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http://unapologetic.wordpress.com/2011/08/24/an-example-part-3/?like=1&source=post_flair&_wpnonce=d0199d491e | # The Unapologetic Mathematician
## An Example (part 3)
Now we can take our differential form and our singular cube and put them together. That is, we can integrate the $1$-form $\omega$ over the circle $c_a$.
First we write down the definition:
$\int\limits_{c_a}\omega=\int\limits_{[0,1]}{c_a}^*\omega$
to find this pullback of $\omega$ we must work out how to push forward vectors from $[0,1]$. That is, we must work out the derivative of $c_a$.
This actually isn’t that hard; there’s only the one basis vector $\frac{d}{dt}$ to consider, and we find
$\displaystyle {c_a}_{*t}\left(\frac{d}{dt}\right)=-2\pi a\sin(2\pi t)\frac{\partial}{\partial x}+2\pi a\cos(2\pi t)\frac{\partial}{\partial y}$
We also have to calculate the composition
$\displaystyle\begin{aligned}\omega(c_a(t))&=-\frac{a\sin(2\pi t)}{(a\cos(2\pi t))^2+(a\sin(2\pi t))^2}dx+\frac{a\cos(2\pi t)}{(a\cos(2\pi t))^2+(a\sin(2\pi t))^2}dy\\&=\frac{1}{a}\left(-\sin(2\pi t)dx+\cos(2\pi t)dy\right)\end{aligned}$
This lets us calculate
$\displaystyle\begin{aligned}\int\limits_{c_a}\omega&=\int\limits_{[0,1]}{c_a}^*\omega\\&=\int\limits_{[0,1]}\left[\omega(c_a(t))\right]\left({c_a}_{*t}\left(\frac{d}{dt}\right)\right)\,dt\\&=\int\limits_{[0,1]}\left[\frac{1}{a}\left(-\sin(2\pi t)dx+\cos(2\pi t)dy\right)\right]\left(2\pi a\left(-\sin(2\pi t)\frac{\partial}{\partial x}+\cos(2\pi t)\frac{\partial}{\partial y}\right)\right)\,dt\\&=2\pi\int\limits_0^1\sin(2\pi t)^2+\cos(2\pi t)^2\,dt=2\pi\int\limits_0^1\,dt=2\pi\end{aligned}$
So, what conclusions can we draw from this? Well, Stokes’ theorem now tells us that the $1$-form $\omega$ cannot be the differential of any $0$-form — any function — on $\mathbb{R}^2\setminus{0}$. Why? Well, if we had $\omega=df$, then we would find
$\displaystyle\int\limits_{c_a}\omega=\int\limits_{c_a}df=\int\limits_{\partial c_a}f=\int\limits_{0}f=0$
which we now know not to be the case. Similarly, $c_a$ cannot be the boundary of any $2$-chain, for if $c_a=\partial c$ then
$\displaystyle\int\limits_{c_a}\omega=\int\limits_{\partial c}\omega=\int\limits_{\partial c}d\omega=\int\limits_{\partial c}0=0$
It turns out that there’s a deep connection between the two halves of this example. Further, in a sense every failure of a closed $k$-form to be the differential of a $k-1$-form and every failure of a closed $k$-chain to be the boundary of a $k+1$-chain comes in a pair like this one.
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Posted by John Armstrong | Differential Topology, Topology
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 25, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9096263647079468, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/23654/is-the-free-energy-bounded | # Is the free energy bounded
Can we say that the free energy of a mixture of a finite number of fluids situated in a bounded container $\Omega$ is bounded?
I was thinking of this as an optional hypothesis for one mathematical theorem I want to prove, and I wondered if the energy of some object can be considered to have an upper bound (maybe very large, but still finite). This seems reasonable if the object is bounded. Still, I was thinking of the black holes as mass concentrated in a point, and their energy is huge (I guess).
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The answer really depends on the physics you put in. If the free energy were unbound, one would have to account for the full physics of GUT scales, which we don't know. Classically, you could think of a GR big bang, which would give positive answer. – Alexey Bobrick Apr 21 '12 at 23:49
## 3 Answers
I will assume you mean the free energy of a classical Newtonian collection of particles at temperature T and volume V, defined as the "free entropy" times the temperature or $T\log(Z)$ where Z is the sum over all configurations of $e^{-\beta E}$. This quantity is bounded for reasonable force-laws (meaning you can't extract too much energy by packing the atoms close) because the volume of the constant E phase-space surface for N particles roughly grows as V^N times the product of N spheres of radius $\sqrt{2mkT}$. So the free energy is bounded above by a small multiple of the ideal gas free energy (the multiple is determined by the detail of the force law).
But at high temperature, the total free entropy is (up to an additive constant) the free gas free entropy
$$\log(Z) = \log({V^N\over N!}) + \log((\sqrt{2mT})^{3N})\approx N\log({N\over V}) + {3\over 2}N \log(T)$$
So that not only is the free energy unbounded, but the free energy divided by the temperature is unbounded at large T.
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Sort of: it depends on whether you take the possibility of black holes seriously.
The energy of "some object", i.e. any particular, bounded system, is always bounded (otherwise you have a perpetual motion machine); the energy a given region can contain is not: you can in principle pack as much energy as you want to in to some fixed volume if you're clever enough.
This is within the limits of reason and non-general-relativistic dynamics, of course. If you pack energy so closely that the associated mass, through $E=mc^2$, exceeds the Scharzschild limit, then whatever it is you've got in there will collapse into a black hole and you will lose that energy behind the event horizon.
I should also mention that while black holes look like they have a lot of energy, in fact they do not, in the following sense: the collapse of a star into a black hole is a spontaneous process (i.e. it does not require any action from outside the system) and as such it has to be exothermic, so it will give out some energy as heat, and the total energy inside any box you place around it will at most remain constant. The energy given off comes from the gravitational potential energy associated with stuff being further apart, which you might not have counted in the first place.
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Only energy differences are measurable, so you would need to define a reference state before you start thinking about what it means to be bounded. That's why most of the time people talk about $\Delta G$, i.e. $G-G_0$, rather than just $G$.
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Yes, but it still can be bounded once you set a datum. The bound depends upon the choice of datum, that's all. – Manishearth♦ Apr 21 '12 at 18:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9614447951316833, "perplexity_flag": "head"} |
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/De_Moivre's_formula | # All Science Fair Projects
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# De Moivre's formula
De Moivre's formula states that for any real number x and any integer n,
(cosx + isinx)n = cos(nx) + isin(nx).
The formula is important because it connects complex numbers (i stands for the imaginary unit) and trigonometry. The expression "cos x + i sin x" is sometimes abbreviated to "cis x".
By expanding the left hand side and then comparing the real and imaginary parts, it is possible to derive useful expressions for cos(nx) and sin(nx) in terms of cos(x) and sin(x). Furthermore, one can use this formula to find explicit expressions for the n-th roots of unity, that is, complex numbers z such that zn = 1.
Abraham de Moivre was a good friend of Newton; in 1698 he wrote that the formula had been known to Newton as early as 1676. It can be derived from (but historically preceded) Euler's formula eix = cos x + i sin x and the exponential law (eix)n = einx (see exponential function).
De Moivre's formula is actually true in a more general setting than stated above: if z and w are complex numbers, then (cos z + i sin z)w is a multivalued function while cos (wz) + i sin (wz) is not, and one can state that
cos (wz) + i sin (wz) is one value of (cos z + i sin z)w.
## Proof
We consider three cases.
For n > 0, we proceed by induction. When n=1, the result is clearly true. For our hypothesis, we assume the result is true for some positive integer k. That is, we assume
$(\cos x + i \sin x)^k = \cos(kx) + i \sin(kx). \,$
Now, considering the case n = k + 1:
$(\cos x+i\sin x)^{k+1}\,$
$= (\cos x+i\sin x)(\cos x+i\sin x)^{k}\,$
$= (\cos(kx)+i\sin(kx))(\cos x+i\sin x)\,$ (by the induction hypothesis)
$= \cos(kx)\cos x - \sin(kx)\sin x + i(\cos(kx)\sin x + \sin(kx)\cos x)\,$
$= \cos(k+1)x + i\sin(k+1)x\,$
We deduce that the result is true for n = k + 1 when it is true for n = k. By the Principle of Mathematical Induction it follows that the result is true for all positive integers n.
When n = 0 the formula is true since cos(0x) + isin(0x) = 1 + i0 = 1, and (by convention) z0 = 1.
When n < 0, we consider a positive integer m such that n = −m. So
$(\cos x + i\sin x)^{n}\, = (\cos x + i\sin x)^{-m}\,$
$=\frac{1}{(\cos x + i\sin x)^{m}} = \frac{1}{(\cos mx + i\sin mx)}\,$, from above
$=\cos(mx) - i\sin(mx)\,$, rationalizing the denominator
$=\cos(-mx) + i\sin(-mx)\, = \cos(nx) + i\sin(nx)\,$
Hence, the theorem is true for all values of n. Q.E.D.
03-10-2013 05:06:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8248329758644104, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/82984/cumulative-distribution-function | # Cumulative Distribution Function
So I know that a cdf F is defined via F(a) = Pr[X ≤ a]. How do I show that the cdf F of a random variable X contains exactly the same information as the function defined via G(a) = Pr[X = a], by expressing F in terms of G and expressing G in terms of F.
-- Compute and plot the cdf for (i) X ∼ Geom(p), (ii) X ∼ Exp(λ ).
-- Identify two key properties that a cdf of any r.v. has to satisfy.
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In general, F cannot be expressed in terms of G. In your example (ii), G=0 identically and there is no way to guess F from that information. – Did Nov 17 '11 at 9:43
I changed the subject line. There is no such thing as a cumulative density function. The word "cumulative" contradicts the word "density". – Michael Hardy Nov 17 '11 at 16:32
## 1 Answer
For the starting period you should develop a clear distinction between discrete random variables and continuous ones:
1. Discrete random variable $X$ takes only finite (or sometimes countable) number of finite real values $\{x_i\}_{i\in I}$ where $I$ is either finite $I = 1,2,...,n$ or countable $I = \mathbb N$. So, for each value we can assign a probability that $X$ will take this single value: $$p_X(x_i) = \mathsf P\{X = x_i\}$$ and $p(x_i)$ is called probability mass function (PMF). PMF uniquely characterizes distribution of a discrete random variable, but there is another way to do it: through cumulative distribution function (CDF). For any real $x$ we define $$F_X(x) := \mathsf P\{X\leq x\} = \sum\limits_{x_i\leq x}p_X(x_i)$$ where the sum is taken over all $i$ such that $x_i\leq x$. You can easily see two (or three :) ) properties of $F_X$: first, $F_X(-\infty) = \lim\limits_{x\to-\infty}F_X(x) = 0$ and $F_X(\infty) = \lim\limits_{x\to\infty}F_X(x)= 1$ and second, $F_X$ is monotonically non-increasing because if $x'\leq x''$ then $\{X\leq x'\}\subseteq \{X\leq x''\}$ and by monotonicity of $\mathsf P$ we get $$F_{X}(x') = \mathsf P\{X\leq x'\}\leq\mathsf P\{X\leq x''\} = F_X(x'').$$ CDF characterizes PMF and hence the whole distribution of $X$ uniquely. We has already shown how to construct CDF from PMF through the sum, let us show the opposite construction: $$p_X(x_i) = \mathsf P\{X\leq x_i\} - \mathsf P\{X<x_i\} = F_{X}(x_i)-\lim\limits_{x\to x_i-0}F(x).$$ It may seem unnecessary complication to use CDF, but in contrast to PMF it can be easily generalized.
2. Continuous real-valued random variable $Y$ takes uncountably many values and due to this complication we cannot assign probability for each single value in an informative way: indeed for any $y\in \mathbb R$ it holds that $$\mathsf P\{Y=y\} = 0$$ so there is no useful information about $Y$ at all from the equality (as Didier has already told you). That's why instead of using PMF which would say nothing to us, the use of CDF still makes sense. As an analogue of PMF one talks about probability density function (PDF) which is a derivative of CDF: $$f_Y(y):=\frac{\mathrm d}{\mathrm dy}F_Y(y)$$ and hence $$F_Y(y) = \int\limits_{-\infty}^y f_Y(t)\mathrm dt.$$ As a minor note, PDF is not a probability of something. In contrast to CDF and PMF it can take value larger than $1$. PDF is rather a likelihood that random variable will take a specific value, not a probability. As an example, let us consider the case of exponential distribution, i.e. $$f_Y(y) = \lambda \mathrm e^{-\lambda y}\cdot 1_{y\geq 0}$$ so for any $y>0$ we have $$F_Y(y) = \int\limits_{-\infty}^y f_Y(t)\mathrm dt = \int\limits_{0}^y \lambda \mathrm e^{-\lambda t}\mathrm dt = -\mathrm e^{-\lambda t}|_{0}^y = 1-\mathrm e^{-\lambda y}.$$
I hope it's now more clear to you, so you just should consider the case when $y\geq 0$ and the case of geometrical distribution. Also, you may be interested in proving the kind of continuity which CDF performs.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9197567105293274, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2011/09/08/switching-orientations/?like=1&_wpnonce=34114a6dfa | # The Unapologetic Mathematician
## Switching Orientations
If we have an oriented manifold $M$, then we know that the underlying manifold has another orientation available; if $\alpha$ is a top form that gives $M$ its orientation, then $-\alpha$ gives it the opposite orientation. We will write $-M$ for the same underlying manifold equipped with this opposite orientation.
Now it turns out that the integrals over the same manifold with the two different orientations are closely related. Indeed, if $\omega$ is any $n$-form on the oriented $n$-manifold $M$, then we find
$\displaystyle\int\limits_{-M}\omega=-\int\limits_M\omega$
Without loss of generality, we may assume that $\omega$ is supported within the image of a singular cube $c$. If not, we break it apart with a partition of unity as usual.
Now, if $c:[0,1]^n\to M$ is orientation-preserving, then we can come up with another singular cube that reverses the orientation. Indeed, let $f(u^1,u^2,\dots,u^n)=(-u^1,u^2,\dots,u^n)$. It’s easy to see that $f^*$ sends $du^1$ to $-du^1$ and preserves all the other $du^i$. Thus it sends $du^1\wedge\dots\wedge du^n$ to its negative, which shows that it’s an orientation-reversing mapping from the standard $n$-cube to itself. Thus we conclude that the composite $\tilde{c}=c\circ f$ is an orientation-reversing singular cube with the same image as $c$.
But then $\tilde{c}:[0,1]\to-M$ is an orientation-preserving singular cube containing the support of $\omega$, and so we can use it to calculate integrals over $-M$. Working in from each side of our proposed equality we find
$\displaystyle\begin{aligned}\int\limits_M\omega&=\int\limits_c\omega=\int\limits_{[0,1]^n}c^*\omega\\\int\limits_{-M}\omega&=\int\limits_{\tilde{c}}\omega=\int\limits_{[0,1]^n}\tilde{c}^*\omega=\int\limits_{[0,1]^n}f^*c^*\omega\end{aligned}$
We know that we can write
$\displaystyle c^*\omega=g(u^1,\dots,u^n)du^1\wedge\dots\wedge du^n$
for some function $g$. And as we saw above, $f^*$ sends $du^1\wedge\dots\wedge du^n$ to its negative. Thus we conclude that
$\displaystyle\tilde{c}^*\omega=-=g(u^1,\dots,u^n)du^1\wedge\dots\wedge du^n$
meaning that when we calculate the integral over $-M$ we’re using the negative of the form on $[0,1]^n$ that we use when calculating the integral over $M$.
This makes it even more sensible to identify an orientation-preserving singular cube with its image. When we write out a chain, a positive multiplier has the sense of counting a point in the domain more than once, while a negative multiplier has the sense of counting a point with the opposite orientation. In this sense, integration is “additive” in the domain of integration, as well as linear in the integrand.
The catch is that this only works when $M$ is orientable. When this condition fails we still know how to integrate over chains, but we lose the sense of orientation.
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Posted by John Armstrong | Differential Topology, Topology
## 2 Comments »
1. [...] if is orientation-reversing from to then it’s orientation-preserving from to , and we already know that integrals over are the negatives of those over . Further, we can assume that the support of [...]
Pingback by | September 12, 2011 | Reply
2. [...] we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the face of a singular -cube to cancel each other [...]
Pingback by | September 16, 2011 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 35, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9126780033111572, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/58068-pointwise-convergence-uniform-convergence.html | # Thread:
1. ## Pointwise convergence to uniform convergence
Does pointwise convergence of continuous functions on a compact set to a continuous limit imply uniform convergence on that set?
I think yes.
Proof.
Suppose that $(M,d)$ is a metric space, let K be a compact subset of M, and define $f_n:K \rightarrow \mathbb {R} \ \ \ \forall n \in N$ such that there exists a continuous function $f:K \rightarrow \mathbb {R}$ with $f_n \rightarrow f$ pointwise.
By definitions, $\forall \epsilon > 0, \exists M_{( \epsilon ,x)}$ such that $d(f_n,f)< \epsilon \ \ \ n \geq M$
Now, what I want to show is that $\exists M_{ \epsilon }$ such that $d(f_n,f) < \epsilon \ \ \ n \geq M$, so an epsilon that doesn't depend on x.
But I find it difficult to work through this, perhaps my understanding of uniform convergence is still poor. Am I on the right track thou?
Thanks!
2. Originally Posted by tttcomrader
Does pointwise convergence of continuous functions on a compact set to a continuous limit imply uniform convergence on that set?
I think yes.
The answer is No. For example, the sequence $f_n(x) = nx^n(1-x)$ converges pointwise but not uniformly to 0 on the unit interval.
3. Okay, here is my formal proof.
Define $f_n(x): [0,1] \rightarrow \mathbb {R}$ by $<br /> f_n(x) = nx^n(1-x) \ \ \ \forall n \in \mathbb {N}<br />$
Claim: This sequence of functions converges pointwise to $f(x)=0$
Proof.
For $x=0$ and $x=1$, we would have $f_n(x)=0$.
Given $\epsilon > 0$, pick $N = ?$
For $x \in (0,1)$, we would have $|f_n(x)-0|=|nx^n(1-x)|<|nx^n|$ I'm stuck here, I know that this would converge to 0, but how would I prove that? What N I should pick to ensure this distance is less than epsilon?
Claim: This sequence of functions do not converge uniformly to 0.
Proof.
Pick $\epsilon = 1$, then $\forall N \in \mathbb {N}$, whenever $n \geq N$ and $x = \frac {1}{n}$ for each index n, we will have $|f_n(x)-0|=|n( \frac {1}{n} ) ^n(1- \frac {1}{n} )| = | \frac {1}{n^{n-1}}(1- \frac {1}{n})|=| \frac {1}{n^{n-1}}- \frac {1}{n^n}| = | \frac {n-1}{n^n}|$ But doesn't this still converges to 0? Did I pick the wrong x?
Thanks!!!
4. Originally Posted by tttcomrader
Okay, here is my formal proof.
Define $f_n(x): [0,1] \rightarrow \mathbb {R}$ by $<br /> f_n(x) = nx^n(1-x) \ \ \ \forall n \in \mathbb {N}<br />$
Both parts of this are actually quite tricky.
Originally Posted by tttcomrader
Claim: This sequence of functions converges pointwise to $f(x)=0$
Proof.
For $x=0$ and $x=1$, we would have $f_n(x)=0$.
Good start. So we now want to show that if 0<x<1 then $nx^n\to0$ as $n\to\infty$. As you can see, this is not obvious. One sneaky way to prove it is to notice that the series $\sum_{n=0}^\infty nx^n$ converges if |x|<1 (easily shown by using the ratio test), and therefore the n'th term of the series must tend to 0.
Originally Posted by tttcomrader
Claim: This sequence of functions do not converge uniformly to 0.[/tex]
To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $x=1-\tfrac 1{n+1}$, at which point the value of the function is $\bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.
As I said, it's a tricky question.
5. Originally Posted by Opalg
To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $x=1-\tfrac n{n+1}$, at which point the value of the function is $\bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.
As I said, it's a tricky question.
I'm a bit unclear about how $<br /> \bigl(\tfrac n{n+1}\bigr)^{n+1}<br /> \rightarrow \frac {1}{e}$ Did we use the natural log here?
6. Originally Posted by tttcomrader
I'm a bit unclear about how $<br /> \bigl(\tfrac n{n+1}\bigr)^{n+1}<br /> \rightarrow \frac {1}{e}$ Did we use the natural log here?
Just 'break it up': $\left( {\frac{n}{{n + 1}}} \right)^{n + 1} = \left( {1 - \frac{1}{{n + 1}}} \right)^n \left( {\frac{n}{{n + 1}}} \right)$
$\lim _{n \to \infty } \left( {1 - \frac{1}{{n + 1}}} \right)^n = e^{ - 1} \,\& \,\lim _{n \to \infty } \left( {\frac{n}{{n + 1}}} \right) = 1$.
In general $\lim _{n \to \infty } \left( {1 + \frac{a}<br /> {{n + b}}} \right)^{cn} = e^{ac}$ (with the correct restrictions on b)
7. I'm just really confuse about here:
So take $x= 1- \frac {n}{n+1}$, then $f_n(x)= | n(1- \frac {n}{n+1})^n[1-(1- \frac {n}{n+1}] |$ $=(1- \frac {n}{n+1})^n( \frac {n^2}{n+1} )$, now, why does this mess equals to $( \frac {n}{n+1})^{n+1}$?
8. Originally Posted by tttcomrader
I'm just really confuse about here:
Now I do agree with that!
Could it be a matter of simple algebra?
I do think you are over doing all of this.
Stop! Think about a simple approach to this problem.
9. Originally Posted by tttcomrader
I'm just really confuse about here:
So take $x= 1- \frac {n}{n+1}$, then $f_n(x)= | n(1- \frac {n}{n+1})^n[1-(1- \frac {n}{n+1}] |$ $=(1- \frac {n}{n+1})^n( \frac {n^2}{n+1} )$, now, why does this mess equals to $( \frac {n}{n+1})^{n+1}$?
Sorry, that's my mistake. In one of the comments above, I wrote $x= 1- \tfrac {n}{n+1}$ when it should have been $x= 1- \tfrac 1{n+1}$. (But you should have spotted that for yourself, because I explained that this is the value of x where the function $nx^n(1-x)$ has its maximum value. If you take the trouble to differentiate, you soon see that this happens when $x=\tfrac n{n+1}$, which can equivalently be written $x=1- \tfrac 1{n+1}$. What I did was to write one of those expressions and then edit it to write the other one, but forgot to alter the n in the numerator.)
10. Originally Posted by Opalg
Both parts of this are actually quite tricky.
Good start. So we now want to show that if 0<x<1 then $nx^n\to0$ as $n\to\infty$. As you can see, this is not obvious. One sneaky way to prove it is to notice that the series $\sum_{n=0}^\infty nx^n$ converges if |x|<1 (easily shown by using the ratio test), and therefore the n'th term of the series must tend to 0.
To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $x=1-\tfrac 1{n+1}$, at which point the value of the function is $\bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.
As I said, it's a tricky question.
I have a question regarding this problem. By using the ratio test, we get the sum of infinite series to be convergent if |x| <1. But how do we know it converges to 0?
11. Originally Posted by anlys
I have a question regarding this problem. By using the ratio test, we get the sum of infinite series to be convergent if |x| <1. But how do we know it converges to 0?
It is not true that the series converges to 0. The series converges (in other words, the terms add up to a finite sum), but its sum is not zero. What is true is that the individual terms of the series converge to 0. That is always the case if a series converges.
12. Originally Posted by Opalg
It is not true that the series converges to 0. The series converges (in other words, the terms add up to a finite sum), but its sum is not zero. What is true is that the individual terms of the series converge to 0. That is always the case if a series converges.
Wouldn't it be easier to apply the L'Hospital rule to show that nx^n converges to 0 as n approaches to infinity? For 0<x<1, we can always write x = 1/(1+a) for some positive number a. So, the limit of nx^n as n goes to infinity equals to the limit of n/(1+a)^n, which is in the infinity/infinity form. So we can use the L'Hospital's rule in this case, which gives us limit of 1/(n*(1+a)^n-1), by differentiating both the numerator and denominator. Thus, as n approaches to infinity, this will approach to 0. So, we get nx^n converges to 0. Is this correct?
13. Originally Posted by anlys
Wouldn't it be easier to apply the L'Hospital rule to show that nx^n converges to 0 as n approaches to infinity? For 0<x<1, we can always write x = 1/(1+a) for some positive number a. So, the limit of nx^n as n goes to infinity equals to the limit of n/(1+a)^n, which is in the infinity/infinity form. So we can use the L'Hospital's rule in this case, which gives us limit of 1/(n*(1+a)^n-1), by differentiating both the numerator and denominator. Thus, as n approaches to infinity, this will approach to 0. So, we get nx^n converges to 0. Is this correct?
There is certainly more than one way to show that $\lim_{n\to\infty}nx^n=0$ when |x|<1. L'Hôpital's rule is one way to approach it, but you need to be more careful in applying it. When you say $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{n(1+a)^{n-1}}$, you are differentiating the numerator as a function of n, and the denominator as a function of a. If you are using n as the variable, then $\tfrac d{dn}(1+a)^n = (1+a)^n\ln (1+a)$. So the calculation should go like this: $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{(1+a)^n\ln (1+a)} = 0$. That looks strange, because it's unusual to see n used as the name for a real variable rather than an integer. But the method is correct.
14. Originally Posted by Opalg
There is certainly more than one way to show that $\lim_{n\to\infty}nx^n=0$ when |x|<1. L'Hôpital's rule is one way to approach it, but you need to be more careful in applying it. When you say $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{n(1+a)^{n-1}}$, you are differentiating the numerator as a function of n, and the denominator as a function of a. If you are using n as the variable, then $\tfrac d{dn}(1+a)^n = (1+a)^n\ln (1+a)$. So the calculation should go like this: $\lim_{n\to\infty}\frac n{(1+a)^n} = \lim_{n\to\infty}\frac 1{(1+a)^n\ln (1+a)} = 0$. That looks strange, because it's unusual to see n used as the name for a real variable rather than an integer. But the method is correct.
Oh, you're right. Since I'm differentiating with respect to n, it should be like how you stated it. I should me more careful on determining which variables I am differentiating next time But, I'm glad the rest is correct. Thank you again for your feedback. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 71, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9465978145599365, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/123594/on-local-rankin-selberg-l-functions-for-steinberg-representations | ## On local Rankin-Selberg L-functions for Steinberg representations
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This is a basic question concerning the local Langlands correspondence for $GL_2(\mathbb{Q}_p)$, in particular the compatibility with tensor products . I apologize if this is too basic but I hope someone with more experience can answer it quickly.
Let $G=GL_2(\mathbb{Q}_p)$, $B \subset G$ the subgroup of upper triangular matrices and $St_G$ be the Steinberg representation of $G$; this is the unique irreducible quotient of the representation $Ind_B^G(|\cdot|^{1/2},|\cdot|^{-1/2})$ (normalized induction).
According to the local Langlands correspondence for $G$, we can attach to $St_G$ a Weil-Deligne representation $(V=\mathbb{C}^2,\sigma:W_F \rightarrow GL(V),N \in End(V))$: in this case the inertia acts trivially, $\sigma(Fr) = \begin{pmatrix} p^{-1/2} & 0 \\ 0 & p^{+1/2} \end{pmatrix}$ as a representation of $W_F$ and $N=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.
Now consider the equality of $L$-factors $L(St_G \times St_G,s)=L(\sigma \otimes \sigma,s)$. According to Jacquet (Automorphic Forms on GL(2), Part II, p. 23), we have $L(St_G \times St_G,s)=(1-q^{-s-1})^{-1}$, an $L$-function of degree $1$. On the RHS, the tensor product of Weil-Deligne representations is defined by $(V,\sigma,N) \otimes (V',\sigma',N')=(V\otimes V',\sigma \otimes \sigma',N \otimes 1 + 1 \otimes N')$. But, if I am not mistaken, $\dim_\mathbb{C} Ker(N \otimes 1 + 1 \otimes N')=2$ so that the RHS seems to be of degree $2$! Where did I go wrong?
-
For what it's worth: if instead you use the "representation of SL_2" language rather than the "N" language then the tensor product of two copies of the standard representation has length 2, so again the RHS sounds like it should have degree 2. – wccanard Mar 5 at 7:28
## 1 Answer
I apologize for answering my own question, but I think that the answer might be of some interest to others.
Since I was getting no answers I asked Professor Jacquet directly today. He explained that the expression for $L(St_G \times St_G,s)$ in the book cited above is not correct. The correct $L$-function appears in Proposition 1.4 of his paper A relation between automorphic forms of GL(2) and GL(3), joint with S. Gelbart. With the expression obtained in that paper we do have the expected equality $L(St_G \times St_G,s)=L(\sigma \otimes \sigma,s)$.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9218358397483826, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2008/04/16/differentiable-convex-functions/ | The Unapologetic Mathematician
Differentiable Convex Functions
We showed that all convex functions are continuous. Now let’s assume that we’ve got one that’s differentiable too. Actually, this isn’t a very big imposition. It turns out that a result called Rademacher’s Theorem will tell us that any Lipschitz function is differentiable “almost everywhere”.
Okay, so what does differentiability mean? Remember our secant-slope function:
$\displaystyle s(\left[a,b\right])=\frac{f(b)-f(a)}{b-a}$
Differentiability says that as we shrink the interval $\left[a,b\right]$ down to a single point $c$ the function has a limit, and that limit is $f'(c)$.
So now take $a<b$. We can pick a $c$ between them and points $x$ and $y$ so that $a<x<c<y<b$. Now we compare slopes to find
$s(\left[a,x\right])\leq s(\left[a,c\right])\leq s(\left[c,b\right])\leq s(\left[y,b\right])$
so as we let $x$ approach $a$ and $y$ approach $b$ we find
$f'(a)\leq s(\left[a,c\right])\leq s(\left[c,b\right])\leq f'(b)$
And so the derivative of $f$ must be nondecreasing.
Let’s look at the statement $f'(a)\leq s(\left[a,x\right])$ a little more closely. We can expand this out to say
$\displaystyle f'(a)\leq\frac{f(x)-f(a)}{x-a}$
which we can rewrite as $f(a)+f'(x)(x-a)\leq f(x)$. That is, while the function lies below any of its secants it lies above any of its tangents. In particular, if we have a local minimum where $f'(a)=0$ then $f(a)\leq f(x)$, and the point is also a global minimum.
If the derivative $f'(x)$ is itself differentiable, then the differential mean-value theorem tells us that $f''(x)\geq0$ since $f'(x)$ is nondecreasing. This leads us back to the second derivative test to distinguish maxima and minima, since a function is convex near a local minimum.
Like this:
Posted by John Armstrong | Analysis, Calculus
4 Comments »
1. Actually, any convex function defined on an interval, is differentiable everywhere except on at most countable set (exercise).
Comment by | April 16, 2008 | Reply
2. I know it’s you, Michael, and you haven’t said anything I haven’t more-or-less already said myself. Yesterday I showed that convex functions are Lipschitz (throwing a sop to you in the process) and today I mentioned Rademacher’s theorem.
Comment by | April 17, 2008 | Reply
3. Yeah, I noticed the sop, thanks, but “more-or-less” is the point here. Rademacher’s theorem is rather involved, and my remark is quite elementary, and makes a nice exercise for an interested reader of your blog. Besides, “except on at most countable set” is stronger than “almost everywhere,” since not all sets of measure zero are countable. In fact, it is true that the derivative of a convex function on an interval is continuous on its definition domain, and it’s another exercise, still rather simple.
Comment by | April 17, 2008 | Reply
4. it really good. it increases me knowledge
Comment by saeed | August 21, 2009 | Reply
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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
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http://crypto.stackexchange.com/questions/6175/how-to-best-obtain-bit-sequences-from-throwing-normal-dice | # How to best obtain bit sequences from throwing normal dice?
Throwing normal dice, one can get sequences of digits in [0,5]. Which is the best procedure in practice to transform such sequences into bit sequences desired?
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– Thomas Jan 30 at 21:40
Perhaps I should add that an example of a "practical" question would be: If a random number in [0, 2^1024-1] is required, how should one best go about to generate it as a bit sequence through throwing normal dice? – Mok-Kong Shen Jan 30 at 22:07
You could always discard die rolls corresponding to 4 and 5. Then each die roll produces two bits of entropy. – Stephen Touset Jan 30 at 23:20
How best to ? Slowly. – Cris Stringfellow Jan 31 at 2:33
## 3 Answers
Thomas' first procedure produces 2 bits per roll with a probability of $\frac23$, i.e. it produces $\frac43 \doteq 1.333$ bits on the average. This can be improved as he describes, but it gets quite complicated soon.
Producing a single bit per roll by taking the result mod two leads to $1$ bit per roll, which is not much worse. Combining the two simple methods like
````1 -> 00
2 -> 01
3 -> 10
4 -> 11
5 -> 0
6 -> 1
````
is quite simple and produces $1+\frac23 = \frac53 \doteq 1.667$ bits per roll. AFAIK, this is the maximum you can get without aggregating the rolls.
You can apply this idea to two rolls and obtain $5$ bits with probability of $\frac{32}{36}$ and $2$ bits otherwise thus getting
$$2 + (5-2) \cdot \frac{32}{36} = 2 + \frac{3 \cdot 8}9 = 2 + \frac83 = \frac{14}3 = 4\frac23$$
bits for two rolls, i.e., $2\frac13 \doteq 2.333$ bits per roll, which is quite close to the theoretical maximum of $\log_2 6 \doteq 2.585$.
For a simple "implementation without computer", note that you can extract 1 bit directly from each roll and need to process the "rest" only (e.g. you can get the bit via $n \bmod 2$ and the rest via $\lfloor \frac{n-1}3 \rfloor$). This makes even combining 3 or 4 rolls using a small table easy.
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This family of procedures is optimal on the criteria of requiring the least dice throws to generate a given number of bits in the worst case. – fgrieu Jan 31 at 20:07
The method you described for both single and two rolls are very fine for the practice, I believe. One just have to look up in a table to write down the bits. For three rolls, however, there is a drop in efficiency, if my computation is not in error. (For the method described by Thomas the efficiency of using 3 rolls is also less than that of using 2 rolls.) – Mok-Kong Shen Jan 31 at 22:35
I have computed the values of efficiencies for different number of rolls in increasing order (in parentheses is value of the method described by Thomas) up to 100 rolls: 1 1.66667 (1.33333), 2 2.33333 (2.22222), 4 2.38272, 7 2.53178 (2.40800), 12 2.57454 (2.54856), 24 2.57598, 36 2.57708, 48 2.57712, 53 2.58454 (2.57951) – Mok-Kong Shen Feb 1 at 11:02
@Mok-Kong Shen: Yes, there's an efficiency drop for 3 rolls. You figures seem to be right. – maaartinus Feb 14 at 2:02
The obvious approach is to consider the following bit extraction algorithm:
1. Roll the die, producing an integer $n$ uniform in $\mathbb{Z}_6$.
2. If $n < 4$, return $n$ as two bits. Otherwise, go to 1.
This will return two uniform bits. The algorithm will always terminate, since the probability of recursion is equal to $\frac{2}{6}$ which is subcritical. A proof of correctness can be found in Dennis' post on this question.
How many dice rolls will the algorithm require? At least one, and arbitrarily many, but on average:
$$1 + \frac{2}{6} + \left ( \frac{2}{6} \right )^2 + \cdots = \sum_{t = 0}^\infty \left ( \frac{2}{6} \right )^t = \frac{3}{2}$$
That is, on average, the algorithm will require one and a half dice rolls. This is, of course, in accordance with information theory, which states that the Shannon entropy of an unbiased dice roll is equal to:
$$\log_2{ \left ( 6 \right )} = \log_2{\left ( 4 \right )} + \log_2{ \left ( \frac{3}{2} \right )}$$
However, this is not optimal. A lot of effort is wasted into compressing this into two uniform bits. Can we do better? Yes. For instance, consider rolling two dice together, drawing a uniform $n$ out of $\mathbb{Z}_{36}$. Then we can apply the same algorithm, but returning five bits instead of four, since $32 < 36$. Similarly, the probability of recursion is also lower at $\frac{4}{36}$ which means we are wasting less time and entropy.
We could also roll three dice, drawing $n$ out of $\mathbb{Z}_{216}$, but this is not optimal, even though we can extract seven bits, the probability of recursion is $\frac{88}{216}$ which is much higher than with even just a single die.
So, clearly, the "trick" is to select an integral number of dice such that $6^k$ is very close to (but greater than) a power of two, to minimize the probability of recursion and maximize our use of entropy. A computer program could be written to find the optimal number of dice to roll according to some efficiency metric, depending on your application's needs (or perhaps a general one exists).
The "intuition" behind this is that you can't use arbitrary amounts of entropy, since the bit is the absolute smallest amount of information possible. Any non-integral amount of entropy remaining must be "left behind" (here we use a recursive procedure to eliminate it, but there are other alternatives*).
However, if you can combine multiple such non-integral amounts of entropy through a non-destructive process (here, throwing multiple dice at the same time) you will naturally get more integral amounts of entropy ($0.5 + 0.5 = 1$) which allows you to make better use, of what you would otherwise have had to throw away.
*The answers to this question also suggest recursion is not an optimal approach in terms of efficiency.
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A quick calculation on my computer shows that throwing the dice 53 times would be best (from 1 to 100 dice throws) ... ah, 12 throws is a bit more practical and quite good too :) – owlstead Jan 31 at 0:21
12 throws gives you 31 bits, nice for an positive signed integer, but 32 would have been more practical :( – owlstead Jan 31 at 0:27
Obviously 359 throws give 1.001066 for 928 bits, unbeaten within 10K throws (unless my precision is off) but you would get a sore arm. – owlstead Jan 31 at 0:33
@owlstead Feel free to edit my answer to insert your results in, they would be a useful and welcome addition :) – Thomas Jan 31 at 0:45
What does "subcritical probability" mean? Wouldn't every probability $< 1$ be good for the expected number of rolls to be finite? – Paŭlo Ebermann♦ Feb 6 at 20:55
show 2 more comments
1 way to get 3 bits of entropy from a single die roll is as follows :
1. Roll the fair die
2. For 2,3,4,5 yield two bits `{01,10,11,00}` and go to 4, for 1 or 6 yield `{1,0}`.
3. For the number on the die appearing right-way-up to the thrower (0-179 degrees) yield `{0}` for the number on the die appearing upside-down to the thrower (180-359 degrees) yield `{1}` and go to 5.
4. For the number on the die being oriented within the first quadrant yield `{00}` for the other quadrants proceeding clockwise yield `{01,10,11}`.
5. Go to 1.
So for example I throw the following sequence `2(12 degrees), 3(111 degrees), 6 (98 degrees)`
```` 2 -> 01, 12 degrees -> 0 , so 2(12) -> 010
3 -> 10, 111 degrees -> 1, so 3(111) -> 101
6 -> 0, 98 degrees -> 01, so 6(98) -> 001
````
Giving `010101001` 169 for 3 throws.
Since number and rotations should both be random variables, the distribution is uniform, unbiased and random. The good thing is rotational distinctions between upper lower half, and all four quadrants are almost as easy to distinguish with the naked eye as the numbers themselves. Though on the boundaries there will be some error and bias, introducing a small additional noise into the signal.
Effectively this procedure amounts to the following : roll the die measuring its number and quadrant of rotation, and discard 1 bit of the quadrant information for the numbers {2,3,4 and 5}.
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Read the statement carefully: the input of the desired algorithm is "digits in [0,5]". – fgrieu Jan 31 at 19:51
Good point. I didn't see that. – Cris Stringfellow Jan 31 at 19:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9121712446212769, "perplexity_flag": "middle"} |
http://www.scholarpedia.org/article/Goos-H%C3%A4nchen_effect | # Goos-Hänchen effect
From Scholarpedia
Paul R. Berman (2012), Scholarpedia, 7(3):11584.
Curator and Contributors
1.00 - Paul R. Berman
The Goos-Hänchen effect refers to the lateral displacement that a wave having finite cross section undergoes when it is totally internally reflected at an interface of two media having different indices of refraction. This lateral shift can be explained in the simplest sense as resulting from the propagation of an evanescent wave parallel to the interface, or as a displacement of the wave in a time interval that can be interpreted as the time delay associated with the scattering process.
## History
The Goos-Hänchen effect is illustrated in Fig. 1 for a beam of light having finite cross section. For an angle of incidence $$\theta _{i}$$ greater than the critical angle required for total internal refection, there is a displacement $$s$$ along the surface. The Goos-Hänchen Shift (GHS) is usually defined as the beam's lateral shift $$D=s\cos \theta _{i}$$ indicated in the diagram. The existence of a lateral shift in total internal reflection is often attributed to Newton, based on Proposition 94 of the Principia or Observation 1 of Book 2, Part 1 of Newton's Optiks. Newton gave both a theoretical basis and experimental evidence for penetration of light into medium 2 under conditions of total internal reflection. However, Newton's picture is associated with a mechanical model, whereas the Goos-Hänchen effect is intrinsically linked to the wave nature of the radiation field or, in the case of quantum mechanics, to the wave nature of matter. In fact, there can be a Goos-Hänchen effect associated with most any form of wave-like phenomena.
Figure 1: An beam of electromagnetic radiation (blue beam) is totally internally reflected (red beam) at the interface between two media having indices of refraction $$n_{1}$$ and $$n_{2}\ ,$$ respectively. There is a displacement $$s$$ along the surface. The Goos-Hänchen shift is the lateral shift $$D$$ indicated in the diagram. The angle of incidence is denoted by $$\theta _{i}\ .$$
An extensive list of references and an historical overview of both theory and experiment related to the Goos-Hänchen effect is given in the four-part article by Lotsch (Lotsch KV, 1970). Puri and Birman (Puri and Birman, 1986) also provide a survey of early work on the Goos-Hänchen effect, as well as more recent extensions involving spatially dispersive media and the role played by surface polaritons. Theories of a lateral shift in total internal reflection of electromagnetic waves were developed by Picht (Picht J, 1929) and by Schaefer and Pich (Schaefer and Pich, 1937). Goos and Hänchen (Goos and Hänchen, 1947) measured this lateral displacement for the first time. Their experimental work inspired new theoretical work by Artmann (Artmann K, 1948) and v. Fragstein (v. Fragstein C, 1949), in which expressions for the lateral shift were obtained, with different shifts predicted for field polarization parallel to or perpendicular to the plane of incidence. Although Goos and Hänchen did not find such a polarization dependence in their original work, they were able to measure the effect in 1949 (Goos and Lindberg-Hänchen, 1949). Wolter (Wolter H, 1950) carried out a measurement of the Goos-Hänchen shift that was in good agreement with theory. The Goos-Hänchen effect continues to attract the interest of researchers. It is but one of a large class of non-specular processes, including the Imbert-Federov effect and reflection by multi-layered media [see, for example, the articles by Tamir (Tamir T, 1986), Li (Li C-F, 2007) , and Krayzel et al. (Krayzel et al., 2010) ].
## Theory
### Goos-Hänchen Shift in Optical Total Internal Reflection
The GHS was first discussed in the context of total internal reflection of electromagnetic radiation. Although the GHS occurs only for beams having finite cross section, the GHS can be related to the phase of the amplitude reflection coefficient calculated for plane waves incident on the interface with an angle of incidence equal to $$\theta _{i}\ .$$ The relationship between the GHS and the phase of the reflection coefficient can be established by considering the scattering of an electromagnetic having finite cross section by the interface. In effect, the GHS is connected with the propagation parallel to the surface of an evanescent wave in medium 2. Alternatively, one can view the GHS in terms of the time delay time $$\tau$$ associated with the scattering of a radiation pulse incident on the interface from medium 1; in this picture, the GHS results from the propagation of the pulse parallel to the interface in medium 2 during the time interval $$\tau$$ (Chiu and Quinn., 1972).
Two distinct cases need be considered, polarization of the electric field perpendicular to the plane of incidence (TE or transverse electric polarization) and polarization parallel to the plane of incidence (TM or transverse magnetic polarization). By matching boundary conditions at the interface, one obtains the standard Fresnel equations for transmission and reflection at an interface. If $$\sin \theta _{i}>n_{2}/n_{1}\equiv n<1,$$ there is total internal reflection; the magnitude of the reflection coefficient is unity and its phase determines the GHS. Assuming that the magnetic permeability constant of each medium is approximately equal to unity, the phase of the (amplitude) refection coefficient is
$\left( \phi _{R}\right) _{\perp }=-2\tan ^{-1}\left( \frac{\sqrt{\sin ^{2}\theta _{i}-n^{2}}}{\cos \theta _{i}}\right) ,$
for electric field polarization perpendicular to the plane of incidence and
$\left( \phi _{R}\right) _{\parallel }=-2\tan ^{-1}\left( \frac{\sqrt{\sin ^{2}\theta _{i}-n^{2}}}{n^{2}\cos \theta _{i}}\right)$
for electric field polarization parallel to the plane of incidence.
If the phase is considered as a function of $$k_{1y}=k_{1}\sin \theta _{i},$$ where $$k_{1}=2\pi /\lambda _{1}$$ is the wavelength in medium 1, the displacement along the surface can be calculated as
$\tag{1} s=-\frac{\partial \phi _{R}}{\partial k_{1y}}=-\frac{1}{k_{1}\cos \theta _{i} }\frac{d\phi _{R}}{d\theta _{i}}.$
For electric field polarization perpendicular to the plane of incidence, the corresponding GHS is $\tag{2} D_{\perp }=s_{\bot }\cos \theta _{i}=-\frac{1}{k_{1}}\frac{d\left( \phi _{R}\right) _{\bot }}{d\theta _{i}}=\frac{\lambda _{1}}{\pi }\frac{\sin \theta _{i}}{\sqrt{\sin ^{2}\theta _{i}-n^{2}}}.$
For electric field polarization parallel to the plane of incidence, the GHS is $\tag{3} D_{\parallel }=s_{\Vert }\cos \theta _{i}=-\frac{1}{k_{1}}\frac{d\left( \phi _{R}\right) _{\Vert }}{d\theta _{i}}= \frac{n^{2}}{\sin ^{2}\theta _{i}\left( 1+n^{2}\right) -n^{2}} D_{\perp},$
These equations were derived by Artmann (Artmann K, 1948) (the form of Eq. (3) given by Artmann was valid only near the critical angle). Near the critical angle, $$D_{\parallel }\approx D_{\perp }/n^{2}>D_{\perp }$$. There is energy flow parallel to the surface in medium 2, but no energy flow perpendicular to the surface; the transmitted wave is an evanescent wave. A numerical simulation of the Goos-Hänchen effect is shown in Fig. 2.
Figure 2: Numerical simulation of the Goos-Hänchen effect. The incident TE beam is taken to have a Gaussian cross-section with waist equal to $$35\lambda _{1}$$. The relative index of refraction is $$n=0.913$$ and the average angle of incidence is $$\theta _{i}=67^{\circ }$$. For these parameters, $$s_{\bot }=2.54\lambda _{1}$$ and $$D_{\bot }=0.99\lambda _{1}$$. The orange line is the interface and there is an evanescent wave in medium 2, above the interface. (Image provided by A. Moreau).
An alternative explanation of the GHS can be given in terms of the time delay associated with the scattering of a radiation pulse at the interface. In this case, the phase of the reflection coefficient is considered to be a function of $k_{1x}=k_{1}\cos \theta _{i}$. The incident radiation pulse is not scattered instantaneously by the surface, but reemerges into medium 1 after a time delay given by
$\tau =\frac{1}{v_{1}\cos \theta _{i}}\frac{\partial \phi _{R}}{\partial k_{1x}}=-\frac{1}{k_{1}v_{1}\sin \theta _{i}\cos \theta _{i}}\frac{d\phi _{R} }{d\theta _{i}},$
where $$v_{1}$$ is the speed of light in medium 1. During this time delay, the pulse propagates parallel to the surface and is displaced by a distance $s=v_{1}\sin \theta _{i}\tau =-\frac{1}{k_{1}\cos \theta _{i}}\frac{d\phi _{R} }{d\theta _{i}},$
in agreement with Eq. (1).
Renard (Renard RH, 1964) questioned the validity of Eqs. (2) and (3) since they lead to a finite shift at grazing incidence, $$\theta _{i}=\pi /2\ .$$ Using a model in which energy is transported from one side of the beam to the other, he arrived at modified expressions that vanished at $$\theta _{i}=\pi /2\ .$$ (Lotsch KV, 1968) also found expressions that vanished at $$\theta _{i}=\pi /2\ ,$$ but these were brought into question by Carnaglia (Carnaglia CK, 1976), who suggested that a corrected version of Lotsch's results agrees with the Artmann result. The existence of a nonvanishing GHS at grazing incidence was further supported by the work of Lai et al. (Lai et al., 1986), McGuirk and Carnaglia (McGuirk and Carnaglia, 1967), and Lai et al. (Lai et al., 2000), who argued that the Artmann result is valid provided the incident beam subtends angles of incidence that are between $$\pi /2-\epsilon$$ and $$\theta _{c}+\epsilon \ ,$$ where $$\epsilon$$ is typically on the order of a hundredth of a radian or so. Further validation of the Artmann result was provided by the numerical simulations of Shi et al. (Shi et al., 2007).
The expressions for the GHS given above diverge at the critical angle where the approximations used in their derivation break down. Early generalizations to include angles near the critical angle were given by Artmann (Artmann K, 1948) and by Wolter (Wolter H, 1950). More recently, Horowitz and Tamir (Horowitz and Tamir, 1971) and Lai et al. (Lai et al., 2000) developed theories involving uniform approximations near the critical angle. In these theories, the maximum GHS actually occurs for $$\theta _{i}>\theta _{c}$$ and there are nonvanishing shifts for $$\theta _{i}<\theta _{c}$$ as well (even though, in the plane wave approximation, there is no GHS for $$\theta _{i}<\theta _{c}$$ since the reflection coefficient is real for $$\theta _{i}<\theta _{c}$$). Both these results reflect the fact that a beam having finite width contains a range of angles of incidence about some average angle of incidence. Hence at angles near the critical angle, there are components in the incident beam that undergo both normal as well as total internal reflection. The maximum value of $$D_{\perp }$$ is typically on the order of a few $$\lambda _{1}$$ for a beam whose width is on the order of a hundred $$\lambda _{1}\ .$$
### Goos-Hänchen Shift in Quantum Scattering
There is a quantum analogue to the GHS, in which medium 1 is replaced by the vacuum, medium 2 by a constant potential, and the optical field pulse by a quantum-mechanical wave packet. The potential energy function in the $$x$$ direction is given by $$V(x)=V_{0}\Theta (x)\ ,$$ where $$V_{0}>0$$ and $$\Theta (x)$$ is the Heaviside step function that vanishes for $$x<0$$ and is equal to unity for $$x\geq 0.$$ If $$E>V_{0}$$ and the angle of incidence is greater than some critical angle, an expression for the GHS can be obtained using standard methods of quantum mechanics. As in the optical case, the GHS can be related to the phase of the reflection coefficient of the corresponding plane wave problem. The eigenfunctions for a plane wave associated with a particle having mass $$m$$ and energy $$E$$ that is incident on the interface with an angle if incidence $$\theta _{i}$$ can be written as
$\psi_{E} \left( x,y\right) =\left\{ \begin{array}{c} e^{ik_{y}y}\left( e^{ik_{x}x}+Re^{-ik_{x}x}\right) \; \; x<0 \\ Te^{ik_{y}y}e^{-\sqrt{\kappa ^{2}-k_{x}^{2}}x} \; \; x>0 \end{array} \right. ,$
where $$k_{x}=k\cos \theta _{i}\ ,$$ $$k_{y}=k\sin \theta _{i}\ ,$$ $$k=\sqrt{ 2mE/\hbar ^{2}}\ ,$$ $$\kappa =\sqrt{2mV_{0}/\hbar ^{2}}\ ,$$ $$\hbar$$ is Planck's constant divided by $$2\pi \ ,$$ $$R$$ is the (amplitude) reflection coefficient, and $$T$$ is the (amplitude) transmission coefficient. It is assumed that $$\left( \kappa ^{2}-k_{x}^{2}\right) >0$$ (or, equivalently, $$\sin ^{2}\theta _{i}>1-V_{0}/E$$), since this corresponds to total internal reflection. The amplitude reflection coefficient is $$R=e^{i\phi _{R}}\ ,$$ where
$\phi _{R}=-2\tan ^{-1}\left( \frac{\sqrt{\kappa ^{2}-k^{2}+k_{y}^{2}}}{\sqrt{ k^{2}-k_{y}^{2}}}\right) ,$
considered as a function of $$k_{y}$$, and the GHS is equal to
$D=s\cos \theta _{i}=-\cos \theta _{i}\frac{\partial \phi _{R}}{\partial k_{y} }=\frac{2\cos \theta _{i}k_{y}}{k_{x}\sqrt{\kappa ^{2}-k_{x}^{2}}}=\frac{2}{k }\frac{\sin \theta _{i}}{\sqrt{\sin ^{2}\theta _{i}-\left( 1-V_{0}/E\right) } }.$
If we define an effective index of refraction for this problem by $$n^{2}=\left( 1-V_{0}/E\right)$$ and use the fact that $$\lambda _{dB}=2\pi /k$$ is the de Broglie wavelength of the particle, we arrive at
$D=\frac{\lambda _{dB}}{\pi }\frac{\sin \theta _{i}}{\sqrt{\sin ^{2}\theta _{i}-n^{2}}},$
a result that was obtained by Hora (Hora H, 1960). Since the GHS is proportional to the de Broglie wavelength $$\lambda _{dB}$$ of the particle, it is a purely wave-like effect. There is a probability current density parallel to the surface in medium 2, but no probability flow perpendicular to the surface; the transmitted wave is an evanescent wave. The quantum GHS has the same form as that of the optical GHS for the case of electric field polarization perpendicular to the plane of incidence.
## Experiment
The definitive experiment establishing the lateral shift was carried out by Goos and Hänchen (Goos and Hänchen, 1947). They compared total internal reflection from the back surface of a prism with the reflection from a silver strip that was deposited on the back of the prism. Since the effect is very small, on the order of several optical wavelengths, they multiplied the relative shift between the light that was totally internally reflected and the light that was reflected from the silver by using an "optical waveguide" (parallel surfaces between which many reflections occurred) that allowed them to increase the relative shift by a factor of 70 or so, limited mainly by losses in reflections from the silver strip. The experiment was further refined in 1949 to the point where they could distinguish the difference in the shift between $$D_{\parallel }$$ and $$D_{\perp }\ .$$ Wolter (Wolter, 1950) repeated the Goos-Hänchen experiment with increased resolution and obtained excellent agreement between his experimental results and theory over a small range of angles about the critical angle. He coined the name Goos-Hänchen Effect for the lateral shift, a label which has remained to this day. Thus, by 1950, the GHS had been firmly established.
The GHS continued to attract attention as new technologies became available. Cowan and Anicin (Cowan and Anicin, 1977) observed the GHS shifts for both TE and TM polarizations for microwave radiation incident on a paraffin prism using a single reflection of the beam. Bretenaker (Bretenaker F, 1992) measured the difference between $$D_{\perp }$$ and $$D_{\parallel }$$ in a single reflection optical experiment using a He-Ne laser cavity field. Schwefel (Schwefel HGL, 2008) used the same method as that employed by Goos and Hänchen in their 1947 experiment, but, by using a glass half-cylinder and a partially coherent LED light source, they were able to measure the effect for a single reflection of both TE and TM waves for all angles of incidence. Their results are consistent with a finite shift at $$\theta _{i}=\pi /2$$ [earlier evidence for a finite shift at $$\pi /2$$ was obtained by Rhodes and Carnaglia (Rhodes and Carnaglia, 1977)].
## Extensions to other Domains
The papers of Lotsch (Lotsch KV, 1970) and Puri and Birman (Puri and Birman, 1986) contain references to a large number of both theoretical and experimental studies of the Goos-Hänchen effect in acoustics, nonlinear optics, absorbing media, spatially dispersive media, plasmas, semiconductors, superlattices, etc. Much of this work is motivated by the possibility that the GHS can serve as a probe of scattering and excitations that occur at and near the interface of two bulk materials. More recently, there have been theoretical proposals for measuring a GHS in neutron scattering (Mâaza and Pardo, 1997, Ignatovitch VK, 2004), in negatively refracting material (Berman PR, 2004), and in graphene (Bretenaker F, 1992). An experiment has been carried out in which evidence for the GHS in neutron scattering was claimed (deHaan et al., 2010). It looks as though the Goos-Hänchen effect will continue to attract the attention of researchers for many years to come.
## References
• Artmann, K., 1948. Ann. Physik 437, 87.
• Beenaker, C. W. J., Sepkhanov, R. A., Akhmerov, A. R., Tworzydlo, J., 2009. Phys. Rev. Lett. 102, 146804.
• Berman, P. R., 2004. Phys. Rev. E 66, 067603; 71:021503.
• Bretenaker, F., Floch, A. L., Dutriaux, L., 1992. Phys. Rev. Lett. 68, 931.
• Carnaglia, C. K., 1976. J. Opt. Soc. Amer. 66, 1425.
• Chiu, K.W. , Quinn, J.J., 1972. Amer. J. Phys. 40, 1847.
• Cowan, J. J., Anicin, B., 1977. J. Opt. Soc. Amer. 67, 1307.
• de Haan, V.-O., Plomp, J., Rekveldt, T.M., Kraan, W.H., van Well, A.A., Dalgliesh, R.M., Langridge, S., 2010. Phys. Rev. Lett. 104, 010401.
• Goos, F., Hänchen, H., 1947. Ann Physik 436, 333.
• Goos, F., Lindberg-Hänchen, H., 1949. Ann Physik 440, 251.
• Hora, H., 1960. Optik 17, 409.
• Horowitz, B. R., Tamir, T., 1971. J. Opt. Soc. Amer. 61, 586.
• Ignatovich, V. K., 2004. Phys. Lett. A 322, 36.
• Krayzel, F., Pollès R., Moreau, A., Mihailovic, M., Granet, G., 2010. J. Eur. Opt. Soc. 5, 10025.
• Lai, H. M., Cheng, F. C., Tang, W. K., 1986. J. Opt. Soc. Amer. A 3, 550.
• Lai, H. M., Kwok, C. W., Loo, Y. W., Tang, B. Y. X. W. K., 2000. Phys. Rev. E 62, 7330.
• Lotsch, K. V., 1968. J. Opt. Soc. Amer. 58, 551.
• Lotsch, H. K. V., 1970. Optik 32, 116; 32:189; 32:299; 32:553.
• Mâaza, M., Pardo, B., 1997. Opt. Comm. 142, 84.
• McGuirk, M., Carnaglia, C. K., 1967. J. Opt. Soc. Amer. 67, 103.
• Picht, J., 1925. Ann. Physik 382, 785.
• Puri, A., Birman, J. L., 1986. J. Opt. Soc. Amer. A 3, 543.
• Renard, R. H., 1964. J. Opt. Soc. Amer 54, 1190.
• Rhodes, D. J., Carnaglia, C. K., 1977. J. Opt. Soc. Amer. 67, 679.
• Schaefer, C., Pich, R., 1937. Ann. Physik 422, 245.
• Schwefel, H. G. L., Köhler, W., Lu, Z. H., Fan, J., Wang, L. J., 2008. Opt. Lett. 33, 794.
• Shi, J.-L., Li, C.-F., Wang Q., 2007. Int. J. Mod. Phys. B21, 2777.
• Tamir, T., 1986. J. Opt. Soc. Amer. A 3, 558.
• v. Fragstein, C., 1949. Ann Physik 439, 271.
• Wolter, H., 1950. Z. Naturforsch 5a, 143.
## Further Reading
Reviews of the Goos-Hänchen effect can be found in the articles by Lotsch (Lotsch KV, 1970), Puri and Birman (Puri and Birman, 1986), and Krayzel et al. (Krayzel et al., 2010) given in the References. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 65, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8803505301475525, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/119689/classification-of-symtrivial-modules-over-a-pid | ## Classification of symtrivial modules over a PID
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Let us call a module $M$ over a commutative ring $R$ symtrivial if the symmetry $M \otimes M \to M \otimes M, a \otimes b \mapsto b \otimes a$ equals the identity (the same notion applies to arbitrary symmetric monoidal categories). Equivalently, every bilinear map on $M \times M$ is symmetric. Obviously $R$, and more generally every invertible $R$-module, is symtrivial, but there are many more symtrivial modules, which follows from the following closure properties.
• A directed colimit of symtrivial modules is symtrivial.
• A direct sum $\oplus_{i \in I} M_i$ is symtrivial iff all $M_i$ are symtrivial and $M_i \otimes M_j = 0$ for $i \neq j$.
• Every quotient of a symtrivial module is symtrivial.
• Every symmetric monoidal functor, possibly non-unital, preserves symtrivial modules. For example, localizations of symtrivial modules are symtrivial.
• If $A$ is a commutative $R$-algebra, then $A$, regarded as an $R$-module, is symtrivial iff $R \to A$ is an epimorphism in the category of commutative rings.
• If $M_{\mathfrak{p}}$ is symtrivial for all $\mathfrak{p} \in \mathrm{Spec}(R)$, then $M$ is symtrivial.
From Nakayama one can deduce that a finitely generated $R$-module is symtrivial iff all $M_{\mathfrak{p}}$ are cyclic $R_{\mathfrak{p}}$-modules.
Question. Is there a classification of symtrivial $\mathbb{Z}$-modules?
Almost equivalent one may ask for a classification of symtrivial $R$-modules, where $R$ is a PID (or even a Dedekind domain?). There is a classification of epimorphisms of commutative rings $R \to A$, see Torsten Schöneberg's answer here. Either $A=R/rR$ for some $0 \neq r \in R$, or
$A \cong A_{n,\widehat{P}} := (\widehat{P} \setminus P)^{-1} R[(x_p)_{p \in P}] / (x_p (1-p x_p),p^{n(p)} (1-p x_p))_{p \in P},$
where $\widehat{P}$ is a set of primes in $R$, $P \subseteq \widehat{P}$ is a subset and $n : P \to \mathbb{N}^+$ is a function. These are in particular symtrivial $R$-modules.
Every locally cyclic $R$-module (:= every finitely generated submodule is cyclic) is symtrivial; these are precisely the submodules of $Q(R)$ and of $Q(R)/R$. Since $Q(R) \otimes Q(R)/R=0$, also $Q(R) \oplus Q(R)/R$ is symtrivial.
We can continue this way and use the closure properties to optain lots of examples of symtrivial modules. Nevertheless, I wonder if any classification is available (similar to the classification of epimorphisms, where in fact every epi can be optained by a canonical order of closure properties).
PS: I invented the notion of symtrivial objects for myself and also wonder if anybody else has worked with them or if this notion is already known under a different name. Any information is appreciated. The background is that for a given cocomplete symmetric monoidal category $C$ one can show that $\mathsf{gr}_{\mathbb{N}}(C)$ is the universal cocomplete symmetric monoidal category over $C$ together with a symtrivial object (namely $1_C[-1]$). Thus, in the context of graded objects symtrivial objects pop out quite naturally, and my question is equivalent to the classification of cocontinuous symmetric monoidal $R$-linear functors $\mathsf{gr}_{\mathbb{N}}(R) \to \mathsf{Mod}(R)$.
-
1
Quick guess is that they are the abelian groups of rank less than $2$, i.e., abelian groups such that the $\mathbb{Q}$-vector space $\mathbb{Q} \otimes_{\mathbb{Z}} A$ is of dimension $0$ or $1$. If so, would that be a satisfactory classification? – Todd Trimble Jan 23 at 20:28
2
Notice that $F: \mathbb{Q} \otimes_{\mathbb{Z}} -$ is a strong symmetric monoidal functor, so if $\sigma \neq 1: F(A) \otimes F(A) \to F(A) \otimes F(A)$, then $\sigma \neq 1: A \otimes A \to A \otimes A$. This means that if $A$ is of rank 2 of greater, then $A$ cannot be symtrivial. Thus being rank $0$ or $1$ is a necessary condition for symtriviality. I think it's also sufficient. – Todd Trimble Jan 23 at 20:34
2
If $A$ is a commutative ring with $\mathbb{Z}$-rank $\leq 1$, then $\mathbb{Z} \to A$ doesn't have to be an epimorphism and therefore $A$ is not symtrivial over $\mathbb{Z}$. For example, $A = \mathbb{Z}/2 \times \mathbb{Z}/2$. – Martin Brandenburg Jan 23 at 20:45
I see, thanks.. – Todd Trimble Jan 23 at 20:46
## 1 Answer
We can describe a symtrivial module over a Dedekind domain $R$ with field of fractions $K$ as follows:
(1) Torsion-free symtrivial modules are submodules of $K$.
(2) The torsion part of a symtrivial module is symtrivial
(3) A torsion module is symtrivial if and only if it is a direct sum of symtrivial $p$-power-torsion modules
(4) A $p$-power-torsion module $M$ is symtrivial if and only if it satisfies either $M/pM=0$ or $M/pM=R/pR$,.
(5) If a $p$-power-torsion module $M$ satisfies $M/pM=R/pR$, then it is the direct sum of a $p$-divisible $p$-torsion module and a cyclic module. (Obviously if $M/pM=0$ then it is a $p$-divisible $p$-torsion module.)
(6) If $0 \longrightarrow T \longrightarrow M \longrightarrow F \longrightarrow 0$ is an exact sequence, with $T$ torsion, $M$ symtrivial, and $F$ torsion-free, then for each prime $p$, either $F$ is $p$-divisible or $T$ has no $p$-power-torsion.
(7) If $T$ is torsion and symtrivial, $F$ is torsion-free and symtrivial, and for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-power-torsion, then $T \oplus F$ is symtrivial.
If none of my proofs are mistaken, the remaining questions are
(Q1) For $T$ torsion symtrivial, $F$ torsion-free symtrivial, such that for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion, what is $\mathrm{Ext}^1(F,T)$?
(Q2) Do all elements of $\mathrm{Ext}^1(F,T)$ give symtrivial modules?
Proofs:
(1) This is immediate, by Todd Trimble's argument.
(2) Indeed, take a module $M$ with torsion submodule $T$ such that $a \otimes b \neq b \otimes a$ in $T\otimes T$, but $a\otimes b = b \otimes a$ in $M\otimes M$. The equality $a \otimes b = b\otimes a$ must be the consequence of finitely many relations, involving finitely many elements. Consider the submodule $M'$ generated by the whole torsion module and those finitely many elements. The torsion-free quotient of $M'$ is a finitely generated submodule of $K$, thus is a fractional ideal, thus projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T\otimes T$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T\otimes T$, they do not equal each other in $M'\otimes M'$ - but a complete set of relations implying that they do are relations of $M' \otimes M'$, a contradiction.
(3) This is immediate from things noted in the original question.
(4) Only if follows from the fact that quotients of symtrivial modules are symtrivial. For if, first note that if $M/pM=0$, then $M$ is $p$-divisible and $p$-torsion, so its tensor product with itself is trivial. If $M/pM=R/pR$, let $x$ be a lift of $1 \in R/pR = M/pM$ to $M$. For any $a \otimes b$, choose $n \geq 0$ such that $p^n a = p^n b = p^n x = 0$. Choose $u,v \in \mathbb{Z}$ with $a \equiv ux \pmod {p^n}$ and $b \equiv vx \pmod {p^n}$. Then $a \otimes b = ux \otimes b = ux \otimes vx = vx \otimes ux = b \otimes ux = b \otimes a$.
(5) Let $\pi$ be a uniformizer of $R_p$. This defines a surjection $p^{n-1}M/p^nM \to p^n M/p^{n+1}M$, so either $p^nM/p^{n+1}M$ is eventually $0$ or we get a nontrivial morphism $M \to \lim_n M/p^n M = \lim_n R/p^nR=\hat{R}_p$ which is torsion-free, so it's eventually $0$. Then we have a surjection $M \to M/p^n M= R/p^nR$ whose kernel is $p$-divisible. Since $p$-divisible $p$-torsion groups are divisible, they are injective, so the exact sequence splits and we get a direct sum.
(6) Localize the exact sequence at $p$. Then $F_p$ is a submodule of $K_p$, so either $R_p$ or $K_p$. In the second case $F$ is $p$-divisible so take the first case. $R_p$ is projective so the exact sequence splits into a direct sum, so $R_p \otimes T_p = 0$ by one of the notes in the question, but $R_p \otimes T_p=T_p$ so $T_p=0$.
(7) It suffices to prove that $T \otimes F=0$, by one of the notes in the question. But we can divide $T$ into $p$-power-torsion modules, and a $p$-divisble module tensor a $p$-power-torsion module is trivial.
Answers to Q1 and Q2:
Q1: We disambiguate: "strongly locally cyclic" means that its localization at each prime is cyclic. "weakly locally cyclic" means that each finitely generated submodule is cyclic. $K_p/R_p$ is in the second class but not the first.
We can write $T$ as the direct sum of divisible torsion module $D$ and a strongly locally cyclic module $C$. Divisible modules are injective, so $Ext^1(F,T)=Ext^1(F,C)$.
$Ext^1(F,C)=\left(\prod_p C_p\right)/C$
Coose any nontrivial homomorphism $R \to F$, then the morphism is injective and the kernel is weakly locally cyclic, say $L$. We have a long exact sequence
$Hom(L,C) \to Hom(F,C) \to Hom(R,C) \to Ext^1(L,C) \to Ext^1(F,C) \to Ext^1(R,C)$
$Hom(R,C)=C$, $Ext^1(R,C) = 0$
$Hom(F,C) = 0$ because $F$ is $p$-divisible for all $p$ such that $C_p$ is nontrivial, so the image of any homomorphism inside $C_p$ is trivial, so the whole homomorphism is trivial.
$Ext^1(L,C) = \prod_p Ext^1(L_p,C_p)$ because $L$ and $C$ are torsion. Whenever $C_p$ is nonzero, $L_p=K_p/R_p$ so $Ext^1(L_p,C_p)=Ext^1(K_p,R_p,R/p^a) = R/p^a=C_p$. So we have a short exact sequence
$0 \to C \to \prod_p C_p \to Ext^1(F,C) \to 0$
If we're interested in classifying extensions abstractly, we just have to mod out by the action of the automorphism group.
Q2: All are symtrivial.
Indeed, let $M$ be any module whose torsion and non-torsion parts satisfy the given conditions, and choose $a \otimes b \in M \otimes M$. Choose $x \in M$ such that $a \in \langle x,t \rangle$, $b\in \langle x,t\rangle$. then $a \otimes b= (k_1 x+ t_1) \otimes (k_2 x + t_2) = k_1k_2 x\otimes x + k_1 x\otimes t_2 + k_2 t_1 \otimes x + t_1 \otimes t_2$. Then $t_1 \otimes t_2 = t_2 \otimes t_1$, so the only thing that remains to check is if $t_1\otimes x=x\otimes t_1$. Suppose $n t_1=0$, then choose $y$ such that $ny \equiv x$ mod $T$, so
$x\otimes t_1=(x-ny)\otimes t_1 =t_1 \otimes (x-ny) = t_1 \otimes x$.
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Thanks a lot for the new version, it is very clear. I still have some problems with (5). I see that $R/p^n R \to M/p^n M$ is surjective, but why is it injective? – Martin Brandenburg Jan 25 at 14:44
The kernel is a submodule of $R/p^nR$, which is a quotient of a local ring of a Dedekind domain, hence a DVR, so all ideals are powers of the maximal ideal, $p$. – Will Sawin Jan 25 at 19:17
(5) Yes, but how can we exclude $R/p^k R \cong M/p^n M$ for $k<n$? (Q2) Why is there some $y$ such that $ny \equiv x$ mod $T$? This is only clear when $M/T = K$. – Martin Brandenburg Jan 25 at 23:41
(5) Because then $p^{n-1}M/p^nM= p^{n-1}(R/p^kR)=0$. (Q2) Because for each prime $p$, either $F$ is $p$-divisble or $T$ has no $p$-torsion – Will Sawin Jan 26 at 0:55
@Will: Is each submodule of $K$ symtrivial ? – TJ Jan 27 at 19:09
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http://mathhelpforum.com/pre-calculus/160974-fibonacci-numbers.html | # Thread:
1. ## Fibonacci numbers.
The Fibonacci numbers are defined by $F_0=0$, $F_1=1$ and $F_n=F_{n-1}+F_{n-2}$
Prove if $2|F_n$, then $3|n$
This is one of those times where it's easy to see intuitively, it's hard to see formally.
Considering the Fibonacci sequence
$0,1,1,2,3,5,8,13,21,36...$
It falls into a pattern of
$E,O,O,E,O,O,E,O,O,E,...$
From, as the first term $F_0$ is even, it's clear to see that every third term beginning from $F_1$ is divisible by 3, as it is the sum of the 2 previous numbers which are odd.
But this just scaffolding, scrap work.
Any tips on how to codify this formally?
2. Given the recursive nature of Fibonacci numbers, a proof by induction seems to be the logical way to go. Have you considered that? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9608275890350342, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2009/06/23/self-adjoint-transformations/?like=1&_wpnonce=76a2d9cd23 | # The Unapologetic Mathematician
## Self-Adjoint Transformations
Let’s now consider a single inner-product space $V$ and a linear transformation $T:V\rightarrow V$. Its adjoint is another linear transformation $T^*:V\rightarrow V$. This opens up the possibility that $T^*$ might be the same transformation as $T$. If this happens, we say that $T$ is “self-adjoint”. It then satisfies the adjoint relation
$\displaystyle\langle v,T(w)\rangle=\langle T(v),w\rangle$
What does this look like in terms of matrices? Since we only have one vector space we only need to pick one orthonormal basis $\left\{e_i\right\}$. Then we get a matrix
$\displaystyle\begin{aligned}t_i^j&=\langle e_j,T(e_i)\rangle\\&=\langle T(e_j),e_i\rangle\\&=\overline{\langle e_i,T(e_j)\rangle}\\&=\overline{t_j^i}\end{aligned}$
That is, the matrix of a self-adjoint transformation is its own conjugate transpose. We have a special name for this sort of matrix — “Hermitian” — even though it’s exactly equivalent to self-adjointness as a linear transformation. If we’re just working over a real vector space we don’t have to bother with conjugation. In that case we just say that the matrix is symmetric.
Over a one-dimensional complex vector space, the matrix of a linear transformation $T$ is simply a single complex number $t$. If $T$ is to be self-adjoint, we must have $t=\bar{t}$, and so $t$ must be a real number. In this sense, the operation of taking the conjugate transpose of a complex matrix (or the simple transpose of a real matrix) extends the idea of conjugating a complex number. Self-adjoint matrices, then, are analogous to real numbers.
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 9 Comments »
1. [...] and Forms I Yesterday, we defined a Hamiltonian matrix to be the matrix-theoretic analogue of a self-adjoint transformation. So why should we separate out [...]
Pingback by | June 24, 2009 | Reply
2. [...] particular, this shows that if we have a symmetric form, it’s described by a self-adjoint transformation . Hermitian forms are also described by self-adjoint transformations . And [...]
Pingback by | July 10, 2009 | Reply
3. [...] this all sort of makes sense. Self-adjoint transformations (symmetric or Hermitian) are analogous to the real numbers sitting inside the complex numbers. Within these, positive-definite matrices are sort of like the positive real numbers. It [...]
Pingback by | July 13, 2009 | Reply
4. [...] off, note that no matter what we use, the transformation in the middle is self-adjoint and positive-definite, and so the new form is symmetric and positive-definite, and thus defines [...]
Pingback by | July 27, 2009 | Reply
5. [...] of the original transformation (or just the same, for a real transformation). So what about self-adjoint transformations? We’ve said that these are analogous to real numbers, and indeed their [...]
Pingback by | August 3, 2009 | Reply
6. [...] All the transformations in our analogy — self-adjoint and unitary (or orthogonal), and even anti-self-adjoint (antisymmetric and [...]
Pingback by | August 5, 2009 | Reply
7. [...] to denote transformations). We also have its adjoint . Then is positive-semidefinite (and thus self-adjoint and normal), and so the spectral theorem applies. There must be a unitary transformation [...]
Pingback by | August 17, 2009 | Reply
8. [...] the underlying space forms a vector space itself. Indeed, such forms correspond to correspond to Hermitian matrices, which form a vector space. Anyway, rather than write the usual angle-brackets, we will write one [...]
Pingback by | November 13, 2010 | Reply
9. This was exactly what I needed. Plowing through Thorpe: Elementary Aspects of Differential Geometry on my own. On p. 58 he notes that the Weingarten map Lp is self-adjoint
Lp(v) dot v = Lp(w) dot v; v,w vectors in a real finite dimension vector space. Elsewhere he notes that the map is linear. So a matrix is definitely involved. This post saved me from plowing through Axler, Strang again (it’s been years) to find what such a matrix must look like.
Luysii
Comment by luysii | January 31, 2013 | Reply
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I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 14, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8956835865974426, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/24256/fields-of-mathematics-that-were-dormant-for-a-long-time-until-someone-revitalized/24292 | ## Fields of mathematics that were dormant for a long time until someone revitalized them
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I thought that the closed question here could be modified to a very interesting question (at least as far as big-list type questions go).
Can people name examples of fields of mathematics that were once very active, then fell dormant for a while (and maybe even were forgotten by most people!), and then were revived and became active again?
Here's an example to show what I mean. In the late 19th and early 20th century, hyperbolic geometry was an active and thriving field, attracting the attention of many of the best mathematicians of the era (for instance, Fricke, Klein, Dehn, etc). Fashions changed, however, and the subject was largely forgotten outside of textbooks. In the late '70's, however, Thurston introduced new ideas and showed that hyperbolic geometry was enormously important for the study of 3-manifolds, and now it and its offshoots have become central topics in low-dimensional topology and geometry.
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Surely somebody here will want to comment on the periodic rebirth of invariant theory? – Jim Humphreys May 11 2010 at 17:44
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Rota has interesting things to say about this phenomenon, but I can't quite remember where he wrote them down... – Qiaochu Yuan May 11 2010 at 20:15
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@Qiaochu: Yes, I was thinking of Gian-Carlo Rota when I wrote that line. He actually wrote things down in a lot of places, having passionate views about invariant theory and its neglect. My own view of the subject has always been more tentative, but his personality is unforgettable for those of us who encountered him. – Jim Humphreys May 11 2010 at 22:24
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Actually, another example just came to me. Dehn proved that the mapping class group of a surface was generated by a finite number of so-called Dehn twists. This was forgotten for many years, and in the '70's Lickorish rediscovered the result (for a while, these mapping classes were known as "Lickorish Twists" <grin>). This was the beginning of an enormous amount of work. However, if you go back to Dehn's papers, you will find many of the things people spent the next 10 or so years rediscovering (for instance, the famous "lantern relation" between mapping classes). – Andy Putman May 12 2010 at 2:49
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+1. I hope that people confronted with potentially interesting but ultimately flawed questions will do this in the future rather than trying to keep the original question open. – Harry Gindi May 12 2010 at 8:05
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## 21 Answers
Work of Julia, Fatou, Montel et al on complex dynamics that was largely forgotten or relegated to complex analysis textbooks until Douady – Hubbard and Mandelbrot revitalized it through the study of the Mandelbrot set supplemented by attractive computer graphics.
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It can be said that the work of Julia, Fatou, Lattes etc. revived earlier work by Schroeder, Koenigs, Boettcher etc. on iteration theory. A good account is Daniel Alexander's book on history of complex dynamics and its follow-up, Early Days in Complex Dynamics: A history of complex dynamics in one variable during 1906-1942 Daniel S. Alexander, Drake University, Des Moines, IA, Felice Iavernaro, Università di Bari, Italy, and Alessandro Rosa A co-publication of the AMS and the London Mathematical Society – Margaret Friedland Nov 3 2011 at 18:27
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The older theory of Hopf algebras, which grew out of algebraic topology as well as some purely algebraic theories, developed to the level of Sweedler's 1969 book and then became something of a backwater (at least as seen from the outside). But a generation later the study of quantum groups by Drinfeld, Jimbo, and their followers aroused interest in Hopf algebras on a far wider scale than the earlier algebraic work. The "dormancy" period here was not all that long, but I think it's fair to say that the earlier theory stayed mostly outside the mainstream (as measured by ICM programs, top journals, big grants, etc.).
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Modular forms were actively studied by number theorists Hecke and Siegel in the 1930s, but it was not widely appreciated. Around the same time Hardy, in a series of lectures on Ramanujan's work delivered at Harvard in 1936, called modular forms -- as represented by Ramanujan's interest in the coefficients of the weight 12 form $\Delta(q) = \sum_{n \geq 1} \tau(n)q^n$ -- "one of the backwaters of mathematics". The study of modular forms basically died off in the 1940s and 1950s. It was revitalized by Weil, Shimura et al. in the 1960s. See the introduction to Lang's book on modular forms for some relevant historical remarks.
[EDIT: As Emerton points out in his comment below, the full quote by Hardy is actually more complimentary, so let me include it here: "We may seem to be straying into one of the backwaters of mathematics, but the genesis of $\tau(n)$ as a coefficient in so fundamental a function compels us to treat it with respect." This is at the start of Chapter X of Hardy's "Ramanjuan: Twelve Lectures on Subjects Suggested by his Life and Work."]
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I think String Theory also kicked new life into modular forms encouraging physicists to revisit a bunch of classical work, reading new physical meaning into it. For example, $\Delta$ plays (part of) the role of a generating function for the number of physical states corresponding to each energy level. – Dan Piponi May 11 2010 at 17:55
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One should be careful in interpreting Hardy's remark, which is often quoted in the spirit that it is here. I think a careful reading of the text will show that Hardy says that "it may seem" that one is entering the backwaters of mathematics, but his intention (in the rest of that chapter, which is the last in his book on Ramanujan) is to show that in fact it is not the backwaters at all. – Emerton May 11 2010 at 17:57
One should also consider the representation theoretic point of view on more general automorphic forms which began to be developed by Gelfand and others in the 50s and 60s, as well as Selberg's work, which together formed the base of Langlands massive work on Eisenstein series. I guess like any deep mathematical subject, the detailed history of its development is complicated. But I don't think one can deny that the connection between modular forms and arithmetic was dormant for quite some time. – Emerton May 11 2010 at 18:01
Matt: Hardy's chapter on Ramanujan's work on tau(n) is not the last one in his book: it is X out of XII. – KConrad May 11 2010 at 18:29
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It dismays me to register my age in this way, but... Around 1970, "automorphic forms and L-functions" was not the booming business it is now. Langlands' results were privately circulated [criticism possible here], Shimura's results about generation of classfields and Hasse-Weil zeta functions were ... difficult, Weil's extension of Hecke's converse theorem was a bit baroque, Gelfand-Fomin-PiatetskiShapiro's understanding of the power of repn theory in automorphic forms was new, ... as was Harish-Chandra's repn theory... Godement's repn theory... Ack!!! It confused me then, as now... :) 8 chrs – paul garrett Jun 22 2011 at 22:59
show 3 more comments
Knot theory. That seems like a canonical example: after a lot of interest up until the 1960s it became mathematical backwater in a way, but experienced an enormous surge in development with the discovery of the Jones polynomial and connections with physics (TQFT).
Related area: braid groups and mapping class groups. Besides connections with knots and physics, needs of low-dimensional topology and solution of some long-standing problems played a major role in the revival.
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Maybe this should go back to Lord Kelvin and Tait – Agol Mar 9 2011 at 22:33
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o.O < ponders the backwaters of 1984--1960 knot theory inhabited by D. Gabai, W. Thurston, D. Rolfsen, H. Hilden, J. Montesinos, J. Birman, C. Gordon, L. Kaufmann, R. Riley, W. Whitten, L. Neuworth, K. Murasugi, R. Fox.......... > – Lee Mosher Apr 1 2012 at 20:59
Symplectic and Contact geometries were invented in the 19th centuries as generalisations of the formalism classical mechanics and geometric optics, respectively. It seems to me both subjects soon died until the 1970-s, when Arnold became interested in the purely topological (as opposed to physics-related) aspects of these subjects and posed a few conjectures. Then, in 1980-s Gromov invented the method of J-holomorphic curves that allowed people to actually solve some problems in these subjects, and now both are very active. Nowadays, people have even invented ways to apply them to the study of general differentiable 3- and 4-manifolds.
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The study of generalized symmetries of differential equations was initiated by Emmy Noether in the context of her famous theorem but by and large the field lay dormant until it was revitalized by the discovery of the equations integrable via the inverse scattering transform (i.e., roughly until 1970s).
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The same theory of integrable systems also picked up on the much earlier work of Jacobi, Neumann and Sofia Kovalevskaya on explicit integration of differential equations by theta functions - see, for example, Mumford's "Tata lectures on Theta". – Victor Protsak May 12 2010 at 2:19
Branko Grunbaum wrote in 1978 lecture notes called "Lectures on Lost Mathematics" Grunbaum talks about areas of geometry that went "underground". The topics discussed there and Grunbaum philosophical comments (e.g. p. 15 of the pdf file where the original manuscript begins) are quite relevant to the topic of the question. Some topics discussed by Grunbaum were "revived" in some cases because of these lecture notes and in other cases independently.
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Hi Gil, thanks for this great link. While many conjectures in the notes are resolved by now, I am afraid some of this material is again "overly neglected" (in Grunbaum's words). That's really too bad... – Igor Pak May 12 2010 at 9:21
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Grünbaum has in two other publications shown the value of taking a "fresh look" at geometrical/combinatorial problems that had been considered by geometers/combinatorialists in the past (19th century), using the power of more recent ideas and methods. These are: Arrangements and Spreads (American Mathematical Society, 1972) and the very recent book: Configurations of Points and Lines (AMS, 2009). Arrangements and Spreads has stimulated many new results in geometry and computational geometry. – Joseph Malkevitch May 12 2010 at 14:03
This might not be exactly what you're asking for but I think it's close: Manjul Bhargava's generalizations of Gauss's composition law to higher composition laws. While Gauss's composition law did not exactly languish in obscurity, it is clear that Bhargava's stunning work has revitalized a classical subject. Perhaps one could argue that this is really a case of a long-standing open problem finally being solved, but my sense is that it's more accurate to say that Bhargava found unsuspected treasures in deceptively familiar territory.
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again related to invariant theory and Cayley's hyperdeterminants... – Abdelmalek Abdesselam May 11 2012 at 20:46
Modal Logic.
This goes back to Aristotle. It was picked up by Medieval and Arab philosophers (often associated with proofs of the existence of God) but I don't think it was taken very seriously by mathematicians until the 20th century. Kripke provided nice semantics for modal logics in terms of possible worlds in the 50s (I think) and since then the subject has blossomed. Nowadays modal logics are a commonplace tool in computer science.
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An interesting point and I'm gonna bring it up to Dr.Kripke in his seminar next semester at the CUNY Graduate Center. – Andrew L May 12 2010 at 0:44
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Leibniz wrote a lot on what (today) is recognized as modal logic. Russell greatly misunderstood some of Leibniz's work because 'modal logic' had not yet been properly invented. – Jacques Carette May 14 2010 at 18:20
@Jacques In fact, I take Russell's comments on the subject as evidence the subject was dormant. – Dan Piponi May 14 2010 at 23:20
Considering the little Aristotle I have read, I find it hard to believe he had anything worthwhile to say about modal logic. It is not sufficient to state that some things are "necessary" and others are "possible", many people have done that, but one must also define at least a rudimentary set of formal rules for manipulating negations and conjunctions of these statements. – Ron Maimon Aug 1 2011 at 13:27
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Aristotle did in fact give some rules for manipulation of propositions with modalities. For example, he gave the rule that "necessarily P" is equivalent to "not possibly not P" as well as a bunch of entailment rules. plato.stanford.edu/entries/aristotle-logic/#Syl – Dan Piponi Aug 1 2011 at 17:16
Some historians have speculated that classical Greek geometers used "hidden" analytic methods to discover results, which they then reconstructed synthetically. Further, it seems that Archimedes was further along towards calculus (though I gather this might be a bit exaggerated) than had been thought. In any case, the gradual decline of Aristotelean finitism and dismissal of empiricist epistemological constraints on geometrical reasoning in the 17th/18th century and the subsequent period of mathematical advance might be an instance of revitalization/rediscovery rather than of completely new developments, though these are claims in need of more careful historical argument than I'm in a position to give.
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The mystery deepens when one considers the discovery of the power series for sine, cosine and inverse tangent (with the corollary that $\pi/4=1-1/3+1/5-1/7+\cdots$) by Madhava and others in India around 1400. – John Stillwell May 11 2010 at 23:16
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John: Do you have a reference for this? – Łukasz Grabowski Oct 29 2010 at 13:06
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Exposition in Mathematics Magazine 63 (1990) 291--306 "The Discovery of the Series Formula for pi by Leibniz, Gregory and Nilakantha" by Ranjan Roy – Gerald Edgar Mar 9 2011 at 14:49
I've heard that finite group theory was pretty big in the late 19th century, lay dormant for a while, and had a big increase in activity around 1960 when cases of the Odd Order Theorem started falling. After the theorem was proved, a full classification of finite simple groups looked (to some experts) like a more reasonable goal than before.
(I am somewhat unqualified to elaborate - edits are welcome.)
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This is probably overstated. The work of Frobenius, Schur, Burnside, Brauer, and others in the first decades of the 20th century took the subject in new directions. This grew out of 19th century invariant theory but opened the door (by the time of the Brauer-Fowler paper in 1955 on centralizers of involutions) to Feit-Thompson and the acceleration of the classification project. Representation theory of finite groups has kept developing in many directions, alongside the structure theory which levelled off more after 1980. – Jim Humphreys May 11 2010 at 17:30
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You should also remember Philip Hall! – Yiftach Barnea May 11 2010 at 19:36
Yes, Hall was a major influence, especially in British group theory. I should emphasize that the first half of the 20th century, with its world wars, depression, and other upheavals, was not a flourishing period for mathematics. There were relatively few people doing real research, no "institutes" before IAS, disbanding of European research centers, few journals, slow communication. Even so, finite group theory, combinatorial group theory, Lie theory all made important progress. But it's true that finite group theory became less visible than other subjects in that period. – Jim Humphreys May 11 2010 at 21:02
Actually, I think that in some sense, (finite) group theory went out of fashion after the announcement of the classification, especially in America. I would say that only now in the last decade or two it is coming back. Is this just my feeling? – Yiftach Barnea May 11 2010 at 22:07
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The structure of finite groups has always been a rather specialized field ignored by most mathematicians. But the ordinary and modular representation theory has continued to have a fairly high profile, though here too it's somewhat of an acquired taste and usually not directly applicable elsewhere. George Lusztig has for example reshaped the agenda for groups of Lie type and attracted new interest in the area. Here the modular theory is still far from being understood. The classification of simple groups is another matter, having reached highest intensity around 1980. – Jim Humphreys May 11 2010 at 22:36
As Jim Humphreys has suggested in the comments, practically all of Gian-Carlo Rota's career could be described as breathing new life into unjustly neglected subjects: Möbius functions of posets, invariant theory, lattice theory, etc. For the purposes of MO, let me single out the umbral calculus as a specific subject that languished and was revived by Rota. For anyone who is skeptical of the power of umbral calculus, I recommend Gessel's paper on applications of the classical umbral calculus. Gessel writes:
When I first encountered umbral notation it seemed to me that this was all there was to it; it was simply a notation for dealing with exponential generating functions, or to put it bluntly, it was a method for avoiding the use of exponential generating functions when they really ought to be used. The point of this paper is that my first impression was wrong: none of the results proved here (with the exception of Theorem 7.1, and perhaps a few other results in section 7) can be easily proved by straightforward manipulation of exponential generating functions.
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Umbral calculus and invariant theory are quite related since the former is one way of making sense of the symbolic method used in the latter. One could surmise that this is why G-C Rota was interested in both subjects at the same time. – Abdelmalek Abdesselam Mar 9 2011 at 16:21
This wikipedia link on dessin d'enfants says the following.
Early proto-forms of dessins d'enfants appeared as early as 1856 in the Icosian Calculus of William Rowan Hamilton in modern terms, Hamiltonian paths on the icosahedral graph. Recognizable modern dessins d'enfants (and Belyi functions) were used by Felix Klein, which he called Linienzüge (German, plural of Linienzug “line-track”, also used as a term for polygon. Dessins d'enfant in their modern form were then rediscovered over a century later and named by Alexander Grothendieck in 1984 in his Esquisse d'un Programme.
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q-special functions (basic hypergeometric functions) were developed between the end of the XIXth and beginning of the XXth century. Then remained somewhat in the background as a very peculiar math gadget. When it turned out that they play a relevant role in the representation theory of quantum groups, towards the end of the 80's, they came back into play and enjoyed a very lively period.
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The lambda calculus was first published in 1933 by Alonzo Church, intended to be an alternative to first-order logic. Two of his students, Kleene and Rosser, proved it inconsistent in 1935. In 1936, another student, Alan Turing, proved that a stripped-down version was equivalent in computational power to the Turing machine. It was pretty much "just another example of a Turing-equivalent language" for about 30 years.
In the late 1960s and 1970s, John McCarthy, Dana Scott, and Peter Landin (among others) revived the lambda calculus: John McCarthy by loosely basing LISP on it, Dana Scott by giving it a set-theoretic interpretation, and Peter Landin by developing a theoretical machine that - unlike Turing's - was similar to actual computers and a transformation into its machine instructions. These things together showed that the lambda calculus was not only good at encoding mathematical procedures as programs, but could be compiled into reasonable programs that run on actual machines. Guy Steele showed shortly after in his series of "Lambda the Ultimate" papers that the programs could be run efficiently, and that the lambda calculus easily models many familiar programming language constructs.
Today, the lambda calculus is THE theory of programming languages. Among the ways to describe algorithms, its mathematical purity is unrivaled.
Compared to other mathematical areas, 30 years isn't a long time for something to lay dormant. But consider that Computer Science has only been around for about 70!
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If I remember right, LISP was invented in the late 50's. Something like 1958 or so. – Asaf Karagila Nov 3 2011 at 17:48
Oh, you're right. 1958 it is. My point still stands, though: programming languages, as a mathematical field, was pretty much dead until Scott, Landin and Steele revived the lambda calculus and connected the mathematics to the machines. That makes John McCarthy a forerunner, a voice crying in the wilderness... "LAMBDA!" – Neil Toronto Nov 6 2011 at 23:27
This example may not be that of a whole field but I think it illustrates an important result that lay dormant for a very long time. A natural question in the theory of graphs is when is a graph the vertex-edge graph of a 3-dimensional convex polyhedron? It turns out that this question was in essence answered by Ernst Steinitz in 1922. However, Steinitz did not use a graph theory framework for his work. As a consequence, almost no one noticed what he had accomplished. Almost no references to Steinitz's work was made until 1962 and 1963 when Branko Grünbaum and Theodore Motzkin wrote two papers where they mentioned what Steinitz had done but reformulated it using graph theory terminology. The result in these terms, now known as Steinitz's Theorem states that a graph is the vertex-edge graph of a convex 3-dimensional polyhedron if and only if the graph is planar and 3-connected. A good place to read about this is in Grünbaum's book: Convex Polytopes (2nd edition). Grünbaum (and others) went on to produce many papers that exploited Steinitz's Theorem in many directions. One way to think of what was accomplished here was that to study the combinatorial properties of 3-dimensional convex polyhedra one does not have to think in 3-dimensions but only in 2 dimensions.
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Hyperdeterminants which are generalisations of determinants to multidimensional hypermatrices were first found and developed by Cayley in the mid 19th century and were actively studied until around 1900. Then general results on invariants such Hilbert's basis theorem made them look redundant. For most of the 20th century few mathematicians would even have recognised Cayley's simplest hyperdeterminants if they came upon them.
Then there were a series of rediscoveries of these objects in areas of mathematics and physics, (e.g. hypergeometric functions, Diophantine equations, qubit entanglement and string theory) Now people are starting to look at them again and realize that they are useful and not yet fully understood.
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Indeed there was renewed interest in these object, especially after the book by Gelfand, Kapranov and Zelevinsky. There was also a series a massive papers by Jouanolou. – Abdelmalek Abdesselam Mar 9 2011 at 16:16
C.S. Peirce was lecturing on what he called the “laws of information” as early as 1865–1866 and later gave a simple form of logarithmic measure for the information content of a logical constraint. Of course, he gave these lectures at those colonial backwaters known as Harvard College and the Lowell Institute, so it's no surprise these seeds of information theory took so long to sprout.
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Polynomial Chaos was developed in the late 30s by N. Wiener, but went more or less unnoticed until Ghanem & Spanos picked up on it for use in finite element analysis in the 80s and 90s. In some ways it still may be an under-utilized approach, given the dominance of the Itō and Stratonovich calculi.
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Did you mean Wiener? – mathphysicist May 11 2010 at 22:57
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@mathphysicist: ain't community wiki great? :) – Willie Wong May 12 2010 at 0:53
My typo rate for mathematicians' names is 2/2 the last couple days... :( – Ed Gorcenski May 12 2010 at 2:05
De Morgan established a calculus of binary relations in 1860. Charles Peirce turned out to the subject in 1870, and found most of the interesting equational laws of relation algebra. The subject fell into neglect between 1900 and 1940, to be revived by Alfred Tarski.
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Point set topological methods are seeing a revivial in a very unexpected place:number theory. The use of ultrafilter methods as well as "tailored" topologies on the integers to produce topological proofs of number theoretic results-pioneered by Kevin Broughan and being pursued by Melvyn Nathanson,myself and several others-are causing a renewed interest in point set methods among number theorists lately.
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Can you give an example of a new number theory result proved in this way? – BCnrd May 12 2010 at 1:33
The key bad word in this answer is methods. The question asks for fields of mathematics, probably because examples of methods that waxed and waned are just too easy to find; it must be commonplace in every field. – Thierry Zell Jan 22 2011 at 1:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9648880958557129, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/55535/list | ## Return to Question
2 added 147 characters in body
Given real numbers $a$ and $b$, and an integer $n \geq 2$, let $f(n,a,b)$ be the minimum of $(nint(ja+b)-nint(ia+b)+1)/(j-i)$ (for $1 \leq i < j \leq n$) minus the maximum of $(nint(ja+b)-nint(ia+b)-1)/(j-i)$ (for $1 \leq i < j \leq n$), where $nint$ is the nearest integer function. What is known about the rate at which $f(n,a,b)$ goes to 0 as $n$ goes to infinity, for "generic" real numbers $a, b$? (That is, real numbers whose continued fractions have convergents that grow as dictated by Khinchin's law.) Experiments suggest that $f(n,a,b) = O(1/n^2)$.
This is related to the question of how accurately one can infer the slope of a line from a digitized version of the line; if there is any literature on this question, I would be very interested in pointers.
I am also interested in knowing how one can most efficiently compute the minimum of $f(n,a,b)$.(nint(ja+b)-nint(ia+b)+1)/(j-i)$(for$1 \leq i < j \leq n$) and the maximum of$(nint(ja+b)-nint(ia+b)-1)/(j-i)$(for$1 \leq i < j \leq n\$).
For context, I'll mention that several recent postings of mine (especially http://mathoverflow.net/questions/54731/sums-of-fractional-parts-of-linear-functions-of-n) approach this question from another angle. Specifically, my posting from last week is related to an estimator of the slope of the line that can be computed in linear (as opposed to quadratic) time but has typical error $O(1/n^{3/2})$.
1
# inferring the slope of a digitized line
Given real numbers $a$ and $b$, and an integer $n \geq 2$, let $f(n,a,b)$ be the minimum of $(nint(ja+b)-nint(ia+b)+1)/(j-i)$ (for $1 \leq i < j \leq n$) minus the maximum of $(nint(ja+b)-nint(ia+b)-1)/(j-i)$ (for $1 \leq i < j \leq n$), where $nint$ is the nearest integer function. What is known about the rate at which $f(n,a,b)$ goes to 0 as $n$ goes to infinity, for "generic" real numbers $a, b$? (That is, real numbers whose continued fractions have convergents that grow as dictated by Khinchin's law.) Experiments suggest that $f(n,a,b) = O(1/n^2)$.
This is related to the question of how accurately one can infer the slope of a line from a digitized version of the line; if there is any literature on this question, I would be very interested in pointers.
I am also interested in knowing how one can most efficiently compute $f(n,a,b)$.
For context, I'll mention that several recent postings of mine (especially http://mathoverflow.net/questions/54731/sums-of-fractional-parts-of-linear-functions-of-n) approach this question from another angle. Specifically, my posting from last week is related to an estimator of the slope of the line that can be computed in linear (as opposed to quadratic) time but has typical error $O(1/n^{3/2})$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 33, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9396283626556396, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/109262/list | ## Return to Answer
2 added 158 characters in body
Using the notation of the question http://mathoverflow.net/questions/87633/construct-the-elliptic-fibration-of-elliptic-k3-surface, one sees that the elliptic curve $C_{[\lambda:\mu]}$ is sent to the curve $C_{[-\lambda: \mu]}$ by the involution $i$. So the surface $S=X/i$ has an elliptic fibration over $\mathbb{P}^1$.
The fixed locus of $i$ is given by the disjoint union of two $(-2)$-curves in $X$, namely the two lines $L_1:\{x=y, \; z=w \}$ and $L_2:\{x=-y, \; z=-w \}$, which are components of the the fibre $C_{[1:0]}$.
Since $i$ has no isolated fixed points, the quotient $S$ is smooth, and the quotient map $\pi \colon X \to S$ is branched over two smooth rational curves, namely the images of $L_1$ and $L_2$.
Using the fact that the topological Euler number of $X$ is $24$ and that the branch locus of the double cover $\pi$ is homeomorphic to the disjoint union of two spheres, one finds that the topological Euler number of $S$ is $\frac{1}{2}(24-4)+4=14$.
On the other hand, by Hurwitz formula one finds $$K_X=\pi^*K_S+L_1+L_2,$$ which yields $K_S^2=\frac{1}{2}(K_X-L_1-L_2)^2=-2$.
Using Noether formula we obtain $\chi(\mathscr{O}_S)=(14-2)/12=1$, i.e. $p_g(S)=q(S)$. In particular $S$ is not birational to a $K3$ surface, hence $i$ must be an anti-symplectic involution, namely $i^* \omega = -\omega$ where $\omega$ is the holomorphic $2$-form on $S$.
By general results, if $i$ is an anti-simplectic involution on a $K3$ surface then $X/i$ is a rational surface or an Enriques surface, and the last case happens exactly when $|\textrm{Fix}(i)|=\emptyset$. Therefore in our case $S$ is a rational surface.
Summing up, the surface $S=X/i$ is a non-minimal rational surface with $K_S^2=-2$ and an elliptic fibration over $\mathbb{P}^1$. Notice that such a fibration is not relatively minimal, since the fibre containing the branch locus also contains two $(-1)$-curves. Contracting those curves, one obtain a non-minimal rational surface $\widetilde{S}$ with $K_{\widetilde{S}}^2=0$ and a relatively minimal elliptic fibration over $\mathbb{P}^1$.
The
By looking at the degenerate fibres on $X$, one checks that the degenerate fibres of $\widetilde{S}$ are two singular fibres of type $I_2$ and two singular fibres of type $I_4$ in Kodaira's classification; the existence of the last two fibres shows in particular that $\widetilde{S}$ is not isomorphic to $\mathbb{P}^2$ blown-up in nine points.
My guess is that $\widetilde{S}$ can be constructed in the following way: take a smooth quadric surface $Q$ and consider two reducible curves $T_1$ and $T_2$ of bidegree $(2,2)$, both composed by two lines in a ruling and two lines in the other ruling. Then $\widetilde{S}$ is obtained by blowing up the $8$ base points of the pencil of elliptic curves generated by $T_1$ and $T_2$. Notice that the $T_i$ are precisely two degenerate fibres of type $I_4$ in that pencil.
1
Using the notation of the question http://mathoverflow.net/questions/87633/construct-the-elliptic-fibration-of-elliptic-k3-surface, one sees that the elliptic curve $C_{[\lambda:\mu]}$ is sent to the curve $C_{[-\lambda: \mu]}$ by the involution $i$. So the surface $S=X/i$ has an elliptic fibration over $\mathbb{P}^1$.
The fixed locus of $i$ is given by the disjoint union of two $(-2)$-curves in $X$, namely the two lines $L_1:\{x=y, \; z=w \}$ and $L_2:\{x=-y, \; z=-w \}$, which are components of the the fibre $C_{[1:0]}$.
Since $i$ has no isolated fixed points, the quotient $S$ is smooth, and the quotient map $\pi \colon X \to S$ is branched over two smooth rational curves, namely the images of $L_1$ and $L_2$.
Using the fact that the topological Euler number of $X$ is $24$ and that the branch locus of the double cover $\pi$ is homeomorphic to the disjoint union of two spheres, one finds that the topological Euler number of $S$ is $\frac{1}{2}(24-4)+4=14$.
On the other hand, by Hurwitz formula one finds $$K_X=\pi^*K_S+L_1+L_2,$$ which yields $K_S^2=\frac{1}{2}(K_X-L_1-L_2)^2=-2$.
Using Noether formula we obtain $\chi(\mathscr{O}_S)=(14-2)/12=1$, i.e. $p_g(S)=q(S)$. In particular $S$ is not birational to a $K3$ surface, hence $i$ must be an anti-symplectic involution, namely $i^* \omega = -\omega$ where $\omega$ is the holomorphic $2$-form on $S$.
By general results, if $i$ is an anti-simplectic involution on a $K3$ surface then $X/i$ is a rational surface or an Enriques surface, and the last case happens exactly when $|\textrm{Fix}(i)|=\emptyset$. Therefore in our case $S$ is a rational surface.
Summing up, the surface $S=X/i$ is a non-minimal rational surface with $K_S^2=-2$ and an elliptic fibration over $\mathbb{P}^1$. Notice that such a fibration is not relatively minimal, since the fibre containing the branch locus also contains two $(-1)$-curves. Contracting those curves, one obtain a non-minimal rational surface $\widetilde{S}$ with $K_{\widetilde{S}}^2=0$ and a relatively minimal elliptic fibration over $\mathbb{P}^1$.
The degenerate fibres of $\widetilde{S}$ are two singular fibres of type $I_2$ and two singular fibres of type $I_4$ in Kodaira's classification; the existence of the last two fibres shows in particular that $\widetilde{S}$ is not isomorphic to $\mathbb{P}^2$ blown-up in nine points.
My guess is that $\widetilde{S}$ can be constructed in the following way: take a smooth quadric surface $Q$ and consider two curves $T_1$ and $T_2$ of bidegree $(2,2)$, both composed by two lines in a ruling and two lines in the other ruling. Then $\widetilde{S}$ is obtained by blowing up the $8$ base points of the pencil of elliptic curves generated by $T_1$ and $T_2$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 117, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9332444667816162, "perplexity_flag": "head"} |
http://mathschallenge.net/full/cuboid_perimeters_to_volume | # mathschallenge.net
## Cuboid Perimeters To Volume
#### Problem
For any given cuboid it is possible to measure up to three different perimeters. For example, one perimeter could be measured this way.
Given that cuboid A has perimeters 12, 16, and 20, and cuboid B has perimeters 12, 16, and 24, which cuboid has the greatest volume?
#### Solution
Let the dimensions of a cuboid be $x, y, z$. The perimeters will be $2(x+y), 2(x+z),$ and $2(y+z)$, so dividing each of the perimeters by 2 will give $x+y, x+z,$ and $y+z$ respectively.
The sum of these three terms will give $2x+2y+2z$, leading to $x+y+z$. Thus by subtracting each of $x+y, x+z,$ and $y+z$ we will be able to obtain $z, y,$ and $x$.
Cuboid$2(x+y)$$2(x+z)$$2(y+z)$$x+y$$x+z$$y+z$
A1216206810
B1216246812
Cuboid$2(x+y+z)$$x+y+z$$x$$y$$z$$xyz$
A241224648
B261315735
Surprising as it may seem, cuboid A has the greater volume.
Problem ID: 357 (26 Aug 2009) Difficulty: 3 Star
Only Show Problem | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8987846374511719, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?p=4032903 | Physics Forums
## Stresses due to change in temperature change
Say you have a member of original length L and after a change in temperature T it changes in length to L'=L(1+αT). If we apply a force so that the length is returned back to L, the change in length is LαT and the strain is ε=LαT/L=αT.
My question is why is the denominator in the strain L and not L' since the new lenth of the member is L' and it is that length that we are shrinking.
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Hello Shawncohen You have a good question, if a slightly garbled title and description. Where did your information come from? You have asked about stresses, but referred to strains in your description. Strictly speaking if we heat something so it expands and then squash it back to its original size it is different from loading it so that does not expand in the first place. We regard the original length as the basis for the strain in the second process. However, you are correct that strictly speaking the basis length for strain in the first example is L'. However we are normally talking about millistrain (parts per thousand) for usual materials so to a first order of approximation the result is the same.
Yeah sorry about that, my original problem in a textbook asked for stress but the problem I had in the solutions was with strain. Thanks a lot though, great help! Can you also recommend anything for me to read on why there is a difference? Thanks again
## Stresses due to change in temperature change
Can you also recommend anything for me to read on why there is a difference?
Not quite sure what you mean - difference in what?
Why if we heat something so it expands and then squash it back to its original size is different from loading it so that it does not expand in the first place.
OK shawn. It is a fundamental principle of elasticity that the behaviour of a body under load does not depend upon its past (stress) history. So if we strain a body by loading it then let go it returns to its original size. We can do this as many times as we like and each time the result is the same. (So long as we remain in the elastic region). Similarly we can heat a body without restraint and it will expand to new size. Then if we cool it it will return to its original size. Again we can do this as many times as we like. Now what happens if we load something or heat it and then wait a thousand years? If after a thousand years someone else comes back and loads or heats it again what will they use as the base size of the body?
they will use the length they see, so the extended length surely? my question is more that when we load the member to prevent it expanding before it is heated it isnt actually subjected to any stress, it only experiences stress when we heat it and I'm finding it hard to get my head around the fact we use an 'extension length' that hasnt actually happened to work out the strain and therefore the stress if this is clearer
that hasnt actually happened to work out the strain and therefore the stress if this is clearer
But we don't do that.
If you heat a body so it expands from L to (L+δL), then walk away and wait five minutes or five centuries you are starting again effectively with a new situation if you then compress it back to L, whilst maintaining its higher temperature.
So your strain on thermal expansion is δL/L
and on physical compression is δL/(L+δL).
However the difference is a second order (δL)2quantity so we ignore it
$$\frac{{\delta L}}{L} - \frac{{\delta L}}{{\left( {L + \delta L} \right)}} = \frac{{{{\left( {\delta L} \right)}^2}}}{{L\left( {L + \delta L} \right)}}$$
However we don't usually consider this.
What we usually consider is partial restraint where we equate the strain in one material to the strain in another withich is joined to the first. Then we equate stains to get the stresses.
so how do you deduce that the strain on thermal expansion is δL/L if it hasnt actually expanded by δL, that's whats confusing me
If you heat a body without restraint it expands from L to (L+δL). Sorry I should have made the without restraint clearer.
so there is strain on a body due to pure heat expansion with no restraint?
so there is strain on a body due to pure heat expansion with no restraint?
Yes of course.
Thermal strain is the main way of obtaining strain without stress.
Of course it may be partly restrained in which case there is some stress and some strain, but neither are as great as they would be if the other were zero.
Thanks a lot, this has helped me a great deal!!
Here is a typical example. Two rods, one of steel and one of copper are fixed between rigid walls as shown The steel rod is heated with a blowtorch so it expands. However it is restrained by the copper rod. The expansion of the steel rod = the contraction of the copper rod But the strains are not equal. Since the x section areas are not equal the stresses will not be equal either. But using the above piece of information we can calculate the the position where the expansion force in the steel just balances the restraint force in the copper. Can you see this? Attached Thumbnails
Yes I did examples like that. The one I had problems with was having just one metal bar fixed by rigid walls and therefore no expansion was allowed what so ever
I suppose the easiest way to regard the fully restrained case would be like this:- Imagine a fully restrained bar between two walls. Heat the bar so that it is subject to a compressive force from the walls as it tries to expand. Now imagine one wall suddenly removed so that the bar extends to L(1+αΔT) ie to L+δL If we now apply a force compressive F so that the bar is recompressed to L the stress is given by stress = F/A = modulus x strain = EδL/(L+δL) ≈ EδL/L By the discussion earlier. Note that really if you look closely at what I said earlier it is not just that (δL)2 is small but that the fraction is really (δL/L)2. For normal materials δL/L is nomally of the order of 1 in 1000 to the square is of the order of 1 in 1 million.
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http://www.reference.com/browse/wiki/Graded_category | Definitions
# Graded category
A graded category is a mathematical concept.
If $mathcal\left\{A\right\}$ is a category, then a $mathcal\left\{A\right\}$-graded category is a category $mathcal\left\{C\right\}$ together with a functor $F:mathcal\left\{C\right\} rightarrow mathcal\left\{A\right\}$.
Monoids and groups can be thought of categories with a single element. A monoid-graded or group-graded category is therefore one in which to each morphism is attached an element of a given monoid (resp. group), its grade. This must be compatible with composition, in the sense that compositions have the product grade. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932572603225708, "perplexity_flag": "head"} |
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Solution to these problems should be postmarked no later than June 30, 2000.
Notes: A set of lines of concurrent if and only if they have a common point of intersection.
7.
Let
$S=\frac{{1}^{2}}{1·3}+\frac{{2}^{2}}{3·5}+\frac{{3}^{2}}{5·7}+\dots +\frac{{500}^{2}}{999·1001} .$
Find the value of $S$.
8.
The sequences $\left\{{a}_{n}\right\}$ and $\left\{{b}_{n}\right\}$ are such that, for every positive integer $n$,
${a}_{n}>0 , {b}_{n}>0 , {a}_{n+1}={a}_{n}+\frac{1}{{b}_{n}} , {b}_{n+1}={b}_{n}+\frac{1}{{a}_{n}} .$
Prove that ${a}_{50}+{b}_{50}>20$.
9.
There are six points in the plane. Any three of them are vertices of a triangle whose sides are of different length. Prove that there exists a triangle whose smallest side is the largest side of another triangle.
10.
In a rectangle, whose sides are 20 and 25 units of length, are placed 120 squares of side 1 unit of length. Prove that a circle of diameter 1 unit can be placed in the rectangle, so that it has no common points with the squares.
11.
Each of nine lines divides a square into two quadrilaterals, such that the ratio of their area is 2:3. Prove that at least three of these lines are concurrent.
12.
Each vertex of a regular 100-sided polygon is marked with a number chosen from among the natural numbers $1,2,3,\dots ,49$. Prove that there are four vertices (which we can denote as $A$, $B$, $C$, $D$ with respective numbers $a$, $b$, $c$, $d$) such that $\mathrm{ABCD}$ is a rectangle, the points $A$ and $B$ are two adjacent vertices of the rectangle and $a+b=c+d$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9226379990577698, "perplexity_flag": "head"} |
http://mathhelpforum.com/geometry/28193-very-interesting-very-challenging-circle-problem.html | # Thread:
1. ## Very Interesting, Very Challenging Circle Problem
There is a circle radius of $\sqrt {50}$
There are two line segments legnths 2 and 6 each with one end point on the circle, the other ends meeting at a 90 degree angle. What is the distance between the center of the circle and the right angle where the two segments meet?
2. Are the line segments inside or outside the circle?
3. Originally Posted by Gauss
Are the line segments inside or outside the circle?
Inside, sorry I wasn't specific enough.
4. Originally Posted by hummeth
There is a circle radius of $\sqrt {50}$
There are two line segments legnths 2 and 6 each with one end point on the circle, the other ends meeting at a 90 degree angle. What is the distance between the center of the circle and the right angle where the two segments meet?
1. Draw a sketch.
2. You are dealing with 2 right triangles. Use Pythagorean theorem:
$\left|\begin{array}{l}y^2 + (6-x)^2 = (\sqrt{50})^2\\ (2+y)^2 + x^2 = 50\end{array} \right.$ ..... Expand the brackets, collect like terms:
$\left|\begin{array}{l}y^2 + 36-12x+x^2 = 50\\ 4+4y+y^2 + x^2 = 50\end{array} \right.$ ..... Subtract the 2nd equation from the first:
$32-4y-12x = 0~\implies~y = -3x+8$ ..... Plug in this term for y into the 1rst equation. You get a quadratic equation in x:
$10x^2-60x+50 = 0~\implies~x = 5~\vee~x=1$ ..... These x-values correspond with the y-values y = -3 (rejected) or y = 5.
Therefore the distance from the point of intersection to the center of the circle is:
$d = \sqrt{5^2+1^2}=\sqrt{26}\approx 5.099$
Attached Thumbnails
5. That's perfect! I figured it out last night with a more algebraic approach. I put the circle on a coordinate plane centerting the circle on the vertex. Then, I set the right angle as point (a,b) and made the two ling segments vertical and horizontal lines on the coordinate plane. The points on the circle were then (a,y) and (x,b) Then I solved these systems.
x^2+y^2=50
a^2+y^2=50
x^2+b^2=50
y-b=6
x-a=2
After some algebra, I got the same answer, roughly 5.1 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8699436187744141, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/476/can-adjoint-linear-transformations-be-naturally-realized-as-adjoint-functors/484 | ## Can adjoint linear transformations be naturally realized as adjoint functors?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Last week Yan Zhang asked me the following: is there a way to realize vector spaces as categories so that adjoint functors between pairs of vector spaces become adjoint linear operators in the usual sense?
It seems as if one needs to declare an inner product by fiat for this to work out. An obvious approach is to take the objects to be vectors and hom(v, w) to be the inner product (so the category should be enriched over C). But I don't see how composition works out here, and Yan says he tried this and it didn't work out as cleanly as he wanted. In this setup I guess we want the category to be additive and the biproduct to be vector addition, but I have no idea whether this actually happens. I think John Baez's ideas about categorified linear algebra, especially categorified Hilbert spaces, are relevant here but I don't understand them well enough to see how they work out.
Anyone who actually knows some category theory care to clear things up?
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" I guess we want the category to be additive and the biproduct to be vector addition, but I have no idea whether this actually happens." Make it happen! Mod out the the messy part so you get what you want. This works remarkably often. Of course, sometimes it gives you the trivial object... – SixWingedSeraph Jan 2 2010 at 1:30
## 6 Answers
There's a canonical way of going the other way, starting with two linear categories with nice finiteness properties, with adjoint functors between them and getting a pair of vector spaces with adjoint linear transformations. The vector spaces are generated by formal symbols for each object in the category, and the inner product between any objects is the dimension of the Hom space (so Hom spaces had better be finite dimensional). Note that this doesn't have to be symmetric.
Functors give linear transformations, and adjoint functors are adjoint in the usual sense.
You can soup up this construction when you have some more structures on your category. For example, if you have a direct sum, then you can impose the relation $[A+B]=[A]+[B]$, and everything will work fine.
If your category is abelian, you can take Grothendieck group, where $[A]+[C]=[B]$ for every short exact sequence $0\to A \to B \to C\to 0$, but then you have to be much more careful about the fact that lots of functors (including Hom with objects in the category!) aren't exact: they don't send short exact sequences to short exact sequences. You need to use derived functors to fix this.
There's no canonical way of going the direction you asked, though in practice we have a very good record of being able to and I don't know of any really good examples of there being two equal natural seeming but different such constructions.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think it's more natural to take advantage of the monoidal structure and regard the vector spaces as functors rather than objects. For simplicity, consider only finite dimensional vector spaces. Given V, we have a functor $F_V: Vect \to Vect$ which sends $W$ to $W \otimes V$. The familiar identification $Hom(U\otimes V, W) = Hom(U, W\otimes V^*)$ shows that the (category theory) adjoint of $F_V$ is `$F_{V^*}$`. (That's $F$ sub $V^*$, in case the font is too small to read.) Chaining together two of these adjunctive identifications of Hom sets, we have
`$Hom(V, X) = Hom(1, X\otimes V^*) = Hom(X^*, V^*)$`.
The above identification sends a linear transformation $g:V\to X$ to the (linear algebra) adjoint `$g^*: X^*\to V^*$`. If $V$ and $X$ are inner product spaces then we can of course identify `$V^*$` with $V$ and `$X^*$` with $X$.
Maybe that's too elementary and not the answer you were looking for. But it seems to me it's the most simple and obvious way to relate linear algebra adjoints to category theory adjoints.
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It just occurred to me that there may be a certain sense in which this is impossible in principle. Every equivalence of categories can be improved to an adjoint equivalence, by modifying either the unit or the counit. This is true for all sorts of categories (internal, enriched, fibered, etc.). So if there were a way to realize all vector spaces (or, say, inner product spaces) as some kind of category such that adjoint linear transformations became adjoint functors, we would expect that any isomorphism of vector spaces would give an equivalence of such categories, and hence could be improved to an adjoint equivalence, i.e. an isomorphism whose adjoint is its inverse. But this is false; not every isomorphism between inner product spaces is unitary/orthogonal.
I can't decide whether this is deep or nonsensical, but I thought I'd throw it out there.
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A neat correspondence between adjoint functions and adjoint functors is possible, if you relax your understanding of what it means for a category to "realize" a Hilbert space a bit. (The adjoint of a linear function only exists if the vector spaces are Hilbert spaces and the function is continuous, so I'll take the question to be about Hilbert space instead of vector spaces.)
Given a Hilbert space $H$, "realize" it as the partially ordered set of closed subspaces $S(H)$, regarded as a category. Then a continuous linear function $f \colon H \to K$ induces a contravariant functor $S(f) \colon S(H)^{\text{op}} \to S(K)$. Now, denoting the adjoint function of $f$ by $f^\dagger \colon K \to H$, we get an adjunction between $S(f)$ and $S(f^\dagger)$. In fact, up to a scalar, any contravariant adjunction between $S(H)$ and $S(K)$ comes from an adjoint pair of functions between $H$ and $K$!
All this comes from a 1974 paper by Paul H. Palmquist, a student of Mac Lane, called "Adjoint functors induced by adjoint linear transformations" in Proceedings of the AMS 44(2):251--254.
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“The adjoint of a linear function only exists if the vector spaces are Hilbert spaces and the function is continuous”—I guess you mean the only situation in which we may naturally view the adjoint of a continuous map $V \to W$ as a map $W \to V$? – L Spice May 9 2011 at 4:15
There a simple way to make this work:
Say `T:V->X` is a map of inner-product vector spaces. You can view `V` as a category, where `Hom(v,w)` is a singleton set containing one real number, the inner product `<v,w>`, and similarly for `X`.
Composition, a binary operation, is defined (stupidly, as in any category with singleton hom-sets) as follows:
`Comp_{uvw} : Hom(u,v)xHom(v,w) -> Hom(u,w) `
by `(<u,v>,<v,w>) |-> <u,w>`
Then the adjoint T*:X->V satisfies
`<Tv,x>=<v,T*x>`, i.e. `Hom(Tv,x)=Hom(v,T*x)`
, meaning it is a right adjoint to T (in a very strong sense: we have equality of these hom-sets instead of just natural isomorphism).
The triviality of this example reflects the fact that that T and T* are called "adjoint" simply because they belong on opposite sides of a comma :)
In general, if H is any function of two variables, we can say that g is right adjoint to f "with respect to H" if H(f(a),b)=H(a,g(b)), and say that "adjoint functors" are "adjoint with respect to Hom" (up to natural isomorphism, of course).
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Okay, but in this setup are the functors between such categories precisely the linear transformations? I don't have much intuition for whether this Hom makes sense. – Qiaochu Yuan Oct 14 2009 at 23:36
this approach forgets the whole vector space structure... – Martin Brandenburg Dec 28 2009 at 12:40
check out
John C. Baez, Higher-Dimensional Algebra II: 2-Hilbert Spaces, online.
"The analogy to adjoints of operators between Hilbert spaces is clear. Our main point here is that that this analogy relies on the more fundamental analogy between the inner product and the hom functor."
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http://physics.stackexchange.com/questions/55512/why-is-a-star-a-pure-state/55548 | # Why Is a star a Pure state?
I am reading some papers about black hole complementarity (Samir D. Mathur. The information paradox: conflicts and resolutions. Proceedings for Lepton-Photon 2011 (expanded). arXiv:1201.2079 [hep-th].) and I found the following sentence;
Thus an initial pure state (the gas making up the collapsing star) evolves to a black hole, and after evaporation, into something that can only be described by a mixed state.
How is a star a pure state? Isn't it a thermal state and as a consequence is mixed? Isnt that why the light coming from the sun is unpolarized?
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2
Do you have a reference for that sentence? – Chris White Mar 1 at 7:23
1
– twistor59 Mar 1 at 13:07
1
The star is represented by a pure state in the sense that the quantum mechanical state of all its components is completely knowable in principle. The paragraph linked (page 7) argues why this has then been treated as evolving into a mixed state - you throw away some information. – twistor59 Mar 1 at 13:13
## 1 Answer
As twistor59 already correctly wrote, a star and every other physical system may be found in a pure state in principle. When a physical system is in a pure state, it just means that we're assuming the maximum possible knowledge about the system – e.g. we know all the eigenvalues of a complete set of commuting observables.
If a physical system is in a mixed (non-pure) state, it just means that our knowledge about the system is smaller than the maximum one and we're creating a probabilistic mixture of pure states – which are mixed in a way that is fully analogous to probabilistic distributions in classical statistical physics.
In quantum mechanics, predictions are always probabilistic but for any pure state, the role for the probabilities is "minimized". Pure states are the closest objects somewhat analogous to "points in the phase space" in classical physics.
In the black hole information puzzle research and many other considerations, we're considering the initial state of the matter to be in a pure state and we can do so without a loss of generality because the evolution of any mixed state is completely determined by the evolution of pure states that belong to a basis of the Hilbert space (space of possible pure states). The previous sentence, easily proved from the evolution equations for density matrices and pure states, is just the quantum counterpart of the classical statement that it's enough to investigate the behavior of well-defined points in the phase space (pure classical states) and the behavior of all probabilistic mixtures is then determined by purely statistical (Hamiltonian-independent) formulae.
A star of a certain temperature may be described by a mixed state but in principle, we may also assume that the actual state of the star is a pure state that just "resembles" the mixed state – a typical pure state in the ensemble, for example – and we may study the behavior of these pure states separately. Indeed, that's necessary if we want the potential information loss to be seen at all. The information loss was the apparent conclusion by Hawking in the 1970s that a pure state will ultimately evolve into a mixed one. There has never been a doubt that a mixed state evolves into a mixed state but the problem was that even a pure state seemed not to be willing to stay pure. This conclusion was later shown to be an artifact of the perturbative approximations in quantum gravity. The full theory of quantum gravity evolves pure states into pure states of the final Hawking radiation.
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Could you give references for your final comments? – Emilio Pisanty Mar 1 at 14:57
Thank you for the answer. I think it answers my doubt. I would also be interested in the reference for your last sentence. – Super Frog Mar 1 at 16:16
Dear Emilio, it's what the thousands of papers on information loss paradox are all about. See some examples at scholar.google.cz/… - Check that all papers allowing pure-to-mixed evolution are before 1995 or surely before 1997 A focused search on newer papers, AdS/CFT, saying that the evolution is unitary, e.g. scholar.google.cz/… – Luboš Motl Mar 1 at 19:03
– Luboš Motl Mar 1 at 19:05
1
Dear @lurscher, a pure (zero von Neumann entropy) state evolves into a pure (zero von Neumann entropy) state. This is the unitarity of the evolution. We ascribe a finite entropy proportional to the area to a black hole but "black hole" either means a mixed state here or the entropy is defined as $\ln N$ where $N$ is the number of macroscopically indistinguishable states from a given microstate, not the von Neumann entropy. A pure black hole (or any other pure) microstate of course carries no von Neumann entropy and no sane physicist has ever claimed otherwise. – Luboš Motl Mar 2 at 7:26
show 2 more comments | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9151276350021362, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/67455/list | ## Return to Question
2 Added follow-up questions
Let $f\colon\thinspace E\to B$ be a Serre fibration whose fibre $F$ is $k-1$-connected, $k\geq 1$. Assume $B$ is a connected CW complex. Then the primary obstruction to the existence of a cross section of $f$ is defined; it is a cohomology class $$\mathfrak{o}(f)\in H^{k+1}(B;\tilde{H}_{k}(F)).$$ Here the coefficients may be twisted by $\pi_1(B)$. The definition involves choosing a section on the $k$-skeleton which you then try to extend, but the class itself is canonical (depends only on the fibration).
Meanwhile, there is the cohomology Leray-Serre spectral sequence of the fibration, with $$E_2^{p,q}=H^p(B;H^q(F))\implies H^*(E),$$ where again the coefficients in the $E_2$ term may be twisted by the action of $\pi_1(B)$.
Here is my question, which I'm a little embarrassed to ask:
Is there a canonical class in the $E_2$ term which relates somehow to $\mathfrak{o}(f)$?
Sorry for being (intentionally) vague.
Edit: As Grigory M points out in his answer, if we work over a field and assume the local system on the base formed from the homology of the fibres is trivial, then the first non-trivial differential $$d_{k+1}\in \mathrm{Hom}(H^k(F),H^{k+1}(B))$$ is the linear dual of an element $$d_{k+1}^\ast\in\mathrm{Hom}(H_{k+1}(B),H_k(F))\cong H^{k+1}(B;H_{k}(F))$$ which should equal the obstruction class.
Has anyone seen a reference for this?
Can anyone give a more general statement when the local coefficient system is non-trivial?
Thanks.
1
# Where does the primary obstruction of a fibration show up in its spectral sequence?
Let $f\colon\thinspace E\to B$ be a Serre fibration whose fibre $F$ is $k-1$-connected, $k\geq 1$. Assume $B$ is a connected CW complex. Then the primary obstruction to the existence of a cross section of $f$ is defined; it is a cohomology class $$\mathfrak{o}(f)\in H^{k+1}(B;\tilde{H}_{k}(F)).$$ Here the coefficients may be twisted by $\pi_1(B)$. The definition involves choosing a section on the $k$-skeleton which you then try to extend, but the class itself is canonical (depends only on the fibration).
Meanwhile, there is the cohomology Leray-Serre spectral sequence of the fibration, with $$E_2^{p,q}=H^p(B;H^q(F))\implies H^*(E),$$ where again the coefficients in the $E_2$ term may be twisted by the action of $\pi_1(B)$.
Here is my question, which I'm a little embarrassed to ask:
Is there a canonical class in the $E_2$ term which relates somehow to $\mathfrak{o}(f)$?
Sorry for being (intentionally) vague. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9356532692909241, "perplexity_flag": "head"} |
http://luckytoilet.wordpress.com/2010/09/03/stepping-stones-solution-with-young-tableaux/ | # Lucky's Notes
Notes on math, coding, and other stuff
## Stepping Stones: solution with Young tableaux
About a year ago or so, I created a math problem and submitted it to CsTutoringCenter. Titled Stepping Stones, my problem statement went like this:
In a certain river, there are a bunch of stepping stones arranged from one side to the other. A very athletic person can cross the river by jumping on these stepping stones, one at a time.
A stepping stone is big enough for only one person, and the gap between two stepping stones is small enough that it is possible to jump between two adjacent stepping stones.
You are an army commander trying to get a group of soldiers across this river (using these stepping stones). Initially your n soldiers are placed on the first n stepping stones. Your task is to get all of them onto the last n stepping stones.
For example, here are the five possible ways to get a group of two soldiers across a river with five stepping stones:
```1) ##--- #-#-- -##-- -#-#- --##- --#-# ---##
2) ##--- #-#-- -##-- -#-#- -#--# --#-# ---##
3) ##--- #-#-- #--#- -#-#- --##- --#-# ---##
4) ##--- #-#-- #--#- -#-#- -#--# --#-# ---##
5) ##--- #-#-- #--#- #---# -#--# --#-# ---##```
Let C(k,n) be the number of ways of which n soldiers can cross a river with k stepping stones. In the example, C(5,2) = 5.
Find C(50,12) mod 987654321.
Of course, small values of $C(k,n)$ may be bruteforced by a computer. But $C(50,12)$ is well out of reach of brute force, and substantial mathematics is needed. Or for the lazy, it is possible to find small values by brute force, then look the sequence up on OEIS to find the formula.
### Bijection to a matrix representation
We find that any instance of the problem can be represented, or bijected to a special matrix, one where each row and column is increasing.
Let us number the soldiers in the following fashion. Let the rightmost soldier, that is, the soldier first to move, be labelled 1. The soldier behind him is labelled 2, and so on, until the last soldier to move is labelled $n$. Since the order of soldiers cannot change, each soldier moves exactly $k-n$ times.
Consider a very simple case, with 4 stones and 2 soldiers. One possible way is the first soldier moving twice, followed by the second moving twice.
This move sequence can be represented by $[1,1,2,2]$. The other sequence, and the only other sequence is $[1,2,1,2]$.
Firstly a sequence like $[1,1,1,2]$ is invalid because in a valid sequence, each soldier has to move the same number of times. Another invalid case is something like $[2,1,1,2]$, since obviously 2 cannot move the first turn. But how can you tell whether $[1,2,1,1,2,1,3,2,3,3,2,3]$ is valid or not?
It isn’t very easy to tell in sequence form. Instead we represent the sequence as a matrix form.
Let’s try some examples first, The sequence $[1,1,2,2]$ in matrix form is:
$\begin{array}{cc} 1&2 \\ 3&4 \end{array}$
The other sequence, $[1,2,1,2]$, is:
$\begin{array}{cc} 1&3 \\ 2&4 \end{array}$
Try a more complex example, $[1,2,1,1,2,1,3,2,3,3,2,3]$:
$\begin{array}{cccc} 1&3&4&6 \\ 2&5&8&11 \\ 7&9&10&12 \end{array}$
To create the matrix, first have a counter and initialize it to 1; when the first soldier moves, place the counter in the first cell that’s unfilled in the first row, and increment the counter. Now if the second soldier moves, we place the counter in the second row (first unfilled cell), and increment it again, and so on. By the time we’re through all of the soldier moves, the matrix should be nice and rectangular.
Perhaps a different explanation is more intuitive. If $A_{3,2} = 7$ (where $A$ is the matrix, $A_{3,2}$ means row 3, column 2), that means on move 7, soldier number 3 makes his move number 2.
From this interpretation, several important facts surface. The rows must be increasing, obviously, since if the row is not increasing, say 7 comes before 5, move 7 happened before move 5, which can’t be!
Less obviously, the column has to be increasing. Suppose that in some matrix, $A_{2,7}=20$, and the cell directly underneath, $A_{3,7} = 19$. In that case soldier 3 made his move 7 before soldier 2 made his move 7. This results in soldier 3 ahead of soldier 2 (or at least on the same stone)!
So with $k$ stones and $n$ soldiers, the matrix should have $n$ rows and $k-n$ columns. The $m \times n$ cells contain the numbers $1 \ldots mn$, while each row and each column is increasing. Our job is to enumerate these matrices, since such a matrix forms a 1-to-1 correspondence to a valid move sequence.
### Enumerating matrices with the hook length formula
A Young tableau is an interesting combinatorial object, based on the Ferrers diagram. From a Ferrers diagram of size $n$, a Young tableau is one where every number from $1 \ldots n$ is filled in it and all rows and all columns are increasing:
From any cell of a Young tableau, a hook is formed by extending all the way down, and all the way to the right:
The hook length of a cell is the length of its hook (including itself). In the above picture, the hook length is 5. Each cell in the tableau has a hook and a hook length.
The number of valid Young tableaux with a given shape $\lambda$ and with $n$ cells is given by the hook length formula:
$N = \frac{n!}{\prod_{x \in \lambda} \mathrm{hook}_x}$
A special case of the hook length formula can be used to enumerate rectangular Young tableaux. For instance, we have a 3*4 Young tableau. If we fill each cell with its hook length we get:
The count would then be
$\frac{12!}{6 \cdot 5 \cdot 4 \cdot 5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$
Or alternatively,
$\frac{12!}{\frac{6!}{3!} \cdot \frac{5!}{2!} \cdot \frac{4!}{1!} \cdot \frac{3!}{0!}}$
Simplifying:
$\frac{12! \cdot 0! \cdot 1! \cdot 2! \cdot 3!}{3! \cdot 4! \cdot 5! \cdot 6!}$
This can be generalized to a formula. If we have $x$ rows and $y$ columns:
$\frac{(xy)! \prod_{i=1}^{y-1}i!}{\prod_{j=x}^{x+y-1} j!}$
For $C(k,n)$, we have $n$ rows and $k-n$ columns, thus by substitution we arrive at our formula:
$C(k,n) = \frac{[n(k-n)]! \prod_{i=0}^{k-n-1}i!}{\prod_{j=n}^{k-1}j!}$
This can be used to compute $C(50,12)$, trivial to implement in Haskell or any other language.
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This entry was posted on Friday, September 3rd, 2010 at 2:38 pm and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### 2 Responses to Stepping Stones: solution with Young tableaux
1. Yessam says:
The amount of math you know blows me away. I’m a grade 12 student in Toronto at the top of my class and you lost me at matrix. Where have you learned all this stuff?
2. luckytoilet says:
Thanks. I’m not sure I know that much math (high school curriculum and some undergraduate topics) but there are plenty of resources out there for a motivated student.
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http://math.stackexchange.com/questions/9023/inconsistent-naming-of-elliptic-integrals?answertab=votes | # Inconsistent naming of elliptic integrals
This may be a question whose answer is lost in the mists of time, but why is the elliptical integral of the first kind denoted as $F(\pi/2,m)=K(m)$ when that of the second kind has $E(\pi/2,m)=E(m)$? It's not very consistent! Aside from convention, is there anything stopping us from rationalising these names a little?
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Of course you're free to use any names you wish for these functions, but the problem is that people might not understand what you mean. There have been attempts to clean up the notation, for example Eagle's book "The Elliptic Functions as They Should Be: An Account, with Applications, of the Functions in a New Canonical Form", but so far the traditional notation has prevailed (as usual). – Hans Lundmark Nov 5 '10 at 7:37
@HansLundmark Thanks, interesting reference. Considering the field I'm in still requires boilerplate like "where E(m) is the second complete elliptic integral" I'm guessing you could sneak in a slight notation change without causing too much angst. But we'll see! – Will Robertson Nov 5 '10 at 7:55
Eh, in my ideal world, we'll all be using Carlson's integrals, but the Legendre-Jacobi ones have a long tradition of use in applications, and I suppose we won't be getting rid of them anytime soon... – J. M. Nov 5 '10 at 11:08
According to Cajori's book, anyway, we have, as expected, Legendre and Jacobi to blame for using $K$ in the complete case and $F$ in the incomplete case for the elliptic integral of the first kind. I have no access to those 19th century articles where they were first introduced, so my guess of why they chose these letters is as good as yours. – J. M. Nov 5 '10 at 11:17
## 1 Answer
This discrepancy seems to come from the various conventions in defining the nome $q$ of complex lattice $\Lambda_{\tau} = \mathbb{Z} \oplus \tau \mathbb{Z}$ as one of the four $e^{2 \pi i \tau}$, $e^{\pi i \tau}$, $e^{2 i \tau}$ or $e^{i \tau}$, where $\tau \in \mathbb{H}$ is the lattice parameter. Investigate the vast literature on theta functions, and you'll see exactly what I mean by the problem of too many "conventions".
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– J. M. Nov 5 '10 at 14:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9521896839141846, "perplexity_flag": "middle"} |
http://www.reference.com/browse/following+ones+nose | Definitions
# Vector space
In mathematics, a vector space (or linear space) is a collection of objects (called vectors) that, informally speaking, may be scaled and added. More formally, a vector space is a set on which two operations, called (vector) addition and (scalar) multiplication, are defined and satisfy certain natural axioms which are listed below. Vector spaces are the basic objects of study in linear algebra, and are used throughout mathematics, science, and engineering.
The most familiar vector spaces are two- and three-dimensional Euclidean spaces. Vectors in these spaces can be represented by ordered pairs or triples of real numbers, and are isomorphic to geometric vectors—quantities with a magnitude and a direction, usually depicted as arrows. These vectors may be added together using the parallelogram rule (vector addition) or multiplied by real numbers (scalar multiplication). The behavior of geometric vectors under these operations provides a good intuitive model for the behavior of vectors in more abstract vector spaces, which need not have a geometric interpretation. For example, the set of (real) polynomials forms a vector space.
A much more extensive idea of what constitutes a vector space is found in the See also subsection for this article, which provides links to more abstract examples of this term.
## Motivation and definition
The space R2 consisting of pairs of real numbers, (x, y), is a common example for a vector space. It is one, because any pair (here a ) can be added:
(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2),
and any vector (x, y) can be multiplied by a real number s to yield another vector (sx, sy). The general vector space notion is a generalization of this idea. It is more general in several ways:
• other fields instead of the real numbers, such as complex numbers or finite fields, are allowed.
• the dimension, which is two above, is arbitrary.
• most importantly, elements of vector spaces are not usually expressed as linear combinations of a particular set of vectors, i.e. there is no preference of representing the vector (x, y) as
(x, y) = x · (1, 0) + y · (0, 1)
over
(x, y) = (−1/3·x + 2/3·y) · (−1, 1) + (1/3·x + 1/3·y) · (2, 1)
The pairs of vectors (1, 0) and (0, 1) or (−1, 1) with (2, 1) are called bases of R2 (see below).
### Definition
Let F be a field (such as the rationals, reals or complex numbers), whose elements will be called . A vector space over the field F is a set V together with two binary operations,
• : V × V → V denoted v + w, where v, w ∈ V, and
• : F × V → V denoted av, where a ∈ F and v ∈ V,
satisfying the axioms below. Let u, v, w be arbitrary elements of V, and a, b be elements of F, respectively.
Associativity of addition u + (v + w) = (u + v) + w
Commutativity of addition v + w = w + v
Identity element of addition There exists an element 0 ∈ V, called the , such that v + 0 = v for all v ∈ V.
Inverse elements of addition For all v ∈ V, there exists an element w ∈ V, called the of v, such that v + w = 0.
Distributivity of scalar multiplication with respect to vector addition a (v + w) = a v + a w
Distributivity of scalar multiplication with respect to field addition (a + b) v = a v + b v
Compatibility of scalar multiplication with field multiplication a (b v) = (ab) v
Identity element of scalar multiplication 1 v = v, where 1 denotes the multiplicative identity in F
### Elementary remarks
The first four axioms can be subsumed by requiring the set of vectors to be an abelian group under addition, and the rest are equivalent to a ring homomorphism f from the field into the endomorphism ring of the group of vectors. Then scalar multiplication a v is defined as (f(a))(v). This can be seen as the starting point of defining vector spaces without referring to a field.
Some sources choose to also include two axioms of closure u + v ∈ V and a v ∈ V for all a, u, and v. When the operations are interpreted as maps with codomain V, these closure axioms hold by definition, and do not need to be stated independently. Closure, however, must be checked to determine whether a subset of a vector space is a subspace.
Expressions of the form “v a”, where v ∈ V and a ∈ F, are, strictly speaking, not defined. Because of the commutativity of the underlying field, however, “a v” and “v a” are often treated synonymously. Additionally, if v ∈ V, w ∈ V, and a ∈ F where vector space V is additionally an algebra over the field F then a v w = v a w, which makes it convenient to consider “a v” and “v a” to represent the same vector.
There are a number of properties that follow easily from the vector space axioms. Some of them derive from elementary group theory, applied to the (additive) group of vectors: for example the zero vector 0 ∈ V and the additive inverse −v of a vector v are unique. Other properties can be derived from the distributive law, for example scalar multiplication by zero yields the zero vector and no other scalar multiplication yields the zero vector.
## History
The notion of a vector space stems conceptually from affine geometry, via the introduction of coordinates in the plane or usual three-dimensional space. Around 1636, French mathematicians Descartes and Fermat found the bases of analytic geometry by tying the solutions of an equation with two variables to the determination of a plane curve.
To achieve a geometric solutions without using coordinates, Bernhard Bolzano introduced in 1804 certain operations on points, lines and planes, which are predecessors of vectors. This work was considered in the concept of barycentric coordinates of August Ferdinand Möbius in 1827. The founding leg of the definition of vectors was the Bellavitis' definition of the bipoint, which is an oriented segment, one of whose ends is the origin and the other one a target.
The notion of vector was reconsidered with the presentation of complex numbers by Jean-Robert Argand and William Rowan Hamilton and the inception of quaternions by the latter mathematician, being elements in R2 and R4, respectively. Treating them using linear combinations goes back to Laguerre in 1867, who defined systems of linear equations.
In 1857, Cayley introduced the matrix notation which allows one to harmonize and simplify the writing of linear maps between vector spaces.
At the same time, Grassmann studied the barycentric calculus initiated by Möbius. He envisaged sets of abstract objects endowed with operations. His work exceeds the framework of vector spaces, since his introduction of multiplication led him to the concept of algebras. Nonetheless, the concepts of dimension and linear independence are present, as well as the scalar product (1844). The primacy of these discoveries was disputed with Cauchy's publication Sur les clefs algébriques.
Italian mathematician Peano, one of whose important contributions was the rigorous axiomatisation of extant concepts, in particular the construction of sets, was one of the first to give the modern definition of vector spaces around the end of 19th century.
An important development of this concept is due to the construction of function spaces by Henri Lebesgue. This was later formalized by David Hilbert and Stefan Banach, in his 1920 PhD thesis.
At this time, algebra and the new field of functional analysis began to interact, notably with key concepts such as spaces of p-integrable functions and Hilbert spaces. Also at this time, the first studies concerning infinite dimensional vector spaces were done.
## Linear maps and matrices
Two given vector spaces V and W (over the same field F) can be related by linear maps (also called linear transformations) from V to W. These are functions that are compatible with the relevant structure—i.e., they preserve sums and scalar products:
f(v + w) = f(v) + f(w) and f(a · v) = a · f(v).
An is a linear map such that there exists an inverse map such that the two possible compositions and are identity maps. Equivalently, f is both one-to-one (injective) and onto (surjective). If there exists an isomorphism between V and W, the two spaces are said to be isomorphic; they are then essentially identical as vector spaces, since all identities holding in V are, via f, transported to similar ones in W, and vice versa via g.
Given any two vector spaces V and W, the set of linear maps V → W forms a vector space HomF(V, W) (also denoted L(V, W): two such maps f and g are added by adding them pointwise, i.e.
(f + g)(v) = f(v) + g(v)
and scalar multiplication is given by
(a·f)(v) = a·f(v).
The case of W = F, the base field, is of particular interest. The space of linear maps from V to F is called the , denoted V∗.
### Matrices
Matrices are a useful notion to encode linear maps. They are written as a rectangular array of scalars, i.e. elements of some field F. Any m-by-n matrix A gives rise to a linear map from Fn, the vector space consisting of n-tuples x = (x1, ..., xn) to Fm, by the following
$\left(x_1, x_2, ..., x_n\right) mapsto left\left(sum_\left\{i=1\right\}^m x_i a_\left\{i1\right\}, sum_\left\{i=1\right\}^m x_i a_\left\{i2\right\}, ..., sum_\left\{i=1\right\}^m x_i a_\left\{in\right\} right\right)$,
or, using the matrix multiplication of the matrix A with the coordinate vector x:
$mathbf x mapsto A mathbf x$.
Moreover, after choosing bases of V and W (see below), any linear map is uniquely represented by a matrix via this assignment.
The determinant of a square matrix tells whether the associated map is an isomorphism or not: to be so it is sufficient and necessary that the determinant is nonzero.
### Eigenvalues and eigenvectors
A particularly important case are endomorphisms, i.e. maps . In this case, vectors v can be compared to their image under f, f(v). Any vector v satisfying λ · v = f(v), where λ is a scalar, is called an eigenvector, with eigenvalue λ. Rephrased, this means that v is an element of kernel of the difference (the identity map In the finite-dimensional case, this can be rephrased using determinants: f having eigenvalue λ is equivalent to
det (f − λ · Id) = 0.
Spelling out the definition of the determinant, the left hand side turns out to be polynomial function in λ, called the characteristic polynomial of f. If the field F is large enough to contain a zero of this polynomial (which automatically happens for F algebraically closed, such as F = C) any linear map has at least one eigenvector. The vector space V may or may not be spanned by eigenvectors, a phenomenon governed by Jordan–Chevalley decomposition.
## Subspaces and quotient spaces
In general, a nonempty subset W of a vector space V that is closed under addition and scalar multiplication is called a subspace of V. Subspaces of V are vector spaces (over the same field) in their own right. The intersection of all subspaces containing a given set of vectors is called its span. Expressed in terms of elements, the span is the subspace consisting of finite sums (called )
a1v1 + a2v2 + ... + anvn,
where the ai and vi (i = 1, ..., n) are scalars and vectors, respectively.
The counterpart to subspaces are quotient vector spaces. Given any subspace W ⊂ V, the quotient space V/W ("V modulo W") is defined as follows: as a set, it consists of v + W = {v + w, w ∈ W}, where v is an arbitrary vector in V. The sum of two such elements v1 + W and v2 + W is (v1 + v2) + W, and scalar multiplication is given by a · (v + W) = (a · v) + W. The key point in this definition is that v1 + W = v2 + W if and only if the difference of v1 and v2 lies in W. This way, the quotient space "forgets" information that is contained in the subspace W.
For any linear map f: V → W, the kernel ker(f) consists of elements v that are mapped to 0 in W. It, as well as the image im(f) = {f(v), v ∈ V}, are linear subspaces of V and W, respectively. There is a fundamental isomorphism
V / ker(f) ≅ im(f).
The existence of kernels and images as above is part of the statement that the category of vector spaces (over a fixed field F) is an abelian category.
## Examples of vector spaces
### Coordinate spaces and function spaces
The first example of a vector space over a field F is the field itself, equipped with its standard addition and multiplication. This is the particular case n = 1 in the vector space usually denoted Fn, known as the where n is an integer. Its elements are n-tuples
(f1, f2, ..., fn), where the fi are elements of F.
Infinite coordinate sequences, and more generally functions from any fixed set Ω to a field F also form vector spaces. The latter applies in particular to common geometric situations, such as Ω being the real line or an interval, open subsets of Rn etc. The vector spaces stemming of this type are called function spaces. Many notions in topology and analysis, such as continuity, integrability or differentiability are well-behaved with respect to linearity, i.e. sums and scalar multiples of functions possessing such a property will still have that property. Hence, the set of such functions are vector spaces. The methods of functional analysis provide finer information about these spaces, see below. The vector space F[x] is given by polynomial functions, i.e.
f (x) = rnxn + rn−1xn−1 + ... + r1x + r0, where the coefficients r0, ..., rn are in F,
or power series, which are similar, except that infinitely many terms are allowed.
### Systems of linear equations
Systems of linear equations also lead to vector spaces. Indeed this source may be seen as one of the historical reasons for developing this notion. For example, the solutions of
a + 3b + c = 0
4a + 2b + 2c = 0
given by triples with arbitrary a, b = a/2, and c = −5a/2 form a vector space. In matrix notation, this can be interpreted as the solution of the equation
Ax = 0,
where x is the vector (a, b, c) and A is the matrix
$begin\left\{bmatrix\right\}$
1 & 3 & 1 4 & 2 & 2end{bmatrix}. Equivalently, this solution space is the kernel of the linear map attached to A (see above).
In a similar vein, the solutions of homogeneous linear differential equations, for example
f ''(x) + 2f '(x) + f (x) = 0
also form vector spaces: since the derivatives of the sought function f appear linearly (as opposed to f ''(x)2, for example) and (f + g)' = f ' + g ', any linear combination of solutions is still a solution. In this particular case the solutions are given by where a and b are arbitrary constants, and e=2.718....
### Algebraic number theory
A common situation in algebraic number theory is a field F containing a smaller field E. Then, by the given multiplication and addition operations of F, F becomes an E-vector space. F is also called a field extension. As such C, the complex numbers are a vector space over R. Another example is Q(z), the smallest field containing the rationals and some complex number z. The dimension of this vector space (see below) is closely tied to z being algebraic or transcendental.
## Basic constructions
In addition to the above concrete examples, there are a number of standard linear algebraic constructions that yield vector spaces related to given ones. In addition to the concrete definitions given below, they are also characterized by universal properties, which determines an object X by specifying the linear maps from X to any other vector space.
### Direct product and direct sum
The direct product $prod_\left\{i in I\right\} V_i$ of a family of vector spaces Vi, where i runs through some index set I, consists of tuples (vi)i ∈ I, i.e. for any index i, one element vi of Vi is given. Addition and scalar multiplication is performed componentwise:
(vi) + (wi) = (vi + wi).
a · (vi) = (a · vi),
A variant of this construction is the direct sum $oplus_\left\{i in I\right\} V_i$ (also called coproduct and denoted $coprod_\left\{i in I\right\}V_i$), where only tuples with finitely many nonzero vectors are allowed. The direct sum is denoted . If the index set I is finite, the two constructions agree, but differ otherwise.
### Tensor product
The tensor product V ⊗F W, or simply V ⊗ W, is a vector space consisting of finite (formal) sums of symbols
v1 ⊗ w1 + v2 ⊗ w2 + ... + vn ⊗ wn,
subject to certain rules mimicking bilinearity, such as
a · (v ⊗ w) = (a · v) ⊗ w = v ⊗ (a · w).
The tensor product—one of the central notions of multilinear algebra—can be seen as the extension of the hierarchy of scalars, vectors and matrices. Via the fundamental isomorphism
HomF (V, W) ≅ V∗ ⊗F W,
matrices, which are essentially the same as linear maps, i.e. contained in the left hand side, translate into an element of the tensor product of the dual of V with W.
In the important case V ⊗ V, the tensor product can be loosely thought of as adding formal "products" of vectors (which, ad hoc, don't exist in vector spaces). In general, there are no relations between the two tensors v1 ⊗ v2 and v2 ⊗ v1. Forcing two such elements to be equal leads to the symmetric algebra, whereas forcing v1 ⊗ v2 = − v2 ⊗ v1 yields the exterior algebra. The latter is the linear algebraic fundament of differential forms: they are elements of the exterior algebra of the cotangent space to manifolds. Tensors, i.e. element of some tensor product have various applications, for example the Riemann curvature tensor encodes all curvatures of a manifold at a time, which finds applications in general relativity, for example, where the Einstein curvature tensor describes the curvature of space-time.
## Bases and dimension
If, in a (finite or infinite) set {vi}i ∈ I no vector can be removed without changing the span, the set is said to be linearly independent. Equivalently, an equation
0 = a1vi1 + ai2v2 + ... + anvin
can only hold if all scalars a1, ..., an equal zero.
A linearly independent set whose span is V is called a basis for V. Hence, every element can be expressed as a finite sum of basis elements, and any such representation is unique (once a basis is chosen). Vector spaces are sometimes introduced from this coordinatised viewpoint.
Using Zorn’s Lemma (which is equivalent to the axiom of choice), it can be proven that every vector space has a basis. It follows from the ultrafilter lemma, which is weaker than the axiom of choice, that all bases of a given vector space have the same cardinality. This cardinality is called the of the vector space. Historically, the existence of bases was first shown by Felix Hausdorff. It is known that, given the rest of the axioms, this statement is in fact equivalent to the axiom of choice.
For example, the dimension of the coordinate space Fn is n, since any element in this space (x1, x2, ..., xn) can be uniquely expressed as a linear combination of n vectors e1 = (1, 0, ..., 0), e2 = (0, 1, 0, ..., 0), to en = (0, 0, ..., 0, 1), namely the sum
$sum_\left\{i=1\right\}^n x_i mathbf\left\{e\right\}_i$.
By the unicity of the decomposition of any element into a linear combination of chosen basis elements vi, linear maps are completely determined by specifying f(vi). Given two vector spaces, V and W, of the same dimension, a choice of bases of V and W and a bijection between the sets of bases gives rise to the map that maps any basis element of V to the corresponding basis element of W. This map is, by its very definition, an isomorphisms. Therefore, vector spaces over a given field are fixed up to isomorphism by the dimension. Thus any n-dimensional vector spaces over F is isomorphic to F0n.
## Vector spaces with additional structures
From the point of view of linear algebra, the vector spaces are completely understood insofar as any vector space is characterized, up to isomorphism, by its dimension. The needs of functional analysis require considering additional structures, especially with respect to convergence of infinite series. On the other hand, the notion of bases as explained above can be difficult to apply to infinite-dimensional spaces, thus also calling for an adapted approach.
Therefore, it is common to study vector spaces with certain additional structures. This is often necessary to recover ordinary notions from geometry or analysis.
### Topological vector spaces
Convergense issues are adressed by considering vector spaces V which also carry a compatible topology, i.e. a structure that allows to talk about elements being close to each other. Compatible here means that addition and scalar multiplication should be continuous maps, i.e. if x and y in V, and a in F vary by a bounded amount (the field also has to carry a topology in this setting), then so do x + y and ax.
Only in such topological vector spaces can one consider infinite sums of vectors, i.e. series, through the notion of convergence. For example, the term
$sum_\left\{i=0\right\}^\left\{infty\right\} f_i$,
where the fi are some elements of a given vector space of real or complex functions means the limit of the corresponding finite sums of functions.
A way of ensuring the existence of limits of infinite series as above is to restrict attention to complete vector spaces, i.e. any Cauchy sequence (which can be thought of as sequences that "should" possess a limit) do have a limit. Roughly, completeness means the absence of holes. E.g. the rationals are not complete, since there are series of rational numbers converging to irrational numbers such as $sqrt 2$. A less immediate example is provided by functions equipped with the Riemann integral.
In the realm of topological vector spaces, such as Banach and Hilbert spaces, all notions should be coherent with the topology. For example, instead of considering all linear maps (also called functionals) V → W, it is useful to require maps to be continuous. For example, the dual space V∗ consists of continuous functionals V → R (or C). If V is some vector space of (well-behaved) functions, this dual space, called space of distributions, which can be thought of as generalized functions, find applications in solving differential equations. Applying the dual construction twice yields the bidual V∗∗. There is always an natural, injective map V → V∗∗. This map may or may not be an isomorphism. If so, V is called reflexive.
#### Banach spaces
Banach spaces, in honor of Stefan Banach, are complete normed vector spaces, i.e. the topology comes from a norm, a datum that allows to measure lengths of vectors.
A common example is the vector space lp consisting of infinite vectors with real entries x = (x1, x2, ...) whose p-norm (1 ≤ p ≤ ∞) given by
$|mathbf x|_p := \left(sum_i |x_i|^p\right)^\left\{1/p\right\}$ for p < ∞ and $|mathbf x|_infty := text\left\{sup\right\}_i |x_i|$
is finite. In the case of finitely many entries, i.e. Rn, the topology does not yield additional insight—in fact, all topologies on finite-dimensional topological vector spaces are equivalent, i.e. give rise to the same notion of convergence. In the infinite-dimensional situation, however, the topologies for different p are inequivalent. E.g. the sequence xn of vectors
xn = (2−n, 2−n, ..., 2−n, 0, 0, ...)—the first 2n components are 2−n, the following ones are 0
yields
$|x_n|_1 = sum_\left\{i=1\right\}^\left\{2^n\right\} 2^\left\{-n\right\} = 1$ and $|x_n|_infty = sup \left(2^\left\{-n\right\}, 0\right) = 2^\left\{-n\right\}$,
i.e. the sequence xn, with n tending to ∞ converges to the zero vector for p = ∞, but does not for p = 1. This is an example for the remark that the study of topological vector spaces is richer than that of vector spaces without additional data.
More generally, it is possible to consider functions endowed with a norm that replaces the sum in the above p-norm by an integral, specifically the Lebesgue integral
$|f|_p := left\left(int |f\left(x\right)|^p dx right\right)^\left\{1/p\right\}$.
The set of integrable functions on a given domain Ω (for example an interval) satisfying |f |p < ∞, and equipped with this norm is denoted Lp(Ω).
Since the above uses the Lebesgue integral (as opposed to the Riemann integral), these spaces are complete. Concretely this means that for any sequence of functions satisfying the condition
$lim_\left\{k, n to infty\right\}int_Omega |\left\{f\right\}_k \left(x\right)-\left\{f\right\}_n \left(x\right)|^p dx = 0 .$
there exists a function f(x) belonging to the vector space Lp(Ω) such that
$lim_\left\{k to infty\right\}int_Omega |\left\{f\right\} \left(x\right)-\left\{f\right\}_k \left(x\right)|^p dx = 0 .$
Imposing boundedness conditions not only on the function, but also on its derivatives leads to Sobolev spaces.
#### Hilbert spaces
Slightly more special, but equally crucial to functional analysis is the case where the topology is induced by an inner product, which allows to measure angles between vectors. This entails, that lengths of vectors can be defined too, namely by $|mathbf v| = sqrt \left\{langle v, v rangle\right\}$. If such a space is complete, it is called Hilbert space, in honor of David Hilbert.
A key case is the Hilbert space L2(Ω), whose inner product is given by
$langle f | g rangle = int_Omega overline\left\{f\left(x\right)\right\} g\left(x\right) dx .$, with $overline\left\{f\left(x\right)\right\}$ being the complex conjugate of f(x).
Reversing this direction of thought, i.e. finding a sequence of functions fn that approximate a given function, is equally crucial. Early analysis, for example, in the guise of the Taylor approximation, established an approximation of differentiable functions f by polynomials. Ad hoc, this technique is local, i.e. approximating f closely at some point x may not approximate the function globally. The Stone-Weierstrass theorem, however, states that every continuous function on [a, b] can be approximated as closely as desired by a polynomial. More generally, and more conceptually, the theorem yields a simple description what "basic functions" suffice to generate a Hilbert space, in the sense that the of their span (i.e. finite sums and limits of those) is the whole space. For distinction, a basis in the linear algebraic sense as above is then called a Hamel basis. Not only does the theorem exhibit polynomials as sufficient for approximation purposes, it, together with the Gram-Schmidt process, also allows the construction of a basis of orthogonal polynomials. Orthogonality means that $langle p | q rangle = 0$, i.e. the polynomials obtained don't interfer. Instead of polynomials, similar statements hold for Legendre polynomials, Bessel functions and Hypergeometric functions.
Resolving general functions into sums of trigonometric functions is known as the Fourier expansion, a technique much used in engineering. It is possible to describe any function f(x) on a bounded, closed interval (or equivalently, any periodic function) as the limit of the following sum
$$
f_N (x) = frac{a_0}{2} + sum_{m=1}^{N}left[a_mcosleft(mxright)+b_msinleft(mxright)right] as N → ∞ , with suitable coefficients am and bm, called Fourier coefficients. This expansion is surprising insofar that countably many functions, namely the rational multiples of sin(mx) and cos(mx), where m takes values in the integers, are enough to express any other function, of which there are uncountably many.
The solutions to various important differential equations can be interpreted in terms of Hilbert spaces. For example, a great many fields in physics and engineering lead to such equations and frequently solutions with particular physical properties are used as basis functions, often orthogonal, that serve as the axes in a corresponding Hilbert space.
As an example from physics, the time-dependent Schrödinger equation in quantum mechanics describes the change of physical properties in time, by means of a partial differential equation determining a wavefunction. Definite values for physical properties such as energy, or momentum, correspond to eigenvalues of an associated (linear) differential operator and the associated wavefunctions are called eigenstates. The spectral theorem describes the representation of linear operators that act upon functions in terms of these eigenfunctions and their eigenvalues.
#### Distributions
Distributions (or "generalized functions") are a powerful instrument to solve differential equations, and are exceeding the cadre of Hilbert spaces. Concretely, a distribution is a map assigning a number to any function in a given vector space. A standard example is given by integrating the function over some domain Ω:
$f mapsto int_Omega f\left(x\right)dx$
The great use of distributions stems from the remark that standard analytic notions such as derivatives can be generalized to distributions. Thus differential equations can be solved in the distributive sense first. This can be accomplished using Green's functions. Then, the found solution can in some cases be proven (e.g. using the Riesz representation theorem) to be actually a true function.
### Algebras over fields
In general, vector spaces do not possess a multiplication operation. (An exceptional case are finite-dimensional vector spaces over finite fields, which turn out to be (finite) fields, as well.) A vector space equipped with an additional bilinear operator defining the multiplication of two vectors is an algebra over a field. An important example is the ring of polynomials F[x] in one variable, x, with coefficients in a field F, or similarly with several variables. In this case the multiplication is both commutative and associative. These rings, and their quotient rings form the basis of algebraic geometry, because they are the rings of functions of algebraic geometric objects.
Another crucial example are Lie algebras, which are neither commutative, nor associative, but the failure to be so is measured by the constraints ([x, y] denotes multiplication of x and y):
[x, y] = −[y, x] and [x, [y, z]] + [y, [x, z]] + [z, [x, y]] = 0
The standard example is the vector space of n-by-n matrices, setting [x, y] to be the commutator xy − yx. Lie algebras are intimately connected to Lie groups. A special case of Lie algebras are Poisson algebras.
### Ordered vector spaces
An ordered vector space is a vector space equipped with an order ≤, i.e. vectors can be compared. Rn can be ordered, for example, by comparing the coordinates of the vectors. Riesz spaces present further key cases.
## Generalizations
### Modules
Modules are to ring (mathematics) what vector spaces are to fields, i.e. the very same axioms, applied to a ring R instead of a field F yield modules. In contrast to the good understanding of vector spaces offered by linear algebra, the theory of modules is in general much more complicated. This is due to the presence of elements r ∈ R that do not possess multiplicative inverses. For example, modules need not have bases as the Z-module (i.e. abelian group) Z/2 shows; those modules that do (including all vector spaces) are known as free modules.
### Vector bundles
A family of vector spaces, parametrised continuously by some topological space X, is a . More precisely, a vector bundle E over X is given by a continuous map
π : E → X,
which is locally a product of X with some (fixed) vector space V, i.e. such that for every point x in X, there is a neighborhood U of x such that the restriction of π to π−1(U) equals the projection V × U → U . The case dim V = 1 is called a line bundle. The interest in this notion comes from the fact that while the situation is simple to oversee locally, there may be global twisting phenomena. For example, the Möbius strip can be seen as a line bundle over the circle S1 (at least if one extends the bounded interval to infinity). The (non-)existence of vector bundles with certain properties can tell something about the underlying space X. For example, over the 2-sphere S2, there is no tangent vector field which is everywhere nonzero, as opposed to the circle S1. The study of all vector bundles over some topological space is known as K-theory. An algebraic counterpart to vector bundles are locally free modules, which—in the guise of projective modules—are important in homological algebra and algebraic K-theory.
### Affine and projective spaces
Affine spaces can be thought of being vector spaces whose origin is not specified. Formally, an affine space is a set with a transitive vector space action. In particular, a vector space is an affine space over itself, by the structure map
V2 → V, (a, b) ↦ a − b.
Sets of the form x + Rm (viewed as a subset of some bigger Rn), i.e. moving some linear subspace by a fixed vector x, yields affine spaces, too.
The set of one-dimensional subspaces of a fixed vector space V is known as projective space, an important geometric object formalizing the idea of parallel lines intersecting at infinity. More generally, the Grassmann manifold consists of linear subspaces of higher (fixed) dimension n. Finally, flag manifolds parametrize flags, i.e. chains of subspaces (with fixed dimension)
0 = V0 ⊂ V1 ⊂ ... ⊂ Vn = V
## See also
• Vector (geometry), for vectors in physics
• Vector field
• Vector spaces without fields
• Coordinates (mathematics)
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http://mathoverflow.net/questions/116051?sort=newest | ## Localizations of non-nilpotent spaces
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For simplicity let's talk about `$p$`-localizations of spaces for a fixed prime `$p$`. Every space `$X$` has a well-defined `$p$`-localization which can be constructed by the small object argument and which becomes a fibrant replacement in the `$p$`-local model structure on the category of spaces. It is well-known that nilpotent spaces have nice enough Postnikov towers and we can localize such spaces by taking the Postnikov tower, localizing step by step and putting it back together by taking the limit of the resulting tower of fibrations. My question is:
Is there an example of a non-nilpotent space `$X$` whose `$p$`-localization we can explicitly describe?
I leave the meaning of "explicitly" ambiguous. I would be interested in any construction not using the small object argument.
Here's my stab at a possible example. For a group `$G$` we define its lower central series by setting `$G_0 = G$` and `$G_{n + 1} = [G_n, G]$` and we can also continue transfinitely by setting `$G_\beta = \bigcap_{\alpha < \beta} G_\alpha$` for limit ordinals `$\beta$`. The group `$G$` is nilpotent if this construction terminates at the trivial subgroup at a finite stage. It is called hypocentral if it terminates at the trivial subgroup at some not necessarily finite stage. According to Wikipedia it is a result of Malcev that there are hypocentral groups with arbitrarily long lower central series.
If we start with a hypocentral group `$G$` and convert its lower central series into a (transfinite) tower of fibrations (whose limit is a `$K(G, 1)$`), `$p$`-localize it step by step and take the limit of the resulting tower of fibrations, do we obtain the `$p$`-localization of `$K(G, 1)$`?
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## 1 Answer
As you know extending $P$-localization functors to non-nilpotent spaces is a very delicate matter. You can do it by homotopical localization techniques. if you do it in this way you get functorial localization which work for non-nilpotent spaces "but" it is the aim of your question it is not "very explicit".
Now let me be very naive and let us go back to D. Sullivan's MIT notes "Geometric topology, localization, periodicity and Galois symmetry" chapter 2 (these notes are really beautiful to read). D. Sullivan first explains how to do $P$-localization when we have a CW-complex, of course he also explains that this CW-procedure gives a "nice localization" when the CW-complex is 1-connected (in that case it is our modern $P$-localization). He proceeds on the skeleton of the CW-complex, localizing cell by cell. Thus you can apply Sullivan's telescope construction to a circle and then to a wedge of circles and any CW-complex you want. The starting point being the localization of spheres through a telescope of maps $S^k\stackrel{\times p}{\rightarrow }S^k$ (you invert degree $p$-maps).
The point is that you can argue that this naive construction on non-nilpotent CW-complexes is not what you expected as a $P$-localization. D. Sullivan also explains how to do it when we have a Postnikov tower.
Let me add that there is another interesting way to $P$-localize which is due to C. Casacuberta and G. Peschke: "Localizing with respect to self-maps of the circle" Trans AMS 339 (1993) 117 – 140.
This $P$-localization is as "explicit" as a homological Bousfield's localization but the homotopy related to it is very interesting. In that setting, a space is $P$-local if and only if the $p$-power map on the loop space $\Omega X$ is a self homotopy equivalence if $p\notin P$.
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http://mathoverflow.net/questions/98881/connectedness-of-the-linear-algebraic-group-so-n | ## Connectedness of the linear algebraic group SO_n
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I apologize in advance if my question is too elementary for MO.
It is a well known fact that the linear algebraic group $G = \mathsf{SO}_n$ is connected, and there exist a few different proofs of this fact. One proof goes by showing that $G$ is generated by unipotent elements, and invoking the theorem that every linear algebraic group with this property is connected.
My question is about a different, more direct proof, involving the Cayley transform $$A \mapsto (I_n + A)^{-1} (I_n - A) ,$$ which maps every matrix $A \in G(k)$ for which $I_n + A$ is invertible (let's write $W$ for the set of such matrices), to a skew-symmetric matrix, and in fact, this map defines an isomorphism of varieties between the non-empty open subset $W$ of $G$, and an open subset of the irreducible variety $V$ of skew-symmetric matrices.
But how does one now conclude that $\mathsf{SO}_n$ is connected? In particular, why is the closure of $W$ equal to $G$? Of course the complement of $W$ is given by the polynomial equation $\det(I_n + A) = 0$, but how is this situation different from, for instance, the fact that $\mathsf{SO}_n$ is the subvariety of $\mathsf{O}_n$ defined by the polynomial equation $\det(A) - 1 = 0$ (or $\det(A) + 1 \neq 0$) whereas $\mathsf{O}_n$ is not connected?
What subtlety am I missing?
added:
1. I'm assuming $\operatorname{char}(k) \neq 2$.
2. The underlying topology is the Zariski toplogy.
3. The proof I'm mentioning in the second paragraph is only valid in characteristic $0$.
4. It seems that the question is not too elementary...
added: I'll add a few more sentences to make clear what I mean by connected in the Zariski topology.
I'm considering $G$ as a group functor (i.e. a functor from the category of commutative $k$-algebras to the category of groups). Then $G$ is connected if and only if its coordinate algebra $k[G] = \mathcal{O}(G)$ has no non-trivial idempotents. In particular, in order to show that $G$ is connected, we may assume w.l.o.g. that $k$ is algebraically closed. [In fact, since $G$ is smooth (since $\operatorname{char}(k) \neq 2$), we may instead consider the group of $\bar{k}$-rational points $G(\bar{k})$, which brings us back to a more classical situation.]
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I am a little confused by the question: Skew matrices form an affine space, and your set of nonsingular matrices is an affine hypersurface (which is presumably easy to show is irreducible). $O_n$ is a much more complicated variety, so this Cayley transform trick seems to have simplified your life. – Igor Rivin Jun 5 at 17:54
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Presumably the OP means connected as a scheme. – Igor Rivin Jun 5 at 20:10
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I'm slightly confused by your second paragraph: the finite group $\mathbf{Z}/p\mathbf{Z}$ may be viewed as a linear algebraic group. Over a field of char. p, it is generated by unipotent elements. But it is not connected. – George McNinch Jun 5 at 23:39
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@Robert, George: I think Tom was starting out in characteristic 0, which is one reason for my attempted clarifications below. – Jim Humphreys Jun 6 at 0:28
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@Tom: I have to confess that I'm still puzzled. It appears that you are only avoiding characteristic $2$, so that you would appear to be claiming that $\mathsf{SO}_2(k)$ is connected in the Zariski topology even when $k$ is finite, and I don't understand this. After all, when $k=\mathbb{Z}_3$, doesn't ${\mathsf{SO}}_2(k)$ have $4$ elements and isn't the complement of any element open? Why then is this group not disconnected? Looking at the answers below, it seems that you at least need $k$ to be infinite for those kinds of arguments to work. – Robert Bryant Jun 6 at 18:53
show 6 more comments
## 2 Answers
Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so the other would have a lower dimension.)
The subvariety where $I_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.
The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO_{n-2}$-bundle on $G_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.
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Thanks a lot! The dimension argument is indeed what I was missing. I'm relieved to see that the answer is not as obvious as I feared. – Tom De Medts Jun 6 at 8:15
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
[EDIT: I've tightened my wording and added a couple of references which I went back to out of curiosity.]
Will's answer has the elements needed for a concrete reply to the question, but the question itself has caused some confusion about the setting and terminology which are worth clarifying. First of all, the underlying field should be of characteristic different from 2, since it gets more subtle to talk about quadratic forms and orthogonal groups in characteristic 2. (This is done however in work of Chevalley and Borel in the algebraic groups framework, where the groups are included in the classification.)
Originally the study of orthogonal groups as Lie groups was carried out by Weyl, Chevalley, and many others. Here the (polynomial) condition on `$n \times n$` matrices is just that the transpose of a matrix must equal its inverse. The orthogonal matrices then form a compact real Lie group `$\mathrm{O}(n)$` or a noncompact complex Lie group `$\mathrm{O}(n, \mathbb{C})$` of dimension `$n(n-1)/2$`. In the euclidean topology, the latter group is homeomorphic to the former group times a vector space. So connectedness questions can be settled in the compact case.
Since eigenvalues of an orthogonal matrix occur along with their inverses, `$\det=\pm 1$` and matrices of det `$-1$` form a closed normal subgroup `$\mathrm{SO}(n)$` or `$\mathrm{SO}(n, \mathbb{C})$` giving in Lie theory the rank `$\ell$` series: `$B_\ell$` with `$n=2\ell+1$` odd, `$D_\ell$` with `$n=2\ell$` even. It's worth following the case `$n=5$` in Will's calculations. To show that the compact group is connected in the topological group setting, Chevalley in Theory of Lie Groups uses induction on `$n$` and the characterization of the successive quotients as spheres.
Now in the Chevalley-Borel setting of linear algebraic groups (over an algebraically closed field `$K$`), much of the previous study carries over with modifications. For linear algebraic groups given the Zariski topology, irreducibility of the underlying variety fortunately coincides with connectedness in that coarse topology; the term "connected" is then preferred. The irreducible (= connected) components of an algebraic group `$G$` are disjoint and equidimensional as well as finite in number (unlike some Lie groups): these are just the cosets of the identity component `$G^\circ$`. We denote the points of the group over `$K$` as `$\mathrm{SO}_n(K)$`, but the scheme language probably adds nothing useful to the study of connectedness here.
The most standard elementary way to show that a linear algebraic group is connected is to show that it is generated by suitable irreducible subsets such as closed connected subgroups. For the classical matrix groups, this is usually done by showing that the group is generated by transvections, hence by connected 1-dimensional unipotent groups. With some care this approach even handles special orthogonal groups in characteristic 2.
On the other hand, the question here raises the possibility of appealing (in characteristic not 2) to a Cayley transform. Here one is able to map isomorphically a nonempty open subset of an affine space (dense in the Zariski topology) onto a nonempty open subset of the matrix group in a concrete way. Then it has to be seen, as Will shows, that none of the hypothetical extra irreducible/connected components of `$\mathrm{SO}_n(K)$` can lie in the excluded hypersurface given by nonvanishing of a determinant. Dimension counting seems necessary here.
The only source I can quote for this slightly esoteric approach is a terse exercise 2.2.2(2) in Springer's book Linear Algebraic Groups, where much is left to the reader's ingenuity. (Are there earlier sources?) Springer himself was attracted to this approach, I think, because he used the Cayley transform for classical groups to realize an isomorphism between unipotent and nilpotent varieties in the group and its Lie algebra.
Earlier arguments appear in at least two places. [Note in each case that for the standard structure theory (over an arbitrary field) involving an isotropic split torus in diagonal form, orthogonal groups are written as matrices using an orthogonal direct sum of hyperbolic planes; over `$K$` this translates to the conventional format above.]
1) Chevalley's 1956-58 seminar Classification des groupes algebriques semi-simples (typeset text, Springer 2005, edited by Cartier). In Expose 22, Chevalley gives an argument for connectedness of some of the linear groups roughly analogous to the inductive argument for compact Lie groups.
2) The second edition of Borel's original notes Linear Algebraic Groups (Springer GTM 126, 1991). In the added Section 23 he discusses examples involving groups of rational points of various classical groups, observing in particulqr that over `$K$` the relevant groups are Zariski-connected. (Characteristic 2 requires as usual extra discussion, as does type `$D_\ell$`.) Here the argument relies on the standard structure theory, showing in effect that a hypothetical coset representative for `$G/G^\circ$` must in fact represent an element of the Weyl group and thus lie in `$G^\circ$`.
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What a great answer! I wish I could accept more than one answer in MO, since the combination of Will's answer and yours is marvelous. One of the sources where I learned about the Cayley transform was indeed Springer's Exercise 2.2.2(2), and the fact that it appeared as an exercise was the main reason for my apologetic first paragraph. – Tom De Medts Jun 6 at 8:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 56, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9477882981300354, "perplexity_flag": "head"} |
http://cs.stackexchange.com/questions/7091/why-is-3n-2on-true | # Why is $3^n = 2^{O(n)}$ true?
$3^n = 2^{O(n)}$ is apparently true. I thought that it was false though because $3^n$ grows faster than any exponential function with a base of 2.
How is $3^n = 2^{O(n)}$ true?
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Beware the abuse of notation! – Raphael♦ Dec 3 '12 at 8:11
Really I can't understand what does $3^n = 2^{O(n)}$ mean? first I changed it to $3^n \in 2^{O(n)}$, after that I saw again this is meaningless. IMO question is meaningless. – Saeed Amiri Dec 3 '12 at 8:40
## 3 Answers
With some algebra (and changing the constant in the $O(n)$), we can actually change the bases.
$$3^n = (2^{\log_2 3})^n = 2^{n\log_2 3}$$
Since $\log_2 3$ is a constant, $n\log_2 3 = O(n)$. So $3^n = 2^{O(n)}$.
I'm not sure what you mean by "$3^n$ grows faster than any exponential function with a base of 2." $2^n = o(3^n)$ of course but it seems you mean something more general. My guess is that your statement applies to something like $O(3^n)$, where you multiply the base by a constant, as opposed to $2^{O(n)}$ where you multiply the number in the exponent by a constant.
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$3^n$ grows faster than any exponential function with a base of $2$.
True. This implies that $3^n = O(2^n)$ cannot be true. But what you have here is $2^{O(n)}$.
Recall that $O(f(n))$ is really a set of functions, and strictly speaking we should be writing $3^n \in 2^{O(n)}$ (or even $(n \mapsto 3^n) \in 2^{O(n \mapsto n)}$). The right-hand side is not the exponential of a function, but the exponential of a set of functions. Expanding the definition of big oh:
$$2^{O(n)} = 2^{\{f \;\mid\; \exists N, \exists p, \forall n \ge N, f(n) \le p n\}} = \{(n \mapsto 2^{f(n)}) \mid \exists N, \exists p, \forall n \ge N, f(n) \le p n\}$$
Since the exponential function $n \mapsto 2^n$ is increasing, we can lift the inequality out of the exponential:
$$2^{O(n)} = \{g \mid \exists N, \exists p, \forall n \ge N, g(n) \le 2^{p \, n}\}$$
Contrast with $$O(2^n) = \{g \mid \exists N, \exists k, \forall n \ge N, g(n) \le k \, 2^{n}\}$$
In $2^{O(n)}$, the multiplicative constant is inside the exponential. In $O(2^n)$, it is multiplied by the exponential. $2^{p \, n} = 2^p 2^n$, so we have (for any $n \ge 0$) $3^n \le 2^{\log_2 3} 2^n$, i.e. we can take $N = 0$ and $p = \log_2 3$, showing that $3^n \in 2^{O(n)}$.
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$3^n=2^{O(n)}$ is in fact true because if you recall $O(n)$ definition you will see that you can add/multiply by any constant. So:
$3^n < 2^{kn}$ // $\log_2$
$n\log_2(3^n) < kn\log_2(2)$
$k > \log_2(3)$
So as you can see $2^{kn}$ is bigger then $3^n \iff k > \log_2(3)$
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http://math.stackexchange.com/questions/273133/calculating-a-continuously-varying-continuously-paid-annuity?answertab=votes | # Calculating a continuously varying, continuously paid annuity
I'm unsure how to calculate a continuously varying, continuously paid annuity. I'll write up my solution (which I suspect is wrong) to one, sample question, and I would greatly appreciate any correction.
The question (problem 29.13 on page 273):
Payments are made to an account at a continuous rate of $(8k+tk)$, where $0\le t\le10$. Interest is credited at a force of interest $\delta_t=\frac1{8+t}$. After 10 years, the account is worth 20,000. Calculate $k$.
My attempt at a solution:
The accumulation function is $e^{\int_0^t(8+s)^{-1}ds}$, which comes to $e^{\left.\ln\left|s+8\right|\right|_0^t}=\frac{t+8}8$, so that the discount (inverse-accumulation) function (to evaluate the present ($t=0$) value) is $\frac8{t+8}$.
Then the present value is the limit of sums of $(\textrm{discount})\times(\textrm{payment})$ i.e. $\int_0^{10}(8k+tk)\frac8{t+8}dt=80k$.
The accumulated value is then $80k\times(\textrm{accumulation(10)})=80k\frac{18}8$; since that's given as $20000$, we have $k=20000/180\approx111.11$.
Could anyone post the right solution or explain what (if anything) is wrong with mine, please?
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I can tell you that Finan has an answer key (no solutions, just answers) and his answer is 111.11. I don't have much time these days to look into it further! – Graphth Jan 12 at 13:58
@Graphth Thank you! I didn't know about his key. Where is it? – user55618 Jan 13 at 0:20
It's not publicly available. I have it because I taught a class from his book. If you are taking a class that uses his book, then your teacher may not want you to have it. So, he doesn't give it out to everyone. – Graphth Jan 13 at 12:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9793162941932678, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/116120/classical-solutions-of-pde-with-mixed-boundary-conditions | # classical solutions of PDE with mixed boundary conditions
Nowadays people usually consider PDEs in weak formulations only, so I have a hard time finding statements about the existence of classical solutions of the Poisson equation with mixed Dirichlet-Neumann boundary conditions.
Maybe someone here can help me and point to a book or article where I can find sufficient conditions on the right hand side that guarantee the existence of a $C^2$ solution.
Note that the unique solvability in $H^1(\Omega)$ follows, e.g., from the paper on discrete maximum principles by Karatson and Korotov in Numer. Math. 99 (2005), 669-698. So (a slightly more precise version of) my question boils down to finding conditions such that this solution is actually C^2 in the interior and continuous differentiable at the boundary.
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WWelcome to MSE. I am adding a reference request tag to your question. – user21436 Mar 3 '12 at 22:08
– Christian Clason Nov 28 '12 at 20:11
## 2 Answers
I believe the papers by Gary Lieberman will answer your question: one is Mixed boundary value problem for elliptic and parabolic differential equations of second order. Another one is Optimal holder regularity for mixed boundary value problems. Thanks
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Thanks. These and especially their reference to Azzam and Kreyszig are very useful! – Arnold Neumaier Dec 2 '12 at 14:11
From my rather primitive knowledge of PDEs, for a well-posed mixed boundary value problem for Poisson equation, I think the effect of the Neumann boundary condition on the regularity of the solution is equivalent to Dirichlet boundary condition of one less differentiability.
Intuitively speaking(most likely I am wrong here), if the uniqueness of the mixed problem can be established(via maximum principle for example), and enough regularity has been assumed for the $u$ to be well-defined on $\Gamma_N$, then it is same that we have $u = H g_N$ on $\Gamma_N$ where $H$ is the Neumann-to-Dirichlet map. If we know all the Dirichlet and Neumann data, the regularity of the solution depends on the regularity of the boundary like the classical theory states.
For reference, in the book Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger, they use an entire chapter six to discuss the classical Schauder estimate for Hölder continuous solutions with various boundary conditions, in Section 6.7 of that book, Lemma 6.27 reads the solution to the Neumann problem of Poisson equation is $C^{2,\alpha}(\bar{\Omega})$-regular if Neumann data $g_N$ is $C^{1,\alpha}$, $f$ is $C^{0,\alpha}$, and the domain is $C^{2,\alpha}$. Now back to the regularity estimate for the Dirichlet problem in Theorem 6.19, if Dirichlet data $g_D$ is $C^{2,\alpha}$, other the conditions are the same, we will get the same $C^{2,\alpha}$-regularity for the solution. Hence I am guessing for mixed boundary problem we should have a similar result that a $C^2(\Omega)\cap C^{1,1}(\bar{\Omega})$ solution should require a $C^2$ domain, $C^1$ Neumann boundary data, and $C^2$ Dirichlet boundary data.
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Thanks. I hadn't realize before that the pure Neumann case is in Gilbarg/Trudinger which I generally know quite well. I still hope to get info on the mixed case from somewhere.... – Arnold Neumaier May 24 '12 at 12:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9005967378616333, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/1263/how-can-you-focus-sound/1501 | # How can you focus sound?
I saw this TED talk and I am curious as to how the sound is focused on the general level. Can anyone explain this or does anyone have any good articles?
-
3
It's worth noting that what he is doing is not sound focusing. He is creating audible sound from high frequency unaudible sound through acoustic non-linear process. An optical similar phenomenon called four-wave mixing does the same. – Bernardo Kyotoku Nov 24 '10 at 11:30
## 4 Answers
I don't think anyone here has really answered your question. In this case, the sound is "focused" using phased arrays. The face of the audio spotlight has multiple transducers:
Flickr
The same signal is output from each of them, but delayed slightly by different amounts, so that the wavefronts all reach the same point in front of the device at the same time. This "virtual focus" is called beamforming.
ref ref
This is how modern radars focus their beams, too. Instead of spinning a satellite dish around, they have lots of little elements that don't move, but the signals are delayed to produce different beam shapes.
-
That's actually diffraction, not focusing. – ptomato Dec 1 '10 at 9:02
Diffraction? Are you thinking of a two-slit experiment? – endolith Dec 1 '10 at 14:36
2
@ptomato- the individual beams diffract as any beam would... but the effect of introducing a variable phase delay between the beams to create a new wavefront is analogous to a wavefront passing through a lens, acquiring phase shift according to the length of each pass – Pete Dec 10 '10 at 5:20
each pass (typo)--> each path – Pete Dec 11 '10 at 4:25
On the general level, you focus sound the same way you focus light -- either by reflecting it from a parabolic surface, or letting it pass through an acoustic lens. An acoustic lens is just like an optical lens in that it consists of a material with a different propagation speed of sound, with varying thickness. See the Wikipedia article on acoustic mirrors.
-
2
You may also focus light using diffractive lenses – belisarius Nov 24 '10 at 12:26
As with sound, but I wanted to keep the answer simple ;-) – ptomato Nov 24 '10 at 14:59
– endolith Apr 19 '12 at 1:45
These ‘audio spotlights’ work by emitting ultrasound at two different frequencies; it is the short wavelength of the ultrasound that causes the beam to be so directed. The two waves interfere and produce sum and different tones at frequencies of $f_1+f_2$ and $f_1-f_2$; if the ultrasound frequencies are, say, $f_1=45\,$kHz and $f_2=44\,$kHz, the difference tone will be at $1$kHz which is in the audible range for humans.
I knew some people who looked at using audio spotlights for noise control some time ago, but (IIRC) the general consensus is that since you're subjecting your victims with very large amounts of ultrasonic noise (greater than 100dB), these devices are probably not too safe for continuous use. (Or, at least, their safety was in no way assured.)
-
Sound is a type of wave, so it has all the wave properties similar to other waves such as light waves. For light waves, you can use a lens to focus the light. A lens has higher refractive index, or lower light speed than the environment. The same is true for sound wave, so what you need is to make a high refractive region [1].
The air surrounding us can be approximated by the ideal gas, so the speed of sound is [2]
$c=\sqrt{\gamma\frac{P}{\rho}}$
where $\gamma$ is the adiabatic index, $p$ is pressure of the air, $\rho$ is density of the air
Here, we want to create a region with high refractive, or equivalently low sound speed. There are few way to achieve this, one is to decrease the pressure, another way is to decrease the temperature (by the ideal gas law $PV=NRT$). However, in both cases, you either need a hard container or a refrigerator near it to keep it cold.
On the other hand, increasing the density can be easily done by using a heavy gas such as carbon dioxide. You just need to fill the gas in a balloon and it can act as a very simple acoustic lens. Note that the size of the balloon or other container must be large compared with the wavelength. There are also other methods to focus sound without using lens. [3]
As said before, the same mechanism can be applied for other wave, for example, a water wave. In a shallow water tank, adding a lens shape obstacle at the bottom can converge water wave because water wave move slowly at the shallow region. This experiment can be easily performed in one's home.
[1] http://hyperphysics.phy-astr.gsu.edu/hbase/sound/refrac.html
[2] http://en.wikipedia.org/wiki/Speed_of_sound#Speed_in_ideal_gases_and_in_air
[3] http://focus.aps.org/story/v14/st3
-
I think building reflectors would be easier than lenses – endolith Apr 18 '12 at 21:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9452102184295654, "perplexity_flag": "middle"} |
http://blog.eqnets.com/2009/08/26/a-minimal-periodic-coloring-theorem-part-3/ | # Equilibrium Networks
Science, networks, and security
## A minimal periodic coloring theorem, part 3
Last time I gave nontrivial examples of periodic colorings of $A_3,$ and demonstrated that the cyclic coloring of $A_{n-1}$ exists for all $n.$ More generally for $n$ composite, let $\{ \sigma_{1k} \}_{k=1}^n \cong G$ for a group $G$ of order $n.$ (By Cayley’s theorem, every finite group can be realized in this way.) Then $\Gamma_\sigma \cong G.$ In other words, every finite group of order $n$ can be realized as a minimal periodic coloring of $A_{n-1}.$
It’s not immediately clear to me that every minimal periodic coloring arises in this way, but I would bet on it. (The first person who can demonstrate this is welcome to claim \$.50 and a Sprite as reward, and an equal reward goes for the first person who can provide some kind of actual picture of what happens with a nonabelian example, say using $S_3$ to color $A_5.$)
Making the connection practically is a bit more involved. I did the simple thing and used permutation matrices to look for faithful representations of finite groups. First up is the code to produce a product of two permutations:
```function y = permprod(s1,s2)
% Product of permutations.
% s1 and s2 represent permutations in $S_n$ as arrays (akin to the two-row
% notation and NOT cycle decomposition).
% This function maps s1 and s2 to permutation matrices, computes their
% product, and returns the corresponding array.
if size(s1,1)*size(s2,1) - 1
'needs row arrays'
return;
elseif size(s1,2)-size(s2,2)
'sizes must match'
return;
else
n = size(s1,2);
end
I = eye(n);
% permutation matrices
m1 = I(s1,:); m2 = I(s2,:);
m = m1*m2;
% now deduce the array
for j = 1:n
s(j) = find(m(j,:),1);
end
y = s;```
and next the inverse of a permutation:
```function y = perminv(s1)
% Inverse permutation.
% s1 represents permutation in $S_n$ as an array (akin to the two-row
% notation and NOT cycle decomposition).
% This function maps s1 to a permutation matrix, computes its inverse,
% and returns the corresponding array.
if size(s1,1) - 1
'needs row array'
return;
else
n = size(s1,2);
end
I = eye(n);
% permutation matrix
m1 = I(s1,:);
m = inv(m1);
% now deduce the array
for j = 1:n
s(j) = find(m(j,:),1);
end
y = s;```
and finally a very crude way to see which choices of permutations produce coherent color shift groups and which don’t:
```function y = colorshiftcheck(varargin)
% Consider elements sjk (or $\sigma_{jk}$) of $S_n$ for 1 ≤ j,k ≤ n and
% subject to the following defining relations:
% 1) sjk*skj = e
% 2) sjk*skl = sjl
% This code starts with $n-1$ permutations s12,...,s1n and proceeds to
% determine whether 1) and 2) hold
% inputs are s12,...,s1n
n = length(varargin)+1;
% get the s1j and check their size, then compute the inverses sj1
for j = 2:n
temp = varargin(j-1);
s{1,j} = temp{1};
if size(s{1,j},2)-n
j-1
'th entry length incorrect'
break;
elseif size(s{1,j},1)-1
j-1
'th entry has too many rows'
break;
end
s{j,1} = perminv(s{1,j});
end
s{1,1} = 1:n;
% form the entries sjk=sj1*s1k
for j = 2:n
for k = 2:n
s{j,k} = permprod(s{j,1},s{1,k});
end
end
% now check to see that the assignments s{j,k} are consistent:
for j = 2:n
for k = 2:n
if sum(abs( perminv(s{k,j}) - s{j,k} ))
'inverse failure at'
[j k]
end
for l = 2:n
if sum(abs( permprod(s{j,l},s{l,k}) - s{j,k} ))
'product failure at'
[j k l]
end
end
end
end
y = s;```
There are probably much more sophisticated ways to do this sort of thing, but the overall idea should be clear: you can specify a symmetry group you want to be reflected in a minimal periodic coloring, and combine these if you want nonminimal periodic colorings (another \$.50 and a Sprite go to the first person who demonstrates that every nonminimal periodic coloring is either obtained in this way or provides a counterexample). For example, combining the two colorings shown last time produces a periodic 8-coloring of $A_3.$
The reason we looked at this sort of thing at Equilibrium is (briefly) that it is possible to make individual packets correspond to vectors of the form $-e_j + e_k$, where the standard basis is indicated, and $j, k$ are respectively one of $n$ source or destination attributes that characterize the packet. These sorts of vectors generate $A_{n-1}.$ (For reasons outlined in a whitepaper of ours [see our website] $n$ can generally be taken to be quite small.) The idea then was to take the induced trajectory (here, the sequence of colors) corresponding to a sequence of observed packets and use various tools from probability to characterize the network traffic (a real explanation of this will have to wait for future posts, as will an explanation of how we can do something along these lines without using the coloring theorem).
### Like this:
This entry was posted on Wednesday, August 26th, 2009 at 22:40 and is filed under Communications security, Equilibrium Networks, Mathematics, Nonrandom bits. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### 2 Responses to A minimal periodic coloring theorem, part 3
1. [...] translating network packets to random walks was similar to the approach described at the end of a previous post, except that the finite space is used instead of the infinite root lattice ), and we seriously [...]
2. Mary says:
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154 views
### Exchange operator in terms of rotation operator
I have studied about exchange operators and rotation operators and I know that an exchange between 2 particles in a combined state is the same as rotating each particle 180 degrees (according to ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.930717945098877, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/69610/list | ## Return to Answer
2 added 88 characters in body
The intuition I have is that relations definable by polynomials is complicated enough that projections can give rise to very complicated sets. The essential point here is not exponential growth, that comes from the unboundedness of the quantifiers, the essential point is that the graph of a very fast growing function can be represented by a polynomial.
On the other hand, if the question is about how polynomial equations can represent graphs of complicated function, then I guess the best explanation is by the people who came up with the idea of how to use these equations to represent the graph of a fast growing function, you which can find about it in My Collaboration with JULIA ROBINSON by Yuri MATIYASEVICH (skip to the line numbered (6) in the text):text) or in his book.
"I saw at once that Julia Robinson had a fresh and wonderful idea. It was connected with the special form of Pell's equation
(6) $$x^2-(a^2-1)y^2 = 1.$$
Solutions $<\chi_0, \psi_0>, <\chi_1, \psi_1>,\cdots, <\chi_n, \psi_n>,\cdots$ of this equation listed in the order of growth satisfy the recurrence relations
(7) $$\chi_{n+1}=2a\chi_n-\chi_{n-1}$$
$$\psi_{n+1}=2a\psi_n-\psi_{n-1}$$
It is easy to see that for any $m$ the sequences $\chi_0,\chi_1,\cdots, \psi_0,\psi_1,\cdots$ are purely periodic modulo $m$ and hence so are their linear combinations. Further, it is easy to check by induction that the period of the sequence
(8) $$\psi_0,\psi_1,\cdots,\psi_n,\cdots (\mod a-1)$$
is
(9) $$0, 1, 2,\cdots, a - 2,$$
whereas the period of the sequence
(10) $$\chi_0-(a-2)\psi_0,\chi_1-(a-2)\psi_1,\cdots, \chi_n-(a-2)\psi_n,\cdots (\mod 4a-5)$$
begins with
(11) $$2^0, 2^1, 2^2,\cdots$$
The main new idea of Julia Robinson was to synchronize the two sequences by imposing a condition $G(a)$ which would guarantee that
(12) $$\text{the length of the period of (8) is a multiple of the length of the period of (10).}$$
If such a condition is Diophantine and is valid for infinitely many values of $a$, then one can easily show that the relation $a = 2^c$ is Diophantine. Julia Robinson, however, was unable to find such a $G$ and, even today, we have no direct method for finding one.
I liked the idea of synchronization very much and tried to implement it in a slightly different situation. When, in 1966, I had started my investigations on Hilbert's tenth problem, I had begun to use Fibonacci numbers and had discovered (for myself) the equation
(13) $$x^2 - xy - y^2=\pm 1$$
which plays a role similar to that of the above Pell's equation; namely, Fibonacci numbers $\phi_n$ and only they are solutions of (13). The arithmetical properties of the sequences $\psi_n$ and $\phi_n$ are very similar. In particular, the sequence
(14) $$0, 1, 3, 8, 21, \cdots$$
of Fibonacci numbers with even indices satisfies the recurrence relation
(15)$$\phi_{n+1}=3\phi_n-\phi_{n-1}$$
similar to (7). This sequence grows like $[(3+\sqrt 5)/2]^n$ and can be used instead of (11) for constructing a relation of exponential growth. The role of (10) can be played by the sequence
(16) $$\psi_0,\psi_1,\cdots,\psi_n,\cdots (\mod a-3)$$
because it begins like (14). Moreover, for special values of a the period can be determined explicitly; namely, if
(17) $$a = \phi_{2k}+\phi_{2k+2},$$
then the period of (16) is exactly
(18) $$0,1,3,\cdots,\phi_{2k},-\phi_{2k},\cdots,-3,1.$$
The simple structure of the period looked very promising.
I was thinking intensively in this direction, even on the night of New Year's Eve of 1970, and contributed to the stories about absentminded mathematicians by leaving my uncle's home on New Year's Day wearing his coat. On the morning of January 3, I believed I had found a polynomial B as in (5) but by the end of that day I had discovered a flaw in my work. But the next morning I managed to mend the construction.
What was to be done next? As a student I had had a bad experience when once I had claimed to have proved unsolvability of Hilbert's tenth problem, but during my talk found a mistake. I did not want to repeat such an embarrassment, and something in my new proof seemed rather suspicious to me. I thought at first that I had just managed to implement Julia Robinson's idea in a slightly different situation. However, in her construction an essential role was played by a special equation that implied one variable was exponentially greater than another. My supposed proof did not need to use such an equation at all, and that was strange. Later I realized that my construction was a dual of Julia Robinson's. In fact, I had found a Diophantine condition $H(a)$ which implied that
(19) $$\text{the length of the period of (16) is a multiple of the length of the period of (8).}$$
This $H$, however, could not play the role of Julia Robinson's $G$, which resulted in an essentially different construction."
1
The intuition I have is that relations definable by polynomials is complicated enough that projections can give rise to very complicated sets. The essential point here is not exponential growth, that comes from the unboundedness of the quantifiers, the essential point is that the graph of a very fast growing function can be represented by a polynomial.
On the other hand, if the question is about how polynomial equations can represent graphs of complicated function, then I guess the best explanation is by the people who came up with the idea, you can find about it in My Collaboration with JULIA ROBINSON by Yuri MATIYASEVICH (skip to the line numbered (6) in the text):
"I saw at once that Julia Robinson had a fresh and wonderful idea. It was connected with the special form of Pell's equation
(6) $$x^2-(a^2-1)y^2 = 1.$$
Solutions $<\chi_0, \psi_0>, <\chi_1, \psi_1>,\cdots, <\chi_n, \psi_n>,\cdots$ of this equation listed in the order of growth satisfy the recurrence relations
(7) $$\chi_{n+1}=2a\chi_n-\chi_{n-1}$$
$$\psi_{n+1}=2a\psi_n-\psi_{n-1}$$
It is easy to see that for any $m$ the sequences $\chi_0,\chi_1,\cdots, \psi_0,\psi_1,\cdots$ are purely periodic modulo $m$ and hence so are their linear combinations. Further, it is easy to check by induction that the period of the sequence
(8) $$\psi_0,\psi_1,\cdots,\psi_n,\cdots (\mod a-1)$$
is
(9) $$0, 1, 2,\cdots, a - 2,$$
whereas the period of the sequence
(10) $$\chi_0-(a-2)\psi_0,\chi_1-(a-2)\psi_1,\cdots, \chi_n-(a-2)\psi_n,\cdots (\mod 4a-5)$$
begins with
(11) $$2^0, 2^1, 2^2,\cdots$$
The main new idea of Julia Robinson was to synchronize the two sequences by imposing a condition $G(a)$ which would guarantee that
(12) $$\text{the length of the period of (8) is a multiple of the length of the period of (10).}$$
If such a condition is Diophantine and is valid for infinitely many values of $a$, then one can easily show that the relation $a = 2^c$ is Diophantine. Julia Robinson, however, was unable to find such a $G$ and, even today, we have no direct method for finding one.
I liked the idea of synchronization very much and tried to implement it in a slightly different situation. When, in 1966, I had started my investigations on Hilbert's tenth problem, I had begun to use Fibonacci numbers and had discovered (for myself) the equation
(13) $$x^2 - xy - y^2=\pm 1$$
which plays a role similar to that of the above Pell's equation; namely, Fibonacci numbers $\phi_n$ and only they are solutions of (13). The arithmetical properties of the sequences $\psi_n$ and $\phi_n$ are very similar. In particular, the sequence
(14) $$0, 1, 3, 8, 21, \cdots$$
of Fibonacci numbers with even indices satisfies the recurrence relation
(15)$$\phi_{n+1}=3\phi_n-\phi_{n-1}$$
similar to (7). This sequence grows like $[(3+\sqrt 5)/2]^n$ and can be used instead of (11) for constructing a relation of exponential growth. The role of (10) can be played by the sequence
(16) $$\psi_0,\psi_1,\cdots,\psi_n,\cdots (\mod a-3)$$
because it begins like (14). Moreover, for special values of a the period can be determined explicitly; namely, if
(17) $$a = \phi_{2k}+\phi_{2k+2},$$
then the period of (16) is exactly
(18) $$0,1,3,\cdots,\phi_{2k},-\phi_{2k},\cdots,-3,1.$$
The simple structure of the period looked very promising.
I was thinking intensively in this direction, even on the night of New Year's Eve of 1970, and contributed to the stories about absentminded mathematicians by leaving my uncle's home on New Year's Day wearing his coat. On the morning of January 3, I believed I had found a polynomial B as in (5) but by the end of that day I had discovered a flaw in my work. But the next morning I managed to mend the construction.
What was to be done next? As a student I had had a bad experience when once I had claimed to have proved unsolvability of Hilbert's tenth problem, but during my talk found a mistake. I did not want to repeat such an embarrassment, and something in my new proof seemed rather suspicious to me. I thought at first that I had just managed to implement Julia Robinson's idea in a slightly different situation. However, in her construction an essential role was played by a special equation that implied one variable was exponentially greater than another. My supposed proof did not need to use such an equation at all, and that was strange. Later I realized that my construction was a dual of Julia Robinson's. In fact, I had found a Diophantine condition $H(a)$ which implied that
(19) $$\text{the length of the period of (16) is a multiple of the length of the period of (8).}$$
This $H$, however, could not play the role of Julia Robinson's $G$, which resulted in an essentially different construction." | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 30, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9758883714675903, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/141219/there-is-no-partial-recursive-function-f-s-t-whenever-n-w-e-has-one-element-f | # there is no partial recursive function f s.t. whenever N-W_e has one element, f converges and N-W_e = f(e)
question is as written in the title:
show that there is no partial recursive function f s.t. whenever N-W_e has one element, f converges and N-W_e = {f(e)}.
W_e is the domain of the program coded by e. So if there is a converging function f such that f(e) = N-W_e, how do I get a contradiction?
-
## 1 Answer
There exists a computable function $g$ such that $$\Phi_{g(e)}(s) = \begin{cases} 1 & \quad \text{ if $s > 0$ and $\Phi_{e,s}(e)\uparrow$} \\ 1 & \quad \text{ if $s = 0$ and $(\exists t)(\Phi_{e,t}(e)\downarrow$)} \\ 1 & \quad \text{ if $(\exists t < s)(\Phi_{e,t}(e)\downarrow$)} \\ \uparrow & \quad \text{ otherwise} \end{cases}$$ where $\uparrow$ means diverging computation and $\downarrow$ means converging computation. Note $\Phi_{e,s}(e)$ means the computation up to $s$ stages. Note that if $\Phi_{e}(e)$ converges then the only divergence occurs on the smallest $s$ that $\Phi_{e,s}(e)\downarrow$. If $\Phi_{e}(e)$ does not converges, then the only element that does not converge is $0$.
Intuitively, $W_{g(e)} = \omega - \{0\}$ if $\Phi_{e}(e)\uparrow$. If $\Phi_{e}(e) \downarrow$, then $W_{g(e)} = \omega - \{t\}$ for some $t > 0$ (not relevant but this $t$ is the smallest stage for which the computation converges). Note that by construction $W_{g(e)}$ is always missing exactly one element.
Even better intuition: As long as $\Phi_{e,s}(e)$ diverges on stage $s > 0$, enumerate s into $W_{g(e)}$. For the first state $t$ (if it exists) such that $\Phi_{e,t}(e)\downarrow$, then enumerate $0$ into $W_{g(e)}$, do not ever enumerate $t$ into $W_{g(e)}$ and enumerate everything bigger than $t$. So if $\Phi_{e}(e)$ diverges, then $W_{g(e)}$ is missing only $0$. If $\Phi_{e}(e)$ converges, then $W_{g(e)}$ is missing only some $t > 0$.
Assume that $f$ exists. Let $K = \emptyset' = \{e : \Phi_{e}(e)\downarrow\}$. Therefore, $e \in \bar{K}$ if and only if $0 \in \omega - W_{g(e)}$ if and only if $f(g(e)) = 0$. Since $W_{g(e)}$ is always missing exactly one element, $f(g(e))$ is a total function. Therefore $\bar{K}$ is computable. Contradiction. Since $\bar{K}$ is actually not even c.e.
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Would you be so kind as to explain what K and K-overscore are? I didn't understand the last part on why f(g(e)) is a total function and why K-overscore is not c.e. – John Lee May 5 '12 at 22:27
$k = \emptyset' = \{e : \Phi_e(e)\downarrow\}$, i.e. the halting set.$\overline{K}$ is the complement of $K$. It is well known that $K$ is not computable but c.e. If $\overline{K}$ was c.e.,then $K$ and $\bar{K}$ is c.e. so $K$ would be computable. Note that only the fact that $\bar{K}$ is not computable is necessary for the proof. By the discussion above, $W_{g(e)}$ always missing exactly 1 element for any $e$. By assumption, whenever $W_x$ is missing exactly one element $f(x)$ is defined and $\omega - W_x = \{f(x)\}$. Hence $f(g(e))$ is total. – William May 5 '12 at 22:53
Thank you very much. I really learned a lot from you. You are an amazing helper. – John Lee May 6 '12 at 5:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 57, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9201788902282715, "perplexity_flag": "head"} |
http://mathematica.stackexchange.com/questions/tagged/equation-solving+algebraic-manipulation | # Tagged Questions
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### How to convert a system of parametric equations to a normal equation?
For example, I have a system of parametric equations (R is a constant number) : ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9200314283370972, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/38379/how-did-einstein-derive-general-relativity | How did Einstein derive general relativity?
How did Einstein derive general relativity (GR)?
Did he use: the equivalence principle? The principle of least action? Anything else?
Note, I'm not looking for a full mathematical derivation of GR! But rather, I'd like to know what Einstein's starting points were.
Note: I found a similar question that has been asked already, but I couldn't find an answer to my question in the answers.
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3 Answers
His starting point was to realize that Newton's gravity didn't satisfy his principles of the (special) theory of relativity because it wasn't Lorentz-invariant and it included action at a distance, faster-than-light effects of gravity that could spread immediately.
So he was looking for a better theory that would be compatible with the principles of relativity. It took him a decade after special relativity was found to find and complete general relativity. Let me completely skip dead ends he had tried although these stories are interesting and one could learn something from them, too. At some point in 1911, in Prague's Viničná Street (see some letters Einstein wrote about Prague), he realized that the equivalence principle was a very special property of gravity – known already to Galileo but not appreciated as an important principle – and it led his final years.
Eventually he realized that the spacetime had to be curved, by arguments based on the equivalence principle, and it must be described by the Riemannian geometry. He was looking for the right equations that could relate the curvature of spacetime and the density of matter in the spacetime and finally in 1915, he found his Einstein's equations.
I think that he found the equations in their explicit form and the Einstein-Hilbert action from which the equations may be derived via the principle of least action were found later – also independently by Hilbert. We may say that the principle of least action wasn't necessary to discover GR; the equivalence principle was essential but Einstein needed (and one needs) more insights than just this principle.
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1
This is not the best way of explaining Einstein's work on GR. I remember telling my uncle that Einstein discovered GR to satisfy Lorentz invariance in gravity. He said that had little or nothing to do with it, and that it is really easy to create a relativistic extension of Newtonian gravity(a non GR extension). What Einstein was really motivated by was extending the principle of relativity to include non-inertial frames, and indeed Einstein was writing as early as 1907 that there was no good reason to believe it couldn't be done. – user7348 Sep 26 '12 at 14:37
1
– Luboš Motl Sep 26 '12 at 14:56
@Motl Okay, that's a pretty good entry, I'll admit. Do you think in the year 2012, we'd have General Relativity without Einstein? – user7348 Sep 26 '12 at 19:14
Yes, I think we would have GR without Einstein much earlier than in 2012. For example, it would have been derived from string theory, a theory of the strong force, in 1974. – Luboš Motl Sep 27 '12 at 5:26
Mach's principle, the equivalence principle and the Ehrenfest paradox all played a part. – Physiks lover Sep 27 '12 at 20:25
I highly recommend reading section 17.7, "A Taste of the History of Einstein's Equation", pages 431 through 434 of MTW's Gravitation
(Click the link to read at Google books).
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Einstein published a book in 1916 called Relativity which he updated in 1952 just a few years before his death. In chapter 25 he discusses Gaussian coordinates and in chapter 28 he gets to the heart of the matter and states:
The Gauss co-ordinate system has to take the place of the body of reference. The following statement corresponds to the fundamental idea of the general principle of relativity: "All Gaussian co-ordinate systems are essentially equivalent for the formulation of the general laws of nature."
Later then elaborates further:
According to the general theory of relativity,...,by application of arbitrary substitutions of the Gauss variables $x_1, x_2, x_3, x_4,$ the equations must pass over into equations of the same form; for every transformation (not only the Lorentz transformation) corresponds to the transition of the on Gauss co-ordinate system into another.
Which essentially gets to his point. Whatever the variables of spacetime are, any relationship between those variables must be respected even when we arbitrarily change variables. Or in other words, our choice of coordinates is arbitrary as long as we include enough variables to describe the underlying spacetime.
It should be noted in the context of discovery that there is a long standing dispute of priority. However, despite uneducated debate, most scholars agree that Einstein developed special and general relativity largely independently.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9685263633728027, "perplexity_flag": "middle"} |
http://physics.aps.org/articles/v4/21 | # Viewpoint: Entangled in a dating game
, School of Computer Science, McGill University, Montreal, Quebec H3A 2A7, Canada
Published March 14, 2011 | Physics 4, 21 (2011) | DOI: 10.1103/Physics.4.21
Entanglement, a crucial resource in quantum information theory, can also improve communication over a classical channel.
#### Entanglement-Enhanced Classical Communication Over a Noisy Classical Channel
R. Prevedel, Y. Lu, W. Matthews, R. Kaltenbaek, and K. J. Resch
Published March 14, 2011 | PDF (free)
In the $90$s, when quantum information was a young field, Charles Bennett and others found that entanglement—a property of quantum mechanics that links different quantum objects so that they cannot, in effect, behave independently—can assist in teleporting quantum bits [1] and in increasing communication rates over both noiseless [2] and noisy quantum channels [3]. However, they also suggested that entanglement would not be particularly helpful [3] when it came to enhancing the capacity of a classical channel. Given this, it is perhaps surprising that Robert Prevedel and his coauthors at the University of Waterloo, Canada, in their paper in Physical Review Letters, show that entanglement can be helpful in a noisy classical channel [4]. They observe this effect because their setup differs from that considered by Bennett and others. Rather than allowing many uses of the classical channel and demanding that communication error rates become negligible in the limit of many uses (as is conventional in quantum information theory), they allow just one use of the classical channel and consider maximizing the chances for a receiver to guess a sender’s transmitted message. The upshot is that entanglement can improve performance here.
Perhaps the best way to explain how entanglement plays a role is with a dating game that we’ll call “Guess That Button.” The game has two contestants, Alice and Bob, who sit in different rooms. The game show host gives Alice the game board (see Fig. 1), and Alice randomly pushes one of the four buttons on the board. Three edges (green, blue, and red) touch the button that Alice pushes. The host then takes the board from Alice, goes to Bob’s room, hands him the board, and one of these three edges randomly lights up. It is then Bob’s goal to “guess that button.” If he does so correctly, he wins a free dinner with Alice at the nicest sushi restaurant in town.
Bob is a bit helpless without more information. Two buttons touch each edge, and he really has to guess randomly which button Alice pushed. His chances of winning are $50/50$, not great odds for a dinner date. It turns out that Alice kind of likes Bob (and she really likes sushi). She realizes that his chances of winning are slim, so she asks the game show host if he’ll allow a strategy with higher chances for a win. The game show host agrees and suggests that Alice randomly pushes either a white or a grey button. He says now that it is Bob’s task to guess the color of the button that she pushes. Alice and Bob agree beforehand (a “code” in the parlance of communication theory) that she’ll randomly pick just one of the buttons on the left.
Bob’s chance to “guess that button” increase significantly with this modified strategy. Suppose Alice pushes the white button on the left. In this case, at random, either the blue, red, or green edge touching it lights up. If the green or red edge lights up, then $2/3$ of the time Bob knows Alice pushed the white button because they agreed she wouldn’t push a button on the right. If the blue edge lights up, then $1/3$ of the time Bob has to guess randomly which button Alice pushed. So the chances of Bob guessing correctly when Alice pushes the white button on the left are $2/3+1/3$ * $1/2=5/6$. Since the game board is symmetric, the chances of winning are the same if Alice pushes the grey button. So, the overall chance for them to win a free sushi dinner is $5/6$. It turns out that this is the best strategy they could use while still keeping the game interesting, that is, a game of chance.
Now enter entanglement. Suppose the host allows Alice and Bob to share two entangled particles before the game begins. Having access to good entanglement increases their chances of winning that dinner if they employ the following strategy: Alice first randomly decides whether she’ll push one of the two white buttons on the top or one of the two grey buttons on the bottom, and she measures her half of the entangled state in a particular way depending on this choice. Each measurement has two possible outcomes. After making this “top or bottom” choice and performing the measurement, she then pushes the button on the left or on the right, depending on the outcome of the measurement. The game show host gives the board to Bob, and one of the green, blue, or red edges touching the button that Alice pushed, randomly lights up. The green edge lights up $1/3$ of the time, and if it does, Bob can determine with certainty whether Alice pushed a white or grey button. If the red or blue edge lights up, Bob employs a different strategy. He measures his half of the shared entanglement in a particular way, depending on certain information that he would like to learn. With an $85%$ chance, he can correctly determine either the red or blue edge that touches the button Alice pushed. This is due to the strong correlations in entanglement—classical correlations would only give him a $75%$ chance of learning this information correctly. So, if the blue edge lights up, it is better for him to determine the red edge, and he chooses the measurement that will give him this information correctly $85%$ of the time. It is better to determine red because he already knows the blue edge, and if he determines the red edge correctly, he can figure out exactly which button Alice pushed, and thus whether it was white or grey. Similarly, if the red edge lights up, he can correctly determine $85%$ of the time that the blue edge touches the button Alice pushed, and then make a good guess at which button Alice pushed. Adding things up, he’ll guess correctly the color of the button Alice pushed $1/3$ of the time (the chance that the lit edge is green), and he’ll guess correctly $2/3$ * ($85%)=57%$ of the time when the lit edge is red or blue. Thus the total chance of guessing the color of the button Alice pushed is about $1/3+57%≈90%$. If they only share classical correlations, the total chance is $1/3+2/3*(75%)=5/6$, which is no better than the classical strategy in the previous paragraph. Thus entanglement gives a significant $7%$ boost to their chances of winning a sushi dinner.
To verify that these results aren’t just magic, Prevedel et al. performed an experimental implementation of the “Guess That Button” game on an optics bench in their quantum optics lab. In place of real contestants in a game show, they simulated the game using polarization entangled photons between two measurement devices, Alice and Bob, and an electronic logic circuit connected to a random number generator that implemented the noisy channel (or, in our analogy, selected randomly which edge would light up). The polarization entanglement was generated using a process called down-conversion, in which a high-energy photon can break into two lower energy photons. One of the photons in the pair was sent to Alice, the other to Bob. In their realization, Alice would randomly choose which of the buttons to push by which way her photon happened to go out of a beam-splitter and the result of a photon counting measurement. Bob’s local decoding action was implemented using fast optical switches and measurement. The experiment was conducted for about $10$ minutes, and, using data from photodetectors, they determined that Alice and Bob were winning about $89%$ of the time when Alice and Bob had access to shared entanglement. This success rate far exceeds what one would expect from the theoretical success probability for a classical strategy, and it is just under the theoretical success probability for the quantum strategy (likely due to slight imperfections in the entangled state).
### References
1. Charles H. Bennett, Gilles Brassard, Claude Crépeau, Richard Jozsa, Asher Peres, and William K. Wootters, Phys. Rev. Lett. 70, 1895 (1993).
2. Charles H. Bennett and Stephen J. Wiesner, Phys. Rev. Lett. 69, 2881 (1992).
3. Charles H. Bennett, Peter W. Shor, John A. Smolin, and Ashish V. Thapliyal, Phys. Rev. Lett. 83, 3081 (1999).
4. R. Prevedel, Y. Lu, W. Matthews, R. Kaltenbaek, and K. J. Resch, Phys. Rev. Lett. 106, 110505 (2011).
### About the Author: Mark M. Wilde
Mark M. Wilde received his B.S. degree in computer engineering from Texas A&M University in 2002, his M.S. degree in electrical engineering from Tulane University in 2004, and his Ph.D. degree in electrical engineering from the University of Southern California in 2008. Currently, he is a Postdoctoral Fellow at the School of Computer Science, McGill University. He has published about 40 articles and preprints in the area of quantum information processing. His current research interests are in quantum error correction and quantum Shannon theory.
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http://en.wikipedia.org/wiki/Skin_effect | # Skin effect
"Skin depth" redirects here. For the depth (layers) of biological/organic skin, see skin.
Distribution of current flow in a cylindrical conductor, shown in cross section. For alternating current, most (63%) of the electrical current flows between the surface and the skin depth, δ, which depends on the frequency of the current and the electrical and magnetic properties of the conductor.
The 3-wire bundles in this power transmission installation act as a single conductor. A single wire using the same amount of metal per kilometer would have higher losses due to the skin effect.
Skin effect is the tendency of an alternating electric current (AC) to become distributed within a conductor such that the current density is largest near the surface of the conductor, and decreases with greater depths in the conductor. The electric current flows mainly at the "skin" of the conductor, between the outer surface and a level called the skin depth. The skin effect causes the effective resistance of the conductor to increase at higher frequencies where the skin depth is smaller, thus reducing the effective cross-section of the conductor. The skin effect is due to opposing eddy currents induced by the changing magnetic field resulting from the alternating current. At 60 Hz in copper, the skin depth is about 8.5 mm. At high frequencies the skin depth becomes much smaller. Increased AC resistance due to the skin effect can be mitigated by using specially woven litz wire. Because the interior of a large conductor carries so little of the current, tubular conductors such as pipe can be used to save weight and cost.
## Cause
Skin depth is due to the circulating eddy currents (arising from a changing H field) cancelling the current flow in the center of a conductor and reinforcing it in the skin.
Conductors, typically in the form of wires, may be used to transmit electrical energy or signals using an alternating current flowing through that conductor. The charge carriers constituting that current, usually electrons, are driven by an electric field due to the source of electrical energy. An alternating current in a conductor produces an alternating magnetic field in and around the conductor. When the intensity of current in a conductor changes, the magnetic field also changes. The change in the magnetic field, in turn, creates an electric field which opposes the change in current intensity. This opposing electric field is called “counter-electromotive force” (back EMF). The back EMF is strongest at the center of the conductor, and forces the conducting electrons to the outside of the conductor, as shown in the diagram on the right.
An alternating current may also be induced in a conductor due to an alternating magnetic field according to the law of induction. An electromagnetic wave impinging on a conductor will therefore generally produce such a current; this explains the reflection of electromagnetic waves from metals.
Regardless of the driving force, the current density is found to be greatest at the conductor's surface, with a reduced magnitude deeper in the conductor. That decline in current density is known as the skin effect and the skin depth is a measure of the depth at which the current density falls to 1/e of its value near the surface. Over 98% of the current will flow within a layer 4 times the skin depth from the surface. This behavior is distinct from that of direct current which usually will be distributed evenly over the cross-section of the wire.
The effect was first described in a paper by Horace Lamb in 1883 for the case of spherical conductors, and was generalised to conductors of any shape by Oliver Heaviside in 1885. The skin effect has practical consequences in the analysis and design of radio-frequency and microwave circuits, transmission lines (or waveguides), and antennas. It is also important even at mains frequencies (50 – 60 Hz) in AC electrical power transmission and distribution systems. Although the term "skin effect" is most often associated with applications involving transmission of electrical currents, the skin depth also describes the exponential decay of the electric and magnetic fields, as well as the density of induced currents, inside a bulk material when a plane wave impinges on it at normal incidence.
## Formula
The AC current density J in a conductor decreases exponentially from its value at the surface JS according to the depth d from the surface, as follows:
$J=J_\mathrm{S} \,e^{-{d/\delta }}$
where δ is called the skin depth. The skin depth is thus defined as the depth below the surface of the conductor at which the current density has fallen to 1/e (about 0.37) of JS. In normal cases it is well approximated as:
$\delta=\sqrt{{2\rho }\over{\omega\mu}}$.
where
ρ = resistivity of the conductor
ω = angular frequency of current = 2π × frequency
μ = absolute magnetic permeability of the conductor[1]
A more general expression for skin depth which is more exact in the case of poor conductors (non-metals) at high frequencies is:[2][3]
$\delta= \left({1 \over \omega}\right) \left\lbrace \left( {{\mu\epsilon} \over 2}\right) \left[ \left(1+\left({1 \over {\rho\omega\epsilon}}\right)^2\right)^{1/2} -1\right]\right\rbrace^{-1/2}$
where $\epsilon$ is the electric permittivity of the material. Note that in the usual form for the skin effect, above, the effect of $\epsilon$ cancels out. This formula is valid away from strong atomic or molecular resonances (where $\epsilon$ would have a large imaginary part) and at frequencies which are much below both the material's plasma frequency (dependent on the density of free electrons in the material) and the reciprocal of the mean time between collisions involving the conduction electrons. In good conductors such as metals all of those conditions are ensured at least up to microwave frequencies, justifying this formula's validity.
This formula can be rearranged as follows to reveal departures from the normal approximation:
$\delta= \sqrt{{2\rho }\over{\omega\mu}} \; \; \sqrt{ \sqrt{1 + \left({\rho\omega\epsilon}\right)^2 } + \rho\omega\epsilon}$
At frequencies much below $1/\rho \epsilon$ the quantity inside the radical is close to unity and the standard formula applies. For instance, in the case of copper this would be true for frequencies much below $10^{18}$ Hz.
However in very poor conductors at sufficiently high frequencies, the factor on the right increases. At frequencies much higher than $1/\rho \epsilon$ it can be shown that the skin depth, rather than continuing to decrease, approaches an asymptotic value:
$\delta \approx {2 \rho} \sqrt{\epsilon \over \mu} \qquad (\omega \gg 1/\rho \epsilon)$
This departure from the usual formula only applies for materials of rather low conductivity and at frequencies where the vacuum wavelength is not much much larger than the skin depth itself. For instance, bulk silicon (undoped) is a poor conductor and has a skin depth of about 40 meters at 100 kHz ($\lambda$=3000m). However as the frequency is increased well into the megahertz range, its skin depth never falls below the asymptotic value of 11 meters. The conclusion is that in poor solid conductors such as undoped silicon, the skin effect doesn't need to be taken into account in most practical situations: any current is equally distributed throughout the material's cross-section regardless of its frequency.
## Resistance
The effective resistance due to a current confined near the surface of a large conductor (much thicker than δ) can be solved as if the current flowed uniformly through a layer of thickness δ based on the DC resistivity of that material. We can therefore assume a cross-sectional area approximately equal to δ times the conductor's circumference. Thus a long cylindrical conductor such as a wire, having a diameter D large compared to δ, has a resistance approximately that of a hollow tube with wall thickness δ carrying direct current. Using a material of resistivity $\rho$ we then find the AC resistance of a wire of length L to be:
$R\approx {{L \rho} \over {\pi (D-\delta) \delta}} \approx {{L \rho} \over {\pi D \delta}}$
The final approximation above assumes $D \gg \delta$.
A convenient formula (attributed to F.E. Terman) for the diameter DW of a wire of circular cross-section whose resistance will increase by 10% at frequency f is:
$D_\mathrm{W} = {\frac{200~\mathrm{mm}}{\sqrt{f/\mathrm{Hz}}}}$
The increase in AC resistance described above is accurate only for an isolated wire. For a wire close to other wires, e.g. in a cable or a coil, the ac resistance is also affected by proximity effect, which often causes a much more severe increase in ac resistance.
## Material effect on skin depth
In a good conductor, skin depth varies as the inverse square root of the conductivity. This means that better conductors have a reduced skin depth. The overall resistance of the better conductor remains lower even with the reduced skin depth. However the better conductor will show a higher ratio between its AC and DC resistance, when compared with a conductor of higher resistivity. For example, at 60 Hz, a 2000 MCM (1000 square millimetre) copper conductor has 23% more resistance than it does at DC. The same size conductor in aluminum has only 10% more resistance with 60 Hz AC than it does with DC. [4]
Skin depth also varies as the inverse square root of the permeability of the conductor. In the case of iron, its conductivity is about 1/7 that of copper. However being ferromagnetic its permeability is about 10,000 times greater. This reduces the skin depth for iron to about 1/38 that of copper, about 220 micrometres at 60 Hz. Iron wire is thus useless for A.C. power lines. The skin effect also reduces the effective thickness of laminations in power transformers, increasing their losses.
Iron rods work well for direct-current (DC) welding but it is impossible to use them at frequencies much higher than 60 Hz. At a few kilohertz, the welding rod will glow red hot as current flows through the greatly increased A.C. resistance resulting from the skin effect, with relatively little power remaining for the arc itself. Only non-magnetic rods can be used for high-frequency welding.
## Mitigation
A type of cable called litz wire (from the German Litzendraht, braided wire) is used to mitigate the skin effect for frequencies of a few kilohertz to about one megahertz. It consists of a number of insulated wire strands woven together in a carefully designed pattern, so that the overall magnetic field acts equally on all the wires and causes the total current to be distributed equally among them. With the skin effect having little effect on each of the thin strands, the bundle does not suffer the same increase in AC resistance that a solid conductor of the same cross-sectional area would due to the skin effect.[5]
Litz wire is often used in the windings of high-frequency transformers to increase their efficiency by mitigating both skin effect and proximity effect. Large power transformers are wound with stranded conductors of similar construction to litz wire, but employing a larger cross-section corresponding to the larger skin depth at mains frequencies.[6] Conductive threads composed of carbon nanotubes[7] have been demonstrated as conductors for antennas from medium wave to microwave frequencies. Unlike standard antenna conductors, the nanotubes are much smaller than the skin depth, allowing full utilization of the thread's cross-section resulting in an extremely light antenna.
High-voltage, high-current overhead power transmission lines often use aluminum cable with a steel reinforcing core; the higher resistance of the steel core is of no consequence since it is located far below the skin depth where essentially no AC current flows. In other applications, solid conductors are replaced by tubes, completely dispensing with the inner portion of the conductor where little current flows. This hardly affects the AC resistance but considerably reduces the weight of the conductor.
Solid or tubular conductors may also be silver-plated to take advantage of silver's higher conductivity. This technique is particularly used at VHF to microwave frequencies where the small skin depth requires only a very thin layer of silver, making the improvement in conductivity very cost effective. Silver or gold plating is similarly used on the surface of waveguides used for transmission of microwaves. This reduces attenuation of the propagating wave due to resistive losses affecting the accompanying eddy currents; the skin effect confines such eddy currents to a very thin surface layer of the waveguide structure. The skin effect itself isn't actually combated in these cases, but the distribution of currents near the conductor's surface makes the use of precious metals (having a lower resistivity) practical.
## Examples
Skin depth vs. frequency for some materials, red vertical line denotes 50 Hz frequency:
Mn-Zn - magnetically soft ferrite
Al - metallic alumium
Cu - metallic copper
steel 410 - magnetic stainless steel
Fe-Si - grain-oriented electrical steel
Fe-Ni - high-permeability permalloy (80%Ni-20%Fe)
We can derive a practical formula for skin depth as follows:
$\delta=\sqrt{{2\rho }\over{(2 \pi f) (\mu_0 \mu_r)}} \approx 503\,\sqrt{\frac{\rho}{\mu_r f}}$
where
$\delta =$ the skin depth in metres
$\mu_r =$ the relative permeability of the medium
$\rho =$ the resistivity of the medium in Ω·m, also equal to the reciprocal of its conductivity: $\rho = 1/ \sigma$ (for Copper, ρ = 1.68×10-8 Ω·m)
$f =$ the frequency of the current in Hz
Gold is a good conductor with a resistivity of 2.44×10-8 Ω·m and is essentially nonmagnetic: $\mu_r =$ 1, so its skin depth at a frequency of 50 Hz is given by
$\delta = 503 \,\sqrt{\frac{2.44 \cdot 10^{-8}}{1 \cdot 50}}= 11.1\,\mathrm{mm}$
Lead, in contrast, is a relatively poor conductor (among metals) with a resistivity of 2.2×10-7 Ω·m, about 9 times that of gold. Its skin depth at 50 Hz is likewise found to be about 33 mm, or $\sqrt{9} = 3$ times that of gold.
Highly magnetic materials have a reduced skin depth owing to their large permeability $\mu_r$ as was pointed out above for the case of iron, despite its poorer conductivity. A practical consequence is seen by users of induction cookers, where some types of stainless steel cookware are unusable because they are not ferromagnetic.[8]
At very high frequencies the skin depth for good conductors becomes tiny. For instance, the skin depths of some common metals at a frequency of 10 GHz (microwave region) are less than a micron:
Conductor Skin depth (μm)
Aluminum 0.80
Copper 0.65
Gold 0.79
Silver 0.64
Thus at microwave frequencies, most of the current flows in an extremely thin region near the surface. Ohmic losses of waveguides at microwave frequencies are therefore only dependent on the surface coating of the material. A layer of silver 3 μm thick evaporated on a piece of glass is thus an excellent conductor at such frequencies.
In copper, the skin depth can be seen to fall according to the square root of frequency:
Frequency Skin depth (μm)
60 Hz 8470
10 kHz 660
100 kHz 210
1 MHz 66
10 MHz 21
100 MHz 6.6
In Engineering Electromagnetics, Hayt points out that in a power station a busbar for alternating current at 60 Hz with a radius larger than one-third of an inch (8 mm) is a waste of copper, and in practice bus bars for heavy AC current are rarely more than half an inch (12 mm) thick except for mechanical reasons.
## Skin effect reduction of the self inductance of a conductor
Refer to the diagram below showing the inner and outer conductors of a coaxial cable. Since the skin effect causes a current at high frequencies to flow mainly at the surface of a conductor, it can be seen that this will reduce the magnetic field inside the wire, that is, beneath the depth at which the bulk of the current flows. It can be shown that this will have a minor effect on the self inductance of the wire itself; see Skilling[9] or Hayt[10] for a mathematical treatment of this phenomenon.
Note that the inductance considered in this context refers to a bare conductor, not the inductance of a coil used as a circuit element. The inductance of a coil is dominated by the mutual inductance between the turns of the coil which increases its inductance according to the square of the number of turns. However when only a single wire is involved, then in addition to the "external inductance" involving magnetic fields outside of the wire (due to the total current in the wire) as seen in the white region of the figure below, there is also a much smaller component of "internal inductance" due to the magnetic field inside the wire itself, the green region in figure B. In a single wire the internal inductance becomes of little significance when the wire is much much longer than its diameter. The presence of a second conductor in the case of a transmission line requires a different treatment as is discussed below.
Due to the skin effect, at high frequencies the internal inductance of a wire vanishes, as can be seen in the case of a telephone twisted pair, below. In normal cases the effect of internal inductance is ignored in the design of coils or calculating the properties of microstrips.
### Inductance per length in a coaxial cable
Let the dimensions a, b, and c be the inner conductor radius, the shield (outer conductor) inside radius and the shield outer radius respectively, as seen in the crossection of figure A below.
Four stages of skin effect in a coax showing the effect on inductance. Diagrams show a cross-section of the coaxial cable. Color code: black=overall insulating sheath, tan=conductor, white=dielectric, green=current into the diagram, blue=current coming out of the diagram, dashed blue lines with arrowheads=magnetic flux (B). The width of the dashed blue lines is intended to show relative strength of the magnetic field integrated over the circumference at that radius. The four stages, A, B, C, and D are non-energized, low frequency, middle frequency and high frequency respectively. There are three regions that may contain induced magnetic fields: the center conductor, the dielectric and the outer conductor. In stage B, current covers the conductors uniformly and there is a significant magnetic field in all three regions. As the frequency is increased and the skin effect takes hold (C and D) the magnetic field in the dielectric region is unchanged as it is proportional to the total current flowing in the center conductor. In C, however, there is a reduced magnetic field in the deeper sections of the inner conductor and the outer sections of the shield (outer conductor). Thus there is less energy stored in the magnetic field given the same total current, corresponding to a reduced inductance. At an even higher frequency, D, the skin depth is tiny: all current is confined to the surface of the conductors. The only magnetic field is in the regions between the conductors; only the "external inductance" remains.
For a given current, the total energy stored in the magnetic fields must be the same as the calculated electrical energy attributed to that current flowing through the inductance of the coax; that energy is proportional to the cable's measured inductance.
The magnetic field inside a coaxial cable can be divided into three regions, each of which will therefore contribute to the electrical inductance seen by a length of cable.[11]
The inductance $L_\text{cen} \,$ is associated with the magnetic field in the region with radius $r < a \,$, the region inside the center conductor.
The inductance $L_\text{ext} \,$ is associated with the magnetic field in the region $a < r < b \,$, the region between the two conductors (containing a dielectric, possibly air).
The inductance $L_\text{shd} \,$ is associated with the magnetic field in the region $b < r < c \,$, the region inside the shield conductor.
The net electrical inductance is due to all three contributions:
$L_\text{total} = L_\text{cen} + L_\text{shd} + L_\text{ext}\,$
$L_\text{ext} \,$ is not changed by the skin effect and is given by the frequently cited formula for inductance L per length D of a coaxial cable:
$L/D = \frac{\mu_0}{2 \pi} \ln \left( \frac {b}{a} \right) \,$
At low frequencies, all three inductances are fully present so that $L_\text{DC} = L_\text{cen} + L_\text{shd} + L_\text{ext}\,$.
At high frequencies, only the dielectric region has magnetic flux, so that $L_\infty = L_\text{ext}\,$.
Most discussions of coaxial transmission lines assume they will be used for radio frequencies, so equations are supplied corresponding only to the latter case.
As the skin effect increases, the currents are concentrated near the outside of the inner conductor (r=a) and the inside of the shield (r=b). Since there is essentially no current deeper in the inner conductor, there is no magnetic field beneath the surface of the inner conductor. Since the current in the inner conductor is balanced by the opposite current flowing on the inside of the outer conductor, there is no remaining magnetic field in the outer conductor itself where $b < r < c \,$. Only $L_\text{ext}$ contributes to the electrical inductance at these higher frequencies.
Although the geometry is different, a twisted pair used in telephone lines is similarly affected: at higher frequencies the inductance decreases by more than 20% as can be seen in the following table.
### Characteristics of telephone cable as a function of frequency
Representative parameter data for 24 gauge PIC telephone cable at 21 °C (70 °F).
Frequency (Hz) R (Ω/km) L (mH/km) G (μS/km) C (nF/km)
1 172.24 0.6129 0.000 51.57
1k 172.28 0.6125 0.072 51.57
10k 172.70 0.6099 0.531 51.57
100k 191.63 0.5807 3.327 51.57
1M 463.59 0.5062 29.111 51.57
2M 643.14 0.4862 53.205 51.57
5M 999.41 0.4675 118.074 51.57
More extensive tables and tables for other gauges, temperatures and types are available in Reeve.[12] Chen[13] gives the same data in a parameterized form that he states is usable up to 50 MHz.
Chen[13] gives an equation of this form for telephone twisted pair:
$L(f) = \frac {l_0 + l_{\infty}(\frac{f}{f_m})^b }{1 + (\frac{f}{f_m})^b} \,$
## Notes
1. The permeability μ can be found from $\mu_r$, the relative permeability of the material by multiplying it by $\mu_0$, the permeability of free space: $\mu= \mu_0 \cdot \mu_r$.
2. Donald G. Fink, H. Wayne Beatty Standard Handbook for Electrical Engineers 11th Edition, McGraw Hill, 1978 table 18-21
3. Xi Nan; Sullivan, C. R. (2005), "An equivalent complex permeability model for litz-wire windings", Industry Applications Conference 3: 2229–2235, doi:10.1109/IAS.2005.1518758, ISBN 0-7803-9208-6, ISSN 0197-2618
4. Central Electricity Generating Board (1982). Modern Power Station Practice. Pergamon Press.
5. "Spinning Carbon Nanotubes Spawns New Wireless Applications". Sciencedaily.com. 2009-03-09. Retrieved 2011-11-08.
6. If the permeability is low, the skin depth is so large that the resistance encountered by eddy currents is too low to provide enough heat
7. ^ a b
## References
• Chen, Walter Y. (2004), Home Networking Basics, Prentice Hall, ISBN 0-13-016511-5
• Hayt, William (1981), Engineering Electromagnetics (4th ed.), McGraw-Hill, ISBN 0-07-027395-2
• Hayt, William Hart. Engineering Electromagnetics Seventh Edition. New York: McGraw Hill, 2006. ISBN 0-07-310463-9.
• Nahin, Paul J. Oliver Heaviside: Sage in Solitude. New York: IEEE Press, 1988. ISBN 0-87942-238-6.
• Ramo, S., J. R. Whinnery, and T. Van Duzer. Fields and Waves in Communication Electronics. New York: John Wiley & Sons, Inc., 1965.
• Ramo, Whinnery, Van Duzer (1994). Fields and Waves in Communications Electronics. John Wiley and Sons.
• Reeve, Whitman D. (1995), Subscriber Loop Signaling and Transmission Handbook, IEEE Press, ISBN 0-7803-0440-3
• Skilling, Hugh H. (1951), Electric Transmission Lines, McGraw-Hill
• Terman, F. E. (1943), Radio Engineers' Handbook, New York: McGraw-Hill . For the Terman formula mentioned above.
• Xi Nan; Sullivan, C. R. (2005), "An equivalent complex permeability model for litz-wire windings", Industry Applications Conference 3: 2229–2235, doi:10.1109/IAS.2005.1518758, ISBN 0-7803-9208-6, ISSN 0197-2618
• Jordan, Edward (1968), Electromagnetic Waves and Radiating Systems, Prentice Hall, ISBN 978-0-13-249995-8 Unknown parameter `|middle=` ignored (help)
• Vander Vorst, Andre; Rosen, Arye; Kotsuka, Youji (2006), RF/Microwave Interaction with Biological Tissues, John Wiley and Sons, Inc., ISBN 978-0-471-73277-8 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 43, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9119794964790344, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?t=220460 | Physics Forums
## infinite quantum wave function
Can a quantum wave function be infinite at a point? For example you could have a radially symmetric wavefunction that's infinite at the center, yet the integrated probability is 1. Is this unphysical somehow?
Laura
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Science Advisor You are describing the Dirac delta function, the wave function of a particle with a precise position.
Mentor
A proper wave function must have a slope that is a continuous function. I don't think this is possible at a point where the function goes to infinity. However, such a function might appear as an idealized limiting case of a sharply "peaked" wave function. For example:
Quote by clem the Dirac delta function, the wave function of a particle with a precise position.
## infinite quantum wave function
Quote by clem You are describing the Dirac delta function, the wave function of a particle with a precise position.
I was thinking about *something* like probability density = exp(-r)/r. Something that goes infinite at the center yet has finite integral. Perfectly well-behaved except at one point. The Dirac delta function wouldn't ever appear in reality although it might as an intermediate step in one's calculations.
Laura
Mentor
Blog Entries: 27
Quote by lark I was thinking about *something* like probability density = exp(-r)/r. Something that goes infinite at the center yet has finite integral. Perfectly well-behaved except at one point. The Dirac delta function wouldn't ever appear in reality although it might as an intermediate step in one's calculations. Laura
What is the fourier transform of a plane wave? This will be the wavefunction in momentum space.
Zz.
now we are talking about dirac deltas, the wavefunction $$D^{n} \delta (x-a)$$ (derivative of delta function) has a meaning ? from Fourier analysis, we could consider the wave fucntion above the Fourier transform of $$x^{n}$$
If the wave function represents probability amplitudes by definition, how could it be greater than 1 at any point?
because its a probability density, you must integrate it to get the probability the particle is in a given range. this situation is unphysical, it corresponds to a wavefunction with infinite expected energy. to see this write the delta in momentum space (neglecting the various constants its e^ipx') now use the momentum space hamiltonian for a free particle ((p^2)/2m)) and take the expectation value. you would get an infinite result, however this is not the only reason why it is unphysical, what the delta wavefunction means in position space is a plane wave in momentum space, this plane wave in momentum space would not be normalizable and thus unphysical.
Have you heard of delta function normalization?
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http://math.stackexchange.com/questions/257373/operators-with-eigenvalue-0-1-that-is-not-projection | # Operators with eigenvalue $\{0,1\}$ that is not projection
Show that there are linear operators T on the Hilbert space H what are not orthogonal projections, but their spectrum consists of the eigenvalues $\{0,1\}.$
I can not come up with an counterexample, but I suppose it must be infinite dimension? Hilbert spaces has orthogonal bases and I did an exersice showing that for projections $T'(T-I) = 0$ So obviously $T \neq TT'$, is zero some kind of accumulation point?
-
## 2 Answers
$$\begin{pmatrix}0 & 0 & 0 \\ 0 & 1 & 57 \\ 0 & 0 & 1\end{pmatrix}$$
-
Oh it was this easy! thanks! – Johan Dec 13 '12 at 12:44
To complement Nate Eldredge's example, it might be more interesting to consider operators that are projections, but not orthogonal projections. We can rephrase this in terms of the SVD: An orthogonal projection is an operator $T$ such that $T^2 = T$, and such that the singular values of $T$ are each either $0$ or $1$.
A non-orthogonal projection, also called an oblique projection, is one for which $T^2 = T$, but the singular values of $T$ may not belong to $\{0,1\}$. Consider this example: $$A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$$ We can check that $A^2 = A$, so this is a projection, but it is not an orthogonal projection, since $$A^* A = \begin{bmatrix} 0 & 0 \\ 0 & 2 \end{bmatrix}$$
-
thanks good example!!! – Johan Dec 13 '12 at 12:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9452934861183167, "perplexity_flag": "head"} |
http://en.wikibooks.org/wiki/Parallel_Spectral_Numerical_Methods/Timestepping | # Parallel Spectral Numerical Methods/Timestepping
## Introduction
We now briefly discuss how to solve initial value problems. For more on this see Bradie (Chap. 7)[1]. A slightly longer but still quick introduction to these ideas can also be found in Boyce and DiPrima.[2].
## Forward Euler
In order to compute solutions to differential equations on computers efficiently, it is convenient to do our calculations at a finite number of specified points and then interpolate between these points. For many calculations it is convenient to use a grid whose points are equally distant from each other.
For the rest of the section $h$ will be our step size, which is assumed to be constant. When solving an ODE or PDE, the choice of $h$ isn't selected at random, but rather requires some intuition and/or theoretical analysis. We are going to start with forward Euler method which is the most basic numerical method. Lets first denote the time at the $n^{\text{th}}$ time-step by $t_n$ and the computed solution at the $n^{\text{th}}$ time-step by $y^n$, where $y^n \equiv y(t=t^n)$. The step size $h$ in terms of $t$ is defined as $h=t^{n+1}-t^n$. Lets first start with a basic ODE with initial conditions, in which $f(t,y)$ is some arbitrary function and $y(t)$ is our solution,
$\frac{dy}{dt} = f(t,y) \qquad y(t^0)=y^0.$
( )
The differential equation can be approximated by finite differences,
$\frac{y^{n+1}-y^n}{h} = f(t^n,y^n).$
Now all we have to do is solve for $y^{n+1}$ algebraically,
$y^{n+1}=y^n+h f(t^n,y^n) \qquad \text{(Forward Euler/Explicit method)}$
( )
If we wanted to calculate $\frac{dy}{dt}$ at time $t^0$, then we could generate an approximation for the value at time $t^{n+1}$ using eq. 2 by first finding $y(t^0)$ and using it to compute $y^{n+1}$. We then repeat this process until the final time is reached.
### An Example Computation
Lets take the ODE in eq. 1 with initial conditions $y(t^0)=1$ where $t^0=0$. Using forward Euler, lets plot two examples where $h=1$ and $h=.1$
A numerical solution to the ODE in eq. 1 with f(t,y) = y demonstrating the accuracy of the Forward Euler method for different choices of timestep.
It should be no surprise that a smaller step size like $h=.1$ compared to $h=1$ will be more accurate. Looking at the line for $h=0.1$, you can see that $y(t)$ is calculated at only 4 points then straight lines are interpolate between each point. This is obviously not very accurate, but gives a rough idea of what the function looks like. The solution for $h=.1$ might require 10 times more steps to be taken, but it is clearly more accurate. Forward Euler is an example of a first order method and approximates the exact solution using the first two terms in the Taylor expansion[footnote 1]
$y(t^n+h)=y(t^n)+h \left.\frac{dy}{dt}\right|_{t^n}+\text{O}(h^2),$
where terms of higher order than O$(h^2)$ are omitted in the approximate solution. Substituting this into eq. 2 we get that
$y^n +h \left.\frac{dy}{dt}\right|_{t^n}+\text{O}(h^2) = y^n +hf(t^n,y^n)$
after cancelling terms and dividing by $h$, we get that
$\left.\frac{dy}{dt}\right|_{t^n}+\text{O}(h) = f(t^n,y^n),$
from which it is clear that the accuracy of the method changes linearly with the step size, and hence it is first order accurate.
## Backwards Euler
A variation of forward Euler can be obtained by approximating a derivative by using a backward difference quotient. Using eq. 1 and applying
$\frac{y^{n}-y^{n-1}}{h}\approx f(t^n,y^n)$
$y^{n}=y^{n-1}+h f(t^n,y^n).$
Stepping the index up from $n$ to $n+1$ we obtain,
$y^{n+1}=y^n+h f(t^{n+1},y^{n+1})\qquad \text{(Backwards Euler/Implicit method)}$
Notice how $y^{n+1}$ is not written explicitly like it was in the forward Euler method. This equation instead implicitly defines $y^{n+1}$ and must be solved to determine the value of $y^{n+1}$. How difficult this is depends entirely on the complexity of the function $f$. For example, if $f$ is just $y^2$, then the quadratic equation could be used, but many nonlinear PDEs require other methods. Some of these methods will be introduced later.
## Crank-Nicolson
By taking an average of the forward and backward Euler methods, we can find the Crank-Nicolson method:
$\frac{y^{n+1}-y^{n}}{h}=\frac{1}{2}f(t^{n+1},y^{n+1})+\frac{1}{2}f(t^n,y^n)$
Rearranging we obtain,
$y^{n+1}=y^n+\frac{h}{2}\left[ f(t^{n+1},y^{n+1})+f(t^n,y^n) \right] \qquad \text{(Crank-Nicolson)}$
Notice again how $y^{n+1}$ is not written explicitly like it was in forward Euler. This equation instead implicitly defines $y^{n+1}$ and so the equation must be solved algebraically to obtain $y^{n+1}$.
## Stability of Forward Euler, Backward Euler and Crank-Nicolson
Let's look at the following ODE
$\frac{dy}{dt}= -\lambda y(t)$
where $\lambda$ is a constant and $y(t^0)=1$ where $t^0=0$. Lets numerically solve this ODE using the forward Euler, backward Euler and Crank-Nicolson time stepping schemes. The results are as follows
$y^{n+1}=y^n-\lambda h y^n \qquad \text{(Forward Euler)}$
$y^{n+1} = \frac{y^n}{(1+\lambda h)} \qquad \text{(Backward Euler)}$
$y^{n+1} = y^n\left(\frac{2-\lambda h}{2+\lambda h}\right) \qquad \text{(Crank-Nicolson)}$
and the exact solution is given by
$y(t)= e^{-\lambda t} \qquad \text{(Exact solution)}$
A numerical solution to the ODE in eq. 2 with λ = 20 and with a timestep of h = 0.1 demonstrating the instability of the Forward Euler method and the stability of the Backward Euler and Crank Nicolson methods.
The figure above shows how both methods converge to the solution, but the forward Euler solution is unstable for the chosen timestep. Listing below is a Matlab program where you can play around with the value of $\lambda$ to see how for a fixed timestep this changes the stability of the method.
( )
```% A program to demonstrate instability of timestepping methods
% when the timestep is inappropriately choosen.
%Differential equation: y'(t)=-y(t) y(t_0)=y_0
%Initial Condition, y(t_0)=1 where t_0=0
clear all; clc; clf;
%Grid
h=.1;
tmax=4;
Npoints = tmax/h;
lambda=.1;
%Initial Data
y0=1;
t_0=0;
t(1)=t_0;
y_be(1)=y0;
y_fe(1)=y0;
y_imr(1)=y0;
for n=1:Npoints
%Forward Euler
y_fe(n+1)=y_fe(n)-lambda*h*y_fe(n);
%Backwards Euler
y_be(n+1)=y_be(n)/(1+lambda*h);
%Crank Nicolson
y_imr(n+1)=y_imr(n)*(2-lambda*h)/(2+lambda*h)
t(n+1)=t(n)+h;
end
%Exact Solution
tt=[0:.001:tmax];
exact=exp(-lambda*tt);
%Plot
figure(1); clf; plot(tt,exact,'r-',t,y_fe,'b:',t,y_be,'g--',t,y_imr,'k-.');
xlabel time; ylabel y;
legend('Exact','Forward Euler','Backward Euler',...
'Crank Nicolson');
```
( )
A Python program to demonstrate instability of different time-stepping methods. Code download
```#!/usr/bin/env python
"""
A program to demonstrate instability of timestepping methods#
when the timestep is inappropriately choosen.################
"""
from math import exp
import matplotlib.pyplot as plt
import numpy
#Differential equation: y'(t)=-l*y(t) y(t_0)=y_0
#Initial Condition, y(t_0)=1 where t_0=0
# Definition of the Grid
h = 0.1 # Time step size
t0 = 0 # Initial value
tmax = 4 # Value to be computed y(tmax)
Npoints = int((tmax-t0)/h) # Number of points in the Grid
t = [t0]
# Initial data
l = 0.1
y0 = 1 # Initial condition y(t0)=y0
y_be = [y0] # Variables holding the value given by the Backward Euler Iteration
y_fe = [y0] # Variables holding the value given by the Forward Euler Iteration
y_imr = [y0] # Variables holding the value given by the Midpoint Rule Iteration
for i in xrange(Npoints):
y_fe.append(y_be[-1]*(1-l*h))
y_be.append(y_fe[-1]/(1+l*h))
y_imr.append(y_imr[-1]*(2-l*h)/(2+l*h))
t.append(t[-1]+h)
print
print "Exact Value: y(%d)=%f" % (tmax, exp(-4))
print "Backward Euler Value: y(%d)=%f" % (tmax, y_be[-1])
print "Forward Euler Value: y(%d)=%f" % (tmax, y_fe[-1])
print "Midpoint Rule Value: y(%d)=%f" % (tmax, y_imr[-1])
# Exact Solution
tt=numpy.arange(0,tmax,0.001)
exact = numpy.exp(-l*tt)
# Plot
plt.figure()
plt.plot(tt,exact,'r-',t,y_fe,'b:',t,y_be,'g--',t,y_imr,'k-.');
plt.xlabel('time')
plt.ylabel('y')
plt.legend(('Exact','Forward Euler','Backward Euler',
'Implicit Midpoint Rule'))
plt.show()
```
## Stability and Accuracy of Forward Euler, Backward Euler and Crank-Nicolson Time Stepping Schemes for $y'=-\lambda y$
We now try to understand these observations so that we have some guidelines to design numerical methods that are stable. The numerical solution to an initial value problem with a bounded solution is stable if the numerical solution can be bounded by functions which are independent of the step size. There are two methods which are typically used to understand stability. The first method is linearized stability, which involves calculating eigenvalues of a linear system to see if small perturbations grow or decay. A second method is to calculate an energy like quantity associated with the differential equation and check whether this remains bounded.
We shall assume that $\lambda\geq0$ so that the exact solution to the ODE does not grow without bound. The forward Euler method gives us
$\frac{y^{n+1}-y^n}{h}=-\lambda y^n$
$y^{n+1}=(1-\lambda h)y^n$
$\Rightarrow | y^{n+1} | \geq | (1-\lambda h) | | y^n | \quad \text{ if } | (1-\lambda h) | > 1$
$\Rightarrow | y^{n+1} | \leq | (1-\lambda h) | | y^n | \quad \text{ if } | (1-\lambda h) | <1.$
We can do a similar calculation for backward Euler to get
$\frac{y^{n+1}-y^n}{h}=-\lambda y^{n+1}$
$y^{n+1}=\frac{y^n}{1+\lambda h}$
$\Rightarrow | y^{n+1} | \leq \left| \frac{y^n}{1+\lambda h} \right| \leq | y^n | \quad \text{ since } \left| \frac{1}{1+\lambda h} \right| <1.$
Thus, the backward Euler method is unconditionally stable, whereas the forward Euler method is not. We leave the analysis of the Crank-Nicolson method as an exercise.
A second method, often used to show stability for partial differential equations is to look for an energy like quantity and show that this bounds the solution and prevents it from becoming too positive or too negative. Usually, the quantity is chosen to be non negative, then all one needs to do is deduce there is an upper bound. We sketch how this is done for an ordinary differential equation so that we can use the same ideas when looking at partial differential equations. Recall that the forward Euler algorithm is given by
$\frac{y^{n+1}-y^n}{h}=-\lambda y^n.$
Multiplying this by $y^{n+1}$ we find that
$(y^{n+1})^2=(1-h\lambda)y^ny^{n+1}.$
Now to obtain a bound on $| y^{n+1}|$ in terms of $| y^n |$, we use the following fact
$(a-b)^2\geq0\Rightarrow a^2+b^2\geq2ab\Rightarrow\frac{(y^{n+1})^2+(y^n)^2}{2}\geq y^ny^{n+1}.$
Hence a sufficient condition for stability if
$(1-h\lambda)>0$
is that
$(y^{n+1})^2\leq(1-h\lambda)\frac{(y^{n+1})^2+(y^n)^2}{2}$
$(y^{n+1})^2\frac{1+h\lambda}{2}\leq\frac{1-h\lambda}{2}(y^n)^2$
$(y^{n+1})^2\leq\frac{1-h\lambda}{1+h\lambda}(y^n)^2,$
thus if $1-h\lambda>0$, then $0<\frac{1-h\lambda}{1+h\lambda}<1$ and so we have stability, we again see that the algorithm is stable provided the timestep is small enough. There are many situations for which $\lambda$ is large and so the timestep, $h$ needs to be very small. In such a situation, the forward Euler method can be very slow on a computer.
Stability is not the only requirement for a numerical method to approximate the solution to an initial value problem. We also want to show that as the timestep is made smaller, the numerical approximation becomes better. For the forward Euler method we have that
$\frac{y^{n+h}-y^{n}}{h}=-\lambda y^n$
now if
$y^n=y(t)$
$y^{n+1}=y(t+h)$
then[footnote 2]
$= y(t)+h\frac{\mathrm{d}y}{\mathrm{d}t} + O(h^2)$
so
$\frac{y^{n+1}-y^n}{h}+\lambda y^n$
$= \frac{y(t+h)-y(t)}{h} +\lambda y(t)$
$=\frac{\mathrm{d}y}{\mathrm{d}t} +O(h)+\lambda y(t)$
$= O(h).$
We can do a similar calculation to show that the Crank-Nicolson method is second order. In this case however, we use Taylor expansions around $y(t+h/2)$.
$\frac{y^{n+1}-y^n}{h}=-\lambda \frac{ y^{n+1} + y^n}{2}$
so
$y^{n+1} =y(t+h)=y(t+h/2)+(h/2)\frac{\mathrm{d}y}{\mathrm{d}t} +(h/2)^2\frac{1}{2}\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + O(h^3)$
$y^{n} =y(t)=y(t+h/2)-(h/2)\frac{\mathrm{d}y}{\mathrm{d}t} +(h/2)^2\frac{1}{2}\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + O(h^3)$
hence
$\frac{y^{n+1}-y^n}{h}+\lambda \frac{ y^{n+1} + y^n}{2}$
$=\frac{\mathrm{d}y}{\mathrm{d}t} + O(h^2) +\lambda \left[y(t+h/2)+O(h^2) \right]$
$= O(h^2).$
Thus this is a second order method.
## Exercises
1. Determine the real values of $\lambda$ and timestep $h$ for which the implicit midpoint rule is stable for the ODE $\frac{\mathrm{d}y}{\mathrm{d}t}=-\lambda y$. Sketch the stable region in a graph of $\lambda$ against timestep $h$.
2. Show that the backward Euler method is a first order method.
## Notes
1. The derivation of the Taylor expansion can be found in most books on calculus.
2. We will use big `Oh' to mean that there exists a constant so that if $f~O(h)$, then for $h\rightarrow0$, we have that $\left|\frac{f}{h}\right|<C$, where $C$ is some constant.
## References
Boyce, W.E.; DiPrima, R.C. (2010). Elementary Differential Equations and Boundary Value Problems. Wiley.
Bradie, B. (2006). A Friendly Introduction to Numerical Analysis. Pearson. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 107, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9260568022727966, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/192728-complex-inequalities-print.html | # complex inequalities
Printable View
• November 26th 2011, 08:03 AM
minicooper58
complex inequalities
if |z|=3 is a circle, how do you know if the region outside the circle is
|z|<3
Many thanks
• November 26th 2011, 08:12 AM
Plato
Re: complex inequalities
Quote:
Originally Posted by minicooper58
if |z|=3 is a circle, how do you know if the region outside the circle is |z|<3 .
$|z-w|$ is the distance between $z~\&~w$.
Thus $\{z:|z|=|z-0|=3\}$ is a circle centered at the origin with radius 3.
BUT the exterior of that circle is $\{z:|z|=|z-0|>3\}$, not it is not $<$.
• November 26th 2011, 08:27 AM
minicooper58
2 Attachment(s)
Re: complex inequalities
Hi , the question and answer is in the attachments . sorry i didn't fully explain the transformation part . I don't understand how the region is outside the circle for (b) .
Many thanks
• November 26th 2011, 12:16 PM
HallsofIvy
Re: complex inequalities
That's not at all the question you originally asked! The set |z|< 3 is the inside of the circle given by |z|= 3. But the function $w= \frac{z}{z+ 1}$ maps the circle |z|= 3, with center at 0 and radius 3, into the circle with center at $\frac{9}{8}$ and radius $\frac{3}{8}$. It also clearly maps 0 to 0. 0 is inside |z|< 3 but outside the circle with center $\frac{9}{8}$ and radius $\frac{3}{8}$. Therefore, the interior of the first circle is mapped to the exterior of the second circle.
All times are GMT -8. The time now is 01:10 AM. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9228304624557495, "perplexity_flag": "middle"} |
http://gilkalai.wordpress.com/2010/05/04/drunken-time-and-drunken-computation/?like=1&source=post_flair&_wpnonce=18b5244a97 | Gil Kalai’s blog
## Drunken Time and Drunken Computation
Posted on May 4, 2010 by
### The problem
We are used to computer programs or models for computations that perform at time $i$ step $U_i$, $i=1,2,\dots,N$. Suppose that time is drunk, so instead of running these steps in their correct order, we apply at time $i$ step $\pi(i)$, where $\pi$ is a random permutation which is substantially correlated with the identity permutation. What then is our computational power?
I suggested this problem in our rump session meeting of last thursday quantum computation seminar. We have these sessions once every several months and usually they are very interesting.
### What are the variables?
The computation is applied to $n$ variables $x_1,x_2,\dots, x_n$. There are three variations to consider regarding the identity of these variables:
a) bits
b) qubits
c) algebraic variables
### What do we allow for a single step $U_i$?
Here also we have three possibilities.
A) a single gate operating on a small number of variables.
B) applying as many gates as we want on disjoint sets of variables.
C) $U_i$ is a transformation where the value of each variable $x_i$ is a function of a bounded set of variables (overlaps are ok).
By a “gate” we refer to the standard notion of gates: Boolean gates for bits for case a), unitary transformation on a small number of qubits for case b) and algebraic gates for case c). (I dont know of an appropriate way to define option C for the quantum case.)
### What is the precise stochastic model?
Write $x(i) = i + Y$, where $Y$ is a normally distributed random variable with standard deviation $\epsilon n$, and define $\pi (i)$ based on the ordering of the $x(i)$s.
### What happens if we consider fully random time evolution, or even adversarial time evolution?
We can ask for the computation power if we talk about completely random permutations. We can also let an adversary choose the permutation. We can also ask, how well can we do if we want to compute something when all the $U_i$‘s are the same? Several participants in the seminar commented that we can have universal classical computation when we run the same step $U_i$ every times. (But this applies for variant C ). One argument raised by Elad Eban was based on known fact regarding cellular automata, and another argument raised by Or Sattath was based on a standard way to move from Turing machines to Boolean circuits. I do not know if we can replace variant C by variant B in these arguments, or, if they apply to the quantum case.
### What do I expect:
a) The classical case: I expect that classic computation prevails (for A and B as well). A rather vague suggestion is to take $n >> N$ and simulate a classical computation by replacing every logical bit with a block of large number of physical bits, performing the computation on pairs of such blocks, and apply “majority” repeatedly within the blocks to guarantee that the physical bits indeed represent the logical bits.
b) The quantum case: I don’t know. This may serve as a (rather wild) model where we cannot go beyond classical computation. (Well, not quite see the update.)
### Update
Oded Regev showed me a very simple argument how drunken time allows the full power of quantum computation. (Perhaps, the arguments by Elad, and Or can be extended to the quamtum case, and maybe the paper quoted in the remarks show it as well, but Oded’s argument is quite clear) Here it is:
”Let’s have a clock register storing the clock in unary encoding. The unitary transformation corresponding to a gate at time t is the following:
* If the clock is at time t, advance it to t+1 and apply the gate
* If the clock is at time t+1, decrease it to t and apply the reverse of the gate
* Otherwise do nothing
Notice that you only need to consider three bits of the clock register (in addition to the two bits of the gate) to do this.
Now duplicate each unitary a sufficiently large number of times. I think what happens is that you get some kind of random walk on the clock register. Most steps do nothing, but with probability 1/T you advance forward, and with probability 1/T you go backwards. Therefore after about T*T^2=T^3 steps you expect the walk to reach time T, at which point you can simply read off the answer. By padding the circuit with identity gates you can increase the probability of measuring the final state of the circuit as much as you wish (so you need not worry about the final state being mixed).”
Further thoughts: Just a constant overhead?
An interesting question is: Is there a linear time algorithm to simulate a an algorithm with T steps on a drunken time machine. This can be relevant for continuous time analogs where we can expect the drunken time effect at all time scales and a superlinear algorithm may blow up.
______
c) Algebraic variables. I don’t know.
### Continuous-time analog?
We described a discrete-time model and it would be interesting to consider continuous time analogs of such drunken time evolution.
I do not know if stochastic perturbations based on “stochastic time-flow” of deterministic evolutions (e.g., those described by differential equations) are equivalent to standard stochastic perturbations.
### Is time really drunk?
Various ”crazy” models regarding time were considered by physicists, mathematicians, and philosophers, and computer scientists for their own sake and also in connection to various models of computations. Itamar Pitowsky showed how undecidable problems can be solved in finite time in certain hypothetical time/space structures. Dave Bacon studied quantum computation with quantum data that can traverse closed time-like curves and Aaronson and Watrous showed that closed time-like curves make quantum and classical computing equivalent (both allowing PSPACE computation). An example I learned from Moshe Vardi is modeling time as a tree that was proposed by Kripke following the introduction of temporal logic by Prior, and these ideas are playing a crucial role in the highly industrial area of automated verification. (Remarks with more examples are most welcome.) (Of course, closed time appeared earlier
I would expect that models with “stochastic time flow” have been considered in the physics and mathematics literature, so I’d be happy to have some pointers. Because of the Law of Large Numbers, small-scale stochastic processes look completely deterministic in larger scales. So if time is drunk in some scales we may not notice it in larger scales. (However, if time is really drunk, I would not expect the amount of drunkeness to be constant but rather to express some measure of non-classical behavior of the dynamics.) Thinking of time as drunk may be lead (or perhaps has already led) to a mathematically useful device.
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### 10 Responses to Drunken Time and Drunken Computation
1. Martin Schwarz says:
You may want to check out the related paper http://arxiv.org/abs/0809.0847, where a model of quantum computation with commuting gates is analyzed, i.e. the order of gate applications is irrelevant. Yet the model seems to be rich enough to create probability distributions which are hard to sample from classically.
2. Jack says:
Do you have a reference for the Itamar Pitowski result?
3. alef says:
Consider n bits placed on a cycle of size n. A bit per vertex.
You don’t see the bits scenery. A walker preforming simple random walk on the cycle broadcast to you at each time step the bit placed where she is.
1. It is possible to reconstruct the scenery.
2. Polynomial many observations are sufficient with high probability.
3. We don’t know an efficient algorithm with polynomially many operations,
to recover the scenery from the polynomial many observations though
4. Gil Kalai says:
Thanks for the comments Martin and Alef,
Jack, I found the following link
http://edelstein.huji.ac.il/staff/pitowsky/papers/Paper14.pdf
for a 1990 paper by Pitowski entitled “The Physical Church’s Thesis and Physical Computational Complexity”.
5. jonas says:
So this would be like the 2001/westley entry in IOCC?
6. a says:
Isn’t distributed computing lower bounds kind of what you want for order of computation being specified by an adversary?
7. Gil Kalai says:
Oded Regev gave a simple argument how to simulate an arbitrary quantum circuit by a quantum process with drunken time. See updated post.
8. JimM says:
It sounds like the problem is solved now, but for C in the quantum case you could consider local Hamiltonians. That might be too artificial.
9. Gil Kalai says:
Right, It looks that for the quantum case even B and most likely A allow universal quantum computing. This is the case even when time is completely random. It is even possible that you can do it with applying the same unitary operator at each step, but I am not sure about it.
Another question is if universal quantum computing prevails (or perhaps we are left with BPP) when we have drunken time plus the most simple noise applying to the qubits. (We can assume that gates are noiseless.)
10. Hola says:
\sum_{i=0}^n {\mathcal{J}_i}
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http://rjlipton.wordpress.com/2011/09/17/can-quantum-machines-do-it-all/ | ## a personal view of the theory of computation
tags: Algorithms, new, P=NP, periods, quantum, SAT
by
What keeps SAT out of BQP?
Dan Boneh is an expert on cryptography, especially the use of advanced number theory, such as the Weil pairing. He is famous for his seminal work on identity based encryption with Matt Franklin, which uses the Weil pairing.
Today I want to recall an old result Dan and I proved years ago and mix in a new result.
When Peter Shor announced in 1994 his instantly famous result that there is a quantum factoring algorithm that runs in polynomial time, many of us were amazed. We quickly read his paper, tried to understand why it worked, and followed the proofs step-by-step. It was—is still—a beautiful result.
Yet there was something about the result that troubled me. Perhaps no one else was troubled, but I did not really understand why the theorem was true. Yes I could follow the steps, yes I could explain the result to my students, but I did not really understand it.
I have discussed this phenomenon before: proofs are not just a mechanical means to move a statement from the conjecture pile to the proved pile. No, proofs are much more—they need to convey why the result is really true. They need to not seem magical, which Peter’s proof seemed to me. Proofs need to explain what is really happening.
Now I will say that Scott Aaronson’s well-noted explanation does this for the central idea, but back then Dan and I had no Scott to guide us. So we did something else.
In order to understand what was going on, Dan and I decided the best way would be to try and generalize Peter’s theorem. This is a standard method in mathematics: one way to really see what is happening is to generalize a theorem. Doing this can often shed new light on why the original theorem is true, and also can be interesting in its own right.
I could give countless examples of this method, it is used so often in math. Here is an example from number theory that is later explained better—I believe—by a classic result from group theory. Consider Fermat’s Little Theorem:
Theorem:
If ${p}$ is a prime and ${a}$ is an integer not divisible by ${p}$, then
$\displaystyle a^{p-1} \equiv 1 \bmod p.$
There are many proofs of this fundamental theorem. See our friends at Wikipedia here for several proofs: They have proofs based on: counting necklaces, using modular arithmetic, using the binomial theorem, and using dynamical systems.
Each proof gives a different insight into why ${p}$ must be prime, why ${p}$ cannot divide ${a}$, and why the theorem is true.
The proof that I personally feel best explains what is happening is the one based on Lagrange’s Theorem:
Theorem: In any finite group, the order of an element divides the order of the group.
As an aside Joseph Lagrange did not prove his theorem, but only discussed special cases in 1771. It was finally proved by Pietro Abbati and published in 1803. The proof of Fermat’s Little Theorem now follows since when ${p}$ is prime, the set of numbers
$\displaystyle \{ 1,2,3,\dots,p-1\}$
form a group under multiplication modulo ${p}$.
## Hidden Linear Functions
Dan and I began to look at what we called the hidden linear problem. A function from ${h: \mathbb{Z} \rightarrow S}$ has period ${q}$ provided
$\displaystyle h(x + q) = h(x)$
for all ${x}$. We say that ${h}$ is m-to-one provided at most ${m}$ integers in the set ${0,1,2,\dots,q-1}$ map to the same value in ${S}$. Note, if ${h}$ is one-to-one, then it is m-to-one with ${m=1}$.
Let ${f(x_{1},\dots,x_{k})}$ be a function from the integers ${\mathbb{Z}^{k}}$ to some arbitrary set ${S}$. Say that f has hidden linear structure over ${q}$ provided there are integers and some function ${h}$ of period ${q}$ so that
$\displaystyle f(x_{1},\dots,x_{k}) = h(x_{1} + \alpha_{2}x_{2} + \cdots + \alpha_{k}x_{k})$
for all integers ${x_{1},\dots,x_{k}}$. Further we say that ${f}$ is m-to-one provided ${h}$ is.
Our conjecture, which eventually became a theorem is:
Theorem:
Suppose that ${f(x_{1},\dots,x_{k})}$ is a function that has a hidden linear structure over ${q}$ which is at most m-to-one. Assume the conditions:
1. Let ${n = \log q}$, so that ${m}$ and ${k}$ are at most polynomial in ${n}$.
2. Let ${p}$ be the smallest prime divisor of ${q}$, so that ${m<p}$.
For such a function ${f}$, in quantum polynomial time in ${n}$ we can recover the values of ${\alpha_{2},\dots,\alpha_{k}}$ modulo ${q}$ from an oracle for ${f}$.
Note that the added conditions are critical. If the function ${h}$ were constant then there would be no way to reconstruct the ${\alpha}$‘s: there must be some limit on how much it collapses. The second condition makes the values of the ${\alpha}$‘s make sense modulo ${q}$.
Look at our paper for details. The above theorem is strong enough to prove both factoring and discrete logarithm are in polynomial quantum time. This already indicated that we had made some progress. In Shor’s paper he uses different but similar arguments to handle factoring and the discrete logarithm. I personally thought that our proof helped me have a greater understanding of Shor’s brilliant work.
By the way, proving this theorem was hard. We used the same type of quantum algorithm that Peter did. But the fact that ${h}$ was not one-to-one created serious difficulties for us. ${ }$ Sometimes relaxing from a constant to `polynomial’ is a walk in the park, but not this time… I recall many times that one of us wanted to give up, but eventually we found the key trick. In the one-to-one case—the case that occurs in Peter’s theorems—there is an exponential sum that must be estimated. This sum is quite easy to handle. However, in the poly-to-one case the sum that arose was much harder to bound. We finally found a trick: when trying to bound a sum, change the sum. More on this another time.
## Quantum Detection of Periods
Dan and I could also prove the following theorem:
Theorem:
Suppose that ${h:\mathbb{Z} \rightarrow S}$ has period ${q}$, and no smaller period. Also let ${h}$ be m-to-one. Add the conditions:
1. Let ${n = \log q}$, so that ${m}$ and ${k}$ are at most polynomial in ${n}$.
2. Let ${p}$ be the smallest prime divisor of ${q}$, so that ${m<p}$.
For such a function ${h}$, in quantum polynomial time in ${n}$ we can recover the the period ${q}$.
In Shor’s paper this theorem is essentially proved with ${m=1}$: the so called bijective case. Obviously our result is a modest generalization to the case where the function can fail to be one-to-one, but a value in the range can only have a polynomial number of pre-images. Since then there have been much better improvements to the period solving problem. The current best results can recover the period provided the function ${h}$ has small auto-correlation—see this for details. Indeed for quantum algorithms that only use ${h}$ as an oracle, the current results are essentially best possible.
## SAT
There is another reason that there may be no quantum algorithm for solving the general period finding problem. That is, given a small circuit—not an oracle—for the function ${h}$, find its period. The reason is that the following is a theorem:
Theorem: The following two problems are equivalent via a random reduction:
1. Finding an assignment to a ${\mathsf{SAT}}$ instance.
2. Finding the period of a Boolean valued function given by a small circuit.
Note, the lower bounds of quantum complexity theory are mostly based on a black box argument—they assume that only oracle access is available to the function whose period you want. There is no way known to show that they apply when the function ${h}$ is given by a circuit. This follows since if ${\mathsf{P} = \mathsf{NP}}$ one could guess the period.
This will be an example of “evolving” a theorem from a trivial beginning to more-meaningful forms. At issue first are: how small is `small’ and what is the time on the reduction? Then the main issue will become, how close are the resulting cases of period-finding to ones that quantum algorithms can handle?
First we prove a version where `small’ means polynomial, the reduction is deterministic polynomial time, and the cases of periodicity involved are trivial. Namely, given an instance formula ${\phi}$ for ${\mathsf{SAT}}$, let ${C}$ be the circuit that on input an assignment ${x}$ outputs ${\phi(x)}$. If ${\phi}$ is not satisfiable then the function ${h}$ computed by ${C}$ is identically zero and so has period ${1}$. If ${\phi}$ is true on the all-${0}$ assignment then we say yes. Else, ${\phi}$ is satisfiable, so ${h(0^n) = 0}$ but ${h(x) = 1}$ for some other ${x}$, so ${h}$ does not have period ${1}$. It may not have any period, which is what makes this trivial. But we certainly know ${\phi \notin \mathsf{SAT} \iff h}$ has period ${1}$.
## Amplifying Periods
Now we want ${h}$ always to have a period, in the case where ${\phi \in \mathsf{SAT}}$. This can be done by a simple padding trick. Let us use some number ${n' > n}$ of variables defining a value ${y}$, and define the circuit ${C'}$ by ${C'(y) = C(y \bmod{2^n})}$. Then the function ${h}$ always has a period dividing ${2^n}$—the question is whether that period is ${1}$ or something larger.
Thus finding the period of a periodic function ${h}$ given by poly-size circuits is ${\mathsf{NP}}$-hard. Of course this ${h}$ is far from bijective—it is just 0-1 valued. If we could make ${h}$ bijective on this period, at least when the formula is satisfiable, then we would put ${\mathsf{SAT}}$ into ${\mathsf{BQP}}$. How close can we come?
We start by arranging that when ${\phi \in \mathsf{SAT}}$, with high probability at least one ${h}$ we run across has a period of some prime order ${q}$, where finding ${q}$ for this ${h}$ is enough to say ${\phi \in \mathsf{SAT}}$. Take the above circuit ${C}$; we wish to see if it has an input ${x}$ so that ${C(x)=1}$. We can assume that ${x}$ is unique, if it exists by using the famous result of Leslie Valiant and Vijay Vazirani.
A further simple random argument shows that we can also assume that the solution ${x}$ is a prime number. Here’s why: Consider flipping each input bit of ${C}$ independently: since the proportion of primes up to ${2^n}$ is order-of ${1/n}$, this will map the ${x}$ to a prime number with reasonable probability, and we only need that our test succeeds on it once. In summary we can assume that ${C(x)}$ is either unsatisfiable or has some prime ${q}$ as the unique solution.
Now let ${h(x)}$ be the function
$\displaystyle \bigvee_{p | x} C(p),$
where ${p}$ is over all prime divisors of ${x}$. Then it is easy to see the following: If ${C(x)}$ is unsatisfiable, then ${h}$ is identically ${0}$ and has period ${1}$. If there is a unique prime ${q}$ so that ${C(q)=1}$, then ${h}$ has period ${q}$. Therefore we have reduced solving ${\mathsf{SAT}}$ to period finding—where we can restrict attention to positive cases where the period has prime length and the function is ${0}$ except for a ${1}$ on the period start.
There is a small point, really a BIG point: the function ${h(x)}$ does not have a poly-size circuit. In order to compute it we must factor ${x}$ into it prime factors and then check each one, ${p}$, to see it makes ${C(p)=1}$. However, factoring, currently, has algorithms that run in time roughly order-of ${\exp(n^\epsilon)}$ for ${\epsilon = 1/3}$. Thus ${h}$ will have circuits of this order of sub-exponential size, where the circuits themselves are also succinct. So long as `small’ means size ${\exp(n^\epsilon)}$ for some known ${\epsilon}$, we are OK.
With that point on hold, we can finally try to work on the period structure we have achieved. For arguments ${w}$ where ${h(w) = 0}$, we would like to provide values in ${\{0,2,\dots,q-1\}}$, or at least provide ${w}$ different values. Not knowing ${q}$ in advance may be a problem, but possibly inside the details of reducing to factoring in the previous paragraph, we may be able to assume that ${q}$ is known.
Here is where we pause and say we’ve at least posed a non-trivial research idea. We may work on it tomorrow and find it doesn’t go any further. We might not put it in a paper yet. But in a blog at least we can see how many ideas get their birth and first steps, where hard theorems successfully raised in the past may give some fostering.
## Open Problems
Is ${\mathsf{SAT}}$ in ${\mathsf{BQP}}$? This is the motivating question.
What other ideas for creating and modifying periods might be relevant?
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from → P=NP, People, Proofs
12 Comments leave one →
1. Dave Bacon
September 17, 2011 5:43 pm
Reminds me a bit of an approach that didn’t quite work: http://arxiv.org/abs/quant-ph/0105005 cool that the archive keeps all versions
2. aram
September 17, 2011 5:54 pm
h(x) does have a poly-size quantum circuit.
But making it injective on Z/q is still going to be hard.
3. September 18, 2011 5:00 pm
This is more rigorous version of the theorem. The rest depends on your free will to understand.
Theorem
For the polynomial $p(x)= \sum x_i^4 - 2 x^T Q x,$ $x \in \mathbb{R}^n, Q \in \mathbb{R}^n \times \mathbb{R}^n,$ $p^*(x)= p(x)-\min p(x)= \sum g_i^2(x),$ where $g_i(x)$ are polynomials at most degree 2.
Proof
$p(x)$ is bounded from below and attains its minimum.
Let $x^*$ be one of the global minima of $p^*(x)$, then ${x^*}^T \nabla p^*(x)= 4 \sum {x^*_i}^4 - 4 {x^*}^T Q x^*= {x^*}^T( {x^*}{x^*}^T-Q) x^* = 0$, from which $\min p(x)= -\sum {x^*_i}^4$, and $p^*(x)= \sum x_i^4 - 2 x^T Q x + \sum {x^*_i}^4$.
Polynomial $\hat p(x)= \sum (x_i^2- {x_i^*}^2)^2$ has $2^n$ zeros, including $x^*$
$p(x)= \sum (x_i^2- {x_i^*}^2)^2+ 2x^T({x^*}{x^*}^T-Q)x$. Matrix $\Delta= {x^*}{x^*}^T-Q$ is positive semi-definite with singular vector $x^*$, since for otherwise there exists $\hat x| \hat x - x^*$ parallel to negative eigenvector of $\Delta$, that $p(\hat x) < p(x^*)$.
By Cholesky decomposition $\Delta= LL^T$, and $p(x)= \sum (x_i^2- {x_i^*}^2)^2+ 2x^T L L^T x$ is a sum of squares. $\square$
$p(x)$ can encode partition problem, max-cut, and may be some more. In decision problems there is polynomial gap between satisfying instances and not satisfying instances. Therefore, semi-definite program decide in polynomial time. Similar thing can be done for the minimization problems like max-cut. There there is a polynomial gap between value of maximal cut and next one. This can be accepted only when some people of sum-of-squares programs (SOS) would be willing to consider it seriously. Lazy skepticism will not work here. My experience with posting is that experts catch idea almost instantly, the most sensible comments I got from people from Parrilo group (see, for example, Scott Aaronson post about “why philosophers … “, around comment #140 or so). The rest are lazy skeptical.
Now, why small fraction of people working on sum of squares programs did not go here. First, people realized that quartic polynomials of certain form encode NP-complete problem and did not look further. For example, Monique Laurent in her review about SOS state quartic polynomial encoding, but does not show anywhere, that it cannot be represented as sum of squares of other polynomials. Second, the major direction there is Positivestelensatz arising as a solution to Hilbert 17th problem, i.e. every positive polynomial can be represented as sum of squared polynomial fractions. If someone find the efficient way to solve it, than there would be an algorithm to find minimum of arbitrary polynomial. This is much more general case (and therefore much more difficult). For establishing P=NP it is sufficient to find a class of polynomials that can be solved by SOS and that encode all instances of some NP-complete problem, that is within framework of NZ Shor 1987 and Parrilo thesis 2000, another relevant reference Parrilo Sturmfels 2003.
The asymmetry arising from the enormous bias toward $P \ne NP$ multiplied by the attraction produced by Clay prize, and all-corner advertisement of the importance of the problem renders collaborative approach toward solving this problem almost impossible. The polymath atmosphere, where everyone can draw random crazy idea is immediately kicked back by the guards of the field. Therefore, the intention of drawing huge attention to the problem with relatively small monetary prize given the commonly assumed risks, and the absence of “polymath” atmosphere, seems to me unethical. Therefore, I consciously will not write any form of formal paper on P=NP, for at least formally be ineligible for the prize. Even if the proof above is not rigorous enough, from this point I think it is easy to patch it. This is cranky, and therefore,
Sincerely yours,
Crackpot Trisector
4. khalik sheikh
September 20, 2011 1:19 pm
this is a great one………………..
5. September 21, 2011 2:02 am
Thanks for patching. The mental illness is not when you talk to yourself, but when then no one understands you. Relax for a moment, the proof is wrong. e.g. $\Delta$ can be negative, since vectors $[\pm x_k]$ are not orthogonal. By subtracting more and more of matrix $\Delta$ it is possible to bring $p^*(x)$ to another global minimum, then repeat that with all the axes that are still negative definite, until there are $2^n$ zeros with $\vec x^*_(1)$ orthogonal to each other. Then one need to construct a sum of squares representation of that minimal polynomial (could we say extreme? I think they are extreme in more narrow sense that Choi and Lam definition), or find counter example. I do not have that construction right now. You can help me finding counter example.
• September 29, 2011 2:33 pm
People, this is my last post. Besides emotions I was trying to be as constructive as my limited ability to think allows. There is no place to discuss this issues, and I’m too stupid to run polymath. The question whether $p^*(x)= p(x)-\min p(x)$ is sum of squares for polynomials $p(x)= \sum x_i^4 - 2 x^T Q x,$ $x \in \mathbb{R}^n, Q \in \mathbb{R}^n \times \mathbb{R}^n -$ real symmetric matrix. This is easy to formulate, and should not require very high-ended math to solve it, it is also very concrete. Positive answer settles down P vs NP question. To my little knowledge, the problem is still open. I’m offering \$200 for the first one who will provide counter-example. I’m giving you some hints to make your life easier. If you ever prove (will not be lazy) there are no counter-examples (e.g. $p^*$ that is not sum of squares), you do not have my permission to use my name in connection to this and related problems (laugh here). I removed all the posts from my blog, in couple of month search engine cash will gone, and you are free to go.
For arbitrary set of linearly independent vectors $\vec y_k \in \mathbb{R}^n, \vec y_i^T \vec y_k= 0, \forall i\neq k$ $\sum_{i=1}^n x_i^4 = \sum_{k=1}^n (\alpha_k x^T \vec y_k \vec y_k^T x)^2+ \sum_m (\beta_m x^T B_m x)^2$ and $\vec y_k^T B_m \vec y_k =0, \forall k,m$, for the set of orthogonal vectors it is easy to find $\alpha_k$
$(x^Ty)^2-t= ( x^T y y^T x)^2-t= 0$ describes a pair of hyperplanes orthogonal to $y$ and distance $\sqrt{t}$ from origin. A set of n vectors defines parallelotop $\sum_k \left(( x^T y_k y_k^T x)^2-t_k\right)^2$.
$p(x)^2+g(x)^2=$ $( p(x) \cos(\phi) + g(x) \sin(\phi))^2 +$ $( - p(x) \sin(\phi) + g(x) \cos(\phi))^2$
$\sum_k \vec x_k \vec x_k^T= [x_1, x_2, ..., x_n] [x_1, x_2, ..., x_n]^T= [x_1, x_2, ..., x_n] R^T R [x_1, x_2, ..., x_n]^T,$ $R^T R= I$.
6. Marc Smith
September 22, 2011 6:27 pm
A friend of mine said that Sakai and Kasahara invented Identity-based Cryptography based on the Weil Pairing. They presented their work on a Japanese conference called SCIS, I guess.
I just think that it is really sad that these guys didn’t had their work recognized. Someone should try to invalidate Boneh’s patent.
That’s just my opinion, the opinion of a student.
• September 23, 2011 11:50 pm
In the references of this paper by Sakai and Kasahara, citation [4] is Sakai-Ohgishi-Kasahara, “Cryptosystems based on pairing over Elliptic Curve”, SCIS Jan 2001, while citation [5] is Boneh-Franklin, CRYPTO August 2001. Citation [2] is Sakai-Ohgishi-Kasahara, “Cryptosystems Based on Pairing” from SCIS 2000, while reference [1] is a TR version from 1999 that mentions elliptic curves.
The Boneh-Franklin paper (2003 journal version) cites the 2000 paper as using the pairing for key-exchange. So does this paper by Boneh and Boyen, in its first paragraph which shows key-exchange as one possible application.
This 2010 paper by Aniket Kate and Ian Goldberg says in its abstract, “In this paper, we design distributed PKG setup and private key extraction protocols for three important IBE schemes; namely, Boneh and Franklin’s BF-IBE, Sakai and Kasahara’s SK-IBE, and Boneh and Boyen’s BB1-IBE.” Perhaps they are regarded today as intrinsically different schemes.
Anyway, this shows the authors recognizing each other and their work being recognized separately by a third party. We have discoursed on independent discovery several times in this blog, where the community has reached equitable conclusions different from your opinion.
• September 24, 2011 8:30 am
Dear KW Regan.
Thanks for the enlightment. It made me feel better about academic ethics.
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http://www.mathplanet.com/education/algebra-1/linear-inequalitites/linear-inequalities-in-two-variables | # Linear inequalities in two variables
The solution of a linear inequality in two variables like Ax + By > C is an ordered pair (x, y) that produces a true statement when the values of x and y are substituted into the inequality.
Example:
Is (1, 2) a solution to the inequality
$\\ 2x+3y>1 \\\\2\cdot 1+3\cdot 2\overset{?}{>}1 \\2+5\overset{?}{>}1 \\7>1 \\$
The graph of an inequality in two variables is the set of points that represents all solutions to the inequality. A linear inequality divides the coordinate plane into two halves by a boundary line where one half represents the solutions of the inequality. The boundary line is dashed for > and < and solid for ≤ and ≥. The half-plane that is a solution to the inequality is usually shaded.
Example:
Graph the inequality
$\\ y\geq -x+1 \\$
Videolesson: Graph the linear inequality
$\\y \geq 2x -3\\$
Next Class: Systems of linear equations and inequalities, Graphing linear systems
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http://math.stackexchange.com/questions/179313/tensor-product-of-hilbert-algebras | # Tensor product of Hilbert Algebras
A Hilbert algebra is an inner product space that is also a *-algebra where the various operations and structures interact according to some axioms. One of those axioms is that the linear operation given by left multiplication by an element must be continuous. That is $x \mapsto yx$ is bounded for all $y$. You can take the algebraic tensor product of hilbert algebras and turn it into something that might be a Hilbert algebra by defining all the structures in the obvious way. (Various arguments are required to show these definitions are consistent, as is always the case with a Tensor product, but I have that under control.) The problem is I don't see why left multiplication is continuous. Without the help of an orthonormal basis of $U$ the Hilbert Algebra, which need not exist, I just don't see how to work this out. Please only use purely algebraic concepts, or general analysis things, but no high-powered machinery from the subject of Hilbert Algebras that could possibly be logically dependent on the fact I'm trying to prove.
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Oh, the essential trick I missed was the recognization that we can get an ON set that spans all the vectors involved as factors in an arbitrary sum of fundamental tensors. – Jeff Aug 6 '12 at 7:55
## 1 Answer
Let $U_i$ be the finite list of Hilbert algebras that I'm tensoring. WLOG I can assume the left multiplication $L$ is by a fundamental tensor $\otimes x_i$. Using the universal property of tensor products, we see that there is a linear map on $\otimes U_i$ to itself that, on fundamental tensors, multiplies by $x_i$ in the $i_{th}$ factor and then fixes the other factors. Since $L$ is a composition of such linear maps, it suffices to show that each such is continuous. Here's the trick: let $\sum_{j} \otimes_i x_{ij}$ be a general tensor, and in each $U_i$ find an ON set that spans all the $x_{ij}$ for that $i$. This shows that $\sum_{j} \otimes_i x_{ij}$ can be recast as a linear combination of mutually orthogonal fundamental tensors, from which the required boundedness estimates are clear.
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http://physics.stackexchange.com/questions/56903/the-interpretation-of-mass-in-quantum-field-theories/56939 | # The interpretation of mass in quantum field theories
Consider a free theory with one real scalar field: $$\mathcal{L}:=-\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -\frac{1}{2}m^2\phi ^2.$$ We write this positive coefficient in front of $\phi ^2$ as $\frac{1}{2}m^2$, and then start calling $m$ the mass (of who knows what at this point) and even interpret it as such. But pretend for a moment that you've never seen any sort of field theory before: if someone were to just write this Lagrangian down, it's not immediately apparent why this should be the mass of anything.
So then, first of all, what is it precisely that we mean when we say the word "mass", and how is our constant related to this physical notion in a way that justifies the interpretation of $m$ as mass?
If it helps to clarify, this is how I think about it. There are two notions of mass involved: the mathematical one that is part of our model, and the physical one which we are trying to model. The physical mass needs to be defined by an idealized thought experiment, and then, if our model is to be any good, we should be able to come up with a 'proof' that our mathematical definition agrees with the physical one.
(Of course, none of this at all has anything to do with this particular field theory; it was just the simplest Lagrangian to write down.)
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"But pretend for a moment that you've never seen any sort of field theory before: if someone were to just write this Lagrangian down, it's not immediately apparent why this should be the mass of anything." - It's a basic feature of science that one is actually allowed - and encouraged - to use the brain and learn things that aren't obvious at the beginning. In the QFT above, one analyzes physics and finds that it contains particles whose mass is $m$. What's the problem? Are you serious that you would want all things in science to be obvious from the beginning and without thinking? – Luboš Motl Mar 14 at 19:35
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@Luboš I think Jonathan is trying to shed some light on the black box represented by "one analyzes physics" in your comment. IMO it's an excellent question. – David Zaslavsky♦ Mar 14 at 19:42
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I agree with David that it's an excellent question. @LubošMotl Your question "Are you serious that you would want all things in science to be obvious from the beginning and without thinking?" is an inaccurate, unproductive manipulation of statements made by the OP. – joshphysics Mar 14 at 19:46
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– John Rennie Mar 14 at 20:02
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I would probably go with QFT=>Rep of Poincare alg.=>Casimir invariant $m^2$. Single particle interpretation: $P^2=m^2$ on shell, and we know that $m$ here is precisely the invariant mass. I don't know anything deeper. The parameters in QFT are given their specific names because they correspond to the appropriate things in the appropriate limits (classical and non relativistic). I think, though, the particle interpretation is key to calling that Lagrangian parameter "mass". – twistor59 Mar 15 at 7:48
show 12 more comments
## 3 Answers
There's two (ultimately related) answers.
For the first answer, just forget about $\hbar$ (but say $c=1$), we are doing a classical relativistic field theory.
The first is that you can consider the field profile around a static, spherically symmetric source of mass $M$ (you need to add a coupling to the action of the form $g \phi J$, where $J$ is an external source, to do this).
So we write the static spherically symmetric source as
$J=J_0 \delta^3(r)$
(if you don't like delta functions you can make it a top hat or a shell and you will find the same conclusions below, this kind of manipulation should be familiar from E/M classes)
You can work out the equations of motion for $\phi$, you will find
$\nabla^2\phi + m^2 \phi = g J_0 \delta^3(r)$
This looks exactly like Poisson's equation, except with this extra $m^2$ term (which I haven't called a mass yet). If you solve it (say using green's functions, or you could just make an explicit ansatz) you will find the solution is
$\phi = g J_0 \frac{e^{-m r}}{4\pi r}$
This is the famous yukawa potential with mass $m$. It describes a force with a range $m^{-1}$, which you may have heard before is the smoking gun for force carrying particle having mass (the weak interactions are so weak at human distances because $m$ is so big)
Now at this point you are still justified in asking why we are calling this quantity $m$, which in the classical analysis above is just an inverse length, a mass.
The second answer addresses this more directly. It comes when we quantize the theory. This is discussed in great detail in many qft textbooks, but the gist is relatively simple. At this point I will say $\hbar=1$, if you really want to keep track of the $\hbar$ dependence you can get it back by dimensional analysis.
We now are no longer If you take the equations of motion for the above system without the source (note we are no longer working with static systems):
$-\partial_t^2\phi + \nabla^2 \phi + m^2 \phi^2 = 0$
and go to fourier space
$\tilde{\phi}(\vec{k},t)\equiv\int\frac{d^3 k}{(2\pi)^3}e^{i\vec{k}\cdot\vec{x}}\phi({\vec{x},t})$
then you will find the equations of motion are
$-\partial_t^2 \tilde{\phi} +\vec{k}^2\tilde{\phi} + m^2\tilde{\phi}=0$
This is just a set of harmonic oscillators, labeled by $\vec{k}$ with frequency
$\omega^2=\vec{k}^2+m^2$
We can think of $\tilde{\phi}(\vec{k},t)$ for a single value of $k$ as being a single quantum variable obeying a harmonic oscillator equation, so there when we quantize these variables we will have one wave function for each value of $\vec{k}$, each of which obeys a schrodinger equation for a harmonic oscillator with the above frequency. Each harmonic oscillator will be quantized and have discrete energy levels. These energy levels are what we call 'particles'. The idea is that for a given momentum, the field $\phi$ can only have certain values of energy, and the interpretation is that these quantized wiggles in $\phi$, or in other words discrete packets of energy, are particles.
Now we remember we are doing quantum mechanics, and we are supposed to identify frequency with energy and momentum with wavelength, using
$E=\omega$, $\vec{p}=\vec{k}$
So we are describing a system with a relationship between energy and momentum given by
$E^2=\vec{p}^2+m^2$
This is einstein's famous formula, and now we see that the parameter $m$ appears in exactly the place the mass of a particle would appear. It is the energy that a particle has when it has $0$ momentum. So we identify the particle like excitations we discovered above with particles of mass $m$.
However I emphasize that there are many, many, many other treatments of this and other ways of looking at it.
A good question would be, what does a particle mean if you don't have a free theory and don't get a harmonic oscillator equation above?
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I've been thinking about this question on and off since it was posted, and I have some thoughts that I hope will shed some light on this stuff. I think it helps to begin with the following:
An Analogy from Mechanics.
Consider the following expression that you'll often encounter in classical mechanics: $$L(x, \dot x) = \frac{1}{2}m\dot x^2$$ and let's say that you've never seen Lagrangian mechanics before. Someone comes and tells you that this Lagrangian "describes" a free particle of mass $m$ moving in one dimension. You could then ask the question "how can we identify the parameter $m$ that appears in this expression with the physical mass of a particle?
How would you answer this question? This isn't just meant to be rhetorical, I think it would help to think about this before reading on!
My thought process in answering this question is as follows. Well, we want to identify the parameter $m$ with the physical definition of mass in classical mechanics. How do we define mass operationally in classical mechanics? Well, we exert a force on the mass, and we see the extent to which it accelerates. In other words, we need to interact with the particle and watch what happens to determine its mass. Next, we return to the Lagrangian above. If it were able to predict how a particle reacts to an interaction, then we could use it to identify the parameter $m$ as the mass of a particle.
Unfortunately, the free particle Lagrangian makes no such dynamical predictions. The corresponding Euler-Lagrange equations lead to the equation of motion $$\ddot x = 0$$ which tells us nothing about how a particle will react to an interaction.
Now, you might try to identify $m$ as the mass by defining the energy and momentum in this mathematical model to be the generators of time and space translations respectively and then say "oh look, these expressions correspond to the expressions for energy and momentum of a non-relativistic particle that we're used to, so $m$ must be the mass!" I don't think there is much content or validity to this argument, because making these "definitions* in no way allows one to identify the parameter in the Lagrangian with the physical mass operationally defined in terms of interactions.
In order to identify $m$ as the mass, we need to put terms in the Lagrangian that correspond to interactions (like a potential term), see what the Euler-Lagrange equations of motion predict about the resulting dynamics, and then compare to experiments to identify $m$ as the mass.
Quantum Field Theory.
Using these analogies, I feel dissatisfied by user20797's response where we obtains the relativistic energy-momentum relationship for a particle because as far as I can tell, that procedure amounts to little more than making definitions for energy and momentum that give you the relation you want. It does not tell you how to relate the parameter $m$ to some empirically defined mass, and that, I feel, was your original (quite excellent/subtle in my opinion) question. I think his first response is probably closer to what is necessary.
However, it seems to me that a proper answer to your question requires an analysis of the following (schematic) form whose details (especially empirically), are quite non-trivial.
1. We note that the mass of particles can be defined by their interactions with one another. In particular, we can use scattering experiments to bring about these interactions.
2. We add interaction terms to the Lagrangian density you wrote down, and we develop a prescription by which the resulting quantized field theory can predict what will happen in scattering experiments.
3. We compare the results of our scattering experiments to the predictions of the quantum field theory, and we find that we can identify the parameter $m$ in the Lagrangian density as the physical mass as it was defined through scattering experiments.
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"In order to identify m as the [m]ass, me need to put terms in the Lagrangian the correspond to interactions (like a potential term), see what the Euler-Lagrange equations of motion predict about the resulting dynamics, and then compare to experiments to identify m as the mass." Agreed that this would be a good operational definition of mass. I suggested this in a comment on the question. You should be able to derive $F=ma$ through some straightforward steps in the appropriate limits. – Michael Brown Mar 26 at 13:10
@MichaelBrown Woops thanks :). The more I thought about this, the more I felt dissatisfied with "the standard dreck" in which one shows that one can reproduce the appropriate free particle spectrum and energy-momentum relationship because it doesn't make contact with some such sensible operational definition of mass via dynamics e.g. scattering. – joshphysics Mar 26 at 13:18
The concept of mass is related to the 4-momentum, or 4-momentum density. It is not just a “quadratic term” in the Lagrangian.
By varying Lagrangian one can obtain the equations of motion. But the momentum, or momentum density need to be defined independently from Lagrangian, in order to enable one to introduce the concept of mass or mass density.
For instance, the momentum can be defined as the sum of two isotropic 4-vectors constructed from spinor and co-spinor:
$\xi = \begin{vmatrix} \xi^1 \\ \xi^2 \end{vmatrix}$ - spinor
$\eta = \begin{vmatrix} \eta_{\dot1} \\ \eta_{\dot2} \end{vmatrix}$ - co-spinor
$p_\mu = \frac{1}{2} (\xi^{+}\sigma_\mu \xi)$ - covariant vector
$\tilde p^\mu = \frac{1}{2} (\eta^{+}\tilde\sigma^\mu \eta)$ - contravariant vector
The vector constructed from spinor is covariant, while the vector constructed from co-spinor is contravariant. This is due to different transformations properties of spinor and co-spinor. Both $p_\mu$ and $\tilde p^\mu$ are isotropic (i.e. $p_\mu p^\mu =\tilde p_\mu \tilde p^\mu =0$), but their sum is not isotropic:
$P_\mu = p_\mu +g_{\mu\nu}\tilde p^\nu$ - momentum or momentum density - depending on the model.
The mass, or mass density, is therefore defined as the "square" of the momentum (density) vector:
$m^2=P_\mu P^\mu$
In general, this definition of mass is independent from the choice of the Lagrangian.
However, the momentum and/or momentum density cannot be defined arbitrarily. It has to be defined in such a way that
1. The total momentum is conserved (as a consequence of the equations of motion),
2. The components of the momentum 4-vector are all real-valued, and
3. The momentum 4-vector is time-like to preserve causality.
In general, we need two ingredients to say that mass in our model is really a mass:
1. Lagrangian, or at least equations of motion, and
2. Appropriate definition of momentum or momentum density (see conditions 2 and 3 above).
The conservation of momentum is required to be the consequence of the equations of motion (except for the Theory of gravity where conservation of momentum is an identity!).
It is also easy to demonstrate that we do not necessarily need the “quadratic” and/or constant mass term in the Lagrangian in order to introduce the concept of mass. Here you can find an example of the model where the “mass term” is variable and complex valued, but still allows for real-valued “mass square” of the momentum density vector.
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We can define mass this way, and I already know how to relate this definition to the term appearing in the Lagrangian. The question is, how do we relate this mathematical definition of mass to a precise, physical, operational definition of mass. Does that make sense? – Jonathan Gleason Mar 15 at 12:48
Please explain what in your opinion is "precise, operational definition of mass". To me the only relativistic physical definition of mass is the square of 4-momentum. – Murod Abdukhakimov Mar 15 at 13:19
– Jonathan Gleason Mar 15 at 13:27
If it helps to clarify, this is how I think about it. There are two notions of mass involved: the mathematical one that is part of our model, and the physical one which we are trying to model. The physical mass needs to be defined by an idealized experiment, and then, if our model is to be any good, we should be able to come up with a 'proof' that our mathematical definition agrees with the physical one. Does that make sense? – Jonathan Gleason Mar 15 at 13:30
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I think I understand your question now. Think of the total energy of two photons resulting from annihilation of particle and its antiparticle in the rest frame of their center of inertia. Energy of photon can be expressed (and measured) via its wavelength in difraction experiment. The total energy will be $2m$. – Murod Abdukhakimov Mar 15 at 14:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9410293102264404, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/26942?sort=newest | ## Is pi a good random number generator?
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Part of what I do is study typical behavior of large combinatorial structures by looking at pseudorandom instances. But many commercially available pseudorandom number generators have known defects, which makes me wonder whether I should just use the digits (or bits) of $\pi$.
A colleague of mine says he "read somewhere" that the digits of $\pi$ don't make a good random number generator. Perhaps he's thinking of the article "A study on the randomness of the digits of $\pi$" by Shu-Ju Tu and Ephraim Fischbach. Does anyone know this article? Some of the press it got (see e.g. http://news.uns.purdue.edu/html4ever/2005/050426.Fischbach.pi.html ) made it sound like $\pi$ wasn't such a good source of randomness, but the abstract for the article itself (see http://adsabs.harvard.edu/abs/2005IJMPC..16..281T ) suggests the opposite.
Does anyone know of problems with using $\pi$ in this way? Of course if you use the digits of $\pi$ you should be careful not to re-use digits you've already used elsewhere in your experiment.
My feeling is, you should use the digits of $\pi$ for Monte Carlo simulations. If you use a commercial RNG and it leads you to publish false conclusions, you've wasted time and misled colleagues. If you use $\pi$ and it leads you to publish false conclusions, you've still wasted time and misled colleagues, but you've also found a pattern in the digits of $\pi$!
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en.wikipedia.org/wiki/… – Steve Huntsman Jun 3 2010 at 18:37
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Are there actually examples where commercial RNGs have led to false conclusions in a published paper? – Qiaochu Yuan Jun 3 2010 at 23:10
I doubt it. I personally am much happier believing a published proof (that I can't find an error in) than the output of some sort of RNG built by hand in the real world. – Robby McKilliam Jun 4 2010 at 1:01
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There are cases where pseudorandom number generators can lead to incorrect simulation results (for example, see "Sensitivity of Ballistic Deposition to Pseudorandom Number Generators" by D'Souza, Bar-Yam, and Kardar (Physical Review E 57 (1998), 5044-5052), mae.ucdavis.edu/dsouza/Pubs/bdrng.final_pre.pdf). These aren't really good PRNGs, certainly not cryptographic ones, but a lot of simulations use whatever lousy PRNG happens to be implemented in their favorite programming language, so this can be a real issue. – Henry Cohn Jan 17 2012 at 13:52
## 10 Answers
Strictly speaking, there are some known patterns in the digits of $\pi$. There are some known results on how well $\pi$ can be approximated by rationals, which imply (for example) that we know a priori that the next $n$ as-yet-uncomputed digits of $\pi$ can't all be zero (for some explicit value of $n$ that I'm too lazy to compute right now). In practice, though, these "patterns" are so weak that they will not affect any Monte Carlo experiments.
The main limitation of using the digits of $\pi$ may be the computational speed. Depending on how many random digits you need, computing fresh digits of $\pi$ might become a computational bottleneck. The further out you go, the harder it becomes to compute more digits of $\pi$.
If you are worried about the quality of random digits that you're getting, then you may want to use cryptographic random number generators. For example, finding a pattern in the Blum-Blum-Shub random number generator would probably yield a new algorithm for factoring large integers! Cryptographic random number generators will run more slowly than the "commercial" random number generators you're talking about but you can certainly find some that will generate digits faster than algorithms for computing $\pi$ will.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Cryptographic PRNG's are the gold standard since if you have a practical way to detect the slightest non-randomness in the output, that is considered a break against the generator, and a significant research result (in cryptanalysis) if the PRNG was considered any good (say if it was based on AES, the Advanced Encryption Standard, in some sensible way). It's easy to make them deterministic: for any key K, just take the encryptions E(0), E(1), E(2), ... where E is the encryption function.
Recent x86 computers have a hardware instruction for AES encryption, so it is very fast. It wouldn't surprise me if AES using the hardware instruction is faster than Mersenne Twister implemented in software.
The first edition of "Numerical Recipes" had some discussion of using DES (predecessor of AES) as an RNG, though they took it out of later editions.
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Why not construct your random numbers with non-sequential digits of pi? For example, digits which are 10 digits apart, on a rolling basis.
And you can buy or get a huge dictionary of the digits of pi which should suffice, along with some clever coding on how you grab your digits.
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How would your suggestions improve the randomness properties or practicality of the digit sequence? – S. Carnahan♦ Jan 1 at 9:15
Please note that the Tu and Fischbach analysis was challenged - I don't know of these concerns are valid. See below
Refutation of claims such as “Pi is less random than we thought”. George Marsaglia Professor Emeritus Florida State University
http://interstat.statjournals.net/YEAR/2006/articles/0601001.pdf
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It is known that $\pi$ doesn't equidistribute very well. I'm not sure what this says (if anything) about the `randomness' of its digits, but it might suggest the use of the golden ratio or Euler-Mascheroni constant over $\pi$.
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... or $\ln 2:$ there is an easy spigot algorithm for it – Victor Protsak Jun 4 2010 at 4:01
mmm I wonder if there is a relationship between ease' of spigot algorithm and goodness' of equidistribution? – Robby McKilliam Jun 4 2010 at 7:52
Are there any rigorous quantitative results which say that $\pi$ equidistributes "slowly"? or just numerics? – unknown (google) Jan 17 2012 at 21:15
Also relevant is the Bailey-Borwein-Plouffe formula
$$\pi = \sum_{i=0}^{\infty} \frac1{16^i}\left( \frac{4}{8i+1}-\frac{2}{8i+4}-\frac{1}{8i+5}-\frac{1}{8i+6}\right),$$
which indicates a certain predictability in the base-16 digits of $\pi$.
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see Steve's comment. – Robby McKilliam Jun 3 2010 at 22:16
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but +1 anyway because I think this formula is really cool. – Robby McKilliam Jun 3 2010 at 22:22
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I wish people wouldn't post naked links. I rarely follow them. – Kevin O'Bryant Jun 4 2010 at 0:17
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I agree that it's a cool formula (and useful, to boot), but how does it indicate "predictability", aside from the fact that the computational complexity of computing a single digit or a string of digits using it is fairly low? – Victor Protsak Jun 4 2010 at 3:59
@Victor: Isn't that what "predictability" means? – Kevin O'Bryant Jan 5 at 16:13
In a technical sense, no. A good pseudorandom number generator would be one that you can plug into any randomized algorithm and expect to see the same behavior that you would from an actual random number generator. One way of making a technical definition out of this is to say that the pseudorandom number generator cannot be distinguished from truly random (with probability bounded away from 1/2) by any polynomial time test.
But the digits of π clearly can be distinguished from random by a polynomial time test, namely a test that computes the digits of π and compares them to your supposedly random sequence.
For the same reason, no fully deterministic sequence can be a good random sequence. Instead, to fit this definition, you need to use a pseudorandom number generator that takes some number n of truly random bits as an input seed and generates from them a longer sequence (polynomial in n) of pseudorandom bits that cannot be distinguished from random by a polynomial time algorithm.
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Yes, all this entropy must come from somewhere! In reverse, it was a big issue with ENIGMA decoding. – Victor Protsak Jun 4 2010 at 3:53
Actually, pi has not been proved to be a normal number, and that is surely the minimum requirement for its use as "random numbers".
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Has ANY specific number been proven to be normal? – Paul Siegel Jun 3 2010 at 21:16
@Paul: Yes. See Wikipedia. – Timothy Chow Jun 3 2010 at 21:45
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@Timothy, you aren't suggesting using .123456789101112131415... as a source of random digits, are you? – Gerry Myerson Jun 4 2010 at 1:54
8
@Gerry: I was suggesting that Paul Siegel consult Wikipedia for the answer to his question as to whether any specific number has been proven to be normal. It was Gerald Edgar, not I, who suggested that normality is a necessary (not sufficient) condition for a "random sequence." For my answer to Jim's question, see my answer to Jim's question. – Timothy Chow Jun 4 2010 at 14:38
In trying to construct a less predictable normal number, are numbers using only a subset of the digits useful? Does Dirichlet''s theorem say that the number whose expansion consists of the last digits of successive primes > 5 gives a number with a normal occurrence of the digits 1,3,7,9? This is a variation on Davidac's comment. – roy smith Jan 12 2011 at 18:45
show 1 more comment
The obvious problem here is that a good pseudorandom number generator will generate a different sequence every time you run it, whereas the digits of pi have never been observed to change.
-
7
Yet ... – Mariano Suárez-Alvarez Jun 3 2010 at 20:13
1
So when you run your simulation the next time, just pick up where you left off. – Nate Eldredge Jun 3 2010 at 21:10
Without a random seeding, it is impossible to answer the question whether the digits of $\pi$ appears random. For a constant seed, the digits of $\pi$ appear constant. – David Harris Jan 17 2012 at 13:44
I think it depends upon your application.
I'd say no if you are using the random numbers to generate cryptographic keys, then you immediately open yourself to attacks, because the attacker can probably mimic your random number generator, and thus you add one weak link into the chain.
-
I agree. If you want to do a Monte Carlo experiment, $\pi$ may work, though it will probably be slower than, say, the Mersenne Twister. If you want to do crypto, it's a very bad idea. – Charles Jun 3 2010 at 18:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9097529649734497, "perplexity_flag": "middle"} |
http://xmds.sourceforge.net/worked_examples.html | # Worked Examples¶
One of the best ways to learn XMDS2 is to see several illustrative examples. Here are a set of example scripts and explanations of the code, which will be a good way to get started. As an instructional aid, they are meant to be read sequentially, but the adventurous could try starting with one that looked like a simulation they wanted to run, and adapt for their own purposes.
The nonlinear Schrödinger equation (partial differential equation)
Kubo Oscillator (stochastic differential equations)
Fibre Noise (stochastic partial differential equation using parallel processing)
Integer Dimensions (integer dimensions)
Wigner Function (two dimensional PDE using parallel processing, passing arguments in at run time)
Finding the Ground State of a BEC (continuous renormalisation) (PDE with continual renormalisation - computed vectors, filters, breakpoints)
Finding the Ground State of a BEC again (Hermite-Gaussian basis)
Multi-component Schrödinger equation (combined integer and continuous dimensions with matrix multiplication, aliases)
All of these scripts are available in the included “examples” folder, along with more examples that demonstrate other tricks. Together, they provide starting points for a huge range of different simulations.
## The nonlinear Schrödinger equation¶
This worked example will show a range of new features that can be used in an XMDS2 script, and we will also examine our first partial differential equation. We will take the one dimensional nonlinear Schrödinger equation, which is a common nonlinear wave equation. The equation describing this problem is:
$\frac{\partial \phi}{\partial \xi} = \frac{i}{2}\frac{\partial^2 \phi}{\partial \tau^2} - \Gamma(\tau)\phi+i|\phi|^2 \phi$
where $$\phi$$ is a complex-valued field, and $$\Gamma(\tau)$$ is a $$\tau$$-dependent damping term. Let us look at an XMDS2 script that integrates this equation, and then examine it in detail.
```<simulation xmds-version="2">
<name>nlse</name>
<author>Joe Hope</author>
<description>
The nonlinear Schrodinger equation in one dimension,
which is a simple partial differential equation.
We introduce several new features in this script.
</description>
<features>
<benchmark />
<bing />
<fftw plan="patient" />
<openmp />
<auto_vectorise />
<globals>
<![CDATA[
const double energy = 4;
const double vel = 0.3;
const double hwhm = 1.0;
]]>
</globals>
</features>
<geometry>
<propagation_dimension> xi </propagation_dimension>
<transverse_dimensions>
<dimension name="tau" lattice="128" domain="(-6, 6)" />
</transverse_dimensions>
</geometry>
<vector name="wavefunction" type="complex" dimensions="tau">
<components> phi </components>
<initialisation>
<![CDATA[
const double w0 = hwhm*sqrt(2/log(2));
const double amp = sqrt(energy/w0/sqrt(M_PI/2));
phi = amp*exp(-tau*tau/w0/w0)*exp(i*vel*tau);
]]>
</initialisation>
</vector>
<vector name="dampingVector" type="real">
<components> Gamma </components>
<initialisation>
<![CDATA[
Gamma=1.0*(1-exp(-pow(tau*tau/4.0/4.0,10)));
]]>
</initialisation>
</vector>
<sequence>
<integrate algorithm="ARK45" interval="20.0" tolerance="1e-7">
<samples>10 100 10</samples>
<operators>
<integration_vectors>wavefunction</integration_vectors>
<operator kind="ex" constant="yes">
<operator_names>Ltt</operator_names>
<![CDATA[
Ltt = -i*ktau*ktau*0.5;
]]>
</operator>
<![CDATA[
dphi_dxi = Ltt[phi] - phi*Gamma + i*mod2(phi)*phi;
]]>
<dependencies>dampingVector</dependencies>
</operators>
</integrate>
</sequence>
<output>
<sampling_group basis="tau" initial_sample="yes">
<moments>density</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
density = mod2(phi);
]]>
</sampling_group>
<sampling_group basis="tau(0)" initial_sample="yes">
<moments>normalisation</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
normalisation = mod2(phi);
]]>
</sampling_group>
<sampling_group basis="ktau(32)" initial_sample="yes">
<moments>densityK</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
densityK = mod2(phi);
]]>
</sampling_group>
</output>
</simulation>
```
Let us examine the new items in the <features> element that we have demonstrated here. The existence of the <benchmark> element causes the simulation to be timed. The <bing> element causes the computer to make a sound upon the conclusion of the simulation. The <fftw> element is used to pass options to the FFTW libraries for fast Fourier transforms, which are needed to do spectral derivatives for the partial differential equation. Here we used the option plan=”patient”, which makes the simulation test carefully to find the fastest method for doing the FFTs. More information on possible choices can be found in the FFTW documentation.
Finally, we use two tags to make the simulation run faster. The <auto_vectorise> element switches on several loop optimisations that exist in later versions of the GCC compiler. The <openmp> element turns on threaded parallel processing using the OpenMP standard where possible. These options are not activated by default as they only exist on certain compilers. If your code compiles with them on, then they are recommended.
Let us examine the <geometry> element.
```<geometry>
<propagation_dimension> xi </propagation_dimension>
<transverse_dimensions>
<dimension name="tau" lattice="128" domain="(-6, 6)" />
</transverse_dimensions>
</geometry>
```
This is the first example that includes a transverse dimension. We have only one dimension, and we have labelled it “tau”. It is a continuous dimension, but only defined on a grid containing 128 points (defined with the lattice variable), and on a domain from -6 to 6. The default is that transforms in continuous dimensions are fast Fourier transforms, which means that this dimension is effectively defined on a loop, and the “tau=-6” and “tau=6” positions are in fact the same. Other transforms are possible, as are discrete dimensions such as an integer-valued index, but we will leave these advanced possibilities to later examples.
Two vector elements have been defined in this simulation. One defines the complex-valued wavefunction “phi” that we wish to evolve. We define the transverse dimensions over which this vector is defined by the dimensions tag in the description. By default, it is defined over all of the transverse dimensions in the <geometry> element, so even though we have omitted this tag for the second vector, it also assumes that the vector is defined over all of tau.
The second vector element contains the component “Gamma” which is a function of the transverse variable tau, as specified in the equation of motion for the field. This second vector could have been avoided in two ways. First, the function could have been written explicitly in the integrate block where it is required, but calculating it once and then recalling it from memory is far more efficient. Second, it could have been included in the “wavefunction” vector as another component, but then it would have been unnecessarily complex-valued, it would have needed an explicit derivative in the equations of motion (presumably dGamma_dxi = 0;), and it would have been Fourier transformed whenever the phi component was transformed. So separating it as its own vector is far more efficient.
The <integrate> element for a partial differential equation has some new features:
```<integrate algorithm="ARK45" interval="20.0" tolerance="1e-7">
<samples>10 100 10</samples>
<operators>
<integration_vectors>wavefunction</integration_vectors>
<operator kind="ex" constant="yes">
<operator_names>Ltt</operator_names>
<![CDATA[
Ltt = -i*ktau*ktau*0.5;
]]>
</operator>
<![CDATA[
dphi_dxi = Ltt[phi] - phi*Gamma + i*mod2(phi)*phi;
]]>
<dependencies>dampingVector</dependencies>
</operators>
</integrate>
```
There are some trivial changes from the tutorial script, such as the fact that we are using the ARK45 algorithm rather than ARK89. Higher order algorithms are often better, but not always. Also, since this script has multiple output groups, we have to specify how many times each of these output groups are sampled in the <samples> element, so there are three numbers there. Besides the vectors that are to be integrated, we also specify that we want to use the vector “dampingVector” during this integration. This is achieved by including the <dependencies> element inside the <operators> element.
The equation of motion as written in the CDATA block looks almost identical to our desired equation of motion, except for the term based on the second derivative, which introduces an important new concept. Inside the <operators> element, we can define any number of operators. Operators are used to define functions in the transformed space of each dimension, which in this case is Fourier space. The derivative of a function is equivalent to multiplying by $$i*k$$ in Fourier space, so the $$\frac{i}{2}\frac{\partial^2 \phi}{\partial \tau^2}$$ term in our equation of motion is equivalent to multiplying by $$-\frac{i}{2}k_\tau^2$$ in Fourier space. In this example we define “Ltt” as an operator of exactly that form, and in the equation of motion it is applied to the field “phi”.
Operators can be explicit (kind="ex") or in the interaction picture (kind="ip"). The interaction picture can be more efficient, but it restricts the possible syntax of the equation of motion. Safe utilisation of interaction picture operators will be described later, but for now let us emphasise that explicit operators should be used unless the user is clear what they are doing. That said, XMDS2 will generate an error if the user tries to use interaction picture operators incorrectly. The constant="yes" option in the operator block means that the operator is not a function of the propagation dimension “xi”, and therefore only needs to be calculated once at the start of the simulation.
The output of a partial differential equation offers more possibilities than an ordinary differential equation, and we examine some in this example.
For vectors with transverse dimensions, we can sample functions of the vectors on the full lattice or a subset of the points. In the <sampling_group> element, we must add a string called “basis” that determines the space in which each transverse dimension is to be sampled, optionally followed by the number of points to be sampled in parentheses. If the number of points is not specified, it will default to a complete sampling of all points in that dimension. If a non-zero number of points is specified, it must be a factor of the lattice size for that dimension.
```<sampling_group basis="tau" initial_sample="yes">
<moments>density</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
density = mod2(phi);
]]>
</sampling_group>
```
The first output group samples the mod square of the vector “phi” over the full lattice of 128 points.
If the lattice parameter is set to zero points, then the corresponding dimension is integrated.
```<sampling_group basis="tau(0)" initial_sample="yes">
<moments>normalisation</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
normalisation = mod2(phi);
]]>
</sampling_group>
```
This second output group samples the normalisation of the wavefunction $$\int d\tau |\phi(\tau)|^2$$ over the domain of $$\tau$$. This output requires only a single real number per sample, so in the integrate element we have chosen to sample it many more times than the vectors themselves.
Finally, functions of the vectors can be sampled with their dimensions in Fourier space.
```<sampling_group basis="ktau(32)" initial_sample="yes">
<moments>densityK</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
densityK = mod2(phi);
]]>
</sampling_group>
```
The final output group above samples the mod square of the Fourier-space wavefunction phi on a sample of 32 points.
## Kubo Oscillator¶
This example demonstrates the integration of a stochastic differential equation. We examine the Kubo oscillator, which is a complex variable whose phase is evolving according to a Wiener noise. In a suitable rotating frame, the equation of motion for the variable is
$\frac{dz}{dt} = i z \;\eta$
where $$\eta(t)$$ is the Wiener differential, and we interpret this as a Stratonovich equation. In other common notation, this is sometimes written:
$dz = i z \;\circ dW$
Most algorithms employed by XMDS require the equations to be input in the Stratonovich form. Ito differential equations can always be transformed into Stratonovich euqations, and in this case the difference is equivalent to the choice of rotating frame. This equation is solved by the following XMDS2 script:
```<simulation xmds-version="2">
<name>kubo</name>
<author>Graham Dennis and Joe Hope</author>
<description>
Example Kubo oscillator simulation
</description>
<geometry>
<propagation_dimension> t </propagation_dimension>
</geometry>
<driver name="multi-path" paths="10000" />
<features>
<error_check />
<benchmark />
</features>
<noise_vector name="drivingNoise" dimensions="" kind="wiener" type="real" method="dsfmt" seed="314 159 276">
<components>eta</components>
</noise_vector>
<vector name="main" type="complex">
<components> z </components>
<initialisation>
<![CDATA[
z = 1.0;
]]>
</initialisation>
</vector>
<sequence>
<integrate algorithm="SI" interval="10" steps="1000">
<samples>100</samples>
<operators>
<integration_vectors>main</integration_vectors>
<dependencies>drivingNoise</dependencies>
<![CDATA[
dz_dt = i*z*eta;
]]>
</operators>
</integrate>
</sequence>
<output>
<sampling_group initial_sample="yes">
<moments>zR zI</moments>
<dependencies>main</dependencies>
<![CDATA[
zR = z.Re();
zI = z.Im();
]]>
</sampling_group>
</output>
</simulation>
```
The first new item in this script is the <driver> element. This element enables us to change top level management of the simulation. Without this element, XMDS2 will integrate the stochastic equation as described. With this element and the option name="multi-path", it will integrate it multiple times, using different random numbers each time. The output will then contain the mean values and standard errors of your output variables. The number of integrations included in the averages is set with the paths variable.
In the <features> element we have included the <error_check> element. This performs the integration first with the specified number of steps (or with the specified tolerance), and then with twice the number of steps (or equivalently reduced tolerance). The output then includes the difference between the output variables on the coarse and the fine grids as the ‘error’ in the output variables. This error is particularly useful for stochastic integrations, where algorithms with adaptive step-sizes are less safe, so the number of integration steps must be user-specified.
We define the stochastic elements in a simulation with the <noise_vector> element.
```<noise_vector name="drivingNoise" dimensions="" kind="wiener" type="real" method="dsfmt" seed="314 159 276">
<components>eta</components>
</noise_vector>
```
This defines a vector that is used like any other, but it will be randomly generated with particular statistics and characteristics rather than initialised. The name, dimensions and type tags are defined just as for normal vectors. The names of the components are also defined in the same way. The noise is defined as a Wiener noise here (kind = "wiener"), which is a zero-mean Gaussian random noise with an average variance equal to the discretisation volume (here it is just the step size in the propagation dimension, as it is not defined over transverse dimensions). Other noise types are possible, including uniform and Poissonian noises, but we will not describe them in detail here.
We may also define a noise method to choose a non-default pseudo random number generator, and a seed for the random number generator. Using a seed can be very useful when debugging the behaviour of a simulation, and many compilers have pseudo-random number generators that are superior to the default option (posix).
The integrate block is using the semi-implicit algorithm (algorithm="SI"), which is a good default choice for stochastic problems, even though it is only second order convergent for deterministic equations. More will be said about algorithm choice later, but for now we should note that adaptive algorithms based on Runge-Kutta methods are not guaranteed to converge safely for stochastic equations. This can be particularly deceptive as they often succeed, particularly for almost any problem for which there is a known analytic solution.
We include elements from the noise vector in the equation of motion just as we do for any other vector. The default SI and Runge-Kutta algorithms converge to the Stratonovich integral. Ito stochastic equations can be converted to Stratonovich form and vice versa.
Executing the generated program ‘kubo’ gives slightly different output due to the “multi-path” driver.
```$ ./kubo
Beginning full step integration ...
Starting path 1
Starting path 2
... many lines omitted ...
Starting path 9999
Starting path 10000
Beginning half step integration ...
Starting path 1
Starting path 2
... many lines omitted ...
Starting path 9999
Starting path 10000
Generating output for kubo
Maximum step error in moment group 1 was 4.942549e-04
Time elapsed for simulation is: 2.71 seconds
```
The maximum step error in each moment group is given in absolute terms. This is the largest difference between the full step integration and the half step integration. While a single path might be very stochastic:
The mean value of the real and imaginary components of the z variable for a single path of the simulation.
The average over multiple paths can be increasingly smooth.
The mean and standard error of the z variable averaged over 10000 paths, as given by this simulation. It agrees within the standard error with the expected result of $$\exp(-t/2)$$.
## Fibre Noise¶
This simulation is a stochastic partial differential equation, in which a one-dimensional damped field is subject to a complex noise. This script can be found in examples/fibre.xmds.
$\frac{\partial \psi}{\partial t} = -i \frac{\partial^2 \psi}{\partial x^2} -\gamma \psi+\beta \frac{1}{\sqrt{2}}\left(\eta_1(x)+i\eta_2(x)\right)$
where the noise terms $$\eta_j(x,t)$$ are Wiener differentials and the equation is interpreted as a Stratonovich differential equation. On a finite grid, these increments have variance $$\frac{1}{\Delta x \Delta t}$$.
```<simulation xmds-version="2">
<name>fibre</name>
<author>Joe Hope and Graham Dennis</author>
<description>
Example fibre noise simulation
</description>
<geometry>
<propagation_dimension> t </propagation_dimension>
<transverse_dimensions>
<dimension name="x" lattice="64" domain="(-5, 5)" />
</transverse_dimensions>
</geometry>
<driver name="mpi-multi-path" paths="8" />
<features>
<auto_vectorise />
<benchmark />
<error_check />
<globals>
<![CDATA[
const real ggamma = 1.0;
const real beta = sqrt(M_PI*ggamma/10.0);
]]>
</globals>
</features>
<noise_vector name="drivingNoise" dimensions="x" kind="wiener" type="complex" method="dsfmt" seed="314 159 276">
<components>Eta</components>
</noise_vector>
<vector name="main" initial_basis="x" type="complex">
<components>phi</components>
<initialisation>
<![CDATA[
phi = 0.0;
]]>
</initialisation>
</vector>
<sequence>
<integrate algorithm="SI" iterations="3" interval="2.5" steps="200000">
<samples>50</samples>
<operators>
<operator kind="ex" constant="yes">
<operator_names>L</operator_names>
<![CDATA[
L = -i*kx*kx;
]]>
</operator>
<dependencies>drivingNoise</dependencies>
<integration_vectors>main</integration_vectors>
<![CDATA[
dphi_dt = L[phi] - ggamma*phi + beta*Eta;
]]>
</operators>
</integrate>
</sequence>
<output>
<sampling_group basis="kx" initial_sample="yes">
<moments>pow_dens</moments>
<dependencies>main</dependencies>
<![CDATA[
pow_dens = mod2(phi);
]]>
</sampling_group>
</output>
</simulation>
```
Note that the noise vector used in this example is complex-valued, and has the argument dimensions="x" to define it as a field of delta-correlated noises along the x-dimension.
This simulation demonstrates the ease with which XMDS2 can be used in a parallel processing environment. Instead of using the stochastic driver “multi-path”, we simply replace it with “mpi-multi-path”. This instructs XMDS2 to write a parallel version of the program based on the widespread MPI standard. This protocol allows multiple processors or clusters of computers to work simultaneously on the same problem. Free open source libraries implementing this standard can be installed on a linux machine, and come standard on Mac OS X. They are also common on many supercomputer architectures. Parallel processing can also be used with deterministic problems to great effect, as discussed in the later example Wigner Function.
Executing this program is slightly different with the MPI option. The details can change between MPI implementations, but as an example:
```$xmds2 fibre.xmds
xmds2 version 2.1 "Happy Mollusc" (r2543)
Copyright 2000-2012 Graham Dennis, Joseph Hope, Mattias Johnsson
and the xmds team
Generating source code...
... done
Compiling simulation...
... done. Type './fibre' to run.
```
Note that different compile options (and potentially a different compiler) are used by XMDS2, but this is transparent to the user. MPI simulations will have to be run using syntax that will depend on the MPI implementation. Here we show the version based on the popular open source Open-MPI implementation.
```$ mpirun -np 4 ./fibre
Found enlightenment... (Importing wisdom)
Planning for x <---> kx transform... done.
Beginning full step integration ...
Rank[0]: Starting path 1
Rank[1]: Starting path 2
Rank[2]: Starting path 3
Rank[3]: Starting path 4
Rank[3]: Starting path 8
Rank[0]: Starting path 5
Rank[1]: Starting path 6
Rank[2]: Starting path 7
Rank[3]: Starting path 4
Beginning half step integration ...
Rank[0]: Starting path 1
Rank[2]: Starting path 3
Rank[1]: Starting path 2
Rank[3]: Starting path 8
Rank[0]: Starting path 5
Rank[2]: Starting path 7
Rank[1]: Starting path 6
Generating output for fibre
Maximum step error in moment group 1 was 4.893437e-04
Time elapsed for simulation is: 20.99 seconds
```
In this example we used four processors. The different processors are labelled by their “Rank”, starting at zero. Because the processors are working independently, the output from the different processors can come in a randomised order. In the end, however, the .xsil and data files are constructed identically to the single processor outputs.
The analytic solution to the stochastic averages of this equation is given by
$\langle |\psi(k,t)|^2 \rangle = \exp(-2\gamma t)|\psi(k,0)|^2 +\frac{\beta^2 L_x}{4\pi \gamma} \left(1-\exp(-2\gamma t)\right)$
where $$L_x$$ is the length of the x domain. We see that a single integration of these equations is quite chaotic:
The momentum space density of the field as a function of time for a single path realisation.
while an average of 1024 paths (change paths="8" to paths="1024" in the <driver> element) converges nicely to the analytic solution:
The momentum space density of the field as a function of time for an average of 1024 paths.
## Integer Dimensions¶
This example shows how to handle systems with integer-valued transverse dimensions. We will integrate the following set of equations
$\frac{dx_j}{dt} = x_j \left(x_{j-1}-x_{j+1}\right)$
where $$x_j$$ are complex-valued variables defined on a ring, such that $$j\in \{0,j_{max}\}$$ and the $$x_{j_{max}+1}$$ variable is identified with the variable $$x_{0}$$, and the variable $$x_{-1}$$ is identified with the variable $$x_{j_{max}}$$.
```<simulation xmds-version="2">
<name>integer_dimensions</name>
<author>Graham Dennis</author>
<description>
XMDS2 script to test integer dimensions.
</description>
<features>
<benchmark />
<error_check />
<bing />
<diagnostics /> <!-- This will make sure that all nonlocal accesses of dimensions are safe -->
</features>
<geometry>
<propagation_dimension> t </propagation_dimension>
<transverse_dimensions>
<dimension name="j" type="integer" lattice="5" domain="(0,4)" />
</transverse_dimensions>
</geometry>
<vector name="main" type="complex">
<components> x </components>
<initialisation>
<![CDATA[
x = 1.0e-3;
x(j => 0) = 1.0;
]]>
</initialisation>
</vector>
<sequence>
<integrate algorithm="ARK45" interval="60" steps="25000" tolerance="1.0e-9">
<samples>1000</samples>
<operators>
<integration_vectors>main</integration_vectors>
<![CDATA[
long j_minus_one = (j-1) % _lattice_j;
if (j_minus_one < 0)
j_minus_one += _lattice_j;
long j_plus_one = (j+1) % _lattice_j;
dx_dt(j => j) = x(j => j)*(x(j => j_minus_one) - x(j => j_plus_one));
]]>
</operators>
</integrate>
</sequence>
<output>
<sampling_group basis="j" initial_sample="yes">
<moments>xR</moments>
<dependencies>main</dependencies>
<![CDATA[
xR = x.Re();
]]>
</sampling_group>
</output>
</simulation>
```
The first extra feature we have used in this script is the <diagnostics> element. It performs run-time checking that our generated code does not accidentally attempt to access a part of our vector that does not exist. Removing this tag will increase the speed of the simulation, but its presence helps catch coding errors.
The simulation defines a vector with a single transverse dimension labelled “j”, of type “integer” (“int” and “long” can also be used as synonyms for “integer”). In the absence of an explicit type, the dimension is assumed to be real-valued. The dimension has a “domain” argument as normal, defining the minimum and maximum values of the dimension’s range. The lattice element, if specified, is used as a check on the size of the domain, and will create an error if the two do not match.
Integer-valued dimensions can be called non-locally. Real-valued dimensions are typically coupled non-locally only through local operations in the transformed space of the dimension, but can be called non-locally in certain other situations as described in the reference. The syntax for calling integer dimensions non-locally can be seen in the initialisation CDATA block:
```x = 1.0e-3;
x(j => 0) = 1.0;
```
where the syntax x(j => 0) is used to reference the variable $$x_0$$ directly. We see a more elaborate example in the integrate CDATA block:
```dx_dt(j => j) = x(j => j)*(x(j => j_minus_one) - x(j => j_plus_one));
```
where the vector “x” is called using locally defined variables. This syntax is chosen so that multiple dimensions can be addressed non-locally with minimal possibility for confusion.
## Wigner Function¶
This example integrates the two-dimensional partial differential equation
$\begin{split}\begin{split} \frac{\partial W}{\partial t} &= \Bigg[ \left(\omega + \frac{U_{int}}{\hbar}\left(x^2+y^2-1\right)\right) \left(x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}\right)\\ &\phantom{=\Bigg[} - \frac{U_{int}}{16 \hbar}\left(x\left(\frac{\partial^3}{\partial x^2 \partial y} +\frac{\partial^3}{\partial y^3}\right)-y\left(\frac{\partial^3}{\partial y^2 \partial x}+\frac{\partial^3}{\partial x^3}\right)\right)\Bigg]W(x,y,t) \end{split}\end{split}$
with the added restriction that the derivative is forced to zero outside a certain radius. This extra condition helps maintain the long-term stability of the integration. The script can be found in examples/wigner_arguments_mpi.xmds under your XMDS2 installation directory.
```<simulation xmds-version="2">
<name>wigner</name>
<author>Graham Dennis and Joe Hope</author>
<description>
Simulation of the Wigner function for an anharmonic oscillator with the initial state
being a coherent state.
</description>
<features>
<benchmark />
<globals>
<![CDATA[
real Uint_hbar_on16;
]]>
</globals>
<arguments>
<argument name="omega" type="real" default_value="0.0" />
<argument name="alpha_0" type="real" default_value="3.0" />
<argument name="absorb" type="real" default_value="8.0" />
<argument name="width" type="real" default_value="0.3" />
<argument name="Uint_hbar" type="real" default_value="1.0" />
<![CDATA[
/* derived constants */
Uint_hbar_on16 = Uint_hbar/16.0;
]]>
</arguments>
<bing />
<fftw plan="patient" />
<openmp />
</features>
<driver name="distributed-mpi" />
<geometry>
<propagation_dimension> t </propagation_dimension>
<transverse_dimensions>
<dimension name="x" lattice="128" domain="(-6, 6)" />
<dimension name="y" lattice="128" domain="(-6, 6)" />
</transverse_dimensions>
</geometry>
<vector name="main" initial_basis="x y" type="complex">
<components> W </components>
<initialisation>
<![CDATA[
W = 2.0/M_PI * exp(-2.0*(y*y + (x-alpha_0)*(x-alpha_0)));
]]>
</initialisation>
</vector>
<vector name="dampConstants" initial_basis="x y" type="real">
<components>damping</components>
<initialisation>
<![CDATA[
if (sqrt(x*x + y*y) > _max_x-width)
damping = 0.0;
else
damping = 1.0;
]]>
</initialisation>
</vector>
<sequence>
<integrate algorithm="ARK89" tolerance="1e-7" interval="7.0e-4" steps="100000">
<samples>50</samples>
<operators>
<operator kind="ex" constant="yes">
<operator_names>Lx Ly Lxxx Lxxy Lxyy Lyyy</operator_names>
<![CDATA[
Lx = i*kx;
Ly = i*ky;
Lxxx = -i*kx*kx*kx;
Lxxy = -i*kx*kx*ky;
Lxyy = -i*kx*ky*ky;
Lyyy = -i*ky*ky*ky;
]]>
</operator>
<integration_vectors>main</integration_vectors>
<dependencies>dampConstants</dependencies>
<![CDATA[
real rotation = omega + Uint_hbar*(-1.0 + x*x + y*y);
dW_dt = damping * ( rotation * (x*Ly[W] - y*Lx[W])
- Uint_hbar_on16*( x*(Lxxy[W] + Lyyy[W]) - y*(Lxyy[W] + Lxxx[W]) )
);
]]>
</operators>
</integrate>
</sequence>
<output>
<sampling_group basis="x y" initial_sample="yes">
<moments>WR WI</moments>
<dependencies>main</dependencies>
<![CDATA[
_SAMPLE_COMPLEX(W);
]]>
</sampling_group>
</output>
</simulation>
```
This example demonstrates two new features of XMDS2. The first is the use of parallel processing for a deterministic problem. The FFTW library only allows MPI processing of multidimensional vectors. For multidimensional simulations, the generated program can be parallelised simply by adding the name="distributed-mpi" argument to the <driver> element.
```$ xmds2 wigner_argument_mpi.xmds
xmds2 version 2.1 "Happy Mollusc" (r2680)
Copyright 2000-2012 Graham Dennis, Joseph Hope, Mattias Johnsson
and the xmds team
Generating source code...
... done
Compiling simulation...
... done. Type './wigner' to run.
```
To use multiple processors, the final program is then called using the (implementation specific) MPI wrapper:
```$ mpirun -np 2 ./wigner
Planning for (distributed x, y) <---> (distributed ky, kx) transform... done.
Planning for (distributed x, y) <---> (distributed ky, kx) transform... done.
Sampled field (for moment group #1) at t = 0.000000e+00
Current timestep: 5.908361e-06
Sampled field (for moment group #1) at t = 1.400000e-05
Current timestep: 4.543131e-06
...
```
The possible acceleration achievable when parallelising a given simulation depends on a great many things including available memory and cache. As a general rule, it will improve as the simulation size gets larger, but the easiest way to find out is to test. The optimum speed up is obviously proportional to the number of available processing cores.
The second new feature in this simulation is the <arguments> element in the <features> block. This is a way of specifying global variables with a given type that can then be input at run time. The variables are specified in a self explanatory way
```<arguments>
<argument name="omega" type="real" default_value="0.0" />
...
<argument name="Uint_hbar" type="real" default_value="1.0" />
</arguments>
```
where the “default_value” is used as the valuable of the variable if no arguments are given. In the absence of the generating script, the program can document its options with the --help argument:
```$ ./wigner --help
Usage: wigner --omega <real> --alpha_0 <real> --absorb <real> --width <real> --Uint_hbar <real>
Details:
Option Type Default value
-o, --omega real 0.0
-a, --alpha_0 real 3.0
-b, --absorb real 8.0
-w, --width real 0.3
-U, --Uint_hbar real 1.0
```
We can change one or more of these variables’ values in the simulation by passing it at run time.
```$ mpirun -np 2 ./wigner --omega 0.1 --alpha_0 2.5 --Uint_hbar 0
Found enlightenment... (Importing wisdom)
Planning for (distributed x, y) <---> (distributed ky, kx) transform... done.
Planning for (distributed x, y) <---> (distributed ky, kx) transform... done.
Sampled field (for moment group #1) at t = 0.000000e+00
Current timestep: 1.916945e-04
...
```
The values that were used for the variables, whether default or passed in, are stored in the output file (wigner.xsil).
```<info>
Script compiled with XMDS2 version 2.1 "Happy Mollusc" (r2680)
See http://www.xmds.org for more information.
Variables that can be specified on the command line:
Command line argument omega = 1.000000e-01
Command line argument alpha_0 = 2.500000e+00
Command line argument absorb = 8.000000e+00
Command line argument width = 3.000000e-01
Command line argument Uint_hbar = 0.000000e+00
</info>
```
Finally, note the shorthand used in the output group
```<![CDATA[
_SAMPLE_COMPLEX(W);
]]>
```
which is short for
```<![CDATA[
WR = W.Re();
WI = W.Im();
]]>
```
## Finding the Ground State of a BEC (continuous renormalisation)¶
This simulation solves another partial differential equation, but introduces several powerful new features in XMDS2. The nominal problem is the calculation of the lowest energy eigenstate of a non-linear Schrödinger equation:
$\frac{\partial \phi}{\partial t} = i \left[\frac{1}{2}\frac{\partial^2}{\partial y^2} - V(y) - U_{int}|\phi|^2\right]\phi$
which can be found by evolving the above equation in imaginary time while keeping the normalisation constant. This causes eigenstates to exponentially decay at the rate of their eigenvalue, so after a short time only the state with the lowest eigenvalue remains. The evolution equation is straightforward:
$\frac{\partial \phi}{\partial t} = \left[\frac{1}{2}\frac{\partial^2}{\partial y^2} - V(y) - U_{int}|\phi|^2\right]\phi$
but we will need to use new XMDS2 features to manage the normalisation of the function $$\phi(y,t)$$. The normalisation for a non-linear Schrödinger equation is given by $$\int dy |\phi(y,t)|^2 = N_{particles}$$, where $$N_{particles}$$ is the number of particles described by the wavefunction.
The code for this simulation can be found in examples/groundstate_workedexamples.xmds:
```<simulation xmds-version="2">
<name>groundstate</name>
<author>Joe Hope</author>
<description>
Calculate the ground state of the non-linear Schrodinger equation in a harmonic magnetic trap.
This is done by evolving it in imaginary time while re-normalising each timestep.
</description>
<features>
<auto_vectorise />
<benchmark />
<bing />
<fftw plan="exhaustive" />
<globals>
<![CDATA[
const real Uint = 2.0;
const real Nparticles = 5.0;
]]>
</globals>
</features>
<geometry>
<propagation_dimension> t </propagation_dimension>
<transverse_dimensions>
<dimension name="y" lattice="256" domain="(-15.0, 15.0)" />
</transverse_dimensions>
</geometry>
<vector name="potential" initial_basis="y" type="real">
<components> V1 </components>
<initialisation>
<![CDATA[
V1 = 0.5*y*y;
]]>
</initialisation>
</vector>
<vector name="wavefunction" initial_basis="y" type="complex">
<components> phi </components>
<initialisation>
<![CDATA[
if (fabs(y) < 3.0) {
phi = 1.0;
// This will be automatically normalised later
} else {
phi = 0.0;
}
]]>
</initialisation>
</vector>
<computed_vector name="normalisation" dimensions="" type="real">
<components> Ncalc </components>
<evaluation>
<dependencies basis="y">wavefunction</dependencies>
<![CDATA[
// Calculate the current normalisation of the wave function.
Ncalc = mod2(phi);
]]>
</evaluation>
</computed_vector>
<sequence>
<filter>
<![CDATA[
printf("Hello world from a filter segment!\n");
]]>
</filter>
<filter>
<dependencies>normalisation wavefunction</dependencies>
<![CDATA[
phi *= sqrt(Nparticles/Ncalc);
]]>
</filter>
<integrate algorithm="ARK45" interval="1.0" steps="4000" tolerance="1e-10">
<samples>25 4000</samples>
<filters where="step end">
<filter>
<dependencies>wavefunction normalisation</dependencies>
<![CDATA[
// Correct normalisation of the wavefunction
phi *= sqrt(Nparticles/Ncalc);
]]>
</filter>
</filters>
<operators>
<operator kind="ip" constant="yes">
<operator_names>T</operator_names>
<![CDATA[
T = -0.5*ky*ky;
]]>
</operator>
<integration_vectors>wavefunction</integration_vectors>
<dependencies>potential</dependencies>
<![CDATA[
dphi_dt = T[phi] - (V1 + Uint*mod2(phi))*phi;
]]>
</operators>
</integrate>
<breakpoint filename="groundstate_break.xsil">
<dependencies basis="ky">wavefunction </dependencies>
</breakpoint>
</sequence>
<output>
<sampling_group basis="y" initial_sample="yes">
<moments>norm_dens</moments>
<dependencies>wavefunction normalisation</dependencies>
<![CDATA[
norm_dens = mod2(phi);
]]>
</sampling_group>
<sampling_group initial_sample="yes">
<moments>norm</moments>
<dependencies>normalisation</dependencies>
<![CDATA[
norm = Ncalc;
]]>
</sampling_group>
</output>
</simulation>
```
We have used the plan="exhasutive" option in the <fftw> element to ensure that the absolute fastest transform method is found. Because the FFTW package stores the results of its tests (by default in the ~/.xmds/wisdom directory), this option does not cause significant computational overhead, except perhaps on the very first run of a new program.
This simulation introduces the first example of a very powerful feature in XMDS2: the <computed_vector> element. This has syntax like any other vector, including possible dependencies on other vectors, and an ability to be used in any element that can use vectors. The difference is that, much like noise vectors, computed vectors are recalculated each time they are required. This means that a computed vector can never be used as an integration vector, as its values are not stored. However, computed vectors allow a simple and efficient method of describing complicated functions of other vectors. Computed vectors may depend on other computed vectors, allowing for spectral filtering and other advanced options. See for example, the Advanced Topics section on Convolutions and Fourier transforms.
The difference between a computed vector and a stored vector is emphasised by the replacement of the <initialisation> element with an <evaluation> element. Apart from the name, they have virtually identical purpose and syntax.
```<computed_vector name="normalisation" dimensions="" type="real">
<components> Ncalc </components>
<evaluation>
<dependencies basis="y">wavefunction</dependencies>
<![CDATA[
// Calculate the current normalisation of the wave function.
Ncalc = mod2(phi);
]]>
</evaluation>
</computed_vector>
```
Here, our computed vector has no transverse dimensions and depends on the components of “wavefunction”, so the extra transverse dimensions are integrated out. This code therefore integrates the square modulus of the field, and returns it in the variable “Ncalc”. This will be used below to renormalise the “phi” field. Before we examine that process, we have to introduce the <filter> element.
The <filter> element can be placed in the <sequence> element, or inside <integrate> elements as we will see next. Elements placed in the <sequence> element are executed in the order they are found in the .xmds file. Filter elements place the included CDATA block directly into the generated program at the designated position. If the element does not contain any dependencies, like in our first example, then the code is placed alone:
```<filter>
<![CDATA[
printf("Hello world from a filter segment!\n");
]]>
</filter>
```
This filter block merely prints a string into the output when the generated program is run. If the <filter> element contains dependencies, then the variables defined in those vectors (or computed vectors, or noise vectors) will be available, and the CDATA block will be placed inside loops that run over all the transverse dimensions used by the included vectors. The second filter block in this example depends on both the “wavefunction” and “normalisation” vectors:
```<filter>
<dependencies>normalisation wavefunction</dependencies>
<![CDATA[
phi *= sqrt(Nparticles/Ncalc);
]]>
</filter>
```
Since this filter depends on a vector with the transverse dimension “y”, this filter will execute for each point in “y”. This code multiplies the value of the field “phi” by the factor required to produce a normalised function in the sense that $$\int dy |\phi(y,t)|^2 = N_{particles}$$.
The next usage of a <filter> element in this program is inside the <integrate> element, where all filters are placed inside a <filters> element.
```<filters where="step end">
<filter>
<dependencies>wavefunction normalisation</dependencies>
<![CDATA[
// Correct normalisation of the wavefunction
phi *= sqrt(Nparticles/Ncalc);
]]>
</filter>
</filters>
```
Filters placed in an integration block are applied each integration step. The “where” flag is used to determine whether the filter should be applied directly before or directly after each integration step. The default value for the where flag is where="step start", but in this case we chose “step end” to make sure that the final output was normalised after the last integration step.
At the end of the sequence element we introduce the <breakpoint> element. This serves two purposes. The first is a simple matter of convenience. Often when we manage our input and output from a simulation, we are interested solely in storing the exact state of our integration vectors. A breakpoint element does exactly that, storing the components of any vectors contained within, taking all the normal options of the <output> element but not requiring any <sampling_group> elements as that information is assumed.
```<breakpoint filename="groundstate_break.xsil">
<dependencies basis="ky">wavefunction</dependencies>
</breakpoint>
```
If the filename argument is omitted, the output filenames are numbered sequentially. Any given <breakpoint> element must only depend on vectors with identical dimensions.
This program begins with a very crude guess to the ground state, but it rapidly converges to the lowest eigenstate.
The shape of the ground state rapidly approaches the lowest eigenstate. For weak nonlinearities, it is nearly Gaussian.
When the nonlinear term is larger ($$U=20$$), the ground state is wider and more parabolic.
## Finding the Ground State of a BEC again¶
Here we repeat the same simulation as in the Finding the Ground State of a BEC (continuous renormalisation) example, using a different transform basis. While spectral methods are very effective, and Fourier transforms are typically very efficient due to the Fast Fourier transform algorithm, it is often desirable to describe nonlocal evolution in bases other than the Fourier basis. The previous calculation was the Schrödinger equation with a harmonic potential and a nonlinear term. The eigenstates of such a system are known analytically to be Gaussians multiplied by the Hermite polynomials.
$\left[-\frac{\hbar}{2 m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}\omega^2 x^2\right]\phi_n(x) = E_n \phi_n(x)$
where
$\phi_n(x,t) = \sqrt{\frac{1}{2^n n!}} \left(\frac{m \omega}{\hbar \pi}\right)^\frac{1}{4} e^{-\frac{m \omega x^2}{2\hbar}} H_n\left(\sqrt{\frac{m \omega}{\hbar}x}\right),\;\;\;\;\;\;E_n = \left(n+\frac{1}{2}\right) \omega$
where $$H_n(u)$$ are the physicist’s version of the Hermite polynomials. Rather than describing the derivatives as diagonal terms in Fourier space, we therefore have the option of describing the entire $$-\frac{\hbar}{2 m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}\omega^2 x^2$$ term as a diagonal term in the hermite-Gaussian basis. Here is an XMDS2 simulation that performs the integration in this basis. The following is a simplified version of the examples/hermitegauss_groundstate.xmds script.
```<simulation xmds-version="2">
<name>hermitegauss_groundstate</name>
<author>Graham Dennis</author>
<description>
Solve for the groundstate of the Gross-Pitaevskii equation using the hermite-Gauss basis.
</description>
<features>
<benchmark />
<bing />
<validation kind="run-time" />
<globals>
<![CDATA[
const real omegaz = 2*M_PI*20;
const real omegarho = 2*M_PI*200;
const real hbar = 1.05457148e-34;
const real M = 1.409539200000000e-25;
const real g = 9.8;
const real scatteringLength = 5.57e-9;
const real transverseLength = 1e-5;
const real Uint = 4.0*M_PI*hbar*hbar*scatteringLength/M/transverseLength/transverseLength;
const real Nparticles = 5.0e5;
/* offset constants */
const real EnergyOffset = 0.3*pow(pow(3.0*Nparticles/4*omegarho*Uint,2.0)*M/2.0,1/3.0); // 1D
]]>
</globals>
</features>
<geometry>
<propagation_dimension> t </propagation_dimension>
<transverse_dimensions>
<dimension name="x" lattice="100" length_scale="sqrt(hbar/(M*omegarho))" transform="hermite-gauss" />
</transverse_dimensions>
</geometry>
<vector name="wavefunction" initial_basis="x" type="complex">
<components> phi </components>
<initialisation>
<![CDATA[
phi = sqrt(Nparticles) * pow(M*omegarho/(hbar*M_PI), 0.25) * exp(-0.5*(M*omegarho/hbar)*x*x);
]]>
</initialisation>
</vector>
<computed_vector name="normalisation" dimensions="" type="real">
<components> Ncalc </components>
<evaluation>
<dependencies basis="x">wavefunction</dependencies>
<![CDATA[
// Calculate the current normalisation of the wave function.
Ncalc = mod2(phi);
]]>
</evaluation>
</computed_vector>
<sequence>
<integrate algorithm="ARK45" interval="1.0e-2" steps="4000" tolerance="1e-10">
<samples>100 100</samples>
<filters>
<filter>
<dependencies>wavefunction normalisation</dependencies>
<![CDATA[
// Correct normalisation of the wavefunction
phi *= sqrt(Nparticles/Ncalc);
]]>
</filter>
</filters>
<operators>
<operator kind="ip" constant="yes" type="real">
<operator_names>L</operator_names>
<![CDATA[
L = EnergyOffset/hbar - (nx + 0.5)*omegarho;
]]>
</operator>
<integration_vectors>wavefunction</integration_vectors>
<![CDATA[
dphi_dt = L[phi] - Uint/hbar*mod2(phi)*phi;
]]>
</operators>
</integrate>
<filter>
<dependencies>normalisation wavefunction</dependencies>
<![CDATA[
phi *= sqrt(Nparticles/Ncalc);
]]>
</filter>
<breakpoint filename="hermitegauss_groundstate_break.xsil" format="ascii">
<dependencies basis="nx">wavefunction</dependencies>
</breakpoint>
</sequence>
<output>
<sampling_group basis="x" initial_sample="yes">
<moments>dens</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
dens = mod2(phi);
]]>
</sampling_group>
<sampling_group basis="kx" initial_sample="yes">
<moments>dens</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
dens = mod2(phi);
]]>
</sampling_group>
</output>
</simulation>
```
The major difference in this simulation code, aside from the switch back from dimensionless units, is the new transverse dimension type in the <geometry> element.
```<dimension name="x" lattice="100" length_scale="sqrt(hbar/(M*omegarho))" transform="hermite-gauss" />
```
We have explicitly defined the “transform” option, which by defaults expects the Fourier transform. The transform="hermite-gauss" option requires the ‘mpmath’ package installed, just as Fourier transforms require the FFTW package to be installed. The “lattice” option details the number of hermite-Gaussian eigenstates to include, and automatically starts from the zeroth order polynomial and increases. The number of hermite-Gaussian modes fully determines the irregular spatial grid up to an overall scale given by the length_scale parameter.
The length_scale="sqrt(hbar/(M*omegarho))" option requires a real number, but since this script defines it in terms of variables, XMDS2 is unable to verify that the resulting function is real-valued at the time of generating the code. XMDS2 will therefore fail to compile this program without the feature:
```<validation kind="run-time" />
```
which disables many of these checks at the time of writing the C-code.
## Multi-component Schrödinger equation¶
This example demonstrates a simple method for doing matrix calculations in XMDS2. We are solving the multi-component PDE
$\frac{\partial \phi_j(x,y)}{\partial t} = \frac{i}{2}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)\phi_j(x,y) - i U(x,y) \sum_k V_{j k}\phi_k(x,y)$
where the last term is more commonly written as a matrix multiplication. Writing this term out explicitly is feasible for a small number of components, but when the number of components becomes large, or perhaps $$V_{j k}$$ should be precomputed for efficiency reasons, it is useful to be able to perform this sum over the integer dimensions automatically. This example show how this can be done naturally using a computed vector. The XMDS2 script is as follows:
```<simulation xmds-version="2">
<name>2DMSse</name>
<author>Joe Hope</author>
<description>
Schroedinger equation for multiple internal states in two spatial dimensions.
</description>
<features>
<benchmark />
<bing />
<fftw plan="patient" />
<openmp />
<auto_vectorise />
</features>
<geometry>
<propagation_dimension> t </propagation_dimension>
<transverse_dimensions>
<dimension name="x" lattice="32" domain="(-6, 6)" />
<dimension name="y" lattice="32" domain="(-6, 6)" />
<dimension name="j" type="integer" lattice="2" domain="(0,1)" aliases="k"/>
</transverse_dimensions>
</geometry>
<vector name="wavefunction" type="complex" dimensions="x y j">
<components> phi </components>
<initialisation>
<![CDATA[
phi = j*sqrt(2/sqrt(M_PI/2))*exp(-(x*x+y*y)/4)*exp(i*0.1*x);
]]>
</initialisation>
</vector>
<vector name="spatialInteraction" type="real" dimensions="x y">
<components> U </components>
<initialisation>
<![CDATA[
U=exp(-(x*x+y*y)/4);
]]>
</initialisation>
</vector>
<vector name="internalInteraction" type="real" dimensions="j k">
<components> V </components>
<initialisation>
<![CDATA[
V=3*(j*(1-k)+(1-j)*k);
]]>
</initialisation>
</vector>
<computed_vector name="coupling" dimensions="x y j" type="complex">
<components>
VPhi
</components>
<evaluation>
<dependencies basis="x y j k">internalInteraction wavefunction</dependencies>
<![CDATA[
// Calculate the current normalisation of the wave function.
VPhi = V*phi(j => k);
]]>
</evaluation>
</computed_vector>
<sequence>
<integrate algorithm="ARK45" interval="2.0" tolerance="1e-7">
<samples>20 100</samples>
<operators>
<integration_vectors>wavefunction</integration_vectors>
<operator kind="ex" constant="yes">
<operator_names>Ltt</operator_names>
<![CDATA[
Ltt = -i*(kx*kx+ky*ky)*0.5;
]]>
</operator>
<![CDATA[
dphi_dt = Ltt[phi] -i*U*VPhi;
]]>
<dependencies>spatialInteraction coupling</dependencies>
</operators>
</integrate>
</sequence>
<output>
<sampling_group basis="x y j" initial_sample="yes">
<moments>density</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
density = mod2(phi);
]]>
</sampling_group>
<sampling_group basis="x(0) y(0) j" initial_sample="yes">
<moments>normalisation</moments>
<dependencies>wavefunction</dependencies>
<![CDATA[
normalisation = mod2(phi);
]]>
</sampling_group>
</output>
</simulation>
```
The only truly new feature in this script is the “aliases” option on a dimension. The integer-valued dimension in this script indexes the components of the PDE (in this case only two). The $$V_{j k}$$ term is required to be a square array of dimension of this number of components. If we wrote the k-index of $$V_{j k}$$ using a separate <dimension> element, then we would not be enforcing the requirement that the matrix be square. Instead, we note that we will be using multiple ‘copies’ of the j-dimension by using the “aliases” tag.
```<dimension name="j" type="integer" lattice="2" domain="(0,1)" aliases="k"/>
```
This means that we can use the index “k”, which will have exactly the same properties as the “j” index. This is used to define the “V” function in the “internalInteraction” vector. Now, just as we use a computed vector to perform an integration over our fields, we use a computed vector to calculate the sum.
```<computed_vector name="coupling" dimensions="x y j" type="complex">
<components>
VPhi
</components>
<evaluation>
<dependencies basis="x y j k">internalInteraction wavefunction</dependencies>
<![CDATA[
// Calculate the current normalisation of the wave function.
VPhi = V*phi(j => k);
]]>
</evaluation>
</computed_vector>
```
Since the output dimensions of the computed vector do not include a “k” index, this index is integrated. The volume element for this summation is the spacing between neighbouring values of “j”, and since this spacing is one, this integration is just a sum over k, as required.
By this point, we have introduced most of the important features in XMDS2. More details on other transform options and rarely used features can be found in the Advanced Topics section.
### Quick search
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http://unapologetic.wordpress.com/2008/10/28/the-category-of-representations/?like=1&source=post_flair&_wpnonce=966197e62c | # The Unapologetic Mathematician
## The Category of Representations
Now let’s narrow back in to representations of algebras, and the special case of representations of groups, but with an eye to the categorical interpretation. So, representations are functors. And this immediately leads us to the category of such functors. The objects, recall, are functors, while the morphisms are natural transformations. Now let’s consider what, exactly, a natural transformation consists of in this case.
Let’s say we have representations $\rho:A:\rightarrow\hom_\mathbb{F}(V,V)$ and $\sigma:A\rightarrow\hom_\mathbb{F}(W,W)$. That is, we have functors $\rho$ and $\sigma$ with $\rho(*)=V$, $\sigma(*)=W$ — where $*$ is the single object of $A$, when it’s considered as a category — and the given actions on morphisms. We want to consider a natural transformation $\phi:\rho\rightarrow\sigma$.
Such a natural transformation consists of a list of morphisms indexed by the objects of the category $A$. But $A$ has only one object: $*$. Thus we only have one morphism, $\phi_*$, which we will just call $\phi$.
Now we must impose the naturality condition. For each arrow $a:*\rightarrow *$ in $A$ we ask that the diagram
$\displaystyle\begin{matrix}V&\xrightarrow{\phi}&W\\\downarrow^{\rho(a)}&&\downarrow^{\sigma(a)}\\V&\xrightarrow{\phi}&W\end{matrix}$
commute. That is, we want $\phi\circ\rho(a)=\sigma(a)\circ\phi$ for every algebra element $a$. We call such a transformation an “intertwiner” of the representations. These intertwiners are the morphisms in the category of $\mathbf{Rep}(A)$ of representations of $A$. If we want to be more particular about the base field, we might also write $\mathbf{Rep}_\mathbb{F}(A)$.
Here’s another way of putting it. Think of $\phi$ as a “translation” from $V$ to $W$. If $\phi$ is an isomorphism of vector spaces, for instance, it could be a change of basis. We want to take a transformation from the algebra $A$ and apply it, and we also want to translate. We could first apply the transformation in $V$, using the representation $\rho$, and then translate to $W$. Or we could first translate from $V$ to $W$ and then apply the transformation, now using the representation $\sigma$. Our condition is that either order gives the same result, no matter which element of $A$ we’re considering.
## 8 Comments »
1. [...] these representations don’t live in a vacuum. No, they’re just the objects of a whole category of representations. We need to consider the morphisms between representations [...]
Pingback by | October 30, 2008 | Reply
2. [...] and Quotient Representations Today we consider subobjects and quotient objects in the category of representations of an algebra . Since the objects are representations we call these [...]
Pingback by | December 5, 2008 | Reply
3. [...] grading day, another straightforward post. It should come as no surprise that the collection of intertwining maps between any two representations forms a vector [...]
Pingback by | December 11, 2008 | Reply
4. [...] Images of Intertwiners The next obvious things to consider are the kernel and the image of an intertwining map. So let’s say we’ve got a representation , a representation , and an intertwiner [...]
Pingback by | December 12, 2008 | Reply
5. [...] Category of Representations is Abelian We’ve been considering the category of representations of an algebra , and we’re just about done showing that is [...]
Pingback by | December 15, 2008 | Reply
6. [...] of . Then since the symmetrizer and antisymmetrizer are elements of the group algebra , they define intertwiners from to itself. The their images are not just subspaces on which the symmetric group acts nicely, [...]
Pingback by | December 22, 2008 | Reply
7. [...] The antisymmetrizer (for today) is an element of the group algebra , and thus defines an intertwiner from to itself. Its image is thus a subrepresentation of acting on [...]
Pingback by | December 23, 2008 | Reply
8. I love ending up in this blog every time I google something. It always answers my questions in such a terse and precise way.
Comment by Ebrahim | December 7, 2012 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
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http://physics.stackexchange.com/questions/9754/dimensonal-analysis-of-damping-constant/9780 | # Dimensonal analysis of damping constant?
What are the units of the damping constant from the following equation by dimensional analysis?
$$\zeta = \frac{c}{2\sqrt{mk}}$$
I'm assuming the units have to be s^-1, as the damping constant is present in the exponential equation which plots damping of y=Ae^kt (which plots amplitude vs time). Is that a correct assumption?
If somebody could do a quick dimensional analysis to confirm it would be great.
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Maybe you could tell, what $c$, $m$, $k$ and $\zeta$ are? (once you know what these other constants are you might be in the position to answer the question yourself) – Fabian May 12 '11 at 8:43
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Is this homework? – Georg May 12 '11 at 8:44
Well mass is kg, k is spring constant (N/m) and ζ is obviously damping constant. I have no idea what the units of c area however. – user3511 May 12 '11 at 9:19
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@user3511 If you know what dimensional analysis is, you can easily perform it yourself. If you do not know what it is, use the internet to learn. If you try to use the internet to learn, and run into trouble because there's something about it you don't understand, ask a question here. As it is, your question makes no sense because you just threw up an equation with undefined quantities, and when someone asked you what the variables meant, you replied that you didn't know. Why are you asking about an equation with quantities you can't even identify? -1 for lack of effort. – Mark Eichenlaub May 12 '11 at 10:26
My apologies, i needed help in confirming the steps and the units of each. Next time I will post them in the question. – user3511 May 12 '11 at 11:43
## 2 Answers
I'm assuming the units have to be s^-1, as the damping constant is present in the exponential equation which plots damping of y=Ae^kt (which plots amplitude vs time). Is that a correct assumption?
No, the damping ratio $\zeta$ is dimensionless:
$$[c] = \frac{[F]}{\left[\frac{dx}{dt}\right]} = \frac{\mathrm{N}}{\mathrm{m}\cdot\mathrm{s}^{-1}} = \frac{\mathrm{kg}\cdot\mathrm{m}\cdot\mathrm{s}^{-2}}{\mathrm{m}\cdot\mathrm{s}^{-1}} = \mathrm{kg}\cdot\mathrm{s}^{-1}$$
$$[\zeta] = \frac{[c]}{\sqrt{[m][k]}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\sqrt{\mathrm{kg}\cdot\mathrm{N}\cdot\mathrm{m}^{-1}}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\sqrt{\mathrm{kg}^2\cdot\mathrm{s}^{-2}}} = \frac{\mathrm{kg}\cdot\mathrm{s}^{-1}}{\mathrm{kg}\cdot\mathrm{s}^{-1}} = 1$$
The solution of the damped harmonic oscillator differential equation (when underdamped) is
$$x(t) = A e^{-\zeta \omega_0 t} \ \sin \left( \sqrt{1-\zeta^2} \ \omega_0 t + \varphi \right)$$
so the exponent is dimensionless (as it must be):
$$[\zeta \omega_0 t] = 1\cdot\mathrm{s}^{-1}\cdot\mathrm{s} = 1$$
Dimensionless and dimensionful parameters
The differential equation for a damped harmonic oscillator is
$$m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$$
We can reduce the number of parameters to 2 just by dividing by $m$
$$\frac{d^2x}{dt^2} + \frac{c}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$$
Then we can transform the two remaining parameters to get a dimensionless one, controlling the shape of the solution, and a dimensionful one, setting the timescale. One way of doing that is to define
$$\omega_0 = \sqrt{\frac{k}{m}}$$
$$\zeta = \frac{\frac{c}{m}}{\omega_0} = \frac{c\sqrt{m}}{m\sqrt{k}} = \frac{c}{\sqrt{k\,m}}$$
so that the differential equation takes the form:
$$\frac{d^2x}{dt^2} + \zeta\omega_0\frac{dx}{dt} + \omega_0^2x = 0$$
The reason to choose $\omega_0$ as the dimensionful parameter is physical: when the system is underdamped, $\omega_0$ is the angular frequency of oscillation.
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I was posting the same thing, but you beat me to it! +1. For a perfect answer, you could add something about the characteristic equation. – Gerben May 12 '11 at 23:01
Thanks, thats great! Can you explain further why the damping constant will be dimensionless in terms of the differential equation? (I don't quite understand the last line). – user3511 May 13 '11 at 8:59
@user3511 I added some more details. – mmc May 13 '11 at 16:43
– user3511 May 15 '11 at 5:57
@user3511 The $\gamma$ shown there is just $\zeta\,\omega_0$. So, while $\zeta$ is dimensionless, $[\gamma] = \mathrm{s}^{-1}$. The same solution can be expressed in many different ways, emphasizing different things :-) – mmc May 15 '11 at 15:04
show 1 more comment
The above is unit less. How? well damping is always force/speed thus $c=[\rm{N}\,\rm{s}\,\rm{m^{-1}}]$, and stiffness is force/distance $k=[\rm{N}\,\rm{m^{-1}}]$, and of course a newton is $[\rm{N}]=[\rm{kg}\,\rm{m}\,\rm{s^{-2}}]$
Combine them all to make
$$\dfrac{\mathrm{N\,/(m/s)}}{\sqrt{{\rm kg\, N/m}}}=\sqrt{\dfrac{{\rm N}\,{\rm s}^{2}}{{\rm kg\, m}}}=\sqrt{\dfrac{{\rm kg\, (m/s^{2})\, s^{2}}}{{\rm kg\, m}}}=1$$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 11, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9353344440460205, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/4934/symmetrical-presentation-of-4-dimensional-rotation-matrix/70260 | ## Symmetrical Presentation of 4-Dimensional Rotation Matrix
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This question is not urgent; just a matter of curiosity...
It is relatively easy to generate an arbitrary 3D or even 4D rotation matrix using conjugation (i.e. YXY−1) of orthogonal rotations. I expect that there are ways to choose the contributing orthogonal angles of rotation in order to get a uniform random distribution of the resulting axis (and angle). (3D rotations are also related to quaternions.)
There's a wonderfully symmetrical 3D rotation matrix presentation, given in the first edition of D. F. Rogers and J. A. Adams book "Mathematical Elements for Computer Graphics". The elements of the matrix are:
u2+(1-u2)c; uv(1-c)+ws; uw(1-c)-vs
uv(1-c)-ws; v2+(1-v2)c; vw(1-c)+us
uw(1-c)+vs; vw(1-c)-us; w2+(1-w2)c
Here, (u,v,w) is a unit vector along the chosen axis of rotation, and s and c are the sine and cosine of the chosen angle of rotation.
[I mention this on my blog page.]
Now, a 4D rotation must be about a "2D-axis", or plane (where a 3D rotation is about a "1D-axis", or line).
I wonder if there's an equally elegant 4x4 matrix, in terms of a pair of mutually orthogonal unit vectors (defining the plane of rotation) and the sine and cosine of the angle of rotation in 4D.
Any on-line references regarding anything mentioned here would we gratefully received.
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## 4 Answers
A 4-d rotation does not have to fix a 2-d axis. For example, (complex) multiplication by $e^{i\theta}$ rotates every vector of unit length in $C^2$ the same way''.
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DC: Thanks; could you please briefly explain multiplication by e^(i theta)? I understand this in 2D, as it applies to complex numbers (pairs of reals) where this may also be written cis(theta) or cos(theta)+i*sin(theta), and thus corresponds to a the 2D matrix [[c,s],[-s,c]]. How is this extended to 4-tuples please? – Rhubbarb Nov 11 2009 at 9:39
DC: Your reply did highlight that I'd made a false assumption: that 4D rotations must have fixed points, as do 3D rotations. However, there is at least one simple counter example: the map (w,x,y,z) --> (-w,-x,-y,-z). Because the dimension is even, this is a true rigid rotation, and does not have a reflection component. This particular map may be thought of as some kind of inversion or reflection through a point. – Rhubbarb Nov 11 2009 at 9:43
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You can think of a pair of complex numbers (ie a vector in $C^2$) as a 4-tuple of real numbers (by taking the real and imaginary part of each complex number). So a transformation which multiplies each complex number by $e^{i\theta}$ "is" a rotation of (real) 4-space in which every vector rotates (through the same angle $\theta$). – Danny Calegari Nov 11 2009 at 14:39
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A nice way to look at the 4-dimensional rotation matrices $SO_4$ is that it's universal cover is isomorphic to $S^3 \times S^3$. The map $S^3 \times S^3 \to SO_4$ is given by left and right quaternionic multiplication by a unit vectors.
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Furthermore, it's "easy" to generate uniform random elements of S^3 (i. e. selected according to the Haar measure, I think). These can be generated by taking a vector of four independent standard normal random variables, and then normalizing to unit length. I suspect that the uniform measure on S^3 x S^3 projects down to uniform measure on SO(4). – Michael Lugo Nov 11 2009 at 2:52
• a 4-d rotation can rotate on two orthogonal planes at the same time. Multiplication by quaternion qx rotates the plane P determined by the unit and non real part of q and the plane orthogonal to that plane. multiplying by q^-1 on the right side gives qxq^-1 gives a rotation about an axis as the multiplication of quaternions is not commutative and the rotations in the P plane cancels out while the rotation in the other plane is doubled. This is related to the Hopf fibration. The symmetry group of the 600-cell is the double cover of the icosahedral group and is called the binary icosahedral group. I think you could represent a rotation as a division into two orthogonal planes and then take two rotations of arbitrary magnitude in each plane.
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http://mathoverflow.net/questions/76325/navier-stokes-equations-in-riemannian-geometry/76452 | ## Navier-Stokes equations in riemannian geometry
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Hello,
The Navier-Stokes equations can be written on a riemannian manifold: $$\dot{u}+\nabla_u u+ \Delta u=(df)^*$$ $$d^* u=0$$ where $\nabla$ is the Levi-Civita connection, $u$ is a vector fields, $\Delta$ is the laplacian, $df$ is the differential of $f$, $(df)^*$ is the dual of $df$ by the metric, $d^*u$ is the divergence of $u$.
The problem is due to Antoine Balan.
Do you have references ?
Thanks in advance.
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Have you looked at the work of Marsden and Weinstein? – Deane Yang Sep 25 2011 at 14:05
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Moreover, googling "Navier-Stokes Riemannian manifold" produces a lot of hits. – Deane Yang Sep 25 2011 at 19:27
From the point of view of extending the results from flat space to compact manifolds, there is no difficulty at all: arxiv.org/abs/0901.4412 – timur Apr 3 2012 at 22:55
## 5 Answers
The answer and comments about Arnold and Marsden papers are a little off side. They concern the equation of inviscid fluids, called Euler equation. This differs from Navier-Stokes by the highest-order derivatives $\Delta u$. This changes completely the functional analysis background. Also, Euler equation has a geometrical interpretation (geodesics on the group of measure-preserving diffeomorphisms), whereas Navier-Stokes has not.
I am not aware of references for Navier-Stokes on manifolds. However, I don't think that this is a real problem. What has been important so far for Navier-Stokes is the space dimension and the embedding theorems we have between functional spaces like Sobolev, Besov and others. For instance, the Cauchy problem must be globally well-posed on every compact surface, and locally well-posed on $3$-manifolds.
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You are missing a $\dot u$ in your equation! We want a dynamic vector field. The sign of your $\nabla_u u$ and $\Delta u$ are usually taken to be opposite, as with the sign of your $df^*$ and $\nabla_u u$. See p. 63 of Arnol'd-Khesin's book `Topological Methods in Fluid Mechanics'. Arnol'd and Khesin definitely knew how to do this. Khesin is still alive!
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Dear Richard Montgomery, I would add a link to the text "Topological Methods in Hydrodynamics" by Arnol'd and Khesin. books.google.com/… – Giuseppe Sep 26 2011 at 5:24
You could look at the paper: Groups of Diffeomorphisms and the motion of an incompressible fluid, by Ebin and Marsden.
About two centuries after Euler, in 1966 Arnold gave a geometric reformulation of the classical equations for an imcompressible fluid in terms of the geodesic spray of left invariant metric on an infinite dimensional Lie Group.
Ebin and Marsden promptly employed this reformulation to obtain existence and uniqueness results for these equations on compact oriented riemannian manifolds.
This circle of ideas is one of the first important application of infinite dimensional manifolds as remarked by Stephen Smale.
By the way, should not the equation contain the time derivative of the unknown $u$?
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For what it's worth, the Navier-Stokes equation on manifolds is also mentioned in this recent paper http://arxiv.org/pdf/1107.2698, see (1.16) there, in connection with another flow for vector fields that the authors define.
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Do you think it would be possible to extend the results of Arnol'd to the Navier-Stokes equation ?
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For instance, considering the Navier-Stokes equation as a small perturbation of the Euler equations, just as it was done by Cruzeiro et al., but on a stochastic point of view – Claire Oct 31 2011 at 20:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.924092710018158, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/6300/what-is-the-angular-momentum-spectrum-of-an-sp3-electron?answertab=votes | What is the angular momentum spectrum of an sp${}^3$ electron?
So, one thing has been annoying me ever since I learned about orbital hybridization: you explain the shape of molecules by postulating that the orbitals of multi-electron atoms are linear combinations of the orbitals of the hydrogen atom. Fine.
Here's what gets me, and I've asked several chemistry professors about this, and I haven't gotten an answer. If I measure the orbital angular momentum of an electron in a hydrogen atom, I get the value $L = \hbar \sqrt{\ell(\ell +1)}$, so an s orbital has angular momentum $0$ and a p orbital has angular momentum $\hbar$.
So, what happens when I measure the orbital angular momentum of a sp${}^{3}$ electron? Do I get $\frac{3\hbar}{4}$ with probability 1? Or do I get $\hbar$ with probability 0.75, and $0$ with probability 0.25? Are the original orbitals really the fundamental thing, and the hybridized orbitals fundamentally just quantum superpositions of these things? Or does the multi-electron potential create a state that just looks like a linear combination of one s and 3 p orbitals?
Or is the actual answer more complicated and this a heuristic picture that I'm taking too seriously?
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The eigenvalue of $L^2$ can never be 3/4 of an allowed eigenvalue: there are no two eigenvalues that differ by a factor of 3/4 (except for two zeros). The allowed spectrum of $L^2$ is always $l(l+1)\hbar^2$ - it follows from basic SO(3) group theory. – Luboš Motl Mar 3 '11 at 20:10
3 Answers
You're definitely taking this picture too seriously. Hybridization of atomic orbitals is just an approximation based on an independent particle model created by Pauling to rationalize some structural trends in chemistry with Quantum Mechanics. We can only assign orbitals unambiguously for one-electron systems, though of course independent particle models were widely employed and still are to explain reactivity trends, etc.
Anyway, if you prepare an electron in an sp3 state, which is an equally weighted linear combination of the px, py pz and s orbitals, the probability to find and electron at a certain angular momentum might be calculated by taking the square of the projection of this angular momentum on the wave function, as QM tells us to do.
Don't worry too much if you find weak spots in general theories developed to explain molecules, reactivity, etc, since they're highly approximated in most cases (unless you do an expensive computation for a system, but then it is not general anymore), and the guys that developed those such as Pauling already knew that. As you know these highly approximated theories (e.g., Molecular Orbital and Valence Bond theory) were and still are very successful, even if they're not very rigorous.
It gets really hard to be rigorous in chemistry beyond a certain point...
@TedBunn
Of course the basic formalism of QM would predict properties of this hypothetical electronic state, I did not say anything that would imply the opposite.
Still, to think about bonding in terms of sp3 hybrid orbitals such as one does in general chemistry is really a very crude picture, unless you're dealing with molecules that have very special properties such as Td symmetry (methane, for example), and still in those cases if you perform a calculation by using VB or MO (Hartree-Fock) theory (and it does not even need to be with a computer, since symmetry is going to make life very easy here) with only the SP3 orbitals of carbon and S of hydrogen in your set of basis functions, you will see a very big quantitative error in predicted ionization energies when you compare to photoelectron spectroscopy measurements of ionization energies. Another good test would be to perform high level quantum chemistry calculation of methane using a software and employing two different basis sets, one containing only s and p functions centered on carbon and s functions centered on hydrogen atoms, and another in which the basis set has functions of several angular momenta centered on each atom. If you compare your results with experiments you will see that the first method will have a much lower accuracy compared to the second one. S and P functions might still be the ones that contribute mostly to the bonding molecular orbitals in the different orbital configurations (set of occupied molecular orbitals) that contribute to a good description of methane, but we can only say that for sure for very special cases. When you go to molecules with very distinct geometries, it is vital to have d functions centered on carbon, for example, in order to determine even qualitatively right chemistry trends.
Quantum chemistry semi-empirical methods widely used in the past often had this problem of only employing valence basis functions centered on an atom in molecule calculations, not giving enough flexibility for the description of molecular geometries, and therefore providing bad results whenever the geometry deviated too much from what the qualitative models (such as the hybridization model) predicted.
In fact, I'd really be surprised (but not too much) if any modern paper that discusses chemical bonding, or that qualitatively discusses bonding orbitals to explain computed trends, do that by employing hybridization theory arguments. Those that I have seen almost always do that by looking at molecular orbitals or some refined valence bond treatments.
Therefore, although hybridization is a deep and important topic that every chemist should dominate, in the way it was developed by Pauling it has severe limitations that people are, or should be, aware of.
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I'm not quite sure I agree that he's taking the picture too seriously. I think it's true, to an excellent approximation, that the electron state in question is a superposition of the given l=0 and l=1 states. There are probably higher-l states in the superposition too, but with low amplitude. The basic formalism of quantum mechanics surely applies to such a state, and it makes sense to talk about the probabilities of getting various results when performing a (hypothetical) orbital angular momentum measurement. (Not that I have a clue how one would perform such a measurement in practice ...) – Ted Bunn Mar 4 '11 at 2:37
@TedBunn I think electron paramagnetic resonance is the tool to study the actual electron charge distribution inside a molecule. From an EPR spectrum you get a justification for the sp3 hybridisation and get a hint on the molecule shape. So an expert for EPR/NMR is needed for the initial question – Alex1167623 Mar 1 '12 at 23:04
More like the latter than the former, although you've got one detail wrong. An $s$ orbital is $\ell=0$, so it corresponds to $L=0$, not $L=\hbar$. A $p$ orbital is $\ell=1$, corresponding to $L=\hbar\sqrt{2}$. If the electron is in a state that is a superposition of $s$ and $p$, then a measurement of angular momentum will yield either 0 or $\hbar\sqrt 2$, with probabilities given by the squares of the amplitudes of the different parts of the superposition. If it's an equally-weighted superposition (and I confess I don't know enough chemistry to know whether that's correct), then the probabilities are 25% and 75% as you say.
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Ugh, right. Correcting the original question. – Jerry Schirmer Mar 3 '11 at 18:24
In the Hydrogen atom $L^2$ commutes with the Hamiltonian so that an eigenstate of $H$ can be at the same time an eigenstate of $L^2$. Additionally, the energy does not depend on the quantum number $l$, so that a superposition of s and p orbitals is still an eigenstate of $H$ with not well-defined angular momentum. In a sp$^3$ state if you measure $L^2$ you will observe $l=0$ 25% of the times, and $l=\sqrt{2}\hbar$ on 75% of the times, as you suggested.
On the other hand, in a molecule, neither $l_i^2$ (the orbital angular momentum of a single electron) nor $L^2$ (the total orbital angular momentum) commute with $H$. (Molecules are not spherical). You typically write your energy eigenstate as an antisymmetrized product of orbitals, which are eigenstates of the Fock operator (the self-consistent one-electron Hamiltonian). Additionally, each orbital can be written as a linear combination of whatever basis (e.g. atomic orbitals, as is usually done in Chemistry). Obviously $l_i^2$ is not well defined for this orbital, nor it shouldn't since you want to obtain faithful representations of the energy eigenstates, nor of the angular momentum operator.
Now if you still want to measure the orbital angular momentum of a single electron in the molecule, you first need to ionize an electron. The angular momentum will depend on the dynamics that led to the ionization. You may still ask what are the angular momentum components of whatever orbital of the molecule, although this information is not physical (I believe there can not be any measurement associated to it as a matter of principle). Then you need to choose an axis first (e.g. whatever bond axis, etc.) There is nothing terribly difficult in doing this, but usually you don't get much interesting information, expect perhaps in diatomic molecules or in studying molecules that are breaking along a particular bond.
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http://mathoverflow.net/users/15334?tab=recent | # Banach
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## Registered User
Name Banach
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| Feb4 | comment | Riesz representation for an infinite-dimensional spaceI looked up projective limit, but I don't see why $C(X)$ is the projective limit of $C(K)$. Would you please indicate what are the coordinate spaces and the bonding maps in the inverse limit? |
| Feb4 | awarded | ● Commentator |
| Feb4 | comment | Riesz representation for an infinite-dimensional spaceAccording to my topology book, there is a restriction. The statement, "a function on X is continuous iff its restrictions on compact subsets are continuous" requires X to be compactly generated. However, in my case X is compactly generated. |
| Feb4 | comment | Riesz representation for an infinite-dimensional space@Yemon, which smaller space do you suggest to consider instead of $C(X)$? Do you know of a reference for Riesz representation for $C(X)$ where $X$ is a metrizable non LCH space? |
| Feb4 | comment | Riesz representation for an infinite-dimensional space What does it mean that $C(X)$ is the projective limit of $C(K)$? Does this require X to be a countable union of compact sets? |
| Feb4 | comment | Riesz representation for an infinite-dimensional space@Yemon, You are misunderstanding the question. I give you a Banach algebra $X$ that is infinite dimensional. I am asking for a topology on $C(X)$ and a characterization of its dual. I am asking if anyone has seen such sort of theorem. The focus is the infinite-dimensionality of $X$. Either you have seen a Riesz-type theorem for infinite-dimensional spaces or not. If you don't like $X$ being an algebra, assume it's just a Banach space, or assume it's just a topological vector space. |
| Feb3 | comment | Riesz representation for an infinite-dimensional space$C(X)$ can be given a topology (most likely not a normable), but I don't want to specify one because that is part of the question. I am asking if there is some sort of Riesz representation theorem for an infinite-dimensional space $X$ where $C(X)$ has some nontrivial topology? |
| Feb3 | comment | Riesz representation for an infinite-dimensional spaceIs $C(X)$ suppose to be a Banach space? |
| Feb3 | comment | Riesz representation for an infinite-dimensional spaceI didn't say $C(X)$ is a Banach space. |
| Feb3 | asked | Riesz representation for an infinite-dimensional space |
| Dec4 | comment | Absolutely Continuous Invariant Measures for Piecewise Convex MapsYou can get away with $C^{1+\alpha}$, but with less regularity, I'm not sure. |
| Dec3 | awarded | ● Editor |
| Dec3 | revised | Absolutely Continuous Invariant Measures for Piecewise Convex Mapsadded 341 characters in body |
| Dec3 | answered | Absolutely Continuous Invariant Measures for Piecewise Convex Maps | | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9005051851272583, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/311027/how-to-put-a-bivariate-normal-distribution-under-standard-form | # How to put a bivariate normal distribution under standard form
If I have the following gaussian integral:
$$\int_{-\infty}^{a}\int_{-\infty}^{b}\exp\left[-\alpha x^2+\beta x-\theta y^2+\gamma y+2\lambda xy\right] dxdy$$ ,where X and Y are standard normal R.V.
Well I can put it under the form:
$$\exp\left(\frac{\beta^2}{4\alpha} +\frac{\lambda^2}{4\theta}\right)\int_{-\infty}^{a}\int_{-\infty}^{b}\exp\left[-(\sqrt\alpha x-\frac{\beta}{2\sqrt\alpha})^2 -(\sqrt\theta y-\frac{\gamma}{2\sqrt\theta})^2+2\lambda xy\right] dxdy$$
But from then... I don't know how to reach a bivariate normal distribution form... and my head starts to hurt.
More precisely, when doing the change of variable. In my second eqn, say I def $\hat{x}:=(\sqrt\alpha x-\frac{\beta}{2\sqrt\alpha})$ and $\hat{y}:=(\sqrt\theta y-\frac{\gamma}{2\sqrt\theta})$, then after shifting limits and adjusting the differentials, I'm left with: $$C\int_{-\infty}^{a*}\int_{-\infty}^{b*}\exp\left[-\hat{x}^2-\hat{y}^2+2\lambda(\hat{x}+...)(\hat{y}+...)\right] d\hat{x}d\hat{y}$$
And now I have some additional x and y double factors that I don't know what to do with...
It looks like this actually http://en.wikipedia.org/wiki/Gaussian_integral#n-dimensional_with_linear_term (n-dimensional with linear term), and I have no problem finding matrix $A$ and vector $B$ that fit my expression but they just go straight to the solution for integration over $R^2$ and give no link to a bivariate...
I still take any hint. Thanks
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Already in dimension 1 there is no general expression except if one defines a new function as the solution... – Did Feb 23 at 6:31
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Some formatting tips: Equations with integrals and displayed fractions look better and are easier to read as displayed equations, which you get by enclosing them in double dollar signs instead of single dollar signs. You can get appropriately sized delimiters (e.g. parentheses) by preceding them with `\left` and `\right`, respectively. Names like exp are interpreted as a juxtaposition of variable names and hence italicized; to get the appropriate font and spacing for them use `\exp`, or `\operatorname{name}` in cases where there's no predefined command. – joriki Feb 23 at 8:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9124037027359009, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/190566/is-this-calculus-of-variations-intuition-justifiable?answertab=oldest | # Is this calculus of variations intuition justifiable?
I'll preface this by saying that I haven't taken an in depth study of the calculus of variations and have only come across it recently in applications; in depth study is on my to do list.
I'm wondering if there is any sense in which the following heuristic for functional optimization is valid. I'm interested in finding a probability density $f$ on $[0, 1]$ with maximum entropy $H[f] = - \int f \log f \ dx$, subject to the constraint that $\int x f(x) \ dx = \mu$. I set this up via Lagrange multipliers as $$\arg \max_f \ -\int f \log f \ dx - \lambda_1 \left(\int f \ dx - 1\right) - \lambda_2 \left(\int x f \ dx - \mu\right).$$ Now, instead I solve an approximate problem using Riemann sums; we partition $[0, 1]$ with $0 = x_0 < x_1 < \cdots < x_n = 1$ and approximate this by $$\arg \max_f \ -\sum f_{x_j} \log f_{x_j} \Delta_{x_j} - \lambda_1 \left(\sum f_{x_j} \Delta_{x_j} - 1\right) - \lambda_2 \left(\sum x_{j} f_{x_j} \ \Delta_{x_j} - \mu \right).$$ This can be solved via multivariate calculus and leads to $f(x) = a e^{bx}$, a truncated exponential distribution, as the density on $[0, 1]$ with maximal entropy for a fixed mean, which I believe is the correct answer.
There are a few problems right off the bat. This makes the assumption of Riemann integrability, which is undesirable. It also ignores measure theoretic details (e.g. the solution to the optimization should only be unique up to a.e. $dx$ equivalence classes). But perhaps if I constrain the optimization to being over continuous functions, this provides an argument which can be patched up and made rigorous?
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http://math.stackexchange.com/questions/58829/a-general-derivation-for-a-combinatoric-problem | # A general derivation for a combinatoric problem
Is there any derivation for the problems:
$1.$Dividing $n$ identical things into $r$ identical groups where some groups may be empty or all $n$ things can go to any one group.
$2.$Dividing $n$ identical items into $r$ identical groups where there is at-least one item in every group.
I am not sure if this is modeled as some other known problem,what I am looking for, is a nice approach/derivation for solving these kind of models.
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## 1 Answer
I think you are refering to partitions and compositions.
Edited: The restriction of "r groups" introduce a restriction. The second problem would correspond to counting $p(n,r)$, number of partitions of a number $n$ in $r$ parts. This is not trivial (see here or here, our counting corresponds to $p(n,k)$ there). The first problem would correspond to the number of partions of size up to $r$, hence it can be expressed as a sum $\sum_{k=1}^r p(n,k)$
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May be related but I think not exactly as there is a restriction of $r$ which needs to be considered explicitly?! – Quixotic Aug 21 '11 at 17:46
YOu are right, added info. – leonbloy Aug 21 '11 at 20:22
I agree that this is not trivial,thanks for the infos. :-) – Quixotic Sep 3 '11 at 15:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9308152198791504, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/65469/is-this-sum-lesser-or-equal-to-2?answertab=active | # Is this sum lesser or equal to 2?
Given any $a, b, c, d \in [0, 1],$ how can the following be proven?
$$a (1-b) + b(1-a) + c(1-d) + d(1-c) \le 2$$
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I assume by symmetry you actually wanted the first term to be $a(1-b)$, right? – anon Sep 18 '11 at 8:40
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One way to find the maximum of a nice expression like this on an interval is to take its gradient and set that to 0. The maximum must occur either on the boundary or where the gradient is 0. If it's on the boundary, then one of the variables is 0 or 1, and the expression simplifies and you can repeat taking the gradient in a lower dimensional space. – Logan Maingi Sep 18 '11 at 8:41
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Furthermore, the LHS is at most $1-ab-cd$. – Did Sep 18 '11 at 9:46
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@anon, yes, $2-ab-cd$. Thanks. – Did Sep 18 '11 at 10:02
## 2 Answers
(To those who originally upvoted: I've changed my answer because this is a much sleeker version, not sure why I didn't see it before.)
This is just a single inequality added to itself with $c,d$ instead of $a,b$ in the second copy. Namely, $$a(1-b)+(1-a)b\le a(1)+(1-a)(1)\le1.$$
Above we used the fact that if $b\in[0,1]$, then $b\le1$ and $1-b\le1$.
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I agree with anon's solution. I'd also like to post mine. Anon's solution is based on a "trick" with inequalities, which is understood, but not so simple to guess from the beginning. OTOH my solution is straightforward.
$$S = a(1−b)+(1−a)b$$
Let's find the derivative of the above with respect to $a$.
$$\frac{dS}{da} = 1-2b.$$
From the above we see that the derivative of $S$ with respect to $a$ is constant (i.e. does not depend on $a$). Hence the extremal values are achieved at the ends of the range.
• Assuming $1-2b > 0, b < 1/2$. The maximum is achieved when $a=1$. Substituting this we get $S = 1-b$. Since $b\ge0$ we get $S \le 1$.
• Assuming $1-2b < 0, b > 1/2$. The maximum is achieved when $a=0$. Substituting this we get $S = b$. Since $b<=1$ we get $S \le 1$.
• If $1-2b = 0$, then $b = 1/2, S = 1/2 \le 1$.
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I've LaTeXified your answer, hope you don't mind. And for the record (since my answer changed), my original trick involved seeing $$[a+(1-a)][b+(1-b)]=1,$$ which can be guessed at by seeing $a$ and $b$ as probabilities for independent events A and B and considering the probability of $$(A\text{ and }B)\text{ or }(A\text{ and not }B)\text{ or }(\text{not }A\text{ and }B)\text{ or }(\text{not }A\text{ and not }B).$$ – anon Sep 18 '11 at 9:43
@anon: I see now. Thanks. – valdo Sep 18 '11 at 11:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9383641481399536, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/314401/fourier-transformation-connection-between-exponential-and-trigonomeric-forms | # Fourier transformation - connection between exponential and trigonomeric forms
On Wikipedia i have come across a Fourier transformation equation in exponential form and its inverse (Wiki):
$$\begin{split} \mathcal{F}(x) &= \int\limits^{\infty}_{-\infty}\mathcal{f}(k) \, e^{2 \pi i kx} \, \textrm{d} k\\ \mathcal{f}(k) &= \int\limits^{\infty}_{-\infty}\mathcal{F}(x) \, e^{2 \pi i kx} \, \textrm{d} x \end{split}$$
but i allso found that there is a trigonometric form of Fourier transformation (PDF, page 2)
$$\begin{split} \mathcal{F}(x) &= \int\limits^{\infty}_{-\infty} f(k) \cos(kx) \, \textrm{d}k\\ \mathcal{f}(k) &= \int\limits^{\infty}_{-\infty} \mathcal{F}(x) \cos(kx) \, \textrm{d}x \end{split}$$
MAIN QUESTION:
Could someone show me, how these pairs of equations are connected?
SUB QUESTION:
(i) I think that $\textrm{d}x$ is used for spatial integration (please correct me if i am wrong).
(ii) I think that $\textrm{d}k$ is used for integration over wave vector (please correct me if i am wrong) .
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## 1 Answer
Do you know about Euler's Formula?
$e^{ix} = cos(x) + isin(x)$
I think that's a hint...
Also, see this formula http://en.wikipedia.org/wiki/Sine_and_cosine_transforms#Fourier_inversion The first one in this column is simply the fourier integral applied over the above relation. Everything else is just a simplification into cosine
And this is even better...a whole justification of the connection http://en.wikipedia.org/wiki/Sine_and_cosine_transforms#Relation_with_complex_exponentials
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Second link you provided is suplying link between time dependant and frequency dependant fourier. I have wave vector ependant and spatial dependant fourier. – 71GA Feb 25 at 23:53
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Call it time or frequency or wave or space --- the formulas are the same, no? – Gerry Myerson Feb 26 at 0:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8498948812484741, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/90227?sort=votes | ## A Kunneth formula for the etale cohomology of the product of (‘simple’) varieties over not (necessarily) algebraically closed field
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If $X$ and $Y$ are varieties over an algebraically closed field, then in the corresponding derived category of complexes we have $RH_{et}(X\times Y,\mathbb{Z}/l^n\mathbb{Z})\cong RH_{et}(X,\mathbb{Z}/l^n \mathbb{Z})\otimes RH_{et}(Y,\mathbb{Z}/l^n\mathbb{Z})$. Where can I found a proof of this fact whose 'idea' would be clear (I don't want to assume that $X$ and $Y$ are smooth, and I don't want to consider cohomology with compact support)? Does the Leray spectral sequence help here? In order to compute the stalks of the corresponding (higher) direct images, one has to consider $Y$ being a strictly henselian scheme; how can this be done?
And what happens when the base field is no longer algebraically closed? In particular, I would like to understand completely the case when $X$ is singular, but $Y$ is just $G_m$ (an affine line minus one point). This particular case is related with the question http://mathoverflow.net/questions/89898/the-gysin-long-exact-sequence-for-the-complement-of-the-zero-section-of-a-line-bu
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## 1 Answer
The only reference I know for Künneth theorems in this generality are SGA4 and SGA4.5. Specifically, SGA4 has theorems for étale cohomology with proper support (which hold in ridiculous generality), but SGA4.5's "Théorèmes de finitude en cohomologie $l$-adique" allows you to get the same results for étale cohomology, provided that the schemes are of finite type over a field (no other assumptions are needed!). The basic Künneth theorem itself is given as Corollaire 1.11 in that exposé. It is stated for schemes of finite type over a separably closed field, but if I'm not mistaken the proof works over any field whatsoever if you replace global sections by pushforwards to the étale topos of the base field. So this only gives you an equivalence between chain complexes with an action of the Galois group. The actual "étale cochains" are the invariants of those, and I don't know how to compute the invariants of a tensor product. But at least this reduces the problem to one in group cohomology.
ADDED: Here's a different approach. I think it may be true that for $X$ and $Y$ schemes of finite type over a field $k$, the cohomology of say $\mathbb{Z}/l^n$ over $X\times_kY$ can be computed as the $\mathbb{Z}/l^n$-cohomology of the homotopy pullback of étale homotopy types $Et(X)\times_{Et(k)}Et(Y)$. My argument is currently very WRONG though. The idea is that (1) $Et(X)$ should be the homotopy fixed points of $Et(\bar X)$ (where $\bar X=X\times_k\bar k$), (2) the result is obviously true over $\bar k$ (by the Künneth theorem), and (3) homotopy fixed points commute with homotopy pullbacks.The usefulness of this is limited since it is not easy to compute the cohomology of a homotopy pullback, but this would give a "closed formula" answer only in terms of $X$, $Y$, and $k$.
ADDED 2: I gave this a bit more thought and it obviously doesn't work. $Et(X)$ is the homotopy orbits of $Et(\bar X)$, and again I don't know how to compute the homotopy orbits of a pullback. The pullback formula is still true if $Y$ is proper and smooth over $k$. In this case you have a fiber sequence of $l$-completed étale homotopy types:
$$Et(\bar Y)\to Et(X\times_kY)\to Et(X)$$
which you can compare with the fiber sequence
$$Et(\bar Y)\to Et(X)\times_{Et(k)}Et(Y)\to Et(X)$$
so that $Et(X\times_kY)$ and $Et(X)\times_{Et(k)}Et(Y)$ are $\mathbb{Z}/l^n$-cohomologically equivalent. These fiber sequences are established in this paper by Friedlander (Corollary 4.8). I wonder if the properness assumption can be removed using SGA 4.5.
ADDED 3: OK, I'm pretty sure the above fiber sequence argument will work for any $X$ and $Y$ of finite type over a field. The only reason Friedlander considers proper and smooth maps is to apply a corollary of the proper/smooth base change theorems from SGA4, exposé XVI, which I think works without that assumption when over a field, thanks to SGA4.5. I need this result myself so I will try to check it thoroughly.
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http://mathhelpforum.com/pre-calculus/166434-period-function-print.html | # Period of function
Printable View
• December 16th 2010, 04:54 PM
horan6
Period of function
Considering the function $u(t,x)=\sum_{h=1}^{\infty}\left(a_{n}\cos\frac{n\p i}{L}ct+b_{n}\sin\frac{n\pi}{L}ct\right)\sin\frac{ n\pi}{L}x$
I know that the period of this function is $2\pi/L$, but I am not sure how to prove it. I tried plugging in the period for t and reducing, but didn't get very far. Is this the right approach? Can anyone help with this?
Thanks in advance!
• December 16th 2010, 05:51 PM
snowtea
If your period is $T$, then $u(t+T,x) = u(t,x)$ for all $t$ and $x$.
• December 18th 2010, 08:43 AM
SammyS
Quote:
Originally Posted by snowtea
If your period is $T$, then $u(t+T,x) = u(t,x)$ for all $t$ and $x$.
Actually, if $\displaystyle T$ is the period, then $\displaystyle u(t+kT,x) = u(t,x),$ for any integer, $\displaystyle k$.
So, you need to show that $\displaystyle T$ is the smallest value for which $\displaystyle u(t+T,x) = u(t,x)$ is true.
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http://mathhelpforum.com/discrete-math/130893-cartesian-product-print.html | Cartesian product
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• February 26th 2010, 10:15 AM
novice
Cartesian product
My book says the converse of the following implication is false:
$A, B, C, D$ are sets such that $A\subseteq C$ and $B \subseteq D$, then $A \times B \subseteq C \times D$.
Although $P \Rightarrow Q \not \equiv Q \Rightarrow P$, I don't understand why since the above implication still can be true under special circumstances.
For exmple, When $A=D$ and $C=B$ or when $A = B = \emptyset$ or $C=D=\emptyset$ , I still can make it true.
I think $P \Rightarrow Q \not \equiv Q \Rightarrow Q$ is not always true. Am I missing something?
• February 26th 2010, 10:20 AM
Drexel28
Quote:
Originally Posted by novice
My book says the converse of the following implication is false:
$A, B, C, D$ are sets such that $A\subseteq C$ and $B \subseteq D$, then $A \times B \subseteq C \times D$.
Although $P \Rightarrow Q \not \equiv Q \Rightarrow P$, I don't understand why since the above implication still can be true under special circumstances.
For exmple, When $A=D$ and $C=B$ or when $A = B = \emptyset$ or $C=D=\emptyset$ , I still can make it true.
I think $P \Rightarrow Q \not \equiv Q \Rightarrow Q$ is not always true. Am I missing something?
Let $(x,y)\in A\times B$ then $x\in A\text{ and }y\in B\implies x\in C\text{ and }y\in D\implies (x,y)\in C\times D$. Done.
• February 26th 2010, 10:59 AM
Plato
Quote:
Originally Posted by novice
My book says the converse of the following implication is false:
$A, B, C, D$ are sets such that $A\subseteq C$ and $B \subseteq D$, then $A \times B \subseteq C \times D$.
Consider: $A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\}$.
• February 26th 2010, 11:08 AM
novice
Quote:
Originally Posted by Plato
Consider: $A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\}$.
You are quite amazing. How could I be so blind?
• February 26th 2010, 11:22 AM
novice
Quote:
Originally Posted by Drexel28
Let $(x,y)\in A\times B$ then $x\in A\text{ and }y\in B\implies x\in C\text{ and }y\in D\implies (x,y)\in C\times D$. Done.
The implication is not a problem, but the converse.
Plato showed the condition for which when it's false. I thought of an empty set on each side of the inclusion, but never thought of having only one emptyset for the leftside while keeping the rightside nonempty.
This is example or over focusing to the point that I loss the entire picture.
• February 26th 2010, 11:30 AM
novice
Quote:
Originally Posted by Plato
Consider: $A = \{ a\} ,\;B = \emptyset ,\;C = \{ c\} ,\;D = \{ d\}$.
Plato,
Now, with the implication, if I declare that $A, B, C$, and $D$ being nonempty sets and that $A \not = B$, I think it will be thought to make the converse false.
Do you agree?
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http://gowers.wordpress.com/2009/03/10/problem-solved-probably/ | # Gowers's Weblog
Mathematics related discussions
## Problem solved (probably)
Without anyone being particularly aware of it, a race has been taking place. Which would happen first: the comment count reaching 1000, or the discovery of a new proof of the density Hales-Jewett theorem for lines of length 3? Very satisfyingly, it appears that DHJ(3) has won. If this were a conventional way of producing mathematics, then it would be premature to make such an announcement — one would wait until the proof was completely written up with every single i dotted and every t crossed — but this is blog maths and we’re free to make up conventions as we go along. So I hereby state that I am basically sure that the problem is solved (though not in the way originally envisaged).
Why do I feel so confident that what we have now is right, especially given that another attempt that seemed quite convincing ended up collapsing? Partly because it’s got what you want from a correct proof: not just some calculations that magically manage not to go wrong, but higher-level explanations backed up by fairly easy calculations, a new understanding of other situations where closely analogous arguments definitely work, and so on. And it seems that all the participants share the feeling that the argument is “robust” in the right way. And another pretty persuasive piece of evidence is that Tim Austin has used some of the ideas to produce a new and simpler proof of the recurrence result of Furstenberg and Katznelson from which they deduced DHJ. His preprint is available on the arXiv.
Better still, it looks very much as though the argument here will generalize straightforwardly to give the full density Hales-Jewett theorem. We are actively working on this and I expect it to be done within a week or so. (Work in progress can be found on the polymath1 wiki.) Better even than that, it seems that the resulting proof will be the simplest known proof of Szemerédi’s theorem. (There is one other proof, via hypergraphs, that could be another candidate for that, but it’s slightly less elementary.)
I have lots of thoughts about the project as a whole, but I want to save those for a different and less mathematical post. This one is intended to be the continuation of the discussion of DHJ(3), and now DHJ(k), into the 1000s. More precisely, it is for comments 1000-1049.
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### 133 Responses to “Problem solved (probably)”
1. Jason Dyer Says:
March 10, 2009 at 1:32 am | Reply
Metacomment.
Appropriate timing of note:
Cantwell.949: A Moser set for n=5 has 124 points
2. Terence Tao Says:
March 10, 2009 at 3:37 am | Reply
1000. DHJ(k)
A small comment: from experience with hypergraph regularity, I would imagine that DHJ(k) will not follow exactly from DHJ(k-1), but from something more like DHJ(k-1,k), and so the induction hypothesis may need to have an additional parameter than just k. (For instance, the simplex removal lemma for 3-uniform hypergraphs requires, at base, something resembling the $K_4$-removal lemma for graphs rather than the triangle removal lemma for graphs. While the $K_4$-removal and triangle removal lemmas have nearly identical proofs, I don’t think one can easily deduce the former from the latter if the latter is a black box.)
• gowers Says:
March 10, 2009 at 10:37 am
1000.1 An $\epsilon^2$ comment: you mean of course DHJ(k-2,k).
I think the jury is still out on whether we need to prove the full DHJ(k-2,k) before doing DHJ(k) or whether passing to subspaces the whole time makes it unnecessary. I’m hoping for the latter, but am ready to switch if we have to.
An $\epsilon^3$ comment: you are now our official powers-of-ten champion …
3. Anonymous Says:
March 10, 2009 at 7:20 am | Reply
1001. Line and equal-slice distributions in $[k]^n$.
It seems to me that the right generalization of the “another useful equivalent definition” of equal-slices on $[k]^n$ is as follows:
Pick $z \in [k]^n$ from the equal-slices distribution. Form $x^i \in [k-1]^n$ by changing all the $k$‘s in $z$ into $i$‘s, for $i = 1...k-1$. Finally, let $y^1, ... y^{k-1}$ be a random permutation of the $x^i$‘s.
Certainly the set $\{y^1, ..., y^{k-1}, z\}$ is a combinatorial line, which is degenerate only if $z$ had no $k$‘s.
I believe that each $y^i$ is individually distributed as equal-slices on $[k-1]^n$, but I can’t quite seem to prove it to myself at this late hour. Anyone want to prove or disprove?
(If it’s true, there must be some slick way to see it, like “break up a necklace of $n$ beads into $k$ pieces, then pretend you did it in one the $k!$ different orders, then…” etc.)
• Ryan O'Donnell Says:
March 10, 2009 at 7:21 am
• gowers Says:
March 10, 2009 at 9:12 am
1001.2 I’m pretty sure it’s false. In fact, isn’t it obviously false, since the expected number of $i$s in $y^i$ is twice the expected number of any j for $j\ne i$?
However, I think what can be proved is that the distribution on each individual $y^i$ is not importantly different from the equal-slices distribution, in that the collection of dense sets is the same in both cases. This is analogous to what happens in the corners problem, where the number of corners containing a point is different from point to point (even if we start off with a triangular grid) but this doesn’t matter because the discrepancy between the uniform measure of a set and the point-in-random-corners measure of a set depends only on the measures and not on $n$.
• gowers Says:
March 10, 2009 at 10:41 am
1001.3 However, I think I may be misunderstanding what you have written, since I don’t quite see why you bother to permute the $x^i$s (given that they were already distributed in a way that is permutation-invariant — or at least that’s how it appears to me from what you say).
• Ryan O'Donnell Says:
March 10, 2009 at 4:41 pm
1001.4. Oops, yes, I think what I meant to say was “Pick a permutation $\pi$ on $[k-1]$ and then form $y^i = \pi(x^i)$, where by $\pi(x^i)$ we mean permute the $j$-sets of $x^i$ according to the permutation $\pi$.”
I will write a new comment on this topic later today.
4. Randall Says:
March 10, 2009 at 7:49 am | Reply
1002. 895 again.
Okay, maybe the book proof of Szemeredi goes through DHJ, but I hope not…I was trying before to get someone to bite on trying something potentially simpler than DHJ (3) that would still prove the corners theorem via the Ajtai-Szem method. (And in particular didn’t use Szemeredi’s theorem as a lemma.) Of course, being the token ergodic theorist, you can probably guess that I would like someone to bite so I don’t have to try to check myself, because I am woefully slow and error prone when it comes to such things. So let me try again…I’ll just think out loud and if none of this makes any sense, everyone can feel free to ignore.
What if I take as my basic object pairs $(\alpha, \beta)$ of $[n]$. A diagonal is a set $D_\gamma= \{ (\alpha,\beta) : \alpha+\beta =\gamma\}$, where $+$ is multi-set union and $\gamma$ is some fixed multi-set (of multiplicity no greater than 2 in each element). A corner is a triple of the form $\{(\alpha,\beta), (\alpha+\eta,\beta), (\alpha,\beta+\eta)\}$, where $\eta$ can be a set plus an anti-set (an antiset is like a set where every element has a multiplicity of -1).
Okay, let $A$ be a set of pairs of density at least $\delta$, i.e. $|A|\geq \delta 4^n$. Now, for some rather small, or at least not-so-big $m$, most of the whole space is covered by diagonals whose “multiplicity 2-or-zero” (for the defining multiset $\gamma$) set of indices has size no bigger than $m$.
Diagonals of this sort have at least $2^{n-m}$ members. Take one of these diagonals $D_\gamma$ on which $A$ is dense, say of relative density at least $\delta /2$. If $A$ is corner-free, then any pair of the form $(\alpha, \beta)$, where $(\gamma -\beta, \beta)\in A$ and $(\alpha, \gamma-\alpha) \in A$ is forbidden. That should be something like $4^{n-m-1} \delta^2$ forbidden pairs, and these pairs lie in a Cartesian product. This should mean that $A$ has a density increment on some other Cartesian product, call it $X\times Y$.
The next step is presumably to use the multidimensional Sperner’s theorem to carve almost all of $B=X\times Y$ up into subgrids of some large but fixed size $r$. (Here a subgrid is a collection $(\zeta\cup FU \{\beta_1,\ldots ,\beta_r\}) \times(\alpha\cup FU \{\beta_1,\ldots ,\beta_r\})$.) I wasn’t really anticipating how hard this would be but maybe you can use the iteration method we use to carve 1-spaces up. Now that I am thinking about this finally, maybe one can do it all in one go. Namely, you can always find $\beta_1,\ldots ,\beta_r$ with combined size at most $M$ such that the probability that $(\zeta\cup FU \{\beta_1,\ldots ,\beta_r\}) \subset X$ and $(\alpha\cup FU \{\beta_1,\ldots ,\beta_r\}) \subset Y$ is above some constant $c$. So for those pairs $(\zeta, \alpha)$ where you get the containment you were shooting for, remove those grids from $B$ and keep them as part of your partition; partition the other pairs into $4^T$ classes according to the value taken on the subgrid associated with that pair. Each of these classes is the intersection of $B$ with a subgrid and we iterate…just like in the proof of DHJ. And, I guess that’s it. Assuming all of the above is correct, you get an increment on a subgrid.
It’s almost 3 here and too late to check so I will apologize in advance if something is amiss.
• gowers Says:
March 10, 2009 at 9:18 am
1002.1 Randall, is your problem the same as Jozsef’s in comment 2? In one of the very early comments I tried to translate that one into set-theoretic language and ended up with something pretty similar. This is mainly to say that I agree that it’s worth seeing whether Szemerédi can be done without going the whole way to DHJ. For now I’m going to concentrate my efforts on DHJ(k) but will follow what you have to say with interest. I suppose there are two issues: can DHJ be avoided, and is the resulting proof substantially simpler?
• Randall Says:
March 10, 2009 at 3:39 pm
1002.2 Yes, it was intended to be the same (not sure whether Jozsef intended to allow inverted corners, but I am here). Oh, yes…I see your comment 7. That looks to be exactly the interpretation I am employing.
Well, perhaps Terry will wake up and say whether the alleged proof I gave is really a proof or can be made into a proof. (And whether it’s simpler than what we have already.) Since he composed and typed the breakthrough post (837?) in eighteen minutes, perhaps he’ll be able to do this despite being very busy with other things.
On a more general philosophical note, Tim, to we ergodic theorists (who have trouble with subscripts appearing on the numbers capital N appearing in proofs), the dynamical proof of van der Waerden (not that of Furstenberg and Weiss but the easy one, due to Blszczyk, Plewik and Turek) is much more pleasant than any proof I have seen of HJ. (Actually, I have long wanted to see a proof, even an unpleasant one, of HJ that made efficacious use of non-metrizable topological dynamics.)
• Randall Says:
March 10, 2009 at 6:58 pm
1002.3 Obviously I meant “efficacious use of *metrizable* topological dynamics”. There exist proofs using non-metrizable topological dynamics (e.g. Furstenberg and Katznelson’s proof). It may not be possible, or it may require some non-commutative stuff like in the FK proof of DHJ. There is a Bergelson-Leibman proof that tried to use recurrence for continuous maps of compact metric spaces, but it was realized after the paper appeared that neither continuity of the transformations nor completeness of the space was actually ever used.
• Randall Says:
March 11, 2009 at 7:34 pm
1002.4 No induction….
I think the idea I had to prove Szemeredi via Ajtai-Szem type argument using Jozsef’s formulation from comment 2 is dead in the water at k=3.5. The problem I don’t see a way around is that there don’t seem to be enough corners at k=3 to forbid enough points at k=4. Obviously not a problem for DHJ since the number of lines at k=3 is 4^k which is equal to the number of points at k=4.
Oh well…if the above is correct (nobody checked?) it may be the simplest proof of the corner’s theorem, at least.
• Randall Says:
March 25, 2009 at 8:57 pm
Hmm…I finally had a look at the above myself, and it seems to be complete garbage for (at least) two reasons…. Perhaps now I am convinced that the DHJ proof is the right one for Szemeredi after all. (Maybe.)
5. gowers Says:
March 10, 2009 at 10:49 am | Reply
Metacomment: because of more spam problems on the wiki, you now have to be registered on it and logged in if you want to edit the main page.
6. Problem solved (probably) « OU Math Club Says:
March 10, 2009 at 4:41 pm | Reply
[...] experiment in doing math research on his blog. It sounds like it was a resounding success! Timothy Gowers announced on his blog that he is pretty convinced that the crowd on his blog has come up with a new proof of the density [...]
7. Terence Tao Says:
March 10, 2009 at 7:10 pm | Reply
Metacomment.
This might be a good time to set up a “meta” thread to deal with issues not directly related to finishing off the proof of DHJ(3)/DHJ(k) [1000-1049] or of computing DHJ-type numbers in small dimension [900-999]. For instance, this seems like a good time to face issues concerning writing up the results, how to perform proper attribution, etc. (At a more frivolous level, there was a suggestion on the wiki to come up with a logo for that wiki; I think that might be a fun topic for the meta thread. More generally, discussion of the strengths and weaknesses of the wiki, or the threaded format, etc. would be apropos at this time, I think.)
• gowers Says:
March 10, 2009 at 7:45 pm
OK what I’ll do is get on and write a post that I’ve been intending to write, in which I discuss general meta-issues that have arisen out of how things have gone. I think there’s a lot to talk about, so comments on that should be a natural place for discussing these issues amongst others.
8. Terence Tao Says:
March 10, 2009 at 7:43 pm | Reply
1003. Austin’s proof
I’ve managed to digest a key ingredient in Austin’s proof (Lemma 6.1 of http://arxiv.org/abs/0903.1633 ) which may be of interest here. Roughly speaking, it says that 1-sets and 2-sets (say) are always “locally independent” of each other. A more precise statement is the following. Let $f: [3]^n \to [-1,1]$ be 02-insensitive (e.g. it is the indicator function of a 1-set), and let $g: [3]^n \to [-1,1]$ be 01-insensitive (e.g. it is the indicator of a 2-set). Suppose also that ${\Bbb E}_{x \in [3]^n} f(x) g(x)$ is large. Then f has large average on large-dimensional spaces, or (equivalently) correlates with a 012-low influence function (a function which is almost invariant under any interchanges between 0, 1, and 2).
Proof. Let x be a random string $[3]^n$, let $A \subset [n]$ be a random medium-sized set of indices, and let $\pi_{A, 02 \to 0}(x) = \pi_{A, 2 \to 0}(x)$ be the string in $[3]^n$ formed by replacing all occurrences of 0 or 2 in A with 0. If A is not too large, then $\pi_{A, 02 \to 0}(x)$ is still roughly uniformly distributed in $[3]^n$, and so
${\Bbb E}_{x \in [3]^n} f( \pi_{A,02 \to 0}(x) ) g( \pi_{A, 02 \to 0}(x) )$ is large.
But as f is 02-insensitive, $f(\pi_{A,02 \to 0}(x)) = f(x)$. As g is 01-insensitive, $g(\pi_{A,02 \to 0}(x)) = g(\pi_{A,02 \to 1}(x)) = g(\pi_{A,012 \to 1}(x))$, where $\pi_{A,012 \to 1}(x)$ is the string formed by sending every digit of x in A to 1. Thus
${\Bbb E}_{x \in [3]^n} f( x ) g( \pi_{A, 012 \to 1}(x) )$ is large.
Freezing A, we thus conclude that f has large mean on many |A|-dimensional subspaces; alternatively, we see that f correlates with the 012-low influence function $G(x) := {\Bbb E}_A g( \pi_{A, 012 \to 1}(x) )$ (if we choose A in a suitably Poisson manner).
A corollary of this is that if one has a 1-set A and a 2-set B which are “strongly stationary” in the sense that their average local density on medium-sized subspaces is close to their global density, then A and B are roughly independent, thus the density of the 12-set $A \cap B$ is roughly the density of A multiplied by the density of B.
As one might imagine, this fact is extremely useful for doing any sort of “counting lemma” involving 12-sets.
To be continued…
• Terence Tao Says:
March 10, 2009 at 7:44 pm
A small correction: where I said that f and g could be indicators of 1-sets and 2-sets respectively, it might be better to think of them as the balanced functions of a 1-set and 2-set, i.e. the indicator minus the density.
• Terence Tao Says:
March 10, 2009 at 7:55 pm
1003.2
Incidentally, this framework also gives a short proof of “lack of lines implies correlation with a local 12-set”. With the notation as before, observe that $x, \pi_{A,01 \to 1}(x), \pi_{A,02 \to 2}(x)$ will form a combinatorial line with high probability if A is not too small. Thus if ${\mathcal A}$ is a set with no lines and density $\delta$ then
${\Bbb E} (1_{\mathcal A}-\delta)(x) 1_{\mathcal A}( \pi_{A,01 \to 1}(x)) 1_{\mathcal A}( \pi_{A,02 \to 2}(x))$ is large.
But if we freeze A and the coordinates outside of A (thus passing to an |A|-dimensional subspace), then $1_{\mathcal A}( \pi_{A,01 \to 1}(x))$, $1_{\mathcal A}( \pi_{A,02 \to 2}(x))$ are indicators of a 2-set and a 1-set respectively, and the claim follows.
• gowers Says:
March 10, 2009 at 11:50 pm
1003.3 Terry, I’m trying to digest 1003.2. It seems to me that the displayed line is true only if there are many x such that $\pi_{A,01\rightarrow 1}(x)$ and $\pi_{A,02\rightarrow 2}(x)$ both belong to $\mathcal{A}.$ In the previous proof this was done with the help of Sperner and that seemed necessary. I’m not sure where the Sperner step is appearing in what you’ve written.
9. Ryan O'Donnell Says:
March 10, 2009 at 9:54 pm | Reply
1004. Random lines/equal slices for DHJ(k+1).
I think what I wrote in 1001 was correct. Let me restate it here for clarity:
Let $z \in [k+1]^n$ be drawn from equal-slices. Form the string $v^j \in [k]^n$ by changing all $(k+1)$‘s in $z$ to $j$‘s, for $j = 1 \dots k$. Finally, pick a random permutation $\pi$ on $[k]$ and define strings $x^i = v^{\pi(i)}$, for $i = 1 \dots k$.
Then each string $x^i$ is distributed as equal-slices on $[k]^n$. Further, since $(v^1, \dots, v^k, z)$ is a combinatorial line (“in order”), we also have that $\{x^1, \dots, x^k, z\}$ is a combinatorial line (possibly “out of order”).
The proof is here; I’ll wikify it soon.
• Ryan O'Donnell Says:
March 10, 2009 at 10:12 pm
1004.1. Wikified here. Hopefully this will help with the generalization of line-free sets correlating with 12-sets.
• gowers Says:
March 10, 2009 at 10:56 pm
1004.2 Ryan, I’m still struggling with the statement of what you are proving. Suppose k=2. Then I pick $z\in[3]^n$ from equal slices and form the strings $u$ and $v$ by turning the 3s of $z$ into 1s and 2s, respectively. Then I randomly decide whether to interchange $u$ to $v$. Now let’s look at the event that the number of 1s in $u$ is between $0.49n$ and $0.51n.$ If I choose $u$ according to equal-slices that is $0.02$. But if I choose it by starting with z, then either I start with about that number of 1s and change the 3s to 2s, or I start with about that number of 2s and change the 3s to 1s. The probability seems to me to be smaller because $z$ is being asked to have rather a lot of 1s or rather a lot of 2s. Since you’ve got a wikified proof, what I’m asking here is for an explanation of where I’m going wrong.
• Ryan O'Donnell Says:
March 10, 2009 at 11:53 pm
1004.3. Hmm, I’m not quite sure what I can say to help.
In your example, when you pick $u$ from equal-slices on $[2]^n$ “in the normal fashion”, then as you say the event $A$ = “number of $1$‘s in $u$ is in $[.49n, .51n]$” occurs with probability exactly $.02$.
When you pick it in the funny way, through, $z$, what is the probability? As you say, half the time $u$ is formed by changing $z$‘s $3$‘s to $2$‘s. In this case, $A$ occurs iff $z$ had between $.49n$ and $.51 n$ $1$‘s. The other half of the time, $u$ is formed by changing $z$‘s $3$‘s to $1$‘s. In this case, $A$ occurs iff $z$ had between $.49n$ and $.51 n$ $1$‘s-and-$3$‘s-together, iff $z$ had between $.49 n$ and $.51 n$ $2$‘s.
Clearly the probability $z$ has between $.49n$ and $.51n$ $1$‘s is the same as the probability $z$ has between $.49n$ and $.51n$ $2$‘s. So we are reduced to asking why this probability is indeed $.02$.
Well, I’m not sure what to say, except that it is. It comes from the fact that $\int_{.49}^{.51} 2(1-p) dp = .02$, where $2(1-p)$ is the density of the first component in a uniform draw from the $3$-simplex. Note that this integral is a bit of a “coincidence”, because it is not true that, e.g., $\int_{0}^{.49} 2(1-p) dp = .49$.
The reason this all works is a bit easier to understand in the case of $k = 2$; see this explanation in the older wiki article on equal-slices.
• gowers Says:
March 11, 2009 at 12:16 am
1004.4 Thanks — I think I’m starting to get it now.
10. Terence Tao Says:
March 10, 2009 at 9:55 pm | Reply
1005. Quantitative bounds?
Is our understanding of the density-increment proof of DHJ(3) good enough that we can venture a probable quantitative bound for n in terms of $1/\delta$? Naively I am expecting tower-exponential behaviour (with the height of the tower being either polynomial or exponential in $1/\delta$).
• gowers Says:
March 10, 2009 at 11:08 pm
1005.1 I’m pretty sure it’s a tower. The rough reason is that when we drop to a subspace we obtain that subspace by applying multidimensional Sperner, which I think restricts its dimension to the log of the previous dimension. But I think the density increase we can get that way is good, so I think the bound should be a tower of height polynomial in $1/\delta$ and therefore not worse than what you get out of the Ajtai-Szemerédi proof of the corners theorem.
Also, if we’ve got the stomach for it, we could think about trying to Shkredovize the argument and beat Shelah’s bound for the colouring DHJ(3). That would be a serious undertaking though, as it would require us to understand the global obstructions rather than just the local ones.
11. Ryan O'Donnell Says:
March 10, 2009 at 10:27 pm | Reply
1006. Equal-slices.
I wonder: can we mostly work in the equal-slices measure for the whole proof (of DHJ(3), say)? What are the advantages of the uniform distribution?
• gowers Says:
March 10, 2009 at 11:12 pm
1006.1 My take on this is that equal-slices measure is very good when you want to mix random points and random lines, but the uniform measure is better when you want to restrict to subspaces. Roughly, the reason for the latter is that if you start with equal-slices and restrict a few coordinates, then the distribution on the resulting subspace becomes much more like a uniform one, or a suitably weighted uniform one. For example, if you’re told that a point’s restriction to the first m coordinates has roughly equal numbers of 1s, 2s and 3s, then with very high probability it belongs to a slice with roughly equal numbers of 1s, 2s and 3s, so the distribution in that (n-m)-dimensional subspace is roughly uniform.
It’s possible that we could argue that we can restrict to subspaces and get an equal-slices density increase, but I think we would still have to go via uniform, so in the end I don’t see a compelling argument for doing that.
• Ryan O'Donnell Says:
March 11, 2009 at 12:08 am
1006.2. That’s a good point. I guess I was idly hoping that the fact that equal-slices is exactly equal to a (fairly natural) mixture of uniform distributions on subspaces might allow us a clever way to stay entirely in the equal-slices world.
12. gowers Says:
March 11, 2009 at 12:55 am | Reply
1007. DHJ(k)
I think I’ve now proved (though with details a bit hazy at a couple of points) the analogue for k of the fact that line-free sets in ${}[3]^n$ correlate locally with 12-sets. The proposed argument can probably be tidied up in a few places. I haven’t yet thought about whether any of the rest of DHJ(k) is going to present problems, but it feels as though we’re closing in on it pretty rapidly. That’s my lot for today though.
13. Terence Tao Says:
March 11, 2009 at 1:20 am | Reply
1008. Re: 1003.3
Yes, you’re right, one also needs Sperner. Without it, what one gets is that if ${\mathcal A}$ is uniformly distributed with respect to 12-sets, then
${\Bbb E} 1_{\mathcal A} 1_{\mathcal A}( \pi_{A,01 \to 1}(x)) 1_{\mathcal A}( \pi_{A,02 \to 2}(x)) \approx {\Bbb E} \delta 1_{\mathcal A}( \pi_{A,01 \to 1}(x)) 1_{\mathcal A}( \pi_{A,02 \to 2}(x))$
and then one has to use DHJ(2) to say that the RHS is large (which one can do, after letting A vary in a sufficiently Poisson-like manner).
I would imagine that the same thing works in higher k, perhaps this is already implicit in 1007.
14. Top Posts « WordPress.com Says:
March 11, 2009 at 1:49 am | Reply
[...] Problem solved (probably) Without anyone being particularly aware of it, a race has been taking place. Which would happen first: the comment count [...] [...]
15. Ryan O'Donnell Says:
March 11, 2009 at 5:38 am | Reply
1009. Equal-slices distribution.
I think the following is an easier explanation of what I wrote.
Claim: Pick $z \in [k]^n$ according to equal-slices. Then pick $J \in [k-1]$ uniformly and form $x$ by changing all the $k$‘s in $z$ to $J$‘s. Then $x$ is distributed according to equal-slices on $[k-1]^n$.
Note: it’s helpful to think of $k$‘s as $\ast$‘s.
Proof sketch via coupling: Draw a circle of $n$ dots. Place bar(1) uniformly at random in the $n+1$ slots. Then place bar(2) uniformly at random in the resulting $n+2$ slots. Etc., placing $k$ bars. Pick $J \in [k-1]$ uniformly. Look at the arc of dots following bar(k) clockwise, up until the next bar. Fill in these dots with $k$‘s. Look also at the arc of dots preceding bar(k) counterclockwise, up until the next bar. Fill in these dots with $J$‘s. Next, for each of the remaining $k-2$ arcs of dots, fill it all with the same digit — using up the remaining digits $[k-1] \setminus \{J\}$ in some arbitrary way. Finally, delete the bars, cut the circle to make it a segment, and then randomly permute the whole sequence of digits. We claim the resulting sequence is distributed as equal-slices on $[3]^n$, and therefore we can take it as $z$.
Now forming $x$ from $z$ is the same as doing the above procedure but using $J$‘s where we were using $k$‘s before. But really, happens under this new procedure? You place $k-1$ bars as if you were doing equal-slices on $[k-1]^n$; then you throw in a phantom extra bar but only use it insofar as deciding to call the arc it lands in the $J$-arc; then you define the other $k-2$ arcs according to the digits in $[k-1] \setminus \{J\}$, etc. Now granted, the arc chosen to be the $J$-arc is not uniformly distributed among arcs — it’s biased towards the longer arcs. But who cares, since $J$ is uniformly random on $[k-1]$?
16. Ryan O'Donnell Says:
March 11, 2009 at 6:01 am | Reply
1010. “Varnavides”-DHJ(k). (Sorry for the scare quotes, but it strikes me as funny somehow to name this concept after a person.)
Tim, I can’t quite follow what you wrote at the end of your proof sketch that line-free sets in $[k]^n$ correlate with lower-complexity sets. I was wondering if you could clarify. In particular, as you say it seems we need to show that DHJ(k) actually implies Varnavides-DHJ(k) under some appropriate measures.
As you suggest, it seems that equal-slices is what to hope for here (it works pretty well for $k = 2$ at least!). In other words, we may hope that if $A \subseteq [k]^n$ has equal-slices measure $\geq \delta$ then if you choose an “equal-slices-combinatorial-line” (i.e., draw from equal-slices on $([k] \cup \{\ast\})^n$) then there is some $c(\delta) > 0$ chance that all $k$ points are in $A$.
Looking at that statement now I worry slightly that it is a bit optimistic. Well, let’s hope it’s true, and let me ask, Tim, about what you wrote.
I guess I don’t quite understand what’s going on with the parameter $M$. On one hand, I assume it’s to be “small” (although still perhaps $\omega_{n \to \infty}(1)$), because otherwise bounding the number of lines in $[k]^M$ by $(k+1)^M$ looks worrisome.
On the other hand, this makes me worry a bit about bullet-point 3: if $\nu$ is supported on $M$-dimensional subspaces then somehow I don’t expect that a set of lines $\mathcal{L}$ being large under $\nu$ should mean it’s large under a more “global” measure on lines such as equal-slices.
Quite likely I’m mistaking your argument or missing something — would you be able to clarify? Thanks!
• gowers Says:
March 11, 2009 at 1:16 pm
1010.1 Ryan, the argument is so sloppily written (little more than a declaration that my gut feeling is that it works) that it is good that you apply a little pressure of this kind. I think I have answers to your specific queries, but that’s not the same as a proof. But maybe it will tip you over to believing that it works.
First of all, my parameter M is not supposed to tend to infinity. As I said in the write-up, it’s meant to be the smallest integer such that we have DHJ(k) with density $\zeta$, where $\zeta$ depends only on the equal-slices density $\theta$ of the set inside which we’re trying to find lots of combinatorial lines.
So what about your second worry? The point here is that the M-dimensional subspaces I’m averaging over are absolutely not local. Instead, they’re supposed to be far more tailored to equal-slices. What I’m not doing is fixing almost all coordinates and taking M small wildcard sets, or something like that. Instead, I’m randomly choosing the sizes of the M wildcard sets (which I allow to add up to 1, so there are in fact no fixed coordinates at all) and then uniformly at random choosing a line from the resulting subspace. That’s why I’m hoping that the resulting measure on lines is sufficiently global in character.
17. jozsef Says:
March 11, 2009 at 8:00 am | Reply
1011. A Shelah-type argument
I’m quite optimistic that a Shelah type argument might work for the general DHJ(k) problem. It’s late evening and I had a hard day, so I’m not sure how far can I go with the argument tonight. Let me give some basic definitions first. In $k^{[n]}$ two points are neighbours if they have same coordinates everywhere but in one position where one is k-2 and the other is k-1. For a subset $S\subset k^{[n]}$ the space $k^{[n]}$ is fliptop if a point p is in S iff any neighbour of p is also in S. (I will wikify it if the argument turns out to be useful) Our goal is to find a large fliptop subspace which is dense. By induction there will be a k-1 combinatorial line which gives a full combinatorial line because of the fliptop property of the subspace. I will follow Shelah’s strategy with some extra conditions to make it (hopefully) work for the density version.
Suppose that $S\subset k^{[n]}$ is c-dense and we already know the subspace version of DHJ for k-1. Partition [n] into m consecutive intervals of lengths $l_1, l_2, \ldots, l_m.$
We will repeat two steps to get a subspace eventually. The first one should keep the density high and the second will provide the fliptop property.
Step one: For any point in $k^{[l_m]}$ there is a certain number of points in S with that tail. Select the points from $k^{[l_m]}$ which are tails of at least $(c-\epsilon_1)k^{n-l_m}$ elements of S. The set of such points is denoted by H1. Applying the induction hypothesis we know that $k^{[l_m]}$ will contain a d1-dimensional subspace where every element avoiding (running) coordinate k-1 are from H1 (so it has many extensions that are in S).
Step two: Consider the elements of the d1-dimensional subspace where every coordinate is k-1 or k-2. We say that two such points are equivalent if they are the tails of the same subset of S. There are no more than $k^{n-l_m}$ such equivalence classes. Therefore if $d1 > k^{n-l_m}$ then we have a Sperner pair, two points where the set of k-1 coordinates of a point contains the other point’s set of k-1 coordinates. It defines a combinatorial line in the subspace and we will work with this line in the following steps.
We will repeat steps one and two in $l_{m-1}$. I will continue, but I would like to read over first what I wrote.
• jozsef Says:
March 11, 2009 at 8:56 am
1011. contd.
Let us denote the k points of the combinatorial line selected previously by $P^1_0, P^1_1,\ldots ,P^1_{k-1}$ (indexed by the running coordinates.)
Step one: select set H2 from $k^{[l_{m-1}]}$.
Select a point if with any $P_i$ it is the tail of at least $(c-\epsilon_2)k^{n-l_m-l_{m-1}}$ elements of S where $0\leq i\leq k-2$. There are two cases; either there are many such points or there is a tail of at least $(c+\delta)k^{n-l_m-l_{m-1}}$ elements of S. (More careful calculations are needed, but let me sketch the argument first.) If there are enough points, the there is a d2-dimensional subspace in $k^{[l_{m-1}]}$ where every point without coordinate k-1 is from H2.
Step two: Consider the elements of the d2-dimensional subspace where every coordinate is k-1 or k-2. We say that two such points, a and b. are equivalent if for any element $t\in k^{[n-l_m-l_{m-1}]}$, $t,a,P_i$ is in S iff $t,b,P_i$ is in S. There are no more than $k^{n-l_m-l_{m_1}}+1$ such equivalence classes. If $d2 > k^{n-l_m-l_{m_1}}+1$ then there is a Sperner pair, two points where the set of k-1 coordinates contains the other point’s set of k-1 coordinates. It defines the second combinatorial line in the subspace and we will work with the two lines in the following steps. Let us denote the k points of the new combinatorial line by $P^2_0, P^2_1,\ldots ,P^2_{k-1}$
• jozsef Says:
March 11, 2009 at 6:18 pm
There are a few typos in the argument. I should fix them eventually, but let me mention her one which might be quite misleading; “There are no more than $k^{n-l_m}$ such equivalence classes.” actually, there are no more than $2^{k^{n-l_m}}$ such equivalence classes. Similarly, in the second iteration There are no more than $2^{k^{n-l_m-l_{m-1}+1}}$ such equivalence classes.
18. jozsef Says:
March 11, 2009 at 9:00 am | Reply
I’ll continue tomorrow morning, but I hope that what I wrote makes sense and maybe someone will tell me if this plan is feasible or not.
• gowers Says:
March 11, 2009 at 9:20 am
1012.1 I’ve just read through it and nothing jumps out at me as wrong, or obviously not going to work. I suppose my one worry is a metaworry, which is that many people must have tried to produce a density version of Shelah’s argument, and there doesn’t seem to be any sign of an obstacle in the above sketch. On the other hand, a balancing metareassurance is that you have a record of finding dazzlingly simple arguments.
I haven’t yet thought about the part where you write “more careful calculations are needed”.
Perhaps another reason to be hopeful is that one way of explaining what Shelah did is to say that he changed an inductive argument from deducing HJ(k) from HJ(2) with the repeated help of HJ(k-1) to deducing HJ(k) from HJ(k-1) with the repeated help of HJ(2), and there is no obvious reason to think that that cannot be done for density as well.
Anyhow, if it works … amazing!
19. gowers Says:
March 11, 2009 at 12:54 pm | Reply
1013. Shelah.
I took the car to be serviced today, which resulted in a long walk to work, during which I pondered Jozsef’s argument. I can’t see a problem, but neither have I thought it through fully. So I want to try to assess its feasibility by seeing if I can sketch it. (Of course, Jozsef has already sketched it, but there’s no harm in having more than one sketch, and if I manage to sketch it without looking any further at what he wrote, then it might make us feel better in the way that people feel better about computer-assisted proofs if more than one person does the computer part and different programs are used.)
What is remarkable about Jozsef’s proposal if it works is that it is so similar to what Shelah did: it would make it very mysterious that the proof had not been discovered earlier (which would add another layer of mystery to the already remarkable fact that Shelah’s proof was itself not discovered earlier).
The idea is this. Suppose that we are trying to prove DHJ(k) with density $\delta$. We begin by choosing some parameters. First, we choose $m$ so large that DHJ(k-1) is true with density $\delta/2$. That is, we assume as an inductive hypothesis that every subset of ${}[k-1]^m$ of density at least $\delta/2$ contains a combinatorial line.
Next, we choose a very rapidly increasing sequence of integers (though in the context of some proofs of this kind the rate of increase is not too bad — this proof should result in a primitive recursive bound for DHJ(k), one level beyond a tower, just like Shelah’s) $r_1<<r_2<<\dots<<r_m$. (Jozsef called them $l_1<<\dots<<l_m$ but on my screen an $l$ is just a vertical line so I’ve changed the notation.)
We shall repeatedly apply the following averaging argument: if $a_1,\dots,a_M$ are integers between 0 and 1, and the average of the $a_i$ is $\delta$, then at least $\epsilon N$ of the $a_i$ are at least ${}\delta-\epsilon.$ The proof is trivial (by contradiction).
Now let $n=r_1+\dots+r_m$ and think of ${}[k]^n$ as ${}[k]^{r_1}\times\dots\times[k]^{r_m}.$ Let $\mathcal{A}$ be a subset of ${}[k]^n$ of density $\delta.$ For each $y\in[k]^{r_m},$ let $\mathcal{A}_y=\{x\in[k]^{n-r_m}:(x,y)\in\mathcal{A}\}.$ Then the average density of $\mathcal{A}_y$ over $y\in[k]^{r_m}$ is $\delta$, so the density of $y$ such that $\mathcal{A}_y$ has density at least $3\delta/4$ is at least $\delta/4$.
Let $\mathcal{B}_m$ be the set of all $y\in[k]^{r_m}$ such that $\mathcal{A}_y$ has density at least $3\delta/4$. By the pigeonhole principle, we can find a subset $\mathcal{C}_m$ of $\mathcal{B}_m$ of density at least $\gamma=\delta/4k^{n-m_r}$ such that $\mathcal{A}_y$ is the same for every $y\in\mathcal{C}_m$.
Let us define a ”binary subspace” to be a set where you fix some of the coordinates and allow the remaining ones to take the values k-1 and k. If t is small enough, and if you choose a random binary subspace of ${}[k]^{r_m}$ by randomly choosing t coordinates to be the variable ones and randomly fixing the others, and if you then choose a random point in one of these random binary subspaces, the result will be very close to the uniform distribution on ${}[k]^{r_m}$. Therefore, by an easy averaging argument we can find a binary subspace inside which the density of $\mathcal{C}_m$ is at least $\gamma/2.$ (Note that for this to work we need $r_m$ to be large enough in terms of $\gamma$ and $t$.) And then by Sperner’s theorem we can find two points in the binary subspace that form a combinatorial line (in the ${}[2]^n$ sense but with $k-1$ and $k$ replacing 1 and 2, or 0 and 1 if you prefer).
We can extend this binary combinatorial line to a full combinatorial line in ${}[k]^{m_r}$, by allowing the wildcards to take the values $1,2,\dots,k-2$ as well. The result is a …
OK, this is the first point at which I’ve got stuck, though I think Jozsef may have dealt with the difficulty I am facing. Time to end this comment and start a new one.
20. gowers Says:
March 11, 2009 at 1:06 pm | Reply
1014. Shelah
The problem with what I was doing just above is this. I’ve just defined a line $L=(y_1,\dots,y_k)$ in ${}[k]^{m_r}$, and I know that the sets $\mathcal{A}_{y_{k-1}}$ and $\mathcal{A}_{y_k}$ are the same. However, I don’t know anything about $\mathcal{A}_{y_i}$ for $i\leq k-2,$ and it would be a serious problem if e.g. they turned out to be empty.
However, I’ve ignored Jozsef’s step 1, which is perhaps his main new idea. He begins by using DHJ(k-1) to pass to a subspace where $\mathcal{A}_y$ has density close to $\delta$ for all sequences $y,$ provided only that they do not have any (variable) coordinate equal to $k$. (This can be done by an averaging argument similar to the one I used above to get dense binary subspaces — this time one wants dense (k-1)-ary subspaces.)
But now I’ve reached the point where I don’t understand Jozsef’s argument. How do I know that it is not the case that $\mathcal{A}_y$ is empty whenever one of the variable coordinates of $y$ is equal to k? I don’t see it at the moment, so I think I’d better wait for Jozsef to wake up.
• jozsef Says:
March 11, 2009 at 5:47 pm
1014.1 Dear Tim, I don’t think that we need that $\mathcal{A}_y$ is dense if one of the variable coordinates is equal to k. This is actually a crucial observation, that’s why we need the fliptop property. At the last step, when only $r_1$ remains, one word of $r_1$ spans a dense subset of the $\{1,2,..,k-1\}^{m-1}$ part of the subspace. By induction there is a line, and by the fliptop property there is an extension to k. Note that the fliptop property was defined by considering running coordinates k as well, however it wasn’t sensitive for density in any ways.
• gowers Says:
March 11, 2009 at 7:15 pm
1014.2 I’m still mystified. In Step 1, let’s choose a subspace such that for every sequence $y$ that avoids running coordinate $k$, there is a set of density almost $\delta$ of $x$ such that $(x,y)\in\mathcal{A}$. Now we go to Step 2. Nothing we’ve done so far stops $\mathcal{A}_y$ being empty for every $y$ that does involve $k$. So we choose our Sperner pair and have two points $y$ and $y'$ (where $y'$ is obtained from $y$ by changing some of the $(k-1)$s to $k$s). How can that help? Indeed, it seems that $\mathcal{A}_y$ could be empty for every point in the combinatorial line you select.
• jozsef Says:
March 11, 2009 at 7:59 pm
Tim – I have to go to the University now, where I will rethink your question again, but I think that the answer is that it is not possible that you switch a running coordinate to k from k-1 and the density drops because the two points are neighbours. I will get back to you soon.
• gowers Says:
March 11, 2009 at 8:02 pm
Jozsef, that gives me time to be more precise about my question. See comment 1016 below.
21. Polymath1: Probable success « Combinatorics and more Says:
March 11, 2009 at 7:21 pm | Reply
[...] northern Israel. Coming back I saw two new posts on Tim Gowers’s blog entitled “Problem solved probably” and “polymath1 and open collaborative mathematics.” It appears [...]
22. Gil Kalai Says:
March 11, 2009 at 7:39 pm | Reply
1015. A bit off-topic
“If this were a conventional way of producing mathematics, then it would be premature to make such an announcement — one would wait until the proof was completely written up with every single i dotted and every t crossed — ”
hmmm, maybe we should regard it premature also in this new mode. What do the rules say?
“but this is blog maths and we’re free to make up conventions as we go along.
So I hereby state that I am basically sure that the problem is solved”
Congratulations! I must admit that even if a full k=3 proof will take 200 more comments and several more weeks the success of the project exceeded my expectations.
“(though not in the way originally envisaged).”
Actually, this is a lovely aspect and a sign of success of the open collaboration idea, isn’t it?
• gowers Says:
March 11, 2009 at 7:49 pm
1015.1 Gil, I feel slightly awkward about it, and won’t feel entirely happy until it’s written up in the conventional way. But if one regards a write-up as having a tree structure, then it’s all there on the wiki down to the small sticks, and only a few leaves are missing. Somehow there don’t seem to be any points where one says “I don’t quite see it but I’m sure it works.” It’s more like “I do see it but it’s slightly tedious to write out in full.” Despite that, I wouldn’t have written what I wrote if it had not been for the fact that other participants seem to share my confidence. But for now I will keep the “(probably)”.
23. gowers Says:
March 11, 2009 at 8:12 pm | Reply
1016 Shelah
Jozsef, I’m still trying to follow your argument in 1011. Let me try to be completely precise about where it seems to me to go wrong. Of course, it could be me who is going wrong.
Consider the following example. I take a random subset S of ${}[k]^{r_m}$ and then take the set $\mathcal{A}$ of all sequences in ${}[k]^{r_1+\dots+r_m}$ such that the final part lies in S. This set has density 1/2. Now I apply your step 1, and it happens that the $d_1$-dimensional subspace I choose has the following property: every point with all running coordinates less than k belong to S, and all other points belong to the complement of S. (With high probability such a subspace exists, and no steps have been taken to avoid it.)
Then the pigeohole argument gives you that all the points with running coordinates less than k can be joined to anything you like in ${}[k]^{r_1+\dots+r_{m-1}}.$ However, if you take any line given by the application of Sperner, then both its top two points must involve a k and give the empty set. Therefore, the whole line gives the empty set (meaning that if your mth part belongs to the line then you don’t belong to $\mathcal{A}$). So at the second stage of the iteration I don’t see how you can do Step 1 (because of the “with any ${}P_i$” requirement).
• jozsef Says:
March 11, 2009 at 8:52 pm
Tim, what you wrote is perfectly right with the exeption that in step one I never consider running coordinates larger than k-1. (with my old notation “larger than k-2″) So, in your example fliptop means that the neighbours given by the last (m-th) k, k-1 pair are simultaneously not in S. I’m still not sure that the argument works, but I don’t see any serious problem yet.
• jozsef Says:
March 11, 2009 at 10:40 pm
I have office hours now, so I will beging to write up the argument and if I won’t have too many students to show up then I will post the sketch in an hour or so.
• gowers Says:
March 11, 2009 at 10:49 pm
Just to try to be more explicit still, let’s imagine that $k=3$ and $r_m=6$. Suppose that all sequences of six 0s and 1s are the tail of every preceding sequence, and that the two points $001122$ and $001222$ are the tails of precisely the same set of sequences, so you choose as your line the line $001*22$. It seems to me that there is no reason for any point in that line to be the tail of any previous sequence.
A separate question: is the aim to construct in each ${}[k]^{r_i}$ a line $L_i$ in such a way that the product of the lines is fliptop (that is, insensitive to changing 1s to 2s when k=3), and $\mathcal{A}$ is dense in that product? My difficulty is that I don’t see the relevance of a statement about $\{0,1\}^{r_m}$ when the line $L_m$ might have 2 as one of its fixed coordinates.
• jozsef Says:
March 11, 2009 at 11:42 pm
I see now. One should be more careful with the selection of the fliptop lines. I still think that it is duable, but it seems that we have to consider more subspaces. I don’t have much time, but I will think about it.
• gowers Says:
March 12, 2009 at 1:36 am
I should stress here that I definitely think this is worth attempting. Even if it turns out not to work 100% straightforwardly, the following question exists and is clearly an interesting one. You have a set $\mathcal{A}\subset[k]^n$ of density $\delta$. How large a subspace $S$ can you find such that $\mathcal{A}$ has density at least $\delta/2$ in the subspace, and membership of $\mathcal{A}$ in the subspace does not change if you flip a $k$ to a $k-1$? I do not see any philosophical reason for that being significantly harder than Sperner plus induction, though such a reason might emerge I suppose.
24. Ryan O'Donnell Says:
March 11, 2009 at 10:43 pm | Reply
1017. Varnavides-DHJ(k).
In 1010.1, Tim wrote “What I’m not doing is fixing almost all coordinates and taking M small wildcard sets, or something like that. Instead, I’m randomly choosing the sizes of the M wildcard sets (which I allow to add up to 1, so there are in fact no fixed coordinates at all) and then uniformly at random choosing a line from the resulting subspace.”
Tim, I think I get it now; as usual I was misinterpreting the word “subspace” and was thinking you were doing what you said you weren’t doing I agree now that the sketch looks relatively solid. Let me see if I can fill in some “leaves”, as you say.
• gowers Says:
March 11, 2009 at 10:50 pm
Great — I’m off for a couple of hours and won’t be able to do much more today, if anything.
25. Ryan O'Donnell Says:
March 11, 2009 at 11:40 pm | Reply
1018. Is Equal-Slices of Equal-Slices equal to Equal-Slices?
Suppose we want to try to prove Varnavides-DHJ(3); as in #1010, let’s try to show that if $A \subseteq [3]^n$ has equal-slices density $\delta > 0$, and we choose a random combinatorial line by drawing its template from equal-slices on $([3] \cup \{\ast\})^n$, then with probability at least \$\eps(\delta) > 0\$ the whole line is in $A$ (assuming $n$ is sufficiently large).
Tim’s strategy for this involves, among other things, picking a combinatorial subspace of dimension $M$ with no free coordinates, according to some distribution $\nu$; then drawing a point from that subspace (thought of as $[k]^M$) from some other distribution, $\sigma$.
It would be particularly helpful if the overall distribution is equal-slices on $[3]^n$. Also helpful would be if \$lambda \nu\$ is a relatively “natural” distribution; also $\sigma$ should have full support. (It’s also not essential that $\nu$ generates subspaces with exactly $M$ dimensions, but never mind that for now.)
Here is perhaps the simplest possible proposal: Let $\nu$ be equal-slices on $[M]^n$, and let $\sigma$ be equal-slices on $[k]^M$.
The question is: might combining these two actually give equal-slices on $[3]^n$ precisely?! I will submit some evidence in a reply to this post.
• Ryan O'Donnell Says:
March 11, 2009 at 11:53 pm
1018.1. Evidence.
Well, my first piece of “evidence” is that in these simple probability matters, sometimes things just turn out nicely
Here is slightly more rigorous evidence… (I would try to simply give a proof, but I haven’t the time to work on it this evening, so I thought I’d quickly throw it out there.)
For my poor brain, let me fix $M = 100$ and fix $k = 3$, $[k] = \{R, G, B\}$ (red, green, blue).
We can think of the “outer” equal slices as choosing $p_1 + \cdots + p_{100} = 1$ uniformly, then drawing from the resulting product distribution on $\{\ast_1, \dots, \ast_{100}\}^n$. We can think of the “inner” equal slices as choosing $q_R + q_G + q_B = 1$ uniformly, then changing each $\ast_i$ to $R$ with probability $q_R$, $G$ with probability $G$, and $B$ with probability $q_B$.
If you think about it a minute, you see that the final is string is drawn from a randomly chosen product distribution $s_R +s_G + s_B = 1$ on $\{R,G,B\}^n$. That’s good! If we could just show that that $(s_R, s_G, s_B)$ is indeed distributed as “uniform on the $3$-simplex” we’d be done.
So how is it chosen? Well, we think of $(q_R, q_G, q_R)$ as being chosen first. Then, think of $p_1, \dots, p_{100}$ as being independent Exponential$(1)$‘s. (Of course, this does not give $100$ numbers adding to $1$, but we use these numbers as proportions, rather than probabilities.)
Then $(s_R, s_G, s_B)$ gets generated by putting each $p_i$ into the “R” bucket with probability $q_R$, into the “G” bucket with probability $q_G$, and into the “B” bucket with probability $q_B$. Finally, you add up all the Exponentials in each bucket, and $(s_R, s_G, s_B)$ is proportional to these totals.
So what is the distribution on the three bucket totals? It shouldn’t be too hard to really work it out, but let me write heuristically here. Once $(q_R, q_G, q_B)$ is fixed, we expect about $100q_R$ of the independent Exponentials to go into the R bucket, and similarly for G and B. (Even moreso do we expect this if $100 = M$ is huge.) And thinking of $100q_R$ as still huge, we’re adding up a huge number of i.i.d. Exponentials, so the CLT will imply that the sum will almost surely be very close to $100q_R$. Similarly, the sum in the G bucket will almost surely be very close to $100q_G$, and again for B.
But in this case, the proportions will be really close to $q_R/q_G/q_B$ I.e., almost surely it seems $(s_R, s_G, s_B)$ will be very nearly $(q_R, q_G, q_B)$. But $(q_R, q_G, q_B)$ is distributed uniformly on the $3$-simplex, as desired!
• Ryan O'Donnell Says:
March 12, 2009 at 6:00 am
1018.2. I don’t think it’s exactly correct, but I still hope some variant is…
26. Ryan O'Donnell Says:
March 12, 2009 at 6:29 am | Reply
1019. ENDS of ENDS = ENDS.
I think I’ve got it now. Define a distribution on $k$-partitions of $n$ called $k$-Equal Non-Degenerate Slices ($k$-ENDS): To get it, take $n$ dots in a line, consider the $n-1$ gaps between them, and place $k-1$ bars in these gaps, uniformly from among the $\binom{n-1}{k-1}$ many possibilities (i.e., “without replacement”). Then the intra-bar blocks of dots correspond to $k$ positive integers $m_1, \dots, m_k$ which add up to $n$. Associating block $m_i$ to character $i \in [k]$, we get a distribution on “Non-Degenerate” slices (each character appears at least once). We can extend this to a distribution on $[k]^n$ as usual, just by randomly permuting the string.
This is actually a kind of convenient variant on equal-slices, because when you’re using equal-slices to pick combinatorial lines, it’s a bit annoying to have to worry about the unlikely case of getting $0$ wildcards.
Anyway, I think it’s now easy to see that if you do $M$-ENDS, and then you do $k$-ENDS on that (as described in #1018), then you just get $k$-ENDS, exactly. Because all that’s really happening is that you’re picking $M-1$ bar-spots uniformly, and then you’re picking $k-1$ of those $M-1$ bars to be the “final bars”, uniformly. So that’s just the same as picking $k-1$ bars uniformly to start with.
I think. Once again, it’s late.
27. Ryan O'Donnell Says:
March 12, 2009 at 6:30 am | Reply
1019.1. “Formula does not parse” above is “(n-1 choose k-1)”. [Fixed]
28. Ryan O'Donnell Says:
March 12, 2009 at 7:13 am | Reply
1019.2. Some googling reveals this paper (among others) which discusses in its Section 2 the proper nomenclature for this stuff, as well as related probability distributions.
k-ENDS -> “random composition into k parts”; the consequent distribution on strings is “random ordered partition into k parts”.
dots -> “balls”
bars -> “walls”
29. gowers Says:
March 12, 2009 at 10:21 am | Reply
1020 DHJ(k)
Ryan, that all looks like good news — it would be very nice if we could get the technicalities to run smoothly so that they don’t obscure the main ideas, and your comments above give me hope that it will be possible.
I’ve got 25 minutes to spare so I’m going to try to explain why I think that DHJ(k) is basically done too. I’m assuming for this that the proof that a line-free set correlates locally with a set of complexity k-2 is going to work out now, which seems likely, as it’s been sketched down to a pretty fine level of detail and you’ve scrutinized the parts that looked most likely to go wrong if anything was going to.
In order to explain how I think the rest of the proof will go, what I need to do is explain the difference between the proof for DHJ(k) and the proof of DHJ(3) outlined on the wiki here.
Analogue of Step 1. We need to replace 1-sets by what I call $\{1,2,\dots,k-2\}$-sets. These are sets \$\latex \mathcal{B}\$ such that $x\in\mathcal{B}$ if and only if $(U_1(x),\dots,U_{k-2}(x))\in\mathcal{Z}$ for some suitable collection $\mathcal{Z}$ of disjoint (k-2)-tuples of sets and $U_i(x)$ stands for the set of places where x takes the value i. To save writing, I’ll call these simple sets. So the next obvious aim is to prove that a simple set can be approximately partitioned into subspaces. Once that’s done, I think it is fair to say that the rest of the proof is exactly as in the k=3 case.
The one other thing that needs to be changed in Step 1 is that the appeal to the multidimensional Sperner theorem has to be replaced by an appeal to multidimensional DHJ(k-1), a proof of which is sketched on the wiki here. Otherwise, the changes are trivial: 3 becomes k, and $\mathcal{U}\otimes[2]^n$ becomes our simple set $\mathcal{B}.$ (I admit that I haven’t been through Step 1 line by line checking what I’m saying — that would use up more than my 25 minutes.)
I’ve looked through Step 2 and I think all the changes needed are of the trivial variety (changing 3s to ks and that kind of thing). It might be nice at some point to write an abstract version of this argument, since what we’re using seems to be very general — that we can partition ${}[k]^n$ into subspaces very conveniently, that the structure of good sets is preserved when we do so, that dense good sets contain lots of subspaces — that kind of thing.
Step 3. And now we do the same trick as we did in DHJ(3) except that we have to apply Step 2 k-1 times instead of twice. The end result is that a set of complexity k-2 can be almost partitioned into subspaces, and that’s what we’re looking for.
30. Anonymous Says:
March 12, 2009 at 4:46 pm | Reply
1021. Varnavides.
Hi Tim, I think the $k$-ary proof that line-free sets correlate with simple sets is now done to about 100% satisfaction in my mind. I’ll try to wikify some details later, but to illustrate the simplicity with which it’s working out, here is a complete proof of Varnavides:
Assume we have DHJ(k).
Technicality: I want to say that a combinatorial line is (also) “degenerate” if the fixed coordinates do not use each character from $[k]$ at least once. Deducing DHJ(k) with this very slightly stronger notion of degeneracy is trivial.
Proof of Varnavides: Assume $A \subseteq [k]^n$ has density $\delta > 0$ under the “uniform $k$-composition” distribution (i.e., equal-slices conditioned on the slice having at least one of each character). Let $M = M(\delta/2)$ denote the least integer such that DHJ(k) guarantees the existence of a nondegenerate line in any subset of $[k]^M$ with uniform $k$-composition density at least $\delta/2$.
Think of choosing $x \in [k]^n$ as first drawing a random $M$-dimensional subspace $V$ from the uniform $M$-composition distribution, and then drawing $x$ randomly from $V \cong [k]^M$ according to the uniform $k$-composition distribution. Since $x \in A$ with probability at least $\delta$, we know that with probability at least $\delta/2$ over the choice of $V$, the $k$-uniform composition density of $A|_V$ is at least $\delta/2$. Call such $V$‘s “good”. By DHJ(k) we know that for any good $V$, there is a (nondegenerate) line inside $A|_V$.
Now suppose we choose a random line $L$ in \$[k]^n\$ according to the uniform $(k+1)$-composition distribution. We claim that it is entirely within $A$ with probability at least $\delta/[2 M^{k/2} (k+1)^M]$ . To see this, again think of $L$ as being drawn by first choosing $V$, and then drawing a random line in $V$ according to uniform $(k+1)$-composition. With probability at least $\delta/2$ $V$ is good, in which case there is at least one nondegenerate line in $A|_V$. The probability $L$ is picked to be it is at least $1/[M^{k/2} (k+1)^M]$, since the least probability under uniform $(k+1)$-composition of $M$ is at least this quantity (er, or something like that).
• Ryan O'Donnell Says:
March 12, 2009 at 4:51 pm
1021.1. That was me again, not logged in. I wanted to add that we can complete the whole “line-free implies correlation with simple set” proof exclusively in this uniform $k$-composition measure rather easily, I think. The only thing to really check is that if you choose a $1 - \epsilon$ fraction of the coordinates $S$ randomly, do uniform-$k$-composition on $S$, and then independently do uniform-$k$-composition OR uniform-$(k-1)$-composition on $\overline{S}$, then the overall probability distribution has total variation distance at most $O(\epsilon \sqrt{n})$ from the usual global uniform $k$-composition distribution. Which is, I’m pretty sure, true.
31. gowers Says:
March 12, 2009 at 10:58 pm | Reply
1022. Shelah
While I wait for reaction about how we should go ahead with writing things up (see this comment), I want to continue thinking a bit about whether the proof we have could be Shelah-ized. Terry said in a comment some time ago that he is a big fan of translating things back and forth between combinatorial and ergodic languages. I (in common with just about every mathematician) am a big fan of “completing the square,” by which I mean completing puzzles of the form “A is to B as C is to ??” For example, “corner is to combinatorial line as large grid is to ??” the answer to which turned out, not completely expectedly, to be “${}[2]^k$ for some large $k$“. And that was just one part of a bigger such puzzle: “the Ajtai-Szemerédi proof of the corners theorem is to the triviality that a dense subset of ${}[n]$ contains two distinct points as ?? is to the proof of Sperner’s theorem.”
So here I would like to think about the puzzle, “The colour-focusing proof of the Hales-Jewett theorem is to Shelah’s proof of the Hales-Jewett theorem as the proof we now have of DHJ(k) is to ??”
The reason I am drawn to this is that I once lectured both proofs of the Hales-Jewett theorem and came to a very clear understanding of the relationship between them, which I’ll have to reconstruct, but I remember it as being natural enough that it will be easy to reconstruct. When I have a spare moment, I’ll write something about this on the wiki, but for now let me just say that Shelah’s basic idea can be interpreted as follows: whereas the old proof of HJ(k) used HJ(k-1) to reduce the problem to the almost trivial HJ(2), Shelah’s new proof used HJ(2) to reduce the problem to HJ(k-1). As I say, I’ll justify that at some point, but here I just want to take that thought and have a stab at applying it to DHJ(k) instead.
Of course, I’m not sure that it’s possible to do so, but just having the goal in mind does, I think, suggest some questions. For instance, here’s a place in the proof where DHJ(k-1) comes into the proof of DHJ(k). One first passes to a subspace, which we’ll notate as though it is the whole space, such that the density of $\mathcal{A}$ is almost $\delta$ and the density of $\mathcal{A}$ in ${}[k-1]^n$ is still positive. Writing $\mathcal{B}$ for $\mathcal{A}\cap[k-1]^n$, we use $\mathcal{B}$ to construct a dense set $\mathcal{C}$ of complexity $k-2$ that is disjoint from $\mathcal{A}$. It doesn’t matter too much what all this means: the main point is that once we have a set of complexity $k-2$ we are able to hit it with DHJ(k-1).
So a natural question, if we are trying to Shelah-ize, is whether we can construct some set of complexity 1 (which means it’s hittable with DHJ(2), aka Sperner) and thereby, somehow, create a fliptop subspace inside which $\mathcal{A}$ is dense. Recall that a fliptop subspace is one inside which if you change any coordinate with value k-1 to a k, then you don’t change whether the point x belongs to the set $\mathcal{A}$. So if your set is dense in such a subspace (or rather, dense in the ${}[k-1]^n$ part of the subspace) then you can hit it with DHJ(k-1) and finish off the proof immediately. Because DHJ(k-1) is used just as the inductive step and not as some part of an iterative procedure for getting to the inductive step, a proof of this kind should be far better quantitatively.
In my next comment I shall think about the following question: are there any complexity-1 sets that can be derived from a set $\mathcal{A}$? For now, I don’t care whether we can actually use them.
32. gowers Says:
March 12, 2009 at 11:40 pm | Reply
1023. Shelah.
Continuing on from 1022, a completely simple-minded idea would be to pass to a subspace such that $\mathcal{A}$ is dense in ${}[2]^n$ and still had density almost $\delta$. (I’m just going to assume that that’s fine.) If we write $\mathcal{B}$ for $\mathcal{A}\cap[2]^n,$ then what set $\mathcal{C}$ can we construct from $\mathcal{B}$?
In the other argument, we looked at all sequences $x$ such that, for every $j$, if you change all $k$s to $j$s then you get a sequence in $\mathcal{B}$. Here, it is clear, we are going to have to do a much more radical collapsing of the sequence. One possibility, which may be nonsense, would be to convert all non-1 coordinates into 2s. In other words, we let $\mathcal{C}$ be the set of all x such that the 1-set of x is equal to the 1-set of some sequence in $\mathcal{B}$.
In the other argument, we then argued that $\mathcal{A}$ is disjoint from $\mathcal{C}$ if it does not contain a combinatorial line. What happens here? Well, bearing in mind that (using the Shelah proof as a guide) we are not trying to go straight for a combinatorial line, but rather for a fliptop subspace, we should perhaps assume as our hypothesis that $\mathcal{A}$ does not contain such a subspace.
Turning that round, we would want to prove that if $\mathcal{A}$ intersects $\mathcal{C}$ then it contains a fliptop subspace. (This doesn’t sound true, but let’s see what’s wrong with it first.)
For $\mathcal{A}$ to intersect $\mathcal{C}$ we need a sequence $x\in\mathcal{A}$ such that if you convert all its non-1s into 2s, then you get a sequence of 1s and 2s that also belongs to $\mathcal{A}$. Why should that be interesting? I don’t know.
But suppose instead that we managed to find quite a large subspace S of such points. Then if x belonged to $\mathcal{A}\cap S$ and had a variable coordinate equal to 2, we would … no, I don’t see where this goes.
Part of the problem is that I seem to be going for the fliptop subspace to be contained in $\mathcal{A}$, whereas what I should be aiming for is merely that $\mathcal{A}$ is dense in the fliptop subspace (which actually seems to be a flipbottom subspace).
I’m going to stop this comment here — I hope I’ve demonstrated the flavour of the argument that I think might conceivably exist.
• gowers Says:
March 12, 2009 at 11:41 pm
1023.1 I should have added the remark that the set $\mathcal{C}$ has complexity 1.
33. Terence Tao Says:
March 12, 2009 at 11:45 pm | Reply
1024. Austin’s proof
I’ve written down an informal (and sketchy) combinatorial translation of the k=3 case of Austin’s argument at
http://michaelnielsen.org/polymath1/index.php?title=Austin%27s_proof
It’s actually very much in the style of the proof of the triangle-removal lemma, especially if one takes an analytic perspective rather than a combinatorial one and talks about decomposing functions rather than regularising graphs. In particular, it’s best to avoid thinking in the classical language of vertices and take a more “point-less” approach in the spirit of probability theory or ergodic theory (or operator algebras, for that matter), thus for instance one should think about averages of products of functions rather than counting triangles etc. One reason for shifting one’s thinking this way is that it allows one to think about strong stationarity in a civilised way, whereas this concept becomes really difficult to work with if one insists on holding on to concrete vertices and edges. (In classical combinatorial language, being strongly stationary is like saying that a certain graph is a “statistically equivalent” to the subgraph formed by restricting of its vertex set from a big space such as [3]^n to a randomly chosen subspace; but defining what “statistically equivalent” means precisely is sort of a pain unless one decides to give up on vertices altogether.)
It’s also clear that this proof will end up being messier than the pure density-increment proof (though cleaner than a finitisation of the Furstenberg-Katznelson proof), and give worse bounds, in large part because one has to apply Graham-Rothschild (with parameters given in turn from a Szemeredi-type regularity lemma!) to get the initial strong stationarity properties. It looks possible to use density increment arguments as a substitute for Graham-Rothschild, but this is still going to have quite lousy bounds.
• Terence Tao Says:
March 12, 2009 at 11:55 pm
1023.1 Information-theoretic approach
Hmm, it occurs to me now that the information-theoretic approach to the regularity lemma that I laid out in
http://front.math.ucdavis.edu/math.CO/0504472
(where I “regularise” random variables using Shannon entropy) is actually well suited to this problem, and may lead to a relatively slick (albeit alien-looking) proof. What I may do at some point is begin by showing how by playing around with conditional Shannon entropies of various random variables, one can establish the triangle removal lemma, and I believe that this proof may extend relatively painlessly to the case under consideration. More on this later…
34. Ryan O'Donnell Says:
March 12, 2009 at 11:47 pm | Reply
1025. Always the uniform-ordered-partition distribution.
Following up on my comment #1006, I feel somewhat sure that we can work entirely within the uniform-ordered-partition distribution (i.e., “equal-nondegenerate-slices”). I’m going to try to write it down now to be sure, but if it’s correct, it should mean that “DHJ(k)=>Varnavides”, “DHJ(k)=>multidimensional-DHJ(k)”, and hopefully also “line-free sets correlate with simple sets” can all be proved with simple arguments.
More precisely, they will be the same simple arguments we have now, but with no annoyances due to all the tedious passing back and forth between various distributions via subspaces.
• gowers Says:
March 12, 2009 at 11:54 pm
1025.1 Wow — I very much hope you’re right about this. I suppose the bit that would be hardest (but not obviously impossible) would be “simple sets can be almost partitioned into subspaces”. In the uniform measure you are then done, but in other measures you can’t instantly argue that a set of density $\alpha$ inside the simple set must have density almost $\alpha$ on one of the subspaces.
But that could well be just another little task to add to your collection. Now the idea would be to prove that the measure on a simple set can be written as a positive linear combination of the measures on subspaces. If that too can be done cleanly in the UOPD/ENS then we are in fantastic shape.
35. gowers Says:
March 13, 2009 at 12:26 am | Reply
1026. Does triangle removal imply DHJ(3)?
A wild thought this one, especially as it’s close to an idea I had several years ago and couldn’t get to work. But with the benefit of a bad memory (something that is very helpful in mathematics, in my view) I now can’t see what is wrong with it.
The idea is sort of ridiculous. Let’s pass to a subspace in which, notating it as ${}[3]^n$, $\mathcal{A}$ has average density at least $\alpha$ in the slices up to $m$, where $m$ is much smaller than $n$. Sorry — by that I mean slices where the 1-set and 2-set are of size at most $m$.
The hope now is that the disjointness graph becomes dense (because two sets of size $m$ are almost always disjoint), so we can just apply the usual triangle-removal lemma. My one reason for not completely dismissing this idea is that when I tried it all those years ago I did not have the concept of equal-slices measure to play with.
Unfortunately, it doesn’t become dense. Only the UV-part of it is dense (because it is just the 1-set and the 2-set that I am assuming to be small).
Let me ask a slightly different question, which probably has a fairly easy answer, and that answer will probably kill off this little idea. Given $n$ points in ${}[3]^N$ (where $N$ can be as big as you like), how many combinatorial lines can they contain?
• jozsef Says:
March 13, 2009 at 1:07 am
Well, if you don’t care about the dimension, then the number of lines might be $cn^2$ Just take 0-1 sequences as i zeros followed by n-i ones (for every i) and 0-2 sequences where the zeros are followed by two-s. For any pair of such 0-1 sequences there is a third 0-2, giving a line. So this number itself won’t show that your plan won’t work.
• gowers Says:
March 13, 2009 at 1:17 am
I don’t understand — large should mean $cn^3$ surely? Or perhaps $Cn^2$ but with the three bipartite graphs dense.
• jozsef Says:
March 13, 2009 at 2:10 am
Probably I don’t understand the question. Isn’t it so that two points define the line uniquely. In any case, my example won’t work.
• jozsef Says:
March 13, 2009 at 3:29 am
As it happened before, I didn’t read your post carefully. It is an extremely busy period for me; heavy teaching, committees, etc. So, I’m just writing those “half baked remarks” as Boris would say. I will try to refrain from that in the future.
36. Ryan O'Donnell Says:
March 13, 2009 at 2:24 am | Reply
1027. A single distribution?
In preparation for trying to justify what I wrote in 1025, I have written this document explaining the distribution. I’ll wikify it if and only if this scheme works out and proves helpful
• Ryan O'Donnell Says:
March 13, 2009 at 7:22 am
1027.1 I wrote some more here but I’ve gotten too sleepy to finish tonight. I’ll try to finish tomorrow and explain it properly in a post, but I have to go out of town tomorrow too…
37. Klas Markström Says:
March 13, 2009 at 9:30 am | Reply
1028.
I will make my first visible visit on this side of the project by asking one of the stupid questions discussed in the metathread.
Regarding the question from 1026. As far as I understand any pair of points specify a unique line. if that is the case n points can specify at most Cn^2 lines.
Now if we have a set of points and a set of lines where every pair of points specify a unique line we have the beginnings of a projective plane, but a projective plane has the same number of points and lines.
So is there another known nice family of combinatorial objects which instead maximises the gap between the number of points and lines?
38. gowers Says:
March 13, 2009 at 9:59 am | Reply
1029
Klas, your “stupid” question has exposed the stupidity (without inverted commas) of mine — it’s not what I meant to ask. Let me have another go.
The way we prove Sperner using equal-slices measure can be thought of like this. We partition the measure on ${}[2]^n$ into measures where the set of combinatorial lines is dense. Each measure is obtained by taking a random permutation of ${}[n]$ and taking all intervals in that permuted set, and then any two points with non-zero measure define a combinatorial line.
So a vague version of what I am trying to ask is whether we can somehow find dense structures inside ${}[3]^n$ and play a similar game, this time exploiting the corners theorem. However, I don’t have a sensible precise version of this question, which perhaps suggests that the answer is no.
39. gowers Says:
March 13, 2009 at 12:47 pm | Reply
1030. Shelah
Another idea. Instead of struggling to preserve the density of $\mathcal{A}$ by constructing just one fliptop subspace, perhaps we can partition the whole of (or almost the whole of) ${}[k]^n$ into fliptop subspaces. Then in at least one of them $\mathcal{A}$ would have density almost $\delta$ and we would be done by induction.
Actually, that’s not quite right unless we’re rather careful, because in order to apply induction we need $\mathcal{A}$ to be dense in the ${}[k-1]^m$ part of the fliptop subspace. Let me not worry about this for the time being.
Instead, I’d like to think about whether some kind of argument like the one we’ve used for partitioning 1-sets into subspaces could do the job here. I can’t see it in advance so I’ll just try to plunge in.
Let m be a medium-sized integer — much smaller than n, but much larger than 1 — and write ${}[k]^n$ as ${}[k]^{n-m}\times[k]^m.$ By the pigeonhole principle we can find a subset $\mathcal{A}_1$ of ${}[k]^{n-m}$ of density $k^{-m}$ such that for every $x\in\mathcal{A}_1$ the set of $y\in[k]^m$ such that $(x,y)\in\mathcal{A}$ is the same. Let that set be $\mathcal{B}_1$, and find a big subspace $S_1$ in ${}[k]^m$ that is fliptop with respect to $\mathcal{B}_1$. Then all the subspaces $\{x\}\times S_1$ such that $x\in\mathcal{B}_1$ are fliptop. (They could be disjoint from $\mathcal{A}$, but we don’t care about this.) Let us include them in our attempted partition and remove them from ${}[k]^n.$ Note that we have removed a positive proportion (depending on $m$) of ${}[k]^n$ as a result of this.
We now partition ${}[k]^n$ according to the part that belongs to ${}[k]^m.$ The result is a number of subspaces of the form $S_y=\{(x,y):x\in[k]^{n-m}\}.$ If $y\notin S_1,$ then we have not thrown away any of $S_y$ and we just repeat the first step. However, if $y\in S_1,$ then things are more complicated, because we have thrown away some of $S_y$ and are therefore not allowed to use that part in any new cell that we want to add to the attempted partition.
At this point I think it helps to imagine that we’ve somehow managed to deal with this problem for several iterations. What position would we expect to be in then? Well, we’d have restricted to some subspace (of codimension $rm,$ where r is the number of iterations completed), and we would want to find a big chunk (which would be a union of fliptop subspaces) to remove. Hmm, it doesn’t seem to work, since by this stage I’m needing to find a big orderly chunk of subspaces inside a set over which I have pretty well no control.
40. Ryan O'Donnell Says:
March 13, 2009 at 3:58 pm | Reply
1031. New distribution.
This writeup is practically the same, but I think it sets things up the best. It’s a bit long, but that was mainly just to convince myself what I was saying is correct.
The gist is this:
The basic objects are length-$m$ lists (ordered) of nonempty sets, where the sets’ union is $[n]$. Such lists are in obvious correspondence with the set of “nondegenerate” points in $[m]^n$ (where a point is degenerate if it does not include each character at least once).
The basic operations are: 1. $\Pi$ = randomly permute the list. 2. $C$ = do a random “coagulation”: pick \$i \in [m-1]\$ uniformly, and merge the $m$th set into the $i$th set (shrinking the list length to $m-1$).
The basic distribution is: Start with the length-$n$ list $\langle \{1\}, \{2\}, \dots, \{n\} \rangle$. If we first apply a $\Pi$ operation, and then apply $n - k$ consecutive $C$ operations, we get a length-$k$ list, which corresponds to a point in $[k]^n$. This is our basic distribution on strings. It is (one can show) the “equal-nondegenerate-slices” distribution.
Finally, note that if we only did $n - (k+1)$ of the $C$ operations, we’d get a string $\lambda \in [k+1]^n$, which we think of as $([k] \cup \{\star\})^n$; i.e., a combinatorial line (template). If we were to further apply the $C$ operation to this, we’d get one of the $k$ points on this line, uniformly at random. And, note that $C \lambda$ is a string in $[k]^n$ distributed according to our basic distribution.
—-
I’m quite sure that at the very least, Varnavides and Multidimensional DHJ(k) become “trivial” now. The proof of line-free sets in $[k]^n$ correlating with simple sets was already pretty simple assuming these two tools, but I’m hoping it also gets a paragraph proof now.
Of course, I further hope that this helps simplify the “other half” of the proof, namely the density-increment argument. I’d like to think more about it except I’m going to visit family for a few days and will likely not get the chance.
• gowers Says:
March 14, 2009 at 9:41 am
Ryan, the link to your write-up isn’t working. I’m looking forward to seeing it …
41. gowers Says:
March 13, 2009 at 4:08 pm | Reply
1032. Shelah/wikification
I’ve now put on the wiki a very slight modification of the usual proof of the (colouring) Hales-Jewett theorem, together with Shelah’s proof. The purpose of the exercise is to show just how similar the two proofs really are: one uses HJ(k-1) to get you down to HJ(2), while the other one uses HJ(2), in pretty well exactly the same way, to get you down to HJ(k-1). If I’d been feeling silly enough, I would have interpolated between the two cases and written a general argument for how to use HJ(s) to get you down to HJ(k-s+1), which can quite clearly be done. (Indeed, it’s so clear that it can be done that maybe what’s already there can be thought of as doing it by giving enough examples to make clear what the general case is.)
I bothered to do this partly because I find it historically interesting that it took such a long time for Shelah’s proof to be discovered, and I think that presenting the two proofs side by side in that way makes it really quite mysterious. But my main motive is to try to get people interested in the process of Shelahification of the proof we (probably) have of DHJ(k). If the induction can be turned upside down so easily in the colouring version, I will need a pretty convincing argument to persuade me that it cannot be done for the density version. And if it can be done, it would be really good to do it because the bounds would drop from Ackermann to primitive recursive.
The proofs are on the page about the colouring Hales-Jewett theorem.
• jozsef Says:
March 13, 2009 at 4:34 pm
Isn’t it so that the iterations give you a k-(k-1) fliptop space, then a (k-1)-(k-2) fliptop space in the smaller cube so on? Then if you check any line, it is fliptop for any consecutive pairs, so it is monochromatic. I was wondering if a statistical variant would work or not; Take an almost fliptop subspace (many neighbours have the same colour) and a smaller almost fliptop subspace so on. At the end if at least one line is pairwise fliptop then we are done.
• gowers Says:
March 13, 2009 at 7:43 pm
Jozsef, I’m not sure if this is what you’re asking, but on the wiki I presented Shelah’s argument in terms of flipbottom subspaces instead of fliptop ones.
• jozsef Says:
March 13, 2009 at 8:18 pm
Tim, I just changed the notation there. I should have followed yours. While I think that I understand Shelah’s proof completely, I’m not sure that I see exactly the limits of the technique. For example, it seems to me that we don’t use that the points of every neighboring pair have the same colour. For example if 11111… is blue and 21111… is red in the first fliptop subspace, it won’t make any trouble later as the line defined by the two elements is outside of the next subspace. In general, every point with coordinate 1 has exactly one “semi-neighbour” which is important in the iteration, where the 1-s are replaced by 2-s. But I don’t see an easy way to consider the important pairs only.
• gowers Says:
March 13, 2009 at 8:29 pm
Yes, I had a similar thought soon after writing up the wiki page. I modified the usual proof so as to ensure, unnecessarily, that it reduced HJ(k) to HJ(2). But the usual colour-focusing proof doesn’t need that full strength. I therefore suspect that it is possible to modify the Shelah proof so that instead of producing a fliptop subspace you produce a subspace with some weaker property. But I haven’t tried to work out any details, or even to convince myself that it’s definitely possible.
42. Gower’s Polymath Experiment - Problem probably solved « Healthy Algorithms Says:
March 13, 2009 at 6:31 pm | Reply
[...] combinatorial proof of the density Hales-Jewitt theorem (DHJ). Big congratulations to them, because the problem is solved, probably. Summarizing why he spent his time on this particular problem, Terry Tao wrote: I guess DHJ is [...]
43. gowers Says:
March 13, 2009 at 8:15 pm | Reply
1033. Shelah
I’m still just throwing out guesses here (and seeing whether I get anywhere with them in real time — no luck so far) and here’s another one.
In the argument we have now, we could present it as saying that we try to find a combinatorial line, and if we can’t then we get correlation with a set of complexity k-2. If we are going to try to Shelah-ize, then we’re very much hoping to find a fliptop subspace (or perhaps, as Jozsef suggests above, something slightly weaker but still sufficient) where $\mathcal{A}$ is still dense. So perhaps we should plunge in and try to construct such a subspace, but instead of being completely determined to succeed, we should hope that either we will succeed or we will get correlation with a set of complexity 1.
The first step of what I was trying to do in 1013, following Jozsef in 1011, was as follows. Write ${}[k]^n$ as ${}[k]^r\times[k]^s$, where $s$ is much bigger than $r$. For each $y\in[k]^s,$ define $\mathcal{A}_y$ to be $\{x\in[k]^r:(x,y)\in\mathcal{A}\}.$ We would like to find $y_1$ and $y_2$ that differ only in a few places where coordinates that were k-1 in $y_1$ are k in $y_2$, such that $\mathcal{A}_{y_1}=\mathcal{A}_{y_2}.$ So far so good, but we would also like $\mathcal{A}$ to have density almost as big as the starting density when restricted to ${}[k]^r\times L$, where $L$ is the unique combinatorial line that contains both $y_1$ and $y_2.$
Now if we can’t do this, then we potentially get an interesting set inside which $\mathcal{A}$ has reduced density. It’s formed as follows: for each pair $(y_1,y_2)$ as above, form the line $L$, and take the union $X$ of all those lines.
Ignoring for now the fact that the union is not an arithmetic operation, so we can’t actually conclude without extra work that if $\mathcal{A}$ has reduced density on every ${}[k]^r\times L$ that it has reduced density on ${}[k]\times X,$ let’s see whether $X$ has any good structure, such as low complexity.
A sequence $z\in[k]^s$ belongs to $X$ if and only if we can find some $j$ and some subset $Z$ of the j-set of $x$ such that when we overwrite $Z$ with (k-1)s or ks, we get two points $y_1$ and $y_2$ with $\mathcal{A}_{y_1}=\mathcal{A}_{y_2}.$ Hmm, that doesn’t seem to have low complexity. Back to the drawing board.
• jozsef Says:
March 13, 2009 at 8:47 pm
Just a technical remark; when you cut $[k]^n$ into two parts the you can play with the densities. The density of A is c, say. If any $A_x$ or $A_y$ had density larger than $c+\epsilon$ then we can go there and repeat. That means that the density of x that $A_x$ is at least $c-\epsilon_2$ is almost 1, I think.
44. gowers Says:
March 13, 2009 at 8:48 pm | Reply
1034. Shelah
Time for a very vague thought indeed. Maybe the mistake in our thoughts so far about a density version of Shelah’s proof is that we’re still using colourings too much. Is there some “density equivalent” of a fliptop subspace?
To get a handle on this question, let us try to focus on the property that’s good about fliptop subspaces: that if you ever find a line with all the points up to k-1 of the same colour, then the whole line must be monochromatic. The obvious density version of that would be that if the first k-1 points belong to $\mathcal{A}$ then so does the kth.
One immediate small observation is that for this to be the case, all we need is that if you change some (k-1)s to ks, you don’t go out of $\mathcal{A}$. We don’t mind if this operation takes you in to $\mathcal{A}$. But is there perhaps some much more general circumstance under which we can deduce that $\mathcal{A}$ contains a line from the fact that $\mathcal{A}\cap[k-1]^n$ does? Or perhaps from the fact that $\mathcal{A}\cap[k-1]^n$ contains lots of combinatorial lines? In the latter case, what is the set of points that $\mathcal{A}$ will be forced to avoid?
I have to go now, but at first glance it still doesn’t look as though there are any super-low-complexity sets floating about.
45. Terence Tao Says:
March 14, 2009 at 4:47 am | Reply
1035. Austin’s proof w/o Ramsey theory or density increment
I’ve written up a more or less complete proof of DHJ(3) using a (rather abstract) triangle-removal approach, based on Austin’s proof, at
http://michaelnielsen.org/polymath1/index.php?title=Austin%27s_proof_II
I found that stationarity is not used as much as I had previously thought, and can in fact be obtained by a simple energy increment argument and the pigeonhole principle rather than more high-powered Ramsey-theoretic tools such as Graham-Rothschild.
The argument is a bit tedious though. It follows the approach to triangle removal used for instance in my reworking of the hypergraph removal lemma, in particular, heavily using lots of conditional expectation computations, rather than the traditional language of cleaning out “bad” cells and then counting both “non-trivial” and “trivial” triangles in the surviving cells. But there are some distracting complications due to the multiplicity of scales, and the presence of some additional random choices (basically, one has to choose some random index sets I to flip from 0 to 2, etc). One has to make the whole argument “relative” or “conditioned” to these random choices, which makes the argument look stranger than it normally does.
The bounds seem to be tower-exponential in nature – not so surprising, being based on triangle removal. Amusingly, if one used uniform measure rather than equal-slices measure, one would be forced to make a double tower-exponential, because of the fact that each new scale would have to essentially be the square root of the previous. Hooray for equal slices measure!
I’m pretty sure the density-increment argument is going to end up being shorter and simpler than this one, but I’m putting the triangle removal proof here for completeness.
I’ll be travelling overseas shortly, and so I’ll probably be contributing very little for the next week or two.
46. Polymath « Maxwell’s Demon Says:
March 14, 2009 at 12:20 pm | Reply
[...] even though it touches on my central topics of maths and communication. Now with its preliminary success feels like a good time to do so. [...]
47. gowers Says:
March 14, 2009 at 12:57 pm | Reply
1036. Graham-Rothschild
A quick pair of closely related questions: is the density version of Graham-Rothschild known, and do we now have the tools to prove it (whether or not it is known)?
For the benefit of anyone who can’t be bothered to look it up on the wiki, the Graham-Rothschild theorem is like the Hales-Jewett theorem except that the basic objects are combinatorial subspaces of some fixed dimension rather than points. For instance, a special case is that if you colour the lines in ${}[3]^n$ with finitely many colours, then there is a combinatorial subspace of dimension m such that all the lines in that subspace have the same colour — provided n is big enough in terms of m and the number of colours.
It may be that the question is easy. For instance, suppose we think of a combinatorial line in ${}[k]^n$ as a point in ${}[k+1]^n$ with coordinates that belong to the set $\{1,2,\dots,k,*\}.$ If we then find in our dense set $\mathcal{A}$ of such lines an m-dimensional subspace, what does that translate into? It gives us some wildcard sets $Z_1,\dots,Z_m,$ each of which can be converted into an element of this alphabet, and it also gives us some fixed coordinates. The problem is that the fixed coordinates can be equal to $*$, so we do not end up with all the lines in some combinatorial subspace.
At first this looks problematic: given a dense subset of $\{1,2,\dots,k,*\}^n,$ we can’t hope to find a subspace inside it with no fixed coordinates equal to $*$, since $\mathcal{A}$ might consist of all points with at least one coordinate equal to $*$. However, that’s not a problem, because our entire space consists of points with at least one $*$ (since they are combinatorial lines).
So here’s a DHJ(3) variant. Suppose you have a dense subset $\mathcal{A}$ of ${}[3]^n.$ Must there be an m-dimensional subspace S such that all the fixed coordinates are equal to 1 or 2, and every point in S that includes at least one 3 is an element of $\mathcal{A}$?
• gowers Says:
March 14, 2009 at 1:28 pm
Just realized that the formulation of that last question was a bit careless, since the usual example of a set with roughly equal numbers of all three coordinates is a counterexample to this when m=2. So it’s not instantly obvious what a density version of Graham-Rothschild would even say, but I hope that with the help of equal-slices measure one might be able to formulate a decent statement.
• Randall Says:
March 14, 2009 at 10:16 pm
Generally, these Ramsey theorems are either “projective” (Schur, Hindman, Ramsey), “affine” (van der Waerden, HJ) or “projective/affine” (Graham-Rothschild, Carlson, CST). Nothing with a projective aspect will typically have a “naive” density version, though I don’t know exactly what you are shooting for here. (Presumably you don’t mean something that the set of words having an odd number of 3s would be a counterexample to.)
• jozsef Says:
March 14, 2009 at 10:58 pm
Hi Randall, It is an interesting question if these problems have density versions or not. Let’s consider Schur’s thm. Ben Green proved that if a subset of [n], S, has only a few, $o(n^2)$, solutions to x+y=z in S then one can remove o(n) elements from S to destroy all solutions. This might be viewed as a density version of Schur’s theorem. On the other hand I don’t see a simple way to deduce Schur’s theorem from this density version. I tried it as it might give a bound on the triangle removal lemma, but without success.
48. Ryan O'Donnell Says:
March 14, 2009 at 5:05 pm | Reply
1037. Uniform ordered partition measure.
Here is the corrected link from 1031; in other words, add “.pdf” to the broken link. I think this measure will work well with the Graham-Rothschild problem described above; again, I’ll try to write more in a few days when I get back home.
• Ryan O'Donnell Says:
March 14, 2009 at 5:06 pm
1037.1 Drat, I keep botching it! It’s here.
49. jozsef Says:
March 14, 2009 at 7:17 pm | Reply
1038. Subspace implies line
A simple argument shows that DHJ(k) implies d-dimensional subspaces in $[k]^n$ for any dense subset (with large enough n). I remember seeing a wiki article about this, but I was unable to find it this morning. It is also true that a subspace theorem implies DHJ. That looks fairly trivial but let me explain it; I want to show that the following statement is equivalent to DHJ(k): “There is a constant, $c < 1$ that for every d there is an n that any c-dense subset of $[k]^n$ contains a d-dimensional subspace.”
I would like to show the direction that the statement above implieas DHJ(k). The other direction is already wikified.
As before, write $[k]^n$ as $[k]^r\times[k]^s$, where s is much bigger than r. For each $y\in [k]^s$, define $\mathcal{A}_y$ to be $\{x\in[k]^r:(x,y)\in\mathcal{A}\}$.
Let Y denote the set of $y\in [k]^s$, that $\mathcal{A}_y$ is empty. Suppose that $\mathcal{A}$ is large, line-free, and its density is $\delta =\Delta-\epsilon$ where $\Delta$ is the limit of density of line-free sets and $\epsilon < (1-c)\Delta$. We can also suppose that no $\mathcal{A}_y$ has density much larger than $\Delta$ as that would guarantee a combinatorial line. But then the density of Y is at most 1-c, so there is a c-dense set Z $Z=[k]^s-Y$ such that any element is a tail of some elements of $\mathcal{A}$. For every $y \in Z$ choose an x $x\in [k]^r:(x,y)\in\mathcal{A}$. This x will be the colour of y. It gives a $[k]^r$ colouring on Z. By the initial condition Z contains arbitrary large subspaces, so by HJ(k) we get a line in $\mathcal{A}$.
• jozsef Says:
March 15, 2009 at 5:06 pm
I’ve find Tim’s post about DHJ(k) implies subspace DHJ(k).
http://michaelnielsen.org/polymath1/index.php?title=DHJ(k)_implies_multidimensional_DHJ(k)
The right title of the post above would be “Weak subspace DHJ(k) implies DHJ(k)”.
50. gowers Says:
March 14, 2009 at 11:33 pm | Reply
1039 Graham-Rothschild
I’ve got a new attempt at a density version of Graham-Rothschild that might have a chance of being true. I realized that equal-slices measure wouldn’t rescue the previous version, since the set of lines with wildcard set of size approximately $\alpha n$ is a counterexample.
As I write, I realize that my new attempt fails too. Basically, if you’ve got a 2D subspace then you must have lines with wildcard sets of sizes x, y and x+y, so a density version looks as though it would have to imply a density version of Schur’s theorem, which is of course false. (I realize that I am sort of repeating what people have already said in their replies to 1036.)
Maybe one could go down the route Jozsef implicitly suggests and try proving that if $\mathcal{A}$ is a set that contains few m-dimensional subspaces, then one can remove a small number of elements from $\mathcal{A}$ and end up with no m-dimensional subspaces. I’m not sure that I feel like trying this though …
51. Klas Markström Says:
March 15, 2009 at 10:52 am | Reply
1040.
Since the density of a subset of [k]^n is quite insensitive to the removal of an element, at least for large n, a set with positive density will contain many lines. How many lines can be guaranteed, as a function of the density and n?
From the equal-slices proof of Sperner it is not hard to get a lower bound for DHJ(2) so I’m curious as to what can be said for DJH(3).
52. jozsef Says:
March 15, 2009 at 6:33 pm | Reply
1041. Line-free sets correlate locally with complexity-1 sets
I would like to go back to one important part of the DHJ(3) proof, to analyze it from a slightly different angle. Let us consider our set $\mathcal{A}$, a dense subset of $[3]^n$, as a subset of $({\Bbb Z}/3{\Bbb Z})^n$. Build a graph on $\mathcal{A}$ as follows; The vertex set is $\mathcal{A}$ and two vertices, a and b, are connected iff for c: a+b+c=0 c is also in $\mathcal{A}$. If $\mathcal{A}$ was dense then there are many edges in any typical subspace. Now DHJ is equivalent to the statement that there are two connected elements of $\mathcal{A}$ with the same set of 3-s and that one’s set of 1-s contains the 1-s of the other. This model leads us to a density Sperner problem.
It just turned out that I have to go somewhere right now. I will come back in a few (4+) hours.
• Ryan O'Donnell Says:
March 15, 2009 at 11:38 pm
1042. The distribution.
Hmm. In fact, it may be even better to view strings in $[k]^n$ being generated in the time-opposite way from the one I described. Specifically, equal-slices and its nondegenerate variant are the same as the following Polya-Eggenberger urn process:
[Non-degenerate version:] Start with one of the $k!$ permutations of the string $123\cdots k$. Repeat the following $n-k$ times: pick a random character in $[k]$ with probability proportional to the number of times it appears already in the string. Now insert that character into a random place in the string.
[Equal-slices version, I think:] Same, except: a) start with the empty string; b) for the purposes of proportionality, pretend there is a phantom copy of each of the $k$ characters.
This yet-another viewpoint on the equal-slices distribution helps with making “subspace arguments” (which the uniform distribution was good for): the point is, if you do a $k$-color Polya urn process for $M + n$ steps and $n \gg M$, then the final distribution hardly depends at all on what happened in the first $M$ steps.
Will write more when I get back in two days.
53. Gil Kalai Says:
March 15, 2009 at 11:57 pm | Reply
1042. Density increasing and an analog problem
(This is a little off topic) Let me mention a problem which I thought of as analogous to Roth/cap set where the gaps between lower and upper bounds are similarly shocking and the current density increasing arguments cannot help; (It is related to old work of mine with Kahn and Linial and also to some more recent work with Shelah that we did not publish.)
you have a subset A of $\{0,1\}^n$ of density c and you would like to find a combinatorial subcube F of dimension 0.9n so that the projection of A to F is of large density say 0.99. In other words, we want to find a set of 0.9n coordinates so that for a fraction 0.99 of the vectors supported on this set we can find a continuation on the other coordinates that is in A. (We usually talk here about restriction to a subcube/subspace and not about projections. But traditionally sections and projections are not unrelated.)
By a density increasing argument doing it one coordinate at a time it was known from the late 80s that this can be achieved if c is $n^{-\alpha}$ for $\alpha=1/5$ or so. A conjecture by Benny Chor asserts that $c=0.999^n$ is good enough!
I think it is a little analogous to Roth (or cap set)
a few points:
1) The proof is also by a slow density increasing argument (you reduce the dimension by one every time) and there are examples the such an argument cannot be improved.
2) There are some conjectures (by Friedgut and others) which may explain why we can get the density down to $n^{-\beta}$ for every $\beta$ maybe even $\beta=\gamma \log n$ but no plans beyond it.
3) There are alarming examples by Ajtai and Linial of Boolean functions descibed by certain random depth 3 circuits (that Ryan already mentioned) that may (or a varian of) give a counter example. It is complicated to check it.
I admit that the analogy with density increasing argument for Roth-type problems is not very strong: this problem is about projection to subcubes and there it is about restrictions to subspaces or similar creatures; But there may be some connecion.
In particular I would try subsets described by low depth small size circuits (with operations over {0,1,2}) as candidates for counter examples for the most ambitious conjectures regarding Roth and cap sets.
(On the positive side maybe more sophisticated density increasing arguments of the kind we talk about here can be used in this problem.)
54. Kristal Cantwell Says:
March 16, 2009 at 3:12 am | Reply
1043. Graham-Rothschild
From 1036: ” A quick pair of closely related questions: is the density version of Graham-Rothschild known, and do we now have the tools to prove it (whether or not it is known)?
For the benefit of anyone who can’t be bothered to look it up on the wiki, the Graham-Rothschild theorem is like the Hales-Jewett theorem except that the basic objects are combinatorial subspaces of some fixed dimension rather than points. For instance, a special case is that if you colour the lines in with finitely many colours, then there is a combinatorial subspace of dimension m such that all the lines in that subspace have the same colour — provided n is big enough in terms of m and the number of colours. ”
I doubt there is density version of the Graham-Rothschild theorem. If one fixes a coordinate and deletes all lines that has a constant coordinate at that point then that will only lower the number of lines by a constant factor but it will prevent the formation of any two dimensional space with all its lines monochromatic as in that case the intersection of the moving coordinates of all of the lines is the null set.
55. Kristal Cantwell Says:
March 16, 2009 at 3:19 am | Reply
1044. Density theorems and Ramsey theorems
Density theorems are stronger than Ramsey theorems. Any density theorem can be converted to a Ramsey theorem as follows. One sets the density less than 1/r and gets a large enough configuration so that the if the density is greater than 1/r there will be the desired result than for any r coloring there will be one color with density 1/r or more and we will have a monochromatic configuration in that color.
56. Kristal Cantwell Says:
March 16, 2009 at 5:49 am | Reply
1045. Density Schur Theorem
I have just remembered the counterexample for a density version of Schur’s theorem. One just takes the odd numbers, they have density 1/2 but since the sum of two odd numbers is even there are no triples a,b,c in the set such that a+b=c.
• gowers Says:
March 16, 2009 at 7:54 am
The fact that the density version of Schur’s theorem is false implies that density Graham-Rothschild is false — see 1038.
57. gowers Says:
March 16, 2009 at 11:36 am | Reply
1046. Bad news and good news.
Bad news: our proof of DHJ(3) is wrong!
Good news: it isn’t too hard to fix.
But it’s still quite amusing that nobody noticed that Substep 2.2 of the write-up on the wiki of the crucial lemma that a dense 1-set can be almost completely partitioned into subspaces was nonsense as presented. I’ve left the old version there as a cautionary tale. Fortunately, the reason for the mistake was that I had translated the corresponding part of Terry’s argument in a sloppy way — the argument itself was sound. But it did give me a bit of a scare …
• Randall Says:
March 16, 2009 at 6:28 pm
1046.1 Good news and polymath….
Well, perhaps it will be somewhat reassuring that the changes you have made seem to be only adding more detail, and that, from the ergodic perspective at least, it was natural to omit said detail; loosely translated (and unless I misunderstand), what you seem not to be checking more carefully is that when you remove the fibers (over some factor) that recur under a subspace, what you have left is still measurable with respect to that subspace. So perhaps the good news is actually that polymath is like math with a net. (In which case the bad news is the danger than everyone involved will internalize that piece of good news to the point where it becomes a problem.)
• Randall Says:
March 16, 2009 at 6:32 pm
I of course meant (a) “now” to be checking more carefully, and (b) measureable with respect to that “factor”.
58. gowers Says:
March 16, 2009 at 1:09 pm | Reply
1047. Wikification
I have now wikified an abstract version of the iterative argument that was wrong before. From the abstract point of view the property that I was forgetting about was the all-important sigma-algebra property. This version should be fine now and is intended to be sufficiently general to deal with DHJ(k) as well. The argument could be sharpened up a bit towards the end — I ran out of energy.
59. gowers Says:
March 16, 2009 at 1:17 pm | Reply
Metacomment. I’ve created a thread for comments 1050-1099, so this one should draw to a close.
60. Michael Nielsen » The Polymath project: scope of participation Says:
March 20, 2009 at 9:29 pm | Reply
[...] week, Gowers announced that the problem was “probably solved”. In fact, if the results hold up, the project [...]
61. Concluding notes on the polymath project — and a challenge « What Is Research? Says:
March 24, 2009 at 5:27 pm | Reply
[...] project, as of February 20, 2009. The project, which began around February 2, 2009, has now been declared successful by Gowers. While the original aim was to find a new proof of a special case of an already proved theorem, the [...]
62. An Open Discussion and Polls: Around Roth’s Theorem « Combinatorics and more Says:
March 25, 2009 at 9:13 am | Reply
[...] (as a little spin off to the polymath1 project, (look especially here and here)) if you have any thoughts about where the truth is for these problems, or about how to [...]
63. A gentle introduction to the Polymath project « The Number Warrior Says:
March 26, 2009 at 1:41 am | Reply
[...] to an online collaborative approach, then kicked things off with a blog post. Six weeks later, the main problem he proposed was declared (essentially) solved. However, the project still continues apace, especially at threads at Terry Tao’s [...]
64. Un blog concentra a cientos de matemáticos en la demostración conjunta de un teorema « Francis (th)E mule Science’s News Says:
April 28, 2009 at 9:27 pm | Reply
[...] (probably),” Nature Physics, 5: 237, April 2009 . Hay una entrada en el blog de Tim con el mismo título. Más allá de la demostración la experiencia muestra a los jóvenes que se inician en la [...]
65. sunee Says:
July 2, 2009 at 10:17 am | Reply
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66. More polymath projects « Algorithmic Game Theory Says:
July 23, 2009 at 9:47 am | Reply
[...] polymath projects After the success of the first polymath project (see also here and here as well as the project wiki), there are now [...]
67. Michael Nielsen » Introduction to the Polymath Project and “Density Hales-Jewett and Moser Numbers” Says:
May 1, 2010 at 7:45 pm | Reply
[...] subprojects of the Polymath Project progressed quickly. On March 10, Gowers announced that he was confident that the polymaths had found a new combinatorial proof of DHJ. Just 37 days [...]
68. Why do we still publish research (via) papers?Research cycle research Says:
September 20, 2010 at 2:18 am | Reply
[...] papers, but platforms like OpenWetWare clearly show that this is doable, and initiatives like the Polymath projects demonstrate that it may even be beneficial to the topic — an aspect that is especially [...]
69. Open Science | Cosmic Variance | Discover Magazine Says:
April 8, 2011 at 8:57 pm | Reply
[...] to the Hales-Jewett theorem — and, several hundred comments later, announced that they had succeeded. Here’s the paper. Buoyed by this success, the group has set up a Polymath Wiki to expedite [...]
70. Drafting proposals in the open – a practical test | Research cycle research Says:
April 26, 2011 at 3:41 pm | Reply
[...] to go because, to paraphrase Michael Nielsen’s account of Tim Gowers’ synopsis of the initial Polymath project, doing science in the open is to traditional science like driving is to pushing a [...]
71. Drafting proposals in the open – a practical test Says:
May 19, 2011 at 4:24 pm | Reply
[...] way to go because, to paraphrase Michael Nielsen’s account of Tim Gowers’ synopsis of the initial Polymath project, doing science in the open is to traditional science like driving is to pushing a [...]
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http://mathhelpforum.com/calculus/60907-question-about-power-series-calc-2-project.html | # Thread:
1. ## Question about a power series for a Calc 2 Project
Greetings all, I'm hoping someone can point me in the right direction for this question. I'd rather not have the answer as I'd like to figure it out myself, however it might not be something that can be answered without giving it away. Well, you'll see what I mean
I have a power series:
This is basically a representation of the number of rabbits born in month n (based off of the Fibonacci numbers). The question asks me to examine this series when n is very large, and it wants to know how many rabbits are born in month n compared to month n-1.
I'm stuck here, I'm not really sure where to start or how to treat this. If anyone could provide some guidance that would be great.
2. Originally Posted by Wintaru
Greetings all, I'm hoping someone can point me in the right direction for this question. I'd rather not have the answer as I'd like to figure it out myself, however it might not be something that can be answered without giving it away. Well, you'll see what I mean
I have a power series:
This is basically a representation of the number of rabbits born in month n (based off of the Fibonacci numbers). The question asks me to examine this series when n is very large, and it wants to know how many rabbits are born in month n compared to month n-1.
I'm stuck here, I'm not really sure where to start or how to treat this. If anyone could provide some guidance that would be great.
Are you asking to sum the series?
3. Originally Posted by Wintaru
Greetings all, I'm hoping someone can point me in the right direction for this question. I'd rather not have the answer as I'd like to figure it out myself, however it might not be something that can be answered without giving it away. Well, you'll see what I mean
I have a power series:
This is basically a representation of the number of rabbits born in month n (based off of the Fibonacci numbers). The question asks me to examine this series when n is very large, and it wants to know how many rabbits are born in month n compared to month n-1.
I'm stuck here, I'm not really sure where to start or how to treat this. If anyone could provide some guidance that would be great.
Im not sure if I follow your question, the infinite power series representation is indexed over n, am I allowed to assume that n is the month 'n' ? and if so does your series represent the number of rabbits as t the time approaches infinity??
Cheers,
David
4. The month is indeed n. Maybe I miscalculated this power series. This is my work to get the equation above.
I was assuming I did this correctly. This equation is supposed to represent:
The question specifically asks us to look at a_n, when n is very large, about how many more rabbits are born in month n compared to month n-1.
Does that help at all to clarify?
5. n is only a summation index and each month will have the same sum. I believed you are confused somewhere.
6. Originally Posted by whipflip15
n is only a summation index and each month will have the same sum. I believed you are confused somewhere.
Right that n is the summation index. So your sum will give the total number of rabbits present in the population at month n. Therefore, each month will not have the same sum..right? I think...I'm rusty on power series.
7. Originally Posted by whipflip15
n is only a summation index and each month will have the same sum. I believed you are confused somewhere.
I'm afraid I don't follow. If n = 2, a_n = 3, if n = 3, a_n = 5, etc. Right?
8. To clear up all this confusion, please give us the original statement of the question.
9. Originally Posted by Wintaru
The month is indeed n. Maybe I miscalculated this power series. This is my work to get the equation above.
I was assuming I did this correctly. This equation is supposed to represent:
The question specifically asks us to look at a_n, when n is very large, about how many more rabbits are born in month n compared to month n-1.
Does that help at all to clarify?
hey mate, I noticed in your expansion of ((1/3)x)^n you have let that equal
x^n/3 which should be x^n/3^n, this may not be the area of concern but something that should be taken into consideration,
Regards,
David
ps - let me know if you need any further help
10. Originally Posted by Wintaru
I have a power series:
This is basically a representation of the number of rabbits born in month n (based off of the Fibonacci numbers).
This is what you said but what you actually mean is that $a_n$ is the number of rabbits born in month n?
11. Originally Posted by o_O
To clear up all this confusion, please give us the original statement of the question.
I'll try, there are 6 questions and this is the last. Questions 1-5 build upon each other and I was trying to avoid putting too much in the post. This is a very summarized version of the project as a whole:
Edit: By the way, I really appreciate the help everyone.
12. Originally Posted by whipflip15
This is what you said but what you actually mean is that $a_n$ is the number of rabbits born in month n?
Then if your series is correct you want to look at
$\begin{array}{c}lim\\n \rightarrow \infty \end{array}\frac{a_{n+1}}{a_n}$
13. Originally Posted by whipflip15
Then if your series is correct you want to look at
$\begin{array}{c}lim\\n \rightarrow \infty \end{array}\frac{a_{n+1}}{a_n}$
Thanks!
I actually was started in that direction, and I got this far:
Again like you said, if my series is correct (a very big IF), then this is a limit of infinity / infinity, do I then use L'Hopital's rule?
14. Ok so I went ahead and tried L'hop, this is what I came up with:
What exactly have I found here? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.974246621131897, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-math-topics/31421-complex-analysis.html | # Thread:
1. ## Complex Analysis
Determine, with motivation, all funtions that is analytic on
{z/ |z| < 4}, with f( 0 ) = i and |f( z ) <= 1 on {z/ |z| < 4}
2. Originally Posted by JacquesRoux
Determine, with motivation, all funtions that is analytic on
{z/ |z| < 4}, with f( 0 ) = i and |f( z ) <= 1 on {z/ |z| < 4}
It seems your problem is faulty somehow. Because the way it is stated $f(z) = i$ is the only such mapping. Because if $f$ is non-constant then by the open mapping theorem $f(0)$ has to be a mapped into an interior point of the range of $f$ but that is not possible since $f(0)=i$ is a boundary point. Thus, we conclude that constant functions are the only such functions, thus, $f(z) = i$ for all $|z|<4$.
3. Thanks for your help!
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http://quant.stackexchange.com/questions/1452/obtaining-characteristics-of-stochastic-model-solution/1453 | # Obtaining characteristics of stochastic model solution
I want to use the following stochastic model
$$\frac{\mathrm{d}S_{t}}{ S_{t}} = k(\theta - \ln S_{t}) \mathrm{d}t + \sigma\mathrm{d}W_{t}\quad (1)$$
using the change in variable $Z_t=ln(S_t)$
we obtain the following SDE
$$\mathrm{d}Z_{t} = k(\theta - \frac{\sigma^2}{2k} - Z_{t}) \mathrm{d}t + \sigma\mathrm{d}W_{t}\quad (2)$$
again we change variable $X_t=e^{kt}Z_t$
we obtain the following SDE
$$\mathrm{d}X_t = k(\theta -\frac{\sigma^2}{2k})e^{kt}\mathrm{d}t + \sigma e^{kt}\mathrm{d}W_{t} \quad (3)$$
This last equation can be integrated easily and we see that
$$(X_{t+1}-X_t)\sim N(\mu, \sigma)$$
My question is: How can I get the ditribution of $\frac{S_{t+1}-S_{t}}{ S_{t}}$ ? (at least its expectation and standard deviation under filtration $F_t$)
If we come back to the first equation it seams that if $\mathrm{d}t$ is small enough then it is a normal distribution with mean $k(\theta - \ln S_{t}) \mathrm{d}t$ and sd $\sigma\mathrm{d}t$, but I want to find mathematically if it is true, and if yes, under what assumptions.
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It seems to me that you found the incorrect SDE for $X_t$, could you please provide your calculations? – Ilya Jul 15 '11 at 8:11
@Gortaur: indeed, there was a $e^{kt}$ missing in the stochastic part – RockScience Jul 15 '11 at 8:24
I think you missed the more important term $\frac{1}{S^2}$, see my answer. Could you refer me to Ito formula you are using? – Ilya Jul 15 '11 at 8:33
can you put some details on your filtration $F_t$? – Beer4All Jul 15 '11 at 9:45
$F_t$ is the natural filtration $F_t=\{S_i\}_{i=0..t}$ – RockScience Jul 18 '11 at 4:54
## 2 Answers
I've edited my answer since Berr4All showed that your equation is right.
What you still can do - is to use Fokker-Planck equation to derive a density.
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Why not considering the empirical way (MC) to get the pdf? – Beer4All Jul 15 '11 at 9:48
@Beer4All: what kind of empirical way do you advise? What is pdf by the way? – Ilya Jul 15 '11 at 9:56
@Beer4All: I see. I am not so happy with randomized methods like Monte-Carlo, so I better use numerical solutions of PDEs if it is possible. – Ilya Jul 15 '11 at 10:18
pdf = probability density function. To get numerically an empiricall estimate of your density function: 1 -- draft sample of $X_t$, 2 -- trace the empirical cdf (cumulative distribution function) of this sample 3 -- then compute (numerically) the differentiated function which is nothing but the empirical density ;) – Beer4All Jul 15 '11 at 10:18
For me MC methods are often the fastest way to get rapid & painless solutions, useful to test your results. – Beer4All Jul 15 '11 at 10:25
show 5 more comments
• 1 ) A first-order Taylor expansion gives $\ln \left(\frac{S_{t+\Delta_t}}{S_t}\right)\approx \frac{S_{t+\Delta_t}-S_{t}}{S_t}+o(\Delta_t)$ , thus unless $\Delta_t$ is not small you can drop the residual term and consider $Z_t\overset{law}{=}\frac{S_{t+\Delta_t}-S_{t}}{S_t}$.
• 2 ) Calculation of the moments: we can proceed by using the classical Dynkin way
we remind that $dZ_t=k(\theta-\frac{\sigma^2}{2k}-Z_t)dt+\sigma dW_t$, so
$d(Z_t^p)\overset{Itô}{=}p\cdot(Z_t)^{p-1}dZ_t+\frac{\sigma^2}{2}dt = \left(\frac{\sigma^2}{2}+ Z_t^{p-1}\cdot{pk(\theta-\frac{\sigma^2}{2k})-Z_t^p\cdot pk} \right)dt+p\sigma Z_t^{p-1}dW_t$
taking the expectation,
$E\left( Z_t^p \right)=E\left( Z_0^p \right)+\int_0^t\left(\frac{\sigma^2}{2}+ E(Z_s^{p-1})\cdot{pk(\theta-\frac{\sigma^2}{2k})-E(Z_s^p)\cdot pk} \right)ds$
finally:
$E\left( Z_t^p \right)=E\left( Z_0^p \right)+\frac{\sigma^2}{2}t+pk(\theta-\frac{\sigma^2}{2k}) \int_0^t E(Z_s^{p-1})ds-pk \int_0^tE(Z_s^p)ds$
so, your p-th moment is the solution of the above deterministic ODE which can be solved step-by-step starting $p=1$.
• 3 ) Exact distribution of the yield: in general it's far to be simple to get exact distribution/ simulation for diffusions (and more for multidimensional SDE's -- example: Heston's joint SDE has an exact pdf which computation requires Malliavin calculus).
So, in my opinion unless you are considering simulations with large time step it would be useless to deal with the exact distribution of your SDE.
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would you like to check out if OP found the right equation for $Z_t$? – Ilya Jul 16 '11 at 19:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9016155004501343, "perplexity_flag": "middle"} |
http://conservapedia.com/Pime_number_theorem | # Prime Number Theorem
### From Conservapedia
(Redirected from Pime number theorem)
The Prime Number Theorem is one of the most famous theorems in mathematics. It states that the number of primes not exceeding n is asymptotic to $\frac{n}{\log(n)}$, where log(n) is the logarithm of (n) to the base e.
$\pi(n)\sim\frac{n}{\ln n}$.
The number of primes not exceeding n is commonly written as π(n), the prime counting function. An asymptotic relationship between a(n) and b(n) is commonly designated as a(n)~b(n). (This does not mean that a(n)-b(n) is small as n increases. It means the ratio of a(n) to b(n) approaches one as n increases.)
The Prime Number Theorem thus states that π(n)~n / log(n) . That is, as n tends to infinity, the relative error between π(n) and n / log(n) tends to zero. This can be expressed using limit notation as
$\lim_{n\to\infty}\frac{\pi(n)}{n/\ln(n)}=1$
In other words, the limit (as n approaches infinity) of the ratio of pi(n) to n/log(n) is one. Put a third way, n / log(n) is a good approximation for π(n).
## Equivalent Statements
Carl Friedrich Gauss conjectured the equivalent statement that π(x) was asymptotic to Li(x) defined as:
$\mbox{Li}(x) = \int_2^x \frac{dt}{\ln t}$.
In fact, for large x this turns out to be a better approximation than π(x). The size of the error Li(x) − π(x) is closely related to the behavior of the Riemann Zeta function
## History of the Theorem
The theorem was first conjectured by Legendre and Gauss (independently) circa 1796. In 1848 and 1850, Chebyshev made significant progress towards proving the theorem using the Riemann zeta function and other non-elementary techniques. His non-elementary proof was completed in 1896 by Hadamard and Charles de la Vallee-Poussin.
The grand culmination of these efforts occurred in 1949 and 1950 when Paul Erdos and Atle Selberg presented an elementary proof of the theorem. This proof earned Selberg the highest prize in math, the Fields Medal. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9170090556144714, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/179495/how-i-can-find-an-optimal-solution-for-a-model-with-concave-convex-objective-fun?answertab=votes | How I can find an optimal solution for a model with concave-convex objective function?
My objective function is sum of three functions, 2 linear functions and a concave function
($1-\exp(x)$); constraints of my model are convex. How can I obtain optimal solution from this problem?
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1 Answer
Solve $\Delta f_{vex}(x^{n+1})=-\Delta g_{cave}(x^n)$ where $f_{vex}(.)$ is the convex part, being the sum of the two linear functions and $g_{cave}$ is the concave part and $x^{n}$ denotes the $n^{th}$iterate of the solution. This is also known as the concave-convex procedure.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9166330695152283, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/10879?sort=oldest | ## Intuition for Group Cohomology
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I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how these are used in the proofs of class field theory.
So far, all I can tell is that the cohomology groups are given by some ad hoc modding out process, then we derive some random properties (like the long exact sequence), and then we compute things like $H^2(\mathrm{Gal}(L/K),I_{L})$, where $I_L$ denotes the group of fractional ideals of a number field $L$, and it just happens to be something interesting for the study of class field theory such as $I_K/\mathrm{N}(I_L)$, where $L/K$ is cyclic and $\mathrm{N}$ denotes the ideal norm. We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups.
What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel modulo the image is interesting), but I don't know why we take these particular kernels modulo these particular images. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture.
For context, if $A$ is a $G$-module where $G$ is cyclic and $\sigma$ is a generator of $G$, then we define the endomorphisms $D=1+\sigma+\sigma^2+\cdots+\sigma^{|G|-1}$ and $N=1-\sigma$ of $A$, and then $H^0(G,A)=\mathrm{ker}(N)/\mathrm{im}(D)$ and $H^1(G,A)=\mathrm{ker}(D)/\mathrm{im}(N)$. Note that this is a slight modification of group cohomology, i.e. Tate cohomology, which the cohomology theory primarily used for Class Field Theory. Group cohomology is the same but with $H^0(G,A) = \mathrm{ker}(N)$. The advantage of Tate cohomology is that it is $2$-periodic for $G$ cyclic.
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1
Some of the answers in this related question deal with group cohomology: mathoverflow.net/questions/640/… – Qiaochu Yuan Jan 6 2010 at 4:30
6
I should point out that this is not ordinary group cohomology, but Tate cohomology, which has some minor alterations to make it 2-periodic. – S. Carnahan♦ Jan 6 2010 at 6:06
1
A similar question has been discussed at mathoverflow.net/questions/8599/… – Leonid Positselski Jan 6 2010 at 12:53
The best treatment of group cohomology which I have seen is the the book "lectures on algebraic geometry 1" by Gunter Harder. Check it out. – Steven Gubkin Aug 31 2010 at 17:12
@Steven: I agree that that book is quite nice, though as far as I can tell, it doesn't seem to give particularly nice intuition for group cohomology. – David Corwin Sep 5 2010 at 17:20
show 1 more comment
## 9 Answers
In my opinion, the best you can do is to see group cohomology in the context of homological algebra. For example, the book by Hilton and Stammbach on homological algebra should be a good introduction.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I'm not sure if this is what you're looking for, but I always think of group (co)homology in terms of the homology of the classifying space for your group. Assuming $G$ is discrete, then there is a topological space $BG$ with the property that $\pi_1 BG=G$ and the higher homotopy groups vanish. By construction, $BG$ has a contractible cover $EG$ so that $EG/G=BG$.
$H_n(BG)$ is the same as the algebraically defined $H_n(G)$ since, if we take the cellular chain complex of $EG$ we end up with a resolution of the integers by $G$-modules because of the action of $G$ on $EG$ passes to the chain groups. Then tensoring by the integral group ring of $G$ just divides out the $G$ action and we get the cellular chain complex of $BG$.
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What is the difference between BG and the Eilenberg-Maclane space K(G,1) mentioned in Allen Hatcher's book? – Anweshi Jan 6 2010 at 12:15
They are the same thing. – Andy Putman Jan 6 2010 at 16:10
5
They are the same thing as long as $G$ has the discrete topology. If $G$ is a topological group, etc, $BG$ won't necessarily be $K(G,1)$. $G \to EG \to BG$ is a fibration so $\pi_n BG = \pi_{n-1} G$. – Josh Roberts Jan 6 2010 at 18:13
1
The standard construction is actually in Hatcher section on $K(G,1)$'s on page 89. A more "hands on" construction exists if you have a finite presentation of your group. The only reference I know of is in an appendix in Knudson's "Homology of Linear Groups". I don't have the book handy and google books doesn't offer a preview of it so I can't give a more specific location, sorry. Basically, you construction a reduced CW complex dimension by dimension killing off higher homotopy as you go. – Josh Roberts Jan 7 2010 at 0:35
1
A very beautiful construction of EG, BG, etc is in Paolo Salvatore's paper: front.math.ucdavis.edu/9907.5073 A pleasant feature of his construction is that it works for any topological monoid, and the "geometric picture" of an element of EG, BG, is very explicit. – Ryan Budney Jan 7 2010 at 6:45
show 3 more comments
The online notes of J.S. Milne on Class Field Theory contain a chapter on group cohomology including Tate cohomology that is easily accessible (and exactly what you need to understand class field theory).
For algebraic intuition the keyword is derived functors. Group cohomology (and homology) are examples of derived functors, which can be considered as a reason why the definitions as are they are, and why they are interesting/useful.
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Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology.
Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176).
If moreover $G$ is cyclic with generator $s$, this implies that an element of $K$ has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$.
Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$ the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff $x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$ .
So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle $x^2+y^2=1$ $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.
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3
Oh, I agree that cohomologists must have known this for ages. But Davidac897's question proves that it might be helpful to have elementary examples explicitly written down to show at an early stage that group cohomology can have applications even before the whole somewhat intimidating machinery is developed. – Georges Elencwajg Jan 6 2010 at 12:25
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I don't like this folklore application because before having done the general proof of $H^1(G,K^*)=0$, where you have to come up with special characters whose linear independancy is used, you have already parameterized the rational circle many times; besides, there are nice geometric approaches to it. – Martin Brandenburg Jan 6 2010 at 16:28
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@Martin: I don't understand your statement about chronology. It is quite possible to prove Theorem 90 without using a parametrization of the rational circle, let alone many of them. Also, I don't see why the existence of a nice geometric proof means one should stop looking for other arguments and points of view. – S. Carnahan♦ Jan 6 2010 at 16:50
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@Elencwajg:This is a very nice observation! I will definitely use it as motivation in my teaching! – SGP Jun 8 2011 at 2:01
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This example was used by Emil Artin in his Hamburg lectures on algebra in I think 1961. – Franz Lemmermeyer Nov 14 at 5:32
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First, let me say beyond knowing that group cohomology comes from derived functors and all the properties that come with it, I don't have much intuition for general group cohomology. However, in number theory there are a number of places in which you can realize the cohomology groups as parametrizing some other objects you are interested in. To understand these other objects (which you could feasibly have very real intuition about) it then suffices to use the machinery of group cohomology. Below are a few examples. I forget to put continuous subscripts on everything, so beware. Mistakes are mine.
First, in the realm of (edit:local) class field theory we have the Brauer group $Br(K)$ which is the abelian group (under the tensor operation) of central simple algebras over $K$ up to the equivalence $A \sim M_n(A)$. It turns out that there is a natural isomorphism between $Br(K)$ and $H^2(G_K, \overline{K}^{\times})$. This isomorphism provides two worlds in which to make important calculations. For instance, the statement that every central simple algebra over $K$ splits over an unramified extension of $K$ may be proved explicitly using the Brauer group or it may be proven by checking that $H^2(G(K^{nr}/K),\overline{K}^\times) = H^2(G_K,\overline{K}^\times)$. From either proof you are able to obtain a proof of the other. Perhaps, also in the realm of class field theory, the local Artin map arises as a map on Tate groups given by a certain cup product but that would take a bit longer to explain. You should look at Milne's notes on CFT for all of this (Ch III and Ch IV for what I have said).
Here is another. Suppose that $B$ is a topological ring and $G$ is a topological group and $G$ acts continuously on $B$. Then, we consider finite free $B$-modules $X$ equipped with a semi-linear action, i.e. $g(bx) = g(b)g(x)$ for all $b \in B$ and $x \in X$. It turns out that all such objects are parametrized by $H^1(G,GL_d(B))$. (Warning: this is non-abelian cohomology, so this is only a pointed set. The point corresponds to the trivial semi-linear representation $B^d$ with the diagonal action.) This comes up very early in the part of $p$-adic Galois representations where one studies the period rings $B_{dR}, B_{HT}$, etc.
Finally, consider the situation where one has a representation $\overline{\rho}:G_{\mathbb Q,S} \rightarrow GL_n(\mathbb F_p)$ and one wants to know whether it has litings $\rho: G_{\mathbb Q,S} \rightarrow GL_n(R)$ where $R$ is some complete DVR with residue field $\mathbb F_p$. If we already have a lifting to $GL_n(R')$ and $R$ and $R'$ are nice enough (there is a surjection $R \rightarrow R'$ whose kernel $I$ is killed by the maximal ideal in $R$) then the obstruction to lifting further lies in the cohomology group $H^2(G_{\mathbb Q,S},I\otimes Ad(\overline{\rho}))$ where $Ad(\overline{\rho})$ is the vectorspace $M_n(\mathbb F_p)$ together with conjugate action by $\overline{\rho}$. This (I've specialized a few things) is written down in Mazur's paper "Deforming Galois Representations".
To finish, I will just say what my adviser told me when I asked him a similar question as you are asking: "Just wait and you will see how much clearer your thought can become with group cohomology."
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I feel like my answer to every question of this kind is "the treatment in Silverman's book on elliptic curves is really nice," but the treatment in Silverman's book on elliptic curves is really nice!
In particular, for H^1 at least I always find it quite nice to think in terms of twists. Quite generally: if X is a variety over a field K, and L/K is a Galois extension, we say that X'/K is a L-twist of X if there exists an isomorphism between X and X' over L. (If we just mean X' is an L-twist for some L, we just call it a twist of X.)
Anyway, it is a nice exercise to check that L-twists of X yield classes in H^1(Gal(L/K),Aut(X/L)). (And in favorable circumstances, the L-twists are actually in bijection with the Galois cohomology set.) This is the basis of the whole story of principal homogeneous spaces, or torsors, which makes up much of the last chapter of Silverman.
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See also this Math.SE post I wrote for some more motivation: http://math.stackexchange.com/a/270266/873. Recall that `$\mathrm H^*(G,M)=\mathrm{Ext}^*(\mathbb{Z},M)$`.
After learning some more math, I've come across the following example of a use of group cohomology which sheds some light on its geometric meaning. (If you want to see a somewhat more concrete explanation of how group cohomology naturally arises, skip the next paragraph.)
We define an elliptic curve to be $E=\mathbb{C}/L$ for a two-dimensional lattice $L$. Note that the first homology group of this elliptic curve is isomorphic to $L$ precisely because it is a quotient of the universal cover $\mathbb{C}$ by $L$. A theta function is a section of a line bundle on an elliptic curve. Since any line bundle can be lifted to $\mathbb{C}$, the universal cover, and any line bundle over a contractible space is trivial, the line bundle is a quotient of the trivial line bundle over $\mathbb{C}$. We can define a function $j(\omega,z):L \times \mathbb{C} \to \mathbb{C} \setminus {0}$. Then we identify $(z,w) \in \mathbb{C}^2$ (i.e. the line bundle over $\mathbb{C}$) with $(z+\omega,j(\omega,z)w)$. For this equivalence relation to give a well-defined bundle over $\mathbb{C}/L$, we need the following: Suppose $\omega_1,\omega_2 \in L$. Then $(z,w)$ is identified with $(z+\omega_1+\omega_2,j(\omega_1+\omega_2,z)w$. But $(z,w)$ is identified with $(z+\omega_1,j(\omega_1,z)w)$, which is identified with $(z+\omega_1+\omega_2,j(\omega_2,z+\omega_1)j(\omega_1,z)w)$. In other words, this forces $j(\omega_1+\omega_2,z) = j(\omega_2,z+\omega_1)j(\omega_1,z)$. This means that, if we view $j$ as a function from $L$ to the set of non-vanishing holomorphic functions $\mathbb{C} \to \mathbb{C}$, with (right) L-action on this set defined by $(\omega f)(z) \mapsto f(z+\omega)$, then $j$ is in fact a $1$-cocyle in the language of group cohomology. Thus $H^1(L,\mathcal{O}(\mathbb{C}))$, where $\mathcal{O}(\mathbb{C})$ denotes the (additive) $L$-module of holomorphic functions on $\mathbb{C}$, classifies line bundles over $\mathbb{C}/L$. What's more is that this set is also classified by the sheaf cohomology $H^1(E,\mathcal{O}(E)^{\times})$ (where $\mathcal{O}(E)$ is the sheaf of holomorphic functions on $E$, and the $\times$ indicates the group of units of the ring of holomorphic functions). That is, we can compute the sheaf cohomology of a space by considering the group cohomology of the action of the homology group on the universal cover! In addition, the $0$th group cohomology (this time of the meromorphic functions, not just the holomorphic ones) is the invariant elements under $L$, i.e. the elliptic functions, and similarly the $0$th sheaf cohomology is the global sections, again the elliptic functions.
More concretely, a theta function is a meromorphic function such that $\theta(z+\omega)=j(\omega,z)\theta(z)$ for all $z \in \mathbb{C}$, $\omega \in L$. (It is easy to see that $\theta$ then gives a well-defined section of the line bundle on $E$ given by $j(\omega,z)$ described above.) Then, note that $\theta(z+\omega_1+\omega_1)=j(\omega_1+\omega_2,z)\theta(z) = j(\omega_2,z+\omega_1)j(\omega_1,z) \theta(z)$, meaning that $j$ must satisfy the cocycle condition! More generally, if $X$ is a contractible Riemann surface, and $\Gamma$ is a group which acts on $X$ under sufficiently nice conditions, consider meromorphic functions $f$ on $X$ such that $f(\gamma z)=j(\gamma,z)f(z)$ for $z \in X$, $\gamma \in \Gamma$, where $j: \Gamma \times X \to \mathbb{C}$ is holomorphic for fixed $\gamma$. Then one can similarly check that for $f$ to be well-defined, $j$ must be a $1$-cocyle in $H^1(\Gamma,\mathcal{O}(X)^\times)$! (I.e. with $\Gamma$ acting by precomposition on $\mathcal{O}(X)^\times$, the group of units of the ring of holomorphic functions on $X$.) Thus the cocycle condition arises from a very simple and natural definition (that of a function which transforms according to a function $j$ under the action of a group). A basic example is a modular form such as $G_{2k}(z)$, which satisfies $G_{2k}(\gamma z) = (cz+d)^{2k} G_{2k}(z)$, where $\gamma = \left(\begin{array}{cc} a & b \ c & d \end{array}\right) \in SL_2(\mathbb{Z})$ acts as a fractional linear transformation. It follows automatically that something as simple as $(cz+d)^{2k}$ is a cocycle in group cohomology, since $G_{2k}$ is, for example, nonzero.
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Topology and Logic as a Source of Algebra by Mac Lane in the AMS Bull. January 1976 may be helpful.
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Seconding some remarks already made: despite the (stodgy) tradition of standard math curricula to ignore these things, it is very useful to understand that group (co)homology consists of the "derived functors" of the functors that take G-[co]fixed vectors on representations spaces. (Unfortunately, the two "co"'s get reversed.) Thus, there is an intrinsic definition. These higher derived functors are the universal/right things to "correct" for the non-exactness of the (co)fixed-vector functors. True, this definition does not explain why we would care so much, but, in the end, it is a large part of what (co) homology is, I think.
It is also very useful to know/prove that any pro/in-jective resolution can be used to compute the (co)homology.
Thus, the particular choices (homogeneous/inhomogeneous bar resolution(s), etc.) are not the definition, despite common assertions that this is so.
The point of clever/insightful choice of resolutions is facilitation of computations about specific situations.
C. Weibel's book on homological algebra does some good sample computations in group (co) homology, in the context that it is an example.
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I think this answer is really good. My point of view is that many things we care about in number theory are acted upon by Galois groups, and often "fixed elements come from below". Somethines, this is sadly false, like for ideals and ideal class group. Group cohomology is then very natural, because you have a tautological exact sequence defining the class group and you would like to compare objects in this sequence (which you care about genuinely because it is tautological) which are fixed by Galois with those coming from below. – Filippo Alberto Edoardo Nov 14 at 2:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 161, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429517388343811, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/64017/eigenvalues-of-laplacian-beltrami-operator/64022 | ## Eigenvalues of Laplacian-Beltrami operator
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I am interested in the first non zero eigenvalue of the Laplace-Beltrami operator in a 2D compact manifold, and if there is a geometric characterization of its value.
I am interested in the case when you fix the volume of the manifold to some value (say $Vol = 1$), and let the other modes of the metric fluctuate. The average curvature of the manifold is imposed by the Gauss-Bonet theorem, but can let the curvature to fluctuate from one point to another.
My intuition says that the first non zero eigenvalue should approach 0 in the limit when the "fluctuations of the curvature" grow, but I can not give a precise meaning to this statement.
so the question is: Is there any characterization of the first eigenvalue(s) of the Laplace-Beltrami operator in a 2D compact riemann manifold as functions of the curvature or its powers (i.g. $\int R^2 \sqrt{g} d^2x$).
So let me be more specific. Imagine a manifold topologically equivalent to a Torus. The metric can be written as
$ds^2=f(x_1,x_2)(dx^2_1+dx^2_2)$
and the scalar curvature
$R=−2\Delta \log(f)$
For the case $f=cte$ we have a flat torus. Now expanding f as a fourier series we will have some regions of the torus with poritive curvature and some regions with negative curvature (this invalid some known theorems, that need wither the curvature to be always positive or always negative). Gauss bonnet says that:
$\int R=0$
So the concrete question is: Is there a characterisation of the first eigenvalue of the Laplace operator in terms of $\int R^2$?
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## 4 Answers
The first eigenvalue of a compact surface can be made arbitrarily small (even for surfaces of fixed genus); see, for example, [1], [2], [3] (and references therefrom).
However, as proved by Sarnak and Xue [4], there are arithmetic examples of (constant negative curvature) compact Riemann surfaces of arbitrarily high genus with the first eigenvalue bounded away from zero (see also [5] for a construction involving Selberg's $3/16$'' theorem).
[1] B. Randol, Small eigenvalues of Laplace operator on compact Riemann surfaces, Bull. AMS 80, 1974 996-1008
[2] R. Schoen, S. Wolpert, S. Yau, Geometric bounds on the low eigenvalues of a compact surface, Proc. Symp. Pure Math, vol. 36, AMS, 1980, 279-285.
[3] P. Buser, On Cheeger’s inequality $λ_1 ≥ \frac{h^2}{4}$. Proc. Symp. Pure. Math. vol. 36, 29–77.
[4] P. Sarnak and X. Xue, Bounds for multiplicites of automorphic representations, Duke Math J. 64, 1991, 207-227.
[5] R. Brooks, E. Makover, Riemann surfaces with large first eigenvalue. J. Anal. Math. 83, 2001, 243–258.
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Yes, This is a classical result due to Lichnerowicz. If the Ricci tensor of a compact Riemannian manifold, is such that $Ric \geq kg$ for some k > 0, then $\lambda_1 \geq \frac{nk}{n - 1}$. In Your case, you just replace the Ricci curvature by $Kg$. You will find a proof in the book of Aubin A Course in Differential Geometry as consequence of the Bochner Formula.
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This inequality is false: there are closed hyperbolic surfaces (constant curvature $=-1$) for which $\lambda_1$ is arbitrarily small. – Agol Jun 22 2011 at 1:51
The inequality only applies to manifolds of positive curvature. – Nate Eldredge Jan 6 2012 at 2:41
Your intuition seems correct: assume that some part of your surface contains a long cylinder with a flat metric. Take a test-function on the surface, supported on that cylinder, equal to +1 roughly on one half of the cylinder and -1 on the other half, and of total integral 0. In this case, the $L^2$-norm of the gradient of your function will be very small compared to the $L^2$-norm of the function. By the variational characterization of the first eigenvalue of the Laplacian (= Rayleigh quotient), you can make the first eigenvalue very small.
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This is a challenging problem. The precise meaning of randomness of the metric ought to play a role. I cannot think of a natural one in a general case but I suggest you consider first the case of the two-torus $T^2$ equipped with a metric of the form
$$g = e^{2 u} d\theta_1^2+ e^{2v} d\theta_2^2$$
where $u,v: T^2\to\mathbb{R}$ are independent random functions. Here you need to specify the nature of the randomness of $u$ and $v$.
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http://unapologetic.wordpress.com/2009/02/05/the-algebra-of-upper-triangular-matrices/?like=1&source=post_flair&_wpnonce=d0199d491e | # The Unapologetic Mathematician
## The Algebra of Upper-Triangular Matrices
Here’s another little result that’s good over any field, algebraically-closed or not. We know that the linear maps from a vector space $V$ (of finite dimension $d$) to itself form an algebra over $\mathbb{F}$. We can pick a basis and associate a matrix to each of these linear transformations. It turns out that the upper-triangular matrices form a subalgebra.
The easy part is to show that matrices of the form
$\displaystyle\begin{pmatrix}\lambda_1&&*\\&\ddots&\\{0}&&\lambda_d\end{pmatrix}$
form a linear subspace of $\mathrm{End}(V)$. Clearly if we add two of these matrices together, we still get zero everywhere below the diagonal, and the same goes for multiplying the matrix by a scalar.
The harder part is to show that the product of two such matrices is again upper-triangular. So let’s take two of them with entries $s_i^j$ and $t_i^j$. To make these upper-triangular, we’ll require that $s_i^j=0$ and $t_i^j=0$ for $i>j$. What we need to check is that the matrix entry of the product $s_i^jt_j^k=0$ for $i>k$. But this matrix entry is a sum of terms as $j$ runs from ${1}$ to $d$, and each term is a product of one matrix entry from each matrix. The first matrix entry can only be nonzero if $i\leq j$, while the second can only be nonzero if $j\leq k$. Thus their product can only be nonzero if $i\leq k$. And this means that all the nonzero entries of the product are on or above the diagonal.
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Posted by John Armstrong | Algebra, Linear Algebra
## 1 Comment »
1. [...] diagonal matrices are themselves diagonal. Thus, diagonal matrices form a further subalgebra inside the algebra of upper-triangular matrices. This algebra is just a direct sum of copies of , with multiplication [...]
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http://math.stackexchange.com/questions/138548/when-do-weak-and-original-topology-coincide/138854 | When do weak and original topology coincide?
Let $X$ be a topological vector space with topology $T$.
When is the weak topology on $X$ the same as $T$? Of course we always have $T_{weak} \subset T$ by definition but when is $T \subset T_{weak}$?
Assume that $X$ is any topological space, not necessarily normed.
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At least for normed space: if and only if the space is finite dimensional (consider the weak closure of $\{x,\lVert x\rVert=1\}$). – Davide Giraudo Apr 29 '12 at 19:33
@DavideGiraudo Thanks Davide, I know : ) (Hence the "not necessarily normed") – Matt N. Apr 29 '12 at 19:33
Did you make a conjecture for the general case? – Davide Giraudo Apr 29 '12 at 19:39
@DavideGiraudo No. ${}{}{}{}{}$ – Matt N. Apr 29 '12 at 19:41
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– Martin Sleziak Apr 30 '12 at 11:52
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3 Answers
Quick Google search led me to the paper Sidney A. Morris: A topological group characterization of those locally convex spaces having their weak topology; Mathematische Annalen, Volume 195, Number 2 (1971), 330–331, DOI: 10.1007/BF01423619. The following result is given in this paper:
Theorem. Let $E$ be a locally convex Hausdorff real topological vector space. Then $E$ has its weak topology if and only if every discrete subgroup (of the additive group structure) of $E$ is finitely generated.
The proof relies on a result from the paper Sidney A. Morris: Locally Compact Abelian Groups and the Variety of Topological Groups Generated by the Reals; Proceedings of the American Mathematical Society, Vol. 34, No. 1 (Jul., 1972), pp. 290–292; DOI: 10.1090/S0002-9939-1972-0294560-4.
Added: (t.b.) Here's a sketch of the argument:
1. Suppose that $E$ has the weak topology, and that $\Gamma \subset E$ is a discrete subgroup of the additive group. Then there exists a neighborhood $U$ of $0\in E$ such that $U \cap \Gamma = \{0\}$. In other words, there are continuous linear functionals $\varphi_1,\dots, \varphi_n$ and $\varepsilon \gt 0$ such that $\{0\} = \Gamma \cap \{x \in E\,:\,\max_{i} |\varphi_i (x)| \lt \varepsilon\}$. But this means that $\gamma \mapsto (\varphi_1(\gamma),\dots,\varphi_n (\gamma))$ is an injective map from $\Gamma$ to a discrete subgroup of $\mathbb{R}^n$, and discrete subgroups of $\mathbb{R}^n$ are finitely generated.
2. Suppose $E$ does not have the weak topology. Then the topology on $E$ is strictly finer than the weak topology, hence there is a symmetric convex open neighborhood $U$ of $0 \in E$ which is not a neighborhood of $0$ in the weak topology, so $U$ cannot contain any linear subspace of finite co-dimension. The Minkowski functional (gauge) of $U$ thus defines a continuous norm $\|\cdot\|$ on some infinite-dimensional subspace $F$ of $E$. A classical argument of Mazur (see e.g. Lindenstrauss–Tzafriri Theorem 1.a.5, and Albiac–Kalton, Theorems 1.4.4, 1.3.9 and 1.3.10) then provides us with a basic sequence in $F$ which we may assume (after changing the norm to an equivalent one if necessary) to be a monotone basic sequence $(x_n)_{n=1}^\infty$. The monotonicity requirement implies that the subgroup generated by $(x_n)_{n=1}^\infty$ is discrete and linear independence implies that it is infinitely generated.
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I've posted this as a CW, in case anyone feels like going through details and elaborate this answer more; feel free to do so. – Martin Sleziak Apr 30 '12 at 14:44
Very nice, thank you! – Matt N. Apr 30 '12 at 14:46
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I added a sketch of the argument. I think the proof I give of part 1 is somewhat simpler than the one in the paper. – t.b. Apr 30 '12 at 23:37
@t.b. Nice, thank you! – Matt N. May 1 '12 at 6:47
I'd like to add that if we assume $X$ to be a normed vector space (over $\mathbb R$) then we have $T_{norm} = T_{weak}$ if and only if $X$ is finite dimensional. To see why this is the case:
$\Longrightarrow$ Assume $X$ is infinite dimensional. To show that then $T_{norm} \neq T_{weak}$ it's enough to find a set that is closed in one of the two topologies but not in the other. Note that $S := \{x \in X \mid \|x\| = 1 \}$ is closed in $T_{norm}$. But it's not closed in $T_{weak}$ since $0$ is in the weak closure of $S$: Let $U$ be any neighbourhood of $0$ in $T_{weak}$. Then there exists an open set $O$ such that $0 \in O \subset U$. We know that $\bigcup_{\varepsilon>0, r_0 \in \mathbb R} \{ \bigcap_{i=1}^n f_i^{-1} (B(r_0, \varepsilon)) \mid n \in \mathbb N \}$ forms a neighbourhood basis of $T_{weak}$. Hence there exist $f_1, \dots f_n \in X^\ast, \varepsilon > 0, r \in \mathbb R$ such that $0 \in O = \bigcap_{i=1}^n f_i^{-1} (B(r, \varepsilon))$.
Now we define a map $\varphi : X \to \mathbb R^n$, $x \mapsto (f_1(x), \dots, f_n(x))$. This map is linear. Hence $\operatorname{dim}{X} = \operatorname{dim}{\ker \varphi} + \operatorname{dim}{\operatorname{im}{\varphi}}$. We know that its image has dimension at most $n$ so since $X$ has infinite dimension we know that its kernel has to have infinite dimension. In particular, we can find an $x$ in $X$ such that $x \neq 0$ and $f_i(x) = 0$ for all $i \in \{1, \dots n\}$, i.e., $\varphi (x) = 0$. Since $\varphi$ is linear we also have $\varphi (\lambda x) = 0$ for all $\lambda \in \mathbb R$, in particular, for $\lambda = \frac{1}{\|x\|}$. Hence we have found a point $\frac{x}{\|x\|}$ that is in $S$ and also in the neighbourhood $U$ of $0$. Since the neighbourhood $U$ was arbitrary we get that $0$ is in the weak closure of $S$.
$\Longleftarrow$ Let $X$ be finite dimensional. Since by definition we always have $T_{weak} \subset T_{norm}$ it's enough to show that every open ball $B_{\|\cdot\|}(x_0, \varepsilon)$ is weakly open. Since $X$ is finite dimensional we can write every $x$ in $X$ as $x = \sum_{i=1}^n x_i e_i$ where $e_i$ is the basis of $X$. Define $f_i : X \to \mathbb R$ as $f_i : x \mapsto x_i$. Then $f_i$ are in $X^\ast$. Also since $X$ is finite dimensional, all norms on $X$ are equivalent and hence it's enough to show that $B_{\|\cdot\|_\infty}(x_0, \varepsilon)$ is weakly open. But that is clear since
$$B_{\|\cdot\|_\infty}(x_0 , \varepsilon ) = \{x \in X \mid \max_i |x_i-x_{0i}|< \varepsilon \} = \{x \in X \mid \max_i |f_i(x-x_0)| < \varepsilon \} = \bigcap_{i=1}^n f_i^{-1} (B(x_{0i}, \varepsilon))$$
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On the implication $\Rightarrow$: what you prove in particular is that every weak neighborhood of $0$ of an infinite-dimensional normed space contains a non-trivial linear subspace. Since the open unit ball is norm-open but contains no subspace, it can't be weakly open. On the implication $\Leftarrow$: One can show that the usual topology on $\mathbb{R}^n$ is the only Hausdorff topology that makes $\mathbb{R}^n$ into a topological vector space. – t.b. May 10 '12 at 17:21
@t.b. Thank you very much for this enlightening comment. – Matt N. May 10 '12 at 18:46
For a finite dimensional vector space there is only one topology that turns it into a topological vector space, so that weak topology and initial topology are necessarily the same.
For an infinite dimensional locally convex topological vector space T with dual T' this is never the case (edit: actually I'm not sure about this, see below), if the initial topology is the Mackey topology.
The Mackey-Arens theorem says that all topologies consistent with a given duality T, T' are comparable:
• The weak topology is the weakest, it is the topology of pointwise convergence.
• The Mackey topology is defined to be the strongest, it is the topology of uniform convergence on every absolutely convex, weakly compact set.
Since T is infinite dimensional, we can pick a sequence $(x_n), n \in \mathbb{N}$, of pairwise different vectors that are part of an algebraic basis of T. This sequence converges to 0 in the weak topology (edit: coming to think of it, I don't know it this is true or how to prove or disprove it. But I will not delete the answer, maybe someone else can improve it :-), but it does not in the Mackey topology. We can check this using the polar $A^°$ of the set $\{x_n, n \in \mathbb{N} \}$, (which is absolutely convex and weakly compact, so that $(x_n)$ would need to converge to 0 on this set, which it obviously does not).
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Since the weak topology is always locally convex by definition, a non-locally convex topological vector space cannot have the weak topology. – t.b. Apr 30 '12 at 13:06
Ugh, right. I deleted my disclaimer about non locally convex spaces. – Tim van Beek May 1 '12 at 10:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 106, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9378662705421448, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/48034/residues-in-qft-propagator?answertab=active | # Residues in QFT propagator
It is a well known fact that the location of the pole of a propagator (in QFT) can be interpreted as the physical mass.
Is there an interpretation for the residue of the propagator?
Note: I´m thinking of generalised propagator, not necessarily a propagator of a fundamental field.
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+1, interesting question. The residue theorem is a special type of Stokes theorem, are you sure that the pole is interpreted as the mass and not the residue? The pole is an infinity, but the residue gives the pole a value (for integrable poles), or any closed closed loop the value of the pole. Intuitively it would seem to give a sort of measure of the mass? I'm not an expert in QFT or QM however... – daaxix Dec 31 '12 at 22:51
@daaxix it's the location of the pole, not the value of the pole, that gives the mass of the particle. – David Zaslavsky♦ Dec 31 '12 at 23:27
@DavidZaslavsky, ah, ok. Well the residue gives the value of the pole itself. Is there a difference between integrable and non-integrable poles in QFT? Are they all integrable? – daaxix Dec 31 '12 at 23:32
There are sometimes branch cuts in propagators, but I couldn't tell you that much about them offhand. Other than that, I can't think of any examples of nonintegrable poles. Generally, a propagator is of the form $\frac{\text{stuff}}{p^2 - m^2}$, and the residue depends on the stuff in the numerator, so it doesn't necessarily correspond to a particular physical quantity that I know of. – David Zaslavsky♦ Jan 1 at 0:08
## 1 Answer
As you say, the location of the pole gives the mass. Higher-order diagrams shift the location of the pole, causing a (often infinite!) renormalization of the mass.
The residue at the pole location simply gives the normalization of the wave function, which, as far as I remember, is just absorbed into scale of the field.
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Thanks Tom. "Happy New year" to you too. But, please don't post these in answers. You could leave a comment below your own post though you can't leave comments in others' post. BTW, please don't leave your user signature. Your user card has your name, rep. score, badges, etc. :-) – Ϛѓăʑɏ βµԂԃϔ Jan 1 at 5:13
Thank you Tom. Thus, just for clarification, the shift of the location of the pole renormalise the mass, whilst the shift to the residue is reabsorbed in the renormalisation of the field. Cheers, and Happy new year. – Dox Jan 2 at 9:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9337446689605713, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/11450/applications-of-artins-holomorphy-conjecture | ## applications of Artin’s holomorphy conjecture
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I wonder why the Artin conjecture is so important. The only reason I could figure out is that one could use the holomorphy of Artin L-series and Weil's converse theorem to show modularity of two-dimensional Galois representations.
Are there other reasons?
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Holomorphy, or meromorphy? – Anweshi Jan 11 2010 at 18:47
2
Very likely s/he means holomorphy. Meromorphy was proved by Brauer. – Pete L. Clark Jan 11 2010 at 18:53
What is the conjecture precisely? – Ryan Budney Jan 11 2010 at 19:03
1
There is a statement available on Wikipedia, although maybe it should be included in the OP: en.wikipedia.org/wiki/… – Qiaochu Yuan Jan 11 2010 at 19:05
## 2 Answers
From a modern view-point, its importance stems from the relationship with modularity/automorphy. Namely, a stronger conjecture, due to Langlands, should be true: if $\rho:G_K \to GL_n({\mathbb C})$ is a continuous irreducible representation, for some number field $K$, then there is a cuspidal automorphic representation $\pi$ of $GL_n({\mathbb A}_n)$ such that $\rho$ and $\pi$ have the same $L$-function; in short, $\rho$ is determined by $\pi$.
This is a non-abelian class field theory. In the case $K = {\mathbb Q}$ and $n = 2$, it says that $2$-dimensional irred. cont. reps. of $G_{\mathbb Q}$ are classified by either weight one holomorphic cuspidal eigenforms (if the rep. is odd) or Maass cuspidal eigenforms with $\lambda = 1/4$ (if the rep. is even). The odd case is now a fully proved theorem (of Deligne--Serre to go from the modular forms to the Galois reps., and of Khare, Wintenberger, and Kisin to go the other way), while the even case is still open. (If the Maass form is dihedral, one can construct a corresponding dihedral Galois rep. --- this is due to Maass himself --- but otherwise that direction is open; if the image of $\rho$ is solvable, then one can construct the corresponding Maass form --- this is due to Langlands and Tunnell.)
As you note, in the two-dimensional case over $\mathbb Q$, Weil's converse theorem shows that Langlands' conjecture is equivalent to the Artin conjecture. (In fact, proving this over any number field $K$ was one of the goals of the famous book of Jacquet--Langlands). In general, the two are also very closely linked, so that no modern number theorist thinks about one without the other. In general, one working principle of modern number theory is that the only way to establish holomorphy of $L$-functions arising from Galois representations is by simultaneously proving automorphy.
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"of Khare, Wintenberger, and Kisin to go the other way". And Emerton. Cf. Berger's recent Bourbaki talk (1017) arxiv.org/abs/1002.4111. – Chandan Singh Dalawat Mar 4 2010 at 4:04
2
Dear Chandan, In fact, going the other way in this instance is a case of 2-dimensional Fontaine--Mazur with Hodge--Tate weights (0,0), and so is not covered by my results on Fontaine--Mazur (which, in addition to other technical caveats, apply only in the distinct Hodge--Tate weights situation). Rather, it follows directly from Serre's conjecture (proved by Khare--Wintenberger--Kisin; see Inventiones 178), with regard to the proof of which I was merely an astonished onlooker! – Emerton Mar 4 2010 at 4:13
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I am waking up an old and already well-answered question, to offer another point of view,
The Artin's conjecture appears very naturally in the context of Chebotarev's density theorem. In fact, we can see Cheobtarev's contribution as a clever trick to circumvent the Artin conjecture by reducing the proof to cases where it is known (by works of Dirichlet and Hecke). But the proof of Chebotarev will be much simpler and more natural if we had the Artin's conjecture, which moreover would give better results as far as the error term is concerned. This is, I think, a good justification of the importance of Artin's conjecture.
To explain the role of Artin's conjecture, let us also assume for simplicity GRH. Then, for $G=Gal(K/\mathbb Q)$ a finite Galois group, and $\rho$ an irreducible Artin representation of $G$, the $L$-function $L(\rho,s)$ has no zero on Re $s>1/2$ (by GRH) and no pole either (by Artin), except for a simple pole at $s=1$ if $\rho$ is trivial. Thus the logarithmic derivative, $L'/L(\rho,s)$ has no pole on Re $s>1/2$ (except perhaps...): this illustrates clearly the symmetric and complementary role played by Artin and Riemann's conjectures; both poles and zeros of $L$ contribute to simple poles of $L'/L$, and Artin eliminates some of them, Riemann the others. Now standard techniques of analytic number theory allows us, by integrating $L'/L$ on a vertical line $2+i \mathbb R$ and moving it near the critical line, to $1/2+\epsilon + i \mathbb R$, to get an estimate of the quantity: $$\pi(\rho,x) = \sum_{p^n < x} \log(p) tr \rho(frob_p)^n$$ where the sum is on prime power less than $x$. This estimate is $O(x^{1/2+\epsilon})$ if $\rho$ is non trivial, and $x + O(x^{1/2+\epsilon})$ if $\rho$ is trivial, because of the pole at $s=1$.
Now let $C$ be subset of $G$ stable by conjugaison, $1_C$ its characteristic function. Since $1_C$ is a central function, it is a linear combinaison of character of irreducible representation of $G$, say $$1_C = \sum_\rho a_\rho tr \rho.$$ Hence $\pi(C,x) := \sum_{p^n< x} \log p 1_C(frob_p) = \sum_\rho a_\rho \pi_\rho(x)$. Since $a_1$ is easily computed as $|C|/|G|$, we get $\pi(C,x)=|C|x/|G| + O(x^{1/2+\epsilon})$, which is up to standard manipulation Chebotarev's density theorem.
Hence we can say that Artin's conjecture play to Chebotarev's density theorem a role analog to the role played by the standard conjectures for the Weil's conjecture proved by Deligne. In both cases, a clever and beautiful trick was used (by Chebotarev and Deligne, respectively) to prove a theorem (Chebotarev's theorem, aka Frobenius' conjecture, and Deligne's theorem, aka the last Weil's conjecture) without proving the conjectural statement that makes the theorem limpid (Artin's conjecture, resp. standard conjectures). This is great, but doesn't make the conjectures any less interesting.
Important addendum A friend of mine made me notice something that kind of weakens significantly the point I was trying to make above. Indeed, for the argument outlined above, one doesn't need GRH + Artin's conjecture: GRH is enough. Or, more precisely, the part of Artin's conjecture that is needed is already known under GRH. Indeed, it is clear that in the argument the absence of poles of $L(\rho,s)$ is only used in the region Re $s>1/2$. But by Brauer's theorem, we know that $L(\rho,s)$ is meromorphic on $\mathbb C$, and by elementary reasoning that $\prod_\rho L(\rho,s) = \zeta_K(s)$, where $\rho$ runs amongst Artin's representations of Gal$(K/\mathbb Q)$. Moreover, by Hecke $\zeta_K$ has no poles (except a simple one at $s=1$). Hence if all the $L(\rho,s)$ have no zero on any given region, then they don't have a polo either on that region -- expected for $L(1,s)$ with its simple pole at $s=1$.
I leave the argument above, because it shows that, the absence of poles for $L(\rho,s)$ is important for questions on the distribution of primes, even if it is a theorem rather than a conjecture. Moreover, this argument has an historical interest, as Artin's conjecture was made in the 1920's, and Brauer's theorem proved in 1946.
I ignore if the question of absence of poles on the critical line for the $L(\rho,s)$ (the only part of Artin's conjecture that is still open) has any direct application on the distribution of primes.
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http://physics.stackexchange.com/questions/45387/why-cant-missing-mass-be-photons/45413 | # Why can't missing mass be photons?
After a star lives and dies, I assume virtually all of it's mass would be photons. If enough stars have already lived and died, couldn't there be enough photon energy out there to account for all the "missing mass" (=dark matter) in the universe?
And if there were enough photons to account for all the missing mass, what would it look like to us?
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3
Only a tiny fraction of a star's mass ever converts into photons. – David Zaslavsky♦ Nov 29 '12 at 7:28
## 4 Answers
As a general rule, zero mass particles which travel with the velocity of light are not good for dark matter, because dark matter concentrates around gravitational attractors. It has to be particles with some mass that can be at rest in order to stay around a galactic center from the beginning . In addition they have to be controlled by weak interactions, if they decay, because the dark matter halo is stable for long periods.
Maybe I should add that very cool and slow photons from the beginning of the formation of the observed universe exist and have been detected as Comsmic Microwave Background radiation , very low frequency photons, uniformly distributed in the cosmos.
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Thanks for your answer. If there were a sea of photons, would they not also concentrate around gravitational attractors? Wouldn't there be more photons in the potential energy well of the gravitational attractor than there would be in flat space? Wouldn't this situation be stable? Are particles with mass really the only candidates? – Tom Fangrow Nov 29 '12 at 6:52
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Aside from the unstable orbit at $r = 3M$ you can't bind photons in a gravitational well. The reason is simply that they always travel at $c$, while particles with mass can take any speed. That's why planets can have stable orbits round the Sun. There is a single distance from a mass where the velocity of light matches it's orbital velocity, but even this orbit is unstable so you'd never find any significant concentration of photons there. – John Rennie Nov 29 '12 at 7:15
As John says; a particle has to have some mass to stay at an orbit.CMB does not concentrate around galaxies, and that is the coldest photons we have observed, because they still travel at the velocity of light. – anna v Nov 29 '12 at 7:31
There is a simple argument why photons emitted by stars can't be dark matter, and that's because there is about ten times more dark matter than normal matter. If all the stars created at the Big Bang had turned into photons there still wouldn't be enough of them.
You might argue that maybe more normal matter than we think was created during the Big Bang, but the theory of Big Bang Nucleosynthesis places a limit on how much normal matter was created, and this limit is four times smaller than the amount of dark matter. The dark matter has to be something odd.
If you're interested in more info this paper is a good review, though harder going than the answers here!
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Thanks for this, I'll check it out. – Tom Fangrow Nov 29 '12 at 7:31
Photons don't have mass. So your assumption's incorrect, although I don't know how much of a (say) main sequence star's mass gets converted into photons over its lifespan.
Photon's can't account for, say, dark matter, because dark matter has mass.
Thermal energy (in the vacuum) is comprised of photons, which can spontaneously form particle-antiparticle pairs. Usually these quickly annihilate, so this is also not a good source of mass.
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1
-1. Photons don't have mass, but they do have energy. It is fully possible that during a star's lifespan, some amount of it's mass is converted into energy (i.e., photons). – Kitchi Nov 29 '12 at 6:32
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The assumption was that all a star's mass is converted to energy in the form of photons. Humor me for a moment. You say dark matter has mass. How do you know? Is it because it affects gravity? Do we know for certain that photons do not affect gravity? – Tom Fangrow Nov 29 '12 at 6:44
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Photons do affect gravity, but that's because gravity is affected by energy, not mass. – David Zaslavsky♦ Nov 29 '12 at 7:28
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Photons do affect and are affected by gravity ( gravitational lensing) they cannot be captured by gravitational wells because of the velocity c. See Johhn's comment in my answer. – anna v Nov 29 '12 at 7:33
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– John Rennie Nov 29 '12 at 10:50
show 3 more comments
Photons are easily detectable. We can count how many photons are there at any distance of us by just counting the photons reaching us from there. It is impossible that the hidden photons ramble the whole universe but mysteriously avoid us.
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http://mathoverflow.net/questions/105586/serre-tate-canonical-lifts-for-finite-fields/105589 | ## Serre-Tate canonical lifts for finite fields
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A result of Serre-Tate states that we can canonically lift an ordinary abelian variety over a perfect field $k$ of positive characteristic to an abelian scheme over the ring of Witt vectors of $k$ and that we can also lift an endomorphism of the variety uniquely to an endomorphism of the canonical lifting.
In the case where $k$ is finite, is the canonical lift a variety? Do we know its dimension? I'd be glad to read your answers and the references to those answers as well.
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## 1 Answer
As you just said, the canonical lift is an abelian scheme over the ring of Witt vectors $W(k)$. Now, if $k$ is finite of characteristic $p$, $W(k)$ is the ring of integers of the unramified extension (say $K$) of $\mathbb{Q}_p$ whose residue field is $k$. The canonical lift has a generic fiber which is an abelian variety over $K$ and has the same dimension as the abelian variety over $k$ you started with. I guess there is a bit of ambiguity in the literature where some people refer to the abelian scheme over $W(k)$ as the canonical lift whereas others refer to its generic fiber over the field of fractions of $W(k)$ as the canonical lift. You can recover the former from the latter by taking Néron models.
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Thank you for clearing things up, this helps a lot. – Marty B. Aug 27 at 1:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9541462063789368, "perplexity_flag": "head"} |
http://proopnarine.wordpress.com/2009/02/ | # Network equivalence
26 Thursday Feb 2009
Posted by in CEG theory, NSF proposal
≈ 3 Comments
The objective here is to establish the degree of similarity between successive communities, in spite of changing taxonomic compositions and diversities, and ecological/guild diversity.
Let $\mathcal{U}$ represent an ecosystem over time (e.g. the Karoo series) comprising a series of chronologically successive communities, $\mathcal{U}=\{ U_{a},U_{b},\ldots ,U_{n}\}.$ The metanetwork representation of $\mathcal{U}$ is identical for all $U$ if guild diversities are omitted. That is, the unparameterized metanetworks are automorphisms $\forall U.$ Species-level networks (slns) are generated from the parameterized metanetworks, a finite set for each one. The question being addressed here is, how many of those slns are isomorphic between metanetworks? In other words, how many of a community’s networks are identical to networks in the preceding and succeeding communities? This is a very important question from the perspective of CEG dynamics, because in the model equivalence of ecological dynamics can transcend taxonomic composition and identity. Consider the two slns in the first figure. Specifically they are different, but dynamically they will respond identically to perturbation. This isomorphism extends to the guild level also. Imagine for a moment that the species in the figure are actually different guilds, and that guilds 2 and 3 are very different organisms. IF guild diversities permit the generation of slns with the same numbers of links, then isomorphic slns will be generated. Furthermore, we can remove any guild and species identity completely from the networks and maintain an equivalence of CEG dynamics under the following condition: If the perturbed species/nodes are part of a connected subgraph, and the connected graph is isomorphic with another subgraph, then CEG dynamics of the two networks will be identical! A minimum measure, therefore, of the continuity of ecosystem dynamics between successive communities, is the number of slns that are isomorphic between the sln sets.
Unfortunately, determining whether two graphs or networks are isomorphic is an NP-complete problem, and slns are very complex graphs. Given the size of the sln sets for each community, it would be impossible to determine the power of their intersection. Given that CEG dynamics are drawn however from the likeliest region of the sln space, that is, stochastic draws are made from defined trophic link distributions, we are really interested in establishing isomorphism of those subsets, not the entire sln space. So, one procedure would be to generate a set of high probability slns from each community, and then test those for isomorphism. The test would have to be one of elimination, i.e. whittling down the number that could be isomorphic, without ever actually arriving at the number that are isomorphic. But that would give us an upper limit of the number that could be isomorphic, and an upper limit on the measure of similarity between the two communities. In the next post I will outline an MCMC approach to generate sets of slns with high likelihood.
It is entirely possible that the result will be that no networks are isomorphic between communities. In that case, it would be obvious that, given similarity of CEG dynamics, the networks are “close enough”. We would then be faced with the even more difficult task of measuring minimum distances between networks, but this is not at all an impossible task. We’d just need some reasonably powerful computers, and acknowledge the fact that our carbon footprints would far outweigh any benefits to be gained from our work.
# Trophic network probability
07 Saturday Feb 2009
Posted by in CEG theory
Given any species level topology $(E,V)\text{,}$ where $E$ represents edges or links, and $V$ represents vertices or species, the probability of that specific topology is the probability of the set of interspecific links specified or composing the topology. Therefore, the probability of the network, given metanetwork $U\text{,}$ is
$p(E,V) = \prod_{x=1}^{\sum \vert G_{u}\vert} \prod _{u=1}^{\vert U \vert}p(r_{x}^{u} \mid r_{x})$
where the right product is the probability of any species $x$ having a particular pattern or topology of links to other species, and the left product is the product of those probabilities for all species in the community. We can see immediately how the probability of any particular species-level food web is built from the probabilities of individual species networks.
The number of species-level networks that can be derived from a metanetwork depends upon permutations of all possible combinations of link topologies of species in the community, and is generally an astronomical number for even a modest number of species and guilds. The network with the greatest probability, or maximum likelihood of occurrence, is one where the probabilities in the formula above are maximized. This can be approximated if one considers that the probability of $x_{i}$ being linked to any $x_{j}$ is equal, regardless of the in-degree of $x_{j}\text{.}$ One can therefore consider the probability of linking to any $x_{j}$ to simply be the proportion of the predator’s set of prey species that is represented by $\vert G_{j}\vert\text{.}$ These are simply maximum likelihood estimates of the metanetwork link probabilities.
The in-links of each species are assigned randomly to species in other guilds for which a metanetwork in-link exists, i.e. those other guilds comprise potential prey of the species, and $a_{iu}=1\text{,}$ where the species belongs to guild $i.$ The resulting network may be represented as a $N\times N$ adjacency matrix, $A_{N}\text{,}$ where $N$ is the total number of species in the community. Binary entries $n_{xy}$ indicate whether species $y$ is prey to species $x\text{.}$ Furthermore, row sums, $\sum n_{xy}$ equal the in-degree of species $x.$
The probability of a species $a\text{'s}$ link topology, $p(a_{i})\text{,}$ or the binary pattern of the row $A_{a}\text{,}$ is the product of the probabilities of each link. This can be calculated efficiently as the multinomial probability of a pattern of links spread among the guilds. Say that the probability of a link between $a$ and a species in guild $G_{j}$ is equal to the fraction of the diversity of $a\text{'s}$ prey that is represented by the species richness of $G_{j}\text{,}$ then the probability of $a$ is
$p(a_{i}) = \frac{r_{a}!}{r_{a}^{1}!r_{a}^{2}!,\ldots,r_{a}^{n}!} \prod_{x=1}^{n}\left( \frac{\vert G_{x}\vert}{\vert R_{i}\vert}\right) ^{r_{a}^{x}} \Rightarrow \frac{r_{a}!}{r_{a}^{1}!r_{a}^{2}!,\ldots,r_{a}^{n}!} \left( \frac{1}{\vert R_{i}\vert}\right)^{r_{a}} \prod_{x=1}^{n}\left( \vert G_{x}\vert\right) ^{r_{a}^{x}}$
where $r_{a}^{x}$ is the number of $a\text{'s}$ links to species in guild $G_{x}\text{,}$ $\vert G_{x}\vert$ is the species richness of $G_{x}\text{,}$ and $\vert R_{i}\vert$ is the total number of prey potentially available to $a$ according to metanetwork $U$, where $a\in G_{i}\text.$ It is important to note that each term in the formula exists iff $a_{ix}=1\text{,}$ i.e. a metanetwork link exists $G_{i}\leftarrow G_{x}.$ This prevents the inclusion of zero probabilities. Finally, the overall probability of the network is calculated as $p(E,V)=\prod_{a=1}^{a=N}p(a)\text{,}$ the product of all species probabilities, which in turn are the products of all link probabilities.
# Another view
05 Thursday Feb 2009
Posted by in Visualization
Tags
food webs, modeling, Network theory, networks
This is another rendering of the San Francisco Bay food web (see below) using a different drawing algorithm. This view arranges guilds hierarchically instead of in a circular fashion. It is very interesting to note that “layers” roughly equivalent to trophic levels emerge naturally from the data. Primary producer guilds are at the bottom, and top predators at the top!
This figure was rendered by Rachel using AT&T’s Graphviz dot algorithm.
# San Francisco Bay community food web
04 Wednesday Feb 2009
Posted by in Visualization
≈ 4 Comments
Tags
connectance, food webs, networks
Now that’s complex! This is a rendering of the metanetwork for the San Francisco Bay food web. The network consists of 163 nodes, each node being a guild. In total, they represent ~1,600 species of invertebrates and fish, as well as four nodes representing various types of autotrophic producers. There are 5,024 links or trophic interactions between the guilds. The dataset currently excludes birds and marine mammals. Those data are being incorporated even as I type! So, when faced with this level of complexity, how does one determine if the system is resilient, or vulnerable to the removal or addition of specific types of species, or can withstand the effects of climate change?
The figure was produced by one of my graduate students, Rachel Hertog, who has done a tremendous amount of work on this project, as well as the Dominican Republican paleocommunities. The data come almost entirely from the collections of the California Academy of Sciences, notably the Dept. of Invertebrate Zoology & Geology, and the Dept. of Ichthyology.
# the written word
01 Sunday Feb 2009
Posted by in Uncategorized
Check out Wordle. It’s fun. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 47, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9331094622612, "perplexity_flag": "middle"} |
http://quant.stackexchange.com/questions/3666/how-do-i-statistically-differentiate-a-series-of-prices-from-a-series-of-returns/3676 | # How do I statistically differentiate a series of prices from a series of returns?
I developed an optimization algorithm that uses returns (among other parameters) as input and basically output an allocation.
As I'm pretty happy with the results, I am in the process of putting the algorithm in production. As a matter of fact, I have to embed some checks into the algorithm to make sure that the input was right and that the output makes sense. For example, I check that the allocation sums to 1, I check that the time series provided are all of the same length and so on...
I would like to add another test which allows me to display a warning if the input time series appear to be prices instead of returns (and people will do this mistake one day I'm sure). So, I would like to setup a statistical test on my set of points $x=x_1, ... , x_n$ to determine whether they are likely to be prices.
Formally the statistical test if a function defined as follows:
$$A(x) \rightarrow \{0,1\}$$
Ideally we would like to find a test that is sufficient and necessary for a time series to be returns (as compared to prices, not to "a random time series").
There are three types of interesting statistical tests:
Type I (necessary and sufficient)
$$A(x)=1 \iff x ~ \text{are returns}$$
Type II (sufficient)
$$A(x)=1 \Longrightarrow x ~ \text{are returns}$$
Type III (necessary)
$$A(x) \neq 1 \Longrightarrow x ~ \text{are prices}$$
The following test is a dummy one:
$$A_{\text{dummy}}(x)=\frac{1}{n} \sum_{i=1}^n x_i > 1$$
This is not good because a Forex time series are prices of a currency expressed in the base currency of the portfolio, and such a series would produce a result of 1.
I came up with a Type II test:
$$A_\text{Type II}(x) = \exists i ~ x_i<0$$
Note that I assume that the input could be either prices or returns.
A series of prices with non-negative returns would fool this test so it is not necessary.
I believe it is impossible to come up with a Type I test, which implies that I can't come up with a Type III test either (otherwise I could construct a Type I test easily).
I would be looking for extra Type II tests to improve the probability of wrong input detection.
Have you ever had to do such test? What method would you recommend?
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You could always compare the input values against the security's closing market price to see if they are similar in magnitude. – chrisaycock♦ Jun 24 '12 at 13:42
I assume that I have no information on what the instruments underlying each time series are. – SRKX♦ Jun 24 '12 at 14:21
## 2 Answers
You could test for whether the input series is I(0) vs the alternative of I(1). Specifically, regress the input series on its own lag, and test whether the coefficient on the lag is significantly different from zero. Price series should have a coefficient close to 1, while return series should have a coefficient close to 0.
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I think that can never be 100% sure, and the most you could do is raise a warning, and your approach makes perfect sense to me.
I want to point out one thing though. While prices cannot be negative, they are sometimes recorded with a negative sign, where negative sign conveys some other information. For example in CRSP:
Price
Usually, the CRSP price is the closing price, the price of the last reported trade on any given day. However, if no trade is recorded on a given trading day, the reported price will be the negative of the average of the bid and asked prices for that day. The price reported in CRSP monthly files is the price of the last trading day of the month. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.947838544845581, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/5759/can-one-build-a-one-way-function-from-aes?answertab=active | # Can one build a one-way function from AES?
We change the AES block cipher encryption:
• we delete the key schedule algorithm
• the user now provides a string of 1408 bits
• we divide the string to 11 sub keys, and use them directly in the encryption part of AES as the round keys.
We call the resulting function $AES' : \mathbb Z_2^{1408} \times \mathbb Z_2^{128} \to \mathbb Z_2^{128}$.
Then we use this function $AES'$ as cryptographic compression function in the following way:
Given text $M$, we divide it to blocks $M_1, \dots M_n$ of size 1408 bits each. Then define $g_i \in \mathbb Z_2^{128}$ as
1. $g_0=0$
2. $g_i = AES'(M_i,g_{i-1})\quad$ (Each $M_i$ is used as the round keys in one AES encryption.)
Then the hash function $h : \mathbb Z_2^{n·1408} \to \mathbb Z_2^{128}$ is defined as follows : $h(M)=g_n$
Is $h$ a one way function? If not, how could an attack reveal $M$, given $h(M)$?
(I don't want you to solve, I just want some hints to solve this problem.)
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Are you only interested in one-wayness or also collision resistance? What do you do if $M$ is not a multiple of 1408 bits long? – mikeazo♦ Dec 18 '12 at 15:31
you may assume that it is a multiple of 1408 – fshtea Dec 18 '12 at 15:37
I edited your question, since this is not about "changing AES encryption", but building an one-way function from parts of AES (quite a different task). – Paŭlo Ebermann♦ Dec 22 '12 at 21:42
Where did you run into this problem? Is this a homework problem, or did you run into it in some practical setting? That may help us give you a more useful answer. – D.W. Dec 23 '12 at 7:46
## 2 Answers
Restrict to $n=1$. Let $M$ and $M'$ be two 1408-bit messages differing in the last 128 bits. What's $h(M)\oplus h(M')$? Can a function $h$ with this property be one-way?
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As Mike asked, it's not clear if you're asking about onewayness, or collision resistance (as you call the function a 'cryptographic compression function').
Assuming you're asking about onewayness, well, given a single 128 bit value $h(M)$, we obviously cannot uniquely deduce the 1408 bit value $M$. However (hint), let us assume that we can ask for the value $h(M \oplus \Delta)$, for 1408 bit values $\Delta$ that we pick; how can we use that to recover $M$? How can we extend this observation to cases where the length of $M$ is a multiple of 1408 bits?
Assuming you're asking about collision resistance, well, how can we pick a value $M$ so that $h(M)$ is a predetermined value? Hint: look at how the AES uses the last round subkeys.
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http://agtb.wordpress.com/2012/07/ | # Turing's Invisible Hand
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## The absolute rule of law at American universities
Posted in Uncategorized on July 27, 2012 | 2 Comments »
American bureaucracy is often nasty, not necessarily because of the excessive love for paperwork (although that certainly plays a role), but rather due to the tendency to follow every rule to the letter even if it doesn’t make sense. Perhaps surprisingly, the bureaucracy at American universities is as unbending as anywhere else.
Harry Lewis, a professor of CS at Harvard and former dean of Harvard College (and an active member of CRCS, where I was a postdoc fellow), recently quoted the following insightful paragraph on his blog:
But morality, not to mention common sense, plays little part in the functioning of our modern universities. The role of academic administrators these days – and this has been true for much of the past 25 years – is to prevent criticism of the institution. This has resulted in the growth of huge public relations infrastructures that team up with fund-raising (“development”) infrastructures. General Counsels have taken charge of much that goes on at the modern university. At their order, universities operate largely on this theory of “risk reduction” – that is to say, things are done not because they are right, nor because they enhance the institution’s educational or scholarly missions. Instead, they are done to protect the institution’s reputation, to protect the jobs of the administrators (for whom general counsel works), to keep the institution from losing government funds, and to keep the institution from getting sued. Truth, principle, and the education and welfare of young people have little to do with it. … The modern administrative university is a business machine without a soul. It is an administrative fiefdom that operates outside of the sights and controls of its governing boards or its alumni, with the primary goal of avoiding criticism.
Though this statement is perhaps too black and white, it does resonate with me. I have a few good examples of how strict bureaucracy can get in the way of doing science, based on my limited experience in Harvard and CMU (which presumably are representative of other good, private American universities), but I don’t want to air my dirty laundry in public. Instead, I’ll give two “harmless” examples from this past year at CMU.
A few weeks after I started working at CMU, CS faculty were required to attend a training session on sexual harassment. I naively assumed that the purpose of such a session would be to create a more supportive and pleasant working environment for men and women alike, thereby indirectly leading to better science and education. However, it quickly became clear that the purpose was to prevent the university from being sued. The discussion was led by a lawyer who explained that, as agents of the university, the university is liable for our actions, and illustrated how dire the situation was using detailed examples of other universities that did get sued.
As a second example, I taught my first lecture at CMU in mid January. It was bitterly cold outside, but it was warm indoors so I went down to class wearing a T-shirt. As I was plugging my laptop in, the fire alarm sounded. It would have taken a while to go to my office and get my three additional layers of clothing, so I figured I would just walk towards the exit and hang out indoors. As soon as I stopped marching towards the door though, I was informed that the university can be sued if I don’t actually leave the building; so I did. I wonder if the university would get sued if, under these circumstances, its employees froze to death.
Of course, Israeli universities take the opposite point of view: there are very few rules or requirements, which are moreover regarded as friendly suggestions. Sometimes things work out amazingly well, and sometimes chaos reigns supreme. Ah, those were the days!
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## Impartial division of scientific credit
Posted in Uncategorized on July 11, 2012 | 6 Comments »
This morning a CMU colleague posted on Google+ a gentlemanly apology about a magazine article that featured his work; in the interview he did not give sufficient credit to a student. I mentioned this to him in the elevator, and we chatted briefly about the difficulty of assigning credit in academia.
This encounter reminded me of Valencia, and I wasn’t thinking about paella. The AAMAS best paper and one of the EC best papers have six authors each, listed in alphabetical order (incidentally, the intersection of the two author lists is nonempty). Now, I am all for alphabetical author ordering, and use it myself whenever possible (in particular, I can’t complain about EC’12 papers with six authors in alphabetical order). But sometimes it is important to know how to assign credit, especially when dealing with award papers or other influential papers that can have an impact on their authors’ careers; doubly so when the influential paper has more than 400 authors (see the last four pages). Of course this issue is typically dealt with through recommendation letters, but it’s not a perfect solution and the information doesn’t always come through.
In some CS communities, such as systems and AI, authors are typically ordered by contribution. In other disciplines such as medicine papers often have fifteen authors or more, but the ninth author’s contribution is typically on the level of catching a lab mouse that was trying to escape. Although this is a more informative way of assigning credit, I’ve heard many stories of huge fights breaking out.
So is there a better way of assigning scientific credit? Yes! In theory… A beautiful JET 2008 paper by de Clippel, Moulin, and Tideman studies the allocation of a (homogeneous) divisible good. The setting fits the assignment of scientific credit perfectly, although as far as I can tell this potential application is not mentioned in the paper.
The main property one would ask for is impartiality: your share of the credit should only depend on your coauthors’ reports (I think this property is equivalent to strategyproofness when your utility is strictly increasing with your share of the credit). Another basic property that de Clippel et al. ask for, which connects the reports with the credit division, is consensuality: if there is a division that agrees with all individual reports then it must be the outcome.
In their model, each player $i$ reports an evaluation $r^i_{jk}$ for every two players $j,k\in N\setminus \{i\}$, where $N$ is the set of players. A report $r^i_{jk}=x$ means that $i$ thinks $j$ deserves $x$ times the credit that $k$ gets. Perhaps a more intuitive way of thinking about the model is that each player reports normalized values that specify how the other players’ share of the credit should be divided among them; the ratios can be obtained by dividing pairs of such reported values.
The case of three players turns out to be rather straightforward. The unique mechanism that is impartial and consensual assigns to player i the share $1/(1+r^j_{ki}+r^k_{ji})$, where $k$ and $j$ are the two other players. The bad news is that this mechanism allocates the entire credit if and only if the reports are consensual (which happens if and only if $r^1_{23}r^2_{31}r^3_{12}=1$). In other words, for three players there is no impartial and consensual mechanism that allocates the entire credit. (Of course I am omitting some details; you can find them in the paper.)
Fortunately, it can be argued that we don’t really need credit division unless there are many authors, and for the case of four and more players de Clippel et al. give a family of (rather complicated) mechanisms that output an exact allocation, and satisfy impartiality and consensuality, as well as two other desiderata: anonymity (in the sense that the mechanism is indifferent to the identity of the players) and continuity.
I toyed for a while with the idea of demonstrating how serious I am by choosing a paper where my share of the credit is small and convincing my coauthors to report evaluations. Ultimately, on top of embarrassing my coauthors, I realized I would have to figure out the gory details of the 4+ player mechanisms. For now it is amusingly self-referential to imagine de Clippel et al. dividing the credit for their paper using the methods therein; with only three authors, I sure hope their evaluations would be consensual!
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http://mathhelpforum.com/algebra/26685-do-i-factor-first-do-i-simplify-first.html | # Thread:
1. ## Do i factor first or do i simplify first?
$\frac{x^2+6x+5}{x^2-25}$
So do i simplify or factor first?
2. Originally Posted by eh501
$\frac{x^2+6x+5}{x^2-25}$
So do i simplify or factor first?
As I can't see a way to simplify this (unless you want to do a division problem) then I would factor first.
-Dan
3. Ok, so i would get $(x+5)(x+1)over(x+5)(x-5)$
Do i just leave it like that?
4. Originally Posted by eh501
Ok, so i would get $(x+5)(x+1)over(x+5)(x-5)$
Do i just leave it like that?
No, now you have the problem in a form where you have a common x + 5 in the numerator and denominator. NOW divide. But remember that you can't divide by 0, so at the end you must write the line $x \neq -5$.
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9426408410072327, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/63561?sort=newest | ## Fibonacci, compositions, history
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
There are three basic families of restricted compositions (ordered partitions) that are enumerated by the Fibonacci numbers (with offsets):
a) compositions with parts from {1,2} (e.g., 2+2 = 2+1+1 = 1+2+1 = 1+1+2 = 1+1+1+1)
b) compositions that do not have 1 as a part (e.g., 6 = 4+2 = 3+3 = 2+4 = 2+2+2)
c) compositions that only have odd parts (e.g., 5 = 3+1+1 = 1+3+1 = 1+1+3 = 1+1+1+1+1)
The connection between (a) & the Fibonacci numbers traces back to the analysis of Vedic poetry in the first millennium C.E., at least (Singh, Hist. Math. 12, 1985).
Cayley made the connection to (b) in 1876 (Messenger of Mathematics).
Who first established the connection with (c), odd-part compositions? It was known by 1969 (Hoggatt & Lind, Fib. Quart.), but I suspect it was done before that. Thanks for any assistance, especially with citations.
BOUNTY! Not sure how much this incentivizes responses, but it would be nice to figure this out. By the way, I have queried Art Benjamin, Neville Robbins, and Doug Lind about this (Doug modestly mentioned of the 1969 article It's even possible this was an original result.'').
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Did you look in Hardy and Wright? I'm not at work so don't have access to my copy, but my memory is that they talk about odd part compositions in their chapter on partitions, and their references (at the end of each section) are sometimes short but often to the point. – Kevin Buzzard Apr 30 2011 at 22:26
Thanks, Kevin. I don't find anything about compositions in Hardy & Wright (I have the fifth edition). All of these restricted compositions are discussed in Heubach & Mansour's 2010 Combinatorics of Compositions and Words, but their history does not go back very far (nor do they claim to trace back to the very first sources). – Brian Hopkins Apr 30 2011 at 23:04
## 1 Answer
I make this an answer instead of a comment so as to bring it to the attention of others.
Using Google Books search for compositions "odd parts" Fibonacci brings up several modern combinatorics texts which might have a reference for the result. It also brings up a 1961 Canadian journal which only has a snippet view, but which might yield useful information. If the poster is still interested in tracking down the source, the search results may prove fruitful, especially as they may not have been available at the time of the first posting. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9498630166053772, "perplexity_flag": "middle"} |
http://mathematica.stackexchange.com/questions/11886/powersrepresentations-algorithm?answertab=active | # PowersRepresentations Algorithm
I'm trying to understand the mathematics behind counting the number of representations of a positive integer by $n$ distinct $k$th powers, i.e. I would really like to know how to do the Mathematica function PowersRepresentations by hand. Any explanations, references, even special cases (I'm particularly interested in $k=2$) would be appreciated. Thanks.
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Have you seen this and this? For the sum of squares case, see this. – J. M.♦ Oct 11 '12 at 2:30
3
I feel like this is more a mathematics question, than it is a Mathematica one. Or do you specifically want to know how Mathematica is calculating power represantations? – jenson Oct 11 '12 at 4:45
lang-mma | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9426831603050232, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/178384/evaluating-int-13-frac-lnx2x22x15-dx?answertab=votes | # Evaluating $\int_1^3\frac{\ln(x+2)}{x^2+2x+15} \ dx$
Could you please give me a hint on how to compute:
$$\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx$$
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2
– Harald Hanche-Olsen Aug 3 '12 at 12:44
1
@KevinCarlson It's possible that they are supposed to use some numerical method, such as Simpson's Rule or some quadrature? If this is the case then the OP needs to confirm. – Shaktal Aug 3 '12 at 12:49
1
:-)) Hi, thanks for the quick replies. Not a student anymore, it's just a hobby of mine (not a wise hobby choice, huh?). I've spent a lot of time on it and I have to admit that I don't like infinite series that much (at least this is what I believe the solution is heading to). – Cristian Bujor Aug 3 '12 at 12:54
1
Hint: Apply partial fractions on the denominator. You'll get complex roots though, and the resulting integrals look easier but still have no elementary solution. They're known though, as the dilogarithm Harald mentioned. – Dario Aug 3 '12 at 13:25
6
There is an elementary solution. After simplifications, Mathematica gives : $$\frac{\pi \log (3)+\log (5) \tan ^{-1}\left(\frac{20 \sqrt{14}}{31}\right)-\log (9) \tan ^{-1}\left(4 \sqrt{14}\right)-2 \log \left(\frac{5}{3}\right) \cot ^{-1}\left(\sqrt{14}\right)}{4 \sqrt{14}}$$ – M. M. Aug 3 '12 at 14:53
show 13 more comments
## 3 Answers
It is not really a simple integral (even if nothing special happens in the range $(1,3)$). Sasha gave a fine approximation (+1) let's provide the dilogarithm answer...
Let's start by factoring the denominator $\ x^2+2x+15$ :
The reduced discriminant is $\Delta=1-1\cdot 15=-14\$ so that that it will have two complex conjugate solutions : $\ a=-1-i\sqrt{14}\$ and $\ \overline{a}=-1+i\sqrt{14}$
Let's rewrite a little the integral : $$I:=\int_1^3\frac{\ln(x+2)}{x^2+2x+15}dx=\int_1^3\frac{\ln(x+2)}{(x-a)(x-\overline{a})}dx$$ $$I=\frac 1{a-\overline{a}}\int_1^3 \left(\frac {\ln(x+2)}{x-a}-\frac {\ln(x+2)}{x-\overline{a}}\right)dx=\frac {I1-I2}{a-\overline{a}}$$
The (promised!) dilogarithm function looks like : $$\operatorname{Li}_2(z)=-\int_0^z \frac {\ln(1-t)}t dt$$
Let's rewrite the first part of our integral the same way : $$I1=\int_1^3 \frac {\ln(x+2)}{x-a} dx=\int_1^3 \frac {\ln(x-a+a+2)}{x-a} dx$$ $$I1=\int_1^3 \frac {\ln((a+2)(\frac{x-a}{a+2}+1)}{x-a} dx=\int_1^3 \frac {\ln(a+2)+\ln(1-\frac{a-x}{a+2})}{x-a} dx$$ $$I1=\left[\ln(x-a)\ln(a+2)\right]_1^3 -\int_1^3 \frac {\ln(1-\frac{a-x}{a+2})}{a-x} dx$$ set $\ \displaystyle t:=\frac{a-x}{a+2}$ (so that $\displaystyle \frac {dt}t=-\frac{dx}{a-x}$) to get : $$I1=\left[\ln(x-a)\ln(a+2)\right]_1^3 -\int_{\frac{a-1}{a+2}}^{\frac{a-3}{a+2}} \frac {\ln(1-t)}{t} (-dt)=\left[\ln(x-a)\ln(a+2)-\operatorname{Li}_2\left(\frac{a-x}{a+2}\right)\right]_1^3$$
Of course the second part of the integral will be : $$I2=\int_1^3 \frac {\ln(x+2)}{x-\overline{a}} dx=\left[\ln(x-\overline{a})\ln(\overline{a}+2)-\operatorname{Li}_2\left(\frac{\overline{a}-x}{\overline{a}+2}\right)\right]_1^3$$
So that your integral should be (with $\ a=-1-i\sqrt{14}\$ and $\ \overline{a}=-1+i\sqrt{14}$) : $$I=\frac {\left[\ln(x-a)\ln(a+2)-\ln(x-\overline{a})\ln(\overline{a}+2)+\operatorname{Li}_2\left(\frac{\overline{a}-x}{\overline{a}+2}\right)-\operatorname{Li}_2\left(\frac{a-x}{a+2}\right)\right]_1^3}{a-\overline{a}}\approx 0.11865036886767$$
EDIT (the last part was corrected $a$ had been replaced by $x$ in $\ln(a+2)$)
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Raymond, your solution is most appreciated, especially for introducing me to the 'Liz'/dilogarithm function :) Thank you – Cristian Bujor Aug 3 '12 at 16:32
@CristianBujor: Thanks Cristian you are welcome! Perhaps that with a little more time you'll see a derivation of Mathematica's result shown by M. Mayrand (it is exact too and elementary!). Anyway each of these efforts represent a 'tour de force' (I think too that something simpler was asked...). Wishing you much fun with Mathematics, – Raymond Manzoni Aug 3 '12 at 16:54
First, let's make a change of variables, $x=3 + u$, which maps $(1,3)$ into $(-1,1)$: $$\int_1^3 \frac{\log(x+2)}{x^2+x+15}\mathrm{d} x = \int_{-1}^1 \frac{\log(4+u)}{u^2 + 6 u+ 23}\mathrm{d} u$$ Now use $\log(4+u) = \log(4) + \log\left(1+\frac{u}{4}\right)$: $$\begin{eqnarray} \int_1^2 \frac{\log(1+2 u)}{2 u^2+7}\mathrm{d} u &=& \int_{-1}^1 \frac{\log(4)}{u^2 + 6 u+ 23}\mathrm{d} u + \int_{-1}^1 \frac{\log\left(1+\frac{u}{4}\right)}{u^2 + 6 u+ 23}\mathrm{d} u \end{eqnarray}$$ The first integral is trivially evaluated by completing the squares in the denominator: $u^2+6 u+23 = (u+3)^2 + 14$, giving: $$I_0 = \int_{-1}^1 \frac{\log(4)}{u^2 + 6 u+ 23}\mathrm{d} u = \frac{\log(4)}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) \approx 0.121478$$ This already gives a good approximation to the correct value ($\approx 0.118650$). The second integral can be done expanding logarithm in a series, and integrating term-wise. The first term: $$\Delta_1 = \int_{-1}^1 \frac{u/4}{u^2 + 6 u+ 23}\mathrm{d} u = -\frac{3}{4} \frac{1}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) + \frac{1}{8} \log\left(\frac{5}{3}\right) \approx -0.001868$$ The second: $$\Delta_2 = -\frac{1}{2} \int_{-1}^1 \frac{(u/4)^2}{u^2 + 6 u+ 23}\mathrm{d} u = -\frac{1}{16} + \frac{5}{32} \frac{1}{\sqrt{14}} \arctan\left(\frac{\sqrt{14}}{11} \right) + \frac{3}{32} \log\left(\frac{5}{3}\right) \approx -0.000918$$ Combining, $I_0 + \Delta_1+\Delta_2 = 0.118692$ which gives a good approximation.
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Hi Sasha, this is solid effort what you've done here. I am very grateful for this and it amazes me how passionate people on this site are. Many thanks, indeed! One question, though: what do you mean by "correct value (~0.118650)"? The "correct value" is what we would like to find out, right? How have you reached to that value? – Cristian Bujor Aug 3 '12 at 14:54
@CristianBujor By correct value, I mean the value obtained by either quadratures, or by numerically approximating the exact expression in terms of dilogarithms that, say, Mathematica produces. – Sasha Aug 3 '12 at 14:58
I really like your clean, practical approach. Thanks for this! – Cristian Bujor Aug 3 '12 at 16:27
This was supposed to be a comment on Raymond's answer, but it got too long. I started with trying to obtain Mayrand's fine expression from the dilogarithmic mess one might obtain through Mathematica or Raymond's route, but wound up with a satisfactorily simple expression.
We start from a version of Raymond's answer with the "elementary portion" already simplified:
$\begin{split} \frac1{2\sqrt{14}}&\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)+\\ &\frac{i}{2\sqrt{14}}\left(\mathrm{Li}_2\left(\frac{2+i\sqrt{14}}{3}\right)+\mathrm{Li}_2\left(\frac{4-i\sqrt{14}}{5}\right)-\right.\\ &\left.\left(\mathrm{Li}_2\left(\frac{2-i\sqrt{14}}{3}\right)+\mathrm{Li}_2\left(\frac{4+i\sqrt{14}}{5}\right)\right)\right)\end{split}$
I grouped the terms in this way, since this allows the easy application of Landen's identity (see this paper for a survey of the various algebraic dilogarithm identities):
$$\mathrm{Li}_2(x)+\mathrm{Li}_2\left(\frac{x}{x-1}\right)=-\frac12\left(\log(1-x)\right)^2$$
Now, we have the relations
$$\frac{\frac{2\pm i\sqrt{14}}{3}}{\frac{2\pm i\sqrt{14}}{3}-1}=\frac{4\mp i\sqrt{14}}{5}$$
which when used with Landen's identity yields
$$\begin{split} &\frac1{2\sqrt{14}}\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)+\\ &\frac{i}{2\sqrt{14}}\left(\left(-\frac12\left(\log\left(1-\frac{2+i\sqrt{14}}{3}\right)\right)^2\right)-\left(-\frac12\left(\log\left(1-\frac{2-i\sqrt{14}}{3}\right)\right)^2\right)\right) \end{split}$$
which, after a few more algebraic manipulations, finally yields
$$\frac1{2\sqrt{14}}\left(\log\frac53\;\arctan\sqrt{14}+\log\,15\;\arctan\frac{\sqrt{14}}{11}\right)-\frac{\arctan\sqrt{14}\log\frac53}{2\sqrt{14}}=\color{blue}{\frac1{2\sqrt{14}}\log\,15\;\arctan\frac{\sqrt{14}}{11}}$$
which is even simpler than Mayrand's original result.
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Very nice. Now, does Landen only apply because the integral ran from 1 to 3? Or does it also apply to the indefinite integral? – Gerry Myerson Aug 4 '12 at 5:49
+1 Nice work J.M. you bet me again ! :-) The quickest way I found to get the same simple result was to start with this WA answer !int2 evaluating this integral from $2$ to $4$ (since $x+2\to x+1$) and using the identity $\rm{Li}_2(x)+\rm{Li}_2(1/x)=\cdots$ to get !rewriting that simplifies as !final – Raymond Manzoni Aug 4 '12 at 7:24
@GerryMyerson: I think that it works only for very specific values (to be able to use a appropriate polylogarithm identity). – Raymond Manzoni Aug 4 '12 at 7:28
@Gerry, the first one; here, it seems the definite integral's limits were fortuitously chosen. It makes me wonder if there's a route that doesn't require a dilogarithmic detour... – J. M. Aug 4 '12 at 7:33
@J.M., that's exactly what I was wondering. But if it really only works because of the limits, that means there's no route to the indefinite integral bypassing dilogs. – Gerry Myerson Aug 4 '12 at 10:37
show 2 more comments | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 20, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9324014782905579, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/170241/when-is-matrix-multiplication-commutative | # When is matrix multiplication commutative?
I know that matrix multiplication in general is not commutative. So, in general:
$A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A$
But for some matrices, this equations holds, e.g. A = Identity or A = Null-matrix $\forall B \in \mathbb{R}^{n \times n}$.
I think I remember that a group of special matrices (was it $O(n)$, the group of orthogonal matrices?) exist, for which matrix multiplication is commutative.
For which matrices $A, B \in \mathbb{R}^{n \times n}$ is $A\cdot B = B \cdot A$?
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4
A sufficient condition is that $A$ and $B$ are simultaneously diagonalizable. – Johannes Kloos Jul 13 '12 at 8:41
– J. M. Jul 13 '12 at 8:57
2
$SO(2)$ is commutative, but that is more of an accident than a general property of (special) orthogonal groups. – Henning Makholm Jul 13 '12 at 10:50
@JohannesKloos That was what I was looking for. If you post it as an answer, I'll accept it. – moose Jul 13 '12 at 13:58
I'm not aware of any general statement in this case. – Johannes Kloos Jul 13 '12 at 15:19
show 5 more comments
## 5 Answers
Two matrices that are simultaneously diagonalizable are always commutative.
Proof: Let $A$, $B$ be two such $n \times n$ matrices over a base field $\mathbb K$, $v_1, \ldots, v_n$ a basis of Eigenvectors for $A$. Since $A$ and $B$ are simultaneously diagonalizable, such a basis exists and is also a basis of Eigenvectors for $B$. Denote the corresponding Eigenvalues of $A$ by $\lambda_1,\ldots\lambda_n$ and those of $B$ by $\mu_1,\ldots,\mu_n$.
Then it is known that there is a matrix $T$ whose columns are $v_1,\ldots,v_n$ such that $T^{-1} A T =: D_A$ and $T^{-1} B T =: D_B$ are diagonal matrices. Since $D_A$ and $D_B$ trivially commute (explicit calculation shows this), we have $$AB = T D_A T^{-1} T D_B T^{-1} = T D_A D_B T^{-1} T D_B D_A = \cdots = BA.$$
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Among the groups of orthogonal matrices $O(n,\mathbb R)$, only the case $n=0$ (the trivial group) and $n=1$ (the two element group) give commutative matrix groups. The group $O(2,\mathbb R)$ consists of plane rotations and reflections, of which the former form an index $2$ commutative subgroup, but reflections do not commute with rotations or among each other in general. The largest commutative subalgebras of square matrices are those which are diagonal on some fixed basis; these subalgebras only have dimension $n$, out of an available $n^2$, so commutation is really quite exceptional among $n\times n$ matrices (at least for $n\geq2$). Nothing very simple can be said that (non-tautologically) characterises all commuting pairs of matrices.
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The orthogonal matrices don't commute, infact there's a subspace of the orthogonals that's non-commutative!
Check that a permutation matrix is an orthogonal matrix (In case you don't know what a permutation matrix is, it's just a matrix $(a_{ij})$ such that a permutation $\sigma$ exists for which $a_{i,\sigma(i)}=1$ and $a_{ij}=0$ for $j\ne\sigma(i)$
Applying to a column vector $x$ the action of the permutation matrices is just permutation of the co-ordinates of $x$. But as we know the symmetry group is non-abelian. So just choose two non-commuting permutations and their corresponding matrices clearly don't commute!
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The only matrices that commutes with all other matrices are the multiples of the identity.
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So there is no group of Matrix pairs that commute. It is $$AB = BA$$ if and only if there is a polynomial $$p \in \mathbb{R}[x]$$ such that $$p(A)=B.$$ This can be proven using Jordan Normalform or by simple computing.
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What does "there is no group of matrix pairs that commute" mean? – user23211 Jul 13 '12 at 10:00
2
There is a small bit of truth here. IIRC, over an algebraically closed field, every matrix that commutes with $A$ is a polynomial in $A$ if and only if every eigenvalue of $A$ has geometric multiplicity $1$. – Robert Israel Jul 13 '12 at 15:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9267473220825195, "perplexity_flag": "head"} |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/D03/d03puc.html | # NAG Library Function Documentnag_pde_parab_1d_euler_roe (d03puc)
## 1 Purpose
nag_pde_parab_1d_euler_roe (d03puc) calculates a numerical flux function using Roe's Approximate Riemann Solver for the Euler equations in conservative form. It is designed primarily for use with the upwind discretization schemes nag_pde_parab_1d_cd (d03pfc), nag_pde_parab_1d_cd_ode (d03plc) or nag_pde_parab_1d_cd_ode_remesh (d03psc), but may also be applicable to other conservative upwind schemes requiring numerical flux functions.
## 2 Specification
#include <nag.h>
#include <nagd03.h>
void nag_pde_parab_1d_euler_roe (const double uleft[], const double uright[], double gamma, double flux[], Nag_D03_Save *saved, NagError *fail)
## 3 Description
nag_pde_parab_1d_euler_roe (d03puc) calculates a numerical flux function at a single spatial point using Roe's Approximate Riemann Solver (see Roe (1981)) for the Euler equations (for a perfect gas) in conservative form. You must supply the left and right solution values at the point where the numerical flux is required, i.e., the initial left and right states of the Riemann problem defined below.
In the functions nag_pde_parab_1d_cd (d03pfc), nag_pde_parab_1d_cd_ode (d03plc) and nag_pde_parab_1d_cd_ode_remesh (d03psc), the left and right solution values are derived automatically from the solution values at adjacent spatial points and supplied to the function argument numflx from which you may call nag_pde_parab_1d_euler_roe (d03puc).
The Euler equations for a perfect gas in conservative form are:
$∂U ∂t + ∂F ∂x = 0 ,$ (1)
with
(2)
where $\rho $ is the density, $m$ is the momentum, $e$ is the specific total energy, and $\gamma $ is the (constant) ratio of specific heats. The pressure $p$ is given by
$p=γ-1 e-ρu22 ,$ (3)
where $u=m/\rho $ is the velocity.
The function calculates the Roe approximation to the numerical flux function $F\left({U}_{L},{U}_{R}\right)=F\left({U}^{*}\left({U}_{L},{U}_{R}\right)\right)$, where $U={U}_{L}$ and $U={U}_{R}$ are the left and right solution values, and ${U}^{*}\left({U}_{L},{U}_{R}\right)$ is the intermediate state $\omega \left(0\right)$ arising from the similarity solution $U\left(y,t\right)=\omega \left(y/t\right)$ of the Riemann problem defined by
$∂U ∂t + ∂F ∂y =0,$ (4)
with $U$ and $F$ as in (2), and initial piecewise constant values $U={U}_{L}$ for $y<0$ and $U={U}_{R}$ for $y>0$. The spatial domain is $-\infty <y<\infty $, where $y=0$ is the point at which the numerical flux is required. This implementation of Roe's scheme for the Euler equations uses the so-called argument-vector method described in Roe (1981).
## 4 References
LeVeque R J (1990) Numerical Methods for Conservation Laws Birkhäuser Verlag
Quirk J J (1994) A contribution to the great Riemann solver debate Internat. J. Numer. Methods Fluids 18 555–574
Roe P L (1981) Approximate Riemann solvers, parameter vectors, and difference schemes J. Comput. Phys. 43 357–372
## 5 Arguments
1: uleft[$3$] – const doubleInput
On entry: ${\mathbf{uleft}}\left[\mathit{i}-1\right]$ must contain the left value of the component ${U}_{\mathit{i}}$, for $\mathit{i}=1,2,3$. That is, ${\mathbf{uleft}}\left[0\right]$ must contain the left value of $\rho $, ${\mathbf{uleft}}\left[1\right]$ must contain the left value of $m$ and ${\mathbf{uleft}}\left[2\right]$ must contain the left value of $e$.
Constraints:
• ${\mathbf{uleft}}\left[0\right]\ge 0.0$;
• Left pressure, $\mathit{pl}\ge 0.0$, where $\mathit{pl}$ is calculated using (3).
2: uright[$3$] – const doubleInput
On entry: ${\mathbf{uright}}\left[\mathit{i}-1\right]$ must contain the right value of the component ${U}_{\mathit{i}}$, for $\mathit{i}=1,2,3$. That is, ${\mathbf{uright}}\left[0\right]$ must contain the right value of $\rho $, ${\mathbf{uright}}\left[1\right]$ must contain the right value of $m$ and ${\mathbf{uright}}\left[2\right]$ must contain the right value of $e$.
Constraints:
• ${\mathbf{uright}}\left[0\right]\ge 0.0$;
• Right pressure, $\mathit{pr}\ge 0.0$, where $\mathit{pr}$ is calculated using (3).
3: gamma – doubleInput
On entry: the ratio of specific heats, $\gamma $.
Constraint: ${\mathbf{gamma}}>0.0$.
4: flux[$3$] – doubleOutput
On exit: ${\mathbf{flux}}\left[\mathit{i}-1\right]$ contains the numerical flux component ${\stackrel{^}{F}}_{\mathit{i}}$, for $\mathit{i}=1,2,3$.
5: saved – Nag_D03_Save *Communication Structure
saved may contain data concerning the computation required by nag_pde_parab_1d_euler_roe (d03puc) as passed through to numflx from one of the integrator functions nag_pde_parab_1d_cd (d03pfc), nag_pde_parab_1d_cd_ode (d03plc) or nag_pde_parab_1d_cd_ode_remesh (d03psc). You should not change the components of saved.
6: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_BAD_PARAM
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_REAL
Left pressure value $\mathit{pl}<0.0$: $\mathit{pl}=〈\mathit{\text{value}}〉$.
On entry, ${\mathbf{gamma}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{gamma}}>0.0$.
On entry, ${\mathbf{uleft}}\left[0\right]=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{uleft}}\left[0\right]\ge 0.0$.
On entry, ${\mathbf{uright}}\left[0\right]=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{uright}}\left[0\right]\ge 0.0$.
Right pressure value $\mathit{pr}<0.0$: $\mathit{pr}=〈\mathit{\text{value}}〉$.
## 7 Accuracy
nag_pde_parab_1d_euler_roe (d03puc) performs an exact calculation of the Roe numerical flux function, and so the result will be accurate to machine precision.
## 8 Further Comments
nag_pde_parab_1d_euler_roe (d03puc) must only be used to calculate the numerical flux for the Euler equations in exactly the form given by (2), with ${\mathbf{uleft}}\left[\mathit{i}-1\right]$ and ${\mathbf{uright}}\left[\mathit{i}-1\right]$ containing the left and right values of $\rho ,m$ and $e$, for $\mathit{i}=1,2,3$, respectively. It should be noted that Roe's scheme, in common with all Riemann solvers, may be unsuitable for some problems (see Quirk (1994) for examples). In particular Roe's scheme does not satisfy an ‘entropy condition’ which guarantees that the approximate solution of the PDE converges to the correct physical solution, and hence it may admit non-physical solutions such as expansion shocks. The algorithm used in this function does not detect or correct any entropy violation. The time taken is independent of the input arguments.
## 9 Example
See Section 9 in nag_pde_parab_1d_cd_ode (d03plc). | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 71, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.580401599407196, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/194735-unique-minimum-simple-function.html | # Thread:
1. ## Unique minimum of a simple function
Hi,
I want to prove that the following function has a unique minimum … and I'm surprised at how hard this is!
$f(x,y) = \frac{\frac{a}{x} + \frac{b}{y} + c}{d - e(x+y)} + \frac{u}{x} + \frac{v}{y}$
The scalars $a,b,c,d,e,u,v$ are all positive. The function is defined on the open triangle $x > 0$, $y > 0$, $x+y < \frac{d}{e}$.
Can you think of some simple proof?
I've tried a few things:
– take the logarithm
– try to show quasi-convexity (which is a stronger result) by several means
– try to rule out saddle points (see my post http://www.mathhelpforum.com/math-he...ts-194122.html and answer by xxp9)
– try to identify a composition of elementary functions
…to no avail
Below is a contour plot of the inverse $1/f$ (because it's nicer to plot than $f$). The function tends to zero on the border of the triangle. You clearly see there's a unique maximum of $1/f$ (minimum of $f$) somewhere inside the triangle.
2. ## Re: Unique minimum of a simple function
Hi again,
I figured out that my function $f$ is actually convex. One can show in a few steps that $f$ is a sum of functions that are all convex on the mentioned triangle.
For some reason (which I won't explain in detail) I had convinced myself that $f$ couldn't be convex. I failed to see the wood for the trees. You can consider this issue as closed (sorry for posting!).
Copyright © 2005-2013 Math Help Forum. All rights reserved. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9435226321220398, "perplexity_flag": "middle"} |
http://floerhomology.wordpress.com/2012/09/22/automatic-transversality-for-dummies/ | ## Automatic transversality for dummies
Posted on September 22, 2012 by
In my last post and in the comments I misremembered and garbled various things about automatic transversality. Let me try to clean things up here.
As Chris Wendl pointed out, a correct statement is the following. Let $X$ be a (completed) four-dimensional strong symplectic cobordism between contact manifolds, and let $J$ be an almost complex structure on $X$ satisfying the usual conditions. If $C$ is a $J$-holomorphic curve in $X$, let $g$ denote the genus of $C$, let $h_+$ denote the number of ends of $C$ at positive hyperbolic orbits, and let $ind$ denote the Fredholm index of $C$.
Theorem. Let $C$ be an immersed $J$-holomorphic curve in $X$ with ends at nondegenerate Reeb orbits. Suppose that $2g-2+h_+<ind$. Then $C$ is automatically transverse (regardless of whether or not $J$ is generic).
Chris Wendl has a paper proving more general versions of this in which $C$ does not have to be immersed and can have ends at Morse-Bott Reeb orbits. (I have used the Morse-Bott version before.) But let me just explain the proof of the above, in my notation.
Proof. To simplify notation, assume all ends of $C$ are positive; it is easy to fix the notation when this is not the case.
Since $C$ is immersed, it has a well-defined normal bundle $N$, and we can regard the linearized Cauchy Riemann operator $D$ as a map from sections of $N$ to sections of $T^{0,1}C\otimes N$. Suppose that $C$ is not transverse. Then the formal adjoint $D^*$ has a nonzero kernel. Let $\psi$ be a section of $T^{0,1}C\otimes N$ which is an element of $Ker(D^*)$. It is a standard fact that every zero of $\psi$ is isolated and has negative multiplicity. (This follows for example from the Carleman similarity principle.) Also the zero set of $\psi$ is compact (this follows from stuff about the asymptotics of $\psi$ which I can review later). So the algebraic count of zeros of $\psi$ is negative:
$\#\psi^{-1}(0) \le 0.$
Now the algebraic count of zeroes of $\psi$ is given in terms of the relative first Chern class of $T^{0,1}C\otimes N$ by the formula
$\#\psi^{-1}(0) = c_1(T^{0,1}C\otimes N,\tau) + wind_\tau(\psi).$
Here $\tau$ is a trivialization of $T^{0,1}C\otimes N$ over the ends of $C$, and $wind_\tau(\psi)$ denotes the sum of the winding numbers of $\psi$ around the ends with respect to the trivialization $\tau$. Indeed, one can regard this formula as the definition of the relative first Chern class $c_1(\cdot,\tau)$.
We can take the trivialization $\tau$ of $T^{0,1}C\otimes N$ over the ends to be induced by a trivialization of the contact plane field $\xi$ over the ends, which we also denote by $\tau$, together with the canonical trivialization of $T^{0,1}C$ over the ends. Then
$c_1(T^{0,1}\Sigma\otimes N,\tau) = \chi(C) + c_1(N,\tau) = c_1(TX|_C,\tau).$
It also follows from the asymptotics of $\psi$ which I didn’t explain here that
$wind_\tau(\psi) \ge \sum_i \lceil CZ_\tau(\gamma_i)/2\rceil.$
Here $i$ indexes the ends of $C$, $\gamma_i$ denotes the Reeb orbit at the $i^{th}$ end of $C$, and $CZ_\tau$ denotes the Conley-Zehnder index with respect to $\tau$. Note that $CZ_\tau(\gamma_i)$ is even if and only if $\gamma_i$ is positive hyperbolic. So if $k$ denotes the total number of ends of $C$, we can rewrite the above inequality as
$2 wind_\tau(\psi) \ge \sum_i CZ_\tau(\gamma_i) + k - h_+.$
Next recall that the Fredholm index is given by the formula
$ind = -\chi(C)+ 2c_1(TX|_C,\tau) + \sum_i CZ_\tau(\gamma_i).$
We now put everything together as follows:
$0 \ge 2\#\psi^{-1}(0)$
$= 2c_1(T^{0,1}C\otimes N,\tau) + 2wind_\tau(\psi)$
$= 2c_1(TX|_C,\tau) + 2wind_\tau(\psi)$
$\ge 2c_1(TX|_C,\tau) + \sum_i CZ_\tau(\gamma_i) + k - h_+$
$= ind + 2 - 2g - h_+.$
So $0 \ge ind + 2 - 2g - h_+$ if $C$ is not transverse. QED.
It is easy to see that automatic transversality can fail for an index $0$ cylinder with both ends at positive hyperbolic orbits (contrary to a foolish claim that I made in the comments to the previous post). Just think about the symplectic fixed point Floer homology for the indentity map on a surface, perturbed by a Morse function. (This is not quite the contact geometry setting but it is close.) It is possible to have a flow line between two index $1$ critical points (e.g. for the height function on a torus standing on its side), and this gives rise to a holomorphic cylinder of index zero whose ends are at positive hyperbolic orbits.
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### 5 Responses to Automatic transversality for dummies
1. Chris Wendl says:
Sorry, I still have a minor quibble.
I don’t think the example at the end shows that automatic transversality fails for index 0 cylinders with two positive hyperbolic orbits. In fact, there’s no situation I can think of in which automatic transversality implies the non-existence of certain holomorphic curves: for the cylinders in question, the necessary criterion is not satisfied so there is simply no conclusion.
But I like the example anyway because it does give a good hint of the fact that hyperbolic orbits cause problems while elliptic orbits do not: there can be no index 0 gradient flow lines between two critical points of index 0 or 2, and if one didn’t already see that from the fact that those critical points are either maxima or minima, one could PROVE it using the fact that index 0 cylinders with two elliptic orbits DO satisfy automatic transversality, i.e. if such a gradient line existed, then the corresponding cylinder would have to be Fredholm regular, which it clearly is not since it is not isolated due to R-invariance.
• floerhomology says:
I’m not sure I understand your quibble, but I think we’re saying the same thing. What I am saying is that nontrivial index 0 cylinders with both ends at positive hyperbolic orbits are not automatically transverse (because sometimes they exist in which case they are necessarily not transverse). Of course the automatic transversality theorem that I reviewed does not apply here because $2g-2+h_+ = ind$.
• Chris Wendl says:
Ah, I see… yes, quite right, I withdraw my quibble!
2. Paolo says:
Hi Michael,
Could you explain why the count of zeroes of $\psi$ is negative? I know that elements of the cokernel are solution of the Cauchy-Riemann type equation on $T^{0,1} C \otimes N$, so shouldn’t the count of zeroes be positive? Sorry for being stupid (or ignorant) here.
• floerhomology says:
In a local complex coordinate $z=s+it$ and a local trivialization of $N$, the operator $D$ has the form $\partial_s + i \partial_t + A(s,t)$ where $A(s,t)$ is a $2\times 2$ matrix. The formal adjoint $D^*$ then has the local form $-\partial_s + i\partial_t + A(s,t)^*$. The symbol of this operator is that of a conjugate Cauchy-Riemann operator, so its zeroes have negative multiplicity instead of positive. That is, if the zeroth order term $A(s,t)$ were not there, then $\psi$ would be in the kernel of $D^*$ if and only if $\overline{\psi}$ is holomorphic; and the Carleman similarity principle basically says that you can locally change the coordinates and trivializations to make the zeroth order term disappear, so that this special case implies the general case.
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http://mathoverflow.net/questions/121297/what-kind-of-subset-is-specr-p-in-specr/121303 | ## What kind of subset is Spec(R_P) in Spec(R)?
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While trying to prove some properties of a subset of a prime spectrum I arrived at the following question: Let $R$ be a commutative ring and let $P \in \mathrm{Spec}(R)$. We can consider $\mathrm{Spec}(R_P)$ as a subset of $\mathrm{Spec}(R)$. My question is: What kind of subset (subspace) is this? I guess it is not true that $\mathrm{Spec}(R_P)$ is open, right? Is it true that a neighborhood of the generic point in $\mathrm{Spec}(R_P)$ is a neighborhood of $P$ in $\mathrm{Spec}(R)$?
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Isn't Spec$(R_P)$ (considered as a subset of Spec$(R)$) the complement of $V(P)$? That would make it open. (This question seems a bit elementary for MO.) – Joe Silverman Feb 9 at 13:08
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Well, V(P) contains all $Q \supset P$, so its complement is the set of all $Q \not\supset P$. But $Spec(R_P)$ consists of all $Q$ with $Q \cap (R \setminus P) = \emptyset$, so $Q \subset P$. This is not the complement of $V(P)$. Am I wrong? This would be great. :-) – Georg S. Feb 9 at 13:13
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@Georg: You are of course right, since spectra are not necessarily totally ordered by inclusion. – Fred Rohrer Feb 9 at 13:14
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$\mathrm{Spec}(R_f)$ is the complement of $V(f)$ (for $f \in R$), but $\mathrm{Spec}(R_P)$ is not the complement of $V(P)$ (for $P \in \mathrm{Spec}(R)$). – Martin Brandenburg Feb 9 at 18:17
## 5 Answers
As already said, in general the image is not open and even not constructible. But it is pro-constructible (that is, locally an intersection of locally constructible subsets, see EGA IV.1.9.4, and 1.9.5(ix) because $X:=\mathrm{Spec}(R)$ is affine hence quasi-compact).
Denote this image by $S$. This is a subset of $X$ having exactly one closed point and which is stable by generization as pointed out by Martin. I claim that these two properties characterize all possibles images $S$ when $X$ is noetherian (this hypothesis could be weakened if we notice that $S$ is always quasi-compact, but I don't know how exactly). Here $X$ needs not be affine.
Indeed, let $s$ be the unique closed point of $S$. Let us show $S$ is the intersection of all open neighborhoods of $s$ in $X$ (then it is easy to show $S$ is the image of $\mathrm{Spec}(O_{X,s})$). Let $U$ be any open neighborhood of $s$ in $X$. Then $S\cap (X\setminus U)$ is noetherian hence admits a closed point if non-empty. But this would be a closed point of $S$ different from $s$. So $S\cap (X\setminus U)=\emptyset$ and $S\subseteq U$. Conversely, any point $x\in X$ in the intersecion of all $U\ni s$ is a generization of $s$ (otherwise the complementary of `$\overline{\{ x\}}$` is an open neighborhood of $s$ not containing $x$).
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Take $R=\mathbb{Z}$ and $P=0$. Then, ${\rm Spec}(R_P)$, considered canonically as a subset of ${\rm Spec}(R)$, consists precisely of the point $0$. In particular, it is not open in ${\rm Spec}(R)$. Hence, it is not a neighbourhood in ${\rm Spec}(R)$ of $0$, despite $0$ being the generic point of ${\rm Spec}(R_P)$ and ${\rm Spec}(R_P)$ being an open neighbourhood of its generic point in itself.
You might want to have a look at Section I.2.5, especially Proposition I.2.5.2, of EGA [1970 edition] (or Section I.2.4 in the first edition) to get some more general information about your situation.
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This is about the fact that $Spec(R_P)$ is homeomorphic to the "generization" of $P$ in $Spec(R)$? Does this perhaps imply an answer about my neighborhood question? – Georg S. Feb 9 at 13:19
If $X$ is a scheme and $x \in X$, then the canonical morphism $j_x : \mathrm{Spec}(\mathcal{O}_{X,x}) \to X$ is a homeomorphism onto its image. The image consists of those $y \in X$ which generalize $x$, i.e. satisfy $x \in \overline{\{y\}}$. Locally, this is just the classification of prime ideals in the localization.
It follows that that the image is closed under generalizations, and contains $x$. In general, I doubt that anything more can be said.
Every open subset is stable under generalizations. The converse is true for constructible subsets of noetherian sober topological spaces (see here). Chevalley's theorem implies that the image of a morphism of finite presentation is constructible. However, $j_x$ is almost never of finite presentation.
For example, when $x$ is a generic point of $X$, this means that the image of $j_x$ equals $\{x\}$, which is usually neither open nor closed.
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$\text{Spec } R_P$ is by the usual correspondences the set of all primes of $R$ which are contained in $P$. In particular, it is somehow dual to $V(P) =$ the set of primes which are contained in $P$. It is not generally equal to the complement of $V(P)$ plus the point $P$ itself (this is very rare although does occur in all valuation rings as Fred pointed out in the comments). It is not open, $\text{Spec } R_P$ as a subset of $R$ is closed under generization.
For example: I like to think of say $\text{Spec }$ of the local ring of the origin in $\mathbb{A}^2$ as the origin itself, some germ of every curve passing through the origin, and the generic point.
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"...does occur in some valuation rings": This is in fact the case in every valuation ring. – Fred Rohrer Feb 9 at 14:48
It is a limit of open subsets (the subsets containing $P$.) Thus, though it is not set-theoretically an open subset, it behaves scheme-theoretically like an open subset.
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What do you mean by "behaves scheme-theoretically like an open subset."? – Martin Brandenburg Feb 9 at 21:39
e.g. if a property holds on every sufficiently small neighborhood of a point, it usually holds on the local ring. – Will Sawin Feb 9 at 21:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 89, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9580367803573608, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/104673?sort=votes | ## Analyticity of the solutions of PDE
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Let's consider a (partial) differential equation with analytic coefficients. The initial conditions may be non-analytic*.
Is it possible a solution $f$, which is non-analytic at any point of the domain?
*Even if the initial conditions are analytic, it is possible that the solution is not [1].
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You certainly need to assume something about the PDE for this question to have a chance of making sense. For example, your PDE could be $\partial_t u =0$, with initial conditions $u(t=0) = f$.... On the other hand, for the heat equation, for example, a solution immediately becomes analytic, even for very rough initial data.. – Otis Chodosh Aug 14 at 17:28
@Otis: I just wanted to know if it is possible to have pde with analytic coefficients and solutions like that in the question. If the answer depends on the particular pde and the initial conditions, I think that the answer should contain the relevant assumptions, not the question. It is more likely that these conditions are known by the person who knows the answer, not by the person who asks the question. – Cristi Stoica Aug 15 at 12:21
You should probably add some motivation as to what sort of PDE's you are interested in, otherwise I doubt there exists a "good" answser. I took a brief glance at the paper you linked to, and if I understand correctly, the paper gives an example of a system of linear PDE's with analytic coefficients and initial data, but non-analytic solutions. This should be contrasted with the Cauchy-Kowalewski theorem, which says that for a linear PDE (for a single function) analytic data and coefficeints imply analyticity of solutions and ( in light of @Bazin's answer) suprisingly existence. – Otis Chodosh Aug 15 at 23:50
In particular, you don't specify if you're interested in systems, single PDE's, linear PDE's, nonlinear PDE's, etc. It seems that the only place that a nice theory might have a hope of existing is in the setting of linear PDEs for one function. Can you give some more precise details, or motivation for the question? – Otis Chodosh Aug 15 at 23:52
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@Otis: Thanks for the comments. I am interested in the most general settings, systems or singles, linear or nonlinear, elliptic, parabolic, hyperbolic etc. My hope was that if somebody has something to say about the question, in a general or in a particular setting, to be free to say, because I am interested in the general question. So my hope was to receive answers of the form "in general, ...", or, "in the particular case of ... equations, ...". I see that Rafe accuses me of being rude, his interpretation of what I said is wrong, but if I made you feel that I was rude with you I apologize. – Cristi Stoica Aug 16 at 6:57
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## 3 Answers
Points off for the rude response to Otis' sincere attempt to help you!
Deanne is correct, but you can see this simply even with the operator $\partial_{x_1}$ in $\mathbb R^n$. If $u(0, x_2, \ldots, x_n)$ is nowhere analytic, then the solution of $\partial u/ \partial_{x_1} = 0$ is constant in $x_1$ but never analytic in the orthogonal directions.
On the other hand, there is a class of operators known as analytic hypoelliptic (which includes both the Laplacian for any analytic metric and the associated heat operator) for which such a statement is true: $L u = f$ and $f$ analytic implies $u$ analytic.
However as the paper you reference indicates, all sorts of wild things can happen in general; that paper gives an example of a constant coefficient system and analytic initial data'' where the solution is not analytic. Actually, that is a rather nonstandard sort of initial value problem where Cauchy data for one component is posed on one hypersurface and Cauchy data for the second component is posed on a transverse hypersurface.
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Rafe, thanks for a much better answer than mine, especially the example of $\partial_1$. – Deane Yang Aug 16 at 2:02
Thanks for the answer. On the other hand, while I agree that Otis's attempt was sincere, I disagree that my reply was rude at all, and I even don't see what may have appeared to you impolite in what I wrote. I tried to explain better what I need, and what I need differs from particularizing the problem and risking, by adding conditions, to miss what I am interested in, and making the problem too localized. – Cristi Stoica Aug 16 at 6:48
My apologies then. – Rafe Mazzeo Aug 16 at 15:58
Rafe, can you detect analytic hypoellipticity via the symbol? (Am I mistaken in remembering that you can detect hypoellipticity via the symbol?) Cool answer, thanks! – Otis Chodosh Aug 16 at 19:19
@Otis: For constant coefficient operators, Petrowsky proved that analytic hypoellipticity is equivalent to ellipticity of the symbol. He also proved that elliptic operators with analytic coefficients are analytic hypoelliptic. – timur Apr 20 at 19:17
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Take the Lewy operator $\mathcal L$ in $\mathbb R^3_{x,y,t}$ $$\mathcal L=\partial_x+i\partial_y+i(x+iy)\partial_t.$$ There exists a set $S$ of second category in $C^\infty(\mathbb R^3)$ such that for all $f\in S$, the equation $\mathcal L u=f$ has no distribution solution. So even with polynomial (even affine here) coefficients, a complex vector field could have terrible pathologies. A good explanation of this phenomenon was given by Lars Hörmander and developed further by several authors. The symbol of the Hans Lewy operator is $a+ib$ and for each point $X\in \mathbb R^3$, you can find $\Xi\in \mathbb R^3$ such that at $(X,\Xi)$, $$a=b=0\quad\text{and the Poisson bracket {$a,b$} is positive}.$$
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Thank you for this example. If I understand well, this is an equation with no solution at all, not a solution which is non-analytic at any point. – Cristi Stoica Aug 15 at 12:23
This can't happen if the PDE is either elliptic or parabolic. If the PDE is hyperbolic and you start with initial data that is nowhere analytic, then it seems plausible that the resulting solution is also nowhere analytic. But I don't have a rigorous proof of this. The idea would be to assume that there is a point where the solution is analytic in a neighborhood and try to derive a contradiction by propagating the analyticity backwards back to the given intial data.
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Thank you for the answer. It makes sense, intuitively. – Cristi Stoica Aug 16 at 6:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9500694870948792, "perplexity_flag": "head"} |
http://physics.aps.org/articles/v2/87 | # Viewpoint: Spins in cold atoms—what a drag!
, NEST-CNR-INFM and Scuola Normale Superiore, I-56126 Pisa, Italy
, Department of Physics and Astronomy, University of Missouri, Columbia, Missouri 65211, USA
Published October 19, 2009 | Physics 2, 87 (2009) | DOI: 10.1103/Physics.2.87
Interactions among noncondensed bosonic atoms in a trap can cause one species of atoms accelerated by a magnetic field to drag along another species of atoms that would normally not interact with the field.
#### Spin Drag in Noncondensed Bose Gases
R. A. Duine and H. T. C. Stoof
Published October 19, 2009 | PDF (free)
Electron-electron interactions between closely spaced nanoelectronic circuits constitute an alternative coupling mechanism to the inductive and capacitive couplings of conventional electronics. It was realized early on [1, 2] that “Coulomb mutual scattering” between spatially separated electronic systems provides a mechanism for momentum relaxation, which then leads to equalization of drift velocities. This intrinsic friction due to electron-electron interactions is presently referred to as “Coulomb drag” [3, 4, 5] and has first been studied experimentally by Gramila et al. [6] and by Sivan, Solomon, and Shtrikman [7] in semiconductor double quantum wells. Now, in a paper in Physical Review Letters, Rembert Duine and Henk Stoof from Utrecht University in the Netherlands add a fresh approach to a different kind of drag physics—the pulling along of spin-aligned ultracold atoms in an optical trap [8].
In order to illustrate the phenomenon of Coulomb drag, consider two neighboring quantum wells, each hosting a two-dimensional electron gas, and drive a constant current on the two-dimensional ($2D$) electron gas in one of the wells (the “active,” or “drive,” layer). If no current is allowed to flow in the other well (the “passive” layer), an electric field develops whose associated force cancels the frictional drag force exerted by the electrons in the active layer on the electrons in the passive one. The transresistance, defined as the ratio of the induced voltage in the passive layer to the applied current in the drive layer, directly measures the rate $ρD$ at which momentum is transferred from the current-carrying $2D$ electron gas to its neighbor. Coulomb drag is ultimately caused by fluctuations in the density of electrons in each layer since two-dimensional layers with uniformly distributed charge will not exert any frictional forces upon each other [3].
This Coulomb drag concept can be extended to other quantum (and classical) fluids with two (or more) distinguishable species that are able to exchange momentum due to mutual collisions. Spin Coulomb drag [9, 10], for example, refers to the friction between two populations of electrons that are not spatially separated as in the bilayer Coulomb drag described above, but that have different spin orientations. Spin Coulomb drag has been recently observed by Weber et al. [11].
Spin drag has also been studied in the context of two-component trapped cold Fermi gases [12, 13, 14, 15]. Ultracold atomic gases are highly tunable and ideally clean systems in which spin drag might be observed in a truly intrinsic regime. Of course, it must be borne in mind that cold gases are profoundly different from electronic systems due to at least four reasons: (i) atoms are neutral objects and thus do not respond directly to conventional electromagnetic forces; (ii) in the case of atoms, “spin” refers to a combination of electronic and nuclear spin; (iii) spin drag is due to short-range interatomic interactions as opposed to long-range Coulomb interactions; (iv) spin dynamics in these systems occur in an ideally ballistic regime because of the absence of the ordinary complications of solid-state environments such as impurities, phonons, and crystalline defects.
In all the above-mentioned cases of spin drag, the actors that collide in a scattering event are fermions, which obey Fermi-Dirac statistics and the Pauli exclusion principle. The rate $ρD$ at which momentum is transferred is typically proportional to $T2$, thus rapidly vanishing as temperature is decreased. The reason for this behavior is essentially Pauli blocking: most of the states in which a fermion with given pseudospin $σ$ might end up after a collision with another fermion of opposite $σ$ are already occupied, and therefore unavailable. Here, the pseudospin $σ$ could be a layer degree of freedom, a real or a hyperfine spin. The rate $ρD$ thus has to vanish as $T$$→$$0$. The surprising ineffectiveness of fermion-fermion scattering is also the physical basis on which Landau’s theory of normal Fermi liquids is built [16]. Coulomb and spin Coulomb drag in electronic devices, or spin drag in trapped cold Fermi gases, are thus “small” effects in this sense.
Duine and Stoof now propose an ingenious scheme to realize spin motive forces for neutral atoms. They show that atoms loaded in an optical trap and in the presence of a slowly varying time-dependent quadrupolar magnetic field experience a fictitious electromagnetic field in the laboratory frame. In the toroidal geometry proposed in their paper, the fictitious magnetic field is zero and the atoms experience only an azimuthal fictitious electric field in the lab frame, $EmF$, which accelerates all atoms with a finite hyperfine spin projection $mF$ along the torus (see Fig. 1).
The authors study specifically the case in which such a trap is loaded with a spinor Bose gas in the $F=1$ hyperfine spin state (a gas of sodium atoms, for example), with equal densities of atoms in the states with hyperfine spin projection $mF=1$ and $mF=0$. The former atoms are accelerated along the torus by the fictitious electric field $EmF$, while the atoms in the $mF=0$ state do not couple to this field. However, they are accelerated indirectly by the spin drag mechanism stemming from the short-range atomic interactions between the two spin species. The key finding of their paper is that the rate of momentum transfer $ρD$ in this system increases as $T-5/2$ as temperature is lowered, at odds with what happens in Fermi systems, as discussed above. The reason for this dramatic difference is quantum statistics: Pauli blocking here is replaced by Bose enhancement. In other words, scattering to states already occupied by other bosons is enhanced rather than suppressed. This leads to a giant spin drag. Indeed, the estimated magnitude of the spin drag relaxation time $τD$ at a temperature of $400nK$ is of the order of $τD∼0.1s$. Considering that the typical density $n$ of a cold atom ensemble is of the order of $1012cm-3$ and using the atomic mass $M$ of sodium, one can estimate a transresistance $ρD=M/(nτDe2)∼10-5$ $Ωm$ (see Ref. [17]), to be compared with the characteristic value of $10-9$ $Ωm$ for ordinary spin Coulomb drag in a bulk semiconductor [9].
Two other aspects distinguish this work from earlier work on electronic systems. Due to the absence of momentum relaxation, a true steady state can never be reached in the trap. Duine and Stoof thus carry out an analysis of spin dynamics as determined by the application of the azimuthal fictitious electric field for a finite period of time. Moreover, the increase of kinetic energy drives the system out of the regime of applicability of linear-response theory, which has been extensively used to describe Coulomb drag and spin drag in electronic systems. The authors are thus forced to go beyond the linear-response regime.
The work by Duine and Stoof lays a serious foundation for the observation of giant spin drag in an intrinsically ballistic regime. Some aspects of this work may be refined. It would be interesting, for example, to study the spin drag relaxation rate in the case of tight transverse confinement: in such a truly one-dimensional regime, many-body effects beyond the Fermi golden rule, i.e., multiple scattering events, might be relevant. A more extensive study of the experimental procedures needed to determine the relaxation rate is also necessary. Finally, it is easy to envision a wealth of possible future developments of this work. It is surely very tempting to investigate spin drag in cold Fermi gases close to a ferromagnetic critical point [18], as these systems parallel electron-hole bilayers close to exciton condensation, which exhibit anomalous Coulomb drag [19].
## Acknowledgments
M.P.’s research in spin transport is currently funded by the European Community’s Seventh Framework Program (FP7/2007-2013) under grant agreement No. 215368. G.V. acknowledges support from DOE Grant No. DE-FG02-05ER46203.
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2. P. J. Price, Physica 117B, 750 (1983).
3. Lian Zheng and A. H. MacDonald, Phys. Rev. B 48, 8203 (1993).
4. A.-P. Jauho and H. Smith, Phys. Rev. B 47, 4420 (1993).
5. For a review see A. G. Rojo, J. Phys. Condens. Matter 11, R31 (1999).
6. T. J. Gramila, J. P. Eisenstein, A. H. MacDonald, L. N. Pfeiffer, and K. W. West, Phys. Rev. Lett. 66, 1216 (1991).
7. U. Sivan, P. M. Solomon, and H. Shtrikman, Phys. Rev. Lett. 68, 1196 (1992).
8. R. A. Duine and H. T. C. Stoof, Phys. Rev. Lett. 103, 170401 (2009).
9. I. D’Amico and G. Vignale, Phys. Rev. B 62, 4853 (2000); 65, 085109 (2002) 68, 045307 (2003).
10. K. Flensberg, T. S. Jensen, and N. A. Mortensen, Phys. Rev. B 64, 245308 (2001).
11. C. P. Weber, N. Gedik, J. E. Moore, J. Orenstein, J. Stephens, and D. D. Awschalom, Nature 437, 1330 (2005).
12. M. Polini and G. Vignale, Phys. Rev. Lett. 98, 266403 (2007).
13. D. Rainis, M. Polini, M. P. Tosi, and G. Vignale, Phys. Rev. B 77, 035113 (2008).
14. Gao Xianlong, M. Polini, D. Rainis, M. P. Tosi, and G. Vignale, Phys. Rev. Lett. 101, 206402 (2008).
15. G. M. Bruun, A. Recati, C. J. Pethick, H. Smith, and S. Stringari, Phys. Rev. Lett. 100, 240406 (2008).
16. See, for example, G.F. Giuliani, and G. Vignale, Quantum Theory of the Electron Liquid (Cambridge University Press, Cambridge, 2005)[Amazon][WorldCat].
17. Note that the electric charge is used just to create an ohmic equivalent of the relaxation time $τD$, i.e., to create the quantity $ρD$ with physical dimensions of resistance out of the other physical quantities $n$, $M$, and $τD$.
18. G.-B. Jo, Y.-R. Lee, J.-H. Choi, C. A. Christensen, T. H. Kim, J. H. Thywissen, D. E. Pritchard, and W. Ketterle, Science 325, 1521 (2009).
19. G. Vignale and A. H. MacDonald, Phys. Rev. Lett. 76, 2786 (1996); A. F. Croxall, K. Das Gupta, C. A. Nicoll, M. Thangaraj, H. E. Beere, I. Farrer, D. A. Ritchie, and M. Pepper, ibid. 101, 246801 (2008); J. A. Seamons, C. P. Morath, J. L. Reno, and M. P. Lilly, 102, 026804 (2009).
### About the Author: Marco Polini
Marco Polini is a CNR-INFM Researcher at the National Enterprise for nanoScience and nanoTechnology (NEST) in Pisa, Italy. He graduated in 1999 from the University of Pisa and received his Ph.D. in January 2003 from the Scuola Normale Superiore di Pisa. Immediately after, he went to work at the University of Texas at Austin as a visiting postdoctoral fellow until July 2003. In July 2003, he was hired at NEST as a tenure-track research scientist until December 2008, when he became permanent staff. Dr. Polini has mainly worked on two-dimensional electron liquids and ultracold atoms trapped in low-dimensional geometries, and he is a coauthor of over 60 publications. Since 2006 his work has mostly focused on the role of electron-electron interactions in single- and bi-layer graphene.
### About the Author: Giovanni Vignale
Giovanni Vignale is Curators’ Professor of Physics at the University of Missouri-Columbia. After graduating from the Scuola Normale Superiore in Pisa in 1979 and gaining his Ph.D. at Northwestern University in 1984, he carried out research at the Max-Planck-Institute for Solid State Research in Stuttgart, Germany, and Oak Ridge National Laboratory. He became a Fellow of the American Physical Society in 1997. Professor Vignale’s main areas of research are many-body theory and density-functional theory. He is coauthor of the book Quantum Theory of the Electron Liquid and has over 150 papers in print.
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Viewpoint | May 13, 2013 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.896674394607544, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/15903/chess-master-problem/15921 | # Chess Master Problem
From Introductory Combinatorics by Richard Brualdi
We have a chess master. He has 11 weeks to prepare for a competition so he decides that he will practice everyday by playing at least 1 game a day. To make sure that he's able to take a couple of breaks, he also decides that he wont play more than 12 games per week.
My question is, how do you prove/dis-prove that there will be a succession of (consecutive) days in which he will play k games, k >= 22?
Specifically, I don't quite understand the use of the pigeonhole principle here.
Thanks in advance! :D
EDIT
I'll try to clarify what I'm asking:
The goal is to prove that for every possible practice schedule that the chess master makes, there will always be a consecutive sequence of days such that he plays exactly k games.
So for example, if k = 22, how do you prove that no matter what his schedule is like, he will always play exactly k games in r days, 1 <= r <= 77.
My main question is: what is the general method for proving different values of k? Are there values of k (from 1 to 132 inclusive) that are impossible to obtain?
Note that he does not necessarily have to play 132 games within 77 days! Also, keep in mind the bounds of k.
-
1
When you say a succession of consecutive days, do you mean two, seven, fourteen, or some other amount of days? – InterestedGuest Dec 30 '10 at 10:03
Like I said in the question: k games, k >= 22. I'm essentially asking for a general proof for the case k >= 22. – K-RAN Dec 30 '10 at 10:06
2
I am not asking the amount of games he will play, I am asking the span of the consecutive days. Two consecutive days for 22 games? 14 consecutive days for 22 games? If it's two consecutive days, it can be easily proven that he won't be able to play 22 games. 14 days (for example), different story. :) – InterestedGuest Dec 30 '10 at 10:07
1
So pretty much any number of consecutive days, as long as he plays more than or equal to 22 games? That's very easy (and probably not what you are looking for), see my answer. – InterestedGuest Dec 30 '10 at 10:11
1
@K-RAN, since he has to play at least a game a day, after 22 days he will have played 22 games, so $k$ holds true. – InterestedGuest Dec 30 '10 at 12:05
show 5 more comments
## 2 Answers
For $\displaystyle 1 \le k \le 24$ you can definitely show that the master must have played exactly $k$ games on some set of consecutive days, using pigeonhole principle.
Suppose the total number of games the master has played till the end of day $\displaystyle j$ is $\displaystyle g_j$.
Now consider the $\displaystyle 154$ numbers: $\displaystyle g_1, g_2, \dots, g_{77}, g_1 + k, g_2 + k, \dots, g_{77} + k$
These are a set of $\displaystyle 154$ numbers between $\displaystyle 1$ and $\displaystyle 132+k$.
For $\displaystyle k \lt 22$, two of these must be equal. Since $\displaystyle g_i \neq g_j$ (at least one game a day) we must have that $\displaystyle g_j + k = g_i$ for some $\displaystyle i,j$.
For $\displaystyle k=22$ we must have that the numbers are $\displaystyle 1,2, \dots, 154$, in which case, the first (and last) $\displaystyle 22$ days, the master must have played $\displaystyle 1$ game everyday.
For $\displaystyle k=23$, we can assume $\displaystyle g_i \neq 23$, and since $g_i \ge 1$, we have $g_i + k \neq 23$.
Thus by an argument similar to above, we must have have $\displaystyle 154$ numbers taking all values in $\displaystyle 1, 2, \dots, 155$, except $\displaystyle 23$ and the master must have played $\displaystyle 1$ game each of the last $\displaystyle 23$ days.
For $\displaystyle k=24$, you can show that the master must have played $\displaystyle 1$ game the first $\displaystyle 23$ days (after eliminating one of the numbers in $\displaystyle 133, 134 \dots$), then a big number of games the next, violating the $\displaystyle 12$ games per week restriction (this is where we actually used that restriction for a specific week).
We might be able to use this kind of argument to show for $\displaystyle k$ close to $\displaystyle 24$, but for larger $\displaystyle k$, I am guessing that you can find a set of games which will miss that (perhaps a computer search will help there).
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The problem is that, while working on this problem, my friend wrote a program to figure out all possible schedules that the chess master can make. Not once did a counter example come up (he started with k = 24 or 25). Either way, thanks a lot! This is the explanation that I was looking for. +1 for clearness :D – K-RAN Dec 30 '10 at 19:43
@K-Ran: It is possible, I was only guessing the non-existence. For instance, if you consider the proof for $k=24$, you will see that there is a lot of scope to increase $k$. Of course, this case-by-case proof is unenlightening, maybe there is a proof which will cover all cases... (the restriction to use pigeonhole might make it harder though). – Aryabhata Dec 30 '10 at 20:05
He has $11$ weeks of training (or $77$ days). The minimum and maximum number of games for this period is $77$ and $11 \cdot 12=132$, respectively.
We have distinct situations:
1) At any period of $k$ consecutive days he plays at least $k$ games. As well you asked.
2) The smallest period of $n$ days such that he plays $k=22$ games occur when he consider a week with a maximum number of games $12$ plus days after and before this week with a maximum possible number of games. Notice that in one day is possible to play $6$ games. So, take one day with $4$ games, one week with $12$ games, and one more day with $6$ games. You get then $22$ games in $9$ days.
Using 1) and 2) we conclude that between $9$ and $k$ days we always can make the choice of $22$ games.
I guess it is very useful!
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http://mathhelpforum.com/calculus/16842-help-w-trig-integrals.html | # Thread:
1. ## Help w/ Trig Integrals
$\int(\sin{x})^3(\cos{x})^3dx$
am i going to use trigonometric or int by parts?
help
2. Hello, ^_^Engineer_Adam^_^!
I have to ask . . .
Are those really $x^3$ ?
We can integrate: $\sin^3\!x\cos^3\!x$ . . . but not $\sin(x^3)\cos(x^3)$
3. oopss
its $(\sin{x})^3 (\cos{x})^3$
4. $\int sin^3(x) cos^3(x) dx$
Let $y = sin(x)$, then $dy = cos(x) dx$
$\int sin^3(x) cos^3(x) dx = \int sin^3(x) cos^2(x) cos(x) dx$
$= \int sin^3(x) (1 - sin^2(x)) cos(x) dx = \int y^3 (1 - y^2) dy$
$= \int (y^3 - y^5)$
I'm sure you can take it from here.
-Dan
5. im sure that either sin(x) or cos(x) is integrable
but the integrator has a different answer
maybe use integration by parts?
6. Originally Posted by Soroban
We can integrate: $\sin^3\!x\cos^3\!x$ . . . but not $\sin(x^3)\cos(x^3)$
Can someone tell me once and for all how can you tell if you can't integrate something, or something is not integrable analytically using elementary functions or whatever...wait, did i ask this question before?
I know we can't integrate $e^{x^2}$ and according to Soroban, we can't integrate $\sin \left( x^3 \right) \cos \left( x^3 \right)$, but how do we know that for sure? What's the proof that we can't integrate those functions by hand?
7. This totally my own opinion: Your confusion is understandable and it comes from the very sad conflating of the words integral and antiderivative. They are not the same. An integral is a number, quite often gotten by way of an antiderivative using the fundamental theorem of integral calculus. An antiderivative is just what is says it is.
Of the antiderivative of $\sin(x^3)$ does exits but we would the series representation for $\sin(x)$ to get it. Therefore, it is proper to say that no elementary representation of the antiderivative of $\sin(x^3)$ exist in the Calculus II sense of the term.
8. Originally Posted by topsquark
$\int sin^3(x) cos^3(x) dx$
Let $y = sin(x)$, then $dy = cos(x) dx$
$\int sin^3(x) cos^3(x) dx = \int sin^3(x) cos^2(x) cos(x) dx$
$= \int sin^3(x) (1 - sin^2(x)) cos(x) dx = \int y^3 (1 - y^2) dy$
Now,
$= \int (y^3 - y^5)$
I'm sure you can take it from here.
-Dan
To continue
$= \frac{1}{4}y^4 - \frac{1}{6}y^6 + C$
$= \frac{1}{4}sin^4(x) - \frac{1}{6}sin^6(x) + C$
Now, my TI-92 comes up with:
$-\frac{sin^2(x) cos^4(x)}{6} - \frac{cos^4(x)}{12}$
and the Integrator comes up with
$\frac{1}{192} ( cos(6x) - 9 cos(2x))$
All of these solutions are correct, despite how it might look. The point is that we are doing indefinite integration, so any solution that differs from another by only a constant are all correct. If you spend the time (or just plug it through on your calculator) you will find that all three solutions differ from each other by some constant. (Neither the TI-92 nor the Integrator remind you to add the arbitrary constant on the end.)
-Dan
9. ## thanks topsquark
ohh ok
10. You can use the fact
$\int\sin^mx\cos^nx~dx={\color{blue}\frac{\sin^{m+1 }(x)\cos^{n-1}(x)}{m+n}+\frac{n-1}{m+n}\int\sin^mx\cos^{n-2}(x)~dx},~m\ne-n$
11. Originally Posted by Jhevon
Can someone tell me once and for all how can you tell if you can't integrate something, or something is not integrable analytically using elementary functions or whatever...wait, did i ask this question before?
I know we can't integrate $e^{x^2}$ and according to Soroban, we can't integrate $\sin \left( x^3 \right) \cos \left( x^3 \right)$, but how do we know that for sure? What's the proof that we can't integrate those functions by hand?
I believe sin(x^3) is done using what is known as a Lommel integral. Don't know much about it though. Just as sin(x^2) is a Fresnel.
I couldn't find reference to Lommel in wiki. Perhaps, that would be a good MathHelpWiki for someone to take on?.
One should be able to use topics from advanced calc to prove sin(x^3) in not integrable by elementary means. Maybe Dirichlet test or something.
It is continuous and differentiable.
I may have to delve into it some more.
12. As far as I know,
$<br /> \int {\sin x^3 ~dx} = - \frac{1}<br /> {2}i\left( {\frac{{x\Gamma\left( {\displaystyle\frac{1}<br /> {3},ix^3 } \right)}}<br /> {{3\sqrt[3]{{ix^3 }}}} - \frac{{x\Gamma\left( {\displaystyle\frac{1}<br /> {3}, - ix^3 } \right)}}<br /> {{3\sqrt[3]{{ - ix^3 }}}}} \right) + k$
13. Yeah. I ran it through Maple and it gave me a horrendous result with LommelSi. It may be equivalent to your result, though. Just a different animal.
14. Originally Posted by DivideBy0
As far as I know,
$<br /> \int {\sin x^3 ~dx} = - \frac{1}<br /> {2}i\left( {\frac{{x\Gamma\left( {\displaystyle\frac{1}<br /> {3},ix^3 } \right)}}<br /> {{3\sqrt[3]{{ix^3 }}}} - \frac{{x\Gamma\left( {\displaystyle\frac{1}<br /> {3}, - ix^3 } \right)}}<br /> {{3\sqrt[3]{{ - ix^3 }}}}} \right) + k$
I know the forum where you got that
15. I can find,
$\int_0^{\infty} \sin x^3 \cos x^3 dx$
And,
$\int_0^{\infty} \sin x^3 dx$
And,
$\int_0^{\infty} \cos x^3 dx$
Eventhough these functions are not elementary.
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http://math.stackexchange.com/questions/248661/use-cauchys-multiplication-theorem-and-the-binomial-theorem-to-prove-expxy | # Use Cauchy's Multiplication Theorem and the Binomial Theorem to prove $\exp(x+y)=\exp(x)\exp(y)$
I am to use Cauchy's Multiplication Theorem and the Binomial Theorem in order to prove
$\exp(x+y)=\exp(x)\exp(y)$
but I have no idea where to begin. All I can think of doing is setting $\exp(x)$ as the sum to infinity of $(x^n)/n!$ and similarly for $\exp(y)$, $(y^n)/n!$
-
## 1 Answer
Well if $x,y\in \mathbb{R}$ then by definition \begin{equation}\exp(x)\exp(y)=(\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}\frac{y^k}{k!}) \end{equation} The Cauchy's Multiplication Theorem tells as that \begin{equation}\sum_{k=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}=\sum_{k=0}^{\infty}a_k\sum_{k=0}^{\infty}b_k\end{equation} when at least one of the two series of the RHS converge absolutely. In our case we have that \begin{equation}\exp(x)\exp(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}\end{equation} Now because $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, \begin{equation}\exp(x)\exp(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}\end{equation} A straightforward application of the binomial theorem yields the resut: \begin{equation}\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{1}{n!}(x+y)^n=\exp(x+y)\end{equation}
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If you could give me some hints and pointers, rather than posting the actual proof, that'd be great. I don't really understand why you're doing what you're doing at each step? – Mathlete Dec 1 '12 at 16:44
Sure I shall add more details if you would like! – Nameless Dec 1 '12 at 16:45
That'd be really useful, thanks! Could you possibly explain where and how you've used Cauchy's multiplication theorem? That's the main part I'm having trouble with. – Mathlete Dec 1 '12 at 16:46
I completely understand that now, thank you very much! – Mathlete Dec 1 '12 at 16:52
@Mathlete No problem. Have a good day – Nameless Dec 1 '12 at 16:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8105238080024719, "perplexity_flag": "middle"} |
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