ k $. Then if $JH$ is nilpotent
and $\deg_z u A \neq \deg_z v B$, for all $ \lambda \neq 0 $ the
polynomial map $ \lambda I + H$ is injective and has inverse polynomial.
\end{teo}
\begin{proof}
According to \cite[Corollary 4.1]{ChE}, we can suppose that the components $ H_1, H_2, H_3 $ of $H$ are
as in (\ref{ache}). Namely,
\begin{eqnarray}\label{ind}
H_1(x, y, z) & = & g( y + b_1 x + v_1 \alpha x^2 ) \, , \\
H_2(x, y, z) & = & v_1z - (b_1 + 2 v_1 \alpha x)g( y + b_1 x + v_1 \alpha x^2 ) \,
, \nonumber \\
H_3(x, y, z) & = & \, \alpha (g( y + b_1 x + v_1 \alpha x^2 ))^2 . \nonumber
\end{eqnarray}
If
$$
\begin{array}{rcl}
u_1 & = & \lambda x + g( y + b_1 x + v_1 \alpha x^2 )\\
u_2 & = & \lambda y + v_1z - (b_1 + 2 v_1 \alpha x)g( y + b_1 x + v_1 \alpha x^2 )\\
u_3 & = & \lambda z + \alpha (g( y + b_1 x + v_1 \alpha x^2 ))^2,
\end{array}
$$
it is easy to obtain
\begin{equation}\label{igualdad para inversa}
\displaystyle \gamma u_2 +\gamma b_1 u_1+ \gamma^2 v_1 \alpha u_1^2 - \gamma^2 v_1 u_3 = y + b_1 x + v_1 \alpha x^2,
\end{equation}
\noindent where $\gamma = {1 \over \lambda}.$ Now, put $\Phi = g( y + b_1 x + v_1 \alpha x^2) = g( \gamma u_2 +\gamma b_1 u_1+ \gamma^2 v_1 \alpha u_1^2 - \gamma^2 v_1 u_3)$ and observe that
$\lambda I + H = (\lambda x + \Phi, \lambda y + v_1 z - (b_1+2 v_1 \alpha x)\Phi, \lambda z +\alpha \Phi^2).$ By using (\ref{igualdad para inversa}), we obtain the inverse of this map which is
$$(\gamma I + P) = (\gamma x + P_1, \gamma y + P_2, \gamma z + P_3)$$
where
\begin{equation}\label{inversa1}
\begin{array}{rcl}
P_1(x,y,z) & = & - \gamma g(\gamma y +\gamma b_1 x+ \gamma^2 v_1 \alpha x^2 - \gamma^2 v_1 z)\\
\\
P_2(x,y,z) & = & - \gamma^2 v_1 z + \gamma(b_1 + 2 \gamma v_1 \alpha x )g(\gamma y +\gamma b_1 x+ \gamma^2 v_1 \alpha x^2 - \gamma^2 v_1 z)\\
\\
& &- \gamma^2 v_1 \alpha (g(\gamma y +\gamma b_1 x+ \gamma^2 v_1 \alpha x^2 - \gamma^2 v_1 z))^2\\
\\
P_3(x,y,z) & = & -\alpha \gamma (g(\gamma y +\gamma b_1 x+ \gamma^2 v_1 \alpha x^2 - \gamma^2 v_1 z))^2.
\end{array}
\end{equation}
\end{proof}
\begin{rem} For $ \lambda < 0 $ (resp. $ \lambda = 1$), the polynomial map $F = \lambda I + H $ of the above Theorem is an example of the WMYC (resp. the Jacobian conjecture) on $\R^3 $.
\end{rem}
\begin{rem}
More examples for the both conjectures in dimension $n \geq 4$ can be constructed consider the following maps of \cite[Proposition 7.1.9]{vE}:
\begin{eqnarray*}
H_1(x_1,\dots,x_n) & = & g(x_2 - a(x_1)) \, , \\
H_i(x_1,\dots,x_n) & = & x_{i+1} + \frac{(-1)^i}{(i-1)!} \,
a^{(i-1)}(x_1) \,
g(x_2 - a(x_1))^{i-1} \, , \; \textrm{if} \; 2 \leq i \leq n-1 \, ,\\
H_{n}(x_1,\dots,x_n) & = & \frac{(-1)^{n}}{(n-1)!} \, a^{(n-1)}(x_1)
\,
g(x_2- a(x_1))^{n-1}
\end{eqnarray*}
where $a(x_1) \in \R[x_1]$ with $\deg a = n-1$ and $ g(t) \in k[t], g(0) = 0$ and
$\deg_t g(t) \geq 1.$ Following the lines of the proof of the Theorem (\ref{11}), we must consider for each fixed $n \geq 4, u_i = \lambda x_i + H_i, \, i = 1, \ldots, n$ and
$$\Phi = g(x_2 - a(x_1)) = g\Big (\frac{1}{\lambda} \, u_2 -a(\frac{1}{\lambda} u) - \frac{1}{\lambda^2} \, u_3 + (-1)^n \sum_{j=4}^{n} \frac{1}{\lambda^{j-1}} \, u_j \Big )$$
for obtain the inverse (polynomial) of $\lambda I + H.$ In fact, the inverse is $\gamma I + P$ where
\begin{eqnarray*}
P_1(x_1, \ldots, x_n) & = & - \gamma \Phi(x_1, \ldots, x_n)\\
\\
P_{i}(x_1, \ldots, x_n) & = & -\gamma(x_{i+1}- \frac{(-1)^i}{(i-1)!} a^{(i-1)}(\gamma (x_1- \Phi(x_1, \ldots, x_n))) (\Phi(x_1, \ldots, x_n))^{i-1}) ,\\\\
P_n(x_1, \ldots, x_n) & = & - \gamma \frac{(-1)^{n}}{(n-1)!} \, a^{(n-1)}(\gamma (x_1- \Phi(x_1, \ldots, x_n))(\Phi(x_1, \ldots, x_n))^{n-1}.
\end{eqnarray*}
with $ 2 \leq i \leq n-1.$ Therefore, for $ \lambda < 0 $ (resp. $ \lambda = 1$), the polynomial map $F = \lambda I + H $ is an example of the WMYC (resp. the Jacobian conjecture) on $\R^n $ with $n \geq 4.$
\end{rem}
\begin{ex}
We consider $n=4,$ the inverse of the map $\lambda I + (H_1,H_2,H_3, H_4)$ is $\gamma I + (P_1,P_2, P_3, P_4)$ where
\begin{eqnarray*}
P_1(x_1, x_2, x_3, x_4) & = & -\gamma \Phi ,\\\\
P_2(x_1, x_2, x_3, x_4) & = & -\gamma^2 x_3 + \gamma^3 x_4 - \gamma (a_1 + \gamma 2 a_2 x_1 + \gamma^2 3 a_3 x_1^2) \Phi + \gamma^2(a_2 + 3 a_3 x_1) \Phi^2 ,\\\\
P_3(x_1, x_2, x_3, x_4) & = & -\gamma^2 x_4 + \gamma (a_2 + \gamma 3 a_3 x_1) \Phi^2 - \gamma 2 a_3 \Phi^3,\\\\
P_4(x_1, x_2, x_3, x_4) & = & - \gamma a_3 \Phi^3,
\end{eqnarray*}
where $\Phi = \Phi(x_1, x_2, x_3, x_4) = (\gamma x_2 - \gamma x_1 - \gamma^2 x_1^2 - \gamma^3 x_1^3 - \gamma^2 x_3 + \gamma^3 x_4).$
\end{ex}
\begin{rem}
A example to WMYC and Jacobian Conjecture, in dimension 4, which does not belong to the above family of maps is $$F(x,y,z,w) = (\lambda x + y, \lambda y+ x^2 - w, \lambda z + y^2, \lambda w + 2y - z) .$$ Is easy to see
that $JH$ is nilpotent and the rows of $JH$ are linearly independent over $\R$ and has inverse
$$F^{-1}(x,y,z,w) = (\gamma (x - \phi), \phi, \gamma(z-\phi^2), \gamma(w- 2 \gamma (x-\phi)\phi+ \gamma (z-\phi^2)) ),$$
where $\phi = \gamma y - \gamma^3 x^2 + \gamma^3 z + \gamma^2 w$ and $\gamma = 1/\lambda.$ Therefore, for $ \lambda < 0 $ (resp. $ \lambda = 1$), the polynomial map $F$ is an example of the WMYC (resp. the Jacobian conjecture) on $\R^4 .$
\end{rem}
\begin{rem} In \cite[Corollary 4.2]{ChE} it is stated that a complete study of maps $H = (u,v, h(u,v))$ with $JH$ is nilpotent will be achieved when considering the case $deg_z (u A) = deg_z (vB).$ In this context, the following result gives a partial progress
to this study since we introduce a large family of maps satisfying $deg_z (u A) = deg_z (vB),$ which through a linear of change of coordinates have the form described in (\ref{ache}).
\end{rem}
\begin{prop}
The map $ \gamma I + P $ with $ P = (P_1,P_2,P_3) $ as in (\ref{inversa1}) has the following properties:
\begin{enumerate}
\item[\textbf{(P0)}] $P(x,y,z)$ has the form $(u(x,y,z), v(x,y,z), h(u(x,y,z), v(x,y,z))).$
\item[\textbf{(P1)}] The Jacobian matrix $JP$ is nilpotent and their rows are linearly independent over $\R.$
\item[\textbf{(P2)}] $deg_z (u A) = deg_z (vB)$.
\item[\textbf{(P3)}] Under the linear change of coordinates $(\widetilde{u},\widetilde{v},\widetilde{w}) = (x, y- \gamma v_1 z, z)$ the map $ \gamma I + P $
is transformed into $ \gamma I + \widehat{P}$ where $\widehat{P}$ has the form $(u,v,h(u,v))$ and $deg_z (u A) \neq deg_z (vB).$
\end{enumerate}
\end{prop}
\begin{proof}
\textbf{(P0)} We have $ P = (u,v,h(u,v)) $ with $ h(u,v) = -\frac{\alpha}{\gamma} \, u^2 $.
\medskip
\textbf{(P1)} By using an algebraic manipulator we see that $(JP)^3 =0 $. We have $ P(0) = 0 $ and $ P = (P_1,P_2,h(P_1,P_2)) $ with $ h(P_1,P_2) = -\frac{\alpha}{\gamma} \, P_1^2 $. Moreover,
if we consider $\omega(x) = b1 + 2 \gamma v_1 \alpha x,$ we have
$$
\begin{array}{rcl}
B & = & \gamma^5 v_1 \omega(x) (g'(t))^2 - 2 \gamma^6 v_1^2 \alpha g(t) (g'(t))^2- \gamma^4 v_1 g'(t)+ \\\\
&&- \gamma^5 v_1 \omega(x) (g'(t))^2 + 2 \gamma^6 v_1^2 \alpha g(t) (g'(t))^2\\\\
&= & -\gamma^4 v_1 g'(t) \neq 0 \, ,
\end{array}
$$
and
$$A = -\gamma^4 v_1 g'(t) (\omega (x) - 2 \gamma v_1 \alpha g(t)) \neq 0,$$
and the result follows from Proposition \ref{independencia}.
\medskip
\noindent\textbf{(P2)} We have
$$u(x,y,z) \cdot A = -2 \gamma^6 v_1^2 \alpha (g(t))^2 g'(t) + \gamma^5 v_1 \omega(x) g(t) g'(t)$$
and
$$v(x,y,z) \cdot B = \gamma^6 v_1^2 z g'(t) -\gamma^5 v_1 \omega(x) g(t) g'(t) + \gamma^6 v_1^2 \alpha (g(t))^2 g'(t).$$
Finally, $\deg_z (u A) = \max \{3k-1, 2k-1\} = 3k-1$ and $\deg_z (v B) = \max \{k,2k-1,3k-1\} = 3k-1.$
\medskip
\noindent\textbf{(P3)} By considering the linear change of coordinate proposed, it is easy to see that the coordinates $\widehat{P_1}, \widehat{P_2}, \widehat{P_3}$ of $\widehat{P}$ are
$$
\begin{array}{rcl}
\widehat{P_1}(x,y,z) & = & g(\gamma y + \gamma b_1 x + \gamma^2 v_1 \alpha x^2)\\\\
\widehat{P_2}(x,y,z) & = & - \gamma^2 v_1 z + \gamma (b_1 + 2 \gamma v_1 \alpha x) g(\gamma y + \gamma b_1 x + \gamma^2 v_1 \alpha x^2)\\\\
\widehat{P_3}(x,y,z) & = & - \alpha \gamma (g(\gamma y + \gamma b_1 x + \gamma^2 v_1 \alpha x^2))^2.
\end{array}
$$
This coordinates $\widehat{P_1}, \widehat{P_2}, \widehat{P_3}$ are the same that (\ref{ache}) with $\gamma = 1,$ thus $deg_z (u A) \neq deg_z (vB).$
\end{proof}
Now we consider, in dimension three, polynomial maps of the form $ F = \lambda I + H $ with $ H(0) = 0 $ and $ JH $ nilpotent such that the components of $ H $ are linearly dependent over $ \R $.
As it was shown in \cite[Proposition 2.1]{CG}, for such vector field $ F = \lambda I + H ,$ there exists $ T \in Gl_3(\R) $ such that $ T^{-1} F T = \lambda I + (P,Q,0)$, where
\begin{eqnarray} \label{normal}
P(x,y,z) & = & -b(z) \, f(a(z) \, x + b(z) \, y) \su c(z) \quad \textrm{and} \nonumber \\
Q(x,y,z) & = & a(z) \, f(a(z) \, x + b(z) \, y) \su d(z)
\,
\end{eqnarray}
with $ a, b, c, d \in \R[z] $ and $ f \in \R[z][t] $. Also it was pointed out in \cite[Proposition 2.3]{CG} that if $ \lambda \neq 0 $, this polynomial map
is injective and thus satisfies the conclusion of WMYC. In fact, if we consider $\gamma = \frac{1}{\lambda},$
we can find explicitly the inverse of $ \lambda I + (P,Q,0) $, with $ P$ and $ Q $ as in (\ref{normal}). Since $ f \in \R[z][t] $, in what follows we replace $ f(t) $ by $ f(t,z) $.
\begin{prop} For $ \lambda \neq 0 $, the inverse of $ \lambda I + (P,Q,0) $, with $ P$ and $ Q $ as in (\ref{normal}), is $ \gamma I + (R,S,0) $, where $\gamma = \frac{1}{\lambda}$ and
$$
\begin{array}{rcl}
R & = & \gamma b(\gamma z) \, f\Big(\gamma \{a(\gamma z) \, x + b(\gamma z) \, y- a(\gamma z) c(\gamma z) - b(\gamma z)d(\gamma z)\},\gamma z\Big) - \gamma c(\gamma z)\\
S & = & -\gamma a(\gamma z) \, f\Big(\gamma \{a(\gamma z) \, x + b(\gamma z) \, y- a(\gamma z) c(\gamma z) - b(\gamma z)d(\gamma z)\},\gamma z\Big) - \gamma d(\gamma z).
\,
\end{array}
$$
\end{prop}
\begin{proof} Putting
$$
\begin{array}{rcl}
u &=&\lambda x -b(z) \, f(a(z) \, x + b(z) \, y,z) \su c(z)\\\\
v &=& \lambda y + a(z) \, f(a(z) \, x + b(z) \, y,z) \su d(z)\\\\
w &=& \lambda z \, ,
\end{array}
$$
we obtain
\begin{equation}\label{inversa}
a(z)u+b(z)v = \lambda(a(z)x+b(z)y) + a(z)c(z) + b(z) d(z) \, ,
\end{equation}
which implies
\begin{displaymath}
a(z)x+b(z)y = \gamma \big( a(z)u+b(z) v - a(z)c(z) - b(z)d(z) \big).
\end{displaymath}
Therefore, by consider that $z = \gamma w,$ we can deduce
$$
\begin{array}{lcl}
u & = & \lambda x - b(\gamma w) \, f(m(u,v,w),\gamma w) \su c(\gamma w)\\\\
v & = & \lambda y + a(\gamma w) \, f(m(u,v,w),\gamma w)\su d(\gamma w),
\end{array}
$$
with $ \; m(u,v,w) = \gamma \big\{ a(\gamma w)u+b(\gamma w) v - a(\gamma w)c(\gamma w) - b(\gamma w)d(\gamma w) \big\} $.
Finally,
$$
\begin{array}{lcl}
x & = & \gamma ( u +b(\gamma w) \, f(m(u,v,w),\gamma w) - c(\gamma w) ) \\\\
y & = & \gamma ( v -a(\gamma w) \, f(m(u,v,w),\gamma w) - d(\gamma w) ).\\\\
\end{array}
$$
\end{proof}
\begin{rem}
Let $ F = \lambda I + H : \R^3 \to \R^3 $ be a polynomial map verifying $ H(0) = 0 $, $ JH $ nilpotent and their components are linearly dependent over $ \R $. If
$ \lambda < 0 $ (resp. $ \lambda = 1 $), then the map $F$ is an example to WMYC (resp. the Jacobian conjecture).
\end{rem}
\section{Examples of almost global stability}
In \cite[Theorem 3.2,Theorem 3.5]{CG} we prove that the vector fields in $ \R^3 $ of the form $ F = \lambda I + H $, with $ \lambda < 0 $, $ H $ as
in (\ref{ind}) and $ g(t) = A_1 \, t + A_2 \, t^2 $, are counterexamples to {\textbf{MYC}} since they have unbounded orbits. Moreover, by following the
respective proofs, we can deduce the existence of an open set of initial states whose trajectories do not converge to the origin.
Thus the origin for these vector fields
is not almost global attractor in the following sense:
\begin{definition} \label{3.1}
Consider the differential equation
\begin{equation} \label{sistema}
\dot{x} = F(x)
\end{equation}
where $F:\R^n \to \R^n$ is a $ C^1-$map and $ F(0) = 0.$ We say the origin is an almost global attractor if all trajectories, except for a set of initial states with zero Lebesgue measure,
converge to the origin.
\end{definition}
From Definition \ref{3.1}, it arises the following question:
\medskip
\noindent {\bf Question 1.} Do there exist counterexamples to the {\textbf{MYC}} with the origin almost global attractor ?
\medskip
In \cite{R}, A. Rantzer introduces a new tool, namely, the density functions in order to obtain sufficient conditions for almost global stability of an equilibrium point for a $ C^1-$vector field in $ \mathbb{R}^n $.
\begin{definition}
\label{density}
A density function of \textnormal{(\ref{sistema})} is a $C^{1}$ map
$\rho\colon \mathbb{R}^{n}\setminus \{0\}\to [0,+\infty)$, integrable outside a ball
centered at the origin that satisfies
\begin{displaymath}
[\triangledown \cdot \rho F](x) >0
\end{displaymath}
almost everywhere with respect to $\mathbb{R}^{n},$ where
$$
\triangledown \cdot [\rho F] = \triangledown \rho \cdot F + \rho[\triangledown\cdot F],
$$
and $\triangledown \rho$, $\triangledown \cdot F$ denote respectively the gradient of $\rho$ and the divergence
of $F$.
\end{definition}
The main result of A. Rantzer \cite{R} is the following.
\begin{teo}
\label{Rantzer}
Given the differential system
$$\dot{x} = F(x),$$
where $F \in C^1, \, F(0) = 0$, suppose there exists a density function $\rho : \R^n \setminus \{0\} \to [0, +\infty)$ such
that $ \rho (x) F(x) / \norm{x}$ is integrable on $\{x \in \R^n: \norm{x} \geq 1\}.$
Then, almost all trajectories converge to the origin, i.e., the
origin is almost globally stable.
\end{teo}
In this context, the next question arises naturally:
\noindent {\bf Question 2.} Do there exist counterexamples to the {\textbf{MYC}} that support density functions ?
In this paper the best answer we give is the following family of almost Hurwitz vector fields; i.e. of vector fields for which the Hurwitz condition hold over $ \R^n - A $ with $ A $ a zero Lebesgue measure set.
\begin{teo}\label{123}
Consider the real numbers $ c \leq a < 0, \, b \in \R, \, k\geq 1,$ and the polynomial $R(z) = \sum\limits_{i=1}^k a_{2i} z^{2i}$ with $a_{2i}>0$ for $i=1, \ldots, k.$ Then
\begin{equation}
\label{casihurwitz}
F(x,y,z) = (y,-x,0) +((ax + by)R(z),(-bx + cy)R(z),-z R(z))
\end{equation} is an almost Hurwitz vector field. Moreover, this vector field has associated the density function $\rho(x,y,z) = (x^2+y^2+R(z))^{-\alpha}$ with $ \; \alpha > \max\{2,
\frac{a+c-1-2k}{2a}, \frac{3-a-c}{2}\} $.
\end{teo}
\begin{proof}
The Jacobian matrix of $F$ in a point $ (x,y,z) $ is
$$
JF(x,y,z) = \left (\begin{array}{ccc}
a R(z) & 1 + b R(z)& *\\
-(1+b R(z)) &c R(z)& *\\
0 & 0 & - (R(z) + zR'(z))
\end{array}
\right ).
$$
Then $ \; \lambda_3 = - (R(z) + zR'(z)) $ is an eigenvalue. The others two eigenvalues $ \lambda_1, \lambda_2 $ verify
$$ \lambda_1 + \lambda_2 = (a + c) \; R(z) \quad \textrm{and} \quad \lambda_1 \lambda_2 = ac R(z)^2 + (1 + b R(z))^2 \, . $$
Therefore, for $ z \neq 0 $ (resp. $ z = 0 $), we have $ \lambda_3 < 0 $ (resp. $ \lambda_3 = 0 $) and $ \lambda_1 $ and $ \lambda_2 $ have negative real part (resp. null real part).
In addition, $F$ verifies the Hurwitz condition except in the invariant plane $z=0.$
In what follows we prove that
the map $$\rho(x,y,z) = (x^2 + y^2+ R(z))^{-\alpha} \, , $$
under the conditions of the theorem is a density function for the vector field $ F $.
The condition $\alpha > 2$ ensures the integrability of $ \rho(x,y,z) $ outside the ball centered at the origin of radius one.
It remains to prove that $\nabla \cdot (\rho F)(x,y,z)$ is positive almost everywhere in $\R^3.$ We have
$$ \triangledown \rho (x,y,z) = \frac{-\alpha}{(x^2+y^2+R(z))^{\alpha + 1}} \; (2 x, 2 y, R'(z)) \, , $$
and
$$ [\triangledown\cdot F] (x,y,z) = (a + c -1) R(z) - z R'(z) \, . $$
Then
\begin{eqnarray*}
[\nabla \cdot \rho F](x,y,z) & = & (\triangledown \rho \cdot F) (x,y,z) + \rho(x,y,z) \; [\triangledown\cdot F] (x,y,z) \\
& = & \frac{-\alpha \, R(z)}{(x^2+y^2+R(z))^{\alpha +1}} \, [2(a x^2 + c y^2) - z \, R'(z)] \\
& & +\frac{1}{(x^2+y^2+R(z))^{\alpha}} \, [(a + c -1) \, R(z) - z \, R'(z)] \\
& = & \frac{1}{(x^2+y^2+R(z))^{\alpha + 1}} \, \left[-2 \alpha (a x^2 + c y^2) R(z) + \alpha z R(z) R'(z) \right. \\
& & + \left. (x^2+y^2+R(z)) \, [(a + c -1) \, R(z) - z \, R'(z)]\right] \\
& = & \frac{1}{(x^2+y^2+R(z))^{\alpha + 1}} \, \left[((a + c - 1) R(z) + (\alpha - 1) z R'(z)) R(z) \right. \\
& & + ((a + c - 1 - 2 \alpha a) R(z) - z R'(z)) x^2 \\
& & + \left.((a + c - 1 - 2 \alpha c) R(z) - z R'(z)) y^2 \right] \, .
\end{eqnarray*}
Since $ R(z) = \sum\limits_{i=1}^k a_{2i} z^{2i}$ and $ z R'(z) = \sum\limits_{i=1}^k 2 i a_{2i} z^{2i}$ with $a_{2i}>0$ for $i=1, \ldots, k $, we obtain $ [\triangledown\cdot F] (x,y,z) > 0 $ for $ z \neq 0 $ and $ [\triangledown\cdot F] (x,y,z) = 0 $ for $ z = 0 $, if
$$ a + c - 1 + 2 (\alpha - 1) i > 0 \, , \quad a + c - 1 - 2 \alpha a - 2 i > 0 \, , \quad a + c - 1 - 2 \alpha c - 2 i > 0 \, , $$
for $i=1, \ldots,k $. Then, the proof is finished due to these inequalities are consequence of our hypothesis $ \; \alpha > \max\{2,
\frac{a+c-1-2k}{2a}, \frac{3-a-c}{2}\} $.
\end{proof}
This family is a generalization of the following example of R. Potrie and P. Monz\'on (\cite{PM})
\begin{equation}\label{PM}
F(x,y,z) = (y -2xz^2,-x-2yz^2, -z^3 ) \, ,
\end{equation}
which has density function $ \rho(x,y,z) = (x^2 + y^2 + z^2)^{-4} $.
\begin{cor} \label{1234} The vector field $ F $ given by (\ref{casihurwitz}) under the conditions of Theorem \ref{123} has the origin as an almost global attractor which is not locally asymptotically stable.
\end{cor}
\begin{proof} We have $ F(x,y,0) = (y,-x,0) $, then the origin is not locally asymptotically stable. On the other hand, to prove that the origin is almost global attractor we use Rantzer's result (Theorem \ref{Rantzer}). Then it is sufficient to show that the condition $\alpha > 2$ ensures the integrability of $ \rho(x,y,z) F(x,y,z)/ \norm{(x,y,z)} $ outside the ball centered at the origin of radius one. In fact, if we consider
$k_0 = \max \{1,a_1\} $, $ k_1 = \max\{1, b^2+c^2 + \abs{b} (a-c)\}$ and $ k_2 = 2 \abs{b} + a-c $, we have that
\begin{eqnarray*}
\norm{F(x,y,z)}^2 & = & x^2 + y^2 + R(z)^2 \, \left[(a^2 + b^2) x^2 + (b^2 + c^2) y^2 + z^2 + 2 b (a - c)xy\right] \\
& & + 2 R(z) \, \left[(a - c) xy + b (x^2 + y^2)\right] \\
& \leq & x^2 + y^2 + R(z)^2 \, \left[(b^2 + c^2) (x^2 + y^2) + z^2 + \abs{b} (a - c) (x^2 + y^2)\right] \\
& & + R(z) (a - c + 2\abs{b}) (x^2 + y^2) \\
& \leq & (x^2+y^2+z^2) + k_1 (x^2+y^2+R(z))^2 (x^2+y^2+z^2) \\
& & + \; k_2 (x^2+y^2+R(z))(x^2+y^2+z^2).
\end{eqnarray*}
and that $x^2 + y^2 + R(z) \geq k_0.$
Thus, this facts combined with the assumption over $\alpha$ imply that
\begin{eqnarray*}
\frac{\norm{F(x,y,z)}^2 \rho(x,y,z)^2}{\norm{(x,y,z)}^2} & \leq & \frac{1}{(x^2+y^2+R(z))^{2 \alpha}} + \frac{k_1}{(x^2+y^2+R(z))^{2 \alpha-2}} \\
& & + \frac{k_2}{(x^2+y^2+R(z))^{2 \alpha-1}} \, \cdot
\end{eqnarray*}
Therefore $ \rho(x,y,z) F(x,y,z)/ \norm{(x,y,z)} $ is integrable outside the ball centered at the origin of radius one.
\end{proof}
\begin{rem}
An alternative --and very simple-- proof of Corollary \ref{1234} is obtained by considering
the fact that $\langle F(x,y,z), (x,y,z) \rangle = (a x^2 + b y^2 - z^2) R(z) < 0 $ for all $ z \neq 0 $ combined with the invariance of the plane $z=0$.
\end{rem}
Motivated by the last remark, we have the following result.
\begin{prop} \label{prop1} Let $ F : \R^3 \to \R^3 $ be a $ C^1-$vector field with $ F(0) = 0 $ such that the plane $ z = 0 $ is invariant and $ [\triangledown\cdot F] (x,y,z) < 0 $ for all $ z \neq 0 $. Suppose that there exists a positive $ C^1-$function $ \rho : \R^3-\{0\} \to \R $ satisfying:
\begin{enumerate}
\item[a)] $ [\triangledown\cdot \rho F] (x,y,z) > 0 $ for all $ z \neq 0 $,
\item[b)] $ \lim_{p \to 0} \, \rho(p) = \infty $.
\item[c)] $ \lim_{\|p\| \to \infty} \, \rho(p) = 0 $.
\end{enumerate}
Then, for any initial state $ \alpha(0) = (x(0),y(0),z(0)) $ verifying $ z(0) \neq 0 $, the trajectory $ \alpha(t) $ exists for all $ t \in [0,\infty[ $ and tends to zero as $ t \to \infty $.
\end{prop}
\begin{proof} Condition a) implies $ [\triangledown\cdot \rho F] (x,y,z) \geq 0 $ everywhere, and
$ (\triangledown \rho \cdot F) (x,y,z) > -\rho(x,y,z) \; [\triangledown\cdot F] (x,y,z) > 0 $ for all $ (x,y,z) $ with $ z \neq 0 $.
\noindent Let us consider the function $ V : \R^3 \to \R $ defined by $ V = \rho^{-1} $ outside the origin and $ V(0) = 0 $. Then $ V $ is continuous by b), it is $C^1 $ outside the origin and
\begin{enumerate}
\item[1)] $ V(p) \geq 0 $ for all $ p $ and $ V(p) = 0 $ if and only if $ p = 0 $.
\item[2)] $ (\triangledown V \cdot F) (x,y,z) = -\rho(x,y,z)^{-2} \, (\triangledown \rho \cdot F) (x,y,z) < 0 $, for $ z \neq 0 $.
\item[3)] $ \lim_{\|p\| \to \infty} \, V(p) = \infty $.
\end{enumerate}
Then, if $ \alpha(0) = (x(0),y(0),z(0)) $ is a initial state with $ z(0) \neq 0 $, we have $ \frac{d}{dt} V(\alpha(t)) < 0 $ by 2). Then the trajectory $ \alpha(t) $ remains over the sublevel $ V(\alpha(0)) $, which is bounded by 3). Therefore, $ \alpha(t) $ exists for all $ t \in [0,\infty[ $ and tend to 0 as $ t \to \infty $.
\end{proof}
\begin{cor} \label{cor1} If in Proposition \ref{prop1} we have $ [\triangledown\cdot \rho F] (x,y,z) > 0 $ for all $ z \in \R $, then the origin is globally asymptotically stable.
\end{cor}
If we perturb the vector field $ F $ defined by (\ref{casihurwitz}) with $ \lambda I $ and $ \lambda < 0 $, then $ G = \lambda I + F $ is a Hurwitz vector field, which is another example of {\bf MYC}, by the following
\begin{prop} \label{prop2} Let $ G = \lambda I + F $ with $ \lambda < 0 $ and $ F $ as in (\ref{casihurwitz}) that verify the conditions of Theorem \ref{123}. Then $ G $ is Hurwitz, admits the same density function $ \rho(x,y,z) = (x^2 + y^2 + R(z))^{\alpha} $ but with $ \alpha > 3 $, and the origin is globally asymptotically stable.
\end{prop}
\begin{proof} $ G $ is Hurwitz due to the fact that $ F $ is almost Hurwitz and $ JG = \lambda I + JF $. Since
$$ [\triangledown\cdot \rho G] (x,y,z) = [\triangledown\cdot \rho F] (x,y,z) + \lambda \frac{(3 - 2\alpha)(x^2 + y^2) + 3 R(z) - \alpha z R'(z)}{(x^2 + y^2 + R(z))^{\alpha + 1}} \; ,$$
under the additional condition $ \alpha > 3 $, $ \rho $ is a density function for $ G $. Finally, the global asymptotic stability of the origin follows from Corollary \ref{cor1}.
\end{proof}
The previous results imply that if we are interested in finding vector fields $F$ with the plane $ z = 0 $ invariant supporting density
functions (but without obvious Lyapunov functions), these fields must verify
$ \triangledown\cdot F >0$ over some open set. The following family of vector fields satisfy these conditions.
\begin{prop} \label{prop3.10}
Consider the vector field $ F = (P,Q,R) $ defined by
\begin{eqnarray*}
P(x,y,z) & = & -x + A_2 \, y + a_1 \, x^3 + 3 \, a_2 \, x^2 \, y + 3 \, a_3 \, x \, y^2 + a_4 \, y^3 + 3 \, a_5 \, x^2 \, z + \\
& & 6 \, a_6 \, x \, y \, z + 3 \, a_7 \, y^2 \, z + 3 \, a_8 \, x \, z^2 + 3 \, a_9 \, y \, z^2 + a_{10} \, z^3 \, , \\
Q(x,y,z) & = & -A_2 \, x - \, y + b_1 \, x^3 + 3 \, b_2 \, x^2 \, y + 3 \, b_3 \, x \, y^2 + b_4 \, y^3 + 3 \, b_5 \, x^2 \, z + \\
& & 6 \, b_6 \, x \, y \, z + 3 \, b_7 \, y^2 \, z + 3 \, b_8 \, x \, z^2 + 3 \, b_9 \, y \, z^2 + b_{10} \, z^3 \, , \\
R(x,y,z) & = & z \, (-1 + 3 \, c_5 \, x^2 +
6 \, c_6 \, x \, y + 3 \, c_7 \, y^2 + 3 \, c_8 \, x \, z + 3 \, c_9 \, y \, z + c_{10} \, z^2) \, ,
\end{eqnarray*}
with
\begin{eqnarray*}
a_8 & = & \frac{2}{3} - a_3 \, , \; a_{10} = \frac{3}{2} \, (a_5 - a_7) \, , \\
b_1 & = & -3 \, a_2 + a_4 + 3 \, b_3 \, , \; b_2 = 1 - a_3 - \frac{c_{10}}{3} \, , \; b_4 = 3 - a_1 - c_{10} \, , \\
b_6 & = & \frac{1}{2} \, (a_5 - a_7) \, , \; b_7 = 2 \, a_6 + b_5 \, , \; b_8 = 2 \, a_2 - a_4 - a_9 - b_3 \, , \\
b_9 & = & -\frac{1}{3} + a_3 + \frac{c_{10}}{3} \, , \; b_{10} = 3 \, a_6 \, , \\
c_5 & = & -\frac{2}{3} + \frac{a_1}{3} + a_3 + \frac{c_{10}}{3} \, , \; c_6 = \frac{2 a_4}{3} - a_2 + b_3 \, , \; c_7 = \frac{4}{3} - \frac{a_1}{3} - a_3 - \frac{c_{10}}{3} \, ,\\
c_8 & = & \frac{1}{2} \, (a_5 + a_7) \, , \; c_9 = a_6 + b_5 \, .
\end{eqnarray*}
Then $ F $ has the density function
$$ \rho(x,y,z) \; = \; \frac{1}{(x^2 + y^2 + z^2)^2} \; \cdot $$
Moreover the plane $ z= 0 $ is invariant and the coefficients of $ F $ can be chosen such that $ \triangledown\cdot F $ is positive over some open set.
\end{prop}
\begin{proof} Clearly the plane $ z = 0 $ is invariant. After straightforward computations we obtain
$$ [\triangledown\cdot \rho F](x,y,z) \; = \; \frac{1 + x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^2} \, ,$$
and that the integrability condition of $ \rho $ holds.
Moreover
\begin{eqnarray*}
\triangledown\cdot F \, (x,y,z) & = & -3 + (1 + 4 a_1) x^2 + (13 - 4 a_1 - 4 c_{10}) y^2 + (24 a_6 + 12 b_5) y z \\
& & + \, (1 + 4 c_{10}) z^2 +
x ((4 a_4 + 12 b_3) y + 12 a_5 z) \, ,
\end{eqnarray*}
assume positive values. For example, if $ y = 0 $, we have $ \; \triangledown\cdot F \, (x,0,z) = -3 + (1 + 4 a_1) x^2 + (1 + 4 c_{10}) z^2 + 12 a_5 x z $ which is positive for $ x, z $ sufficient large and $ a_1 > 0 $ and $ a_5 < 0 $.
\end{proof}
\begin{rem} The coefficients of the vector field $ F = (P,Q,R) $ of Proposition \ref{prop3.10} were obtained from the equation
$$ [\triangledown\cdot \rho F](x,y,z) \; = \; \frac{1 + x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^2} \, ,$$
which is equivalent to
$$ -4 \, (x P + y Q + z R) \; + \; (x^2 + y^2 + z^2) \, [\triangledown\cdot F] \; = \; (1 + x^2 + y^2 + z^2) \, (x^2 + y^2 + z^2) \, . $$
\end{rem}
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