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"The greatest gifts we can give our children are the roots of responsibility and the wings of independence."
- Maria Montessori
8 Principles of Montessori Education
Philosophy
Movement and cognition are closely intertwined,.
The Montessori-inspired distance learning program will embody these evidence-based outcomes throughout all of the grade levels.
Montessori is a comprehensive educational approach that is based on self-directed activity and hands-on learning. Students work individually or in groups to explore and discover the knowledge of the world around them and to develop their maximum potential.
Key principles of the Montessori method include: Observing the child and fostering learning experiences based upon their interests and needs, helping the child develop independence and self-direction through freedom within limits, fostering natural curiosity and an innate love for learning, collaboration within a multi-age, family-like setting, and creating a prepared environment where the students can actively engage in experiential learning.
27740 Jefferson Ave.
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TITLE: Roots of 1 in $\mathbb Q_p$
QUESTION [4 upvotes]: How to prove, that all roots of 1
in $\mathbb Q_p$ are roots of $x^{p-1}-1$?
If we consider the ring homomorphism
$$
\mathbb Z_p \to \mathbb F_p^*,
$$
then we see, that all the roots in power $p-1$ are
equal to 0 modulo $p$. Using Hensel's lemma
we can construct a solusion to $x^{p-1}-1=0$,
which is the same as the first modulo $p$.
REPLY [1 votes]: Chapter 2, $\S 1.4$ of my notes on local fields contains a discussion of roots of unity in local fields. In particular there is a complete proof that the group of roots of unity in $\mathbb{Q}_p$ is cyclic of order $p-1$.
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Responsible Business
Setting the industry standard as a responsible business for best practice and beyond.
WJ UK is committed to conducting business with the highest levels of integrity, honesty and responsibility. At WJ UK we aim to attract the best and brightest people, whatever their background, to bring new perspectives to some of the most challenging and inspiring projects around the UK & Ireland.
Through delivering in the areas of health & safety, well-being, environment & sustainability and corporate social responsibility we aim to be the best and most responsible business in our sector.
Our Mission
To deliver confidence through excellence.
Our Vision
To be the water management partner of choice.
Our Values
Our services detail what we do, but it’s our values that describe how we do it – who we are, how we work and what we stand for. If you ask us to describe what it’s like to work with WJ UK, our values are what we talk about.
Confidence
Thanks to our experience and longevity, we are justifiably assured in what we do and how we do it; but we are never arrogant. We instill confidence in our customers through our technical expertise, knowledge and service – proven in our enviable record of repeat business.
Pride
This reflects the long history our company enjoys and the enviable heritage we have within the civil engineering sector. UK employee is efficient and organised with the overriding personality traits of being thorough, careful and vigilant. This value also confirms WJ UK’s strive to complete any task we undertake to the best of our abilities.
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Olympic finalist and World Championship gold medallist Shanaze Reade has been brought in by on-demand restaurant food delivery service, Deliveroo, to help the current Milton Keynes based riders improve their riding skills.
Shanaze has been giving expert guidance on training, diet and how to handle different types of terrain.
Reade’s competitive cycling career spans 14 years to date, combining BMX riding, in which she has won the UCI BMX World Championships three times, and track cycling.
Reade partnered Victoria Pendleton in the Team Sprint at the World Championships in 2007, with the pair winning gold, before racing in the final of the 2012 Summer Olympic Games in London.
Milton Keynes riders get the most out of themselves,” said Shanaze.
Deliveroo riders in Milton Keynes can cycle more than 150km a week, delivering high-quality meals to thousands of homes and offices across the city in an average of 32 minutes, so are reliant on their endurance and stamina.
“Keeping fit whilst earning money is one of the key benefits of being a Deliveroo rider,” said Dan Warne, Managing Director of UK & IE at Deliveroo.
“Our riders cover big distances each week in cities across the UK, so we wanted to bring someone in to give them some guidance on endurance cycling and who better than an Olympian cyclist?”
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"Environmentally Friendly Solutions"
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Management Profiles
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Chem-usa.com News
Client Log In
Members of Congress and other Bush Administration officials are delaying EPA from releasing a long-awaited health risk reassessment for dioxin, due out later this year, industry and environmental sources say. will be raising $2 million to finance the further development and expansion of its email marketing division. This offer will be taking place in the first quarters of 2002.
press release
EPA Administrator Christine Todd Whitman says the agency will soon propose a rule to grant the regulatory flexibility that it has promised to industrial participants under the National Environmental Performance Track program. The voluntary program was set up in 2000 to reward companies that pledge to reduce emissions beyond federal mandates.
press release
EPA has no plans to change drinking water standards that were set before 1997, it says in a draft report released late last month. Standards for a total of 68 chemicals do not need to be revised, because existing data indicates that they are adequate to protect public health and the environment, says EPA. Another review is scheduled to be completed in 2008.
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| 286,656
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TITLE: What is the sum $\sum_{m} e^{i (U_m k + \beta_m)} $ when $U$ and $\beta$ follow different distributions
QUESTION [1 upvotes]: I have the following function.
$$ x(k) = \sum_{m} e^{i (U_m k + \beta_m)} $$
$i = \sqrt{-1}$
Here, $U_m$ are samples drawn from a Gaussian random distribution.
$$ U_m \sim \mathcal{N}(\mu, \sigma) $$ and $\beta_i$ are samples drawn from an uniform distribution.
$$ \beta_m \sim \mathcal{U}[-\pi, +\pi] $$.
I want to write $x$ as a function of $\mu$ and $\sigma$ like this,
$$ x(k) = f(k, \mu, \sigma) $$ when the number of samples summed in $x$ tends to $\infty$.
The solution I tried so far:
I used the expected value principles,
First I took the first term in the sum that is
$$ \sum_{m} e^{i U_m k} = ( \mathbb{E}[e^{i U}] ) \times ( \mathbb{E[U]} ) N = N \mu e^{-\sigma^2 k^2 /2} e^{i \mu k} $$ where $N$ is the number of samples.
The same I did for the second term and it turns out to be $0$.
$$ \sum_{m} e^{i \beta_i} = ( \mathbb{E}[e^{i \beta}] ) \times ( \mathbb{E[\beta]} ) N = 0$$
The first term suggests that the signal is decaying with $k \sigma$ and the second one suggests that the sum should be $0$. When I remove the second term from the original numerical sum and plot the function with $k$, I see in the simulation that the signal is indeed decaying.
However, when I add the second term, it becomes like a periodic signal. So, it doesn't hold the properties of both analysis I did. I guess I am missing something. The numerical sum I am getting is the expected one I believe as it is periodic and finite as $k$ increases.
============ THE SIMULATION ===============================
clear;
close all;
Mu = 7.5 .* 0.4189;
Sigma = 1 .* 0.4189;
Nt = 128;
K = 0:1:Nt-1;
x = zeros(1, Nt);
Nu = 100000;
beta = -pi + 2 * pi .* rand([1 Nu]);
U = normrnd(Mu, Sigma, [1 Nu]);
for k = 1:Nt
x(k) = [sum(exp(1j .* K(k) .* U + 1j .* beta) )];
end
figure; plot(real(x)); hold on; plot(imag(x)); grid on;
The result looks like this:
With respect to $k$, this sum is still a periodic signal. I can agree with this because after all, it is a sum of periodic signals. I have used the number of points in the sum to be $100000$. The number of $k$ points is $128$. How can I correctly write the original sum as a function of $\mu$ and $\sigma$ ?
========================== EDIT =========================================
I did some more analysis and found the distribution of $e^{i (U_m k + \beta)}$. I don't know how to approach the sum after this to get a statistical analysis of $x$ when some finite number of samples are taken in $k$ domain.
Understanding the function inside the sum:
I deduced the distribution of $U_m k + \beta_m$. It looks like the following. It is a convolution of both distributions.
$$ p(x) = \int_{0}^{2\pi} \frac{1}{2\pi \sqrt{2 \pi k^2 \sigma^2}} e^{-(kx - k\mu - \tau)/(2k^2\sigma^2)} d\tau $$
$$ p(x) = \frac{1}{4\pi} \Big[ \operatorname{erf}\Big(\frac{k\mu-x+2\pi}{\sqrt{2}k\sigma}\Big) - \operatorname{erf}\Big(\frac{k\mu-x}{\sqrt{2}k\sigma}\Big) \Big] $$
I used the CDF technique to find the distribution of $\cos((U_m k + \beta_m))$.
$$ F(y) = p(Y \leq y) = p(\cos(X) \leq y) = p(X \leq \cos^{-1}(y))$$
$F(y) = \int_{-\infty}^{\cos^{-1}(y)} \frac{1}{4\pi} \Big[ \operatorname{erf}\Big(\frac{k\mu-x+2\pi}{\sqrt{2}k\sigma}\Big) - \operatorname{erf}\Big(\frac{k\mu-x}{\sqrt{2}k\sigma}\Big) \Big] dx $
I have seen that the function inside the integral is $0$ at $-\infty$ so the expression becomes,
$$ F(y) = \frac{\left(k \mu-\cos ^{-1}(y)\right) \text{erf}\left(\frac{k \mu-\cos ^{-1}(y)}{\sqrt{2} k \sigma}\right)+\left(-k \mu+\cos ^{-1}(y)-2 \pi \right) \text{erf}\left(\frac{k \mu-\cos ^{-1}(y)+2 \pi }{\sqrt{2} k \sigma}\right)+\sqrt{\frac{2}{\pi }} k \sigma \left(e^{-\frac{\left(\cos ^{-1}(y)-k \mu\right)^2}{2 k^2 \sigma^2}}-e^{-\frac{\left(k \mu-\cos ^{-1}(y)+2 \pi \right)^2}{2 k^2 \sigma^2}}\right)}{4 \pi } $$
Then I took the derivative in terms of $y$ of this expression to find the pdf of $\cos(x)$, that is the pdf of $\cos(U_m k + \beta_m)$
$$ g(y) = \frac{\frac{\text{erf}\left(\frac{k \mu-\cos ^{-1}(y)}{\sqrt{2} k \sigma}\right)}{\sqrt{1-y^2}}-\frac{\text{erf}\left(\frac{k \mu-\cos ^{-1}(y)+2 \pi }{\sqrt{2} k \sigma}\right)}{\sqrt{1-y^2}}+\sqrt{\frac{2}{\pi }} k \sigma \left(\frac{\left(k \mu-\cos ^{-1}(y)+2 \pi \right) e^{-\frac{\left(k \mu-\cos ^{-1}(y)+2 \pi \right)^2}{2 k^2 \sigma^2}}}{k^2 \sigma^2 \sqrt{1-y^2}}+\frac{\left(\cos ^{-1}(y)-k \mu\right) e^{-\frac{\left(\cos ^{-1}(y)-k \mu\right)^2}{2 k^2 \sigma^2}}}{k^2 \sigma^2 \sqrt{1-y^2}}\right)+\frac{\sqrt{\frac{2}{\pi }} \left(k \sigma-\cos ^{-1}(y)\right) e^{-\frac{\left(k \mu-\cos ^{-1}(y)\right)^2}{2 k^2 \sigma^2}}}{k \sigma \sqrt{1-y^2}}+\frac{\sqrt{\frac{2}{\pi }} \left(-k \mu+\cos ^{-1}(y)-2 \pi \right) e^{-\frac{\left(k \mu-\cos ^{-1}(y)+2 \pi \right)^2}{2 k^2 \sigma^2}}}{k \sigma \sqrt{1-y^2}}}{4 \pi } $$
$$ -1< y < 1 $$
It looks like a cosine inverse distribution. Numerically also the distribution of $\cos(U_m k + \beta_m)$ looked like a cosine inverse one.
The $g(y)$ can be simplified to:
$$ g(y) = \frac{\text{erf}\left(\frac{k \mu-\cos ^{-1}(y)}{\sqrt{2} k \sigma}\right)-\text{erf}\left(\frac{k \mu-\cos ^{-1}(y)+2 \pi }{\sqrt{2} k \sigma}\right)}{4 \pi \sqrt{1-y^2}} $$
REPLY [1 votes]: If I understand correctly your question, you are trying to find:
\begin{align}
\lim\limits_{M\to \infty} \frac{1}{M}\sum_{m=1}^{M}e^{i\left(kU_m + \beta_m\right)}
\end{align}
where $U_m \sim \mathcal N(\mu, \sigma^2)$ and $\beta_m \sim \mathcal U(-\pi, \pi)$. By the law of big number it is equivalent to compute the expectation of $e^{i(kU + \beta)}$. So let's do that:
\begin{align}
\mathbb E\left[e^{i\left(kU+\beta\right)}\right] &= \mathbb E\left[e^{ikU}\right]\mathbb E\left[e^{i\beta}\right]\\
&= e^{ik\mu-\frac{1}{2}\sigma^2k^2}\frac{e^{i\pi} - e^{-i\pi}}{i(\pi - (-\pi))} = 0.
\end{align}
However if you had $e^{ik(U+\beta)} in that case the expectation would be:
\begin{align}
\mathbb E\left[e^{ik\left(U+\beta\right)}\right] &= \mathbb E\left[e^{ikU}\right]\mathbb E\left[e^{ik\beta}\right]\\
&= e^{ik\mu-\frac{1}{2}\sigma^2k^2}\frac{e^{ik\pi} - e^{-ik\pi}}{ik(\pi - (-\pi))} = \frac1{k\pi}e^{ik\mu-\frac12\sigma^2k^2} \sin\left(k\pi\right).
\end{align}
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tag:blogger.com,1999:blog-55174437702089415742017-02-08T20:50:47.296-08:00Bath and Beauty Product Reviewsjburkett ColorStay Ultimate Liquid Lipstick - A Review<b>Colorstay Ultimate Liquid Lipstick</b><br /><br /.<br /><br /.<br /><br /><b>My own Experience with this lipstick:</b><br /><br /. <br /><br /. <br /><br /? <br /><br /. <br /><br />As it is, this is my favorite for now, thank you so very much, Revlon! This is a winner! jburkett Day Scrub Cream by Clinique - Review. <br /><br />I recall learning that some people will need to use it more, like two to three times a week as a gentle scrub, and some only every seven days. Regardless, it is nice, and it rinses off leaving skin feeling much nicer than before. <br />! <br /><br />Love this product. It is like many of the other favorite Clinique products that I love, in that its a good thing to get when I don't know what else to add onto a purchase for a gift!<br /><br />Highly recommended.jburkett Fraichelle by Lancome - A ReviewMy Favorite Skin Scrub!<br /><br /.<br /><br />The fragrance, as I shared before is wonderful. It is a true bath product treat, if I can put it like that. If this is a gift for someone, they will love it! If it is a little treat for yourself, you won't regret it. <br /><br /.jburkett Hour Cream by Elizabeth Arden - Review. <br /><br />One nice thing about this also, is that you can use it as a lip balm for your lips as needed. Use like a vaseline or petroleum jelly for your lips. It smells nice too! <br /><br /. <br /><br />Elizabeth Arden is known for having a makeup gift on occasion also. Try it sometime, I think you will like it!jburkett Liquid Facial Soap Mild - ReviewHere I am again sharing one of my favorite Clinique products with you. It is their liquid facial soap, the mild variety. Here is what I like about it. <br /><br /.<br /><br /.<br /><br />This liquid mild facial soap comes recommended. I see nothing wrong with it. It doesn't small as pretty as some, but that is part of the appeal to others sometimes.jburkett - NutriMin C - Review - An Enzyme Peel<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 248px;" src="" border="0" alt=""id="BLOGGER_PHOTO_ID_5732209965562640690" /></a><br /><span style="font-weight:bold;">REgain - Illuminating Enzyme Peel</span><br /><br /!<br /><br />I was not looking to drop money on skin care that night. However, this one thing caught my eye. It was an enzyme peel, made of fruit and extracts, etc. I learned more about it, and bought it.<br /><br /. <br /><br /.<br /><br />So, this product comes recommended. I like it, and how natural it is.jburkett - Dramatically Different Moisturizing Gel - Review. <br /><br />First, it is ideal for all skin, in my opinion. It is especially helpful though, to the skin that might have an occasional acne flare up or break out. All skin needs moisture. Skin that has the occasional break out also needs it, but doesn't need added oils. <br /><br />To put on makeup, or even have just nice smooth looking skin, one needs to use a product like this. It adds moisture, but no oil. I love it. I have gone through bottles of it, and it is a lifesaver.<br /><br />For hot summers, or humid situations, your skin will love this moisturizing gel. It is fragrance free, hypoallergenic, and makes skin happy.<br /><br />So I can't recommend it enough. It comes highly recommended. If you are wanting a Clinique gift at the makeup counter at the mall, and don't know what else to get, this is a winner for sure.jburkett of Vanilla Bean Noel Body Lotion, Bath and Shower Gel, and Spray by Bath and Body Works. <br /><br /. <br /><br /! <br /><br />I would love to know what others have experienced in this way. Is it just that right after someone puts it on, that people can smell it so much? Or is the fragrance that strong that it lingers in the room where they are? <br /><br /! <br /><br />For a day when I don't feel like flowers or a "cleaner" fragrance, I like this one a lot. Its a great vanilla fragrance and I would recommend trying it at least once. <br /><br />Their natural skin soothing agents are shea butter, Vitamin E, and Jojoba oil. I love the extra additions like this that are so good for the skin.jburkett<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;width: 320px; height: 230px;" src="" border="0" alt=""id="BLOGGER_PHOTO_ID_5700459206874598578" /></a>jburkett Citrus Lotion, Body Lotion, Creme, Bubble Bath and Shower Gel. <br /><br /.<br /><br /. <br /><br />I would purchase this again. I hope they make it in the future.jburkett Kisses Bubble Bath, Shower Gel, Hand Creme, and Lotion by Victoria's Secret. <br /><br />It is hard to describe this scent, but it is a very sensual and sexy scent, without being overpowering or too musky or anything like that. <br /><br /. <br /><br />For ID, the bottle had what looks to be little purple cosmos flowers on it in the picture. A favorite fragrance!<br /><br />Highly recommended product.jburkett Soap - Sole-Vita - With Colavita Extra Virgin Olive OilI. <br /><br /. <br /><br /.jburkett Strawberry Lip Gloss or Lip Balm Review6224214320207378" /></a><br />This lip gloss was part of a lovely gift set from my mother, and came with some great hand and body lotion as well. <br />! <br /><br /.<br /><br />I can recommend this very highly, and I hope Watkins still makes it!jburkett Strawberry Hand and Body Lotion Review. <br /><br />This strawberry lotion is wonderful, and I would highly recommend it. I currently put it on every night at least on my hands and lower arms, before I go to bed to keep from getting dry in these Winter months. <br /><br /. <br /><br />The directions recommend using with the Strawberry Body Wash and Bubble Bath, and finishing off with the lip gloss.jburkett Story.<br /><br /. <br /><br />I say, wear whatever you love, and that others compliment you on. You can't lose that way.jburkett and Beauty Products and Our LivesIts hard to find a person in this world that doesn't make use of some kind of bathing product or beauty product. I mean even shampoos and soaps here, even if they are homemade. They are simply a part of our lives.<br /><br /.<br /><br />This blog is partly dedicated to just remembering some of those tried and true favorites that you can't help but want to go back to again and again. Its easy to forget some of the names sometimes, so this will help in that regard.<br /><br /. <br /><br />My commitment is to honesty, so you can know that if I am even raving about a product, its not because I am paid to do so! I always welcome feedback and any thoughts. Do share with me your experiences also.jburkett
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TITLE: Algebraic Topology-Explanation required for the following definition
QUESTION [3 upvotes]: I am currently reading the book A combinatorial introduction to topology by Michael Henle.
Under "Compactness and Connectedness" there is the following definition which I didn't understand at all. I do know the definition of nearness.
Let $P=\{P_1,P_2,P_3,\ldots \}$ be a sequence of points. The point $Q$ is near the sequence $P=\{P_n\}$ if either $Q=P_n$ for an infinite number of terms of the sequence or $Q=P_n$ only a finite number of times and $Q$ is near the set of other values of $P$.
Any help would be greatly appreciated.
REPLY [3 votes]: The definition of "$Q$ is near the set $A$" is "every neighborhood of $P$ contains a point of $A$".
Now, a sequence is really a function $P$ from the set of positive integers to the space. We write $P_n$ instead of $P(n)$, but $P$ is really a function. The author uses set notation, and writes $P=\{P_1,P_2,\dots\}$, but this is not quite correct, since there is a difference between the sequence, which is a function, and its range, which is a set. So, for example, if $P$ is the sequence $P_1=1,P_2=-1,P_3=1,P_4=-1,\dots$ and if $P'$ is the sequence $P'_1=P'_2=1,P'_3=P'_4=P'_5=\dots=-1$, then both are very different as sequences, but have the same range, namely $\{-1,1\}$.
To say that $Q$ is near a sequence $P$ means that, either for infinitely many values of $n$, $Q$ is the value taken by the function $P$ at $n$, $Q=P_n$, or else this fails, but $Q$ is near the set resulting from excluding $Q$ from the range of $P$.
To illustrate, for the sequence $P$ given by $P_1=1,P_2=-1,P_3=1,P_4=-1,\dots$ Here, the point $Q=1$ is near $P$, because $Q=P_1=P_3=P_5=\dots$ Also, the point $R=-1$ is near $P$, because $R=P_2=P_4=\dots$ On the other hand, no other point $S$ is near $P$, because if $S\ne1$ and $S\ne-1$, then there is a neighborhood of $S$ so small that neither $1$ nor $-1$ is in it.
Now, look at the sequence $P'$ given by $P'_1=P'_2=1,P'_3=P'_4=P'_5=\dots=-1$. As before, if $S\ne1$ and $S\ne -1$, then $S$ is not near $P'$. Also, $R=-1$ is near $P'$ as $R=P'_3=P'_4=P'_5=\dots$ However, $Q=1$ is not near $P'$, because there are only two values of $n$ for which $Q=P'_n$, namely, $n=1$ or $n=2$. The set of other values of $P'$ is just $-1$, and there is a neighborhood of $Q$ that misses $-1$, so $Q$ is not near the set of other values.
All this being said, in practice, most sequences $P$ we are interested in will not repeat values, so a point $Q$ will be near $P$ iff $Q$ is not in the range of $P$, but $Q$ is near the range of $P$. For example, $0$ is near the sequence $P$ given by $P_n=1/n$ for $n=1,2,\dots$
It is in order to avoid having to phrase arguments awkwardly, and to avoid unnecessarily splitting them into two cases, that the definition is presented to cover both possibilities, when $Q$ is repeatedly listed in the sequence $P$, and when is actually being ``approached'' by the sequence.
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My courses
Biology 30S
The theme of gr.11 biology revolves around the big picture of wellness and homeostasis in our human bodies.
Biology 40S
In Biology 12, the course explores the genetic makeup of organisms and the diversity of life.
Chemistry 30S
Fundamentals of Chemistry
In Chemistry 11, the course explores the fundamentals of qualitative and quantitative chemistry.
Chemistry 40S
Applications of Chemistry
In Chemistry 12, the course continues where chemistry 11 left off and ventures into a more theoretical look of chemistry.
Mr.Yeung’s Wonderments
L.O.L Surprise dolls
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ECSU Athletic Department Summer Camps
The Elizabeth City State University Department of Athletics will hold several camps this summer. A 7-on-7 football camp will be held on June 13th from 8 am to 4 pm for ages 14 to 18. On June 14, there will be an individual/position camp for ages 7 to 12. There will also be a Lil’ Viking Day Camp held on June 16th from 9 am to 3 pm for ages 6 to 14. For more information about the camps and fees, contact Coach Monterio Hand (252)-335-3827 or Graham Hobbs (252)335-3700.
A women’s basketball team camp will be held June 17th and 18th from 9 am to 5 pm and a Lil’ Vikings Basketball Camp will be held June 21st through 23rd from 8 am to noon for ages 8 to 13. There will also be a Guard/Post position camp held on June 24th and 25th from 9 am to 2 pm for ages 12 to 17. For more information about the camps and fees, contact Coach Laquanda Dawkins at lddawkins@ecsu.edu or 252-335-8772.
Individual camps for men’s basketball will be held August 1st through August 3rd for ages 7 to 12 from 9 am to noon. A guard/post position camp will be held August 3rd and August 4th for ages 13 to 18 from 2 pm to 5 pm. For more information about the camps and fees, contact ajdunk@ecsu.edu or 252-335-3461.
A Viking Sports Camp will be held July 11th through July 22nd, Monday to Friday, for ages 5 to 15. For more information about the camp and fees, contact Camp Director Richard Harfst rfharfst@ecsu.edu or 252-335-3396.
Athletic Department Summer Camps:
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TITLE: Is the Weierstrass function given in Counterexamples in Analysis a typo?
QUESTION [1 upvotes]: Let $0 < a < 1$; let $b$ be an odd integer; let $ab > 1 + \frac{3\pi}{2}$; let
$f: x \mapsto \sum_{n \geq 0}a^{n}\cos (b^{n}\pi x): \Bbb{R} \to \Bbb{R}$. Then $f$ is everywhere continuous and nowhere differentiable.
On page 39 of Counterexamples in Analysis (Dover edition), the authors write the Weierstrass function $f$ as $f(x) = \sum_{n \geq 0}b^{n}\cos (a^{n} \pi x)$; then I do not see how it can converge (of this "typo" series)...
If this is really an obvious typo, then there should already be someone that pointed it out; but so far I did not find any source doing this, so I ask here to clarify this point.
REPLY [1 votes]: The typo is the constraints on $a$ and $b$. In the book it says that $b$ is an odd integer, it should actually be $a$. And $b$ should be $0<b<1$. That, or the $a$ and $b$ should be switched in the series.
To see this converges, just use the Weierstrass M-test. $|a^n\cos(b^n \pi x)|\leq a^n$ and $\sum a^n$ converges.
| 145,665
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High speed, flexible and compact palletizing machineThe Flex palletizer has been developed to respond to the market requirements for high speed, flexibility and compactness of the palletizing section of the modern production lines. The heart of the machine is the robot (4 or 6 axis, according to the application) which picks up the incoming cases and places them on the pallet, according to a pre-set pattern already stored in the machine memory.
The robot is flexible and very fast: picking up individual cases, ink-jet, labelling or weighing them before placing them on the pallet. All this can be done even at high speed. The machine can be equipped with large capacity and empty pallet magazine with automatic handling of empty and full pallets. Due to its modular design, the pallet handling section can be customized according to the requirements of the end user.
A simpler version is also available, with a dual position empty pallet loading station. The machine can be supplied as a stand alone unit or connected to an IMA side loading case-packer, particularly where space is at a premium.
The machine also features a standard PLC unit which interacts with the robot processor, thereby making operation straightforward. A large and comprehensible operator interface makes change-overs simple and assists with diagnostics. The machine is fully guarded by transparent panels interlocked with safety switches and light curtains in the pallet removal area.
| 198,868
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OK, resolved this myself (see below), but I thought my little adventure here might be of value to others, so I decided to post it anyway. Also, I'm open to other solutions anyone might have that would be more elegant than what I have done.
OK, so... getting into 2013, now. I updated our Standard.ipt, and initially converted the "Colors" to the new "Appearance" global standard. Then, I found out that the new Appearances makes Rubber reflective... and any metal surface just... frankly, for my money... stupidly reflective. THEN, I found that I could NOT get the reflectiveness to turn off. It looks like the Brass material gets this "Brass - Satin" Appearance tied to it, and under the Appearance Properties-Metal settings, you can change the "Finish", but all of them... super reflecty. Looks stupid, IMO. Don't want it. So, this all happened because I updated to the new Appearances format. So THEN, I try using an older .ipt to create a new Standard.ipt, this time NOT converting to "Appearances", and instead retaining the old Color Styles... which produced the not-so-bloody-reflective effect desired.
However, NOW, when I start a new part, I get the "Style conflict" msg for "Color:Rubber - Black - Version 1.1, Color: Titanium, Material:RUBBER, material
ILICONE..." and so on. OK, so lets go to the Style Library Manager... oh, but wait... "Colors" and "Materials" are no longer IN the Style Library Manager, so I can't delete them. It's as if these Styles have gotten pushed out to the Style Library, but now in a format that the Style Library Manager interface no longer recognizes...
OK, in the course of writing this all out, I found it. The conflicting "Colors" and "Materials" can be deleted from the "Inventor Material Library" of the Appearance Browser, and the "Inventor Material Library" of the Material Browser, respectively... NOT through the "Style Library Manager" as some (read: I) might expect. Now, I may well push the document styles out to the style library, and manage them there... but that's for another thread...
seen a number of posts from you recently concerning the Style Library, maybe this link will help;
Thanks for the link... some good night reading there.
not a problem. I'm still learning about the Style Library and found that recently. Bookmarked it right away!
Here is an article about the Color and Appearances Libraries in 2013. It's pretty informative.
But does it help in reality?
Creating unique materials and colors for all members of the Autodesk CAD family is a big and useful step for the future. On the other hand, I'd like to see same colors and appearances for 2012 parts now in 2013. Legacy colors and materials should be available, until the user decides to use the new ones.
For me, this is a big startup problem for 2013. The sooner a solution is available, the better ..
Walter
There are definitely some pluses and minuses to this new set up. For those users who work across many of Autodesk's products, this can be a very welcome tool. From here going forward it will be quite useful. But the obvious drawback, as you stated, was the issues with legacy parts and their colors or styles no longer being available. It would be nice to be able to have all of those still available, but it looks like the frustrating work around is going to be re-creating those styles yourself. This is reminiscent of a similar issue when Styles were introduced in Inventor 9.
What are your results with transparent materials like glass in 2013, John?
Walter
I am actually quite impressed with the ability to control the transparency and index of refraction for any material in the appearance library. It makes for much clearer and life like parts. We are currently in the process of updating our training courses for the 2013 release of Inventor, and there is going to be quite a bit of information on this topic in our 2D Drafting and Customization course. We will also probably be adding a free mini-course about dealing with the Colors and Appearance libraries to our YouTube channel in the coming weeks.
| 402,429
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Best Six Sigma and Lean Management Training in Atlanta
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9th February 2022
Looking at the intraday performance of the FTSE 100, the index hit a post-pandemic high this week on the news of BP profits hitting an eight-year high after oil and gas prices surged last year.
Whilst the FTSE 100 slumped slightly lower by the end of the day, it has been one of the strongest global equity markets year to date, reflecting the strong presence of sectors within the index which can be described as value – banks, oil and gas, miners…
Turning to the Eurozone, the European Central Bank (ECB) chose not to alter monetary stimulus in their first meeting of 2022 last week, but gave a narrative suggesting that policy could potentially tighten. Following this, markets rushed to price in a hawkish stance. However, on Monday this week the President of the ECB spoke, revealing a more cautious narrative.
Christine Lagarde allayed fears of dramatic interest rate increases and entrenched inflation, as the ECB believe that Eurozone inflation is set to fall back, and stabilise around their 2% target.
Additionally, she highlighted that they will be accommodative in their use of monetary policy, bolstering our sentiment that interest rate rises will be measured, gradual and data dependent.
Data to look out for this week includes UK GDP, UK industrial production, US CPI and the University of Michigan consumer sentiment index.
Hannah Owen, Portfolio Specialist.
| 280,403
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TITLE: Counting problem, nickels and quarters.
QUESTION [1 upvotes]: A machine selects five coins at random from a purse containing ten quarters and four nickels.
1) Given that at least one nickel is selected, find the probability that there are at least two nickels selected.
I already solve (event A: the probability that there is at least one nickel selected) which is $0.8741$ and (event B:the probability that there are at least two nickels selected) which is $0.454$. The question asks about
$$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$
The answer key said is $=0.519$. But I have no idea what I need to do to reach that number.
Thanks for your help.
REPLY [2 votes]: Note that in this specific case, $A\cap B = B$ since the event "Has at least one nickel" is a superset to the event "Has at least two nickels." I.e. $A\supseteq B$, which is the same thing as saying $B\subseteq A$.
That is to say, whenever you have at least two nickels you also always have at least one nickel.
You say you calculated $P(A)\approxeq 0.8741$ and $P(B)\approxeq 0.454$ and you know that $P(B|A)=\frac{P(A\cap B)}{P(A)}$
Now, taking into account what I mentioned before, you have:
$$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(B)}{P(A)}\approxeq \frac{0.454}{0.8741}\approxeq 0.519$$
| 144,140
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Description
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- Brown wooden print case with piano finish
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- Requires four C batteries included
- 18.5 in tall x 17.25 in wide x 4.25 in deep
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In the past we've talked about which retirement accounts are best for different situations, and we looked at reasons why you might want to invest in a Roth IRA, 401k, Roth 401k or A Traditional IRA. … [Read more...]
Personal finance topics including budgeting, debt reduction, investing, income creation and saving.
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\begin{document}
\begin{flushleft}
{\bf\Large {Quaternion Offset Linear Canonical Transform in One-dimensional Setting}}
\parindent=0mm \vspace{.2in}
{\bf{M. Younus Bhat$^{1},$ and Aamir H. Dar$^{2}$ }}
\end{flushleft}
{{\it $^{1}$ Department of Mathematical Sciences, Islamic University of Science and Technology Awantipora, Pulwama, Jammu and Kashmir 192122, India.E-mail: $\text{ gyounusg@gmail.com}$}}
{{\it $^{2}$ Department of Mathematical Sciences, Islamic University of Science and Technology Awantipora, Pulwama, Jammu and Kashmir 192122, India.E-mail: $\text{ahdkul740@gmail.com}$}}
\begin{quotation}
\noindent
{\footnotesize {\sc Abstract.} In this paper, we introduce quaternion offset linear canonical transform of integrable and square integrable functions. Moreover, we show that the proposed transform satisfies all the respective properties like inversion formula, linearity, Moyal's formula , product theorem and the convolution theorem. \\
{ Keywords:} Offset linear canonical transform; Quaternion Offset linear canonical transform; Moyal's formulla; Convolution. \\
\noindent
\textit{2000 Mathematics subject classification: } }
\end{quotation}
\section{\bf Introduction}
\noindent
The classical Integral transform has been generalized to the six-parameter $(A, B, C, D, p, q)$ transform called the offset linear canonical transform (OLCT).
For a matrix parameter $\Lambda =\left[\begin{array}{cccc}A & B &| & p\\C & D &| & q \\\end{array}\right] $, the OLCT of any signal $f$ is defined as
\begin{equation}\label{eqn 1}
\mathcal O_\Lambda[f](w)=\int f(t)\mathcal K_{\Lambda}(t,w)dt,
\end{equation}
where $\mathcal K_{\Lambda}(t,w)$ denotes the kernel of the OLCT and is given by
\begin{equation}
\mathcal K_{\Lambda}(t,w) =\frac{1}{\sqrt{i2\pi B}}\exp\left\{\frac{i}{2B}\left(At^2-2t(w-p)-2w(Dp-Bq)+D(w^2+p^2)\right)\right\}
\end{equation}
with $AD-BC=1.$
It is here worth to mention that if $ B=0$ then the OLCT defined by (\ref{eqn 1}) is simply a time
scaled version off multiplied by a linear chirp. Hence, without loss of generality, in this paper we
assume $B\ne0$.
Looking at the core of OLCT, one can derive it as a time-shifted and frequency-modulated child of the parent linear canonical transform (LCT) \cite{xulct, 16l}. On the application side OLCT is similar to LCT but due to its two extra parameters $p$ and $q$, it is more general and flexible than parental LCT. Hence it has gained more popularity in optics, signal and image processing. For more details, we refer to \cite{15l, 14, xiang}.
In the prospect of signal processing, one can consider that any
signal processing tool converts the time-domain signals into frequency-domain . Further in signal processing, convolution of two
functions \cite{17l,18l,ownwvd} is a most useful tool in constructing a filter for denoising the given noisy signals(see\cite{19l}).\\
In past decades, hypercomplex algebra has become a leading area of research with its applications in color image processing, image filtering, watermarking, edge detection and pattern recognition(see \cite{1,2,3,10,4,5,6,7}). The Cayley-Dickson algebra of order four is labeled as quaternions which has wide applications in optical and signal processing. The extension of Fourier transform in quaternion algebra is known as quaternion Fourier transform(QFT) \cite{8} which is said to be the substitute of the commonly used two-dimensional Complex Fourier Transform (CFT). The QFT has wide range of applications see\cite{11,12}.
The quaternionic offset linear canonical transform (QOLCT) can be defined as a generalization of the quaternionic linear canonical transform (QLCT) and has been studied in \cite{bah2}. Here the authors derive the relationship between the QOLCT and the quaternion Fourier transform (QFT). Moreover, they proved the Plancherel formula, and some properties related to the QOLCT. For more details we refer to \cite{bia,13,ownqgolct,ownd}.
But to the best of our knowledge, theory about one-dimensional quaternion OLCT(1D-QOLCT) is still in its infancy. Therefore it is worthwhile to study the theory of 1D-QOLCT which can be productive for signal processing theory and applications.
In this paper, our main objectives are to introduce the novel integral transform called the one-dimensional quaternion offset linear canonical transform(1D-QOLCT) and study its properties, such as inversion formula, linearity,
Moyal's formula, convolution theorem and product theorem 1D-QOLCT.\\
This paper is organized as follows: In Section 2, we summarize the general definitions and basic properties of quaternions. In Section 3, we introduce 1D-QOLCT and obtain various properties linearity, Moyal's formula, convolution and product theorem of the proposed transform.
\section{\bf Preliminaries}
\subsection{Quaternions}\ \\
Let $\mathbb R$ and $\mathbb C$ be the usual set of real numbers and set of complex numbers, respectively. The division ring of quaternions in the honor of Hamilton, is denoted by $\mathbb H$ and is defined as
\begin{align*}
\mathbb H&=\{h_0+e_1h_1+e_2h_2+e_3h_3:h_0,h_1,h_2,h_3\in\mathbb R \}\\
&=\{z_1+e_2z_2:z_1,z_2\in\mathbb C\}\quad(Cayley Dickson form)
\end{align*}
where $e_1,e_2,e_3$ satisfy Hamilton’s multiplication rule\\
$$ e_1e_2=-e_2e_1=e_3,\quad e_2e_3=-e_3e_2=e_1,\quad e_3e_1=-e_1e_3=e_2$$
and\\
$$e^2_1=e^2_2=e^2_3=1$$
\parindent=0mm \vspace{.2in}
Every member of $\mathbb H$ is known as quaternion. In quaternion algebra addition, multiplication, conjugate
and absolute value of quaternions are defined by \\
$$(a_1+e_2a_2)+(b_1+e_2b_2)=(a_1+b_1)+e_2(a_2+b_2),$$
$$(a_1+e_2a_2)(b_1+e_2b_2)=(a_1b_1-\overline a_2b_2)+e_2(a_2b_1+\overline a_1 b_2), $$
$$(a_1+e_2a_2)^c=\overline a_1-e_2a_2,$$
$$|a_1+e_2a_2|=\sqrt{|a_1|^2+|a_2|^2},$$\\
here $\overline a_k$ is the complex conjugate of $a_k$ and $|a_k|$ is the modulus of the
complex number $a_k, k = 1, 2.$
For all $a=a_1+e_2a_2,\quad b=b_1+e_2b_2 \in \mathbb H,$ the following properties of conjugate and modulus and multiplicative inverse
are well known.\\
$$(a^c)^c=a,\quad (a+b)^c=a^c+b^c,\quad(ab)^c=b^ca^c,$$ $$\quad|a|^2=aa^c=|a_1|^2+|a_2|^2,\quad|ab|=|a||b|,$$
$$a^{-1}=\frac{\overline a}{|a|^2}.$$\\
We denote $L^p(\mathbb R,\mathbb H),$ the Banach space of all quaternion-valued functions $f$
satisfying $$\|f\|_p=\left(\int|f_1(t)|^p+|f_2(t)|^pdt\right)^{1/p}<\infty,\quad p=1,2.$$ And on $L^2(\mathbb R,\mathbb H)$ the inner product $\langle f,g\rangle=\int f(t)[g(t)]^cdt,$ where integral of a quaternion valued function is defined by $\int(f_1+e_2f_2)(t)dt=\int f_1(x)dt+e_2\int f_2(x)dt,$ whenever the integral exists.
\section{\bf Quaternion one-dimensional offset linear canonical transform }
In this section we will introduce the definition of quaternion one-dimensional offset linear canonical transform(1D-QOLCT) by using \cite{1D-QFT,1D-QFrFT,1D-QLCT}. Prior to that we note $e_1,e_2$ and $e_3$ ((or equivalently $i,j,k$) denote the three imaginary units in the quaternion algebra.
\begin{definition}\label{def 1D-QOLCT} The 1D-QOLCT of any signal $f\in L^1(\mathbb R,\mathbb H)$ with respect a matrix parameter $\Lambda=(A,B,C,D,p,q)$ is defined by
\begin{equation}\label{eqn 1D-QOLCT}
\mathbb Q^{\mathbb H}_{\Lambda}[f(t)](w)=\int f(t)\mathcal K^{ e_2}_{\Lambda}(t,w)d{ t}
\end{equation}
where
\begin{equation}
\mathcal K_{\Lambda}(t,w) =\frac{1}{\sqrt{i2\pi B}}\exp\left\{\frac{i}{2B}\left(At^2-2t(w-p)-2w(Dp-Bq)+D(w^2+p^2)\right)\right\}
\end{equation}
With $AD-BC=1.$
Now we can find that if $f(t)$ is real-valued signal in (\ref{eqn 1D-QOLCT}), then we can interchange the kernel in Definition \ref{def 1D-QOLCT}.
\end{definition}
By appropriately choosing parameters in $\Lambda=(A,B,C,D,p,q)$ the 1D-QOLCT(\ref{eqn 1D-QOLCT})
gives birth to the following existing time-frequency transforms:
\begin{itemize}
\item For $\Lambda=(0,1,-1,0,0,0)$, the 1D-QOLCT (\ref{def 1D-QOLCT}) boils down to the quaternion one-dimensional Fourier Transform\cite{1D-QFT}
\item For $\Lambda=(A,B,C,D,0,0)$ the 1D-QOLCT(\ref{def 1D-QOLCT}) reduces to the Quaternion one-dimensional Linear Canonical Transform\cite{1D-QLCT}.\\
\item For $\Lambda=(\cos\theta,\sin\theta,-\sin\theta,\cos\theta,0,0)$ the 1D-QOLCT (\ref{def 1D-QOLCT}) reduces to the Quaternion one-dimensional fractional Fourier Transform\cite{1D-QFrFT}.
\end{itemize}
\begin{definition}[Inversion]\label{def inverse}
The inverse of a 1D-QOLCT with parameter $\Lambda =\left[\begin{array}{cccc}A & B &| & p\\C & D &| & q \\\end{array}\right] $ is given by a 1D-QOLCT with parameter $\Lambda^{-1} =\left[\begin{array}{cccc}D& -B &| &Bq-Dp\\-C & A &| &Cp-Aq \\\end{array}\right] $ as
\begin{equation}\label{eqn inverse}
f(t)=\{\mathcal O^{\mathbb H}_{\Lambda}\}^{-1}[\mathcal O^{\mathbb H}_{\Lambda}[f]](t)
=\int\mathcal O^{\mathbb H}_{\Lambda}[f](w){\mathcal K^{ e_2}_{\Lambda^{-1}}(w,t)}dw\\
\end{equation}
where ${\mathcal K^{ e_2}_{\Lambda^{-1}}(w,t)}={\mathcal K^{ -e_2}_{\Lambda}(t,w)}=\overline{\mathcal K^{ e_2}_{\Lambda}(t,w)}$
\end{definition}
\begin{definition}\label{def2 1D-QOLCT}
Let $f=f_1+e_2f_2$ be a quaternion valued signal in $L^1(\mathbb R,\mathbb H),$ then the quaternion quadratic-phase Fourier transform is defined as
\begin{equation}\label{eqn2 1D-QOLCT }
\mathcal O^{\mathbb H}_{\Lambda}[f(t)]( w)=\mathcal O^{\mathbb H}_{\Lambda}[f_1(t)]( w)+e_2\mathcal O^{\mathbb H}_{\Lambda}[f_2(t)]( w).
\end{equation}
By above definition, it is consistent with the offset linear canonical transform on $L^1(\mathbb R,\mathbb C).$ Now it is clear from the definition of quaternion offset linear canonical transform and the properties of offset linear canonical transform on $L^1(\mathbb R,\mathbb H),$ that $\mathcal O^{\mathbb H}_{\Lambda}\left(\mathcal O^{\mathbb H}_{\Gamma}[f]\right)=\mathcal O^{\mathbb H}_{\Lambda\Gamma}[f]$ and $\{\mathcal O^{\mathbb H}_{\Lambda}[f]\}^{-1}=\mathcal O^{\mathbb H}_{\Lambda^{-1}}[f]$ for every signal $f\in L^1(\mathbb R,\mathbb H).$
\end{definition}
\begin{theorem}\label{linear} The quaternion quadratic-phase Fourier transform $\mathcal O^{\mathbb H}_{\Lambda}$ is $\mathbb H-$linear on $L^1(\mathbb R,\mathbb H).$
\begin{proof}
Let us consider two quaternion signals $f=f_1+e_2f_2$ and $g=g_1+e_2g_2$ in $L^1(\mathbb R,\mathbb H),$ now by the linearity of $\mathcal O^{\mathbb H}_{\Lambda}$ on $L^1(\mathbb R,\mathbb C),$ we obtain
\begin{align*}
\mathcal O^{\mathbb H}_{\Lambda}[f+g]&=\mathcal O^{\mathbb H}_{\Lambda}[(f_1+e_2f_2)+(g_1+e_2g_2)]\\
&=\mathcal O^{\mathbb H}_{\Lambda}[(f_1+g_1)+e_2(f_2+g_2)]\\
&=\mathcal O^{\mathbb H}_{\Lambda}[f_1]+\mathcal O^{\mathbb H}_{\Lambda}[g_1]+e_2\left(\mathcal O^{\mathbb H}_{\Lambda}[f_2]+\mathcal O^{\mathbb H}_{\Lambda}[g_2]\right)\\
&=\left(\mathcal O^{\mathbb H}_{\Lambda}[f_1]+e_2\mathcal O^{\mathbb H}_{\Lambda}[f_2]\right)+\left(\mathcal O^{\mathbb H}_{\Lambda}[g_1]+e_2\mathcal O^{\mathbb H}_{\Lambda}[g_2]\right)\\
&=\mathcal O^{\mathbb H}_{\Lambda}[f]+\mathcal O^{\mathbb H}_{\Lambda}[g].
\end{align*}
Now to prove $\mathbb H-$linearity, we let $q=q_1+e_2q_2 \in\mathbb H$ and $f=f_1+e_2f_2\in L^1(\mathbb R,\mathbb H)$ be arbitrary,then we have
\begin{align*}
\mathcal O^{\mathbb H}_{\Lambda}[e_2f]&=\mathcal O^{\mathbb H}_{\Lambda}[e_2(f_1+e_2f_2)]\\
&=\mathcal O^{\mathbb H}_{\Lambda}[e_2f_1-f_2]\\
&=e_2\mathcal O^{\mathbb H}_{\Lambda}[f_1]-\mathcal O^{\mathbb H}_{\Lambda}[f_2]\\
&=e_2\left(\mathcal O^{\mathbb H}_{\Lambda}[f_1]+e_2\mathcal O^{\mathbb H}_{\Lambda}[f_2]\right)\\
&=e_2\mathcal O^{\mathbb H}_{\Lambda}[f].
\end{align*}
Therefore,
\begin{align*}
\mathcal O^{\mathbb H}_{\Lambda}[qf]&=\mathcal O^{\mathbb H}_{\Lambda}[q_1f]+\mathcal O^{\mathbb H}_{\Lambda}[e_2q_2f]\\
&=q_1\mathcal O^{\mathbb H}_{\Lambda}[f]+e_2q_2\mathcal O^{\mathbb H}_{\Lambda}[f]\\
&=(q_1+e_2q_2)\mathcal O^{\mathbb H}_{\Lambda}[f]\\
&=q\mathcal O^{\mathbb H}_{\Lambda}[f].\\
\end{align*}
Which completes proof.
\end{proof}
\end{theorem}
\begin{theorem}[Moyal's formula ]\label{moyal} Let $f,g\in L^1(\mathbb R,\mathbb H)\cap L^2(\mathbb R,\mathbb H)$ be two signals functions with $\mathcal O^{\mathbb H}_{\Lambda}[f]\in L^1(\mathbb R,\mathbb H)$, $\langle f,g \rangle=\langle\mathcal O^{\mathbb H}_{\Lambda}[f],\mathcal O^{\mathbb H}_{\Lambda}[g] \rangle$ .
\begin{proof}
For $f,g\in L^1(\mathbb R,\mathbb H)\cap L^2(\mathbb R,\mathbb H)$ with $\mathcal O^{\mathbb H}_{\Lambda}[f]\in L^1(\mathbb R,\mathbb H)$,
\begin{align*}
\langle f,g \rangle &=\int f(t)[g(t)]^cdt\\\\
&=\int\int\mathcal O^{\mathbb H}_{\Lambda}[f](w){\mathcal K^{ e_2}_{\Lambda^{-1}}(w,t)}dw[g(t)]^cdt\quad(by \ref{eqn inverse})\\\\
&=\int\int\mathcal O^{\mathbb H}_{\Lambda}[f](w){\mathcal K^{ -e_2}_{\Lambda}(t,w)}[g(t)]^cdwdt\\\\
&=\int\mathcal O^{\mathbb H}_{\Lambda}[f](w)\left\{\int{\overline{\mathcal K^{ e_2}_{\Lambda}(t,w)}}[g(t)]^cdw\right\}dt\\\\
&=\int\mathcal O^{\mathbb H}_{\Lambda}[f](w)\left\{\int{g(t)\mathcal K^{e_2}_{\Lambda}(t,w)}dw\right\}^cdt\\\\
&=\int\mathcal O^{\mathbb H}_{\Lambda}[f](w)\{\mathcal O^{\mathbb H}_{\Lambda}[g](w)\}^cdt\\\\
&=\langle\mathcal O^{\mathbb H}_{\Lambda}[f],\mathcal O^{\mathbb H}_{\Lambda}[g] \rangle.
\end{align*}
Which completes proof.
\end{proof}
\end{theorem}
\begin{lemma}\label{lem conju}
For $f\in L^p(\mathbb R ,\mathbb C ),p=1,2$; we have $\mathcal O^{\mathbb H}_{\Lambda}[\overline{f}](w)=\overline{\mathcal O^{\mathbb H}_{\Lambda^{-1}}[f](w)}.$
\begin{proof}
It follows from definition \ref{def 1D-QOLCT} that
\begin{align*}
\mathcal O^{\mathbb H}_{\Lambda}[\overline{f}](w)&=\int\overline{f(t)}\mathcal K^{e_2}_\Lambda(t,w)dt\\\\
&=\int\overline{f(t)\overline{\mathcal K^{e_2}_\Lambda(t,w)}}dt\\\\
&=\int\overline{f(t){\mathcal K^{-e_2}_{\Lambda}(t,w)}}dt\\\\
&=\overline{\int{f(t){\mathcal K^{e_2}_{\Lambda^{-1}}(w,t)}}dt}\\\\
&=\overline{\mathcal O^{\mathbb H}_{\Lambda^{-1}}[f](w)}.
\end{align*}
Which completes the proof.
\end{proof}
\end{lemma}
\begin{remark}The Lemma \ref{lem conju} can also be written as $\mathcal O^{\mathbb H}_{\Lambda^{-1}}[\overline{f}](w)=\overline{\mathcal O^{\mathbb H}_{\Lambda}[f](w)}.$
\end{remark}
\begin{definition}\label{def con}
For $f\in L^2(\mathbb R,\mathbb H)$ and $g\in L^1(\mathbb R,\mathbb H),$ define
\begin{equation}
(f\ast g) =(f_1\ast g_1 -\mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_2\ast g_2])+e_2(f_2\ast g_1+\mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_1\ast g_1]),
\end{equation}
where $ \ast$ is the proposed definition of convolution.
\end{definition}
\begin{lemma} \label{lemma con} Under the assumptions of definition \ref{def con}, we have
\begin{equation}
\mathcal O^{\mathbb H}_{\Lambda}[f\ast g(t)](u)=\mathcal O^{\mathbb H}_{\Lambda}[f](u)\mathcal O^{\mathbb H}_{\Lambda}[ g](u)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}
\end{equation}
\end{lemma}
\begin{theorem}[Convolution theorem]\label{thm con} Let $f,g$ be two given signal functions such that $f\in L^2(\mathbb R,\mathbb H)$ and $g\in L^1(\mathbb R,\mathbb H),$ , then for all $w\in \mathbb R$ we have
\begin{equation}
\mathcal O^{\mathbb H}_{\Lambda}[f\ast g](w)=\mathcal O^{\mathbb H}_{\Lambda}[f](w)\mathcal O^{\mathbb H}_{\Lambda}[ g](w)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}
\end{equation}
\begin{proof}
By applying Definition \ref{def con} and Lemma \ref{lemma con}, we have
\begin{align*}
\mathcal O^{\mathbb H}_{\Lambda}[f\ast g](w)&= \mathcal O^{\mathbb H}_{\Lambda}\left[(f_1\ast g_1 -\mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_2\ast g_2])\right](w)+e_2\mathcal O^{\mathbb H}_{\Lambda}\left[(f_2\ast g_1+\mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_1\ast g_1])\right](w)\\\\
&=\mathcal O^{\mathbb H}_{\Lambda}[f_1](w)\mathcal O^{\mathbb H}_{\Lambda}[g_1](w)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}-\mathcal O^{\mathbb H}_{\Lambda}\mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_2](w)\mathcal O^{\mathbb H}_{\Lambda}[g_2](w)\\\\
&\qquad\times\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}\\\\
&+e_2\left\{\mathcal O^{\mathbb H}_{\Lambda}[f_2](w)\mathcal O^{\mathbb H}_{\Lambda}[g_1](w)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}\right.\\\\
&\qquad\left.+\mathcal O^{\mathbb H}_{\Lambda}\mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_1](w)\mathcal O^{\mathbb H}_{\Lambda}[g_1](w)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}\right\}\\\\
&=\left\{\big[\mathcal O^{\mathbb H}_{\Lambda}[f_1]\mathcal O^{\mathbb H}_{\Lambda}[g_1]- \mathcal O^{\mathbb H}_{\Lambda^{-2}}[\overline f_2]\mathcal O^{\mathbb H}_{\Lambda}[g_2]\big](w)\right.\\\\
&\qquad\left.e_2\big[\mathcal O^{\mathbb H}_{\Lambda}[f_2]\mathcal O^{\mathbb H}_{\Lambda}[g_1]- \mathcal O^{\mathbb H}_{\Lambda^{-1}}[\overline f_1]\mathcal O^{\mathbb H}_{\Lambda}[g_1]\big](w) \right\}\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}\\\\
&=\left\{\big[\mathcal O^{\mathbb H}_{\Lambda}[f_1]\mathcal O^{\mathbb H}_{\Lambda}[g_1]- \overline{\mathcal O^{\mathbb H}_{\Lambda}[ f_2]}\mathcal O^{\mathbb H}_{\Lambda}[g_2]\big](w)\right.\\\\
&\qquad\left.e_2\big[\mathcal O^{\mathbb H}_{\Lambda}[f_2]\mathcal O^{\mathbb H}_{\Lambda}[g_1]+\overline{\mathcal O^{\mathbb H}_{\Lambda}[ f_1]}\mathcal O^{\mathbb H}_{\Lambda}[g_1]\big](w) \right\}\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}\\\\
&=\mathcal O^{\mathbb H}_{\Lambda}[f](w)\mathcal O^{\mathbb H}_{\Lambda}[g](w)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}.
\end{align*}
Which completes the proof.
\end{proof}
\end{theorem}
\begin{definition}\label{def pro} For $f\in L^2(\mathbb R,\mathbb H)$ and $g\in L^1(\mathbb R,\mathbb H),$ define
\begin{equation}
(f\otimes g) =(f_2\otimes g_2 +\overline{\mathcal O^{\mathbb H}_{\Lambda^{-2}}[ f_1]}\otimes [g_1])+e_2(\overline{\mathcal O^{\mathbb H}_{\Lambda^{-2}}[ f_1]}\otimes[ g_2]-f_2\otimes g_1),
\end{equation}
where $ \otimes$ is the proposed definition of convolution.
\end{definition}
\begin{lemma}\label{lem pro}Under the assumptions of definition \ref{def pro}, we have
\begin{equation}
\left(\mathcal O^{\mathbb H}_{\Lambda}[f(t)]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ g(t)]\right)(u)=\mathcal O^{\mathbb H}_{\Lambda}\left[f(t) g(t)\exp\left\{\frac{e_2}{2B}\left(2w(Dp-Bq)-Dw^2\right)\right\}\right]
\end{equation}
where $ \mathcal O^{\mathbb H}_{\Lambda}[f]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ g]=\mathcal O^{\mathbb H}_{\Lambda}[fg]$
\end{lemma}
\begin{theorem}[Product theorem]\label{thm pro}Let $f,g$ be two given signal functions such that $f\in L^2(\mathbb R,\mathbb H)$ and $g\in L^1(\mathbb R,\mathbb H),$ , then for all $w\in \mathbb R$ we have
\begin{equation}\label{eqn pro}
\mathcal O^{\mathbb H}_{\Lambda}[\overline {f}g]=\mathcal O^{\mathbb H}_{\Lambda}[f]\otimes\mathcal O^{\mathbb H}_{\Lambda}[g].
\end{equation}
\begin{proof}
By Definition \ref{def pro} and Lemma \ref{lem pro}, we have
\begin{align*}
\mathcal O^{\mathbb H}_{\Lambda}[\overline {f}g]&=\mathcal O^{\mathbb H}_{\Lambda}[(\overline {f_1}-e_2f_2)(g_1+e_2g_2)]\\
&=\mathcal O^{\mathbb H}_{\Lambda}[\overline {f_1}g_1]+\mathcal O^{\mathbb H}_{\Lambda}[ {f_2}g_2]+e_2\left(\mathcal O^{\mathbb H}_{\Lambda}[\overline {f_1}g_2]-\mathcal O^{\mathbb H}_{\Lambda}[ {f_2}g_1]\right)\\
&=\mathcal O^{\mathbb H}_{\Lambda}[\overline {f_1}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_1}]+\mathcal O^{\mathbb H}_{\Lambda}[ {f_2}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_2}]\\
&\qquad\qquad +e_2\left(\mathcal O^{\mathbb H}_{\Lambda}[\overline {f_1}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_2}]- \mathcal O^{\mathbb H}_{\Lambda}[ {f_2}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_1}]\right)\\
&=\overline{\mathcal O^{\mathbb H}_{\Lambda^{-1}}[{f_1}]}\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_1}]+\mathcal O^{\mathbb H}_{\Lambda}[ {f_2}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_2}]\\
&\qquad\qquad +e_2\left(\overline{\mathcal O^{\mathbb H}_{\Lambda^{-1}}[ {f_1}]}\otimes\mathcal O^{\mathbb H}_{\Lambda}[{g_2}]- \mathcal O^{\mathbb H}_{\Lambda}[ {f_2}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_1}]\right)\\
&=\overline{\mathcal O^{\mathbb H}_{\Lambda^{-2}}\mathcal O^{\mathbb H}_{\Lambda}[{f_1}]}\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_1}]+\mathcal O^{\mathbb H}_{\Lambda}[ {f_2}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_2}]\\
&\qquad\qquad +e_2\left(\overline{\mathcal O^{\mathbb H}_{\Lambda^{-2}}\mathcal O^{\mathbb H}_{\Lambda}[ {f_1}]}\otimes\mathcal O^{\mathbb H}_{\Lambda}[{g_2}]- \mathcal O^{\mathbb H}_{\Lambda}[ {f_2}]\otimes\mathcal O^{\mathbb H}_{\Lambda}[ {g_1}]\right)\\
&=\mathcal O^{\mathbb H}_{\Lambda}[f]\otimes\mathcal O^{\mathbb H}_{\Lambda}[g].
\end{align*}
Which completes the proof.
\end{proof}
\end{theorem}
\parindent=0mm \vspace{.4in}
\section*{ Conclusion }
In this paper, we have proposed the definition of the novel integral transform known as the one-dimensional quaternion offset linear canonical transform (1D-QOLCT) which is embodiment of several well known signal processing tools. We then obtained Moyal's formula, convolution theorem and product theorem for proposed transform. Our future work about convolution and corellation theorems for two-sided short-time offset linear canonical transform and uncertainty principles for short-time quaternion offset linear canonical transform is in progress.
\section*{Declarations}
\begin{itemize}
\item Availability of data and materials: The data is provided on the request to the authors.
\item Competing interests: The authors have no competing interests.
\item Funding: No funding was received for this work
\item Author's contribution: Both the authors equally contributed towards this work.
\item Acknowledgements: This work is supported by the UGC-BSR Research Start Up Grant(No. F.30-498/2019(BSR)) provided by UGC, Govt. of India.
\end{itemize}
| 30,245
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TITLE: Why are my calculations wrong? picking balls from urns, probability
QUESTION [0 upvotes]: Three urns contain balls. Urn A contains 2 black balls and 2 red balls. Urn B contains 3 white balls and 1 black ball. Urn C contains 1 black, 1 white and 1 red ball.
An urn is chosen, and then two balls are removed one after the other.
The probability to chose urn A is 0.5, the probability to choose urn B is 0.25 and the probability to chose urn C is 0.25. in each urn any ball is equally likely to be removed.
Now I have to calculate the probability that at least one black ball is removed and the second ball removed is white.
In the former parts of this question, I already calculated the probability that at least one black ball is removed, and the probability is $ \frac{17}{24} $
and I calculated the probability that the second ball removes is white, the probability is $ \frac{13}{48} $.
Now I want to use the conditional probability formula and denote $ E_1 $ as the event that at least one balck ball removed, and $ E_2 $ as the event that the second ball removed is white.
Thus, $ \mathbb{P}\left(E_{1}\cap E_{2}\right)=\mathbb{P}\left(E_{1}|E_{2}\right)\cdot\mathbb{P}\left(E_{2}\right) $
I already calculated $ \mathbb{P}\left(E_{2}\right) $ so all I have to do is to calculate $ \mathbb{P}\left(E_{1}|E_{2}\right) $. given that $ E_2 $ occured, the only way at least one black ball is removed is if the first ball removed is black, also since we know that the second ball removed was white, it cannot be that urn A was picked. therefore the probability we want is the probability that urn B was picked multiplied by the probability that the first ball is black, plus the probability that urn C was picked multiplied by the probability that the first ball is black. That is :
$ \frac{1}{4}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{3} = \frac{7}{47} $
now if we'll mulyiply it by the probability of $ E_2 $ we'll get $ \frac{7}{47}\cdot\frac{13}{48} $ which leads to the wrong answer.
I'll now present the right answer, but I'd like to know where the first way falls, Im new to this and I dont understand what I did wrong.
This is the right way:
without using the conditional probability formula, assuming the second ball is white, the only possibilities we are interested in are $\left(B,black,white\right),\left(C,black,white\right) $
That is the probability:
$ \frac{1}{4}\cdot\frac{1}{4}\cdot1+\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{16}+\frac{1}{24}=\frac{5}{48} $.
Thanks in advance.
REPLY [0 votes]: Your first approach was wrong, because although you
have correctly identified the probability of white being
chosen (2nd) as 13/48, under this scenario, it is significantly
more likely that the white ball came from urn B than urn C.
Intuitively, this is because when urn B is selected, your chance
of a white ball is 75% vs the 33.33% from urn C. Intuiting to
the extreme, suppose that urn C had 100 balls, only 1 of which
was white. Then if a white ball is picked, it is much more
likely that it would have come from urn B.
This is the main problem with your first approach.
There is also an auxiliary problem.
Suppose that urn B is chosen, and it is assumed that the 2nd ball
chosen is white. Then the chance that the 1st ball chosen is black
is 1/3, not 1/4. This is because you have to eliminate one of the
white balls in urn B from contention.
An easier way to intuit this is to pretend that the white ball
is to be chosen 1st and the black ball is to be chosen 2nd. Then,
the consideration re the last paragraph becomes more obvious.
Further, an easy way to see that this type of intuition is required
is to pretend that urn B had two balls - 1 white and 1 black.
Then, under the assumption that Urn B was selected, and that the 1st ball chosen was white,
the chances would be certainty rather than 1/2 that the 2nd ball chosen
is black.
A similar error occurred with respect to your temporary consideration
of Urn C being chosen.
| 70,554
|
Property Attributes
- MLS#6323700
- TypeBackup
- CountyMARICOPA
- CityPhoenix
- NeighborhoodMoon Valley Gardens Plat B
- Zip85029
- StyleRanch
- Year Built1970
- Taxes$ 1665
- Price$ 475,000
- Bedrooms4
- Full Bathrooms2
- Half Bathrooms0
- Sqr Footage1823
- Lot Size10,020 SqFt
Listing Provided Courtesy Of:
A.Z. & Associates
Data Source:
Arizona Regional MLS (ARMLSN) - 22 - AZ
Property Description
Beautifully Remodeled Home with meticulous desert landscape and great curb appeal. Stunning NEW kitchen features self closing doors, granite counters, all SS appliances, and white wood cabinetry with stylish hardware. 20 x 20 tile floor, carpet, white wooden blinds, canned lighting, 6 ceiling fans, switches, outlets, doors,. Brick fireplace and patio access from great room. Refreshing pool with backyard covered patio, swaying palm trees, landscape w/irrigation, and endless blue skies! 2 RV Gates. A must see ''gem'' in the desert! It wont last long
General Features
Interior Features
Exterior Features
Amenities
- Foreclosure
- Views
- Short Sale
- New Construction
- Adult 55+
- Lease To Own
- No HOA Fees
- Furnished
- Primary On Main
- Air Conditioning
- Seller Finance
- Green
- Fixer Upper
- Horse
- Golf
- Fireplace
- Deck
- Garage
- Basement
- Pool
| 65,300
|
25 I received a dividend of € 14.85 from Agrium Inc.
- On July 25 I received a dividend of € 28.05 from General Electric Co.
- On July 27 I received a dividend of € 49.69 from Bank of Nova Scotia.
- On July 27 I received a dividend of € 20.14 from Cisco Systems Inc.
Dividend income reported after the deduction of taxes. I have also updated my Monthly Dividend sheet.
Transactions of My Portfolio:
- On July 26, 2016 I bought 50 shares of Omega Healthcare Investors Inc. (OHI)
News from My Portfolio Holdings:
July 25, 2016
- Golar LNG: Golar and Schlumberger Form OneLNG Joint Venture: OneLNG targets development of low cost gas reserves to LNG
- Gilead Sciences, Inc.: Announces Second Quarter 2016 Financial Results
July 26, 2016
- UPM-Kymmene Corporation: Interim report Q2/2016: UPM's comparable EBIT increased by 21%, cash flow reaching newhighs
- McDonald's Corporation: Reports Second Quarter 2016 Results
- T. Rowe Price Group, Inc.: Reports Second Quarter 2016 Results
- Baxter International Inc.: Reports Second Quarter 2016 Results and Raises Financial Outlook for Full-Year 2016
July 27, 2016
- Altria Group, Inc.: Reports 2016 Second-Quarter and First-Half Results; Raises Full-Year Earnings Guidance
- The Coca-Cola Company: Reports Second Quarter 2016 Results
- Agrium Inc.: Agrium Announces Investment in Finistere AgTech Venture Fund
- Archer Daniels Midland Company: ADM Continues to Expand Starch Business with New Products
July 28, 2016
- Neste Corporation: Interim Report for January-June 2016
- ConocoPhillips: Reports Second-Quarter 2016 Results; Continued Strong Operational Performance
- Digital Realty Trust, Inc.: Reports Second Quarter 2016 Results
July 29, 2016
- Chevron Corporation: Reports Second Quarter Loss of $1.5 Billion
Articles that caught my attention:
- ConocoPhillips (COP) Dividend Stock Analysis by Ken Faulkenberry at Dividend Value Builder
- Undervalued Dividend Growth Stock of the Week: Amgen, Inc. (AMGN) by Jason Fieber at Daily Trade Alert
- Warren Buffett’s Top 10 Pieces of Investment Advice by Simply Safe Dividends
- Wal-Mart Stores Is A Rare Deal On Sale In A Toppy Market by Passive Income Pursuit at Seeking Alpha
- Return Obsession by Ben Reynolds at Sure Dividend
- College Dorm Dividends by DivHut
- Taking Aim At Target by Passive Income Pursuit at Seeking Alpha
- Dear Vodafone: I Like You And Your 5% Yield, But Cash Flow Concerns Keep Me Away (For Now) by Dividend Drive at Seeking Alpha
- Emerson Electric Co. (EMR) Dividend Stock Analysis by D4L at Dividend Growth Stocks
- Reading Between the Lines by Roadmap2Retire
- Phillips 66: Value & Yield In An Overpriced Market by Ben Reynolds at Sure Dividend
- Four Attractively Valued Dividend Growth Stocks For Further Research by Dividend Growth Investor
- Is Caterpillar’s Dividend Still Safe? by Simply Safe Dividends
- 2 Avoids And 1 Potential Buy by Income Surfer at Seeking Alpha
- Duke Energy (DUK) Dividend Stock Analysis by D4L at Dividend Growth Stocks
- 5 Reasons why BT’s shares could be riskier than you think by John Kingham at UK Value Investor
- 5 High Yielding Dividend Investments To Buy And Hold For The Next 30 Years by Dividends4Me at Dividend Yield - Stock, Capital, Investment
- Axfood, dyr men stadig utdelningshöjare by petrusko
- Dividend Safety Analysis: Cisco (CSCO) by Simply Safe Dividends
- Dividend Discount Model Calculation Explained by DivGuy at Dividend Monk
Thanks for stopping by, have a nice next week!
Pretty good week for dividends this week DividendHawk, congrats :) Thanks for sharing these articles.
Tristan
Thanks for the mentions DH! And for sharing the rest of these great articles. Hope you have a great weekend.
Thanks for the mention DH. Your monthly dividend sheet looks like things are going along swimmingly for you! I hope you have a great week ahead.
-Bryan
Thank you for including my ConocoPhillips (COP) Dividend Stock Analysis!
| 100,993
|
So, as far as old vs. new posts, I decided that I'm going to do both. If I just try to do the old stuff, I'll feel like I'm just falling further behind & it would be true. Plus, with Halloween today & the holidays right around the corner, no way I'd get caught up in time to share this year's themed manis with you. So, I'll just try to state whether what you're looking at is old or new ... although in the end, it probably doesn't matter that much, right? Nails are nails & a mani that was beautiful a few months ago is still beautiful today.
This post is a new mani, for this year's Halloween. I wasn't sure at first if I was going to do any Halloween manis this year, I just didn't have much inspiration, and then I decided to do these ... they weren't my favorite ever, because as you know I'm a perfectionist, but it was fun to do & good practice at freehanding (which I need). Base colors were OPI Black Onyx & OPI Alpine Snow, bones done with AS & the skull done with black acrylic:
Here's the tutorial:
And I actually ended up doing another, even free-handier, Halloween design - stay tuned for all of the little Goodwill Halloween faces from the commercial!
Thanks for reading.
~Colette
I love this mani!! and I miss seeing your post on here :)
That is so awesome. Absolutely love it!
| 49,246
|
\begin{document}
\title[Monadicity theorem and weighted projective lines of tubular type]
{Monadicity theorem and weighted projective lines of tubular type}
\author[Jianmin Chen, Xiao-Wu Chen, Zhenqiang Zhou] {Jianmin Chen, Xiao-Wu Chen$^*$, Zhenqiang Zhou}
\thanks{$^*$ the corresponding author}
\subjclass[2010]{16W50, 14A22, 14H52}
\date{\today}
\keywords{weighted projective line, elliptic curve, equivariantization, monad, graded ring}
\maketitle
\dedicatory{}
\commby{}
\begin{abstract}
We formulate a version of Beck's monadicity theorem for abelian categories, which is applied to the equivariantization of abelian categories with respect to a finite group action. We prove that the equivariantization is compatible with the construction of quotient abelian categories by Serre subcategories. We prove that the equivariantization of the graded module category over a graded ring is equivalent to the graded module category over the same ring but with a different grading. We deduce from these results two equivalences between the category of (equivariant) coherent sheaves on a weighted projective line of tubular type and that on an elliptic curve, where the acting groups are cyclic and the two equivalences are somehow adjoint to each other.
\end{abstract}
\section{Introduction}
The close relationship between weighted projective lines of tubular type and elliptic curves is known to experts since the creation of weighted projective lines in \cite{GL87}. Roughly speaking, the category of coherent sheaves on a weighted projective line of tubular type is equivalent to the category of equivariant coherent sheaves on an elliptic curve with respect to a finite abelian group action. This result is implicitly exploited in \cite{Po} using ramified elliptic Galois coverings over the field of complex numbers; compare \cite{Ploog}. Moreover, the somehow converse result holds true by \cite{Len, Lentalk}: the category of coherent sheaves on an elliptic curve is equivalent to the category of equivariant coherent sheaves on a weighted projective line of tubular type with respect to a different finite abelian group action, which contains the action given by the Auslander-Reiten translation.
The goal of this paper is to re-exploit the above mentioned equivalences in an explicit form. We emphasize that our treatment of the four tubular types is in a uniform manner, and that we deduce these equivalences from general results on the equivariantization of abelian categories. These general results are direct applications of the famous Beck's monadicity theorem which characterizes the module category over a monad; see \cite[Chapter VI]{McL}. We mention that monads appear naturally, since the category of equivariant objects is isomorphic to the module category over a certain monad; see \cite{Ch,El2014}.
Let us describe the content of this paper. In Section 2, we collect some basic facts on monads and modules over monads. We formulate a version of Beck's monadicity theorem for abelian categories, which is convenient for applications; see Theorem \ref{thm:Beck}. We give a self-contained proof to Theorem \ref{thm:Beck}. In Section 3, we recall the notions of a group action on a category and the category of equivariant objects. The observation we need is that the category of equivariant objects is isomorphic to the category of modules over a certain monad; see Proposition \ref{prop:monadic}. In Section 4, we apply Theorem \ref{thm:Beck} to prove that the equivariantization of an abelian category is compatible with quotient abelian categories by Serre subcategories; indeed, a slightly more general result on the category of modules over an exact monad is obtained; see Proposition \ref{prop:app1} and Corollary \ref{cor:equiv}. In Section 5, we prove that the equivarianzation of the graded module category over a graded ring with respect to a degree-shift action is equivalent to the graded module category over the same ring but with a coarser grading; see Proposition \ref{prop:app2}. This enables us to recover \cite[Theorem 6.4.1]{NV} on the graded module category over the corresponding graded group ring; see Corollary \ref{cor:NV}. For the case that the acting group is finite abelian, we generalize slightly a result in \cite{Len} and prove that the equivariantization of the graded module category over a graded ring with respect to a twisting action is equivalent to the graded module category over the same ring but with a refined grading; see Proposition \ref{prop:recover}.
In Section 6, we recall some basic facts on the homogeneous coordinate algebra of a weighted projective line. We prove that the quotient graded module category over a restriction subalgebra of the homogeneous coordinate algebra is equivalent to a quotient of the category of coherent sheaves on the weighted projective line by a certain Serre subcategory; see Proposition \ref{prop:eff}. This motivates the notion of an effective subgroup of the grading group for the homogeneous coordinate algebra; see Definition \ref{defn:1}. For these subgroups, Proposition \ref{prop:eff} claims an equivalence between the quotient graded module category over the restriction subalgebra and the category of coherent sheaves on the weighted projective line.
The final section is devoted to the study of the relationship between weighted projective lines of tubular type and elliptic plane curves. We prove that the homogeneous coordinate algebra of an elliptic plane curve is isomorphic to the restriction subalgebra of the homogeneous coordinate algebra of the weighted projective line with respect to a suitable effective subgroup; see Theorem \ref{thm:R-S}. Applying Theorem \ref{thm:R-S} and the results in Sections 4 and 5, we prove the main result: the equivariantization of the category of coherent sheaves on a weighted projective line of tubular type with respect to a certain degree-shift action is equivalent to the category of coherent sheaves on an elliptic plane curve; the equivariantization of the category of coherent sheaves on an elliptic plane curve with respect to a certain twisting action is equivalent to the category of coherent sheaves on a weighted projective line of tubular type; see Theorem \ref{thm:main}. We mention that here the acting groups are cyclic, and that we need two different kinds of group actions to have these two equivalences. On the other hand, the two equivalences obtained are somehow adjoint to each other; see Remark \ref{rem:main}(1).
We emphasize that the two equivalences in Theorem \ref{thm:main} are contained in \cite{GL87} and \cite{Len, Lentalk}, which are presented in a slightly different form with sketched proofs; compare \cite{Hille} and Remark \ref{rem:main}(3). As pointed out in \cite{GL87}, these equivalences link the classification of indecomposable coherent sheaves on elliptic curves in \cite{At} with the classification of indecomposable modules of tubular algebras in \cite{Rin}.
\section{Modules over monads}
In this section, we recall some basic facts from \cite[Chapter VI]{McL} on monads and modules over monads. We formulate a version of Beck's monadicity theorem for abelian categories, for which we provide a self-contained proof.
Let $\mathcal{C}$ be a category. Recall that a \emph{monad} on $\mathcal{C}$ is a triple $(M, \eta, \mu)$ consisting of an endofunctor $M\colon \mathcal{C}\rightarrow \mathcal{C}$ and two natural transformations, the \emph{unit} $\eta\colon{\rm Id}_\mathcal{C}\rightarrow M$ and the \emph{multiplication} $\mu \colon M^2\rightarrow M$, subject to the relations $\mu \circ M\mu =\mu \circ \mu M$ and $\mu\circ M\eta={\rm Id}_M=\mu\circ \eta M$. We sometimes suppress the unit and the multiplication, and denote the monad simply by $M$.
Monads arise naturally in adjoint pairs. Assume that $F\colon \mathcal{C}\rightarrow \mathcal{D}$ is a functor which admits a right adjoint $U\colon \mathcal{D}\rightarrow \mathcal{C}$. We denote by $\eta\colon {\rm Id}_\mathcal{C}\rightarrow UF$ the unit and $\epsilon\colon FU\rightarrow {\rm Id}_\mathcal{D}$ the counit; they satisfy $\epsilon F\circ F\eta ={\rm Id}_F$ and $U\epsilon\circ \eta U={\rm Id}_U$. We denote this adjoint pair on $\mathcal{C}$ and $\mathcal{D}$ by the quadruple $(F, U; \eta, \epsilon)$. We suppress the unit and the counit when they are clear from the context.
Each adjoint pair $(F, U;\eta, \epsilon)$ on two categories $\mathcal{C}$ and $\mathcal{D}$ defines a monad $(M, \eta, \mu)$ on $\mathcal{C}$, where $M=UF\colon \mathcal{C}\rightarrow \mathcal{C}$ and $\mu=U\epsilon F\colon M^2=UFUF\rightarrow U{\rm Id}_\mathcal{D}F=M$. The resulting monad
$(M, \eta, \mu)$ on $\mathcal{C}$ is said to be \emph{defined} by the adjoint pair $(F, U; \eta, \epsilon)$. Indeed, as recalled now every monad is defined by some adjoint pair.
For a monad $(M, \eta, \mu)$ on $\mathcal{C}$, an $M$-\emph{module} is a pair $(X, \lambda)$ consisting of an object $X$ in $\mathcal{C}$ and a morphism $\lambda \colon M(X)\rightarrow X$ subject to the conditions $\lambda\circ M\lambda =\lambda\circ \mu_X$ and $\lambda\circ \eta_X={\rm Id}_X$; the object $X$ is said to be the \emph{underlying object} of the module. A morphism $f\colon (X, \lambda)\rightarrow (X', \lambda')$ between two $M$-modules is a morphism $f\colon X\rightarrow X'$ in $\mathcal{C}$ satisfying $f\circ \lambda=\lambda'\circ M(f)$. This gives rise to the category $M\mbox{-Mod}_\mathcal{C}$ of $M$-modules; moreover, we have the \emph{forgetful functor} $U_M\colon M\mbox{-Mod}_\mathcal{C}\rightarrow \mathcal{C}$. The functor $U_M$ is faithful.
We observe that each object $X$ in $\mathcal{C}$ yields an $M$-module $F_M(X)=(M(X), \mu_X)$, the \emph{free $M$-module} generated by $X$. Indeed, this gives rise to the \emph{free module functor} $F_M\colon \mathcal{C}\rightarrow M\mbox{-Mod}_\mathcal{C}$ sending $X$ to the free module $F_M(X)$, and a morphism $f\colon X\rightarrow Y$ to the morphism
$M(f)\colon F_M(X)\rightarrow F_M(Y)$.
We have the adjoint pair $(F_M, U_M; \eta, \epsilon_M)$ on $\mathcal{C}$ and $M\mbox{-Mod}_\mathcal{C}$, where for an
$M$-module $(X, \lambda)$, the counit $\epsilon_M$ is given such that $$(\epsilon_M)_{(X, \lambda)}=\lambda\colon F_M U_M(X,\lambda)=(M(X), \mu_X)\longrightarrow (X, \lambda).$$ We observe that this adjoint pair $(F_M, U_M; \eta, \epsilon_M)$ defines the given monad $(M, \eta, \mu)$.
The above adjoint pair $(F_M, U_M; \eta, \epsilon_M)$ enjoys the following universal property: for any adjoint pair $(F, U; \eta, \epsilon)$ on $\mathcal{C}$ and $\mathcal{D}$ that defines the monad $M$, there is a unique functor $$K\colon \mathcal{D}\longrightarrow M\mbox{-Mod}_\mathcal{C}$$ satisfying $KF=F_M$ and $U_MK=U$; see \cite[VI.3]{McL}. This unique functor $K$ will be referred as the \emph{comparison functor} associated to the adjoint pair $(F, U; \eta, \epsilon)$. For its construction, we have
\begin{align}\label{equ:K}
K(D)=(U(D), U\epsilon_D), \quad K(f)=U(f)
\end{align}
for an object $D$ and a morphism $f$ in $\mathcal{D}$. Here, we observe that $M=UF$ and that $(U(D), U\epsilon_D)$ is an $M$-module.
Following \cite[VI.3]{McL} the adjoint pair $(F, U)$ is \emph{monadic} (\emph{resp.} \emph{strictly monadic}) provided that the associated comparison functor $K\colon \mathcal{D}\rightarrow M\mbox{-Mod}_\mathcal{C}$ is an equivalence (\emph{resp.} an isomorphism). In these cases, we might identify $\mathcal{D}$ with $M\mbox{-Mod}_\mathcal{C}$.
The famous monadicity theorem of Beck gives intrinsic characterizations on (strictly) monadic adjoint pairs; see \cite[VI.7]{McL}. However, the conditions in Beck's monadicity theorem on coequalizers seem rather technical.
In the following result, we formulate a version of Beck's monadicity theorem for abelian categories, which is quite convenient for applications.
\begin{thm}\label{thm:Beck}
Let $F\colon \mathcal{A}\rightarrow \mathcal{B}$ be an additive functor on two abelian categories. Assume that $F$ admits a right adjoint $U\colon \mathcal{B}\rightarrow \mathcal{A}$ which is exact. Denote by $M$ the defined monad on $\mathcal{A}$. Then the comparison functor $K\colon \mathcal{B}\rightarrow M\mbox{-{\rm Mod}}_\mathcal{A}$ is an equivalence if and only if the functor $U\colon \mathcal{B}\rightarrow \mathcal{A}$ is faithful.
\end{thm}
The ``only if" part of Theorem \ref{thm:Beck} is trivial, since $U=U_MK$ and the forgetful functor $U_M$ is faithful. We will present two proofs for the ``if" part. The first one is an application of the original version of Beck's monadicity theorem, while the second one is more self-contained and seems to be of independent interest.
The following observation is well known.
\begin{lem}\label{lem:cok}
Let $F\colon \mathcal{A}\rightarrow \mathcal{B}$ be an exact functor on two abelian categories, which is faithful. Then $F$ preserves and reflects cokernels, that is, for two morphisms $f\colon X\rightarrow Y$ and $c\colon Y\rightarrow C$ in $\mathcal{A}$, $c$ is a cokernel of $f$ if and only if $F(c)$ is a cokernel of $F(f)$.
\end{lem}
\begin{proof}
The ``only if" part follows from the right exactness of $F$.
For the ``if" part, we assume that $F(c)$ is a cokernel of $F(f)$. In particular, $F(c\circ f)=0$, and thus $c\circ f=0$ since $F$ is faithful. By the exactness of $F$, we have the following fact: for any complex $\xi\colon X\rightarrow Y\rightarrow Z$ in $\mathcal{A}$ with cohomology $H$ at $Y$, the complex $F(\xi)\colon F(X)\rightarrow F(Y)\rightarrow F(Z)$ in $\mathcal{B}$ has cohomology $F(H)$ at $F(Y)$. Since $F$ is faithful, we infer that the complex $F(\xi)$ is exact at $F(Y)$, that is, $F(H)\simeq 0$, if and only if $H\simeq 0$, that is, the complex $\xi$ is exact at $Y$. We apply this fact to the complexes $X\stackrel{f}\rightarrow Y\stackrel{c}\rightarrow C$ and $Y\stackrel{c}\rightarrow C\rightarrow 0$. We infer that $c$ is a cokernel of $f$.
\end{proof}
\vskip 5pt
\noindent \emph{The first proof of the ``if" part of Theorem \ref{thm:Beck}.}\quad Recall that a coequalizer of two parallel morphisms in an additive category equals a cokernel of their difference. Then we apply Lemma \ref{lem:cok} to the functor $U$, and infer that $U$ preserves and reflects coequalizers. Then the comparison functor $K$ is an equivalence by the ``equivalence" version of Beck's monadicity theorem; see \cite[VI.7, Exercise 6]{McL}. \hfill $\square$
\vskip 5pt
The second proof is an application of a slightly more general result, which seems to be standard in relative homological algebra.
Let $\mathcal{X}$ be a full subcategory of a category $\mathcal{C}$. Let $C$ be an object in $\mathcal{C}$. Recall that a \emph{right $\mathcal{X}$-approximation} of $C$ is a morphism $f\colon X\rightarrow C$ with $X\in \mathcal{X}$ such that any morphism $t\colon T\rightarrow C$ with $T\in \mathcal{X}$ factors through $f$, that is, there exists a morphism $t'\colon T\rightarrow X$ with $t=f\circ t'$; compare \cite[Section 3]{AS}. In case that $\mathcal{C}$ is abelian,
an \emph{$\mathcal{X}$-presentation} of $C$ means an exact sequence $X_1\stackrel{g}\rightarrow X_0\stackrel{f}\rightarrow C\rightarrow 0$ such that $f$ is a right $\mathcal{X}$-approximation of $C$ and that the induced morphism $\bar{g}\colon X_1\rightarrow {\rm Ker}\; f$ is a right $\mathcal{X}$-approximation of ${\rm Ker}\; f$.
For a functor $F\colon \mathcal{C}\rightarrow \mathcal{D}$ and a full subcategory $\mathcal{X}\subseteq \mathcal{C}$, we denote by $F(\mathcal{X})$ the full subcategory of $\mathcal{D}$ consisting of objects that are isomorphic to $F(X)$ for some object $X$ in $\mathcal{X}$. In particular, we write ${\rm Im}\; F=F(\mathcal{C})$, the \emph{essential image} of $F$.
\begin{exm}\label{exm:1}
Let $(F, U; \eta, \epsilon)$ be an adjoint pair on two categories $\mathcal{C}$ and $\mathcal{D}$. Then for any object $D$ in $\mathcal{D}$, the counit $\epsilon_D\colon FU(D)\rightarrow D$ is a right ${\rm Im}\; F$-approximation. Indeed, any morphism $t\colon F(C)\rightarrow D$ factors through $\epsilon_D$ as $t=\epsilon_D\circ F(U(t)\circ \eta_C)$.
Let us state a special case explicitly. For a monad $M=(M, \eta, \mu)$ on $\mathcal{C}$, we apply the above to the adjoint pair $(F_M, U_M; \eta, \epsilon_M)$. Then for any $M$-module $(X, \lambda)$, the counit
\begin{align}\label{equ:epi}
(\epsilon_M)_{(X, \lambda)}=\lambda\colon F_MU_M(X, \lambda)=(M(X), \mu_X)\longrightarrow (X, \lambda)
\end{align}
is a right ${\rm Im}\; F_M$-approximation of $(X, \lambda)$ in $M\mbox{-{\rm Mod}}_\mathcal{C}$; it is an epimorphism, since $\lambda\circ \eta_X={\rm Id}_X$.
\end{exm}
The following general result seems to be standard in relative homological algebra, whose proof is also standard.
\begin{prop}\label{prop:equi}
Let $F\colon \mathcal{A}\rightarrow \mathcal{B}$ be a right exact functor on two abelian categories. Let $\mathcal{X}$ be a full subcategory of $\mathcal{A}$. We assume that the following conditions are satisfied.
\begin{enumerate}
\item[(i)] For any object $C$ in $\mathcal{A}$, there exists an epimorphism $f\colon X\rightarrow C$ which is a right $\mathcal{X}$-approximation such that $F(f)$ is a right $F(\mathcal{X})$-approximation of $F(C)$.
\item[(ii)] The restricted functor $F|_\mathcal{X}\colon \mathcal{X}\rightarrow \mathcal{B}$ is fully faithful.
\end{enumerate}
Then the functor $F\colon \mathcal{A}\rightarrow \mathcal{B}$ is fully faithful. Assume that in addition the following condition is satisfied.
\begin{enumerate}
\item[(iii)] For any object $B$ in $\mathcal{B}$, there is an epimorphism $F(A)\rightarrow B$ for some object $A$ in $\mathcal{A}$.
\end{enumerate}
Then the functor $F\colon \mathcal{A} \rightarrow \mathcal{B}$ is an equivalence.
\end{prop}
\begin{proof}
Let $A, A'$ be two objects in $\mathcal{A}$. By the right exactness of $F$ and (i), we have two $\mathcal{X}$-presentations $\xi\colon X_1\stackrel{d}\rightarrow X_0\stackrel{p}\rightarrow A\rightarrow 0$ and $\xi'\colon X'_1\stackrel{d'}\rightarrow X'_0\stackrel{p'}\rightarrow A'\rightarrow 0$ such that $F(\xi)$ and $F(\xi')$ are $F(\mathcal{X})$-presentations of $F(A)$ and $F(A')$, respectively. To show that $F$ is full, take a morphism $t\colon F(A)\rightarrow F(A')$. Since $F(\xi')$ is an $F(\mathcal{X})$-presentation, we have the following commutative diagram
\[\xymatrix{
F(X_1) \ar@{.>}[d]^-{t_1}\ar[r]^{F(d)} & F(X_0) \ar@{.>}[d]^-{t_0} \ar[r]^{F(p)} & F(A)\ar[d]^-{t} \ar[r] & 0\\
F(X'_1) \ar[r]^-{F(d')} & F(X'_0) \ar[r]^{F(p')} & F(A') \ar[r] & 0.
}\]
By (ii), there exist morphisms $s_i\colon X_i\rightarrow X'_i$ such that $F(s_i)=t_i$ and $s_0\circ d=d'\circ s_1$. The identity $s_0\circ d=d'\circ s_1$ implies the existence of a morphism $s\colon A\rightarrow A'$ such that $s \circ p=p'\circ s_0$. It follows that $F(s)=t$ by comparing commutative diagrams.
The faithfulness of $F$ follows by a converse argument. Indeed, given a morphism $s\colon A\rightarrow A'$ with $F(s)=0$, we have morphisms $s_i\colon X_i\rightarrow X'_i$ such that $s \circ p=p'\circ s_0$ and $s_0\circ d=d'\circ s_1$; here, we use the $\mathcal{X}$-presentation $\xi'$. Then in the above diagram, $t=F(s)=0$ and thus $t_0=F(s_0)$ factors through $F(d')$. By (ii), $s_0$ factors through $d'$, and thus $s\circ p=p'\circ s_0=0$. We infer that $s=0$, since $p$ is epic.
We assume that (iii) is satisfied. Then any object $B$ fits into an exact sequence $F(A')\stackrel{g}\rightarrow F(A)\rightarrow B\rightarrow 0$. By the fully faithfulness of $F$, there exists a morphism $s\colon A'\rightarrow A$ satisfying $F(s)=g$. It follows that $F({\rm Cok}\; s)\simeq B$. This proves the denseness of $F$, and thus $F$ is an equivalence.
\end{proof}
\vskip 5pt
\noindent \emph{The second proof of the ``if" part of Theorem \ref{thm:Beck}.}\quad Since $F$ has a right adjoint, it is right exact. Then the endofunctor $M=UF$ on $\mathcal{A}$ is right exact. It follows that the category $M\mbox{-Mod}_\mathcal{A}$ is abelian; indeed, a sequence of $M$-modules is exact if and only if the corresponding sequence of the underlying objects in $\mathcal{A}$ is exact.
Recall from (\ref{equ:K}) that $K(B)=(U(B), U\epsilon_B)$ and that $K(f)=U(f)$ for an object $B$ and a morphism $f$ in $\mathcal{B}$. Since the functor $U$ is exact and faithful, it follows that the comparison functor $K\colon \mathcal{B}\rightarrow M\mbox{-Mod}_\mathcal{A}$ is exact and faithful. Consider $\mathcal{X}={\rm Im}\; F$ as a full subcategory of $\mathcal{B}$. We claim that the pair $K$ and $\mathcal{X}$ satisfy the three conditions in Proposition \ref{prop:equi}. Then we infer that $K$ is an equivalence.
For the claim, we verify the conditions (i)-(iii). For (i), take any object $B$ in $\mathcal{B}$ and consider the counit $\epsilon_B\colon FU(B)\rightarrow B$, which is a right $\mathcal{X}$-approximation by Example \ref{exm:1}. We observe that $K(\epsilon_B)$ equals $(\epsilon_M)_{K(B)}$, the counit $\epsilon_M$ applied on the $M$-module $K(B)$. In particular, by Example \ref{exm:1} $K(\epsilon_B)$ is a right ${\rm Im}\; F_M$-approximation of $K(B)$ in $M\mbox{-Mod}_\mathcal{A}$, which is an epimorphism. Recall that $F_M=KF$ and thus ${\rm Im}\; F_M=K(\mathcal{X})$. Hence, the epimorphism $K(\epsilon_B)$ is a right $K(\mathcal{X})$-approximation of $K(B)$. By Lemma \ref{lem:cok} the functor $K$ reflects epimorphisms. It follows that $\epsilon_B$ is an epimorphism. This is the required morphism in (i).
The condition (ii) is well known, since the restriction of $K$ on $\mathcal{X}={\rm Im}\; F$ is fully faithful; see \cite[Lemma 3.3]{Ch} and compare \cite[VI.3 and VI.5]{McL}. The condition (iii) follows from the epimorphism (\ref{equ:epi}) for any $M$-module $(X, \lambda)$, since $F_MU_M=KFU_M$ and thus $(M(X), \mu_X)=KF(X)$. Set $B=F(X)$. In particular, we have by (\ref{equ:epi}) a required epimorphism $K(B)\rightarrow (X, \lambda).$
\hfill $\square$
\vskip 5pt
\section{Equivariant objects as modules}
In this section, we recall the notions of a group action on a category and the category of equivariant objects. We give a direct proof of the fact that in the additive case, the category of equivariant objects is isomorphic to the category of modules over a certain monad.
Let $G$ be an arbitrary group. We write $G$ multiplicatively and denote its unit by $e$. Let $\mathcal{C}$ be an arbitrary category.
The notion of a group action on a category is well known; compare \cite{De,RR,DGNO}. An \emph{action} of $G$ on $\mathcal{C}$ consists of the data $\{F_g, \varepsilon_{g, h}|\; g, h\in G\}$, where each $F_g\colon \mathcal{C}\rightarrow \mathcal{C}$ is an auto-equivalence and each $\varepsilon_{g, h}\colon F_gF_h\rightarrow F_{gh}$ is a natural isomorphism such that a $2$-cocycle condition holds, that is,
\begin{align}\label{equ:2-coc}
\varepsilon_{gh, k}\circ \varepsilon_{g,h}F_k=\varepsilon_{g, hk}\circ F_g\varepsilon_{h,k}
\end{align}
for all $g, h, k\in G$. We observe that there exists a unique natural isomorphism $u\colon F_e\rightarrow {\rm Id}_\mathcal{C}$, called the \emph{unit} of the action, satisfying $\varepsilon_{e,e}=F_eu$; moreover, we have $F_{e}u=uF_e$ by (\ref{equ:2-coc}).
The given action is \emph{strict} provided that each $F_g$ is an automorphism and each isomorphism $\varepsilon_{g, h}$ is the identity, in which case the unit is also the identity. Therefore, a strict action coincides with a group homomorphism from $G$ to the automorphism group of $\mathcal{C}$.
Let $G$ act on $\mathcal{C}$. A \emph{$G$-equivariant object} in $\mathcal{C}$ is a pair $(X, \alpha)$, where $X$ is an object in $\mathcal{C}$ and $\alpha$ assigns for each $g\in G$ an isomorphism $\alpha_g\colon X\rightarrow F_g(X)$ subject to the relations
\begin{align}\label{equ:rel}
\alpha_{gg'}=(\varepsilon_{g,g'})_X \circ F_g(\alpha_{g'}) \circ \alpha_g.\end{align}
These relations imply that $\alpha_e=u^{-1}_X$. A morphism $\theta\colon (X, \alpha)\rightarrow (Y, \beta)$ between two $G$-equivariant objects is a morphism $\theta\colon X\rightarrow Y$ in $\mathcal{C}$ such that
$\beta_g\circ \theta=F_g(\theta)\circ \alpha_g$ for all $g\in G$. This gives rise to
the category $\mathcal{C}^G$ of $G$-equivariant objects, and the \emph{forgetful functor}
$U\colon \mathcal{C}^G\rightarrow \mathcal{C}$ defined by $U(X, \alpha)=X$. The process forming the category $\mathcal{C}^G$ of equivariant objects is known as the \emph{equivariantization} with respect to the group action; see \cite{DGNO}.
In what follows, we assume that the group $G$ is finite and that $\mathcal{C}$ is an additive category. In this case the forgetful functor $U$
admits a left adjoint $F\colon \mathcal{C}\rightarrow \mathcal{C}^G$, which is known as the \emph{induction functor}; see \cite[Lemma 4.6]{DGNO}.
The functor $F$ is defined as follows: for an object $X$, set $F(X)=(\oplus_{h\in G} F_h(X), \varepsilon)$, where for each $g\in G$, the isomorphism ${\rm \varepsilon}_g \colon \oplus_{h\in G} F_h(X)\rightarrow F_g(\oplus_{h\in G} F_h(X))$ is diagonally induced by the isomorphism $(\varepsilon_{g, g^{-1}h})_X^{-1}\colon F_h(X)\rightarrow F_g(F_{g^{-1}h}(X))$. Here, to verify that $F(X)$ is indeed an equivariant object, we need the $2$-cocycle condition (\ref{equ:2-coc}). The functor $F$ sends a morphism $\theta\colon X\rightarrow Y$ to $F(\theta)=\oplus_{h\in G} F_h(\theta)\colon F(X)\rightarrow F(Y)$.
For an object $X$ in $\mathcal{C}$ and an object $(Y, \beta)$ in $\mathcal{C}^G$,
a morphism $F(X)\rightarrow (Y, \beta)$ is of the form $\sum_{h\in G}\theta_h\colon \oplus_{h\in G} F_h(X)\rightarrow Y$
satisfying $F_g(\theta_h)=\beta_g\circ \theta_{gh}\circ (\varepsilon_{g, h})_X$ for any $g, h\in G$. The adjunction of $(F, U)$ is given by the
following natural isomorphism
\begin{align*}
{\rm Hom}_{\mathcal{C}^G} (F(X), (Y, \beta))\stackrel{\sim}\longrightarrow {\rm Hom}_\mathcal{C}(X, U(Y, \beta))
\end{align*}
sending $\sum_{h\in G} \theta_h$ to $\theta_e \circ u_X^{-1}$. The corresponding unit $\eta\colon {\rm Id}_\mathcal{C}\rightarrow UF$ is
given such that $\eta_X=(u_X^{-1}, 0, \cdots, 0)^t$, where `t' denotes the transpose;
the counit $\epsilon\colon FU\rightarrow {\rm Id}_{\mathcal{C}^G}$ is given such that
$\epsilon_{(Y, \beta)}=\sum_{h\in G} \beta_h^{-1}$.
Let us compute the monad $M=(UF, \eta, \mu)$ defined by the adjoint pair $(F, U; \eta, \epsilon)$. This monad is said to be \emph{defined} by the group action. The endofunctor $M\colon \mathcal{C}\rightarrow \mathcal{C}$ is given by $M(X)=\oplus_{h\in G} F_h(X)$ and $M(\theta)=\oplus_{h\in G} F_h(\theta)$ for a morphism $\theta$ in $\mathcal{C}$. The multiplication $\mu\colon M^2\rightarrow M$ is given by
$$\mu_X=U\epsilon_{F(X)}\colon M^2(X)=\oplus_{h, g\in G} F_hF_g(X) \longrightarrow M(X)=\oplus_{h\in G} F_h(X)$$
with the property that the corresponding entry
$F_hF_g(X)\rightarrow F_{h'}(X)$ is $\delta_{hg, h'}(\varepsilon_{h, g})_X$; here, $\delta$ is the Kronecker symbol.
The following result shows that the category of equivariant objects is isomorphic to the category of $M$-modules. Roughly speaking, equivariant objects are modules. We mention that the result is an immediate consequence of Beck's monadicity theorem in \cite[VI.7]{McL}; see also \cite[Lemma 4.3]{Ch} and compare \cite[Proposition 3.10]{El2014}. We provide a direct proof for completeness.
\begin{prop}\label{prop:monadic}
Let $\mathcal{C}$ be an additive category and $G$ be a finite group acting on $\mathcal{C}$. Keep the notation as above.
Then the adjoint pair $(F,U; \eta, \epsilon)$ is strictly monadic, that is, the associated comparison functor $K\colon \mathcal{C}^G\rightarrow M\mbox{-{\rm Mod}}_\mathcal{C}$ is an isomorphism of categories.
\end{prop}
\begin{proof}
We have just computed explicitly the monad $M$. Let us take a closer look at $M$-modules. An $M$-module is a pair $(X, \lambda)$ with $\lambda=\sum_{h\in G}\lambda_h\colon M(X)=\oplus_{h\in G} F_h(X)\rightarrow X$.
The condition $\lambda\circ \eta_X={\rm Id}_X$ is equivalent to $\lambda_e=u_X$,
and $\lambda\circ M(\lambda)=\lambda\circ \mu_X$ is equivalent to
$\lambda_{hg}\circ \varepsilon_{h,g}=\lambda_h\circ F_h(\lambda_g)$ for any $h, g\in G$. Hence, if we
set $\alpha_h\colon X\rightarrow F_h(X)$ to be $(\lambda_h)^{-1}$ and compare (\ref{equ:rel}), we obtain an object $(X, \alpha)\in \mathcal{C}^G$.
Roughly speaking, the maps $\lambda_h$'s carry the same information as $\alpha_h$'s.
Recall from (\ref{equ:K}) that the associated comparison functor $K\colon \mathcal{C}^G\rightarrow M\mbox{-Mod}_\mathcal{C}$ is constructed such that $K(X, \alpha)=(U(X, \alpha), U\epsilon_{(X, \alpha)})$, which equals the $M$-module $(X, \lambda)$ with $\lambda_h=(\alpha_h)^{-1}$ by the explicit form of the counit $\epsilon$. It follows immediately that $K$ induces a bijection on objects, and is fully faithful. We infer that $K$ is an isomorphism of categories.
\end{proof}
\begin{rem}\label{rem:monadic}
Proposition \ref{prop:monadic} can be extended slightly. Let $G$ be a group whose cardinality $|G|$ might be infinite. Assume that the additive category $\mathcal{C}$ has coproducts with any index set of cardinality less or equal to $|G|$. For a $G$-action on $\mathcal{C}$, we define the induction functor $F$ and the monad $M$ as above by replacing finite coproducts by coproducts indexed by a possibly infinite index set. The same argument proves that the comparison functor $K\colon \mathcal{C}^G\rightarrow M\mbox{-{\rm Mod}}_\mathcal{C}$ is an isomorphism of categories.
\end{rem}
\section{Exact monads and quotient abelian categories}
In this section, we give the first application of Theorem \ref{thm:Beck}, which states that the formation of the module category of an exact monad is compatible with quotient abelian categories by Serre subcategories. This result applies to the category of equivariant objects in an abelian category.
\subsection{}
Let $\mathcal{A}$ be an abelian category. A \emph{Serre subcategory} $\mathcal{N}$ of $\mathcal{A}$ is by definition a full subcategory which is closed under subobjects, quotient objects and extensions. In other words, for an exact sequence $0\rightarrow X\rightarrow Y\rightarrow Z\rightarrow 0$ in $\mathcal{A}$, $Y$ lies in $\mathcal{N}$ if and only if both $X$ and $Z$ lie in $\mathcal{N}$. It follows that a Serre subcategory $\mathcal{N}$ is abelian and the inclusion functor $\mathcal{N}\rightarrow \mathcal{A}$ is exact.
For a Serre subcategory $\mathcal{N}$ of $\mathcal{A}$, we denote by $\mathcal{A}/\mathcal{N}$ the \emph{quotient abelian category}; see \cite{Ga62}. The objects of $\mathcal{A}/\mathcal{N}$ are the same as $\mathcal{A}$, and for two objects $X$ and $Y$, a morphism in $\mathcal{A}/\mathcal{N}$ is represented by a morphism $X'\rightarrow Y/Y'$, where $X'\subseteq X$ and $Y'\subseteq Y$ are subobjects with both $X/X'$ and $Y'$ in $\mathcal{N}$. The quotient functor $q\colon \mathcal{A}\rightarrow \mathcal{A}/\mathcal{N}$ sends an object $X$ to $X$, a morphism $f\colon X\rightarrow Y$ to the morphism $q(f)$, which is represented by $f\colon X\rightarrow Y$; here, the corresponding $X'$ and $Y'$ are $X$ and $0$, respectively. The functor $q$ is exact with its essential kernel ${\rm Ker}\; q=\mathcal{N}$. In particular, for a morphism $f\colon X\rightarrow Y$ in $\mathcal{A}$, $q(f)=0$ if and only if its image ${\rm Im}\; f$ lies in $\mathcal{N}$.
Let $F\colon \mathcal{A}\rightarrow \mathcal{A}'$ be an exact functor. Then the essential kernel ${\rm Ker}\; F$ is a Serre subcategory of $\mathcal{A}$, and $F$ induces a unique exact functor $F'\colon \mathcal{A}/{{\rm Ker}\; F}\rightarrow \mathcal{A}'$. We say that $F$ is a \emph{quotient functor} provided that $F'$ is an equivalence.
Let $F\colon \mathcal{A}\rightarrow \mathcal{A}'$ be an exact functor. Assume that $\mathcal{N}\subseteq \mathcal{A}$ and $\mathcal{N}'\subseteq \mathcal{A}'$ are Serre subcategories such that $F(\mathcal{N})\subseteq \mathcal{N}'$. Then there is a uniquely induced exact functor $\bar{F}\colon \mathcal{A}/\mathcal{N}\rightarrow \mathcal{A}'/{\mathcal{N}'}$.
\begin{lem}\label{lem:quoinduced}
Keep the notation as above. Assume that $F\colon \mathcal{A}\rightarrow \mathcal{A}'$ is a quotient functor. Then the induced functor $\bar{F}\colon \mathcal{A}/\mathcal{N}\rightarrow \mathcal{A}'/{\mathcal{N}'}$ is also a quotient functor.
\end{lem}
\begin{proof}
Denote by $\mathcal{C}$ the inverse image of $\mathcal{N}'$ under $F$. Then $\mathcal{C}$ is a Serre subcategory of $\mathcal{A}$ containing $\mathcal{N}$ and ${\rm Ker}\; F$. The essential kernel ${\rm Ker}\; \bar{F}$ of $\bar{F}$ equals $\mathcal{C}/\mathcal{N}$.
We observe the following isomorphisms of abelian categories
$$(\mathcal{A}/\mathcal{N})/(\mathcal{C}/\mathcal{N})\simeq \mathcal{A}/\mathcal{C}\simeq (\mathcal{A}/{{\rm Ker}\; F})/(\mathcal{C}/{{\rm Ker}\; F}).$$
By the assumption, we have the equivalence $F'\colon \mathcal{A}/{{\rm Ker}\; F}\stackrel{\sim}\longrightarrow \mathcal{A}'$, which induces an equivalence $(\mathcal{A}/{{\rm Ker}\; F})/(\mathcal{C}/{{\rm Ker}\; F}) \stackrel{\sim}\longrightarrow \mathcal{A}'/\mathcal{N}'$. Consequently, we have the required equivalence $(\mathcal{A}/\mathcal{N})/{\rm Ker}\; \bar{F}\stackrel{\sim}\longrightarrow \mathcal{A}'/\mathcal{N}'$.
\end{proof}
Let $M\colon \mathcal{A}\rightarrow \mathcal{A}$ be an \emph{exact} monad on $\mathcal{A}$, that is, $M=(M, \eta, \mu)$ is a monad on $\mathcal{A}$ and the endofunctor $M$ is exact. Recall that the category $M\mbox{-Mod}_\mathcal{A}$ of $M$-modules is abelian, where a sequence of $M$-modules is exact if and only if the corresponding sequence of underlying objects in $\mathcal{A}$ is exact. It follows that both the free module functor $F_M\colon \mathcal{A}\rightarrow M\mbox{-Mod}_\mathcal{A}$ and the forgetful functor $U_M\colon M\mbox{-Mod}_\mathcal{A}\rightarrow \mathcal{A}$ are exact.
Let $\mathcal{N}\subseteq \mathcal{A}$ be a Serre subcategory such that $M(\mathcal{N})\subseteq \mathcal{N}$. Then we have the induced endofunctor $\bar{M}\colon \mathcal{A}/\mathcal{N}\rightarrow \mathcal{A}/\mathcal{N}$; moreover, the natural transformations $\eta$ and $\mu$ induce natural transformations $\bar{\eta}\colon {\rm Id}_{\mathcal{A}/\mathcal{N}}\rightarrow \bar{M}$ and $\bar{\mu}\colon \bar{M}^2\rightarrow \bar{M}$. Indeed, we obtain a monad $\bar{M}=(\bar{M}, \bar{\eta}, \bar{\mu})$ on $\mathcal{A}/\mathcal{N}$, called the \emph{induced monad} of $M$.
We will need the following standard fact.
\begin{lem}\label{lem:quotientAbel}
Let $F\colon \mathcal{A}\rightarrow \mathcal{A}'$ be an exact functor between two abelian categories, which has an exact right
adjoint $U\colon \mathcal{A}'\rightarrow \mathcal{A}$. Assume that $\mathcal{N}\subseteq \mathcal{A}$ and $\mathcal{N}'\subseteq \mathcal{A}'$ are Serre subcategories such that $F(\mathcal{N})\subseteq \mathcal{N}'$ and $U(\mathcal{N}')\subseteq \mathcal{N}$. Then the following statements hold.
\begin{enumerate}
\item The induced functor $\bar{F}\colon \mathcal{A}/\mathcal{N}\rightarrow \mathcal{A}'/{\mathcal{N}'}$ is left adjoint to the induced functor $\bar{U}\colon \mathcal{A}'/\mathcal{N}'\rightarrow \mathcal{A}/\mathcal{N}$.
\item The monad on $\mathcal{A}/\mathcal{N}$ defined by the adjoint pair $(\bar{F}, \bar{U})$ coincides with the induced monad of the one on $\mathcal{A}$ defined by the adjoint pair $(F, U)$.
\item Denote by $U^{-1}(\mathcal{N})$ the inverse image of $\mathcal{N}$. Assume that $U^{-1}(\mathcal{N})=\mathcal{N}'$. Then the induced functor $\bar{U}$ is faithful.
\end{enumerate}
\end{lem}
\begin{proof}
Denote the given adjoint pair by $(F, U; \eta, \epsilon)$ and its defined monad by $(M=UF, \eta, \mu)$.
We observe that the unit $\eta$ (\emph{resp.},
the counit $\epsilon$) induces naturally a natural transformation
$\bar{\eta}\colon {\rm Id}_{\mathcal{A}/\mathcal{N}}
\rightarrow \bar{U}\bar{F}$ (\emph{resp.}, $\bar{\epsilon}\colon
\bar{F} \bar{U} \rightarrow {\rm Id}_{\mathcal{A}'/{\mathcal{N}'}}$). Then we apply
\cite[IV.1, Theorem 2(v)]{McL} to deduce the adjoint pair $(\bar{F}, \bar{U}; \bar{\eta}, \bar{\epsilon})$. It follows that the monad defined by this adjoint pair coincides with $(\bar{M}, \bar{\eta}, \bar{\mu})$, the induced monad of $M$.
For (3), take any morphism $\theta\colon X\rightarrow Y$ in $\mathcal{A}'/\mathcal{N}'$ with $\bar{U}(\theta)=0$. We assume that $\theta$ is represented by a morphism $f\colon X'\rightarrow Y/Y'$, where $X'\subseteq X$ and $Y'\subseteq Y$ are subobjects satisfying that both $X/X'$ and $Y'$ lie in $\mathcal{N}'$. Then $\bar{U}(\theta)$ is represented by $U(f)\colon U(X')\rightarrow U(Y/Y')$, and thus $q(U(f))=0$ in $\mathcal{A}/\mathcal{N}$, or equivalently, the image ${\rm Im}\;U(f)$ lies in $\mathcal{N}$. We observe that ${\rm Im}\; U(f)\simeq U({\rm Im}\; f)$ and then ${\rm Im}\; f$ lies in $\mathcal{N}'$, since $U^{-1}(\mathcal{N})=\mathcal{N}'$. We infer that $\theta$ equals zero in $\mathcal{A}'/{\mathcal{N}'}$. We are done. \end{proof}
\subsection{}
Let $\mathcal{A}$ be an abelian category, and $M\colon \mathcal{A} \rightarrow \mathcal{A}$ be an exact monad. We will apply Lemma \ref{lem:quotientAbel} to the adjoint pair $(F_M, U_M)$.
Let $\mathcal{N}\subseteq \mathcal{A}$ be a Serre subcategory with $M(\mathcal{N})\subseteq \mathcal{N}$. Then we have the restricted
exact monad $M\colon \mathcal{N}\rightarrow \mathcal{N}$ and the category $M\mbox{-Mod}_\mathcal{N}$ of $M$-modules in $\mathcal{N}$. We view $M\mbox{-Mod}_\mathcal{N}$ as a full subcategory of $M\mbox{-Mod}_\mathcal{A}$, in other words, an $M$-module $(X, \lambda)$ lies in $M\mbox{-Mod}_\mathcal{N}$ if and only if the underlying object $X$ lies in $\mathcal{N}$. We observe that $M\mbox{-Mod}_\mathcal{N}\subseteq M\mbox{-Mod}_\mathcal{A}$ is a Serre subcategory.
Consider the induced monad $\bar{M}$ on $\mathcal{A}/\mathcal{N}$. We observe that an $M$-module $(X, \lambda)$ yields an $\bar{M}$-module $(X, q(\lambda))$ on $\mathcal{A}/\mathcal{N}$, where $q(\lambda)\colon \bar{M}(X)=qM(X)\rightarrow X$. This gives rise to an exact functor $$\Phi\colon M\mbox{-Mod}_\mathcal{A}\longrightarrow \bar{M}\mbox{-Mod}_{\mathcal{A}/\mathcal{N}}, \quad (X, \lambda)\mapsto (X, q(\lambda)).$$
The functor $\Phi$ sends a morphism $\theta$ to $q(\theta)$. We observe that $\Phi$ vanishes on the Serre subcategory $M\mbox{-Mod}_\mathcal{N}$, and induces an exact functor $\bar{\Phi}\colon M\mbox{-{\rm Mod}}_\mathcal{A}/{M\mbox{-{\rm Mod}}_\mathcal{N}}\rightarrow \bar{M}\mbox{-{\rm Mod}}_{\mathcal{A}/\mathcal{N}}$.
\begin{prop}\label{prop:app1}
Let $M\colon \mathcal{A}\rightarrow \mathcal{A}$ be an exact monad on an abelian category $\mathcal{A}$, and let $\mathcal{N}\subseteq \mathcal{A}$ be a Serre subcategory satisfying $M(\mathcal{N})\subseteq \mathcal{N}$. Keep the notation as above. Then the induced functor
$$\bar{\Phi}\colon M\mbox{-{\rm Mod}}_\mathcal{A}/{M\mbox{-{\rm Mod}}_\mathcal{N}}\stackrel{\sim}\longrightarrow \bar{M}\mbox{-{\rm Mod}}_{\mathcal{A}/\mathcal{N}}$$
is an equivalence of categories.
\end{prop}
\begin{proof}
Consider the free module functor $F_M\colon \mathcal{A}\rightarrow M\mbox{-{\rm Mod}}_\mathcal{A}$ and the forgetful functor $U_M\colon M\mbox{-{\rm Mod}}_\mathcal{A}\rightarrow \mathcal{A}$, both of which are exact. We observe that $F_M(\mathcal{N})\subseteq M\mbox{-{\rm Mod}}_\mathcal{N}$, and $U_M^{-1}(\mathcal{N})=M\mbox{-{\rm Mod}}_\mathcal{N}$. We apply Lemma \ref{lem:quotientAbel}(1) and (3), and obtain the adjoint pair $\overline{F_M}$ and $\overline{U_M}$ between quotient abelian categories $\mathcal{A}/\mathcal{N}$ and $M\mbox{-{\rm Mod}}_\mathcal{A}/{M\mbox{-{\rm Mod}}_\mathcal{N}}$, where $\overline{U_M}$ is faithful. Moreover, by Lemma \ref{lem:quotientAbel}(2) the monad defined by this adjoint pair coincides with the induced monad $\bar{M}$ of $M$. We apply Theorem \ref{thm:Beck} to the adjoint pair $(\overline{F_M}, \overline{U_M})$ and infer that the associated comparison functor $$K\colon M\mbox{-{\rm Mod}}_\mathcal{A}/{M\mbox{-{\rm Mod}}_\mathcal{N}}\longrightarrow \bar{M}\mbox{-Mod}_{\mathcal{A}/\mathcal{N}}$$ is an equivalence.
It remains to observe that $K=\bar{\Phi}$. Indeed, by (\ref{equ:K}) the comparison functor $K$ sends an $M$-module $(X, \lambda)$ to an $\bar{M}$-module $(\overline{U_M}(X, \lambda), \overline{U_M}(\bar{\epsilon}_M)_{(X, \lambda)})$, which equals $(X, q(\lambda))$. Hence, the functors $K$ and $\bar{\Phi}$ agree on objects. For the same reason, they agree on morphisms.
\end{proof}
We now apply Proposition \ref{prop:app1} to the category of equivariant objects. Let $G$ be a finite group which acts on an abelian category $\mathcal{A}$ by the data $\{F_g, \varepsilon_{g,h}\; |\; g,h\in G\}$. Then the category $\mathcal{A}^G$ of $G$-equivariant objects is abelian; indeed, a sequence of equivariant objects is exact if and only if the corresponding sequence of the underlying objects in $\mathcal{A}$ is exact.
Let $\mathcal{N}\subseteq \mathcal{A}$ be a Serre subcategory which is \emph{invariant} under this action, that is, $F_g(\mathcal{N})\subseteq \mathcal{N}$ for any $g\in G$. Then the quotient category $\mathcal{A}/\mathcal{N}$ inherits a $G$-action, which is given by the data $\{\bar{F_g}, \bar{\varepsilon}_{g,h}\; |\; g,h\in G\}$. The quotient functor $q\colon \mathcal{A}\rightarrow \mathcal{A}/\mathcal{N}$ induces an exact functor
$$\Psi\colon \mathcal{A}^G\longrightarrow (\mathcal{A}/\mathcal{N})^G.$$
More precisely, $\Psi$ sends a $G$-equivariant object $(X, \alpha)$ to $(X, q(\alpha))$, where $q(\alpha)_g=q(\alpha_g)\colon X\rightarrow \bar{F_g}(X)=qF_g(X)$ for each $g\in G$, and $\Psi$ sends a morphism $\theta\colon (X, \alpha)\rightarrow (Y, \beta)$ to $q(\theta)\colon (X, q(\alpha)) \rightarrow (Y, q(\beta))$. We observe that the functor $\Psi$ is exact and that its essential kernel equals $\mathcal{N}^G$. Therefore, we have the induced functor $\bar{\Psi}\colon \mathcal{A}^G/{\mathcal{N}^G}\rightarrow (\mathcal{A}/\mathcal{N})^G$.
\begin{cor}\label{cor:equiv}
Let $G$ be a finite group acting on an abelian category $\mathcal{A}$, and let $\mathcal{N}\subseteq \mathcal{A}$ be a Serre subcategory which is invariant under the action. Keep the notation as above. Then the induced functor
$$\bar{\Psi}\colon \mathcal{A}^G/{\mathcal{N}^G}\stackrel{\sim}\longrightarrow (\mathcal{A}/\mathcal{N})^G$$
is an equivalence of categories.
\end{cor}
\begin{proof}
Denote by $M$ the monad on $\mathcal{A}$ that is defined by the group action. Then $M$ is exact and by the invariance of $\mathcal{N}$, we have $M(\mathcal{N})\subseteq \mathcal{N}$. We observe that the induced monad $\bar{M}$ on $\mathcal{A}/\mathcal{N}$ coincides with the monad defined by the induced $G$-action on $\mathcal{A}/\mathcal{N}$.
By Proposition \ref{prop:monadic} we identify $\mathcal{A}^G$ with $M\mbox{-Mod}_\mathcal{A}$, and $(\mathcal{A}/\mathcal{N})^G$ with $\bar{M}\mbox{-Mod}_{\mathcal{A}/\mathcal{N}}$. Then the equivalence follows from Proposition \ref{prop:app1}. Here, one observes that the functor $\bar{\Phi}$ in Proposition \ref{prop:app1} corresponds to the functor $\bar{\Psi}$.
\end{proof}
\section{Graded module categories and graded group rings}
In this section, we give the second application of Theorem \ref{thm:Beck}, and prove that the equivarianzation of the graded module category over a graded ring with respect to a certain degree-shift action is equivalent to the graded module category over the same ring but with a coarser grading; this result is essentially due to \cite[Theorem 6.4.1]{NV} in a different setup. On the other hand, we prove that the equivariantization of the graded module category over a graded ring with respect to a certain twisting action is equivalent to the graded module category over the same ring but with a refined grading. All modules are right modules.
\subsection{}
Let $G$ be an arbitrary group. Let $R=\oplus_{g\in G} R_g$ be a $G$-graded ring with a unit. Here, the unit $1_R$ lies in $R_e$, and the subgroups $R_g$ of $R$ are called the homogeneous components of degree $g$.
A $G$-graded $R$-module $X$ is an $R$-module with a decomposition $X=\oplus_{g\in G}{X_g}$ into homogeneous components $X_g$ such that $X_g.R_{g'}\subseteq X_{gg'}$, where the dot ``." means the right action. An element $x$ in $X_g$ is said to be \emph{homogeneous of degree} $g$, denoted by $|x|=g$ or ${\rm deg}\; x=g$. The \emph{grading support} of $X$ is by definition ${\rm gsupp}(X)=\{g\in G\; |\; X_g\neq 0\}$, which is a subset of $G$.
We denote by ${\rm Mod}^G\mbox{-}R$ the abelian category of $G$-graded $R$-modules, where the homomorphism between graded modules are module homomorphisms that preserve the degrees. More precisely, a homomorphism $f\colon X\rightarrow X'$ is an $R$-module homomorphism satisfying $f(X_g)\subseteq X'_{g}$, and we denote by $f_g\colon X_g\rightarrow X'_g$ its restriction to the corresponding homogeneous component. We denote by ${\rm mod}^G\mbox{-}R$ the full subcategory formed by finitely presented graded modules.
For a $G$-graded $R$-module $X$ and $g'\in G$, the \emph{shifted module} $X(g')$ is defined such that $X(g')=X$ as an ungraded $R$-module and that its grading is given by $X(g')_g=X_{g'g}$ for each $g\in G$. This gives rise to an automorphism $(g')\colon {\rm Mod}^G\mbox{-}R\rightarrow {\rm Mod}^G\mbox{-}R$, called the \emph{degree-shift functor}. For example, we consider $R(g')$ as a $G$-graded $R$-module with $1_R$ having degree $g'^{-1}$. Indeed, the set $\{R(g)\;|\; g\in G\}$ is a set of projective generators in the category ${\rm Mod}^G\mbox{-}R$.
Each subgroup $G'\subseteq G$ has a strict action on ${\rm Mod}^G\mbox{-}R$ by assigning to each $g'\in G'$ the automorphism $F_{g'}=(g'^{-1})$. Such a $G'$-action on ${\rm Mod}^G\mbox{-}R$ is referred as a \emph{degree-shift action}; in this case, $G'$ acts also on ${\rm mod}^G\mbox{-}R$.
Let $\pi\colon G\rightarrow H$ be a homomorphism of groups. We define an $H$-graded ring $\pi_*(R)$ as follows: as an ungraded ring
$\pi_*(R)=R$, while its homogeneous component is given by $\pi_*(R)_h=\oplus_{g\in \pi^{-1}(h)} R_g$ for each $h\in H$; compare \cite[Subsection 1.2]{NV}. Then we have the abelian category ${\rm Mod}^H\mbox{-}\pi_*(R)$ of $H$-graded $\pi_*(R)$-modules and its full subcategory ${\rm mod}^H\mbox{-}\pi_*(R)$ consisting of finitely presented modules, both of which carry a natural strict $H$-action by degree-shift.
We define a functor $\pi_*\colon {\rm Mod}^G\mbox{-} R\rightarrow {\rm Mod}^H\mbox{-} \pi_*(R)$ as follows: for a $G$-graded $R$-module $X=\oplus_{g\in G}X_g$, we assign an $H$-graded $\pi_*(R)$-module $\pi_*(X)$ such that $\pi_*(X)=X$ as an ungraded $R$-module and that its homogeneous component is given by $\pi_*(X)_h=\oplus_{g\in \pi^{-1}(h)} X_g$. The functor $\pi_*$ acts on homomorphisms by the identity. We observe that $\pi_*$ sends $R$ to $\pi_*(R)$, and more generally, $\pi_*$ sends $R(g)$ to $\pi_*(R)(h)$ for each $g\in G$ and $h=\pi(g)$. We observe also that the functor $\pi_*$ is exact.
The functor $\pi_*$ has a right adjoint $\pi^*\colon {\rm Mod}^H\mbox{-} \pi_*(R)\rightarrow {\rm Mod}^G\mbox{-}R$, which is defined as follows. For an $H$-graded $\pi_*(R)$-module $Y=\oplus_{h\in H} Y_h$, we define a $G$-graded abelian group $\pi^*(Y)$ such that its homogeneous component $\pi^*(Y)_{g}=Y_{\pi(g)}$ for each $g\in G$. A homogeneous element $r\in R_{g'}$ acts on an element $y\in \pi^*(Y)_g$ as the given action $y.r$ on $Y$, where the resulting element $y.r\in Y_{\pi(g'g)}$ is viewed now as an element in $\pi^*(Y)_{gg'}$. This defines a $G$-graded $R$-module $\pi^*(Y)$. For an $H$-graded $\pi_*(R)$-module homomorphism $f\colon Y\rightarrow Y'$, the corresponding homomorphism $\pi^*(f)\colon \pi^*(Y)\rightarrow \pi^*(Y')$ sends an element $y\in \pi^*(Y)_g=Y_{\pi(g)}$ to $f(y)\in Y'_{\pi(g)}=\pi^*(Y')_g$ for each $g\in G$. We observe that the functor $\pi^*$ is exact.
The adjoint pair $(\pi_*, \pi^*)$ is given by the following natural isomorphism
\begin{align*}
{\rm Hom}_{{\rm Mod}^H\mbox{-}\pi_*(R)} (\pi_*(X), Y)\stackrel{\sim}\longrightarrow {\rm Hom}_{{\rm Mod}^G\mbox{-}R} (X, \pi^*(Y))
\end{align*}
which sends $f\colon \pi_*(X)\rightarrow Y$ to $f'\colon X\rightarrow \pi^*(Y)$ such that $f'_g\colon X_g\rightarrow \pi^*(Y)_g=Y_{\pi(g)}$ is the restriction of $f_{\pi(g)}\colon \pi_*(X)_{\pi(g)}\rightarrow Y_{\pi(g)}$ to the direct summand $X_g$. It follows that the unit $\eta\colon {\rm Id}_{{\rm Mod}^G\mbox{-}R}\rightarrow \pi^*\pi_*$ is given such that $(\eta_X)_g\colon X_g\rightarrow \pi^*\pi_*(X)_g=\oplus_{g'\in \pi^{-1}(\pi(g))} X_{g'}$ is the inclusion of $X_g$. The counit $\epsilon\colon \pi_*\pi^*\rightarrow {\rm Id}_{{\rm Mod}^H\mbox{-}\pi_*(R)}$ is given such that $(\epsilon_Y)_h\colon \pi_*\pi^*(Y)_h=\oplus_{g\in \pi^{-1}(h)} Y_{\pi(g)}\rightarrow Y_h$ maps each direct summand $Y_{\pi(g)}$ identically to $Y_h$ if $h\in \pi(G)$, and that $(\epsilon_Y)_h=0$ otherwise.
\begin{lem}\label{lem:monadN}
Let $N$ be the kernel of $\pi\colon G\rightarrow H$. Then the monad defined by the adjoint pair $(\pi_*, \pi^*; \eta, \epsilon)$ coincides with the monad defined by the degree-shift action of $N$ on ${\rm Mod}^G\mbox{-}R$.
\end{lem}
\begin{proof}
Denote by $(M, \eta', \mu)$ the monad that is defined by the degree-shift action of $N$ on ${\rm Mod}^G\mbox{-}R$. As we computed above, $\pi^*\pi_*(X)_g=\oplus_{g'\in \pi^{-1}(\pi(g))} X_{g'}=\oplus_{n\in N} X(n^{-1})_g$ for each $g\in G$. Indeed, this proves that the endofunctor $\pi^*\pi_*$ equals $M$, which is by definition $\oplus_{n\in N} F_n=\oplus_{n\in N}(n^{-1})$. It is direct to verify that $\eta=\eta'$ and $\mu=\pi^* \epsilon \pi_*$.
\end{proof}
We consider the degree-shift action of $N$ on ${\rm Mod}^G\mbox{-}R$, and thus the category $({\rm Mod}^G\mbox{-}R)^N$ of $N$-equivariant objects. Consider the following functor
\begin{align}\label{equ:theta}
\Theta\colon {\rm Mod}^H\mbox{-}\pi_*(R)\longrightarrow ({\rm Mod}^G\mbox{-}R)^N
\end{align}
sending $Y$ to $\Theta(Y)=(\pi^*(Y), {\rm Id})$, where ${\rm Id}_n\colon \pi^*(Y)\rightarrow \pi^*(Y)(n^{-1})$ is the identity for each $n\in N$; here, we observe that $\pi^*(Y)(n^{-1})=\pi^*(Y)$. The functor $\Theta$ sends a homomorphism $f$ to $\pi^*(f)$.
The main observation is as follows. We mention that it is implicitly due to \cite[Theorem 6.4.1]{NV}; see Corollary \ref{cor:NV}.
\begin{prop}\label{prop:app2}
Keep the notation as above. Then the functor $\Theta$ is an equivalence if and only if $\pi\colon G\rightarrow H$ is surjective. In this case, if in addition $N$ is finite, the equivalence $\Theta$ restricts to an equivalence
$$\Theta\colon {\rm mod}^H\mbox{-}\pi_*(R)\stackrel{\sim}\longrightarrow ({\rm mod}^G\mbox{-}R)^N.$$
\end{prop}
\begin{proof}
We claim that the functor $\pi^*$ is faithful if and only if $\pi$ is surjective. Recall that $\pi^*(Y)_g=Y_{\pi(g)}$ for each $g\in G$, and that for a morphism $f\colon Y\rightarrow Y'$, $\pi^*(f)_g=f_{\pi(g)}$. In particular, $\pi^*(f)=0$ if and only if $f_{h}=0$ for all $h\in \pi(G)$. This proves the ``if" part of the claim. On the other hand, if $\pi$ is not surjective, we take $h\in H$ that is not contained in $\pi(G)$. Consider $Y=\pi_*(R)(h)$ in ${\rm Mod}^H\mbox{-}\pi_*(R)$. Then $\pi^*(Y)=0$, or equivalently, $\pi^*({\rm Id}_Y)=0$. This shows that $\pi^*$ is not faithful in this case.
Set $\mathcal{A}={\rm Mod}^G\mbox{-}R$. Denote by $M$ the monad on $\mathcal{A}$ defined by the adjoint pair $(\pi_*, \pi^*)$. We consider the associated comparison functor $K\colon {\rm Mod}^H\mbox{-}\pi_*(R)\rightarrow M\mbox{-Mod}_{\mathcal{A}}$. By Lemma \ref{lem:monadN}, Proposition \ref{prop:monadic} and Remark \ref{rem:monadic}, we identify $ M\mbox{-Mod}_{\mathcal{A}}$ with $\mathcal{A}^N$, with which the functor $\Theta$ is identified with this comparison functor $K$. Here, one compares the definition of $\Theta$ in (\ref{equ:theta}) with the construction of $K$ in (\ref{equ:K}). Recall that the functor $\pi^*$ is exact. Then we apply Theorem \ref{thm:Beck} to the adjoint pair $(\pi_*, \pi^*)$, and infer that $\Theta$ is an equivalence if and only if $\pi^*$ is faithful, which by the above claim is equivalent to the surjectivity of $\pi$.
The restricted equivalence follows from Lemma \ref{lem:fp}(2).
\end{proof}
The following fact is well known.
\begin{lem}\label{lem:fp}
Assume that the homomorphism $\pi\colon G\rightarrow H$ is surjective with its kernel $N$ finite. Let $X$ be a $G$-graded $R$-module and $Y$ an $H$-graded $\pi_*(R)$-module. Keep the notation as above. Then the following statements hold.
\begin{enumerate}
\item The $G$-graded $R$-module $X$ is finitely presented if and only if so is $\pi_*(X)$.
\item The $H$-graded $\pi_*(R)$-module $Y$ is finitely presented if and only if so is $\pi^*(Y)$.
\end{enumerate}
\end{lem}
\begin{proof}
Recall that $\pi_*(R(g))=\pi_*(R)(h)$ for each $g\in G$ and $h=\pi(g)$. Thus the exact functor $\pi_*$ preserves finitely generated projective modules. Then the ``only if" part of (1) follows.
We observe that $\pi^*(\pi_*(R)(h)) \simeq \oplus_{n\in N} R(ng)$ for $h=\pi(g)$, and that $N$ is finite. It follows that the exact functor $\pi^*$ preserves finitely generated projective modules. The ``only if" part of (2) follows.
For the ``if" part of (1), we assume that $\pi_*(X)$ is finitely presented, and thus so is $\pi^*\pi_*(X)$. The unit $\eta_X\colon X\rightarrow \pi^*\pi_*(X)$ is a split monomorphism. It follows that $X$ is finitely presented.
It remains to prove the ``if" part of (2). We claim that for each $H$-graded $\pi_*(R)$-module $Z$, if $\pi^*(Z)$ is finitely generated, so is $Z$. Indeed, we recall that the exact functor $\pi_*$ preserves finitely generated projective modules. It follows that $\pi_*\pi^*(Z)$ is finitely generated. Since the counit $\epsilon_Z\colon \pi_*\pi^*(Z)\rightarrow Z$ is surjective, we infer that $Z$ is finitely generated. Here, for the surjectivity of $\epsilon_Z$ we use the surjectivity of the homomorphism $\pi\colon G\rightarrow H$.
We assume that $\pi^*(Y)$ is finitely presented. By the claim $Y$ is finitely generated. Take an exact sequence $0\rightarrow K\rightarrow P\rightarrow Y\rightarrow 0$ in ${\rm Mod}^H\mbox{-}\pi_*(R)$ with $P$ finitely generated projective. Applying the exact functor $\pi^*$ and the fact that $\pi^*(P)$ is finitely generated projective, we infer that $\pi^*(K)$ is finitely generated. By the claim again, we have that $K$ is finitely generated. This shows that $Y$ is finitely presented.
\end{proof}
\subsection{} We will relate Proposition \ref{prop:app2} to \cite[Theorem 6.4.1]{NV}.
Let $N$ be a normal subgroup of a group $G$. Let $R=\oplus_{g\in G} R_g$ be a $G$-graded ring. The \emph{graded group ring} $R^{\rm gr}[N]$ is defined as follows: $R^{\rm gr}[N]=\oplus_{n\in N} Ru_n$ is a free left $R$-module with a basis $\{u_{n}\;|\; n\in N\}$, which is $G$-graded by means of $|ru_n|=|r| \cdot n$ for a homogeneous element $r$ in $R$; the multiplication is given as follows $$(ru_n) (r'u_{n'})=rr'u_{(|r'|^{-1}\cdot n\cdot |r|')n'}.$$
We mention that the elements $u_n$ are invertible.
We observe that $R^{\rm gr}[N]$ is a $G$-graded ring, and that the canonical map $R\rightarrow R^{\rm gr}[N]$, sending $r$ to $ru_{e}$, identifies $R$ as a $G$-graded subring of $R^{\rm gr}[N]$. We refer for the details to \cite[Subsection 6.1]{NV}.
The following result establishes the link between the graded group ring $R^{\rm gr}[N]$ and the category of $N$-equivariant objects in ${\rm Mod}^G\mbox{-}R$. We emphasize that here we consider the degree-shift action of $N$ on ${\rm Mod}^G\mbox{-}R$.
\begin{prop}\label{prop:iso}
Keep the notation as above. Then there is an isomorphism of categories
$$({\rm Mod}^G\mbox{-}R)^N\stackrel{\sim}\longrightarrow {\rm Mod}^G\mbox{-}(R^{\rm gr}[N]).$$
\end{prop}
\begin{proof}
Recall that an $N$-equivariant object $(X, \alpha)$ is a $G$-graded $R$-module $X$ with isomorphisms $\alpha_n\colon X\rightarrow F_nX=X(n^{-1})$ in ${\rm Mod}^G\mbox{-}R$, subject to the relations $F_n(\alpha_{n'})\circ \alpha_n=\alpha_{nn'}$. We also view $\alpha_n$ as an automorphism on $X$ of degree $n^{-1}$.
We define a functor $\Delta \colon ({\rm Mod}^G\mbox{-}R)^N\stackrel{\sim}\longrightarrow {\rm Mod}^G\mbox{-}(R^{\rm gr}[N])$ as follows. To an $N$-equivariant object $(X, \alpha)$, we associate a $G$-graded $R^{\rm gr}[N]$-module $\Delta(X, \alpha)=X$ such that the $R$-action is the same as $X$ and that $$x.u_n=(\alpha_{|x|\cdot n\cdot |x|^{-1}})^{-1}(x)$$
for each homogeneous element $x$ in $X$ and $n\in N$. It is rountine to verify that this defines a $G$-graded $R^{\rm gr}[N]$-module structure on $X$. The functor $\Delta$ acts by the identity on morphisms. Roughly speaking, the functor $\Delta$ rearranges the isomorphisms $\alpha_n$'s into the action of the invertible elements $u_n$'s.
The inverse functor of $\Delta$ associates to a $G$-graded $R^{\rm gr}[N]$-module $Y$ the $N$-equivaraint object $(Y, \beta)$, where the isomorphism $\beta_{n}\colon Y\rightarrow Y(n^{-1})$ is given by $\beta_n(y)=y.(u_{|y|^{-1}\cdot n^{-1}\cdot |y|})$ for each homogeneous element $y\in Y$.
\end{proof}
We combine Propositions \ref{prop:app2} and \ref{prop:iso} to recover the following result.
\begin{cor}\label{cor:NV} {\rm (\cite[Theorem 6.4.1]{NV})} Let $R=\oplus_{g\in G}R_g$ be a $G$-graded ring and let $N\subseteq G$ be a normal subgroup. Consider the canonical projection $\pi\colon G\rightarrow G/N$. Then there is an equivalence of categories
$${\rm Mod}^{G/N}\mbox{-}\pi_*(R)\stackrel{\sim}\longrightarrow {\rm Mod}^G\mbox{-}(R^{\rm gr}[N]).$$
\end{cor}
\subsection{} We will show that in the case that $N$ is a finite abelian group which is a direct summand of $G$, one might recover the category of $G$-graded $R$-modules from the category of $G/N$-graded $\pi_*(R)$-modules via the equivariantization with respect to a certain twisting action; see Proposition \ref{prop:recover}.
Throughout this subsection, we assume that $A$ is a finite abelian group. Let $k$ be a splitting field of $A$. In other words, the group algebra $kA$ is isomorphic to a direct product of copies of $k$; in particular, the characteristic of the field $k$ does not divide the order of $A$. Denote by $\widehat{A}$ the \emph{character group} of $A$, that is, the group of linear characters of $A$ over $k$.
Let $V$ be a vector space over $k$. By a \emph{linear $\widehat{A}$-action} on $V$ we mean a group homomorphism from $\widehat{A}$ to the general linear group of $V$. By an \emph{$A$-gradation} on $V$, we mean a decomposition $V=\oplus_{a\in A} V_a$ into subspaces indexed by $A$. We have the following well-known one-to-one correspondence:
\begin{align}\label{equ:corres}
\{A\mbox{-gradations on }V\} \longleftrightarrow \{\mbox{linear }\widehat{A}\mbox{-actions on }V\}.
\end{align}
The correspondence identifies an $A$-gradation $V=\oplus_{a\in A} V_a$ with the linear $\widehat{A}$-action given by $\chi.v=\chi(a)v$ for any $\chi \in \widehat{A}$ and $v\in V_a$. Here, $\chi.v$ denotes the left action of $\chi$ on $v$.
We will consider a restriction of this correspondence. Let $H$ be an arbitrary group and let $R=\oplus_{h\in H} R_h$ be an $H$-graded algebra over $k$. By an $\widehat{A}$-\emph{action as graded automorphisms} on $R$ we mean a group homomorphism from $\widehat{A}$ to the automorphism group of $R$ as an $H$-graded algebra.
Consider the canonical projection $\pi\colon H\times A\rightarrow H$ which sends $(h, a)$ to $h$. An $A$-\emph{refinement} of $R$ means an $(H\times A)$-graded algebra $\bar{R}$ such that $\pi_*(\bar{R})=R$, or equivalently, each homogeneous component $R_h=\oplus_{a\in A} \bar{R}_{(h, a)}$ has an $A$-gradation such that $rr'\in \bar{R}_{(hh', aa')}$ for any elements $r\in \bar{R}_{(h, a)}$ and $r'\in \bar{R}_{(h', a')}$.
The following result is well known; compare \cite[Proposition 1.3.13 and Remarks 1.3.14]{NV}.
\begin{lem}\label{lem:corres}
Let $R$ be an $H$-graded algebra. Then there is a one-to-one correspondence
\begin{align}\label{equ:corres2}
\{A\mbox{-refinements of }R\} \longleftrightarrow \{\widehat{A}\mbox{-actions as graded automorphisms on }R\},
\end{align}
which identifies an $A$-refinement $\bar{R}$ with the $\widehat{A}$-action given by $\chi.r=\chi(a)r$ for $\chi\in \widehat{A}$ and $r\in \bar{R}_{(h, a)}$.
\end{lem}
\begin{proof}
We apply the correspondence (\ref{equ:corres}) to the homogeneous component $R_h$ for each $h\in H$. It suffices to show that the decomposition $R=\oplus_{(h, a)\in H\times A} \bar{R}_{(h, a)}$ makes $R=\bar{R}$ an $(H\times A)$-graded algebra if and only if the corresponding $\widehat{A}$-action on $R$ is given by $H$-graded algebra automorphisms. Take the ``only if" part for example. For any two elements $r\in \bar{R}_{(h, a)}\subseteq R_{h}$ and $r\in \bar{R}_{(h', a')}\subseteq R_{h'}$, we have
$$\chi.(rr')=\chi(aa')rr'=(\chi(a)r)(\chi(a')r')=(\chi.r)(\chi.r'),$$
where the leftmost equality uses the $(H\times A)$-gradation on $R=\bar{R}$. This proves that $\chi$ acts on $R$ by an $H$-graded algebra automorphism.
\end{proof}
Let $\sigma\colon R\rightarrow R$ be an automorphism as an $H$-graded algebra. For each $X\in \mbox{Mod}^H\mbox{-}R$, the \emph{twisted module} $X^\sigma$ is defined such that $X^\sigma=X$ as an $H$-graded space and that it has a new $R$-action ``$_\circ$" given by $x_\circ r=x.\sigma(r)$. This gives rise to an automorphism $(-)^\sigma\colon \mbox{Mod}^H\mbox{-}R\rightarrow \mbox{Mod}^H\mbox{-}R$ . In this way, each subgroup $G'$ of the automorphism group of $R$ as an $H$-graded algebra has a strict action on $\mbox{Mod}^H\mbox{-}R$ by assigning to each $g\in G'$ the automorphism $F_g=(-)^{g^{-1}}$. Such an action is referred as a \emph{twisting action} of $G'$ on $\mbox{Mod}^H\mbox{-}R$.
Let $A$ be a finite abelian group. Consider an $A$-refinement $\bar{R}$ of $R$, and the corresponding $\widehat{A}$-action on $R$. Then there is a strict $\widehat{A}$-action on $\mbox{Mod}^H\mbox{-}R$ by assigning to each $\chi\in \widehat{A}$ the automorphism $F_\chi=(-)^{\chi^{-1}}$; here, we identify $\chi$ as an element in the automorphism group of $R$. We refer to this action as the \emph{twisting action} of $\widehat{A}$ on $\mbox{Mod}^H\mbox{-}R$ that \emph{corresponds to the $A$-refinement $\bar{R}$}. We will consider the category $(\mbox{\rm Mod}^H\mbox{-}R)^{\widehat{A}}$ of $\widehat{A}$-equivariant objects in $\mbox{Mod}^H\mbox{-}R$.
By Proposition \ref{prop:app2} there is an equivalence of categories
\begin{align}\label{equ:inverseequi}
\Theta\colon {\rm Mod}^H\mbox{-}R\stackrel{\sim}\longrightarrow ({\rm Mod}^{(H\times A)}\mbox{-}\bar{R})^A
\end{align}
where we consider the degree-shift action of $A$ on ${\rm Mod}^{(H\times A)}\mbox{-}\bar{R}$. The following is a different equivalence, which is somehow adjoint to (\ref{equ:inverseequi}); see Remark \ref{rem:Len1}.
Recall that the canonical projection $\pi\colon H\times A\rightarrow H$ satisfies $\pi_*(\bar{R})=R$. We define the following functor
$$\Gamma \colon \mbox{\rm Mod}^{(H\times A)}\mbox{-}\bar{R} \longrightarrow (\mbox{\rm Mod}^H\mbox{-}R)^{\widehat{A}}$$
which sends an $(H\times A)$-graded $\bar{R}$-module $\bar{X}$ to $\Gamma(\bar{X})=(\pi_*(\bar{X}), \gamma)$, where each isomorphism $\gamma_\chi\colon \pi_*(\bar{X})\rightarrow F_\chi(\pi_*(\bar{X}))=\pi_*(\bar{X})^{\chi^{-1}}$ is given by $\gamma_\chi(x)=\chi^{-1}(a)x$ for $x\in \bar{X}_{(h, a)}\subseteq \pi_*(\bar{X})_h$. The action of $\Gamma$ on morphisms is given by $\pi_*$.
The following result generalizes slightly a result in \cite{Len}.
\begin{prop}\label{prop:recover}
Let $R$ be an $H$-graded algebra over $k$, and let $A$ be a finite abelian group such that $k$ is a splitting field of $A$. Consider an $A$-refinement $\bar{R}$ of $R$ and the corresponding twisting action of $\widehat{A}$ on $\mbox{\rm Mod}^H\mbox{-}R$. Then the above functor $\Gamma$ is an isomorphism of categories. Moreover, we have a restricted isomorphism
$$ \Gamma\colon \mbox{\rm mod}^{(H\times A)}\mbox{-}\bar{R} \stackrel{\sim}\longrightarrow (\mbox{\rm mod}^H\mbox{-}R)^{\widehat{A}}.$$
\end{prop}
\begin{proof}
Let us indicate the inverse of $\Gamma$. For an $\widehat{A}$-equivariant object $(X, \alpha)$ in $\mbox{\rm Mod}^H\mbox{-}R$, we have for each $\chi\in \widehat{A}$ the isomorphism $\alpha_\chi\colon X\rightarrow F_\chi(X)=X^{\chi^{-1}}$ in $\mbox{\rm Mod}^H\mbox{-}R$; in particular, $\alpha_\chi$ induces a $k$-linear automorphism on each homogeneous component $X_h$ for each $h\in H$. In view of (\ref{equ:rel}), we observe that these isomorphisms $(\alpha_\chi)^{-1}=\alpha_{\chi^{-1}}$ induce a linear $\widehat{A}$-action on $X_h$. By the correspondence (\ref{equ:corres}) each homogeneous component $X_h$ has an $A$-gradation $X_h=\oplus_{a\in A}\bar{X}_{(h, a)}$ such that $\alpha_{\chi^{-1}}(x)=\chi(a)x$ for $x\in \bar{X}_{(h, a)}$. This makes $\bar{X}=X=\oplus_{(h, a)\in H\times A} \bar{X}_{(h, a)}$ an $(H\times A)$-graded $\bar{R}$-module; moreover, we have $\alpha_\chi=\gamma_\chi$ for each $\chi\in \widehat{A}$. In other words, $(X, \alpha)=\Gamma(\bar{X})$. Then the assignment $(X, \alpha)\mapsto \bar{X}$ defines the inverse functor of $\Gamma$.
The restricted isomorphism follows from Lemma \ref{lem:fp}(1).
\end{proof}
\begin{rem}\label{rem:Len1}
We mention that one might infer the equivalence (\ref{equ:inverseequi}) from Proposition \ref{prop:recover} and a general result \cite[Theorem 7.2]{El2014}. Here, we identify $A$ with the character group of $\widehat{A}$. In other words, these two equivalences are somehow \emph{adjoint} to each other. We thank Helmut Lenzing for this remark.
\end{rem}
\section{Weighted projective lines and quotient abelian categories}
In this section, we recall from \cite{GL87, GL90} some basic facts on weighted projective lines. For a weighted projective line, we study the relationship between the restriction subalgebras of its homogeneous coordinate algebra and certain quotient categories of the category of coherent sheaves on it. Here, we work on an arbitrary field $k$.
\subsection{}
Let $t\geq 1$ be an integer. A \emph{weight sequence} $\mathbf{p}=(p_1, p_2, \cdots, p_t)$ with $t$ weights consists of $t$ positive integers satisfying $p_i\geq 2$. We will assume that $p_1\geq p_2\geq \cdots \geq p_t$.
The \emph{string group} $L(\mathbf{p})$ associated to a weight sequence $\mathbf{p}$ is an abelian group with generators $\vec{x}_1, \vec{x}_2, \cdots, \vec{x}_t$ subject to the relations $p_1\vec{x}_1=p_2\vec{x}_2=\cdots =p_t\vec{x}_t$. This common element is denoted by $\vec{c}$, called the \emph{canonical element}. The abelian group $L(\mathbf{p})$ is of rank one, where the canonical element $\vec{c}$ has infinite order. We observe an isomorphism $L(\mathbf{p})/{\mathbb{Z}\vec{c}}\simeq \prod_{i=1}^t \mathbb{Z}/p_i\mathbb{Z}$ of abelian groups. From these facts, we infer that each element $\vec{x}$ in $L(\mathbf{p})$ is uniquely expressed in its \emph{normal form}
\begin{align}
\vec{x}=l\vec{c}+\sum_{i=1}^t l_i \vec{x}_i
\end{align}
with $l\in \mathbb{Z}$ and $0\leq l_i< p_i$. Set $\phi(\vec{x})=l$; this gives rise to a map $\phi\colon L(\mathbf{p})\rightarrow \mathbb{Z}$. For each $1\leq i\leq t$, we define a homomorphism $\pi_i\colon L(\mathbf{p})\rightarrow \mathbb{Z}/{p_i\mathbb{Z}}$ such that $\pi_i (\vec{x}_j)=\delta_{i, j}\bar{1}$; it is surjective.
We denote by $\mathbb{N}=\{0, 1, 2\cdots\}$ the set of natural numbers. Define the following map
\begin{align*}{\rm mult}\colon L(\mathbf{p})\longrightarrow \mathbb{N}, \quad \vec{x}\mapsto {\rm max}\{\phi(\vec{x})+1, 0\}.
\end{align*}
We set $p={\rm lcm}(\mathbf{p})={\rm lcm}(p_1, p_2, \cdots, p_t)$ to be the least common multiple of $\mathbf{p}$. The following homomorphism of abelian groups, called the \emph{degree map}, is well defined
\begin{align*}
\delta\colon L(\mathbf{p})\longrightarrow \mathbb{Z}, \quad \vec{x}_i\mapsto \frac{p}{p_i}.\end{align*}
The degree map is surjective and its kernel coincides with the torsion subgroup of $L(\mathbf{p})$. We observe that $\delta(\vec{c})=p$.
Recall that the \emph{dualizing element} $\vec{\omega}$ in $L(\mathbf{p})$ is defined as $\vec{\omega}=(t-2)\vec{c}-\sum_{i=1}^t\vec{x}_i$; its normal form is $\vec{\omega}=-2\vec{c}+\sum_{i=1}^t(p_i-1)\vec{x}_i$. One computes that $\delta(\vec{\omega})=p((t-2)-\sum_{i=1}^t\frac{1}{p_i})$.
The string group $L(\mathbf{p})$ is partially ordered such that $\vec{x}\leq \vec{y}$ if and only if $\vec{y}-\vec{x}\in \mathbb{N}\{\vec{x}_1, \vec{x}_2, \cdots, \vec{x}_t\}$, which is further equivalent to $\phi(\vec{y}-\vec{x})\geq 0$. Here, for a subset $S$ of an abelian group $A$, we denote by $\mathbb{N}S$ the submonoid of $A$ generated by $S$. We denote by $L(\mathbf{p})_+$ the positive cone of $L(\mathbf{p})$, which by definition equals the submonoid $\mathbb{N}\{\vec{x}_1, \vec{x}_2, \cdots, \vec{x}_t\}$.
\subsection{} Let $k$ be an arbitrary field. A \emph{weighted projective line} $\mathbb{X}=\mathbb{X}(\mathbf{p}, \lambda)$ is given by a \emph{parameter sequence} $\lambda=(\lambda_1, \lambda_2, \cdots, \lambda_t)$, that is, a collection of pairwise distinct rational points on the projective line $\mathbb{P}^1_k$, together with a weight sequence $\mathbf{p}=(p_1, p_2, \cdots, p_t)$. We will assume that the parameter sequence $\lambda$ is \emph{normalized} such that $\lambda_1=\infty$, $\lambda_2=0$ and $\lambda_3=1$.
The \emph{homogeneous coordinate algebra} $S=S(\mathbf{p}, \lambda)$ of a weighted projective line $\mathbb{X}$ is by definition $k[X_1, X_2, \cdots, X_t]/I$, where $I$ is the ideal generated by $X_i^{p_i}-(X_2^{p_2}-\lambda_i X_1^{p_1}), 3\leq i\leq t$. We write $x_i=X_i+I$ in $S$.
The algebra $S$ is $L(\mathbf{p})$-graded by means of ${\rm deg}\; x_i=\vec{x}_i$. Hence, $S=\oplus_{\vec{x}\in L(\mathbf{p})} S_{\vec{x}}$, where $S_{\vec{x}}$ is the homogeneous component of degree $\vec{x}$. We observe that $S_{\vec{x}}\neq 0$ if and only if $\vec{x}\geq \vec{0}$; in this case, if $\vec{x}=l\vec{c}+\sum_{i=1}^t l_i\vec{x}_i$ is in its normal form, then the subset $\{x_1^{ap_1}x_2^{bp_2}x_1^{l_1}x_2^{l_2}\cdots x_t^{l_t}\; |\; a+b=l, a, b\geq 0\}$ is a basis of $S_{\vec{x}}$. We infer that \begin{align}\label{equ:mult}
{\rm dim}_k \; S_{\vec{x}}={\rm mult}(\vec{x})\end{align}
for any $\vec{x}\in L(\mathbf{p})$.
For an irreducible monic polynomial $f(X)\in k[X]$, we define its \emph{homogenization} $f^h(X, Y)=X^{d}f(Y/X)\in k[X, Y]$, where $d$ is the degree of $f$. The set of all homogeneous prime ideals of $S$ is given by ${\rm Spec}^{L(\mathbf{p})}(S)=\{(0), (x_i), (f^h(x_1^{p_1}, x_2^{p_2})), \mathfrak{m}=(x_1, x_2, \cdots, x_t)\; |\; f\neq X-\lambda_i, 1\leq i\leq t\}$; compare \cite[Proposition 1.3]{GL87}. Here, $\mathfrak{m}$ is the unique homogeneous maximal ideal of $S$.
The following observation is direct. Recall that for an $L(\mathbf{p})$-graded $S$-module $X$, we denote by ${\rm gsupp}(X)$ its grading support; see Subsection 5.1.
\begin{lem}\label{lem:gsupp}
Consider the graded $S$-modules $S/\mathfrak{p}(\vec{x})$ for $\mathfrak{p}\in {\rm Spec}^{L(\mathbf{p})}(S)$ and $\vec{x}\in L(\mathbf{p})$. Write $\vec{x}=l\vec{c}+\sum_{j=1}^t l_j\vec{x}_j$ in its normal form. Then the following statements hold.
\begin{enumerate}
\item Assume that $\mathfrak{p}=(0)$ or $(f^h(x_1^{p_1}, x_2^{p_2}))$. Then $\mbox{\rm gsupp}\; S/\mathfrak{p}(\vec{x})=L(\mathbf{p})_+-\vec{x}$. In particular, $\mathbb{N}(m_0\vec{c})\subseteq \mbox{\rm gsupp}\; S/\mathfrak{p}(\vec{x})$ for sufficiently large $m_0>0$.
\item Assume that $\mathfrak{p}=(x_i)$. Then $\mbox{\rm gsupp}\; S/\mathfrak{p}(\vec{x})=\sum_{j\neq i}\mathbb{N}\vec{x}_j-\vec{x}$. In particular, $\mathbb{N}(m_0\vec{c})-l_i\vec{x}_i\subseteq \mbox{\rm gsupp}\; S/\mathfrak{p}(\vec{x})\subseteq \pi_i^{-1}(-\bar{l_i})$ for sufficiently large $m_0>0$.
\item Assume that $\mathfrak{p}=\mathfrak{m}$. Then $\mbox{\rm gsupp}\; S/\mathfrak{p}(\vec{x})=\{-\vec{x}\}$.
\end{enumerate}
\end{lem}
\begin{proof}
We use the facts that $\mbox{\rm gsupp}(X(\vec{x}))=\mbox{\rm gsupp}(X)-\vec{x}$, and that the quotient algebra $S/\mathfrak{p}$ is a graded domain which is generated as an algebra by $x_i$'s that are not contained in $\mathfrak{p}$. It follows that $\mbox{\rm gsupp}(S/\mathfrak{p})=L(\mathbf{p})_+$ for $\mathfrak{p}=(0)$ or $(f^h(x_1^{p_1}, x_2^{p_2}))$, and that $\mbox{\rm gsupp}( S/(x_i))=\sum_{j\neq i} \mathbb{N}\vec{x}_j$. The remaining statements are evident.
\end{proof}
Let $H\subseteq L(\mathbf{p})$ be an infinite subgroup. Consider the \emph{restriction subalgebra} $S_H=\oplus_{\vec{x}\in H}S_{\vec{x}}$ of $S$, which will be viewed as an $H$-graded algebra. For example, $S_{\mathbb{Z}\vec{c}}$ is isomorphic to the polynomial algebra $k[X, Y]$ by identifying $X$ with $x_1^{p_1}$ and $Y$ with $x_2^{p_2}$. More generally, for $m\geq 1$ the restriction subalgebra $S_{\mathbb{Z}(m\vec{c})}$ is isomorphic to the subalgebra of $k[X, Y]$ generated by monomials of degree $m$.
\begin{lem}
Let $H\subseteq L(\mathbf{p})$ be an infinite subgroup. Then the restriction subalgebra $S_H$ is a finitely generated $k$-algebra, and as an $S_H$-module, $S$ is finitely generated.
\end{lem}
\begin{proof}
Assume that $H\cap \mathbb{Z}\vec{c}=\mathbb{Z}(m\vec{c})$ for some $m\geq 1$. Recall from \cite[Proposition 1.3]{GL87} that $S$ is a finitely generated $S_{\mathbb{Z}\vec{c}}$-module, and thus a finitely generated $S_{\mathbb{Z}(m\vec{c})}$-module. Since the algebra $S_{\mathbb{Z}(m\vec{c})}$ is noetherian, the algebra $S_H$, viewed as an $S_{\mathbb{Z}(m\vec{c})}$-submodule of $S$, is finitely generated. Then the statements follow immediately.
\end{proof}
Recall that ${\rm mod}^{L(\mathbf{p})}\mbox{-}S$ denotes the abelian category of finitely presented $L(\mathbf{p})$-graded $S$-modules. Consider the \emph{restriction functor} $${\rm res}\colon {\rm mod}^{L(\mathbf{p})}\mbox{-}S\rightarrow {\rm mod}^H\mbox{-}S_H,$$ which sends an $L(\mathbf{p})$-graded $S$-module $X$ to $X_H=\oplus_{h\in H} X_h$, which is naturally an $H$-graded $S_H$-module. The functor ``res" is well defined by the following fact.
\begin{lem}
Let $H\subseteq L(\mathbf{p})$ be an infinite subgroup, and let $X$ be a finitely generated $L(\mathbf{p})$-graded $S$-module. Then the $H$-graded $S_H$-module $X_H$ is finitely generated.
\end{lem}
\begin{proof}
Assume that $H\cap \mathbb{Z}\vec{c}=\mathbb{Z}(m\vec{c})$ for some $m\geq 1$. We observe that $X$, viewed as an $S_{\mathbb{Z}(m\vec{c})}$-module, is finitely generated. Thus as its submodule, $X_H$ is a finitely generated $S_{\mathbb{Z}(m\vec{c})}$-module. In particular, $X_H$ is a finitely generated $S_H$-module.
\end{proof}
The following fact is well known.
\begin{lem}\label{lem:res}
Let $H\subseteq L(\mathbf{p})$ be an infinite subgroup. Then the restriction functor ${\rm res}\colon {\rm mod}^{L(\mathbf{p})}\mbox{-}S\rightarrow {\rm mod}^H\mbox{-}S_H$ is a quotient functor.
\end{lem}
\begin{proof}
Indeed, the functor ``res" admits a left adjoint $S\otimes_{S_H}-\colon {\rm mod}^H\mbox{-}S_H\rightarrow {\rm mod}^{L(\mathbf{p})}\mbox{-}S$. Moreover, the counit ${\rm res}\circ S\otimes_{S_H}-\longrightarrow {\rm Id}_{{\rm mod}^H\mbox{-}S_H}$ is an isomorphism. It follows from \cite[1.3 Proposition (iii)]{GZ} that the restriction functor is a quotient functor.
\end{proof}
\subsection{}
Let $S=S(\mathbf{p}, \lambda)$ be the homogenous coordinate algebra of a weighted projective line $\mathbb{X}=\mathbb{X}(\mathbf{p}, \lambda)$. Set $L=L(\mathbf{p})$. Denote by ${\rm mod}_0^{L}\mbox{-}S$ the full subcategory of ${\rm mod}^{L}\mbox{-}S$ consisting of finite dimensional modules; it is a Serre subcategory. The corresponding quotient category is denoted by ${\rm qmod}^L\mbox{-}S={\rm mod}^{L}\mbox{-}S/{{\rm mod}_0^{L}\mbox{-}S}$, and the quotient functor is denoted by $q\colon {\rm mod}^{L}\mbox{-}S\rightarrow {\rm qmod}^L\mbox{-}S$.
For each $\vec{x}\in L$, the degree-shift functor $(\vec{x})$ on ${\rm mod}^{L}\mbox{-}S$ induces an automorphism on the quotient category ${\rm qmod}^L\mbox{-}S$, which is also denoted by $(\vec{x})$. Consequently, each subgroup of $L$ has a strict action on ${\rm qmod}^L\mbox{-}S$. Set $\mathcal{S}_i=q(S/(x_i))$ for $1\leq i\leq t$.
We mention that the quotient category ${\rm qmod}^L\mbox{-}S$ is equivalent to the category ${\rm coh}\mbox{-}\mathbb{X}$ of coherent sheaves on $\mathbb{X}$, and the object $\mathcal{S}_i$ corresponds to a simple sheaf concentrated at the homogeneous prime ideal $(x_i)$; see \cite[Subsections 1.8 and 1.7]{GL87}.
Let $H\subseteq L$ be an infinite subgroup. Consider the $H$-graded restriction subalgebra $S_H$. We have the quotient category ${\rm qmod}^H\mbox{-}S_H={\rm mod}^H\mbox{-}S_H/{{\rm mod}_0^H\mbox{-}S_H}$. The restriction functor ${\rm res}\colon {\rm mod}^{L}\mbox{-}S\rightarrow {\rm mod}^{H}\mbox{-}S_H$ induces the corresponding functor betwen the quotient categories $${\rm res}\colon {\rm qmod}^L\mbox{-}S\rightarrow {\rm qmod}^H\mbox{-}S_H,$$ which will also be referred as the \emph{restriction functor}.
We will consider the following Serre subcategory of ${\rm qmod}^L\mbox{-}S$
$$\mathcal{N}_H=\langle \mathcal{S}_i(l\vec{x}_i)\; |\; 1 \leq i\leq t, 0\leq l\leq p_i-1 \mbox{ with } -\bar{l}\notin \pi_i(H)\rangle.$$
That is, $\mathcal{N}_H$ is the Serre subcategory generated by these simple sheaves $ \mathcal{S}_i(l\vec{x}_i)$.
Recall that for each $1\leq i\leq t$, the surjective homomorphism $\pi_i\colon L\rightarrow \mathbb{Z}/{p_i\mathbb{Z}}$ is defined by $\pi_i(\vec{x}_j)=\delta_{i, j}\bar{1}$. The following concept is related to the general notion of a wide subcategory of the category formed by all the line bundles on a weighted projective line in \cite{KLM}.
\begin{defn}\label{defn:1}
Let $\mathbf{p}$ be a weight sequence and $L(\mathbf{p})$ be the string group. An infinite subgroup $H\subseteq L(\mathbf{p})$ is said to be \emph{effective} provided that for each $1\leq i\leq t$, $\pi_i(H)=\mathbb{Z}/{p_i\mathbb{Z}}$, or equivalently, $\bar{1}$ lies in $\pi_i(H)$.
\end{defn}
For example, any infinite subgroup of $L(\mathbf{p})$ containing the dualizing element $\vec{\omega}$ is effective, while the subgroup $\mathbb{Z}\vec{c}$ generated by $\vec{c}$ is not effective for $t\geq 1$.
The following result justifies the terminology ``effective subgroup". We mention that in spirit it is close to \cite[Proposition 8.5]{GL90}; compare the argument in \cite[Example 5.8]{GL87}. The equivalence part of the following proposition is contained in a more general result in \cite{KLM}.
\begin{prop}\label{prop:eff}
Let $L=L(\mathbf{p})$. Keep the notation as above. Then the restriction functor ${\rm res}\colon {\rm qmod}^L\mbox{-}S\rightarrow {\rm qmod}^H\mbox{-} S_H$ induces an equivalence of categories
$${\rm qmod}^L\mbox{-}S/\mathcal{N}_H\stackrel{\sim}\longrightarrow {\rm qmod}^H\mbox{-}S_H. $$
In particular, the restriction functor ${\rm res}\colon {\rm qmod}^L\mbox{-}S\rightarrow {\rm qmod}^H\mbox{-}S_H$ is an equivalence if and only if the subgroup $H\subseteq L$ is effective.
\end{prop}
\begin{proof}
The restriction functor ${\rm res}\colon {\rm mod}^L\mbox{-}S\rightarrow {\rm mod}^H\mbox{-}S_H$ is a quotient functor by Lemma \ref{lem:res}. Then by Lemma \ref{lem:quoinduced} the induced functor ${\rm res}\colon {\rm qmod}^L\mbox{-} S\rightarrow {\rm qmod}^H\mbox{-}S_H$ is also a quotient functor. It suffices to calculate its essential kernel. Instead, we will compute the essential kernel of the composite $F\colon {\rm mod}^L\mbox{-}S\stackrel{q}\rightarrow {\rm qmod}^L\mbox{-} S \stackrel{\rm res}\rightarrow {\rm qmod}^H\mbox{-} S_H$. Indeed, for an $L$-graded $S$-module $X$, $F(X)=0$ if and only if $X_H$ is a finite dimensional $S_H$-module, which is equivalent to the condition that $\mbox{\rm gsupp}(X)\cap H$ is a finite set.
We observe that $\mathbb{Z}(m\vec{c})\subseteq H$ for a sufficiently large $m>0$. Let $\vec{x}=l\vec{c}+\sum_{i=1}^tl_i\vec{x}_i$ an arbitrary element in $L$ that is written in its normal form. It follows from Lemma \ref{lem:gsupp} that $\mbox{\rm gsupp}(S/\mathfrak{p}(\vec{x})) \cap H $ is finite if and only if $\mathfrak{p}=\mathfrak{m}$, or $\mathfrak{p}=(x_i)$ and $-\bar{l_i}\notin \pi_i(H)$.
Recall that a finitely generated $L$-graded $S$-module $X$ has a finite filtration with factors isomorphic to $S/\mathfrak{p}(\vec{x})$ for some $\mathfrak{p}\in {\rm Spec}^L(S)$ and $\vec{x}\in L$. Recall that ${\rm Ker}\; F$ is a Serre subcategory of ${\rm mod}^L\mbox{-}S$. It follows that
$${\rm Ker}\; F=\langle S/\mathfrak{m}(\vec{y}), S/(x_i)(\vec{x}) \; |\; \vec{y}\in L, 1\leq i\leq t, \vec{x} \mbox{ satisfying that } -\bar{l_i}\notin \pi_i(H)\rangle.$$
Then we are done by using the fact that the essential kernel of ``res" equals $q({\rm Ker}\; F)$, which equals $\mathcal{N}_H$. Here, we recall from \cite[Subsection 1.6]{GL87} the well-known fact that $\mathcal{S}_i(\vec{x})=\mathcal{S}_i(l_i\vec{x}_i)$ for each $1\leq i\leq t$.
\end{proof}
\section{Weighted projective lines of tubular type}
In this section, we study weighted projective lines of tubular type. They are explicitly related to elliptic plane curves via two different equivariantizations. We emphasize that the idea goes back to \cite[Example 5.8]{GL87} and more explicitly to \cite{Len, Lentalk}; compare \cite{Po,Hille,Ploog}.
\subsection{}
We recall that a weight sequence $\mathbf{p}=(p_1, p_2, \cdots, p_t)$ is of \emph{tubular type} provided that the dualizing element $\vec{\omega}$ in $L(\mathbf{p})$ is torsion, or equivalently, $\delta(\vec{\omega})=0$. An elementary calculation yields a complete list of weight sequences of tubular type: $(2,2,2,2)$, $(3,3,3)$, $(4,4,2)$ and $(6,3,2)$. We observe that in each case the order of $\vec{\omega}$ equals $p={\rm lcm}(\mathbf{p})$, which further equals $p_1$.
We have the following easy observation.
\begin{lem}\label{lem:mult}
Let $\mathbf{p}$ be a weight sequence of tubular type and let $n\geq 1$. Recall that $p={\rm lcm}(\mathbf{p})$. Then the following equation holds
$$\sum_{j=0}^{p-1} {\rm mult}(n\vec{x}_1+j\vec{\omega})=n.$$
\end{lem}
\begin{proof}
We observe that ${\rm mult}(\vec{x}+\vec{c})={\rm mult}(\vec{x})+1$ provided that $\phi(\vec{x})\geq -1$. For a weight sequence $\mathbf{p}$ of tubular type, we recall that $p=p_1$ and thus $p\vec{x}_1=\vec{c}$. We observe that $\phi(n\vec{x}_1+j\vec{\omega})\geq -1$ for each $n\geq 1$ and $0\leq j< p-1$. Denote the left hand side of the equation by $f(n)$. It follows that $f(n+p)=f(n)+p$ for each $n\geq 1$.
Therefore, to show the required equation, it suffices to prove that $f(i)=i$ for $1\leq i\leq p$. These $p$ equations are easy to verify for each of the four cases. Take the case $\mathbf{p}=(6,3,2)$ for example. We have that $f(4)=\sum_{j=0}^{5} {\rm mult}(4\vec{x}_1+j\vec{\omega})=1+0+1+1+1+0=4$ and that $f(5)=\sum_{j=0}^{5} {\rm mult}(5\vec{x}_1+j\vec{\omega})=1+0+1+1+1+1=5$. We omit the details.
\end{proof}
The following consideration is inspired by \cite[Example 5.8]{GL87}. Let $\mathbf{p}$ be a weight sequence of tubular type. We consider the subgroup $H(\mathbf{p})=\mathbb{Z}(3\vec{x}_1)\oplus \mathbb{Z}\vec{\omega}$ of $L(\mathbf{p})$ generated by $3\vec{x}_1$ and $\vec{\omega}$. It inherits the partial order from $L(\mathbf{p})$ such that its positive cone equals $H(\mathbf{p})_+=H(\mathbf{p})\cap L(\mathbf{p})_+$, which is a submonoid of $H(\mathbf{p})$. We consider the subset $3\vec{x}_1+\mathbb{Z}\vec{\omega}=\{3\vec{x}_1+j\vec{\omega}\; |\; j=0, 1,\cdots, p-1\}$ of $H(\mathbf{p})$. We list in Table 1 all the positive elements in $3\vec{x}_1+\mathbb{Z}\vec{\omega}$ explicitly.
\begin{table}[h]
\caption{The list of generators of $H(\mathbf{p})_+$}
\begin{tabular}{|c|c|c|c|}
\hline
the weight sequence $\mathbf{p}$ & \multicolumn{2}{c}{ \quad the elements in $(3\vec{x}_1+\mathbb{Z}\vec{\omega})\cap H(\mathbf{p})_+$} & \\
\hline
$(2,2,2,2)$ & \; $3\vec{x}_1$ \; & \multicolumn{1}{c}{\; $3\vec{x}_1+\vec{\omega}=\vec{x}_2+\vec{x}_3+\vec{x}_4$} &\\
\hline
$(3,3,3)$ & \; $3\vec{x}_1$ \; & \multicolumn{1}{c}{\; $3\vec{x}_1+2\vec{\omega}=\vec{x}_1+\vec{x}_2+\vec{x}_3$} & \\
\hline
$(4,4,2)$ & \; $3\vec{x}_1$ \; & $3\vec{x}_1+2\vec{\omega}=\vec{x}_1+2\vec{x}_2$ & $3\vec{x}_1+3\vec{\omega}=\vec{x}_2+\vec{x}_3$ \\
\hline
$(6,3,2)$ & \; $3\vec{x}_1$ \; & $3\vec{x}_1+2\vec{\omega}=\vec{x}_1+\vec{x}_2$& $3\vec{x}_1+3\vec{\omega}=\vec{x}_3$ \\
\hline
\end{tabular}
\end{table}
We have the following elementary observation, for which we include an elementary proof here. For a conceptual reasoning, we refer to Remark \ref{rem:thm}.
\begin{lem}\label{lem:gene}
Let $\mathbf{p}$ be a weight sequence of tubular type. Then we have $$H(\mathbf{p})_+=\mathbb{N}((3\vec{x}_1+\mathbb{Z}\vec{\omega})\cap H(\mathbf{p})_+).$$
In other words, the monoid $H(\mathbf{p})_+$ is generated by the subset $(3\vec{x}_1+\mathbb{Z}\vec{\omega})\cap H(\mathbf{p})_+$.
\end{lem}
\begin{proof}
Recall that $p={\rm lcm}(\mathbf{p})=p_1$ equals the order of $\vec{\omega}$. For each $0\leq j\leq p-1$, we define $m(j)$ to be the least natural number $m$ such that $m(3\vec{x}_1)+j\vec{\omega}\geq \vec{0}$. We observe that $m(0)=0$ and $m(j)\geq 1$ for $j\neq 0$.
We claim that $m(j)(3\vec{x}_1)+j\vec{\omega}$ lies in $\mathbb{N}((3\vec{x}_1+\mathbb{Z}\vec{\omega})\cap H(\mathbf{p})_+)$ for each $0\leq j\leq p-1$. Then we are done. Indeed, for any positive element $\vec{x}=m(3\vec{x}_1)+j\vec{\omega}$, we have $m\geq m(j)$, and thus $\vec{x}=(m-m(j))(3\vec{x}_1)+(m(j)(3\vec{x}_1)+j\vec{\omega})$, which lies in $\mathbb{N}((3\vec{x}_1+\mathbb{Z}\vec{\omega})\cap H(\mathbf{p})_+)$.
The claim is easily verified case by case.
\end{proof}
\subsection{}
We recall that a weighted projective line $\mathbb{X}=\mathbb{X}(\mathbf{p}, \lambda)$ over a field $k$ is of \emph{tubular type} provided that the weight sequence $\mathbf{p}$ is of tubular type. Recall that $S=S(\mathbf{p}, \lambda)$ is the homogeneous coordinate algebra of $\mathbb{X}$.
We fix the notation for $S$ in the four tubular types. For the weight type $(2,2,2,2)$, there exists a parameter $\lambda \neq 0, 1$ in the presentation of the corresponding homogeneous coordinate algebra $S$. In other three cases, we omit the normalized parameter sequences. Hence, we refer to $(2,2,2,2; \lambda)$, $(3,3,3)$, $(4,4,2)$ and $(6,3,2)$ as the \emph{type} of $S$. We emphasize that $\lambda\in k$, which is not equal to $0$ or $1$.
We list the homogeneous coordinate algebras $S$ explicitly, according to their types. We define the \emph{associated} $\mathbb{Z}$-graded algebras $R$ of the same type on the right hand side, where all the variables are of degree one.
{\tiny
\begin{align*}
&S(2,2,2,2; \lambda)=\frac{k[X_1, X_2, X_3, X_4]}{(X_3^2-(X_2^2-X_1^2), X_4^2-(X_2^2-\lambda X_1^2))}, \quad R(2,2,2,2;\lambda)= \frac{k[X,Y, Z]}{(Y^2Z-X(X-Z)(X-\lambda Z))};
\\ &S(3,3,3)=\frac{k[X_1, X_2, X_3]}{(X_3^3-(X_2^3-X_1^3))}, \hskip 98pt R(3,3,3)=\frac{k[X, Y, Z]}{(X^3-(Y^2Z-Z^2Y))};\\
&S(4,4,2)=\frac{k[X_1, X_2, X_3]}{(X_3^2-(X_2^4-X_1^4))}, \hskip 98pt R(4,4,2)=\frac{k[X, Y, Z]}{(Y^2Z-X(X-Z)(X+Z))};\\
&S(6,3,2)=\frac{k[X_1, X_2, X_3]}{(X_3^2-(X_2^3-X_1^6))}, \hskip 98pt R(6,3,2)=\frac{k[X, Y, Z]}{(Y^2Z-(X^3-Z^3))}.
\end{align*}
}
We denote by $\mathbb{E}$ the projective plane curve defined by $R$. By Serre's theorem, there is an equivalence between the category ${\rm coh}\mbox{-}\mathbb{E}$ of coherent sheaves on $\mathbb{E}$ and the quotient category ${\rm qmod}^\mathbb{Z}\mbox{-}R$ of ${\rm mod}^\mathbb{Z}\mbox{-}R$ by the Serre subcategory ${\rm mod}_0^\mathbb{Z}\mbox{-}R$ consisting of finite dimensional modules.
\begin{rem}\label{rem:j}
We mention that $\mathbb{E}(2,2,2,2;\lambda)$ is an elliptic curve for $\lambda \neq 0, 1$. For other types, we have requirement on the characteristic of the field $k$; see also \cite{Len}. If ${\rm char}\; k\neq 3$, $\mathbb{E}(3,3,3)$ is an elliptic curve with $j$-invariant $0$; if ${\rm char}\; k\neq 2$, $\mathbb{E}(4,4,2)$ is an elliptic curve with $j$-invariant $1728$; if ${\rm char}\; k\neq 2, 3$, $\mathbb{E}(6,3,2)$ is an elliptic curve with $j$-invariant $0$.
\end{rem}
We define three homogeneous elements $x$, $y$ and $z$ in $S$ according to Table 2. By comparing with Table 1, all their degrees lie in $(3\vec{x}_1+\mathbb{Z}\vec{\omega})\cap H(\mathbf{p})_+$. For an explanation of the term ``extra degree" in Table 2, we refer to Remark \ref{rem:extra}.
\begin{table}[h]
\caption{The elements $x$, $y$ and $z$ in $S$ and their extra degrees}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
the types of $S$ and $R$ & $x$ & $y$ & $z$ & ${\rm deg}\; x$ & ${\rm deg}\; y$ & ${\rm deg}\; z$ \\
\hline
$(2,2,2,2; \lambda)$ & $x_1x_2^2$ & $x_2x_3x_4$ & \; $x_1^3$ \; & {\small $ (1, \bar{0})$} &{\small $(1, \bar{1})$} & {\small $(1, \bar{0})$} \\
\hline
$(3,3,3)$ & $x_1x_2x_3$ & $x_2^3$ & \; $x_1^3$ \; & {\small $(1, \bar{2})$} & {\small $(1, \bar{0})$} & {\small $(1, \bar{0})$} \\
\hline
$(4,4,2)$ & $x_1x_2^2$ & $x_2x_3$ & \; $x_1^3$ \; & {\small $(1, \bar{2})$} & {\small $(1, \bar{3})$} & {\small $(1, \bar{0})$} \\
\hline
$(6,3,2)$ & $x_1x_2$ & $x_3$ & \; $x_1^3$ \; & {\small $(1, \bar{2})$} & {\small $(1, \bar{3})$} & {\small $(1, \bar{0})$} \\
\hline
\end{tabular}
\end{table}
We consider the subalgebra $k[x, y, z]$ of $S$ generated by $x$, $y$ and $z$. We observe that $k[x, y, z]\subseteq S_{H(\mathbf{p})}$. Consider the surjective homomorphism $\pi\colon H(\mathbf{p})\rightarrow \mathbb{Z}$ defined by $\pi(3\vec{x}_1)=1$ and $\pi(\vec{\omega})=0$.
The central technical result of this section is as follows.
\begin{thm}\label{thm:R-S}
Let $S$ be the homogeneous coordinate algebra of a weighted projective line $\mathbb{X}$ of tubular type and let $R$ be the associated $\mathbb{Z}$-graded algebra of the same type. Let $H=H(\mathbf{p})$. Keep the above notation. Then we have $k[x, y, z]=S_H$, and that there is an isomorphism of $\mathbb{Z}$-graded algebras
$$R \stackrel{\sim}\longrightarrow \pi_*(S_H).$$
\end{thm}
\begin{proof}
We claim that $k[x, y, z]\subseteq S$ is an integral extension. Recall that an integral extension preserves Krull dimensions and that the Krull dimension of the algebra $S$ is two. Hence the Krull dimension of $k[x, y, z]$ is also two.
Recall that $S_{\mathbb{Z}\vec{c}}=k[x_1^{p_1}, x_2^{p_2}]$ and that $S$ is a finitely generated $S_{\mathbb{Z}\vec{c}}$-module. Hence $S$ is integral over $S_{\mathbb{Z}\vec{c}}$ and thus over the subalgebra $k[x_1, x_2]$. For the claim, it suffices to show that both $x_1$ and $x_2$ are integral over $k[x, y, z]$. Since $x_1^3=z$, the element $x_1$ is integral over $k[x, y, z]$.
We consider the element $x_2$ case by case. We take $\mathbf{p}=(2,2,2,2)$ for example, while other cases are similar and even easier. Observe in Table 2 that $y^2=x_2^2(x_2^2-x_1^2)(x_2^2-\lambda x_1^2)$. This implies that $x_2$ is integral over $k[x, y, z, x_1]$ and thus over $k[x, y,z]$. We are done with the claim.
We define a homomorphism $\theta\colon R\rightarrow S$ of algebras by sending $X$ to $x$, $Y$ to $y$ and $Z$ to $z$; see Table 2. The relation of $R$ is satisfied by an elementary calculation. The image of $\theta$ is $k[x, y, z]$. Recall that $R$ is an integral domain of Krull dimension two. Comparing the Krull dimensions of $R$ and the image of $\theta$, we infer that $\theta$ is injective. Consequently, we have an injective homomorphism of $\mathbb{Z}$-graded algebras $\theta \colon R\rightarrow \pi_*(S_H)$. Here, we recall that $k[x, y,z]\subseteq S_H$.
We claim that the injective homomorphism $\theta \colon R\rightarrow \pi_*(S_{H})$ is an isomorphism. Then the two required statements follow. Indeed, it suffices to show that for each $n\geq 1$, ${\rm dim}_k\; R_n={\rm dim}_k\; \pi_*(S_H)_n$. Recall that $\pi_*(S_H)_n=\oplus_{j=0}^{p-1} S_{3n\vec{x}_1+j\vec{\omega}}$, and thus has dimension $\sum_{j=0}^{p-1} {\rm mult}(3n\vec{x}_1+j\vec{\omega})$; see (\ref{equ:mult}). By Lemma \ref{lem:mult} we have ${\rm dim}_k\; \pi_*(S_H)_n=3n$. On the other hand, it is well known that ${\rm dim}_k\; R_n=3n$. Then we are done.
\end{proof}
\begin{rem}\label{rem:thm}
Recall that $\vec{x}\in H(\mathbf{p})$ is positive if and only if $(S_{H(\mathbf{p})})_{\vec{x}} =S_{\vec{x}} \neq 0$. We deduce from the equality $k[x, y, z]=S_{H(\mathbf{p})}$ that each $\vec{x}\in H(\mathbf{p})_+$ lies in the submonoid generated by the degrees of $x$, $y$ and $z$. This gives another proof of Lemma \ref{lem:gene}.
\end{rem}
\begin{rem} \label{rem:extra}
We now explain the ``extra degrees" of $x$, $y$ and $z$ in $S_{H(\mathbf{p})}$. We recall that $p={\rm lcm}(\mathbf{p})$ equals the order of $\vec{\omega}$. So we have the following isomorphism of abelian groups
\begin{align}
\psi\colon H(\mathbf{p})=\mathbb{Z}(3\vec{x}_1)\oplus \mathbb{Z}\vec{\omega}\stackrel{\sim}\longrightarrow \mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}}
\end{align}
such that $\psi(3\vec{x}_1)=(1, \bar{0})$ and $\psi(\vec{\omega})=(0, \bar{1})$. By this isomorphism, the $H(\mathbf{p})$-graded algebra $S_{H(\mathbf{p})}$ is $\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$-graded. This yields the \emph{extra degrees} of $x$, $y$ and $z$, which are computed by applying $\psi$ to the degrees of $x$, $y$ and $z$; compare Table 1.
We observe that the algebras $R$ are naturally $\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}}$-graded according to the degrees in Table 2; here, we abuse $X$, $Y$, $Z$ with $x$, $y$ and $z$, respectively. We denote the resulting $\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}}$-graded algebras by $\bar{R}$; it is a $\mathbb{Z}/{p\mathbb{Z}}$-refinement of $R$.
Then the isomorphism $\theta$ in the above proof, which is given according to Table 2, yields an isomorphism of $\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}}$-algebras
\begin{align}\label{equ:iso1}
\theta\colon \bar{R}\stackrel{\sim}\longrightarrow S_{H(\mathbf{p})}.
\end{align}
\end{rem}
\vskip 10pt
Let $\mathbf{p}$ be a weight sequence of tubular type. Recall that $p={\rm lcm}(\mathbf{p})$. Denote by $C_p=\langle g\; |\; g^p=1\rangle $ the cyclic group of order $p$.
We assume that $k$ is a splitting field of $C_p$. Equivalently, $p$ is invertible in $k$ which contains a primitive $p$-th root of unity. Fix $\omega\in k$ a primitive $p$-th root of unity. For example, if $\mathbf{p}=(2,2,2,2)$, then the characteristic of $k$ is not equal to $2$ and $\omega=-1$.
We identify $C_p$ with the character group $\widehat{\mathbb{Z}/{p\mathbb{Z}}}$ of $\mathbb{Z}/{p\mathbb{Z}}$ such that $g(\bar{1})=\omega$. Applying the correspondence (\ref{equ:corres2}) to the above $\mathbb{Z}/{p\mathbb{Z}}$-refinement $\bar{R}$ of the $\mathbb{Z}$-graded algebra $R$, we obtain in Table 3 the corresponding $C_p$-action on $R$ as $\mathbb{Z}$-graded algebra automorphisms.
\begin{table}[h]
\caption{The cyclic group action on $R$}
\begin{tabular}{|c|c|c|}
\hline
the types of $R$ & the action of $C_p=\langle g\; |\; g^p=1\rangle $ on generators & the order of $\omega\in k$ \\
\hline
$(2,2,2,2; \lambda)$ & $g.x=x, \; g.y=\omega y,\; g.z=z$ & 2\\
\hline
$(3,3,3)$ & $g.x=\omega^2 x, \; g.y=y, \; g.z=z$ & 3\\
\hline
$(4,4,2)$ & $g.x=\omega^2 x, \;g.y=\omega^3 y, \; g.z=z$ & 4\\
\hline
$(6,3,2)$ & $g.x=\omega^2 x,\; g.y=\omega^3 y, \; g.z=z$ & 6\\
\hline
\end{tabular}
\end{table}
We mention that the cyclic group action on $R$ corresponds to a ramified Galois covering from the elliptic curve $\mathbb{E}$ to the projective line $\mathbb{P}_{k}^1$, where the Galois group is isomorphic to the acting cyclic group; compare \cite[Subsection 1.3]{Po}.
\subsection{} We will formulate the main result of this paper. As mentioned earlier, the result is implicitly contained in \cite[Example 5.8]{GL87} and later explicitly in \cite{Len} with more details; see also \cite{Lentalk}. The treatment here is slightly different from \cite{Len, Lentalk}, and we strengthen slightly the results therein; compare Remark \ref{rem:main}(3). We mention that related results appear in \cite{Hille}.
Let $k$ be an arbitrary field. Let $S$ be the homogeneous coordinate algebra of a weighted projective line $\mathbb{X}$ of tubular type, and $R$ be the associated $\mathbb{Z}$-graded algebra of the same type. Let $\mathbf{p}$ be the weight sequence and denote $p={\rm lcm}(\mathbf{p})$. Recall that the dualizing element $\vec{\omega}$ in $L=L(\mathbf{p})$ has order $p$. The subgroup $\mathbb{Z}\vec{\omega}$ acts on ${\rm mod}^L\mbox{-}S$ by the degree-shift action, which induces the \emph{degree-shift action} of $\mathbb{Z}\vec{\omega}$ on ${\rm qmod}^L\mbox{-}S$. We consider the category $({\rm qmod}^L\mbox{-}S)^{\mathbb{Z}\vec{\omega}}$ of equivariant objects.
Assume that $k$ is a splitting field of the cyclic group $C_p$ of order $p$. Consider the $C_p$-action on $R$ as $\mathbb{Z}$-graded algebra automorphisms in Table 3. Such an action induces the twisting action of $C_p$ on ${\rm mod}^\mathbb{Z}\mbox{-}R$, which further induces the \emph{twisting action} of $C_p$ on ${\rm qmod}^\mathbb{Z}\mbox{-}R$. We consider the category $({\rm qmod}^\mathbb{Z}\mbox{-}R)^{C_p}$ of equivariant objects.
\begin{thm}{}\rm (Lenzing-Meltzer)\label{thm:main}
Let $k$ be a field. Let $S$ be the homogeneous coordinate algebra of a weighted projective line $\mathbb{X}$ of tubular type, and let $R$ be the associated $\mathbb{Z}$-graded algebra of the same type, where we keep the notation as above. Then the following statements hold.
\begin{enumerate}
\item There is an equivalence of categories
$$({\rm qmod}^L\mbox{-}S)^{\mathbb{Z}\vec{\omega}} \stackrel{\sim}\longrightarrow {\rm qmod}^\mathbb{Z}\mbox{-}R.$$
\item Assume that $k$ is a splitting field of $C_p$. Then there is an equivalence of categories
$${\rm qmod}^L\mbox{-}S\stackrel{\sim}\longrightarrow ({\rm qmod}^\mathbb{Z}\mbox{-}R)^{C_p}.$$
\end{enumerate}
\end{thm}
\begin{proof}
Recall the subgroup $H=H(\mathbf{p})=\mathbb{Z}(3\vec{x}_1)\oplus \mathbb{Z}\vec{\omega}$ of $L$ from Subsection 7.1; it is an effective subgroup in the sense of Definition \ref{defn:1}. By Proposition \ref{prop:eff} the restriction functor is an equivalence $${\rm res}\colon {\rm qmod}^L\mbox{-}S \stackrel{\sim}\longrightarrow {\rm qmod}^H\mbox{-}S_H.$$
Recall from Remark \ref{rem:extra} the $\mathbb{Z}/{p\mathbb{Z}}$-refinement $\bar{R}$ of the $\mathbb{Z}$-graded algebra $R$. By the isomorphism $\theta$ in (\ref{equ:iso1}) we have an isomorphism of categories $$\theta^*\colon {\rm qmod}^H\mbox{-}S_H\stackrel{\sim}\longrightarrow {\rm qmod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R}.$$
Consider the following composite equivalence
\begin{align}\label{equ:comp}
\theta^*\circ {\rm res}\colon {\rm qmod}^L\mbox{-}S \stackrel{\sim}\longrightarrow {\rm qmod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R}.\end{align}
We observe that this equivalence transfers the degree-shift action of $\mathbb{Z}\vec{\omega}$ on ${\rm qmod}^L\mbox{-}S$ to the degree-shift action of $\mathbb{Z}/{p\mathbb{Z}}$ on ${\rm qmod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R}$. Consequently, this equivalence induces an equivalence of categories
\begin{align}\label{equ:ZZ}
({\rm qmod}^L\mbox{-}S)^{\mathbb{Z}\vec{\omega}} \stackrel{\sim}\longrightarrow ({\rm qmod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R})^{\mathbb{Z}/{p\mathbb{Z}}}.
\end{align}
By Proposition \ref{prop:app2} we have an equivalence of categories
$$({\rm mod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R})^{\mathbb{Z}/{p\mathbb{Z}}} \stackrel{\sim}\longrightarrow {\rm mod}^\mathbb{Z} \mbox{-}R,$$
which restricts to an equivalence
$({\rm mod}_0^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R})^{\mathbb{Z}/{p\mathbb{Z}}} \stackrel{\sim}\longrightarrow {\rm mod}_0^\mathbb{Z} \mbox{-}R$
of full subcategories consisting of finite dimensional modules. Applying Corollary \ref{cor:equiv}, we have an equivalence of categories
$$({\rm qmod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R})^{\mathbb{Z}/{p\mathbb{Z}}}\stackrel{\sim}\longrightarrow {\rm qmod}^\mathbb{Z} \mbox{-}R.$$
We combine this equivalence with (\ref{equ:ZZ}) to obtain (1).
For (2), we assume that $k$ is a splitting field of $C_p$. Recall that $C_p$ is identified with the character group of $\mathbb{Z}/{p\mathbb{Z}}$. By Proposition \ref{prop:recover} we have an isomorphism of categories $${\rm mod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R}\stackrel{\sim}\longrightarrow ({\rm mod}^\mathbb{Z}\mbox{-}R)^{C_p},$$
which restricts to an isomorphism ${\rm mod}_0^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R}\stackrel{\sim}\longrightarrow ({\rm mod}_0^\mathbb{Z}\mbox{-}R)^{C_p}$ of full subcategories consisting of finite dimensional modules. Applying Corollary \ref{cor:equiv}, we have an equivalence of categories
$${\rm qmod}^{(\mathbb{Z}\times \mathbb{Z}/{p\mathbb{Z}})}\mbox{-}\bar{R}\stackrel{\sim}\longrightarrow ({\rm qmod}^\mathbb{Z}\mbox{-}R)^{C_p}.$$
We combine this equivalence with (\ref{equ:comp}) to obtain (2).
\end{proof}
\begin{rem}\label{rem:main}
(1) We recall that $\mathbb{X}$ is the weighted projective line defined by $S$ and that $\mathbb{E}$ is the projective plane curve defined by $R$. If $k$ is a splitting field of $C_p$, then $\mathbb{E}$ is an elliptic plane curve. Then Theorem \ref{thm:main} relates the categories of coherent sheaves on $\mathbb{X}$ and $\mathbb{E}$ via two equivariantizations.
We emphasize that the two group actions on relevant categories are completely different: one is the degree-shift action and another is the twisting action. However, as in Remark \ref{rem:Len1} the two equivalences obtained in Theorem \ref{thm:main} are somehow \emph{adjoint} to each other. We mention that by Remark \ref{rem:j} the assignment $\mathbb{X}\mapsto\mathbb{E}$ is not a bijection up to isomorphism; see also \cite{Lentalk}.
(2) For the $L$-graded algebra $S$ and the associated $\mathbb{Z}$-graded $R$ one might consider the quotient categories ``${\rm QMod}$" of arbitrary graded modules, which are equivalent to the categories of quasi-coherent sheaves on $\mathbb{X}$ and $\mathbb{E}$, respectively. Then a similar result as Theorem \ref{thm:main} holds for theses two quotient categories ``${\rm QMod}$".
(3) Let us mention the version of Theorem \ref{thm:main} in \cite{Lentalk}.
Consider the degree map $\delta\colon L\rightarrow \mathbb{Z}$ whose kernel equals the torsion subgroup ${\rm tor}(L)$ of $L$; moreover, we have a decomposition $L={\rm tor}(L)\oplus \mathbb{Z}\vec{x}_1$. Consider the $\mathbb{Z}$-graded algebra $\underline{S}=\delta_*(S)$; see Subsection 5.1. We view the $L$-graded algebra $S$ as a ${\rm tor}(L)$-refinement of $\underline{S}$. Observe that $\mathbb{Z}\vec{\omega}\subseteq {\rm tor}(L)$.
If the field $k$ is a splitting field of $C_p$, then it is also a splitting field of ${\rm tor}(L)$. In this case, we denote by $A$ the character group of ${\rm tor}(L)$. Then $A$ acts on $\underline{S}$ as $\mathbb{Z}$-graded algebra automorphisms by Lemma \ref{lem:corres}. We consider the \emph{degree-shift action} of ${\rm tor}(L)$ on ${\rm qmod}^L\mbox{-}S$, and the \emph{twisting action} of $A$ on ${\rm qmod}^\mathbb{Z}\mbox{-}\underline{S}$. Then under the same assumptions as Theorem \ref{thm:main}, the same argument yields the following two adjoint equivalences:
\begin{align}\label{equ:LM}
({\rm qmod}^L\mbox{-}S)^{{\rm tor}(L)} \stackrel{\sim}\longrightarrow {\rm qmod}^\mathbb{Z}\mbox{-}\underline{S}, \quad {\rm qmod}^L\mbox{-}S\stackrel{\sim}\longrightarrow ({\rm qmod}^\mathbb{Z}\mbox{-}\underline{S})^{A}.\end{align}
Indeed, the argument here is easier since it does not rely on Proposition \ref{prop:eff} and Theorem \ref{thm:R-S}.
The major difference between the two equivalences (\ref{equ:LM}) and the ones in Theorem \ref{thm:main} is the fact that the acting groups in (\ref{equ:LM}) are not cyclic except the case $\mathbf{p}=(6,3,2)$, while all the acting groups in Theorem \ref{thm:main} are cyclic.
We mention that for the cases $\mathbf{p}=(4,4,2)$ and $(6,3,2)$, the variables of the $\mathbb{Z}$-graded algebra $\underline{S}$ are not all of degree one. Therefore, to verify that $ {\rm qmod}^\mathbb{Z}\mbox{-}\underline{S}$ is equivalent to the category of coherent sheaves on an elliptic curve, one might need the general fact on Veronese subalgebras \cite[Proposition 5.10]{AZ}, or the axiomatic description of the category of coherent sheaves on an elliptic curve in \cite[Theorem C]{RV}. If the characteristic of $k$ is not equal to $2$, the $\mathbb{Z}$-graded algebra $\underline{S}$ for the case $\mathbf{p}=(2,2,2,2)$ defines an elliptic space curve in its Jacobi form. If the characteristic of $k$ is not equal to $3$, the $\mathbb{Z}$-graded algebra $\underline{S}$ for the case $\mathbf{p}=(3,3,3)$ defines an elliptic curve as a Fermat curve of degree $3$.
\end{rem}
\vskip 15pt
\noindent {\bf Acknowledgements}\; J. Chen is very grateful to Helmut Lenzing for giving her the manuscript \cite{Len} during his visit at Xiamen University in 2011. The authors thank Helmut Lenzing very much for enlightening and helpful discussions. The authors are grateful to Henning Krause and Hagen Meltzer for helpful comments.
J. Chen and Z. Zhou are supported by National Natural Science Foundation of China (No. 11201386) and Natural Science Foundation of Fujian Province (No. 2012J05009); X.W. Chen is supported by NCET-12-0507, National Natural Science Foundation of China (No. 11201446) and the Alexander von Humboldt Stiftung.
\bibliography{}
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TITLE: Connection between Fourier transform and Taylor series
QUESTION [152 upvotes]: Both Fourier transform and Taylor series are means to represent functions in a different form.
What is the connection between these two? Is there a way to get from one to the other (and back again)? Is there an overall, connecting (geometric?) intuition?
REPLY [11 votes]: Taylor series at $t=0$ of some function $\textrm{f}(t)$ is defined as
$$ \textrm{f}\left(t\right) =\sum_{j=0}^{\infty}
{
h_j\cdotp\frac{d^{j}}{dt^{j}}\textrm{f}(0)\cdotp t^{j}
}
$$
where $ h_j=1/{j!}$ and $\frac{d^0}{dt^0}\textrm{f}\left(t\right)=\textrm{f}\left(t\right)$
Fourier series is defined as
$$
\textrm{f}\left(t\right) =
\sum_{n=1}^{\infty}
{ \left(
a_n\cdot\cos \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)+
b_n\cdot\sin \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)
\right)
}
$$
with coefficients:
$$
\begin{align}
a_n&=\frac{2}{T}\cdotp\int_{t_1}^{t_2}{\textrm{f}(t)\cdotp\cos\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)\,dt}\\
b_n&=\frac{2}{T}\cdotp\int_{t_1}^{t_2}{\textrm{f}(t)\cdotp\sin\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)\,dt}
\end{align}
$$
For full-wave function $t_1=-{T}/{2}$ and $t_2=+{T}/{2}$ for any positive period $T$.
Let find Taylor series of cosine and sine functions:
$$
\begin{align}
\cos\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)&=\sum_{k=0}^{\infty}{\frac{(-1)^k}{\left(2\cdotp k \right)!}
\cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{2\cdotp k}}\\
\sin\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)&=\sum_{k=0}^{\infty}{\frac{(-1)^k}{\left(2\cdotp k + 1\right)!}
\cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{\left(2\cdotp k +1\right)}}
\end{align}
$$
and substitute this expansion to Fourier coefficients:
$$
\begin{align}
a_n&=\frac{2}{T}\cdotp
\int_{t_1}^{t_2}
{
\underbrace
{
\left(
\sum_{j=0}^{\infty}
{
h_j\cdotp\frac{d^{j}}{dt^{j}}\textrm{f}(0)\cdotp t^{j}
}
\right)\cdotp
\left(\sum_{k=0}^{\infty}
{
\frac{(-1)^k}{\left(2\cdotp k \right)!}
\cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{2\cdotp k}
}\right)
}_{\mbox{$\textrm{Tc}(t)$}}
\,dt
}\\
b_n&=\frac{2}{T}\cdotp
\int_{t_1}^{t_2}
{
\underbrace
{
\left(
\sum_{j=0}^{\infty}
{
h_j\cdotp\frac{d^{j}}{dt^{j}}\textrm{f}(0)\cdotp t^{j}
}
\right)
\cdotp
\left(\sum_{k=0}^{\infty}
{\frac{(-1)^k}{\left(2\cdotp k + 1\right)!}
\cdotp \left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)^{\left(2\cdotp k +1\right)}
}
\right)
}_{\mbox{$\textrm{Ts}(t)$}}
\,dt
}
\end{align}
$$
Now consider $\textrm{Tc}(t)$:
For some first indicies $j$ and $k$ brackets can be disclosured and terms multiplied in sequence as shown:
$$
\begin{align*}
\textrm{Tc}(t)&=&\textrm{f}(0)+\frac{d}{dt}\textrm{f}(0)\cdotp t + \left(-2\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{2}\cdotp\frac{d^2}{dt^2}\textrm{f}(0)\right)\cdotp t^{2}+\\&&
+\left(-2\cdotp\frac{d}{dt}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{6}\cdotp\frac{d^3}{dt^3}\textrm{f}(0)\right)\cdotp t^{3}+\\&&
+\left(\frac{2}{3}\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{4}-\frac{d^2}{dt^2}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{24}\cdotp\frac{d^4}{dt^4}\textrm{f}(0)\right)\cdotp t^{4}+\dots
\end{align*}
$$
Now integrate this function for each term:
$$
\begin{align*}
\int_{t_1}^{t_2}{\textrm{Tc}(t)\,dt}&=&\textrm{f}(0)\left(t_2-t_1\right)+\frac{1}{2}\cdotp\frac{d}{dt}\textrm{f}(0)\cdotp\left(t_2^2-t_1^2\right) + \\&& + \frac{1}{3}\cdot\left(-2\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{2}\cdotp\frac{d^2}{dt^2}\textrm{f}(0)\right)\cdotp\left(t_2^3-t_1^3\right) +\\&&
+\frac{1}{4}\cdotp\left(-2\cdotp\frac{d}{dt}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{6}\cdotp\frac{d^3}{dt^3}\textrm{f}(0)\right)\cdotp\left(t_2^4-t_1^4\right)+\\&&
+\frac{1}{5}\left(\frac{2}{3}\cdotp\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{4}-\frac{d^2}{dt^2}\textrm{f}(0)\cdotp\left(\frac{\pi\cdotp n}{T}\right)^{2}+\frac{1}{24}\cdotp\frac{d^4}{dt^4}\textrm{f}(0)\right)\cdotp\left(t_2^5-t_1^5\right)+\dots
\end{align*}
$$
Collect coefficients at previous integral above $\frac{d^i}{dt^i}\textrm{f}(0)$:
$$
\begin{align*}
\int_{t_1}^{t_2}{\textrm{Tc}(t)\,dt}&=&\textrm{f}(0)\cdotp\left( \left(t_2-t_1\right)-
\frac{2}{3}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^3-t_1^3\right)+\frac{2}{15}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^5\cdotp\left(t_2^5-t_1^5\right)+\dots\right) +
\\&&
+\frac{d}{dt}\textrm{f}(0)\cdotp\left(\frac{1}{2}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^2-t_1^2\right)-
\frac{1}{2}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^4-t_1^4\right)+\frac{1}{9}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^4\cdotp\left(t_2^6-t_1^6\right)+\dots\right)+
\\&&
+\frac{d^2}{dt^2}\textrm{f}(0)\cdotp\left(\frac{1}{6}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^3-t_1^3\right)-
\frac{1}{5}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^5-t_1^5\right)+\frac{1}{21}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^4\cdotp\left(t_2^7-t_1^7\right)+\dots\right)+
\\&&
+\frac{d^3}{dt^3}\textrm{f}(0)\cdotp\left(\frac{1}{24}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^4-t_1^4\right)-
\frac{1}{18}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^6-t_1^6\right)+\frac{1}{72}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^4\cdotp\left(t_2^8-t_1^8\right)+\dots\right)+\dots
\end{align*}
$$
Now it is easy to recognize sequences at brackets (rhs expression is multiplied by $2/T$).
For $\textrm{f}(0)$:
$$
\begin{align*}
\left(t_2-t_1\right)-
\frac{2}{3}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^3-t_1^3\right)+\dots &:& \frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ 1\cdot\left(2\cdot i+1 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+1 \right) -t_1^ \left(2\cdot i+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
For $\frac{d}{dt}\textrm{f}(0)$:
$$
\begin{align*}
\frac{1}{2}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^2-t_1^2\right)-\frac{1}{2}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^4-t_1^4\right)+\dots &:&\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ 1\cdot \left(2\cdot i+2 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+2 \right) -t_1^ \left(2\cdot i+2 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
For $\frac{d^2}{dt^2}\textrm{f}(0)$:
$$
\begin{align*}
\frac{1}{6}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^3-t_1^3\right)-\frac{1}{5}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^5-t_1^5\right)+\dots &:&\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ 1\cdot2\cdot \left(2\cdot i+3 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+3 \right) -t_1^ \left(2\cdot i+3 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
For $\frac{d^3}{dt^3}\textrm{f}(0)$:
$$
\begin{align*}
\frac{1}{24}\cdotp\frac{\pi\cdotp n}{T}\cdotp \left(t_2^4-t_1^4\right)-
\frac{1}{18}\cdotp\left(\frac{\pi\cdotp n}{T}\right)^2\cdotp\left(t_2^6-t_1^6\right)+\dots&:&\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ 1\cdot 2\cdot 3\cdot\left(2\cdot i+4 \right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+4 \right) -t_1^ \left(2\cdot i+4 \right)\right) }{T^ \left(2\cdot i+1 \right)}
\end{align*}
$$
and so on.
Finally overall sequence for $\frac{d^m}{dt^m}$ is computed as:
$$
\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ m!\cdot\left(1+m+2\cdot i\right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+m+1 \right) -t_1^ \left(2\cdot i+m+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}
$$
Now we can find sum using CAS:
$$
\textrm{Ct}(n,m)=\sum_{i=0}^{\infty}{\left(\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ m!\cdot\left(1+m+2\cdot i\right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+m+1 \right) -t_1^ \left(2\cdot i+m+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}\right)}
$$
and $\textrm{Ct}(n,m)$ becomes quite complex expression containing Lommel's or hypergeometric functions.
In particular, when $m=0$ function becomes
$$
\textrm{Ct}(n,0)=\frac{\sin\left(\frac{2\pi n\cdot t_2}{T})\right)-\sin\left(\frac{2\pi n \cdot t_1}{T}\right)}{\pi n}
$$
for $m=1$:
$$
\textrm{Ct}(n,1)= \frac
{2\pi n
\left(
\sin\left(\frac{2\pi n\cdot t_2}{T})\right)\cdot t_2-\sin\left(\frac{2\pi n\cdot t_1}{T}\right)\cdot t_1
\right)
+T\cdot
\left(
\cos\left(\frac{2\pi n\cdot t_2}{T})\right)-\cos\left(\frac{2\pi n\cdot t_1}{T}\right)
\right)
}
{ 2\cdot(\pi n)^2 }
$$
and so on.
So we can write expression for $a_n$:
$$
a_n=\sum_{m=0}^{\infty}{\frac{d^m}{dt^m}\textrm{f}(0)\cdot\textrm{Ct}(n,m)}
$$
or
$$
a_n=\sum_{m=0}^{\infty}{\frac{1}{m!}\cdot\frac{d^m}{dt^m}\textrm{f}(0)\cdot\left(\sum_{i=0}^{\infty}{\left(\frac{\left(-1 \right)^i\cdot 2^ \left(2\cdot i+1 \right)\cdot n^ \left(2\cdot i \right)}{ \left(1+m+2\cdot i\right)\cdot \left(2\cdot i \right)! }\cdot \frac {\pi^ \left(2\cdot i \right)\cdot \left(t_2^ \left(2\cdot i+m+1 \right) -t_1^ \left(2\cdot i+m+1 \right)\right) }{T^ \left(2\cdot i+1 \right)}\right)}\right)}
$$
We can easy see that $\frac{1}{m!}$ is Taylor's coefficient, thus relationship between the Fourier coefficients and the coefficient in the Taylor expansion using special functions is established.
In the particular case of full wave function $t_1=-{T}/{2}$ and $t_2=+{T}/{2}$ we can write for $\textrm{Ct}(n,m)$ more simple closed form:
$$
\frac{
\left(-1 \right)^n T^m
\left(
\left(\pi n\right)^ \left(-m-\frac{1}{2} \right)\left(\textrm{I}\cdot\textrm{L}_{\textrm{S1}}\left(m+\frac{3}{2}, \frac{1}{2}, -\pi n \right)-\textrm{L}_{\textrm{S1}}\left(m+\frac{3}{2}, \frac{1}{2}, \pi n \right)\right)+1+\left(-1 \right)^m \right)
}
{2^ {m}\cdot\left(m+1 \right)!}
$$
where $\textrm{L}_{\textrm{S1}}(\mu,\nu,z)=\textrm{s}_{\mu,\nu}(z)$ is first Lommel function and $\textrm{I}=(-1)^{2^{-1}}$ (Complex $a_n$ coefficient).
For example let consider parabolic signal with period $T$: $\textrm{g}(t)=A\cdot t^2+B\cdot t + C$.
Coefficients $a_n$ can be found using Fourier formula:
$$
a_n=\frac{2}{T}\cdotp\int_{-T/2}^{T/2}{\left(A\cdot t^2+B\cdot t + C\right)\cdotp\cos\left({\frac{2\pi\cdotp n \cdotp t}{T}}\right)\,dt} = A\cdot\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n
$$
For $\textrm{g}(t)$ function: $\textrm{g}'(t)=2A\cdot t + B$ , $\textrm{g}''(t)=2A$, derivatives greater than two is zero. So we can use $\textrm{Ct}(n,m)$ for $m=0,1,2$. It is easy to check if $t_1=-{T}/{2}$ and $t_2=+{T}/{2}$ then $\textrm{Ct}(n,0) = 0,\textrm{Ct}(n,1)=0$ we have zero values for odd and non zero for even values, in particular, whenfor $m=2$:
$$
\textrm{Ct}(n,2) = \frac{1}{2}\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n
$$
for $m=4$ as example $\textrm{Ct}(n,4) = \frac{1}{48}\left(\frac{T}{\pi n}\right)^4\cdot\left((\pi n)^2-6\right)\cdot(-1)^n
$.
Finally we can obtain for example for $m$ up to 4:
$$
\begin{align*}
a_n=\sum_{m=0}^{4}{\frac{d^m}{dt^m}\textrm{g}(0)\cdot\textrm{Ct}(n,m)}=&\\
=\left(A\cdot 0^2+B\cdot 0 + C\right)\cdot 0+\left(2A\cdot 0+B\right)\cdot 0+\frac{1}{2}\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n \cdot 2A+0\cdot0+...\cdot 0=\\
=A\cdot\left(\frac{T}{\pi n}\right)^2\cdot(-1)^n
\end{align*}
$$
the same result using Fourier integral. There is interesting result for non-integer harmonic for $\textrm{g}(t)$ is:
$$
a_n=\frac{1}{120}A\cdot T^2 \cdot (-1)^{n+1} \left({}_{2}\textrm{
F}_1\left( 1, 3; \frac{7}{2}; -\left(\frac{\pi n}{2}\right)^2\right)\cdot(\pi n)^2-20\right)
$$
where ${}_{2}\textrm{
F}_1\left(a,b;c;z\right)$ is hypergeometric function.
So we can plot coefficients calculated from Fourier integral and for this special result (for $T=1,\,A=1$):
Cosine series. Red circles and red line is Fourier cosine coefficients (for real part of non-integer power of -1), solid blue line is real part of obtained expression with hypergeometric function and dash green line is imaginary. For integer $n$ imaginary part is zero and $a_n$ is real.
Similar expressions can be obtained for the sine $b_n$ series.
| 184,940
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City description
A city of extremes, this urban metropolis of 7 million inhabitants, one and a half times bigger than New York, sits on a plateau at the foot of the eastern chain of the Andes. It is a gateway to the Colombian soul. Let the bright colors of the La Candelaria district surprise you and be intoxicated by its stylish, colonial buildings, small, Iberian-inspired homes and cobbled streets. A bohemian neighborhood where the streets echo with children's laughter and salsa rhythms and where Baroque churches stand alongside the Botero Museum and houses with wooden balconies. Even more amazing will be your visit to the largest square in Bogotá: plaza de bolivar, with its incredible dimensions, rows of windows and archways. Why not plan a sumptuous picnic using local products such as fritanga (meat and fried vegetables) and almojabana (cheese bread) and take a trip out to laguna de Guatavita, the birthplace of the legend of Eldorado?
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Ashlee Holmes Gets A Dramatic Hair Transformation
I’ve been dying to get my hair done for a while now but I’m super picky when it comes to trusting someone with my hair. I heard about Johnny Ramirez through word of mouth and I decided to give him a try. Some of Johnny’s most famous clients include Miranda Kerr and Kirsten Dunst. I figured if he’s good enough for a Victoria’s Secret model or two –he’s good enough for me haha I decided to check out his spot conveniently located in Beverly Hills.
Watching Johnny work is like watching a true artist. He really knows what he’s doing and he cares about your hair as much as you do. He puts a lot if time and energy into each client’s hair. I really respect him for taking so much time to make sure you’re just as happy with the results as he is. He actually listens to what you want.
I trusted Johnny with my hair not just because of his reputation –but because Johnny also happens to have amazing fashion sense. He even designed the interior of his salon –which has a very vintage yet modern vibe… I can’t fully explain it.. but I LOVE it!
Check out this gallery of the INSANE transformation of my hair –>
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\begin{document}
\centerline{\LARGE\bf On the~peripheral spectrum of positive elements}
\medskip
\begin{center}\large \fbox{\it\kern1pt Egor A. Alekhno}\\[9pt]
Belarusian State University, Minsk, Belarus
\end{center}
\bigskip
{\small \noindent\textbf{Abstract.} Let $A$ be an~ordered Banach algebra with a~unit ${\bf e}$ and
a~cone $A^+$. An~element $p$ of $A$ is said to be an~order idempotent if $p^2=p$ and $0\le p\le{\bf
e}$. An~element $a\in A^+$ is said to be irreducible if the~relation $({\bf e}-p)ap=0$, where
$p$~is an~order idempotent, implies $p=0$ or $p={\bf e}$. For an~arbitrary element $a$ of~$A$
the~peripheral spectrum $\sigma_{\rm per}(a)$ of $a$ is the~set $\sigma_{\rm
per}(a)=\{\lambda\in{\Bbb C}:|\lambda|=r(a)\}$, where $r(a)$ is the~spectral radius of $a$. We
investigate properties of the~peripheral spectrum of an~irreducible element $a$. The~conditions
under which $\sigma_{\rm per}(a)$ contains of or coincides with $r(a)H_m$, where $H_m$ is the~group
of all $m^{\rm th}$ roots of unity, and the~spectrum $\sigma(a)$ is invariant under the~rotation on
angle $\frac{2\pi}m$ for some $m\in{\Bbb N}$, are given. The~correlation between these results and
the~existence of a~cyclic form of $a$ is considered. The~conditions under which $a$ is primitive,
i.e., $\sigma_{\rm per}(a)=\{r(a)\}$, are studied. The~necessary assumptions on the~algebra $A$
which imply the~validity of these results, are discussed. In~particular, the~Lotz-Schaefer axiom is
introduced and finite-rank elements of $A$ are defined. Other approaches to the~notions of
irreducibility and primitivity are discussed. The~conditions under which the~inequalities $0\le
b<a$ imply $r(b)<r(a)$, are studied. The closedness of the~center $A_{\bf e}$, i.e., of~the~order
ideal generated by ${\bf e}$ in $A$, is proved.}
\bigskip
{\small \noindent\textbf{Mathematical Subject Classification.} 47A10, 46B40, 46H30, 46H10}
\smallskip
{\small \noindent\textbf{Keywords.} Ordered Banach algebra, Irreducible element, Peripheral
spectrum, Primitivity, Lotz-Schae\-fer axiom, Finite-rank element, Center.}
\medskip
\section{Introduction and preliminaries}
Let $A$ be a~(complex) Banach algebra with an~algebraic unit ${\bf e}$ and $A^+$ a~(closed, convex)
cone in $A$. As usual, for elements $a, b \in A$ the~symbol $a\ge b$ (or $b\le a$) means $a - b \in
A^+$. Under this ordering, $A$ is an~ordered linear space. From the~definition of the~cone, it
follows that the~inequalities $a\ge b$ and $b\ge a$ imply $a = b$ for all $a, b \in A$ and $\alpha
x + \beta y\ge 0$ for all elements $x, y\in A^+$ and all scalars $\alpha, \beta\in{\Bbb R}^+$.
The~elements of $A^+$ are called {\it positive}. If ${\bf e}\ge 0$ and the~inequalities $a, b \ge
0$ imply $ab\ge 0$ then $A$ is called an~{\it ordered Banach algebra}. An~important example of
an~ordered Banach algebra is the~algebra of (linear, bounded) operators on an~ordered Banach space.
Namely, if $E$ is an~ordered Banach space with (closed) cone $E^+$ then the~algebra $B(E)$ of all
operators on $E$ is an~ordered Banach algebra under the~natural order if and only if the~linear
space $E^+ - E^+$ is dense in $E$. In~particular, if $E$ is a~(complex) Banach lattice then
the~algebra $B(E)$ is an~ordered Banach algebra.
The~study of ordered Banach algebras was initiated in \mbox{\cite{MR, RR}}. In these papers and in
a~number of subsequent ones the~main emphasis was on the~study of spectral properties of positive
elements. Nevertheless, in spite of the~considerable progress in this direction, several aspects of
the~theory have received almost no attention. In~particular, the~properties of the~peripheral
spectrum of positive elements in ordered Banach algebras were not enough studied. The~main purpose
of this note is to take a~step in this direction.
\smallskip
We recall some well-known notions which will be necessary later on. Let $B$ be a~Banach algebra
with a~unit. The~{\it spectrum} \mbox{\cite[p.~19]{BD}} of an~element $b$ of $B$ is the~set
$$\sigma(b;B)=\{\lambda\in{\Bbb C} : \lambda-b \ \text{is an~invertible element of} \ B\}.$$
When no confusion can occur, we write $\sigma(b)$ to denote $\sigma(b;B)$. The~{\it resolvent
set}~$\rho(b;B)$ \mbox{\cite[p.~245]{AbrAl}} of $b$ is the~complement of the~spectrum, i.e.,
$\rho(b;B)={\Bbb C}\setminus\sigma(b;B)$. As~is well known, the~spectrum $\sigma(b;B)$ is
a~non-empty compact set (we assume $B\neq\{0\}$). The~{\it spectral radius} $r(b)$
\mbox{\cite[p.~23]{BD}} of an~element $b$ is defined via the~formula $r(b)=\max{\{|\lambda| :
\lambda\in\sigma(b;B)\}}$. By the~Gelfand formula \mbox{\cite[pp. 11,~23]{BD}}, the~equality
$r(b)=\lim\limits_{n\to\infty}\|b^n\|_B^{\frac1n}$ holds. The~{\it peripheral spectrum}
$\sigma_{\rm per}(b;B)$ (see, e.g., \mbox{\cite[Section~4]{MR}}) of an~element $b$ is the~set
$$\sigma_{\rm per}(b;B)=\{\lambda\in\sigma(b;B) : |\lambda|=r(b)\}.$$
Obviously, $\sigma_{\rm per}(b;B)$ is also a~non-empty compact set. Next, the~{\it resolvent
function} \mbox{\cite[p.~245]{AbrAl}} $R(\cdot,b):\rho(b;B)\to B$ of $b$ is defined by
$R(\lambda,b)=(\lambda-b)^{-1}$ and is an~analytic function on the~open set $\rho(b;B)$.
In~particular, if $\lambda_0$ is an~isolated point of $\sigma(b;B)$ then in a~sufficiently small
punctured neighbourhood of this point a~Laurent series expansion
\begin{equation}\label{1}
R(\lambda,b)=\sum\limits_{j=-\infty}^{+\infty}b_{\lambda_0,j}(\lambda-\lambda_0)^j
\end{equation}
holds, where $b_{\lambda_0,j}\in B$ for all $j\in{\Bbb Z}$. Of course, this expansion also holds
when the~point $\lambda_0\notin\sigma(b;B)$; in this case, $b_{\lambda_0,j}=0$ for $j\le-1$.
If~$\lambda_0=r(a)$, we will simply write $b_j$ instead of $b_{\lambda_0,j}$. A~point $\lambda_0$
is said to be a~{\it pole} \mbox{\cite[p.~264]{AbrAl}} of the~resolvent $R(\cdot,b)$ of
order~$k\in{\Bbb N}$ whenever $\lambda_0$ is an~isolated point of $\sigma(b;B)$,
$b_{\lambda_0,j}=0$ for $j<-k$, and $b_{\lambda_0,-k}\neq0$; in this case, the~identities
$bb_{\lambda_0,-k}=b_{\lambda_0,-k}b=\lambda_0b_{\lambda_0,-k}$ hold. If $k=1$ then $\lambda_0$ is
called a~{\it simple pole}.
\smallskip
The~considerable progress in the~study of the~peripheral spectrum was attained in the~case of
ordered spaces. The~results concerning to the~spectral theory of non-negative matrices, i.e., to
the~spectral theory in finite dimensional (Archimedean) Riesz spaces, and, in~particular, to
properties of the~peripheral spectrum of non-negative matrices can be found in \mbox{\cite[Chapter
1]{Sch}} and \mbox{\cite[Chapter 8]{AbrAl}}. Properties of the~peripheral spectrum of positive
operators on Banach lattices are considered, e.g., in \cite[Chapter~V and, in~particular, Sections
V.4 and~V.5]{Sch} (see also \mbox{\cite{Al,AbrAl,Gl}}). Some results concerning to the~peripheral
spectrum of positive elements in ordered Banach algebras can be found in \mbox{\cite[Section
4]{MR}}. Below, as far as necessary, we recall some results in these directions. Now we mention
only the~following result~\cite{RR}. If $A$ is an~ordered Banach algebra such that the~spectral
radius function in $A$ is monotone, i.e., the~inequalities $0\le a\le b$ in~$A$ imply $r(a)\le
r(b)$, then $r(a)\in\sigma_{\rm per}(a)$; in particular, this is true when the~cone $A^+$ is normal
and, hence, when the~ordered linear space $A$ is Dedekind complete.
\medskip
The~paper is organized as follows. In the~second section properties of the~peripheral spectrum of
irreducible elements are studied (see, in~particular, Theorem~\ref{thm12}) and some assumptions on
the~algebra $A$ which allow obtaining nice spectral properties, are considered. In~the~next
section, using a~special assumption (the~Lotz-Schaefer axiom), the~results of the~preceding section
will be made more precisely and, moreover, the~notion of finite-rank elements in ordered Banach
algebras and the~conditions of the~primitivity of irreducible elements are discussed. Another
approach to the~notion of irreducibility is considered in the~fourth section. In the~next section,
conditions under which the~inequa\-lities $0\le b<a$ imply $r(b)<r(a)$ are studied. In~the~last
section, the~closedness of the~center $A_{\bf e}$ of an~ordered Banach algebra $A$ is proved.
\medskip
For any unexplained terminology, notations, and elementary properties of ordered Banach spaces, we
refer the~reader to~\cite{AlT}. For information on the~theory of Riesz spaces, Banach lattices, and
operators on these spaces, we suggest~\mbox{\cite{AbrAl, AlB}} (see also \cite{Sch}). More details
on elementary properties of Banach algebras can be found in~\cite{BD} (see also~\cite{BMSmW}).
Throughout the~note, unless stated otherwise, $A$ will stand for an~arbitrary ordered Banach
algebra with a~unit ${\bf e}\neq0$.
\section{The~peripheral spectrum of irreducible elements}
An~important result in the~spectral theory of positive elements is the~{\it theorem about
the~Frobenius normal form}. For the~case of matrices it means that (see, e.g., \mbox{\cite[p.
31]{Sch}}) an~arbitrary non-negative matrix $A$ can be transformed into a~block lower-triangular
matrix
$$\left(\begin{array}{cccc} A_{11} & 0 & \ldots & 0 \\ A_{21} & A_{22} & \ldots & 0 \\ \ldots &
\ldots & \ldots & \ldots \\ A_{n1} & A_{n2} & \ldots & A_{nn}
\end{array}\right)$$
using a simultaneous permutation of rows and columns, where the~matrices $A_{11},\ldots,A_{nn}$ are
irreducible. We recall that a~matrix $D$ is called {\it irreducible} \mbox{\cite[p. 19]{Sch}}
whenever it can not be transformed to the~form $D=\left(\begin{smallmatrix} D_{11} & 0 \\ D_{21} &
D_{22}\end{smallmatrix}\right)$ using a~simultaneous permutation of rows and columns, where
$D_{ii}\neq D$ for $i=1,2$ (the~zero $1\times1$ matrix is irreducible). Obviously,
$\sigma(A)=\bigcup\limits_{i=1}^n\sigma(A_{ii})$. Therefore, we obtain the~dependence of
the~spectrum of an~arbitrary non-negative matrix $A$ on the~spectrum of irreducible matrices.
Now we consider the~case of an~ordered Banach algebra $A$. An~element $p \in A$ is called an~{\it
order idempotent} \cite{Al2} if $0\le p\le{\bf e}$ and $p^2 = p$. Under the~partial ordering
induced by~$A$, the~set of all order idempotents ${\bf OI}(A)$ of $A$ is a~Boolean algebra and its
lattice operations satisfy the~identities $p\wedge q = pq$ and $p\vee q = p + q - pq$ for all $p, q
\in {\bf OI}(A)$ (see~\cite{Al2}). For $p \in {\bf OI}(A)$ and $a\in A$, we put $p^{\rm d}={\bf
e}-p$ and $a_p=pap$. Obviously, $p^{\rm d}\in{\bf OI}(A)$. A~positive element $a\in A$ is said to
be {\it order continuous} \cite{Al2} if $p_\alpha a\downarrow0$ and $ap_\alpha\downarrow0$ in~$A$
whenever $p_\alpha\downarrow0$ in ${\bf OI}(A)$.\footnote{In~\cite{Al3} the~notion of order
continuity was extended to the~case of an~arbitrary element $a\in A$, not necessarily positive.
However, for our purposes, we can limit oneself to this special situation only.} The~collection of
all order continuous element of $A$ will be denoted by $A_{\rm n}$. The~algebra $A$ is called {\it
order regular} if $A_{\rm n}$ is a~subsemi-group of $A$, i.e., $a,b\in A_{\rm n}$ for every $a,b\in
A_{\rm n}$. Next, an~order idempotent $p\in A$ is called $a$-{\it invariant}~\cite{Al2}, where
$a\in A$, if $p^{\rm d}ap=0$. A~positive element $a\in A$ is said to be {\it
irreducible}~\cite{Al2} whenever $a$ has no non-trivial (i.e., $p\neq0$ and $p\neq{\bf e}$)
invariant order idempotents; all other elements of $A$ are called {\it reducible}. An~element $a\in
A^+$ is said to be {\it irreducible with respect to an~order idempotent} $p\in{\bf OI}(A)$ whenever
there exists no $q\in{\bf OI}(A)$ such that $0<q<p$ and $(p-q)aq=0$. An~element $b\in A$ is called
a~{\it block}~\cite{Al2} of an~element $a\in A^+$ if there exists \mbox{$a$-invariant} order
idempotents $p_1$~and~$p_2$ satisfying the~relations $p_2<p_1$ and $b=a_{p_2p_1^{\rm d}}$.
An~element $b$ is called a~{\it spectral block} of an~element $a\in A^+$~\cite{Al2} if $b$ is
a~block of $a$ and $r(b)=r(a)$. An~element $a\in A^+$ is said to {\it have the~Frobenius normal
form}~\cite{Al2} if there exists \mbox{$a$-invariant} order idempotents $p_0,p_1,\ldots,p_n$, which
determine this form, such that ${\bf e}=p_n\ge p_{n-1}\ge\ldots\ge p_0=0$ and if the~relation
$r(a_{p_ip_{i-1}^{\rm d}})=r(a)$ holds for some $i=1,\ldots,n$ then $a_{p_ip_{i-1}^{\rm d}}$ is
irreducible with respect to $p_ip_{i-1}^{\rm d}$. In this case, we have the~inclusion~\cite{Al2}
\begin{equation}\label{2}
\sigma(a)\subseteq\bigcup\limits_{i=1}^n\sigma(a_{p_ip_{i-1}^{\rm d}}).
\end{equation}
An~order continuous element $a\in A$ is said to be {\it spectrally order continuous}~\cite{Al2} if
for every spectral block $b$ of $a$ the~condition that $r(a)$ is a~pole of $R(\cdot,b)$ of order
$k$ implies the~order continuity of the~element $b_{-k}$ (see~(\ref{1})). The~spectral radius
$r(a)$ is called a~{\it finite-rank pole} (abbreviated an~$f$-{\it pole}) of the~resolvent
$R(\cdot,a)$ of a~positive element $a\in A$ if the~inequalities $0\le b\le a$ imply $r(b)\le r(a)$
and if $r(b)=r(a)$ then $r(a)$ is a~pole of $R(\cdot,b)$.
The~theorem about the~Frobenius normal form has the~following form~\cite{Al3}:
\begin{thm}\label{thm1}\
Let an~ordered Banach algebra $A$ be Dedekind complete and let an~ele\-ment~$a$ of $A$ be
spectrally order continuous. Assume that $r(a)>0$ and $r(a)$ is an~$f$-pole of $R(\cdot,a)$. Then
the~element $a$ has the~Frobenius normal form.
\end{thm}
\begin{exm}\label{exm2}\
{\rm {\bf (a)} Let us consider the~ordered Banach algebra $B(E)$, where $E$ is an~arbitrary Banach
lattice. If $T\in B(E)$ then, on the~one hand, the~notion of order continuity can be defined for
the~{\it operator} $T$ on $E$, i.e., \mbox{\cite[p. 46]{AlB}} if a~net
$x_\alpha\stackrel{o}{\longrightarrow}0$ then $Tx_\alpha\stackrel{o}{\longrightarrow}0$.
The~collection of all order continuous operators on $E$ is denoted by $B_{\rm n}(E)$. On the~other
hand, the~notion of order continuity can be defined for $T$ as of an~{\it element} in the~ordered
Banach algebra $B(E)$ which is defined only for a~positive operator. These two notions may differ.
Moreover~\cite{Al3}, the~inclusion $(B_{\rm n}(E))^+\subseteq(B(E))_{\rm n}$ and the~converse one
do not hold in general. Nevertheless, in~the~case of a~Dedekind complete Banach lattice $E$ these
two notions coincide. i.e., we have $(B_{\rm n}(E))^+=(B(E))_{\rm n}$, and, in~particular,
the~algebra $B(E)$ is order regular.
A~positive operator $T$ on a~Banach lattice $E$ is called ({\it band}) {\it
irreducible}~\mbox{\cite[p. 349]{AbrAl}} if $T$~has no non-trivial invariant bands. Obviously,
${\bf OI}(B(E))$ is the~collection of all order projections on $E$. If $E$ is Dedekind complete
then every band $B$ in $E$ is a~projection band and we have the~one-to-one correspondence between
the~set of all bands in $E$ and the~set of all order projections on $E$. Therefore, an~operator $T$
on a~Dedekind complete Banach lattice~$E$ is an~irreducible operator if and only if $T$ is
an~irreducible element of the~ordered Banach algebra $B(E)$. Next, if $E$ is a~Dedekind complete
Banach lattice, the~Lorenz seminorm $\|x\|_L=\inf{\{\sup{\|y_\alpha\|_E:0\le
y_\alpha\uparrow|x|}\}}$ is a~norm on $E$ (e.g., the~order continuous dual $E_{\rm n}^\sim$
separates points of $E$ or $E$ is an~$AM$-space with an~order unit), and $T$ is a~positive order
continuous operator on $E$ with $r(T)>0$ then \cite{Al2} $r(T)$~is an~$f$-pole of the~resolvent
$R(\cdot,T)$ if and only if $r(T)$ is a~finite-rank pole of $R(\cdot,T)$, i.e., $r(T)$ is a~pole of
$R(\cdot,T)$ and the~residue $T_{-1}$ of $R(\cdot,T)$ at $r(T)$ is a~finite-rank operator. Various
assumptions under which an~operator $T$ has the~Frobenius normal form in a~special case of
the~algebra $B(E)$ can be found in~\cite{Al2} .
{\bf (b)} If $E$ is an~ordered linear space and an~element $x\in E^+$ then the~{\it order ideal
$E_x$ generated by} $x$ is the~set \mbox{\cite[p. 103]{AlT}}
\begin{equation}\label{9}
E_x=\{y\in E:-\lambda x\le y\le\lambda x \ \text{for some} \ \lambda\in{\Bbb R}^+\}.
\end{equation}
Under the~algebraic operations and the~ordering induced be $E$, $E_x$ is a~real ordered linear
space satisfying $E_x\subseteq E^+-E^+$. If $A$ is an~ordered Banach algebra and $b\in A_{\rm n}$
then $A_b^+\subseteq A_{\rm n}$. The~order ideal $A_{\bf e}$ is called \cite{Al3} the~{\it center}
of $A$. As will be shown in Section~\ref{sec6}, $A_{\bf e}$ is closed in~$A$ and, hence, is a~{\it
real} ordered Banach algebra. Again, if ${\bf e}\in A_{\rm n}$ then $A_{\bf e}^+\subseteq A_{\rm
n}$. However, in general, the~inclusion ${\bf e}\in A_{\rm n}$ does not hold (see~\cite{Al3}).
Nevertheless, in every case, the~algebra $A_{\bf e}$ is order regular. Indeed, let $a\in(A_{\bf
e})_{\rm n}$ and $b\in A_{\bf e}^+$. If $p_\alpha\downarrow0$ in ${\bf OI}(A_{\bf e})$ then for
some $\lambda\ge0$, we have $0\le p_\alpha ab\le\lambda p_\alpha a\downarrow0$ in $A_{\bf e}$;
analogously, $abp_\alpha\downarrow0$. Thus, $ab,ba\in(A_{\bf e})_{\rm n}$.
The~author does not know an~example of an~ordered Banach algebra which is not order
regular.}\hfill$\Box$
\end{exm}
It follows from the~theorem about the~Frobenius normal form (see Theorem~\ref{thm1}) and
the~inclusion~(\ref{2}) that the~spectrum of a~wide class of positive elements of an~ordered Banach
algebra~$A$ is determined by the~spectra of irreducible elements. The~next result and
Corollary~\ref{cor5} make more precisely this result for the~case of the~peripheral spectrum.
\begin{lem}\label{lem3}\
Let $B$ be a~Banach algebra with a~unit ${\bf u}$. Let elements $b,p_0,p_1,\ldots,p_n\in B$ with
$n\in{\Bbb N}$ satisfy $({\bf u}-p_j)bp_j=0$, $p_0=0$, $p_n={\bf u}$, and
$p_ip_j=p_{\min{\{i,j\}}}$ for all $i,j=0,\ldots,n$. Then $r(b_{q_j})\le r(b)$ and
$$\sigma_{\rm per}(b)=\bigcup\{\sigma_{\rm per}(b_{q_j}):r(b_{q_j})=r(b)\},$$
where $q_j=p_j-p_{j-1}$ and $b_{q_j}=q_jbq_j$ for $j=1,\ldots,n$.
\end{lem}
{\bf Proof.} We mention first that for an~arbitrary scalar $\lambda$ belonging to the~unbounded
connected component $\rho_\infty(b)$ of the~resolvent set $\rho(b)$ the~identity $({\bf
u}-p_j)R(\lambda,b)p_j=0$ holds for all $j=0,\ldots,n$. Indeed, using an~elementary induction, we
get $({\bf u}-p_j)b^np_j=0$ for all $n\in{\Bbb N}$. Whence, in view of the~expansion
$R(\lambda,b)=\frac1\lambda+\frac1{\lambda^2}b+\ldots$ with $\lambda>r(b)$, we have the~identity
$({\bf u}-p_j)R(\lambda,b)p_j=0$. Taking the~unique extension of the~analytic function $({\bf
u}-p_j)R(\lambda,b)p_j$ to $\lambda\in\rho_\infty(b)$, we conclude that the~last identity holds for
all $\lambda\in\rho_\infty(b)$; in~particular,
\begin{equation}\label{3}
({\bf u}-p_j)R(\lambda,b)q_j=0.
\end{equation}
Now let us consider a~non-zero scalar $\lambda\in\rho_\infty(b)$. The~proof will be completed if we
will check the~invertibility of the~element $\lambda-b_{q_j}$ for every $j=1,\ldots,n$. To this
end, we define the~element $z$ via the~formula $z=\frac1\lambda({\bf
u}-p_j)+q_jR(\lambda,b)q_j+\frac1\lambda p_{j-1}$. Using (\ref{3}) and the~equality
$q_jbp_{j-1}=0$, we have
$$(\lambda-b_{q_j})z={\bf u}-p_j+(\lambda-b_{q_j})q_jR(\lambda,b)q_j+p_{j-1}=
{\bf u}-q_j+q_j(\lambda-b_{q_j})R(\lambda,b)q_j=$$
$$={\bf u}-q_j+q_j(\lambda-b+b({\bf u}-q_j))R(\lambda,b)q_j=
{\bf u}+q_jb({\bf u}-q_j)R(\lambda,b)q_j=$$
$$={\bf u}+q_jb({\bf u}-p_j+p_j-q_j)R(\lambda,b)q_j=
{\bf u}+(q_jb({\bf u}-p_j)+q_jbp_{j-1})R(\lambda,b)q_j=$$
$$={\bf u}+q_jb({\bf u}-p_j)R(\lambda,b)q_j={\bf u}.$$
Analogously, $z(\lambda-b_{q_j})={\bf u}$.\hfill$\Box$
\smallskip
As the~next example shows, the~inclusion~(\ref{2}) cannot make more precisely in general.
\begin{exm}\
{\rm If $a\in A$, $q$ is an~idempotent of $A$, and $q\neq {\bf e}$ then $0\in\sigma(a_q)$. Thus,
the~inclusion~(\ref{2}) is proper in general. We shall show that the~identity
$\sigma(a)\setminus\{0\}=(\sigma(a_{p^{\rm d}})\cup\sigma(a_p))\setminus\{0\}$ also does not hold,
where $p$ is an~\mbox{$a$-invariant} order idempotent. To see this, we consider the~space
$\ell_\infty$ of all bounded sequences and define the~operator $S$ on $\ell_\infty$ via the~formula
$$Sx=(x_1,0,0,0,x_3,0,0,0,x_5,\ldots)+(0,x_2,x_4,x_6,0,x_8,x_{10},x_{12},0,\ldots),$$
where $x=(x_1,x_2,\ldots)\in\ell_\infty$. As is easy to see, $S$ is invertible. Put $T=I+S$.
Obviously, $1\notin\sigma(T)$. The~band $B=\{x\in\ell_\infty:x_{2k}=0 \ {\rm for \ all} \ k\in{\Bbb
N}\}$ is \mbox{$T$-invariant} and
$$P_BTP_Bx=(x_1,0,x_3,0,x_5,0,x_7,0,x_9,\ldots)+(x_1,0,0,0,x_3,0,0,0,x_5,\ldots),$$
where $P_B$ is the~order projection on $B$. In~particular, $1\in\sigma(P_BTP_B)$.}\hfill$\Box$
\end{exm}
\begin{cor}\label{cor5}\
Let order idempotents $p_0,p_1,\ldots,p_n$ determine the~Frobenius normal form of an~element $a\in
A$. Then
$$\sigma_{\rm per}(a)=\bigcup\{\sigma_{\rm per}(a_{q_j}):r(a_{q_j})=r(a) \ \text{and} \ a_{q_j} \
\text{is irreducible with respect to} \ q_j\},$$ where $q_j=p_jp_{j-1}^{\rm d}$ for $j=1,\ldots,n$.
\end{cor}
Keeping the~preceding corollary in mind, we now turn to the~study of the~peripheral spectrum of
irreducible elements. Firstly, let us recall some spectral properties of irreducible elements. It
should be mentioned at once that there exists an~ordered Banach algebra $A$ such that every
positive element of $A$ is irreducible; the~latter is equivalent to the~identity ${\bf
OI}(A)=\{0,{\bf e}\}$. For example, the~ordered Banach algebra $A=A_0\otimes{\Bbb C}$ obtained from
an~ordered Banach algebra~$A_0$ by adjoining a~unit or the~algebras $C(K)$ of all continuous
functions on $K$ and $B(C(K))$, where $K$ is a~connected (Hausdorff) compact. Therefore, in
general, one cannot expect any distinctive spectral properties of irreducible elements and, hence,
a~special class of ordered Banach algebras should be distinguished. We shall say that an~ordered
Banach algebra $A$ has a~{\it disjunctive product}~\cite{Al2} if for any $a,b\in A_{\rm n}$ with
$ab=0$ there exists an~element $p\in{\bf OI}(A)$ satisfying $ap=p^{\rm d}b=0$. The~algebra $B(E)$,
where $E$ is a~Dedekind complete Banach lattice, has~\cite{Al2} a~disjunctive product. Next,
an~element $a\in A$ is said to be {\it algebraically strictly positive}, in symbols $a\gg0$,
whenever $p_1ap_2>0$ for all $0<p_1,p_2\in{\bf OI}(A)$. The~next result holds~\cite{Al2}.
\begin{thm}\label{thm6}\
Let $A$ be an~ordered algebra with a~disjunctive product and with the~Boo\-lean algebra ${\bf
OI}(A)$ Dedekind complete. Assume that a~non-zero element $a\in A$ is order continuous, that $r(a)$
is a~pole of $R(\cdot,a)$ of order $k$, and that $a_{-k}$ is also order continuous. If the~element
$a$ is irreducible then the~following statements hold:
\begin{description}
\item[(a)] The~spectral radius $r(a)>0$;
\item[(b)] The~point $r(a)$ is a~simple pole of $R(\cdot,a)$;
\item[(c)] The~residue $a_{-1}\gg0$ and the~resolvent $R(\lambda,a)\gg0$ for all
$\lambda>r(a)$;
\item[(d)] If $0\le b<a$ and if some element $c\in A_{\rm n}$ satisfies $0< r(b)c\le bc$ then $r(b)<r(a)$.
\end{description}
\end{thm}
The~following lemma and Corollary~\ref{cor8} show that, under additional assumptions on~$A$, the
condition about the~order continuity of the~coefficient $a_{-k}$ of the~Laurent series expansion of
$R(\,\cdot\,, a)$ around $r(a)$ in the~preceding theorem can be rejected.
\begin{lem}\
Let an~ordered Banach algebra $A$ be order regular and let $a\in A_{\rm n}$. Then the~resolvent
$R(\lambda,a)\in A_{\rm n}$ for all $\lambda>r(a)$.
\end{lem}
{\bf Proof.} For an~arbitrary number $n\in{\Bbb N}$, we define the~elements $b_n$ and $c_n$ as
follows
$$
b_n=\frac1\lambda{\bf e}+\ldots+\frac1{\lambda^n}a^{n-1} \ \ \text{and} \ \
c_n=\frac1{\lambda^{n+1}}a^n+\frac1{\lambda^{n+2}}a^{n+1}+\ldots.
$$
Obviously, $b_n+c_n=R(\lambda,a)$, $c_n\to0$ as $n\to\infty$, and, since $A$ is order regular,
$b_n\in A_{\rm n}$. Let $p_\alpha\downarrow0$ in ${\bf OI}(A)$ and let $p_\alpha R(\lambda,a)\ge x$
with $x\in A$. Fix an~index $\alpha_0$. For every $\alpha\ge\alpha_0$, the~inequality $x\le
p_\alpha b_n+p_{\alpha_0}c_n$ holds and, hence, $x-p_{\alpha_0}c_n\le p_\alpha
b_n\downarrow_{\alpha\ge\alpha_0}0$. Therefore, $x\le p_{\alpha_0}c_n\to0$ as $n\to\infty$.
Finally, $x\le0$ or $p_\alpha R(\lambda,a)\downarrow0$. Analogously, $R(\lambda,a)p_\alpha
\downarrow0$.\hfill$\Box$
\begin{cor}\label{cor8}\
Let an~ordered Banach algebra $A$ be order regular and let $A_{\rm n}$ be closed in~$A$. Let
an~element $a\in A_{\rm n}$ such that $r(a)$ is a~pole of the~resolvent $R(\cdot,a)$ of order~$k$.
Then $a_{-k}\in A_{\rm n}$.
\end{cor}
{\bf Proof.} The~identity $a_{-k}=\lim\limits_{\lambda\downarrow r(a)}(\lambda-r(a))^kR(\lambda,a)$
holds. It only remains to recall the~preceding lemma and the~closedness of $A_{\rm n}$.\hfill$\Box$
\smallskip
If $E$ is a~Dedekind complete Banach lattice such that the~Lorenz seminorm~\mbox{$\|\cdot\|_L$} is
a~norm on $E$ (see Example~\ref{exm2}{\bf (a)}) then \cite{KW} the~set $(B(E))_{\rm n}$ is closed
in $B(E)$. Consequently, the~wide class of ordered Banach algebras of the~form $B(E)$
auto\-ma\-ti\-cally satisfies the~assumption about the~closedness of $A_{\rm n}$.
Thus, it follows from Theorem~\ref{thm6} and the~preceding corollary that, under the~next
assumptions on $A$, order continuous irreducible elements have nice spectral properties:
\begin{description}
\item[{\bf (A$_1$)}] An~ordered Banach algebra $A$ is order regular and has a~disjunctive product,
the~Boolean algebra ${\bf OI}(A)$ is Dedekind complete, and the~set $A_{\rm n}$ is closed in $A$.
\end{description}
If $E$ is a~Dedekind complete Banach lattice and $P$ is a~non-zero order continuous projection on
$E$ such that $P\gg0$ as an~element of $B(E)$ then \cite{Al2} $\dim{R(P)}=1$, where $R(P)$ is
the~range of the~operator $P$. In~particular, in view of part~{\bf (c)} of Theorem~\ref{thm6}, ``as
a rule", the~residue $T_{-1}$ of the~resolvent $R(\cdot,T)$ of an~irreducible operator $T$ at
$r(T)$ is a~rank-one operator. On the~other hand, as is well known, if $Z$ is a~Banach space and
a~projection $Q\in B(Z)$ then $Q$ is a~rank-one operator if and only if $Q$ is a~minimal idempotent
of the~algebra $B(Z)$, i.e., $QB(Z)Q={\Bbb C}Q$. In the~case of an~arbitrary ordered Banach algebra
$A$ and of an~idempotent $b\in A$, the~condition $b\gg0$ does not imply the~minimality of
the~idempotent $b$. For this reason, we must axiomatize this property and make the~following
assumption:
\begin{description}
\item[{\bf (A$_2$)}] Every algebraically strictly positive idempotent $b$ of $A$ is minimal.
\end{description}
As usual, through $L_a$ and $R_a$, we will denote the~operators on an~algebra $B$ defined by
\begin{equation}\label{7}
L_ab=ab \ \ \text{and} \ \ R_ab=ba,
\end{equation}
where $a,b\in B$, and through $N(S)$, we will denote the~null space of the~operator $S$ acting
between two linear spaces.
\begin{prop}\label{prop9}\
An~ordered Banach algebra $A$ satisfies Axiom~{\bf (A$_2$)} if and only if for an~arbitrary
idempotent $b\gg0$ of the~following identity holds
\begin{equation}\label{4}
\dim{N(I-L_b)\cap N(I-R_b)}=1.
\end{equation}
\end{prop}
In general, the~next result is true: {\it If $B$ is a~Banach algebra and an~idempotent $b\in B$
then $b$ is minimal if and only if {\rm (\ref{4})} holds}.
\smallskip
{\bf Proof.} Let $A$ satisfy Axiom~{\bf (A$_2$)} and let $a\in A$ such that $a=ba=ab$. Then
$a=bab=\lambda b$ for some $\lambda\in{\Bbb C}$. For the~converse, if $b^2=b\gg0$ then for every
$a\in A$ the~element $bab$ belongs to $N(I-L_b)\cap N(I-R_b)$. On the~other hand, this space
contains $b$ and, hence, $bab=\mu b$ for some $\mu\in{\Bbb C}$.\hfill$\Box$
\smallskip
For the~study of the~peripheral spectrum of a~positive operator on a~Banach lattice,
the~possibility of the~restriction of a~problem to the~case of operators on the~space $C(K)$ of all
continuous functions on a~compact space $K$ which is simpler for the~study, is important. Recall
that if $E$ is a~Riesz space satisfying Axiom~{\bf (OS)}, i.e., \mbox{\cite[p. 54]{Sch}} if
the~inequalities $0\le z_n\le\lambda_n z$, where $z_n,z\in E$ and
$\lambda=(\lambda_1,\lambda_2,\ldots)\in\ell_1$, imply the~existence of
$\sup\limits_n\sum\limits_{j=1}^nz_j$, then the~order ideal $E_x$ generated by $x$ is, under
the~Minkowski norm $\|\cdot\|_x$ defined by
\begin{equation}\label{28}
\|y\|_x=\inf{\{\lambda\in{\Bbb R}^+:-\lambda x\le y\le\lambda x\}}
\end{equation}
with $y\in E_x$, an~$AM$-space with order unit $x$~\mbox{\cite[p. 102]{Sch}}. Therefore, by
the~Kakutani-Bohnenblust-M.-S.Krein theorem~\mbox{\cite[p. 201]{AlB}}, $E_x$ is lattice isometric
onto a~spa\-ce~$C(K)$ and, moreover, under this isomorphism, $x$ is mapped onto the~constant-one
function~${\rm 1}\!\!{\rm 1}_K$. Evidently, every Dedekind complete Riesz space $E$ satisfies
Axiom~{\bf (OS)}. Now let $E$~be a~Dedekind complete Banach lattice. Recall that an operator $T$ on
$E$ is said to be {\it regular}~\mbox{\cite[p. 12]{AlB}} if it can be written as a~difference of
two positive operators. As is well known, every regular operator is bounded. By
the~Riesz-Kantorovich theorem~\mbox{\cite[p. 14]{AlB}}, the~space $L_{\rm r}(E)$ of all regular
operators on $E$ is a~Dedekind complete Riesz space. Hence, if $T\in L_{\rm r}(E)$ then the~order
ideal $(B(E))_T$ is lattice isometric onto a~space $C(K)$. We axiomatize this property and make
the~next assumption:
\begin{description}
\item[{\bf (A$_3$)}] The~order
ideal $A_b$ generated by the~non-zero element $b\in A^+$ is lattice isomorphic onto a~space $C(K)$
and, under this isomorphism, $x$ is mapped onto~${\rm 1}\!\!{\rm 1}_K$.
\end{description}
Every complex Riesz space $E$ is the~complexification of the~real Riesz space $E_{\Bbb R}$
satisfying Axiom~{\bf (OS)} (see \mbox{\cite[Section~II.11]{Sch}}). In this case, the~algebra
$L(E)$ of all operators on $E$ is isomorphic onto the~complexification of the~algebra $L(E_{\Bbb
R})$ and, in~particular, every $T\in L(E)$ has a~unique decomposition $T=T_1+iT_2$, where $T_j$ are
real maps on $E_{\Bbb C}$, i.e., $T_j(E_{\Bbb R})\subseteq E_{\Bbb R}$ for $j=1,2$. Defining
the~set $A_{\rm r}$ of all {\it regular}~\cite{Al3} elements of an~ordered Banach algebra $A$ by
$A_{\rm r}=A^+-A^+$, we axiomatize this property and make the~next assumption:
\begin{description}
\item[{\bf (A$_4$)}] The~equality $a+ib=0$ with $a,b\in A_{\rm r}$ implies $a=b=0$.
\end{description}
We continue our discussion with two auxiliary results.
\begin{lem}\label{lem10}\
Let $B$ be a~Banach algebra with a~unit, let $b\in B$, and let $m,k\in{\Bbb N}$. If
$\lambda_0\in\sigma(b)$ and the~set
$\{\lambda_0,\lambda_0e^{i\frac{2\pi}m},\ldots,\lambda_0e^{i\frac{2\pi}m(m-1)}\}\cap\sigma(b)$
consists entirely of poles of $R(\cdot,b)$ of orders which are not greater than $k$, then
$\lambda_0^m$ is a~pole of $R(\cdot,b^m)$ of order which is not greater than~$km$.
\end{lem}
{\bf Proof.} Put $\omega_j=e^{i\frac{2\pi}mj}$ with $j=0,\ldots,m-1$. We claim first that
$\lambda_0^m$ is an~isolated point of $\sigma(b^m)$. Indeed, if a~sequence $\{\xi_n\}$ in
$\sigma(b^m)$ satisfies $\xi_n\neq\lambda_0^m$ for all $n$ and $\xi_n\to\lambda_0^m$ as
$n\to\infty$ then, taking into account the~identity $\sigma(b^m)=f(\sigma(b))$ with $f(z)=z^m$, we
find a~sequence~$\{\mu_n\}$ in $\sigma(b)$ with the~property $\mu_n^m=\xi_n$. Let $\{\mu_{n_r}\}$
be an~arbitrary convergent subsequence of~$\{\mu_n\}$. If $\mu_{n_r}\to\mu_0\in\sigma(b)$ then
$\mu_0^m=\lambda_0^m$ and, hence, $\mu_0=\lambda_0\omega_j$ for some $j=0,\ldots,m-1$. Therefore,
$\mu_0$ is an~isolated point of~$\sigma(b)$. Thus, $\mu_{n_r}=\mu_0$ for sufficiently large $r$ and
for such~$r$, we have $\xi_{n_r}=\lambda_0^m$, a~contradiction.
For arbitrary numbers $\lambda,z\in{\Bbb C}$ the~identity
$\lambda^m-z^m=\prod\limits_{j=0}^{m-1}(\lambda-\omega_jz)$ holds. The~latter implies
\begin{equation}\label{5}
\lambda^m-b^m=\prod\limits_{j=0}^{m-1}(\lambda-\omega_jb).
\end{equation}
If ${\cal U}_{\lambda_0^m}$ is a~punctured neighbourhood of the~point $\lambda_0^m$ satisfying
${\cal U}_{\lambda_0^m}\subseteq\rho(b^m)$ then the~set ${\cal V}=\{\lambda\in{\Bbb
C}:\lambda^m\in{\cal U}_{\lambda_0^m}\}$ is open and the~inclusion ${\cal
V}\subseteq\rho(\omega_jb)$ holds for all $j=0,\ldots,m-1$. Now, using~(\ref{5}), we obtain
\begin{equation}\label{6}
R(\lambda^m,b^m)=\prod\limits_{j=0}^{m-1}R(\lambda,\omega_jb)
\end{equation}
for all $\lambda\in{\cal V}$. Let us consider an~arbitrary sequence $\{\lambda_n\}$ in ${\Bbb C}$
satisfying $\lambda_n\neq\lambda_0^m$ for all $n$ and $\lambda_n\to\lambda_0^m$. It is not
difficult to show that there exists a~sequence $\{\theta_n\}$ in ${\Bbb C}$ with the~properties
$\theta_n\neq\lambda_0^m$, $\theta_n^m=\lambda_n$ for all $n$, and $\theta_n\to\lambda_0$ as
$n\to\infty$. Obviously, $\theta_n\in{\cal V}$ for all sufficiently large $n$. In view of
the~identity~(\ref{6}), we have
$$(\lambda_n-\lambda_0^m)^{km}R(\lambda_n,b^m)=
(\theta_n^m-\lambda_0^m)^{km}R(\theta_n^m,b^m)=
(\theta_n^m-\lambda_0^m)^{km}\prod\limits_{j=0}^{m-1}R(\theta_n,\omega_jb)=$$
$$=\prod\limits_{j=0}^{m-1}(\theta_n^m-\lambda_0^m)^kR(\theta_n,\omega_jb)=
(-1)^{m+1}\prod\limits_{j=0}^{m-1}(\theta_n^m-\lambda_0^m)^kR(\theta_n\omega_j^{-1},b)=$$
$$=(-1)^{m+1}\prod\limits_{j=0}^{m-1}(\theta_n^m-\lambda_0^m)^kR(\theta_n\omega_j,b)=$$
$$=(-1)^{m+1}
\prod\limits_{j=0}^{m-1}\Big(\sum\limits_{l=0}^{m-1}\theta_n^{m-1-l}\lambda_0^l\Big)^k\cdot
\prod\limits_{j=0}^{m-1}(\theta_n-\lambda_0)^kR(\theta_n\omega_j,b)=$$
$$=(-1)^{(m+1)(k+1)}
\prod\limits_{j=0}^{m-1}\Big(\sum\limits_{l=0}^{m-1}\theta_n^{m-1-l}\lambda_0^l\Big)^k
\cdot\prod\limits_{j=0}^{m-1}(\theta_n\omega_j-\lambda_0\omega_j)^kR(\theta_n\omega_j,b)\to 0$$ as
$n\to\infty$.\hfill$\Box$
\medskip
Let $m\in{\Bbb N}$. Consider a~nonempty subset $J$ of the~set $\{1,\ldots,m\}$. Let
$J=\{j_1,\ldots,j_r\}$ with $r=1,\ldots,m$. We define the~shift $J-1$ of $J$ via the~formula
$J-1=\{j_1-1,\ldots,j_r-1\}$, where in the~case of $j_k=1$ for some $k=1,\ldots,r$, we put
$j_k-1=m$. Now, using the~elementary induction, the~set $J-l$ can be defined easily for every
$l\in{\Bbb N}$. Obviously, $J-m=J$.
\begin{lem}\label{lem11}\
Let $m\in{\Bbb N}$ and let $J$ be a~nonempty subset of $\{1,\ldots,m\}$. If $l$ is a~minimal
natural number satisfying $J-l=J$ then $l$ is a~divisor of $m$.
\end{lem}
{\bf Proof.} Evidently, $J-kl=J$ for all $k\in{\Bbb N}$ and $l\le m$. The~representation $m=nl+r$
holds with $n,r\in{\Bbb N}$ and $0\le r<l$. Therefore, $J=J-(nl+r)=J-r$. Whence, using
the~minimality of $l$, we infer $r=0$.\hfill$\Box$
\smallskip
Let be a~Banach algebra with a~unit. As is well known, if an~element $a\in B$ then the~identities
$\sigma(L_a;B(A))=\sigma(R_a;B(A))=\sigma(a;A)$ hold (see~(\ref{7})). We shall say that a~point
$\lambda\in\sigma(a)$ is the~{\it joint eigenvalue} of $a$ whenever there exists a~non-zero element
$c\in B$ satisfying
\begin{equation}\label{8}
ac=ca=\lambda c.
\end{equation}
Now let $A$ be an~ordered Banach algebra, let $a\in A$, and let $b\in A^+$. The~{\it joint
spectrum} of an~element $a$ with respect of $b$ is the set $\sigma_{\rm j}(a;b;A)$ of all complex
numbers $\lambda$ such that there exists an~element $c\in A$ which is not nilpotent,
satisfies~(\ref{8}), and has the~representation in the~form $c=c'+ic''$, where $c'$ and $c''$
belong to the~order ideal $A_b$ (see~(\ref{9})). Again, when no confusion can occur, we write
$\sigma_{\rm j}(a;b)$ to denote $\sigma_{\rm j}(a;b;A)$.
\smallskip
Below, through $H_m$ with $m\in{\Bbb N}$, we will denote the~group of all $m^{\rm th}$ roots of
unity.
\smallskip
Now we are ready to state and to prove the~main result of this paper.
\begin{thm}\label{thm12}\
Let an~ordered Banach algebra $A$ satisfy Axioms {\bf (A$_1$)}-{\bf (A$_4$)}. Let $a$ be
a~non-zero, order continuous, and irreducible element of $A$ such that $r(a)$ is a~pole
of~$R(\cdot,a)$. Let $m\in{\Bbb N}$, $m>1$. The~following statements are equivalent:
\begin{description}
\item[(a)] $a^m$ is reducible;
\item[(b)] For some divisor $m'$ of $m$, $m'>1$, there exist non-zero order idempotents
$p_1,\ldots,p_{m'}$ of $A$ satisfying $\sum\limits_{j=1}^mp_j={\bf e}$, $p_{j'}p_{j''}=0$ for
$j'\neq j''$, and $p_ja=ap_{j+1}$ for all $j=1,\ldots,m'$, where for $j=m'$, we put $j+1=1$;
\item[(c)] For some divisor $m''$ of $m$, $m''>1$, and for all $j=0,\ldots,m''-1$ the~points
$r(a)e^{i\frac{2\pi}mj}$ are poles of $R(\cdot,a)$.
\end{description}
If the~condition~{\bf (b)} holds then the~spectrum $\sigma(a)$ is invariant under the~rotation on
angle~$\frac{2\pi}{m'}$, i.e., $\sigma(a)=\frac{2\pi}{m'}\sigma(a)$.
Moreover, $\sigma_{\rm per}(a)\cap\sigma_{\rm j}(a;a_{-1})=r(a)H_{m_0}$ for some $m_0\in{\Bbb N}$.
\end{thm}
Below, we shall say that an~arbitrary element $a\in A$ has the~{\it cyclic form} whenever there
exist order idempotents $p_1,\ldots,p_{m'}$ with $m'>1$ satisfying the~conditions of part~{\bf
(b)}. In this case, $p_1,\ldots,p_{m'}$ is said to determine the~cyclic form of $a$.
\medskip
{\bf Proof.} {\bf (a)}~$\Longrightarrow$~{\bf (b)} Taking into account the~reducibility of $a^m$,
we find a~non-trivial order idempotent $q_1$ of $A$ satisfying $q_1^{\rm d}a^mq_1=0$. Obviously,
$q_1^{\rm d}aa^{m-1}q_1=0$. Since the~algebra $A$ has a~disjunctive product (Axiom~{\bf (A$_1$)}),
there exists an~order idempotent~$q_2$ of $A$ such that $q_1^{\rm d}aq_2=q_2^{\rm d}a^{m-1}q_1=0$.
Using the~elementary induction, we find $q_1,\ldots,q_m\in{\bf OI}(A)$ satisfying $q_1^{\rm
d}aq_2=q_2^{\rm d}aq_3=\ldots=q_m^{\rm d}aq_1=0$. The~identity
\begin{equation}\label{10}
\prod\limits_{j=1}^mq_j=0
\end{equation}
holds. To see this, keeping the~relation
\begin{equation}\label{11}
aq_j=q_{j-1}aq_j
\end{equation}
for $j=1,\ldots,m$ (for $j=1$, we put $j-1=m$) in mind, we have
$$\Big({\bf e}-\prod\limits_{j=1}^mq_j\Big)a\prod\limits_{j=1}^mq_j=
\Big(q_m-\prod\limits_{j=1}^mq_j\Big)a\prod\limits_{j=1}^mq_j=$$
$$=\Big(q_mq_1-\prod\limits_{j=1}^mq_j\Big)a\prod\limits_{j=1}^mq_j=\ldots=
\Big(q_mq_1\ldots q_{m-1}-\prod\limits_{j=1}^mq_j\Big)a\prod\limits_{j=1}^mq_j=0.$$
The~irreducibility of $a$ and the~relation $q_1\neq{\bf e}$ imply~(\ref{10}).
For $r\in\{1,\ldots,m\}$, we put ${\cal P}_r=\{J\subseteq\{1,\ldots,m\}:{\rm card}\! \ J=r\}$. Now
we assume the~validity of the~identity
\begin{equation}\label{12}
\prod\limits_{k\in J}q_k=0
\end{equation}
for all $J\in{\cal P}_r$, where $r=2,\ldots,m$ and for $r=m$, we obtain~(\ref{10}). Let us show
that in this case either there exists the~required collection of idempotents or
the~identity~(\ref{12}) is valid for all $J\in{\cal P}_{r-1}$. To this end, let (\ref{12}) hold for
all $J\in{\cal P}_r$, $r\ge2$. Then for all $J_1,J_2\in{\cal P}_{r-1}$, $J_1\neq J_2$, we have
\begin{equation}\label{13}
\prod\limits_{j\in J_1}q_j\cdot\prod\limits_{k\in J_2}q_k=0.
\end{equation}
We shall say that two subsets $J_1,J_2\in{\cal P}_{r-1}$ is {\it equivalent} whenever $J_1-l=J_2$
for some $l\in{\Bbb N}$ (see the~remarks before Lemma~\ref{lem11}). As is easy to see, the~relation
introduced above is an~equivalence relation on ${\cal P}_{r-1}$ really. Thus, the~set ${\cal
P}_{r-1}$ is the~union of (disjoint) equivalence classes ${\cal I}_1,\ldots,{\cal I}_t$ with
$t\in{\Bbb N}$. Clearly, if $J\in{\cal I}_s$, where $s\in1,\ldots,t$, then $J-1\in{\cal I}_s$.
By~Lemma~\ref{lem11}, $m_s={\rm card}\! \ {\cal I}_s$ is a~divisor of $m$. If $J\in{\cal I}_s$ then
\begin{equation}\label{14}
{\cal I}_s=\{J,J-1,\ldots,J-(m_s-1)\}
\end{equation}
Fix $s$. In view of (\ref{11}), (\ref{12}), (\ref{13}), and~(\ref{14}), for an~arbitrary set
$J_0\in{\cal I}_s$, we have
$$\Big({\bf e}-\sum\limits_{J\in{\cal I}_s}\prod\limits_{j\in J}q_j\Big)
a\prod\limits_{j\in J_0}q_j=$$
$$=\Big({\bf e}-\sum\limits_{J\in{\cal I}_s}\prod\limits_{j\in J}q_j\Big)
\prod\limits_{k\in J_0-1}q_j\cdot a\prod\limits_{j\in J_0}q_j= \Big(\prod\limits_{k\in
J_0-1}q_j-\prod\limits_{k\in J_0-1}q_j\Big)a\prod\limits_{j\in J_0}q_j=0.$$ Since $J_0$ is
arbitrary, we obtain $\Big({\bf e}-\sum\limits_{J\in{\cal I}_s}\prod\limits_{j\in J}q_j\Big)a
\sum\limits_{J\in{\cal I}_s}\prod\limits_{j\in J_0}q_j=0$. Thus, for every $s\in1,\ldots,t$ either
$\sum\limits_{J\in{\cal I}_s}\prod\limits_{j\in J_0}q_j=0$ or $\sum\limits_{J\in{\cal
I}_s}\prod\limits_{j\in J_0}q_j={\bf e}$. Moreover, either the~former of these equalities hold for
all indexes $s$ or, in view of~(\ref{13}), the~second one holds for the~unique index~$s_0$ (if
$r=m$ then $s_0=t=1$ and $m_{s_0}=m$). We consider the~second case. Let ${\cal
I}_{s_0}=\{J_1,\ldots,J_{m_{s_0}}\}$ and let $J_{j-1}=J_j-1$ for all $j=1,\ldots,m_{s_0}$. Now we
define required order idempotents $p_1,\ldots,p_{m_{s_0}}$ via the~formula $p_j=\prod\limits_{r\in
J_j}q_r$ for $j=1,\ldots,m_{s_0}$. Obviously, $\sum\limits_{j=1}^{m_{s_0}}p_j={\bf e}$ and
$p_{j'}p_{j''}=0$ for $j'\neq j''$. Next, for an~arbitrary index $j=1,\ldots,m_{s_0}$, we have
$$ap_{j+1}=a\prod\limits_{r\in J_j}q_r=
\sum\limits_{J\in{\cal I}_{s_0}}\prod\limits_{r\in J}q_r\cdot a\cdot\prod\limits_{r\in J_{j+1}}q_r
=$$
$$=\Big(\sum\limits_{J\in{\cal I}_{s_0}}\prod\limits_{r\in J}q_r\Big)
\prod\limits_{r\in J_{j+1}-1}q_r\cdot a= \Big(\sum\limits_{J\in{\cal I}_{s_0}}\prod\limits_{r\in
J}q_r\Big)\prod\limits_{k\in J_j}q_k\cdot a= \prod\limits_{r\in J_j}q_r\cdot a=p_ja.$$ The~order
idempotents $p_j$ are non-zero. Indeed, if $p_j=0$ for a~index $j=1,\ldots,m_{s_0}$ then
$ap_{j+1}=0$ and, hence, $p_{j+1}^{\rm d}ap_{j+1}=0$. Thus, $p_{j+1}=0$. Using the~elementary
induction, we conclude $p_j=0$ for all $j$, a~contradiction (we have shown that if
$p_1,\ldots,p_{m'}$ satisfies the~conditions of part~{\bf (b)} then $p_j\neq0$ for all
$j=1,\ldots,m'$ and, hence, $p_j\neq{\bf e}$ for all~$j$). Now we assume that there exists no such
index $s_0$. The~latter implies the~equality $\prod\limits_{k\in J_j}q_k=0$ for all $J\in{\cal
I}_s$ and all $s$, i.e., the~validity of the~relation~(\ref{12}) for all $J\in{\cal P}_{r-1}$.
Now, using the~identity~(\ref{10}), i.e., (\ref{12})~for $r=m$, and the~construction above and
taking the~finite number of steps, either we will construct the~required collection of order
idempotents $p_1,\ldots,p_{m'}$ or we will reduce the~problem to the~case when
the~identity~(\ref{10}) holds with $r=1$, i.e., to the~case of $q_1,\ldots,q_m=0$. The~latter is
impossible as $q_1\neq0$.
{\bf (b)}~$\Longrightarrow$~{\bf (a)} As was shown above, if $p_1,\ldots,p_{m'}$ satisfy
the~condition of part~{\bf (b)} then $p_j$ is not-trivial for all $j=1,\ldots,m'$. Now, using
the~elementary induction, we have
$$p_ja^{m'}=ap_{j+1}a^{m'-1}=a^2p_{j+2}a^{m'-2}=\ldots=a^{m'}p_{j+m'}=a^{m'}p_j$$
for an~arbitrary index~$j$, whence
$$p_ja^m=p_ja^{m'\frac{m}{m'}}=a^{m'}p_ja^{(m'-1)\frac{m}{m'}}=\ldots=a^mp_j.$$
Finally, $p_ja^mp_j^{\rm d}=0$ and, in~particular, $a^m$ is reducible.
{\bf (b)}~$\Longrightarrow$~{\bf (c)} We define the~element $d$ via the~formula
$d=\sum\limits_{j=1}^{m'}e^{i\frac{2\pi}{m'}j}p_j$, where the~order idempotents $p_1,\ldots,p_{m'}$
satisfy the~condition of part~{\bf (b)}. As is easy to see, the~element~$d$ is invertible and
$d^{-1}=\sum\limits_{r=1}^{m'}e^{-i\frac{2\pi}{m'}j}p_r$. We have the~equalities
$$e^{i\frac{2\pi}{m'}}dad^{-1}=
e^{i\frac{2\pi}{m'}}\Big(\sum\limits_{j=1}^{m'}e^{i\frac{2\pi}{m'}j}p_j\Big)a
\Big(\sum\limits_{r=1}^{m'}e^{-i\frac{2\pi}{m'}j}p_r\Big)=$$
$$=e^{i\frac{2\pi}{m'}}\sum\limits_{j,r=1}^{m'}e^{i\frac{2\pi}{m'}(j-r)}p_jap_r=
e^{i\frac{2\pi}{m'}}\sum\limits_{j=1}^{m'}e^{i\frac{2\pi}{m'}(j-(j+1))}ap_{j+1}=a.$$ Therefore,
$\sigma(a)=e^{i\frac{2\pi}{m'}}\sigma(dad^{-1})=e^{i\frac{2\pi}{m'}}\sigma(a)$. Consequently,
part~{\bf (b)} implies the~inva\-riance of the~spectrum $\sigma(a)$ under the~rotation on angle
$\frac{2\pi}{m'}$. In~particular, the~points $r(a)e^{i\frac{2\pi}{m'}j}$ belong to $\sigma_{\rm
per}(a)$ for $j=0,\ldots,m'-1$ as $r(a)\in\sigma(a)$. Let us show that these points are simple
poles of $R(\cdot, a)$. Using the~identity $\lambda-a=e^{i\frac{2\pi}{m'}}d(\lambda
e^{-i\frac{2\pi}{m'}}-a)d^{-1}$ for all $\lambda\in{\Bbb C}$, we obtain
$R(\lambda,a)=e^{-i\frac{2\pi}{m'}}dR(\lambda e^{-i\frac{2\pi}{m'}},a)d^{-1}$ for all
$\lambda\in\rho(a)$. Let $\lambda_0$ be a~pole of $R(\cdot,a)$ of order $k$. Then
$\lambda_0e^{i\frac{2\pi}{m'}}$ is an~isolated point of $\sigma(a)$ and we have the~equalities
$$\lim\limits_{\lambda\to\lambda_0e^{i\frac{2\pi}{m'}}}
(\lambda-\lambda_0e^{i\frac{2\pi}{m'}})^kR(\lambda,a)=
\lim\limits_{\lambda\to\lambda_0e^{i\frac{2\pi}{m'}}} (\lambda-\lambda_0e^{i\frac{2\pi}{m'}})^k
e^{-i\frac{2\pi}{m'}}dR(\lambda e^{-i\frac{2\pi}{m'}},a)d^{-1}=$$
$$=\lim\limits_{\lambda\to\lambda_0e^{i\frac{2\pi}{m'}}}
(\lambda e^{-i\frac{2\pi}{m'}}-\lambda_0)^ke^{i\frac{2\pi}{m'}k} e^{-i\frac{2\pi}{m'}}dR(\lambda
e^{-i\frac{2\pi}{m'}},a)d^{-1}=$$
$$=\lim\limits_{\mu\to\lambda_0}(\mu-\lambda_0)^ke^{i\frac{2\pi}{m'}(k-1)}dR(\mu,a)d^{-1}=0.$$
Thus, the point $\lambda_0e^{i\frac{2\pi}{m'}}$ is a~pole of $R(\cdot,a)$ of order which is not
greater than $k$. On the~other hand, $r(a)$ is a~simple pole of $R(\cdot,a)$ and so the~points
$r(a)e^{i\frac{2\pi}{m'}j}$, $j=0,\ldots,m'-1$, are also simple poles of $R(\cdot,a)$.
{\bf (c)}~$\Longrightarrow$~{\bf (a)} For $j=0,1,\ldots,m''-1$, we put
$\omega_j=e^{i\frac{2\pi}{m''}j}$. Using the~identity \mbox{\cite[p. 22]{BD}}
$\frac1{1-z^{m''}}=\frac1{m''}\sum\limits_{j=0}^{m''-1}\frac1{1-\omega_j^{-1}z}$ which is valid for
all complex numbers $z\notin\{\omega_0,\ldots,\omega_{m''-1}\}$, we have
$\frac1{\lambda^{m''}-z^{m''}}=
\frac1{m''\lambda^{m''-1}}\sum\limits_{j=0}^{m''-1}\frac{\omega_j}{\lambda\omega_j-z}$ for all
$\lambda,z\in{\Bbb C}$ satisfying $\frac{z}\lambda\notin\{\omega_0,\ldots,\omega_{m''-1}\}$.
Therefore, for $\lambda$ from a~sufficiently small punctured neighbourhood of $r(a)$, we obtain
\begin{equation}\label{15}
m''\lambda^{m''-1}R(\lambda^{m''},a^{m''})= \sum\limits_{j=0}^{m''-1}\omega_jR(\lambda\omega_j,a).
\end{equation}
In view of our condition and Lemma~\ref{lem10}, $r(a)^{m''}$ is a pole of $R(\cdot,a^{m''})$.
Assume that the~element $a^m$ is irreducible. Then the~element $a^{m''}$ is also irreducible.
Consequently, in view of Axiom~{\bf (A$_1$)}, $r(a)>0$, the~point $r(a)^m$ is a~simple pole of
the~resolvent $R(\cdot,a^{m''})$, and the~residue $(a^{m''})_{-1}\gg0$. The~point $r(a)$ is also
a~simple pole of the~function $\lambda\to R(\lambda^{m''},a^{m''})$. Indeed, $r(a)$ is an~isolated
singular point of this function and for every natural $k\in{\Bbb N}$ the~relation
\begin{equation}\label{16}
(\lambda-r(a))^kR(\lambda^{m''},a^{m''})=
\frac{(\lambda^{m''}-r(a)^{m''})^kR(\lambda^{m''},a^{m''})}
{(\lambda^{m''-1}+\lambda^{m''-2}r(a)+\ldots+r(a)^{m''-1})^k}
\end{equation}
holds. Therefore, for every $k>1$, we have
\begin{equation}\label{17}
\lim\limits_{\lambda\to r(a)}(\lambda-r(a))^kR(\lambda^{m''},a^{m''})=0.
\end{equation}
Let $k_j$ be the~order of the~pole of $R(\cdot,a)$ at the~point $r(a)\omega_j$, where
$j=0,1,\ldots,m''-1$. We claim equality $k_j=1$ (this fact is not assumed in part~{\bf (c)};
moreover, we mention that the~cone $A^+$ is not assumed to be normal). Put $l=\max\limits_{0\le
j\le m''-1}k_j$. Proceeding by contradiction, we suppose $l>1$. Therefore, using the~identities
(\ref{15}) and~(\ref{17}), we obtain
$$0=\sum\limits_{j=0}^{m''-1}\omega_j\lim\limits_{\lambda\to r(a)}
(\lambda-r(a))^lR(\lambda\omega_j,a)=$$
$$=\sum\limits_{j=0}^{m''-1}\lim\limits_{\lambda\to r(a)}
(\lambda\omega_j-r(a)\omega_j)^l\frac1{\omega_j^{l-1}}R(\lambda\omega_j,a)=
\sum\limits_{j=0}^{m''-1}\frac1{\omega_j^{l-1}}a_{-l,r(a)\omega_j}$$ and so
\begin{equation}\label{18}
\sum\limits_{j=0}^{m''-1}\frac1{\omega_j^{l-1}}a_{-l,r(a)\omega_j}=0.
\end{equation}
Taking into account the~integral representation of the~coefficients of the~Laurent series expansion
of the~resolvent $R(\cdot,a)$ around the~point $\omega_j$ and using the~functional calculus, for
arbitrary~$j_0$ satisfying $k_{j_0}=l$, we get
$a_{-1,r(a)\omega_{j_0}}a_{-l,r(a)\omega_{j_0}}=a_{-l,r(a)\omega_{j_0}}$ and
$a_{-1,r(a)\omega_{j_0}}a_{-l,r(a)\omega_j}=0$ for $j\neq j_0$. Now, from the~(\ref{18}), it
follows that
$$0=a_{-1,r(a)\omega_{j_0}}\sum\limits_{j=0}^{m''-1}\frac1{\omega_j^{l-1}}a_{-l,r(a)\omega_j}=
\frac1{\omega_{j_0}^{l-1}}a_{-l,r(a)\omega_{j_0}}.$$ Therefore, $a_{-l,r(a)\omega_{j_0}}=0$,
a~contradiction. Thus, $l=1$. Next, according to~(\ref{16}),
$$\lim\limits_{\lambda\to r(a)}(\lambda-r(a))R(\lambda^{m''},a^{m''})=
\frac{(a^{m''})_{-1}}{m''r(a)^{m''-1}}.$$ Using the~last equality and the~identity~(\ref{16}) once
more, we obtain
$$(a^{m''})_{-1}=\lim\limits_{\lambda\to r(a)}(\lambda-r(a))
\sum\limits_{j=0}^{m''-1}\omega_jR(\lambda\omega_j,a)=
\sum\limits_{j=0}^{m''-1}a_{-1,r(a)\omega_j}$$ and, hence,
$$(a^{m''})_{-1}a_{-1,r(a)\omega_j}=a_{-1,r(a)\omega_j}(a^{m''})_{-1}=a_{-1,r(a)\omega_j}.$$
In other words, for all $j=0,1,\ldots,m''-1$ the idempotents $a_{-1,r(a)\omega_j}$ belong to
the~space $N(I-R_{(a^{m''})_{-1}})\cap N(I-L_{(a^{m''})_{-1}})$. In view of Axiom~{\bf (A$_2$)}
(see Proposition~\ref{prop9} and, in~particular, the~identity~(\ref{4})), this space is
one-dimensional. Therefore, taking into account the~condition $m''>1$, we get
$a_{-1,r(a)\omega_j}=0$ for all $j$. This contradiction establishes that $a^m$ is reducible.
Now we suppose that Axioms {\bf (A$_3$)} and~{\bf (A$_4$)} hold and show that for some $m_0\in{\Bbb
N}$, we have the~equality $\sigma_{\rm per}(a)\cap\sigma_{\rm j}(a;a_1)=r(a)H_{m_0}$. As is easy to
see, in~view of Axiom~{\bf (A$_4$)}, the~complexification $(A_{a_{-1}})_{\Bbb C}$ of the~order
ideal $A_{a_{-1}}$ generated by $a_{-1}$ is isomorphic onto the~complex linear subspace
$A_0=\{b+ic:b,c\in A_{a_{-1}}\}$ of $A$. On~the~other hand, in view of Axiom~{\bf (A$_3$)},
the~order ideal $A_{a_{-1}}$ is lattice isomorphic onto a~space $C(K)$ and, under this isomorphism,
$a_{-1}$~is mapped onto~${\rm 1}\!\!{\rm 1}_K$. In~particular, $A_{a_{-1}}$ is a~Riesz space and so
the~space $(C(K))_{\Bbb C}$ can be identified with~$A_0$, i.e., with the~complexification
$(A_{a_1})_{\Bbb C}$ of $A_{a_{-1}}$. If for a~number $\lambda\ge0$ and $b\in A_{a_{-1}}$,
the~inequalities $-\lambda a_{-1}\le b\le\lambda a_{-1}$ hold then, taking into account
the~identity $aa_{-1}=r(a)a_{-1}$, we have $-\lambda r(a)a_{-1}\le L_ab\le\lambda r(a)a_{-1}$.
The~latter implies the~\mbox{$L_a$-invariance} of $A_{a_{-1}}$ and the~relation
$\|L_ab\|_{a_{-1}}\le r(a)\|b\|_{a_{-1}}$. Whence $\|\widehat{L_a}\|_{B(A_{a_{-1}})}=r(a)$, where
the~positive operator $\widehat{L_a}$ is a~restriction of $L_a$ to $A_{a_{-1}}$. Thus,
$r(\widehat{L_a})=r(a)$. Since $a_{-1}$ is a~minimal idempotent (Axiom~{\bf (A$_2$)}) for arbitrary
$b\in A$, we find a~scalar $f(b)$ satisfying $a_{-1}ba_{-1}=f(b)a$. The~function $f$ on
$A_{a_{-1}}$ defined above is linear and positive and satisfies the~inequality
$|f(b)|\le\|b\|_{a_{-1}}$ for all $b\in A_{a_{-1}}$. In~particular, the~functional~$f$ is bounded.
Since $a_{-1}\gg0$, $f$ is strictly positive. For arbitrary $a,b\in A_{a_{-1}}$, we have
the~equalities
$$f(ab)a_{-1}=a_{-1}aba_{-1}=r(a)a_{-1}ba_{-1}=r(a)f(b)a_{-1}.$$
Whence for the~adjoint operator $\widehat{L_a}^*$, we get $(\widehat{L_a}^*f)(b)=f(ab)=r(a)f(b)$ or
$\widehat{L_a}^*f=r(a)f$.
Fix $\lambda_0\in\sigma_{\rm per}(a)\cap\sigma_{\rm j}(a;a_1)$. There exists an~element $x_0$ of
$A$ which is not nilpotent, satisfies the~relations $L_ax_0=R_ax_0=\lambda_0x_0$, and has
the~representation in the~form $x_0=x_0'+ix_0''$, where $x_0',x_0''\in A_{a_{-1}}$. In~particular,
$x_0\in(A_{a_{-1}})_{\Bbb C}$ and $\widehat{L_a}x_0=\lambda_0x_0$. For the~element $x_0$
the~modulus $|x_0|$ exists, belongs to $A_{a_{-1}}$, and is given by
$|x_0|=\sup\limits_{\varphi\in[0,2\pi]}{|(\cos{\varphi})x_0'+(\sin{\varphi})x_0'|}$. We can assume
$\||x_0|\|_{a_{-1}}=1$. For the~element $ax_0=ax_0'+iax_0''\in(A_{a_{-1}})_{\Bbb C}$ and for
the~modulus of this element in $(A_{a_{-1}})_{\Bbb C}$, we have
$$|ax_0|=\sup\limits_{\varphi\in[0,2\pi]}{|(\cos{\varphi})ax_0'+(\sin{\varphi})ax_0'|}\le
a\sup\limits_{\varphi\in[0,2\pi]}{|(\cos{\varphi})x_0'+(\sin{\varphi})x_0'|}=a|x_0|.$$ Whence
$r(a)|x_0|=|\lambda_0x_0|=|ax_0|\le a|x_0|$ or $0\le(a-r(a))|x_0|$. On the~other hand,
$0=a_{-1}(a-r(a))|x_0|$ and $a_{-1}\gg0$. Thus, $r(a)|x_0|=a|x_0|$; analogously,
$r(a)|x_0|=|x_0|a$. Therefore, for $\lambda>r(a)$, we have
$$R(\lambda,a)|x_0|=\frac1\lambda|x_0|+\frac1{\lambda^2}r(a)|x_0|+\ldots=
\frac{|x_0|}\lambda\Big(1+\frac{r(a)}\lambda+\Big(\frac{r(a)}\lambda\Big)^2+\ldots\Big)=
\frac{|x_0|}{\lambda_0-r(a)}$$ and, hence, $a_{-1}|x_0|=|x_0|$; analogously, $|x_0|a_{-1}=|x_0|$.
Consequently, we have the~equality $|x_0|=a_{-1}|x_0|=a_{-1}|x_0|a_{-1}=\mu a_{-1}$ for some
$\mu\in{\Bbb R}$. Since $\||x_0|\|_{a_{-1}}=1$, we infer $\mu=1$ and so $|x_0|=a_{-1}$. There
exists (see \mbox{\cite[Lemma 5.1(I)]{NS}}) an operator~$S$ on~$(A_{a_{-1}})_{\Bbb C}$ depending
upon~$x_0$ and satisfying the~identity $\widehat{L_a}=r(a)\lambda_0^{-1}S^{-1}\widehat{L_a}S$;
analogously, $\widehat{R_a}=r(a)\lambda_0^{-1}S^{-1}\widehat{R_a}S$ for the~same operator $S$,
where $\widehat{R_a}$ is a~restriction of~$R_a$ to $A_{a_{-1}}$. Now, if $\lambda'\in\sigma_{\rm
per}(a)\cap\sigma_{\rm j}(a;a_1)$ and $L_ax=R_ax=\lambda'x$ for $x\in(A_{a_{-1}})_{\Bbb C}$ then
$\widehat{L_a}Sx=\frac{\lambda_0\lambda'}{r(a)}Sx$ and
$\widehat{R_a}Sx=\frac{\lambda_0\lambda'}{r(a)}Sx$. Whence $\frac{\lambda_0\lambda'}{r(a)}$ belongs
to the~set $\sigma_{\rm per}(a)\cap\sigma_{\rm j}(a;a_1)$. Finally, since $\lambda_0$ and
$\lambda'$ are arbitrary, the set $\frac1{r(a)}\sigma_{\rm per}(a)\cap\sigma_{\rm j}(a;a_1)$ is
the~group of all $m_0^{\rm th}$~roots of unity for some $m_0\in{\Bbb N}$. We used Axiom~{\bf
(A$_4$)} where $b$ is an~algebraically strictly positive idempotent only. We didn't use
the~assumption that $x_0$ is not nilpotent (this assumption from the~definition of $\sigma_{\rm
j}(a;b)$ will be needed for the~validity of Lemma~\ref{lem15} and Corollary~\ref{cor16}).
The~proof of the~theorem is now complete.\hfill$\Box$
\smallskip
In the~proof of the~preceding theorem, using a~number $m$ of part~{\bf (a)}, a~number~$m'$
satisfying the~conditions of part~{\bf (b)} was found. A~number $m'$ is not uniquely
de\-ter\-mined. Indeed, for the~cyclic matrix $\left(\begin{smallmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1
& 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{smallmatrix}\right)$ two situations $m'=2$ and $m'=4$
are possible (then the~required order idempotents of part~{\bf (b)} correspond, in~the~former, to
the~coordinates $\{1,3\}$ and $\{2,4\}$ and, in the~second case, to $\{1\},\ldots,\{4\}$). For
the~case of the~analogous cyclic $6\times6$ matrix three situations $m'=2$ ($\{1,3,5\}$ and
$\{2,4,6\}$), $m'=3$ ($\{1,4\}$, $\{2,5\}$, and $\{3,6\}$), and $m'=6$ ($\{1\}$,\ldots,$\{6\}$) are
possible.
As is easy to see from the~proof of Theorem~\ref{thm12}, if $p_1,\ldots,p_{m'}$ satisfying
the~conditions of part~{\bf (b)} exist then part~{\bf (c)} holds with $m''=m'$. In is not known if
the~converse is valid. I.e., does part~{\bf (b)} hold with $m''=m'$ if part~{\bf (c)} holds? In
other words, the~author does not know the~direct proof of the~implication {\bf
(c)}~$\Longrightarrow$~{\bf (b)} of the~preceding theorem. Nevertheless, in the~following
particular case such a~proof is possible. Let the~condition of part~{\bf (c)} of
Theorem~\ref{thm12} hold and let the~number $m''$ have the~representation in the~form of
the~product $m''=m_1\ldots m_n$, where~$n\in{\Bbb N}$, $m_k$ are prime numbers for all
$k=1,\ldots,n$, and $m_{k'}\neq m_{k''}$ for all $k'\neq k''$. Obviously, for arbitrary
$k=1,\ldots,n$ and all $j=0,1,\ldots,m_k-1$ the~points $r(a)e^{i\frac{2\pi}{m_k}j}$ are poles of
$R(\cdot,a)$ and so (see the~proof of the~implication {\bf (c)}~$\Longrightarrow$~{\bf (a)})
$a^{m_k}$ is reducible. In view of the~validity of the~implication {\bf (a)}~$\Longrightarrow$~{\bf
(a)} and since $m_k$ is a~prime number, there exists non-zero order idempotents
$p_{m_k,1},\ldots,p_{m_k,m_k}$ determining the~cyclic form of the~element $a$. For an~arbitrary
$n$-tuple $(j_1,\ldots,j_n)$, where $j_k=1,\ldots,m_k$ for all $k=1,\ldots,n$, we put
$p_{j_1,\ldots,j_n}=p_{m_1,j_1}\ldots p_{m_n,j_n}$. If $(j_1',\ldots,j_n')\neq(j_1'',\ldots,j_n'')$
then $p_{j_1',\ldots,j_n'}\perp p_{j_1'',\ldots,j_n''}$ and, hence, the~number of order idempotents
$p_{j_1,\ldots,j_n}$ is equal to~$m''$. Next,
$\sum\limits_{j_1=1}^{m_1}\ldots\sum\limits_{j_n=1}^{m_n}p_{j_1,\ldots,j_n}={\bf e}$,
$p_{j_1,\ldots,j_n}a=ap_{j_1-1,\ldots,j_n-1}$, and $p_{j_1-m'',\ldots,j_n-m''}=p_{j_1,\ldots,j_n}$.
If for some $r\in{\Bbb N}$ the~equality $p_{j_1-r,\ldots,j_n-r}=p_{j_1,\ldots,j_n}$ is valid then
for all $k=1,\ldots,n$ the~number $m_k$ is a~divisor of $r$. Therefore, $r=m_1\ldots m_kl=m''l$
with~$l\in{\Bbb N}$. Finally, the~collection $\{p_{j_1,\ldots,j_n}\}$ determines the~cyclic form of
$a$.
\smallskip
The~next result makes more precisely correlations between the~values $m$, $m'$, and~$m''$.
\begin{cor}\
Under the~assumptions of Theorem~{\rm \ref{thm12}}, the~following equalities hold
$$\min{\{m\in{\Bbb N}:a^m \ \text{is reducible}\}}=$$
$$=\min{\{m\in{\Bbb N}: \text{there exist} \ p_1,\ldots, p_m \ \text{in} \ {\bf OI}(A) \
\text{determining the~cyclic form of} \ a\}}$$
$$=\min{\{m\in{\Bbb N}:
r(a)e^{i\frac{2\pi}mj} \ \text{is a~pole of} \ R(\cdot,a) \ \text{for} \ j=0,1,\ldots,m-1\}};$$ if
$a^m$ is irreducible for all $m$ or such $p_1,\ldots, p_m$ does not exist then we suppose that
the~respective minimum is equal to one.
\end{cor}
We continue with the~following two auxiliary results.
\begin{lem}\label{lem14}\
Let $c\in A^+$, let $p_0,p_1,\ldots,p_n$, where $n\in{\Bbb N}$, be a~collection of
\mbox{$c$-invariant} order idempotents satisfying the~relations ${\bf e}=p_n\ge\ldots\ge p_1\ge
p_0$, and let $q_j=p_jp_{j-1}^{\rm d}$ for all $j=1,\ldots,n$. If for every $j$ the~elements
$c_{q_j}$ are nilpotent then the~element $c$ is also nilpotent.
\end{lem}
{\bf Proof.} For $n=1$ the~assertion is obvious. Let us assume $n>1$. We consider first the~case of
$c_{q_j}=0$ for all $j$ and will show the~identity $c^n=0$. To this end, using the~induction on
$k=1,\ldots,n$, we shall prove the~validity of the~equality $q_sc^kq_j=0$ for all $j=1,\ldots,n$
and $s=\max{\{1,j-k+1\}},\ldots,n$ which, for $k=n$, implies $q_sc^nq_j=0$ for all $s$ and $j$. In
view of the~relation $\sum\limits_{j=1}^nq_j={\bf e}$, we obtain $c^n=0$. If $j<s$ then $0\le
q_scq_j\le p_{s-1}^{\rm d}cp_j\le p_{j-1}^{\rm d}cp_j=0$. Taking into account our condition, we
have the~equality $q_scq_j=0$ for $j\le s$. Therefore, for $k=1$ our induction hypothesis is true
and the~identity $c=\sum\limits_{s=1}^{n-1}\sum\limits_{j=s+1}^nq_scq_j$ holds. Assume the~validity
of our assertion for some $k<m$. Since $n-k+1\ge\max{\{1,j-k+1\}}$ for all $j$, we obtain
$q_sc^kq_j=0$ for $s\ge n-k+1$ and this equality also holds for $j<s+k$. Therefore,
$c^k=\sum\limits_{s=1}^{n-k}\sum\limits_{j=s+k}^nq_sc^kq_j$. Evidently, $q_mc^{k+1}q_j=0$ for all
$j$ and $q_sq_j=0$ for $s\neq j$. Now, for arbitrary indexes $l$ and $t$ with $l\le m-1$, we have
$$q_lc^{k+1}q_t=q_lcc^kq_t=
\sum\limits_{j=l+1}^nq_lcq_j\cdot\sum\limits_{s=1}^{n-k}\sum\limits_{j=s+k}^nq_sc^kq_jq_t=
\sum\limits_{j=l+1}^nq_lcq_j\cdot\sum\limits_{s=1}^{t-k}q_sc^kq_t.$$ Consequently, if $t-k\le l$
then $q_lc^{k+1}q_t=0$, as desired.
In a~general case, we mention first the~validity of the~equality $(c^k)_{q_j}=(c_{q_j})^k$ for all
$k\in{\Bbb N}$ and $j=1,\ldots,n$. Indeed, the~case of $k=1$ is obvious. If the~required equality
is true for some~$k$ then
$$(c_q)^{k+1}=(c_q)^kc_q=(c^k)_qc_q=qc^kqcq=qc^{k+1}q-qc^kq^{\rm d}cq=(c^{k+1})_q.$$
Now we choose $k_j\in{\Bbb N}$ satisfying $c_{q_j}^k=0$ for all $j=1,\ldots,n$. Putting
$k=\max\limits_{1\le j\le n}{k_j}$, we have $(c^k)_{q_j}=(c_{q_j})^k=0$. As was showed above,
$c^{kn}=0$.\hfill$\Box$
\smallskip
For a~non-zero element $q\in{\bf OI}(A)$, we put $A(q)=\{a\in A:qaq=a\}$. Under the~linear
operations, the~multiplication, the~norm, and the~order induced by $A$, the~linear space $A(q)$ is
an~ordered Banach algebra with unit $q$ being a~closed subalgebra of $A$. We mention at once
the~following properties of $A(q)$~\cite{Al2}: {\bf (a)} For the~order interval $[0,q]$ in $A$
the~identity ${\bf OI}(A(q))={\bf OI}(A)\cap[0,q]$ is valid; {\bf (b)} An~element $z\in A(q)$ is
irreducible with respect $q$ if and only if $z$ is irreducible in $A(q)$; {\bf (c)} The~identity
$(A(q))_{\rm n}=A_{\rm n}\cap A(q)$ holds; {\bf (d)} If $A$ has a~disjunctive product then $A(q)$
also has a~disjunctive product. From this remarks, it follows that if $A$ satisfies Axiom~{\bf
(A$_j$)} for some $j=1,3,4$ then $A(q)$ also satisfies this axiom.
\begin{lem}\label{lem15}\
Let $A$ satisfy Axiom~{\bf (A$_1$)}. Let an~element $a\in A_{\rm n}$ such that $r(a)>0$ is a~pole
of $R(\cdot,a)$ and the~residue $a_{-1}$ possesses a~modulus $|a_{-1}|$. Let the~collection of
order idempotents $p_0,p_1,\ldots,p_n$ determine the~Frobenius normal form of the~element~$a$. Then
the~inclusions
\begin{eqnarray}\label{19}
& \sigma_{\rm per}(a;A)\cap\sigma_{\rm j}(a;|a_{-1}|;A)\subseteq &\nonumber\\
& \subseteq \bigcup\{\sigma_{\rm per}(a_{q_j};A(q_j))\cap\sigma_{\rm
j}(a_{q_j};(a_{q_j})_{-1};A(q_j)): r(a_{q_j})=r(a)\}\subseteq\sigma_{\rm per}(a;A) &
\end{eqnarray}
hold with $q_j=p_jp_{j-1}^{\rm d}$ for $j=1,\ldots,n$.
\end{lem}
{\bf Proof.} We denote the~middle part of~(\ref{19}) by ${\cal S}$. By Corollary~\ref{cor5}, to
check the~left inclusion in~(\ref{19}), it suffices to establish the~relation
$$\bigcup\{\sigma_{\rm per}(a_{q_j};A)\cap\sigma_{\rm j}(a;|a_{-1}|;A): r(a_{q_j})=r(a)\}\subseteq
{\cal S}.$$ To this end, let $\lambda\in\sigma_{\rm j}(a;|a_{-1}|;A)$ and let $|\lambda|=r(a)$.
Then there exists an~element $c\in A$ which is not nilpotent, satisfies $ac=ca=\lambda c$, and has
a~representation in the~form $c=c'+ic''$ with $c',c''\in A_{|a_{-1}|}$. For arbitrary
$j=0,1,\ldots,n$, the~order idempotent~$p_j$ is \mbox{$|a_{-1}|$-invariant} and, hence, is
\mbox{$c$-invariant}. Therefore, for $j=1,\ldots,n$, we have
$$q_j^{\rm d}cq_j=(p_j^{\rm d}+p_{j-1})cp_jp_{j-1}^{\rm d}=p_{j-1}cq_j;$$
analogously, $q_jaq_j^{\rm d}=q_jcp_j^{\rm d}$. Thus, $q_jaq_j^{\rm d}cq_j=0$. Consequently,
$$\lambda c_{q_j}=q_jacq_j=q_jaq_jcq_j+q_jaq_j^{\rm d}cq_j=a_{q_j}c_{q_j};$$
analogously, $\lambda c_{q_j}=c_{q_j}a_{q_j}$. In view of Lemma~\ref{lem14}, there exists an~index
$j_0$ such that $c_{q_{j_0}}$~is not nilpotent. In~particular, $c_{q_{j_0}}\neq0$ and, hence,
$\lambda\in\sigma(a_{q_{j_0}})$. On the~other hand, Lemma~\ref{lem3} yields $r(a_{q_{j_0}})\le
r(a)$. Therefore, $r(a_{q_{j_0}})=r(a)$, $a_{q_{j_0}}$ is irreducible with respect~$q_{j_0}$, and
$\lambda\in\sigma_{\rm per}(a_{q_{j_0}};A(q_{j_0}))$. Then \cite{Al2} $r(a)$ is a~simple pole of
$R(\cdot,a_{q_{j_0}})$ and for $\lambda$ from a~sufficient small punctured neighbourhood of $r(a)$
the~identity
$$R(\lambda,a_{q_{j_0}})=
\frac1\lambda(p_{j_0}^{\rm d}+p_{j_0-1})+q_{j_0}R(\lambda,a)q_{j_0}$$ holds. Thus,
$0\le(a_{q_{j_0}})_{-1}=(a_{-1})_{q_{j_0}}$. The~last equality implies
$(c')_{q_{j_0}},(c'')_{q_{j_0}}\in A_{(a_{q_{j_0}})_{-1}}$ and, hence, $\lambda\in\sigma_{\rm
j}(a_{q_{j_0}};(a_{q_{j_0}})_{-1};A(q_{j_0}))$.
To check the~right inclusion in~(\ref{19}), mention the~next inclusion $\sigma_{\rm
per}(a_q;A(q))\subseteq\sigma_{\rm per}(a_q;A)$ for arbitrary $q\in{\bf OI}(A)$. In fact, if
$(\lambda-a_q)b=b(\lambda-a_q)={\bf e}$ for an~element $b\in A$ then
$(\lambda-a_q)b_q=b_q(\lambda-a_q)=q$. Now it only remains to recall Lemma~\ref{lem3}.\hfill$\Box$
\smallskip
Theorem~\ref{thm12} and the~preceding lemma imply the~next consequence which charac\-terizes
the~peripheral spectrum of an~arbitrary positive element of $A$.
\begin{cor}\label{cor16}\
Let the~assumptions of Lemma~{\rm \ref{lem15}} satisfy, let $A$ satisfy Axioms {\bf (A$_3$)}
and~{\bf (A$_4$)}, and let the~algebra $A(q)$ satisfy Axiom~{\bf (A$_2$)} for every non-zero
$q\in{\bf OI}(A)$. Then the~inclusions
$$\sigma_{\rm per}(a;A)\cap\sigma_{\rm j}(a;|a_{-1}|;A)\subseteq
r(a)\bigcup\limits_{s=1}^nH_{m_s}\subseteq\sigma_{\rm per}(a;A)$$ hold with some
$m_1,\ldots,m_n\in{\Bbb N}$.
\end{cor}
We close this section with the~next assertion about elements having the~cyclic form.
\begin{prop}\
Let an~element $a\in A$ have the~cyclic form. Suppose that there exists at least one pole of
the~resolvent $R(\cdot,a)$ among points of $\sigma_{\rm per}(a)$. Then the~decomposition
$a^m=\sum\limits_{j=1}^mb_j$ holds, where $1<m\in{\Bbb N}$, $b_{j'}b_{j''}=0$ for $j'\neq j''$, and
$r(b_j)=r(a)^m$ for all $j=1,\ldots,m$. If, in~addition, $a\in A^+$ then there exists such
a~decomposition that $b_{j'}\wedge b_{j''}=0$ for $j'\neq j''$.
\end{prop}
{\bf Proof.} Using our condition, we find order idempotents $p_1,\ldots,p_m$ of $A$, where $m>1$,
satisfying $p_ja=ap_{j+1}$ for $j=1,\ldots,m$. Evidently, $p_ja^m=ap_{j+1}a^{m-1}=\ldots=a^mp_j$,
whence $p_ja^mp_j=a^mp_j=p_ja^m$ for all $j$ and $p_{j'}a^mp_{j''}=0$ for $j'\neq j''$. Therefore,
putting $b_j=p_ja^mp_j$, we obtain $a^m=\sum\limits_{j=1}^mb_j$. Fix an~index $j$. Let $\lambda_0$
be a~pole of~$R(\cdot,a)$ and let $|\lambda_0|=r(a)$. Since the~equality $p_jR(\lambda,a)=0$ is
impossible $\lambda$ for sufficiently close to $\lambda_0$, $\lambda\neq\lambda_0$, there exists
a~number $s\in{\Bbb N}$ satisfying $p_ja_{\lambda_0,s}\neq0$ and $p_ja_{\lambda_0,t}=0$ for $t<s$.
Keeping the~identity $aa_{\lambda_0,s}=r(a)a_{\lambda_0,s}+a_{\lambda_0,s-1}$ in mind and using
the~elementary induction, it is not difficult to check the~validity of the~relation
$a^na_{\lambda_0,s}=\sum\limits_{t=0}^n\lambda_0^tC_n^ta_{\lambda_0,s-(n-t)}$ for all $n\in{\Bbb
N}$, where $C_n^t$ are binomial coefficients. The~latter implies the~equality
$p_ja^ma_{\lambda_0,s}=\lambda_0^mp_ja_{\lambda_0,s}$ or
$b_ja_{\lambda_0,s}=\lambda_0^ma_{\lambda_0,s}$. Thus, $\lambda_0^m\in\sigma(b_j)$. Taking into
account the~inequality $r(b_j)\le r(a)^m$, we have $r(b_j)=r(a)^m$. In view of the~definition of
$b_j$, the~last assertion is clear.\hfill$\Box$
\section{The~Lotz-Schaefer axiom}
In the~spectral theory of positive operators the~next Lotz-Schaefer theorem is one of the~most
significant results (see \mbox{\cite[pp. 351-352]{Sch}}). In some situations, e.g., in~the~case of
irreducible operators (see~\cite{Al}), this theorem allows the~study of points of the~peripheral
spectrum of an~operator to reduce to the~case of poles only.
\begin{thm}\
Let $T$ be a~positive operator on a~Banach lattice $E$ and let $r(T)$ be a~finite-rank pole of
$R(\cdot,T)$. Then the~peripheral spectrum $\sigma_{\rm per}(T)$ of $T$ consists entirely of poles
of $R(\cdot,T)$.
\end{thm}
As was mentioned above (see Axiom~{\bf (A$_2$)} and the~remarks before this axiom), algebraically
strictly positive projections in an~ordered Banach algebra $A$ can be considered as
a~generalization of rank-one operators. Moreover, the~residue $a_{-1}$ of the~resolvent
$R(\cdot,a)$ of the~irreducible element $a$ at the~point $r(a)$ satisfies this condition (see
Theorem~\ref{thm6}{\bf (c)}). An~element $b\in A$ is said to be {\it relatively algebraically
strictly positive} whenever there exists a~non-zero order idempotent~$q$ of $A$ such that $b\in
A(q)$ and $q_1aq_2>0$ for all $0<q_1,q_2\in{\bf OI}(A)\cap[0,q]$; in this case, we write $b\gg_q0$.
Obviously, $b\gg0$ if and only if $b\gg_{\bf e}0$. As is easy to see, if $P$ is a~non-zero order
continuous projection on a~Dedekind complete Banach lattice $E$ being relatively algebraically
strictly positive element in the~ordered Banach algebra $B(E)$ then $\dim{R(P)}=1$. On~the~other
hand, the~collection $F(E)$ of finite-rank operators on $E$ is an~algebraic ideal of $B(E)$ and if
a~non-zero operator $T\in F(E)$ then the~algebraic ideal generated by~$T$ in $B(E)$ coincides
with~$F(E)$. Keeping these remarks in mind, we define the~set ${\cal F}(A)$ of {\it finite-rank
elements} of an~ordered Banach algebras $A$ as the~(two-sided) algebraic ideal generated by the~set
of all relatively algebraically strictly positive order continuous projectors of $A$; if such
projectors do not exist, we put ${\cal F}(A)=\emptyset$. If $a\in A(q)$ is a~relatively
algebraically strictly positive element of $A(q)$ then $a$ is such an~element of $A$, and, hence,
${\cal F}(A(q))\subseteq{\cal F}(A)$. Now, axiomatizing the~respective theorem, we can introduce
the~Lotz-Schaefer axiom in the~following manner:
\begin{description}
\item[{\bf (A$_{\rm LS}$)}] If $a$ is a~positive element of $A$, $r(a)$ is a~pole of $R(\cdot,a)$,
and the~residue $a_{-1}$ is a~finite-rank element then the~peripheral spectrum $\sigma_{\rm
per}(a)$ of $a$ consists entirely of poles of $R(\cdot,a)$.
\end{description}
If $E$ is a~Dedekind complete Banach lattice admitting a~weak order unit $x>0$ and a~strictly
positive order continuous functional $f$ then the~projection $\frac1{f(x)}f\otimes x\gg0$ in
the~ordered Banach algebra $B(E)$ and, hence, ${\cal F}(B(E))=F(E)$. Therefore, the~Lotz-Schaefer
theorem implies the~validity of Axiom~{\bf (A$_{\rm LS}$)} for a~wide class of ordered Banach
algebras of the~form $B(E)$. If we want the~validity of a~similar axiom for a~wider class of
algebras of the~form $B(E)$ then we must introduce the~next weaker axiom:
\begin{description}
\item[{\bf (A$_{\rm LS}'$)}] If $a$ is a~positive element of $A$, $r(a)$ is a~pole of $R(\cdot,a)$,
the~residue $a_{-1}$ is an~order continuous element, and $a_{-1}\gg0$ then the~peripheral spectrum
$\sigma_{\rm per}(a)$ of $a$ consists entirely of poles of $R(\cdot,a)$.
\end{description}
Now if $E$ is an~arbitrary Dedekind complete Banach lattice then the~ordered Banach algebra~$B(E)$
satisfies Axiom~{\bf (A$_{\rm LS}'$)}. Moreover, the~last two conditions of this axiom
automatically hold when the~element $a$ is irreducible (see Theorem~\ref{thm6}{\bf (c)}). However,
as the~results below show (see, e.g., Theorem~\ref{thm23}), in the~case of an~arbitrary positive
elements Axiom~{\bf (A$_{\rm LS}$)} is more useful than Axiom~{\bf (A$_{\rm LS}'$)}.
Nevertheless, there are other cases when an~algebra $A$ need not satisfy Axiom~{\bf (A$_{\rm
LS}$)}. If ${\bf OI}(A)=\{0,{\bf e}\}$ then ${\bf e}\gg0$ and, hence, ${\cal F}(A)=A$. Next, if
$A=L_\infty(\Omega,\mu)$, with $\mu$~a~\mbox{$\sigma$-finite} diffuse measure on
a~$\sigma$-algebra, then there exist no algebraically relatively strictly positive elements in $A$
and, hence, ${\cal F}(A)=\{0\}$.
If we will assume that the~residue~$a_{-1}$ is a~minimal idempotent of $A$ instead of
$a_{-1}\in{\cal F}(A)$ then Axiom~{\bf (A$_{\rm LS}$)} is valid for the~algebra $B(E)$, where $E$
is an~arbitrary Banach lattice. Unfortunately, the~residue $a_{-1}$ of the~resolvent $R(\cdot,a)$
of an~irreducible element $a$ which belongs to an~ordered Banach algebra $A$ satisfying Axiom~{\bf
(A$_1$)} need not be a~minimal idempotent (see~\cite{Al2}). Therefore, under this assumption,
Axiom~{\bf (A$_{\rm LS}$)} is employed only for a~narrow class of ordered Banach algebras.
Moreover, as the~next example shows, Axiom~{\bf (A$_{\rm LS}'$)} in its present form and Axiom~{\bf
(A$_{\rm LS}$)} with the~assumption about the~minimality of $a_{-1}$ do not hold in general for
an~arbitrary ordered Banach algebra $A$. This example also shows that {\it the~peripheral spectrum
$\sigma_{\rm per}(a)$ of an~irreducible element $a$ need not be cyclic} while $r(a)$ is a~pole of
$R(\cdot,a)$ and the~residue $a_{-1}\gg0$.
\begin{exm}\
{\rm Consider the~space $\ell_\infty$ of all bounded sequences $x=(x_1,x_2,\ldots)$. Under
the~natural algebraic operations, multiplication, and $\sup$-norm, this space is a~commutative
Banach algebra with unit ${\bf e}=(1,1,\ldots)$. Fix an~arbitrary number $\lambda_0\in{\Bbb C}$ and
a~sequence $\{z_n\}$ in ${\Bbb C}$ satisfying $|z_n|=1$, $z_n\neq\lambda_0$ for all $n$,
$\lambda_0\neq1$ , $z_1=1$, $z_n\neq z_m$ for all $n\neq m$, and $z_n\to\lambda_0$ as $n\to\infty$.
Define the~sequence $z\in\ell_\infty$ by $z=(z_1,z_2,\ldots)$ and consider the~algebraic wedge
$K_0$ generated by ${\bf e}$ and~$z$, i.e.,
$$K_0=\Big\{\sum\limits_{j=0}^n\alpha_jz^j:
n\in{\Bbb N} \ \text{and} \ \alpha_j\in{\Bbb R}^+ \ \text{for all} \ j=0,1,\ldots,n\Big\}.$$ We
claim that the~closure $\overline{K_0}$ of $K_0$ is a~normal cone. Indeed, for arbitrary
$y=\sum\limits_{j=0}^n\alpha_jz^j$ with $\alpha_j\in{\Bbb R}$ the~inequalities
$\Big|\sum\limits_{j=0}^n\alpha_j\Big|\le\|y\|_{\ell_\infty}\le\sum\limits_{j=0}^n|\alpha_j|$ holds
as $z_1=1$. Therefore, if $\alpha_j\ge0$ for $j=0,1,\ldots,n$ then
\begin{equation}\label{20}
\|y\|_{\ell_\infty}=\sum\limits_{j=0}^n\alpha_j.
\end{equation}
Let $x\in\ell_\infty$ and let $\pm x\in\overline{K_0}$. There exist two sequences $\{x_n\}$ and
$\{y_n\}$ in $K_0$ satisfying $x_n\to x$ and $y_n\to-x$ and, hence, $x_n+y_n\to0$ as $n\to\infty$.
Let $x_n=\sum\limits_{j=0}^{k_n}\alpha_{nj}z^j$ and $y_n=\sum\limits_{j=0}^{k_n}\beta_{nj}z^j$,
where $\alpha_{nj},\beta_{nj}\ge0$ for all $n$ and $j=1,\ldots,k_n$. Then
$\sum\limits_{j=0}^{k_n}(\alpha_{nj}+\beta_{nj})z^j\to0$ and so
$0\le\sum\limits_{j=0}^{k_n}\alpha_{nj}\le\sum\limits_{j=0}^{k_n}(\alpha_{nj}+\beta_{nj})\to0$ as
$n\to\infty$. Taking into account the~identity~(\ref{20}), we have
$\|x_n\|_{\ell_\infty}=\sum\limits_{j=0}^n\alpha_{nj}\to0$. Therefore, $x=0$. Thus,
$K=\overline{K_0}$ is a~cone and, under the~order induced by $K$, $\ell_\infty$ is an~ordered
Banach algebra. In view of the~relation $z_n\neq z_m$ with $n\neq m$, the~system $\{{\bf
e},z,z^2,\ldots\}$ is linearly independent and, in~particular, every element $w\in K_0$ has
a~unique representation in the~form $w=\sum\limits_{j=0}^n\omega_jz^j$ with $\omega_j\ge0$. Now if
$u=\sum\limits_{j=0}^m\mu_jz^j\in K_0$ and $0\le_{K_0}w\le_{K_0}u$ then
$\|w\|_{\ell_\infty}=\sum\limits_{j=0}^n\omega_j\le\sum\limits_{j=0}^n\mu_j=\|u\|_{\ell_\infty}$.
Thus, $K_0$ is a~normal cone and, hence \mbox{\cite[p. 81, Exercise 9]{AlT}}, $K$ is normal. We
mention at once that the~relation $\overline{K-K}\neq\ell_\infty$ holds as the~Banach space
$\ell_\infty$ is not separable; in~particular, $K$~is not generating. Obviously, $r(z)=1$ and
$R(\lambda,z)=(\frac1{\lambda-z_1},\frac1{\lambda-z_2},\ldots)$ for all
$\lambda\notin\sigma(z)=\overline{\{z_1,z_2,\ldots\}}$. In~particular, $\xi=1$ is an~isolated point
of $\sigma(z)$. For $\lambda$ close to this point and $\lambda\neq1$ the~inequality
$|\lambda-1|\le|\lambda-z_n|$ holds for all $n$. Whence
$\Big|\frac{(\lambda-1)^2}{\lambda-z_n}\Big|\le|\lambda-1|$ and so
$(\lambda-1)^2\|R(\lambda,z)\|_{\ell_\infty}\le|\lambda-1|\to0$ as~$\lambda\to1$. Thus, $\xi=1$ is
a~simple pole of $R(\cdot,z)$. Let $r=\inf\limits_{n>1}{|1-z_n|}>0$. For arbitrary
$\epsilon$~and~$\lambda$ satisfying $0<\epsilon<r$ and $|\lambda-1|<\epsilon$, we have
$|\lambda-z_n|\ge|1-z_n|-|\lambda-1|\ge r-\epsilon$ and, hence,
$\Big|\frac{\lambda-1}{\lambda-z_n}\Big|\le\frac{\epsilon}{r-\epsilon}\to0$ as $\epsilon\to0$.
Consequently, $\lim\limits_{\lambda\to1}\Big|\frac{\lambda-1}{\lambda-z_n}\Big|=0$ uniformly in
$n=2,3,\ldots$ and so the~residue $R(\cdot,z)$ at the~point $\xi=1$ is equal to ${\bf
e_1}=(1,0,0,\ldots)$. If we can verify the~identity ${\bf OI}(\ell_\infty)=\{0,{\bf e}\}$ (under
the~order induced by $K$) then this means the~validity of the~relation ${\bf e_1}\gg0$ while
the~point $\lambda_0$ is not an~isolated point of the~spectrum $\sigma(z)$. To this~end, let
the~sequence $p\in{\bf OI}(\ell_\infty)$. Then $p$ is a~characteristic function $\chi_A$ of
a~subset $A$ of ${\Bbb N}$ which has a~representation in the~form
$p=\sum\limits_{j=0}^n\theta_jz^j$ with $\theta_j\ge0$. If $1\notin A$ then $(\chi_A)_1=0$, whence
$\sum\limits_{j=0}^n\theta_j=0$ or $p=0$. If $1\in A$ then $1\notin{\Bbb N}\setminus A$, whence
$p={\bf e}$, as required. Obviously, the~residue ${\bf e_1}$ is a~minimal idempotent of the~algebra
$\ell_\infty$. Moreover, as is easy to see, we can choose the~sequence $\{z_n\}$ such that
the~peripheral spectrum $\sigma_{\rm per}(z)$ of the~irreducible element~$z$ is not
cyclic.}\hfill$\Box$
\end{exm}
Now, assuming the~Lotz-Schaefer axiom~{\bf (A$_{\rm LS}$)}, we continue the~study of
the~perip\-heral spectrum and, in~particular, will obtain some consequences of Theorem~\ref{thm12}.
\smallskip
Before, we discuss the~following property of the~algebra $B(E)$, where $E$ is a~complex Banach
lattice being the~complexification of the~real Banach lattice $E_{\Bbb R}$. Let
$\{S_n\}$~and~$\{T_n\}$ be two sequences in the~space $B_{\rm r}(E)=B_{\rm r}(E_{\Bbb R})$ of all
regular operators on $E$ such that the~sequence $\{S_n+iT_n\}$ converges in $B(E)$. Then
$\{S_n\}$~and~$\{T_n\}$ are also convergent. Indeed, fix a~number $\epsilon>0$ and find an~index
$k\in{\Bbb N}$ satisfying $\|S_n-S_m+i(T_n-T_m)\|_{B(E)}<\epsilon$ for all $n,m\ge k$. Using
the~condition $S_n\in B_{\rm r}(E)$ and the~inequality $\|y+iz\|_E\ge\max{\{\|y\|_E,\|z\|_E\}}$ for
all $y,z\in E_{\Bbb R}$, we have
$$\|(S_n-S_m)x\|_E\le\|(S_n-S_m)x+i(T_n-T_m)x\|_E<\epsilon$$
for an~arbitrary element $x\in E_{\Bbb R}$ with $\|x\|_E=\|x\|_{E_{\Bbb R}}=1$. Whence
$\|S_n-S_m\|_{B(E_{\Bbb R})}<\epsilon$ and so $\|S_n-S_m\|_{B(E)}<2\epsilon$. Thus, the~sequence
$\{S_n\}$ is convergent; the~case of $\{T_n\}$ is analogous. We axiomatize this property and make
the~next assumption:
\begin{description}
\item[{\bf (A$_5$)}] The~convergence of the~sequence $\{b_n+ic_n\}$, where $\{b_n\}$ and
$\{c_n\}$ are two sequences in~$A_{\rm r}$, implies the~convergence of $\{b_n\}$ and $\{c_n\}$.
\end{description}
Evidently, if an~ordered Banach algebra $A$ satisfies Axiom~{\bf (A$_5$)} then the~algeb\-ra~$A(q)$
also satisfies this axiom for every non-zero order idempotent $q$ of $A$. Next, as the~example of
the~ordered Banach algebra $C^1[a,b]$ of all complex functions $x$ represented in the~form
$x=x_1+ix_2$, where the~functions $x_1,x_2:[a,b]\to{\Bbb R}$ are continuously differentiable, under
the~natural algebraic operations, multiplication, order, and norm
$\|x\|_{C^1[a,b]}=\max\limits_{t\in[a,b]}|x(t)|+\max\limits_{t\in[a,b]}|{\dot x}(t)|$, shows,
Axiom~{\bf (A$_5$)} does not imply the~normality of a~cone~$A^+$.
\begin{thm}\label{thm20}\
Let an~ordered Banach algebra $A$ satisfy Axioms~{\bf (A$_1$)}-{\bf (A$_5$)} and~{\bf (A$_{\rm
LS}'$)}. Let an~element $a$ of $A$ be non-zero, order continuous, and irreducible. Let the~point
$r(a)$ be a~pole of $R(\cdot,a)$. Then the~peripheral spectrum $\sigma_{\rm per}(a)$ consists
entirely of poles of $R(\cdot,a)$ and has the~form $\sigma_{\rm per}(a)=r(a)H_m$ for some
$m\in{\Bbb N}$.
\end{thm}
{\bf Proof.} In view of Theorem~\ref{thm12}, it suffices to establish the~inclusion $\sigma_{\rm
per}(a)\subseteq\sigma_{\rm j}(a;a_{-1})$. To this end, let $\lambda_0\in\sigma_{\rm per}(a)$.
Since $A$ satisfies Axiom~{\bf (A$_{\rm LS}'$)}, $\lambda_0$ is a~pole of $R(\cdot,a)$ of order
$k$. The~relations $aa_{\lambda_0,-k}=a_{\lambda_0,-k}a=\lambda_0a$ hold. We claim that $k=1$ and
the~residue $a_{\lambda_0,-1}$ has the~representation in the~form $a_{\lambda_0,-1}=z'+iz''$, where
the~elements $z',z''\in A_{a_{-1}}$. The~last two assertions imply the~inclusion
$\lambda_0\in\sigma_{\rm j}(a;a_{-1})$. To check them, let
$\lambda_0=r(a)(\cos{\varphi}+i\sin{\varphi})$ with $\varphi\in[0,2\pi)$. We have the~equalities
$$a_{\lambda_0,-k}=\lim\limits_{\lambda\to\lambda_0}(\lambda-\lambda_0)^kR(\lambda,a)=
\lim\limits_{t\downarrow1}(t\lambda_0-\lambda_0)^kR(t\lambda_0,a)=
\lim\limits_{t\downarrow1}(t-1)^k\lambda_0^k\sum\limits_{j=0}^\infty\frac1{(t\lambda_0)^{j+1}}a^j
=$$
$$=\lim\limits_{t\downarrow1}(t-1)^k\sum\limits_{j=0}^\infty
\frac{\cos{((k-j+1)\varphi)}+i\sin{((k-j+1)\varphi)}}{t^{j+1}}a^j.$$ Fix an~arbitrary sequence
$\{t_n\}$ in ${\Bbb R}$ satisfying $t_n\downarrow1$ and put
$$b_n=(t_n-1)^k\sum\limits_{j=0}^\infty\frac{\cos{((k-j+1)\varphi)}}{t_n^{j+1}}a^j \ \ {\rm and} \ \
c_n=(t_n-1)^k\sum\limits_{j=0}^\infty\frac{\sin{((k-j+1)\varphi)}}{t_n^{j+1}}a^j.$$ Obviously,
$b_n+ic_n\to a_{\lambda_0,-k}$ as $n\to\infty$ and $\pm
b_n\le(t_n-1)^k\sum\limits_{j=0}^\infty\frac1{t_n^{j+1}}a^j\to a_{-k}$; analogously, for $\{c_n\}$.
Taking into account Axiom~{\bf (A$_5$)}, we conclude the~convergence of $\{b_n\}$ and $\{c_n\}$ and
the~validity of the~representation $a_{\lambda_0,-k}=x'+ix''$, where $-a_{-k}\le x',x''\le a_{-k}$.
If $k>1$ then $a_{-k}=0$ and, hence, $x',x''=0$ or $a_{\lambda_0,-k}=0$, a~contradiction. Thus,
$k=1$, as required.\hfill$\Box$
\begin{lem}\label{lem21}\
If $A$ satisfies Axiom~{\bf (A$_{\rm LS}$)} then $A(q)$ also satisfies this axiom for every
non-zero $q\in{\bf OI}(A)$.
\end{lem}
{\bf Proof.} Let $0\le a\in A(q)$, let $r(a)$ be a~pole of $R_q(\cdot,a)$, where $R_q(\cdot,a)$ is
a~resolvent of $a$ in~$A(q)$, and let the~residue $a_{-1}$ of $R_q(\cdot,a)$ at the~point $r(a)$
satisfy the~condition $a_{-1}\in{\cal F}(A(q))$. We can assume $r(a)>0$. The~inclusions~\cite{Al2}
\begin{equation}\label{21}
\rho_\infty(a;A)\subseteq\rho(a;A(q)) \ \ {\rm and} \ \
\rho(a;A(q))\setminus\{0\}\subseteq\rho(a;A)
\end{equation}
hold, where $\rho_\infty(a;A)$ is the~unbounded connected component of $\rho(a;A)$. Moreover, for
arbitrary $\lambda\in\rho(a;A(q))\setminus\{0\}$, we have the~identity~\cite{Al2}
\begin{equation}\label{22}
R(\lambda,a)=R_q(\lambda,a)+\frac1\lambda q^{\rm d}.
\end{equation}
Thus, $r(a)$ is a~pole of $R(\cdot,a)$ and the~residues of $R(\cdot,a)$ and $R_q(\cdot,a)$ at
$r(a)$ coincide. In view of the~inclusion ${\cal F}(A(q))\subseteq{\cal F}(A)$, $a_{-1}\in{\cal
F}(A)$. If $\lambda\in\sigma_{\rm per}(a;A(q))$ then $\lambda\in\sigma_{\rm per}(a;A)$ and, hence,
$\lambda$ is a~pole of $R(\cdot,a)$. Taking into account the~first inclusion of~(\ref{21}), we
infer that $\lambda$ is an~isolated point of $\sigma(a;A(q))$. Consequently, in view of~(\ref{22}),
$\lambda$ is a~pole of $R_q(\cdot,a)$.\hfill$\Box$
\smallskip
The~next result which follows easily from Corollary~\ref{cor5}, Theorem~\ref{thm20},
the~inclu\-sions~(\ref{21}), and the~preceding lemma, characterizes the~peripheral spectrum of
a~wide class of positive elements.
\begin{cor}\
Let $A$ satisfy Axioms {\bf (A$_1$)}, {\bf (A$_3$)}-{\bf (A$_5$)}, and~{\bf (A$_{\rm LS}$)}, let
the~algebra $A(q)$ satisfy Axiom~{\bf (A$_2$)} for every non-zero $q\in{\bf OI}(A)$, and let
an~element $a\in A_{\rm n}$ such that it has the~Frobenius normal form and $r(a)$ is a~pole of
$R(\cdot,a)$. Then the~identity $\sigma_{\rm per}(a)=r(a)\bigcup\limits_{s=1}^nH_{m_s}$ holds with
some $m_1,\ldots,m_n\in{\Bbb N}$.
\end{cor}
As the~next theorem shows, Axiom~{\bf (A$_{\rm LS}$)} can be employed to positive elements having
the~Frobenius normal form. Moreover, this result, Theorem~\ref{thm20}, and the~pre\-ceding
corollary illustrate the~importance of the~notion of finite-rank element as we define it above. It
also shows that our definition in the~abstract case is the~right one to use.
\begin{thm}\label{thm23}\
Let an~ordered Banach algebra $A$ satisfy Axiom~{\bf (A$_1$)}. Let an~element $a\in A_{\rm n}$ have
the~Frobenius normal form. If $r(a)$ is a~pole of $R(\cdot,a)$ then the~residue $a_{-1}$ is
a~finite-rank element.
\end{thm}
{\bf Proof.} The~idea is borrowed from the~proof of the~implication {\bf
(d)}~$\Longrightarrow$~{\bf (c)} of Theorem~2.14 in \cite{Al2}. Let order idempotents
$p_0,p_1,\ldots,p_n$, ${\bf e}=p_n\ge\ldots\ge p_0=0$, determine the~Frobenius normal form of $a$.
Put $q_j=p_jp_{j-1}^{\rm d}$ for $j=1,\ldots,n$. If $r(a)=0$ then $r(a_{q_j})=0$ and $a_{q_j}$ is
irreducible with respect $q_j$ for all $j$, whence $a_{q_j}=0$. Thus, ${\bf
OI}(A)\cap[0,q_j]=\{0,q_j\}$ and so $q_j\gg_{q_j}0$. On the~other hand, $a_{-1}={\bf
e}=\sum\limits_{j=1}^nq_j\in{\cal F}(A)$.
Now we can suppose $r(a)>0$. For the proof, we use induction on $n$. For $n=1$ the~element $a$ is
irreducible and it remains to use Theorem~\ref{thm6}{\bf (c)}. Next, assume that the~desired
assertion is proved if a~parameter of the induction lies between $1$ and $n-1\ge1$. Let us verify
our assertion for $n$.
We consider first the~case of the~identity $r(a_{p_{n-1}})=r(a)$ and show the~inclusion
\begin{equation}\label{23}
(a_{-1})_{p_{n-1}}\in{\cal F}(A).
\end{equation}
The~relation \cite{Al2} $(a_{p_{n-1}})_{-1}=(a_{-1})_{p_{n-1}}$ holds in $A$. If
$r(a)\notin\sigma(a_{p_{n-1}})$ then $(a_{p_{n-1}})_{-1}=0$ and (\ref{23}) is obvious. Let
$r(a)\notin\sigma(a_{p_{n-1}})$. Then \cite{Al2} $r(a)$ is a~pole of
$R_{p_{n-1}}(\cdot,a_{p_{n-1}})$ in $A(p_{n-1})$ (see the~proof of Lemma~\ref{lem21}). The~order
idempotents $p_{n-1},\ldots,p_0$ are \mbox{$a_{p_{n-1}}$-invariant}. If $r(a_{q_j})=r(a_{p_{n-1}})$
then $a_{q_j}$ is irreducible with respect $q_j$ in $A$ and so in $A(p_{n-1})$. By the~induction
hypothesis, the~residue $(a_{p_{n-1}})_{-1}\in{\cal F}(A(p_{n - 1}))$ (see the~proof of
Lemma~\ref{lem21} once more) and, hence, $(a_{p_{n-1}})_{-1}\in{\cal F}(A)$.
Consider the~case of the~identity $r(a_{p_{n-1}^{\rm d}})=r(a)$ and show the~inclusion
\begin{equation}\label{24}
(a_{-1})_{p_{n - 1}^{\rm d}}\in{\cal F}(A).
\end{equation}
Assuming without loss of generality that $r(a)\in\sigma(a_{p_{n-1}}^{\rm d})$, we have
the~relations
$$(a_{-1})_{p_{n - 1}^{\rm d}}=(a_{p_{n - 1}^{\rm d}})_{-1}\in{\cal F}(A(p_{n - 1}^{\rm d}))\subseteq
{\cal F}(A).$$
Both inequalities $r(a_{p_{n-1}})\le r(a)$ and $r(a_{p_{n-1}^{\rm d}})\le r(a)$ cannot be strict
simulta\-neously. To~complete the proof, we consider three possible cases.
{\sf Case~1:} $r(a_{p_{n-1}})=r(a_{p_{n-1}^{\rm d}})=r(a)$. As was shown above, the inclusions
(\ref{23})~and~(\ref{24}) hold. For $\lambda\in\rho(a;A)\setminus\{0\}$, we have
the~identity~\cite{Al2}
\begin{equation}\label{25}
p_{n-1}R(\lambda,a)p_{n-1}^{\rm d}= R(\lambda,a_{p_{n - 1}})p_{n-1}ap_{n-1}^{\rm
d}R(\lambda,a_{p_{n-1}^{\rm d}})
\end{equation}
which implies
$$p_{n - 1}ap_{n-1}^{\rm d}=(a_{p_{n - 1}})_{-1}p_{n-1}ap_{n-1}^{\rm d}(a_{p_{n-1}^{\rm d}})_0+
(a_{p_{n-1}})_0p_{n-1}ap_{n-1}^{\rm d}(a_{p_{n-1}^{\rm d}})_{-1}\in{\cal F}(A)$$ as ${\cal F}(A)$
is an~algebraic ideal. Using the~$a_{-1}$-invariance of $p_{n-1}$, we obtain $a_{-1}\in{\cal
F}(A)$.
{\sf Case~2:} $r(a_{p_{n-1}})=r(a)$ and $r(a_{p_{n-1}^{\rm d}})<r(a)$. Then (\ref{23}) holds.
Moreover, we have~\cite{Al2} $p_{n-1}^{\rm d}a_{-1}=0$. Hence, taking into account~(\ref{25}), we
conclude $a_{-1}\in{\cal F}(A)$.
{\sf Case~3:} $r(a_{p_{n-1}})<r(a)$ and $r(a_{p_{n-1}^{\rm d}})=r(a)$. Then (\ref{24}) holds.
Moreover, we have~\cite{Al2} $a_{-1}p_{n-1}=0$. Hence, taking into account~(\ref{25}), we conclude
$a_{-1}\in{\cal F}(A)$.\hfill$\Box$
\smallskip
We mention the~following important consequence of the~preceding theorem which shows once more that
the~definition of an~$f$-pole is the~right and natural one to use (see~\cite{Al2}, where the~detail
discussion of this notion can be found).
\begin{cor}\label{cor24}\
Let $A$ be a~Dedekind complete and let $a\in A$ be a~spectrally order continuous element with
$r(a)>0$. If $r(a)$ is a~finite-rank pole of $R(\cdot,a)$ then the~residue $a_{-1}$ is
a~finite-rank element.
\end{cor}
{\it It is not known if the~point $r(a)$ is an~$f$-pole of $R(\cdot,a)$ of an~arbitrary irreducible
element $a$ of $A$ such that $r(a)$ is a~pole of $R(\cdot,a)$} (of course, under the~assumptions of
Theorem~\ref{thm6}). As can be shown (see~\cite{Al2}), the~affirmative answer to this question is
equivalent to: {\it $0\le b< a$ implies $r(b)<r(a)$}, which will be discussed in
Section~\ref{sec5}. In~particular, it is not known if the~converse to the~preceding corollary is
true.
\medskip
We now turn our attention to the~conditions of the~primitivity of irreducible element~$a$ in
an~ordered Banach algebra $A$. Recall that an~element $b$ of an~arbitrary Banach algebra $B$ is
called {\it primitive} if the~peripheral spectrum $\sigma_{\rm per}(b)$ contains at most one point;
all other elements of $B$ are called {\it imprimitive}.
We begin with the~next criteria of the~primitivity.
\begin{thm}\label{thm25}\
Let an~ordered Banach algebra $A$ satisfy Axioms {\bf (A$_1$)}-{\bf (A$_5$)} and~{\bf (A$_{\rm
LS}$)}. Let $a\in A$ be a~non-zero irreducible element such that $r(a)$ is a~pole of $R(\cdot,a)$.
The~following statements are equivalent:
\begin{description}
\item[(a)] The~element $a$ is primitive;
\item[(b)] The~element $a^m$ is primitive for all $m\in{\Bbb N}$;
\item[(c)] The~element $a^m$ is irreducible for all $m\in{\Bbb N}$;
\item[(d)] The~sequence $\{(\frac{a}{r(a)})^n\}$ converges to an~algebraically strictly positive
element.
\end{description}
\end{thm}
{\bf Proof.} The~implication {\bf (a)}~$\Longrightarrow$~{\bf (b)} follows at once from
the~identity $\sigma(a^m)=f(\sigma(a))$ with $f(z)=z^m$ and the~implication {\bf
(b)}~$\Longrightarrow$~{\bf (a)} is obvious.
{\bf (a)}~$\Longrightarrow$~{\bf (c)} If $a^m$ is reducible for some $m\in{\Bbb N}$ then $m>1$ and,
in view of Theorem~\ref{thm12}, the~points $r(a)e^{i\frac{2\pi}{m''}j}\in\sigma(a)$ for all
$j=0,1,\ldots,m''-1$ with $m''>1$, a~contradiction.
{\bf (c)}~$\Longrightarrow$~{\bf (a)} Proceeding by contradiction and using Axiom~{\bf (A$_{\rm
LS}$)} and Theorem~\ref{thm20}, we conclude the~validity of the~identity $\sigma_{\rm
per}(a)=r(a)H_k$ with $k>1$. Taking into account Theorem~\ref{thm12} once more, we obtain
the~reducibility $a^m$ of for some $m>1$, which is impossible.
{\bf (a)}~$\Longrightarrow$~{\bf (d)} We shall prove first the~next assertion: {\it If $b$ is
a~primitive positive element of an~arbitrary ordered Banach algebra $A_0$ such that $r(b)>0$ is
a~simple pole of $R(\cdot,b)$ then $(\frac{b}{r(b)})^n$ converges to the~residue $b_{-1}$ of
$R(\cdot,b)$ at $r(b)$}. Indeed, we define the~element~$c$ by $c=b-r(b)b_{-1}$. The~Spectral
Mapping Theorem yields $\sigma(c)=(\sigma(c)\cup\{0\})\setminus\{r(b)\}$. Taking into account
the~primitivity of $b$, we infer $r(c)<r(b)$ or $r(\frac{c}{r(b)})<1$. By the~Gelfand formula,
the~equality $r(\frac{c}{r(b)})=\lim\limits_{n\to\infty}\frac{\|c^n\|_{A_0}}{r(b)^n}$ holds and,
in~particular, $(\frac{c}{r(b)})^n\to0$ as $n\to\infty$. Using the~identities
$bb_{-1}=b_{-1}b=r(b)b_{-1}$, we obtain the~relations $b_{-1}c=cb_{-1}=0$. Therefore,
$b^n=c^n+r(b)^nb_{-1}$ for all $n$. Consequently, the~relations
$(\frac{b}{r(b)})^n=(\frac{c}{r(b)})^n+b_{-1}\to b_{-1}$ hold, as~required. Now it only remains to
remember Theorem~\ref{thm6}{\bf (b)},{\bf (c)}. According to it, $r(a)$ is a~simple pole of
$R(\cdot,A)$ and $a_{-1}\gg0$ as $a$ is irreducible.
{\bf (d)}~$\Longrightarrow$~{\bf (c)} If $a^m$ is reducible for some number $m>1$ then $p^{\rm
d}a^mp=0$ for some non-trivial $p\in{\bf OI}(A)$. Evidently, $p^{\rm d}a^{mn}p=0$ for all
$n\in{\Bbb N}$. Thus, $0=p^{\rm d}(\frac{a}{r(a)})^{mn}p\to p^{\rm d}a_{-1}p$ as $n\to\infty$ and,
hence, $p^{\rm d}a_{-1}p=0$. The~latter contradicts to the~algebraic strict positivity of
$a_{-1}$.\hfill$\Box$
\smallskip
A~non-zero order idempotent $p$ of $A$ is called {\it order minimal} if the~equality
$\sum\limits_{j=1}^np_j={\bf e}$, where $p_j\in{\bf OI}(A)$ for $j=1,\ldots,n$ and
$p_{j'}p_{j''}=0$ for $j'\neq j''$, implies the~existence of an~index $j_0$ satisfying $p\le
p_{j_0}$.
\begin{cor}\
Under the~assumptions of Theorem~{\rm \ref{thm25}}, each of the~following conditions guarantees
the~primitivity of the~element $a$:
\begin{description}
\item[(a)] The~element $a\gg0$;
\item[(b)] The~element $a_p>0$ for some order minimal $p\in{\bf OI}(A)$.
\end{description}
\end{cor}
{\bf Proof.} {\bf (a)} We shall show the~irreducibility of $a^m$ for all $m\in{\Bbb N}$. In view of
part~{\bf (a)} of the~preceding theorem, the~latter implies the~desired assertion. To this end, we
assume $q^{\rm d}a^mq=0$ for some $m>1$ and $q\in{\bf OI}(A)$. Since $A$ has a~disjunctive product,
there exists an~order idempotent $q_1$ satisfying $q^{\rm d}aq_1=q_1^{\rm d}a^{m-1}q=0$. We can
suppose $q\neq{\bf e}$. Then $q_1=0$ and so $a^{m-1}q=0$. If $m>2$ then there exists an~order
idempotent $q_2$ satisfying $aq_2=q_2^{\rm d}a^{m-2}q=0$ and so $a^{m-2}q=0$. Finally, using
a~reverse finite induction, we obtain $aq=0$ or $q=0$, as required.
{\bf (b)} Proceeding by contradiction and taking into account Theorems \ref{thm25} and~\ref{thm12},
we find $p_1,\ldots,p_m\in{\bf OI}(A)$ with $m>1$ determining the~cyclic form of the~element~$a$
and, in~particular, $p_ja=ap_{j+1}$ for all $j=1,\ldots,m$. In view of the~order minimality of $p$,
we choose an~index~$j_0$ satisfying $p\le p_{j_0}$. Then the~equalities $pa=pp_{j_0}a=pap_{j_0+1}$
hold and, hence, $a_p=pap_{j_0+1}p=0$, a~contradiction.\hfill$\Box$
\smallskip
If $b$ is an~element of a~Banach algebra $B$ with a~unit ${\bf u}$ such that $r(b)\in\sigma(b)$
then $b+\lambda{\bf u}$ is primitive for all numbers $\lambda>0$. The~next result makes more
precisely this fact.
\begin{cor}\
Let $A$ satisfy Axioms {\bf (A$_1$)}-{\bf (A$_5$)} and let $a,b\in A^+$. If $a+b$ is irreducible,
$b_p>0$ for all $0<p\in{\bf OI}(A)$, and $r(a+b)$ is a~pole of $R(\cdot,a+b)$ then $a+b$ is
primitive.
\end{cor}
{\bf Proof.} Again, proceeding by contradiction and taking into account Theorems \ref{thm25}
and~\ref{thm12}, we find $p_1,\ldots,p_m\in{\bf OI}(A)$ with $m>1$ determining the~cyclic form
of~$a+b$ and, in~particular, $p_j(a+b)=(a+b)p_{j+1}$ for all $j=1,\ldots,m$. Therefore,
$0=p_j(a+b)p_j=a_{p_j}+b_{p_j}>0$, a~contradiction.\hfill$\Box$
\smallskip
An~element $a$ of $A$ is said to be {\it symmetric} whenever $paq=qap$ for all $p,q\in{\bf OI}(A)$.
\begin{cor}\
Suppose that all assumptions of Theorem~{\rm \ref{thm25}} are satisfied and, in~addi\-tional,
the~element $a$ is symmetric. Then the~element $a$ is primitive if and only if $a+r(a){\bf e}$ is
invertible.
\end{cor}
{\bf Proof.} The~necessity is obvious. We shall prove the~sufficiency. Proceeding by
contra\-diction, we find $p_1,\ldots,p_m\in{\bf OI}(A)$ with $m>1$ determining the~cyclic form
of~$a$. In~view of the~validity of the~implication {\bf (b)}~$\Longrightarrow$~{\bf (c)} of
Theorem~\ref{thm12} and our condition, the~natural number $m$ is odd. Then $p_1ap_2=ap_2>0$. Since
$a$ is symmetric, $p_2ap_1>0$. On the~other hand, $p_2ap_1=ap_3p_1=0$, a~contradiction.\hfill$\Box$
\smallskip
As~follows from Theorem~\ref{thm25}, if an~element $a$ of $A$ is irreducible and primitive then
$a^m$ is irreducible for all $m\in{\Bbb N}$. Unfortunately, the~relation $a^m\gg0$ need not hold
for any $m$ in the~case of an~integral operator even (see~\cite{Al}). Nevertheless, we have
the~next result.
\begin{cor}\
Suppose that all assumptions of Theorem~{\rm \ref{thm25}} are satisfied and, in~addi\-tional,
the~element $a$ is primitive. If $a^{m_0}\gg0$ for some $m_0\in{\Bbb N}$ then $a^m\gg0$ for all
natural $m\ge m_0$.
\end{cor}
{\bf Proof.} Proceeding by contra\-diction, we find $m>m_0$ and non-zero $p,q\in{\bf OI}(A)$
satisfying $pa^mq=0$. Obviously, $pa^{m_0}a^{m-m_0}q=0$. Since the~algebra $A$ has a~disjunctive
product, we have $pa^{m_0}p_1=p_1^{\rm d}a^{m-m_0}q=0$ for some $p_1\in{\bf OI}(A)$. According to
our condition, $p_1=0$ and so $a^{m-m_0}q=0$. In view of Theorem~\ref{thm25}, the~element
$a^{m-m_0}$ is irreducible and, hence, $q=0$, a~contradiction.\hfill$\Box$
\smallskip
For the~case of an~arbitrary (not necessarily irreducible) element $a\in A^+$, we have the~next.
\begin{prop}\
Let $A$ be an~ordered Banach algebra. Let $a\in A$ be a~positive element such that $r(a)$ is a~pole
of $R(\cdot,A)$ and every point $\alpha\in\sigma_{\rm per}(a)$ is an~eigenvalue of either
the~operator $L_a$ or the~operator $R_a$ on $A$. The~following statements hold:
\begin{description}
\item[(a)] The~sequence $a^n\to0$ as $n\to\infty$ if and only if $r(a)<1$;
\item[(b)] If $r(a)=1$ then the~sequence $\{a^n\}$ is convergent if and only if $r(a)$ is a~simple
pole of $R(\cdot,a)$ and the~element $a$ is primitive;
\item[(c)] If $r(a)>1$ then the~sequence $\{a^n\}$ is not convergent.
\end{description}
\end{prop}
{\bf Proof.} {\bf (b)} Let $r(a)=1$ and let $\{a^n\}$ be convergent. Then $\|a^n\|_A\le c$ for all
$n$ and some constant $c\in{\Bbb R}^+$. For all $\lambda\in{\Bbb R}$, $\lambda>1$, we have
$$(\lambda-1)^2\|R(\lambda,a)\|_A=
(\lambda-1)^2\Big\|\sum\limits_{j=0}^\infty\frac1{\lambda^{j+1}}a^j\Big\|_A\le$$
$$(\lambda-1)^2\sum\limits_{j=0}^\infty\frac1{\lambda^{j+1}}\|a^j\|_A\le
C(\lambda-1)^2\sum\limits_{j=0}^\infty\frac1{\lambda^{j+1}}=
C(\lambda-1)^2\frac\lambda{\lambda-1}=C\lambda(\lambda-1)\to0$$ as $\lambda\downarrow1$. Thus,
$r(a)$ is a~simple pole of $R(\cdot,a)$. Now we consider $\alpha\in\sigma_{\rm per}(a)$. In view of
our condition, $\alpha$ is an~eigenvalue of $L_a$ (the~case of $R_a$ is analogous), i.e.,
$ab=\alpha b$ for some non-zero $b\in A$. Obviously, $a^nb=\alpha^nb$ for all $n\in{\Bbb N}$ and
the~sequence $\{\alpha^nb\}$ converges. Therefore, $\{\alpha^n\}$ converges. The~latter is possible
for the~case of $\alpha=1$ only and, hence, $a$ is primitive. The~converse assertion, namely,
the~relation $a^n\to a_{-1}$ as $n\to\infty$, was shown in the~proof of the~implication {\bf
(a)}~$\Longrightarrow$~{\bf (d)} of Theorem~\ref{thm25}.
{\bf (a)} The~sufficiency is clear. We shall check the~necessity. Assume that $a^n\to0$ as
$n\to\infty$. This implies $r(a)\le1$. If $r(a)=1$ then, as was mentioned above, $a^n\to a_{-1}$
and, hence, $a_{-1}=0$, a~contradiction.
{\bf (c)} If $r(a)=1$ then $\{a^n\}$ is unbounded and, in~particular, is not
convergent.\hfill$\Box$
\section{Other viewpoints on the~irreducibility and\newline the~primitivity}
Recall that an~element $a\in A^+$ is said to be irreducible whenever the~equality $p^{\rm d}ap=0$,
where $p\in{\bf OI}(A)$, implies $p=0$ or $p={\bf e}$. Irreducible elements were introduced
in~\cite{BGr} for the~case of Banach lattice algebras and in~\cite{Al2} for the~case of ordered
Banach algebras. Under the~natural assumptions, these elements have nice spectral properties and,
under such a~notion of irreducibility, the~theorem about the~Frobenius normal form holds.
Nevertheless, the~purpose of this section is to establish some results which allow us in a~new
fashion to take a~glance at the~algebraic nature of such notions as the~irreducibility and
the~primitivity and, thus, to break some ground for further research of these notions in an~ordered
Banach algebras.
\medskip
Let $E$ be a~Dedekind complete Banach lattice. As usual, we denote the~set of all opera\-tors on
$E$ of the~form $\sum\limits_{j=1}^nf_j\otimes x_j$, where $f_j\in E_{\rm n}^\sim$, $x_j\in E$, and
$(f_j\otimes x_j)x=(f_j(x))x_j$ for all $j=1,\ldots,n$ and $x\in E$, by $E_{\rm n}^\sim\otimes E$.
The~band $(E_{\rm n}^\sim\otimes E)^{\rm dd}$ generated by $E_{\rm n}^\sim\otimes E$ in the~Banach
lattice $L_{\rm r}(E)$ with the~$r$-norm $\|T\|_r=\|T\|_{B(E)}$ is called \mbox{\cite[p.
193]{AbrAl}} the~{\it band of abstract integral operators}. As is easy to see, $(E_{\rm
n}^\sim\otimes E)^{\rm dd}\subseteq L_{\rm n}(E)$ and if $T\in L_{\rm r}(E)$ and $S\in(E_{\rm
n}^\sim\otimes E)^{\rm dd}$ then $TS,ST\in(E_{\rm n}^\sim\otimes E)^{\rm dd}$. In~particular, under
the~$r$-norm, $(E_{\rm n}^\sim\otimes E)^{\rm dd}$ is an~ordered Banach algebra (possibly, without
a~unit). If $E$~is a~function space and $E_{\rm n}^\sim$ separates points of $E$ then, by
the~Lozanovsky theorem \mbox{\cite[p. 199]{AbrAl}}, $(E_{\rm n}^\sim\otimes E)^{\rm dd}$ coincides
with the~band of regular integral operators on $E$. Next, as can be shown (see~\cite{Sch2}), if
an~arbitrary Banach lattice $E$ possesses the~non-trivial band $E_{\rm c}^\sim$ of all
$\sigma$-order continuous functionals, i.e., $E_{\rm c}^\sim\neq\{\emptyset\}$, and admitts
a~$\sigma$-order continuous irreducible operator $T$ then $E_{\rm n}^\sim=E_{\rm c}^\sim$ and
$E_{\rm n}^\sim$ separates the~points of $E$. We also recall that in a~Banach algebra $B$ with or
without a~unit the~{\it wedge operator}~$W_b$ on $B$ \mbox{\cite[pp. 17, 70]{BMSmW}}, where $b\in
B$, is defined by $W_b=bab$ for $a\in B$. As is easy to see, $r(W_b)\le r(b)^2$ (we put
$r(b)=\lim\limits_{n\to\infty}\|b^n\|_B^{\frac1n}$ if $B$ does not have a~unit).
\begin{thm}\label{thm31}\
Let $E$ be a~Dedekind complete Banach lattice such that $E_{\rm n}^\sim$ separates points of $E$.
Let $T$ be a~non-zero positive order continuous operator on $E$ such that $r(T)$ is a~pole of
$R(\cdot,T)$. If the~restriction of the~wedge operator $W_T$ to $(E_{\rm n}^\sim\otimes E)^{\rm
dd}$ is an~irreducible operator then two operators $T$ and $T'$ are also irreducible and, moreover,
primitive, where $T'$ is the~restriction of the~adjoint operator $T^*$ to $E_{\rm n}^\sim$.
\end{thm}
{\bf Proof.} Consider \mbox{a~$T$-invariant} band $B\neq\{0\}$. Fix $\lambda>r(T)$. Using
\mbox{the~$R(\lambda,T)$-invariance} of this band and the~inequality $TR(\lambda,T)\le\lambda
R(\lambda,T)$, we find a~non-zero element $z_0\in B^+$ satisfying $Tz_0\le\lambda z_0$. For
an~arbitrary non-zero functional $h_0\in(E^*)^+$ such that $T^*h_0\le\lambda h_0$, we have
$W_T(h_0\otimes z_0)\le\lambda^2h_0\otimes z_0$. As is easy to see, the~operator $W_T$ is order
continuous on~$(E_{\rm n}^\sim\otimes E)^{\rm dd}$. Therefore, $h_0\otimes z_0$ is a~weak order
unit in $(E_{\rm n}^\sim\otimes E)^{\rm dd}$. If $z\in E^+$ and $z_0\perp z$ then $(h\otimes
z_0)\wedge(h\otimes z)=h\otimes(z_0\wedge z)=0$, whence $h\otimes z=0$ or $z=0$. Thus, $z_0$ is
a~weak order unit. Consequently, $B=E$ and the~irreducibility of $T$ has been proved.
In~particular, $T'$~is (see~\cite{Al}, the~proof of Theorem~1) also irreducible. Moreover, there
exist a~weak order unit $x_0\in E$ and a~strictly positive functional $f\in E_{\rm n}^\sim$
satisfying $Tx_0=r(T)x_0$ and $T^*f_0=r(T)f_0$.
Now let us verify the~primitivity of the~operator $T$. In view of the~inclusion \mbox{\cite[p.
256]{AbrAl}} $\rho_\infty(T^*;B(E^*))\subseteq\rho(T';B(E_{\rm n}^\sim))$, the~latter implies
the~primitivity of $T'$. Proceeding by contradiction, we find (\cite{Al}; see also part~{\bf (b)}
of Theorem~\ref{thm12}) elements $y_1,\ldots,y_m$, where $m>1$, satisfying
$\sum\limits_{j=1}^my_j=x_0$, $y_{j'}\wedge y_{j''}=0$ for $j'\neq j''$, and $Ty_{j+1}=r(T)y_j$ for
$j=1,\ldots,m$. Define the~functionals $g_j\in E^*$ via the~formula $g_j=P_{y_j}^*f_0$, where
$P_{y_j}$ is the~order projection from $E$ onto the~projection band $B_{y_j}$ generated by $y_j$.
Obviously, $\sum\limits_{j=1}^mg_j=f_0$, $g_{j'}\perp g_{j''}$ for $j'\neq j''$, and $g_j\in E_{\rm
n}^\sim$. Taking into account the~identities $T^*g_j=T^*P_{y_j}^*f_0=P_{y_{j+1}}^*T^*f_0$, we have
the~relation $T^*g_{j'}\perp T^*g_{j''}$ for $j'\neq j''$. Moreover, $(T^*g_j)y_k=g_j(y_{k-1})=0$
for $k\neq j+1$, whence $(T^*g_j)(x_0-y_{j+1})=g_ky_{j+1}=0$ and so $(T^*g_j\wedge g_k)x_0=0$.
Since $x_0$ is a~weak order unit, the~last equality implies $T^*g_j\perp g_k$ for $k\neq j+1$.
Consequently,
$$T^*g_j-r(T)g_{j+1}\perp r(T)\sum\limits_{\stackrel{n=1}{n\neq j+1}}^mg_n-
\sum\limits_{\stackrel{n=1}{n\neq j}}^mT^*g_n.$$ On the~other hand,
$\sum\limits_{n=1}^mT^*g_n=r(T)\sum\limits_{n=1}^mg_n$ or
$$T^*g_j-r(T)g_{j+1}=r(T)\sum\limits_{\stackrel{n=1}{n\neq j+1}}^mg_n-
\sum\limits_{\stackrel{n=1}{n\neq j}}^mT^*g_n,$$ whence $T^*g_j=r(T)g_{j+1}$ for all
$j=1,\ldots,m$. Put $f_j=g_{m-j+1}$. Evidently,
$$T^*f_{j+1}=T^*g_{m-j}=r(T)g_{m-j+1}=r(T)f_j.$$
Thus, $W_T\sum\limits_{n=1}^mf_n\otimes y_n=r(T)^2\sum\limits_{n=1}^mf_n\otimes y_n$. Therefore,
$\sum\limits_{n=1}^mf_n\otimes y_n$ is a~weak order unit in the~band~$(E_{\rm n}^\sim\otimes
E)^{\rm dd}$. This contradicts to the~relation $f_j\otimes y_k\perp\sum\limits_{n=1}^mf_n\otimes
y_n$ for all $j\neq k$.\hfill$\Box$
\smallskip
It is not known if the~converse to the~assertion of the~preceding theorem holds. Nevertheless, as
the~next result shows, we can assert the~converse in the~case of the~Banach algebra $M_n({\Bbb C})$
of all $n\times n$ matrices with complex entries and the~natural multiplication and order. We
mention first that a~matrix $A\in M_n({\Bbb C})$ is irreducible if and only if the~transpose $A^t$
of $A$ is irreducible.
\begin{thm}\
Let $T$ be an~$n\times n$~positive matrix. Then $T$ is irreducible and primitive if and only if
the~wedge operator $W_T$ on $M_n({\Bbb C})$ is irreducible.
\end{thm}
{\bf Proof.} In view of Theorem~\ref{thm31}, it is enough to verify the~necessity. Clearly, we can
assume~$n>1$. Let the~inequality $W_TS\le\lambda S$ hold for a~number $\lambda\ge0$ and a~non-zero
positive matrix $S\in M_n({\Bbb C})$. Since the~matrix $T$ is primitive, $T^k$ is strongly positive
for some $k\in{\Bbb N}$ (see Theorem~\ref{thm25}), i.e., all its entries are strictly positive.
Therefore, $\lambda^kS\ge W_T^kS=T^kST^k$ and so $S$ is also strongly positive. Finally, $W_T$ is
irreducible.\hfill$\Box$
\medskip
The~following result suggests another approach to the~notion of irreducibility in ordered Banach
algebras.
\begin{thm}\
Let $T$ be an~$n\times n$~matrix. Then $T$ is irreducible if and only if the~operator $L_T+R_T$ on
$M_n({\Bbb C})$ is irreducible.
\end{thm}
{\bf Proof.} Necessity. We recall first that (see the~equalities~(\ref{7})) $(L_T+R_T)Q=TQ+QT$,
where $Q\in M_n({\Bbb C})$. Let the~inequality $L_TS\le\lambda S$ holds for a~number $\lambda\ge0$
and a~non-zero positive matrix $S\in M_n({\Bbb C})$, $S=[s_{ij}]$. There exist indexes
$i_0,j_0=1,\ldots,n$ such that the~entry $s_{i_0j_0}>0$. If $s^{j_0}$ is the~$j_0^{\rm th}$ column
of $S$ then, in view of the~inequality $TS\le\lambda S$, we have $Ts^{j_0}\le\lambda s^{j_0}$.
Taking into account the~irreducibility of $T$, we obtain $s_{ij_0}>0$ for all $i=1,\ldots,n$. On
the~other hand, the~inequality $S^tT^t\le\lambda S^t$ holds, $T^t$~is irreducible, and
$s_{j_0i}^{t_0}>0$ for all $i$, where $S^t=[s_{ij}^t]$. As was shown above, $s_{ji}^t>0$ or
$s_{ij}>0$ for all $i,j$, and we are done.
Sufficiency. Let the~operator $L_T+R_T$ be irreducible. In~particular, $L_T+R_T\ge0$. Then
$0\le(L_T+R_T)I=2T$ or $T\ge0$. Assume that $T$ is reducible. Then for some $k=1,\ldots,n-1$ there
exist indexes $j_1,\ldots,j_k$ satisfying $t_{ij}=0$ for all $i\notin J$ and $j\in J$, where
$J=\{j_1,\ldots,j_k\}$. Consider the~band $B$ in $M_n({\Bbb C})$ defined by
$$B=\{S\in M_n({\Bbb C}):s_{ij}=0 \ \text{for all} \ i\notin J \ \text{and all} \ j\}.$$
Let $S\in B$. If $TS=[(ts)_{ij}]$ then for $i\notin J$, we have
$$(ts)_{ij}=\sum\limits_{m=1}^nt_{im}s_{mj}=
\sum\limits_{m\notin J}t_{im}s_{mj}+\sum\limits_{m\in J}t_{im}s_{mj}=0.$$ If $ST=[(st)_{ij}]$ then
$(st)_{ij}=0$ for $i\notin J$. Thus, the~band $B$ is \mbox{$L_T+R_T$-invariant}, which is
impossible.\hfill$\Box$
\section{When does $0\le b< a$ imply $r(b)<r(a)$?}\label{sec5}
Let $T$ be an~order continuous irreducible operator on a~Banach lattice $E$, let $r(T)$~be a~pole
of $R(\cdot,T)$ of order $k$, and let the~coefficient $T_{-k}$ of the~Laurent series expansion
of~$R(\cdot,T)$ around~$r(T)$ also be order continuous (the~latter holds if, e.g., the~Lorenz
seminorm~$\|\cdot\|_L$ on~$E$ is~a~norm). Then \cite{Al2} the~operator inequalities $0\le S< T$,
where $S,T\in B(E)$, imply the~spectral radius inequality $r(S)<r(T)$. In~particular, if $S,T\in
M_n({\Bbb C})$, $T$ is irreducible, and $0\le S< T$ then $r(S)<r(T)$. As was mentioned above (see
remarks after Corollary~\ref{cor24}), the~analogous question, i.e., the~validity of the~inequality
$r(b)<r(a)$ where $0\le b< a$ and the~element $a$ is irreducible, remains open in the~case of
an~ordered Banach algebra. The~purpose of the~present section is to discuss a~number of additional
conditions under which the~answer to this question is affirmative. We mention at once that in
research of this problem the~assumption that $r(a)$ is a~pole of $R(\cdot,a)$ is natural absolutely
and cannot be even reject in the~case of operators (see~\mbox{\cite{Al, Al4}}).
\begin{thm}\label{thm34}\
Let $A$ be a~finite-dimensional ordered Banach algebra with a~disjunctive product and let $a,b\in
A$. If the~element $a$ is irreducible then $0\le b<a$ implies $r(b)<r(a)$.
\end{thm}
{\bf Proof.} We claim that all assumptions of Theorem~\ref{thm6} hold. Indeed, if elements
$q_j\in{\bf OI}(A)$ with $j=1,\ldots,k$ and $q_{j'}q_{j''}=0$ for $j'\neq j''$ then
$q_1,\ldots,q_k$ are linearly independent. Therefore, there exists a~maximal collection of pairwise
disjoint elements $\{p_1,\ldots,p_n\}$ in ${\bf OI}(A)$. Obviously, $\sum\limits_{j=1}^np_j={\bf
e}$. We have the~identity
\begin{equation}\label{26}
{\bf OI}(A)=\Big\{\sum\limits_{j\in J}p_j:J\subseteq\{1,\ldots,n\}\Big\}.
\end{equation}
Actually, if an~order idempotent $p$ of $A$ satisfies $0<pp_j<p_j$ for some $j=1,\ldots,n$ then
$0<pp_j<p_j$ and the~collection $\{p_1,\ldots,p_{j-1},pp_j,p^{\rm d}p_j,p_{j+1},\ldots,p_n\}$
consists of pairwise disjoint elements. The~latter contradicts to the~maximality of
$\{p_1,\ldots,p_n\}$ and, hence, either $pp_j=0$ or $pp_j=p_j$. Putting $J=\{j:pp_j>0\}$, we obtain
$p=p\sum\limits_{j=1}^np_j=\sum\limits_{j\in J}pp_j=\sum\limits_{j\in J}p_j$, and (\ref{26})~has
been checked. In~particular, the~Boolean algebra ${\bf OI}(A)$ is Dedekind complete. Consider a~net
$\{b_\alpha\}$ satisfying $b_\alpha\downarrow0$ in ${\bf OI}(A)$. As~was shown above,
$b_\alpha=\sum\limits_{j\in J_\alpha}p_j$ for every $\alpha$, where
$J_\alpha\subseteq\{1,\ldots,n\}$. Evidently, if $b_{\alpha'}<b_{\alpha''}$ then
$J_{\alpha'}\subsetneq J_{\alpha''}$. From the~latter, we conclude easily the~existence of
an~index~$\alpha_0$ such that $b_{\alpha_0}=0$. Thus, the~equality $A^+=A_{\rm n}$ holds. On
the~other hand, as is well known, the~spectrum $\sigma(x;B)$ of every element $x$ of an~arbitrary
finite-dimensional Banach algebra $B$ with a~unit is finite and consists of poles of
the~resolvent~$R(\cdot,x)$. Now it only remains to use part~{\bf (d)} of
Theorem~\ref{thm6}.\hfill$\Box$
As the~next example shows, the~preceding theorem is not valid without the~assump\-tion about
a~disjunctive product.
\begin{exm}\
{\rm Consider the~ordered Banach algebra $A_0={\Bbb C}^n$, where $n\in{\Bbb N}$ and $n\ge2$, under
the~natural algebraic operations, multiplication, and order and under some algebra-norm. Then
the~ordered Banach algebra $A$ obtained from $A_0$ by adjoining a~unit gives the~required example.
Another example is the~ordered Banach algebra $A={\Bbb C}^2$ under the~natural algebraic operations
and order, the~multiplication given by $(x_1,x_2)(y_1,y_2)=(x_1y_1,x_1y_2+x_2y_1)$, and the~norm
$\|(x_1,x_2)\|_A=|x_1|+|x_2|$. Indeed, as is easy to see, the~element ${\bf e}=(1,0)$ is a~unit of
$A$, ${\bf OI}(A)=\{0,{\bf e}\}$, and the~spectrum $\sigma(x;A)=\{x_1\}$, where $x=(x_1,x_2)$. Then
every element of the~algebra $A$ is irreducible, $0\le(1,0)<(1,1)$, and $r((1,0))=r((1,1))=1$.
The~situation does not change in the~case of the~algebra of the~form $B(E)$. Indeed, let $H$~be
an~arbitrary real Hilbert space and let $z\in H$ with $\|z\|_H=1$. Under the~order generated by
the~ice cream cone $K=\{y\in H:\langle y,z\rangle\ge\frac1{\sqrt{2}}\|y\|_H\}$, the~space $H$ is
a~real ordered Banach space. Since the~cone $K$ is generating, $B(H)$ is a~real ordered Banach
algebra and, hence, $B(H_{\Bbb C})$ is a~complex ordered Banach algebra, where $H_{\Bbb C}$ is
the~complexification of $H$. If $\dim{H}\ge3$ then \cite{Al3} the~center $(B(H_{\Bbb
C}))_I=\{\lambda I:\lambda\in{\Bbb R}\}$ and, in~particular, we have ${\bf OI}(B(H_{\Bbb
C}))=\{0,I\}$. There exists (see~\cite{Al3} once more) a~non-zero positive operator $T$ satisfying
$T^2=0$. Obviously, $I+T$ is an~irreducible element of $B(H_{\Bbb C})$, $0\le I<I+T$, and
$r(I)=r(I+T)=1$. On the~other hand, if $E$ is a~two-dimensional ordered Banach space with
generating cone $E^+$ then, as is well known, $E^+$~is a~lattice cone. From the~latter follows
easily that the~ordered Banach algebra $B(E)$ has a~disjunctive product and, hence,
Theorem~\ref{thm34} can be applied in this case.}\hfill$\Box$
\end{exm}
An~arbitrary Banach algebra $B$ with a~unit is said to be {\it Fredholm} if the~following three
conditions hold:
\begin{description}
\item[(a)] The~open subset $\Phi(B)$ of $B$ is determined and $\Phi(B)=-\Phi(B)$.
Elements of~$\Phi(B)$ are called {\it Fredholm};
\item[(b)] Two functions ${\rm nul}, {\rm def}:B\to{\Bbb N}\cup\{0,\pm\infty\}$ are determined
such that ${\rm nul}\! \ b={\rm nul}\! \ (-b)$ and ${\rm def}\! \ b={\rm def}\! \ (-b)$ for all
$b\in B$ and the~set ${\rm Inv}\! \ B$ of invertible elements of $B$ satisfies ${\rm Inv}\! \
B=\{b\in B:{\rm nul}\! \ b={\rm def}\! \ b=0\}$;
\item[(c)] The~{\it punctured neighbourhood property} holds: if $b\in\Phi(B)$ then there exists
a~number $\epsilon>0$ such that ${\rm nul}\! \ (\lambda-b)$ and ${\rm def}\! \ (\lambda-b)$ are
constants on the~set $\{\lambda\in{\Bbb C}:0<|\lambda|<\epsilon\}$.
\end{description}
As is well known, the~Banach algebra $B(Z)$, where $Z$ is an~arbitrary Banach space, is Fredholm
(see~\mbox{\cite[Section~4.4]{AbrAl}}). In this case, $\Phi(B(Z))$ coincides with the~class of all
Fredholm operators on $Z$ and ${\rm nul}\! \ T=\dim{N(T)}$ and ${\rm def}\! \ T={\rm codim}\! \
R(T)$ for all $T\in\Phi(B(Z))$. Moreover, every Banach algebra $B$ with a~unit ${\bf u}$ is
Fredholm (see~\mbox{\cite[Sections F2 anf F3]{BMSmW}}), i.e., the~set $\Phi(B)$ and two functions
${\rm nul}$ and ${\rm def}$ on $B$ satisfying the~required properties can be defined. Nevertheless,
in~\cite{BMSmW} for the~deter\-mi\-nation of these objects the~notion of {\it inessential ideal}
$J$ was used. That is, $J$ is an~algebraic ideal and zero is the~only possible accumulation point
of $\sigma(b;B)$ for each $b\in J$. In this case, the~set $\Phi(B)$ is defined by
\begin{equation}\label{27}
\Phi(B)=\{b\in B:{\bf u}-ab,{\bf u}-ba\in J \ \text{for some} \ a\in B\}.
\end{equation}
In the~definition of the~Fredholm algebra given above the~notion of inessential ideal is not
required. Moreover, in some cases, e.g., of the~algebra $C(K)$, where the~compact space $K$ does
not contain unisolated points, the~zero ideal is a~unique inessential ideal.
\begin{thm}\
Let an~ordered Banach algebra $A$ satisfy Axiom~{\bf (A$_1$)} and let the~cone $A^+$ be normal. Let
$a,b\in A$ be such that $0\le b<a$, the~element $a$ is irreducible, $r(a)$~is a~pole of
$R(\cdot,a)$. Then each of the~following conditions guarantees the~inequality $r(b)<r(a)$:
\begin{description}
\item[(a)] $A$ is a~Fredholm algebra and $r(a)-a\in\Phi(A)$;
\item[(b)] There exists an~inessential ideal $J$ of $A$ which contains the~residue $a_{-1}$;
\item[(c)] The~ideal ${\cal F}(A)$ of finite-rank elements is inessential.
\end{description}
\end{thm}
{\bf Proof.} {\bf (a)} Proceeding by contradiction, we assume $r(b)=r(a)$. For arbitrary
$\epsilon\in[0,1)$, we define the~element $a_\epsilon$ by $a_\epsilon=(1-\epsilon)a+\epsilon b$.
Obviously, $0\le b\le a_\epsilon\le a$. Therefore, in~view of the~normality of the~cone $A^+$,
$r(a_\epsilon)=r(a)$. Moreover, $r(a)-a_\epsilon\to r(a)-a$ as $\epsilon\to0$ and, hence,
$r(a)-a_\epsilon\in\Phi(A)$ for sufficiently small $\epsilon$. Fix such a~number $\epsilon$. Then
${\rm nul}\! \ (\lambda-a_\epsilon)$ and ${\rm def}\! \ (\lambda-a_\epsilon)$ are constants on
the~set $\{\lambda\in{\Bbb C}:0<|\lambda-r(a)|<\delta\}$ for some $\delta>0$. On the~other hand,
for $\lambda$ close to $r(a)$, we have the~inclusion $\lambda-a\in{\rm Inv}\! \ A$ or ${\rm nul}\!
\ (\lambda-a)={\rm def}\! \ (\lambda-a)=0$. Since the~element $\lambda-a_\epsilon$ is invertible
for $|\lambda|>r(a)$, the~identities ${\rm nul}\! \ (\lambda-a_\epsilon)={\rm def}\! \
(\lambda-a_\epsilon)=0$ hold for $|\lambda|>r(a)$ and so for all numbers $\lambda$ satisfying
$0<|\lambda-r(a)|<\delta$. Therefore, for such $\lambda$ the~element $\lambda-a_\epsilon$ is
invertible and, in~particular, $r(a)$ is an~isolated point of $\sigma(a_\epsilon)$. Using
the~normality cone $A^+$ once more, we conclude that $r(a)$ is a~simple pole of
$R(\cdot,a_\epsilon)$. The~residue $(a_\epsilon)_{-1}$ satisfies the~relations
$$
r(a)(a_\epsilon)_{-1}=a_\epsilon(a_\epsilon)_{-1}\le a(a_\epsilon)_{-1} \ \ \text{or} \ \
0\le(a-r(a))(a_\epsilon)_{-1}.
$$
On the~other hand, $a_{-1}(a-r(a))(a_\epsilon)_{-1}=0$. Taking into account the~relation
$a_{-1}\gg0$, we get
$$a(a_\epsilon)_{-1}=r(a)(a_\epsilon)_{-1}=a_\epsilon(a_\epsilon)_{-1}=
((1-\epsilon)a+\epsilon b)(a_\epsilon)_{-1}$$ and, hence, $(a-b)(a_\epsilon)_{-1}=0$. The~element
$a_\epsilon$ is also irreducible and so $(a_\epsilon)_{-1}\gg0$. Now the~last equality yields
$a=b$, a~contradiction.
{\bf (b)} As was mentioned above, $A$ is a~Fredholm algebra and the~set $\Phi(A)$ of Fredholm
elements can be defined via the~formula~(\ref{27}). The~coefficients $a_{-1}$ and $a_0$ of
the~Laurent series expansion of $R(\cdot,a)$ around $r(a)$ satisfies ${\bf e}-(r(a)-a)a_0={\bf
e}-a_0(r(a)-a)=a_{-1}$ and, hence, $r(a)-a\in\Phi(A)$. Now the~required assertion follows at once
from part~{\bf (a)}.
{\bf (c)} It suffices to observe the~inclusion $a_{-1}\in{\cal F}(A)$ and to use part~~{\bf
(b)}.\hfill$\Box$
\section{The~closedness of the~center}\label{sec6}
A~normed algebra with a~unit ${\bf e}$ and with a~(closed, convex) cone $A^+$ is called an~{\it
ordered normed algebra}~\cite{Al3} if ${\bf e}\ge0$ and the~inequalities $a,b\ge0$ imply $ab\ge0$.
The~{\it center}~\cite{Al3} of an ordered normed algebra $A$ is called the~order ideal $A_{\bf e}$
generated by ${\bf e}$ (see Example~\ref{exm2}{\bf (b)}), i.e.,
$$A_{\bf e}=\{a\in A:-\lambda{\bf e}\le a\le\lambda{\bf e} \ \text{for some} \
\lambda\in{\Bbb R}^+\}.$$ The~aim of this section is to prove the~closedness of the~center $A_{\bf
e}$ in $A$ (Theorem~\ref{thm38}).
As was shown in \cite{Al3}, if an~element $a\in A_{\bf e}$ then $a^2\ge0$. The~next result makes
more precisely this fact.
\begin{lem}\
In an~arbitrary ordered normed algebra $A$ the~following identity holds
$$A_{\bf e}=\{a\in A:(\lambda+a)^2\ge0 \ \text{for every} \ \lambda\in{\Bbb R}\}.$$
\end{lem}
{\bf Proof.} In view of the~remarks above, it suffices to show that if $(\lambda+a)^2\in A^+$ for
all real $\lambda$ then $a\in A_{\bf e}$. To this end, we consider the~Banach algebra $B$ being
a~completion of $A$ and $K=\overline{A^+}$, where the~closure was taken in $B$. Obviously, $K$ is
a~wedge and $K\cdot K\subseteq K$. The~relations $a^2\in A^+$ and $(\lambda^2-a^2)^2\in A^+$ with
$\lambda\in{\Bbb R}$ are valid. For $\lambda>r(a)$ the~element $(\lambda^2-a^2)^{-1}$ is well
defined and belongs to $K$. Whence $\lambda^2-a^2\in A^+\cdot K\subseteq K$. On the~other hand,
$\lambda^2-a^2\in A$. In view of the~closedness of~$A^+$ in $A$, $\lambda^2-a^2\in K\cap A=A^+$.
Therefore, $0\le_{A^+}a^2\le_{A^+}\le\lambda^2{\bf e}$ and, in~particular, $a^2\in A_{\bf e}$.
Next, using the~relation $({\bf e}\pm a)^2\in A^+$, we obtain $\pm2a\le_{A^+}{\bf e}+a^2$. Finally,
$a\in A_{\bf e}$.\hfill$\Box$
\begin{thm}\label{thm38}\
The~center $A_{\bf e}$ is a~closed subset of an~arbitrary ordered normed algebra~$A$.
\end{thm}
{\bf Proof.} Consider a~sequence $\{a_n\}$ in the~center $A_{\bf e}$ satisfying $a_n\to a$ in $A$
as $n\to\infty$. Then $(\lambda+a_n)^2\ge0$ for all $\lambda\in{\Bbb R}$ and, hence,
$(\lambda+a)^2\ge0$. In view of the~preceding lemma, the~element $a\in A_{\bf e}$.\hfill$\Box$
\smallskip
The~Minkowski norm $\|\cdot\|_{\bf e}$ can be defined on the~center $A_{\bf e}$ (see~(\ref{28})).
Under this norm, $A_{\bf e}$ is a~(real) ordered normed algebra. The~next result follows
immediately from the~equality $A_{\bf e}=\bigcup\limits_{n=1}^\infty n[-{\bf e},{\bf e}]$,
the~Baire category theorem, and the~preceding theorem.
\begin{cor}\
In an~ordered Banach algebra $A$ the~embedding $(A_{\bf e},\|\cdot\|_A)\to(A_{\bf e},\|\cdot\|_{\bf
e})$ is continuous. In~particular, there exists a~constant $c>0$ satisfying $\|a\|_{\bf e}\le
c\|a\|_A$ for all $a\in A_{\bf e}$.
\end{cor}
\vspace{1.0cm}
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In The Mouth Of The Lion by Frank Gates
An on the run, former South African Special Forces soldier is recruited by the CIA - Michael Memphis wanted to leave his old life behind him.
But the world needs men like him.
It was to be a simple mission. Meet with a top secret scientist in the employ of the hostile and secretive South African Apartheid regime. The scientist is a desperate man attempting to trade sensitive intelligence to the highest bidder. But within hours of landing both men are hunted down by the brutal South African security police, fearful a terrible state-secret might be exposed jeopardizing South Africa’s relations with their Western allies. Allies that consider the South African government absolutely vital in eliminating the Cold War threat of communism on the African continent.
A fact that threatens to overshadow the missing men, leaving their lives hanging in the balance.
That is, unless the two men can escape.
Though faced with extraordinarily few choices, Memphis has been carefully plotting their freedom since their capture - while also planning to deliver and expose an even bigger, far more dangerous evil to the Americans and the world.
High risk, high reward, “In The Mouth Of The Lion” is a nonstop, page-turning high-speed thriller that will have you right in the middle of the action…!
*Looking for familiar ground? You’re on it, if you enjoy reading David Baldacci, Dan Brown’s Robert Langdon, Lee Child’s Jack Reacher, Tom Clancy’s Jack Ryan, Nelson DeMille’s John Corey, Mark Dawson’s John Milton, Vince Flynn’s Mitch Rapp, Mark Greaney’s Gray Man, Gregg Hurwitz’s Orphan X, John Le Carre, Robert Ludlum’s Jason Bourne, John Sandford’s Lucas Davenport, Daniel Silva’s Gabriel Allon, Wayne Stinnett’s Jesse McDermitt, Brad Taylor’s Pike Logan, Brad Thor’s Scot Harvath and Tim Tigner’s Kyle Achilles.
| 20,404
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How to Save on Grilling This Summer
Summer is about to kick off with the upcoming three-day weekend, which means the start of grilling season. If you're planning a cookout or grilling outdoors, it can be easy to scramble last minute and spend too much money because you're not armed with the right tools. These tips will help you grill on the cheap and start your summer right:
Find the right grill for you. If you're in the market for a grill, there are a few things to consider before you make your purchase. There is a big debate on charcoal grills versus gas grills. One benefit of gas grills is they are far more economical long-term, especially if you grill quite frequently. Of course, there are other factors that may be important to you when deciding which grill to buy, including where it will be placed outside, how much you entertain and your preferred taste. Whichever grill you decide to buy, do some research in advance to find coupons and sales -- stores will periodically have sales on patio and outdoor equipment, even in the peak season. If you buy a grill online, use a cash back program to earn money back on your purchase.
Buy the right amount of fuel. Depending on what type of grill you own, you'll need either propane or charcoal for fuel. If you estimate how much fuel you'll need ahead of a cookout, and then fire up your grill for the right amount of time, you can avoid some unnecessary spending.
%VIRTUAL-article-sponsoredlinks%For a charcoal grill, you only need approximately three pounds of charcoal for four to six people. Most charcoal is ready after heating for about 15 minutes. Time your cooking to start right when your grill is ready, and then turn the grill off as soon as you are done. Charcoal is an expensive fuel and this can help reduce your consumption by quite a bit.
If you have a gas grill, look around before you decide where to purchase a propane tank and its future refills; don't settle for the first option, as you might be able to find it cheaper somewhere else. While propane is the less expensive fuel, still be considerate of when you turn your grill on and off to cook. Once started a gas grill will be ready to cook in about 10 minutes.
Choose your meat wisely. You have your grill and your fuel -- now it's time to choose your meat. It's tempting to select meats that are premium cuts and known to be delicious, but there are a few secrets that will cost you a lot less money and still leave you reigning as king or queen of the grill.
When cooking chicken, buying legs and thighs instead of breasts can save money and even add more flavor to your meal. For steak, consider buying "in bulk" and get larger cuts of meat at the butcher, then cut it up into smaller pieces on your own. When grilling fish, the "catch of the day" is always priciest, so look to other seafood options and don't be afraid to ask an expert what would be tastiest when cooked on the grill.
If you're looking to spice up a cut of meat that could use some more flavor, find a recipe online to make your own rub or marinade. Many can be made with ingredients you already have in the house.
Think ahead to future meals. When you're cooking out, whether it's a casual weekday night with the family or a full-fledged party, consider throwing on extra while the grill is still hot. Grilled meat works well in leftovers such as salads and sandwiches, and it's a smart way to make the most of your fuel while getting a few lunches and dinners ready for the week ahead. Just be sure not to make so much that you won't be able to eat it all in time.
Maintain and care for your grill. Proper grill maintenance is essential for keeping your grill around for many years to come. Clean it thoroughly before and after each grilling session, including disposing of charcoal ashes after they cooled, or changing the catch-pan liner in your gas grill. Rub down the grate with oil or cooking spray before and after each use, which will keep meat from sticking. Wipe up any spills with a damp paper towel when grill is cool, as grease and salt can accelerate corrosion.
Happy grilling!
Visit Kitchen Daily for more grilling tips and recipes.
Jon Lal is the founder of coupons and cash back website BeFrugal.com, which saves shoppers an average of $27 per order thanks to coupons plus an average of 7 percent cash back at more than 3,000 stores.
11 PHOTOS
10 Strange and Sneaky Supermarket Savings Strategies
How to Save on Grilling This Summer.
Most stores with bakeries bake more than customers will buy. One store near me always has a section of not-as-fresh breads and sweet items 50 percent off. At these prices, those are often more cost-effective than homemade..
Store usually just want to get rid of these unpopular items, and they may never been seen again. Sometimes, they are products discontinued by the manufacturer. They seem to be more frequent in the frozen food aisle, in my experience.
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\section{rodier type structure theorem of generalized principal series}
Recall that an irreducible supercuspidal representation $\tau$ of $M$ is called \underline{regular} in $G$ if the only element $w\in W_M$ such that $\tau^w\simeq \tau$ is the identity element. A parabolic induction $Ind^G_{P=MN}(\rho)$ is called a generalized principal series if the inducing data $\rho$ is a supercuspidal representation of the Levi subgroup $M$ of $P$. Furthermore, if our inducing data is a regular supercuspidal representation, we call the associated induced representation a regular generalized principal series. In this section, we will first recall Rodier's structure theorem of the constituents of regular principal series of split groups (see \cite[Theorem, Pg.418]{rodier1981decomposition}), then extend it to regular generalized principal series of arbitrary connected reductive group and its finite central covering group.
Recall that in \cite{rodier1981decomposition}, for a regular character $\chi$ of the torus $T$ of the Borel subgroup $B=TU$ of a connected split reductive group $G$, let $S$ be the set of coroots $\alpha^\vee$ such that $\chi_\alpha=|\cdot|$, and $-S:=\{-\alpha^\vee:~\alpha\in S \}$. Then
\begin{thm}[Rodier structure theorem] (see \cite[Theorem, Pg. 418]{rodier1981decomposition})\label{rps}
The constituents $\pi_\Gamma$ of the regular principal series $Ind^G_B(\chi)$ are parameterized by the connected components $\Gamma$ of $$\mathfrak{a}_T^\star-\bigcup_{\alpha^\vee\in S}Ker(\alpha^\vee)$$
satisfying the following property:
the Jacquet module $r_B(\pi_\Gamma)$ of $\pi_\Gamma$ with respect to $B$ is equal to
\[\bigoplus_{wC^+\subset \Gamma}\chi^w. \]
\end{thm}
Before turning to our Rodier type structure theorem for regular generalized principal series of arbitrary connected reductive groups, we first investigate the main ideas behind Rodier structure theorem which has not been pointed out explicitly in \cite{rodier1981decomposition}. Once those ideas are streamlined clearly, it would be readily to see how simple yet beautiful our Rodier type structure theorem is.
For regular principal series $Ind^G_B(\chi)$, we have that, as representations of $T$,
\[r_B(Ind^G_B(\chi))=r_B(Ind^G_B(\chi^w))=\bigoplus_{w'\in W_T}\chi^{w'}, \]
for any $w\in W_T$. Applying Frobenius reciprocity, we know that
$Ind^G_B(\chi^w)$ has a unique irreducible subrepresentation, and
$$Hom_G(Ind^G_B(\chi^w),~Ind^G_B(\chi^{w'}))\simeq \mathbb{C}$$
for any $w,~w'\in W_T$.
As the Jacquet module functor is exact (cf. \cite{bernstein1977induced,casselman1995introduction,waldspurger2003formule}), so for any
$\pi\in JH(Ind^G_B(\chi))$ and any $w\in W_T$, we know that $\pi$ is of multiplicity at most one in $Ind^G_B(\chi^w)$ and is uniquely determined by its Jacquet module $r_B(\pi)$ with respect to $B$. Moreover, for any $\chi^w\in r_B(\pi)$,
\[Ind^G_B(\chi^w)\simeq Ind^G_B(\chi^{w'})\mbox{ if and only if }\chi^{w'}\in r_B(\pi). \]
Therefore, the determination of the set $JH(Ind^G_B(\chi))$ of the constituents of $Ind^G_B(\chi)$ is equivalent to determining the orbits $\mathcal{O}$ of the set $\{\chi^w:~w\in W_T \}$ under the equivalent relation $\sim$: for $w,~w'\in W_T$,
\[\chi^w\sim \chi^{w'} \mbox{ if and only if } A(w,w')\mbox{ is an isomorphism,} \]
where $A(w,w')$ is the unique, up to scalar, non-zero $G$-equivalent homomorphism in $$Hom_G(Ind^G_B(\chi^w),~ Ind^G_B(\chi^{w'}))\simeq \mathbb{C}.$$
For simplicity, we will abbreviate those intertwining operators $A(w,w')$ as $A$ in what follows.
The next step is to give a characterization of those pairs $(w,w')\subset W_T$ satisfying
\[Ind^G_B(\chi^w)\simeq Ind^G_B(\chi^{w'}). \]
Let us first take a look at the simple basic case, i.e. the pairs $(w,w')$ with $w'=ww_\alpha$ for some simple root $\alpha$. That is to say
\[wC^+\mbox{ and }w'C^+\mbox{ share the same wall }Ker(w.\alpha^\vee). \]
For such a pair, via the induction by stage property of parabolic inductions and the uniqueness property of the intertwining operators $A$, we have also the induction by stage property of $A$, i.e. the following diagram commutes:
\[\xymatrix{Ind^G_{B}(\chi^{w})\ar[r]^A\ar@{=}[d]&Ind^G_{B}(\chi^{ww_{\alpha}})\ar@{=}[d]\\
Ind^G_{P_\alpha}\circ Ind^{M_\alpha }_{B\cap M_\alpha}(\chi^{w})\ar[r]^-{Ind(A)}&Ind^G_{P_\alpha}\circ Ind^{M_\alpha }_{B\cap M_\alpha}(\chi^{ww_{\alpha}}), } \]
where $P_\alpha=M_\alpha N_\alpha$ is the co-rank one parabolic subgroup associated to the simple root $\alpha$ (cf. \cite{silberger2015introduction}). Whence
\[Ind^G_B(\chi^w)\simeq Ind^G_B(\chi^{ww_\alpha}) \]
if and only if
\[Ind^{M_\alpha}_{B\cap M_\alpha}(\chi^{w})\simeq Ind^{M_\alpha }_{B\cap M_\alpha}(\chi^{ww_{\alpha}})\]
if and only if
\[(\chi^w)_\alpha=\chi_{w.\alpha}\neq |\cdot|^{\pm 1}.\tag*{($\star$)} \]
Moreover, if $(\star)$ does not hold, i.e. $\chi_{w.\alpha}=|\cdot|^{\pm 1}$, which is to say that $Ind^{M_\alpha}_{B\cap M_\alpha}(\chi^w)$ is reducible, then the Jacquet modules of $Ker(A)$ and $Im(A)$ with respect to $B$ are as follows:
\[r_B(Ker(A))=r_B\circ Ind^G_{P_\alpha}(Ker(A))=\bigoplus_{w''\in W_T:~w''^{-1}.\alpha>0}\chi^{ww''}\tag*{$(\star\star)$} \]
and
\[r_B(Im(A))=r_B\circ Ind^G_{P_\alpha}(Im(A))=\bigoplus_{w''\in W_T:~w''^{-1}.\alpha>0}\chi^{ww_\alpha w''}, \]
which follow from Bernstein--Zelevinsky geometrical lemma (see \cite{bernstein1977induced,casselman1995introduction,waldspurger2003formule}). Notice that any coroot $\beta'^\vee$ corresponding to a positive root $\beta'$ satisfies the following relations:
\[\left<w''.\beta,~w''.\beta'^\vee\right>=\left<\beta,~\beta'^\vee\right>>0\mbox{ for any }\beta\in C^+\mbox{ and any }w''\in W_T. \]
In particular,
\[\left<\beta,~\alpha^\vee \right>>0\mbox{ and }\left<\beta,~w''^{-1}.\alpha^\vee \right>=\left<w''.\beta,~\alpha^\vee \right>>0\mbox{ for any }\beta\in C^+\mbox{ and any } w''\in W_T\mbox{ s.t. }w''^{-1}.\alpha>0. \]
Thus we obtain a geometrical characterization of the double coset \[ P_\alpha\backslash G/B=\big\{w''\in W_T:~w''^{-1}.\alpha>0 \big\}=\big\{w''\in W_T:~w''C^+\mbox{ and }C^+ \mbox{ are on the same side of }Ker(\alpha^\vee) \big\}. \]
Applying the same argument as above after conjugating $C^+,~\alpha$ and $w''$ by $w$, we know that the Jacquet module $r_B(Ker(A))$ of $Ker(A)$ in $(\star\star)$ is equal to
\[r_B(Ker(A))=\bigoplus_{w''}\chi^{w''}, \]
where $w''$ runs over those Weyl elements in $W_T$ such that
\[wC^+\mbox{ and }w''C^+\mbox{ are on the same side of }Ker(w.\alpha^\vee). \]
Analogously,
\[r_B(Im(A))=\bigoplus_{w''}\chi^{w''},\tag{$SB$} \]
where $w''$ runs over those Weyl elements in $W_T$ such that
\[w'C^+\mbox{ and }w''C^+\mbox{ are on the same side of }Ker(w.\alpha^\vee). \]
At last, we would like to emphasize that the condition $(\star)$ is equivalent to saying that
\[\mbox{the co-rank one parabolic induction associated to $w.\alpha$ is irreducible.} \]
Such a condition is the right language we need in our statement of Rodier type structure theorem later on.
Now we turn to the discussion of the general case, i.e. the pairs $(w,w')$ with $w'=ww_{\alpha_1}\cdots w_{\alpha_s}$ for some simple roots $\alpha_1,\cdots,\alpha_s$. We require that such a decomposition is minimal in the sense that it gives rise to a minimal gallery between $wC^+$ and $w'C^+$ (see \cite[Section 1.2]{casselman1995introduction}), i.e. $w^{-1}w'=w_{\alpha_1}\cdots w_{\alpha_s}$ is a reduced decomposition. Thus we have (see \cite[Lemma 8.3.2]{springer2010linear})
\[R(w^{-1}w'):=\big\{\alpha\in \Phi^+:~(w^{-1}w').\alpha<0 \big\}=\big\{\alpha_s, ~w_{\alpha_s}.\alpha_{s-1},~\cdots, ~(w_{\alpha_s}\cdots w_{\alpha_2}).\alpha_1 \big\}.\tag{$RD$} \]
The key observation for the general case is to show that
\[A(w,w')=A(ww_{\alpha_1}\cdots w_{\alpha_{s-1}},w')\circ\cdots\circ A(ww_{\alpha_1},ww_{\alpha_1}w_{\alpha_2})\circ A(w,ww_{\alpha_1}),\tag{$KO$} \]
where $A(w_1,w_2)$ is the unique, up to scalar, nonzero intertwining operator in $$Hom_G(Ind^G_B(\chi^{w_1}),~Ind^G_B(\chi^{w_2}))\simeq \mathbb{C},$$ for any $w_1,~w_2\in W_T$.
Before moving to the proof of the claim $(KO)$, we first discuss its implication to our previous question, i.e. when is $A(w,w')$ an isomorphism?
Given $(KO)$, it is easy to see that $A(w,w')$ is an isomorphism if and only if
\[\big\{w.\alpha_1^\vee, ww_{\alpha_1}.\alpha_2^\vee,\cdots,ww_{\alpha_1}\cdots w_{\alpha_{s-1}}.\alpha_s^\vee \big\}\bigcap (S\bigcup -S)=\emptyset. \]
Thus Rodier structure theorem, i.e. Theorem \ref{rps} holds.
In what follows, we finish the proof of our claim $(KO)$. To show the equality $(KO)$ is to show that the composition map on the right hand side is nonzero, i.e. the Jacquet module of its image is nonzero. Notice that the Jacquet module of its image on the right hand side of $(KO)$ with respect to $B$ is equal to
\[Jim:=r_B(Im(A(w,ww_{\alpha_1})))\bigcap r_B(Im(A(ww_{\alpha_1},ww_{\alpha_1}w_{\alpha_2})))\bigcap\cdots\bigcap r_B(Im(A(ww_{\alpha_1}\cdots w_{\alpha_{s-1}},w'))), \]
which follows from the simple fact that Jacquet module is invariant under isomorphism. Then it reduces to show that $Jim\neq 0$. Recall that $(SB)$ says:
\begin{align*}
r_B&(Im(A(w,ww_{\alpha_1})))=\\
&\big\{w''\in W_T:~ww_{\alpha_1}C^+\mbox{ and }w''C^+\mbox{ are on the same side of }Ker(w.\alpha_1^\vee) \big\}, \\
r_B&(Im(A(ww_{\alpha_1},ww_{\alpha_1}w_{\alpha_2})))=\\
&\big\{w''\in W_T:~ww_{\alpha_1}w_{\alpha_2}C^+\mbox{ and }w''C^+\mbox{ are on the same side of }Ker(ww_{\alpha_1}.\alpha_2^\vee) \big\},\\
&\vdots\\
r_B&(Im(A(ww_{\alpha_1}\dots w_{\alpha_{s-1}},w')))=\\
&\big\{w''\in W_T:~w'C^+\mbox{ and }w''C^+\mbox{ are on the same side of }Ker(ww_{\alpha_1}\dots w_{\alpha_{s-1}}.\alpha_s^\vee) \big\}.
\end{align*}
From the expression of $R(w^{-1}w')$ in $(RD)$, it is readily to check that
\[\chi^{w'}\in Jim, \]
whence the claim $(KO)$ holds.
At last, we would like to see what explicit form of the set $Jim$ is for its own sake. As $$Jim\bigcup Jer=\big\{w''\in W_T:~\chi^{w''} \big\}\mbox{ and }Jim\bigcap Jer=\emptyset$$ where $Jer:=$
\[r_B(Ker(A(w,ww_{\alpha_1})))\bigcup r_B(Ker(A(ww_{\alpha_1},ww_{\alpha_1}w_{\alpha_2})))\bigcup\cdots\bigcup r_B(Ker(A(ww_{\alpha_1}\cdots w_{\alpha_{s-1}},w'))), \]
it is equivalent to determine the set $Jer\overset{(KO)}{=}r_B(Ker(A(w,w')))$.
From the geometrical description of $r_B(Ker(A))$ in $(SB)$, we obtain that
\[Jer=r_B(Ker(A(w,w')))=\bigoplus_{w''\in Y}\chi^{w''} \]
where $Y$ is the set of $w''\in W_T$ for which there exists a coroot $\alpha^\vee\in S$ such that the chambers $wC^+$ and $w''C^+$ are on the same side of the wall $Ker(\alpha^\vee)$, and the chambers $wC^+$ and $w'C^+$ are separated by the wall. Moreover,
\[Jim=r_B(Im(A(w,w')))=\bigoplus_{w''\in W_T-Y}\chi^{w''}. \]
Now we turn to the discussion of extending Rodier structure theorem, i.e. Theorem \ref{rps} to regular generalized principal series for arbitrary connected reductive group and its finite central covering group. Recall that the key ideas in the proof of Rodier structure theorem for regular principal series analyzed as above are as follows:
\begin{enumerate}[(i)]
\item The double coset $B\backslash G/B=W_T=\left<w_\alpha:~\alpha\in \Delta\right>$ is a coxeter group.
\item For a reduced decomposition of $w^{-1}w'=w_{\alpha_1}w_{\alpha_2}\cdots w_{\alpha_{s-1}}w_{\alpha_s}$, we have
\[A(w,w')=A(ww_{\alpha_1}\cdots w_{\alpha_{s-1}},w')\circ\cdots\circ A(ww_{\alpha_1},ww_{\alpha_1}w_{\alpha_2})\circ A(w,ww_{\alpha_1}).\]
\end{enumerate}
For general standard parabolic group $P=MN$ of $G$ and supercuspidal representation $\rho$ of $M$, even though Bernstein--Zelevinsky geometrical lemma says that only elements in the relative Weyl group $W_M:=N_G(M)/M$ appear in $r_P\circ Ind^G_P(\rho)$ instead of the whole double coset $P\backslash G/P$, $W_M$ is NOT a coxeter group in general. To overcome such a difficulty, we discover two novel observations in what follows.
The first observation is about the structure of the relative Weyl group $W_M$. Recall that $\Phi^0_M$ (resp. $(\Phi_M^0)^+$) is the set of those relative (resp. positive) roots which contribute reflections in $W_M$ and $W_M^0$ is the ``small'' relative Weyl group $W_M^0:=\left<w_\alpha:~\alpha\in \Phi^0_M\right>$. Notice that $w_{w.\alpha}=ww_\alpha w^{-1}$ for any $w\in W_M$ and $\alpha\in \Phi_M$, we know that $W_M$ preserves $\Phi_M^0$, whence $W_M^0\lhd W_M$ and $\Phi_M^0$ is a root system. From the previous analysis of regular principal series, it is readily to see that all previous arguments works perfectly for the ``small'' relative Weyl group $W_M^0$ which is a coxeter group.
Analogous to the definition of Knapp--Stein $R$-group, we define
\[W_M^1:=\big\{w\in W_M:~w.(\Phi_M^0)^+>0 \big\}. \]
It is easy to see that $W_M^1$ is a subgroup of $W_M$, and $W_M^1\cap W_M^0=\{1\}$. Then it is natural to guess that $W_M=W_M^0 W_M^1$ which is stated as the following lemma.
\begin{lem}\label{key2}
Keep the notation as above. Then we have
\[W_M=W_M^0\rtimes W_M^1. \]
\end{lem}
\begin{proof}
It suffices to show
\[W_M=W_M^0. W_M^1. \]
Recall that $\Phi_M^0$ is a root system with the set of simple roots denoted by $\Delta_M^0$, and $W_M$ preserves $\Phi_M^0$. Therefore for $w\in W_M$, we may consider the action of $w$ on $\Delta_M^0$. If
\[w.\Delta_M^0>0, \qquad i.e.\quad w.\Phi_M^0>0, \]
then by definition, we have $w\in W_M^1$. Otherwise, there exists some $\alpha\in \Delta_M^0$ such that
\[w.\alpha<0. \]
By the decomposition structure of $w$ as reflections as in \cite{moeglin_waldspurger_1995}, we know that
\[w=w_\alpha w'\mbox{ with }\mathfrak{l}(w)>\mathfrak{l}(w'), \]
where $\mathfrak{l}(-)$ is the length function. It is easy to see that $w'\in W_M$. So the Lemma follows by induction on the length of $w$.
\end{proof}
The second observation is about the structure of those ``mysterious'' intertwining operators $A(w,ww_1)$ with $w\in W_M^0$ and $w_1\in W_M^1$.
As pointed out previously, we define $S$ to be the set of those relative positive coroots $\alpha^\vee$ such that the co-rank one parabolic induction $Ind^{M_\alpha}_{P\cap M_\alpha}(\rho)$ associated to $\alpha\in \Phi^0_M$ is reducible. Notice that $W_M\neq W_M^0$in general, thus the following key Lemma is needed to claim a similar result as Theorem \ref{rps} for regular generalized principal series.
\begin{lem}\label{key1}Keep the notions as above.
For any $w\in W_M^0$ and any $w_1\in W_M^1$, we have that
\[A(w,ww_1):~Ind^G_P(\rho^w)\longrightarrow Ind^G_P(\rho^{ww_1})\mbox{ is an isomorphism.} \]
\end{lem}
\begin{proof}
it reduces to show
\[Ind^G_P(\rho)\simeq Ind^G_P(\rho^{w_1}), \]
which follows from the associativity property of intertwining operators (please refer to \cite[IV.3.(4)]{waldspurger2003formule} for the notions). To be precise, up to non-zero scalar, the non-trivial intertwining operator
\[A: ~Ind_P^G(\rho)\longrightarrow Ind_P^G(\rho^{w_1}) \]
is equal to
\[J_{P|w_1^{-1}Pw_1}(\rho^{w_1})\circ\lambda(w_1):~Ind_P^G(\rho)\longrightarrow Ind^G_{w_1^{-1}Pw_1}(\rho^{w_1})\longrightarrow Ind_P^G(\rho^{w_1}). \]
By \cite[IV.3.(4)]{waldspurger2003formule}, we have
\[J_{P|w_1^{-1}Pw_1}(\rho^{w_1})J_{w_1^{-1}Pw_1|P}(\rho)=\prod j_\alpha(\rho) J_{P|P}(\rho), \]
where $\alpha$ runs over $\Phi_M(P)\bigcap \Phi_M(\overline{w_1^{-1}Pw_1})$ with $\overline{w_1^{-1}Pw_1}$ the opposite parabolic subgroup of $w_1^{-1}Pw_1$. Notice that
\[w_1.C^+_M=C^+_M, \]
so we have
\[\Phi_M(P)\bigcap \Phi_M(\overline{w_1^{-1}Pw_1})\bigcap (S\bigcup -S)=\emptyset .\]
Thus, in view of \cite[Corollary 1.8]{silberger1980special},
\[\prod j_\alpha(\rho) J_{P|P}(\rho)=\prod j_\alpha(\rho)\neq 0,~\infty. \]
Whence $A$ is an isomorphism.
\end{proof}
Combining the previous analysis and our key Lemma \ref{key1}, we could now claim our Rodier type structure theorem for the general case as follows:
\begin{thm}(Rodier Type Structure Theorem)\label{mthm}
Keep the notation as previous. The constituents $\pi_\Gamma$ of the regular generalized principal series $Ind^G_P(\rho)$ are parameterized by the connected components $\Gamma$ of $$^{0}\mathfrak{a}_M^\star-\bigcup_{\alpha^\vee\in S}Ker(\alpha^\vee)$$
satisfying the following property:
the Jacquet module $r_P(\pi_\Gamma)$ of $\pi_\Gamma$ with respect to $P$ is equal to
\[\bigoplus_{wC^+_M\subset \Gamma}\rho^w. \]
\end{thm}
A direct corollary of Rodier type structure theorm, i.e. Theorem \ref{mthm} is the well-known Harish-Chandra--Silberger's irreducibility criterion for regular generalized principal series $Ind^G_P(\rho)$ as follows:
\begin{thm}(see \cite[Theorem 5.4.3.7]{silberger2015introduction})\label{irredcrit}
If $\rho$ is a regular supercuspidal representation of the Levi subgroup $M$ of $P=MN$ in $G$, then the following are equivalent
\begin{enumerate}[(i)]
\item $Ind^G_P(\rho)$ is irreducible.
\item $S$ is an empty set, i.e. no co-rank one reducibility.
\end{enumerate}
\end{thm}
Well, to sum up, we would like to emphasis that the previous argument works in a broad sense if the following two analogous ingredients exist in general
\begin{enumerate}[(i)]
\item Bernstein--Zelevinsky geometrical lemma (cf. \cite{bernstein1977induced,casselman1995introduction}).
\item Harish-Chandra Plancherel formula theory, especially intertwining operator theory (cf. \cite{waldspurger2003formule}).
\end{enumerate}
A direct example is that the finite central covering group $\widetilde{G}$ of $G$ enjoys those properties listed as above (cf. \cite{BJ,luo2017R}). Thus we know that
\begin{thm}\label{covering}
Rodier type structure theorem, i.e. Theorem \ref{mthm} and Theorem \ref{irredcrit} hold for finite central covering group $\widetilde{G}$.
\end{thm}
Another direct corollary of the above structure theorems, i.e. Theorem \ref{mthm} and Theorem \ref{covering}, is as follows:
\begin{cor}\label{univirred}
Keep the notions as before. If all co-rank one reducibility conditions lie in a Levi subgroup $L$ of a parabolic subgroup $Q=LV$ in $G$, then we have
\[Ind^G_Q(\sigma)\mbox{ is always irreducible for any }\sigma\in JH(Ind^L_{L\cap P}(\rho)). \]
\end{cor}
Indeed, such a natural universal irreducibility structure is a special case of a general phenomenon \cite[Theorem 2.1]{luo2018C}.
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Sometimes us photographers can be fussy. I mean real fussy. We often stew about a certain color or position of something in our photograph and forget to look at the whole picture.
I am no different. Often times I sit and can’t decide which version of a particular shot is better. I look at each discriminating detail trying to sway my decision one way or the other.
One such situation are these recent photos of a male red winged blackbird singing and displaying while perched on a cattail stem. Both are similar shots of the same bird taken at the same spot, one quickly after the other.
One portrays the bird on the left while the other one portrays the one on the right. Both have similar backgrounds and positions of the bird. Both, in my view, are great photographs of this beautufil species of bird. Shown separately each photo would be a great display of this birds springtime song and displaying ritual but shown together, for us fussy photographers, it is little more than a decision making headache.
We photographers often forget to step back and stop looking at each minute detail and look at the overall shot and what it portrays. Most people looking at photographs, either to buy for their wall or just to casually observe, don’t fret over small, ever so small details between two similar pictures. They look at which photograph grabs them immediately so why do us photographers do it? Why do we sit and stew over such small details?
I don’t have an answer. I wish I did but I don’t. I will say, however, I have finally decided which one of these I do like best, a decision which didn’t come easily or quickly. A decision which I have finally made after hours of comparison and critiquing. Wait a minute. Now it’s the other one. See, I told you us photographers are fussy creatures.
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Microsoft co-founder Paul Allen and his group the Underthinkers are releasing an album, Everywhere at Once, in August. Allen is the not-so-famous-as-Bill Gates co-founder of the technology company Microsoft, but is a huge guitar-head and music aficionado.
Everywhere at Once features some stellar guests. Joe Walsh, Derek Trucks, Heart’s Ann and Nancy Wilson, Chrissie Hynde, Doyle Bramhall II, Los Lobos’ David Hidalgo and Ivan Neville all feature on the album. Allen himself co-wrote all of the songs.
Seattle-native Allen saw Jimi Hendrix perform when he was 16 and was key in founding the EMP Museum in Seattle. Allen paid a reported $2m at auction to buy the guitar Hendrix played in ’69 at Woodstock.
“I've rarely gone a week without picking up a guitar,” he wrote in his memoir. “It's more than a hobby; it gives me balance and keeps me in the moment, which can be a challenge with all the projects I'm pursuing at any one time… I take music with me wherever I go.”
Proceeds from Everywhere at Once’s sales will go toward the EMP Museum‘s educational programs. Which is good, as in March 2013 Allen was estimated by Forbes to be the 53rd-richest person in the world.
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\begin{document}
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\author{Erez Lapid}
\thanks{E.L. was partially supported by Grant \# 711733 from the Minerva Foundation.}
\address{Department of Mathematics, Weizmann Institute of Science, Rehovot 7610001, Israel}
\email{erez.m.lapid@gmail.com}
\author{Alberto M{\'{\i}}nguez}
\thanks{A.M. was partially supported by P12-FQM-2696}
\address{Institut de Math\'ematiques de Jussieu, Universit\'e Paris VI, Paris, France}
\email{alberto.minguez@imj-prg.fr}
\title[Geometric conditions for $\square$-irreducibility]{Geometric conditions for $\square$-irreducibility of certain representations
of the general linear group over a non-archimedean local field}
\begin{abstract}
Let $\pi$ be an irreducible, complex, smooth representation of $\GL_n$ over a local non-archimedean (skew) field.
Assuming $\pi$ has regular Zelevinsky parameters, we give a geometric necessary and sufficient criterion for the irreducibility
of the parabolic induction of $\pi\otimes\pi$ to $\GL_{2n}$.
The latter irreducibility property is the $p$-adic analogue of a special case of the notion of ``real representations'' introduced by Leclerc and studied recently by
Kang--Kashiwara--Kim--Oh (in the context of KLR or quantum affine algebras).
Our criterion is in terms of singularities of Schubert varieties of type $A$ and admits a simple combinatorial description.
It is also equivalent to a condition studied by Geiss--Leclerc--Schr\"oer.
\end{abstract}
\maketitle
\setcounter{tocdepth}{1}
\tableofcontents
\section{Introduction}
In this paper, which is a sequel to \cite{MR3573961}, we study several questions arising from the problem of characterizing reducibility of parabolic induction
for smooth, complex representations of the general linear group over a non-archimedean local field $F$.\footnote{In the body of the paper, we also consider
skew fields but for the introduction we stick to the commutative case.}
We connect this representation-theoretic question to combinatorics and geometry.
As customary, we consider all groups $\GL_n(F)$, $n\ge0$ at once and denote simply by $\Irr$ the set of equivalence classes of irreducible representations of $\GL_n(F)$, $n\ge0$.
By the Zelevinsky classification \cite{MR584084}, $\Irr$ is in one-to-one correspondence with the monoid of multisegments, which are certain essentially combinatorial objects.
We write $\zele{\m}$ for the irreducible representation corresponding to the multisegment $\m$ and denote by $\times$ normalized parabolic induction.
Then $\zele{\m+\n}$ occurs with multiplicity one in the Jordan--H\"older sequence of $\zele{\m}\times\zele{\n}$. Consequently,
\[
\zele{\m}\times\zele{\n}\text{ is irreducible }\iff\LI(\m,\n)\text{ and }\LI(\n,\m)
\]
where $\LI(\m,\n)$ is the condition $\soc(\zele{\m}\times\zele{\n})=\zele{\m+\n}$ and $\soc$ denotes the socle.
This was the point of departure of \cite{MR3573961} which led us to study the property $\LI(\m,\n)$ and characterize it purely combinatorially
in special cases.
In general, $\soc(\pi\times\sigma)$ is not necessarily irreducible for $\pi,\sigma\in\Irr$.
In fact, as was shown by Leclerc \cite{MR1959765}, there are examples of $\pi\in\Irr$ for which $\pi\times\pi$ is semisimple of length $2$.
However, it turns out that if $\pi\times\pi$ is irreducible (in which case we say that $\pi$ is \LM)\footnote{In general, an object $M$ in a ring category
is called ``real'' if $M\otimes M$ is simple. In the $p$-adic case at hand we opted for a different terminology, for obvious reasons.}
then for any $\sigma\in\Irr$, $\soc(\pi\times\sigma)$ is irreducible and
occurs with multiplicity one in the Jordan--H\"older sequence of $\pi\times\sigma$.
This is an analogue of a recent result of Kang--Kashiwara--Kim--Oh, originally proved in the context of finite-dimensional modules
of either quiver Hecke (a.k.a. KLR) algebras or quantum affine algebras \cite{MR3314831}.
The argument can be adapted to the $p$-adic setting without much difficulty -- see \S\ref{sec: KKKO}.
Granted this result, two natural interrelated problems arise. The first is to characterize (combinatorially or otherwise) the
$\square$-irreducibility of $\pi$. The second is to characterize the condition $\LI(\m,\n)$ or more generally determine
$\soc(\pi\times\sigma)$, at least when $\pi$ is \LM.
We focus on the first question in this paper, leaving the second one for a future work.
Let us briefly recall the geometry pertaining to the Zelevinsky classification \cite{MR617466, MR783619, MR863522}.
Consider pairs $(V,A)$ where $V=\oplus_{n\in\Z}V_n$ is a finite-dimensional $\Z$-graded $\C$-vector space
and $A$ is in the space $E_+(V)$ of $\C$-linear (nilpotent) endomorphisms of $V$ such that $A(V_n)\subset V_{n+1}$ for all $n$.
The isomorphism types of such pairs are parameterized by (certain) multisegments in a simple way.
Similarly if $E_+(V)$ is replaced by $E_-(V)$, with the obvious meaning.
Given $V$ as before, the group $\GL(V)$ of grading preserving linear automorphisms of $V$ acts with finitely many orbits on each of the spaces $E_\pm(V)$,
which are in duality with respect to the $\GL(V)$-invariant pairing $A,B\mapsto\tr AB=\tr BA$.
Consider the algebraic set
\[
\commvar(V)=\{(A,B)\in E_+(V)\times E_-(V):AB=BA\}.
\]
By a well-known result of Pyasetskii \cite{MR0390138}, the set of $\GL(V)$-orbits in $E_+(V)$ (or $E_-(V)$) is in canonical bijection with the
set of irreducible components of $\commvar(V)$.
The work of Geiss--Leclerc--Schr\"oer (in a more general context) highlighted the property that an irreducible component of $\commvar(V)$
admits an open $\GL(V)$-orbit. The following is a stronger variant of a special case of their beautiful conjecture. \nocite{MR2242628}
\begin{conjecture}(cf. \cite[Conjecture 18.1]{MR2822235}, \cite{LecChev}) \label{conj: GLSi}
Let $\Comp_\m$ be the irreducible component in $\commvar(V_\m)$ (for suitable $V_\m$) corresponding to a multisegment $\m$.
Then $Z(\m)$ is \LM\ if and only if $\Comp_\m$ admits an open $\GL(V_\m)$-orbit.
\end{conjecture}
The pertinent geometric condition admits an even more down-to-earth interpretation (see \S\ref{sec: GLSconj}).
Our main result is a proof of this conjecture in the so-called regular case, where we link the above condition to another geometric criterion.
Before stating it, let us introduce some more notation.
\begin{itemize}
\item For any integers $a\le b+1$ let $Z([a,b])$ be the character $\abs{\det\cdot}^{(a+b)/2}$ of $\GL_{b-a+1}(F)$.
\item For any permutation $\sigma\in S_k$, $k>0$ we denote by $\cell_\sigma$ (resp., $X_\sigma$) the corresponding Schubert cell (resp., variety)
in the flag variety of type $A_{k-1}$.
Thus, $\cell_\sigma$ is Zariski open in its closure $X_\sigma$ and $X_\sigma\supset\cell_{\sigma'}$ if and only if $\sigma'\le\sigma$ in the Bruhat order.
\item For $N>1$ let $U_q(\widehat{\sll}_N)$ be the quantum affine algebra pertaining to the affine Lie algebra $\widehat{\sll}_N$ where $q\in\C^*$ is not a root of unity.
The finite-dimensional simple modules of $U_q(\widehat{\sll}_N)$ are parameterized by Drinfeld polynomials,
or what amounts to the same, by monomials in the formal variables $Y_{i,a}$, $i=1,\dots,N-1$, $a\in\C^*$ (e.g., \cite{MR2642561}).
\end{itemize}
\begin{theorem} \label{thm: maini}
Let $\m=[a_1,b_1]+\dots+[a_k,b_k]$ where $a_1,\dots,a_k,b_1,\dots,b_k$ are integers such that $a_i\le b_i$ for all $i$.
Assume that $b_1>\dots>b_k$ and that $a_1,\dots,a_k$ are distinct.
Then Conjecture \ref{conj: GLSi} holds for
\[
\pi=Z(\m)=\soc(Z([a_1,b_1])\times\dots\times Z([a_k,b_k])).
\]
Moreover, let $\sigma,\sigma_0\in S_k$ be the permutations such that $a_{\sigma(1)}<\dots<a_{\sigma(k)}$ and for all $i$
\[
\sigma_0^{-1}(i)=\max\{j\le x_i:j\notin\sigma_0^{-1}(\{i+1,\dots,k\})\}\text{ where }x_i=\#\{j:a_j\le b_i+1\}.
\]
For $N>1+\max_i(b_i-a_i)$ let $L_N$ be the finite-dimensional simple module of $U_q(\widehat{\sll}_N)$ corresponding to the monomial $\prod_{i=1}^kY_{b_i-a_i+1,q^{a_i+b_i}}$.
Then $\sigma_0\le\sigma$ and the following conditions are equivalent.
\begin{enumerate}
\item \label{cond: pisqrirred} $\pi$ is \LM.
\item \label{cond: Lreal} $L_N$ is real, i.e., $L_N\otimes L_N$ is irreducible, for $N\gg1$.
\item \label{cond: GLS} $\Comp_\m$ admits an open $\GL(V_\m)$-orbit. (See Conjecture \ref{conj: GLSi}.)
\addtocounter{enumi}{1}
\begin{enumerate}[leftmargin=0pt, label*=\alph*]
\item \label{cond: smlocus} The smooth locus of $X_\sigma$ contains $\cell_{\sigma_0}$.
\item \label{cond: rsmlocus} $X_\sigma$ is rationally smooth at any point of $\cell_{\sigma_0}$.
\item \label{cond: ntrans} The number of transpositions $\tau\in S_k$ such that $\sigma_0\tau\le\sigma$ is equal to the length of $\sigma$.
\item \label{cond: KL} The Kazhdan--Lusztig polynomial $P_{\sigma_0,\sigma}$ with respect to $S_k$ is $1$.
\item \label{cond: KLall} $P_{\sigma',\sigma}\equiv1$ for every $\sigma'\in S_k$ such that $\sigma_0\le\sigma'\le\sigma$.
\end{enumerate}
\item \label{cond: detform} In the Grothendieck group we have
\[
\pi=\sum_{\sigma'\in S_k:\sigma_0\le\sigma'\le\sigma}\sgn\sigma\sigma'
\ Z([a_{\sigma(1)},b_{\sigma'(1)}])\times\dots\times Z([a_{\sigma(k)},b_{\sigma'(k)}]).
\]
\item \label{cond: sseq} There does not exist a sequence $1\le n_1<\dots<n_r\le k$, $r\ge4$ such that if $a'_i=a_{n_i}$ and $b'_i=b_{n_i}$ then either
\[
a'_{i+1}<a'_i\le b'_{i+1}+1,\ i=3,\dots,r-1,\ a'_3<a'_1\le b'_3+1\text{ and }a'_r<a'_2<a'_{r-1}
\]
or
\[
a'_{i+1}<a'_i\le b'_{i+1}+1,\ i=4,\dots,r-1,\ a'_4<a'_2\le b'_4+1\text{ and }a'_3<a'_r<a'_1<a'_l
\]
where $l=2$ if $r=4$ and $l=r-1$ otherwise.
\end{enumerate}
\end{theorem}
The equivalence of conditions \ref{cond: pisqrirred} and \ref{cond: Lreal} follows from the quantum Schur--Weyl duality \cite{MR1405590}.
The equivalence of conditions \ref{cond: smlocus}, \ref{cond: rsmlocus}, \ref{cond: ntrans}, \ref{cond: KL} and \ref{cond: KLall}
(for any $\sigma,\sigma_0\in S_k$) is well known (\cite{MR788771}).
The equivalence of conditions \ref{cond: KLall} and \ref{cond: detform} follows from the properties of the Arakawa--Suzuki functor \cite{MR1652134}
and the Kazhdan--Lusztig conjecture \cite{MR610137, MR632980} -- see \S\ref{sec: KLid}.
If $\max_ia_i\le b_k+1$ then $\sigma_0$ is the identity and condition \ref{cond: smlocus} simply becomes the smoothness of $X_\sigma$.
In this case condition \ref{cond: sseq} is tantamount to the well-known smoothness criterion of Lakshmibai--Sandhya \cite{MR1051089} that $\sigma$ avoids the patterns $3412$ and $4231$.
In the general case, the equivalence of conditions \ref{cond: smlocus} and \ref{cond: sseq} follows from the description of the maximal singular loci of $X_\sigma$ due (independently) to
Billey--Warrington, Cortez, Kassel--Lascoux--Reutenauer and Manivel \cite{MR1990570, MR1994224, MR2015302, MR1853139}, as explained in \S\ref{sec: smth pairs}
and \S\ref{sec: combi}. Incidentally, $\sigma_0$ is a stack-sortable permutation in the sense of Knuth. (Roughly speaking, it encodes how the sets
$\{a_1,\dots,a_k\}$ and $\{b_1,\dots,b_k\}$ are interleaved.)
Thus, the main innovative part of the paper is the equivalence of the conditions \ref{cond: pisqrirred}, \ref{cond: GLS} and \ref{cond: sseq}.
(See Theorem \ref{thm: main}.)
The ensuing equivalence of conditions \ref{cond: GLS} and \ref{cond: smlocus} is striking since at first glance, the two geometric conditions are seemingly of a different nature.
Indeed, at present we do not have a good geometric insight for this equivalence. Instead, we prove it combinatorially.
The case where $a_1>\dots>a_k$ (i.e., where $\sigma$ is the longest element of $S_k$) is especially important.
It was considered in \cite{MR3573961} under the name ``ladder representations''. In other contexts it has been known under different names.
Let us say a few words about the proof.
The implication \eqref{cond: sseq}$\implies$\eqref{cond: pisqrirred} is proved in \S\ref{sec: main} by induction on $k$.
For the induction step we use the simple observation
that if $\pi\hookrightarrow\pi_1\times\pi_2$ and $\pi\times\pi_1$ is irreducible then $\pi$ is \LM\ provided that $\pi_2$ is \LM.
(See Lemma \ref{lem: albertoidea}.)
In the case at hand we take $\pi_1$ to be a ladder representation and use the results of \cite{MR3573961}
to check the required properties combinatorially.
A parallel argument yields the implication \eqref{cond: sseq}$\implies$\eqref{cond: GLS}.
For the inverse direction $\lnot$\eqref{cond: sseq}$\implies\lnot$\eqref{cond: pisqrirred}, i.e., to prove non-$\square$-irreducibility,
we make several reductions to certain basic cases for which we use the following ``double socle'' strategy.
Given $\pi=\zele{\m}$, we construct $\square$-irreducible representations $\pi_1,\pi_2$ such that $\pi\hookrightarrow\pi_1\times\pi_2$.
Then $\Pi:=\soc(\pi_1\times\soc(\pi_2\times\pi))\hookrightarrow\pi_1\times\pi_2\times\pi$ is irreducible and hence
$\Pi\hookrightarrow\omega\times\pi$ for some irreducible subquotient $\omega$ of $\pi_1\times\pi_2$.
We then show that this is not possible unless $\omega=\pi$. This ensures that $\Pi\hookrightarrow\pi\times\pi$ and hence that $\pi$ is not \LM\
provided that $\Pi\ne\zele{\m+\m}$.
The proof is rather technical and once again, uses the results of \cite{MR3573961}.
The same reductions apply to the implication $\lnot$\eqref{cond: sseq}$\implies\lnot$\eqref{cond: GLS}, for which the basic cases are easy to verify.
As far as we know, Theorem \ref{thm: maini} is the first instance where a non-trivial infinite family of non-\LM\ representations is exhibited.
We remark that Theorem \ref{thm: maini} is proved more generally for Zelevinsky's segment representations
$Z([a,b])=\soc(\rho\abs{\det\cdot}^a\times\rho\abs{\det\cdot}^{a+1}\times\dots\times\rho\abs{\det\cdot}^b)$
for any fixed supercuspidal $\rho\in\Irr$.
Theorem \ref{thm: maini} implies the following curious identity of Kazhdan--Lusztig polynomials with respect to $S_{2k}$.
\begin{theorem}[Corollary \ref{cor: KLidnt}]
Let $\sigma,\sigma_0\in S_k$ be such that the equivalent conditions \ref{cond: smlocus}--\ref{cond: KLall} of Theorem \ref{thm: maini} are satisfied and
$\sigma_0$ is $213$-avoiding.
Let $\tilde\sigma\in S_{2k}$ be given by $\tilde\sigma(2i-j)=2\sigma(i)-j$, $i=1,\dots,k$, $j=0,1$ and
let $H\simeq S_k\times S_k$ be the parabolic subgroup of $S_{2k}$ of type $(k,k)$. Then
\[
\sum_{w\in H}\sgn w\ P_{\widetilde{\sigma'}w,\widetilde{\sigma}}(1)=1
\]
for any $\sigma'$ such that $\sigma_0\le\sigma'\le\sigma$.
\end{theorem}
See also Theorem \ref{thm: higherKL} for a generalization and \cite{1705.06517} for a follow-up conjecture.
It would be interesting to have a geometric interpretation of these identities.
At the moment, it is not clear what would replace the smoothness condition \ref{cond: smlocus} of Theorem \ref{thm: maini} in the non-regular case.
At any rate, it seems that in order to attack Conjecture \ref{conj: GLSi} in the general case with our approach, it is imperative to generalize
the results of \cite{MR3573961} to a broader class of representations. We hope to pursue this in a forthcoming work.
The determination of $\soc(\pi\times\sigma)$ is a very useful tool in representation theory.
Already in the case where $\pi$ is supercuspidal, partial results in this direction were used by M\oe glin--Waldspurger
to explicate the Zelevinsky involution \cite{MR863522}. The analysis in this case was completed independently in \cite{MR2306606} and \cite{MR2527415}
and was a key ingredient for explicating the theta correspondence for dual pairs of type II \cite{MR2504432}.
These results were extended in \cite{MR3573961} to ladder representations and yielded
a new and simplified proof of the classification of the unitary dual of $\GL_m(F)$ and its inner forms.
The problem makes sense for classical groups as well.
This approach was used by Gan--Takeda \cite{MR3454380} in their new proof of the theta correspondence for dual pairs of type I.
A better understanding in this case is a prerequisite for determining the (still unknown) unitary dual of classical groups.
(See \cite{1703.09475} for some preliminary results in this direction.)
As alluded to above, the real modules of quantum affine algebras and KLR algebras play a crucial role in the monoidal categorification of
certain (quantum) cluster algebras \cite{MR3077685, MR2682185, 1412.8106, 1502.06714}.
In particular, they are expected to represent the cluster monomials.
It remains to be seen whether our results shed any light on this procedure, or whether they can be extended beyond type $A$.
We caution however that although the notions of quantum affine algebras and KLR algebras make sense for any Cartan datum,
the link to representation theory of $p$-adic groups works well only for type $A$.
Thus, the study of reducibility questions of parabolic induction (say, for classical groups) may lead to different questions.
The contents of this paper is as follows.
In \S\ref{sec: KKKO} we translate some of the results and proofs of Kang--Kashiwara--Kim--Oh into the language of $p$-adic groups (mostly $\GL(n,F))$.
Essentially, the role of $R$-matrices is played by the usual intertwining operators.
We then recall in \S\ref{sec: classification} the Zelevinsky classification (extended to division algebras) and the irreducibility criteria of \cite{MR3573961},
which are the principal tools for the proof of the main result.
In \S\ref{sec: GLSconj} we explicate the openness criterion of Geiss--Leclerc--Schr\"oer in the case at hand and state equivalent forms
of Conjecture \ref{conj: GLSi}.
We also give some consistency checks which will be used in the proof of the main result.
This concludes the first part of the paper.
In the second part we focus on the case of irreducible representations with regular parameters, for which our main result applies.
In \S\ref{sec: smth pairs} we recall some well-known facts about singularities of Schubert varieties of type $A$.
In \S\ref{sec: combi} we introduce the main combinatorial criterion for multisegments and reinterpret it using the results of
\cite{MR1990570, MR1994224, MR2015302, MR1853139}.
The recent thesis of Deng Taiwang \cite{1603.06387} sheds more light on some of the material of this section as well as on \S\ref{sec: KLid}.
The main result is stated in \S\ref{sec: main} where the $\square$-irreducibility part is proved.
Exemplars of non-\LM\ representations are constructed in \S\ref{sec: basicases}.
The reduction to these special cases is accomplished in \S\ref{sec: comproof} where the proof of the main result is completed.
Finally, in \S\ref{sec: KLid} we interpret the main result in terms of an identity of Kazhdan--Lusztig polynomials via the Arakawa--Suzuki functor.
\subsection*{Acknowledgement}
We are grateful to a number of mathematicians and institutes for their help in preparing this manuscript.
First and foremost we are grateful to Guy Henniart for suggesting the formulation of Lemma \ref{lem: Henniart} and for several other suggestions.
We thank Max Gurevich for his input on Lemma \ref{lem: LM}.
We are thankful to David Hernandez and Bernard Leclerc for very useful correspondence and to Pascal Boyer for communicating to us the thesis of Taiwang Deng.
We are indebted to Greg Warrington for clarifying some points in \cite{MR1990570} and most importantly for
his C code for computing the Kazhdan--Lusztig polynomials for the symmetric group which was pivotal for coming up with
the formulation of our main result.
Finally, we thank Karim Adiprasito, Arkady Berenstein, Joseph Bernstein, Michel Brion, Tobias Finis, Masaki Kashiwara,
David Kazhdan, Eric Opdam, Gordan Savin and Toshiyuki Tanisaki for useful correspondence and discussions at various stages of this project.
Part of this work was done while the first-named author was visiting the Institute for Mathematical Sciences, National University of Singapore in spring 2016.
He thanks the IMS for their kind support.
The second-named author would like to thank the CNRS for the one year of ``d\'el\'egation'' where part of this work was completed and the
University of Seville for its hospitality during the academic year 2017-2018.
Both authors would like to thank Instituto de Matem\'aticas Universidad de Sevilla for providing good working conditions during
a visit of the first-named author.
\subsection*{Notation}
Throughout we fix a non-archimedean local division algebra $D$ with center $F$.
We denote by $\abs{\cdot}$ the normalized absolute value on $F$ and by $\# A$ the cardinality of a finite set $A$.
For any reductive group $G$ over $F$ we denote by $\Reps(G)$ the category of complex, smooth representations
of $G(F)$ of finite length (hence admissible) and by $\Irr G$ the set of irreducible objects of $\Reps(G)$ up to equivalence.
We have a well-known decomposition $\Reps(G)=\oplus_{\mathcal{D}}\Reps_{\mathcal{D}}(G)$ according to supercuspidal data
(i.e., pairs $(M,\sigma)$ where $M$ is a Levi subgroup of $G$ and $\sigma$ is a supercuspidal representation of $M(F)$, up to conjugation by $G(F)$).
We will mostly consider the groups $G_n=\GL_n(D)$, $n=0,1,2,\dots$, the multiplicative group of the ring of $n\times n$ matrices over $D$.
If $\pi_i\in\Reps(G_{n_i})$ $i=1,2$, we denote by $\pi_1\times\pi_2\in\Reps(G_{n_1+n_2})$ the representation
parabolically induced from $\pi_1\otimes\pi_2$ (normalized induction).
This functor (and the isomorphism of induction by stages) endow $\Reps=\oplus_{n\ge0}\Reps(G_n)$ with the structure of a ring category\footnote{i.e.,
a locally finite $\C$-linear abelian monoidal category in which $\End(1)=\C$ and the tensor product bifunctor is bilinear and biexact -- cf. \cite[Ch. 4]{MR3242743}.}
where the identity (which we denote by $1$) is the one-dimensional representation of $G_0$.
Let $\Gr_n$ (resp., $\Gr$) be the Grothendieck group of $\Reps(G_n)$ (resp., $\Reps$).
Even though $\times$ is not commutative in $\Reps$, $\Gr=\oplus_{n\ge0}\Gr_n$ is nevertheless a commutative graded ring under $\times$.
Set $\Irr=\cup_{n\ge0}\Irr G_n$ and let $\Cusp\subset\Irr$ be the subset of supercuspidal representations of $G_n$, $n>0$.
(Note that by convention we exclude $1\in\Irr G_0$ from $\Cusp$.)
Let $\pi,\pi'\in\Reps(G_n)$ and $\chi$ a character of $F^*$. We use the following notation and terminology.
\begin{itemize}
\item $\deg(\pi)=n$.
\item $\pi^\vee\in\Reps(G_n)$ is the contragredient of $\pi$.
\item $\pi\chi\in\Reps(G_n)$ is the representation obtained from $\pi$ by twisting by the character $\chi\circ\Nrd$ where $\Nrd$ is the reduced norm on $G_n$.
\item $\JH(\pi)$ is the Jordan--H\"older sequence of $\pi$ (i.e., the image of $\pi$ in $\Gr_n$), viewed as a finite multiset of $\Irr G_n$.
\item We write $\pi'\le\pi$ if $\JH(\pi')\subset\JH(\pi)$ (including multiplicities).
\item $\soc(\pi)$ (resp., $\coss(\pi)$) is the socle (resp., cosocle) of $\pi$, i.e., the largest semisimple subrepresentation (resp., quotient) of $\pi$.
\item We say that $\pi$ is \SI\ if $\soc(\pi)$ is irreducible and occurs with multiplicity one in $\JH(\pi)$.
\item $\supp\pi$ is the supercuspidal support of $\pi$ considered as a finite subset of $\Cusp$ (without multiplicity).
\item $\jac_{(m,n-m)}(\pi)\in\Reps(G_m\times G_{n-m})$ is the (normalized) Jacquet module of $\pi$ with respect to the standard (upper triangular)
parabolic subgroup of $G_n$ of type $(m,n-m)$, $0\le m\le n$. Often we simply write $\jac(\pi)$ if $m$ and $n$ are clear from the context.
\item For $\rho_1,\dots,\rho_k\in\Cusp$ with $m=\deg\rho_1+\dots+\deg\rho_k\le n$, $\jac_{(m,n-m)}(\pi)_{\rho_1+\dots+\rho_k;*}$
denotes the maximal subrepresentation $\sigma$ of $\jac_{(m,n-m)}(\pi)$
with the property that any supercuspidal irreducible subquotient of a Jacquet module of $\sigma$ is of the form
$\rho_{\tau(1)}\otimes\dots\otimes\rho_{\tau(k)}\otimes\rho'_1\otimes\dots\otimes\rho'_l$ for some permutation $\tau$ of $\{1,\dots,k\}$
and $\rho'_i\in\Cusp$. Similarly for $\jac_{(n-m,m)}(\pi)_{*;\rho_1+\dots+\rho_k}$.
\end{itemize}
\part{}
\section{Some results of Kang--Kashiwara--Kim--Oh} \label{sec: KKKO}
The purpose of this section is to translate some of the results of \cite{MR3314831} and \cite{1412.8106} to the language of
representations of reductive groups (mostly, $G_n$) over $F$.
\subsection{}
For the next lemma let $G$ be a (connected) reductive group defined over $F$ and let $P,Q$ be parabolic subgroups of $G$
(defined over $F$) such that $R=P\cap Q$ is also a parabolic subgroup. We denote (normalized) parabolic induction by $I_P^G$.
It is a functor from $\Reps(M_P)$ to $\Reps(G)$ where $M_P$ denotes the Levi part of $P$. Similarly for $I_R^P$ and $I_R^Q$.
We are very grateful to Guy Henniart for suggesting to us the following neat formulation.
\begin{lemma} \label{lem: Henniart}
Let $\pi\in\Reps(M_R)$ and let $\sigma$ (resp., $\tau$) be a subrepresentation of $I_R^P(\pi)$ (resp., $I_R^Q(\pi)$).
Assume that $I_P^G(\sigma)\subset I_Q^G(\tau)$ (as subrepresentations of $I_R^G(\pi)$).
Then there exists a subrepresentation $\kappa$ of $\pi$ such that $\sigma\subset I_R^P(\kappa)$ and $I_R^Q(\kappa)\subset\tau$.
\end{lemma}
\begin{proof}
Let $\kappa$ be the space $\{f(e):f\in\sigma\}\subset\pi$ where we view $\sigma$ as a subrepresentation of $I_R^P(\pi)$.
It is clear that $\kappa$ is a subrepresentation of $\pi$ and $\sigma\subset I_R^P(\kappa)$
(and in fact, $\kappa$ is the smallest subrepresentation with this property).
It remains to show that $I_R^Q(\kappa)\subset\tau$.
We first recall that there exists a compact open subgroup $K_0$ of $G$ and basis $\bss$ of neighborhoods of $1$ in $G$ consisting of
normal subgroups of $K_0$ such that $PK\cap Q=R(K\cap Q)$ for every $K\in\bss$.
Indeed, fix an opposite parabolic $\bar Q$ to $Q$ with unipotent radical $\bar V$.
By \cite[p. 16]{MR771671} $G$ admits a compact open subgroup $K_0$ and a basis of neighborhoods of $1$
consisting of normal subgroups of $K_0$ satisfying $K=(K\cap \bar V)(K\cap Q)$.
On the other hand, we have $P\bar Q\cap Q=R$.
Hence, $PK\cap Q=(P(K\cap\bar V)\cap Q)(K\cap Q)\subset (P\bar V\cap Q)(K\cap Q)=R(K\cap Q)$ as required.
For any $v\in\pi$ and a compact open subgroup $K$ of $G$ such that $v\in\pi^{K\cap R}$ denote by $\varphi_{v;K}$
the element of $I_R^Q(\pi)$ which is supported in $R(K\cap Q)$ and takes the value $v$ on $K\cap Q$.
We claim that for any $K\in\bss$, $\tau$ contains $\{\varphi_{f(e);K}:f\in\sigma^{P\cap K}\}$.
Indeed, let $f\in\sigma^{P\cap K}$ and consider the element $\varphi'_f\in I_P^G(\sigma)$
which is supported in $PK$ and has constant value $f$ on $K$. We can view $\varphi'_f$ as an element of
$I_R^G(\pi)\simeq I_Q^G(I_R^Q(\pi))$. Let $\phi_f$ (resp., $\psi_f$) be the image of $\varphi'_f$ in $I_R^G(\pi)$ (resp., $I_Q^G(I_R^Q(\pi))$).
Then for any $g\in G$, $\phi_f(g)=\varphi'_f(g)(1)$ and $\psi_f(g)\in I_R^Q(\pi)$ is given by $q\mapsto\varphi'_f(qg)(1)$.
Since $I_P^G(\sigma)\subset I_Q^G(\tau)$ we infer that $\psi_f(g)\in\tau$ for all $g\in G$ and in particular, $\psi_f(e)\in\tau$.
On the other hand, since $\varphi'_f\rest_Q$ is supported in $PK\cap Q = R(K\cap Q)$, $\psi_f(e)$ coincides with $\varphi_{f(e);K}$. Our claim follows.
Fix a compact set $\Omega\subset Q$ such that $R\Omega=Q$ and let $\psi\in I_R^Q(\kappa)$. We need to show that $\psi\in\tau$.
Since $\psi$ takes only finitely many values on $\Omega$, there exists $K\in\bss$ and for any $g\in\Omega$ there exists
$f_g\in\sigma^{P\cap K}$ such that $f_g(e)=\psi(g)$ (and in particular $\psi(g)\in\pi^{K\cap R}$). By the discussion above we therefore have
\begin{equation} \label{eq: intau}
\varphi_{\psi(g);K_1}\in\tau\text{ for any $g\in\Omega$ and $K_1\in\bss$ contained in $K$}.
\end{equation}
Let $K_2=\cap_{g\in\Omega}g^{-1}Kg$. Then
\[
\psi=\sum_{\eta}I_R^Q(\pi,\eta^{-1})\varphi_{\psi(\eta);K_2^\eta}
\]
where $\eta$ ranges over a set of representatives of $R\bs Q/(K'\cap Q)$ contained in $\Omega$ and $K_2^\eta=\eta K_2\eta^{-1}$.
It remains to show that $\varphi_{\psi(\eta);K_2^\eta}\in\tau$ for all $\eta\in\Omega$.
However, if $K_1\in\bss$ is any subgroup of $K_2^\eta$ then
\[
\varphi_{\psi(\eta);K_2^\eta}=\sum_{\gamma\in(K_2^\eta\cap R)(K_1\cap Q)\bs K_2^\eta\cap Q}I_R^Q(\pi,\gamma^{-1})\varphi_{\psi(\eta);K_1}
\]
and the claim therefore follows from \eqref{eq: intau}.
\end{proof}
Specializing to the general linear groups we obtain
\begin{corollary}(cf. \cite[Lemma 3.1]{MR3314831}) \label{cor: threetensor}
Let $\pi_i\in\Reps(G_{n_i})$, $i=1,2,3$.
Let $\sigma$ (resp., $\tau$) be a subrepresentation of $\pi_1\times\pi_2$ (resp., $\pi_2\times\pi_3$).
Assume that $\sigma\times\pi_3\subset\pi_1\times\tau$. Then there exists a subrepresentation $\omega$ of $\pi_2$ such that
$\sigma\subset\pi_1\times\omega$ and $\omega\times\pi_3\subset\tau$.
In particular, if $\pi_2$ is irreducible and $\sigma\ne0$ then $\tau=\pi_2\times\pi_3$.
\end{corollary}
\begin{proof}
Indeed, let $P$ (resp., $Q$, $R$) be the standard parabolic subgroup of $G_{n_1+n_2+n_3}$ of type $(n_1+n_2,n_3)$ (resp., $(n_1,n_2+n_3)$, $(n_1,n_2,n_3)$)
so that $R=P\cap Q$. For brevity we denote by $I^{1,2}$ the functor of parabolic induction from $G_{n_1+n_2}\cap R$ to $G_{n_1+n_2}$;
similarly for $I^{2,3}$.
By Lemma \ref{lem: Henniart}, there exists a subrepresentation $\kappa$ of $\pi_1\otimes\pi_2\otimes\pi_3$ such that
$\sigma\otimes\pi_3\subset I_R^P(\kappa)$ and $I_R^Q(\kappa)\subset\pi_1\otimes\tau$.
Let $\alpha$ be the smallest subrepresentation of $\pi_2\otimes\pi_3$ such that $\kappa\subset\pi_1\otimes\alpha$.
Namely, $\alpha$ is the sum of the subrepresentations $\alpha_\lambda:=\lambda^*(\kappa)$
where $\lambda$ varies in $\pi_1^*$ and $\lambda^*$ is the map $\lambda\otimes\id_{\pi_2\otimes\pi_3}:\pi_1\otimes\pi_2\otimes\pi_3\rightarrow\pi_2\otimes\pi_3$.
Note that $I^{2,3}(\alpha)\subset\tau$ since for any $\lambda\in\pi_1^*$ we have
\[
I^{2,3}(\alpha_\lambda)=(\lambda\otimes\id_{\pi_2\times\pi_3})(I_R^Q(\kappa))\subset\tau.
\]
Now for any $w\in\pi_3$ let $\omega_w$ be the subrepresentation
\[
\omega_w=\{v\in\pi_2:v\otimes w\in\alpha\}
\]
of $\pi_2$ and let $\omega=\cap_{w\in\pi_3}\omega_w$.
Since $I^{2,3}(\alpha)\subset\tau$ we have $\omega\times\pi_3\subset\tau$. It remains to show that $\sigma\subset\pi_1\times\omega$.
By assumption $\sigma\otimes\pi_3\subset I_R^P(\kappa)\subset I_R^P(\pi_1\otimes\alpha)$.
Thus, for any $w\in\pi_3$ we have
\[
\sigma\otimes\C w\subset I_R^P(\pi_1\otimes\alpha)\cap (\pi_1\times\pi_2\otimes\C w).
\]
As a representation of $G_{n_1+n_2}$ the latter is
\begin{multline*}
I^{1,2}(\pi_1\otimes\alpha)\cap I^{1,2}(\pi_1\otimes\pi_2\otimes\C w)
=\\I^{1,2}(\pi_1\otimes(\alpha\cap\pi_2\otimes\C w))=
I^{1,2}(\pi_1\otimes\omega_w\otimes\C w)=\pi_1\times\omega_w\otimes\C w.
\end{multline*}
It follows that $\sigma\subset\pi_1\times\omega_w$ for all $w$ and hence $\sigma\subset\pi_1\times\omega$ as required.
\end{proof}
\subsection{}
Let $\pi_i\in\Reps(G_{n_i})$, $i=1,2$.
We write $M_{\pi_1,\pi_2}(s)$ for the standard intertwining operator
\[
M_{\pi_1,\pi_2}(s):\pi_1\abs{\cdot}^s\times\pi_2\rightarrow\pi_2\times\pi_1\abs{\cdot}^s
\]
(see e.g. \cite[\S IV]{MR1989693}). (It depends on a choice of a Haar measure, but this will be immaterial for us.)
If $\pi_1,\pi_2\ne0$ let $r_{\pi_1,\pi_2}\ge0$ be the order of the pole of $M_{\pi_1,\pi_2}(s)$ at $s=0$ and let
\[
R_{\pi_1,\pi_2}=\lim_{s\rightarrow0}s^{r_{\pi_1,\pi_2}}M_{\pi_1,\pi_2}(s).
\]
Thus, $R_{\pi_1,\pi_2}$ is a non-zero intertwining operator from $\pi_1\times\pi_2$ to $\pi_2\times\pi_1$.
The following result is standard.
\begin{lemma} \label{lem: elemM}
Let $0\ne\pi_i\in\Reps(G_{n_i})$, $i=1,2,3$ and
\[
0\rightarrow\tau\rightarrow\pi_1\rightarrow\tau'\rightarrow0
\]
a short exact sequence. Then
\begin{enumerate}
\setcounter{enumi}{-1}
\item $r_{1,\pi_1}=r_{\pi_1,1}=0$ and $R_{1,\pi_1}=R_{\pi_1,1}=\id_{\pi_1}$.
\item We have a commutative diagram of short exact sequences
\[
\xymatrix{
0\ar[r] & \tau\abs{\cdot}^s\times\pi_2 \ar[r] \ar[d]^{M_{\tau,\pi_2}(s)} & \pi_1\abs{\cdot}^s\times\pi_2\ar[r] \ar[d]^{M_{\pi_1,\pi_2}(s)} &
\tau'\abs{\cdot}^s\times\pi_2\ar[r] \ar[d]^{M_{\tau',\pi_2}(s)} & 0\\
0\ar[r] & \pi_2\times\tau\abs{\cdot}^s \ar[r] & \pi_2\times\pi_1\abs{\cdot}^s \ar[r] & \pi_2\times\tau'\abs{\cdot}^s\ar[r] & 0}
\]
\item \label{part: sqle} $r_{\tau,\pi_2}\le r_{\pi_1,\pi_2}$ if $\tau\ne0$; $r_{\tau',\pi_2}\le r_{\pi_1,\pi_2}$ if $\tau'\ne0$.
\item \label{part: nonzeroR} $R_{\pi_1,\pi_2}$ restricts to an intertwining operator $\tau\times\pi_2\rightarrow\pi_2\times\tau$. More precisely,
\[
R_{\pi_1,\pi_2}\rest_{\tau\times\pi_2}=\begin{cases}R_{\tau,\pi_2}&\text{if }\tau\ne0\text{ and }r_{\pi_1,\pi_2}=r_{\tau,\pi_2},\\
0&\text{otherwise.}\end{cases}
\]
\item $M_{\pi_1\times\pi_2,\pi_3}(s)=(M_{\pi_1,\pi_3}(s)\times\id_{\pi_2\abs{\cdot}^s})\circ(\id_{\pi_1\abs{\cdot}^s}\times M_{\pi_2,\pi_3}(s))$.
\item \label{part: ifisom} $r_{\pi_1\times\pi_2,\pi_3}\le r_{\pi_1,\pi_3}+r_{\pi_2,\pi_3}$ and
\[
(R_{\pi_1,\pi_3}\times\id_{\pi_2})\circ(\id_{\pi_1}\times R_{\pi_2,\pi_3})=
\begin{cases}R_{\pi_1\times\pi_2,\pi_3}&\text{if }r_{\pi_1\times\pi_2,\pi_3}=r_{\pi_1,\pi_3}+r_{\pi_2,\pi_3}\\0&\text{otherwise.}\end{cases}
\]
Moreover, $r_{\pi_1\times\pi_2,\pi_3}=r_{\pi_1,\pi_3}+r_{\pi_2,\pi_3}$ if at least one of $R_{\pi_i,\pi_3}$, $i=1,2$ is an isomorphism or if $\pi_3$ is irreducible
\cite[Lemma 2.8]{1412.8106}.
\item $M_{\pi_2\abs{\cdot}^s,\pi_1\abs{\cdot}^s}(-s)\circ M_{\pi_1,\pi_2}(s)=c(s)\id_{\pi_1\abs{\cdot}^s\times\pi_1}$ for some meromorphic function $c(s)$.
\item \label{part: compR} Suppose that $R_{\pi_1,\pi_2}$ is an isomorphism. Then $R_{\pi_2,\pi_1}\circ R_{\pi_1,\pi_2}$ is a non-zero scalar.
\end{enumerate}
\end{lemma}
\begin{proof}
The only non-evident part is that if $\pi_3$ is irreducible then $r_{\pi_1\times\pi_2,\pi_3}=r_{\pi_1,\pi_3}+r_{\pi_2,\pi_3}$.
Suppose that this is not the case. Then $(R_{\pi_1,\pi_3}\times\id_{\pi_2})\circ(\id_{\pi_1}\times R_{\pi_2,\pi_3})=0$
and therefore $\pi_1\times\operatorname{Im} R_{\pi_2,\pi_3}\subset\Ker R_{\pi_1,\pi_3}\times\pi_2$.
This contradicts an obvious analogue of Corollary \ref{cor: threetensor} since both $R_{\pi_2,\pi_3}$ and $R_{\pi_1,\pi_3}$ are non-zero.
\end{proof}
\begin{corollary}(cf. \cite[Theorem 3.2]{MR3314831}) \label{cor: maincor}
Let $0\ne\pi\in\Reps(G_n)$ be such that $R_{\pi,\pi}$ is a (non-zero) scalar.
Then for any $\sigma\in\Irr G_m$, $\soc(\pi\times\sigma)$ is irreducible and is equal to the image of $R_{\sigma,\pi}$.
In particular, $\pi$ and $\pi\times\pi$ are irreducible.
Similarly, $\soc(\sigma\times\pi)$ is irreducible and is equal to the image of $R_{\pi,\sigma}$.
Finally, $\soc(\pi\times\sigma)\simeq\coss(\sigma\times\pi)$ and $\soc(\sigma\times\pi)\simeq\coss(\sigma\times\pi)$.
\end{corollary}
\begin{proof}
By assumption, $R_{\pi,\pi}=\lambda^{-1}\id_{\pi\times\pi}$ for some non-zero scalar $\lambda$.
Therefore, by Lemma \ref{lem: elemM} part \ref{part: ifisom}
\begin{equation} \label{eq: radd}
r_{\pi\times\sigma,\pi}=r_{\pi,\pi}+r_{\sigma,\pi}\text{ and }\lambda R_{\pi\times\sigma,\pi}=\id_{\pi}\times R_{\sigma,\pi}.
\end{equation}
Let $\tau$ be a non-zero subrepresentation of $\pi\times\sigma$.
Thus, we have a commutative diagram
\[
\xymatrix{
\tau\times\pi \ar[rr]^{\lambda R_{\pi\times\sigma,\pi}} \ar@{^{(}->}[d] &&\pi\times\tau \ar@{^{(}->}[d]\\
\pi\times\sigma\times\pi \ar[rr]^{\id_{\pi}\times R_{\sigma,\pi}} && \pi\times\pi\times\sigma}
\]
Hence, $\tau\times\pi\subset\pi\times R_{\sigma,\pi}^{-1}(\tau)$.
It follows from Corollary \ref{cor: threetensor} that $R_{\sigma,\pi}^{-1}(\tau)=\sigma\times\pi$, i.e., the image of $R_{\sigma,\pi}$ is contained in $\tau$.
Since $\tau$ was arbitrary, we conclude that the image of $R_{\sigma,\pi}$ is irreducible and is equal to $\soc(\pi\times\sigma)$.
Applying this with $\sigma=1$ we obtain that $\pi$ is irreducible. Taking $\sigma=\pi$ we conclude that $\pi\times\pi$ is irreducible as well.
The other part is proved in a similar way. Analogously for the irreducibility of $\coss(\pi\times\sigma)$ and $\coss(\sigma\times\pi)$.
Finally, since both $\coss(\sigma\times\pi)$ and the image of $R_{\sigma,\pi}$ are irreducible, they coincide.
\end{proof}
\begin{corollary} (Cf. \cite[Corollary 3.3]{MR3314831}) \label{cor: LMconds}
The following conditions are equivalent for $0\ne\pi\in\Reps(G_n)$.
\begin{enumerate}
\item $\pi\times\pi$ is \SI.
\item $\pi\times\pi$ is irreducible.
\item $\End_{G_{2n}}(\pi\times\pi)=\C$.
\item $R_{\pi,\pi}$ is a scalar.
\end{enumerate}
\end{corollary}
\begin{proof}
Trivially, 2$\implies$1 and 3$\implies$4. By Schur's lemma 2$\implies$3. By Corollary \ref{cor: maincor} 4$\implies$2. It remains to show that 1$\implies$3.
Suppose that $\pi\times\pi$ is \SI\ and let $\pi_0=\soc(\pi\times\pi)$.
Let $A\in\End(\pi\times\pi)$. Since $\pi_0$ is irreducible, $A$ acts as a scalar $\lambda\in\C$ on $\pi_0$. Let $A'=A-\lambda\id_{\pi\times\pi}$.
Then $\Ker A'\supset\pi_0$. On the other hand $\operatorname{Im}(A')$, if non-zero, must contain $\pi_0$.
This would contradict the assumption that $\pi_0$ occurs with multiplicity one in $\JH(\pi\times\pi)$.
Hence $A\equiv\lambda\id_{\pi\times\pi}$ as required.
\end{proof}
We say that $\pi$ is \LM\ if it satisfies the conditions of Corollary \ref{cor: LMconds}.
We denote by $\IrrS\subset\Irr$ the set of \LM\ representations.
Note that $\pi\in\IrrS$ if and only if $\pi^\vee\in\IrrS$.
\begin{remark}
We do not know whether $\pi\times\pi$ is semisimple for every $\pi\in\Irr$, or even whether
$R_{\pi,\pi}$ is always an isomorphism (or equivalently, whether $R_{\pi,\pi}\circ R_{\pi,\pi}$ is a non-zero scalar).
\end{remark}
We write $\pi^{\times n}=\pi\times\dots\times\pi$ ($n$ times).
\begin{corollary} (cf. \cite[Corollary 3.4 and p. 391]{MR3314831}) \label{cor: pi1pi2LM}
Suppose that $\pi_1,\pi_2\in\IrrS$ and $R_{\pi_1,\pi_2}$ is an isomorphism. Then $\pi_1\times\pi_2\in\IrrS$ (and in particular $\pi_1\times\pi_2\in\Irr$).
If $\pi\in\IrrS$ then $\pi^{\times n}\in\IrrS$ for all $n\ge1$.
\end{corollary}
\begin{proof}
Let $\pi_3=\pi_1\times\pi_2$. We have
\[
M_{\pi_3,\pi_3}(s)=(\id_{\pi_1}\times M_{\pi_1,\pi_2}(s)\times\id_{\pi_2\abs{\cdot}^s})\circ(M_{\pi_1,\pi_1}(s)\times M_{\pi_2,\pi_2}(s))
\circ(\id_{\pi_1\abs{\cdot}^s}\times M_{\pi_2,\pi_1}(s)\times\id_{\pi_2}).
\]
By assumption, $R_{\pi_1,\pi_1}$, $R_{\pi_2,\pi_2}$ and $R_{\pi_1,\pi_2}$ are isomorphisms and therefore
(cf. Lemma \ref{lem: elemM} part \ref{part: ifisom}) $r_{\pi_3,\pi_3}=\sum_{i,j=1,2}r_{\pi_i,\pi_j}$ and
\[
R_{\pi_3,\pi_3}=(\id_{\pi_1}\times R_{\pi_1,\pi_2}\times\id_{\pi_2})\circ(R_{\pi_1,\pi_1}\times R_{\pi_2,\pi_2})
\circ(\id_{\pi_1}\times R_{\pi_2,\pi_1}\times\id_{\pi_2}).
\]
By assumption, this is proportional to $\id_{\pi_1}\times (R_{\pi_1,\pi_2}\circ R_{\pi_2,\pi_1})\times\id_{\pi_2}$
which is a scalar by Lemma \ref{lem: elemM} part \ref{part: compR}. The first part follows.
Similarly, the second part follows from Corollary \ref{cor: LMconds} and the fact that $R_{\pi^{\times n},\pi^{\times n}}$ is a non-zero scalar if $R_{\pi,\pi}$ is.
\end{proof}
We can slightly strengthen Corollary \ref{cor: maincor}. We are grateful to Max Gurevich for this observation.
\begin{lemma} (cf. \cite[Theorem 3.1]{1412.8106}) \label{lem: LM}
Suppose that $\pi\in\IrrS$.
Then for any $\sigma\in\Irr$, $\pi\times\sigma$ and $\sigma\times\pi$ are \SI.
\end{lemma}
\begin{proof}
We prove that $\pi\times\sigma$ is \SI. The other assertion is proved similarly.
Let $\tau=\soc(\pi\times\sigma)$ and $\tau'=\Ker R_{\sigma,\pi}\subset\sigma\times\pi$. We already know that $\tau$ is the image of $R_{\sigma,\pi}$
and $\tau$ is irreducible. Since $\JH(\sigma\times\pi)=\JH(\pi\times\sigma)$, it remains to show that $\tau$ does not occur in $\JH(\tau')$.
First note that $R_{\pi\times\sigma,\pi}\rest_{\tau\times\pi}\ne 0$, that is (Lemma \ref{lem: elemM} part \ref{part: nonzeroR}),
$r_{\pi\times\sigma,\pi}=r_{\tau,\pi}$. For otherwise, we would have (by \eqref{eq: radd}) $\tau\times\pi\subset\Ker(R_{\pi\times\sigma,\pi})=\pi\times\tau'$
which contradicts Corollary \ref{cor: threetensor} since $\tau'\subsetneq\sigma\times\pi$.
Since $\lambda R_{\sigma\times\pi,\pi}=R_{\sigma,\pi}\times\id_{\pi}$
(where as before $R_{\pi,\pi}=\lambda^{-1}\id_{\pi\times\pi}$),
the restriction of $R_{\sigma\times\pi,\pi}$ to $\tau'\times\pi$ vanishes.
Hence, by Lemma \ref{lem: elemM} parts \ref{part: nonzeroR} and \ref{part: ifisom},
\[
r_{\tau',\pi}<r_{\sigma\times\pi,\pi}=r_{\sigma,\pi}+r_{\pi,\pi}=r_{\pi\times\sigma,\pi}=r_{\tau,\pi}
\]
by the above. Thus, by Lemma \ref{lem: elemM} part \ref{part: sqle} $\tau$ cannot occur in $\JH(\tau')$ as required.
\end{proof}
\begin{remark}
In \cite{MR3573961} we defined a ``left multiplier'' to be an irreducible representation such that $\pi\times\sigma$ is \SI\ for any irreducible $\sigma$.
In view of Lemma \ref{lem: LM} this is equivalent to the conditions of Corollary \ref{cor: LMconds}.
\end{remark}
The following result gives a recursive way to deduce $\square$-irreducibility.
\begin{lemma} \label{lem: albertoidea}
Suppose that $\pi\hookrightarrow\pi_1\times\pi_2$ and $\pi\times\pi_1$ is irreducible.
Then $\pi\in\IrrS$ provided that $\pi_2\in\IrrS$.
\end{lemma}
\begin{proof}
Since $\pi\times\pi_1\in\Irr$ it follow from Lemma \ref{lem: LM} that $\Pi:=\pi\times\pi_1\times\pi_2$ is \SI\ provided that $\pi_2\in\IrrS$.
Since $\pi\times\pi\hookrightarrow\Pi$ we infer that $\pi\times\pi$ is \SI.
Hence $\pi\in\IrrS$ by Corollary \ref{cor: LMconds}.
\end{proof}
\subsection{}
One of the most fundamental facts in the representation theory of the groups $G_n$ is that $\Cusp\subset\IrrS$ \cite{MR0499010}.
For any $\pi\in\Reps(G_n)$ and $\rho\in\Cusp$ with $d=\deg(\rho)$ define
\begin{gather*}
\lmlt_\rho(\pi):=\max\{i\ge0:\jac_{di,n-di}(\pi)_{i\rho;*}\ne0\}=
\max\{i\ge0:\rho^{\times i}\otimes\sigma\le\jac_{di,n-di}(\pi)\text{ for some }\sigma\ne0\},\\
\rmlt_\rho(\pi):=\max\{i\ge0:\jac_{n-di,di}(\pi)_{*;i\rho}\ne0\}=\max\{i\ge0:\sigma\otimes\rho^{\times i}\le\jac_{n-di,di}(\pi)\text{ for some }\sigma\ne0\},
\end{gather*}
and let
\[
\lnrset(\pi)=\{\rho\in\Cusp:\lmlt_\rho(\pi)>0\},
\rnrset(\pi)=\{\rho\in\Cusp:\rmlt_\rho(\pi)>0\}
\subset\supp\pi.
\]
If $\pi\in\Irr$ then $\rho\in\lnrset(\pi)$ (resp., $\rho\in\rnrset(\pi)$) if and only if there exists $\pi'\in\Irr$, necessarily unique,
such that $\pi\hookrightarrow\rho\times\pi'$ (resp., $\pi\hookrightarrow\pi'\times\rho$).
More generally, for $d=\deg\rho$ and $m=\lmlt_\rho(\pi)$ (resp., $m=\rmlt_\rho(\pi)$) the representation
$\jac_{(md,n-md)}(\pi)_{m\cdot\rho;*}$ (resp., $\jac_{(n-md,md)}(\pi)_{*;m\cdot\rho}$) is irreducible, i.e.
\[
\jac_{(md,n-md)}(\pi)_{m\cdot\rho;*}=\rho^{\times m}\otimes\pi'
\text{ (resp., $\jac_{(n-md,md)}(\pi)_{*;m\cdot\rho}=\pi'\otimes\rho^{\times m}$)}
\]
where $\pi'\in\Irr G_{n-md}$ (\cite{MR2306606}). In particular,
\[
\pi\hookrightarrow\rho^{\times m}\times\pi'
\text{ (resp., $\pi\hookrightarrow\pi'\times\rho^{\times m}$)}.
\]
Moreover, $\lmlt_\rho(\pi')=0$ (resp., $\rmlt_\rho(\pi')=0$). We write $\lderiv_\rho(\pi)=\pi'$ (resp., $\rderiv_\rho(\pi)=\pi'$).
The following result easily follows from the geometric lemma of Bernstein--Zelevinsky \cite{MR0579172} and Frobenius reciprocity.
\begin{lemma} \label{lem: lnrprop}
Let $\pi,\pi'\in\Reps(G_n)$, $\pi_i\in\Irr$, $i=1,2$ and $\rho\in\Cusp$. Then
\begin{enumerate}
\item If $\pi'\le\pi$ then $\lnrset(\pi')\subset\lnrset(\pi)$ and $\rnrset(\pi')\subset\rnrset(\pi)$.
\item $\lnrset(\pi_1\times\pi_2)=\lnrset(\pi_1)\cup\lnrset(\pi_2)$ and $\rnrset(\pi_1\times\pi_2)=\rnrset(\pi_1)\cup\rnrset(\pi_2)$.
\item $\lmlt_\rho(\pi_1\times\pi_2)=\lmlt_\rho(\pi_1)+\lmlt_\rho(\pi_2)$ and similarly for $\rmlt_\rho$.
\item $\rho^{\times \lmlt_\rho(\pi_1\times\pi_2)}\otimes\lderiv_\rho(\pi_1)\times\lderiv_\rho(\pi_2)\le\jac(\pi_1\times\pi_2)$ and
$\rderiv_\rho(\pi_1)\times\rderiv_\rho(\pi_2)\otimes\rho^{\times \rmlt_\rho(\pi_1\times\pi_2)}\le\jac(\pi_1\times\pi_2)$.
\item If $\pi\hookrightarrow\pi_1\times\pi_2$ then $\lnrset(\pi)\supset\lnrset(\pi_1)$ and $\rnrset(\pi)\supset\rnrset(\pi_2)$.
\end{enumerate}
\end{lemma}
\begin{corollary} (cf. \cite[Proposition 4.20]{1502.06714})
Suppose that $\pi_1\times\pi_2$ is irreducible. Then for any $\rho\in\Cusp$ we have
\[
\lderiv_\rho(\pi_1\times\pi_2)=\lderiv_\rho(\pi_1)\times\lderiv_\rho(\pi_2).
\]
In particular, $\lderiv_\rho(\pi_1)\times\lderiv_\rho(\pi_2)$ is irreducible. Similarly for $\rderiv_\rho$.
\end{corollary}
\begin{corollary} (cf. \cite[Corollary 4.21]{1502.06714})
If $\pi\in\IrrS$ then so are $\lderiv_{\rho}(\pi)$ and $\rderiv_\rho(\pi)$.
\end{corollary}
We say that $\pi'\in\Irr$ is a descendant of $\pi\in\Irr$ if there exists a sequence $\pi_0,\pi_1,\dots,\pi_n\in\Irr$, $n>0$
such that $\pi_0=\pi$, $\pi_n=\pi'$ and for all $i=1,\dots,n$, $\pi_i=\lderiv_{\rho_i}(\pi_{i-1})$ or $\pi_i=\rderiv_{\rho_i}(\pi_{i-1})$
for some $\rho_i\in\lnrset(\pi)$ (resp. $\rho_i\in\rnrset(\pi)$).
\begin{corollary} \label{cor: derisLM}
If $\pi\in\IrrS$ then so is any descendant of $\pi$.
\end{corollary}
By \cite[Lemma 2.5]{MR3573961} we also conclude
\begin{corollary} \label{cor: extractrho}
Suppose that $\pi_1\in\IrrS$, $\pi_2\in\Irr$ and $\rho\in\Cusp$. Let $m_i=\lmlt_\rho(\pi_i)$, $\pi'_i=\lderiv_\rho(\pi_i)$, $i=1,2$ and $m=m_1+m_2$.
Then $\rho^{\times m}\times\pi'_1\times\pi'_2$ is \SI. Hence, if moreover $m_2=0$ or $\rho\times\pi_1$ is irreducible, so that
\[
\pi_1\times\pi_2\hookrightarrow\rho^{\times m_2}\times\pi_1\times\pi_2'\hookrightarrow\rho^{\times m}\times\pi_1'\times\pi'_2,
\]
then
\[
\soc(\pi_1\times\pi_2)=\soc(\rho^{\times m}\times\soc(\pi'_1\times\pi'_2)).
\]
\end{corollary}
\section{Classification} \label{sec: classification}
We recall the classification of $\Irr$ which goes back to Bernstein--Zelevinsky and Zelevinsky in the case where $D=F$ \cite{MR0579172, MR584084}.
We refer the reader to \cite{MR3573961} and the references therein for more details and the history. Here we just record the facts and set the notation.
\subsection{}
For any $\rho\in\Cusp$ there exists a unique positive real number\footnote{In fact, $s_\rho$ is an integer, but this will not play any role here.} $s_\rho$ such that
$\rho\abs{\cdot}^{s_\rho}\times\rho$ is reducible.
(If $D=F$ then $s_\rho=1$.)
We write $\nu_\rho=\abs{\cdot}^{s_\rho}$, $\rshft{\rho}=\rho\nu_\rho$, $\lshft{\rho}=\rho\nu_\rho^{-1}$.
Note that $\nu_{\rho^\vee}=\nu_\rho$ and $\nu_{\rho\chi}=\nu_\rho$ for any character $\chi$ of $F^*$.
Moreover, if $\rho_1, \rho_2\in\Cusp$ then $\rho_1\times\rho_2$ is reducible if and only if $\rho_2$ is equal
to either $\rshft\rho_1$ or $\lshft\rho_1$.
A \emph{segment} is a finite non-empty subset of $\Cusp$ of the form $\Delta=\{\rho_1,\dots,\rho_k\}$ where $\rho_{i+1}=\rshft{\rho}_i$, $i=1,\dots,k-1$.
We write $b(\Delta)=\rho_1$, $e(\Delta)=\rho_k$ and $\deg\Delta=\sum_{i=1}^k\deg\rho_i=k\cdot\deg\rho_1$.
Since $\Delta$ is determined by $b(\Delta)$ and $e(\Delta)$ we often write $\Delta$ as $[b(\Delta),e(\Delta)]$.
Let $\Delta=\{\rho_1,\dots,\rho_k\}$ be a segment as before.
Then the representation $\rho_1\times\dots\times\rho_k\in\Reps(G_{\deg\Delta})$ is \SI.
We denote $\zele{\Delta}=\soc(\rho_1\times\dots\times\rho_k)\in\Irr G_{\deg\Delta}$.
For convenience, we also set $\zele{\emptyset}=1$.
We have
\[
\jac_{(i\deg\rho_1,(k-i)\deg\rho_1)}(\zele{\Delta})=\zele{\{\rho_1,\dots,\rho_i\}}\otimes\zele{\{\rho_{i+1},\dots,\rho_k\}},\ \ 0\le i\le k.
\]
Also, $\zele{\Delta}^\vee=\zele{\Delta^\vee}$ where $\Delta^\vee=\{\rho_k^\vee,\dots,\rho_1^\vee\}$.
We set
\begin{gather*}
\lshft{\Delta}=\{\lshft{\rho}_1,\dots,\lshft{\rho}_k\},\ \ \rshft{\Delta}=\{\rshft{\rho}_1,\dots,\rshft{\rho}_k\},\\
\Delta^+=[b(\Delta),e(\rshft{\Delta})],\ ^+\Delta=[b(\lshft\Delta),e(\Delta)],\ \Delta^-=[b(\Delta),e(\lshft{\Delta})],\ ^-\Delta=[b(\rshft\Delta),e(\Delta)].
\end{gather*}
Note that $\Delta^-$ or $^-\Delta$ can be empty.
Given two segments $\Delta_1$, $\Delta_2$ we write $\Delta_1\prec\Delta_2$ if $b(\Delta_1)\notin\Delta_2$, $b(\lshft{\Delta}_2)\in\Delta_1$ and
$e(\Delta_2)\notin\Delta_1$. In this case $\soc(\zele{\Delta_1}\times\zele{\Delta_2})=\zele{\Delta_1'}\times\zele{\Delta'_2}$ where
$\Delta_1'=\Delta_1\cup\Delta_2$, $\Delta_2'=\Delta_1\cap\Delta_2$ (the latter is possibly empty).
If either $\Delta_1\prec\Delta_2$ or $\Delta_2\prec\Delta_1$ then we say that $\Delta_1$ and $\Delta_2$ are linked.
In this case we say that $(\Delta'_1,\Delta'_2)$ as above is the offspring of $(\Delta_1,\Delta_2)$.
Note that $\{b(\Delta_1'),b(\Delta'_2)\}=\{b(\Delta_1),b(\Delta_2)\}$ and $\{e(\Delta_1'),e(\Delta'_2)\}=\{e(\Delta_1),e(\Delta_2)\}$.
(By convention, if $\Delta'_2=\emptyset$ in the case at hand, we write $b(\Delta_2')=b(\Delta_{3-j})$ and $e(\Delta_2')=e(\Delta_j)$ if $\Delta_j\prec\Delta_{3-j}$.)
A multisegment is a formal sum $\m=\Delta_1+\dots+\Delta_k$ of segments.
(We omit empty segments from this sum.)
In other words, the set $\Mult$ of multisegments is the free commutative monoid generated by all segments.
Write $\supp\m=\cup_{i=1}^k\Delta_i$ and $\deg\m=\sum_{i=1}^k\deg\Delta_i$.
Assume that $\Delta_1,\dots,\Delta_k$ is a sequence of segments such that $\Delta_i\not\prec\Delta_j$ for all $i<j$.
(Any multisegment can be ordered this way.)
Then the representation
\[
\std(\m):=\zele{\Delta_1}\times\dots\times\zele{\Delta_k}\in\Reps(G_{\deg\m})
\]
is \SI\ and depends only on $\m=\Delta_1+\dots+\Delta_k$.
The main result of the classification is that the map
\[
\m\in\Mult\mapsto\zele{\m}:=\soc(\std(\m))\in\Irr G_{\deg\m}
\]
defines a bijection between $\Mult$ and $\Irr$.
We write the inverse bijection as $\pi\mapsto\m(\pi)$.
Following Zelevinsky, we write $\m\adj\n$ if $\m$ is obtained from $\n$ by replacing a pair of linked segments in $\n$ by its offspring.
The transitive closure of this relation is denoted by $\obt$. (In particular, $\m\obt\m$.)
We recall some basic properties of the Zelevinsky classification. Let $\m,\n\in\Mult$. Then
\begin{enumerate}
\setcounter{enumi}{-1}
\item $\zele{0}=1$.
\item $\supp\zele{\m}=\supp\m$.
\item $\zele{\m}\le\std(\n)$ if and only if $\m\obt\n$.
\item $\std(\m)$ is irreducible, i.e. $\zele{\m}=\std(\m)$, if and only if $\m$ is \emph{pairwise unlinked}, that is,
no two segments in $\m$ are linked.
\item $\zele{\m+\n}$ occurs with multiplicity one in $\JH(\zele{\m}\times\zele{\n})$.
\item In particular, if $\zele{\m}\times\zele{\n}$ is irreducible then $\zele{\m}\times\zele{\n}=\zele{\m+\n}$.
\item We write $\LI(\zele{\m},\zele{\n})$ (resp., $\RI(\zele{\m},\zele{\n})$) for the condition
$\soc(\zele{\m}\times\zele{\n})=\zele{\m+\n}$ (resp., $\coss(\zele{\m}\times\zele{\n})=\zele{\m+\n}$).
Thus, $\zele{\m}\times\zele{\n}$ is irreducible if and only if both $\LI(\zele{\m},\zele{\n})$ and $\RI(\zele{\m},\zele{\n})$.
\item The condition $\LI(\zele{\m},\zele{\n})$ is satisfied if $\Delta\not\prec\Delta'$ for any segment $\Delta$ of $\m$ and $\Delta'$ of $\n$.
\end{enumerate}
As a ring, $\Gr$ is freely generated by $\zele{\Delta}$ as $\Delta$ ranges over all segments.
Equivalently, $\Gr$ is freely generated as an abelian group by $\std(\m)$, $\m\in\Mult$ (as well as by $\zele{\m}$, $\m\in\Mult$).
The change of basis matrix is unitriangular with respect to $\obt$ and its coefficients are given by values of Kazhdan--Luzstig polynomials
with respect to the symmetric group -- see \S\ref{sec: KLid}.
\subsection{Auxiliary results}
\begin{definition} \label{def: detachable}
Let $\m=\Delta_1+\dots+\Delta_k\in\Mult$.
We say that $\Delta_i$ is a detachable segment of $\m$ if at least one of the following conditions holds:
\begin{subequations}
\begin{equation} \label{eq: deltafirst}
\Delta_i\not\prec\Delta_j\text{ and }\lshft{\Delta}_i\not\prec\Delta_j\text{ for all }j\ne i
\end{equation}
or,
\begin{equation} \label{eq: deltalast}
\Delta_j\not\prec\Delta_i\text{ and }\lshft{\Delta}_j\not\prec\Delta_i\text{ for all }j\ne i.
\end{equation}
\end{subequations}
\end{definition}
\begin{lemma} \label{lem: 1stred}
Suppose that $\Delta$ is a detachable segment of $\m\in\Mult$ and let $\m'=\m-\Delta$.
Assume that $\zele{\m}\in\IrrS$. Then $\zele{\m'}\in\IrrS$.
\end{lemma}
\begin{proof}
Suppose that \eqref{eq: deltafirst} holds.
Let $\pi=\zele{\m}$ and $\pi'=\zele{\m'}$. Then $\pi\hookrightarrow\zele{\Delta}\times\pi'$ by the first condition on $\Delta$.
Thus, by Frobenius reciprocity $\jac(\pi)\twoheadrightarrow\zele{\Delta}\otimes\pi'$.
Hence, by the geometric lemma
\begin{equation} \label{eq: occurs}
\zele{\Delta+\Delta}\otimes\pi'\times\pi'=\zele{\Delta}\times\zele{\Delta}\otimes\pi'\times\pi'\le\jac(\pi\times\pi).
\end{equation}
Assume that $\pi$ is \LM. Then $\pi\times\pi=\zele{\m+\m}\hookrightarrow\zele{\Delta+\Delta}\times\zele{\m'+\m'}$.
On the other hand, it is easy to see using the geometric lemma that the condition $\lshft{\Delta}\not\prec\Delta'$ for any segment $\Delta'$ of $\m'$ guarantees that
\[
\jac(\zele{\Delta+\Delta}\times\zele{\m'+\m'})_{\Delta+\Delta;*}=\zele{\Delta+\Delta}\otimes\zele{\m'+\m'}.
\]
It follows from \eqref{eq: occurs} that $\pi'\times\pi'=\zele{\m'+\m'}$, i.e., that $\pi'$ is \LM\ as required.
The argument with the condition \eqref{eq: deltalast} is similar.
\end{proof}
For $\rho\in\Cusp$ let $f_\rho:\Cusp\rightarrow\Cusp$ be the function given by
\[
f_\rho(\rho')=\begin{cases}\lshft{\rho'}&\text{if }\rho'=\rho\nu_\rho^l\text{ for some }l\in\Z_{>0},\\\rho'&\text{otherwise.}\end{cases}
\]
Thus, for any segment $\Delta$, $f_\rho(\Delta)$ is either $\Delta$, $\Delta^-$ or $\lshft{\Delta}$.
\begin{definition} \label{def: contractible}
Let $\m=\Delta_1+\dots+\Delta_k\in\Mult$ and $\rho\in\Cusp$. We say that $\m$ is $\rho$-contractible if for every $i$,
$\#(\Delta_i\cap\{\rho,\rshft\rho\})\ne1$, i.e., either $\{\rho,\rshft{\rho}\}\subset\Delta_i$ or $\Delta_i\cap\{\rho,\rshft{\rho}\}=\emptyset$.
In this case, we say that the $\rho$-contraction of $\m$ is $f_\rho(\Delta_1)+\dots+f_\rho(\Delta_k)$.
We call $\m$ contractible if it is $\rho$-contractible for some $\rho\in\supp\m$.
\end{definition}
The following assertion follows from Corollary \ref{cor: fzele} of \S\ref{sec: KLid}.
\begin{proposition} \label{prop: contract}
Suppose that $\m$ is $\rho$-contractible for some $\rho\in\Cusp$ and let $\m'$ be the $\rho$-contraction of $\m$.
Then $\zele{\m}\in\IrrS$ if and only if $\zele{\m'}\in\IrrS$.
\end{proposition}
For any $\m\in\Mult$ we write $\lnrset(\m)=\lnrset(\zele{\m})\subset\supp\m$ and for any $\rho\in\Cusp$ we write
$\lderiv_\rho(\m)=\m(\lderiv_\rho(\zele{\m}))$ and similarly for $\rderiv_\rho(\m)$.
We recall the following combinatorial recipe for $\lnrset(\m)$, $\lderiv_\rho(\m)$ and $\m(\soc(\rho\times\zele{\m}))$.
\begin{lemma} (\cite{MR2306606, MR2527415}) \label{lem: desclder}
For $\rho'\in\Cusp$ let $I_{\rho'}=\{i:b(\Delta_i)=\rho'\}$. Then there exists a subset $I\subset I_\rho$ and an injective function
$f:I\rightarrow I_{\rshft\rho}$ such that if $J=I_\rho\setminus I$ then we have the following properties.
\begin{enumerate}
\item $\Delta_i\prec\Delta_{f(i)}$ for all $i\in I$.
\item If $\Delta_i\prec\Delta_j$ with $i\in I$ and $j\notin f(I)$ then $^+\Delta_j\not\prec\Delta_{f(i)}$.
\item If $\Delta_j\prec\Delta_{j'}$ with $j\in J$ and $j'\in I_{\rshft\rho}$ then $j'\in f(I)$ and $\Delta_{f^{-1}(j')}\not\prec\,^-\Delta_j$.
\end{enumerate}
Moreover, we have the following.
\begin{enumerate}
\item $\rho\in\lnrset(\m)$ if and only if $J\ne\emptyset$.
\item $\lderiv_\rho(\m)=\m+\sum_{j\in J}(^-\Delta_j-\Delta_j)$. In particular, $\sum_{i\in I}\Delta_i$
is independent of $I$ and $f$.
\item $\soc(\rho\times\zele{\m})=\zele{\m+\{\rho\}}$ if $f(I)=I_{\rshft\rho}$ and otherwise,
$\soc(\rho\times\zele{\m})=\zele{\m-\Delta_j+\,^+\Delta_j}$ where $j\in I_{\rshft\rho}\setminus f(I)$ is such that
$\Delta_j\not\prec\,^+\Delta_r$ for all $r\in I_{\rshft\rho}\setminus f(I)$.
\end{enumerate}
\end{lemma}
For convenience we record the following special cases.
\begin{lemma} \label{lem: spclcaseextrho}
Let $\m=\Delta_1+\dots+\Delta_k$ and $\rho\in\Cusp$. Let $n_\rho=\#\{i:b(\Delta_i)=\rho\}$ and similarly for $n_{\rshft\rho}$.
\begin{enumerate}
\item If $n_\rho=0$ then $\rho\notin\lnrset(\zele{\m})$.
\item If $n_\rho>n_{\rshft\rho}$ then $\rho\in\lnrset(\zele{\m})$.
\item Suppose that $n_\rho=1$ and let $\Delta$ be the segment of $\m$ such that $b(\Delta)=\rho$.
Then $\rho\in\lnrset(\zele{\m})$ if and only if there does not exist $\Delta'$ in $\m$ such that $b(\Delta')=\rshft\rho$ and $\Delta\prec\Delta'$.
In this case $\lderiv_\rho(\m)=\m-\Delta+\,^-\Delta$.
\item Suppose that $n_\rho=2$ and $n_{\rshft\rho}=1$. Let $s$ and $l$ be the indices such that $b(\Delta_s)=b(\Delta_l)=\rho$ with
$\Delta_s\subset\Delta_l$ and let $j$ be such $b(\Delta_j)=\rshft\rho$. Assume that $\Delta_l\prec\Delta_j$.
Then $\rho\in\lnrset(\m)$ and $\lderiv_\rho(\m)=\m-\Delta_s+\,^-\Delta_s$.
\end{enumerate}
\end{lemma}
\begin{lemma} \label{lem: soctimesrho}
Let $\m=\Delta_1+\dots+\Delta_k$ and $\rho\in\Cusp$. Let $n_\rho$ and $n_{\rshft\rho}$ be as before.
\begin{enumerate}
\item Suppose that $n_{\rshft\rho}=1$ and let $\Delta$ be the segment of $\m$ such that $b(\Delta)=\rshft\rho$.
Then
\[
\soc(\rho\times\zele{\m})=\begin{cases}\zele{\m+\{\rho\}}&\text{if $\exists\Delta'$ in $\m$ such that $b(\Delta')=\rho$ and }\Delta'\prec\Delta,\\
\zele{\m-\Delta+\,^+\Delta}&\text{otherwise.}\end{cases}
\]
\item Suppose that $n_\rho\le1$ and $n_{\rshft\rho}=2$. Let $s$ and $l$ be the indices such that $b(\Delta_s)=b(\Delta_l)=\rshft\rho$ with
$\Delta_s\subset\Delta_l$. If there exists $j$ such $b(\Delta_j)=\rho$ then assume that $\Delta_j\prec\Delta_s$.
Then $\soc(\rho\times\zele{\m})=\zele{\m-\Delta_l+\,^+\Delta_l}$.
\end{enumerate}
\end{lemma}
\subsection{Reduction to cuspidal lines} \label{sec: fixrho}
An equivalence class for the equivalence relation on $\Cusp$ generated by $\rho\sim\rshft\rho$ is called a \emph{cuspidal line}.
Thus, the cuspidal line containing $\rho\in\Cusp$ is $\Z_\rho:=\{\rho\nu_\rho^n:n\in\Z\}$.
For any cuspidal line $\cspline$ consider the Serre ring subcategory $\Reps_\cspline$ of $\Reps$ consisting of the representations whose supercuspidal support
is contained in $\cspline$. Let $\Gr_\cspline$ be the Grothendieck ring of $\Reps_\cspline$.
The following assertions are consequences of Zelevinsky classification:
\begin{enumerate}
\item As a commutative ring, $\Gr$ (resp., $\Gr_\cspline$) is freely generated (over $\Z$) by the images of $\zele{\Delta}$,
where $\Delta$ varies over all segments (resp., the segments contained in $\cspline$).
\item If $\pi_i\in\Irr\Reps_{\cspline_i}$ with $\cspline_1,\dots,\cspline_r$ distinct then $\pi_1\times\dots\times\pi_r$ is irreducible.
\item Conversely, any $\pi\in\Irr$ can be written uniquely (up to permutation) as $\pi=\pi_1\times\dots\times\pi_r$ where $\pi_i\in\Irr\Reps_{\cspline_i}$ and
$\cspline_1,\dots,\cspline_r$ distinct.
\item $\Gr$ is the coproduct (in the category of commutative rings) over all cuspidal lines of $\Gr_\cspline$, i.e.,
$\Gr$ is the inductive limit over finite sets $\{\cspline_1,\dots,\cspline_r\}$ of $\otimes_{i=1}^r\Gr_{\cspline_i}$.
\end{enumerate}
In practice, this enables us to reduce questions about $\Irr$ to $\Irr\Reps_\cspline$.
For instance, if $\pi=\pi_1\times\dots\times\pi_r$ as above then $\pi\in\IrrS$ if and only if $\pi_i\in\IrrS$ for all $i$.
Let $\rho\in\Cusp$ and denote by $\Mult_\rho$ the submonoid of multisegments supported in $\Z_\rho$.
Let $D'$ be another local non-archimiedean division algebra (not necessarily with center $F$) and let $\rho'$ be an irreducible supercuspidal
representation of some $\GL_m(D')$, $m>0$.
Define $\phi_{\rho,\rho'}:\Z_\rho\rightarrow\Z_{\rho'}$ by $\phi_{\rho,\rho'}(\rho\nu_\rho^n)=\rho'\nu_{\rho'}^n$.
(Thus, $\phi_{\rho,\rho'}$ is the unique bijection between $\Z_\rho$ and $\Z_{\rho'}$ which commutes with $\rshft{}$ and which maps $\rho$ to $\rho'$.)
It induces a bijection $\phi_{\rho,\rho'}:\Mult_\rho\rightarrow\Mult_{\rho'}$.
Sometimes it will be convenient to use the following fact.
\begin{theorem} \label{thm: indepcspline}
There is an equivalence of ring categories between $\Reps_\rho$ and $\Reps_{\rho'}$ taking
$\rho$ to $\rho'$ and $\rshft\rho$ to $\rshft{\rho'}$, hence taking
$\zele{\m}$ to $\zele{\phi_{\rho,\rho'}(\m)}$ and $\std(\m)$ to $\std(\phi_{\rho,\rho'}(\m))$ for any $\m\in\Mult_\rho$.
\end{theorem}
This follows from the explication of the Bernstein components of $\Reps$ as categories of finite-dimensional representations of Hecke algebras which in turn follows
either by the results of \cite{MR2827179} or by type theory \cite{MR1204652, MR2081220, MR2188448, MR2216835, MR2427423, MR2889743, MR2946230}.
In principle, one can circumvent the use of Theorem \ref{thm: indepcspline} for the purpose of the paper.
However, we will use it sporadically in \S\ref{sec: basicases} in order to simplify some inessential aspects of the argument.
\begin{remark}
Let $I$ be a finite set of cuspidal lines and let $\Reps_I$ be the Serre ring subcategory of $\Reps$ consisting of the representations whose supercuspidal support
is contained in $\cup I$. Clearly, $\Reps$ is the inductive limit of the $\Reps_I$'s as $I$ varies over the directed set of finite sets of cuspidal lines
(with respect to inclusion). Once can show that if $I=\{\cspline_1,\dots,\cspline_r\}$ then parabolic induction
gives rise to an equivalence of categories of the tensor product of $\Reps_{\cspline_1},\dots,\Reps_{\cspline_r}$ in the sense of \cite[\S5]{MR1106898}
with $\Reps_I$. We will not use this fact here.
\end{remark}
From now on we fix $\rho\in\Cusp$ and for simplicity write $\Reps_\rho=\Reps_{\Z_\rho}$, $\Gr_\rho=\Gr_{\Z_{\rho}}$, $\Irr_\rho=\Irr\Reps_\rho$.
We will only consider multisegments in $\Mult_\rho$.
We identify segments supported in $\Z_\rho$ with sets of integers of the form
$[a,b]=\{n\in\Z:a\le n\le b\}$ (with $a,b\in\Z$) by $[a,b]_\rho=\{\rho\nu_\rho^n:n\in [a,b]\}$.
We will also write $[a]=[a,a]$ for brevity.
If $\rho$ is clear from the context (which will often be the case) then we suppress it from the notation.
It will be convenient to use the convention that
\begin{equation} \label{eq: convention}
\zele{[a_1,b_1]+\dots+[a_k,b_k]}=\std([a_1,b_1]+\dots+[a_k,b_k])=0\text{ if $a_i>b_i+1$ for some $i$}.
\end{equation}
We order the segments supported in $\Z_\rho$ right-lexicographically, namely we write $[a_1,b_1]<_e[a_2,b_2]$ if either
$b_1<b_2$ or $b_1=b_1$ and $a_1<a_2$. Similarly for the left-lexicographic relation $<_b$.
Given $\m,\n\in\Mult_\rho$ we write $\n<_b\m$ if $\Delta'<_b\Delta$ for any segment $\Delta$ of $\m$ and $\Delta'$ of $\n$.
This implies that $\soc(\zele{\m}\times\zele{\n})=\zele{\m+\n}$.
For later use, we mention the following result which follows from \cite[Lemma 4.11]{MR3573961}.
\begin{lemma} \label{lem: extractsegment}
Let $\m_1,\m_2\in\Mult_\rho$ and $\pi_i=\zele{\m_i}$, $i=1,2$. Assume that the maximal segment $\Delta$ of $\m_1$
with respect to $<_b$ occurs with multiplicity one in $\m_1$ and $\Delta'<_b\Delta$ for any segment $\Delta'$ of $\m_2$.
Assume that $\pi'_1\times\pi_2$ is \SI\ where $\pi'_1=\zele{\m_1-\Delta}$. Then $\zele{\Delta}\times\pi_1'\times\pi_2$ is \SI\ and hence
\[
\m(\soc(\pi_1\times\pi_2))=\Delta+\m(\soc(\pi'_1\times\pi_2)).
\]
The same holds if $<_b$ is replaced by $<_e$. Dually, suppose that the minimal segment $\Delta$ of $\m_2$
with respect to $<_b$ occurs with multiplicity one in $\m_2$ and that $\Delta<_b\Delta'$ for any segment $\Delta'$ of $\m_1$.
Assume that $\pi_1\times\pi'_2$ is \SI\ where $\pi'_2=\zele{\m_2-\Delta}$. Then $\pi_1\times\pi_2'\times\zele{\Delta}$ is \SI\ and hence
\[
\m(\soc(\pi_1\times\pi_2))=\Delta+\m(\soc(\pi_1\times\pi'_2)).
\]
Similarly if $<_b$ is replaced by $<_e$.
\end{lemma}
Recall that a \emph{ladder} is a multisegment of the form $\m=[a_1,b_1]+\dots+[a_k,b_k]$ where
$a_1>\dots>a_k$ and $b_1>\dots>b_k$. The corresponding irreducible representation $\zele{\m}$ is called a ladder representation.
It is known that a ladder representation is \LM\ \cite{MR3573961}.
We will also need the following result which follows from Frobenius reciprocity and the description of the Jacquet modules
of a ladder representation \cite{MR2996769}.
\begin{lemma} \label{lem: chopladders}
Let $\m$ be a ladder as above and let $c_1,\dots,c_k\in\Z$ be such that $a_i\le c_i\le b_i+1$ for all $i$ and $c_1>\dots>c_k$. Then
\[
\zele{\m}=\soc(\zele{\sum_i[a_i,c_i-1]}\times\zele{\sum_i[c_i,b_i]}).
\]
\end{lemma}
One of the main results of \cite{MR3573961} is the description of $\soc(\pi\times\sigma)$ when $\pi$ is a ladder representation and $\sigma$ is irreducible.
We will recall an important consequence of this description but we first make a definition which makes sense
for any pair of multisegments and which we will revisit in the next section.
\begin{definition} \label{def: XandY}
Let $\m=\Delta_1+\dots+\Delta_k$ and $\n=\Delta'_1+\dots+\Delta'_l$ be two multisegments.
Let $\X_{\m;\n}=\{(i,j):\Delta_i\prec\Delta'_j\}$, $\Y_{\m;\n}=\{(i,j):\lshft{\Delta}_i\prec\Delta'_j\}$
and let $\rltn$ be the relation between $\X_{\m;\n}$ and $\Y_{\m;\n}$ given by
\[
(i_1,j_1)\rltn(i_2,j_2)\text{ if either }\begin{cases}i_1=i_2\text{ and }\Delta'_{j_2}\prec\Delta'_{j_1},\text{ or }\\
j_1=j_2\text{ and }\Delta_{i_1}\prec\Delta_{i_2}.\end{cases}
\]
A $\rltn$-matching is an injective function $f:\X_{\m;\n}\rightarrow \Y_{\m;\n}$ such that $x\rltn f(x)$ for all $x\in \X_{\m;\n}$.
We write $\LC(\m,\n)$ for the condition that there exists a $\rltn$-matching from $\X_{\m;\n}$ to $\Y_{\m;\n}$.
\end{definition}
\begin{theorem}\cite{MR3573961} \label{thm: laddercomb}
Suppose that $\pi=\zele{\m}$ is a ladder and $\sigma=\zele{\n}\in\Irr$.
Then $\LI(\pi,\sigma)$ if and only if $\LC(\m,\n)$. Similarly, $\RI(\pi,\sigma)$ if and only if $\LC(\n,\m)$.
Thus $\pi\times\sigma$ is irreducible if and only if both $\LC(\m,\n)$ and $\LC(\n,\m)$.
\end{theorem}
\subsection{The Zelevinsky involution} \label{sec: zeleinvo}
The combinatorial analogue $\m\mapsto\m^\#$ of the Zelevinsky involution was defined by M\oe glin--Waldspurger \cite{MR863522}.
(See also \cite{MR1371654} for an alternative description.)
For $0\ne\m=\Delta_1+\dots+\Delta_k$ with $\Delta_1\ge_e\dots\ge_e\Delta_k$ define $l>0$ and indices $1=i_1<\dots<i_l$ recursively by
\[
i_{j+1}=\min\{i:\Delta_i\prec\Delta_{i_j}\text{ and }e(\Delta_i)=e(\lshft{\Delta}_{i_j})\}
\text{ if such an index exists, otherwise $l=j$.}
\]
Set $\del(\m)=[e(\Delta_{i_l}),e(\Delta_1)]$ and
\[
\m^-=\m+\sum_{j=1}^l(\Delta_{i_j}^--\Delta_{i_j}).
\]
We also write $\zele{\m}^-=\zele{\m^-}$ and $\del(\zele{\m})=\del(\m)$.
\begin{remark}
The multisegment $\m$ is uniquely determined by $\m^-$ and $\del(\m)$. Indeed, writing $\m^-=\Delta'_1+\dots+\Delta'_l$
with $\Delta'_1\ge_e\dots\ge_e\Delta'_l$ and $\del(\m)=(\rho_1,\dots,\rho_s)$, define $0\le r\le s$ and $j_1>\dots>j_r$ by
\begin{gather*}
j_1=\max\{j:e(\Delta'_j)=\lshft{\rho}_1\}\text{ if defined, otherwise }r=0,\\
j_{i+1}=\max\{j:\Delta'_{j_i}\prec\Delta'_j\text{ and }e(\Delta'_j)=\lshft{\rho}_{i+1}\}\text{ if defined, otherwise }r=i.
\end{gather*}
Then
\[
\m=\m^-+\sum_{i=1}^r({\Delta'_{j_i}}^+-\Delta'_{j_i})+\sum_{i=r+1}^s\{\rho_i\}.
\]
\end{remark}
The map $\m\mapsto\m^\#$ is defined recursively by $0^\#=0$ and
\[
\m^\#=(\m^-)^\#+\del(\m),\ \ \m\ne0.
\]
We may then define $\zele{\m}^t=\zele{\m^\#}$.
This definition extends by linearity to $\Gr$ and determines an involution of graded rings \cite{MR863522, MR1285969, MR1390967, MR2349436}.
In particular,
\begin{proposition} \label{prop: ZIred}
Suppose that $\pi\in\IrrS$. Then $\pi^t\in\IrrS$.
\end{proposition}
We refer the reader to \cite{1701.07329} for a recent, more categorical point of view of Zelevinsky involution.
\begin{lemma} (\cite[Lemma 4.13]{MR3573961}) \label{lem: suppn>suppm}
Let $\m,\n\in\Mult_\rho$. Assume that $\max\supp\m<\max\supp\n$ and that $\zele{\m}\in\IrrS$.
Then $\soc(\zele{\m}\times\zele{\n})$ is the irreducible representation $\pi$ satisfying
\[
\pi^-=\soc(\zele{\m}\times\zele{\n^-})\text{ and }\del(\pi)=\del(\n).
\]
\end{lemma}
\subsection{Regular multisegments}
In the second part of the paper we will specialize to a certain class of multisegments.
Namely, we say that a multisegment $\m=\Delta_1+\dots+\Delta_k$ is \emph{regular} if $b(\Delta_1),\dots,b(\Delta_k)$ are distinct and
$e(\Delta_1),\dots,e(\Delta_k)$ are distinct. Note that if $\m$ is regular and $\n\obt\m$ then $\n$ is also regular.
A sub-multisegment of a multisegment $\m$ is a multisegment $\m_1$ for which there exists a multisegment $\m_2$ such that $\m=\m_1+\m_2$.
Clearly, a sub-multisegment of a regular multisegment is also regular.
The same is true for the $\rho$-contraction of a $\rho$-contractible regular multisegment. However, the Zelevinsky involution
does not preserve regularity.
\section{A variant of a conjecture of Geiss--Leclerc--Schr\"oer} \label{sec: GLSconj}
\subsection{}
There is a more geometric way, also due to Zelevinsky, to think about the Zelevinsky classification \cite{MR617466, MR783619, MR863522, MR1648174}.
Namely, consider pairs $(V,A)$ where $V$ is a finite-dimensional $\Cusp$-graded $\C$-vector space $V=\oplus_{\rho\in\Cusp}V_\rho$
and $A$ is a (necessarily nilpotent) $\C$-linear endomorphisms of $V$ such that $A(V_\rho)\subset V_{\rshft{\rho}}$ for all $\rho\in\Cusp$.
(We denote by $E_\rightarrow(V)$ the space of such endomorphisms.)
The isomorphism types of such pairs $(V,A)$ are parameterized by multisegments.
Namely, for any segment $\Delta$ let $V_\Delta$ be the $\Cusp$-graded vector space $\C^\Delta$ with basis $\{\xbasis_\rho:\rho\in\Delta\}$ and let
$\rshft A_\Delta\in E_\rightarrow(V_\Delta)$ be given by $\rshft A_\Delta \xbasis_\rho=\xbasis_{\rshft\rho}$
where by convention $\xbasis_\rho=0$ if $\rho\notin\Delta$.
To any multisegment $\m=\Delta_1+\dots+\Delta_k$ define $V_\m=\oplus_{i=1}^kV_{\Delta_i}$ with basis $\{x^i_\rho:i=1,\dots,k, \rho\in\Delta_i\}$, and
$\rshft A_\m=\oplus_{i=1}^k\rshft A_{\Delta_i}\in E_\rightarrow(V_\m)$.
Then $\{(V_\m,\rshft A_\m):\m\in\Mult\}$ is a set of representatives for the isomorphism types of pairs $(V,A)$ as above.
The previous discussion applies verbatim equally well if we change $\lshft{}$ with $\rshft{}$ throughout.
For any finite-dimensional $\Cusp$-graded vector space $V$,
the group $\GL(V)$ of grading preserving linear automorphisms of $V$ acts with finitely many orbits on each of the spaces $E_\leftrightarrows(V)$.
Note that these spaces are in duality with respect to the $\GL(V)$-invariant pairing $A,B\mapsto\tr AB=\tr BA$.
Consider the algebraic set
\[
\commvar(V)=\{(A,B)\in E_\rightarrow(V)\times E_\leftarrow(V):AB=BA\}
\]
with the canonical $\GL(V)$-equivariant projection maps $p_{\leftrightarrows}:\commvar(V)\rightarrow E_\leftrightarrows(V)$.
The following is a special case of a result of Pyasetskii.
\begin{theorem} \cite{MR0390138} (cf. \cite{MR1371654})
\begin{enumerate}
\item $\dim\commvar(V)=\dim E_\rightarrow(V)=\dim E_\leftarrow(V)$ and the irreducible components of $\commvar(V)$ are equi-dimensional.
\item If $C$ is an irreducible component of $\commvar(V)$ then $p_\rightarrow(C)$ admits a (unique) open $\GL(V)$-orbit which we denote by $p_\rightarrow(C)^{\gen}$.
\item The map $C\mapsto p_\rightarrow(C)^{\gen}$ is a bijection
between the set of irreducible components of $\commvar(V)$ and the set of $\GL(V)$-orbits in $E_\rightarrow(V)$.
\item The inverse map is given by $\OO\mapsto\overline{p_{\rightarrow}^{-1}(\OO)}$ (Zariski closure).
\item Similar statements hold for $p_\leftarrow$.
\item For any $\GL(V)$-orbit $\OO$ of $E_\rightarrow(V)$, $p_{\leftarrow}(p_\rightarrow^{-1}(\OO))$ contains a unique open $\GL(V)$-orbit
$\OO^\#$. Thus, $p_\rightarrow^{-1}(\OO)\cap p_\leftarrow^{-1}(\OO^\#)$ is non-empty and open in both $p_\rightarrow^{-1}(\OO)$
and $p_\leftarrow^{-1}(\OO^\#)$.
\item The map $\OO\rightarrow\OO^\#$ is a bijection between the sets of $\GL(V)$-orbits in $E_\leftrightarrows(V)$.
\end{enumerate}
\end{theorem}
We denote by $\rshft{\OO}_\m$ the $\GL(V_\m)$-orbit of $\rshft{A}_\m$ in $E_\rightarrow(V)$ and similarly for $\lshft{\OO}_\m$.
Then $(\rshft\OO_\m)^\#=\lshft{\OO}_{\m^\#}$ where $\m^\#$ is as in \S\ref{sec: zeleinvo}. (We can identify $V_\m$ and $V_{\m^\#}$.)
The following is a variant (of a special case) of a beautiful conjecture of Geiss--Leclerc--Schr\"oer.
\begin{conjecture}(cf. \cite[Conjecture 18.1]{MR2822235}, \cite{LecChev}) \label{conj: GLS}
An irreducible representation $\pi=Z(\m)$ is \LM\ if and only if $p_\rightarrow^{-1}(\rshft{\OO}_\m)$ admits an open $\GL(V_\m)$-orbit.
(Clearly, such an orbit would necessarily be contained in $p_\leftarrow^{-1}(\lshft\OO_{\m^\#})$.)
\end{conjecture}
We emphasize however that a counterexample to Conjecture \ref{conj: GLS} would not necessarily invalidate the conjecture made in [ibid.].
The pertinent openness condition admits a homological interpretation.
Alternatively, we can rephrase it by saying that the stabilizer $G_\m$ of $\rshft{A}_\m$ in $\GL(V_\m)$ admits an open orbit in
the space $C_\m=\{B\in E_{\leftarrow}(V_\m):B\rshft{A}_\m=\rshft{A}_\m B\}$.
The advantage is that this is a linear action and by passing to the Lie algebra
the condition becomes the existence of $\lambda\in C_\m$ such that $[\Lieg_\m,\lambda]=C_\m$ where $\Lieg_\m=\operatorname{Lie} G_\m$,
viewed as a subalgebra of the Lie algebra of $\Cusp$-grading preserving endomorphisms of $V_\m$.
It is easy to explicate $\Lieg_\m$ and its action on $C_\m$ (cf. \cite[Lemmas II.4 and II.5]{MR863522}).
Let $\X_\m=\{(i,j):\Delta_i\prec\Delta_j\}$ and $\Y_\m=\{(i,j):\lshft{\Delta}_i\prec\Delta_j\}$.
Then $C_\m$ has a basis $\alpha_{i,j}$, $(i,j)\in \X_\m$ given by
\[
\alpha_{i,j}(\xbasis_\rho^l)=\delta_{j,l}\xbasis_{\lshft{\rho}}^i,\ \ \rho\in\Cusp, l=1,\dots,k,
\]
while as a vector space, $\Lieg_\m$ has a basis $\beta_{i,j}$, $(i,j)\in \Y_\m$ given by
\[
\beta_{i,j}(\xbasis_\rho^l)=\delta_{j,l}\xbasis_\rho^i,\ \ \rho\in\Cusp, l=1,\dots,k.
\]
Moreover, we have
\[
[\beta_{i,j},\alpha_{l,m}]=\delta_{j,l}\alpha_{i,m}-\delta_{i,m}\alpha_{l,j},\ \ (i,j)\in \Y_\m,\ (l,m)\in \X_\m
\]
where for convenience we set $\alpha_{i,j}=0$ if $(i,j)\notin \X_\m$.
In other words, any $\lambda\in C_\m$ is determined by its coordinates $\lambda_{i,j}$, $(i,j)\in \X_\m$ satisfying
\[
\lambda(\xbasis_\rho^j)=\sum_{i:(i,j)\in \X_\m,\lshft\rho\in\Delta_i}\lambda_{i,j}\xbasis_{\lshft\rho}^i.
\]
Similarly, any $g\in G_\m$ is determined by its coordinates $g_{i,j}$, $(i,j)\in \Y_\m$ satisfying
\[
g(\xbasis_\rho^j)=\sum_{i:(i,j)\in \Y_\m,\rho\in\Delta_i}g_{i,j}\xbasis_\rho^i.
\]
Thus, we have the following characterization of the condition that $p_\rightarrow^{-1}(\rshft{\OO}_\m)$ admits an open $\GL(V_\m)$-orbit.
(The surjectivity of the map $g\in\Lieg_\m\mapsto [g,\xi]\in C_\m$ is rephrased by the injectivity of the dual map.)
\begin{definition} \label{def: GLS}
Let $\m=\Delta_1+\dots+\Delta_k$ be a multisegment.
Consider the $\C$-vector space $\C^{\Y_\m}$ with basis $\{e_{i,j}:(i,j)\in \Y_\m\}$).
We say that $\m$ satisfies the condition \GLS\ if there exists $\lambda\in C_\m$ such that the vectors
\begin{equation} \label{eq: cxcy}
\xvec_{i,j}(\lambda):=\sum_{r:(r,j)\in \X_\m,(i,r)\in \Y_\m}\lambda_{r,j}e_{i,r}-\sum_{s:(s,j)\in \Y_\m,(i,s)\in \X_\m}\lambda_{i,s}e_{s,j},\ \ (i,j)\in \X_\m
\end{equation}
are linearly independet in $\C^{\Y_\m}$.
\end{definition}
This condition is easy to check (at least probabilistically) on a computer.
All in all, we get the following equivalent reformulation of Conjecture \ref{conj: GLS}.
\begin{conjecture}\label{conj: GLS2}
$Z(\m)$ is \LM\ if and only if $\m$ satisfies (GLS).
\end{conjecture}
\begin{remark} \label{rem: Gmorb}
Clearly, the linear independence of $\{\xvec_{i,j}(\lambda):(i,j)\in\X_\m\}$ is a Zariski open $G_\m$-invariant condition on $\lambda\in C_\m$.
\end{remark}
\begin{remark} \label{rem: leftrightarrow}
In the formulation of Conjecture \ref{conj: GLS} we could have used $p_\leftarrow^{-1}(\lshft{\OO}_\m)$ instead of $p_\rightarrow^{-1}(\rshft{\OO}_\m)$.
Indeed, an analogous argument would yield the restatement made in Conjecture \ref{conj: GLS2}.
\end{remark}
\begin{remark} \label{rem: neighij}
Note that $\X_\m=\X_{\m;\m}$ and $\Y_\m=\Y_{\m;\m}$ in the notation of Definition \ref{def: XandY}.
We continue to write $\rltn$ for the relation defined there. Thus, we get a bipartite graph $\Graph_\m$ whose vertices are $\X_\m\coprod\Y_\m$
(disjoint union) and whose edges are given by $\rltn$.
For any $(i,j)\in \X_\m$ and $\lambda\in C_\m$ we denote by $N_{i,j}(\lambda)\subset\Y_\m$ the set of non-zero coordinates of $\xvec_{i,j}(\lambda)$
and define $N_{i,j}:=\{y\in \Y_\m:(i,j)\rltn y\}$ (the neighbors of $(i,j)$ in $\Graph_\m$). Then $N_{i,j}(\lambda)\subset N_{i,j}$ with equality if $\lambda_{i,j}\ne0$ for all $(i,j)\in \X_\m$.
Note that $N_{i,j}\supset\{(i,i),(j,j)\}$ for all $(i,j)\in \X_\m$.
\end{remark}
\begin{remark} \label{rem: matching}
Clearly, \GLS\ implies the existence of a $\rltn$-matching from $\X_\m$ to $\Y_\m$.
(An example of a multisegment without such a matching is $[4,6]+[1,5]+[2,4]+[3,3]+[0,2]$.)
However, the latter condition is not sufficient. (See Remark \ref{rem: kirred} below.)
\end{remark}
It will be convenient to attach labels to the edges of the graph $\Graph_\m$.
Namely, we write $(i,j)\lrltn{(j',j)}(i,j')$ if $(i,j),(j',j)\in\X_\m$ and $(i,j')\in\Y_\m$; similarly $(i,j)\lrltn{(i,i')}(i',j)$ if $(i,j),(i,i')\in\X_\m$, $(i',j)\in\Y_\m$.
\begin{remark} \label{rem: strongmatch}
Suppose that $f:\X_\m\rightarrow\Y_\m$ is a $\rltn$-matching and let
\[
L_f=\{r\in\X_\m:x\lrltn rf(x)\text{ for some }x\in\X_\m\}.
\]
We say that $f$ is strong (resp., extra-strong) if there exists an enumeration
$r_1,\dots,r_n$ of $\X_\m$ such that $r_i\not\lrltn rf(r_j)$ (resp., $r_i\not\rltn f(r_j)$) for any $i<j$ and $r\in L_f$.
Clearly, if there exists a strong $\rltn$-matching $f$ then $\m$ satisfies \GLS.
Indeed, taking $\lambda_{i,j}\ne0$ if $(i,j)\in L_f$ and $0$ otherwise, the $f(\X_\m)$-coordinates of $\{\xvec_{i,j}(\lambda):(i,j)\in\X_\m\}$
form a lower triangular matrix with non-zero diagonal entries.
We do not know whether \GLS\ implies the existence of a strong $\rltn$-matching.
\end{remark}
\subsection{Some examples}
\begin{example}
Let $\m=[2,2]+[2,2]+[1,1]+[1,1]$.
Here $\X_\m=\{(3,1),(3,2),(4,1),((4,2)\}$ and $\Y_\m=\{(1,1),(1,2),(2,1),(2,2),(3,3),(3,4),(4,3),(4,4)\}$.
The $\xvec_{i,j}(\lambda)$'s are given by the following table:
\begin{table}[h!]
\begin{tabular}{ c || c | c | c | c | c | c | c | c}
& $1,1$ & $1,2$ & $2,1$ & $2,2$ & $3,3$ & $3,4$ & $4,3$ & $4,4$ \\
\hline\hline
$3,1$ & $-\lambda_{3,1}$ & & $-\lambda_{3,2}$ & & $\lambda_{3,1}$ & $\lambda_{4,1}$ & \\
$3,2$ & & $-\lambda_{3,1}$ & & $-\lambda_{3,2}$ & $\lambda_{3,2}$ & $\lambda_{4,2}$ & \\
$4,1$ & $-\lambda_{4,1}$ & & $-\lambda_{4,2}$ & & & & $\lambda_{3,1}$ & $\lambda_{4,1}$ \\
$4,2$ & & $-\lambda_{4,1}$ & & $-\lambda_{4,2}$ & & & $\lambda_{3,2}$ & $\lambda_{4,2}$
\end{tabular}
\end{table}
Thus, $(3,1)\mapsto (1,1)$, $(3,2)\mapsto (1,2)$, $(4,1)\mapsto (4,3)$, $(4,2)\mapsto (4,4)$ is a strong $\rltn$-matching, hence $\m$ is \GLS.
However, there does not exist an extra-strong $\rltn$-matching.
\end{example}
Next, we make a simple general observation.
\begin{remark} \label{rem: genrem}
Suppose for simplicity that $b(\Delta_i)\le e(\rshft\Delta)$ for all $i,j$. Denote by $E_{i,j}$ the $k\times k$-matrix whose $(i,j)$-th entry is $1$
and all other entries vanish. The linear span $\yy$ of
\[
\{E_{i,j}:b(\Delta_i)\le b(\Delta_j)\text{ and }e(\Delta_i)\le e(\Delta_j)\}
\]
is a Lie subalgebra of the Lie algebra of $k\times k$ matrices and
$\{E_{i,j}:b(\Delta_j)=e(\rshft\Delta_i)\}$ spans a Lie ideal $\zz$ of $\yy$.
We can identify $\Lieg_\m$ with the quotient $\yy/\zz$ and
$C_\m$ with the linear span $\xx$ of $\{E_{i,j}:(i,j)\in \X_\m\}$ which is a nilpotent Lie ideal of $\yy$ whose center contains $\zz$.
The condition \GLS\ is that there exists $h\in\xx$ such that $[\yy,h]=\xx$. Equivalently, if $G$ (resp., $H$) is the subgroup of $\GL_k(\C)$
corresponding to $\yy$ (resp., $\xx$) the condition is that $G$ acts (by conjugation) with an open orbit on $H$.
\end{remark}
\begin{example}
Suppose that $\m=\Delta_1+\dots+\Delta_k$ is a ladder. Assume for simplicity that $\Delta_k\prec\Delta_1$.
Then $\m$ satisfies \GLS. Indeed, in view of Remark \ref{rem: genrem} this reflects the fact that the Borel subgroup of $\GL_k(\C)$
acts (by conjugation) on its unipotent radical with an open orbit (given by $u_{1,2}\dots u_{k-1,k}\ne0$).
Alternatively, $\X_\m=\{(i,j):1\le j<i\le k\}$, $\Y_\m\cup\{(k,1)\}=\{(i,j):1\le j\le i\le k\}$ and the map $f((i,j))=(i-1,j)$ is a strong $\rltn$-matching.
(We enumerate $\X_\m$ as $(k,k-1),\dots,(2,1),(k,k-2),\dots,(3,1),\dots,(k,2),(k-1,1),(k,1)$.)
Note that in this case $p_\rightarrow^{-1}(\rshft\OO_\m)\cap p_\leftarrow^{-1}(\lshft\OO_{\m^\#})$ itself is a $\GL(V_\m)$-orbit.
\end{example}
\begin{example}
For $k>2$ consider
\begin{equation}
\m=[k-1,2k-2]+[k,2k-3]+\sum_{i=1}^{k-2}[k-1-i,2k-3-i].
\end{equation}
Here is a drawing for $k=7$:
\[
\xymatrix@=0.6em{
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ}
\]
We claim that $\m$ satisfies \GLS.
Let $P=M\ltimes U$ be the standard parabolic subgroup of $\GL_k(\C)$ of type $(2,1,\dots,1)$
and let $P'$ be the subgroup $T\ltimes U$ of $P$ of codimension $2$ where $T$ is the diagonal torus.
Thus, $P'$ is the inverse image of $T$ under the projection $\pr_M:P\rightarrow M$.
In view of Remark \ref{rem: genrem} we need to check that
\begin{equation} \label{eq: P'openorb}
\text{$P'$ has an open orbit (by conjuagtion) on $U$.}
\end{equation}
Indeed, the element $y=I_k+\sum_{i=2}^{k-1}E_{i,i+1}$ is a Richardson element with respect to $P$ with centralizer
\[
P_y=\{g\in P:g_{1,i}=0\text{ for }1<i<k,\ g_{i,j}=g_{i+1,j+1}\text{ if }1<i\le j<k\},
\]
a group of dimension $\dim M=k+2$.
For instance, for $k=7$
\[
P_y=\{\begin{pmatrix}
a&&&&&&b\\
c&d&e&f&g&h&i\\
&&d&e&f&g&h\\
&&&d&e&f&g\\
&&&&d&e&f\\
&&&&&d&e\\
&&&&&&d
\end{pmatrix}:a,b,c,d,e,f,g,h,i\in\C,ad\ne0\}.
\]
Note that the dimension of $\pr_M(P_y)$ is $3$. Suppose that $x=pyp^{-1}$ for $p\in P$.
Then
\begin{multline*}
\pr_M(P'_x)=\pr_M(P_x\cap P')=\pr_M(pP_yp^{-1}\cap P')=\pr_M(pP_yp^{-1})\cap T\\=\pr_M(p)\pr_M(P_y)\pr_M(p)^{-1}\cap T
\end{multline*}
which is the group of scalar matrices provided that $p_{1,2}p_{2,2}\ne0$.
Hence, under this condition, $P'_x$ is of codimension $2$ in $P_x$ and the assertion \eqref{eq: P'openorb} follows.
Note that the element of $C_\m$ corresponding to $E_{1,3}+\sum_{i=3}^{k-1}E_{i,i+1}$ belongs to $\OO_\m^\#$ but
its orbit under $G_\m$ is not open in $C_\m$. Hence, in this case $p_\rightarrow^{-1}(\rshft\OO_\m)\cap p_\leftarrow^{-1}(\lshft\OO_{\m^\#})$ is not a $\GL(V_\m)$-orbit.
Alternatively, we can deduce that $\m$ satisfies \GLS\ by observing that
\[
\X_\m=\{(i,j):1\le j<i\le k\}\setminus\{(2,1)\},\ \ \Y_\m=\{(i,j):1\le j\le i\le k\}\setminus\{(k,2),(2,1)\}
\]
and the map $f((j,i))=(j-1,i)$ if $(j,i)\ne (3,1)$ and $f((3,1))=(1,1)$ is a strong $\rltn$-matching.
(We enumerate $\X_\m$ as $(k,k-1),\dots,(3,2),(k,k-2),\dots,(3,1),\dots,(k,2),(k-1,1),(k,1)$.)
\end{example}
\begin{example} \label{exam: 34*12}
For $k>4$ consider
\begin{equation} \label{eq: mincase34*12}
\m=[k-1,2k-2]+[k,2k-3]+\sum_{i=1}^{k-4}[k-1-i,2k-3-i]+[1,k]+[2,k-1].
\end{equation}
Here is a drawing for $k=7$:
\[
\xymatrix@=0.6em{
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ}
\]
We show that $\m$ does not satisfy \GLS.
Let $P=M\ltimes U$ be the standard parabolic subgroup of $\GL_k(\C)$ of type $(2,1,\dots,1,2)$
and let $P'$ be the subgroup $T\ltimes U$ of $P$ of codimension $4$.
As before, $P'$ is the inverse image of $T$ under the projection $\pr_M:P\rightarrow M$.
In view of Remark \ref{rem: genrem} we need to show that
\begin{equation} \label{eq: P'nopenorb}
\text{$P'$ does not have an open orbit (by conjuagtion) on $U$.}
\end{equation}
Suppose on the contrary that $P'$ admits an open orbit $\OO$.
Clearly, $\OO$ is contained in the Richardson orbit of $P$. Fix a representative $x\in\OO$.
Then the centralizer $P'_x$ of $x$ in $P'$ is of codimension $4$ in the centralizer $P_x$ of $x$ in $P$
and hence (since $P'\supset U$) $\pr_M(P'_x)$ is of codimension $4$ in $\pr_M(P_x)$.
However, the element $y=I_k+\sum_{i=2}^{k-2}E_{i,i+1}$ is a Richardson element with respect to $P$ and
\begin{multline*}
P_y=\{g\in P:g_{1,i}=g_{k+1-i,k}=0\text{ for }1<i<k-2,\\g_{2,1}=g_{3,k},\ g_{k,k-1}=g_{1,k-2},\ g_{1,1}=g_{k,k},\
g_{i,j}=g_{i+1,j+1}\text{ for }1<i\le j<k-1\},
\end{multline*}
a group of dimension $\dim M=k+4$.
For instance, for $k=7$
\[
P_y=\{\begin{pmatrix}
a&&&&b&c&d\\
e&p&q&r&s&t&f\\
&&p&q&r&s&e\\
&&&p&q&r&\\
&&&&p&q&\\
&&&&&p&\\
&&&&&b&a
\end{pmatrix}:a,b,c,d,e,f,p,q,r,s,t\in\C,ap\ne0\}.
\]
Note that the dimension of $\pr_M(P_y)$ is $4$. Suppose that $x=pyp^{-1}$ for $p\in P$.
Then as before
\[
\pr_M(P'_x)=\pr_M(p)\pr_M(P_y)\pr_M(p)^{-1}\cap T
\]
and the latter contains the group of scalar matrices. In particular, $\pr_M(P'_x)$ cannot be $0$-dimensional.
The assertion \eqref{eq: P'nopenorb} follows.
\end{example}
We can also give a simple necessary condition for a multisegment to satisfy \GLS.
\begin{remark} \label{rem: kirred}
We say that a pair $(i,j)\in \X_\m$ is irreducible if $N_{i,j}=\{(i,i),(j,j)\}$.
In this case $\xvec_{i,j}(\lambda)=\lambda_{i,j}(e_{i,i}-e_{j,j})$, and in particular $\xvec_{i,j}(\lambda)$ belongs to the $k-1$-dimensional space
\[
\{\sum_{i=1}^k\alpha_ie_{i,i}:\sum_{i=1}^k\alpha_i=0\}.
\]
Thus, if the number of irreducible pairs in $\X_\m$ is at least $k$ then $\m$ does not satisfy \GLS.
For instance for Leclerc's example $\m=[3,4]+[1,3]+[2,2]+[0,1]$ the set $\X_\m=\{(2,1),(3,1),(4,2),(4,3)\}$ consists entirely
of irreducible pairs. Hence $\m$ does not satisfy \GLS\ even though there is a $\rltn$-matching from $\X_\m$ to $\Y_\m$.
More examples of this kind will be given later. (See Remark \ref{rem: types not GLS}.)
On the other hand, in Example \ref{exam: 34*12} above, the only irreducible pairs are $(3,1)$, $(3,2)$, $(k-1,k-2)$, $(k,k-2)$ so the argument
does not apply in this case.
\end{remark}
\subsection{}
Finally, we provide a few sanity checks for Conjecture \ref{conj: GLS}.
\begin{remark} \label{rem: contractGLS}
Under the assumptions of Proposition \ref{prop: contract} the conditions \GLS\ for $\m$ and $\m'$ are equivalent (in fact identical).
\end{remark}
\begin{lemma} \label{lem: lderivm}
If $\m$ is \GLS\ then so are $\lderiv_{\rho}(\m)$ and $\rderiv_{\rho}(\m)$ for any $\rho\in\Cusp$.
\end{lemma}
\begin{proof}
We prove it for $\m'=\lderiv_{\rho}(\m)$. (The argument for $\rderiv_{\rho}(\m)$ is similar.)
Write $\m=\Delta_1+\dots+\Delta_k$.
Let $I$, $J$ and $f$ be as in Lemma \ref{lem: desclder}. Then
\[
\m'=\Delta_1'+\dots+\Delta'_k\text{ where }\Delta'_j=\begin{cases}^-\Delta_j\text{ (possibly empty)}&\text{if }j\in J,\\
\Delta_j&\text{otherwise.}\end{cases}
\]
In the course of the proof we will freely use the properties of $f$ described in Lemma \ref{lem: desclder} without further notice. Let
\[
A=\X_\m\cap (J\times f(I)),\ \ \tilde A=\Y_\m\cap (J\times I).
\]
Then $(j,i)\mapsto (j,f(i))$ is a bijection between $\tilde A$ and $A$ and in particular $\#A=\#\tilde A$.
For simplicity we order the $\Delta_i$'s by $\ge_e$ and write $I_{\ge i}=\{j\in I:j\ge i\}$.
We first show that there exists $\lambda\in C_\m$ for which $\{\xvec_{i,j}(\lambda):(i,j)\in\X_\m\}$ (as in \eqref{eq: cxcy})
are linearly independent and for all $i\in I$ we have
\begin{equation} \label{eq: speclambda}
\lambda_{j,f(i)}=\delta_{i,j}\text{ for any $j\in J\cup I_{\ge i}$ such that }(j,f(i))\in \X_\m.
\end{equation}
To that end, we prove by induction on $l\ge0$ that there exists $\lambda\in C_\m$ for which $\{\xvec_{i,j}(\lambda):(i,j)\in\X_\m\}$
are linearly independent and \eqref{eq: speclambda} is satisfied for the first $l$ elements of $I$.
The base of the induction is the assumption that $\m$ satisfies \GLS.
For the induction step, let $r\in I$ and suppose that $\tilde\lambda\in C_\m$ is such that $\{\xvec_{i,j}(\tilde\lambda):(i,j)\in\X_\m\}$ are linearly independent
and \eqref{eq: speclambda} holds for $\tilde\lambda$ for all $i\in I$ with $i<r$.
We may assume in addition that $\tilde\lambda_{r,f(r)}\ne0$.
Let $g\in G_\m$ be the element given by
\[
g\xbasis_{\rho'}^i=\begin{cases}\sum_{j\in J\cup I_{\ge r}:(j,r)\in \Y_\m\text{ and }\rho'\in\Delta_j}\tilde\lambda_{j,f(r)}\xbasis_{\rho'}^j&\text{if }i=r\\
\xbasis_{\rho'}^i&\text{otherwise.}\end{cases}
\]
Thus,
\[
g_{j,i}=\begin{cases}\tilde\lambda_{j,f(r)}&\text{if }i=r\text{ and }j\in J\cup I_{\ge r},\\\delta_{i,j}&\text{otherwise,}\end{cases}\ \ (j,i)\in \Y_\m.
\]
Clearly, $g$ is invertible since $\tilde\lambda_{r,f(r)}\ne0$.
Let $\lambda=g^{-1}\tilde\lambda g$. By Remark \ref{rem: Gmorb} $\{\xvec_{i,j}(\lambda):(i,j)\in\X_\m\}$ are linearly independent.
We show that \eqref{eq: speclambda} is satisfied for all $i\in I$ with $i\le r$.
Fix $i\in I$. Clearly, $g\xbasis_{\rshft{\rho}}^{f(i)}=\xbasis_{\rshft{\rho}}^{f(i)}$.
If $i<r$ then by induction hypothesis, the coordinate of $\tilde\lambda \xbasis_{\rshft{\rho}}^{f(i)}$ at $x_\rho^i$ is $1$
and the coordinates at $x_\rho^j$, $j\in J\cup I_{>i}$, and in particular at $j=r$, vanish. Thus,
$\lambda \xbasis_{\rshft{\rho}}^{f(i)}=\tilde\lambda\xbasis_{\rshft{\rho}}^{f(i)}$ and \eqref{eq: speclambda} is satisfied.
On the other hand, we can write $\tilde\lambda\xbasis_{\rshft\rho}^{f(r)}=\xi_1+\xi_2$ where
\[
\xi_1=\sum_{j\in J\cup I_{\ge r}:(j,f(r))\in \X_\m}\tilde\lambda_{j,f(r)}\xbasis_\rho^j\ \ \ \text{ and }\ \ \
\xi_2=\sum_{j\notin J\cup I_{\ge r}:(j,f(r))\in \X_\m}\tilde\lambda_{j,f(r)}\xbasis_\rho^j=g\xi_2.
\]
Note that if $j\in J\cup I_{\ge r}$ then $(j,r)\in \Y_\m\iff(j,f(r))\in \X_\m$. Thus, $\xi_1=g\xbasis_\rho^i$ and
hence $\lambda\xbasis_{\rshft\rho}^{f(r)}=\xbasis_\rho^r+\xi_2$. It follows that $\lambda$ satisfies \eqref{eq: speclambda} for $i=r$ as well,
completing the induction step.
Now let
\begin{gather*}
A'=\X_{\m'}\cap (I\times J),\ \ \widetilde{A'}=\Y_{\m'}\cap(f(I)\times J),\\
B=\X_\m\cap(\{i:e(\Delta_i)=\lshft\rho\}\times J), \ \ \tilde B=\Y_\m\cap(\{i:e(\Delta_i)=\rho\}\times J).
\end{gather*}
As before, $(i,j)\mapsto (f(i),j)$ is a bijection between $A'$ and $\widetilde{A'}$.
In particular, $\#A'=\#\widetilde{A'}$. It is also easy to see that
\[
\X_{\m'}\setminus\X_\m=A',\ \X_\m\setminus\X_{\m'}=A\cup B,\ \Y_{\m'}\setminus\Y_\m=\widetilde{A'},\ \ \Y_\m\setminus\Y_{\m'}=\tilde A\cup\tilde B.
\]
Suppose that $\lambda$ satisfies \eqref{eq: speclambda} for all $i\in I$. We claim that
\begin{enumerate}
\item \label{part: BB'} $N_{i,j}(\lambda)\subset\tilde A$ for any $(i,j)\in A$.
\item $N_{i,j}(\lambda)\cap\tilde B=\emptyset$ for any $(i,j)\in \X_\m\setminus B$.
\item If $(i',j')\lrltn{(i,j)}(i'',j'')$ with $(i',j')\in \X_\m\setminus B$ and $(i,j)\in B$ then $i'=i$
and $(i'',j'')=(j,j')\in\tilde A$.
\end{enumerate}
The first part follows from \eqref{eq: speclambda}.
For the second part, assume on the contrary that $(i,j)\in \X_\m\setminus B$ and
$(i',j')\in N_{i,j}(\lambda)\cap\tilde B$. In particular, $(i,j)\rltn(i',j')$.
By \eqref{eq: speclambda} we cannot have $i'=i$. Hence $j'=j$ which is also impossible since $(i,j)\notin B$.
The third part is also easy.
Now let $\lambda'$ be the element of $C_{\m'}$ whose coordinates are given by
\[
\lambda'_{i,j}=\begin{cases}\lambda_{i,j}&(i,j)\in \X_\m\cap\X_{\m'},\\0&(i,j)\in A',\end{cases}\ \ \ (i,j)\in \X_{\m'}
\]
and let
\[
\xvec'_{i,j}(\lambda')=\sum_{r:(r,j)\in \X_{\m'},(i,r)\in \Y_{\m'}}\lambda'_{r,j}e'_{i,r}-\sum_{s:(s,j)\in \Y_{\m'},(i,s)\in \X_{\m'}}\lambda_{i,s}e'_{s,j},\ \ (i,j)\in \X_{\m'}
\]
where $\{e'_{i,j}:(i,j)\in\Y_{\m'}\}$ is the standard basis for $\C^{\Y_{\m'}}$.
We show that $\{\xvec'_{i,j}(\lambda'):(i,j)\in X_{\m'}\}$ are linearly independent in $\C^{\Y_{\m'}}$.
For any $(i,j)\in \X_\m\cap\X_{\m'}=X_\m\setminus(A\cup B)$ the coordinates of $\xvec_{i,j}(\lambda)$ at $\tilde B$ vanish while the coordinates
at $\Y_\m\cap\Y_{\m'}=\Y_\m\setminus (\tilde A\cup\tilde B)$ coincide with those of $\xvec'_{i,j}(\lambda')$.
On the other hand, the non-zero coordinates of $\xvec'_{i,j}(\lambda')$, $(i,j)\in A'$ are confined to $\widetilde{A'}$.
Moreover, by the assumption on $\lambda$, the square submatrix pertaining to the rows in $A'$ and the columns in $\widetilde{A'}$ is
lower unitriangular for a suitable enumeration of the rows and columns:
the entry of $\xvec'_{i,j}(\lambda')$, $(i,j)\in A'$ in the $(f(i'),j')$-column is $\lambda_{i,f(i')}$ if $j'=j$, $i\le i'$
and $(i,f(i'))\in X_\m$ and $0$ otherwise.
Thus, the linear independence of $\{\xvec'_{i,j}(\lambda'):(i,j)\in X_{\m'}\}$ follows from the fact that the non-zero coordinates of $\xvec_{i,j}(\lambda)$, $(i,j)\in A$
are confined to $\tilde A$ and $\#A=\#\tilde A$. The lemma follows.
\end{proof}
\begin{table}[h!]
\begin{tabular}{ c || c | c | c }
& $\Y_\m\cap\Y_{\m'}$ & $\tilde A$ & $\tilde B$ \\
\hline\hline
$\X_\m\cap\X_{\m'}$ & $M_1$ & $*$ & $0$ \\ \hline
$A$ & $0$ & $M_2$ & $0$ \\ \hline
$B$ & $*$ & $*$ & $*$ \\
\end{tabular}
\quad
\begin{tabular}{ c || c | c }
& $\Y_\m\cap\Y_{\m'}$ & $\tilde A'$ \\
\hline\hline
$\X_\m\cap\X_{\m'}$ & $M_1$ & $*$ \\ \hline
$A'$ & $0$ & $M_3$ \\
\end{tabular}
\caption{Comparing $\xvec_{i,j}(\lambda)$ (left) and $\xvec'_{i,j}(\lambda')$ (right) for $\lambda$, $\lambda'$ as above. The full-rank matrix $M_1$ is common to both
while $M_2$ and $M_3$ are unitriangular up to permutation}
\end{table}
\begin{remark} \label{rem: GLS^t}
The condition \GLS\ is clearly invariant under $\m\mapsto\m^\vee$. It is also invariant under $\m\mapsto\m^\#$.
(This follows from Remark \ref{rem: leftrightarrow} and the fact that $p_\rightarrow^{-1}(\rshft{\OO}_\m)\cap p_\leftarrow^{-1}(\lshft{\OO}_{\m^\#})$
is non-empty and open in $p_\rightarrow^{-1}(\rshft{\OO}_\m)$.)
\end{remark}
\begin{remark} \label{rem: GLScomb}
Suppose that $\Delta$ is a detachable segment of a multisegment $\m$. (See Definition \ref{def: detachable}.)
It is clear that if $\m$ satisfies \GLS\ then so does $\m-\Delta$. Indeed, suppose that $\m=\Delta_1+\dots+\Delta_k$ with $\Delta=\Delta_k$
and let $\m'=\m-\Delta$. Thus, $\X_{\m'}=\{(i,j)\in \X_\m:i,j\ne k\}$ and similarly for $\Y_{\m'}$.
Since $\Delta$ is detachable, $N_{i,j}\subset \Y_{\m'}$ for all $(i,j)\in \X_{\m'}$.
The claim follows.
\end{remark}
\part{}
In the second part of the paper we state and prove our main result, which is to characterize, for regular multisegments $\m$,
the condition that $\zele{\m}\in\IrrS$ and in particular to show that Conjecture \ref{conj: GLS} holds in this case.
This will involve both geometry and combinatorics. In the next couple of sections we recall the interplay between the two in the context
of Schubert varieties and interpret it for the case of regular multisegments.
\section{Smooth pairs} \label{sec: smth pairs}
In this section we recall some well-known facts about singularities of Schubert varieties of type A.
\subsection{}
Let $B_k$ be the Borel subgroup of upper triangular matrices in $\GL_k$ over $\C$ and consider the $B_k$ action on the flag
variety $B_k\bs\GL_k$. For any $\sigma\in S_k$
let $\cell_\sigma$ be the corresponding Schubert cell, i.e., the $B_k$-orbit of the permutation matrix corresponding to $\sigma$,
and let $X_\sigma$ be the corresponding Schubert variety (the Zariski closure of $\cell_\sigma$).
Recall that $\cell_\sigma$ is open in $X_\sigma$ and has dimension $\ell(\sigma)=\#\{i<j:\sigma(i)>\sigma(j)\}$.
Also, $\cell_{\sigma_0}\subset X_\sigma$ if and only if $\sigma_0\le\sigma$ in the Bruhat order. We write $[\sigma_0,\sigma]$ for the Bruhat interval
\[
[\sigma_0,\sigma]=\{\sigma_1\in S_k:\sigma_0\le\sigma_1\le\sigma\}.
\]
\begin{definition}
Let $\sigma,\sigma_0\in S_k$ with $\sigma\ge\sigma_0$.
We say that $(\sigma,\sigma_0)$ is a smooth pair if
$\cell_{\sigma_0}$ is contained in the smooth locus $X_\sigma^{\smth}$ of $X_\sigma$.
\end{definition}
In particular, if $\sigma_0=e$ then $(\sigma,\sigma_0)$ is a smooth pair if and only if $X_\sigma$ is smooth.
In this case we simply say that $\sigma$ is smooth.
Since $X_\sigma^{\smth}$ is open in $X_\sigma$,
\begin{equation} \label{eq: redrelsmth}
\text{if $(\sigma,\sigma_0)$ is a smooth pair and $\sigma_1\in [\sigma_0,\sigma]$ then $(\sigma,\sigma_1)$ is also a smooth pair.}
\end{equation}
It is known that $(\sigma,\sigma_0)$ is a smooth pair if and only if the Kazhdan--Lusztig polynomial
$P_{\sigma_0,\sigma}$ with respect to $S_k$ is $1$. (See \cite{MR1782635}.)
However, there is a much simpler well-known combinatorial criterion for smoothness which we recall next.
Denote by $\rflx$ the set of reflexions in the symmetric group $S_k$.
The elements of $\rflx$ are the transpositions $t_{i,j}$, $1\le i<j\le k$.
Recall that for any $\sigma\in S_k$ and $i<j$ we have $\sigma t_{i,j}>\sigma$ if and only if
$\sigma(i)<\sigma(j)$, otherwise $\sigma t_{i,j}<\sigma$. Thus,
\[
\#\{t\in\rflx:\sigma t<\sigma\}=\ell(\sigma).
\]
For $\sigma,\sigma_0\in S_k$ with $\sigma\ge\sigma_0$ we define
\[
\rsig(\sigma_0,\sigma)=\{t\in\rflx:\sigma_0t\in [\sigma_0,\sigma]\}
\]
and
\[
\asig(\sigma_0,\sigma)=\{t\in\rflx:\sigma_0t\le\sigma\}=\rsig(\sigma_0,\sigma)\cup\{t_{i,j}:i<j\text{ and }\sigma_0(i)>\sigma_0(j)\}.
\]
The following result due to Lakshmibai--Seshadri is well known and admits many generalizations.
(Cf.~\cite{MR1653040} or \cite{MR1782635} for more details.)
\begin{proposition}[\cite{MR752799}] \label{prop: LS}
The dimension of the tangent space of $X_\sigma$ at any point of $\cell_{\sigma_0}$ is equal to $\#\asig(\sigma_0,\sigma)$.
Thus, $\#\asig(\sigma_0,\sigma)\ge\ell(\sigma)$ (or equivalently, $\#\rsig(\sigma_0,\sigma)\ge\ell(\sigma)-\ell(\sigma_0)$
and equality holds if and only if $(\sigma,\sigma_0)$ is a smooth pair.
\end{proposition}
This proposition provides an efficient algorithm for deciding whether a given pair is smooth
since the sets $\rsig(\sigma_0,\sigma)$ and $\asig(\sigma_0,\sigma)$ are easily computable.
We also remark that the inequality $\#\asig(\sigma_0,\sigma)\ge\ell(\sigma)$ and the fact that equality implies that
$\#\asig(\sigma_1,\sigma)=\ell(\sigma)$ for any $\sigma_1\in[\sigma_0,\sigma]$ can also be proved combinatorially
using \cite[Lemma 2.2]{MR1827861} and induction on $\ell(\sigma)-\ell(\sigma_0)$.
\subsection{} \label{sec: minsingBL}
In \cite{MR1051089} a combinatorial criterion for the smoothness of Schubert variety of type $A_n$ was given in terms of pattern avoidance.
Namely, $\sigma$ is smooth if and only if $\sigma$ avoids the pattern $4231$ and $3412$.
Moreover, in the non-smooth case, a conjectural description of the irreducible components of the smooth locus was given as well.
This conjecture was solved independently in \cite{MR1990570, MR1994224, MR2015302, MR1853139}
(with an important earlier contribution in \cite{MR1827861}).
In order to state the main result of these papers we first recall certain permutations introduced in \cite[\S9]{MR1990570}.
For $r,s\ge2$ and $t=1,2,3$, with $s=2$ if $t=3$, let $\tau_{r,s}^{(t)},\delta_{r,s}^{(t)}\in S_k$ with $k=r+s$ be the pairs of permutations given by
\begin{subequations}
\begin{equation} \label{eq: taurs1}
\tau_{r,s}^{(1)}(i)=\begin{cases}k&i=1,\\
r+2-i&1<i\le r,\\
r+k-i&r<i<k,\\
1&i=k.\end{cases}\ \
\delta_{r,s}^{(1)}(i)=\begin{cases}r+1-i&i\le r,\\r+k+1-i&i>r.\end{cases}
\end{equation}
\begin{equation} \label{eq: taurs2}
\tau_{r,s}^{(2)}(i)=\begin{cases}r+1&i=1,\\
r+1-i&1<i<r,\\
k&i=r,\\
1&i=r+1,\\
r+k+1-i&r+1<i<k,\\
r&i=k,
\end{cases}\ \
\delta_{r,s}^{(2)}(i)=\begin{cases}r-i&i<r,\\
r+1&i=r,\\
r&i=r+1,\\
r+k+2-i&i>r+1.\end{cases}
\end{equation}
\begin{equation} \label{eq: taurs3}
\tau_{r,2}^{(3)}(i)=\begin{cases}
r+1&i=1,\\
k&i=2,\\
k+1-i&2<i\le r,\\
1&i=r+1,\\
2&i=k.
\end{cases}\ \
\delta_{r,2}^{(3)}(i)=\begin{cases}
1&i=1,\\
k+1-i&1<i<k,\\
k&i=k.\end{cases}
\end{equation}
\end{subequations}
In the notation of \cite[\S9]{MR1990570} we have $\tau_{r,s}^{(1)}=w_{r,s}$, $\tau_{r,s}^{(2)}=w_{r-1,2,s-1}$,
$\tau_{r,2}^{(3)}=w_{1,r,1}$ and similarly $\delta_{r,s}^{(1)}=x_{r,s}$, $\delta_{r,s}^{(2)}=x_{r-1,2,s-1}$,
$\delta_{r,2}^{(3)}=x_{1,r,1}$.\footnote{Note the following typo in \cite[(9.2)]{MR1990570}: the last entry of $w_{k,m}$, which is $1$, is missing.}
In particular, $\tau_{2,2}^{(2)}=\tau_{2,2}^{(3)}$ and $\delta_{2,2}^{(2)}=\delta_{2,2}^{(3)}$.
It follows from \cite{MR1827861} (cf.~\cite[Theorem 37]{MR1990570}) that
\begin{equation} \label{eq: pairs not smooth}
\text{the pairs $(\tau_{r,s}^{(1)},\delta_{r,s}^{(1)})$, $(\tau_{r,s}^{(2)},\delta_{r,s}^{(2)})$, $(\tau_{r,2}^{(3)},\delta_{r,2}^{(3)})$ are not smooth.}
\end{equation}
For a subset $I\subset\{1,\dots,k\}$ of size $l$ we write $\rmv_I$ for the ``flattened'' permutation in $S_{k-l}$ obtained from $\sigma$ by removing the entries
$(i,\sigma(i))$, $i\in I$ and keeping the relative order of all other entries. In other words, $\rmv_I(\sigma)=\jmath\circ\sigma\circ\imath$ where
$\imath:\{1,\dots,k-l\}\rightarrow\{1,\dots,k\}\setminus I$
and $\jmath:\{1,\dots,k\}\setminus\sigma(I)\rightarrow\{1,\dots,k-l\}$ are the monotone bijections.
\begin{theorem}\cite{MR1990570, MR1994224, MR2015302, MR1853139} \label{thm: BW}
Suppose that $\sigma_0\le\sigma$ but $(\sigma,\sigma_0)$ is not a smooth pair. Then
there exist $\sigma_1\in[\sigma_0,\sigma]$, a subset $I\subset\{1,\dots,k\}$ and integers $r,s\ge2$ and $t=1,2,3$, with $s=2$ if $t=3$, such that
\begin{enumerate}
\item $\sigma_1(i)=\sigma(i)$ for all $i\in I$.
\item $\rmv_I(\sigma)=\tau_{r,s}^{(t)}$ and $\rmv_I(\sigma_1)=\delta_{r,s}^{(t)}$.
\item The Bruhat intervals $[\sigma_1,\sigma]$ and $[\delta_{r,s}^{(t)},\tau_{r,s}^{(t)}]$ are isomorphic as posets.
(Equivalently, $\ell(\sigma)-\ell(\sigma_1)=\ell(\tau_{r,s}^{(t)})-\ell(\delta_{r,s}^{(t)})$.)
\end{enumerate}
(In the terminology of \cite{MR2422304} this means that $[\delta_{r,s}^{(t)},\tau_{r,s}^{(t)}]$ interval pattern embeds into $[\sigma_1,\sigma]$.)
\end{theorem}
Note that the case $\sigma_0=e$ is essentially a reformulation of the original result of \cite{MR1051089}.
\subsection{} \label{sec: rmvi}
Given $\sigma\in S_k$ and an index $i$, we write for simplicity $\rmv_i(\sigma)=\rmv_{\{i\}}(\sigma)$.
The following result is probably well known. For convenience we include the proof.
\begin{lemma} \label{lem: rmvi}
Suppose that $(\sigma,\sigma_0)$ is a smooth pair in $S_k$ and $i$ is an index such that $\sigma(i)=\sigma_0(i)$.
Then $(\rmv_i(\sigma),\rmv_i(\sigma_0))$ is a smooth pair in $S_{k-1}$.
\end{lemma}
Before giving the proof, we first recall the following elementary fact.
\begin{lemma} (\cite[Lemma 17]{MR1990570}) \label{lem: elemi}
Suppose that $\sigma_0,\sigma\in S_k$ and $i$ is an index such that $\sigma(i)=\sigma_0(i)$. Then $\sigma_0\le\sigma$ if and only if
$\rmv_i(\sigma_0)\le\rmv_i(\sigma)$.
\end{lemma}
Next, we introduce some notation.
For any index $i$ let
\[
\rflx_i=\{t\in\rflx:t(i)\ne i\}
\]
(so that $\#\rflx_i=k-1$).
For any $\sigma\in S_k$ let
\[
\ell_i(\sigma)=\#\{t\in\rflx_i:\sigma t<\sigma\}=\#\{r<i:\sigma(r)>\sigma(i)\}+\#\{r>i:\sigma(r)<\sigma(i)\}
\]
so that $2\ell(\sigma)=\sum_i\ell_i(\sigma)$.
For any $\sigma_0\le\sigma$ let $\rsig_i(\sigma_0,\sigma)=\rsig(\sigma_0,\sigma)\cap\rflx_i$
and $\asig_i(\sigma_0,\sigma)=\asig(\sigma_0,\sigma)\cap\rflx_i$.
We need another (probably well-known) result.
\begin{lemma} \label{lem: lcldeod}
Suppose that $\sigma_0\le\sigma$ and let $i$ be an index such that $\sigma(i)=\sigma_0(i)$. Then
\[
\#\asig_i(\sigma_0,\sigma)\ge\ell_i(\sigma).
\]
Equivalently, $\#\rsig_i(\sigma_0,\sigma)\ge\ell_i(\sigma)-\ell_i(\sigma_0)$.
\end{lemma}
\begin{proof}
We prove the result by induction on $\ell(\sigma)-\ell(\sigma_0)$.
If $\sigma=\sigma_0$ the assertion is trivial.
Otherwise, it follows from \cite[Proposition 14]{MR1990570} and the fact that $\sigma_0(i)=\sigma(i)$
that $\rsig(\sigma_0,\sigma)\ne\rsig_i(\sigma_0,\sigma)$.
For the induction step, take any $t\in\rsig(\sigma_0,\sigma)\setminus\rflx_i$ and let
\[
\phi_t=\phi_t^{\sigma_0t,\sigma}:\asig(\sigma_0t,\sigma)\rightarrow\asig(\sigma_0,\sigma)
\]
be the injective map defined in \cite[Lemma 2.2]{MR1827861} (cf.~\cite[\S6]{MR1990570}).
It is easy to see from the definition that $\phi_t(\asig_i(\sigma_0t,\sigma))\subset\asig_i(\sigma_0,\sigma)$.
Thus, $\#\asig_i(\sigma_0,\sigma)\ge\#\asig_i(\sigma_0t,\sigma)$ and the assertion follows from the induction hypothesis.
\end{proof}
Finally, we can prove Lemma \ref{lem: rmvi}.
\begin{proof}[Proof of Lemma \ref{lem: rmvi}]
By Proposition \ref{prop: LS} and the assumption we have
\[
\#\{t\in\rflx:\sigma_0t\le\sigma\}=\ell(\sigma).
\]
For simplicity, write $\tilde\sigma_0=\rmv_i(\sigma_0)$ and $\tilde\sigma=\rmv_i(\sigma)$.
By Lemma \ref{lem: elemi} we have $\tilde\sigma_0\le\tilde\sigma$.
If $t\in\rflx\setminus\rflx_i$ then $\sigma_0t(i)=\sigma_0(i)=\sigma(i)$ and $\rmv_i(\sigma_0t)=
\rmv_i(\sigma_0)t'$ where $t'=\rmv_i(t)$ is a reflexion in $S_{k-1}$.
Thus, again by Lemma \ref{lem: elemi},
$\sigma_0 t\le\sigma$ if and only if $\tilde\sigma_0t'\le\tilde\sigma$.
Clearly $\rmv_i$ induces a bijection between $\rflx\setminus\rflx_i$ and the set $\rflx'$ of reflexions in $S_{k-1}$.
It follows from Lemma \ref{lem: lcldeod} that
\begin{multline*}
\#\{t'\in\rflx':\tilde\sigma_0t'\le\tilde\sigma\}=
\#\{t\in\rflx:\sigma_0t\le\sigma\}-\#\{t\in\rflx_i:\sigma_0t\le\sigma\}\\
=\ell(\sigma)-\#\{t\in\rflx_i:\sigma_0t\le\sigma\}\le\ell(\sigma)-\ell_i(\sigma)=\ell(\tilde\sigma).
\end{multline*}
By Proposition \ref{prop: LS} once again it follows that $(\tilde\sigma,\tilde\sigma_0)$ is a smooth pair as required.
\end{proof}
\section{Balanced multisegments} \label{sec: combi}
In this section we introduce the main combinatorial condition on multisegments for which our main result applies
and relate it to the results of the previous section.
\subsection{} \label{sec: biseq}
Henceforth we fix an integer $k\ge1$.
Let $\tseq$ be a pair of two sequences of integers $a_1\le\dots\le a_k$ and $b_1\ge\dots\ge b_k$ such that
$a_{k+1-i}\le b_i+1$ for all $i$.
We write $\tseq$ as $\bitmplt$ and refer to it simply as a \emph{\biseq}.
By \cite[Lemma 15]{MR3163355}, there exists a unique $\sigma_0=\sigma_0(\tseq)\in S_k$ such that for any $\sigma\in S_k$
we have
\begin{equation} \label{def: propsigma0}
a_{\sigma^{-1}(i)}\le b_i+1\text{ for all $i$ if and only if }\sigma\ge\sigma_0
\end{equation}
(in the Bruhat order).\footnote{This was stated in \cite{MR3163355} under the assumption that $a_1<\dots<a_k$
and $b_1>\dots>b_k$, but the proof, which is in any case elementary, works under the weaker assumption.}
Moreover, $\sigma_0$ is $213$-avoiding, i.e. there do not exist indices $a<b<c$ such that
$\sigma_0(b)<\sigma_0(a)<\sigma_0(c)$. In other words, $i\mapsto k+1-\sigma_0(i)$ is s stack-sortable permutation in the sense of Knuth \cite[\S2.2.1]{MR3077152}.
The permutation $\sigma_0$ is defined recursively as follows. Given $\sigma_0^{-1}(k),\dots,\sigma_0^{-1}(i+1)$ we set
\begin{equation} \label{def: sigma0}
\sigma_0^{-1}(i)=\max\{j\notin\sigma_0^{-1}(\{i+1,\dots,k\}):a_j\le b_i+1\}
\end{equation}
which is well-defined since $a_{k+1-i}\le b_i+1$.
It follows that $\sigma_0(i)<\sigma_0(i+1)$ whenever $a_i=a_{i+1}$ and $\sigma_0^{-1}(i)<\sigma_0^{-1}(i+1)$ whenever $b_i=b_{i+1}$.
\begin{example}
For $l\ge0$ let
\begin{equation} \label{def: akl}
\tseq_{k,l}=\begin{pmatrix}1&2&\dots&k\\k+l-1&k+l-2&\dots&l\end{pmatrix}.
\end{equation}
Then
\[
\sigma_0(\tseq_{k,l})(i)=\begin{cases}i&i\le l,\\k+l+1-i&\text{otherwise.}\end{cases}
\]
The condition $\sigma\ge\sigma_0(\tseq)$ becomes $\sigma(i)\le k+l+1-i$ for all $i>l+1$.
\end{example}
For any $\sigma\in S_k$ let
\[
\m_\sigma=\m_\sigma(\tseq)=\sum_{i=1}^k[a_{\sigma^{-1}(i)},b_i]
\]
and $\pi_\sigma=\zele{\m_\sigma}$. Thus, $\m_\sigma\in\Mult$ (i.e., $\pi_\sigma\ne0$) if and only if $\sigma\ge\sigma_0$.
Note that $\m_{\sigma_0}$ is pairwise unlinked.
If we want to stress the dependence on $\rho$ we will write $\m_\sigma^{(\rho)}(\tseq)$.
\begin{remark}
Clearly, as we vary $\tseq$ and $\sigma\in S_k$ (with $\sigma\ge\sigma_0(\tseq)$), $\m_\sigma(\tseq)$ range over all
multisegments with $\le k$ segments.
More precisely, given $\m=\Delta_1+\dots+\Delta_k$, we may assume that $e(\Delta_1)\ge\dots\ge e(\Delta_k)$.
We take $\sigma\in S_k$ such that $b(\Delta_{\sigma(i)})\le b(\Delta_{\sigma(j)})$ whenever $i>j$ and set
$b_i=e(\Delta_i)$ and $a_i=b(\Delta_{\sigma(i)})$. Finally, if we sort the sequence $b(\Delta_1),\dots,b(\Delta_k),e(\Delta_1)+\frac32,\dots,e(\Delta_k)+\frac32$
as $c_1\le\dots\le c_{2k}$ and replace $c_i$ by the letter $X$ if $c_i\in\{b(\Delta_1),\dots,b(\Delta_k)\}$ and by the letter $Y$ otherwise
then we get a Dyck word $\word$ of length $2k$ and $\sigma_0(\tseq)$ is the $213$-avoiding permutation corresponding to $\word$ (see Lemma \ref{lem: comb12}
below).
\end{remark}
Note that
\begin{equation} \label{eq: contrasigma}
\m_\sigma^{(\rho)}(\tseq_{k,l})^\vee=\m_{\sigma^{-1}}^{(\rho^*)}(\tseq_{k,l})\text{ where }\rho^*=(\rho\nu_\rho^{k+l})^\vee.
\end{equation}
We say that a \biseq\ $\tseq=\bitmplt$ is regular if $a_1<\dots<a_k$ and $b_1>\dots>b_k$.
In this case the multisegments $\m_\sigma(\tseq)$, $\sigma \geq \sigma_0$, are regular and distinct.
Moreover, $\m_{\sigma_1}(\tseq)\obt\m_{\sigma_2}(\tseq)$ if and only if $\sigma_1\le\sigma_2$.
(In the non-regular case it is still true that $\m_{\sigma_1}(\tseq)\obt\m_{\sigma_2}(\tseq)$ if $\sigma_1\le\sigma_2$.)
For completeness we recall a standard combinatorial result (cf. \cite[Exercise 6.19]{MR1676282}).
A Dyck word of length $2k$ is a string composed of the letters $X$ and $Y$, each appearing $k$ times, such that in any initial segment,
the number of $Y$'s does not exceed the number of $X$'s. It is well-known that the number of Dyck words of length $2k$ is
the Catalan number $C_k={2k\choose k}-{2k\choose k+1}$.
\begin{lemma} \label{lem: comb12}
The following sets are in natural bijections:
\begin{enumerate}
\item Dyck words of length $2k$.
\item Regular \biseq s $\tseq=\bitmplt$ such that $a_1=2$ and $b_1=2k-1$.
\item $213$-avoiding permutations in $S_k$.
\end{enumerate}
\end{lemma}
\begin{proof}
Given a Dyck word $\word$ of length $2k$ let $a_i'$ (resp., $b_i'$) $i=1,\dots,k$ be the position of the $i$-th $X$ (resp., $Y$)
from the left (resp., from the right). Then $a'_i<b'_{k+1-i}$.
Letting $a_i=a_i'+1$ and $b_i=b'_i-1$, $i=1,\dots,k$, we get a regular \biseq\ $\tseq(\word)=\bitmplt$ with $a_1=2$, $b_1=2k-1$.
To any \biseq\ $\tseq=\bitmplt$
we assign the $213$-avoiding permutation $\sigma_0(\tseq)$ defined by \eqref{def: propsigma0} and \eqref{def: sigma0}.
Finally, given any permutation $\sigma_0\in S_k$ we assign the Dyck word $\word(\sigma_0)$ such that for any $i$, the number of $X$'s to the left of the
$i$-th $Y$ from the right is $x_i=\max_{j\ge i}\sigma^{-1}(j)$. (Clearly, $x_1\ge\dots\ge x_k$ and $x_i\ge k+1-i$ for all $i$.)
It is easy to see that $\tseq(\word(\sigma_0(\tseq')))=\tseq'$, $\word(\sigma_0(\tseq(\word')))=\word'$ and
$\sigma_0(\tseq(\word(\sigma)))=\sigma$ for any regular \biseq\ $\tseq'=\bitmplt$ with $a_1=2$ and $b_1=2k-1$,
a Dyck word $\word'$ of length $2k$ and a $213$-avoiding permutation $\sigma\in S_k$.
\end{proof}
\begin{remark}
For any permutation $\sigma\in S_k$, the permutation $\sigma_0(\tseq(\word(\sigma)))$ is the unique
maximal (with respect to Bruhat order) $213$-avoiding permutation $\le\sigma$.
It is obtained from $\sigma$ by repeatedly interchanging $\sigma(i)$ and $\sigma(j)$ whenever $i<j<k$ and $\sigma(j)<\sigma(i)<\sigma(k)$.
\end{remark}
\subsection{}
We now introduce the key combinatorial property.
\begin{definition}
Let $\m$ be a multisegment.
\begin{enumerate}
\item We say that $\m$ is almost pairwise unlinked (\APU) if there exists a pairwise unlinked multisegment $\m'$ such that $\m'\adj\m$.
\item The complexity of $\m$ (denoted $\cmplx(\m)$) is the maximal integer $l\ge0$ for which there exists a chain of multisegments
$\m_l\adj\dots\adj\m_1\adj\m$.
\item The depth of $\m$ (denoted $\depth(\m)$) is the number of \APU\ multisegments $\obt\m$.
\item If $\m$ is regular, we say that $\m$ is balanced $\depth(\m)=\cmplx(\m)$.
\end{enumerate}
\end{definition}
Note that $\cmplx(\m)=0$ if and only if $\depth(\m)=0$ if and only if $\m$ is pairwise unlinked.
\begin{example} \label{ex: lecorig}
Let $\tseq=\tseq_{4,2}$. Here $\sigma_0(\tseq)=(1243)$ (where we use the notational convention $\sigma=(\sigma(1)\dots\sigma(k))$)
and $\m_{\sigma_0}(\tseq)=[1,5]+[2,4]$. In the following table we list the \APU\ multisegments $\m$ such that $\m_{\sigma_0}\adj\m$
and the corresponding permutation $\sigma$ such that $\m=\m_\sigma(\tseq)$.
\begin{tabular}{c|l}
$\sigma$ & $\m_\sigma(\tseq)$\\
\hline
$(1342)$ & $[1,5]+[4,4]+[2,3]$\\
$(3241)$ & $[4,5]+[2,4]+[1,3]$\\
$(1423)$ & $[1,5]+[3,4]+[2,2]$\\
$(4213)$ & $[3,5]+[2,4]+[1,2]$\\
$(2143)$ & $[2,5]+[1,4]$
\end{tabular}
Let
\[
\m=\m_\sigma(\tseq)=[4,5]+[2,4]+[3,3]+[1,2]
\]
where $\sigma=(4231)$.
All the \APU\ multisegments in the table above are $\obt\m$ and therefore $\depth(\m)=5$.
On the other hand, the chain
\begin{align*}
[1,5]+[2,4]&\adj\\ [2,5]+[1,4]&\adj\\ [2,5]+[3,4]+[1,2]&\adj\\ [2,5]+[4,4]+[3,3]+[1,2]&\adj\m
\end{align*}
is of maximal length and therefore $\cmplx(\m)=4$. In conclusion, $\m$ is not balanced.
\end{example}
In general, it is easy to see by induction on the number of segments that if $\m=\Delta_1+\dots+\Delta_k$ is regular then
$\cmplx(\m)=\#\X_\m$. Indeed, if $\X_{\m}=\emptyset$ then $\m$ is pairwise unlinked and the assertion is clear.
Otherwise, let $(i,j)\in\X_\m$ and let $\m'\adj\m$ be the multisegment obtained from $\m$
by replacing the pair $(\Delta_i,\Delta_j)$ with its offspring. It is easy to see that $\# \X_{\m'}\le\# \X_{\m}-1$
with an equality if there does not exist an index $l$ such that $(i,l),(l,j)\in\X_\m$. The induction step follows.
Let $\tseq=\bitmplt$ be a regular \biseq\ and let $\sigma_0=\sigma_0(\tseq)$.
Then for any $\sigma\ge\sigma_0$ we have
\[
\depth(\m_\sigma(\tseq))=\#\{t\in\rflx:\sigma_0t\in [\sigma_0,\sigma]\}
\]
and
\[
\cmplx(\m_\sigma(\tseq))=\ell(\sigma)-\ell(\sigma_0).
\]
Thus, we get the following consequence of Proposition \ref{prop: LS}.
\begin{corollary}\label{cor: fgd and smth}
For any regular multisegment $\m$ we have $\depth(\m)\ge\cmplx(\m)$.
Moreover, if $\tseq=\bitmplt$ is a regular \biseq\ and $\sigma\ge\sigma_0(\tseq)$ then
$\m_\sigma(\tseq)$ is balanced if and only if $(\sigma,\sigma_0(\tseq))$ is a smooth pair
\end{corollary}
In Proposition \ref{prop: nonsmth} below we will give a simpler combinatorial characterization of balanced multisegments,
using the results of \cite{MR1990570, MR1994224, MR2015302, MR1853139}.
The following is an immediate consequence of \eqref{eq: redrelsmth}.
\begin{corollary} \label{cor: si0le}
Let $\m$ and $\m'$ be two regular multisegments. Write $\m=\Delta_1+\dots+\Delta_k$ and $\m'=\Delta_1'+\dots+\Delta'_{k'}$
with $e(\Delta_1)>\dots>e(\Delta_k)$, $e(\Delta'_1)>\dots>e(\Delta'_{k'})$. Assume that
\begin{enumerate}
\item $k'=k$.
\item $e(\Delta_i')=e(\Delta_i)$ for all $i$.
\item $b(\Delta'_i)\ge b(\Delta_i)$ for all $i$.
\item For all $i\ne j$ we have $b(\Delta_i)<b(\Delta_j)$ if and only if $b(\Delta'_i)<b(\Delta'_j)$.
\item $\m$ is balanced.
\end{enumerate}
Then $\m'$ is balanced.
\end{corollary}
Indeed, if we write $\m=\m_\sigma(\tseq)$ for a \biseq\ $\tseq=\bitmplt$ and $\sigma\in S_k$
then $\m'=\m_\sigma(\tseq)$ for some \biseq\ $\tseq'$ such that $\sigma_0(\tseq')\ge\sigma_0(\tseq)$.
Hence, the corollary follows from \eqref{eq: redrelsmth}.
Similarly, we can infer the following from the results of \S\ref{sec: rmvi}.
\begin{lemma} \label{lem: submlt}
A sub-multisegment of a balanced multisegment is balanced.
\end{lemma}
\begin{proof}
By induction, it is enough to check that if $\m$ is balanced then $\m'=\m-\Delta$ is balanced for any segment $\Delta$ in $\m$.
We may assume that $\m=\m_\sigma(\tseq)$ for a \biseq\ $\tseq=\bitmplt$ and let $\sigma_0=\sigma_0(\tseq)$.
Then for some $i$, $\m'=\m_{\sigma'}(\tseq')$ where $\sigma'=\rmv_i(\sigma)$ and $\tseq'$ is obtained from $\tseq$ by removing $a_i$ and $b_{\sigma(i)}$.
Let $\sigma_0'=\sigma_0(\tseq')\in S_{k-1}$ and let $\tilde\sigma$ be ``unflattening'' of $\sigma_0'$, namely the (unique) permutation in $S_k$ such that
$\tilde\sigma(i)=\sigma(i)$ and $\rmv_i(\tilde\sigma)=\sigma_0'$.
By Lemma \ref{lem: elemi} $\tilde\sigma\le\sigma$. It is also easy to see that $\sigma_0\le\tilde\sigma$.
Indeed, (cf.~\eqref{def: propsigma0}) the relations $a_{\tilde\sigma^{-1}(j)}\le b_j+1$, $j\ne\sigma(i)$ amount to the property $\sigma_0'\le\sigma'$,
while the corresponding inequality for $j=\sigma(i)$ also holds because $\tilde\sigma(i)=\sigma(i)$.
Thus, $(\sigma,\tilde\sigma)$ is a smooth pair (by \eqref{eq: redrelsmth}), and hence $(\sigma',\sigma_0')$ is a smooth pair by Lemma \ref{lem: rmvi}.
The lemma follows.
\end{proof}
\subsection{}
Corollary \ref{cor: fgd and smth} gives an efficient way to detect whether a given regular multisegment is balanced.
However, it will be useful to have another combinatorial criterion for balanced (regular) multisegments.
\begin{definition}
We say that a regular multisegment $\m=\Delta_1+\dots+\Delta_k$, $k\ge4$
with $e(\Delta_1)>\dots>e(\Delta_k)$ is of type $4231$ (resp., $3412$) if
\[
\Delta_{i+1}\prec\Delta_i,\ i=3,\dots,k-1,\
\Delta_3\prec \Delta_1\text{ and }b(\Delta_k)<b(\Delta_2)<b(\Delta_{k-1})
\]
(resp.,
\[
\Delta_{i+1}\prec\Delta_i,\ i=4,\dots,k-1,\
\Delta_4\prec\Delta_2,\text{ and }b(\Delta_3)<b(\Delta_k)<b(\Delta_1)<b(\Delta_l)
\]
where $l=2$ if $k=4$ and $l=k-1$ otherwise).
\end{definition}
\begin{example} \label{ex: bexam}
The ``minimal'' examples of multisegments of type $4231$ and $3412$ for $k\ge4$ are given by
\begin{gather*}
\m=[k,k+1]+[2,k]+[k-1]+[k-2]+\dots+[3]+[1,2]=\m_\sigma(\tseq_{k,2})\\
\text{where }\sigma(i)=\tau_{2,k-2}^{(1)}(i)=\begin{cases}k&i=1,\\2&i=2,\\k-i+2&i=3,\dots,k-1,\\1&i=k,\end{cases}
\end{gather*}
and
\begin{gather*}
\m=[3,k+2]+[k,k+1]+[1,k]+[k-1]+[k-2]+\dots+[4]+[2,3]=\m_\sigma(\tseq_{k,3})\\
\text{ where }\sigma(i)=\tau_{2,k-2}^{(2)}(i)=\begin{cases}3&i=1,\\k&i=2,\\1&i=3,\\
k-i+3&i=4,\dots,k-1,\\2&i=k,\end{cases}
\end{gather*}
respectively.
The corresponding drawings for $k=6$ are
\[
\xymatrix@=0.6em{
&&&&&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&\circ&&&\\
&&&\circ&&&&\\
&&\circ&&&&&\\
\circ\ar@{-}[r]&\circ&&&&&}
\text{ and }
\xymatrix@=0.6em{
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&\circ\ar@{-}[r]&\circ&\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&&&\\
&&&&\circ&&&\\
&&&\circ&&&&\\
&\circ\ar@{-}[r]&\circ&&&&&}
\]
\end{example}
More generally, using the notation of \S\ref{sec: minsingBL}, for any \biseq\ $\tseq=\bitmplt$ we have
\begin{multline} \label{eq: 4231 or 3412}
\text{$\m_\sigma(\tseq)$ is of type $4231$ (resp., $3412$) if and only if $\sigma=\tau_{2,k-2}^{(t)}$
and $\sigma_0(\tseq)\le\delta_{2,k-2}^{(t)}$}
\\\text{where $t=1$ (resp., $t=2$).}
\end{multline}
This easily follows from the defining property \eqref{def: propsigma0} of $\sigma_0(\tseq)$.
Note that since $\sigma_0(\tseq)$ is $213$-avoiding, $\sigma_0(\tseq)\le\delta_{2,k-2}^{(t)}$ if and only if $\sigma_0(\tseq)\le\sigma_0(\tseq_{k,t+1})$.
\begin{remark} \label{rem: types not GLS}
If $\m$ is of type $4231$ or $3412$ then $\m$ does not satisfy \GLS. Indeed, in the terminology of Remark \ref{rem: kirred},
$\X_\m$ contains the $k$ irreducible pairs $(2,1)$, $(3,1)$, $(i+1,i)$, $i=3,\dots,k-1$, $(k,2)$ (resp.,
$(3,1)$, $(3,2)$, $(4,2)$, $(i+1,i)$, $i=4,\dots,k-1$, $(k,1)$) in the $4231$ (resp., $3412$) case.
\end{remark}
\begin{proposition} \label{prop: nonsmth}
A regular multisegment $\m$ is balanced if and only if $\m$ does not admit a sub-multisegment of type $4231$ or $3412$.
\end{proposition}
\begin{proof}
For the `only if' direction we may assume, by Lemma \ref{lem: submlt}, that $\m$ itself is of type $4231$ or $3412$.
In this case the claim follows from \eqref{eq: 4231 or 3412}, \eqref{eq: redrelsmth}, \eqref{eq: pairs not smooth} and Corollary \ref{cor: fgd and smth}.
For the converse direction, assume that $\m$ is regular but not balanced. Write $\m=\m_\sigma(\tseq)$ for a \biseq\ $\tseq=\bitmplt$
and let $\sigma_0=\sigma_0(\tseq)$. By Corollary \ref{cor: fgd and smth}, $(\sigma,\sigma_0)$ is not a smooth pair.
Let $\sigma_1,I,r,s,t$ be as in Theorem \ref{thm: BW}.
By passing to the sub-multisegment determined by $I$ we may assume that $I=\emptyset$.
Removing the segments $\Delta_i$, $2<i\le r$ (if $r>2$) we obtain a sub-multisegment of type $4231$
if $t=1$ and of type $3412$ if $t$ is either $2$ or $3$.
\end{proof}
\begin{remark} \label{rem: contrblncd}
Suppose that $\m$ is a regular multisegment which is $\rho$-contractible for some $\rho\in\supp\m$.
(See Definition \ref{def: contractible}.) Then $\m$ is balanced if and only if the $\rho$-contraction of $\m$ is balanced.
\end{remark}
\section{The main result} \label{sec: main}
Our main result is the following.
\begin{theorem} \label{thm: main}
Suppose that $\m$ is a regular multisegment. Then the following conditions are equivalent:
\begin{enumerate}
\item \label{item: mbal} $\m$ is balanced.
\item \label{item: zmLM} $\zele{\m}$ is \LM.
\item \label{item: mGLS} $\m$ satisfies \GLS.
\end{enumerate}
\end{theorem}
In this section we will prove the implications \ref{item: mbal}$\implies$\ref{item: zmLM} and \ref{item: mbal}$\implies$\ref{item: mGLS}.
In view of Propositions \ref{prop: LS} and \ref{prop: nonsmth}, Corollary \ref{cor: fgd and smth} and formula \eqref{def: sigma0},
Theorem \ref{thm: main} implies Theorem \ref{thm: maini} of the introduction, except for conditions \ref{cond: Lreal} and \ref{cond: detform}
which will be dealt with in the remark below and in \S\ref{sec: KLid} respectively.
(Recall that conditions \ref{cond: smlocus}--\ref{cond: KLall} are equivalent by \cite{MR788771}.)
\begin{remark}
Using the quantum Schur--Weyl duality \cite{MR1405590} we may translate Theorem \ref{thm: main} to the language of representation theory
of the quantum affine algebra $U_q(\widehat{\sll}_N)$ when $q$ is not a root of unity.
Recall that the finite-dimensional irreducible representations of $U_q(\widehat{\sll}_N)$ are parameterized by Drinfeld polynomials,
or what amounts to the same, by monomials in the formal variables $Y_{i,a}$, $i=1,\dots,N-1$, $a\in\C^*$ (e.g., \cite{MR2642561}).
We write $L(M)$ for the irreducible representation corresponding to a monomial $M$.
\begin{corollary} \label{cor: mainq}
Suppose that $\m=\sum_{i=1}^k[a_i,b_i]$ is a regular multisegment and $N>b_i-a_i+1$ for all $i$.
Let $M=\prod_{i=1}^kY_{b_i-a_i+1,q^{a_i+b_i}}$.
Then $L(M)$ is real (i.e., $L(M)\otimes L(M)$ is irreducible) if $\m$ is balanced. The converse also holds provided that
$N>2\sum_{i=1}^k(b_i-a_i+1)$.\footnote{This lower bound is far from optimal. See Remark \ref{rem: lwrbndN}.}
\end{corollary}
\end{remark}
\begin{proof}[Proof of the implication \ref{item: mbal}$\implies$\ref{item: zmLM} of Theorem \ref{thm: main}]
We argue by induction on the number of segments $k$ in $\m$. The base of the induction is the trivial case $k=0$.
For the induction step we use Lemma \ref{lem: albertoidea}.
Write $\m=\m_\sigma(\tseq)$ where $\tseq=\bitmplt$ is a \biseq\ and $\sigma\in S_k$.
Write $\Delta_i=[a_{\sigma^{-1}(i)},b_i]$ so that $\m=\Delta_1+\dots+\Delta_k$ and let $\sigma_0=\sigma_0(\tseq)\in S_k$.
By assumption $(\sigma,\sigma_0)$ is smooth.
For convenience write $a'_i=a_{k+1-i}$ and set $\sigma'(i)=\sigma(k+1-i)$ so that $b(\Delta_{\sigma'(1)})=a'_1>\dots>b(\Delta_{\sigma'(k)})=a'_k$.
We construct $\pi_1$ and $\pi_2$ as follows. Let $m\ge1$ be the largest integer such that $\sigma'(1)<\dots<\sigma'(m)$.
We define indices $n_1<\dots<n_m$ with $n_i\ge\sigma'(i)$ for all $i$ recursively as follows.
We take $n_m=\sigma'(m)$ and given $n_{i+1}$, $1\le i<m$ we define $n_i$ to be the largest index $\sigma'(i)\le j<n_{i+1}$
such that $\lshft{\Delta}_j\prec\Delta_{\sigma'(i)}$.
Let $\Delta_i'=[a'_i,b_{n_i}]\subset\Delta_{\sigma'(i)}$, $i=1,\dots,m$ and let $l>1$ be the largest integer $\le m$ such that
$\Delta'_l\not\prec\Delta'_{l-1}$ (i.e., such that $a'_{l-1}>b_{n_l}+1$) if such an index exists; otherwise let $l=1$.
We take $\m_1=\sum_{i=l}^m\Delta_i'$ and $\m_2=\sum_{i=1}^k\Delta''_i$ where $\Delta''_{\sigma'(i)}=
\Delta_{\sigma'(i)}\setminus\Delta'_i=[b_{n_i}+1,b_{\sigma'(i)}]$, $i=l,\dots,m$ and $\Delta''_j=\Delta_j$
if $j\notin\sigma'(\{l,\dots,m\})$.
Let $\pi_i=\zele{\m_i}$, $i=1,2$. Clearly $\pi_1$ is a ladder and $\m_2$ is regular.
The induction step will follow from Lemma \ref{lem: albertoidea} and the lemma below.
\end{proof}
\begin{lemma} \label{lem: indstep}
We have
\begin{enumerate}
\item $\pi\hookrightarrow\pi_1\times\pi_2$.
\item $\m_2$ is balanced. Hence, by induction hypothesis $\pi_2$ is \LM.
\item $\pi\times\pi_1$ is irreducible.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\m_3=\sum_{i=1}^{l-1}\Delta_{\sigma'(i)}$, $\m_4=\sum_{i=l}^m\Delta''_{\sigma'(i)}$ and
$\m_5=\sum_{j\notin\sigma'(\{1,\dots,m\})}\Delta_j$ so that $\m_2=\m_3+\m_4+\m_5$ and $\m_5<_b\m_4<_b\m_3$.
Set $\pi_i=\zele{\m_i}$, $i=3,4,5$. Thus,
\[
\pi_2\hookrightarrow\pi_3\times\pi_4\times\pi_5
\]
and therefore
\[
\pi_1\times\pi_2\hookrightarrow\pi_1\times\pi_3\times\pi_4\times\pi_5.
\]
Note that $\pi_1\times\pi_3$ is irreducible since no segment in $\m_1$ in linked with any segment in $\m_3$
(by the definition of $l$). Thus, $\pi_1\times\pi_3\simeq\pi_3\times\pi_1$ is a ladder, and in particular, \LM.
Thus, $\pi_1\times\pi_3\times\pi_4$ is \SI. Since $\lnrset(\pi_5)\cap\supp(\pi_i)=\emptyset$, $i=1,3,4$, it follows that
$\pi_1\times\pi_3\times\pi_4\times\pi_5$ is \SI\ (\cite[Lemma 1.5]{MR3573961}). Thus,
\[
\soc(\pi_1\times\pi_2)=\soc(\pi_1\times\pi_3\times\pi_4\times\pi_5)
\simeq\soc(\pi_3\times\pi_1\times\pi_4\times\pi_5)=\soc(\soc(\pi_3\times\soc(\pi_1\times\pi_4))\times\pi_5).
\]
Let $\m_6=\sum_{i=l}^m\Delta_{\sigma'(i)}$ and $\pi_6=\zele{\m_6}$. Note that $\pi_6$ is a ladder.
Moreover, by Lemma \ref{lem: chopladders} we have $\pi_6=\soc(\pi_1\times\pi_4)$. (Note that $\pi_4$ is a ladder.)
Since $\m_5<_b\m_6<_b\m_3$ it follows that
\[
\soc(\pi_1\times\pi_2)=\zele{\m_3+\m_6+\m_5}=\zele{\m}
\]
and the first part follows.
Next, we show that $\m_2$ is balanced.
Note that if $i>l$ is such that $\Delta'_i\ne\Delta_{\sigma'(i)}$, i.e., $n_i>\sigma'(i)$, then it follows from the definition of $n_{i-1}$
and $l$ that $n_{i-1}\ge\sigma'(i)>\sigma'(i-1)$ and therefore $\Delta'_{i-1}\ne\Delta_{\sigma'(i-1)}$.
Let $l' \leq m$ be the smallest index $\ge l$ such that $\Delta'_{l'}=\Delta_{\sigma'(l')}$, i.e., such that $n_{l'}=\sigma'(l')$.
By the above, $n_i=\sigma'(i)$ for all $i\ge l'$.
Let $a''_i=b(\Delta_i)$ if $i\notin\sigma'(\{l,\dots,m\})$ and $a''_{\sigma'(i)}=b_{n_i}+1$ for $l\le i\le m$.
Thus, $\Delta''_i=[a''_i,b_i]$ for all $i$. It is easy to see that $a''_i<a''_j$ if and only if
$b(\Delta_i)<b(\Delta_j)$ (i.e., if and only if $\sigma'^{-1}(i)>\sigma'^{-1}(j)$).
It follows from Lemma \ref{lem: submlt} and Corollary \ref{cor: si0le} that $\m_2$ is balanced.
Finally, we show the irreducibility of $\pi\times\pi_1$ using the combinatorial condition given by Theorem \ref{thm: laddercomb}.
The condition $\LI(\pi_1,\pi)$ is easy to check and does not depend on the condition on $\m$.
Indeed, recall that (cf.~Definition \ref{def: XandY})
\[
\X_{\m_1;\m}=\{(i,j)\in\{l,\dots,m\}\times\{1,\dots,k\}:\Delta'_i\prec\Delta_j\}
\]
and
\[
\Y_{\m_1;\m}=\{(i,j)\in\{l,\dots,m\}\times\{1,\dots,k\}:\lshft{\Delta'_i}\prec\Delta_j\}.
\]
If $(i,j)\in \X_{\m_1;\m}$ then $j=\sigma'(i')$ for some $l<i'<i$ and hence $(i-1,j)\in \Y_{\m_1;\m}$.
Thus $(i,j)\mapsto(i-1,j)$ is a matching from $\X_{\m_1;\m}$ to $\Y_{\m_1;\m}$.
The condition $\RI(\pi_1,\pi)$ is more delicate and relies on the assumption on $\m$.
Recall
\[
\X_{\m;\m_1}=\{(i,j)\in\{1,\dots,k\}\times\{l,\dots,m\}:\Delta_i\prec\Delta'_j\}
\]
and
\[
\Y_{\m;\m_1}=\{(i,j)\in\{1,\dots,k\}\times\{l,\dots,m\}:\lshft{\Delta}_i\prec\Delta'_j\}.
\]
Let $l'$ be as above.
Consider first the case where $l'=l$, i.e., $\Delta'_i=\Delta_{\sigma'(i)}$ for all $i=l,\dots,m$.
For any $(i,j)\in \X_{\m;\m_1}$ let $h(i,j)$ be the largest index $r$ such that
$\Delta_i\prec\Delta_r$ and $\lshft{\Delta}_r\prec\Delta'_j$. Since $\Delta'_j=\Delta_{\sigma'(j)}$,
$h(i,j)$ is well-defined and $\sigma'(j)\le h(i,j)<i$.
We claim that the function $f:\X_{\m;\m_1}\rightarrow \Y_{\m;\m_1}$ given by $f(i,j)=(h(i,j),j)$ is a matching.
The only issue is injectivity.
Suppose on the contrary that $i_1<i_2$, $(i_1,j),(i_2,j)\in \X_{\m;\m_1}$ and $h(i_1,j)=h(i_2,j)$.
We cannot have $\Delta_{i_2}\prec\Delta_{i_1}$ since otherwise $h(i_2,j)\ge i_1$ while $h(i_1,j)<i_1$.
On the other hand, $b(\Delta_{i_1})<b(\Delta'_j)\le e(\Delta_{i_2})+1$, hence $b(\Delta_{i_1})\le e(\Delta_{i_2})$.
Since $i_1<i_2$ we must therefore have $b(\Delta_{i_1})<b(\Delta_{i_2})$.
Let $j'$ be the largest index $\ge j$ such that $\Delta_{i_1}\prec\Delta'_{j'}$ and $\Delta_{i_2}\prec\Delta'_{j'}$.
If $j'=m$ then $\Delta_{\sigma'(m+1)}+\Delta_{\sigma'(m)}+\Delta_{i_1}+\Delta_{i_2}$ forms a submultisegment of type $3412$
in contradiction with the assumption that $\m$ is balanced (Proposition \ref{prop: nonsmth}).
Thus, $j'<m$. We cannot have $\sigma'(j'+1)<i_1$ since otherwise $\Delta_{i_1},\Delta_{i_2}\prec\Delta'_{j'+1}$,
rebutting the maximality of $j'$. Also, we cannot have $\sigma'(j'+1)=i_1$ since $b(\Delta_{i_1})<b(\Delta_{i_2})$
and $i_2>\sigma'(j')$. Thus, $\sigma'(j'+1)>i_1$ and the definition of $n_{j'}$ would give $n_{j'}\ge i_1$, controverting our assumption.
Suppose now that $l'>l$. We claim that $(i,j)\mapsto(i,j+1)$ is a matching from $\X_{\m;\m_1}$ to $\Y_{\m;\m_1}$.
That is, for any $(i,j)\in \X_{\m;\m_1}$ we have $j<m$ and $\lshft{\Delta}_i\prec\Delta'_{j+1}$.
Suppose on the contrary that $(i,j)\in \X_{\m;\m_1}$ and either $j=m$ or $\lshft{\Delta}_i\not\prec\Delta'_{j+1}$.
Assume first that $j<l'$, i.e., that $n_j>\sigma'(j)$. Since $\Delta_i\prec\Delta'_j$ we must have $n_j<i$.
In particular, $j<m$. Since $\lshft{\Delta}_i\not\prec\Delta'_{j+1}$ we also have $n_{j+1}>i$.
From the definition of $n_j$ and the fact that $n_j<i$ it follows that $\lshft{\Delta}_i\not\prec\Delta'_j$.
Therefore, $e(\Delta_i)+1=b(\Delta'_j)$. However, then
$\Delta'_{j+1}\not\prec\Delta'_j$ since $e(\Delta'_{j+1})<e(\Delta_i)=b(\Delta'_j)-1$, repudiating the assumption that $j'\ge l$.
Assume now that $j\ge l'$, so that $\Delta'_j=\Delta_{\sigma'(j)}$.
For simplicity write $i_0=n_{l'-1}$ so that $\sigma'(l'-1)<i_0<\sigma'(l')$ and $\lshft{\Delta}_{i_0}\prec\Delta_{\sigma'(l'-1)}$.
If $b(\Delta_{i_0})>b(\Delta_i)$ (i.e., if $\sigma'^{-1}(i_0)>\sigma'^{-1}(i)$) then
\[
\Delta_{\sigma'(l'-1)}+\Delta_{i_0}+\Delta_{\sigma'(l')}+\dots+\Delta_{\sigma'(j)}+\Delta_i
\]
is a submultisegment of type $4231$ which violates the assumption that $\m$ is balanced by Proposition \ref{prop: nonsmth}.
Therefore $b(\Delta_{i_0})<b(\Delta_i)$.
Assume first that $j=m$ and let $i_1=\sigma'(m+1)$. by the definition of $m$, $i_1<\sigma'(m)$. By the definition of $l'$
we also have $i_1<\sigma'(l')$. Suppose first that $i_1>\sigma'(l'-1)$. Then as before,
\[
\Delta_{\sigma'(l'-1)}+\Delta_{i_1}+\Delta_{\sigma'(l')}+\dots+\Delta_{\sigma'(j)}+\Delta_i
\]
is a submultisegment of type $4231$, denying the assumption that $\m$ is balanced.
On the other hand, if $i_1<\sigma'(l'-1)$ then
\[
\Delta_{i_1}+\Delta_{\sigma'(l'-1)}+\Delta_{i_0}+\Delta_{\sigma'(l')}+\dots+\Delta_{\sigma'(j)}+\Delta_i
\]
is a submultisegment of type $3412$,
which once again violates the assumption on $\m$.
Thus $j<m$. It is now clear that $\lshft{\Delta}_i\prec\Delta'_{j+1}$, that is, $\sigma'(j+1)\le i$
for otherwise $n_j\le i<\sigma'(j+1)$, gainsaying the assumption $j\ge l'$.
This finishes the proof of the lemma, and hence the implication \ref{item: mbal}$\implies$\ref{item: zmLM} of Theorem \ref{thm: main}.
\end{proof}
\begin{example}
Consider the balanced multisegment
\[
\m=[12,14]+[9,13]+[6,12]+[3,10]+[1,9]+[2,8]+[5,6]+[4,5]
\]
\[
\xymatrix@=0.6em{
&&&&&&&&&&&\bullet\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&\bullet\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&\bullet\ar@{-}[r]&\bullet\ar@{-}[r]&\bullet\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&\bullet\ar@{-}[r]&\bullet\\
&&&\bullet\ar@{-}[r]&\bullet}
\]
Here, in the notation of the proof of Theorem \ref{thm: main} we have
$\sigma'=(12378465)$, $m=5$, $n_1=3$, $n_2=5$, $n_3=6$, $n_4=7$, $n_5=8$, $l=2$, $l'=4$.
We marked by solid dots the part of $\Delta_{\sigma'(i)}$ which belongs to $\Delta'_i$.
Thus, $\m_1=[9,9]+[6,8]+[5,6]+[4,5]$ and
\[
\m_2=[12,14]+[10,13]+[9,12]+[3,10]+[1,9]+[2,8]
\]
\[
\xymatrix@=0.6em{
&&&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ}
\]
\end{example}
\begin{proof}[Proof of the implication \ref{item: mbal}$\implies$\ref{item: mGLS} of Theorem \ref{thm: main}]
We will show in fact that there exists an extra-strong $\rltn$-matching from $\X_\m$ to $\Y_\m$.
(See Remark \ref{rem: strongmatch}.)
The argument parallels the one above for the implication \ref{item: mbal}$\implies$\ref{item: zmLM}.
In particular, we argue by induction on $k$.
We use the same notation as in the proof above.
Recall that $\Delta''_i=[a''_i,b_i]$ where $b(\Delta_i)\le a''_i$ for all $i$ and $a''_i<a''_j$ if and only if $b(\Delta_i)<b(\Delta_j)$.
It easily follows that
\begin{enumerate}
\item $\X_{\m_2}\subset \X_\m$; $\Y_{\m_2}\subset \Y_\m$.
\item If $(i,r)\in \X_\m$, $(r,j)\in \Y_\m$ and $(i,j)\in \X_{\m_2}$ then $(i,r)\in \X_{\m_2}$ and $(r,j)\in \Y_{\m_2}$.
\item Similarly, if $(r,j)\in \X_\m$, $(i,r)\in \Y_\m$ and $(i,j)\in \X_{\m_2}$ then necessarily $(r,j)\in \X_{\m_2}$ and $(i,r)\in \Y_{\m_2}$.
\end{enumerate}
Thus, the non-zero coordinates of $\{\xvec_{i,j}(\lambda):(i,j)\in \X_{\m_2}\}$ are confined to $\Y_{\m_2}$ and the entries coincide with those
of $\xvec_{i,j}(\lambda)$ with respect to $\m_2$ (and the same $\lambda_{i,j}$).
We have already shown in Lemma \ref{lem: indstep} that $\m_2$ is balanced. Hence, by induction hypothesis it suffices to check that
there exists a strong $\rltn$-matching $f:\X'\rightarrow\Y'$ where $\X'=\X_\m\setminus \X_{\m_2}$ and $\Y'=\Y_\m\setminus \Y_{\m_2}$.
Consider first the case $l'=l$. Thus, $\X'=\{(i,\sigma'(j))\in \X_\m:l\le j\le m\}$.
Let $f:\X'\rightarrow\Y'$ be defined by $f(i,\sigma'(j))=(h(i,j),\sigma'(j))$ where $h$ is as in Lemma \ref{lem: indstep}. Then $f$ is injective.
Let $\somele$ be the skewed lexicographic order on $\Y_\m$ given by $(i,j)\somele(i',j')$ if either $j'>j$ or ($j=j'$ and $i'\le i$).
It follows from the definition of $f$ that if $(i,j)\rltn(i',j')\in \Y_\m$ with $(i,j)\in\X'$ then $f(i,j)\somele(i',j')$.
Suppose now that $l'>l$. It is easy to see that
\[
\X'=\{(i,\sigma'(j))\in \X_\m:l\le j\le m,\Delta_i\prec\Delta'_j\}.
\]
Thus, the proof of Lemma \ref{lem: indstep} shows that the rule $(i,\sigma'(j))\mapsto (i,\sigma'(j+1))$ defines an injective function $f:\X'\rightarrow\Y'$.
Let $\somele$ be the skewed lexicographic order on $\Y_\m$ given by $(i,j)\somele(i',j')$ if either $i'<i$ or ($i'=i$ and $j'\ge j$).
Clearly, for any $(i,\sigma'(j))\in\X'$ and $(i,\sigma'(j))\rltn(i',\sigma'(j'))$ with $l\le j,j'\le m$ we have $(i,\sigma'(j+1))\somele(i',\sigma'(j'))$.
In both cases the proof is complete.
\end{proof}
\begin{remark}
Once proved, Theorem \ref{thm: main}, together with Lemma \ref{lem: submlt}, imply that if $\m$ is regular and $\zele{\m}\in\IrrS$ then
$\zele{\n}\in\IrrS$ for any sub-multisegment $\n$ of $\m$.
However, this is no longer true in the non-regular case. For instance, we can take $\n=[4,5]+[2,4]+[3,3]+[1,2]$ and $\m=\n+[2,3]$.
(It can be proved using Lemma \ref{lem: albertoidea} that $\zele{\m}$ is \LM, but in the next section we show that $\zele{\n}$ is not \LM.)
\end{remark}
\begin{remark}
Lemma \ref{lem: albertoidea}, though simple, provides a powerful tool for proving $\square$-irreducibility.
While we do not yet have sufficient evidence to make precise conjectures we may ask the following question.
Given $\pi\in\IrrS$ which is not supercuspidal, do there always exist $1\ne\pi_1,\pi_2\in\IrrS$ such that
$\pi\hookrightarrow\pi_1\times\pi_2$ and $\pi\times\pi_1$ is irreducible?
\end{remark}
\section{Basic cases} \label{sec: basicases}
It remains to prove the other implications of Theorem \ref{thm: main}.
In this section we carry out the main step by showing that for certain ``basic'' unbalanced regular multisegments $\m$
(generalizing Example \ref{ex: bexam}) which are introduced below, $\zele{\m}\notin\IrrS$ and $\m$ is not \GLS.
\subsection{}
The idea of the proof is the following.
Suppose that $\pi=\zele{\m}=\soc(\pi_1\times\pi_2)$ with $\pi_1,\pi_2\in\IrrS$. Then the \wtness\
$\Pi=\soc(\pi_1\times\soc(\pi_2\times\pi))\hookrightarrow\pi_1\times\pi_2\times\pi$ is irreducible, and hence
$\Pi\hookrightarrow\omega\times\pi$ for some $\omega\in\JH(\pi_1\times\pi_2)$.
If we can show that this is not possible unless $\omega=\pi$ then necessarily $\Pi\hookrightarrow\pi\times\pi$ and hence $\pi\notin\IrrS$
provided that $\Pi\ne\zele{\m+\m}$.
To facilitate the argument it is useful to introduce the following concept.
Let $\pi\in\Irr$ and $\pi_1,\pi_2,\Pi$ as before. We say that $(\pi_1,\pi_2)$ is a \emph{\spltng}\ for $\pi$ with \wtness\ $\Pi$ if in addition
\begin{enumerate}
\item $\m(\pi)=\m(\pi_1)+\m(\pi_2)$.
\item $\rnrset(\pi)=\rnrset(\pi_1)\cup\rnrset(\pi_2)$.
\item $\lnrset(\Pi)=\lnrset(\pi)$.
\end{enumerate}
In practice, determining $\Pi$ is the most technically involved step.
The properties above limit the possible $\omega$'s in $\JH(\pi_1\times\pi_2)$ such that $\Pi\hookrightarrow\omega\times\pi$.
Before making this more precise, we recall that the ascent set of a permutation $\sigma\in S_k$ is defined by
\[
\desl(\sigma):=\{1\le i<k:\sigma(i)<\sigma(i+1)\}\cup\{k\}.
\]
The following is an immediate consequence of Lemma \ref{lem: spclcaseextrho}. (Recall the notation of \S\ref{sec: biseq}.)
\begin{lemma} \label{lem: lnrdes}
Let $l\ge0$ and $\sigma\in S_k$ such that $\sigma(i)\le k+l+1-i$ for all $i>l+1$.
Let $\m=\m_\sigma(\tseq_{k,l})$ (see \eqref{def: akl}).
Then $\lnrset(\m)=\{\rho\nu_\rho^i:i\in\desl(\sigma)\}$ and $\rnrset(\m)=\{\rho\nu_\rho^{k+l-i}:i\in\desl(\sigma^{-1})\}$.
\end{lemma}
\begin{corollary} \label{cor: sigmarest}
Let $k,l\ge1$ and $\tau\in S_k$ be such that $\tau(i)\le k+l+1-i$ for all $i>l+1$.
Let $\pi=\zele{\m_\tau(\tseq_{k,l})}$ and assume that $(\pi_1,\pi_2)$ is a \spltng\ for $\pi$ with \wtness\ $\Pi\in\Irr$.
Let $\omega$ be an irreducible subquotient of $\pi_1\times\pi_2$ such that $\Pi\hookrightarrow\omega\times\pi$.
Then $\omega=\m_\sigma(\tseq_{k,l})$ where $\sigma\le\tau$, $\desl(\sigma)\subset\desl(\tau)$ and $\desl(\sigma^{-1})\subset\desl(\tau^{-1})$.
\end{corollary}
\begin{proof}
Since $\pi_1\times\pi_2\le\std(\m_\tau(\tseq_{k,l}))$ we have $\omega=\m_\sigma(\tseq_{k,l})$ for some $\sigma\le\tau$.
By Lemma \ref{lem: lnrprop} we have $\lnrset(\omega)\subset\lnrset(\Pi)=\lnrset(\pi)$ and $\rnrset(\omega)\subset\rnrset(\pi_1)\cup\rnrset(\pi_2)=\rnrset(\pi)$.
Thus, by Lemma \ref{lem: lnrdes}, $\desl(\sigma)\subset\desl(\tau)$ and $\desl(\sigma^{-1})\subset\desl(\tau^{-1})$ as required.
\end{proof}
It will be convenient to use the following notation: given a segment $\Delta$ and $k\ge0$ let
$\speh{\Delta}k=\Delta_1+\dots+\Delta_k$ where $\Delta_1=\Delta$ and $\Delta_{i+1}=\lshft{\Delta}_i$, $i=1,\dots,k-1$.
The ``basic'' multisegments come in three families which are introduced and analyzed in the following subsections.
\subsection{Basic multisegments of type $423{*}1$}
As in example \ref{ex: bexam} for $k\ge4$ let $\pi=\zele{\m}$ with
\begin{equation} \label{eq: basic4231}
\m=[k,k+1]+[2,k]+\speh{[k-1]}{k-3}+[1,2].
\end{equation}
By Remark \ref{rem: types not GLS}, $\m$ does not satisfy \GLS.
Let
\[
\Pi=\zele{\speh{[2,k+1]}2+\speh{[k,k+1]}k}=\zele{\speh{[2,k+1]}2}\times\zele{\speh{[k,k+1]}k}.
\]
\begin{proposition} \label{prop: basictype4231}
We have
\[
\Pi\hookrightarrow\pi\times\pi.
\]
In particular, $\pi\notin\IrrS$.
\end{proposition}
\begin{remark}
The case $k=4$ (where $\m=[4,5]+[2,4]+[3]+[1,2]$, see Example \ref{ex: lecorig}) is the original example given by Leclerc for an ``imaginary'' representation \cite{MR1959765}.
\end{remark}
Following the above-mentioned strategy we first show the following.
\begin{lemma} \label{lem: ws4231}
Let
\[
\pi_1=\zele{[k,k+1]+[2,k]},\ \ \pi_2=\zele{\speh{[k-1]}{k-3}+[1,2]}.
\]
Then $(\pi_1,\pi_2)$ is a \spltng\ for $\pi$ with \wtness\ $\Pi$.
\end{lemma}
\begin{proof}
Note that $\pi_1$ and $\pi_2$ are ladders and in particular $\pi_1,\pi_2\in\IrrS$.
Also, $\pi=\soc(\pi_1\times\pi_2)$, $\rnrset(\pi_1)\cup\rnrset(\pi_2)=\{[k]\}\cup\{[2]\}=\rnrset(\pi)$ and $\lnrset(\Pi)=\lnrset(\pi)=\{[2],[k]\}$.
Let
\[
\pi_3=\zele{[k-1,k+1]+[1,k]+\speh{[k-2,k-1]}{k-2}}.
\]
By Lemma \ref{lem: extractsegment} (applied repeatedly) we have
\[
\soc(\pi_1\times\pi_3)=\zele{[k,k+1]+[1,k]+\speh{[k-2,k-1]}{k-2}+\m(\soc(\zele{[2,k]}\times\zele{[k-1,k+1]}))}=\Pi.
\]
It remains to show that
\[
\soc(\pi_2\times\pi)=\pi_3.
\]
Note that $\del(\pi)=\del(\pi_3)=[k,k+1]$ and $\del(\pi^-)=\del(\pi_3^-)=[2,k]$.
Hence, by Lemma \ref{lem: suppn>suppm} (applied twice) it suffices to show that
\[
\soc(\pi_2\times(\pi^-)^-)=(\pi_3^-)^-.
\]
This is straightforward. Indeed,
\[
(\pi^-)^-=\zele{[2,k-1]+[1]}
\]
and by Lemma \ref{lem: extractsegment} we have
\begin{multline*}
\soc(\pi_2\times(\pi^-)^-)=\zele{\speh{[k-1]}{k-3}+[1]+\m(\soc(\zele{[1,2]}\times\zele{[2,k-1]}))}
\\=\zele{[1,k-1]+\speh{[k-1]}{k-1}}=(\pi_3^-)^-
\end{multline*}
as required.
\end{proof}
In order to conclude Proposition \ref{prop: 3412basic} it remains to show that
\[
\Pi\not\hookrightarrow\omega\times\pi
\]
for any irreducible subquotient $\omega$ of $\pi_1\times\pi_2$, other than $\pi$.
Recall that $\m=\m_{\sigma_1}(\tseq_{k,2})$ where
\[
\sigma_1(i)=\begin{cases}k&i=1,\\2&i=2,\\k+2-i&i=3,\dots,k-1,\\1&i=k.\end{cases}
\]
Note that $\sigma_1^{-1}=\sigma_1$ and $\desl(\sigma_1)=\{2,k\}$.
\begin{lemma} \label{lem: sigma14231}
Suppose that $\sigma\le\sigma_1$ and $\desl(\sigma)\cup\desl(\sigma^{-1})\subset\desl(\sigma_1)$.
Then either $\sigma=\sigma_1$ or
\[
\sigma(i)=\begin{cases}3-i&i=1,2,\\k+3-i&i=3,\dots,k.\end{cases}
\]
\end{lemma}
\begin{proof}
Let $i=\sigma(1)$ and $j=\sigma(2)$. Note that $i>j$ and $i,j$ determine $\sigma$ uniquely since $\sigma(3)>\sigma(4)>\dots>\sigma(k)$.
Suppose first that $i\ne k$. Then since $\sigma^{-1}(i)=1<\sigma^{-1}(i+1)$, we have $i=2$, in which case $j=1$.
On the other hand, if $i=k$ then $j\le\sigma_1(2)=2$. If $j=1$ then $2=\sigma^{-1}(1)>\sigma^{-1}(2)$
and therefore $i=2$, a contradiction. Hence, $j=2$ and $\sigma=\sigma_1$.
\end{proof}
It follows from Corollary \ref{cor: sigmarest} and Lemma \ref{lem: sigma14231} that if $\omega$ is an irreducible subquotient of $\pi_1\times\pi_2$,
other than $\pi$, such that $\Pi\hookrightarrow\omega\times\pi$ then necessarily $\omega=\zele{\speh{[2,k+1]}2}$. However, in this case
$\omega\times\pi$ is irreducible (e.g., \cite[Proposition 6.6]{MR3573961}) and is not equal to $\Pi$ (since $\m(\omega)+\m(\pi)\ne\m(\Pi)$).
We get a contradiction. This finishes the proof of Proposition \ref{prop: basictype4231}.
\begin{remark}
In the notation of Proposition \ref{prop: basictype4231}, we expect that $\pi\times\pi$ decomposes as a direct sum of $\Pi$
and $\zele{\m+\m}$. We will not say more about that here.
\end{remark}
\subsection{Basic multisegments of type $3{*}41{*}2$} \label{sec: 3*41*2}
Next, for $k>l>2$ consider $\pi=\zele{\m}$ where
\begin{equation} \label{eq: basic3412}
\m=[l,l+k-1]+\speh{[l-2,l+k-2]}{l-3}+[k,k+1]+[1,k]+\speh{[k-1]}{k-l-1}+[l-1,l].
\end{equation}
This is a generalization of Example \ref{ex: bexam} (in which $l=3$). Here is a drawing for $k=8$, $l=5$:
\[
\xymatrix@=0.6em{
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&\circ\\
&&&&&&\circ\\
&&&&\circ\ar@{-}[r]&\circ}
\]
By Remark \ref{rem: kirred} $\m$ does not satisfy \GLS. Indeed, $\X_\m$ contains the $k$ irreducible pairs
$(i+1,i)$, $i=1,\dots,l-3$, $(l,l-2)$, $(l,l-1)$, $(l+1,l-1)$, $(i+1,i)$, $i=l+1,\dots,k-1$, $(k,1)$.
Note that $\lnrset(\pi)=\{[l-2],[l],[k]\}$ and $\rnrset(\pi)=\{[l],[k],[k+2]\}$.
\begin{proposition} \label{prop: 3412basic}
Let
\begin{multline*}
\Pi=\zele{\speh{[l-2,l+k-1]}{l-2}+\speh{[l,l+k-1]}{l}+\speh{[k,k+1]}{k+2-l}}=\\
\zele{\speh{[l-2,l+k-1]}{l-2}}\times\zele{\speh{[l,l+k-1]}{l}}\times\zele{\speh{[k,k+1]}{k+2-l}}.
\end{multline*}
Then
\[
\Pi\hookrightarrow\pi\times\pi.
\]
In particular, $\pi\notin\IrrS$.
\end{proposition}
As before, the main step is the following.
\begin{lemma} \label{lem: weakerbasic3412}
Let
\begin{gather*}
\pi_1=\zele{[l,l+k-1]+\speh{[l-2,l+k-2]}{l-3}}\\
\pi_2=\zele{[k,k+1]+[1,k]+\speh{[k-1]}{k-l-1}+[l-1,l]}.
\end{gather*}
Then $(\pi_1,\pi_2)$ is a \spltng\ for $\pi$ with \wtness\ $\Pi$.
\end{lemma}
\begin{proof}
By the ``if'' part of Theorem \ref{thm: main} $\pi_1,\pi_2\in\IrrS$.
It is clear that $\pi=\soc(\pi_1\times\pi_2)$, $\rnrset(\pi_1)\cup\rnrset(\pi_2)=\{[k+2]\}\cup\{[l],[k]\}=\rnrset(\pi)$
and $\lnrset(\Pi)=\lnrset(\pi)$. Let
\[
\pi_3=\zele{\speh{[l-1,l+k-1]}{l-3}+\speh{[2,k+1]}2+[1,k+2]+\speh{[k,k+1]}{k+2-l}}.
\]
We first show that $\soc(\pi_1\times\pi_3)=\Pi$.
Let $\pi_4=\zele{\speh{[l-2,l+k-2]}{l-3}}$ and $\pi_5=\soc(\pi_4\times\pi_3)$. By Lemma \ref{lem: extractsegment} we have
\[
\soc(\pi_1\times\pi_3)=\zele{[l,l+k-1]+\m(\pi_5)}.
\]
Note that $\del(\pi_3)=[k+2,k+l-1]$ and
\[
\pi_3^-=\zele{\speh{[l-1,l+k-2]}{l-1}+[1,k+1]+\speh{[k,k+1]}{k+2-l}}.
\]
Hence, by Lemma \ref{lem: suppn>suppm} $\del(\pi_5)=[k+2,k+l-1]$ and $\pi_5^-=\soc(\pi_4\times\pi_3^-)$.
By Theorem \ref{thm: laddercomb} we have $\LI(\pi_4,\pi_3^-)$ and therefore
\[
\pi_5^-=\zele{\speh{[l-2,l+k-2]}{l-2}+\speh{[l-1,l+k-2]}{l-1}+\speh{[k,k+1]}{k+2-l}}.
\]
Thus,
\[
\pi_5=\zele{\speh{[l-2,l+k-1]}{l-2}+\speh{[l-1,l+k-2]}{l-1}+\speh{[k,k+1]}{k+2-l}}.
\]
All in all, $\soc(\pi_1\times\pi_3)=\Pi$ as claimed.
Ir remains to show that
\begin{equation} \label{eq: tau2tausigma}
\soc(\pi_2\times\pi)=\pi_3.
\end{equation}
Assume first that $l>3$. Then, since $[1],[l-1]\notin\lnrset(\pi)$ we have by Corollary \ref{cor: extractrho}
\begin{equation} \label{eq: soctau2tau}
\soc(\pi_2\times\pi)=\soc([l-1]\times\soc([1]\times\soc(\pi_6\times\pi)))
\end{equation}
where (see Lemma \ref{lem: spclcaseextrho})
\[
\pi_6=\lderiv_{[l-1]}(\lderiv_{[1]}(\pi_2))=\zele{[k,k+1]+[2,k]+\speh{[k-1]}{k-l}}.
\]
Note that by Theorem \ref{thm: laddercomb} $\pi_6\times\pi_1$ is irreducible. Therefore,
\begin{multline*}
\soc(\pi_6\times\pi)=\soc(\pi_6\times\pi_1\times\pi_2)=
\soc(\pi_1\times\soc(\pi_6\times\pi_2))\\=
\zele{[l,l+k-1]+\speh{[l-2,l+k-2]}{l-3}+\m(\soc(\pi_6\times\pi_2))}.
\end{multline*}
Let $\pi_7=\zele{[k,k+1]+\speh{[k-1]}{k-l-1}}$. By Lemma \ref{lem: extractsegment} we have
\[
\soc(\pi_6\times\pi_2)=\zele{[l-1,l]+[1,k]+\m(\soc(\pi_6\times\pi_7))}
\]
and by Corollary \ref{cor: extractrho}
\[
\soc(\pi_6\times\pi_7)=\soc(\soc(\pi_6\times\zele{\speh{[k]}{k-l}})\times[k+1]).
\]
Moreover, by Lemma \ref{lem: extractsegment} and Lemma \ref{lem: chopladders}
\begin{multline*}
\soc(\pi_6\times\zele{\speh{[k]}{k-l}})=\zele{[k,k+1]+\m(\soc(\zele{[2,k]+\speh{[k-1]}{k-l}}\times \zele{\speh{[k]}{k-l}}))}
\\=\zele{[2,k]+\speh{[k,k+1]}{k+1-l}}.
\end{multline*}
Using Lemma \ref{lem: soctimesrho} it follows that
\[
\soc(\pi_6\times\pi_7)=\zele{[2,k+1]+\speh{[k,k+1]}{k+1-l}}
\]
and hence
\[
\soc(\pi_6\times\pi_2)=\zele{\speh{[2,k+1]}2+\speh{[k,k+1]}{k+2-l}}
\]
and
\[
\soc(\pi_6\times\pi)=\zele{[l,l+k-1]+\speh{[l-2,l+k-2]}{l-3}+\speh{[2,k+1]}2+\speh{[k,k+1]}{k+2-l}}.
\]
The relation \eqref{eq: tau2tausigma} now follows from \eqref{eq: soctau2tau} using Lemma \ref{lem: soctimesrho}.
Consider now the remaining case $l=3$. We first write using Corollary \ref{cor: extractrho}
\[
\soc(\pi_2\times\pi)=\soc([2]\times\soc(\pi_8\times\pi))
\]
where
\[
\pi_8=\lderiv_{[2]}(\pi_2)=\zele{[k,k+1]+[1,k]+\speh{[k-1]}{k-3}}.
\]
By Lemma \ref{lem: extractsegment} we have
\[
\soc(\pi_8\times\pi)=\zele{[2,3]+\m(\soc(\pi_8\times\pi_9))}
\]
where
\[
\pi_9=\zele{[3,k+2]+[k,k+1]+[1,k]+\speh{[k-1]}{k-4}}.
\]
Now, since $[1]\times\pi_8$ is irreducible we have by Corollary \ref{cor: extractrho}
\[
\soc(\pi_8\times\pi_9)=\soc([1]\times[1]\times\soc(\pi_{10}\times\pi_{11}))
\]
where
\[
\pi_{10}=\lderiv_{[1]}(\pi_8)=\zele{[k,k+1]+[2,k]+\speh{[k-1]}{k-3}}
\]
and
\[
\pi_{11}=\lderiv_{[1]}(\pi_9)=\zele{[3,k+2]+[k,k+1]+[2,k]+\speh{[k-1]}{k-4}}.
\]
Again by Corollary \ref{cor: extractrho}
\[
\soc(\pi_{10}\times\pi_{11})=\soc([2]\times\soc(\pi_{12}\times\pi_{11}))
\]
where (by Lemma \ref{lem: spclcaseextrho})
\[
\pi_{12}=\lderiv_{[2]}(\pi_{10})=\zele{[k,k+1]+[3,k]+\speh{[k-1]}{k-3}}.
\]
By Lemma \ref{lem: extractsegment} we have
\[
\soc(\pi_{12}\times\pi_{11})=\zele{[2,k]+\m(\soc(\pi_{12}\times\pi_{13}))}
\]
where
\[
\pi_{13}=\zele{[3,k+2]+[k,k+1]+\speh{[k-1]}{k-4}}=\zele{[3,k+2]}\times\zele{[k,k+1]+\speh{[k-1]}{k-4}}.
\]
Clearly,
\[
\soc(\pi_{12}\times\pi_{13})=\zele{[3,k+2]+\m(\soc(\pi_{12}\times\zele{[k,k+1]+\speh{[k-1]}{k-4}}))}
\]
while by Corollary \ref{cor: extractrho}
\[
\soc(\pi_{12}\times\zele{[k,k+1]+\speh{[k-1]}{k-4}})=\soc(\soc(\pi_{12}\times\zele{\speh{[k]}{k-3}})\times[k+1]).
\]
Since
\begin{multline*}
\soc(\pi_{12}\times\zele{\speh{[k]}{k-3}})=\zele{[k,k+1]+[3,k]+\m(\soc(\zele{\speh{[k-1]}{k-3}}\times\zele{\speh{[k]}{k-3}}))}\\
=\zele{[3,k]+\speh{[k,k+1]}{k-2}}
\end{multline*}
we obtain (using Lemma \ref{lem: soctimesrho})
\begin{gather*}
\soc(\pi_{12}\times\zele{[k,k+1]+\speh{[k-1]}{k-4}})=\zele{[3,k+1]+\speh{[k,k+1]}{k-2}},\\
\soc(\pi_{12}\times\pi_{13})=\zele{[3,k+2]+[3,k+1]+\speh{[k,k+1]}{k-2}},\\
\soc(\pi_{12}\times\pi_{11})=\zele{[2,k]+[3,k+2]+[3,k+1]+\speh{[k,k+1]}{k-2}},\\
\soc(\pi_{10}\times\pi_{11})=\zele{[2,k]+[2,k+2]+[3,k+1]+\speh{[k,k+1]}{k-2}},\\
\soc(\pi_8\times\pi_9)=\zele{[1,k]+[1,k+2]+[3,k+1]+\speh{[k,k+1]}{k-2}},\\
\soc(\pi_8\times\pi)=\zele{[1,k]+[1,k+2]+[3,k+1]+\speh{[k,k+1]}{k-1}},
\end{gather*}
and finally
\[
\soc(\pi_2\times\pi)=\zele{[1,k+2]+\speh{[2,k+1]}2+\speh{[k,k+1]}{k-1}}=\pi_3
\]
as required.
\end{proof}
In order to conclude Proposition \ref{prop: 3412basic} it remains to show that
\[
\Pi\not\hookrightarrow\omega\times\pi
\]
for any irreducible subquotient $\omega$ of $\pi_1\times\pi_2$, other than $\pi$.
Recall that $\m(\pi)=\m_{\sigma_1}(\tseq_{k,l})$ where
\[
\sigma_1(i)=\begin{cases}l&i=1,\\l-i&i=2,\dots,l-2,\\k&i=l-1,\\1&i=l,\\l+k-i&i=l+1,\dots,k-1,\\l-1&i=k.\end{cases}
\]
Note that $\sigma_1^{-1}=\sigma_1$ and $\desl(\sigma_1)=\{l-2,l,k\}$.
\begin{lemma} \label{lem: sigma3412}
Suppose that $\sigma\le\sigma_1$ and $\desl(\sigma)\cup\desl(\sigma^{-1})\subset\desl(\sigma_1)$.
Then $\sigma$ is one of the following four permutations:
\begin{gather*}
\sigma=\sigma_1,\\
\sigma(i)=\begin{cases}l&i=1,\\l-i&i=2,\dots,l-2,\\l-1&i=l-1,\\1&i=l,\\k+l+1-i&i=l+1,\dots,k,\end{cases}\\
\sigma(i)=\begin{cases}l-1-i&i=1,\dots,l-2,\\k&i=l-1,\\l&i=l,\\k+l-i&i=l+1,\dots,k-1,\\l-1&i=k,\end{cases}\\
\sigma(i)=\begin{cases}l-1-i&i=1,\dots,l-2,\\l&i=l-1,\\l-1&i=l,\\k+l+1-i&i=l+1,\dots,k.\end{cases}
\end{gather*}
In particular, $\sigma=\sigma^{-1}$ and if $\sigma\ne\sigma_1$ then $\sigma$ is smooth.
\end{lemma}
\begin{proof}
We have $\sigma(1)\le\sigma_1(1)=l$ and $\sigma(i)>\sigma(i+1)$ for all $1\le i<l-2$.
Also $\sigma(1)\ne l-1$ since $\sigma^{-1}(l-1)>\sigma^{-1}(l)$. Thus, either
$\sigma(i)=l-1-i$ for all $1\le i\le l-2$ or $\sigma(1)=l$. In the latter case, if $l>3$ then $\sigma(2)\le\sigma_1(2)=l-2$ and
therefore there exists $1<j<l$ such that $\sigma(i)=l-i$ for all $1<i<j$ and $\sigma(i)=l-1-i$ for all $j\le i\le l-2$.
In fact, $j=l-1$ for otherwise $\sigma^{-1}(l-j)\ge l-1>j=\sigma^{-1}(l-j-1)$ in contradiction with the assumption on $\sigma$.
Thus, either $\sigma(i)=l-1-i$ for all $1\le i\le l-2$ or $\sigma(1)=l$ and $\sigma(i)=l-i$ for all $1<i\le l-2$.
By a similar reasoning, either $\sigma(i)=k+l+1-i$ for all $l<i\le k$ or $\sigma(k)=l-1$ and $\sigma(i)=k+l-i$ for all $l<i<k$.
Taking into account the condition $\sigma(l-1)>\sigma(l)$, $\sigma$ must be one of the four possibilities listed in the statement of the lemma.
\end{proof}
Assume for simplicity that $\rho^\vee=\rho\nu_\rho^{k+l}$, or equivalently, that $\pi^\vee=\pi$.
(We may do so since by Theorem \ref{thm: indepcspline}, the validity of Theorem \ref{thm: main} is independent of the choice of $\rho$.)
It follows from Corollary \ref{cor: sigmarest}, Lemma \ref{lem: sigma3412}, the ``if'' direction of Theorem \ref{thm: main} and \eqref{eq: contrasigma}
that if $\omega$ is an irreducible subquotient of $\pi_1\times\pi_2$, other than $\pi$, such that $\Pi\hookrightarrow\omega\times\pi$
then $\omega\in\IrrS$ and $\omega=\omega^\vee$.
However, in this case, it would follow from Lemma \ref{lem: selfdualcases} below\footnote{Alternatively, we could compute
$\soc(\omega\times\pi)$ directly. This will unnecessitate the assumption that $\rho$ is essentially self-dual.}
that $\omega\times\pi$ is irreducible and we obtain a contradiction
(since $\m(\pi)$ is not a sub-multisegment of $\m(\Pi)$). This finishes the proof of Proposition \ref{prop: 3412basic}.
\begin{lemma} \label{lem: selfdualcases}
Suppose that $\pi_1$, $\pi_2$ and $\pi$ are irreducible and self-dual and $\pi\hookrightarrow\pi_1\times\pi_2$.
If at least one of $\pi_1$ and $\pi_2$ is \LM\ then $\pi_1\times\pi_2$ is irreducible. In particular, $\m(\pi)=\m(\pi_1)+\m(\pi_2)$.
\end{lemma}
\begin{proof}
Since $\pi$ is a subrepresentation of $\pi_1\times\pi_2$, $\pi^\vee$ is a quotient of $\pi_1^\vee\times\pi_2^\vee$.
Thus, by the self-duality assumption, $\pi$ is a quotient of $\pi_1\times\pi_2$ as well.
Since by assumption $\pi_1\times\pi_2$ is \SI, it must be irreducible.
\end{proof}
\subsection{Basic multisegments of type $34{*}12$}
Finally, as in Example \ref{exam: 34*12} consider for $k>4$
\[
\pi=\zele{\m},\ \m=[k-1,2k-2]+[k,2k-3]+\speh{[k-2,2k-4]}{k-4}+[1,k]+[2,k-1].
\]
\begin{proposition} \label{prop: lastcase3412}
We have $\Pi\hookrightarrow\pi\times\pi$ where
\begin{multline*}
\Pi=\zele{\speh{[k,2k-3]}{k-1}+[1,2k-2]+\speh{[k-1,2k-2]}{k-1}}=\\
\zele{\speh{[k,2k-3]}{k-1}}\times\zele{[1,2k-2]}\times\zele{\speh{[k-1,2k-2]}{k-1}}.
\end{multline*}
In particular, $\pi$ is not \LM.
\end{proposition}
To that end we first show
\begin{lemma} \label{lem: weakspec2}
Let
\[
\pi_1=\zele{[k-1,2k-2]},\ \ \pi_2=\zele{[k,2k-3]+\speh{[k-2,2k-4]}{k-4}+[1,k]+[2,k-1]}.
\]
Then $(\pi_1,\pi_2)$ is a \spltng\ for $\pi$ with \wtness\ $\Pi$.
\end{lemma}
\begin{proof}
Clearly $\pi_1$ is \LM\ and the same is true for $\pi_2$ by the ``if'' part of Theorem \ref{thm: main}.
It is also clear that $\pi=\soc(\pi_1\times\pi_2)$, $\rnrset(\pi_1)\cup\rnrset(\pi_2)=\{[2k-2]\}\cup\{[k-1],[k]\}=\rnrset(\pi)$
and $\lnrset(\Pi)=\lnrset(\pi)=\{[1],[k-1],[k]\}$.
Let
\[
\pi_3=\zele{\speh{[k,2k-3]}{k-1}+[1,2k-2]+\speh{[k-2,2k-3]}{k-2}}.
\]
Clearly, $\Pi=\soc(\pi_1\times\pi_3)$.
It remains to show that
\[
\soc(\pi_2\times\pi)=\pi_3.
\]
Since $\del(\pi_3)=\del(\pi)=[2k-2]$ it is enough to show by Lemma \ref{lem: suppn>suppm} that $\soc(\pi_2\times\pi^-)=\pi_3^-$.
For brevity set $\Delta=[k,2k-3]$.
We have
\[
\pi^-=\zele{\Delta+\speh{^+\Delta}{k-3}+[1,k]+[2,k-1]}.
\]
Since $\lnrset(\pi^-)=\{[1],[k-1],[k]\}$ we have by Corollary \ref{cor: extractrho}
\[
\soc(\pi_2\times\pi^-)=\soc([k-2]\times\soc([k-3]\times\dots\times\soc([2]\times\soc(\pi_4\times\pi^-))\dots))
\]
where (using Lemma \ref{lem: spclcaseextrho})
\[
\pi_4=\lderiv_{[2]}(\dots\lderiv_{[k-3]}(\lderiv_{[k-2]}(\pi_2))\dots)=\zele{\speh{\Delta}{k-3}+[1,k]+[3,k-1]}.
\]
Since $[1]\times\pi_4$ is irreducible and $\lmlt_{[1]}(\pi_4)=\lmlt_{[1]}(\pi^-)=1$ we have by Corollary \ref{cor: extractrho}
\[
\soc(\pi_4\times\pi^-)=\soc([1]\times[1]\times\soc(\pi_5\times\pi_6))
\]
where
\[
\pi_5=\lderiv_{[1]}(\pi_4)=\zele{\speh{\Delta}{k-3}+[2,k]+[3,k-1]}
\]
and
\[
\pi_6=\lderiv_{[1]}(\pi^-)=\zele{\Delta+\speh{^+\Delta}{k-2}+[2,k-1]}.
\]
By Lemma \ref{lem: extractsegment} we have
\[
\soc(\pi_5\times\pi_6)=\zele{[2,k-1]+\m(\soc(\pi_5\times\pi_7))}
\]
where
\[
\pi_7=\zele{\Delta+\speh{^+\Delta}{k-2}}=\zele{\Delta}\times\zele{\speh{^+\Delta}{k-2}}.
\]
We have
\[
\soc(\pi_5\times\pi_7)=\zele{\speh{^+\Delta}{k-2}}\times\soc(\pi_5\times\zele{\Delta})
\]
and since $^-\Delta\cap\rnrset(\pi_5)=\emptyset$,
\[
\soc(\pi_5\times\zele{\Delta})=\soc(\soc(\pi_5\times[k])\times\zele{^-\Delta}).
\]
Also,
\[
\soc(\pi_5\times[k])=\zele{\speh{\Delta}{k-2}+[2,k]}.
\]
Thus,
\[
\soc(\pi_5\times\zele{\Delta})=\zele{\speh{\Delta}{k-2}+[2,2k-3]}
\]
and hence by Lemma \ref{lem: soctimesrho}
\[
\soc(\pi_5\times\pi_7)=\zele{\speh{^+\Delta}{k-2}+\speh{\Delta}{k-2}+[2,2k-3]},
\]
\[
\soc(\pi_5\times\pi_6)=\zele{\speh{^+\Delta}{k-2}+\speh{\Delta}{k-1}+[2,2k-3]},
\]
\[
\soc(\pi_4\times\pi^-)=\zele{[1,2k-3]+[1,k]+\speh{^+\Delta}{k-3}+\speh{\Delta}{k-1}},
\]
and finally (again by Lemma \ref{lem: soctimesrho})
\[
\soc(\pi_2\times\pi^-)=\zele{[1,2k-3]+\speh{[k-2,2k-3]}{k-2}+\speh{\Delta}{k-1}}=\pi_3^-
\]
as required.
\end{proof}
We can write $\pi=\m_{\sigma_1}(\tseq_{k,k-1})$ where
\[
\sigma_1(i)=\begin{cases}i+k-2&i=1,2,\\k+1-i&i=3,\dots,k-2,\\i-k+2&i=k-1,k.\end{cases}
\]
Note that $\sigma_1^{-1}=\sigma_1$ and $\desl(\sigma_1)=\{1,k-1,k\}$.
Proposition \ref{prop: lastcase3412} is concluded from Lemma \ref{lem: weakspec2} exactly as before using the following elementary lemma.
\begin{lemma}
Suppose that $\sigma\le\sigma_1$ and $\desl(\sigma)\cup\desl(\sigma^{-1})\subset\desl(\sigma_1)$.
Then $\sigma$ is one of the following four permutations:
\begin{gather*}
\sigma=\sigma_1,\\
\sigma(i)=\begin{cases}1&i=1,\\k+2-i&i=2,\dots,k,\end{cases}\\
\sigma(i)=\begin{cases}k-i&i=1,\dots,k-1,\\k&i=k,\end{cases}\\
\sigma(i)=\begin{cases}i&i=1,k,\\k+1-i&i=2,\dots,k-1.\end{cases}
\end{gather*}
In particular, $\sigma=\sigma^{-1}$ and if $\sigma\ne\sigma_1$ then $\sigma$ is smooth.
\end{lemma}
\begin{proof}
Let $i=\sigma(1)$ and $j=\sigma(k)$.
Then $i\le\sigma_1(1)=k-1$ and $\sigma^{-1}(i)=1<\sigma^{-1}(i+1)$, hence $i\in\{1,k-1\}$.
Similarly $j\in\{2,k\}$. Since $\sigma(2)>\sigma(3)>\dots>\sigma(k-1)$, $\sigma$ must be one of the four
possibilities above.
\end{proof}
\section{End of proof of Theorem \ref{thm: main}} \label{sec: comproof}
In this section we complete the proof of the remaining parts of Theorem \ref{thm: main}.
Namely, we show that if $\m$ is a regular unbalanced multisegment then $\zele{\m}$ is not \LM\ and $\m$ does not satisfy \GLS.
We will achieve this by reducing the statement to the cases considered in the previous section.
The first reduction uses Lemma \ref{lem: 1stred}. It motivates the following definition.
\begin{definition}
Let $\m=\Delta_1+\dots+\Delta_k$ be a regular unbalanced multisegment.
We say that $\m$ is \emph{minimal unbalanced} if $\m-\Delta$ is balanced for every detachable segment $\Delta$ of $\m$. (See Definition \ref{def: detachable}.)
\end{definition}
We can explicate the minimal unbalanced multisegments as follows.
\begin{lemma} \label{lem: MU}
Suppose that $\m=\Delta_1+\dots+\Delta_k$ is a regular multisegment with $e(\Delta_1)>\dots>e(\Delta_k)$.
Then $\m$ is minimal unbalanced if and only if precisely one of the following three conditions holds.
\begin{enumerate}
\item (case $4{*}23{*}1$)
\begin{enumerate}
\item $b(\Delta_k)<b(\Delta_i)<b(\Delta_1)$ for all $1<i<k$.
\item There do not exist $1<i,j<k-1$ such that $b(\Delta_{i+1})<b(\Delta_j)<b(\Delta_i)$.
\item There exists $i$ such that $\Delta_{i+1}\not\prec\Delta_i$.
\item Let $r=\max\{i:\Delta_{i+1}\not\prec\Delta_i\}$.
Then $\Delta_{r+1}\prec\Delta_1$ and $r<k-1$.
\end{enumerate}
\item (case $3{*}41{*}2$) There exists $1<r<k-1$ such that if $\tau$ is the transposition $r\leftrightarrow r+1$ then
\begin{enumerate}
\item $\Delta_{\tau(i+1)}\prec\Delta_{\tau(i)}$, $i=1,\dots,r-1,r+1,\dots,k-1$.
\item $b(\Delta_{\tau(2)})<b(\Delta_k)<b(\Delta_1)<b(\Delta_{\tau(k-1)})$.
\end{enumerate}
\item (case $34{*}12$) $k>4$, $\Delta_2\subset\Delta_1$, $\Delta_3\prec\Delta_1$, $\Delta_{i+1}\prec\Delta_i$, $i=3,\dots,k-3$,
$\Delta_k\prec\Delta_{k-2}$, $\Delta_k\subset\Delta_{k-1}$, $\Delta_k\prec\Delta_2$.
\end{enumerate}
\end{lemma}
Here is an example of the case $4{*}23{*}1$ with $k=8$ and $r=5$:
\[
\xymatrix@=0.6em{
&&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&&\circ\ar@{-}[r]&\circ&\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&}
\]
Next is an example of the case $3{*}41{*}2$ with $k=8$ and $r=4$:
\[
\xymatrix@=0.6em{
&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&&\circ\ar@{-}[r]&\circ\\
&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ}
\]
Finally, here is an example of the case $34{*}12$ with $k=7$:
\[
\xymatrix@=0.6em{
&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ}
\]
\begin{proof}
Let $S=\{i:\Delta_i\text{ is detachable in }\m\}$.
Note that since $\m$ is regular, $i\in S$ if and only if $\Delta_i\not\prec\Delta_j$ for all $j<i$
or $\Delta_j\not\prec\Delta_i$ for all $j>i$. In particular, $\{1,k\}\subset S$.
Moreover, by Proposition \ref{prop: nonsmth}, $\m$ is minimal unbalanced if and only if there exists a submultisegment
$\m_A=\sum_{i\in A}\Delta_i$, $A\subset\{1,\dots,k\}$ of $\m$ which is either of type $4231$ or $3412$ and
\begin{equation} \label{eq: equmin}
\text{for any $A$ such that $\m_A$ is of type $4231$ or $3412$ we have $A\supset S$}
\end{equation}
(and in particular, $1,k\in A$).
Consider the families above.
In the $4{*}23{*}1$ case $S=\{1,k\}$ and $\m_{\{1,r,r+1,\dots,k\}}$ forms a sub-multisegment of type $4231$.
On other hand, for any sub-multisegment $\m_A$ of type $4231$ we have $\{1,k\}\subset A$ and there is no sub-multisegment of type $3412$.
In the $3{*}41{*}2$ case $S=\{1,r,r+1,k\}$ and $\m_{\{1,r,r+1,\dots,k\}}$ is a sub-multisegment of type $3412$.
Any sub-multisegment of type $4231$ necessarily contains $S$ and there is no sub-multisegment of type $4231$.
In the $34{*}12$ case, $S=\{1,2,k-1,k\}$ and $\m_S$ is the unique sub-multisegment of type $3412$;
there is no sub-multisegment of type $4231$.
Thus in all cases $\m$ is a minimal unbalanced multisegment.
It is also clear that the three cases are disjoint.
Conversely, suppose that $\m$ is minimal unbalanced and let $\m_A$ be a sub-multisegment of type $4231$ or $3412$.
By the minimality assumption $1,k\in A$.
Let $j_{\min}$ (resp., $j_{\max}$) be the index $j$ for which $b(\Delta_j)$ is minimal (resp., maximal). Then $j_{\min},j_{\max}\in S\subset A$.
Assume first that $\m_A$ is of type $4231$. In this case $j_{\min}=k$ and $j_{\max}=1$.
In other words $b(\Delta_k)<b(\Delta_i)<b(\Delta_1)$ for all $1<i<k$.
Note that for any $i<k$ there exists $j>i$ such that $\Delta_j\prec\Delta_i$. Indeed, if $i\in A$, we can choose
$j\in A$ as well, while if $i\notin A$ then $i\notin S$ and the claim is clear.
It follows that for any $i$ there exists a sequence $i_0<\dots<i_m$, $m\ge0$ such that $i_0=i$, $i_m=k$
and $\Delta_{i_{j+1}}\prec\Delta_{i_j}$, $j=0,\dots,m-1$.
Similarly, for any $i$ there exists a sequence $i_0<\dots<i_m$, $m\ge0$ such that $i_0=1$, $i_m=i$
and $\Delta_{i_{j+1}}\prec\Delta_{i_j}$, $j=0,\dots,m-1$.
Next we show that we cannot have $b(\Delta_l)<b(\Delta_j)<b(\Delta_i)$ for any $1<i<l<j$.
Assume otherwise, and consider a counterexample with $j$ minimal and with $b(\Delta_i)$ minimal (with respect to $j$).
We first claim that $\Delta_j\prec\Delta_i$.
Indeed, by the above, there exists $s<j$ such that $\Delta_j\prec\Delta_s$.
If $b(\Delta_s)>b(\Delta_i)$ then $\Delta_j\prec\Delta_i$ as required.
Otherwise, $s>1$ and by the minimality of $j$ we have $s<l$ for otherwise we could replace $j$ by $s$.
However, this contradicts the minimality of $b(\Delta_i)$, since we can now replace $i$ by $s$.
Let $j_0,\dots,j_m$ be a sequence such that $j_0=j$, $j_m=k$ and $\Delta_{j_{l+1}}\prec\Delta_{j_l}$ for $l=0,\dots,m-1$.
Let $s>0$ be the first index such that $b(\Delta_{j_s})<b(\Delta_l)$.
Then $\m_{\{i,l,j_0,\dots,j_s\}}$ is a sub-multisegment of type $4231$, which repudiates \eqref{eq: equmin} and the assumption that $i>1$.
By passing to the contragredient we also conclude that we cannot have $b(\Delta_j)<b(\Delta_i)<b(\Delta_l)$ for any $i<l<j<k$.
Clearly, there exists some $i<k$ such that $\Delta_{i+1}\not\prec\Delta_i$. Let $r$ be the maximal such index.
We have $b(\Delta_{r+1})>b(\Delta_r)$ for otherwise $r\in S\setminus A$.
In particular $r<k-1$. Also by what we showed before we have $b(\Delta_{r+1})>b(\Delta_i)$ for all $1<i<r$.
Thus, $\Delta_{r+1}\prec\Delta_1$ for otherwise $r+1\in S\setminus A$. This concludes the case where $\m_A$ is of type $4231$.
Assume now that $\m_A$ is of type $3412$. Write $A=\{1,r,s,t,\dots,\}$ where $1<r<s<t<\dots$. In this case $j_{\min}=s$ and $j_{\max}=r$.
In particular, $S\supset\{1,r,s,k\}$.
We first claim that $\Delta_{i+1}\prec\Delta_i$ for all $i<r-1$. Assume on the contrary that $i$ is a minimal counterexample.
Since $i+1\notin A\supset S$ there exists $j<i$ such that $\Delta_{i+1}\prec\Delta_j$. Let $j$ be maximal with respect to this property.
Then $\m_{\{j,j+1,i+1,s\}}$ is of type $4231$, a contradiction to \eqref{eq: equmin}.
Moreover, if $r>2$ then $b(\Delta_2)<b(\Delta_k)$ for otherwise $\m_{A\cup\{2\}\setminus\{1\}}$ is a sub-multisegment of type $3412$
in violation of \eqref{eq: equmin}.
Consider first the case where $r=2$, $\#A=4$ (i.e., $t=k$) and $b(\Delta_k)<b(\Delta_3)<b(\Delta_1)$.
We claim that in this case we have $s=k-1$. Otherwise, $k-1\notin A\supset S$ and therefore $\Delta_k\prec\Delta_{k-1}$.
Necessarily $b(\Delta_{k-1})<b(\Delta_2)$ (since $j_{\max}=2$) and hence $\Delta_{k-1}\prec\Delta_2$ (since $\Delta_k\prec\Delta_2$).
If $b(\Delta_{k-1})>b(\Delta_3)$ then $\m_{\{2,3,k-1,k\}}$ is of type $4231$.
Otherwise $\m_{\{1,2,s,k-1\}}$ is of type $3412$. Both cases rebut \eqref{eq: equmin}. Hence $s=k-1$ as claimed.
Suppose that $i$ is such that $b(\Delta_1)<b(\Delta_i)<b(\Delta_2)$. Then $\m_{\{1,i,k-1,k\}}$
would be a sub-multisegment of type $3412$ which is excluded by \eqref{eq: equmin}.
By a similar reasoning we conclude that $b(\Delta_k)<b(\Delta_i)<b(\Delta_1)$ for all $2<i<k-1$.
Finally, we have $b(\Delta_{i+1})<b(\Delta_i)$ (and hence $\Delta_{i+1}\prec\Delta_i$) for all $2<i<k-2$ otherwise
$\Delta_{\{1,i,i+1,k-1\}}$ would be a sub-multisegment of type $4231$.
Thus, we are in the case $34{*}12$ of the lemma.
From now on we assume that $\#A>4$ or $r>2$ or $b(\Delta_k)>b(\Delta_3)$ or $b(\Delta_1)<b(\Delta_3)$.
We show that $r=s-1$. Assume on the contrary that $r<s-1$ and let $i$ be the index in $A\setminus\{1\}$ such that $b(\Delta_i)<b(\Delta_{r+1})$
and $b(\Delta_i)$ is maximal with respect to this property. If $i=s$ then $\m_{A\cup\{r+1\}\setminus\{s\}}$ is of type $3412$.
If $i=t$ and either $\#A>4$ or $\#A=4$ and $b(\Delta_1)<b(\Delta_{r+1})$ then $\m_{A\cup\{r+1\}\setminus\{r\}}$ is of type $3412$.
If $i=t$, $\#A=4$, $b(\Delta_{r+1})<b(\Delta_1)$ and $r>2$ then $\m_{\{1,2,r+1,s\}}$ is of type $4213$.
If $i>t$ then $\m_{\{r,r+1,t,\dots,i\}}$ is of type $4231$. All these cases lead to a contradiction to \eqref{eq: equmin}.
Thus $s=r+1$.
Assume that $A$ is a maximal subset of $\{1,\dots,k\}$ with respect to inclusion such that $\m_A$ is of type $3412$.
It remains to show that $A\supset\{s,\dots,k\}$. Assume on the contrary that this is not the case and let $s<j<k$ be the maximal element not in $A$.
Suppose first that $b(\Delta_j)<b(\Delta_{j+1})$. If $b(\Delta_j)<b(\Delta_k)$ then $j\in S\setminus A$ and we get a contradiction.
Otherwise, let $l$ be the first index $>j$ such that $b(\Delta_l)<b(\Delta_j)$. Let $j^-=r$ if $j=r+2$ and $j^-=j-1$ otherwise.
Then $\m_{\{j^-,j,j+1,\dots,l\}}$ is a multisegment of type $4231$, controverting \eqref{eq: equmin}.
Thus $b(\Delta_j)>b(\Delta_{j+1})$.
Assume now that $b(\Delta_j)>b(\Delta_{j^-})$. Since $j\notin A\supset S$ we must have $\Delta_{j+1}\prec\Delta_j$
and there exists $l$ such that $\Delta_j\prec\Delta_l$. Necessarily $r\le l<j^-$ and $l\ne s$. Now
$\m_{\{l,j^-,j,j+1\}}$ is of type $4231$, gainsaying \eqref{eq: equmin}.
Thus $b(\Delta_j)<b(\Delta_{j^-})$.
Since $\Delta_{j+1}\prec\Delta_{j^-}$ and $b(\Delta_{j+1})<b(\Delta_j)<b(\Delta_{j^-})$
we infer that $\Delta_j\prec\Delta_{j^-}$ and $\Delta_{j+1}\prec\Delta_j$.
By the maximality of $A$ we necessarily have $j=k-1$ and $b(\Delta_{k-1})<b(\Delta_1)$.
But then, $\m_{A\cup\{k-1\}\setminus\{k\}}$ is of type $3412$, denying \eqref{eq: equmin}.
This concludes the proof of the lemma.
\end{proof}
Next, we will use Corollary \ref{cor: derisLM} (and its terminology) to motivate the following definition.
\begin{definition}
Let $\m$ be a minimal unbalanced (regular) multisegment. We say that $\m$ is absolutely minimal unbalanced if no descendant $\m'$ of $\m$ is regular unbalanced.
\end{definition}
Recall that two segments $\Delta'\prec\Delta$ are juxtaposed if $e(\Delta')=b(\lshft{\Delta})$.
We say that the segments $\Delta_1,\dots,\Delta_k$ are back-to-back juxtaposed if $e(\Delta_{i+1})=b(\lshft\Delta_i)$ for all $i=1,\dots,k-1$.
\begin{lemma} \label{lem: absmincls}
Let $\m=\Delta_1+\dots+\Delta_k$ be an absolutely minimal unbalanced multisegment with $e(\Delta_1)>\dots>e(\Delta_k)$.
Then exactly one of the following conditions holds.
\begin{enumerate}
\item (case $4{*}23{*}1$) There exists $1<r<k-1$ such that
\begin{enumerate}
\item $\Delta_{i+1}=\lshft{\Delta}_i$ for all $1<i<r$.
\item $\Delta_1,\Delta_{r+1},\Delta_{r+2},\dots,\Delta_k$ are back-to-back juxtaposed.
\item $b(\Delta_k)=b(\lshft{\Delta}_r)$, $e(\Delta_2)=e(\lshft{\Delta}_1)$, $b(\Delta_2)=b(\lshft{\Delta}_{k-1})$, $e(\Delta_{r+1})=e(\lshft{\Delta}_r)$.
\end{enumerate}
\item (case $3{*}41{*}2$)
There exists $1<r<k-1$ such that if $\tau$ is the transposition $r\leftrightarrow r+1$ then
\begin{enumerate}
\item $\Delta_{\tau(r+1)},\Delta_{\tau(r+2)},\dots,\Delta_{\tau(k)}$ are back-to-back juxtaposed.
\item $b(\Delta_{\tau(2)})=b(\lshft{\Delta}_k)$, $b(\Delta_k)=b(\lshft{\Delta}_1)$, $b(\Delta_1)=b(\lshft{\Delta}_{\tau(k-1)})$
and $b(\Delta_{\tau(i+1)})=b(\lshft{\Delta}_{\tau(i)})$, $i=2,\dots,r-1$.
\item $e(\Delta_{i+1})=e(\lshft{\Delta}_i)$, $i=1,\dots,r+1$.
\end{enumerate}
\item (case $34{*}12$) $\m$ is of the form \eqref{eq: mincase34*12}.
\end{enumerate}
\end{lemma}
An example of the case $4{*}23{*}1$ with $k=8$ and $r=4$ is:
\[
\xymatrix@=0.6em{
&&&&&&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ&\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&&&\circ\ar@{-}[r]&\circ&\\
&&&&&&\circ\ar@{-}[r]&\circ&\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&}
\]
An example of the case $3{*}41{*}2$ with $k=8$ and $r=4$ is:
\[
\xymatrix@=0.6em{
&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&&&&\circ\ar@{-}[r]&\circ\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\\
&&&&&\circ\\
&&&\circ\ar@{-}[r]&\circ}
\]
\begin{remark}
One can show that the converse to the lemma holds as well, but we will not need this fact.
\end{remark}
\begin{proof}
We separate into the cases provided by Lemma \ref{lem: MU}.
Consider first the $4{*}23{*}1$ case. As before, let $r>1$ be the maximal index such that $\Delta_{r+1}\not\prec\Delta_r$.
If $\Delta_{i+1}\prec\Delta_i$ for some $1<i<r$ then $b(\Delta_{i+1})=b(\lshft{\Delta}_i)$, for otherwise $\lderiv_{b(\Delta_{i+1})}(\m)$
is regular unbalanced, in contradiction to the assumption on $\m$. Similarly, $e(\Delta_{i+1})=e(\lshft{\Delta}_i)$. Thus,
$\Delta_{i+1}=\lshft{\Delta}_i$ for all $1<i<r$ such that $\Delta_{i+1}\prec\Delta_i$.
Next, we show that $\Delta_1,\Delta_{r+1},\Delta_{r+2},\dots,\Delta_k$ are back-to-back juxtaposed.
If $e(\Delta_{r+1})\ne b(\lshft{\Delta}_1)$ then $\lderiv_{b(\Delta_1)}(\m)$ is regular unbalanced and we get a contradiction.
Suppose on the contrary that $e(\Delta_{i+1})\ne b(\lshft{\Delta}_i)$ for some $r<i<k$ and let $i$ be the minimal such index.
Then $\Delta_i$ is not a singleton, i.e. $b(\Delta_i)\ne e(\Delta_i)$ and by the minimality of $i$, this amounts to
$b(\Delta_i)\ne b(\lshft{\Delta}_j)$ where $j=1$ if $i=r+1$ and $j=i-1$ otherwise. Thus, $b(\Delta_i)\in\lnrset(\m)$ and
$\lderiv_{b(\Delta_i)}(\m)$ is regular unbalanced, a contradiction.
Suppose now that the set $\{i:\Delta_{i+1}\not\prec\Delta_i\}$ is not a singleton and let $s>1$ be the
penultimate element of this set. Then $\Delta_1+\Delta_s+\Delta_r+\Delta_k$ is a sub-multisegment of type $4231$.
For every $r<i<k$ $\Delta_i$ is a singleton for otherwise $b(\Delta_i)\in\lnrset(\m)$ and $\lderiv_{b(\Delta_i)}$ is regular
unbalanced, a contradiction. Write $\Delta_i=\{\rho_i\}$, $i=r+1,\dots,k-1$ and set $\rho_k=\lshft{\rho}_{k-1}$.
We have $b(\Delta_{s+1})=\rho_k$ since otherwise $b(\Delta_{s+1})\in\lnrset(\m)$ and $\lderiv_{b(\Delta_{s+1})}$ is regular unbalanced.
Let $\m_k=\m$ and define inductively $\m_{i-1}=\lderiv_{\rho_i}(\m_i)$, $i=k,\dots,r+1$.
It easily follows from Lemma \ref{lem: spclcaseextrho} that $\rho_i\in\lnrset(\m_i)$, $i=r+1,\dots,k$
and $\m_r$ is obtained from $\m$ by removing $\Delta_{r+1}$ and replacing $\Delta_{s+1}$ by $[\rho_{r+1},e(\Delta_{s+1})]$.
Thus, $\m_r$ is regular unbalanced and we get a contradiction. In conclusion $\Delta_{i+1}\prec\Delta_i$ for all $i<r$.
Finally, $e(\Delta_{j+1})=e(\lshft{\Delta}_j)$, $j=1,r$ for otherwise $\rderiv_{e(\Delta_j)}(\m)$ is regular unbalanced.
Similarly, $b(\Delta_k)=b(\lshft{\Delta}_r)$ and $b(\Delta_2)=b(\lshft{\Delta}_{k-1})$
In the $3{*}41{*}2$ case, we have $e(\Delta_{i+1})=e(\lshft{\Delta}_i)$ for all $i\le r+1$, otherwise $\rderiv_{e(\Delta_i)}(\m)$ is unbalanced.
By passing to the contragredient we get the analogous relations for the $b(\Delta_i)$'s.
Also, $e(\Delta_{r+2})=b(\lshft{\Delta}_r)$ for otherwise $\lderiv_{b(\Delta_r)}(\m)$ would be a regular unbalanced multisegment.
Suppose on the contrary that $e(\Delta_{i+1})\ne b(\lshft{\Delta}_i)$ for some $i>r+1$ and let $i$ be the minimal such index.
Then $\Delta_i$ is not a singleton, that is $b(\Delta_i)\ne e(\Delta_i)$. Equivalently, by the minimality of $i$,
$b(\Delta_i)\ne b(\lshft{\Delta}_{\tau(i-1)})$. Hence, $\lderiv_{b(\Delta_i)}(\m)$ is regular unbalanced
in contradiction with the absolute minimality assumption. In conclusion, $\Delta_{\tau(i)}$, $i=r+1,\dots,k$ are back-to-back juxtaposed.
In the $34{*}12$ case, we have $e(\Delta_{i+1})=e(\lshft{\Delta}_i)$ for all $1\le i\le k-1$ since otherwise $\rderiv_{e(\Delta_i)}(\m)$ is regular unbalanced.
Analogously, by passing to the contragredient, we have $b(\Delta_{i+1})=b(\lshft{\Delta_i})$ for all $2<i<k-2$,
$b(\rshft{\Delta}_{k-1})=b(\Delta_k)=b(\lshft{\Delta}_{k-2})$ and $b(\rshft{\Delta}_3)=b(\Delta_1)=b(\lshft{\Delta}_2)$.
Finally, $e(\Delta_k)=b(\lshft{\Delta}_2)$, for otherwise $\lderiv_{b(\Delta_2)}(\m)$ is regular unbalanced.
Thus, $\m$ is of the form \eqref{eq: mincase34*12}.
\end{proof}
\begin{corollary} \label{cor: absmin}
Suppose that $\m$ is absolutely minimal unbalanced multisegment. Then at least one of the following conditions holds.
\begin{enumerate}
\item $\m$ is of the form \eqref{eq: basic4231}, \eqref{eq: basic3412} or \eqref{eq: mincase34*12}.
\item $\m$ is contractible.
\item $\m^\#$ is regular unbalanced but not minimal unbalanced.
\end{enumerate}
\end{corollary}
\begin{proof}
This is trivial if $\m$ is of type $34{*}12$.
If $\m$ is of type $3{*}41{*}2$ then $\m$ is contractible unless every $\Delta_i$, $r+1<i<k$ is a singleton,
in which case $\m$ is of the form \eqref{eq: basic3412}.
Finally, if $\m$ is of type $4{*}23{*}1$ then once again, $\m$ is contractible unless every $\Delta_i$, $r<i<k$ is a singleton, in which case
\begin{equation} \label{eq: minproof}
\m=[k,k+r-1]+\speh{[r,k+r-2]}{r-1}+\speh{[k-1]}{k-r-1}+[1,r].
\end{equation}
It is then easy to see from the recipe of $\m^\#$ (\S\ref{sec: zeleinvo}) that
\[
\m^\#=\speh{[k,k+r-1]}{r-1}+[r,k]+\speh{[k-r+1,k-1]}{k-2r+1}+\speh{[r-1,2r-2]}{r-1}
\]
if $k\ge 2r$ and
\[
\m^\#=\speh{[k,k+r-1]}{k-r-1}+\speh{[r,2r]}{2r-k}+[r+1,k]+\speh{[k-r,k-1]}{k-r}
\]
otherwise. Thus, $\m^\#$ is regular unbalanced but upon removing its last segment we remain with an unbalanced multisegment unless $r=2$
in which case $\m$ is of the form \eqref{eq: basic4231}.
The result follows.
\end{proof}
Here is a drawing for $\m$ as in \eqref{eq: minproof} for $k=8$ and $r=4$:
\[
\xymatrix@=0.6em{
&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&\circ&\\
&&&&&\circ&\\
&&&&\circ&\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&}
\]
$\m^\#$ is given by
\[
\xymatrix@=0.6em{
&&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&\\
\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ\ar@{-}[r]&\circ&}
\]
Finally, we can prove the remaining parts of Theorem \ref{thm: main}, namely that if $\m$ is an unbalanced multisegment then $\zele{\m}$ is not \LM\ and $\m$ is not \GLS.
Indeed, assume on the contrary that $\m$ is an unbalanced multisegment with minimal $\deg\m$ such that $\pi=\zele{\m}$ is \LM.
In view of Lemma \ref{lem: 1stred}, Corollary \ref{cor: derisLM}, Proposition \ref{prop: contract} and Remark \ref{rem: contrblncd},
the minimality of $\deg\m$ implies that $\m$ is absolutely minimal unbalanced and not contractible.
Moreover, by Proposition \ref{prop: ZIred} if $\m^\#$ is regular then it is necessarily minimal unbalanced.
By Corollary \ref{cor: absmin} $\pi$ is therefore one of the representations considered in Propositions
\ref{prop: basictype4231}, \ref{prop: 3412basic} and \ref{prop: lastcase3412} of the last section.
These propositions yield the required contradiction.
By a similar reasoning, using Lemma \ref{lem: lderivm} and Remarks \ref{rem: contractGLS}, \ref{rem: GLS^t} and \ref{rem: GLScomb}
no unbalanced multisegment can be \GLS.
\begin{remark}
The use of Propositions \ref{prop: contract} (whose proof depends on the material of the next section) is not indispensable.
The ideas of \S\ref{sec: basicases} work slightly more generally for all multisegments listed in Lemma \ref{lem: absmincls}.
However, the additional reduction alleviates the bookkeeping. Similarly, the use of the Zelevinsky involution is not essential.
\end{remark}
\begin{remark} \label{rem: lwrbndN}
In principle, one can explicate the argument of this section to weaken the lower bound on $N$ stated in the converse part of Corollary \ref{cor: mainq}.
However, we will not pursue this matter here.
\end{remark}
\section{An identity of Kazhdan--Lusztig polynomials} \label{sec: KLid}
Using the Arakawa--Suzuki equivalence \cite{MR1652134} we may reinterpret Theorem \ref{thm: main} in terms of the
Kazhdan--Lusztig polynomials for the symmetric group $S_{2k}$ (Corollary \ref{cor: KLidnt}).
\subsection{The Arakawa--Suzuki functor}
We sketch the setup, referring the reader to \cite{MR2320806} and \cite{MR3495794} and the references therein for more details.
Consider the category $\CO$ with respect to $\mathfrak{gl}_k$.
For any $\mu\in\Z^k$ let $M(\mu)$ (resp., $L(\mu)$) be the Verma (resp., simple) module with highest weight $\mu$.
Suppose that $\mu=(\mu_1,\dots,\mu_k)\in\Z^k$ with $\mu_1\ge\dots\ge\mu_k$ and let $S_\mu$ be the stabilizer of $\mu$ in $S_k$,
a parabolic subgroup of $S_k$.
As is well-known, for any $\mu'\in\Z^k$ and $w\in S_k$, $L(\mu')$ occurs in $\JH(M(w\mu))$ if and only if $\mu'$
is of the form $w'\mu$ with $w'\ge w$ in the Bruhat order of $S_k$.
In the latter case, if we take $w'$ to be of maximal length in its coset $w'S_\mu$ then the multiplicity of $L(\mu)$ in $\JH(M(w\mu))$
is $P_{w,w'}(1)$ where $P_{w,w'}(q)$ denotes the Kazhdan--Lusztig polynomial with respect to $S_k$ (\cite{MR560412, MR610137, MR632980, MR1237825, MR1802178}).
In other words, denoting by $\grimg{\cdot}$ the image of an object of a locally finite abelian category in its Grothendieck group, we have
\begin{equation} \label{eq: KLconj}
\grimg{M(w\mu)}=\sum_{w'\in S_k:w'\text{ of maximal length in }w'S_\mu}P_{w,w'}(1)\grimg{L(w'\mu)}.
\end{equation}
Equivalently, for any $w\in S_k$ of maximal length in $wS_\mu$ we have
\[
\grimg{L(w\mu)}=\sum_{w'\in S_k}\sgn ww'\ P_{w'w_0,ww_0}(1)\grimg{M(w'\mu)}
\]
where $w_0$ is the longest element of $S_k$. Recall that $P_{w,w'}\equiv0$ unless $w\le w'$.
Fix $\lambda=(\lambda_1,\dots,\lambda_k)\in\Z^k$ with $\lambda_1\ge\dots\ge\lambda_k$.
For any integer $l\ge0$ let $F_{\lambda,l}$ be the exact functor of Arakawa--Suzuki from category $\CO$ to the category of
finite-dimensional representations of the graded affine Hecke algebra $\GH_l$ of $\GL_l$ (\cite{MR1652134}).
Let $\chi$ be an integral infinitesimal character of the center $\mathfrak{z}$ of the universal enveloping algebra of $\mathfrak{gl}_k$
and let $\CO_\chi$ be the full subcategory of $\CO$ on which $\mathfrak{z}$ acts by $\chi$.
There is at most one $l$ for which $F_{\lambda,l}$ is non-zero on $\CO_{\chi}$.
For this $l$ (if exists) $Z(\GH_l)$ acts by an integral character $\chi'$ (depending on $\chi$) on the image of $F_{\lambda,l}$.
Denote by $J_\chi$ the maximal ideal of $Z(\GH_l)$ corresponding to $\chi'$ (i.e., which annihilates the image of $F_{\lambda,l}$).
Let now $\IH_l$ be the Iwahori--Hecke algebra of $\GL_l(F)$.
The category of finite-dimensional representations of $\IH_l$ is equivalent to the category $\Reps_I(\GL_l(F))$ of finite-length representations of $\GL_l(F)$
which are generated by the vectors which are fixed under the Iwahori subgroup.
To $\chi'$ corresponds a character $\tilde\chi$ of $Z(\IH_l)$.
Let $J_{\tilde\chi}$ be the corresponding maximal ideal of $Z(\IH_l)$.
Then the algebras $\GH_l/J_\chi\GH_l$ and $\IH_l/J_{\tilde\chi}\IH_l$ are isomorphic \cite{MR991016}.
Thus, we may view $F_{\lambda,l}$ as an exact functor from $\CO_{\chi}$ to the full subcategory of $\Reps_I(\GL_l(F))$ on which $Z(\IH_l)$ acts by $\tilde\chi$.
We will omit $\chi$ from the notation since it will be generally clear from the context.
Taking $D=F$ and $\rho$ to be the trivial one-dimensional character of $\GL_1(F)=F^*$, the functor $F_{\lambda,l}$ satisfies
\[
F_{\lambda,l}(M(\mu))=\std(\m_{\mu,\lambda})\text{ and }
F_{\lambda,l}(L(\mu))=\begin{cases}\zele{\m_{\mu,\lambda}}&
\text{if $\mu_i\le\mu_{i+1}$ whenever $\lambda_i=\lambda_{i+1}$,}\\0&\text{otherwise,}\end{cases}
\]
where $\m_{\mu,\lambda}=\sum_{i=1}^k[\mu_i,\lambda_i]$ and $l=\sum_{i=1}^k(\lambda_i-\mu_i+1)$.
(Recall the notational convention \eqref{eq: convention}.)
Let $\mu=(\mu_1,\dots,\mu_k)\in\Z^k$ with $\mu_1\ge\dots\ge\mu_k$.
Note that $\m_{w\mu,\lambda}$ depends only on the double coset $S_\lambda wS_\mu$ of $w$.
Moreover, $\m_{w\mu,\lambda}\obt\m_{w'\mu,\lambda}$
(i.e., $\zele{\m_{w'\mu,\lambda}}$ occurs in $\JH(\std(\m_{w\mu,\lambda}))$) if and only if
$S_\lambda wS_\mu\le S_\lambda w'S_\mu$ with respect to the partial order on the double coset set $S_\lambda\bs S_k/S_\mu$ induced by
the Bruhat order of $S_k$ (by passing to the representatives of minimal length).
Applying $F_{\lambda,l}$ to \eqref{eq: KLconj} we get that for any $\mu=(\mu_1,\dots,\mu_k)\in\Z^k$ with $\mu_1\ge\dots\ge\mu_k$ and $w\in S_k$ we have
\[
\grimg{\std(\m_{w\mu,\lambda})}=\sum_{w'\in S_k:w'\text{ is of maximal length in }S_\lambda w'S_\mu}P_{w,w'}(1)\grimg{\zele{\m_{w'\mu,\lambda}}}.
\]
Equivalently, for $w$ of maximal length in $S_\lambda wS_\mu$ we have
\begin{equation} \label{eq: invgen}
\grimg{\zele{\m_{w\mu,\lambda}}}=\sum_{w'\in S_k}\sgn w'w\ P_{w'w_0,ww_0}(1)\ \grimg{\std(\m_{w'\mu,\lambda})}.
\end{equation}
In other words, the coefficients of $\zele{\m_{w\mu,\lambda}}$ in the basis
$\std(\m_{w'\mu,\lambda})$, $w'\in S_\lambda\bs S_k/S_\mu$ (ignoring $0$ terms) are
\[
\sgn w\sum_{x\in S_\lambda w'S_\mu}\sgn x\ P_{xw_0,ww_0}(1),\ \ w'\in S_\lambda\bs S_k/S_\mu.
\]
By Theorem \ref{thm: indepcspline}, this relation holds for arbitrary $\rho\in\Cusp$ and $D$.
Going back to the setup of \S\ref{sec: smth pairs} and \S\ref{sec: combi} we infer:
\begin{corollary} \label{cor: detsmth}
Let $\tseq=\bitmplt$ be a \biseq, $\sigma_0=\sigma_0(\tseq)$ and let $\sigma\in S_k$ be such that $\sigma(i)<\sigma(i+1)$ whenever $a_i=a_{i+1}$
and $\sigma^{-1}(i)<\sigma^{-1}(i+1)$ whenever $b_i=b_{i+1}$. Then
\begin{equation} \label{eq: geninv}
\grimg{\zele{\m_{\sigma}(\tseq)}}=\sum_{\sigma'\in [\sigma_0,\sigma]}\sgn \sigma'\sigma\ P_{\sigma',\sigma}(1)\ \grimg{\std(\m_{\sigma'}(\tseq))}.
\end{equation}
In particular, if $(\sigma,\sigma_0)$ is a smooth pair then
\[
\grimg{\zele{\m_{\sigma}(\tseq)}}=\sum_{\sigma'\in [\sigma_0,\sigma]}\sgn \sigma'\sigma\ \grimg{\std(\m_{\sigma'}(\tseq))}.
\]
The converse also holds in the case where $\tseq$ is regular.
\end{corollary}
This follows from \eqref{eq: invgen} by taking $\lambda=(b_1,\dots,b_k)$, $\mu=(a_k,\dots,a_1)$, $w=\sigma w_0$ and noting that
$\m_{\sigma}(\tseq)=\m_{\sigma w_0\mu,\lambda}$.
\begin{remark}
The corollary suggests that if $(\sigma,\sigma_0)$ is a smooth pair then the semisimplification of the Jacquet module of $\zele{\m_{\sigma}(\tseq)}$
is relatively easy to compute. (See \cite{MR2996769} for a special case.)
In view of Theorem \ref{thm: main} this is in accordance with Conjectures 1 and 2 of \cite{LecChev}.
\end{remark}
Next, we go back to Proposition \ref{prop: contract}.
Let $f_b,f_e:\Z\rightarrow\Z$ be the strictly monotone maps
\[
f_b(n)=\begin{cases}n+1&\text{if }n>0,\\n&\text{otherwise,}\end{cases}\ \ \
f_e(n)=\begin{cases}n+1&\text{if }n\ge0,\\n&\text{otherwise.}\end{cases}
\]
Note that $f_b(x)=f_e(y)+1$ if and only if $x=y+1$ so that
\begin{equation} \label{eq: aleb+1}
\text{$x\le y+1$ if and only if $f_b(x)\le f_e(y)+1$.}
\end{equation}
Define an injective endofunction $f$ on the set of segments by $f([x,y])=[f_b(x),f_e(y)]$.
We extend $f$ by additivity to an injective endomorphism (also denoted by $f$) of $\Mult_\rho$.
On the other hand, $f$ also defines an injective (non-graded) ring endomorphism $\phi$ of $\Gr_\rho$
determined by $\phi(\grimg{\zele{[a,b]}})=\grimg{\zele{f([a,b])}}$ for any segment $[a,b]$.
Thus $\phi(\grimg{\std(\m)})=\grimg{\std(f(\m))}$ for any $\m\in\Mult_\rho$.
\begin{corollary} \label{cor: fzele}
Under the notation above we have $\phi(\grimg{\zele{\m}})=\grimg{\zele{f(\m)}}$ for any $\m\in\Mult_\rho$, i.e., $\phi$ preserves irreducibles.
In particular, if $\m_1,\m_2\in\Mult_\rho$ then $\zele{\m_1}\times\zele{\m_2}$ is irreducible if and only if $\zele{f(\m_1)}\times\zele{f(\m_2)}$ is irreducible.
\end{corollary}
\begin{proof}
Given $\tseq=\bitmplt$ let $f(\tseq)=\begin{pmatrix}f_b(a_1)&\dots&f_b(a_k)\\f_e(b_1)&\dots&f_e(b_k)\end{pmatrix}$.
By \eqref{eq: aleb+1} $\sigma_0(f(\tseq))=\sigma_0(\tseq)$ and $f(\m_\sigma(\tseq))=\m_\sigma(f(\tseq))$ for any $\sigma\in S_k$.
Therefore, $\phi(\grimg{\std(\m_\sigma(\tseq))})=\grimg{\std(\m_\sigma(f(\tseq)))}$ for any $\sigma\in S_k$.
It follows from \eqref{eq: geninv} that $\phi(\grimg{\zele{\m_\sigma(\tseq)}})=\grimg{\zele{\m_\sigma(f(\tseq))}}=\grimg{\zele{f(\m_\sigma(\tseq))}}$.
The corollary follows.
\end{proof}
\begin{remark}
It would be interesting to have a more functorial proof of Corollary \ref{cor: fzele}.
\end{remark}
\begin{remark} \label{rem: maini}
Now that Proposition \ref{prop: contract} is proved,
Corollary \ref{cor: detsmth}, together with the statement \eqref{eq: redrelsmth} provides the last missing part of
Theorem \ref{thm: maini} of the introduction. (See the discussion following Theorem \ref{thm: main}.)
\end{remark}
\subsection{}
Let $H$ be the parabolic subgroup of $S_{2k}$
\[
H=\{w\in S_{2k}:\{w(2i-1),w(2i)\}=\{2i-1,2i\}\ \forall i\}\simeq S_2^k.
\]
As is well known, the map
\[
w\mapsto M_w=\#(w(\{2i-1,2i\})\cap\{2j-1,2j\})_{i,j=1,\dots,k}
\]
is bi-$H$-invariant and defines a bijection between $H\bs S_{2k}/H$
and the set $\Matk$ of $k\times k$ matrices with entries in $\{0,1,2\}$ such that the sum of the entries in each row and each column is $2$.
In turn, by the Birkhoff--von-Neumann theorem, these are precisely the matrices that can be written as the sum of two
$k\times k$ permutation matrices. (We will write $\permat{\sigma}$ for the permutation matrix corresponding to $\sigma\in S_k$.)
The corresponding permutations in $S_k$ (say $\sigma_1$, $\sigma_2$) are not uniquely determined (even up to interchanging).
However, the conjugacy class of $\sigma_2^{-1}\sigma_1$ in $S_k$, which will be denote by $[w]$, is uniquely determined by the double coset.
More precisely, we have the following.
\begin{lemma} \label{lem: RHS}
For any $M\in\Matk$ let $C_1,\dots,C_s$ be the equivalence classes for the equivalence relation
generated by $i\sim j$ if there exists $l$ such that $M_{i,l}=M_{j,l}=1$.
Then the set
\[
\{(\sigma_1,\sigma_2)\in S_k\times S_k:\permat{\sigma_1}+\permat{\sigma_2}=M\}
\]
has cardinality $2^r$ where $r=\{i:\abs{C_i}>1\}$.
Moreover, for any $(\sigma_1,\sigma_2)\in S_k\times S_k$ such that $\permat{\sigma_1}+\permat{\sigma_2}=M$,
the cycles of $\sigma_2^{-1}\sigma_1$ are the $C_i$'s. In particular,
the conjugacy class of $\sigma_2^{-1}\sigma_1$ in $S_k$ is determined by $M$ only.
\end{lemma}
\begin{proof}
The symmetric $k\times k$ matrix $MM^t-2I_k$ has non-negative integer entries and the sum along each row and column is two.
Therefore, it is the adjacency matrix of an
undirected $2$-regular graph $G$ (possibly containing loops and double edges), with vertex set $\{1,\dots,k\}$.\footnote{As usual,
a loop counts twice for the degree of a vertex.}
Hence, the connected components of $G$ are cycles (including loops and $2$-cycles) and their underlying vertex sets are the $C_i$'s.
Note that the loops in $G$ correspond to the indices $i$ for which there exists $l$ such that $M_{i,l}=2$, while the $2$-cycles in $G$
(i.e., the double edges) correspond to the pairs of indices $i\ne j$ for which there exist $l\ne m$ such that $M_{i,l}=M_{j,l}=M_{i,m}=M_{j,m}=1$.
If $M=\permat{\sigma_1}+\permat{\sigma_2}$ then the edges of $G$ (counted with multiplicities) are given by $\{\sigma_1(i),\sigma_2(i)\}$, $i=1,\dots,k$.
Moreover, any such presentation gives rise to an orientation of $G$, given by $\sigma_1(i)\rightarrow\sigma_2(i)$ such that $G$ is the union
of directed cycles, i.e., such that the indegree and the outdegree of each vertex is one. Conversely, any such orientation
arises from a presentation $M=\permat{\sigma_1}+\permat{\sigma_2}$ where $\sigma_1$ and $\sigma_2$ are uniquely determined
and the cycles of $\sigma_2^{-1}\sigma_1$ are the $C_i$'s.
Clearly, the number of such orientations is $2^r$ where $r$ is the number of non-trivial connected components of $G$.
\end{proof}
Given a \biseq\ $\tseq=\bitmplt$ we write $\tilde{\tseq}$ for the duplicated \biseq\
$\begin{pmatrix}a_1&a_1&\dots&a_k&a_k\\b_1&b_1&\dots&b_k&b_k\end{pmatrix}$ of length $2k$.
Similarly, for any $\sigma\in S_k$ we write $\tilde\sigma$ for the permutation in $S_{2k}$ given by
$\tilde\sigma(2i-j)=2\sigma(i)-j$, $i=1,\dots,k$, $j=0,1$.
Clearly, $\tilde\sigma$ normalizes the subgroup $H$ of $S_{2k}$.
It easily follows from \eqref{def: sigma0} that
\begin{equation} \label{eq: dbltseq}
\sigma_0(\tilde\tseq)=\widetilde{\sigma_0(\tseq)}.
\end{equation}
Let $\iota:S_k\times S_k\rightarrow S_{2k}$ be the embedding
\[
\iota(\sigma_1,\sigma_2)(2(i-1)+j)=2(\sigma_j(i)-1)+j,\ \ i=1,\dots,k,\ j=1,2.
\]
In particular, $\iota(\sigma,\sigma)=\tilde\sigma$.
Clearly, if $\sigma_1'\le\sigma_1$ and $\sigma_2'\le\sigma_2$ then $\iota(\sigma_1',\sigma_2')\le\iota(\sigma_1,\sigma_2)$.
Also, for any $w\in S_{2k}$ and $\sigma_1,\sigma_2\in S_k$
\begin{equation} \label{eq: iotasum}
\iota(\sigma_1,\sigma_2)\in HwH\text{ if and only if }\permat{\sigma_1}+\permat{\sigma_2}=M_w.
\end{equation}
For any $\sigma\in S_k$ let $\rk_\sigma:\{1,\dots,k\}^2\rightarrow\Z_{\ge0}$ be the rank function
\[
\rk_\sigma(i,j)=\#\{u=1,\dots,i:\sigma(u)\le j\}.
\]
It is well known that for any $\sigma,\tau\in S_k$ we have $\tau\le\sigma$ if and only if $\rk_\sigma\le\rk_\tau$ on $\{1,\dots,k\}^2$.
We will use the following combinatorial result.
\begin{proposition}[\cite{1710.06115}] \label{prop: tight}
Let $(\sigma,\sigma_0)$ be a smooth pair and let $\tau\in S_k$.
Suppose that $\rk_\tau(i,j)=\rk_\sigma(i,j)$ for all $(i,j)\in\{1,\dots,k\}^2$ such that $\rk_{\sigma_0}(i,j)=\rk_\sigma(i,j)$.
Then $\tau\le\sigma$.
\end{proposition}
Note that for $\sigma_0=\id$, i.e., when $\sigma$ itself is smooth, this is a classical result. (See \cite{MR1934291} and the references therein.)
We obtain the following consequence.
\begin{corollary} \label{cor: dblsame}
Suppose that $(\sigma,\sigma_0)$ is a smooth pair. Let $\sigma_1,\sigma_2\in S_k$ be such that
$\sigma_0\le\sigma_1,\sigma_2$ and $H\iota(\sigma_1,\sigma_2)H\le H\tilde\sigma$. Then $\sigma_1,\sigma_2\le\sigma$.
\end{corollary}
\begin{proof}
Indeed, the condition $H\iota(\sigma_1,\sigma_2)H\le H\tilde\sigma$ means that
\[
\rk_{\sigma_1}(i,j)+\rk_{\sigma_2}(i,j)\ge 2\rk_{\sigma}(i,j),\ \ i,j=1,\dots,k.
\]
On the other hand, $\rk_{\sigma_1}(i,j),\rk_{\sigma_2}(i,j)\le\rk_{\sigma_0}(i,j)$ since $\sigma_0\le\sigma_1,\sigma_2$.
Hence, whenever $\rk_{\sigma_0}(i,j)=\rk_\sigma(i,j)$ we also have $\rk_{\sigma_1}(i,j)=\rk_{\sigma_2}(i,j)=\rk_{\sigma}(i,j)$.
By Proposition \ref{prop: tight} we conclude that $\sigma_1,\sigma_2\le\sigma$ as required.
\end{proof}
Let $\clsf$ be the class function on $S_k$ given by $\clsf(\sigma)=\sgn\sigma\ 2^r$ where $r$ is the number of non-trivial cycles of $\sigma$.
We now interpret Theorem \ref{thm: main} in terms of an identity of Kazhdan--Lusztig polynomials.
\begin{corollary} (of Theorem \ref{thm: main}) \label{cor: KLidnt}
Suppose that $(\sigma,\sigma_0)$ is a smooth pair with $\sigma_0$ $213$-avoiding. Then for any $x\in [\widetilde{\sigma_0},\tilde\sigma]$ we have
\begin{equation} \label{eq: allxrltn}
\sum_{w\in HxH}\sgn w\ P_{w,\widetilde{\sigma}}(1)=\clsf([x]).
\end{equation}
In particular,
\begin{equation} \label{eq: parkl}
\sum_{w\in H}\sgn w\ P_{\widetilde{\sigma'}w,\widetilde{\sigma}}(1)=1.
\end{equation}
for any $\sigma'\in[\sigma_0,\sigma]$.
\end{corollary}
\begin{proof}
Let $\tseq$ be a regular \biseq\ such that $\sigma_0=\sigma_0(\tseq)$ (see Lemma \ref{lem: comb12}) and let $\m=\m_\sigma(\tseq)$.
By Theorem \ref{thm: main} $\zele{\m}\times\zele{\m}$ is irreducible, i.e., $\zele{\m}\times\zele{\m}=\zele{\m+\m}$.
We will deduce the corollary from Corollary \ref{cor: detsmth} by computing the coefficient of $\std(\m_x(\tilde\tseq))$ in the expansion of
$\zele{\m}\times\zele{\m}=\zele{\m+\m}$ in terms of standard modules in two different ways.
On the one hand,
\[
\grimg{\zele{\m}}=\sum_{\sigma'\in S_k:\sigma'\le\sigma}\sgn\sigma\sigma'\ \grimg{\std(\m_{\sigma'}(\tseq))}
\]
where of course only the terms $\sigma'\ge\sigma_0$ give a non-zero contribution.
Note that for any $\sigma_1,\sigma_2\in S_k$ we have
\[
\std(\m_{\sigma_1}(\tseq))\times\std(\m_{\sigma_2}(\tseq))=\std(\m_{\sigma_1}(\tseq)+\m_{\sigma_2}(\tseq))=
\std(\m_{\iota(\sigma_1,\sigma_2)}(\tilde\tseq))
\]
and this is non-zero (i.e., by \eqref{eq: dbltseq}, $\iota(\sigma_1,\sigma_2)\ge\widetilde{\sigma_0}$) if and only if $\sigma_0\le\sigma_1,\sigma_2$.
Thus,
\begin{multline*}
\grimg{\zele{\m}\times\zele{\m}}=\sum_{\sigma_1,\sigma_2\in S_k:\sigma_1,\sigma_2\le\sigma}\sgn\sigma_1\sigma_2\ \grimg{\std(\m_{\sigma_1}(\tseq)+\m_{\sigma_2}(\tseq))}\\=
\sum_{\sigma_1,\sigma_2\in S_k:\sigma_1,\sigma_2\le\sigma}\sgn\sigma_1\sigma_2\ \grimg{\std(\m_{\iota(\sigma_1,\sigma_2)}(\tilde\tseq))}.
\end{multline*}
On the other hand, by \eqref{eq: geninv} we have
\[
\grimg{\zele{\m+\m}}=\sum_{w\in S_{2k}}\sgn w\ P_{w,\tilde\sigma}(1)\ \grimg{\std(\m_w(\tilde\tseq))}.
\]
Comparing coefficients, for any $x\in H\bs S_{2k}/H$ such that $x\ge\widetilde{\sigma_0}$ we get
\begin{equation} \label{eq: intprtirred}
\sum_{w\in HxH}\sgn w\ P_{w,\tilde\sigma}(1)=\sum_{\sigma_1,\sigma_2\in S_k:\sigma_1,\sigma_2\le\sigma\text{ and }\iota(\sigma_1,\sigma_2)\in HxH}\sgn\sigma_1\sigma_2.
\end{equation}
Recall that $\iota(\sigma_1,\sigma_2)\ge\widetilde{\sigma_0}$ (or equivalently, $H\iota(\sigma_1,\sigma_2)H\ge\widetilde{\sigma_0}H$)
if and only if $\sigma_0\le\sigma_1,\sigma_2$.
Thus, by Corollary \ref{cor: dblsame}, if $x\in [\widetilde{\sigma_0},\tilde\sigma]$ then the condition $\sigma_1,\sigma_2\le\sigma$ on the
right-hand side of \eqref{eq: intprtirred} is superfluous.
Hence, by Lemma \ref{lem: RHS} and \eqref{eq: iotasum} the right-hand side of \eqref{eq: intprtirred} is $\clsf([x])$, proving our claim.
\end{proof}
\begin{remark}
The (computer-assisted) example $\sigma=(4231)$, $\sigma_0=(1324)$ shows that the condition that $\sigma_0$ is $213$-avoiding is essential
for the relation \eqref{eq: allxrltn}.
On the other hand, in \cite{1705.06517} we conjecture among other things that for any smooth pair $(\sigma,\sigma_0)$ we have
\[
\sum_{w\in H}\sgn w\ P_{\widetilde{\sigma_0}w,\widetilde{\sigma}}(q)=q^{\ell(\sigma)-\ell(\sigma_0)}
\]
(and in particular \eqref{eq: parkl} holds)
and prove it in the case where $\sigma$ is a product of distinct simple reflexions (i.e., a Coxeter element in a parabolic subgroup of $S_n$).
We also remark that for the relation \eqref{eq: parkl} (assuming $\sigma_0$ is $213$-avoiding) we haven't used the result of \cite{1710.06115}
since Corollary \ref{cor: dblsame} is trivial if $H\iota(\sigma_1,\sigma_2)H=H\widetilde{\sigma'}$
(in which case $\sigma_1=\sigma_2=\sigma'$).
\end{remark}
\subsection{}
More generally, let $m>1$ and consider the parabolic subgroup $H\simeq S_m\times\dots\times S_m$ of $S_{mk}$ of type $(m,\dots,m)$
($k$ times) and the subgroup $K\simeq S_k\times\dots\times S_k$ ($m$ times) of $S_{mk}$ given by
\[
K=\{\sigma\in S_{mk}:\sigma(i)\equiv i\pmod m\text{ for all }i\}.
\]
Thus, $H\cap K=1$ and the normalizer of $H$ is $H\rtimes\{\tilde\sigma:\sigma\in S_k\}$ where as before
\[
\tilde\sigma(mi-j)=m\sigma(i)-j,\ i=1,\dots,k,\ j=0,\dots,m-1.
\]
We have
\begin{theorem} \label{thm: higherKL}
Suppose that $(\sigma,\sigma_0)$ is a smooth pair with $\sigma_0$ $213$-avoiding. Then
\begin{equation} \label{eq: gencaseirredcoseq}
\sum_{w\in HxH}\sgn w\ P_{w,\tilde\sigma}(1)=\sum_{\tau\in HxH\cap K}\sgn\tau
\end{equation}
for any $x\in[\widetilde{\sigma_0},\tilde\sigma]$. In particular,
\begin{equation} \label{eq: spclcasesigma'}
\sum_{w\in H}\sgn w\ P_{\widetilde{\sigma'}w,\widetilde{\sigma}}(1)=1
\end{equation}
for any $\sigma'\in[\sigma_0,\sigma]$.
\end{theorem}
Indeed, if $\tseq$ is a regular \biseq\ such that $\sigma_0=\sigma_0(\tseq)$ and $\m=\m_\sigma(\tseq)$ then as in the proof of Corollary \ref{cor: KLidnt}
(using an obvious analog of Corollary \ref{cor: dblsame}), the left-hand side (resp., right-hand side) of
\eqref{eq: gencaseirredcoseq} is $\sgn\tilde\sigma$ times the coefficient of $\std(\m_x(\tilde\tseq))$ in the expansion of
$\zele{m\cdot\m}$ (resp., $\zele{\m}^{\times m}$) in terms of standard modules.
The theorem therefore follows from Theorem \ref{thm: main} and Corollary \ref{cor: pi1pi2LM} which imply that $\zele{\m}^{\times m}=\zele{m\cdot\m}$.
(Once again, for \eqref{eq: spclcasesigma'} we do not need to use \cite{1710.06115}.)
Note that as before, the double cosets $H\bs S_{mk}/H$ correspond to matrices of size $k\times k$ with non-negative integer entries,
whose sums along each row and each column are all equal to $m$. Each such matrix can be written as a sum of $m$ permutation matrices.
Thus, $HxH\cap K\ne\emptyset$ for all $x\in S_{mk}$.
However, for $m>2$ it is no longer true that $\sgn$ is constant on $HxH\cap K$.
In fact, the right-hand side of \eqref{eq: gencaseirredcoseq} is much more mysterious for $m>2$.
For instance, in the case where $m=k$ and the double coset $HxH$ corresponds to the matrix all of whose entries are 1,
the right-hand side of \eqref{eq: gencaseirredcoseq} is $(-1)^{m\choose 2}$ times the difference $\partial_m$ between the number
of even and odd Latin squares of size $m\times m$. Clearly $\partial_m=0$ if $m$ is odd but it is still an open question,
known as the Alon--Tarsi conjecture, whether $\partial_m\ne0$ for all even $m$ \cite{MR1179249}.
This conjecture is related to other problems in linear algebra. Some progress on it was made by Janssen, Drisko, Zappa and others \cite{MR1309160, MR1451417, MR1624999, MR1453404}.
In particular, $\partial_m\ge0$. An upper bound for $\partial_m$ was given by Alpoge \cite{MR3638338}.
| 20,131
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TITLE: Linear Programming Transformations
QUESTION [2 upvotes]: What is the process of performing a transformation from a given problem to another linear programming problem such that the transformed problem has an optimal solution iff the initial problem has a solution. I've learned about reductions (working with complexity), and the question seems quite similar, but now that the optimal keyword is included, I know that it can't be exactly like a reduction.
For reference, here is the question that has spurred my own question:
Consider the linear inequalities $L0$ in a finite number of unknowns
${x_1,\ldots,x_n}$ given below:
$$\sum_{j=1}^{n}a_{ij}x_j \le b_i \tag{L0}$$ for $i=1,2,\ldots,m$.
Show how to transform this into a linear program $L1$ so that $L1$ has
an optimum solution if and only if $L0$ has a solution. Argue the
correctness of your transformation.
REPLY [3 votes]: I found this PDF really helpful. Look at the second system of equations.
The basic idea is to change it to a minimization problem. It's clear that for all $b_i \geq 0$ the original system $L0$ is solvable: just set all $x_i$ to $0$ for a feasible (not optimal, but feasible) solution.
But for $b_i$ < $0$ it is possible that $L0$ is not solvable. So for every i s.t. $b_i<0$ , introduce a new variable $w_i \geq 0$ Then the equation becomes:
$$\sum_{j=1}^{n}a_{ij}x_j - w_i \le b_i \tag{L0}$$ for all i s.t. $b_i$ < $0$, and
$$\sum_{j=1}^{n}a_{ij}x_j \le b_i \tag{L0}$$ for all i s.t. $b_i \ge 0$.
So the minimization problem is to minimize $$Z=\sum_{i \text{ s.t. } b_i < 0}w_i$$
If minimum $Z=0$ (the optimum solution), $L0$ has a feasible solution (because you get back the original system.)
| 148,646
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MILWAUKEE — The Milwaukee Urban League is looking to help at least 303 people get their driver's license, free of charge.
It’s all through a program they’ve been operating for more than 20 years. Now, they’re pushing harder to get more people into the courses than ever before.
The monthly course is free to everyone who qualifies. That includes anyone enrolled in the State of Wisconsin’s Food Share Employment and Training (FSET) program. Drivers must also have no unpaid citations or fines.
“We need programs like this. We do,” said Douglas Harris.
Harris got his license through the four-day course over the summer. He finally got it after spending when his license expired when he was homeless.
Now, he’s back on his feet.
“Now, I just got my driver’s license. So it’s going to take me a minute to get a car, but at least I’m heading in the right direction,” Harris said.
The program is taught by an instructor. A Wisconsin Department of Transportation employee comes in on the fourth and final day of classes to offer the learner’s permit tests.
“It’s such a blessing because I had no idea up until last Thursday,” said Alaina Applewhite, a current student of the class.
Applewhite took the bus the class. She also takes the bus with her 2-month-old daughter Naomi. With temperatures dropping, she said it’s time to get on the road.
“It means more availability to jobs and not having my daughter out in this bad weather,” she said.
The average cost-per-person is estimated around $350. Each person in the course will not have to pay.
According to Milwaukee Police:
1,966 people have been cited this year for operating without a license.
7,309 people have been cited with a suspended or revoked license.
The Urban League said they are looking to have at least 303 students by the end of the contract year that began on Oct. 1.
| 114,492
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Goblet
(c. 1760)
- Medium
- glass, enamel (wheel-engraved, enamel-twist stem)
- Measurements
- 17.3 × 8.4 cm diameter
- Place/s of Execution
- England
- Accession Number
- 1657-D5
- Department
- International Decorative Arts
- Credit Line
- National Gallery of Victoria, Melbourne
William and Margaret Morgan Endowment, 1968
- Gallery location
- Not on display
- Description
- Ovoid bowl, engraved with sunflower and bee, opaque twist stem, plain foot.
| 297,061
|
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Wiki
It appears that and a similar fix for the ctype errors with the compaq
extended math library has resolved this issue.
Thanks again.
FYI.
I've found the following CFLAGS very useful on my Alpha:
CFLAGS=-O3 -mcpu=ev56 -mtune=ev56 -mieee -mno-soft-float -mfp-regs -pipe
-pthread
(the -mno-soft-float -mfp-regs attempt to use the internal FP components of
the Alpha processor and the -mieee corrects for NAN (not-a-number, e.g.
divide by zero) issues).
Anyone have other/better optimizations for ev56 class?
-----Original Message-----
From: Brian Parkhurst [mailto:brianp@...]
Sent: Tuesday, July 12, 2005 22:49
To: gentoo-alpha@g.o
Subject: Re: [gentoo-alpha] Recent Glibc breaks CPQ-Java?
I@...
----- Original Message -----
From: "Jose Luis Rivero (YosWinK)" <yoswink@g.o>
To: <gentoo-alpha
--
gentoo-alpha@g.o mailing list
Updated Jun 17, 2009
Summary:
Archive of the gentoo-alpha mailing list.
Donate to support our development efforts.
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| 333,998
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\begin{document}
\maketitle
\begin{abstract}
In this paper we study elements $\beta \in \fqn$ having normal $\alpha$-depth $b$; that is, elements for which $\beta, \beta - \alpha, \ldots, \beta-(b-1)\alpha$ are simultaneously normal elements of $\fqn$ over $\fq$. In~\cite{GoveGary}, the authors present the definition of normal $1$-depth but mistakenly present results for normal $\alpha$-depth for some fixed normal element $\alpha \in \fqn$. We explain this discrepancy and generalize the given definition of normal $(1-)$depth from~\cite{GoveGary} as well as answer some open questions presented in~\cite{GoveGary}.
\end{abstract}
{\bf Keywords:} finite fields, normal bases, primary decomposition
{\bf MSC:} 11T30, 11T71, 12Y05
\maketitle
\section{Introduction and notation}
Throughout this document, we use the following standard notation. Let $p$ be a prime and let $q$ be a power of $p$, the finite field of $q$ elements is denoted $\fq$, and the finite degree $n$ extension of $\fq$ is denoted $\fqn$. The (relative) trace function is denoted $\Tr_{\fqn:\fq}\colon \fqn\to\fq$. We remark that the trace function is onto, and for any $k\not\equiv0\pmod{p}$, the element $k\alpha$ is also normal. For any positive integer $n$, denote by $e = v_p(n)$, the \emph{$p$-ary valuation} of $n$; that is the largest integer $e$ such that $p^e$ divides $n$ but $p^{e+1}$ does not divide $n$. We also denote by $\tau = p^e$; specifically, $\tau = 1$ ($e=0$) if $\gcd(p,n) = 1$.
In Section~\ref{sec:Frob}, we derive conditions for elements to be normal that we will use later in the paper. In Section~\ref{sec:depthb}, we correct and generalize the notion of normal elements of depth $b$ from~\cite{GoveGary}. Also motivated by~\cite{GoveGary}, in Section~\ref{sec:conjugates} we observe that depth is not necessarily invariant under conjugation, and further analyze the depth of the conjugates of normal elements.
\section{Finite fields as Frobenius modules}\label{sec:Frob}
In this section, we follow~\cite{LS, prim-1-normal} and introduce finite fields as Frobenius modules. Let $\sigma_q\colon \overline{\fq}\to\overline{\fq}$ denote the Frobenius $q$-automorphism. Clearly, $\sigma_q$ fixes $\fq$ and for any $n > 0$ and $\alpha \in \overline{\fq}$, $\sigma_q^n(\alpha) = \alpha$ if and only $\alpha \in \fqn$. Moreover, the Galois group of $\fqn$ over $\fq$ is cyclic of order $n$ and generated by $\sigma_q$.
Let $\alpha \in \fqn$ and let $\cB$ consist of the Galois orbit of $\alpha$; that is, $\cB = \{\alpha, \alpha^q, \ldots, \alpha^{q^{n-1}}\}$. If $\cB$ is a linearly independent set, then $\alpha$ is a \emph{normal element} of $\fqn$ and $\cB$ is a \emph{normal basis} of $\fqn$ over $\fq$. We also call $\alpha$ a \emph{cyclic vector} for $\fqn$ as a vector space over $\fq$.
For $f(x) = \sum_{i=0}^{m} a_i x^i$, denote the action of $f$ on $\overline{\fq}$ by
\[ f \circ \alpha = f(\sigma_q)(\alpha) = \sum_{i=0}^{m} a_i \alpha^{q^i}. \]
Clearly, $(f + g)\circ \alpha = f\circ\alpha + g\circ\alpha$ for any $f,g \in \fq[x]$, and $(x^n-1) \circ \alpha = 0$ if and only if $\alpha \in \fqn$. Moreover, $(fg)\circ \alpha = f\circ(g\circ \alpha)$, so that if $f \circ \alpha = 0$ for any $\alpha \in \fqn$, then $f$ divides $x^n-1$.
\begin{definition} \ \\
\begin{enumerate}
\item For any $\alpha \in \fqn$, define the \emph{annihilator} of $\alpha$ as the polynomial $\ann_\alpha \in \fq[x]$ of smallest degree such that $\ann_\alpha \circ \alpha = 0$.
\item For any $f \in \fq[x]$, define $\ker(f) = \{\alpha \in \fqn\colon \ann_\alpha = f\}$, the set of elements of $\fqn$ annihilated by $f$ under $\circ$.
\end{enumerate}
\end{definition}
Observe that $\ann_\alpha$ annihilates any linear combination of Galois conjugates of $\alpha$. We have $\ker(x^n-1) = \fqn$ and $\ann_\alpha(x)$ divides $x^n-1$ for any $\alpha$. Moreover, $\alpha$ is a normal element of $\fqn$ over $\fq$ if and only if $\ann_\alpha(x) = x^n-1$ by linear independence of the conjugates of $\alpha$. We summarize these observations in Proposition~\ref{prop:numnormal}.
\begin{proposition}\label{prop:numnormal}
For any prime power $q$, the number of normal elements of $\fqn$ over $\fq$ is given by $\Phi_q(x^n-1)$, where $\Phi_q$ is Euler's totient function over $\fq$; that is, $\Phi_q(x^n-1)$ is the number of polynomials in $\fq[x]$ of degree less than $n$ that are relatively prime with $x^n-1$.
\end{proposition}
Existence of normal elements can be gleaned directly from Proposition~\ref{prop:numnormal}, since $\Phi_q(x^n-1)$ is nonzero for all $n \geq 1$.
We now introduce a map central to the remainder of this work. Suppose $\alpha \in \fqn$ is normal and define the map $\phi_\alpha\colon \fq[x]\to \fqn$ by $\phi(f) = f\circ\alpha$. Then $\ker(\phi_\alpha) = ( x^n-1 )$, since $\alpha$ is normal; similarly $\phi_\alpha$ is onto since the set $\cB_{\alpha} = \{\alpha, \alpha^q, \ldots, \alpha^{q^{n-1}}\}$ is a basis. Hence $\fqn\cong \fq[x]/(x^n-1)$ as Frobenius modules. We will abuse notation and refer to this isomorphism also as $\phi_\alpha$.
Let $g(x) = \sum_{i=0}^{n-1} g_i x^{i} \in \fq[x]$, and $\beta = \phi_\alpha(g) =\sum_{i=0}^{n-1} g_i \alpha^{q^i}$. Then $\beta^q = \sum_{i=0}^{n-1} g_{i-1} \alpha^{q^i}$. Thus $\phi_\alpha^{-1}(\beta^q) = x\phi_\alpha^{-1}(\beta) \mod (x^n-1)$. Thus the Frobenius action on $\fq[x]/(x^n-1)$ is induced by $\overline{\sigma_q}(g):= xg(x)$, with $\overline{\sigma_q} = \phi_\alpha \sigma_q \phi_\alpha^{-1}$.
We exploit the decomposition of $\fq[x]/(x^n-1)$ as a Frobenius module. We follow the treatment in~\cite{Steel}. Let $e = \nu_p(n)$ be the valuation of $n$ at $p$ and let $x^n-1 = f_1^{e_1} \cdots f_r^{e_r}$ be the primary factorization of $x^n-1$, then $e_i = p^{e} = \tau$ for all $i = 1, \ldots, r$. In particular, $\tau = 1$ if $\gcd(p,n) = 1$. Denote by $\overline{V_i} = \fq[x]/(f_i^{\tau})$, then
\begin{equation}
\fq[x]/(x^n-1) \cong \bigoplus_{i=1}^r \overline{V_i}.\label{eqn:primary_decomposition}
\end{equation}
Explicitly, we write the image of $g$ in $\bigoplus_{i=1}^r \overline{V_i}$ as $(g \mod f_1^{\tau},\ldots,g \mod f_r^{\tau})$. We abuse notation slightly and write $V_i=\phi_\alpha(\overline{V_i})$.
\begin{equation}
\bigoplus_{i=1}^r V_i\cong \fqn \cong \fq[x]/(x^n-1) \cong \bigoplus_{i=1}^r \overline{V_i}.\label{eqn:primary_decomposition2}
\end{equation}
Equation~\eqref{eqn:primary_decomposition2} is the \emph{primary decomposition} of $\fqn$ as a Frobenius module. Moreover, we observe that each $V_i$ is stable under $\sigma_q$.
\begin{proposition}\label{prop:normalelt}
Let $\alpha$ be a normal element of $\fqn$, and suppose $\beta=\phi_\alpha(g(x))$. Then $\ann_\beta = \frac{x^n-1}{\gcd(x^n-1,g(x))}$, and $\beta$ is normal if and only if $\gcd(x^n-1,g(x))=1$. Furthermore, $V_i=\ker(f_i^{\tau})$.
\end{proposition}
\begin{proof}
Let $f(x) = \sum_{i=0}^{m} a_i x^i$. Then $f\circ \beta = 0$ if and only if $f(x)g(x)\in (x^n-1)$. The smallest degree polynomial satisfying $f(x)g(x)\in (x^n-1)$ is clearly $\frac{x^n-1}{\gcd(x^n-1,g(x))}$, as claimed. Since $\beta$ is normal if and only if $\ann_\beta(x) = x^n-1$, $\beta$ is normal if and only if $\gcd(x^n-1,g(x))=1$.
Now $\beta \in \phi_\alpha(\overline{V_i})$ if and only if $f_j^{\tau}$ divides $g(x)$ for all $j\ne i$, which occurs if and only if $f_i(x)^{\tau}g(x)\in (x^n-1)$, if and only if $f_i^{\tau}\circ \beta =0$, if and only if $\beta \in \ker(f_i^{\tau})$.
\end{proof}
We summarise the characterisations of normal elements here.
\begin{proposition}\label{prop:normalequiv}
Let $\alpha$ be a normal element of $\fqn$, and suppose $\beta=\phi_\alpha(g(x))$. Let $x^n-1 = f_1^{\tau} \cdots f_r^{\tau}$, with the $f_i$ being distinct irreducible polynomials in $\fq[x]$. Let $g_i=g\mod f_i^{\tau}$, and $\beta=\sum_{i=1}^r \beta_i$ for $\beta_i\in V_i$. Then the following are equivalent:
\begin{enumerate}
\item
$\beta$ is normal,
\item
$\gcd(x^n-1,g(x))=1$,
\item
$\gcd(f_i,g_i)=1$ for each $i$.
\item
$\ann_{\beta_i}=f_i^{\tau}$ for each $i$,
\item
$\beta_i\in \ker(f_i^{\tau})\backslash \ker(f_i^{\tau-1})$ for each $i$,
\end{enumerate}
\end{proposition}
\begin{proof}
($1. \iff 2.$) This is Proposition~\ref{prop:normalelt}.
($2. \iff 3.$) Let $g_i = g\mod f_i^\tau$, then $g = h f_i^\tau + g_i$ for some $h \in \fq[x]$. Then (the irreducible) $f_i$ divides $g_i$ for some $1 \leq i \leq r$ if and only if $f_i$ divides $g$, contradicting $\gcd(x^n-1, g(x)) = 1$.
($3. \iff 4.$) Let $g_i = g \mod f_i^\tau$ with $\beta_i = \phi_\alpha(g_i) \in V_i$. Clearly $\beta_i \in V_i$ if and only if $\beta_i \in \ker(f_i^\tau)$, so $\ann_{\beta_i} = f_i^k$ for $1 \leq k \leq \tau$. Now, $f_i^k\circ \beta_i = 0$ if and only if $f_i^k g_i \circ \alpha = 0$ if and only if $f_i^k g_i \in (x^n-1)$. Now $\gcd(f_i, g_i) = 1$ if and only if $k = \tau$ for all $i$.
($4. \iff 5.$) By the minimality of $\ann_{\beta_i}$, we have $\ann_{\beta_i} = f_i^\tau$ if and only if $\beta_i \in \ker(f_i^\tau)$ and $\beta_i \notin \ker(f_i^{\tau-1})$. \qedhere
\end{proof}
If $\gcd(p,n) =1$, then $\tau=1$, and thus we get the following.
\begin{corollary}\label{cor:normalPD}
Let $\gcd(p,n) =1$ and let $\beta = \beta_1 + \beta_2 + \cdots + \beta_r$ with $\beta_i \in V_i$, then $\beta$ is a normal element if and only if $\prod_{i=1}^r \beta_i \neq 0$.
\end{corollary}
\section{Depth-$b$ normal elements}\label{sec:depthb}
\begin{definition}\label{def:depth}
Let $b\in \NN$ with $b\leq p$. If $\beta \in \fqn$ is such that $\beta, \beta - \alpha, \ldots, \beta-(b-1)\alpha$ are normal elements of $\fqn$ over $\fq$ for some $\alpha\in \fqn$, then we say that $\beta$ has \emph{normal $\alpha$-depth $b$}.
\end{definition}
In \cite{GoveGary}, the authors introduced normal depth, where the definition was for $\theta=1$. However the results in \cite{GoveGary} are in fact referring to normal $\alpha$-depth, for some fixed normal element $\alpha$. We will explain the discrepancy below, and consider the more general problem.
We remark that Defintion~\ref{def:depth} can be extended for $b \geq p$ when $q$ is a power of $p$ by imposing an ordering on the elements of $\fq$ (or even further still, on $\fqn$). Since~\cite{GoveGary} and Section~\ref{sec:conjugates} are mostly concerned with depth $2$, we will not treat these sorts of extensions in this work.
We recap (and generalize) the main question from~\cite{GoveGary}.
\begin{question}\label{question}
To what extent do the conjugates of an element $\beta$ having normal $\alpha$-depth $b$ also have normal $\alpha$-depth $b$?
\end{question}
In particular in \cite{GoveGary}, they focus on normal depth $2$ and search for \emph{lonely elements}: that is, normal elements of depth $2$ having a conjugate that fails to have normal depth $2$.
\begin{lemma}
Without loss of generality, fix a normal element $\alpha$ of $\fqn$ satisfying $\Tr_{\fqn:\fq}(\alpha)=n/\tau$, since if $\alpha'$ is any normal element with $\Tr_{\fqn:\fq}(\alpha') = k \neq 0$, the element $\alpha = \alpha' \frac{\tau}{nk}$ is normal (since $\tau/n \not\equiv 0\pmod{p}$). Then,
\begin{enumerate}
\item
$\phi_\alpha^{-1}(\alpha^{q^i})=x^i$ and $\phi_\alpha^{-1}(1) = (\tau/n)\frac{x^n-1}{x-1}$,
\item
the image of $\alpha$ in $\bigoplus_{i=1}^r \overline{V_i}$ is $(1,1,\ldots,1)$, and the image of $1$ is $((x-1)^{\tau-1},0,\ldots,0)$.
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item For $f(x) = \sum_{i=0}^{m} a_i x^i$, we have $\phi_\alpha(f) = f \circ \alpha = \sum_{i=0}^{m} f_i \alpha^{q^i}$. Hence, $x^i \circ \alpha = \alpha^{q^i}$ or $\phi_\alpha^{-1}(\alpha^{q^i}) = x^i$. Similarly,
\[
\Tr_{\fqn:\fq}(\alpha) = \left(\sum_{i=0}^{n-1} x^i\right)\circ \alpha = n/\tau,
\]
so by linearity, $\phi_\alpha^{-1}(1) = (\tau/n) \sum_{i=0}^{n-1} x^i = (\tau/n) \frac{x^n-1}{x-1}$.
\item Since $\alpha = \phi_\alpha(1)$, we have $g_i = g\mod f_i^\tau = 1$ for all $1 \leq i \leq r$. Similarly, $1 = (\tau/n)\phi_\alpha (\frac{x^n-1}{x-1})$, and with $\tau = p^{\nu_p(n)}$ and by linearity of Frobenius,
\begin{align*}
\frac{x^n-1}{x-1} &= \sum_{i=0}^{n-1} x^i \equiv (n/\tau) \sum_{i=0}^{\tau-1} x^i \mod (x^\tau -1) \\
&= (n/\tau) \frac{x^\tau-1}{x-1} = (n/\tau)(x-1)^{\tau-1}. \qedhere
\end{align*}
\end{enumerate}
\end{proof}
\begin{proposition}
Let $\alpha \in \fqn$ be normal. An element $\beta=\phi_\alpha(g(x))$ has normal $\alpha$-depth $b$ if and only if $\gcd(x^n-1,g(x)-c)=1$ for all $c\in \{0,\ldots,b-1\}$.
\end{proposition}
\begin{proof}
The proof is immediate from the linearity of $\phi$ and from Proposition~\ref{prop:normalequiv}, Remark 2.
\end{proof}
In \cite{GoveGary}, the number $\#\{g:\gcd(x^n-1,g(x)-c)=1~\forall~c\in \{0,\ldots,b-1\}\}$ was defined as $\Phi_b(x^n-1)$.
\begin{theorem}
Let $\alpha \in \fqn$ be normal with $\Tr_{\fqn:\fq}(\alpha) = \tau/n$, let $e = \nu_p(n)$, and let $\beta=\phi_\alpha(g(x))$ also be normal. Then
\begin{enumerate}
\item if $e > 0$, then $\beta$ has normal $1$-depth $p$; moreover, $\beta - c$ is normal for all $c \in \fq$,
\item if $e = 0$, then $\beta$ has normal $1$-depth $b$ if and only if $g(1) \geq b$ (under a suitable implicit ordering of the elements of $\fq$). In particular, $\beta - c$ is normal if and only if $g(1) \neq c$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $g_i=g\mod f_i^{\tau}$. Then the image of $\beta-c$ in $\bigoplus_{i=1}^r \overline{V_i}$ is $(g_1 -c (x-1)^{\tau-1},g_2,\ldots,g_r)$.
If $e>0$ and $\beta$ is normal, then $\gcd(g_1,(x-1)^\tau)=1$. If $\beta-c$ is not normal, then $(x-1)$ divides $g_1 -c (x-1)^{\tau-1}$, implying $(x-1)$ divides $g_1$, a contradiction. Thus $\beta-c$ is normal for all $c\in \fq$.
If $e=0$, then $g_1=g(1)$, and the image of $g$ is $(g(1) -c ,g_2,\ldots,g_r)$. By Corollary~\ref{cor:normalPD}, $\beta$ is normal if and only if $g_i\ne0$ for each $i$. Hence, $\beta-c$ is {\it not} normal if and only if $g(1)= c$.
\end{proof}
In \cite{GoveGary} the authors mistakenly state that the number of elements having normal $1$-depth $b$ is equal to $\Phi_b(x^n-1)$. This assumably arose by the erroneous assumption that $\phi_\alpha(1)=1$. Instead, since $\phi_\alpha(1) = \alpha$, $\Phi_b(x^n-1)$ refers to the number of elements having normal $\alpha$-depth $b$, and so for the remainder of this paper we focus on this case as well.
\section{Conjugates: Lonely and Sociable elements}\label{sec:conjugates}
Throughout this section, we use the notation from Section~\ref{sec:depthb}; in particular, $x^n-1 = (f_1 \cdots f_r)^{\tau}$ where $n = \tau m$ with $\gcd(m,\tau) = 1$, and $f_i$ is irreducible for $1 \leq i \leq r$. Suppose $\beta=\phi_\alpha(g(x))$ has normal $\alpha$-depth $b$. We consider the normal $\alpha$-depth of its conjugates. Recall that $\beta^{q^i}=\phi_\alpha(x^i g(x))$. Thus we need to consider the common divisors of $x^ig(x)-c$ with $x^n-1$, or equivalently $g(x)-cx^i$ with $x^n-1$.
\begin{definition}\label{def:lonely-social}
An element $\beta\in \fqn$ is \emph{$(\alpha,b)$-lonely} if $\beta$ has normal $\alpha$-depth $b$, but $\beta^{q^i}$ does not have normal $\alpha$-depth $b$ for some $i$. If $\beta^{q^i}$ has normal $\alpha$-depth $b$ for all $i$, we say that $\beta$ is \emph{$(\alpha,b)$-sociable}.
\end{definition}
Similar to Proposition~\ref{prop:normalequiv}, we have a number of equivalent characterizations of sociable elements.
\begin{theorem}\label{thm:tfae-sociable}
Let $x^n-1 = f_1^\tau f_2^\tau \cdots f_r^\tau$ with $f_i$ irreducible, $1 \leq i \leq r$. Let $\beta\in \fqn$ with $g(x)=\phi_\alpha^{-1}(\beta)$ and let $g_i = g\mod f_i^\tau$. Then the following are equivalent:
\begin{enumerate}
\item
$\beta$ is $(\alpha,b)$-sociable,
\item
$\gcd(x^n-1,g(x)-cx^j)=1$ for all $j\in \{0,\ldots,n-1\}$, $c\in \{0,\ldots,b-1\}$,
\item
$\gcd(f_i,g_i - cx^j)=1$ for all $i\in \{1,\ldots,r\}$, $j\in \{0,\ldots,n-1\}$, $c\in \{0,\ldots,b-1\}$,
\item
$g(\theta)\notin \{c\theta^j:c \in \{0,\ldots,b-1\}, j\in \{0,\ldots,n-1\}\}$ and $\theta$ a root of $x^n-1$.
\end{enumerate}
\end{theorem}
\begin{proof}
The equivalence of items $1.$, $2.$, $3.$ come directly from applying Proposition~\ref{prop:normalequiv} to Definition~\ref{def:lonely-social}. Here we prove only $3. \iff 4.$
Suppose $\gcd(f_i, g_i-cx^j) \neq 1$ for some $1 \leq i \leq r$, $0 \leq j \leq n-1$, which occurs if and only if $f_i(\theta_i) = g_i(\theta_i) - c\theta_i^j = 0$ for some $\theta_i \in \F_{q^{\deg(f_i)}}$; that is, $g_i(\theta_i) = c\theta_i^j$. The fourth equivalence follows, since $g_i = g\mod f_i^\tau$, so $g(\theta_i) = g_i(\theta_i).$
\end{proof}
The number of $\beta$ that are $(\alpha,b)$-sociable is the number of $g$ satisfying the conditions on their roots given in the fourth equivalence of Theorem~\ref{thm:tfae-sociable}.
\begin{lemma}\label{lem:trivial}
Let $x^n-1 = f_1^\tau \cdots f_r^{\tau}$ and let $\theta_i$ be a root of $f_i$, $1 \leq i \leq r$. Then there are exactly $q^{\deg(f_i)}$ possible values for $g(\theta_i)$ for $g\in \fq[x]$. Furthermore, let $\theta_{ij} = \theta_i^{q^j}$ for $j = 0, 1, \ldots, \deg(f_i)$ be the roots of $f_i$ in $\fq(\theta_i)$, and fix $\gamma_{i} \in \fq(\theta_i)$, $1 \leq i \leq r$. Then there exist precisely $q^{\frac{n(\tau-1)}{\tau}}$ polynomials $g$ of degree at most $n$ with $g(\theta_{ij}) = \gamma_{i}^{q^j}$ for all $1 \leq i \leq r$, $0 \leq j \leq \deg(f_i)-1$.
\end{lemma}
\begin{proof}
Clearly $g(\theta_i)\in \fq(\theta_i)=\mathbb{F}_{q^{\deg(f_i)}}$, and so there are at most $q^{\deg(f_i)}$ possible values for $g(\theta_i)$. As $g$ has coefficients in $\fq$, we have that $g(\theta_i^{q^j})=g(\theta_i)^{q^j}$ for any $j$.
With $n = n_0\tau$, two polynomials $g$ and $h$ in $\fq[x]$ agree on all $n_0$-th roots of unity if and only if $f_1f_2 \cdots f_r$ divides $g-h$. As $\deg(f_1f_2 \cdots f_r) = n_0$, there are $q^{n-n_0} = q^{n_0(\tau-1)}$ such polynomials $h$ of degree at most $n$.
\end{proof}
For $\beta$ that are $(\alpha, b)$-sociable, Theorem~\ref{thm:tfae-sociable} provides a number of forbidden values for $g(\theta_i)$. The precise number of forbidden values that ensure that $\beta$ is $(\alpha,b)$-sociable is complicated in general, but we can solve it completely in some cases left open in \cite{GoveGary}.
\begin{proposition}
The number of elements in $\fqn$ that are $(\alpha,b)$-sociable is {\it at most}
\[
q^{\frac{n(\tau-1)}{\tau}}\prod_{i=1}^r (q^{\deg( f_i)} - n(b-1)-1)
\]
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:trivial}, there are at most $q^{\deg(f_i)}$ choices for $g(\theta_i)$ for each $i = 1, \ldots, r$. By the final assertion of Theorem~\ref{thm:tfae-sociable}, an upper bound on the number of forbidden choices of $g(\theta_i)$ occurs when all of $c\theta_i^j$ are distinct for all $c \in \{1, \ldots, b-1\}$ and $j \in \{0,\ldots, n-1\}$. This gives $n(b-1)$ forbidden values for $g(\theta_i)$, and the further restriction $g(\theta_i)\ne 0$ together with Lemma \ref{lem:trivial} completes the proof.
\end{proof}
\begin{proposition}
Suppose $(n,q-1)=1$. Then the number of elements that are $(\alpha,b)$-sociable is
\[
q^{\frac{n(\tau-1)}{\tau}}\prod_{i=1}^r (q^{\deg(f_i)} - (b-1)\ord(\theta_i)-1),
\]
where $\theta_i$ is a root of $f_i$, $1 \leq i \leq r$.
\end{proposition}
\begin{proof}
Since $(n,q-1)=1$, $x^n-1$ has only one root in $\fq$, namely $1$. Thus as each $\theta_i^j$ is an $n$-th root of $1$ (in some extension field), we have that $\theta_i^j\notin \fq$ for all $i$ and all $1<j<\ord(\theta_i)$. Therefore $\#\{c\theta_i^j:c \in \{1,\ldots,b-1\}, j\in \{0,\ldots,n-1\}\} = (b-1)\ord(\theta_i)$. As $g(\theta_i)\ne 0$, there are $q^{\deg(f_i)} - (b-1)\ord(\theta_i)-1$ choices for $g(\theta_i)$ for each $i$ for which $\phi_\alpha(g)$ is $(\alpha,b)$-sociable. The factor $q^{\frac{n(\tau-1)}{\tau}}$ follows from Lemma \ref{lem:trivial}.
\end{proof}
\begin{corollary}\label{cor:n=qs}
Suppose $n=q^s$. Then the number of elements that are $(\alpha,b)$-sociable is
\[
q^{q^s-q^{s-1}}(q - b).
\]
\end{corollary}
For a specific example of Corollary~\ref{cor:n=qs}, taking $q=n$, $b=2$, we get that there are $q^{q-1}(q-2)$ elements which are $(\alpha,2)$-sociable in $\mathbb{F}_{q^q}$.
\begin{corollary}
Suppose $n$ is prime, $n\notin \{p,q-1\}$, and let $x^n-1 = (x-1)f_2 \cdots f_r$. Then the number of elements that are $(\alpha,b)$-sociable is
\[
(q-b)\prod_{i=2}^r (q^{\deg(f_i)} - (b-1)n-1).
\]
\end{corollary}
In \cite{GoveGary}, focus is applied to the case $b=2$, the case of $(\alpha,2)$-lonely/sociable elements. We now apply Theorem~\ref{thm:tfae-sociable} to this situation.
\begin{proposition}\label{prop:linearsplit}
Suppose $n|(q-1)$. Then the number of elements that are $(\alpha,2)$-sociable is
\begin{equation}
\prod_{i=1}^n \left(q- \frac{n}{(i,n)}-1\right).\label{eqn:coprime}
\end{equation}
\end{proposition}
\begin{proof}
As $n|(q-1)$, $x^n-1$ factorises in to a product of distinct linear factors over $\fq$. Let $f_i = x-\theta_i$. Then $\beta$ is $(\alpha,2)$-sociable if and only if $g(\theta_i)\ne 0, \theta_i^j$ for any $j$. Thus the number of forbidden choices for $g(\theta_i)$ is $\ord(\theta_i)+1$. Letting $\theta$ be a primitive $n$-th root of unity in $\fq$, and letting $\theta_i=\theta^i$, then $\ord(\theta_i)=\frac{n}{(i,n)}$ and the result follows.
\end{proof}
\begin{remark}
Note that Formula~\eqref{eqn:coprime} is not true in general. Issues arise when there exist $c_1,c_2\in \{0,\ldots,b-1\}$ such that $c_1=c_2\theta_i^j$, in which case $\#\{c\theta_i^j:c \in \{0,\ldots,b-1\}, j\in \{0,\ldots,n-1\}\}$ is more difficult to calculate. The conditions of the previous two theorems were chosen to avoid this possibility.
\end{remark}
The following example of Proposition~\ref{prop:linearsplit} provides an answer to the first open question left in~\cite{GoveGary}.
\begin{example}
Suppose $n=3$, and suppose $x^3-1$ factors into distinct linear factors over $\fq$, say $x^3-1=(x-1)(x-\lambda)(x-\mu)$; equivalently, if $q \equiv 1\pmod{3}$. Then $\phi_\alpha(g)$ has normal $\alpha$-depth $2$ if and only if $\{0,1\}\cap\{g(1),g(\lambda),g(\mu)\}=\emptyset$. Similarly, $\phi_\alpha(g)^{q^i}$ has normal $\alpha$-depth $2$ if and only if $\{0,1\}\cap\{g(1),\lambda^i g(\lambda),\mu^i g(\mu)\}=\emptyset$. Since a polynomial of degree at most three is uniquely determined by its evaluation at three different elements of $\fq$, then there are $(q-2)^3$ elements of $\alpha$-depth $2$, of which $(q-2)(q-4)^2$ are not lonely. Thus there are $4(q-2)(q-3)$ lonely elements.
\end{example}
We can also apply Proposition~\ref{prop:linearsplit} to provide a partial answer to the second open question in~\cite{GoveGary}.
\begin{example}\label{ex:p=n+1}
Suppose $n = 4$, $q=5$, then $x^4-1 = (x-1)(x-2)(x-3)(x-4)$, with $\ord(1) = 1$, $\ord(4) = 2$ and $\ord(2)=\ord(3)=4$. A direct application of Proposition~\ref{prop:linearsplit} shows that there are no $(\alpha,2)$-sociable elements.
\end{example}
Example~\ref{ex:p=n+1} generalizes in an obvious way.
\begin{proposition}
Let $q = n+1$, then there are no $(\alpha, 2)$-sociable elements of $\fqn$.
\end{proposition}
\begin{proof}
Let $\theta$ be a primitive element in $\fq$. Then $x^n-1=x^{q-1}-1=\prod_{\lambda\in \fq^*}(x-\lambda)=\prod_{i=0}^{q-2}(x-\theta^i) $. Hence for $\beta=\phi_\alpha(g)$ to be $(\alpha, 2)$-sociable it would require that $g(\theta)\ne 0$ and $g(\theta)\ne \theta^i$ for any $0\leq i\leq q-2$, which is impossible as $g(\theta)\in \fq$.
\end{proof}
The following was proved in \cite[Proposition 4.3]{GoveGary}. We include an alternative proof here.
\begin{proposition}
Suppose $\frac{x^n-1}{x-1}$ is irreducible over $\fq$. Then the number of elements that are $(\alpha,2)$-sociable is
\[
(q-2)(q^{n-1}-n-1),
\]
and the number of elements that are $(\alpha,2)$-lonely is
\[
(q-2)(n-1).
\]
\end{proposition}
\begin{proof}
Recall that $\frac{x^n-1}{x-1}$ is irreducible over $\fq$ if and only if $q$ is primitive modulo $n$. Then $\{\theta^{q^i}\colon i = 0, \ldots, n-2\} = \{\theta^i\colon i = 1, \ldots, n-1\}$ is the set of distinct roots of $\frac{x^n-1}{x-1}$. Thus, an element $\phi_\alpha(g)$ is $(\alpha,2)$-sociable if and only if $g(1) \neq 0,1$ and $g(\theta) \neq \theta^i$ for $i = 1,2, \ldots, n-1$. Hence, there are $(q-2)(q^{n-1}-n+1)$ elements that are $(\alpha,2)$-sociable and $(q-2)(n-1)$ lonely elements in $\fqn$.
\end{proof}
The following example examines two cases of $(\alpha, 3)$-sociable elements, giving the first directions towards the third open problem in~\cite{GoveGary}.
\begin{example}
\begin{enumerate}
\item Consider the case $q=7$, $n=3$, $b=3$. Then $x^n-1=(x-1)(x-2)(x-4)$, and $2^3=1$. Now the set $\{c \theta^j:c\in \{0,1,2\},j\in \{0,1,2\}\}$ is equal to $\{0,1,2\}$ for $\theta=1$ and $\{0,1,2,4\}$ for $\theta=2,4$. Thus the number of $(\alpha,3)$-sociable elements is $(7-3)(7-4)^2=36$.
\item In the case $q=13$, $n=3$, $b=3$, we have $x^n-1=(x-1)(x-3)(x-9)$. Now the set $\{c \theta^j:c\in \{0,1,2\},j\in \{0,1,2\}\}$ is equal to $\{0,1,2\}$ for $\theta=1$ and $\{0, 1, 2, 3, 5, 6, 9\}$ for $\theta=3,9$. Thus the number of $(\alpha,3)$-sociable elements is $(13-3)(13-7)^2=36$.
\end{enumerate}
\noindent These two examples illustrates how extra care must be taken when an element of $\{0,\ldots,b-1\}$ is a nontrivial $n$-th root of unity.
\end{example}
\section{Conclusions and future directions}
In this paper, we study a generalization of normal elements of depth $b$, as presented in~\cite{GoveGary}. Since depth is not invariant under conjugation, we further analyze the depth of the conjugates of normal elements.
The notion of depth readily lends itself to further generalization. One such ``natural'' generalization is as follows. Given some total ordering $\mathcal{O}$ of the elements of $\fqn$, say $\mathcal{O} = \{o_0, o_1, \ldots, o_{q^n-1}\}$, an element $\beta \in \fqn$ has normal $(\mathcal{O},\alpha)$-depth $b$ if $\beta - o_0\alpha, \beta - o_1\alpha, \ldots, \beta - o_{b-1}\alpha$ are simultaneously normal. Here, $\beta$ has $\alpha$-depth $b$ if $o_i = i$ for $i = 0, \ldots, b-1$. Some interesting questions here occur when $\alpha$ is a normal element of $\fqn$ over $\fq$ and $\mathcal{O}_\zeta = (0, \zeta, \zeta^2, \ldots, \zeta^{q^n-2})$, for a primitive element $\zeta \in \fqn$. Determining conditions for which $\beta$ has $(\mathcal{O}_\zeta, \alpha)$-depth $2$, or statistics on the possible values of $b$ for which $\beta$ has $(\mathcal{O}_\zeta, \alpha)$-depth $b$ is the subject of future work.
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Nandamuri Kalyan Ram, Aditi Arya in the lead roles.
Jagapati Babu plays an important supporting role.
The first look poster is released on 4th July 2016. Kalyan Ram is featured in an action packed role in this movie.
Junaid Siddiqui and Mukesh G handled the editing and cinematography respectively.
Music composed byAnup Rubens.
Music released on ADITYA Music Company.
Audio was launched on 5th October 2016, held atHyderabad.
Nandamuri Harikrishna, NTR, Dil Raju and many other celebrities attended the function.
The film is expected to release on 7 October 2016 on the occasion of Dussera.
| 65,376
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99pp., paperback, No Place, 2020
A new collection of poems by Cas Vos.
Poet and theologian Cas Vos has published 10 previous volumes of poetry, as well as translations of Homer's lliad and Odyssey and several theological works. He was Dean of the Faculty of Theology at the University of Pretoria from 2000 to 2010, when he retired. He was awarded the Andrew Murray Prize for Theology in 1999.
| 214,981
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TITLE: Find k such that a piecewise function is continuous
QUESTION [0 upvotes]: Find k such that $f(x) = x^2$ if $x \leq 2$ and $f(x) = k - x^2$ if $x > 2$.
My first thought was well, since from the left side $f(x) \to 4$ as $x \to 2$ it seems to me the other part should pick up at 4 as well otherwise we would have a 'hole'. Which leads me to think $k = 8$, as $8 - 2^2 = 4$.
Is it correct that although it has a kink in the graph, the function is continuous because it's defined at $x = 2$
REPLY [1 votes]: Yes, your answer is correct. The kink in the graph means the function is not differentiable at 2, but has no bearing on whether it is continuous. It's continuous if there are no breaks in the graph, and a kink is not a break. So your function is continuous if $k=8$. Note that it's not enough that the function be defined. It has to also have no break. It's defined at $x=2$ but not continuous if you choose $k=7.$ Only continuous if $k=8.$
Awesome username btw.
| 29,167
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TITLE: Lagrangian mechanics formulation of a simple free motion of two masses in uniform gravity field
QUESTION [2 upvotes]: As a part of larger project, I decided to test my Lagrangian formulation of simple system of two rigidly connected point masses as indicated below.
I introduce the generalized coordinates vector $\bar{q} = [x, y, q]^T$, where $x,y$ are coordinates of position of mass $M$ and $q$ is the angular deflection from horizontal of the massless rod of length $l$, in the middle of which mass $m$ is attached. Then, the position coordinates of $m$ are given by
$$ x_m = x + \dfrac{l}{2}\cos(q) $$
$$ y_m = y + \dfrac{l}{2}\sin(q) $$
Having this, I defined kinetic ($T$) and potential ($V$) energies
$$ T = \dfrac{1}{2}M(\dot{x}^2 + \dot{y}^2) + \dfrac{1}{2}m(\dot{x}_m^2 + \dot{y}_m^2)$$
$$ V = g \left( My + m\left(y+\dfrac{l}{2}\sin(q)\right) \right) $$
Then, the Lagrangian $L$ and mechanical energy of the system $E$ are
$$ L = T-V$$
$$ E = T+V$$
Further, the derivations were done by my by hand (a few times to double check) and using SymPy (symbolic math package of Python) and they match. Upon integrating the EOM, I receive a nicely looking plots for state vector variables $\bar{Q} = [x,y,q,\dot{x},\dot{y},\dot{q}]^T$, if my initial conditions involve null $q$ nd $\dot{q}$. However, if they are non-zero, the motion looks highly non-physical visually (I created a Matlab animation of the motion) and additionally the mechanical energy of the system is not adding up to be constant (see pic below for $\bar{Q}_0 = [1,1000,1,1,0,-0.1]^T$).
My question is therefore, are my assumptions and initial formulation correct for the given situation? I am quite new to Lagrangian formalism and afters hours of tackling this seemingly simple problem, started to think there might be some rule of Lagrangian mechanics I violate. All classical mechanics problems I was able to find online are not free motion of multibody systems, therefore I was not able to check with actual solution for problem like this.
To give some more context, my intention is to expand this problem to put the multibody system in orbital flight (central point-mass gravity field) and increase complexity of structure, leading to n-link pendulum attached to bus mass $M$. If there is a fundamental error in the proceedings I presented above here, I assume I would not be able to extend the formulation to more complex system I just described.
I will be eternally grateful for any help / advice! I will be also happy to provide more details on my solution.
EDIT
To clarify, I post my EL equations:
$$ M \frac{d^{2}}{d t^{2}} x{\left(t \right)} - 0.5 m \left(l \sin{\left(q{\left(t \right)} \right)} \frac{d^{2}}{d t^{2}} q{\left(t \right)} + l \cos{\left(q{\left(t \right)} \right)} \left(\frac{d}{d t} q{\left(t \right)}\right)^{2} - 2 \frac{d^{2}}{d t^{2}} x{\left(t \right)}\right)= 0 $$
$$ M \frac{d^{2}}{d t^{2}} y{\left(t \right)} + g \left(M + m\right) + 0.5 m \left(- l \sin{\left(q{\left(t \right)} \right)} \left(\frac{d}{d t} q{\left(t \right)}\right)^{2} + l \cos{\left(q{\left(t \right)} \right)} \frac{d^{2}}{d t^{2}} q{\left(t \right)} + 2 \frac{d^{2}}{d t^{2}} y{\left(t \right)}\right) = 0 $$
$$ l m \left(0.5 g \cos{\left(q{\left(t \right)} \right)} + 0.25 l \frac{d^{2}}{d t^{2}} q{\left(t \right)} - 0.5 \sin{\left(q{\left(t \right)} \right)} \frac{d^{2}}{d t^{2}} x{\left(t \right)} + 0.5 \cos{\left(q{\left(t \right)} \right)} \frac{d^{2}}{d t^{2}} y{\left(t \right)}\right) = 0$$
$$ \begin{bmatrix} M+m & 0 & -\dfrac{l}{2}m\sin(q) \\ 0 & M+m & \dfrac{l}{2}m\cos(q) \\ -\dfrac{l}{2}m\sin(q) & \dfrac{l}{2}m\cos(q) & m\dfrac{l^2}{4} \end{bmatrix} \begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{q} \end{bmatrix} = \begin{bmatrix} \dfrac{l}{2}m\cos(q)\dot{q}^2 \\ -g(M+m) + \dfrac{l}{2}m\sin(q)\dot{q}^2 \\ -\dfrac{l}{2}gm\cos(q) \end{bmatrix} $$
REPLY [1 votes]: Remark. This case generalizes to your real goal to have a system of more than two mass-points attached to each other by rods where the rods can rotate relative to each other. Basically, in a uniform force field like the gravitational force field near the surface of the earth the problem becomes equivalent to the geodesic motion of a point onto a torus of dimension equal to the number of rods.
Lagrangian approach. First, you place an inertial coordinate frame (say a frame attached to the ground) in the plane of the system. Denote it by $O\,\vec{e}_x\,\vec{e}_y$, where the coordinate vectors are of unit length and perpendicular to each other. Moreover, $O\,\vec{e}_x\,\vec{e}_y$ is oriented so that $\vec{e}_y$ is the vertical vector along which gravitational acceleration is $\vec{g} = - \, g \, \vec{e}_y$. The position vectors of the mass-points, originating from point $O$ and ending at the either of the mass points, are
$\vec{r}_M$ and $\vec{r}_m$. The Lagrangian is then
$$L\, =\, \frac{M}{2}\,\left|\frac{d\vec{r}_M}{dt}\right|^2
\, + \, \frac{m}{2}\,\left|\frac{d\vec{r}_m}{dt}\right|^2
\, - \, Mg\,\big(\vec{r}_M \cdot \vec{e}_y\big)
\, - \, mg\,\big(\vec{r}_m \cdot \vec{e}_y\big)$$
with holonomic constriant
$$\big|\vec{r}_M \, - \, \vec{r}_m\big|^2 \, = \, \frac{l^2}{4}$$
Perform the linear chainge of coordinates:
\begin{align}
&\vec{r}_G \, =\, \frac{M}{M+m} \, \vec{r}_M \, + \, \frac{m}{M+m} \, \vec{r}_m\\
&\vec{r}_l \, = \, \vec{r}_M \, - \, \vec{r}_m
\end{align}
with inverse
\begin{align}
&\vec{r}_M \, =\, \vec{r}_G \, + \, \frac{m}{M+m} \, \vec{r}_l\\
&\vec{r}_m \, = \, \vec{r}_G \, - \, \frac{M}{M+m} \, \vec{r}_l
\end{align}
In the new coordinates, the Lagrangian becomes
\begin{align}
L \, =& \, \frac{M}{2}\,\left|\frac{d\vec{r}_G}{dt} \, + \, \frac{m}{M+m} \, \frac{d\vec{r}_l}{dt}\right|^2
\, + \, \frac{m}{2}\,\left|\frac{d\vec{r}_G}{dt} \, - \, \frac{M}{M+m} \, \frac{d\vec{r}_l}{dt}\right|^2 \\
&\, - \,g\,\Big(\big(M\,\vec{r}_M \, + \, m \, \vec{r}_m \big)\cdot \vec{e}_y\Big) \\
\, =&\, \frac{M}{2} \left|\frac{d\vec{r}_G}{dt}\right|^2 \, + \, \frac{M\,m}{(M+m)}\left(\frac{d\vec{r}_G}{dt}\cdot\frac{d\vec{r}_l}{dt}\right) \, + \, \frac{M\,m^2}{2(M+m)^2}\, \left|\frac{d\vec{r}_l}{dt}\right|^2\\
&\, + \, \frac{m}{2} \left|\frac{d\vec{r}_G}{dt}\right|^2 \, - \, \frac{M\,m}{(M+m)}\left(\frac{d\vec{r}_G}{dt}\cdot\frac{d\vec{r}_l}{dt}\right) \, + \, \frac{M^2\,m}{2(M+m)^2} \left|\frac{d\vec{r}_l}{dt}\right|^2\\
&\, - \,(M+m)\,g\,\Big(\vec{r}_G \cdot \vec{e}_y\Big)
\end{align}
and after some regrouping and simplification the Lagrangian and the holonomic constraint become:
$$L \, = \, \frac{M\,m}{2(M+m)}\, \left|\frac{d\vec{r}_l}{dt}\right|^2
\, + \, \frac{(M+m)}{2}\, \left|\frac{d\vec{r}_G}{dt}\right|^2 \, - \, (M+m)\,g\,\Big(\vec{r}_G \cdot \vec{e}_y\Big) $$
$$\big|\vec{r}_l\big|^2 = \frac{l^2}{4}$$
Introduce the generalized coordinates:
$$\vec{r}_l \, = \, \frac{l}{2}\, \cos(q) \, \vec{e}_x \, + \, \frac{l}{2}\, \sin(q) \, \vec{e}_y$$
$$\vec{r}_G \, = \, x_G \, \vec{e}_x \, + \, y_G\, \vec{e}_y$$ which parametrixe the holonimc constrints (they tirivialize it). Then, the Lagrnagian simplifies to
$$L \, = \, \frac{M\,m\ l^2}{8(M+m)}\, \left(\frac{dq}{dt}\right)^2
\, + \, \frac{(M+m)}{2}\, \left(\frac{dx_G}{dt}\right)^2 \, + \, \frac{(M+m)}{2}\, \left(\frac{dy_G}{dt}\right)^2 \, - \, (M+m)\,g\,y_G $$
Consequently, the Euler-Lagrange equations are
\begin{align}
&\frac{d^2{x}_G}{dt^2} \, = \, 0 \\
&\frac{d^2{y}_G}{dt^2} \, = \, - \, g\,\\
& \frac{d^2q}{dt^2} \, = \, 0
\end{align}
You can solve them right away
\begin{align}
&x_G \, = \, x_G(0) \, + \, v_x\, t \\
&y_G \, = \, y_G(0) \, + \, v_y\, t \, - \, \frac{g}{2} \, t^2\,\\
&q \, = \, q(0) \, + \, \omega_0 \, t
\end{align}
where $x_G(0), \, y_G(0), \, q(0)$ are constants describing the initial configuration of the rod and $v_x, \, v_y, \, \omega_0$ are constants describing the initial velocities of the rod.
The rigid body interpretation. The special case you are modelling in your OP can be thought of as a solid body, basically a rigid rod, which moves in a plane and whose mass is concentrated at its ends. First, you place an inertial coordinate frame (say a frame attached to the ground). Denote it by $O\,\vec{e}_x\,\vec{e}_y\,\vec{e}_z$, where the coordinate vectors are of unit length and perpendicular to each other. Moreover, $O\,\vec{e}_x\,\vec{e}_y\,\vec{e}_z$ is oriented so that $\vec{e}_y$ is the vertical vector along which gravitational acceleration is $\vec{g} = - \, g \, \vec{e}_y$, the system moves in the coordinate plane $O\,\vec{e}_x\,\vec{e}_y$, just like on your picture, and $\vec{e}_z$ is pointing perpendicularly to the picture.
Denote by $\vec{r}_M$ the position vector, pointing from the origin $O$ of the coordinate frame to the mass-point $M$ and by $\vec{r}_m$ the position vector, pointing from the origin $O$ of the coordinate frame to the mass-point $m$. The condition that the two mass points are connected by a mass-less rod of length $\frac{l}{2}$, can be written as the quadratic equation.
$$\big|\vec{r}_M - \vec{r}_m\big|^2 = \frac{l^2}{4}$$ Let $G$ be the center of gravity of the rod, i.e. the center of gravity, which is
$$\vec{r}_G = \frac{M}{M+m} \, \vec{r}_M \, + \, \frac{m}{M+m} \, \vec{r}_m$$
The force that acts on the system of the two mass points on the rod is gravity and is applied to the center of mass $\vec{r}_G$ of the rod. The general equations of motion of a rigid body can be written as
\begin{align}
&\frac{d^2\vec{r}_G}{dt^2} \, = \, - \, g\, \vec{e}_y\\
& \frac{d}{dt}\,I \,\omega\, \vec{e}_z \, = \, \vec{T}
\end{align} where $\omega \, \vec{e}_z$ is the angular velocity of the rod and $$I = \frac{M\,m\,l^2}{4\,(M+m)}$$ is its moment of inertia along the $O\, \vec{e}_z$ axis with respect to the center of mass $G$ of the rod. The other moments of inertia are zero because this is a one dimensional rod. The vector $\vec{T}$ is the sum of the torques of all forces acting on the rod, calculated with respect to the center of mass $G$. However, there is only one force, the gravity force $-(M+m)\, g \, \vec{e}_y$, acting on the rod and it is applied to the center of gravity $G$. Since $$\vec{T} = \vec{GG} \times \big(-(M+m)\, g \, \vec{e}_y\big) = \vec{0}\times \big(-(M+m)\, g \, \vec{e}_y\big) = \vec{0}$$ i.e. the torque is zero, the equations of motion become
\begin{align}
&\frac{d^2\vec{r}_G}{dt^2} \, = \, - \, g\, \vec{e}_y\\
& \frac{d}{dt}\,I \,\omega\, \vec{e}_z \, = \, \vec{0}
\end{align}
As you can see, these equations are decoupled and can be written as
\begin{align}
&\frac{d^2\vec{r}_G}{dt^2} \, = \, - \, g\, \vec{e}_y\\
& \,I \, \frac{d\omega}{dt} \, = \, {0}
\end{align}
But, the angular velocity, in your notations, is simply $\omega = \frac{dq}{dt}$ and also
$$\vec{r_G} = x_G \, \vec{e}_x + y_G\, \vec{e}_y$$ so
\begin{align}
&\frac{d^2{x}_G}{dt^2} \, = \, 0 \\
&\frac{d^2{y}_G}{dt^2} \, = \, - \, g\,\\
& \,I \, \frac{d^2q}{dt^2} \, = \, {0}
\end{align}
You can solve them right away
\begin{align}
&x_G \, = \, x_G(0) \, + \, v_x\, t \\
&y_G \, = \, y_G(0) \, + \, v_y\, t \, - \, \frac{g}{2} \, t^2\,\\
&q \, = \, q(0) \, + \, \omega_0 \, t
\end{align}
where $x_G(0), \, y_G(0), \, q(0)$ are constants describing the initial configuration of the rod and $v_x, \, v_y, \, \omega_0$ are constants describing the initial velocities of the rod.
If I have time, I can get to these equations from Lagrangian point of view too. Basically you just have to change coordinates of your Lagrangian and the holonomic constrant $\big|\vec{r}_M - \vec{r}_m\big|^2 = \frac{l^2}{4}$ like this:
\begin{align}
&\vec{r}_G \, =\, \frac{M}{M+m} \, \vec{r}_M \, + \, \frac{m}{M+m} \, \vec{r}_m\\
&\vec{r}_l \, = \, \vec{r}_M \, - \, \vec{r}_m
\end{align}
Remark. This case does not generalize to your real goal to have a system of more than two mass-points attached to each other by rods where the rods can rotate relative to each other.
| 147,422
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{\bf Problem.} What is the remainder when $2001 \cdot 2002 \cdot 2003 \cdot 2004 \cdot 2005$ is divided by 19?
{\bf Level.} Level 4
{\bf Type.} Number Theory
{\bf Solution.} Reducing each factor modulo 19 first, we see that $2001 \cdot 2002 \cdot 2003 \cdot 2004 \cdot 2005 \equiv 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \equiv 30240 \equiv \boxed{11} \pmod{19}$.
| 110,026
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| 225,702
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VDPAU
Template:Article summary start Template:Article summary text Template:Article summary heading Template:Article summary wiki Template:Article summary wiki Template:Article summary end
Video Decode and Presentation API for Unix is an open source library and API to offload portions of the video decoding process and video post-processing to the GPU video-hardware.
Contents
Supported hardwareAUR.
Supported formats
1Supported by the libva-driver-intel-g45-h264 package. See H.264 decoding on GMA 4500 for instructions and caveats.
Configuration
The libvdpau-va-glAUR driver (for Intel Graphics or AMD Catalyst) needs to be enabled manually. To enable it, create the following file:
/etc/profile.d/vdpau_vaapi.sh
#!/bin/sh export VDPAU_DRIVER=va_gl
make it executable:
# chmod +x /etc/profile.d/vdpau_vaapi.sh
and reboot or relogin.
In order to check what features are supported by your GPU, run the following command, which provided by the vdpauinfoAUR package:
$ vdpauinfo
Supported software
Adobe Flash Player
In order to enable hardware based video decoding, add the following line to the
/etc/adobe/mms.cfg file:
EnableLinuxHWVideoDecode=1
MPlayer
Install mplayer or mplayer2AUR package, available in the official repositories.
$ mplayer -vo vdpau, -vc ffmpeg12vdpau,ffwmv3vdpau,ffvc1vdpau,ffh264vdpau,ffodivxvdpau, foobar.mpeg
- -vo - Select vdpau video output driver
- -vc - Select vdpau video codecs
MPlayer based players:
- gnome-mplayer: to enable hardware acceleration:
Edit -> Preferences -> Player, then set Video Output to "vdpau".
- smplayer: to enable hardware acceleration:
Options -> Preferences -> General -> Video, then set Output driver to "vdpau".
VLC media player
Install vlc package, available in the official repositories.
To enable hardware acceleration:
Tools -> Preferences -> Input & Codecs, then set Hardware-accelerated decoding to "Video Decode and Presentation API for Unix (VDPAU)".
| 50,140
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A »
GMRF Welcomes 4 Tanzanian Albinos In US To Help With Medical Treatment
Founder of New York-based Global Medical Relief Fund (GMRF), Elissa Montanti, welcomed back four Tanzanian albino children on Saturday at the John F. Kennedy International Airport to help them with medical treatment. Some two years ago they had undergone prosthetics after being attacked in the country for witchcraft benefits. Children with albinism live in fear and danger in Tanzania. One ...Read More »
World Bank Announces Huge Financial Package For Africa
About $57 billion in financing has been announced by the World Bank for sub-Saharan Africa for the period of next three fiscal years. It is learned out of the total assured the International Development Association will be supporting with $45 billion. It will support 448 projects too which are underway in the region. The association is World Bank fund and .. »
East Africa Facing Hunger On Massive Scale, DEC Appeals For Fund
About sixteen million people in east Africa are facing hunger on a massive scale and Disasters Emergency Commission called on Wednesday a major appeal for funds to scale up the response to drought. The areas struck with drought most include Somalia, southern and south eastern Ethiopia and northern as well as coastal Kenya. There had been rianfall shortage of about ...Read More »
Former President Jakaya Kikwete Inaugurates JMKF Foundation Aimed In Healthcare, Education
Former Tanzanian President Jakaya Kikwete is looking ahead to do good for his countrymen. He inaugurated his foundation in Dar es Salaam on Monday that is learned to be focusing in healthcare, education, youth and governance sectors. Dubbed as Jakaya Mrisho Kikwete Foundation (JMKF), the newly formed foundation is aimed to serve fellow countrymen. During the launching event Kikwete said, ...Read More »
| 235,549
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TITLE: References: Infinite dimensional Lie algebras
QUESTION [10 upvotes]: What I really want are properties (if it is abelian, nilpotent, solvable, simple, or semisimple; Cartan subalgebras...) of the Lie algebra of smooth functions on a symplectic manifold $(M,\omega)$; the Lie bracket being the Poisson bracket $\{ \cdot , \cdot \}$.
The symplectic form induces a surjective Lie algebra homomorphism between $(C^\infty(M), \{ \cdot , \cdot \} )$ and the hamiltonian vector fields on $M$, which is a Lie subalgebra of the Lie algebra of vector fields. Therefore the properties of $(C^\infty(M), \{ \cdot , \cdot \} )$ are related to the Lie algebra of vector fields on $M$.
Are there any references (in English) on infinite dimensional Lie algebras treating the examples above?
Any reference dealing with a specific symplectic manifold will be very useful (specially to rule out general statements).
Kac's book on Infinite dimensional Lie algebras deals with Kac-Moody algebras, and "E. Cartan, Les groups de transformations continus, infinis, simples, C. R. Acad. Sc., t.144 (1907) 1094." is in French (I cannot read it).
REPLY [1 votes]: One should certainly mention
A. Lichnerowicz, L'Algèbre de Lie des automorphisme infinitésimaux symplectiques, Symp. Math. XIV 11-24, Academic Press 1974.
A. Lichnerowicz Les variétés de Poisson et leurs algèbres de Lie associées, J. Differential Geom. 12 (1977), 253–300.
I do not have the chance to do it at present, but I remember some other refs by Lichnerowicz at around the same years. Being in French they're often forgot in the literature.
| 108,792
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All children at the Lord Derby Academy can report concerns or worries to any member of staff. For the designated Safeguarding Officers, please click the link below:
To report a concern, click here
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If you feel you are unable to speak with a member of staff please see below for a list of alternative agencies that can offer support:
| 279,382
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Each week we talk to HK's brightest and best
Name Alan To Yiu-lun
Sport Lion and Dragon Dance
Achievements Alan To Yiu-lun, 20, is a powerful force in the Hong Kong Chan Ka Fai Dragon and Lion Association.
He joined the association more than seven years ago, and has helped them win several titles.
To was in the squad that represented the association at the New Year Lion and Dragon Dance Competition when they successfully defended their title.
He also helped the association to win third place at the International Luminous Dragon Dance Competition in 2006.
Soundbite 'I've been practising lion and dragon dance for more than seven years,' To said.
'When I was studying in Form Two, we had to join an extra- curricular activity at school. I chose to take lion and dragon dance lessons because I thought they were cool and different.
'Thanks to my coach Chan, I found lion and dragon dance very attractive. It always revitalises me.
'Dragon dance requires great team spirit. It is a good feeling when nine people co-operate with each other closely to win.
'Practising lion and dragon dance requires lots of energy. I'm constantly exhausted yet I always enjoy it.
'I've also learned how to deal with my teammates, which has helped me a lot in my daily life.' From his coach 'Alan's fundamental skills are good, due to his hard work,' said coach Chan Ka-fai.
'He's willing to learn and to try new things. Although the practice is hard, he never gives up.
'Alan has always remained humble, which has impressed me a lot.'
Ambitions 'I still have a great interest in learning more about lion and dragon dance,' To said.
'It has brightened up my life. Even though I've left school and I'm working full time now, I still try my best to find more time to practise.
'I hope I can keep on performing and taking part in competitions for as long as possible.'
| 383,666
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chandlerhallmark@gmail.com
Part 1: The Ultimate Roadtripping Guide to Southern Utah and Northern Arizona
The Great Outdoors…
As I was watching Forrest Gump as a small child the scene with the orange desert and pointy rocks stuck out to me. I knew I wanted to visit one day, but it seemed it would stay a dream as I was living in Nashville, a 22-hour drive away. The Great American Roadtrip was always something I wanted to do, but living closer to the East coast I knew it would be expensive and possibly out of the question.
Flash forward to 2019. We had been living in Colorado for about a year when we finally decided to just get out there and explore. We wanted to drive out west with a vague plan and see our country, gosh darn it!
The drive from Denver to Monument Valley is a long 8-hours with plenty of empty desert in between and the occasion tumbleweed, yes they’re a real thing. We arrived in Monument Valley around sunset which put us in the perfect mood to set up camp. We stayed at Goulding’s Resort RV & Campground for, I believe, about $30 a night for a camping spot. The views were great, but we thought the price was too high and we were placed right next to another couple in an empty campground. Rex cooked dinner while I set up camp and we settled into a routine that we would soon adopt for the rest of the week.
The next morning we packed up camp, bought our $20 pass to enter the park and explored Monument Valley. (Sadly, the National Parks pass won’t work here). There are strict rules on what you can and cannot do during your time at Monument Valley because it is sacred Najavo land so hiking was out of the question. We, in turn, decided to drive the 17-mile loop to view all the historic buttes, canyons, and wildlife. There are so many great photo opportunities everywhere you look!
A little after lunchtime and a full morning of jumping out of the car to take a million pictures we drove 4 hours farther down south to Grand Canyon National Park.
There is nothing quite like the drive to the South Rim of the Grand Canyon let me tell you. What seems to be a large crack in the ground slowly starts growing larger, but then, BAM, it disappears. Until… you enter the park. Whoever says the Grand Canyon is boring is out of their mind, because this was nothing like I had seen before! It was God’s country.
This time we could use our National Parks Pass which was great to have because it gives you so much freedom to come and go as you please, especially if you’re trying to avoid the National Park food prices. The National Parks Pass is $80 for a full year and it paid for itself many times over for us.
I honestly did little research on the Grand Canyon before we got there so we drove around for hours just in awe of the beauty and going with the flow. We climbed up the stairs of the Desert View Watchtower, which I highly recommend visiting, and saw tourists attempting to pet the elk (smh).
As the sun was setting we realized we didn’t have a spot to camp. I had two locations that I researched before, but because it was so late in the day they were completely full. With daylight dwindling I started to get anxious. I am a planner and I like to know where I am sleeping at night. We drove around for two hours without any luck when we decided to chat up a park ranger. He was very informative and helped us find a disperse camping spot because we were relatively new to the whole disperse camping world. We followed his directions and found an amazing spot that was secluded and better than the campgrounds I had planned on staying at earlier.
Fun Fact: You can camp anywhere in a National Forest! These spots are usually found off dirt roads with limited signs and zero bathrooms. Rex and I now only disperse camp because we end up saving a ton of money because it’s free! Another perk is that your only neighbors are usually wildlife. Read more on dispersed camping here .
The next morning we chose to hike the South Kaibab Trail and I highly recommend it, but please make sure you have plenty of water and sunscreen because there is no shade whatsoever. The hike is 3.1 miles long with a 1,177 elevation gain and you see that gain the whole time hiking back up so save some energy. What I found interesting is that some people paid for a tour and road horses or hire alpacas to carry their supplies. That’s when it hit me. I was in the Wild West.
This is Part 1 of my Great American Roadtrip. Stay tuned to read Part 2 where I visit Antelope Canyon, find the coolest campsite I’ve ever stayed at, and meet up with my friends!
4 thoughts on “Part 1: The Ultimate Roadtripping Guide to Southern Utah and Northern Arizona”
Okay love this so much
Thank you!
Great article. Had never heard the term “ dispersed” camping.
Terry and I made some trips to Blanding, Utah and went to several of the parks near there. Then down to southwest Utah for a few days. We had been to Salt Lake City and out to Antelope Island, which was a fun trip. Can’t wait to go back. So much to see – keep exploring and camping!
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This article was posted by CrystalWind.ca.
A Few Of My Favorite Things
Category: Inspired Mind Written by Cheryl Richardson Views: 937
Neighborhood holiday lights are sparkling and candles are flickering in the windows of the houses in our small town. The Christmas tree farm across the lake is lined with cars filled with families waiting to pick out the perfect tree. And our quaint downtown Market Square is already filled with shoppers in search of the perfect gifts.
In the spirit of gift-giving, I thought I’d share a few of my favorite things in case it gives you some ideas for the loved ones in your life (and makes your shopping easier).
First, I wanted you to know that my new card deck, Self Care Wisdom Cards, is now available and Hay House is having a 50% off sale this week. It’s an illustrated deck with important messages that help you build a stronger, more loving relationship with yourself. You can find them here.
One of my favorite unusual gifts is a milk frother because it makes any cup of coffee or tea extra special. You can use it with oat and almond milk and, in less than a minute, you’ll add a creamy, whipped topping to your favorite drink! You can find it here.
Okay, this might seem strange, but I also love giving practical gifts as stocking stuffers and Trader Joe’s sponges are the perfect example. My friend Beth turned me on to these and I think of her every time I wash dishes. They’re soft and squishy and mold to any surface. You can find them here.
Another practical gift idea comes from my assistant Lisa who grew tired of watching me struggle to open boxes and packages. This handy little gadget will save your hands and a whole lot of frustration. You can find it here.
Because I love supporting writers, I’m including two books on my list. The first is by Joan Borysenko, a woman whom I adore. With her signature wisdom, good sense, and laugh-out-loud humor, Joan has created a new version of her popular book, Pocketful of Miracles. It’s filled with daily prayers, affirmations, and meditations that will rest well at your bedside table or on your bathroom counter. It’s a beautiful hardcover book and the daily messages are such a blessing for those in need of comfort and spiritual support. You can find it here.
For the fiction lovers on your list, I just finished a novel by Deepak Malhotra called “The Peacemaker’s Code” and it’s a compelling story about what happens when a professor who teaches negotiation strategies is brought in by the government to negotiate the world out of a disastrous fate. It’s a wild ride that will also teach you about history and the art of negotiation (and it offers hope for humanity!). You can find it here.
Finally, as a gift to yourself, you might want to pick up the new sunscreen from AnnMarieGianni.com. It’s called Sun Love and the unscented, silky new formula glides on and allows you to use just a small amount for good coverage (and it’s on sale this week!). It’s my go-to favorite self-care addition to my daily walks. You can find it here.
Happy gift-giving!45 guests and no members online
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TITLE: $\sigma$-field generated by random variables
QUESTION [1 upvotes]: Consider a probability space with infinitely many coin flips $X_1, X_2, X_3, \dots$. Consider the sequence of $\sigma$-fields $\{\mathcal{F_n}\}$ generated by these random variables. Describe the elements of $\mathcal{F}_1$, $\mathcal{F_2}$, and $\mathcal{F_3}$.
My intial attempt for $\mathcal{F}_1$: $$\mathcal{F}_1 = \big\{ \emptyset, \Omega, \{ (a_1,a_2, \dots): X_1 = 0 \}, \{ (a_1,a_2, \dots): X_1 = 1 \} \big\}.$$
I'm not sure if this is correct, and I'm not quite sure how to proceed with $\mathcal{F_2}$ and $\mathcal{F_3}$. I'm having a difficult time thinking about $\sigma$-fields generated by random variables.
REPLY [1 votes]: I guess that $\mathscr F_1$ is the $\sigma $-algebra generated by $X_1$. So it is
$$\{\{H,T\}, \{H\}, \{T\},\emptyset\}.$$
Then $\mathscr F_2$ is the $\sigma $-algebra generated by $X_1,X_2$. That is, $$\mathscr F_2=\mathscr F_1\times\mathscr F_1$$
containing $\Omega_2$, the set of the possible pairs of $H $ and $T $, all the outcomes of double coin flipping, and all the subsets of this set:
$$\mathscr F_2=\{\Omega_2\}\cup 2^{\Omega_2}.$$
And for the $\sigma $-algebra generated by $X_1,X_2, X_3$ you will have to do the same with $\Omega_3$, the set of the possible triplets of $H$ and $T$.
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TITLE: Derivative at $x=0$ for $\frac{1}{x\ln(2)} - \frac{1}{2^x - 1}$
QUESTION [0 upvotes]: A function $f$ is defined on $\Bbb R$ as $\frac{1}{x\ln 2 } - \frac{1}{2^x -1}$ for $x \neq 0$ and $\frac{1}{2}$ for $x = 0$. Show that $f$ is continuous and find $f'(0)$
From my knowledge
$$f'(0) = \lim_{x\to 0}\frac{f(x) - f(0)}{x}\\
=\frac{\frac{1}{x\ln(2)} - \frac{1}{2^x-1} - \frac{1}{2}}{x}\\
= \frac{\frac{x \ln(2)}{2(2^x-1)} - \frac{1}{2}}{x}\\
= \frac{x \ln(2) - (2^x -1)}{2x(2^x - 1)}\\
= \frac{x^2 \ln^2(2)}{4x(2^x - 1)}$$
Which on using mclaurin series of $2^x$ gives $\frac{-\ln 2}{4}$ but the correct answer is $\frac{-\ln 2}{12}$
REPLY [1 votes]: We have $$\frac{1}{x\ln(2)}-\frac{1}{2^x-1}-\frac{1}{2}=\frac{(2^x-1)2-2x\ln(2)-(2^x-1)x\ln(2)}{x\ln(2)(2^x-1)\cdot 2}$$
REPLY [1 votes]: You did not include all terms in the series. In intermediate steps, there is also a term $\frac{x^3\ln^3(2)}{3!}$ which is missed.
$$\frac{\frac{1}{x\ln(2)} - \frac{1}{2^x-1} - \frac{1}{2}}{x} = \frac{2(2^{x}-1)-2(x \ln(2))-x\ln(2)(2^{x}-1)}{2x^2 \ln(2)(2^{x}-1)} \\
= \frac{2(x\ln(2)+\tfrac{x^2 \ln^2(2)}{2}+\tfrac{x^3 \ln^3(2)}{6})-2x\ln 2 - x \ln(2)(x\ln(2)+\tfrac{x^2 \ln^2(2)}{2})}{2x^2 \ln(2)(2^{x}-1)}\\
= \frac{-\tfrac{x^3 \ln^3(2)}{6}}{2x^2 \ln(2)(2^{x}-1)}$$
which will give $ \frac{-\ln(2)}{12}$ as the limit.
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Item Details
Tests of Anthracite Coal in House Heating Boilers
by Charles Bayard Baxter, Fred Terrill
- Format
- Thesis/Dissertation; Online
- Published
- 1908.
- Language
- Description
- 40 leaves : ill., charts, photos. ; anthracite coal in house heating boilers / c| by Charles Bayard Baxter, Fred Terrill.
▾See more
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Henry Alken Senior
The meet; Drawing cover; Full cry; The death
Oil on canvas: 11(h) x 14.8(w) in /
27.9(h) x 37.5(w) cm
A set of four, the third and fourth signed lower right: H. Alken
This framed painting is for sale.
Please contact us on:
+44 (0)20 7493 3939
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. The hybrid architecture will be the building blocks of future Hyundai HEVs that includes that 2010 Sonata scheduled to be introduced to the US market.
The Santa Fe has the Theta 2.4L engine coupled with a six speed automatic gear shifting system under its hood. A 30Kw motor is also installed to provide power for its regenerative braking. The design of the engine was focus on decreasing friction along its cooling system and drivelines so fuel economy will be at its optimal.
The software of the engine was enhanced to improve handling of injection pressure, exhaust system, and cycle timing. The engine is also automatically shut off instead of being left on idle to save more fuel. With this stop-start function, the driver can easily start the car by applying pressure on the gas pedal which in turn will trigger the integrated starter generator.
The fuel saving features of the Santa Fe places it at 38 miles per gallon of fuel. Top speed was recorded at 105 mph.
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Adler, Rose M. (Nee Forman) Sept. 6, 1997, age 87 years of Milwaukee. Beloved wife of Max. Dear mother of Susan (Dr. Michael) Rontal, Bloomfield Hills, MI and Dr. Richard (Elyce) Lernor, Bayside. Loving grandmother of Dr. Matthew Rontal, Daniel Rontal and Andrew Rontal, Bloomfield Hills, MI., Lauie (Robert) Lazarus, Cynthia (Michael) Marks, Deborah (Scot) Eisendrath, San Diego, CA., Robyn (Lior) Zorea, San Francisco, CA and great-grandmother of Mitchell Shiner, San Diego, CA. Fond sister of Laura (Julius) Levin, Evansville, IN and sister-in-law of Gertrude (David) Winters, Chicago, IL. Dear aunt of Donna Cooper, Milwaukee, Connie (George) Helmbock, Chicago, IL and Judy (Carlos) Neu, Boston, MA. Further survived by many beloved family members and friends. Rose was active in Beth El Ner Tamid Sisterhood, Mt. Sinai Hospital Aux., Hadassah, Brandeis University and ORT. Services as the funeral home 3 PM Tues. Not in state. Entombment Spring Hill Cemetery. Memorials to a charity of your choice preferred. GOODMAN - BENSMAN WHITEFISH BAY FUNERAL HOME 4750 N. Santa Monica Blvd. 964-3111 Allen, Nancy M. Suddenly, age 46, of Waukesha, WI. Formerly of Chicago's North-West side. Beloved daughter of the late Richard and Helen (Nee Nagy) Williamson. Dear niece of many aunts and uncles. Loved by her many cousins. Resting at the Columbian Chapels Funeral Home, 6901 W. Belmont Avenue, Chicago. Visitation Tuesday, 3PM until the time of service, 7PM. In lieu of flowers, memorials may be given to Nancy's favorite charities: AIDS Foundation and Battered Womens' Shelter. For information Call 708-848-3010. Andreska, Joseph F. "Oiler" Of South Milwaukee. Passed away Sept. 6, 1997 at the age of 80 years. Beloved husband of the late Eleanore Andreska. Dear father of Kenneth (Diann), Joan (Harold) Gosh, and James (Beverly) Andreska. Proud grandfather of 15 grandchildren and 19 great-grandchildren. Fond brother of Steve (Mary), Josephine (Millard) Walker, Lorraine (John "Bill") Nunn, and Agnes Fernholz. Further survived by nieces, nephews, other relatives and friends. Mass of Christian Burial Tues. Sept 9 at 11:00 AM at St. Adalbert Church 1611 Manitoba Ave. South Milwaukee. Interment Holy Sepulcher Cemetery. In State at the church only Tues. 10:00 AM until time of Mass. MOLTHEN-BELL & SONS Serving the Family South Milwaukee 762-0154 Blau, Martha "Mania" (Nee Ostry) Sept. 6, 1997, age 75 years, of Mequon. Beloved wife of Tibor. Dear mother of Fred (Harriett) Blau, Germantown and Robert (Marybeth) Blau, Boston, MA. Fond sister of Jack Ostry, Short Hills, NJ and Sophie (David) Orzech, New Milford, NJ. Further survived by 5 grandchildren. Services at the Funeral Home 11 AM Tues. Not in state. Interment Greenwood Cemetery. Memorials to the American Cancer Society preferred. GOODMAN - BENSMAN WHITEFISH BAY FUNERAL HOME 4750 N. Santa Monica Blvd. 964-3111 Byrne, Eulail Therese (Nee Flanagan) of Madison, WI, formerly of Whitefish Bay. Please see Tues. newspaper for complete notice or call SCHMIDT & BARTELT William R. Feerick Associate Whitefish Bay 964-3040 Charles, Frank M. Found Eternal Peace Sun. Sept 7, 1997 surrounded by his loving family at Froedtert Memorial Hospital after a short illness. Long time member of St. Jude the Apostle Parish, Wauwatosa. Completed notice with time and day of services to be given later. SCHMIDT & BARTELT Chudada, Joseph A. Of Pewaukee, Sat. Sept. 6, 1997 at the age of 58 years. Beloved husband of Donna (Nee Porter). Dear father of Chris and Ryan at home and Jolene (Michael) Miller of Bremerton, WA. Fond brother of Catherine (Brad) Spencer of Muskego. Loving grandfather of Jennifer and Jessica Miller. Further survived by other relatives and friends. Funeral Mass of Christian Burial Wed. Sept. 10 at 7:00 PM at St. Anthony on the Lake, W280 N2101 Prospect Ave., Pewaukee. Visitation after 4:00 PM Wed. at Church. Interment Salem Cemetery, Wales. Joe was a resident of Pewaukee for over 24 years and enjoyed hunting and fishing in the area. He was employed at Medusa Cement for 35 years where he managed daily operations. YONKE & SON FUNERAL HOME Serving Colla, Rose Please refer to Sun. paper. SCHAFF FUNERAL HOME 541-7533 Curtis, Stephen J. Age 40. See Tues. Journal Sentinel or call. KRAUSE FUNERAL HOME 9000 W. Capitol Dr. 464-4640 Czerwinski, James A. Fri. Sept. 5, 1996, age 65 years. Beloved husband of Patricia (nee Verhein). Loving father of James, Bruce Venne, Mark (Candy Meyers) Venne and Dawn (Brian) Lucas. Dear Grandpa Bear of Bruce, Beau, Daniel Venne and Amanda Meyers. Dear brother of Harold (Rose), Ronald (Harriet), Richard (Frieda), Gerald (Julie) and Laurine Struck. Preceded in death by his parents Amanda and Alexander, and sisters Eleanor (Rodman) Forbes, Caroline (Geno) Hudzinski and Mable (Kenneth) Leinwe ber. Further survived by brothers and sisters in law, nieces, nephews, other relatives and friends. Visitation at the Funeral Home MON 4-7 PM. Funeral Services Mon. 7 PM. Military Rites Tues. 11 AM at Forest Home Cemetery. James was a lifetime member of the South Milwaukee Yacht Club and a member of Muskego Rod and Gun Club as well as a Veteran of the Korean War, honorably serving our country. PRASSER-KLECZKA BAY VIEW CHAPEL 3275 S. Howell Ave. 483-2322 Daniels, Terry A. Of Menomonee Falls Sept 6, 1997 at the age of 56 years. Beloved wife of Cathy (nee Roskopf). For time and place of Visitation and Mass please refer to Tues. Journal Sentinel or call. SCHMIDT & BARTELT A.A. SCHMIDT & SONS Menomonee Falls 251-3630 Drefahl, Wilbur H. Sat. Sept. 6, 1997 age 79 years. Beloved husband of Daisy (nee Pollard). Dear father of Gerald and Thomas (Judith) Drefahl. Dear grandfather of Nicholas and John. Also survived by other relatives and many friends. Visitation at the Funeral Home TUES. 10 AM to 12 Noon. MASONIC SERVICES at 12 Noon by WEST ALLIS LODGE NO. 291 F&AM. Interment Arlington Park Cemetery. Wilbur was a member of the Bay View High School 1936 City Football Championship Team, Tripoli Temple, Wisconsin Scottish Rite Bodies and retiree of Nordberg after 41 years of service. Memorials suggested to the Shrine Hospitals for crippled children. PRASSER-KLECZKA SOUTH SUBURBAN CHAPEL 6080 S. 27th Street 282-6080 Ferlindes, Joseph September 6, 1997, age 66. Beloved husband of Julia. Dear father of Frank (Tina), Steve (Maria) and Ana Ferlindes. Proud grandfather of Maria, Jessica, Stephen and Matthew. Fond brother of Ana (Mitch) Selovic. Further survived by nieces, nephews, other relatives and friends. Funeral Services Tuesday at 9:30 AM from Schaff St. Augustines Church, 69th and W. Rogers St. for Mass at 10:00 AM. Interment Mount Olivet Cemetery. Visitation Mon. at Schaff 4:30-8:00 PM. SCHAFF FUNERAL HOME 5920 W. Lincoln Ave. 541-7533 Geisel, Arthur F. Of Pewaukee Lake. Died Sun. Sept. 7, 1997 at the age of 85 years. Dear husband of Ann, formerly Goede (nee Pieper). Dear father of Paul Geisel, Jack (Diane) Geisel and William (Maureen) Goede. Dear father-in-law of Karen Geisel. Preceded in death by his first wife Marcella and his son Jim. Also survived by 9 grandchildren, 3 great-grandchildren, nieces, nephews, other relatives and friends. Family will receive friends and relatives Tues. at the Funeral Home from 3-5 PM. Private interment. Art's family would like to thank Dr. Adel Korkor and the nursing staff at Franciscan Woods. Memorials to the American Kidney Foundation would be appreciated. BECKER-RITTER ELMBROOK Michael Feerick, Associate 14075 West North Ave. 782-5330 Gresl, Robert F. Age 79 of Inverness, FL, formerly of Wind Lake, WI. Beloved husband of Lorraine S. Gresl of Inverness. Further survived by daughters Susan, Patricia, Eileen, Theresa, Marie, Barbara and Joanne and a son Robert. Also survived by 6 grandchildren, 1 great-grandson, 3 brothers and 4 sisters. Funeral Mass was held in Palm Bay, FL. Mr. Gresl was a life long member of the Amateur Trap Shooting Assn. and was a retired sod farmer. Donations appreciated to the National Stroke Assn, 96 Inverness Dr., East, Ste I, Englewood, CO 80112. Gutkowski, Raymond D. Born to eternal life September 6, 1997, age 80 years. Beloved father of Dale (Janice) and Linda (Robert) Plicka. Also survived by grandchildren, many loving family members and friends. Mass of Christian Burial will be held Wednesday, September 10, at St. Augustine's Catholic Church, West Allis, at 10AM. PLEASE MEET AT CHURCH. Visitation Tuesday at the funeral home from 4-8PM. Interment St. Adalbert Cemetery. In lieu of flowers, memorials to the charity of your choice appreciated. BORGWARDT FUNERAL HOMES 1603 S. 81st ST West Allis 476-2010 Harmeyer, Earl C. Of West Allis, Fri., Sept. 5, 1997, age 70 years. Please see Sun. paper for complete notice. BEVSEK-VERBICK West Allis 546-4342 10210 W. Lincoln Ave. Ivkovich, Frank J. Age 83, died Sept. 3, 1997 in Vista, CA. Please see Sunday Journal Sentinel or call MAX A. SASS & SONS MISSION HILLS CHAPEL 8989 W. Loomis Rd. 427-0767 Joerres, George A. September 6, 1997, age 66 years. Beloved husband of Helen. Dear father of Joseph, Debra (Thomas) Whalen, Barbara (Timothy) Reynolds, Laura (Mark) Miller and Craig. Loving grandfather of 13 grandchildren. Son of Sophie Joerres. Brother of Art (Betty), Jeannette (Bernie) Schoen and Shirley (Warren) Allhands. Further survived by nieces, nephews, other relatives and friends. Complete Funeral Service will be held Tuesday at 7 PM at Schaff. Visitation Tuesday 4:30 PM to time of service. Private interment. SCHAFF FUNERAL HOME 5920 W. Lincoln Ave. 541-7533 Joslin, Robert G. Services pending. CHURCH AND CHAPEL Milwaukee 827-0659 Waukesha 549-0659 Kachelmeyer, Joyce L. (Nee Scott) Age 66 years. Sept. 7, 1997. Beloved wife of Joseph. Dear mother of Mary Jo (Ralph) Sundling, Joseph Jr. (Laura), John (Kathleen) and Jerry (Annette). Loving grandmother of Bob and Tom Sundling, David, Kristen, Jimmy, Laura, Jaclyn, Katie, Danny and Christopher. Fond sister of Florence (Richard) Kass and Richard (Audrey) Scott. Also survived by many other relatives and friends. In state Tues. 1 PM until time of services at 3 PM all at Wisconsin Memorial Park Chapel of the Chimes, 132nd and W. Capitol Dr. Interment to follow. In lieu of flowers, donations to the American Cancer Society preferred. JELACIC FUNERAL HOME Serving the Family 466-2134 Knippel, Frances J. (Nee Caggio) Saturday, September 6, 1997, age 73 years. Beloved wife of the late Daniel. Dear mother of Donna Frodsham, Angela Calvert, Barbara (Lyn) Dombrowski, Joann Schultz, Daniel, and Marc (Cheryl) Knippel. Dear sister of Anthony (Marge) Caggio, Virginia King, Mary Ferro, Josephine (Fred) Krenz and Joann (John) Asdigan. Loving grandmother of Marie (Mark) Zingler, Kimberly, Bryan, Randy, Dannie, Kelly, Bill, Andrew, Eric, Jackie, Mandy, Melissa and Matthew. Loving grandmother of Calvin Kuckes. Fond guardian of Marie Strain and dear friend of Jane, Lee, Thuy, Evan, Georgia, Betty, Rich and Karen. Also survived by nieces, nephews, other relatives and many other dear friends. FUNERAL SERVICES TUESDAY, SEPTEMBER 9, 10:30AM AT SACRED HEART OF JESUS CHURCH, 3635 S. Kinnickinnic. Interment Wood National Cemetery. Frances was a retiree of the Milwaukee County DHSS, member of Christian Womens' Society, member of VFW Womens' Auxiliary. VISITATION TUESDAY AT CHURCH ONLY, from 9AM until the time of Mass. If desired, memorials to Sacred Heart of Jesus Church appreciated. SCHRAMKA-REMBOWSKI Kosnicki, Benedict J. Of Milwaukee. Age 91. Passed away Saturday, September 6, 1997 at the South Shore Manor. He was born in Chicago on February 29, 1906, son of the late John and Anna (nee Krajerz) Kosnicki. On January 18, 1932 Ben was united in marriage to Adeline Kamasinski who preceded him in death in 1982. Ben was employed by Ladish Drop Forge for many years until his retirement in 1971. He was a member of St. James Catholic Church. Surviving are 2 daughters and sons-in-law, Lorrie (Ted) Kroes of Racine, Corinne Kuras (Ken) Skurulsky of Bay View, 1 son Michael Kosnicki of Milwaukee, 1 daughter-in-law Joan Kosnicki of Shawano, 14 grandchildren, 28 great-grandchildren, other relatives and many dear friends. Also preceded in death by his son, Ronald and his sister, Florence Badke. A Mass of Christian Burial will be celebrated on Tuesday, September 9 at 11:00 AM at St. James Catholic Church, 7219 S. 27th St. Interment will follow at St. Adalbert Cemetery. Relatives and friends are welcome to meet with the family on Tuesday 10:00 AM until time of Service at the church. A very special thank you to the staff at South Shore Manor. MARESH-MEREDITH FUNERAL HOME 803 Main St. 634-7888 Kucharski, Florence C. See Sunday paper Lambrecht, Estelle D. (Nee Machajewski) age 90 years. Passed away Friday Sept. 5, 1997. Beloved wife of the late Arthur. Fond aunt of Florence Schaefer and Erwin Piczkowski. Further survived by other relatives and friends. A Mass of Christian Burial will be celebrated Wed. 9:30 AM at St. Mary's Church, Hales Corners, 9520 W. Forest Home Ave. PLEASE MEET AT CHURCH. Entombment St. Adalbert Cemetery. Not in state. BRUSKIEWITZ FOREST LANE CHAPEL Serving the Family 321-1700 LaPlante, Ovid John Services pending. CHURCH AND CHAPEL Milwaukee 827-0659 Waukesha 549-0659 Marx, Charles A. Sat. Sept. 6, 1997 age 95 years. Beloved husband of the late Evelyn (nee Brill). Father of James (Audrey) Marx and Kay (Edward) Cieslik, the late Charles (Patricia) Marx and the late Donald Marx. Loving grandfather of 7 and great-grandfather of 5. Further survived by nieces, nephews, other relatives and friends. Mass of Christian Burial will be held Tues. 11 AM at Blessed Sacrament Church, 3100 S. 41st St. Entombment St. Adalbert's Cemetery. In state Tues. at the Church from 10 AM until the time of Mass (11 AM). Charles was a retired supervisor at American Motors Co. HERITAGE FUNERAL HOMES Tebo Peppey Klemmer & Scheuerell Chapel 321-7440 McConnell, Joseph Age 68 years. Sept. 4, 1997. Survived by his loving wife Lee Anna; 1 son, Johnny Wade; 4 daughters, Gloria (Tommy) May, Ida, Diana Marie McConnell and Jennefier Joiner; 4 brothers, Richard (Cecelia), Jr., James (Mary), John and William (Ida) McConnell; 2 sisters, Mary Sutton and Bonita (Tommy) Judd; 3 grandchildren; 1 uncle Issac Matthew; and a host of other fond relatives and friends. Funeral Wed. 1 PM, Chapel of the Chimes (Wisconsin Memorial Park) 13235 W. Capitol Dr. Visitation Tues. 2-8 PM. Family hour 6-7 PM at: NEW GOLDEN GATE 5665 N. Teutonia Ave. 358-0538 Miller, Golden Sept. 4, 1997, age 64 years. Visitation Tues. 10 AM-1 PM. Funeral 1 PM. All services at Mt. Ararat Baptist Church, 3118 W. Atkinson Ave. NEW GOLDEN GATE FUNERAL HOME 5665 N. Teutonia 358-0538 Parsons, Hildegarde E. (nee Bachmann) of Cedarburg. Born to Eternal Life of Sept. 6, 1997 at the age of 79 years. Beloved wife of the late Bill Parsons. Dear mother of Don (Lani) Parsons, Kathe (Dave) Cochrane and Carol (Fred) Schultz. Loving grandmother of Lisa (Todd) LeRoy, Christopher (Sandra) Parsons and 5 great-grandchildren. Fond sister of Elsie Dietrich and Ruth Shully. Sister in law of Ruth Fraser, Bob Ellmer, Ria Bachmann and John Hayburn. Also survived by nieces, nephews, other relatives and many dear friends. Friends may greet the family on Tues. Sept. 9 from 10 AM to time of Memorial Services at 11 AM at Wisconsin Memorial Park Chapel of the Flowers. In lieu of flowers memorials to M.S. Society or Care United Hospice appreciated. SCHMIDT & BARTELT William R. Feerick, Associate Mequon 241-8085 Patrinos, Bessie S. (Nee Pantelis) of Brookfield died September 6, 1997 at age 102 years. Beloved mother of Mary (Peter) Karter, Ft. Lauderdale, FL and Amelia (Chris) Karter, Brookfield; 7 grandchildren, 20 great-grandchildren, other relatives and friends. Preceded by daughter Anne Trott, great-grandson Christopher John, and by sisters Eleni, Chloe, Chris and Anne. Visitation Tues. 10 AM until 10:30 AM services at funeral home. Burial at Prairie Home Cemetery, Waukesha, will follow. Mrs. Patrinos was a member of the Greek Orthodox Church of the Annunciation, Milwaukee. Memorials to the church appreciated. RANDLE-DABLE 1110 S. Grand 547-4035 Waukesha Petermann, Julia L. Phillips-Gosetti (nee Ring) Age 85 years. Born to Eternal Life Sept. 6, 1997. Beloved mother of Albert Phillips, Coretia Cira, Cindy Downs and John Petermann. Dear grandmother of Albert II, Greg, Brian (Vicky), Chris, Jeffrey and Jay. Further survived by other relatives and friends. Visitation Wed. after 10 AM at MOTHER OF GOOD COUNSEL CHURCH (6924 W. Lisbon Ave.). Mass of Christian Burial 12 noon. Interment Holy Cross Cemetery. CHURCH AND CHAPEL FUNERAL SERVICE Serving the Family Milwaukee 827-0659 Waukesha 549-0659 Popp, Dennis R. Sept. 3, 1997, age 55 years. Father of Joseph and Nancy. Private services were held. JOHN J. WALLOCH FUNERAL HOME 4309 S. 20th St. 281-7145 Powers, Lawrence Please see Sunday's paper. Bruskiewitz Forest Lane Chapel 321-1700 Russell, Cornelius Age 84. Sept. 4, 1997. Visitation Tues., Sept. 9, 1997, 12 noon-8 PM. Funeral Wed. 11 AM at PITTS FUNERAL HOME, 2031 W. Capitol Dr. Interment Graceland Cemetery. PITTS FUNERAL HOME 2031 W. Capitol Dr. 447-6000 Schaff, Anna (Nee Nikolaus) Sept. 6, 1997. Age 90 years. Beloved wife of the late Paul. Dear mother of Mathias (Anna) and Michael (Tonita). Grandmother of Harry (Mary) and Robert. Sister of Johan (Elizabeth) Nickolaus. Further survived by nieces, nephews other relatives and friends. Funeral Service Wed. at 9 AM from Schaff to Sacred Heart Church, 49th and W. Wells St. for Mass at 9:30 AM. Interment Holy Cross Cemetery. Visitation Tues. at Schaff 4:30-8:00 PM. SCHAFF FUNERAL HOME 5920 W. Lincoln Ave. 541-7533 Schultz, Franklin H. Age 84 years of Mequon, formerly of Jackson. Sept. 5, 1997. Dear grandfather of Jackie (David) Kelsey, Jeff (Renee) Schultz, Shary (Mark) Kraft, James (Kerri) Schultz, Daniel Schultz and Susan Schultz. Further survived by a daughter-in-law Deanna Schultz, other relatives and friends. Funeral services will be held Mon. Sept. 8, at 7:00 PM at the funeral home. Burial will be Tues. at 9:30 AM at Valhalla Memorial Park, Milwaukee. Visitation will be Mon. at the Funeral Home from 5 PM until the time of services. SCHMIDT FUNERAL HOMES "A Golden Rule Funeral Home" N168 W20135 Main St. Jackson Semrau, Joseph Age 86 yrs. Sept. 5, 1997. Beloved and dearest brother of Helen Semrau, Pearl Franecki, the late Regina Kazmierczak and the late Victor Semrau. Brother-in-law of Hedwig Semrau. Further survived by many nieces, nephews, grand-nieces, grand-nephews, and cousins, Bruno, Pearl and Alice Kasprowicz and Gertrude Kuchta. Joseph was a army veteran of 5 years and lifetime member of the American Legion. Funeral service Wed. at 9:30am from the Max A. Sass & Sons Funeral Home to ST. HELEN CHURCH at 10am. Entombment St. Adalbert Cemetery. Parish Vigil Tues. 6pm. Visitation Tues. 4-8pm at the Funeral Home. MAX A. SASS & SONS 1515 W. Oklahoma Ave. 645-4992 Smith, Helen L. (nee Bradley), Fri., Sept. 5, 1997, age 73 years. Beloved wife of Russell J. Smith. Dearest mother of Judith Ann (Santiago) Maldonado and Carole Jean Smith. Sister-in-law of Myron (Carolyn), Clayton (Evelyn) and Allen (Nancy) Smith. Also survived by other relatives and friends. Visitation at the Funeral Home TUES., 4-8 PM. Prayer service at 7 PM. Funeral WED. 9:30 AM from the Funeral Home to IMMACULATE CONCEPTION CHURCH. Mass of Christian Burial at 10 AM. Entombment Arlington Park Cemetery. PRASSER-KLECZKA BAY VIEW CHAPEL 3275 S. Howell Ave. 483-2322 Sobanski, Neal Sept. 5, 1997 age 52 years. Beloved son of Ted J. and Marion (nee Tuchel). Dearest brother of Joanne (Charles) Goeb of Florida, Tom of California and James (Gwen) Sobanski. Loving friend of Barbara Socha. Uncles, aunts, nieces, nephews, many other relatives and friends also survive. Visitation Tues. 4-8 PM with evening services at 7 PM. Complete Military Services Wed. 10 AM. Retired Lt. Colonel US Army and Viet Nam Veteran. Neal was Captain of the Dept. of Defense Police at 440th Air Force Reserve. JOHN J. WALLOCH FUNERAL HOME 4309 S. 20th St. 281-7145 Stawski, Dorothy Services pending. LARSEN BROS. NEW BERLIN FUNERAL HOME Stieg, Viola L. (Nee Reuter) PORT WASHINGTON September 6, 1997, age 90 years. Widow of the late Oliver. Survived by 3 grandchildren, 5 great-grandchildren, 3 sisters Evelyn (Len) Ruffing, Dorothy Haensgen and Florence Pein, other relatives and friends. Funeral Wednesday, September 10, 2 PM at Friedens Evangelical Church, 454 N. Milwaukee St., Port Washington. Interment Union Cemetery. In state at the CHURCH Wednesday from 12 noon until time of services. Memorials to Friedens Evangelical Church appreciated. EERNISSE FUNERAL HOME Port Washington 284-2601 Stys, Richard Sept. 7, 1997 age 73 years. Beloved husband of Mabel (nee Borlick). See Tues. paper for complete notice. JOHN J. WALLOCH FUNERAL HOME 4309 S. 20th St. 281-7145 Thompto, Virginia F. of Waukesha, died Sept. 6, 1997 at age 70 years. Dear mother of Cynthia R. (Patrick) Hopfensperger, Cedarburg and Robert A. (Mare) Younger, Kentwood, MI. Stepmother of Ann (Salvatore) San Felipo, Cedarburg, Harry (Joanne) Thompto, Wind Lake and Guy (Dianna) Thompto, Saukville. Proud grandmother and step-grandmother of 11 grandchildren, nieces, nephews, other relatives and friends. Preceded by husbands Robert R. Younger and Harry Thompto. Memorial Service 1 PM Wed. Sept. 10 at First Presbyterian Church, 810 N. East Ave., Waukesha. Memorials to the Church or to the American Cancer Society appreciated. RANDLE-DABLE Waukesha 547-4035 Vingleman, Ann See Sunday Journal Journal Sentinel or call. CHURCH AND CHAPEL Milwaukee 827-0659 Waukesha 549-0659 Weis, Christine H. See Sun. Jouranl Sentinel or call: CHURCH AND CHAPEL Milwaukee 827-0659 Waukesha 549-0659 Welton, Josephine W. Of Brookfield, age 89. Sept. 5, 1997. Dear mother of Lloyd (Allergra) and Richard (Arlene). Sister of Delores Stafford, 3 grandchildren, other relatives and friends. In lieu of flowers, Masses appreciated. In State 1:00 pm Tues. at MOTHER OF GOOD COUNSEL CHURCH with the Mass at 1:30pm. Burial Holy Cross. CHURCH AND CHAPEL FUNERAL SERVICE Milwaukee 827-0659 Waukesha 549-0659 Wolowicz, Alice (Nee Sidney) PORT WASHINGTON, Sept. 5, 1997, age 87 years. Dear mother of James (Janice) Wolowicz. Fond sister of Bill (Betty) Sidney, Esther Davidson and Gloria Nowakowski. Further survived by 4 grandchildren, 8 great-grandchildren, other relatives and friends. Funeral service 7 PM Tues. Sept 9 at Portview Christian Center, 3457 Cty Trk LL, Port Washington. Interment Highland Memorial Park, New Berlin. Visitation at 4-7 PM. Portview Christian Center. POOLE FUNERAL HOME (414) 284-4431 Port Washington Serving the Family Zellmer, Delbert Carl Sept. 6, 1997 age 58 years. Beloved husband of Rebecca Abraham. Dear father of Darryl (Kelly) Zellmer, Connie (Scott) Miller and Amihan. Fond grandfather of Phillip and Kevin Miller and Kyle Zellmer. Brother of Marvin and Robert Zellmer. Further survived by other relatives. Visitation Tues. Sept. 9, 1997 at ST. SEBASTIAN'S CHURCH from 4 PM until time of Mass at 7 PM. Interment Resurrection Cemetery 11 AM Wed., meet at the funeral home at 10 AM. SCHMIDT & BARTELT 10121 W. North Ave. 774-5010
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TITLE: Evaluating Infinite Series with Binomial Coefficient
QUESTION [1 upvotes]: I am not sure whether my argument works. I think I'm mostly concerned with how I'm working with the binomial coefficient. So I am wondering if I can get feed back as to whether my argument is correct/ the series is divergent or not.
Evaluate the following series:
$$\sum_{k=2}^{\infty} {3k \choose k} \cdot 7!$$
Let us begin by considering the sequence of partial sums:
$$ S_k = {3k \choose k} \cdot 7!$$
for $ k \in \mathbb{N}: k >1$. This expression can be expanded and rewritten to give:
$$ S_k = {3k \choose k} \cdot 7! = \left( \frac {(3k)!}{k! \cdot (2k)!} \right) \cdot 7! = \left( \frac{(3k) \cdot (3k - 1) \cdot ... \cdot (2k + 1)}{k!} \right) \cdot 7! $$
Now, in order to evaluate the convergence, or lack thereof, of the infinite series we will consider the limit of the ratio $ \frac {S_{k + 1}}{S_k}$. In order to do this, we must first find an expression for $S_{k + 1}$:
$$S_{k+1} = {3(k+1) \choose k+1} \cdot 7! = \left( \frac {(3k + 3)!}{(k + 1)! \cdot (2 k + 2)!} \right) \cdot 7! = \left( \frac{(3k + 3) \cdot (3k + 2) \cdot ... \cdot (2k + 3)}{(k + 1)!} \right) \cdot 7!$$
Finally, denote the aforementioned limit of the ratio $ \frac {S_{k + 1}}{S_k}$ by $Q$ and observe that:
$$Q = \lim_{k\to\infty} \frac{S_{k + 1}}{S_k} = \lim_{k\to\infty} \left( \frac{3k + 3) \cdot (3k + 2) \cdot ... \cdot (2k + 3)}{(k + 1)!} \cdot \frac{k!}{(3k) \cdot (3k - 1) \cdot ... \cdot (2k + 1)} \right)$$
$$ = \lim_{k\to\infty} \left( \frac {(3k + 3) \cdot (3k + 2) \cdot (3k + 1)}{(k + 1) \cdot (2k + 2) \cdot (2k + 1)} \right) = \frac {27}{4}$$
We have found $Q = \frac{27}{4}$ and thus, clearly, have that $Q > 1$. It follows from the Quotient test for absolute convergence that the series $\sum_{k=2}^{\infty} {3k \choose k} \cdot 7!$ is divergent, or:
$$\sum_{k=2}^{\infty} {3k \choose k} \cdot 7! = \infty$$
REPLY [1 votes]: This argument works, but here are the ways to make it simpler:
you can immediately drop $7!$ as constant factors have no bearing on convergence
argument itself: so the series cannot converge, you can argue either that
$\binom{3k}{k} > 1$ or
$\binom{3k+3}{k+1} > \binom{3k}{k}$
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David McMahon
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Left!
1 comment:
I have to admit that when I saw that the memorial was to be held at St. Michael's, I didn't think of the current church, which I have never been inside, but of the little green chapel on the hill above our house that we attended many Sundays.
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TITLE: Can someone explain to me why this pattern appears when consolidating numbers within multiplication tables to single digits?
QUESTION [3 upvotes]: This is something I came across about a year ago while playing with numbers in my head, I shared it with a couple friends who couldn't explain why it happens either, and since forgot about it until recently.
Two examples of consolidating a number to a single digit:
27 becomes 2 + 7 = 9, so 27 consolidates to 9.
47 becomes 4 + 7 = 11, then 11 becomes 1 + 1 = 2, so 47 consolidates to 2.
Pretty arbitrary, but when we apply this to the different multiplication tables a pattern emerges for each of them, and the patterns display a very interesting symmetry/negative correlation with each other. Also note that once complete every sequence loops through itself again infinitely.
An example of a table with it's consolidations beneath:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24...
2, 4, 6, 8, 1, 3, 5, 7, 9, 2, 4, 6...
Okay, that should be all the pretext needed.
I will now write the main sequences down with my observations beneath.
Multiples of 0:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0...
(Multiples of 0 remain the same number)
Multiples of 1:
1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5...
(Multiples of 1 ascend through each digit)
Multiples of 2:
2, 4, 6, 8, 1, 3, 5, 7, 9, 2, 4, 6, 8, 1...
(Multiples of 2 ascend through the even numbers, then the odd numbers)
Multiples of 3:
3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6, 9, 3, 6...
(Multiples of 3 ascend through the single digit multiples of 3)
Multiples of 4:
4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, 7, 2...
(Multiples of 4 have two alternating sequences of descending numbers. This can also be seen as one ascending sequence, ascending through even numbers then odd numbers, always skipping one number that should logically be next)
Multiples of 5:
5, 1, 6, 2, 7, 3, 8, 4, 9, 5, 1, 6, 2, 7...
(Multiples of 5 have two alternating sequences of ascending numbers. This can also be seen as one descending sequence, descending through odd numbers then even numbers, always skipping one number that should logically be next)
Multiples of 6:
6, 3, 9, 6, 3, 9, 6, 3, 9, 6, 3, 9, 6, 3...
(Multiples of 6 descend through the single digit multiples of 3)
Multiples of 7:
7, 5, 3, 1, 8, 6, 4, 2, 9, 7, 5, 3, 1, 8...
(Multiples of 7 descend through the odd numbers, then the even numbers)
Multiples of 8:
8, 7, 6, 5, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4...
(Multiples of 8 descend through each digit)
Multiples of 9
9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9...
(Multiples of 9 remain the same number)
This is the end of our pattern of sequences. The more observant of you will have spotted the negative correlations between sequences I mentioned earlier.
0 with 9
1 with 8
2 with 7
3 with 6
4 with 5
These sequences are a reflection of each other. Continuing our sequences with higher tables will give the exact same patterns repeating each 9 multiples.
eg.
9 with 18
10 with 17
11 with 16
12 with 15
13 with 14
The only anomaly is 0, based on it's nature, always gives 0.
9 will always give 9.
18 will always give 9.
27 will always give 9 etc.
These are the only multiples without a negative correlation to their counterpart, but a positive correlation.
So finally, can anyone explain to me why we find these patterns? I've found absolutely no mention of this on the internet when I looked, I think because of how arbitrary the question is, but there must be a solid reasoning as to why this happens this way. Thank you for your time!
(One final footnote for those interested, I tried this with the prime numbers to see if I could find a pattern but to no avail, here are some of them:)
2, 3, 5, 7, 2, 4, 8, 1, 5, 2, 4, 1, 5, 7, 2, 8, 5, 7, 4, 8...
REPLY [2 votes]: $\color {green}{\text{ Definition :}}$
What you mean is to find the Digital Root of the numbers . We calculate digital roots by repeated summation of the digits of a number until we reach a single value .
The more mathematical way to look at digital roots is by defining them as the remainder of a number $\mod 9$. (in Base $10$)
For eg .
$$47=4+7=11=1+1=2$$
which is equivalent to
$$\color{blue}{47 \,\equiv \,2\mod 9}$$
$\color{red}{ \text { Patterns in the numbers:}}$
Multiples of 1:
It is clear that in each multiple you add $1$ , to get other . Hence the order increases from $1$ to $9$ and then reaches back to $1.$
Multiples of 2:
It is clear that in each multiple you add $2$ , to get other . Hence the order increases from $2$ to $8$ with a step of $2$ and then reaches to $1.$ and again continues with a step of $2$.
Multiples of 3:
It is clear that in each multiple you add $3$ , to get other . Hence the order increases from $3$ to $9$ with a step of $3$ and then reaches back to $3.$
Multiples of 4:
In Case of $4$ , you add $4$ or in terms of modular arithmetics , you add $-5 \quad (\, 4\, \equiv -5 \, \equiv \mod 9)$. Hence the order decreases from $4$ to $8$ to $3$ with a step of $-5$
Multiples of 5:
In Case of $5$ , you add $5$ or in terms of modular arithmetics , you add $-4 \quad (\, 5\, \equiv -4 \, \equiv \mod 9)$. Hence the order decreases from $5$ to $1$ to $6$ with a step of $-4$
Multiples of 6:
In Case of $6$ , you add $6$ or in terms of modular arithmetics , you add $-3 \quad (\, 6\, \equiv -3 \, \equiv \mod 9)$ Hence the order decreases from $6$ to $3$ to $9$ with a step of $-3$
Multiples of 7:
In Case of $7$ , you add $7$ or in terms of modular arithmetics , you add $-2 \quad (\, 7\, \equiv -2 \, \equiv \mod 9)$ . Hence the order decreases from $7$ to $5$ to $3$ with a step of $-2$
Multiples of 8:
In Case of $8$ , you add $8$ or in terms of modular arithmetics , you add $-1 \quad (\, 8\, \equiv -1 \, \equiv \mod 9)$ . Hence the order decreases from $8$ to $7$ to $6$ with a step of $-1$.
Multiples of 9:
For $9$ it is trivial to see that since all the terms are divisible by $ 9 $, the digital roots would be $9$ only.
Some helpful links :
Number Theory
Modular Arithmetic
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\begin{document}
\title{Equivariant $\D$-modules on binary cubic forms}
\author{Andr\'as C. L\H{o}rincz}
\address{Department of Mathematics, Purdue University, West Lafayette, IN 47907}
\email{alorincz@purdue.edu}
\author{Claudiu Raicu}
\address{Department of Mathematics, University of Notre Dame, 255 Hurley, Notre Dame, IN 46556\newline
\indent Institute of Mathematics ``Simion Stoilow'' of the Romanian Academy}
\email{craicu@nd.edu}
\author{Jerzy Weyman}
\address{Department of Mathematics, University of Connecticut, Storrs, CT 06269}
\email{jerzy.weyman@uconn.edu}
\subjclass[2010]{Primary 13D45, 14F10, 16G20}
\date{\today}
\keywords{Binary forms, equivariant $\D$-modules, quivers, local cohomology}
\begin{abstract}
We consider the space $X=\Sym^3\bb{C}^2$ of binary cubic forms, equipped with the natural action of the group $\GL_2$ of invertible linear transformations of $\bb{C}^2$. We describe explicitly the category of $\GL_2$-equivariant coherent $\D_X$-modules as the category of representations of a quiver with relations. We show moreover that this quiver is of tame representation type and we classify its indecomposable representations. We also give a construction of the simple equivariant $\D_X$-modules (of which there are $14$), and give formulas for the characters of their underlying $\GL_2$-representations. We conclude the article with an explicit calculation of (iterated) local cohomology groups with supports given by orbit closures.
\end{abstract}
\maketitle
\section{Introduction}\label{sec:intro}
Let $W=\bb{C}^2$ denote a $2$-dimensional complex vector space, and let $X = \Sym^3 W$ denote the corresponding space of binary cubic forms. There is a natural action of the group $\GL(W) = \GL_2(\bb{C})$ (which we'll simply denote by $\GL$) on the space $X$, with four orbits:
\begin{itemize}
\item The zero orbit $O_0 = \{0\}$.
\item The orbit $O_2 = \{ l^3 : 0\neq l \in W\}$ of cubes of linear forms, whose closure $\ol{O_2}$ is the affine cone over the twisted cubic curve.
\item The orbit $O_3 = \{ l_1^2 \cdot l_2 : 0 \neq l_1,l_2 \in W \mbox{ distinct up to scaling}\}$ whose closure $\ol{{O}_3}$ is the hypersurface defined by the vanishing of the cubic discriminant (and it is also the affine cone over the tangential variety to the twisted cubic curve).
\item The dense orbit $O_4 = \{ l_1 \cdot l_2 \cdot l_3 : 0\neq l_1,l_2,l_3 \in W \mbox{ distinct up to scaling}\}$.
\end{itemize}
Note that our indexing of the orbits is chosen so that $O_i$ has dimension $i$ for $i=0,2,3,4$. Letting $\D = \D_X$ denote the Weyl algebra of differential operators on $X$ with polynomial coefficients, our goal is to completely classify and give concrete constructions of the simple $\GL$-equivariant $\D$-modules, as well as to give an explicit quiver description of the category of equivariant coherent $\D$-modules and to analyze its structure.
By the Riemann--Hilbert correspondence, the simple equivariant $\D$-modules are known to be indexed by irreducible equivariant local systems on the orbits of the group action, but their explicit realization is in general difficult to obtain (see Open Problem 3 in \cite[Section~6]{mac-vil}). In the case of binary cubic forms the classification is summarized as follows.
\begin{thm-simples*}
There exist 14 simple $\GL$-equivariant $\D$-modules on $X=\Sym^3 W$, which can be classified according to their support as follows:
\begin{itemize}
\item $6$ of the modules, denoted $G_{-1},G_0=S,G_1,G_2,G_3,G_4$, have full support and correspond to irreducible rank one equivariant local systems on $O_4$.
\item $3$ of the modules, denoted $Q_0,Q_1,Q_2$, have full support and correspond to irreducible rank two equivariant local systems on $O_4$.
\item one of the modules, denoted $P$, has support $\ol{O_3}$.
\item $3$ of the modules, denoted $D_0,D_1,D_2$, have support $\ol{O_2}$.
\item one of the modules, denoted $E$, has support $O_0$.
\end{itemize}
All the modules have a concrete description, starting with $G_0=S$ which is the affine coordinate ring of $X$. They are permuted by the Fourier transform, which fixes $P$ and $G_3$ and swaps the modules in each of the pairs
\[(S,E),\ (G_{-1},D_1),\ (G_1,D_2),\ (G_2,G_4),\ (Q_1,Q_2),\ (D_0,Q_0),\]
and are also permuted by the holonomic duality functor, which fixes $S,Q_0,P,D_0,E,G_3$ and swaps the modules in each of the pairs
\[(D_1,D_2),\ (Q_1,Q_2),\ (G_2,G_4),\ (G_1,G_{-1}).\]
\end{thm-simples*}
We will analyze in more detail each of the $14$ simple modules in the theorem above, but before doing so we discuss the category $\opmod_{\GL}(\D_X)$ of $\GL$-equivariant coherent $\D_X$-modules. It is known quite generally that categories of $\D$-modules (or perverse sheaves) admit a quiver interpretation \cites{gel-mac-vil,vilonen,lor-wal}, but explicit descriptions of such categories are hard to come by \cites{galligo-granger-1,galligo-granger-2,mac-vil-cusp,braden-grinberg,lor-wal}. In the case of binary cubic forms we have the following.
\begin{thm-category*}
There is an equivalence of categories
\[\opmod_{\GL}(\D_X) \simeq \rep(\Q,\I),\]
where $\rep(\Q,\I)$ is the category of finite-dimensional representations of a quiver $\Q$ with relations $\I$, described as follows. The vertices and arrows of the quiver $\Q$ are depicted in the diagram
\[\xymatrix@=2.3pc@L=0.2pc{
s \ar@<0.5ex>[dr]^{\alpha_1} & & d_0 \ar@<0.5ex>[dl]^{\alpha_2} & & & g_{1} \ar@<0.5ex>[rr]^{\gamma_1} & & d_1 \ar@<0.5ex>[ll]^{\delta_1} & \\
& p \ar@<0.5ex>[ul]^{\beta_1} \ar@<0.5ex>[ur]^{\beta_2}\ar@<0.5ex>[dl]^{\beta_4} \ar@<0.5ex>[dr]^{\beta_3} & & & \overset{q_1}{\bullet} & \overset{q_2}{\bullet} & \overset{g_2}{\bullet} & \overset{g_3}{\bullet} & \overset{g_4}{\bullet} \\
q_0 \ar@<0.5ex>[ur]^{\alpha_4} & & e \ar@<0.5ex>[ul]^{\alpha_3} & & & g_{-1} \ar@<0.5ex>[rr]^{\gamma_{-1}} & & d_2 \ar@<0.5ex>[ll]^{\delta_{-1}} &
}\]
and the set of relations $\I$ is given by all 2-cycles and all non-diagonal compositions of two arrows:
\[\alpha_i \beta_i \mbox{ and } \beta_i\alpha_i \mbox{ for } i=1,2,3,4,\quad \gamma_i\delta_i\mbox{ and }\delta_i\gamma_i\mbox{ for }i=1,-1,\mbox{ and}\]
\[\alpha_1 \beta_2, \alpha_1\beta_4 , \alpha_2\beta_1, \alpha_2\beta_3, \alpha_3\beta_2,\alpha_3\beta_4,\alpha_4\beta_1,\alpha_4\beta_3.\]
Moreover, the bijection between the simple $\GL$-equivariant $\D_X$-modules and the nodes of the above quiver is given by replacing upper case symbols with the corresponding lower case symbols.
\end{thm-category*}
Once a quiver description of the category $\opmod_{\GL}(\D_X)$ is given, there are powerful tools available to understand its structure, such as Auslander--Reiten theory or tilting theory. We give an indication of this at the end of Section~\ref{sec:category-modDx} without going into the details of the theory. Instead we focus on proving that the quiver of $\opmod_{\GL}(\D_X)$ is of tame representation type, and to classify its indecomposable objects by relating them to the indecomposables of the extended Dynkin quiver of type $\hat{D}_4$. These latter indecomposables are well-understood, as their classification constitutes the solution of the celebrated \defi{four subspace problem} (see \cite{gel-pon} and \cite[Section XIII.3]{sim-sko2}).
In commutative algebra, the importance of the study of equivariant $\D$-modules comes from the desire to understand local cohomology modules with support in orbit closures \cites{raicu-weyman-witt,raicu-weyman,raicu-weyman-loccoh,raicu-veronese}. Our investigations owe a great deal to the work of Gennady Lyubeznik, who pioneered the use of $\D$-module techniques in commutative algebra. In the seminal paper \cite{lyubeznik}, Lyubeznik studies iterated local cohomology groups with respect to general families of supports, which lead to invariants known to be notoriously hard to compute in concrete examples. We show in Section~\ref{sec:loccoh} how basic knowledge of the structure of $\opmod_{\GL}(\D_X)$ can help to compute explicitly local cohomology groups associated with the space of binary cubics.
What is missing (and we promised to clarify) from the statements of the theorems above is the concrete construction of the simple equivariant $\D$-modules. In order to achieve this, we begin by setting up some notation. We let $V = W^*$ denote the dual vector space to $V$ so that $\Sym^3 V$ is naturally identified with the space of linear forms on $X$, and let $S = \Sym(\Sym^3 V)$ denote the ring of polynomial functions on $X$. If $\{w_0,w_1\}$ is a basis for $W$ then $\{w_0^3,3w_0^2 w_1,3w_0 w_1^2, w_1^3\}$ is a basis for $\Sym^3 W$. Choosing $\{x_0,x_1,x_2,x_3\}$ to be the dual basis of $\Sym^3 V$ we obtain an identification $S = \bb{C}[x_0,x_1,x_2,x_3]$. The condition for a cubic form
\begin{equation}\label{eq:generic-cubic}
f = x_0\cdot w_0^3 + 3x_1 \cdot w_0^2 w_1 + 3x_2 \cdot w_0 w_1^2 + x_3\cdot w_1^3
\end{equation}
to be in $\ol{O_2}$ is equivalent to the vanishing of the $2\times 2$ minors of
\[ A=\begin{bmatrix}
x_0 & x_1 & x_2 \\
x_1 & x_2 & x_3 \\
\end{bmatrix}\]
while the condition that $f\in \ol{O_3}$ is equivalent to the vanishing of the discriminant
\begin{equation}\label{eq:discriminant}
\Delta = 3 x_1^2 x_2^2 - 4x_0 x_2^3 - 4 x_1^3 x_3 - x_0^2 x_3^2 + 6 x_0x_1x_2x_3.
\end{equation}
We write $S_{\Delta}$ for the localization of $S$ at $\Delta$ and for every integer $i\in\bb{Z}$ we define
\begin{equation}\label{eq:Fi}
F_i = S_{\Delta} \cdot \Delta^{i/6}
\end{equation}
which is a $\D$-module. Moreover, we have $F_i = F_j$ if and only if $i-j$ is divisible by $6$. For $i=-1,0,1,2,3,4$ we define $G_i$ to be the $\D$-submodule of $F_i$ generated by $\Delta^{i/6}$, and part of the Theorem on Simple $\D$-modules is the fact that each $G_i$ is a simple $\D$-module. It is easy to see that $G_0=S$ is simple, and we will show later that in fact $G_i = F_i$ for $i=2,3,4$. Moreover, $F_1$ and $F_{-1}$ have $\D$-module length two so that the quotients $F_1/G_1$ and $F_{-1}/G_{-1}$ are simple $\D$-modules, which we label $D_1$ and $D_2$ respectively.
The modules $D_0$ and $E$ arise as local cohomology modules, for instance $D_0 = H^2_{\ol{O_2}}(S)$ and $E = H^4_{O_0}(S)$, while in order to construct $P$ we proceed as follows. The $\D$-module $F_0 = S_{\Delta}$ is generated by $\Delta^{-1}$, and therefore the same is true about the quotient $S_{\Delta}/S$. There exist a surjective $\D$-module map
\[ \pi: S_{\Delta}/S \lra E\]
sending $\Delta^{-1}$ to the socle generator of $E$, and $P$ can be realized as $\ker(\pi)$.
To construct the remaining modules $Q_0,Q_1,Q_2$ we use the Fourier transform, which may be defined as follows. If we write $\D = S \langle \pd_0,\pd_1,\pd_2,\pd_3\rangle$ with $\pd_i = \frac{\pd}{\pd x_i}$, then for every $\D$-module $M$ its \defi{Fourier transform $\mc{F}(M)$} is the $\D$-module with the same underlying abelian group, but with the \emph{new} action of the generators $x_i,\pd_i$ of $\D$ given in terms of the \emph{old} action by
\[ x_i \cdot^{new} m = - \pd_i \cdot^{old} m,\quad \pd_i \cdot^{new} m = x_i \cdot^{old} m\mbox{ for every }m\in M.\]
We use this definition only for the purpose of this Introduction, but in order to preserve equivariance more care is needed in defining the Fourier transform, as we explain in Section~\ref{subsec:Dmods}. The most basic example of Fourier transform is exhibited by the identification of $E$ with $\mc{F}(S)$, but this extends to other pairs of simple $\D$-modules as indicated in our first theorem. We have $Q_0 = \mc{F}(D_0)$ and
\[ Q_1 = (Q_0)_{\Delta} \oo_{S_{\Delta}} F_2 = (Q_0)_{\Delta} \cdot \Delta^{1/3},\quad Q_2 = (Q_0)_{\Delta} \oo_{S_{\Delta}} F_4 = (Q_0)_{\Delta} \cdot \Delta^{2/3},\]
finalizing the list of simple objects in $\opmod_{\GL}(\D_X)$.
The article is organized as follows. In Section~\ref{sec:prelim} we record some preliminaries and basic notation regarding $\GL$-representations, (equivariant) $\D$-modules, and quivers. In Section~\ref{sec:simple-Dmods} we analyze the simple $\GL$-equivariant $\D$-modules on the space of binary cubic forms, and describe their characters. In Section~\ref{sec:category-modDx} we study the category of equivariant $\D$-modules by providing its quiver description and classifying its indecomposable objects. We finish with a number of explicit computations of local cohomology modules in Section~\ref{sec:loccoh}.
\section{Preliminaries}\label{sec:prelim}
\subsection{Admissible representations and their characters}\label{subsec:repthy}
Throughout this paper we let $W$ denote a complex vector space of dimension two, we let $V = W^{\vee}$ be its dual, and write $\GL$ for the group $\GL(W)\simeq\GL_2(\bb{C})\simeq\GL(V)$ of invertible linear transformations of $W$. We let $\Lambda$ denote the set of (isomorphism classes of) finite dimensional irreducible $\GL$-representations, which are classified by \defi{dominant weights} $\ll=(\ll_1,\ll_2)\in \bb{Z}^2$, $\ll_1\geq\ll_2$. Concretely, we write $\S_{\ll}V$ for the irreducible $\GL$-representation corresponding to $\ll$, which is determined by the following conventions:
\begin{itemize}
\item If $\ll=(0,0)$ then $\S_{\ll}V = \bb{C}$ with the trivial $\GL$-action.
\item If $\ll=(a,0)$ with $a\geq 0$ then $\S_{\ll}V = \Sym^a V$ is the space of homogeneous degree $a$ polynomial functions on $W$.
\item For every dominant $\ll=(\ll_1,\ll_2)$ we have $\S_{(\ll_1+1,\ll_2+1)}V = \S_{\ll}V \oo \bw^2 V$, and in particular if $\ll=(1,1)$ then $\S_{(1,1)}V = \bw^2 V$. We will often refer to $\bw^2 V$ as the \defi{determinant of $V$} and denote it $\det(V)$.
\end{itemize}
More generally, if $U$ is any $k$-dimensional vector space (or a $\GL$-representation), we define the \defi{determinant of $U$} to be $\det(U) = \bw^k U$, so for instance we get
\begin{equation}\label{eq:det-sym}
\det(\Sym^a V) = \bw^{a+1}\Sym^a V = \S_{(b,b)}V\mbox{ where }b=\frac{a(a+1)}{2}.
\end{equation}
We write $\S_{\ll}W$ for the $\GL$-representation dual to $\bb{S}_{\ll}V$, so that if we let $\ll^{\vee} = (-\ll_2,-\ll_1)$ then
\[ \S_{\ll}W = \Hom_{\bb{C}}(\S_{\ll}V,\bb{C}) = \S_{\ll^{\vee}}V.\]
We next consider \defi{admissible $\GL$-representations} to be representations that decompose into a (possibly infinite) direct sum
\begin{equation}\label{eq:admissible}
M = \bigoplus_{\ll} (\S_{\ll}V)^{\oplus a_{\ll}},\mbox{ where }a_{\ll}\geq 0.
\end{equation}
The \defi{Grothendieck group $\Gamma(\GL)$ of virtual admissible $\GL$-representations} is defined as a direct product of copies of $\bb{Z}$ indexed by the set $\Lambda$
\[ \Gamma(\GL) = \bb{Z}^{\Lambda} = \{\mbox{maps } f: \Lambda \lra \bb{Z} \}.\]
We will often represent elements of $\Gamma(\GL)$ as formal sums
\begin{equation}\label{eq:formalsum}
a=\sum_{\ll} a_{\ll} \cdot e^{\ll}
\end{equation}
where $e^{\ll}$ corresponds to $\S_{\ll}V$ and $a_{\ll}\in\bb{Z}$ is the \defi{multipicity} of $\S_{\ll}V$. The correspondence between maps $f:\Lambda\lra\bb{Z}$ and the formal sum $a$ in (\ref{eq:formalsum}) is given by $a_{\ll} = f(\ll)$ for all $\ll\in\bb{Z}^2$ dominant. Given an element $a\in\Gamma(\GL)$ as in (\ref{eq:formalsum}) we will write
\[ \langle a,e^{\ll} \rangle = a_{\ll}.\]
Besides the additive group operation, $\Gamma(\GL)$ also has a partially defined multiplication given by convolution of functions:
\begin{equation}\label{eq:convolution}
(f\cdot g)(\ll) = \sum_{\mu+\delta = \ll} f(\mu) \cdot g(\delta)
\end{equation}
which is defined precisely when all the sums in (\ref{eq:convolution}) involve finitely many non-zero terms. Using this multiplication, we can make sense of inverting some virtual representations such as $(1-e^{\ll})$:
\[ \frac{1}{1-e^{\ll}} = 1 + e^{\ll} + e^{2\ll} + \cdots\]
Given an admissible representation $M$ as in (\ref{eq:admissible}) we write $[M]$ for the class of $M$ in $\Gamma(\GL)$. For example, when $\Sym(V) = \bigoplus_{i\geq 0}\Sym^i V$ is the ring of polynomial functions on $W$ then $[\Sym(V)] = 1/(1-e^{(1,0)})$. The following will be important in our study of $\D$-modules on binary cubics:
\begin{lemma}[{\cite[Section~6.1]{landsberg-manivel}}]\label{lem:charS} If $S = \Sym(\Sym^3 V)$ is the ring of polynomial functions on the space $X=\Sym^3 W$ of binary cubic forms then
\begin{equation}\label{eq:character-S}
[S] = \frac{1+e^{(6,3)}}{(1-e^{(3,0)})(1-e^{(4,2)})(1-e^{(6,6)})}.
\end{equation}
\end{lemma}
We say that a sequence $(f_n)_{n}$ of elements in $\Gamma(\GL)$ \defi{converges to $f$}, and write $\lim_{n\to\infty} f_n = f$, if for each dominant weight $\ll$ we have that the sequence of integers $(f_n(\ll))_n$ is eventually constant, equal to $f(\ll)$. A typical example of convergent sequence arises from localization: if $\Delta$ is the discriminant of the binary cubic form (see (\ref{eq:discriminant})) then it spans a one-dimensional $\GL$-representation with $[\bb{C}\cdot\Delta] = e^{(6,6)}$. We get based on Lemma~\ref{lem:charS} that
\begin{equation}\label{eq:character-Sdelta}
[S_{\Delta}] = \lim_{n\to\infty} [\Delta^{-n}\cdot S] = \frac{1+e^{(6,3)}}{(1-e^{(3,0)})(1-e^{(4,2)})} \cdot e^{(6,6)\bb{Z}}
\end{equation}
where $e^{(6,6)\bb{Z}} = \sum_{i\in\bb{Z}} e^{(6i,6i)}$. We have more generally the following.
\begin{lemma}\label{lem:localization}
Suppose that $M$ is a $\GL$-equivariant $S$-module which is an admissible representation, and suppose that $\Delta$ is a non-zerodivisor on $M$. If for each dominant weight $\ll$ we have that the sequence of multiplicities $\langle [M],e^{\ll+(6n,6n)}\rangle$ is eventually constant, with stable value equal to $m_{\ll}^{stab}$, then
\[ [M_{\Delta}] = \lim_{n\to\infty} [\Delta^{-n}\cdot M] = \sum_{\ll} m_{\ll}^{stab}\cdot e^{\ll}.\]
\end{lemma}
\begin{proof}
The condition that $\Delta$ is a non-zerodivisor on $M$ insures that $M$ embeds into $M_{\Delta}$ and that one can write $M_{\Delta} = \bigcup_{n\geq 1} \Delta^{-n}\cdot M$. It then suffices to show that for each dominant weight $\ll$ we have that $\langle \Delta^{-n}\cdot M,e^{\ll} \rangle = m_{\ll}^{stab}$ for $n\gg 0$. Since $[\bb{C}\cdot\Delta] = e^{(6,6)}$, it follows that $[\bb{C}\cdot\Delta^n] = e^{(6n,6n)}$ and therefore
\[\langle \Delta^{-n}\cdot M,e^{\ll} \rangle = \langle M,e^{\ll}\cdot[\Delta^n] \rangle = \langle [M],e^{\ll+(6n,6n)}\rangle = m_{\ll}^{stab} \mbox{ for }n\gg 0.\qedhere\]
\end{proof}
The main examples of admissible representations in this work come from $\GL$-equivariant coherent $\D_X$-modules (see \cite[Theorem~2.4]{lor-wal},
where admissible representations are called \defi{multiplicity-finite}). If $M$ is a $\GL$-equivariant $\D_X$-module then the same is true about the localization $M_{\Delta}$, hence Lemma~\ref{lem:localization} applies. In the next section we record some more generalities on equivariant $\D$-modules.
\subsection{Equivariant $\D$-modules {\cite{hot-tak-tan}}}\label{subsec:Dmods}
Given a smooth complex algebraic variety $Y$ we denote by $\D_Y$ the sheaf of differential operators on $Y$. If $G$ is a connected affine algebraic group acting on $Y$, with Lie algebra $\lie$, we get by differentiating the $G$-action on $Y$ a map from $\lie \oo_{\bb{C}} \mc{O}_Y$ to the sheaf of algebraic vector fields on $Y$, which in turn extends to a map of sheaves of algebras $U(\lie) \oo_{\bb{C}} \mc{O}_Y \to \D_Y$, where $U(\lie)$ denotes the universal enveloping algebra of $\lie$. We will always assume that $G$ acts on $Y$ with finitely many orbits and we will be interested in the study of
\[ \opmod_G(\D_Y) = \mbox{the category of }G\mbox{-equivariant coherent }\D_Y\mbox{-modules},\]
as defined in \cite[Definition 11.5.2]{hot-tak-tan}: a $\D_Y$-module $M$ is \defi{equivariant} if we have a $\D_{G\times Y}$-isomorphism $\tau: p^*M \rightarrow m^*M$, where $p: G\times Y\to Y$ denotes the projection and $m: G\times Y\to Y$ the map defining the action, with $\tau$ satisfying the usual compatibility conditions. When $Y$ is affine, this amounts to $M$ admitting an algebraic $G$-action whose differential coincides with the $\lie$-action induced by the natural map $\lie \to \D_Y$. It follows from \cite[Proposition~3.1.2]{VdB:loccoh} that a morphism of $\D_Y$-modules between objects in $\opmod_G(\D_Y)$ is automatically $G$-equivariant, so $\opmod_G(\D_Y)$ is equivalent to a full subcategory of the category $\opmod(\D_Y)$ of all coherent $\D_Y$-modules, which is closed under taking submodules and quotients. If $G$ is semi-simple, then if follows from \cite[Proposition~1.4]{lor-wal} that this subcategory is also closed under taking extensions. Moreover, it follows from \cite[Theorem~11.6.1]{hot-tak-tan} that every module in $\opmod_G(\D_Y)$ is regular and holonomic, and via the Riemann--Hilbert correspondence $\opmod_G(\D_Y)$ is equivalent to the category of $G$-equivariant perverse sheaves on $Y$. This category was shown in \cite[Theorem 4.3]{vilonen} to be equivalent to the category of finitely generated modules over a finite dimensional $\bb{C}$-algebra, which in turn is equivalent to the category of representations of a quiver with relations (see \cite[Corollary~I.6.10,~Theorems~II.3.7~and~III.1.6]{ass-sim-sko}). A more direct, $\D$-module theoretic approach to express $\opmod_G(\D_Y)$ as a category of quiver representations is described in \cite{lor-wal}, and it is this approach that we will follow in our study.
In what follows it will be important to consider several other categories of $\D$-modules, as well as the functors between them. If $Z\subset Y$ is a $G$-stable closed subvariety of $Y$, we define $\opmod_G^Z(\D_Y)$ to be the full subcategory of $\opmod_G(\D_Y)$ consisting of equivariant $\D_Y$-modules with support contained in $Z$. Associated to any $G$-equivariant local system $\mc{S}$ on some open subset contained in the smooth locus of $Z$ we have (see \cite[Remark~7.2.10]{hot-tak-tan})
\[\mc{L}(Z,\mc{S};Y) = \mbox{the \defi{intersection homology} }\D_Y\mbox{-module corresponding to }\mc{S}\]
which is a simple object in $\opmod_G^Z(\D_Y)$ (and hence also simple in $\opmod_G(\D_Y)$ and in $\opmod(\D_Y)$). In the case when $\mc{S}$ is the constant sheaf $\bb{C}$, we simply write $\mc{L}(Z;Y)$ for the corresponding intersection homology $\D_Y$-module. One important construction of objects in $\opmod_G^Z(\D_Y)$ comes from considering local cohomology functors $\mc{H}^i_Z(Y,\bullet)$: for each $i\geq 0$ and each $M\in\opmod_G(\D_Y)$ we have
\[ \mc{H}^i_Z(Y,M) = \mbox{the }i\mbox{-th local cohomology module of }M\mbox{ with support in }Z,\]
which is an element of $\opmod_G^Z(\D_Y)$. If we write $c = \dim(Y) - \dim(Z)$ for the codimension of $Z$ in $Y$ and take $M=\mc{O}_Y$ to be the structure sheaf of $Y$, then
\begin{itemize}
\item $\mc{H}^i_Z(Y,\mc{O}_Y) = 0$ for all $i<c$.
\item $\mc{H}^c_Z(Y,\mc{O}_Y)$ contains $\mc{L}(Z;Y)$ as a $\D_Y$-submodule, and the quotient $\mc{H}^c_Z(Y,\mc{O}_Y) / \mc{L}(Z;Y)$ has support contained in the singular locus of $Z$ (in particular $\mc{H}^c_Z(Y,\mc{O}_Y) = \mc{L}(Z;Y)$ when $Z$ is smooth).
\end{itemize}
More generally, if $Z' \subset Z$ is closed, and if we let $V = Z \setminus Z'$ denote the complement, then we can talk about the local cohomology functors $\mc{H}^i_V(Y,\bullet)$ with support in $V$, which are related to the usual ones via a long exact sequence for every $\D_Y$-module $M$ (see \cite[(1.2)]{lyubeznik}):
\[ \cdots \lra \mc{H}^i_{Z'}(Y,M) \lra \mc{H}^i_Z(Y,M) \lra \mc{H}^i_V(Y,M) \lra \mc{H}^{i+1}_{Z'}(Y,M) \lra \cdots\]
Moreover, if $M\in\opmod_G(\D_Y)$ then we have that $\mc{H}^i_Z(Y,M)\in\opmod_G^Z(\D_Y)$.
When $Z$ is itself smooth we can talk about various categories of $\D_Z$-modules. As a consequence of Kashiwara's equivalence \cite[Theorem~1.6.1]{hot-tak-tan} we get an equivalence of categories
\begin{equation}\label{eq:Kashiwara-equiv}
\opmod_G(\D_Z) \simeq \opmod^Z_G(\D_Y),
\end{equation}
under which $\mc{O}_Z \in \opmod_G(\D_Z)$ corresponds to $\mc{L}(Z;Y) = \mc{H}^c_Z(Y,\mc{O}_Y) \in \opmod^Z_G(\D_Y)$.
We next consider a non-empty $G$-stable open subset $U\subset Y$, and write $j:U \to Y$ for the corresponding open immersion. The direct and inverse image functors $j_*$ and $j^*$ for quasi-coherent sheaves restrict to functors between the corresponding categories of $\D$-modules:
\begin{equation}\label{eq:adjoint-pairs}
\opmod(\D_U) \overset{j_*}{\underset{j^*}{\rightleftarrows}} \opmod(\D_Y), \qquad \opmod_G(\D_U) \overset{j_*}{\underset{j^*}{\rightleftarrows}} \opmod_G(\D_Y), \qquad \opmod_G^{W\cap U}(\D_U) \overset{j_*}{\underset{j^*}{\rightleftarrows}} \opmod_G^W(\D_Y),
\end{equation}
where in the last pair of categories $W$ denotes a closed $G$-stable subvariety of $Y$. In each of the three cases, we have that $j_*$ is right adjoint to $j^*$, and $j^*$ is (left) exact, so $j_*$ takes injective objects to injective objects. For any $\D_Y$-module $M$, the adjunction between $j_*$ and $j^*$ gives a natural map $M \to j_* j^*M$. If we let $Z = Y\setminus U$ we get an exact sequence
\begin{equation}\label{eq:4term-loccoh}
0 \lra \mc{H}_Z^0(Y,M) \lra M \lra j_* j^*M \lra \mc{H}_Z^1(Y,M) \lra 0
\end{equation}
and for every $k\geq 1$ we have isomorphisms (where $R^k j_*$ denotes the $k$-th derived functor of $j_*$)
\[ R^k j_*(j^* M) \simeq \mc{H}_Z^{k+1}(Y,M).\]
As a consequence of the Riemann--Hilbert correspondence and \cite[Theorem~11.6.1]{hot-tak-tan} we have the following.
\begin{theorem}\label{thm:simples}
Suppose that $Y$ is a smooth complex algebraic variety, and that $G$ is an algebraic group acting on $Y$.
(a) If $G$ acts transitively on $Y$, we identify $Y$ with $G/K$ for some algebraic subgroup $K$ of $G$ and let $H = K/K^0$ denote the component group of $K$ (where $K^0$ is the connected component of the identity element in $K$). The category $\opmod_G(\D_Y)$ is equivalent to the category of finite dimensional complex representations of $H$ and is therefore semisimple (since $H$ is finite).
(b) If $G$ acts with finitely many orbits on $Y$ then we have a bijective correspondence
\[\left\{(O,\mc{S}) : \begin{array}{c}
O\mbox{ is a }G\mbox{-orbit and } \\ \mc{S}\mbox{ is an equivariant irreducible local system on }O
\end{array}
\right\}
\llra \{\mbox{simple objects of }\opmod_G(\D_Y)\}\]
\[\mbox{given by }\qquad(O,\mc{S}) \llra \mc{L}(O,\mc{S};Y).\]
Moreover, if we fix $O = G/K$ and let $H = K/K^0$ as in part (a), then we have a bijective correspondence
\[\{\mbox{equivariant irreducible local systems on }O\} \llra \{\mbox{irreducible representations of }H\}.\]
\end{theorem}
The following will play a key role in our description of the category of equivariant $\D$-modules in Section~\ref{sec:category-modDx}.
\begin{lemma}\label{lem:inj}
Let $G,Y$ be as in Theorem~\ref{thm:simples}, and let $O$ denote one of the orbits. Consider $Z=\ol{O}\setminus O$, $U=Y\setminus Z$, and let $j: U \to Y$ be the corresponding open immersion. If $M$ is a simple equivariant $\D_Y$-module with support $\ol{O}$ then $j_* j^* M$ is the injective envelope of $M$ in the category $\opmod^{\ol{O}}_G (\D_Y)$.
\end{lemma}
\begin{proof}
Since $M$ is simple with support $\ol{O}$ and $\mc{H}_Z^0(Y,M)$ is a $\D$-submodule with support contained in $Z\subsetneq\ol{O}$, it follows that $\mc{H}_Z^0(Y,M)=0$. Using (\ref{eq:4term-loccoh}) we get a natural injective map $M\hra j_* j^* M$, so it suffices to verify that $j_*j^*M$ is an injective object in $\opmod^{\ol{O}}_G (\D_Y)$ and that it is indecomposable.
If we let $W = \ol{O}$ then $W\cap U = O$, so based on (\ref{eq:adjoint-pairs}) we can think of $j_*,j^*$ as a pair of adjoint functors between the categories $\opmod^{{O}}_G (\D_U)$ and $\opmod^{\ol{O}}_G (\D_Y)$. Using Kashiwara's equivalence (\ref{eq:Kashiwara-equiv}) and Theorem~\ref{thm:simples}(a) we get that $\opmod^{O}_G (\D_U)\simeq \opmod_G (\D_O)$ is semisimple and in particular $j^*M$ is an injective object. As remarked earlier, $j_*$ takes injectives to injectives, hence $j_*j^*M$ is injective in $\opmod^{\ol{O}}_G (\D_Y)$.
To show that $j_* j^* M$ is indecomposable, we note that $j^*M$ is simple and therefore
\[\Hom_{\D_Y} (j_* j^* M, j_* j^* M) = \Hom_{\D_U} (j^*j_* j^* M, j^* M)=\Hom_{\D_U} (j^* M, j^* M)=\bb{C}.\]
If $j_* j^* M$ decomposed as a direct sum of two submodules then $\Hom_{\D_Y} (j_* j^* M, j_* j^* M)$ would have dimension at least two, concluding our proof.
\end{proof}
We end this section by recalling two important functors on $\opmod_G(\D_Y)$:
\begin{itemize}
\item The \defi{holonomic duality functor} $\bb{D}: \opmod_G(\D_Y) \lra \opmod_G(\D_Y)$ is a \defi{duality of categories} (i.e. an equivalence of categories between $\opmod_G(\D_Y)$ and its opposite category $\opmod_G(\D_Y)^{op}$). Translated via the Riemann--Hilbert correspondence this functor corresponds to Verdier duality \cite[Corollary~4.6.5]{hot-tak-tan}, and in particular it restricts to a duality on $\opmod_G^Z(\D_Y)$ for any $G$-stable subvariety $Z$. At the level of $\D_Y$-modules it is defined via (see \cite[Section~2.6]{hot-tak-tan})
\[ \bb{D}(M) = \ShExt^{\ n}_{\D_Y}(M,\D_Y) \oo_{\mc{O}_Y} \omega_Y^{-1}\]
where $n=\dim(Y)$, $\omega_Y$ is the canonical line bundle on $Y$, and where $\bullet \oo_{\mc{O}_Y} \omega_Y^{-1}$ is the functor that transforms right $\D_Y$-modules (of which $\ShExt^{\ n}_{\D_Y}(M,\D_Y)$ is one) into left $\D_Y$-modules \cite[Proposition~1.2.12]{hot-tak-tan}. What will be important for us is that $\bb{D}$ interchanges injective and projective objects in $\opmod_G^Z(\D_Y)$, and that it permutes the collection of simple objects.
\item The \defi{Fourier transform}. If $Y$ is an affine space, corresponding to a finite dimensional $G$-representation $U$, we let $Y^{\vee}$ denote the affine space corresponding to the dual representation $U^{\vee}$. As explained in \cite[Section~2.5]{raicu-matrices} we get an equivalence of categories
\[\mc{F}^{\circ}:\opmod_G(\D_Y) \lra \opmod_G(\D_{Y^{\vee}}),\quad \mc{F}^{\circ}(M) = M \oo_{\bb{C}} \det(U^{\vee}).\]
In the case when $G = \GL(W)$ and $Y=\S_{\ll}W$, we can further use the natural isomorphism $\tau:\GL(W) \simeq \GL(V)$ given by $\tau(\phi) = (\phi^{\vee})^{-1}$ (the inverse transpose), which is compatible with the respective actions of $\GL(V)$ and $\GL(W)$ on $\S_{\ll}V$, to obtain an equivalence of categories
\[ \opmod_{\GL(W)}(\D_{\S_{\ll}V}) \overset{\tau}{\lra} \opmod_{\GL(V)}(\D_{\S_{\ll}V})\]
Fixing a vector space isomorphism $T:V\to W$ we obtain a group isomorphism $\tau':\GL(V) \simeq \GL(W)$ given by $\tau'(\phi) = T\circ\phi\circ T^{-1}$, and because of the functoriality of $\bb{S}_{\ll}$ an isomorphism $\tl{T}:\S_{\ll}V \lra S_{\ll}W$ compatible with the respective actions of $\GL(V)$ and $\GL(W)$ on the source and the target. This provides a further equivalence of categories
\[ \opmod_{\GL(V)}(\D_{\S_{\ll}V}) \overset{(\tau',\tl{T})}{\lra} \opmod_{\GL(W)}(\D_{\S_{\ll}W}).\]
Putting all of these equivalences together, we obtain a self equivalence of categories
\begin{equation}\label{eq:def-Fourier}
\mc{F}:\opmod_{\GL(W)}(\D_{\S_{\ll}W}) \overset{\mc{F}^{\circ}}{\lra} \opmod_{\GL(W)}(\D_{\S_{\ll}V}) \overset{\tau}{\lra} \opmod_{\GL(V)}(\D_{\S_{\ll}V}) \overset{(\tau',\tl{T})}{\lra} \opmod_{\GL(W)}(\D_{\S_{\ll}W})
\end{equation}
which we will refer to (by abuse of language) as the Fourier transform on $Y=\S_{\ll}W$.
\end{itemize}
\begin{remark}\label{rem:Fourier} The construction of a Fourier transform as a self-equivalence of $\opmod_G(\D_{Y})$ can be done more generally when $G$ is a connected linear reductive group. Let $T\subset G$ be a maximal torus and let $B\subset G$ be a Borel subgroup containing $T$. There exists an involution $\theta \in \Aut(G)$ such that $\theta(t)=t^{-1}$ for all $t\in T$ and $B\cap \theta(B)=T$ (for example, see \cite[II. Corollary 1.16]{jantzen}). If $V$ is any $G$-module, we can twist the action of $G$ by $\theta$ to obtain a $G$-module $V^*$, which is isomorphic to the usual dual representation $V^{\vee}$ of $V$. Twisting the action of $G$ on $Y$ by $\theta$ gives then an equivalence of categories $\tau_{\theta}:\opMod_G(\D_{Y}) \simeq \opMod_G(\D_{Y^*})$. Using a $G$-isomorphism $Y^*\simeq Y^{\vee}$ together with the usual Fourier transform $\mathcal{F}^{\circ} : \opMod_G(\D_{Y^{\vee}}) \xrightarrow{\sim} \opMod_G(\D_{Y})$ we obtain an involutive self-equivalence of $\opmod_G(\D_Y)$ (which we also call the Fourier transform)
\[\mathcal{F} : \opmod_G(\D_Y) \overset{\tau_{\theta}}{\lra} \opmod_G(\D_{Y^*}) \xrightarrow{\sim} \opmod_G(\D_{Y^{\vee}}) \overset{\mc{F}^{\circ}}{\lra} \opMod_G(\D_{Y}).\]
\end{remark}
\subsection{The Fourier transform of admissible representations}\label{subsec:Fourier-admissible}
We let $U = \Sym^3 W$ and recall from (\ref{eq:det-sym}) that
\[ \det(U) = \S_{(6,6)}W.\]
We define the Fourier transform (relative to $U$) $\mc{F}:\Lambda\to\Lambda$ via
\[ \mc{F}(\ll) = \ll^{\vee} - (6,6) = (-\ll_2-6,-\ll_1-6).\]
This in turn induces a Fourier transform $\mc{F}:\Gamma(\GL) \lra \Gamma(\GL)$ given by
\[\mc{F}\left(\sum_{\ll} a_{\ll} e^{\ll}\right) = \sum_{\ll} a_{\ll} e^{\mc{F}(\ll)}.\]
The motivation behind this definition is as follows. If $X = \Sym^3 W$ is the affine space corresponding to $U$, and if $M\in\opmod_{\GL}(\D_X)$ then it follows from \cite[Theorem~2.4]{lor-wal} that $[M]$ is an admissible $\GL$-representation, and therefore
\[ M \simeq \bigoplus_{\ll} \S_{\ll}V^{\oplus a_{\ll}},\mbox{ for }a_{\ll}\in\bb{Z}.\]
Applying the Fourier transform $\mc{F}:\opmod_{\GL}(\D_X)\lra \opmod_{\GL}(\D_X)$ as constructed in (\ref{eq:def-Fourier}) we obtain
\[\mc{F}(M) \simeq \bigoplus_{\ll} \S_{\ll+(6,6)}W^{\oplus a_{\ll}} = \bigoplus_{\ll} \S_{\ll^{\vee}-(6,6)}V^{\oplus a_{\ll}},\mbox{ that is }[\mc{F}(M)]=\mc{F}([M]).\]
We rewrite the above conclusion as follows:
\begin{equation}\label{eq:wts-Fourier}
\langle [\mc{F}(M)], e^{\ll} \rangle = \langle [M], e^{\ll^{\vee} - (6,6)} \rangle \mbox{ for every dominant weight }\ll.
\end{equation}
We record two basic instances of the Fourier transform which will be used later. Using the fact that the simple $\D$-module $E = \mc{L}(O_0;X)$ supported at the origin can be realized as $\mc{F}(S)$ it follows from (\ref{eq:character-S}) that
\begin{equation}\label{eq:character-E}
[E] = [\mc{F}(S)] = \frac{1+e^{(-3,-6)}}{(1-e^{(0,-3)})(1-e^{(-2,-4)})(1-e^{(-6,-6)})}\cdot e^{(-6,-6)}.
\end{equation}
We can also use (\ref{eq:wts-Fourier}) in conjunction with (\ref{eq:character-Sdelta}) to show that the character of $S_{\Delta}$ doesn't change under taking Fourier transform:
\[ [\mc{F}(S_{\Delta})] = \frac{1+e^{(-3,-6)}}{(1-e^{(0,-3)})(1-e^{(-2,-4)})} \cdot e^{(6,6)\bb{Z}} \cdot e^{(-6,-6)} = [S_{\Delta}].\]
This is a reflection of the fact that $\mc{F}(S_{\Delta})$ has three composition factors, $S,E$ and $P$, and the Fourier transform exchanges $S$ and $E$, and it preserves $P$.
\subsection{Quivers and their representations}\label{subsec:quivers}
We begin by establishing some notation and reviewing some basic results regarding the representation theory of quivers, following \cite{ass-sim-sko}. A \defi{quiver} $\Q$ is an oriented graph, i.e. a pair $\Q=(\Q_0,\Q_1)$ formed by a finite set of vertices $\Q_0$ and a finite set of arrows $\Q_1$. An arrow $\a\in \Q_1$ has a \defi{head} (or a \defi{target}) $h(\a)$ and a \defi{tail} (or a \defi{source}) $t(\a)$ which are elements of $\Q_0$:
\[\xymatrix{
t(\a) \ar[r]^{\a} & h(\a)
}\]
A \defi{(directed) path $p$ in $\Q$ from $a$ to $b$ of length $l$} is a sequence of arrows
\begin{equation}\label{eq:pathQ}
p:\quad\xymatrix{
a=a_0 \ar[r]^{\a_1} & a_1 \ar[r]^{\a_2} & a_2 \ar[r]^{\a_3} & \ \cdots\ \ar[r]^{\a_l} & a_l = b \\
}
\end{equation}
where $a_i\in \Q_0$ and $\a_i\in \Q_1$. We call $a$ (resp. $b$) the \defi{source} (resp. \defi{target}) of the path, and write $p=\a_1\a_2\cdots\a_l$. The complex vector space with basis given by the paths in $\Q$ has a natural multiplication induced by concatenation of paths (where $pq=0$ if the source of $q$ is different from the target of $p$). The corresponding $\bb{C}$-algebra is called the \defi{path algebra of the quiver $\Q$} and is denoted $\bb{C}\Q$. A \defi{relation} in $\Q$ is a $\bb{C}$-linear combination of paths of length at least two having the same source and target. We define a \defi{quiver with relations} $(\Q,\mc{I})$ to be a quiver $\Q$ together with a finite set of relations $\mc{I}$. The \defi{quiver algebra of $(\Q,\mc{I})$} is the quotient $\mc{A}=\bb{C}\Q/\langle\mc{I}\rangle$ of the path algebra by the ideal generated by the relations. We will often use the word quiver to refer to a quiver with relations $(\Q,\I)$, and talk about $\sum c_i\cdot p_i = 0$ as being a relation in the quiver if $\sum c_i\cdot p_i \in \langle\mc{I}\rangle$, or equivalently if the relation $\sum c_i\cdot p_i = 0$ holds in the quiver algebra $\mc{A}$. We will moreover always assume that the ideal of relations $\langle\mc{I}\rangle$ contains any path whose length is large enough, so that the corresponding quiver algebra $\mc{A}$ is finite dimensional (see \cite[Section II.2]{ass-sim-sko}).
A \defi{(finite-dimensional) representation} $V$ of a quiver $(\Q,\mc{I})$ is a family of (finite-dimensional) vector spaces $\{V_x\,|\, x\in Q_0\}$ together with linear maps $\{V(\a) : V_{t(\a)}\to V_{h(\a)}\, | \, \a\in Q_1\}$ satisfying the relations induced by the elements of $\mc{I}$. More precisely, for every path $p:a\to b$ as in (\ref{eq:pathQ}) we consider the composition
\[ V(p) = V(\a_l) \circ V(\a_{l-1}) \circ \cdots \circ V(\a_1)\]
and we ask that for every element $\sum_i c_i\cdot p_i\in\mc{I}$ with $c_i\in\bb{C}$ and $p_i$ a path from $a$ to $b$, we have that
\[ \sum_i c_i\cdot V(p_i) = 0.\]
A morphism $\phi:V\to V'$ of two representations $V,V'$ of $(\Q,\mc{I})$ is a collection of linear maps
\[\phi = \{\phi(x) : V_x \to V'_x\,| \,x\in Q_0\},\]
with the property that for each $\a\in Q_1$ we have
\[\phi(h(\a))\circ V(\a)=V'(\a)\circ \phi(t(\a)).\]
We note that the data of a representation of $(\Q,\mc{I})$ is equivalent to that of a module over the quiver algebra $\mc{A}$, and a morphism of representations corresponds to an $\mc{A}$-module homomorphism. In other words, the category $\rep(\Q,\mc{I})$ of finite-dimensional representations of $(\Q,\mc{I})$ is equivalent to that of finitely generated $\mc{A}$-modules \cite[Section~III.1, Thm.~1.6]{ass-sim-sko}. Moreover, this is an abelian category with enough projectives and injectives, and having finitely many simple objects, as we explain next.
The (isomorphism classes of) simple objects in $\rep(\Q,\mc{I})$ are in bijection with the vertices of $\Q$. For each $x\in \Q_0$, the corresponding simple $\sS^x$ is the representation with
\begin{equation}\label{eq:defS}
(\sS^x)_x=\bb{C},\ (\sS^x)_y=0\mbox{ for all }y\in \Q_0\setminus\{x\},\mbox{ and }\sS^x(\a)=0\mbox{ for all }\a\in\Q_1.
\end{equation}
A representation of $(\Q,\mc{I})$ is called \defi{indecomposable} if it is not isomorphic to a direct sum of two non-zero representations. Just like the simple objects, the indecomposable projectives (resp. injectives) in $\rep(\Q,\mc{I})$ are in bijection with the vertices of $\Q$. For each $x\in \Q_0$, we let $\sP^x$ (resp. $\sI^x$) denote the \defi{projective cover} (resp. \defi{injective envelope}) of~$\sS^x$, as constructed in \cite[Section III.2]{ass-sim-sko}. For $y\in\Q_0$, the dimension of $(\sP^x)_y$ (resp. $(\sI^x)_y$) is given by the number of paths from $x$ to $y$ (resp. from $y$ to $x$), considered up to the relations in $\mc{I}$. More precisely
\begin{equation}\label{eq:defP-I}
(\sP^x)_y = \mbox{Span}\{ p\in\mc{A}:p\mbox{ is a path from }x\mbox{ to }y\},\ (\sI^x)_y = \mbox{Span}\{ p\in\mc{A}:p\mbox{ is a path from }y\mbox{ to }x\}^{\vee},
\end{equation}
where $^{\vee}$ denotes as usual the dual vector space. For an arrow $\a\in\Q_0$, thought of as an element of $\mc{A}$, we have that $\sP^x(\a)$ is right multiplication by $\a$, while $\sI^x(\a)$ is the dual to left multiplication by $\a$.
Unlike the classification of simple objects, and that of indecomposable injective and projective objects, the classification of general indecomposable objects in $\rep(\Q,\mc{I})$ is significantly more involved. A quiver $(\Q,\mc{I})$ is said to be of \defi{finite representation type} if it has finitely many (isomorphism types of) indecomposable representations. It is of \defi{tame representation type} if all but a finite number of indecomposable representations of $(\Q,\mc{I})$ of a given dimension belong to finitely many one-parameter families \cite[XIX.3, Definition 3.3]{sim-sko3}, and it is of \defi{wild representation type} otherwise. As the name suggests, in the wild case the classification of indecomposables is essentially intractable, while in the finite and tame cases it is more manageable. A special instance of tame representation type, and the one that will concern us in this paper, is when the number of one-parameter families of indecomposables is bounded as the dimension of the representations grows. We say in this case that the quiver $(\Q,\mc{I})$ is of \defi{domestic tame representation type} \cite[XIX.3, Definitions 3.6, 3.10 and Theorem 3.12]{sim-sko3}. One example of such a quiver, which will appear again in Section~\ref{subsec:indecomp}, is given by the following.
\begin{example}\label{ex:dee4hat}
Let $Q$ be the extended Dynkin quiver $\hat{D}_4$ (with no relations):
\[\hspace{-0.2in} \hat{D}_4: \hspace{0.15in} \begin{aligned} \xymatrix@R=0.7pc@C=2.5pc{
1 \ar[dr]^{\alpha_1} & & 2 \ar[dl]_{\alpha_2}\\
& 5 & \\
4 \ar[ur]_{\alpha_4} & & 3\ar[ul]^{\alpha_3}
}\end{aligned}\]
The representation theory of the the quiver $\hat{D}_4$ is well-understood. The quiver is of (domestic) tame representation type, and all its indecomposable representations have been classified -- this is, in the language of quivers, the famous four subspace problem solved in \cite{gel-pon}. For the complete description of the indecomposables of $\hat{D}_4$ together with the Auslander-Reiten quiver, we refer the reader to \cite[Section XIII.3]{sim-sko2}. In the following we illustrate only the 1-parameter families of indecomposable representations (they are taken from \cite[XIII.3. Table 3.14]{sim-sko2}): for each $n>0$ there exists precisely one $1$-parameter family of indecomposable representations $V$ of $\hat{D}_4$ with dimension vector $(n,n,n,n,2n)$ (i.e. with $\dim V_1 = \dim V_2 = \dim V_3 = \dim V_4 = 1$ and $\dim V_5 = 2n$), and moreover these are all the $1$-parameter families occuring in the classification of indecomposable of $\hat{D}_4$.
Let $I_n$ denote the $n\times n$ identity matrix, and for any $\lambda\in\bb{C}$ we denote by $J_n(\lambda)$ the $n\times n$ Jordan block
\[J_n(\lambda) = \begin{bmatrix}
\lambda & 1 & 0 & \dots & 0 \\
0 & \lambda & 1 & \dots & 0 \\
\vdots & \vdots & \ddots &\ddots & \vdots\\
0 & 0 & \dots & \lambda & 1 \\
0 & 0 & \dots & 0 & \lambda
\end{bmatrix}.\]
Then for any $n>0$, we have the following 1-parameter family of indecomposables $R_n(\lambda)$, where $\lambda \in \bb{C}$:
\[\hspace{-0.2in} R_n(\lambda) : \hspace{0.15in} \begin{aligned} \xymatrix@R=2pc@C=4pc{
\bb{C}^n \ar[dr]^{\footnotesize\begin{bmatrix} I_n\\ 0 \end{bmatrix}} & & \bb{C}^n \ar[dl]_{\footnotesize\begin{bmatrix} 0\\I_n \end{bmatrix}}\\
& \bb{C}^{2n} & \\
\bb{C}^n \ar[ur]_{\footnotesize\begin{bmatrix} I_n\\ J_n(\lambda) \end{bmatrix}} & & \bb{C}^n\ar[ul]^{\footnotesize\begin{bmatrix} I_n\\ I_n \end{bmatrix}}
}\end{aligned}\]
\end{example}
In Section~\ref{sec:category-modDx} we will be concerned with the quiver associated with the category of $\GL$-equivariant $\D$-modules on the space of binary cubic forms. We will prove that the said quiver is of (domestic) tame representation type and we will classify its indecomposables by directly relating them to indecomposables of~$\hat{D}_4$. The following reduction lemmas will play a key role in describing this relationship.
Consider a vertex $x$ in a quiver $(\Q,\mc{I})$:
\[\xymatrix@C=4pc@R=1pc{
\cdots \circ \ar@<0.6ex>[ddr]^(0.4){\a_1} & & \circ \cdots\\
\cdots \circ \ar[dr]^(0.4){\a_2} & & \circ \cdots\\
\vdots & x \ar@<0.6ex>[uur]^(0.6){\b_1} \ar[ur]^(0.6){\b_2} \ar[dr]^{\b_n} & \vdots \\
\cdots \circ \ar[ur]^{\a_m} & & \circ \cdots
}\]
Following \cite{martinez}, we say $x$ is a \defi{node} of $(\Q,\mc{I})$ if we have $\a_i\b_j = 0$ for all $1\leq i \leq m, 1\leq j \leq n$. If $x$ is a node of $(\Q,\mc{I})$, we can ``separate the node" to obtain a new quiver $(\Q', \I')$:
\[\xymatrix@C=3pc@R=1pc{
\cdots \circ \ar@<0.7ex>[ddr]^(0.4){\a_1} & & & \circ \cdots\\
\cdots \circ \ar@<0.2ex>[dr]^(0.4){\a_2} & & & \circ \cdots\\
\vdots & x' & x \ar@<0.7ex>[uur]^(0.6){\b_1} \ar@<0.2ex>[ur]^(0.6){\b_2} \ar[dr]^{\b_n} & \vdots \\
\cdots \circ \ar[ur]^{\a_m} & & & \circ \cdots
}\]
The relations in $\I'$ are the natural ones obtained from $\I$ by removing the relations $\a_i \b_j = 0$. If $V$ is a representation of $(\Q,\I)$, then we can induce a representation $V'$ of $(\Q',\I')$ by letting
\[V'_{x'} = \sum_{i=1}^{m} \operatorname{Im} V(\a_i) \subset V_x \mbox{ and } V'_{x} = V_x / V'_{x'}.\]
The following is a consequence of \cite[Theorem 2.10(b)]{martinez} (albeit formulated in a slightly different language):
\begin{lemma}\label{lem:node}
The functor $V\mapsto V'$ induces a bijection between the (isomorphism classes of) non-simple indecomposable objects in $\rep(\Q,\I)$ and those in $\rep(\Q',\I')$.
\end{lemma}
\vspace{0.1in}
Consider next the following situation. Suppose that $(\Q,\I)$ is the quiver
\[\xymatrix@C=4pc@R=1pc{
y \ar@<0.6ex>[ddr]^(0.4){\a} & & z\\
\cdots \circ \ar[dr]^(0.4){\a_1} & & \circ \cdots\\
\vdots & x \ar@<0.6ex>[uur]^(0.6){\b} \ar[ur]^(0.6){\b_1} \ar[dr]^{\b_n} & \vdots \\
\cdots \circ \ar[ur]^{\a_m} & & \circ \cdots
}\]
with relations $\a_i\cdot\b= 0$ and $\a\cdot\b_j=0$, for all $1\leq i \leq m,1\leq j \leq n$, where the diagram above is indicative of the fact that the only arrow connected to the vertex $y$ (resp. $z$) is $\a$ (resp. $\b$). Let $\sP^y$ be the indecomposable projective corresponding to $y$, and let $\sI^z$ be the indecomposable injective corresponding to $z$. It follows from the explicit description in (\ref{eq:defP-I}) that $\sP^y\simeq\sI^z$ with
\[(\sP^y)_y = (\sP^y)_x = (\sP^y)_z = \bb{C},\ \sP^y(\a) = \sP^y(\b) = \mbox{id}_{\bb{C}},\mbox{ and }(\sP^y)_t = 0\mbox{ for }t\neq x,y,z.\]
We have moreover the following.
\begin{lemma}\label{lem:addrel}
If $(\Q,\I)$ is a quiver as above and if $V$ is an indecomposable representation of $(\Q,\I)$ with $V\not\simeq \sP^y$, then $V(\b) \circ V(\a) = 0$.
\end{lemma}
\begin{proof}
We let $A= \sum_{j=1}^m \operatorname{im} V(\a_j)$ and denote $i: A\hookrightarrow V_x$ the inclusion. We consider the Dynkin quiver of type $D_4$ (with no relations)
\[
\xymatrix@R=0.2pc{
y \ar[dr]^{\a} & \\
& x \ar[r]^{\b} & z \\
a \ar[ur]_{\gamma} & &
}
\]
and its representation $V'$ given by
\[V': \hspace{0.2in} \begin{aligned}\xymatrix@R=0.2pc{
V_y \ar[dr]^{V(\a)} & \\
& V_x \ar[r]^{V(\b)} & V_z \\
A \ar[ur]_i & &
}\end{aligned}\]
If we assume that $V(\b) \circ V(\a) \neq 0$, then $V'$ has an indecomposable summand $X$ with the property that $X(\b) \circ X(\a) \neq 0$. Moreover, it follows from the relations in $(\Q,\I)$ that $V'(\b) \circ V'(\gamma) = V(\b) \circ i = 0$, hence $X(\b) \circ X(\gamma) = 0$. The classification in \cite[Chapter VII,~Example~5.15(b)]{ass-sim-sko} of indecomposables of the Dynkin quiver of type $D_4$ shows that there is only one indecomposable $X$ with $X(\b) \circ X(\a) \neq 0$ and $X(\b) \circ X(\gamma) = 0$, namely
\[X: \hspace{0.2in} \begin{aligned}\xymatrix@R=0pc{
\bb{C} \ar[dr]^{1} & & \\
& \bb{C} \ar[r]^{1} & \bb{C} \\
0 \ar[ur] & &
}\end{aligned}\]
We can then write $V'\simeq X^{\oplus k} \oplus Y$, for some $k\geq 1$, where $Y$ is a representation of the quiver of type $D_4$ with $Y(\b)\circ Y(\a)=0$. Due to the relations $\a \cdot \b_j=0$, we can lift this isomorphism to a decomposition $V\simeq(\sP^y)^{\oplus k} \oplus Z$ in $\rep(\Q,\I)$, where $Z$ is a representation of $(\Q,\I)$ with $Z(\a) = Y(\a)$, $Z(\b)=Y(\b)$.
Since $V$ was indecomposable, we conclude that $k=1$ and $Z=0$, i.e. $V\simeq\sP^y$ which is a contradiction. It follows that we must have $V(\b) \circ V(\a)=0$, concluding the proof.
\end{proof}
\section{The simple equivariant $\D$-modules on binary cubics}\label{sec:simple-Dmods}
In this section we give a classification of the simple $\GL$-equivariant holonomic $\D$-modules on the space $X = \Sym^3 W$ of binary cubic forms. For each of these simple modules we give an explicit description of the characters of the underlying admissible representations.
\subsection{The classification of simple equivariant $\D$-modules}\label{subsec:classification}
We will use Theorem~\ref{thm:simples} in order to obtain a classification of the simple equivariant $\D$-modules. We choose a basis $\{w_0,w_1\}$ for $W$ and identify $\GL\simeq\GL_2(\bb{C})$ relative to this basis. The non-zero elements of $X = \Sym^3 W$ are homogeneous cubic polynomials in the variables $w_0,w_1$ as in (\ref{eq:generic-cubic}), and as such they factor into a product of three linear forms. The $\GL$-orbit structure of $X$ is then described by the different types of factorizations: a non-zero cubic form may have three distinct (up to scaling) linear factors, or precisely one factor that is repeated twice, or it could be the cube of a linear form. The following table records representatives of each orbit $O_i$ ($i=0,2,3,4$), together with their stabilizers $K_i$ and the component groups $H_i = K_i/K_i^0$ of these stabilizers (we write $1$ for the trivial group, $C_k$ for the cyclic group of order $k$, and $\rtimes$ for a semidirect product).
\begin{center}
\renewcommand{\arraystretch}{1.5}
\begin{tabular}{c|c|c|c}
Orbit & Representative & Stabilizer & Component group \\
\hline
$O_0$ & $0$ & $\GL$ & 1 \\
\hline & & &\\[-10pt]
$O_2$ & $w_0^3$ & $\left\{\begin{pmatrix} \xi & \b \\ 0 & \a\end{pmatrix}:\xi^3 = 1,\a\in\bb{C}^{\times},\b\in\bb{C}\right\}$ & $C_3$ \\[15pt]
\hline & & &\\
$O_3$ & $w_0^2\cdot w_1$ & $\left\{\begin{pmatrix} \a & 0 \\ 0 & \a^{-2} \end{pmatrix}:\a\in\bb{C}^{\times}\right\}$ & 1 \\[15pt]
\hline & & &\\
$O_4$ & $w_0^3 + w_1^3$ & $\left\{\begin{pmatrix} \xi_1 & 0 \\ 0 & \xi_2 \end{pmatrix}:\xi_1^3 = \xi_2^3 = 1\right\} \bigcup \left\{\begin{pmatrix} 0 & \xi_1 \\ \xi_2 & 0 \end{pmatrix}:\xi_1^3 = \xi_2^3 = 1\right\}$ & $(C_3\times C_3)\rtimes C_2$ \\[10pt]
& & & \\
\end{tabular}
\end{center}
We note that the semidirect product $H_4 = K_4 = (C_3\times C_3)\rtimes C_2$ (where $C_2$ acts on $C_3\times C_3$ by interchanging the factors) is in fact isomorphic to $C_3 \times S_3$ (where $S_3$ denotes the group of permutations on $3$ letters) if we make the identifications
\[ C_3 \simeq \left\{\begin{pmatrix} \xi & 0 \\ 0 & \xi \end{pmatrix}:\xi^3 = 1\right\} \mbox{ and }S_3\simeq\left\{\begin{pmatrix} \xi & 0 \\ 0 & \xi^{-1} \end{pmatrix}:\xi^3 = 1\right\} \bigcup \left\{\begin{pmatrix} 0 & \xi \\ \xi^{-1} & 0 \end{pmatrix}:\xi^3 = 1\right\}.\]
The irreducible representations of $C_3$ are all $1$-dimensional (and there are three of them), while $S_3$ has one $2$-dimensional irreducible representation, and two $1$-dimensional. By tensoring the irreducible representations of $C_3$ with those of $S_3$ we conclude that $H_4\simeq C_3\times S_3$ has three $2$-dimensional irreducible representations, and six $1$-dimensional ones. According to Theorem~\ref{thm:simples}, we obtain the following classification for the simple $\GL$-equivariant $\D$-modules on $X$:
\begin{itemize}
\item $6$ modules with full support, corresponding to the $1$-dimensional representations of~$H_4$.
\item $3$ modules with full support, corresponding to the $2$-dimensional irreducible representations of $H_4$.
\item $1$ module with support $\ol{O_3}$.
\item $3$ modules with support $\ol{O_2}$, corresponding to the $1$-dimensional representations of~$H_2$.
\item $1$ module with support $O_0$.
\end{itemize}
Our next goal is to describe the characters of the $14$ simple modules listed above and establish some basic links between them. We have discussed the character of $E=\mc{L}(O_0;X)$ in (\ref{eq:character-E}). The characters of the $3$ modules with support $\ol{O_2}$ have been computed in \cite{raicu-veronese} and we start by recalling their description below.
\subsection{$\D$-modules supported on $\ol{O_2}$ (the cone over the twisted cubic)}\label{subsec:Dmods-on-twistedcubic}
We consider the collection of integers $(\nu_i)_{i\in\bb{Z}}$ encoded by the generating function
\[ \sum_{i\in\bb{Z}} \nu_i \cdot t^i = \frac{1}{(1-t^2)(1-t^3)} = 1 + t^2 + t^3 + t^4 + t^5 + 2t^6 + t^7 + 2t^8 + 2t^9 + 2t^{10} + \cdots\]
and define for each dominant weight $\ll=(\ll_1,\ll_2)$ the integers
\begin{equation}\label{eq:mll-ell}
m_{\ll} = \nu_{\ll_1-5} - \nu_{\ll_2-6}\mbox{ and }e_{\ll} = \langle [E],e^{\ll^{\vee}} \rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [S],e^{\ll-(6,6)} \rangle.
\end{equation}
For each $j=0,1,2$ we set
\begin{equation}\label{eq:allj=0}
a_{\ll}^j = 0\mbox{ if }\ll_1 + \ll_2 \not\equiv j\ (\mbox{mod }3)
\end{equation}
and otherwise let
\begin{equation}\label{eq:allj-general}
a_{\ll}^j=
\begin{cases}
m_{\ll}+e_{\ll} & \mbox{if }j=0; \\
m_{\ll} & \mbox{if }j=1,2. \\
\end{cases}
\end{equation}
With this notation, \cite[Thm.~1.2]{raicu-veronese} asserts that the characters of the simple $\GL$-equivariant $\D$-modules $D_0,D_1,D_2$ supported on the twisted cubic are
\begin{equation}\label{eq:chars-Dj}
[D_j] = \sum_{\ll} a_{\ll}^j \cdot e^{\ll^{\vee}} \mbox{ for }j=0,1,2.
\end{equation}
Notice the use of the dual weights $\ll^{\vee}$ in the above formula, which accounts for the fact that $e^{\ll}$ denotes the class of $\S_{\ll}V$ in $\Gamma(\GL)$ and not that of $\S_{\ll}W$! Notice also that (\ref{eq:allj=0}) and (\ref{eq:chars-Dj}) imply that
\begin{equation}\label{eq:wts-D012}
\mbox{if }\langle [D_j],e^{\ll} \rangle \neq 0\mbox{ then }\ll_1 + \ll_2 + j \equiv 0\ (\mbox{mod }3).
\end{equation}
\begin{example}\label{ex:wts-D0}
It will be useful later to know that
\begin{equation}\label{eq:wts-D0}
\langle [D_0],e^{(-1,-5)} \rangle = \langle [D_0],e^{(-6,-9)} \rangle = 1 \mbox{ and } \langle [D_0], e^{(3,0)} \rangle = \langle [D_0], e^{(-2,-4)} \rangle = 0.
\end{equation}
This is equivalent to proving that $a^0_{(5,1)} = a^0_{(9,6)} = 1$ and $a^0_{(0,-3)} = a^0_{(4,2)} = 0$. We have that
\[m_{(5,1)} = \nu_0 - \nu_{-5} = 1,\ m_{(9,6)} = \nu_4 - \nu_0 = 0,\ m_{(0,-3)} = \nu_{-5} - \nu_{-9} = 0\mbox{ and }m_{(4,2)} = \nu_{-1} - \nu_{-4} = 0.\]
It follows from (\ref{eq:character-E}) that if $\langle [E],e^{(a,b)} \rangle \neq 0$ then $a,b\leq -6$, so
\[ e_{(5,1)} = \langle [E],e^{(-1,-5)} \rangle = 0,\ e_{(0,-3)} = \langle [E],e^{(3,0)} \rangle = 0\mbox{ and }e_{(4,2)} = \langle [E],e^{(-2,-4)} \rangle = 0.\]
We have moreover that
\[ e_{(9,6)} = \langle [S],e^{(3,0)} \rangle = \left\langle \frac{1+e^{(6,3)}}{(1-e^{(3,0)})(1-e^{(4,2)})(1-e^{(6,6)})}, e^{(3,0)} \right\rangle = 1.\]
We conclude that
\[a^0_{(5,1)} = m_{(5,1)} + e_{(5,1)} = 1,\ a^0_{(9,6)} = m_{(9,6)} + e_{(9,6)} = 1,\mbox{ and }\]
\[a^0_{(0,-3)} = m_{(0,-3)} + e_{(0,-3)} = 0,\mbox{ and }a^0_{(4,2)} = m_{(4,2)} + e_{(4,2)}=0,\]
as desired.
\end{example}
\begin{example}\label{ex:wts-D12}
It will also be useful later to know that
\begin{equation}\label{eq:wts-D12}
\langle [D_2], e^{(-5,-9)} \rangle = 1 \mbox{ and } \langle [D_1], e^{(3,-1)}\rangle = 0.
\end{equation}
This is equivalent to proving that $a_{(9,5)}^1 = 1$ and $a_{(1,-3)}^2=0$, which in turn follow from
\[ m_{(9,5)} = \nu_4 - \nu_{-1} = 1\mbox{ and }m_{(1,-3)} = \nu_{-5} - \nu_{-9} = 0.\]
\end{example}
\begin{lemma}\label{lem:maa}
The values of $m_{(a,a)}$ when $a\in\bb{Z}$ are as follows:
\begin{equation}\label{eq:maa}
m_{(a,a)} = \begin{cases}
-1 & \mbox{if }a \equiv 0\ (\mbox{mod }6),\ a\geq 6 \\
1 & \mbox{if }a \equiv \pm 1\ (\mbox{mod }6),\ a\geq 5 \\
0 & \rm{otherwise}.
\end{cases}
\end{equation}
\end{lemma}
\begin{proof} Since $m_{(a,a)} = \nu_{a-5}-\nu_{a-6}$, it follows that $m_{(a,a)}=0$ for $a<5$. To compute $m_{(a,a)}$ for $a\geq 5$ we consider the generating function
\[\sum_{i\geq 0} (\nu_i - \nu_{i-1})\cdot t^i = (1-t)\cdot \sum_{i\geq 0} \nu_i\cdot t^i = \frac{1}{(1+t)(1-t^3)} = \frac{1-t+t^2}{1-t^6} = (1-t+t^2) + (t^6-t^7+t^8) + \cdots \]
from which the formula for $m_{(a,a)}$ follows.
\end{proof}
Combining Lemma~\ref{lem:maa} with (\ref{eq:chars-Dj}) we conclude that for $a\in\bb{Z}$ we have
\begin{equation}\label{eq:Dj-sl2-inv}
\langle D_1,e^{(a,a)}\rangle = \begin{cases}
1 & \mbox{if }a \equiv 1\ (\mbox{mod }6),\ a\leq -5 \\
0 & \rm{otherwise}.
\end{cases}
\qquad
\langle D_2,e^{(a,a)}\rangle = \begin{cases}
1 & \mbox{if }a \equiv -1\ (\mbox{mod }6),\ a\leq -7 \\
0 & \rm{otherwise}.
\end{cases}
\end{equation}
In particular, both $D_1, D_2$ contain invariant sections for the action of the special linear group $\SL \subset \GL$. In contrast, $D_0$ has no $\SL$-invariant sections as seen by the following.
\begin{lemma}\label{lem:D0-sl2-inv}
For every integer $a\in\bb{Z}$ we have
\begin{equation}\label{eq:D0-sl2-inv}
\langle [D_0], e^{(a,a)} \rangle = 0.
\end{equation}
\end{lemma}
\begin{proof}
Using (\ref{eq:allj=0}), it suffices to consider the case when $a\in 3\bb{Z}$. Using (\ref{eq:allj-general}), and replacing $a$ by $-a$, it is then enough to show that $m_{(a,a)}+e_{(a,a)}=0$ for all $a\in 3\bb{Z}$ which follows from
\[ m_{(a,a)} + e_{(a,a)} \overset{(\ref{eq:mll-ell})}{=} m_{(a,a)} + \langle [S],e^{(a-6,a-6)}\rangle \overset{(\ref{eq:character-S}),(\ref{eq:maa})}{=} \begin{cases}
-1 + 1 = 0& \mbox{if }a \equiv 0\ (\mbox{mod }6),\ a\geq 6 \\
0 + 0 = 0 & \mbox{otherwise}.
\end{cases}
\qedhere\]
\end{proof}
Equipped with the description of the characters of $D_0,D_1,D_2$, we next prove that none of them appears as a composition factor in the localization $S_{\Delta}$. This will be important later in our analysis of the remaining simple $\D$-modules.
\begin{lemma}\label{lem:D12-notin-Sdelta}
$D_1,D_2$ do not appear as composition factors of $S_{\Delta}$.
\end{lemma}
\begin{proof}
It follows from (\ref{eq:character-Sdelta}) that if $\langle [S_{\Delta}],e^{\ll}\rangle \neq 0$ then $\ll_1 + \ll_2$ is divisible by $3$, so (\ref{eq:wts-D012}) implies that $\langle [D_1],e^{\ll}\rangle = \langle [D_2],e^{\ll}\rangle = 0$ showing that $D_1,D_2$ cannot occur as subrepresentations of $S_{\Delta}$ and therefore they are not composition factors of $S_{\Delta}$.
\end{proof}
\begin{lemma}\label{lem:D0-notin-Sdelta}
$D_0$ does not appear as a composition factor of $S_{\Delta}$.
\end{lemma}
\begin{proof}
As shown in Example~\ref{ex:wts-D0}, we have $\langle [D_0], e^{(-1,-5)} \rangle = 1$, so to prove the result it suffices to verify that
\begin{equation}\label{eq:wt-1-5-notSdelta}
\langle [S_{\Delta}], e^{(-1,-5)}\rangle = 0.
\end{equation}
Assuming this isn't the case, it follows from (\ref{eq:character-Sdelta}) that we can find integers $0\leq a\leq 1$, $b,c\geq 0$ and $t\in\bb{Z}$ such that
\begin{equation}\label{eq:wt-1-5}
(-1,-5) = a\cdot(6,3) + b\cdot(3,0) + c\cdot(4,2) + t\cdot(6,6).
\end{equation}
Considering the difference between the first and second components of the above weights we get that
\[ 4 = 3a + 3b + 2c\]
which has a unique solution satisfying $0\leq a\leq 1$, $b,c\geq 0$, namely $a=b=0$ and $c=2$. The equation (\ref{eq:wt-1-5}) becomes $(-1,-5) = (8+6t,4+6t)$ which has no integer solution. We get a contradiction, showing (\ref{eq:wt-1-5-notSdelta}) and concluding our proof.
\end{proof}
\subsection{The remaining simple equivariant $\D$-modules}
We are left with describing the character of $P$, as well as those of the simple $\D$-modules with full support. We first identify $P$ as a composition factor of $S_{\Delta}$.
\begin{lemma}\label{lem:P-in-Sdelta}
The $\D$-module $P$ appears with multiplicity one as a composition factor of $S_{\Delta}$.
\end{lemma}
\begin{proof}
Since $S_{\Delta}/S$ does not have full support, $S$ is the only composition factor of $S_{\Delta}$ with full support. It follows from the classification and Lemma~\ref{lem:D0-notin-Sdelta} that the remaining composition factors of $S_{\Delta}/S$ must be among $P,D_1,D_2,E$. Consider the dominant weight $(3,-3)$ and note that in order to compute $\langle [S_{\Delta}],e^{(3,-3)} \rangle$ we need to solve, as in the proof of Lemma~\ref{lem:D0-notin-Sdelta}, the equation
\begin{equation}\label{eq:wt3-3}
(3,-3) = a\cdot(6,3) + b\cdot(3,0) + c\cdot(4,2) + t\cdot(6,6),\mbox{ where }0\leq a\leq 1,\ b,c\geq 0\mbox{ and }t\in\bb{Z}.
\end{equation}
Every solution has $6=3a+3b+2c$, which in turn implies that $(a,b,c)$ is one of $(1,1,0)$, $(0,2,0)$, $(0,0,3)$. It follows that there is only one integer solution of (\ref{eq:wt3-3}), namely $(a,b,c,t) = (1,1,0,-1)$, and therefore
\[\langle [S_{\Delta}],e^{(3,-3)} \rangle = 1.\]
Using (\ref{eq:character-S}), (\ref{eq:character-E}) and (\ref{eq:wts-D012}) we conclude that
\[ \langle [S],e^{(3,-3)} \rangle = \langle [E],e^{(3,-3)} \rangle = \langle [D_1],e^{(3,-3)} \rangle = \langle [D_2],e^{(3,-3)} \rangle = 0,\]
so the only composition factor of $S_{\Delta}$ that may contain $\S_{(3,-3)}V$ is $P$. Since the multiplicity of $\S_{(3,-3)}V$ in $S_{\Delta}$ is $1$, the same must be true about the multiplicity of $P$ as a composition factor of $S_{\Delta}$.
\end{proof}
\begin{remark}\label{rem:P-in-Sdelta}
Since $S_{\Delta}/S = H^1_{\ol{O_3}}(X,\mc{O}_X)$ it follows that $P=\mc{L}(\ol{O}_3,X)$ appears as a submodule of $S_{\Delta}/S$ and that the quotient $(S_{\Delta}/S)/P$ is supported on a proper closed subset of $\ol{O_3}$. Using Lemma~\ref{lem:D0-notin-Sdelta} and combining (\ref{eq:character-Sdelta}) with (\ref{eq:wts-D012}) we get that $D_0,D_1,D_2$ are not composition factors of $S_{\Delta}$, hence $(S_{\Delta}/S)/P$ must have support in $O_0$. It is not clear a priori that $(S_{\Delta}/S)/P\neq 0$, but as we'll see shortly we have that $(S_{\Delta}/S)/P=E$.
\end{remark}
\begin{lemma}\label{lem:FD0-full}
The Fourier transform $\mc{F}(D_0)$ is different from $P,D_0,D_1,D_2,E$ and hence it is a simple $\GL$-equivariant $\D$-module with full support.
\end{lemma}
\begin{proof}
Since the weight $\ll=(-1,-5)$ satisfies $\ll = \ll^{\vee} - (6,6)$, it follows that
\[\langle [D_0],e^{(-1,-5)}\rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [\mc{F}(D_0)],e^{(-1,-5)}\rangle \overset{(\ref{eq:wts-D0})}{=} 1.\]
Equation (\ref{eq:wt-1-5-notSdelta}) shows that $\mc{F}(D_0)$ does not appear as a composition factor of $S_{\Delta}$, hence $\mc{F}(D_0)\neq P$. We have
\[\langle [\mc{F}(D_0)],e^{(3,0)}\rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [D_0],e^{(-6,-9)}\rangle \overset{(\ref{eq:wts-D0})}{=} 1,\]
and since $\langle [D_0],e^{(3,0)}\rangle = 0$ by (\ref{eq:wts-D0}), we conclude that $\mc{F}(D_0) \neq D_0$. The fact that $\mc{F}(D_0) \neq D_1,D_2$ follows from (\ref{eq:wts-D012}), and the fact that $\mc{F}(D_0)\neq E$ follows from $\langle [E],e^{(3,0)}\rangle = 0$ which is a consequence of (\ref{eq:character-E}).
\end{proof}
\begin{lemma}\label{lem:FD0-delta}
The $\D$-module composition factors of $\mc{F}(D_0)_{\Delta}$ are among $D_0, P, \mc{F}(D_0)$. Moreover, both~$\mc{F}(D_0)$ and~$P$ appear as composition factors with multiplicity one.
\end{lemma}
\begin{proof}
We start by noting that since $\mc{F}(D_0)$ is simple and has full support, it is torsion free as an $S$-module. In particular $\Delta$ is a non-zerodivisor on $\mc{F}(D_0)$ and we get using Lemma~\ref{lem:localization} that
\[ [\mc{F}(D_0)_{\Delta}] = \lim_{n\to\infty} \left([\mc{F}(D_0)] \cdot e^{(-6n,-6n)}\right).\]
We also note that $\mc{F}(D_0)$ is a submodule of $\mc{F}(D_0)_{\Delta}$ and that the quotient $\mc{F}(D_0)_{\Delta}/\mc{F}(D_0)$ has support contained in $\ol{O_3}$, which means that the composition factors of $\mc{F}(D_0)_{\Delta}/\mc{F}(D_0)$ are among $P,D_0,D_1,D_2,E$, and in particular $\mc{F}(D_0)$ appears with multiplicity one.
Since $D_0$ has no $\SL$-invariant sections by Lemma~\ref{lem:D0-sl2-inv}, the same must be true about $\mc{F}(D_0)$ and thus also about $\mc{F}(D_0)_{\Delta}$. Since $E,D_1,D_2$ have $\SL$-invariant sections by (\ref{eq:character-E}) and (\ref{eq:Dj-sl2-inv}), it follows that they cannot appear as composition factors of $\mc{F}(D_0)_{\Delta}$, so the only remaining potential composition factors are~$P$ and~$D_0$. To conclude it remains to show that $P$ does appear as a composition factor and that it has multiplicity one.
We have
\[ \langle [\mc{F}(D_0)],e^{(-2,-4)} \rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [D_0],e^{(-2,-4)} \rangle \overset{(\ref{eq:wts-D0})}{=} 0, \]
so $D_0$ and $\mc{F}(D_0)$ do not contain $\S_{(-2,-4)}V$ as a subrepresentation. Moreover, using Lemma~\ref{lem:localization} we get
\[
\begin{aligned}
\langle [\mc{F}(D_0)_{\Delta}],e^{(-2,-4)} \rangle &= \lim_{n\to\infty} \langle [\mc{F}(D_0)],e^{(6n-2,6n-4)} \rangle = \lim_{n\to\infty} \langle [D_0],e^{(-2-6n,-4-6n)} \rangle \\
&= \lim_{n\to\infty} (m_{(6n+4,6n+2)} + e_{(6n+4,6n+2)})
\end{aligned}
\]
We have $e_{(6n+4,6n+2)} = \langle [S], e^{6n-2,6n-4} \rangle = 1$ for all $n\geq 1$, and $m_{(6n+4,6n+2)} = \nu_{6n-1} - \nu_{6n-4}$ which can be computed from the generating function
\[ \sum_{i\geq 0} (\nu_i - \nu_{i-3})\cdot t^i = \frac{1-t^3}{(1-t^2)(1-t^3)} = \frac{1}{1-t^2} = 1+t^2 + t^4 + t^6 \cdots\]
Taking $i=6n-1$ we get that $\nu_i - \nu_{i-4} = 0$ and therefore $m_{(6n+4,6n+2)} = 0$ for all $n$. We conclude that
\begin{equation}\label{eq:FD0-delta-2-4}
\langle [\mc{F}(D_0)_{\Delta}],e^{(-2,-4)} \rangle=1,
\end{equation}
so there is a composition factor of $\mc{F}(D_0)_{\Delta}$ containing $\S_{(-2,-4)}V$ as a subrepresentation. Since $D_0$ and $\mc{F}(D_0)$ do not have this property, this composition factor must then be $P$, occurring with multiplicity one as desired.
\end{proof}
\begin{corollary}\label{cor:factors-Sdelta}
The module $P$ has no $\SL$-invariant sections, and is therefore not isomorphic to $S_{\Delta}/S$. In fact, $S_{\Delta}/S$ has length two as a $\D$-module, with composition factors $P$ and $E$.
\end{corollary}
\begin{proof}
It follows from Lemma~\ref{lem:FD0-delta} that $P$ is a composition factor of $\mc{F}(D_0)_{\Delta}$. Since $\mc{F}(D_0)_{\Delta}$ has no $\SL$-invariant sections, the same must be true about $P$, but
\[\langle [S_{\Delta}/S], e^{(-6,-6)}\rangle = \langle [S_{\Delta}], e^{(-6,-6)}\rangle - \langle [S], e^{(-6,-6)}\rangle \overset{(\ref{eq:character-S}),(\ref{eq:character-Sdelta})}{=} 1 - 0 = 1,\]
and therefore $P\neq S_{\Delta}/S$. Using Remark~\ref{rem:P-in-Sdelta} and the fact that $\langle [S_{\Delta}/S], e^{(-6,-6)}\rangle = \langle [E], e^{(-6,-6)}\rangle = 1$ we conclude that $(S_{\Delta}/S)/P = E$.
\end{proof}
As a consequence of Corollary~\ref{cor:factors-Sdelta} and using (\ref{eq:character-S}), (\ref{eq:character-Sdelta}), (\ref{eq:character-E}), we determine the character of $P$ via
\[[P] = [S_{\Delta}] - [S] - [E].\]
\begin{corollary}\label{cor:factors-FD0-delta}
The $\D$-module $\mc{F}(D_0)_{\Delta}$ has length three, with composition factors $D_0, P, \mc{F}(D_0)$, each appearing with multiplicity one.
\end{corollary}
\begin{proof} We already know that $P$ and $\mc{F}(D_0)$ appear as composition factors with multiplicity one, so we need to show that the same is true for $D_0$. We consider the weight $\ll=(-6,-9)$ and observe using (\ref{eq:character-S}), (\ref{eq:character-Sdelta}) and (\ref{eq:character-E}) that
\[ \langle [S], e^{(-6,-9)} \rangle = 0\quad\mbox{and}\quad\langle [S_{\Delta}], e^{(-6,-9)} \rangle = \langle [E], e^{(-6,-9)} \rangle = 1\]
which based on Corollary~\ref{cor:factors-Sdelta} implies that
\[ \langle [P], e^{(-6,-9)} \rangle = \langle [S_{\Delta}], e^{(-6,-9)} \rangle - \langle [S], e^{(-6,-9)} \rangle - \langle [E], e^{(-6,-9)} \rangle = 0.\]
We have
\[ \langle [D_0],e^{(-6,-9)} \rangle \overset{(\ref{eq:wts-D0})}{=} 1 \mbox{ and } \langle [\mc{F}(D_0)],e^{(-6,-9)} \rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [D_0],e^{(3,0)} \rangle \overset{(\ref{eq:wts-D0})}{=} 0.\]
Since $P$ and $\mc{F}(D_0)$ do not contain $\S_{(-6,-9)}V$ as a subrepresentation, and $D_0$ contains it with multiplicity one, it follows that the multiplicity of $D_0$ as a composition factor of $\mc{F}(D_0)_{\Delta}$ is equal to
\[\langle [\mc{F}(D_0)_{\Delta}],e^{(-6,-9)} \rangle = \lim_{n\to\infty} \langle [\mc{F}(D_0)],e^{(6n-6,6n-9)} \rangle = \lim_{n\to\infty} \langle [D_0],e^{(3-6n,-6n)} \rangle = \lim_{n\to\infty} (m_{(6n,6n-3)} + e_{(6n,6n-3)})\]
We have $e_{(6n,6n-3)} = \langle [S], e^{6n-6,6n-9} \rangle = 1$ for all $n\geq 1$, and $m_{(6n,6n-3)} = \nu_{6n-5} - \nu_{6n-9}$ which can be computed from the generating function
\[ \sum_{i\geq 0} (\nu_i - \nu_{i-4})\cdot t^i = \frac{1-t^4}{(1-t^2)(1-t^3)} = \frac{1+t^2}{1-t^3} = (1+t^2) + (t^3 + t^5) + (t^6 + t^8) + \cdots\]
Taking $i=6n-5$ we get that $\nu_i - \nu_{i-4} = 0$ and therefore $m_{(6n,6n-3)} = 0$ for all $n$. We conclude that $\langle [\mc{F}(D_0)_{\Delta}],e^{(-6,-9)} \rangle = 1$, so that $D_0$ appears with multiplicity one as a composition factor of $\mc{F}(D_0)_{\Delta}$, concluding the proof.
\end{proof}
\begin{lemma}\label{lem:FD12-full}
The Fourier transforms $\mc{F}(D_1)$ and $\mc{F}(D_2)$ are simple $\GL$-equivariant $\D$-modules with full support.
\end{lemma}
\begin{proof}
Just as in the proof of Lemma~\ref{lem:FD0-full}, it suffices to verify that $\mc{F}(D_1)$ and $\mc{F}(D_2)$ are not among the modules $P,D_0,D_1,D_2,E$. It follows from (\ref{eq:wts-Fourier}) and (\ref{eq:wts-D012}) that
\[ \mbox{if }\langle [\mc{F}(D_j)],e^{\ll} \rangle \neq 0\mbox{ then }\ll_1 + \ll_2 \equiv j \ (\mbox{mod }3),\]
so we only need to verify that $\mc{F}(D_1)\neq D_2$ and $\mc{F}(D_2)\neq D_1$. Since $\mc{F}$ is an involution, these two assertions are equivalent, so we only verify the first one. We have
\[ \langle [\mc{F}(D_2)], e^{(3,-1)}\rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [D_2], e^{(-5,-9)} \rangle \overset{(\ref{eq:wts-D12})}{=} 1\mbox{ and }\langle [D_1], e^{(3,-1)} \rangle \overset{(\ref{eq:wts-D12})}{=} 0\]
so $\mc{F}(D_2)\neq D_1$, concluding the proof.
\end{proof}
We write $Q_0 = \mc{F}(D_0)$ and let $Q_j = (Q_0)_{\Delta}\cdot \Delta^{j/3}$ for $j=1,2$. We have that $Q_0,Q_1,Q_2$ are $\D$-modules with full support and no $\SL$-invariant sections, and $Q_0$ is simple because the Fourier transform preserves simplicity. Our aim is to show that $Q_1$ and $Q_2$ are also simple, but for now we only show the following.
\begin{lemma}\label{lem:Q12-factors}
All of the composition factors of $Q_1,Q_2$ have full support, or equivalently, none of the modules $E,P,D_0,D_1,D_2$ appear as composition factors in $Q_1,Q_2$.
\end{lemma}
\begin{proof} Since $[\bb{C}\cdot\Delta^{j/3}] = e^{(2j,2j)}$ it follows that
\begin{equation}\label{eq:wts-Qj}
\mbox{if }\langle [Q_j],e^{\ll} \rangle \neq 0\mbox{ then }\ll_1 + \ll_2 \equiv j \ (\mbox{mod }3),
\end{equation}
so it suffices to check that $D_1$ is not a composition factor of $Q_2$, and that $D_2$ is not a composition factor of~$Q_1$. To see this it suffices to observe that by (\ref{eq:Dj-sl2-inv}) the modules $D_1,D_2$ have non-zero $\SL$-invariant sections, whereas $Q_1$ and $Q_2$ do not.
\end{proof}
We define as in the introduction $F_i = S_{\Delta}\cdot \Delta^{i/6}$ for $i=-1,0,1,2,3,4$, and let $G_i \subseteq F_i$ denote the $\D$-submodule generated by $\Delta^{i/6}$.
\begin{lemma}\label{lem:G0234}
We have that $G_i = F_i$ for $i=2,3,4$. Moreover, $G_i$ is a simple $\D$-module for $i=0,2,3,4$.
\end{lemma}
\begin{proof}We have that $G_0 = S = \mc{L}(X;X)$ is a simple $\D$-module, so we only need to focus on the cases $i=2,3,4$. Recall from \cite[Example~9.1]{b-functions} that the $b$-function of $\Delta$ is
\[b_{\Delta}(s) = (s+1)^2\cdot(s+5/6)\cdot(s+7/6)\]
hence no integer translate of $i/6$ is a root of $b_{\Delta}(s)$ for $i=2,3,4$. The argument in \cite[Section~7]{raicu-matrices} with $f=\Delta$ applies to show that $G_i = F_i$ for $i=2,3,4$, and that each such $G_i$ is a simple $\D$-module.
\end{proof}
\begin{corollary}\label{cor:simples-full-support}
The 9 simple equivariant $\D$-modules on $X = \Sym^3 W$ having full support are
\[G_0,G_2,G_3,G_4,\mc{F}(D_1),\mc{F}(D_2), Q_0,Q_1,Q_2.\]
Moreover, we have that $G_1=\mc{F}(D_2)$, $G_{-1}=\mc{F}(D_1)$, $D_1 = F_1/G_1$ and $D_2 = F_{-1}/G_{-1}$.
\end{corollary}
\begin{proof}
We have already seen that $G_0,G_2,G_3,G_4,\mc{F}(D_1),\mc{F}(D_2),Q_0=\mc{F}(D_0)$ are simple. For $j=1,2$ let $Q_j'$ denote any of the composition factors of $Q_j$, which by Lemma~\ref{lem:Q12-factors} has full support. It follows from (\ref{eq:wts-Qj}) that the modules $Q_0,Q_1',Q_2'$ are mutually distinct. Moreover, they are distinct from $G_0,G_2,G_3,G_4,\mc{F}(D_1),\mc{F}(D_2)$ since the latter contain non-zero $\SL$-invariant sections, whereas $Q_0,Q_1',Q_2'$ do not. It follows that
\begin{equation}\label{eq:simples-full-support}
G_0,G_2,G_3,G_4,\mc{F}(D_1),\mc{F}(D_2), Q_0,Q_1',Q_2'
\end{equation}
are all the 9 simple equivariant $\D$-modules on $X = \Sym^3 W$ having full support. Since for $j=1,2$ the modules $Q_j'$ were chosen arbitrarily among the composition factors of $Q_j$, and since the list (\ref{eq:simples-full-support}) is independent on these choices, we conclude that in fact each $Q_j$ only has one composition factor (possibly with multiplicity bigger than one). We have seen in (\ref{eq:FD0-delta-2-4}) that $\langle [Q_0],e^{(-2,-4)}\rangle = 1$, which implies that $\langle [Q_1],e^{(0,-2)}\rangle = 1$ and $\langle [Q_2],e^{(2,0)}\rangle = 1$, and thus the composition factors of $Q_1$ and $Q_2$ cannot appear with multiplicity bigger than one. This means that $Q_1' = Q_1$ and $Q_2'=Q_2$, proving our first assertion.
The $\D$-modules $F_1$ and $F_{-1}$ are torsion free, so they contain a simple submodule with full support. Looking at the congruence of $\ll_1 + \ll_2$ modulo $3$ for the representations $\S_{\ll}V$ contained in $F_1$ and $F_{-1}$, and comparing with the list of simple equivariant $\D$-modules that we obtained, we conclude that
\begin{itemize}
\item $\mc{F}(D_2)$ and $Q_2$ are the only possible submodules of $F_1$ with full support, and $D_1$ is the only possible composition factor of $F_1$ that does not have full support.
\item $\mc{F}(D_1)$ and $Q_1$ are the only possible submodules of $F_{-1}$ with full support, and $D_2$ is the only possible composition factor of $F_{-1}$ that does not have full support.
\end{itemize}
Since $F_1 = S_{\Delta}\cdot\Delta^{1/6}$ it follows from (\ref{eq:character-Sdelta}) that
\begin{equation}\label{eq:character-F1}
[F_1] = \frac{1+e^{(6,3)}}{(1-e^{(3,0)})(1-e^{(4,2)})} \cdot e^{(6,6)\bb{Z}+(1,1)}
\end{equation}
and in particular $\langle [F_1],e^{(a+2,a)} \rangle \neq 0$ if and only if $a \equiv 3\ (\mbox{mod }6)$. This shows that $\langle [F_1],e^{(2,0)} \rangle = 0$, and since $\langle [Q_2],e^{(2,0)}\rangle = 1$ we conclude that $Q_2$ is not a submodule of $F_1$. It follows that $\mc{F}(D_2)$ is a submodule of $F_1$, and in fact $\mc{F}(D_2)$ has multiplicity one as a composition factor of $F_1$, due to the fact that
\[\langle [\mc{F}(D_2)], e^{(1,1)} \rangle \overset{(\ref{eq:wts-Fourier}),(\ref{eq:Dj-sl2-inv})}{=} 1\quad\mbox{ and }\quad \langle [F_1], e^{(1,1)} \rangle = 1.\]
Since $\bb{C}\cdot\Delta^{1/6}$ is the unique copy of $\S_{(1,1)}V$ inside $F_1$, it follows that $\Delta^{1/6}\in \mc{F}(D_2)$. Since $\mc{F}(D_2)$ is simple, it is generated by $\Delta^{1/6}$ as a submodule of $F_1$, hence $\mc{F}(D_2) = G_1$. Now the quotient $F_1/G_1$ can only have $D_1$ as a composition factor, and $D_1$ will appear with multiplicity one since
\[ \langle [F_1], e^{(-5,-5)} \rangle \overset{(\ref{eq:character-F1})}{=} 1 \overset{(\ref{eq:Dj-sl2-inv})}{=} \langle [D_1], e^{(-5,-5)} \rangle\quad\mbox{and}\quad\langle [G_1], e^{(-5,-5)} \rangle = \langle [\mc{F}(D_2)], e^{(-5,-5)} \rangle \overset{(\ref{eq:wts-Fourier}),(\ref{eq:Dj-sl2-inv})}{=} 0.\]
This shows that $F_1/G_1 = D_1$, as desired. The proof that $G_{-1} = \mc{F}(D_1)$ and that $F_{-1}/G_{-1}$ is completely analogous, and we leave the details to the interested reader.
\end{proof}
\begin{remark}
Since the modules $G_i$, $i=-1,0,1,2,3,4$, have holonomic rank one (when restricted to the dense orbit $O_4$ they are isomorphic to $F_i$, which are free $S_{\Delta}$-modules of rank one), they correspond via the classification in Section~\ref{subsec:classification} to the $1$-dimensional representations of the component group $H_4\simeq C_3\times S_3$. The remaining simple $\D$-modules with full support, $Q_0,Q_1,Q_2$, correspond to the $2$-dimensional representations of $H_4$, and in particular they have holonomic rank two, which is perhaps not immediately clear from their construction.
\end{remark}
At this point we can describe the characters of all the remaining simple $\D$-modules. Using (\ref{eq:chars-Dj}) and (\ref{eq:wts-Fourier}) we obtain from Corollary~\ref{cor:simples-full-support} explicit formulas for $[G_1] = \mc{F}[D_2]$ and $[G_{-1}] = \mc{F}[D_1]$. Since $[G_i] = [S_{\Delta}] \cdot e^{(i,i)}$ for $i=2,3,4$, they can be computed from (\ref{eq:character-Sdelta}), while the character of $G_0=S$ has been determined in (\ref{eq:character-S}). The character of $Q_0$ is $[Q_0]=\mc{F}[D_0]$, while the characters of $Q_1,Q_2$ are determined by $[Q_i] = [Q_0]\cdot e^{(2i,2i)}$ for $i=1,2$.
We end this section by analyzing some non-simple $\D$-modules. In Section~\ref{sec:category-modDx} we will study from a quiver perspective the indecomposable objects in the category of equivariant $\D$-modules. Two such example is given below.
\begin{lemma}\label{lem:indQ0-del}
The module $D_0$ is not isomorphic to a submodule of $(Q_0)_\Delta / Q_0$, and in particular $(Q_0)_\Delta / Q_0$ is indecomposable.
\end{lemma}
\begin{proof}
Recall from Corollary \ref{cor:factors-FD0-delta} that $(Q_0)_\Delta/Q_0$ has length two with composition factors $P$ and $D_0$. If it were decomposable, it would be isomorphic to $P\oplus D_0$, and in particular $D_0$ would be a submodule of $(Q_0)_\Delta / Q_0$. It is then enough to prove that this is not the case.
Assume by contradiction that $D_0\subset (Q_0)_\Delta/Q_0$, let $\lambda= (-6,-9)$ and recall that $\langle [D_0],e^{\lambda} \rangle \overset{(\ref{eq:wts-D0})}{=} 1$. Using (\ref{eq:allj-general}) we obtain
\begin{equation}\label{eq:wts-3-6D0}
\langle [D_0], e^{(-3,-6)}\rangle= 0 \mbox{ and } \langle [D_0], e^{(0,-3)} \rangle = 0.
\end{equation}
Let $v_\lambda\in D_0 \subset (Q_0)_\Delta/Q_0$ be a highest weight vector of weight $\lambda$ and choose a lift of $v_\lambda$ to a highest weight vector $\tilde{v}_\lambda \in (Q_0)_\Delta$. Since $\Delta$ has weight $(6,6)$ and $\langle [D_0], e^{\ll+(6,6)} \rangle = \langle [D_0], e^{(0,-3)} \rangle = 0$, we conclude that $\Delta \cdot v_\lambda = 0$ and therefore $\Delta \cdot \tilde{v}_\lambda \in Q_0$. We have however that
\[\langle [Q_0], e^{\ll+(6,6)} \rangle = \langle [Q_0], e^{(0,-3)} \rangle \overset{(\ref{eq:wts-Fourier})}{=} \langle [D_0], e^{(-3,-6)} \rangle \overset{(\ref{eq:wts-3-6D0}) }{=} 0\]
which implies that $\Delta \cdot \tilde{v}_{\ll}=0$. This is a contradiction since $\Delta$ is a non-zero divisor on $(Q_0)_\Delta$, concluding our proof.
\end{proof}
\begin{lemma}\label{lem:indS-del}
The module $E$ is not isomorphic to a submodule of $S_\Delta / S$, and in particular $S_\Delta / S$ is indecomposable.
\end{lemma}
\begin{proof}
By Corollary~\ref{cor:factors-Sdelta}, $S_{\Delta}/S$ has length two, with composition factors $P$ and $E$, so we can prove our result using the same strategy as in Lemma~\ref{lem:indQ0-del}. For that it suffices to take again $\ll = (-6,-9)$ and use
\[ \langle [E], e^{\ll}\rangle = 1,\mbox{ and } \langle [E], e^{\ll+(6,6)}\rangle = \langle [S], e^{\ll+(6,6)}\rangle = 0.\qedhere\]
\end{proof}
\section{The category of equivariant coherent $\D$-modules}\label{sec:category-modDx}
In this section we continue to denote by $W$ a 2-dimensional complex vector space, by $\GL=\GL(W)$ the group of invertible linear transformations of $W$, and by $X=\Sym^3 W$ the space of binary cubic forms which is equipped with a natural action of $\GL$. The goal of this section is to describe the category of $\GL$-equivariant coherent $\D_X$-modules as the category of finite-dimensional representations of a quiver (as mentioned in Section \ref{subsec:Dmods}). Then we identify (up to isomorphism) all the indecomposable representations of the quiver, so implicitly, all the indecomposable equivariant $\D_X$-modules.
\subsection{The quiver description of the category of equivariant $\D$-modules}
We now proceed to determine the quiver $(\Q,\I)$ for the category of $\GL$-equivariant coherent $\D_X$-modules. We will use freely the following general properties of the correspondence between the category $\opmod_{\GL}(\D_X)$ and the quiver $(\Q,\I)$ representing~it:
\begin{itemize}
\item There is a bijective correspondence between simples $M\in\opmod_{\GL}(\D_X)$ and nodes $m$ of $\Q$ -- using notation (\ref{eq:defS}), the representation of $(\Q,\I)$ corresponding to $M$ is $\sS^m$.
\item If $D\in\opmod_{\GL}(\D_X)$ corresponds to $\sV^D\in\rep(\Q,\I)$ then for every simple $M\in\opmod_{\GL}(\D_X)$ corresponding to a node $m$ we have that the multiplicity of $M$ as a composition factor of $D$ is equal to $\dim\sV^D_m$.
\item Specializing the remark above to the case when $D$ is the injective envelope of the simple $M$, we have that $\sV^D = \sI^m$. Using (\ref{eq:defP-I}), it follows that for every simple $N$ with corresponding node $n$, the multiplicity of $N$ as a composition factor of $D$ is equal to the number of paths in $\Q$ (modulo relations) from $n$ to $m$.
\item Dually, if $D$ is the projective cover of a simple $M$ then the multiplicity of a simple $N$ as a composition factor of $D$ is equal to the number of paths from $m$ to $n$.
\item If $m,n$ are nodes of $\Q$ corresponding to the simples $M,N$, then the number of arrows from $m$ to $n$ equals $\dim_{\bb{C}} \Ext^1(M,N)$ \cite[Chapter III, Lemma 2.12]{ass-sim-sko}.
\item The holonomic duality functor $\bb{D}$ (resp. the Fourier transform $\mc{F}$) described in Section~\ref{subsec:Dmods} induces a self-duality (resp. a self-equivalence) of the category $\rep(\Q,\I)$. If $m,n$ are nodes of $\Q$ corresponding to the simples $M,N$, then we write $n=\bb{D}(m)$ (resp. $n=\mc{F}(m)$) when $N=\bb{D}(M)$ (resp. $N = \mc{F}(M)$). The number of arrows from $m$ to $n$ is equal to the number of arrows from $\bb{D}(n)$ to $\bb{D}(m)$, as well as to the number of arrows from $\mc{F}(m)$ to $\mc{F}(n)$. An analogous relationship holds for the number of paths from $m$ to $n$ (considered as always modulo the relations in $\I$).
\end{itemize}
Based on these general remarks we can prove the following.
\begin{lemma}\label{lem:MN'N''}
Suppose that we have an exact sequence in $\opmod_{\GL}(\D_X)$
\begin{equation}\label{eq:sesM-IM-N}
0 \lra M \lra I^M \lra N \lra 0
\end{equation}
where $I^M$ denotes the injective envelope of the simple module $M$, and $N$ has length two, with composition factors $N',N''$. We assume that $M,N',N''$ are pairwise distinct and let $m,n',n''$ denote the corresponding nodes of $\Q$. If $N''$ is not isomorphic to a submodule of $N$ then there exists a unique arrow in $\Q$ from $n'$ to $m$, and no arrow from $n''$ to $m$.
\end{lemma}
\begin{proof} Since the socle of $I^M$ is $M$, we get $\Hom(N',I^M) = \Hom(N'',I^M) = 0$. Moreover, the assumptions on $N$ show that $N'$ must be a submodule of $N$, with $N'' = N/N'$, which in turn implies that $\Hom(N',N) = \bb{C}$. The long exact sequences associated to $\Hom(N',\bullet)$ and $\Hom(N'',\bullet)$ applied to (\ref{eq:sesM-IM-N}) yield
\[0 \lra \Hom(N',N) \lra \Ext^1(N',M) \lra \Ext^1(N',I^M) = 0,\]
\[0 \lra \Hom(N'',N) \lra \Ext^1(N'',M) \lra \Ext^1(N'',I^M) = 0,\]
where the vanishing of $\Ext^1(\bullet,I^M)$ follows from the fact that $I^M$ is injective. We get that $\Ext^1(N',M)=\bb{C}$ and $\Ext^1(N'',M)=0$, i.e. there exists exactly one arrow from $n'$ to $m$, and no arrow from $n''$ to $m$.
\end{proof}
We are now ready to prove the main theorem regarding the quiver description of $\opmod_{\GL}(\D_X)$.
\begin{theorem}\label{thm:quiver}
There is an equivalence of categories
\[\opmod_{\GL}(\D_X) \simeq \rep(\Q,\I),\]
where $\rep(\Q,\I)$ is the category of finite-dimensional representations of a quiver $\Q$ with relations $\I$, described as follows. The vertices and arrows of the quiver $\Q$ are depicted in the diagram
\[\xymatrix@=2.3pc@L=0.2pc{
s \ar@<0.5ex>[dr]^{\alpha_1} & & d_0 \ar@<0.5ex>[dl]^{\alpha_2} & & & g_{1} \ar@<0.5ex>[rr]^{\gamma_1} & & d_1 \ar@<0.5ex>[ll]^{\delta_1} & \\
& p \ar@<0.5ex>[ul]^{\beta_1} \ar@<0.5ex>[ur]^{\beta_2}\ar@<0.5ex>[dl]^{\beta_4} \ar@<0.5ex>[dr]^{\beta_3} & & & \overset{q_1}{\bullet} & \overset{q_2}{\bullet} & \overset{g_2}{\bullet} & \overset{g_3}{\bullet} & \overset{g_4}{\bullet} \\
q_0 \ar@<0.5ex>[ur]^{\alpha_4} & & e \ar@<0.5ex>[ul]^{\alpha_3} & & & g_{-1} \ar@<0.5ex>[rr]^{\gamma_{-1}} & & d_2 \ar@<0.5ex>[ll]^{\delta_{-1}} &
}\]
and the set of relations $\I$ is given by all 2-cycles and all non-diagonal compositions of two arrows:
\[\alpha_i \beta_i \mbox{ and } \beta_i\alpha_i \mbox{ for } i=1,2,3,4,\quad \gamma_i\delta_i\mbox{ and }\delta_i\gamma_i\mbox{ for }i=1,-1,\mbox{ and}\]
\[\alpha_1 \beta_2, \alpha_1\beta_4 , \alpha_2\beta_1, \alpha_2\beta_3, \alpha_3\beta_2,\alpha_3\beta_4,\alpha_4\beta_1,\alpha_4\beta_3.\]
\end{theorem}
\begin{proof}
Let $M$ be any of the simples $G_2,G_3,G_4,Q_1,Q_2$, and consider the open embedding $j:O_4 \to X$. We have by construction that $j_*j^*M =M_{\Delta}= M$, hence $M$ is injective by Lemma \ref{lem:inj}. The holonomic duality functor $\bb{D}$ takes a simple $\D$-module corresponding to a local system on $U$ to the simple corresponding to the dual local system and therefore
\[G_2 \overset{\bb{D}}{\llra}G_4,\ \bb{D}(G_3)=G_3,\ Q_1 \overset{\bb{D}}{\llra}Q_2.\]
Since $\bb{D}$ takes injectives to projectives, this shows that $M$ is both injective and projective, hence $\Ext^1(M,N) = \Ext^1(N,M)=0$ for all $N\in\opmod_{\GL}(\D_X)$. We conclude that $g_2,g_3,g_4,q_1,q_2$ are isolated nodes of the quiver~$\Q$.
Consider next the modules $G_i$ with $i=\pm 1$. We have by Lemma~\ref{lem:inj} that $j_*j^* G_{i} = F_{i}$ is the injective envelope of $G_i$. Combining this with Corollary \ref{cor:simples-full-support} we get non-split exact sequences
\[0 \to G_{1} \to F_{1} \to D_1\to 0 \mbox{ and } 0 \to G_{-1} \to F_{-1} \to D_2\to 0\]
so $\Ext^1(D_2,G_{-1}) = \Ext^1(D_1,G_{1}) = \bb{C}$. It follows that in $\Q$ there exist unique arrows $\delta_1$ from $d_1$ to $g_1$, and $\delta_{-1}$ from $d_2$ to $g_{-1}$. Moreover, since $F_i$ is the injective envelope of $G_i$ there are no other paths of positive length (and in particular no arrows) with target $g_1$ or $g_{-1}$. Using
\[D_1 \overset{\bb{D}}{\llra}D_2,\ G_1 \overset{\bb{D}}{\llra}G_{-1},\]
we get unique arrows $\gamma_1$ from $g_1=\bb{D}(g_{-1})$ to $d_1=\bb{D}(d_2)$, and $\gamma_{-1}$ from $g_{-1}=\bb{D}(g_1)$ to $d_2=\bb{D}(d_1)$, and moreover there are no other paths with source in $g_1$ or $g_{-1}$. Applying next the Fourier transform
\[G_1 \overset{\mc{F}}{\llra}D_2,\ G_{-1} \overset{\mc{F}}{\llra}D_1,\]
we conclude that there are unique arrows in and out of $d_1,d_2$ (namely $\gamma_{\pm 1},\delta_{\pm 1}$) and no other paths of positive length in or out of $d_1,d_2$. This forces the relations $\delta_i\gamma_i=\gamma_i\delta_i=0$ for $i=\pm 1$.
We are left to consider the restriction of the quiver to the nodes $e,d_0,p,q_0,s$. Since the modules $E,D_0,P,Q_0,S$ are preserved by holonomic duality, it follows that for $m,n\in\{e,d_0,p,q_0,s\}$, the number of arrows from $m$ to $n$ is equal to the number of arrows from $n$ to $m$.
Applying Lemma~\ref{lem:inj} with $Y=X$, $O=O_3$ and $M=S$ we find that $S_{\Delta}$ is the injective envelope of $S$, and combining Lemmas~\ref{lem:MN'N''} and~\ref{lem:indS-del} we conclude that there exists a unique arrow from $p$ to $s$, which we denote by $\b_1$, and no arrow from $e$ to $s$. Applying the duality $\bb{D}$ we find a unique arrow $\a_1$ from $s$ to $p$. Since $D_0$ and $Q_0$ do not appear as composition factors of $S_{\Delta}$, there is no path (and thus no arrow) from either $d_0$ or $q_0$ to $s$. We can next argue in the similar fashion by taking $M=Q_0$ and appealing to Lemma~\ref{lem:indQ0-del} to conclude that there are unique arrows $\b_4$ from $p$ to $q_0$ and $\a_4$ from $q_0$ to $p$, there is no arrow from $d_0$ to $q_0$, and no path from either $e$ or $s$ to $q_0$.
Since the Fourier transform fixes $P$ and swaps the elements in each pair $(E,S)$ and $(Q_0,D_0)$, it follows from the conclusions of the preceding paragraph that there exists a pair of arrows $\a_3,\b_3$ between $p$ and $e$, and another pair of arrows $\a_2,\b_2$ between $p$ and $d_0$. Moreover, there are no paths from either $d_0$ or $q_0$ to $e$, and no paths from either $e$ or $s$ to $d_0$, so in particular we have obtained a complete list of arrows in $\Q$.
The relations $\a_1\b_1=\a_4\b_4=0$ follow from the fact that $S$ and $Q_0$ appear with multiplicity one inside their respective injective envelopes. Applying the duality $\bb{D}$ we conclude that $\b_1\a_1=\b_4\a_4=0$, and further applying the Fourier transform we get $\a_2\b_2=\b_2\a_2=\a_3\b_3=\b_3\a_3=0$. Since there are no paths between either of $e,s$ and either $q_0,d_0$ we conclude that
\[\alpha_1 \beta_2 = \alpha_1\beta_4 = \alpha_2\beta_1 =\alpha_2\beta_3 = \alpha_3\beta_2 = \alpha_3\beta_4 = \alpha_4\beta_1 = \alpha_4\beta_3 = 0.\]
Since $E$ appears in the injective envelope of $S$, there has to be a path from $e$ to $s$ which is necessarily $\a_3\b_1$. By holonomic duality, there is a path from $s$ to $e$, namely $\a_1\b_3$. Similarly, since $D_0$ appears in the injective envelope of $Q_0$ we get the path $\a_2\b_4$ from $d_0$ to $q_0$, and by duality we have the path $\a_4\b_2$ from $q_0$ to $d_0$. This shows that there are no more relations in the quiver $\Q$, concluding our proof.
\end{proof}
\begin{remark}
As any $\D$-module in $\opmod_{\GL}(\D_X)$ has a projective (resp. injective) resolution, the indecomposable projective (resp. injective) $\D$-modules play a special role. We can construct all the indecomposable projective (resp. injective) $\D$-modules in $\opmod_{\GL}(\D_X)$ using $\D$-module-theoretic tools as follows. Up to duality $\bb{D}$, it is enough to describe the injectives. We already obtained in the proof of Theorem \ref{thm:quiver} the injective envelopes for the simples with full support. Namely, if $M$ is a simple equivariant $\D$-module with full support, then its injective envelope is $M_\Delta$ (see Lemma \ref{lem:inj}). By taking Fourier transform (which preserves injectives), the injective envelopes of $E,D_0,D_1,D_2$ are $\mc{F}(S_\Delta), \mc{F}((Q_0)_\Delta),\mc{F}(F_{-1}),\mc{F}(F_1)$, respectively. We are left to identify the injective envelope of $P$. One possible description is as follows. Let $U=X\backslash \ol{O_2}$ and $j: U \to X$ the open embedding. Since $O_3$ is smooth and closed in $U$ of codimension 1, the sheaf $\mc{H}^1_{O_3} (\mc{O}_{U})$ is the simple $\D_{U}$-module $j^*P$, and $\mc{H}^i_{O_3}(\mc{O}_U)=0$ for $i\neq 1$. We have $j_{*} \mc{H}^1_{O_3} (\mc{O}_{U})= H^0 \mc{H}^1_{O_3} (\mc{O}_{U}) = H^1_{O_3}(X,\mc{O}_X)$, since the spectral sequence $H^p(\mc{H}^q_{O_3}(U,\mc{O}_U))\Rightarrow H^{p+q}_{O_3}(X,\mc{O}_X)$ degenerates. By Lemma \ref{lem:inj}, this shows that $H=H^1_{O_3}(X,\mc{O}_X)$ is the injective envelope of $P$ in the subcategory $\opmod_G^{\ol{O_3}}(\D_X)$. By Theorem \ref{thm:quiver}, the $\D$-module $H$ and its Fourier transform $\mc{F}(H)$ will correspond to the following representations of the quiver $(\Q,\I)$:
\[H: \begin{aligned}\xymatrix@=1.3pc@L=0.1pc{
0 \ar@<0.5ex>[dr] & & \bb{C} \ar@<0.5ex>[dl]^{1} \\
& \bb{C} \ar@<0.5ex>[ul] \ar@<0.5ex>[ur]^{0}\ar@<0.5ex>[dl] \ar@<0.5ex>[dr]^{0} & \\
0 \ar@<0.5ex>[ur] & & \bb{C}\ar@<0.5ex>[ul]^{1}
}\end{aligned} \hspace{0.6in}
\mc{F}(H): \begin{aligned}\xymatrix@=1.5pc@L=0.1pc{
\bb{C} \ar@<0.5ex>[dr]^{1} & & 0 \ar@<0.5ex>[dl] \\
& \bb{C} \ar@<0.5ex>[ul]^{0} \ar@<0.5ex>[ur]\ar@<0.5ex>[dl]^{0} \ar@<0.5ex>[dr] & \\
\bb{C} \ar@<0.5ex>[ur]^{1} & & 0\ar@<0.5ex>[ul]
}\end{aligned}\]
Note that $P$ is a submodule of both $H$ and $\mc{F}(H)$. Due to the description in (\ref{eq:defP-I}), we obtain the injective envelope $I^P$ of $P$ by the exact sequence
\[0\lra P \lra H \oplus \mc{F}(H) \lra I^P \lra 0,\]
where the first map is the diagonal inclusion.
\end{remark}
\subsection{Description of the indecomposables}\label{subsec:indecomp}
In this section we describe the indecomposable representations of the quiver $\rep(\Q,\I)$ obtained in Theorem \ref{thm:quiver}. We consider each connected component of $\Q$. Clearly, the only indecomposable representations of the isolated vertices are the simples.
Next, consider the quiver $\xymatrix{1 \ar@<0.5ex>[r] & 2 \ar@<0.5ex>[l]}$ with all 2-cycles zero. There are only 4 indecomposable representations (see also \cite{lor-wal}): the 2 simples $\xymatrix{\bb{C} \ar@<0.5ex>[r] & 0 \ar@<0.5ex>[l]}$ and $\xymatrix{0 \ar@<0.5ex>[r] & \bb{C} \ar@<0.5ex>[l]}$, together with their respective projective covers $\xymatrix{\bb{C} \ar@<0.5ex>[r]^1 & \bb{C} \ar@<0.5ex>[l]^0}$ and $\xymatrix{\bb{C} \ar@<0.5ex>[r]^0 & \bb{C} \ar@<0.5ex>[l]^1}$.
Denote by $(\tilde{\Q},\tilde{\I})$ the largest connected component of $(\Q,\I)$, that is, the quiver
\[\hspace{-0.2in}(\tilde{\Q},\tilde{\I}): \hspace{0.1in} \begin{aligned}\xymatrix@=2.3pc@L=0.1pc{
1 \ar@<0.5ex>[dr]^{\alpha_1} & & 2 \ar@<0.5ex>[dl]^{\alpha_2} \\
& 5 \ar@<0.5ex>[ul]^{\beta_1} \ar@<0.5ex>[ur]^{\beta_2}\ar@<0.5ex>[dl]^{\beta_4} \ar@<0.5ex>[dr]^{\beta_3} & \\
4 \ar@<0.5ex>[ur]^{\alpha_4} & & 3\ar@<0.5ex>[ul]^{\alpha_3}
}\end{aligned}\]
with relations $\tilde{\I}$ as in Theorem \ref{thm:quiver}. In contrast with the quivers obtained in \cite{lor-wal}, the quiver $(\tilde{\Q},\tilde{\I})$ is not representation-finite, that is, the category $\rep(\tilde{\Q},\tilde{\I})$ has infinitely many indecomposable representations. This follows by realizing the extended Dynkin quiver $\hat{D}_4$ as a subquiver $\Q(\alpha)$ (resp. $\Q(\beta)$) of $(\tilde{\Q},\tilde{\I})$ by considering only the arrows $\alpha_i$ (resp. the arrows $\beta_i$), where $i=1,2,3,4$. Nevertheless, we are able to describe the indecomposables due to the fact that $(\tilde{\Q},\tilde{\I})$ is of tame representation type (Theorem \ref{thm:tame}).
More precisely, consider the quiver $\hat{D}_4$ as in Example \ref{ex:dee4hat}. We construct two embeddings of categories
\[ \alpha: \rep(\hat{D}_4) \to \rep(\tilde{\Q},\tilde{\I}) \, \mbox{ and } \, \beta : \rep(\hat{D}_4) \to \rep(\tilde{\Q},\tilde{\I})\]
as follows. For a representation $V\in \rep(\hat{D}_4)$, set $\alpha(V),\beta(V)$ to be the representations
\[\alpha(V): \hspace{0.1in} \begin{aligned}\xymatrix@=2pc@L=0.1pc{
V_1 \ar@<0.5ex>[dr]^{V_{\alpha_1}} & & V_2 \ar@<0.5ex>[dl]^{V_{\alpha_2}} \\
& V_5 \ar@<0.5ex>[ul]^{0} \ar@<0.5ex>[ur]^{0}\ar@<0.5ex>[dl]^{0} \ar@<0.5ex>[dr]^{0} & \\
V_4 \ar@<0.5ex>[ur]^{V_{\alpha_4}} & & V_3\ar@<0.5ex>[ul]^{V_{\alpha_3}}
}\end{aligned} \hspace{0.4in}
\beta(V): \hspace{0.1in} \begin{aligned}\xymatrix@=2pc@L=0.1pc{
V^*_1 \ar@<0.5ex>[dr]^{0} & & V_2^* \ar@<0.5ex>[dl]^{0} \\
& V_5^* \ar@<0.5ex>[ul]^{V^*_{\alpha_1}} \ar@<0.5ex>[ur]^{V^*_{\alpha_2}}\ar@<0.5ex>[dl]^{V^*_{\alpha_4}} \ar@<0.5ex>[dr]^{V^*_{\alpha_3}} & \\
V^*_4 \ar@<0.5ex>[ur]^{0} & & V^*_3\ar@<0.5ex>[ul]^{0}
}\end{aligned}\]
Clearly, $\alpha,\beta$ send indecomposables to indecomposables. For example, for an arbitrary $n>0$, we have in $\rep(\tilde{\Q},\tilde{\I})$ two 1-parameter families of indecomposables $\alpha(R_n(\lambda))$ and $\beta(R_n(\lambda))$ in the dimension vector $(n,n,n,n,2n)$ induced by the representations $R_n(\lambda)\in\rep(\hat{D}_4)$ described in Example \ref{ex:dee4hat}.
Note that we have 4 projective-injective indecomposable representations of $(\tilde{\Q},\tilde{\I})$ that are not obtained from $\alpha,\beta$, namely: $\sP^1 = \sI^2$, $\sP^2 = \sI^1$, $\sP^3 = \sI^4$ and $\sP^4=\sI^3$.
As explained in Example \ref{ex:dee4hat}, the indecomposable representations of $\hat{D}_4$ are classified. Using this, we obtain the classification for $(\tilde{\Q},\tilde{\I})$:
\begin{theorem}\label{thm:tame}
Let $X$ be an indecomposable representation in $\rep(\tilde{\Q},\tilde{\I})$. Assume that $X$ is not isomorphic to either of the projective-injectives $\sP^1,\sP^2,\sP^3,\sP^4$. Then there is an indecomposable $V$ in $\rep(\hat{D}_4)$ such that $X$ is isomorphic to either $\alpha(V)$ or $\beta(V)$. In particular, the quiver $(\tilde{\Q},\tilde{\I})$ is of (domestic) tame representation type.
\end{theorem}
\begin{proof}
Due to the relations $\beta_i\alpha_i = 0$, the vertices $1,2,3,4$ of $(\tilde{\Q},\tilde{\I})$ are nodes as defined Section \ref{subsec:quivers}. After separating each of these nodes, we obtain the following quiver $(\Q',\I')$
\[\xymatrix@=2.3pc@L=0.15pc{
1' & \ar[dr]^{\alpha_1} 1 & & 2' & 2 \ar[dll]^{\alpha_2}\\
& & 5 \ar[ull]^{\beta_1} \ar[ur]^{\beta_2} \ar[dl]^{\beta_4} \ar[drr]^{\beta_3} & & \\
4 \ar[urr]^{\alpha_4} & 4' & & 3 \ar[ul]^{\alpha_3} & 3'
}\]
with relations generated by $\alpha_i \beta_i$, where $i=1,2,3,4$, and $\alpha_1\beta_2,\alpha_1\beta_4,\alpha_2\beta_1,\alpha_2\beta_3,\alpha_3\beta_2,\alpha_3\beta_4,\alpha_4\beta_1,\alpha_4\beta_3$. By Lemma \ref{lem:node}, we can assume $X$ is an indecomposable of $(\Q',\I')$.
Now we apply Lemma \ref{lem:addrel} four consecutive times. Since $X$ is not isomorphic to $\sP^1,\sP^2,\sP^3,\sP^4$, we obtain that the indecomposable $X$ must be a representation of the quiver $(\Q',\I'')$, where $\I''$ is obtained from $\I'$ by adding the relations $\alpha_1\beta_3, \alpha_3\beta_1,\alpha_2\beta_4,\alpha_4\beta_2$. But now the vertex $5$ is a node of $(\Q',\I'')$, and by separating the node, we obtain two copies of $\hat{D}_4$ (one copy with the middle vertex a sink and one copy with the middle vertex a source). Using Lemma \ref{lem:node} again, we obtain the result.
\end{proof}
\begin{remark}
It can be shown that the quiver $(\Q',\I')$ obtained by separating the 4 nodes has global dimension 2, and any each indecomposable $X \in \rep(\Q',\I')$, either the projective dimension or the injective dimension of $X$ is at most $1$. Hence $(\Q',\I')$ is quasi-tilted by \cite[Chapter XX, Theorem 3.10]{sim-sko3}, which suggests that an alternative approach could be used to classify its indecomposables via tilting theory (for more on quasi-tilted algebras see \cite[Section XX.3]{sim-sko3}; for more on tilting theory, see \cite[Chapter VI]{ass-sim-sko}).
\end{remark}
\begin{remark}
One can understand the morphisms between the indecomposable representations via Auslander-Reiten theory -- see \cite[Chapter VI]{ass-sim-sko} for the basic theory and relevant terminology. We can describe the Auslander-Reiten quiver of $(\tilde{\Q},\tilde{\I})$ by standard methods as follows. There are two ${\hat D}_4$ type sub-quivers of $(\tilde{\Q},\tilde{\I})$: the quiver $\Q(\alpha)$ with arrows $\alpha_i$ and the quiver $\Q(\beta)$ with arrows $\beta_j$. The Auslander-Reiten quivers for both of these sub-quivers have three parts: preprojective, preinjective and the regular one consisting of so-called tubes (see \cite[Section XIII.3]{sim-sko2}).
Now we describe the components of the Auslander-Reiten quiver of $(\tilde{\Q},\tilde{\I})$. First, the tubes of both sub-quivers $\Q(\alpha)$ and $\Q(\beta)$ appear as separate components of the Auslander-Reiten quiver of $(\tilde{\Q},\tilde{\I})$. Next, the preinjective component of the quiver $\Q(\alpha)$ and the preprojective component of $\Q(\beta)$ are glued together along the four simple representations $\sS^1, \sS^2, \sS^3, \sS^4$ to become one component of the Auslander-Reiten quiver of $(\tilde{\Q},\tilde{\I})$. Finally, the last component of the Auslander-Reiten quiver of $(\tilde{\Q},\tilde{\I})$ is obtained by gluing together the preinjective component of $\Q(\beta)$ and the preprojective component of $\Q(\alpha)$ along the simple module $\sS^5$. The four projective-injective modules $\sP^1,\sP^2, \sP^3, \sP^4$ are also part of this component, and they occur in the middle along with $\sS^5$. They are part of four almost split sequences, described as follows. Denote by $M_{\alpha_i}$ the two dimensional module with the nonzero linear map $\alpha_i$ and by $M_{\beta_j}$ the two dimensional module with the nonzero linear map $\beta_j$. Then we have an almost split sequence
\[
\begin{matrix}&&\sP^3&&\\
&\nearrow&&\searrow&\\
M_{\beta_1}&&&&M_{\alpha_3}\\
&\searrow&&\nearrow\\
&&\sS^5&&
\end{matrix}
\]
and three more analogous ones by symmetry.
\end{remark}
\section{Some local cohomology calculations}\label{sec:loccoh}
We conclude this article with a couple of examples of local cohomology computations for equivariant $\D$-modules, with support in orbit closures. Many of these calculations are standard, but some require an understanding of the structure of the category $\opmod_{\GL}(\D_X)$. We begin with the following general observation. Suppose that $M$ is a $\GL$-equivariant $\D_X$-module with support contained in $\ol{O_i}$. It follows that for $j\geq i$ we have
\[H^k_{\ol{O_j}}(M) = \begin{cases}
M & \mbox{if }k = 0, \\
0 & \mbox{otherwise}.
\end{cases}\]
We will thus only be interested in studying the local cohomology groups $H^k_{\ol{O_j}}(M)$ when $j<i$. For iterated local cohomology groups of $S$ with respect to a family of $\GL$-invariant closed subsets we have the following.
\begin{theorem}\label{thm:iteratedloccoh}
The non-zero iterated local cohomology groups $H^{i_1}_{\ol{O_{j_1}}}(\cdots H^{i_k}_{\ol{O_{j_k}}}(S))$ with $j_1<\cdots<j_k$ are:
\begin{equation}\label{eq:HS}
H^1_{\ol{O_3}}(S)=\frac{S_{\Delta}}{S},\quad H^2_{\ol{O_2}}(S)= D_0,\quad H^4_{O_0}(S) = E,
\end{equation}
\begin{equation}\label{eq:HHS}
H^1_{\ol{O_2}}(H^1_{\ol{O_3}}(S))=D_0,\quad H^3_{O_0}(H^1_{\ol{O_3}}(S))= E,\quad H^2_{O_0}(H^2_{\ol{O_2}}(S))= E,
\end{equation}
\begin{equation}\label{eq:HHHS}
H^2_{O_0}(H^1_{\ol{O_2}}(H^1_{\ol{O_3}}(S)))=E.
\end{equation}
\end{theorem}
The formulas (\ref{eq:HHS}--\ref{eq:HHHS}) follow from (\ref{eq:HS}) and the fact that the orbit closures are cohomological complete intersections (see \cite[Example~4.2]{lyub-nmbs}). In particular, the only non-vanishing Lyubeznik number for each orbit closure is the highest one. Based on Theorem~\ref{thm:iteratedloccoh} we can also determine the local cohomology groups of simple $\GL$-equivariant $\D_X$-modules with support in the orbit closures as follows.
\begin{theorem}\label{thm:loccoh-simples}
The only non-zero local cohomology modules $H^k_{\ol{O_j}}(M)$ when $M$ is a simple $\GL$-equivariant $\D_X$-module with support $\ol{O_i}$ and $j<i$ are the ones in (\ref{eq:HS}), as well as the following:
\begin{equation}\label{eq:HD0}
H^2_{O_0}(D_0)=E,
\end{equation}
\begin{equation}\label{eq:HP}
H^1_{\ol{O_2}}(P)=D_0 \oplus E,\quad H^1_{O_0}(P) = H^3_{O_0}(P) = E,
\end{equation}
\begin{equation}\label{eq:HQ0}
H^1_{\ol{O_3}}(Q_0)=\frac{(Q_0)_{\Delta}}{Q_0},\quad H^2_{\ol{O_2}}(Q_0)= H^2_{O_0}(Q_0) = E,
\end{equation}
\begin{equation}\label{eq:HG}
H^1_{\ol{O_3}}(G_1) = H^1_{\ol{O_2}}(G_1) = D_1,\quad H^1_{\ol{O_3}}(G_{-1}) = H^1_{\ol{O_2}}(G_{-1}) = D_2.
\end{equation}
\end{theorem}
Perhaps the most interesting computations in Theorem~\ref{thm:loccoh-simples} are that of $H^1_{\ol{O_2}}(P)=D_0$, as well as that of the local cohomology groups of $Q_0$, particularly since they require some of the structure theory that we developed for the category of $\GL$-equivariant $\D_X$-modules.
\begin{proof}[Proof of Theorem~\ref{thm:iteratedloccoh}]
Since $\ol{O_3}$ is the hypersurface defined by $\Delta=0$ it follows that $H^i_{\ol{O_3}}(S) = S_{\Delta}/S$ for $i=0$, and it is $0$ otherwise. The calculation of $H^{\bullet}_{\ol{O_2}}(S)$ is a special case of \cite[Theorem~1.2]{raicu-veronese}, while that of $H^{\bullet}_{O_0}(S)$ is known since the ideal of $O_0$ is the maximal homogeneous ideal of $S$ \cite[Corollary~A1.6]{geom-syzygies}.
For $i<j$ we have a spectral sequence $H^p_{\ol{O_i}}(H^q_{\ol{O_j}}(S)) \Rightarrow H^{p+q}_{\ol{O_i}}(S)$. We get using (\ref{eq:HS}) that this spectral sequence degenerates, since $H^q_{\ol{O_j}}(S)$ is non-zero for a unique value of $q$. This proves (\ref{eq:HHS}), as well as the vanishing of the remaining modules $H^p_{\ol{O_i}}(H^q_{\ol{O_j}}(S))$. The equation (\ref{eq:HHHS}), as well as the vanishing of $H^{i_1}_{O_0}(H^{i_2}_{\ol{O_2}}(H^{i_3}_{\ol{O_3}}(S)))$ for $(i_1,i_2,i_3)\neq(2,1,1)$ follows from (\ref{eq:HS}--\ref{eq:HHS}) since $H^1_{\ol{O_2}}(H^1_{\ol{O_3}}(S))= D_0 = H^2_{\ol{O_2}}(S)$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:loccoh-simples}]
The equation (\ref{eq:HD0}) and the vanishing of $H^i_{O_0}(D_0)$ for $i\neq 2$ follows from (\ref{eq:HS}--\ref{eq:HHS}). To prove (\ref{eq:HP}), consider the non-split exact sequence
\begin{equation}\label{eq:PSdE}
0 \lra P \lra S_{\Delta}/S \lra E \lra 0
\end{equation}
coming from Corollary~\ref{cor:factors-Sdelta} and Lemma~\ref{lem:indS-del}. Since $E$ is supported at $O_0$ we have $H^i_{\ol{O_j}}(E) = E$ for $i=0$ and it is $0$ otherwise, while the local cohomology modules of $S_{\Delta}/S = H^1_{\ol{O_3}}(S)$ have been computed in (\ref{eq:HHS}). The long exact sequence obtained by applying $H^0_{\ol{O_2}}(\bullet)$ to (\ref{eq:PSdE}) yields then a short exact sequence
\[0\lra E \lra H^1_{\ol{O_2}}(P) \lra D_0 \lra 0\]
and the vanishing of $H^i_{\ol{O_2}}(P)$ for $i\neq 1$. By Theorem~\ref{thm:quiver}, any extension between $D_0$ and $E$ is split, hence $H^1_{\ol{O_2}}(P) = D_0 \oplus E$. We next consider the long exact sequence obtained by applying $H^0_{O_0}(\bullet)$ to (\ref{eq:PSdE}) and get
\[ E = H^0_{O_0}(E) \simeq H^1_{O_0}(P),\quad H^3_{O_0}(P) \simeq H^3_{O_0}(S_{\Delta}/S) = E,\]
and $H^i_{O_0}(P)=0$ for $i\neq 1,3$, proving (\ref{eq:HP}).
The calculation of $H^i_{\ol{O_3}}(Q_0)$ follows from the fact that $\ol{O_3}$ is defined by $\Delta=0$. Since the only non-vanishing is obtained for $i=1$, the spectral sequence $H^p_{\ol{O_2}}(H^q_{\ol{O_3}}(Q_0)) \Rightarrow H^{p+q}_{\ol{O_2}}(Q_0)$ degenerates and we get $H^i_{\ol{O_2}}(Q_0) = H^{i-1}_{\ol{O_2}}(H^1_{\ol{O_3}}(Q_0))$. We have from Corollary~\ref{cor:factors-FD0-delta} and Lemma~\ref{lem:indQ0-del} a non-split exact sequence
\[0\lra P \lra H^1_{\ol{O_3}}(Q_0) \lra D_0 \lra 0,\]
which yields by applying $H^0_{\ol{O_2}}(\bullet)$, and using (\ref{eq:HP}) and the fact that $D_0$ has support in $\ol{O_2}$, an exact sequence
\[0\lra H^0_{\ol{O_2}}(H^1_{\ol{O_3}}(Q_0)) \overset{\a}{\lra} D_0 \overset{\b}{\lra} D_0\oplus E \lra H^1_{\ol{O_2}}(H^1_{\ol{O_3}}(Q_0)) \lra 0,\]
as well as the vanishing $H^i_{\ol{O_2}}(H^1_{\ol{O_3}}(Q_0))=0$ for $i\geq 2$. If the map $\a$ is non-zero, then it must be an isomorphism since $D_0$ is simple, which would show that $D_0$ is a submodule of $H^1_{\ol{O_3}}(Q_0)=(Q_0)_{\Delta}/Q_0$ and contradict Lemma~\ref{lem:indQ0-del}. It follows that $\a=0$, hence $H^1_{\ol{O_2}}(Q_0)=H^0_{\ol{O_2}}(H^1_{\ol{O_3}}(Q_0)) = 0$, and moreover we have $\coker(\b)=E$ which implies $H^2_{\ol{O_2}}(Q_0)=H^1_{\ol{O_2}}(H^1_{\ol{O_3}}(Q_0)) = E$. To finish the proof of (\ref{eq:HQ0}) we note that $H^q_{\ol{O_2}}(Q_0)$ is non-zero only for $q=2$, and therefore it yields the degeneration of the spectral sequence $H^p_{O_0}(H^q_{\ol{O_2}}(Q_0)) \Rightarrow H^{p+q}_{O_0}(Q_0)$. We get that $H^q_{\ol{O_2}}(Q_0)=E$ for $q=2$ and is $0$ otherwise.
Since $F_1/G_1 = D_1$ we have $(G_1)_{\Delta} = (F_1)_{\Delta} = F_1$ and therefore
\[H^1_{\ol{O_3}}(G_1) = (G_1)_{\Delta}/G_1 = F_1/G_1 = D_1\]
is the only non-vanishing group $H^i_{\ol{O_3}}(G_1)$. The standard (by now) spectral sequence argument shows that $H^i_{\ol{O_2}}(G_1) = D_1$ for $i=1$ and is $0$ otherwise. A similar argument, based on $F_{-1}/G_{-1}=D_2$, applies to show the rest of (\ref{eq:HG}). To conclude the determination of the local cohomology modules of $G_{\pm 1}$ it remains to prove that $H^i_{O_0}(G_{\pm 1})=0$ for all $i$, which in turn is a consequence of the fact that $H^i_{O_0}(D_j)=0$ for all~$i$ and $j=1,2$. This vanishing result follows since on the one hand the modules $H^i_{O_0}(D_j)$ are direct sums of copies of $E$ (beging supported on $O_0$), and on the other hand their characters may only involve terms of the form~$e^{\ll}$ with $|\ll| \equiv 1,2\ \mbox{(mod 3)}$ by (\ref{eq:wts-D012}). Comparing this with (\ref{eq:character-E}) we get the desired vanishing.
Finally, if $M$ is any of the modules $G_2,G_3,G_4,Q_1,Q_2$ we have that $M_{\Delta}=M$, so $H^i_{\ol{O_3}}(M)=0$ for all~$i$. Using the spectral sequence, this shows that $H^i_{\ol{O_2}}(M)=H^i_{O_0}(M)=0$ for all $i$, concluding the proof.
\end{proof}
\section*{Acknowledgements}
Experiments with the computer algebra software Macaulay2 \cite{M2} have provided numerous valuable insights. Raicu acknowledges the support of the Alfred P. Sloan Foundation, and of the National Science Foundation Grant No.~1600765. Weyman acknowledges partial support of the Sidney Professorial Fund and of the National Science Foundation grant No.~1400740.
\end{document}
| 137,628
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TITLE: How to prove that $\vdash\phi\rightarrow\neg\Box\neg\phi$ is a theorem in S5?
QUESTION [0 upvotes]: I want to prove that $\vdash\phi\rightarrow\neg\Box\neg\phi$ is a theorem in S5
I have S5 definition :
\begin{align*}
T&:\Box p\rightarrow p\\
5&:\Diamond p\rightarrow \Box\Diamond p\\
K&:\Box (p\rightarrow q)\rightarrow(\Box p\rightarrow \Box \phi)\\
Nec&:\frac{\phi}{\Box\phi}
\end{align*}
And theorem definition :
A formula is a theorem of S5 $\Leftrightarrow\phi$ is valid in all frames where R is an equivalence relation
\begin{align}
&\neg (\phi\rightarrow\neg\Box\neg\phi),w_i\mbox{ as an hypothesis}\\
&\phi,w_i \mbox{ from }R_\neg \mbox{ on } 1\\
&\Diamond\phi,w_i\mbox{ from }R_\neg \mbox{ on } 1\\
\end{align}
I'm not sure I have here an equivalence relation to conclude that
Therefore $\phi$ is valid in all frames and $\vdash\phi\rightarrow\neg\Box\neg\phi$ is a theorem in S5
Third attempt
\begin{align}
\neg\phi\rightarrow\Box\neg\phi,w_i \mbox{ contrapositive of }T\\
\neg\neg\phi\rightarrow\neg\Box\neg\phi,w_i \mbox{ negation of }1
\end{align}
Therefore $\phi$ is valid and $\vdash\phi\rightarrow\neg\Box\neg\phi$ is a theorem in S5
REPLY [2 votes]: 1) $\Box\neg\phi\to\neg\phi$, obtained by substituting $\neg\phi$ for $p$ in the $\mathbf{T}$ axiom.
2) $\neg\neg\phi\to\neg\Box\neg\phi$, by contraposition from (1).
3) $\phi\to\neg\neg\phi$, by classical logic (a tautology).
4) $\phi\to\neg\Box\neg\phi$, by syllogism from (3) and (2).
That's it.
| 88,116
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Fha Loan Interest Rate Today
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Let’s say you want to buy that mid-priced home using a 30-year fixed-rate mortgage at this week’s average interest rate,
3.19%, 4.53%. FHA 30 Year Fixed, 3.38%, 3.37%, +0.01. 30 Yr. FHA, 3.83%, 0.28, +0.04. 5/1 ARM, 3.43%. Loan Calculator | Compare Rates | Daily Email Update. Mortgage rates held steady today after moving sharply lower yesterday .
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Use this calculator to see the kind of monthly payment you can expect from today’s low mortgage rates. end to the.
Check today’s best mortgage rates where you are. Americans who refinanced in the spring of this. Rates on home loans tend.
Rate And Unit Rate Calculator An Intrinsic Calculation For Ju Teng International Holdings Limited (HKG:3336) Suggests It’s 50% Undervalued – the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC).
Mortgage interest rates fell on all five loan types the MBA tracks. On an unadjusted basis, the MBA’s composite index increased by 5% in the past week. The seasonally adjusted purchase index dipped by.
While the decline in rates has prompted many home owners to refinance their loans, it may not be enough to create a major uptick in home-buying activity So far the drop in mortgage rates has mostly.
| 294,158
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In my regular YouTube wanderings W2AEW’s video “#90: Measure Capacitors and Inductors with an Oscilloscope” showed up. My current study has been related to building an Arduino based LC meter to measure Capacitors and Inductors, but that is a bit of a chicken and egg problem because you need to have precise accounting of the inductors and capacitors in the circuit for the LC meter because the measurements come from noting change when adding another capacitor or inductor to the circuit. (Yeah, you can purchase ones with a 1% tolerance, but I’m looking around for how much I can build and keep my costs down.) Also, I had a fun conversation in G+’s Ham Radio Homebrewing group where Sporadic-Z was encouraging me to try wire wound inductors and this peaked my interest.
So this method W2AEW is talking about requires a “Fast Edge Pulse Generator” circuit to generate a sharp square wave which is another tool I don’t have at this point, but I realized the Siglent oscilloscope has a 1khz square wave generator so I hooked that up to the breadboard to provide my power and throw together the capacitor test circuit. The first capacitors I grabbed to try this out on were big electrolytic capacitors. That was pointless and therefore a failure/learning opportunity. These big guys took the wave and made it a straight line, but eventually I realized the problem with that and went with some smaller capacitors I’d pulled out of something ages ago and I found success in the form of a sharktooth wave. I was able to find a few other similar capacitors and see variations. The pictures below are of the most aesthetically pleasing wave. I’m not sure what the values of the capacitors are at this point, but he provided his notes in the video on PDF. I thought that pretty awesome of him so printed out a copy and will try to do the math later.
So that wave isn’t in his suggested form. He recommended doing 8 squares per division high and I need to remember that when I go back for the math portion of this exercise. There were some other good hints about how to line things up to make measurements of stuff easier which is one of the reasons I intend to watch again. That seemed a skill worth collecting to make my life easier. In most videos they move things around without talking about why they are doing it. I appreciate W2AEW assuming I don’t know much.
I’m hoping I can get to trying out the inductor stuff tomorrow. If I can measure out a few of these capacitors maybe I can find the 10pF and 1nF ones that are suggested in the notes or understand the math enough to substitute something I do have. I’m also expecting fully understanding this will be good for understanding the LC meter I’m contemplating so when I have trouble with that I’ll have a clue about how to troubleshoot it.
If you’re “following along at home”, here is the video. I’d recommend subscribing to his channel if this interests you. His videos seem to be growing more interesting as I learn more about how this stuff works. He’s up to 230-some videos now. I enjoyed his more recent one on panel meters. The 6 video series on Capacitors and Inductors also looks good.
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Chili Lime Corn on The Cob
Warm weather and sweet corn on the cob go hand in hand. Grilled Chili Lime Corn On The Cob gives this summer staple a twist that’s hard to resist. Tart lime, spicy chili powder, and cilantro butter on sweet charred corn offer a vibrant flavor perfect for any BBQ.
We love serving this chili lime corn on the cob alongside Hawaiian Burgers, Lemon Rosemary Grilled Chicken, Glazed Turkey Burgers with Creamed Feta and Grilled Flank Steak Sliders.
Be sure to sign up for my email… to get new recipes and ideas in your inbox!
Chili Lime Corn on the Cob
Guys! We are finally in the thick of prime fresh corn season! I might be the weird one here, but the moment I start seeing homegrown corn stands everywhere I get probably a little too excited.
If you’re the same as me, you’ll go crazy over this chile lime corn on the cob with homemade chili lime butter. With this recipe, we can take the perfectly selected and shucked corn on the cob and lather it with a mix of seasonings and spices, mixed into melted butter.
If you thought you liked corn, just wait until you try it coated in this mixture, it will blow your mind. This easy recipe can just be thrown on the grill and ready in minutes, to make getting dinner on the table even easier.
What Kind of Corn Should I Buy?
Let’s be honest for a second. The produce aisle at the grocery store can be intimidating, with all of the options these days. From organic to homegrown, to white or yellow, who knows what to choose?
Well, I’m here to hopefully ease some of that produce aisle pain, for your next trip to the store with a few things to keep in mind for this recipe.
White, Yellow or Multicolored?
You might have always been told that one kernel color is sweeter than the other, meaning you should give attention to which color corn you are buying.
I’ll let you in on a little secret, there’s actually no difference in the taste of white corn vs. yellow corn. Both have the same sweetness and overall flavor. The color really serves no purpose, so feel free to choose whatever cob color you wish.
Fresh or Frozen?
I highly suggest buying fresh corn on the cob, because the flavor and texture will be a lot better, than that of frozen corn. While I do suggest fresh, frozen will work too if that’s all you have to work with.
Just know that it might not come out to the same quality level as that of fresh corn.
What Do I Need To Make Chili Lime Butter?
This is a pretty straight forward recipe, with a lot of common ingredients you probably already have on hand. You’ll also want to be sure you have a few of the necessary tools that will allow you to get this recipe made in no time.
Ingredients:
- Butter, softened or melted
- Fresh cilantro, chopped
- lime, zested and juiced
- Chili powder
- Seasoned salt
- Fresh cracked pepper
- Sugar
- Crushed red pepper
Tools:
Chili Lime Corn – Great for Summer Grilling
Is it even really summer if you’re not firing up the grill as much as you possibly can? What better excuse to get the grill going than to try out a brand new grilled corn on the cob recipe?
This chili lime corn is the epitome of the taste of summer, and great for any occasion. Make a bulk batch when you need some quick sides for a party or a small batch for when you just need to spice up your weeknight dinner routine.
Tips for the Best Chili Lime Corn on the Cob
- Don’t peel back the husk too early: When shopping for you fresh corn, you want to be sure to get a good ear, but don’t peel back the husk while shopping. It will deplete the shelf life of the corn, and dry it out in no time.
- Use a basting brush: When applying your chile lime butter, use a basting brush. It makes spreading the butter easy as pie and gets you an even coat.
- Wrap it in foil: For a juicer end result, wrap the buttered corn in tin foil, and cook for 15-20 minutes, turning every few minutes. Either way is delicious, but the corn tends to turn out a little juicier when cooked in the foil.
Sides That Will Steal the Show at Your Next BBQ
Easy Classic Bean Salad – A great side dish in the summer but yet hearty enough for a fall side as well! Quick to make, packed with flavor, and travels well making it perfect for potlucks!
Grilled Chipotle Peach Salsa – A fresh spicy-sweet fruit salsa recipe perfect for dipping. Spoon peach salsa over fish or chicken for a perky topping!
Grilled Carrots with Chipotle Lime Dip – A surprisingly easy and healthy way to serve carrots and boy are they full of flavor. They would make a wonderful snack or a great addition to your dinner table tonight!
Rainbow Veggie Kabobs – The perfect summer side dish! Serve them with my delicious and easy-to-make Grilled Lemon Garlic Sauce.
Caprese Pasta Salad – a fresh, easy and simple summer pasta salad perfect for BBQs. It has all the flavors of a traditional Caprese salad in pasta form!
Blueberry Watermelon Feta Mint Salad – Made with a simple balsamic dressing for a perfect sweet and savory salad for any summer celebration.
Like This Grilled Chili Lime Corn? Pin It!
Your New Summer Staple!
There’s no lack of places to find corn in the summer, so you have no excuses to not make this recipe. Trust me once you make it once, you’ll find yourself stocking up on new ears of corn every week.:
Warm weather and Grilled Chili Lime Corn On The Cob go hand in hand. Tart lime, spicy chili powder, and cilantro butter on sweet charred corn offer a vibrant flavor perfect for any BBQ. .
Chili Lime Corn on The Cob
Ingredients
Instructions
Notes
Recommended Products
Warm weather and Grilled Chili Lime Corn On The Cob go hand in hand. Tart lime, spicy chili powder, and cilantro butter on sweet charred corn offer a vibrant flavor perfect for any BBQ.
- 4 ears of corn, husked
- 6 tablespoons butter, softened or melted
- 1-2 tablespoons fresh cilantro, chopped
- 1 medium lime, zested and juiced
- 1 teaspoon chili powder
- 1/2 teaspoon seasoned salt
- 1/4 teaspoon fresh cracked pepper
- 1/4 teaspoon sugar
- 1 pinch crushed red pepper
- Preheat outdoor grill to medium or broiler.
- In a small bowl, combine butter, lime zest, lime juice and all the seasonings. Lightly brush corn with butter mixture.
- Place buttered corn on grill*, cover and cook until tender and slightly charred, about 10-12 minutes, turning occasionally during cooking. Remove from the grill and slather each ear with remaining butter.
.
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\begin{document}
\begin{abstract}
Let $E/\Q$ be an elliptic curve and $p$ a rational prime of good ordinary reduction. For every imaginary quadratic field $K/\Q$ satisfying the Heegner hypothesis for $E$ we have a corresponding line in $E(K)\otimes \Qp$, known as a shadow line. When $E/\Q$ has analytic rank $2$ and $E/K$ has analytic rank $3$, shadow lines are expected to lie in $E(\Q)\otimes \Qp$. If, in addition, $p$ splits in $K/\Q$, then shadow lines can be determined using the anticyclotomic $p$-adic height pairing. We develop an algorithm to compute anticyclotomic $p$-adic heights which we then use to provide an algorithm to compute shadow lines. We conclude by illustrating these algorithms in a collection of examples.
\end{abstract}
\subjclass [2010] {11G05, 11G50, 11Y40.}
\keywords{Elliptic Curve, Universal Norm, Anticyclotomic $p$-adic Height, Shadow Line.}
\thanks{The authors are grateful to the organizers of the conference ``WIN3: Women in Numbers 3'' for facilitating this collaboration\\
\indent and acknowledge the hospitality and support provided by the Banff International Research Station. \\
\indent During the preparation of this manuscript: the second author was partially supported by NSA grant H98230-12-1-0208 and \\
\indent NSF grant DMS-1352598; the third author was partially supported by NSF grant DGE-1144087.}
\maketitle
\section*{Introduction} Fix an elliptic curve $E/\Q$ of analytic rank $2$ and an odd prime $p$ of good ordinary reduction. Assume that the $p$-primary part of the Tate-Shafarevich group of $E/\Q$ is finite. Let $K$ be an imaginary quadratic field such that the analytic rank of $E/K$ is $3$ and the Heegner hypothesis holds for $E$, i.e., all primes dividing the conductor of $E/\Q$ split in $K$. We are interested in computing the subspace of $E(K) \otimes \Qp$ generated by the anticyclotomic universal norms. To define this space, let $K_\infty$ be the anticyclotomic $\Zp$-extension of $K$ and $K_n$ denote the subfield of $K_\infty$ whose Galois group over $K$ is isomorphic to $\Z/p^n\Z$. The module of \textit{universal norms} is defined by
\[
\mathcal{U} = \bigcap_{n \geq 0} N_{K_n/K}(E(K_n) \otimes \Zp),
\]
where $N_{K_n/K}$ is the norm map induced by the map $E(K_n) \to E(K)$ given by $P \mapsto \underset{\sigma\in \mathrm{Gal}(K_n/K)}\sum P^\sigma$.
Consider
\[
L_K := \cU \otimes \Qp \subseteq E(K) \otimes \Qp.
\]
By work of Cornut \cite{Cornut}, and Vatsal \cite{Vatsal} our assumptions on the analytic ranks of $E/\Q$ and $E/K$ together with the assumed finiteness of the $p$-primary part of the Tate-Shafarevich group of $E/\Q$ imply that $\dim L_K \geq 1$. Bertolini \cite{Bertolini} showed that $\dim L_K = 1$ under certain conditions on the prime $p$. Wiles and \c Ciperiani \cite{CW}, \cite{Ciperiani} have shown that Bertolini's result is valid whenever $\Gal (\Q(E_p)/\Q)$ is not solvable; here $E_p$ denotes the full $p$-torsion of $E$ and $\Q(E_p)$ is its field of definition.
The $1$-dimensional $\Qp$-vector space $L_K$ is known as the \emph{shadow line} associated to the triple $(E,K, p)$.
Complex conjugation acts on $E(K) \otimes \Qp$, and we consider its two eigenspaces $E(K)^+\otimes \Qp$ and $E(K)^-\otimes \Qp$.
Observe that $E(K)^+\otimes \Qp =E(\Q) \otimes \Qp$. By work of Skinner-Urban \cite{SU}, Nekov\'a\v r \cite{Nekovar}, Gross-Zagier \cite{GZ}, and Kolyvagin \cite{Kolyvagin} we know that
\[
\dim E(K)^+\otimes \Qp \geq 2 \indent \text{and}\indent \dim E(K)^-\otimes \Qp =1.
\]
Then by the Sign Conjecture \cite{MRgrowth} we expect that
\[
L_K \subseteq E(\Q) \otimes \Qp.
\]
Our main motivating question is the following:
\begin{Quest}[Mazur, Rubin]
As $K$ varies, we presumably get different shadow lines $L_K$ -- what are these lines, and how are they distributed in $E(\Q) \otimes \Qp$ ?
\end{Quest}
In order to gather data about this question one can add the assumption that $p$ splits in $K/\mathbb{Q}$ and then make use of the \emph{anticyclotomic $p$-adic height pairing} on $E(K)\otimes\mathbb{Q}_p$. It is known that $\mathcal{U}$ is contained in the kernel of this pairing \cite{MTpairing}. In fact, in our situation we expect that $\mathcal{U}$ equals the kernel of the anticyclotomic $p$-adic height pairing. Indeed we have $\dim E(K)^-\otimes \Qp =1$ and the weak Birch and Swinnerton-Dyer Conjecture for $E/\Q$ predicts that $\dim E(\Q) \otimes \Qp=2$ from which the statement about $\cU$ follows by the properties of the anticyclotomic $p$-adic height pairing and its expected non-triviality. (This is discussed in \S \ref{sec:Shadow} in further detail.) Thus computing the anticyclotomic $p$-adic height pairing allows us to determine the shadow line $L_K$.
Let $\Gamma(K)$ be the Galois group of the maximal $\Zp$-power extension of $K$, and let $I(K) = \Gamma(K) \otimes_{\Zp} \Qp$. Identifying $\Gamma(K)$ with an appropriate quotient of the idele class group of $K$, Mazur, Stein, and Tate \cite[\S 2.6]{MST} gave an explicit description of the universal $p$-adic height pairing
\[
( \, , ) : E(K) \times E(K) \to I(K).
\]
One obtains various $\Qp$-valued height pairings on $E$ by composing this universal pairing with $\Qp$-linear maps $I(K) \to \Qp$. The kernel of such a (non-zero) $\Qp$-linear map corresponds to a $\Zp$-extension of $K$.
In particular, the anticyclotomic $\Zp$-extension of $K$ corresponds to a $\Qp$-linear map $\rho : I(K) \to \Qp$ such that $\rho \circ c = -\rho$, where $c$ denotes complex conjugation. The resulting anticyclotomic $p$-adic height pairing is denoted by $( \, , )_\rho$. One key step of our work is an explicit description of the map $\rho$, see \S \ref{sec:character}. As in \cite{MST}, for $P \in E(K)$ we define the anticyclotomic $p$-adic height of $P$ to be $h_\rho(P) = -\frac{1}{2}(P, P)_\rho$. Mazur, Stein, and Tate \cite[\S2.9]{MST} provide the following formula\footnote{The formula appearing in \cite[\S2.9] {MST} contains a sign error which is corrected here.} for the anticyclotomic $p$-adic height of a point $P \in E(K)$:
\[
h_\rho(P ) = \rho_\pi(\sigma_\pi(P)) - \rho_\pi(\sigma_\pi(P^c)) + \sum_{w \nmid p\infty} \rho_w(d_w(P )),
\]
where $\pi$ is one of the prime divisors of $p$ in $K$ and the remaining notation is defined in \S \ref{sec:Height}. An algorithm for computing $\sigma_\pi$ was given in \cite{MST}. Using our explicit description of $\rho$, in \S \ref{sec:Height} we find a computationally feasible way of determining the contribution of finite primes $w$ which do not divide $p$. This enables us to compute anticyclotomic $p$-adic height pairings.
We then proceed with a general discussion of shadow lines and their identification in $E(\Q) \otimes \Qp$, see \S \ref{sec:Shadow}. In \S \ref{Algs} we present the algorithms that we use to compute anticyclotomic $p$-adic heights and shadow lines. We conclude by displaying in \S\ref{Examples} two examples of the computation of shadow lines $L_K$ on the elliptic curve ``389.a1" with the prime $p=5$ and listing the results of several additional shadow line computations.
\section{Anticyclotomic character}\label{sec:character}
Let $K$ be an imaginary quadratic field with ring of integers $\cO_K$ in which $p$ splits as $p \cO_K=\pi\pi^c$, where $c$ denotes complex conjugation on $K$. Let $\A^\times$ be the group of ideles of $K$. We also use $c$ to denote the involution of $\A^\times$ induced by complex conjugation on $K$. For any finite place $v$ of $K$, denote by $K_v$ the completion of $K$ at $v$, $\cO_v$ the ring of integers of $K_v$, and $\mu_v$ the group of roots of unity in $\cO_v$. Let $\Gamma(K)$ be the Galois group of the maximal $\Zp$-power extension of $K$. As in \cite{MST}, we consider the idele class $\Qp$-vector space $I(K) =\Gamma(K) \otimes_{\Zp} \Qp $. By class field theory $\Gamma(K)$ is a quotient of $J' := \A^\times/\overline{K^\times \C^\times \prod_{w \nmid p} \cO_w^\times}$ by its finite torsion subgroup $T$, see the proof of Theorem 13.4 in \cite{Washington}. The bar in the definition of $J'$ denotes closure in the idelic topology, and the subgroup $T$ is the kernel of the $N$-th power map on $J'$ where $N$ is the order of the finite group
\[
\A^\times/\overline{K^\times \C^\times \prod_{w \nmid p} \cO_w^\times}(1 + \pi\cO_\pi) (1 + \pi^c\cO_{\pi^c}).
\]
Thus we have
\begin{equation}\label{I(K)}
I(K) = J'/T \otimes_{\Zp}\Qp.
\end{equation}
We shall use this idelic description of $\Gamma(K)$ in what follows.
\begin{definition}[Anticyclotomic $p$-adic idele class character] \label{anticyclotomic}
An \emph{anticyclotomic} $p$-adic idele class character is a continuous homomorphism
\[
\rho: \A^\times/K^\times\rightarrow\Zp
\]
such that $\rho\circ c = -\rho$.
\end{definition}
\begin{lemma}\label{factoring characters}
Every $p$-adic idele class character
\[
\rho: \A^\times/K^\times\rightarrow\Zp
\]
factors via the natural projection
\[
\A^\times/K^\times\twoheadrightarrow \A^\times / \Bigl(K^\times\C^\times\prod_{w\nmid p}{\cO_w^\times}\prod_{v\mid p}{\mu_v}\Bigr).
\]
\end{lemma}
\begin{proof}
This is an immediate consequence of the fact that $\Zp$ is a torsion-free pro-$p$ group.
\end{proof}
The aim of this section is to define a non-trivial anticyclotomic $p$-adic idele class character. By the identification \eqref{I(K)}, such a character will give rise to a $\Qp$-linear map $I(K)\rightarrow \Qp$ which cuts out the anticyclotomic $\Zp$-extension of $K$.
\subsection{The class number one case}
We now explicitly construct an anticyclotomic $p$-adic idele class character $\rho$ in the case when the class number of $K$ is $1$.
Recall our assumption that $p$ splits in $K/\Q$ as $p\cO_K=\pi\pi^c$ and let
\[
U_\pi = 1 + \pi\cO_\pi \indent \text{and} \indent U_{\pi^c} = 1 + \pi^c\cO_{\pi^c}.
\]
Define a continuous homomorphism
\[
\varphi : \A^\times \to U_\pi \times U_{\pi^c}
\]
as follows. Let $(x_v)_v \in \A^\times$. Under our assumption that $K$ has class number $1$, we can find $\alpha \in K^\times$ such that
\[
\alpha x_v \in \cO_v^\times \indent \text{ for all finite }v.
\]
Indeed, the ideal $\fa_v$ corresponding to the place $v$ is principal, say generated by $\varpi_v \in \cO_K$. Then take $\alpha = \prod_{v} \varpi_v^{-\ord_v(x_v)}$, where the product is taken over all finite places $v$ of $K$. We define
\begin{equation} \label{varphi}
\varphi((x_v)_v) = ((\alpha x_\pi)^{p - 1}, (\alpha x_{\pi^c})^{p - 1}).
\end{equation}
Note that since $p$ is split in $K$ we have $\cO_\pi^\times \cong \Zp^\times \cong \mu_{p - 1} \times U_\pi$, and similarly for $\pi^c$. To see that $\varphi$ is independent of the choice of $\alpha$, we note that any other choice $\alpha' \in K^\times$ differs from $\alpha$ by an element of $\cO_K^\times$. Since $K$ is an imaginary quadratic field, $\cO_K^\times$ consists entirely of roots of unity. In particular, under the embedding $K \hookrightarrow K_\pi$ we see that $\cO_K^\times \hookrightarrow \mu_{p - 1}$. Thus, any ambiguity about $\alpha$ is killed when we raise $\alpha$ to the $(p - 1)$-power. Therefore, $\varphi$ is well-defined. The continuity of $\varphi$ is easily verified.
\begin{proposition}
Suppose that $K$ has class number $1$. Then the map $\varphi$ defined in \eqref{varphi} induces an isomorphism of topological groups
\[
\A^\times/\Bigl(K^\times\C^\times\prod_{w\nmid p}{\cO_w^\times}\prod_{v\mid p}{\mu_v}\Bigr)\rightarrow U_\pi \times U_{\pi^c}.
\]
\end{proposition}
\begin{proof}
For $v\in\{\pi,\pi^c\}$, the $p$-adic logarithm gives an isomorphism $U_v\cong 1+p\Zp\rightarrow p\Zp$. Hence, raising to the power $(p-1)$ is an automorphism on $U_v$ for $v\in\{\pi,\pi^c\}$ and consequently $\varphi$ is surjective. It is easy to see that $K^\times \C^\times \prod_{w \nmid p} \cO_w^\times \subset \ker \varphi$. Since $\mu_v\cong \F_p^\times$ for $v\in\{\pi,\pi^c\}$, we have $\prod_{v\mid p}{\mu_v}\subset \ker\varphi$.
We claim that $\ker \varphi = K^\times\C^\times\prod_{w \nmid p} \cO_w^\times\prod_{v\mid p}{\mu_v}$. Let $(x_v)_v \in \ker \varphi$ and let $\alpha \in K^\times$ be such that $\alpha x_v \in \cO_v^\times$ for all finite $v$.
It suffices to show that $(\alpha x_v)_v\in\C^\times\prod_{w \nmid p} \cO_w^\times\prod_{v\mid p}{\mu_v}$. This is clear: since $(x_v)_v\in\ker\varphi$, we have $\alpha x_v\in\mu_v$ for $v\in\{\pi,\pi^c\}$.
Finally, since $\varphi$ is a continuous open map, it follows that $\varphi$ induces the desired homeomorphism.
\end{proof}
By Lemma \ref{factoring characters} we have reduced the problem of constructing an anticyclotomic $p$-adic idele class character to the problem of constructing a character
\begin{equation}
\chi : U_\pi \times U_{\pi^c}\to \Zp
\end{equation}
satisfying $\chi \circ c = -\chi$. Note that this last condition implies that $\chi(x, y) = \chi(x/y^c, 1)$. Explicitly:
\begin{equation}\label{anticyc property}
\chi(x, y) = -\chi \circ c(x, y) = -\chi(y^c, x^c) = -\chi(y^c, 1) - \chi(1, x^c) = -\chi(y^c, 1) + \chi(x, 1) = \chi(x/y^c, 1).
\end{equation}
In other words, $\chi$ factors via the surjection
\begin{eqnarray*}
f_\pi: U_\pi \times U_{\pi^c}\twoheadrightarrow U_\pi\\
(x,y)\mapsto x/y^c.
\end{eqnarray*}
Therefore, it is enough to define a character $U_\pi \to \Zp$. Fixing an isomorphism of valued fields $\psi: K_\pi\to \Qp$ gives an identification $U_\pi\cong 1+p\Zp$. Now, up to scaling, there is only one choice of character, namely $\log_p : 1 + p \Zp \to p\Zp$. We write $\log_p$ for the unique group homomorphism $\log_p:\Qp^\times\to (\Qp,+)$ with $\log_p(p)=0$ extending $\log_p : 1 + p \Zp \to p\Zp$. The extension to $\Zp^\times$ of the map $\log_p$ is explicitly given by
\[
\log_p(u)=\frac{1}{p-1}\log_p(u^{p-1}).
\]
We choose the normalization $\rho = \frac{1}{p(p - 1)}\log_p\circ\psi \circ f_\pi \circ \varphi$. We summarize our construction of the anticyclotomic $p$-adic idele class character $\rho$ in the following proposition.
\begin{proposition}\label{charOne}
Suppose that $K$ has class number 1. Fix a choice of isomorphism $\psi:K_\pi\to \Qp$. Consider the map $\rho:\A^\times/K^\times\rightarrow \Zp$ such that
\begin{eqnarray*}
\rho((x_v)_v)
=\frac{1}{p}\log_p\circ\psi\left(\frac{\alpha x_\pi}{\alpha^cx_{\pi^c}^c} \right)
\end{eqnarray*}
where $\alpha\in K^\times$ is such that $\alpha x_v \in \cO_v^\times$ for all finite $v$. Then $\rho$ is the unique (up to scaling) non-trivial anticyclotomic $p$-adic idele class character.
\end{proposition}
\begin{proof}
Let $\alpha\in K^\times$ be such that $\alpha x_v \in \cO_v^\times$ for all finite $v$. By our earlier discussion and the definition of the extension of $\log_p$ to $\Zp^\times$, we have
\begin{eqnarray*}
\rho((x_v)_v) &=& \frac{1}{p(p - 1)}\log_p\circ \psi\left(\frac{(\alpha x_\pi)^{p - 1}}{(\alpha^cx_{\pi^c}^c)^{p - 1}} \right)\\
&=&\frac{1}{p}\log_p\circ\psi\left(\frac{\alpha x_\pi}{\alpha^cx_{\pi^c}^c} \right).
\end{eqnarray*}
\end{proof}
\subsection{The general case}
There is a simple generalization of the construction of $\rho$ to the case when the class number of $K$ may be greater than one. Let $h$ be the class number of $K$. We can no longer define the homomorphism $\varphi$ of \eqref{varphi} on the whole of $\A^\times$ because $\cO_K$ is no longer assumed to be a principal ideal domain. However, we can define
\[
\varphi_h: (\A^\times)^h\to U_\pi\times U_{\pi^c}
\]
in a similar way, as follows. Let $\fa_v$ be the ideal of $K$ corresponding to the place $v$. Then $\fa_v^h$ is principal, say generated by $\varpi_v \in \cO_K$. For $(x_v)_v\in\A^\times$ we set $\alpha(v) = {\varpi_v}^{-\ord_v(x_v)}$. Then $\alpha(v)x_v^h \in \cO_v^\times$ and $\alpha(v) \in \cO_w^\times$ for all $w \neq v$. Note that $\alpha(v) = 1$ for all but finitely many $v$. Set $\alpha = \prod_v \alpha(v)$ and observe that $\alpha x_v^h \in \cO_v^\times$ for all $v$. Then we define $\varphi_h$ by
\begin{equation}\label{varphi_h}
\varphi_h((x_v)_v^h)=((\alpha x_\pi^h)^{p - 1}, (\alpha x_{\pi^c}^h)^{p - 1}).
\end{equation}
Fix an isomorphism $\psi:K_\pi\to \Qp$. As before, we can now use the $p$-adic logarithm to define an anticyclotomic character $\rho:(\A^\times)^h\to \Zp$ by setting
\[
\rho = \frac{1}{p(p - 1)}\log_p \circ \psi \circ f_\pi \circ \varphi_h.
\]
We extend the definition of $\rho$ to the whole of $\A^\times$ by setting $\rho((x_v)_v)=\frac{1}{h}\rho((x_v)_v^h)$.
As in Proposition \ref{charOne}, we now summarize our construction of the anticyclotomic $p$-adic idele class character in this more general setting.
\begin{proposition} \label{general character}
Let $h$ be the class number of $K$. Fix a choice of isomorphism $\psi:K_\pi\to \Qp$. Consider the map $\rho:\A^\times/K^\times\rightarrow \frac{1}{h}\Zp$ such that
\begin{eqnarray*}
\rho((x_v)_v)= \frac{1}{hp}\log_p\circ\psi\left(\frac{\alpha x_\pi^h}{\alpha^cx_{\pi^c}^{ch}} \right)
\end{eqnarray*}
where $\alpha\in K^\times$ is such that $\alpha x_v^h \in \cO_v^\times$ for all finite $v$. Then $\rho$ is the unique (up to scaling) non-trivial anticyclotomic $p$-adic idele class character.
\end{proposition}
\begin{remark}
Note that $\rho:\A^\times/K^\times\rightarrow \frac{1}{h}\Zp$, so if $p\mid h$ then $\rho$ is not strictly an anticyclotomic idele class character in the sense of Definition \ref{anticyclotomic}. However, the choice of scaling of $\rho$ is of no great importance since our purpose is to use $\rho$ to define an anticyclotomic height pairing on $E(K)$ and compute the kernel of this pairing.
\end{remark}
\begin{remark} \label{alpha}
The ideal $\prod_v \fa_v^{-h\, \ord_v(x_v)}$ is principal and a generator of this ideal is the element $\alpha \in K$ that we use when evaluating the character $\rho$ defined in Proposition \ref{general character}.
\end{remark}
\section{Anticyclotomic $p$-adic height pairing}\label{sec:Height}
We wish to compute the anticyclotomic $p$-adic height $h_\rho$ using our explicit description of the anticyclotomic idele class character $\rho$ given in Proposition \ref{general character}. For any finite prime $w$ of $K$, the natural inclusion $K_w^\times\hookrightarrow \A^\times$ induces a map $\iota_w : K_w^\times \rightarrow I(K)$, and we write $\rho_w = \rho \circ \iota_w$. For every finite place $w$ of $K$ and every non-zero point $P \in E(K)$ we can find $d_w(P) \in \cO_w$ and $a_w(P), b_w(P) \in \cO_w$, each relatively prime to $d_w(P)$, such that
\begin{equation}\label{denom}
\left(\iota_w(x(P)), \iota_w(y(P)) \right) = \left(\frac{a_w(P)}{d_w(P)^2}, \frac{b_w(P)}{d_w(P)^3} \right).
\end{equation}
We refer to $d_w(P)$ as a \textit{local denominator} of $P$ at $w$. The existence of $d_w(P)$ follows from the Weierstrass equation for $E$ and the fact that $\cO_w$ is a principal ideal domain. Finally, we let $\sigma_\pi$ denote the $\pi$-adic $\sigma$-function of $E$.
Given a non-torsion point $P \in E(K)$ such that
\begin{itemize}
\item $P$ reduces to $0$ modulo primes dividing $p$, and
\item $P$ reduces to the connected component of all special fibers of the Neron model of $E$,
\end{itemize}
we can compute its anticyclotomic $p$-adic height using the following formula\footnote{ The formula appearing in \cite[\S2.9] {MST} contains a sign error which is corrected here.} \cite[\S2.9] {MST} :
\begin{equation}\label{eq:height}
h_\rho(P) = \rho_\pi(\sigma_\pi(P)) - \rho_\pi(\sigma_\pi(P^c)) + \sum_{w \nmid p\infty} \rho_w(d_w (P )).
\end{equation}
In the following lemmas, we make some observations which simplify the computation of $h_{\rho}(P)$.
\begin{lemma} \label{units die}
Let $w$ be a finite prime such that $w\nmid p$. Let $x_w\in K_w^\times$. Then $\rho_w(x_w)$ only depends on $\ord_w(x_w)$. In particular, if $x_w\in\cO_w^\times$, then $\rho_w(x_w)=0$.
\end{lemma}
\begin{proof}
This follows immediately from Lemma \ref{factoring characters}. Alternatively, note that the auxiliary element $\alpha$ used in the definition of $\rho$ only depends on the valuation of $x_w$.
\end{proof}
\begin{lemma} \label{char-prime}
Let $w$ be a finite prime of $K$. Then $\rho_{w^c}=-\rho_{w}\circ c$. In particular, if $w=w^c$, then $\rho_w=0$.
\end{lemma}
\begin{proof}
This is an immediate consequence of the relations $\rho \circ c = -\rho$ and $c \circ \iota_{\lambda^c} = \iota_\lambda \circ c$.
\end{proof}
Lemma \ref{char-prime} allows us to write the formula \eqref{eq:height} for the anticyclotomic $p$-adic height as follows:
\begin{equation}\label{simplified height}
h_{\rho}(P) = \rho_\pi\left(\frac{\sigma_\pi(P)}{\sigma_\pi(P^c)} \right) +\sum_{\substack{\ell = \lambda\lambda^c\\ \ell \neq p}} \rho_\lambda\left(\frac{d_\lambda(P)}{d_{\lambda^c}(P)^c} \right).
\end{equation}
\begin{remark} In order to implement an algorithm for calculating the anticyclotomic $p$-adic height $h_\rho$, we must determine a finite set of primes which includes all the split primes $\ell = \lambda \lambda^c \nmid p$ for which $\rho_\lambda\left(\frac{d_\lambda(P)}{d_{\lambda^c}(P)^c} \right) \neq 0$. Let $k_\lambda$ be the residue field of $K$ at $\lambda$ and set $\cD (P )= \prod_{\lambda \nmid p\infty} (\#k_\lambda)^{\ord_\lambda(d_\lambda(P ))}.$
It turns out that $\cD (P )$ can be computed easily from the leading coefficient of the minimal polynomial of the $x$-coordinate of $P$ \cite[Proposition 4.2]{BCS}.
Observe that $\rho_\lambda\left(\frac{d_\lambda(P)}{d_{\lambda^c}(P)^c} \right) \neq 0$ implies that $\ord_\lambda(d_\lambda(P ))\neq 0$ or $\ord_{\lambda^c}(d_{\lambda^c}(P)) \neq 0$. Hence, the only primes $\ell\neq p$ which contribute to the sum in \eqref{simplified height} are those that are split in $K/\Q$ and divide $\cD( P)$. However, in the examples that we have attempted, factoring $\cD( P)$ is difficult due to its size.
\end{remark}
\medskip
We now package together the contribution to the anticyclotomic $p$-adic height coming from primes not dividing $p$. Consider the ideal $x(P ) \cO_K$ and denote by $\delta(P)\subset \cO_K$ its denominator ideal. Observe that by \eqref{denom} we know that all prime factors of $\delta(P )$ appear with even powers. Fix $\dd(P ) \in \cO_K$ as follows:
\begin{equation} \label{denom-replacement}
\dd(P )\cO_K =\prod_{\mathfrak{q}}{\mathfrak{q}^{h\, \ord_{\mathfrak{q}}(\delta(P))/2}}
\end{equation}
where $h$ is the class number of $K$, and the product is over all prime ideals $\mathfrak{q}$ in $\cO_K$.
\begin{proposition} \label{packaging}
Let $P \in E(K)$ be a non-torsion point which reduces to $0$ modulo primes dividing $p$, and to the connected component of all special fibers of the Neron model of $E$.
Then the anticyclotomic $p$-adic height of $P$ is
\[
h_\rho(P)= \frac{1}{p}\log_p\left(\psi\left(\frac{\sigma_\pi(P)}{\sigma_\pi(P^c)} \right)\right)+\frac{1}{hp}\log_p\left(\psi\left(\frac{\dd(P )^c}{\dd( P)}\right)\right),
\]
where $\psi:K_\pi\to \Qp$ is the fixed isomorphism.
\end{proposition}
\begin{proof}
By \eqref{eq:height} we have
\begin{equation}\label{eq:height2}
h_\rho(P) = \rho_\pi\left(\frac{\sigma_\pi(P)}{\sigma_\pi(P^c)} \right) + \sum_{w \nmid p\infty} \rho_w(d_w (P )).
\end{equation}
Let $P=(x,y) \in E(K)$. Since $P$ reduces to the identity modulo $\pi$ and $\pi^c$, we have
\begin{align*}
\ord_{\pi}(x)=-2e_\pi,& \indent \ord_{\pi}(y)=-3e_\pi, \\
\ord_{\pi^c}(x)=-2e_{\pi^c},& \indent \ord_{\pi^c}(y)=-3e_{\pi^c},
\end{align*}
for positive integers $e_\pi$ and $e_{\pi^c}$. Since the $p$-adic $\sigma$ function has the form $\sigma(t)=t+\cdots\in t\Zp[[t]]$, we see that
\begin{eqnarray*}
\ord_\pi(\sigma_\pi(P))=\ord_\pi\left(\sigma_\pi\left(\frac{-x}{y}\right)\right)=\ord_\pi\left(\frac{-x}{y}\right)=e_\pi
\end{eqnarray*}
and similarly
\[\ord_\pi(\sigma_\pi(P^c))=\ord_\pi\left(\frac{-x^c}{y^c}\right)=\ord_{\pi^c}\left(\frac{-x}{y}\right)=e_{\pi^c}.\]
Thus,
\begin{equation}
\label{eq:order at pi}
\ord_{\pi}\left(\frac{\sigma_\pi(P)}{\sigma_\pi(P^c)} \right)=e_\pi-e_{\pi^c}.
\end{equation}
Let $\alpha\in K^\times$ generate the principal ideal $\pi^{h}$. By \eqref{eq:order at pi} and the definition of the anticyclotomic $p$-adic idele class character, we have
\begin{eqnarray*}
\rho_\pi\left(\frac{\sigma_\pi(P)}{\sigma_\pi(P^c)} \right)&=& \frac{1}{hp}\log_p\circ\psi\left(\frac{\alpha^{e_{\pi^c}-e_\pi}\sigma_\pi(P)^h}{(\alpha^c)^{e_{\pi^c}-e_\pi}\sigma_\pi(P^c)^h }\right)\\
&=&\frac{1}{p}\log_p\left(\psi\left(\frac{\sigma_\pi(P)}{\sigma_\pi(P^c)}\right)\right)+\frac{1}{hp}\log_p\left(\psi\left(\frac{\alpha}{\alpha^c}\right)^{e_{\pi^c}-e_\pi}\right).
\end{eqnarray*}
Now it remains to show that
\begin{equation}
\label{eq:cancellation}
\sum_{w \nmid p\infty} \rho_w(d_w (P ))=\frac{1}{hp}\log_p\left(\psi\left(\frac{\dd ( P)^c}{\dd ( P)}\right)\right)-\frac{1}{hp}\log_p\left(\psi\left(\frac{\alpha}{\alpha^c}\right)^{e_{\pi^c}-e_\pi}\right).
\end{equation}
By the definition of $\rho$, we have
\begin{equation} \label{awayfromp}
\sum_{w \nmid p\infty} \rho_w(d_w(P))=\frac{1}{h} \sum_{w \nmid p\infty} \rho_w(d_w(P)^h).
\end{equation}
Since $\ord_w(d_w(P)^h)=\ord_w(\dd ( P))$, Lemma \ref{units die} gives $\rho_w(d_w(P)^h)=\rho_w(\dd ( P))$ for every $w\nmid p\infty$. Substituting this into \eqref{awayfromp} gives
\begin{eqnarray*}
\sum_{w \nmid p\infty} \rho_w(d_w(P))&=&\frac{1}{h} \sum_{w \nmid p\infty}\rho_w(\dd ( P))\\
&=&\frac{1}{h} \sum_{w \nmid p\infty}\rho\circ \iota_w(\dd ( P))\\
&=&\frac{1}{h} \rho\Bigl(\prod_{w \nmid p\infty}\iota_w(\dd ( P))\Bigr).
\end{eqnarray*}
Now $\prod_{w \nmid p\infty}\iota_w(\dd ( P))$ is the idele with entry $\dd ( P)$ at every place $w\nmid p\infty$ and entry $1$ at all other places. Define $\beta\in\cO_K$ by $\dd ( P)=\alpha^{e_\pi}(\alpha^c)^{e_{\pi^c}}\beta$. Thus, by Proposition \ref{general character} and Remark \ref{alpha}, we get
\begin{eqnarray*}
\frac{1}{h} \rho\Bigl(\prod_{w \nmid p\infty}\iota_w(\dd ( P))\Bigr)&=&\frac{1}{hp}\log_p\left(\psi\left(\frac{\beta^c}{\beta} \right)\right)\\
&=&\frac{1}{hp}\log_p\left(\psi\left(\frac{\dd ( P)^c}{\dd ( P)} \right)\right)-\frac{1}{hp}\log_p\left(\psi\left(\frac{\alpha}{\alpha^c}\right)^{e_{\pi^c}-e_\pi}\right)
\end{eqnarray*}
as required. This concludes the proof.
\end{proof}
In \cite{MST}, the authors describe the ``universal'' $p$-adic height pairing $(P,Q)\in I(K)$ of two points $P,Q\in E(K)$. Composition of the universal height pairing with any $\Qp$-linear map $\rho:I(K)\rightarrow \Qp $ gives rise to a canonical symmetric bilinear pairing
\[(\ ,\ )_\rho:E(K)\times E(K)\rightarrow \Qp\]
called the \emph{$\rho$-height pairing}. The $\rho$-height of a point $P\in E(K)$ is defined to be $-\frac{1}{2}(P,P)_\rho$.
Henceforth, we fix $\rho$ to be the anticyclotomic $p$-adic idele class character defined in \S\ref{sec:character}. The corresponding $\rho$-height pairing is referred to as the \emph{anticyclotomic $p$-adic height pairing}, and it is denoted as follows:
\[
\langle\ , \ \rangle=(\ ,\ )_\rho: E(K)\times E(K)\rightarrow \Qp
\]
Observe that
\[
\langle P,Q\rangle=h_\rho(P)+h_\rho(Q)-h_\rho(P+Q).
\]
Let $E(K)^+$ and $E(K)^-$ denote the $+1$-eigenspace and the $-1$-eigenspace, respectively, for the action of complex conjugation on $E(K)$. Since $\sigma_\pi$ is an odd function, using \eqref{simplified height} we see that the anticyclotomic height satisfies
\[
h_\rho(P)=0 \indent \text{ for all }P\in E(K)^+\cup E(K)^-.
\]
Therefore, the anticyclotomic $p$-adic height pairing satisfies
\begin{equation} \label{restriction vanishes}
\langle E(K)^+, E(K)^+\rangle=\langle E(K)^-,E(K)^-\rangle=0.
\end{equation}
Consequently, if $P\in E(K)^+$ and $Q\in E(K)^-$, then
\begin{align} \label{pairing-height}
\langle P,Q\rangle & = h_\rho(P)+h_\rho(Q)-h_\rho(P+Q)\\
&= -\frac{1}{2}\langle P,P\rangle -\frac{1}{2}\langle Q,Q\rangle -h_\rho(P+Q) \nonumber \\
&= -h_\rho(P+Q).\nonumber
\end{align}
\bigskip
\section{The shadow line} \label{sec:Shadow}
Let $E$ be an elliptic curve defined over $\Q$ and $p$ an odd prime of good ordinary reduction. Fix an imaginary quadratic extension $K/\Q$ satisfying the Heegner hypothesis for $E/\Q$ (i.e., all primes dividing the conductor of $E/\Q$ split in $K$). Consider the anticyclotomic $\Zp$-extension $K_\infty$ of $K$. Let $K_n$ denote the subfield of $K_\infty$ whose Galois group over $K$ is isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$. The module of \textit{universal norms} for this $\Zp$-extension is defined as follows:
\[
\cU := \bigcap_{n \geq 0} N_{K_n/K}(E(K_n) \otimes \Zp) \subseteq E(K)\otimes \Zp,
\]
where $N_{K_n/K}$ is the norm map induced by the map $E(K_n) \to E(K)$ given by $P \mapsto \underset{\sigma\in \mathrm{Gal}(K_n/K)}\sum P^\sigma$.
By work of Cornut \cite{Cornut} and Vatsal \cite{Vatsal} we know that for $n$ large enough, we have a non-torsion Heegner point in $E(K_n)$. Since $p$ is a prime of good ordinary reduction, the trace down to $K_{n-1}$ of the Heegner points defined over $K_n$ is related to Heegner points defined over $K_{n-1}$, see \cite[\S 2]{BCS} for further details. Due to this relation among Heegner points defined over the different layers of $K_\infty$, if the $p$-primary part of the Tate-Shafarevich group of $E/K$ is finite then these points give rise to non-trivial universal norms. Hence, if the $p$-primary part of the Tate-Shafarevich group of $E/K$ is finite then $\cU$ is non-trivial whenever the Heegner hypothesis holds. By \cite{Bertolini}, \cite{CW}, and \cite{Ciperiani} we know that if $\Gal (\Q(E_p)/\Q)$ is not solvable then $\cU\simeq \Zp$.
Consider
\[
L_K := \cU \otimes \Qp.
\]
If the $p$-primary part of the Tate-Shafarevich group of $E/K$ is finite then $L_K$ is a line in the vector space $E(K) \otimes \Qp$ known as the \emph{shadow line} associated to the triple $(E, K, p)$. The space $E(K) \otimes \Qp$ splits as the direct sum of two eigenspaces under the action of complex conjugation
\[
E(K) \otimes \Qp= E(K)^+ \otimes \Qp \oplus E(K)^- \otimes \Qp.
\]
Observe that
\[
E(K)^+ \otimes \Qp= E(\Q) \otimes \Qp \indent\text{and}\indent E(K)^- \otimes \Qp\simeq E^K(\Q) \otimes \Qp,
\]
where $E^K$ denotes the quadratic twist of $E$ with respect to $K$. Since the module $\cU$ is fixed by complex conjugation, the shadow line $L_K$ lies in one of the eigenspaces:
\[
L_K \subseteq E(\Q) \otimes \Qp \indent\text{or}\indent L_K \subseteq E(K)^- \otimes \Qp.
\]
The assumption of the Heegner hypothesis forces the analytic rank of $E/K$ to be odd, and hence the dimension of $E(K) \otimes \Qp$ is odd by the Parity Conjecture \cite{Nekovar} and our assumption of the finiteness of the $p$-primary part of the Tate-Shafarevich group of $E/K$. Hence, $\dim E(K)^- \otimes \Qp \neq \dim E(\Q) \otimes \Qp$. The Sign Conjecture states that $L_K$ is expected to lie in the eigenspace of higher dimension \cite{MRgrowth}.
Our main motivating question is the following:
\begin{question}[Mazur, Rubin] \label{Quest}
Consider an elliptic curve $E/\Q$ of positive even analytic rank $r$, an imaginary quadratic field $K$ such that $E/K$ has analytic rank $r+1$, and a prime $p$ of good ordinary reduction such that the $p$-primary part of the Tate-Shafarevich group of $E/\Q$ is finite. By the Sign Conjecture, we expect $L_K$ to lie in $E(\Q)\otimes \Qp$. As $K$ varies, we presumably get different shadow lines $L_K$. What are these lines and how are they distributed in $E(\Q)\otimes \Qp$?
\end{question}
Note that in the statement of the above question we make use of the following results:
\begin{enumerate}
\item Since $E/\Q$ has positive even analytic rank we know that $\dim E(\Q)\otimes \Qp \geq 2$ by work of Skinner-Urban \cite[Theorem 2]{SU} and work of Nekovar \cite{Nekovar} on the Parity Conjecture. \label{Qrank}
\item Since our assumptions on the analytic ranks of $E/\Q$ and $E/K$ imply that the analytic rank of $E^K/\Q$ is $1$, by work of Gross-Zagier \cite{GZ} and Kolyvagin \cite{Kolyvagin} we know that
\begin{enumerate}
\item $\dim E(K)^- \otimes \Qp=1$; \label{Krank}
\item the $p$-primary part of the Tate-Shafarevich group of $E^K/\Q$ is finite, and hence the finiteness of the $p$-primary part of the Tate-Shafarevich group of $E/K$ follows from the finiteness of the $p$-primary part of the Tate-Shafarevich group of $E/\Q$. \label{Sha}
\end{enumerate}
\end{enumerate}
Thus by \eqref{Sha} we know that $L_K \subseteq E(K)\otimes \Qp$, while \eqref{Qrank} and \eqref{Krank} are the input to the Sign Conjecture.
It is natural to start the study of Question \ref{Quest} by considering elliptic curves $E/\Q$ of analytic rank $2$. In this case, assuming that
\begin{equation}\label{algRank}
\rank_\Z E(\Q) =2,
\end{equation}
we identify $L_K$ in $E(\Q)\otimes \Qp$ by making use of the anticyclotomic $p$-adic height pairing, viewing it as a pairing on $E(K)\otimes \Zp$. This method forces us to restrict our attention to quadratic fields $K$ where $p$ splits. It is known that $\cU$ is contained in the kernel of the anticyclotomic $p$-adic height pairing \cite[Proposition 4.5.2]{MTpairing}. In fact, in our situation, the properties of this pairing and \eqref{algRank} together with the fact that $\dim E(K)^- \otimes \Qp=1$ imply that either $\cU$ is the kernel of the pairing or the pairing is trivial. Thus computing the anticyclotomic $p$-adic height pairing allows us to verify the Sign Conjecture and determine the shadow line $L_K$.
In order to describe the lines $L_K$ for multiple quadratic fields $K$, we fix two independent generators $P_1, P_2$ of $E(\Q)\otimes \Qp$ (with $E$ given by its reduced minimal model) and compute the slope of $L_K\otimes \Qp$ in the corresponding coordinate system. For each quadratic field $K$ we compute a non-torsion point $R \in E(K)^-$ (on the reduced minimal model of $E$). The kernel of the anticyclotomic $p$-adic height pairing on $E(K)\otimes \Zp$ is generated by $aP_1+bP_2$ for $a,b\in\Zp$ such that $\langle aP_1+bP_2, R\rangle=0$. Then by \eqref{pairing-height} the shadow line $L_K\otimes \Qp$ in $E(\Q)\otimes \Qp$ is generated by $h_\rho(P_2+R) P_1-h_\rho(P_1+R) P_2$ and its slope with respect to the coordinate system induced by $\{P_1, P_2\}$ equals
\[
-h_\rho(P_1+R)/ h_\rho(P_2+R).
\]
\bigskip
\section{Algorithms} \label{Algs}
Let $E/\Q$ be an elliptic curve of analytic rank $2$; see \cite[Chapter 4]{Bradshaw} for an algorithm that can provably verify the non-triviality of the second derivative of the $L$-function. Our aim is to compute shadow lines on the elliptic curve $E$. In order to do this using the method described in \S \ref{sec:Shadow} we need to
\begin{itemize}
\item verify that $\rank_\Z E(\Q)=2$, and
\item compute two $\Z$-independent points $P_1, P_2 \in E(\Q)$.
\end{itemize}
By work of Kato \cite[Theorem 17.4]{Kato}, computing the $\ell$-adic analytic rank of $E/\Q$ for any prime $\ell$ of good ordinary reduction gives an upper bound on $\rank_\Z E(\Q)$ (see \cite[Proposition 10.1]{SW}). Using the techniques in \cite[$\S 3$]{SW}, which have been implemented in \texttt{Sage}, one can compute an upper bound on the $\ell$-adic analytic rank using an approximation of the $\ell$-adic $L$-series, thereby obtaining an upper bound on $\rank_\Z E(\Q)$. Since the analytic rank of $E/\Q$ is $2$, barring the failure of standard conjectures we find that $\rank_\Z E(\Q) \leq 2$. Then using work of Cremona \cite[Section 3.5]{Cremona} implemented in \texttt{Sage}, we search for points of bounded height, increasing the height until we find two $\Z$-independent points $P_1, P_2\in E(\Q)$. We have thus computed a basis of $E(\Q)\otimes \Qp$.
We will now proceed to describe the algorithms that allow us to compute shadow lines on the elliptic curve $E/\Q$.
\bigskip
\begin{algorithm}\label{Kgenerator} Generator of $E(K)^-\otimes \Qp$.\\
\textit{Input}:
\begin{itemize}
\item an elliptic curve $E/\Q$ (given by its reduced minimal model) of analytic rank $2$,
\item an odd prime $p$ of good ordinary reduction;
\item an imaginary quadratic field $K$ such that
\begin{itemize}
\item the analytic rank of $E/K$ equals $3$, and
\item all rational primes dividing the conductor of $E/\Q$ split in $K$.
\end{itemize}
\end{itemize}
\textit{Output}: A generator of $E(K)^-\otimes \Qp$ (given as a point on the reduced minimal model of $E/\Q$).\\
\begin{enumerate}
\item Let $d\in \Z$ such that $K= \Q(\sqrt d)$. Compute a short model of $E^K$, of the form $y^2 = x^3 + ad^2x + bd^3$.
\item Our assumption on the analytic ranks of $E/\Q$ and $E/K$ implies that the analytic rank of $E^K/\Q$ is $1$. Compute a non-torsion point\footnote{Note that by \cite{GZ} and \cite{Kolyvagin} the analytic rank of $E^K/\Q$ being $1$ implies that the algebraic rank of $E^K/\Q$ is $1$ and the Tate-Shafarevich group of $E^K/\Q$ is finite. Furthermore, in this case, computing a non-torsion point in $E^K(\Q)$ can be done by choosing an auxiliary imaginary quadratic field $F$ satisfying the Heegner hypothesis for $E^K/\Q$ such that the analytic rank of $E^K/F$ is $1$ and computing the corresponding basic Heegner point in $E^K(F )$.} of $E^K(\Q)$ and denote it $(x_0,y_0)$. Then $(\frac{x_0}{d}, \frac{y_0\sqrt{d}}{d^2})$ is an element of $E(K)$ on the model $y^2 = x^3 + ax + b$.
\item Output the image of $(\frac{x_0}{d}, \frac{y_0\sqrt{d}}{d^2})$ on the reduced minimal model of $E$.
\end{enumerate}
\end{algorithm}
\bigskip
\begin{algorithm} \label{height} Computing the anticyclotomic $p$-adic height associated to $(E,K,p)$.\\
\textit{Input}:
\begin{itemize}
\item elliptic curve $E/\Q$ (given by its reduced minimal model);
\item an odd prime $p$ of good ordinary reduction;
\item an imaginary quadratic field $K$ such that $p$ splits in $K/\Q$;
\item a non-torsion point $P\in E(K)$.
\end{itemize}
\textit{Output}: The anticyclotomic $p$-adic height of $P$.\\
\begin{enumerate}
\item Let $p\cO_K= \pi \pi^c$. Fix an identification $\psi: K_{\pi} \simeq \Qp$. In particular, $v_p(\psi(\pi)) = 1$.
\item Let $m_0 = \lcm \{c_{\ell}\}$, where $\ell$ runs through the primes of bad reduction for $E/\Q$ and $c_{\ell}$ is the Tamagawa number at $\ell$. Compute\footnote{Note that Step \ref{bad-red} and Step \ref{p-reduction} are needed to ensure that the point whose anticyclotomic $p$-adic height we will compute using formula \eqref{eq:height} satisfies the required conditions. } $R = m_0 P$. \label{bad-red}
\item Determine the smallest positive integer $n$ such that $nR$ and $nR^c$ reduce to $0\in E(\F_p)$ modulo $\pi$. Note that $n$ is a divisor of $\#E(\F_p)$. Compute\ $T= nR.$ \label{p-reduction}
\item Compute $\dd( R) \in \cO_K$ defined in \eqref{denom-replacement} as a generator of the ideal
\[
\prod_{\mathfrak{q}}{\mathfrak{q}^{h\, \ord_{\mathfrak{q}}(\delta( R))/2}}
\]
where $h$ is the class number of $K$, the product is over all prime ideals $\mathfrak{q}$ of $\cO_K$, and $\delta ( R)$ is the denominator ideal of $x(R )\cO_K$.
\item Let $f_n$ denote the $n$th division polynomial associated to $E$. Compute $\dd(T) = \dd(nR) = f_n(R )^h \dd(R )^{n^2}$.
Note that by Step \eqref{bad-red} and Proposition 1 of Wuthrich \cite{Wuthrich} we see that $f_n(R )^h \dd(R )^{n^2} \in \cO_K$ since $\dd(T)$ is an element of $K$ that is integral at every finite prime.
\item Compute $\sigma_{\pi}(t) := \sigma_p(t)$ as a formal power series in $t\Zp[[t]]$ with sufficient precision. This equality holds since our elliptic curve $E$ is defined over $\Q$.
\item We use Proposition \ref{packaging} to determine the anticyclotomic $p$-adic height of $T$: compute
\begin{align*}
h_\rho(T)&= \frac{1}{p}\log_p\left(\psi\left(\frac{\sigma_\pi(T)}{\sigma_\pi(T^c)} \right)\right)+\frac{1}{hp}\log_p\left(\psi\left(\frac{\dd(T )^c}{\dd( T)}\right)\right)\\
&= \frac{1}{p}\log_p\left(\psi\left(\frac{\sigma_p\left(\frac{-x(T)}{y(T)}\right)}{\sigma_p\left(\frac{-x(T)^c}{y(T)^c}\right)}\right)\right) + \frac{1}{hp}\log_p\left(\psi\left(\frac{\dd(T )^c}{\dd( T)}\right)\right)\\
&= \frac{1}{p}\log_p
\left( \frac {\sigma_p\left(\psi\left(\frac{-x(T)}{y(T)}\right)\right)} {\sigma_p\left(\psi\left(\frac{-x(T)^c}{y(T)^c}\right)\right)}\right)
+ \frac{1}{hp}\log_p\left(\psi\left(\frac{\dd(T )^c}{\dd( T)}\right)\right).\\
\end{align*}
\item Output the anticyclotomic $p$-adic height of $P$: compute\footnote{As a consistency check we compute the height of $nP$ and verify that $h_{\rho}(nP) = \frac{1}{n^2}h_{\rho}( P)$ for positive integers $n\leq 5$.}
\[
h_\rho( P) = \frac{1}{n^2m_0^2}h_\rho(T).
\]
\end{enumerate}
\end{algorithm}
\bigskip
\begin{algorithm} \label{shadow} Shadow line attached to $(E,K,p)$.\\
\textit{Input}:
\begin{itemize}
\item an elliptic curve $E/\Q$ (given by its reduced minimal model) of analytic rank $2$ such that $\mathrm {rank}_\Z E(\Q) =2$;
\item an odd prime $p$ of good ordinary reduction such that the $p$-primary part of the Tate-Shafarevich group of $E/\Q$ is finite;
\item two $\Z$-independent points $P_1, P_2 \in E(\Q)$;
\item an imaginary quadratic field $K$ such that
\begin{itemize}
\item the analytic rank of $E/K$ equals $3$, and
\item $p$ and all rational primes dividing the conductor of $E/\Q$ split in $K$.
\end{itemize}
\end{itemize}
\textit{Output}: The slope of the shadow line $L_K \subseteq E(\Q)\otimes\Q_p$ with respect to the coordinate system induced by $\{P_1, P_2\}$. \\
\begin{enumerate}
\item Use Algorithm \ref{Kgenerator} to compute a non-torsion point $S \in E(K)^-$. We then have generators $P_1, P_2, S$ of $E(K)\otimes \Qp$ such that $P_1, P_2 \in E(\Q)$ and $S \in E(K)^-$ (given as points on the reduced minimal model of $E/\Q$) .
\item Compute $P_1 + S$ and $P_2 +S$.
\item Use Algorithm~\ref{height} to compute\footnote{We compute the height of $P_1 + P_2 + S$ as a consistency check.} the anticyclotomic $p$-adic heights: $h_\rho(P_1+ S)$ and $h_\rho(P_2+S)$.
Finding that at least one of these heights is non-trivial implies that the shadow line associated to $(E,K,p)$ lies in $E(\Q)\otimes \Q_p$, i.e., the Sign Conjecture holds for $(E, K, p)$.
\item The point $h_\rho(P_2+S) P_1-h_\rho(P_1+S) P_2$ is a generator of the shadow line associated to $(E,K,p)$. Output the slope of the shadow line $L_K \subseteq E(\Q)\otimes\Q_p$ with respect to the coordinate system induced by $\{P_1, P_2\}$: compute
\[
-h_\rho(P_1+S)/ h_\rho(P_2+S) \in \Q_p.
\]
\end{enumerate}
\end{algorithm}
\bigskip
\section{Examples}\label{Examples}
Let $E$ be the elliptic curve $\mlq\mlq 389.a1\mrq\mrq$ \cite[\href{http://www.lmfdb.org/EllipticCurve/Q/389.a1}{Elliptic Curve 389.a1}]{lmfdb} given by the model
\[
y^2 + y = x^3 + x^2 - 2x.
\]
We know that the analytic rank of $E/\Q$ equals $2$ \cite[\S6.1]{Bradshaw} and $\rank_\Z E(\Q) = 2$, see \cite{Cremona}. In addition, $5$ and $7$ are good ordinary primes for $E$. We find two $\Z$-independent points
\[
P_1 = (-1,1), P_2 = (0,0) \in E(\Q).
\]
We will now use the algorithms described in \S\ref{Algs} to compute the slopes of two shadow lines on $E(\Q)\otimes \Q_5$ with respect to the coordinate system induced by $\{P_1, P_2\}$.
\bigskip
\subsection{Shadow line attached to $\big(\mlq\mlq 389.a1\mrq\mrq, \Q(\sqrt{-11}),5\big)$}\label{ex1}
The imaginary quadratic field $K = \Q(\sqrt{-11})$ satisfies the Heegner hypothesis for $E$ and the quadratic twist $E^K$ has analytic rank $1$. Moreover, the prime $5$ splits in $K$.
We use Algorithm \ref{Kgenerator} to find a non-torsion point $S = (\frac{1}{4} , \frac{1}{8}\sqrt{-11} - \frac{1}{2}) \in E(K)^-$. We now proceed to compute the anticyclotomic $p$-adic heights of $P_1 + S$ and $P_2 + S$ which are needed to determine the slope of the shadow line associated to the triple $\big(\mlq\mlq389.a1\mrq\mrq, \Q(\sqrt{-11}),5\big)$.
We begin by computing
\begin{align*}
A_1 &:= P_1 + S = \left(-\frac{6}{25}\sqrt{-11} + \frac{27}{25}, -\frac{62}{125}\sqrt{-11} + \frac{29}{125}\right),\\
A_2 &:= P_2 + S = (-2\sqrt{-11}, -4\sqrt{-11}- 12).
\end{align*}
We carry out the steps of Algorithm \ref{height} to compute $h_{\rho}(A_1)$:
\begin{enumerate}
\item Let $5 \cO_K = \pi \pi^c$, where $\pi = (\frac{1}{2}\sqrt{-11} + \frac{3}{2})$ and $\pi^c = (-\frac{1}{2}\sqrt{-11} + \frac{3}{2})$. This allows us to fix an identification
\[
\psi: K_{\pi} \rightarrow \Q_5
\]
that sends
\[
\frac{1}{2}\sqrt{-11} + \frac{3}{2} \mapsto 2 \cdot 5 + 5^{2} + 3 \cdot 5^{3} + 4 \cdot 5^{4} + 4 \cdot 5^{5} + 3 \cdot 5^{7} + 5^{8} + 5^{9} + O(5^{10}).
\]
\item Since the Tamagawa number at 389 is trivial, i.e., $c_{389} = 1$, we have $m_0 = 1$. Thus $R = A_1$.
\item We find that $n = 9$ is the smallest multiple of $R$ and $R^c$ such that both points reduce to $0$ in $E(\cO_K/\pi)$. Set $T = 9R$.
\item Note that the class number of $K$ is $h = 1$. We find $\dd( R) = \frac{1}{2} \sqrt{-11} - \frac{3}{2}$.
\item Let $f_9$ denote the $9$th division polynomial associated to $E$. We compute
\begin{align*}
\dd(T) &= \dd(9R) \\
&= f_9(R ) \dd(R )^{9^2}\\
&= 24227041862247516754088925710922259344570 \sqrt{-11} \\
&\qquad - 147355399895912034115896942557395263175125.
\end{align*}
\item We compute
\begin{align*}\sigma_{\pi}(t) :&= \sigma_5(t)\\
&=t + \left(4 + 5 + 3 \cdot 5^{2} + 5^{3} + 2 \cdot 5^{4} + 3 \cdot 5^{5} + 2 \cdot 5^{6} + O(5^{8})\right)t^{3} \\
&\quad+ \left(3 + 2 \cdot 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 2 \cdot 5^{4} + 2 \cdot 5^{5} + 2 \cdot 5^{6} + O(5^{7})\right)t^{4} \\
&\quad+ \left(1 + 5 + 5^{2} + 5^{3} + 3 \cdot 5^{4} + 3 \cdot 5^{5} + O(5^{6})\right)t^{5}\\
&\quad + \left(4 + 2 \cdot 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 3 \cdot 5^{4} + O(5^{5})\right)t^{6}\\
&\quad + \left(4 + 3 \cdot 5 + 4 \cdot 5^{2} + O(5^{4})\right)t^{7} + \left(3 + 3 \cdot 5^{2} + O(5^{3})\right)t^{8}\\
&\quad + \left(3 \cdot 5 + O(5^{2})\right)t^{9} + \left(2 + O(5)\right)t^{10} + O(t^{11}).
\end{align*}
\item We use Proposition \ref{packaging} to determine the anticyclotomic $p$-adic height of $T$: we compute
\begin{align*}
h_\rho(T)&= \frac{1}{p}\log_p\left(\psi\left(\frac{\sigma_\pi(T)}{\sigma_\pi(T^c)} \right)\right)+\frac{1}{hp}\log_p\left(\psi\left(\frac{\dd(T )^c}{\dd( T)}\right)\right)\\
&= \frac{1}{p}\log_p
\left( \frac {\sigma_p\left(\psi\left(\frac{-x(T)}{y(T)}\right)\right)} {\sigma_p\left(\psi\left(\frac{-x(T)^c}{y(T)^c}\right)\right)}\right)
+ \frac{1}{hp}\log_p\left(\psi\left(\frac{\dd(T )^c}{\dd( T)}\right)\right)\\
&= 3 + 5 + 5^{2} + 4 \cdot 5^{4} + 3 \cdot 5^{5} + 4 \cdot 5^{7} + 3 \cdot 5^{8} + 5^{9} +O(5^{10}).
\end{align*}
\item We output the anticyclotomic $p$-adic height of $A_1$:
\begin{align*}
h_\rho(A_1) &= \frac{1}{9^2}h_\rho(T) \\
&= 3 + 3 \cdot 5 + 3 \cdot 5^{2} + 2 \cdot 5^{4} + 4 \cdot 5^{5} + 4 \cdot 5^{6} + 3 \cdot 5^{8} + O(5^{10}).
\end{align*}
\end{enumerate}
Repeating Steps (1) -- (8) for $A_2$ yields
\[
h_{\rho}(A_2) = 3 + 2 \cdot 5 + 4 \cdot 5^{2} + 2 \cdot 5^{5} + 5^{6} + 4 \cdot 5^{7} + 4 \cdot 5^{9} +O(5^{10}).
\]
As a consistency check, we also compute
\[
h_\rho(P_1 + P_2 + S) = 1 + 5 + 3 \cdot 5^{2} + 5^{3} + 2 \cdot 5^{4} + 5^{5} + 5^{6} + 4 \cdot 5^{8} + 4 \cdot 5^{9} + O(5^{10}).
\]
Observe that, numerically, we have
\[
h_\rho(P_1 + P_2 + S) = h_\rho(P_1 + S) + h_\rho(P_2 + S).
\]
The slope of the shadow line $L_K \subseteq E(\Q)\otimes\Q_p$ with respect to the coordinate system induced by $\{P_1, P_2\}$ is thus
\[
-\frac{h_\rho(P_1+S)}{h_\rho(P_2+S)}=4 + 2 \cdot 5 + 5^{2} + 3 \cdot 5^{3} + 5^{4} + 5^{6} + 5^{7} + O(5^{10}).
\]
\bigskip
\subsection{Shadow line attached to $\big(\mlq\mlq 389.a1\mrq\mrq, \Q(\sqrt{-24}),5\big)$}
Consider the imaginary quadratic field $K = \Q(\sqrt{-24})$. Note that $K$ satisfies the Heegner hypothesis for $E$, the twist $E^K$ has analytic rank $1$, and the prime $5$ splits in $K$.
Using Algorithm \ref{Kgenerator} we find a non-torsion point $S = \left(\frac{1}{2} ,\frac{1}{8} \sqrt{-24} - \frac{1}{2}\right) \in E(K)^-$. We then compute
\begin{align*}P_1 + S &= \left(-\frac{1}{6} \sqrt{-24} + \frac{1}{3}, -\frac{5}{18} \sqrt{-24} - 1\right)\\
P_2 + S &= \left(-\frac{1}{2} \sqrt{-24} - 2, -6\right).\end{align*}
Many of the steps taken to compute $h_\rho(P_1 + S)$ and $h_\rho(P_2 + S)$ are quite similar to those in $\S\ref{ex1}$. One notable difference is that in this example the class number $h$ of $K$ is equal to $2$. We find that
\begin{align*}
h_\rho(P_1 + S) &= 4 + 2 \cdot 5 + 3 \cdot 5^{4} + 2 \cdot 5^{5} + 4 \cdot 5^{6} + 2 \cdot 5^{7} + 5^{8} + 2 \cdot 5^{9} + O(5^{10}),\\
h_\rho(P_2 + S ) &=1 + 5 + 5^{3} + 5^{5} + 2 \cdot 5^{6} + 4 \cdot 5^{7} + 2 \cdot 5^{8} + 3 \cdot 5^{9} + O(5^{10}).\\
\end{align*}
In addition, we compute $h_\rho(P_1 + P_2 + S)$ and verify that
\begin{align*}
h_\rho(P_1 + P_2 + S) &= 4 \cdot 5 + 5^{3} + 3 \cdot 5^{4} + 3 \cdot 5^{5} + 5^{6} + 2 \cdot 5^{7} + 4 \cdot 5^{8} + O(5^{10})\\
&= h_\rho(P_1 +S) + h_\rho(P_2 + S).
\end{align*}
This gives that the slope of the shadow line $L_K \subseteq E(\Q)\otimes\Q_p$ with respect to the coordinate system induced by $\{P_1, P_2\}$ is
\[
-\frac{h_\rho(P_1+S)}{h_\rho(P_2+S)}= 1 + 5 + 3 \cdot 5^{2} + 3 \cdot 5^{5} + 3 \cdot 5^{6} + 3 \cdot 5^{7} + 2 \cdot 5^{8} + 5^{9} +O(5^{10}).
\]
\subsection{Summary of results of additional computations of shadow lines}
The algorithms developed in \S \ref{Algs} enable us to compute shadow lines in many examples which is what is needed to initiate a study of Question \ref{Quest}. We will now list some results of additional computations of slopes of shadow lines on the elliptic curve $\mlq\mlq 389.a1\mrq\mrq$. In the following two tables we fix the prime $p=5, 7$ respectively, and vary the quadratic field.
\smallskip
\begin{table}[h]
\begin{center}
\caption{Slopes of shadow lines for $\big(\mlq\mlq 389.a1\mrq\mrq,K, 5\big)$}
\begin{tabular}{||c | l||}
\hline
$K$ & slope \\ \hline
$\Q(\sqrt{-11})$ & $4 + 2 \cdot 5 + 5^{2} + 3 \cdot 5^{3} + 5^{4} + 5^{6} + 5^{7} + O(5^{10})$ \\
$\Q(\sqrt{-19})$ & $1 + 4 \cdot 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 2 \cdot 5^{5} + 5^{6} + 4 \cdot 5^{7} + 3 \cdot 5^{8} + 4 \cdot 5^{9} + O(5^{10})$ \\
$\Q(\sqrt{-24})$ & $1 + 5 + 3 \cdot 5^{2} + 3 \cdot 5^{5} + 3 \cdot 5^{6} + 3 \cdot 5^{7} + 2 \cdot 5^{8} + 5^{9} + O(5^{10})$ \\
$\Q(\sqrt{-59})$ & $4 + 5 + 4 \cdot 5^{2} + 5^{3} + 2 \cdot 5^{4} + 2 \cdot 5^{5} + 3 \cdot 5^{7} + 4 \cdot 5^{8} + 2 \cdot 5^{9} + O(5^{10})$\\
$\Q(\sqrt{ -79})$ & $2 + 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 4 \cdot 5^{4} + 4 \cdot 5^{5} + 3 \cdot 5^{6} + 3 \cdot 5^{7} + 3 \cdot 5^{8} + 2 \cdot 5^{9} + O(5^{10})$\\
$\Q(\sqrt{ -91})$ & $4 + 3 \cdot 5 + 5^{2} + 5^{4} + 2 \cdot 5^{5} + 4 \cdot 5^{6} + 5^{7} + 2 \cdot 5^{9} + O(5^{10})$\\
$\Q(\sqrt{ -111})$ & $5^{-2} + 4 \cdot 5^{-1} + 4 + 4 \cdot 5 + 2 \cdot 5^{2} + 2 \cdot 5^{3} + 4 \cdot 5^{4} + 2 \cdot 5^{5} + 3 \cdot 5^{6} + 5^{7} + 2 \cdot 5^{8} + 5^{9} + O(5^{10})$\\
$\Q(\sqrt{ -119})$ & $4 \cdot 5^{-1} + 2 + 2 \cdot 5 + 2 \cdot 5^{2} + 4 \cdot 5^{3} + 4 \cdot 5^{4} + 2 \cdot 5^{5} + 5^{6} + 4 \cdot 5^{7} + 4 \cdot 5^{8} + 4 \cdot 5^{9} + O(5^{10})$\\
$\Q(\sqrt{ -159})$ & $2 \cdot 5 + 4 \cdot 5^{4} + 4 \cdot 5^{5} + 5^{6} + 5^{7} + 4 \cdot 5^{8} + 5^{9} + O(5^{10})$\\
$\Q(\sqrt{ -164})$ & $3 + 2 \cdot 5 + 4 \cdot 5^{2} + 5^{3} + 4 \cdot 5^{4} + 3 \cdot 5^{5} + 3 \cdot 5^{6} + 3 \cdot 5^{8} + 4 \cdot 5^{9} + O(5^{10})$\\
\hline
\end{tabular}
\end{center}
\end{table}
\smallskip
\begin{table}[h]
\begin{center}
\caption{Slopes of shadow lines for $\big(\mlq\mlq 389.a1\mrq\mrq,K, 7\big)$}
\begin{tabular}{||c | l||}
\hline
$K$ & slope \\ \hline
$\Q(\sqrt{-19})$ & $3 + 2 \cdot 7 + 2 \cdot 7^{2} + 3 \cdot 7^{3} + 7^{4} + 7^{5} + 4 \cdot 7^{7} + 6 \cdot 7^{9} + O(7^{10})$\\
$\Q(\sqrt{ -20})$ & $1 + 5 \cdot 7 + 6 \cdot 7^{2} + 6 \cdot 7^{3} + 2 \cdot 7^{4} + 3 \cdot 7^{5} + 3 \cdot 7^{6} + 3 \cdot 7^{7} + O(7^{10})$ \\
$\Q(\sqrt{ -24})$ & $1 + 3 \cdot 7 + 3 \cdot 7^{2} + 2 \cdot 7^{3} + 6 \cdot 7^{4} + 2 \cdot 7^{5} + 2 \cdot 7^{6} + 6 \cdot 7^{7} + 2 \cdot 7^{8} + O(7^{10})$\\
$\Q(\sqrt{ -52})$ & $1 + 5 \cdot 7 + 7^{2} + 3 \cdot 7^{3} + 3 \cdot 7^{4} + 2 \cdot 7^{5} + 5 \cdot 7^{6} + 3 \cdot 7^{9} + O(7^{10})$\\
$\Q(\sqrt{ -55})$ & $1 + 7 + 6 \cdot 7^{2} + 3 \cdot 7^{3} + 5 \cdot 7^{4} + 3 \cdot 7^{5} + 7^{7} + 4 \cdot 7^{9} + O(7^{10})$ \\
$\Q(\sqrt{ -59})$ & $2 + 7 + 3 \cdot 7^{2} + 3 \cdot 7^{3} + 5 \cdot 7^{4} + 5 \cdot 7^{5} + 2 \cdot 7^{6} + 4 \cdot 7^{7} + 7^{8} + 6 \cdot 7^{9} + O(7^{10})$ \\
$\Q(\sqrt{ -68})$ & $4 + 4 \cdot 7 + 2 \cdot 7^{3} + 5 \cdot 7^{4} + 5 \cdot 7^{5} + 7^{6} + 7^{7} + 5 \cdot 7^{8} + 5 \cdot 7^{9} + O(7^{10})$ \\
$\Q(\sqrt{ -87})$ & $3 \cdot 7 + 4 \cdot 7^{2} + 7^{3} + 2 \cdot 7^{4} + 2 \cdot 7^{5} + 7^{6} + 5 \cdot 7^{7} + 7^{9} + O(7^{10})$ \\
$\Q(\sqrt{ -111})$ & $7^{-2} + 2 \cdot 7^{-1} + 5 + 2 \cdot 7 + 7^{2} + 2 \cdot 7^{3} + 6 \cdot 7^{4} + 5 \cdot 7^{5} + 7^{6} + 2 \cdot 7^{7} + 2 \cdot 7^{9} + O(7^{10})$\\
$\Q(\sqrt{ -143})$ & $5 + 5 \cdot 7 + 2 \cdot 7^{3} + 4 \cdot 7^{4} + 3 \cdot 7^{5} + 3 \cdot 7^{6} + 2 \cdot 7^{7} + 2 \cdot 7^{9} + O(7^{10})$ \\
\hline
\end{tabular}
\end{center}
\end{table}
| 65,688
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Everything you need to understand or teach The Exile by William Kotzwinkle.
The Exile brings the old world of decadence, corruption, and infinite cynicism into direct juxtaposition with William Kotzwinkle's new world of comic absurdity so that "past and present collide; fascist Germany and contemporary Los Angeles interpenetrate." The problem that Kotzwinkle had to solve was how to join his continuously evolving comic vision of an absurd urban society—the Los Angeles of the spaced-out 1980s—to his sinister sense of Berlin as the symbolic city of death. The linkage between Los Angeles and Berlin is developed further by Kotzwinkle's mordant satire of the film industry (which he calls "accurate reporting") and by his exceptionally vivid evocation of the atmosphere of wartime Germany, a familiar subject that is pictured in the novel with an intensity and clarity that exposes the horror and terror anew.
(read more from the Short Guide)
| 48,234
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Did you know that November is Diabetes Awareness Month? I honestly had no idea. I am embarrassed to say that, but I just didn’t know. I have been very fortunate to not have diabetes touch my life, or anyone close to me. I recently met someone who has it, but he manages it very quietly, so you would never know. My friend Carolyn, from All Day I Dream About Food is my closest real connection to diabetes. She is constantly in the kitchen doing amazing things, making over the most decadent dish to make it diabetic friendly.
Today many of us are getting together to help spread the word about Diabetes. Did you know that 25.8 million people (adults and children) in the United States have diabetes. In 2010, 1.9 million new cases of diabetes were diagnosed in people 20 years of age and under. It is a wide spread disease that can affect anyone.
Today Carolyn challenged us to share a recipe that was gluten free, lower in carbs, and didn’t use any type of sugar. I came up with this Greek yogurt dip. It is a little spicy, and you can never go wrong with chili and lime together. Serve with veggies, or your favorite gluten free crackers for a super healthy snack. You can easily serve this at any party, and no one will ever know that you are serving something healthy!
Recent Comments
| 202,225
|
TITLE: Example of non-finitely generated $R$-algebra
QUESTION [2 upvotes]: By definition, an $R$-algebra is a ring homomorphism $f: R \to S$. For example, if $R=\mathbb Z$ and $S= \mathbb Z / n \mathbb Z$ then the projection $k \mapsto k \mod n$ is a ring homomorphism so that $\mathbb Z / n \mathbb Z$ is a $\mathbb Z$-algebra. I think the point of an algebra is that it's a bit like a module in that we extend its structure by adding a ring that is acting on it. In the case of modules, we start with an abelian group and in the case of algebras we start with a ring.
Now for my question: I've been trying to come up with a non-finitely generated $R$-algebra but couldn't. Can someone help me and give me an example? Thank you.
REPLY [3 votes]: You may find the following example illustrative of the questions involved. If $K$ is a field, then the polynomial ring $K[X]$, although it is a (countably) infinite dimensional vector space over $K$, is finitely generated (by $X$ alone) as a $K$-algebra. However the field $K(X)$ of rational functions (quotients of polynomials) is not finitely generated as a $K$-algebra, since any finite set of generators can only produce finitely many irreducible factors in the denominators. The argument is similar to $\mathbf Q$ being non finitely generated as a $\mathbf Z$-algebra (see the comment by Dylan Moreland), because of the inifiniteness of the set of prime factors one needs for denominators.
One can also prove (can you do it?) that the ring $K[[X]]$ of formal power series is not finitely generated as a $K$-algebra. Nor is $\mathbf R$ finitely generated as $\mathbf Q$-algebra; in these examples the sheer (uncountable) dimension as a vector space already excludes finite generation.
| 66,529
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\begin{document}
\maketitle
\begin{abstract}
In this article, we compute the Buchsbaum-Rim function of two variables associated to
a direct sum of cyclic modules and give a formula for the last positive associated Buchsbaum-Rim multiplicity
in terms of the ordinary Hilbert-Samuel multiplicity of an ideal.
This is a generalization of a formula for the last positive Buchsbaum-Rim multiplicity given
by Kirby and Rees.
\end{abstract}
\section{Introduction}
Let $(R, \fkm)$ be a Noetherian local ring with the maximal ideal $\fkm$ of dimension
$d>0$ and let $C$ be a nonzero $R$-module of finite length.
Let $\varphi: R^n \to R^r$ be an $R$-linear map of free modules with $C=\Coker \varphi$,
and put $M:=\Im \varphi \subset F:=R^r$. Then
one can consider the function
$$\lambda_C(p):=\ell_R([\Coker \Sym_R(\varphi)]_{p})=\ell_R(S_{p}/M^{p}), $$
where $S_p$ (resp. $M^p$) is a homogeneous component of
degree $p$ of $S=\Sym_R(F)$ (resp. $R[M]=\Im \Sym_R(\varphi)$). The function of this type was introduced by
Buchsbaum-Rim \cite{BR2} and they proved that $\lambda_C(p)$ is eventually a polynomial of degree $d+r-1$.
Then they defined a multiplicity of $C$ as
$$e(C):=\mbox{(The coefficient of} \ p^{d+r-1} \ \mbox{in the polynomial})\times (d+r-1)!, $$
which is now called the {\it Buchsbaum-Rim multiplicity} of $C$. They also proved that it
is independent of the choice of $\varphi$.
Note that the Buchsbaum-Rim multiplicity $e(R/I)$ of a cyclic module $R/I$ defined
by an $\fkm$-primary ideal $I$ in $R$
coincides with the ordinary Hilbert-Samuel multiplicity $e(I)$ of the ideal $I$.
More recently, Kleiman-Thorup \cite{KT1, KT2} and Kirby-Rees \cite{KR1, KR2} introduced another kind of multiplicities
which is related to the Buchsbaum-Rim multiplicity. They considered the function of two variables
$$\Lambda(p, q):={\ell}_R(S_{p+q}/M^{p}S_{q}), $$
and proved that it is eventually a polynomial of total degree $d+r-1$.
Then they defined a sequence of multiplicities of $C$ as, for $j=0, 1, \dots , d+r-1$,
$$e^j(C):=(\mbox{The coefficient of} \ p^{d+r-1-j}q^j \ \mbox{in the polynomial})\times (d+r-1-j)!j! $$
and proved that it is independent of the choice of $\varphi$. Moreover they proved that
$$e(C) = e^0(C) \geq e^1(C) \geq \dots \geq e^{r-1}(C)>e^r(C)= \dots = e^{d+r-1}(C)=0$$
where $r=\mu_R(C)$ is the minimal number of generators of $C$. Namely,
the first multiplicity $e^0(C)$ is just the classical Buchsbaum-Rim multiplicity $e(C)$, and the sequence is always
a descending sequence of non-negative integers with $e^j(C)=0$ if $j \geq r$, and $e^{r-1}(C)$ is the last positive
multiplicity. Thus the multiplicity $e^j(C)$ is now called {\it $j$-th Buchsbaum-Rim multiplicity} of $C$ or
the {\it associated Buchsbaum-Rim multiplicity} of $C$.
In this article, we investigate the detailed relation between the classical Buchsbaum-Rim multiplicity $e(C)=e^0(C)$
and the other one $e^j(C)$ for $j=1, 2, \dots , r-1$ by computing these invariants in a certain concrete case.
There are some computation of the classical Buchsbaum-Rim multiplicity
(see \cite{Bi, CLU, J, KR1, KR2} for instance).
However, it seems that the computation of the other associated Buchsbaum-Rim multiplicities is done only
for very special cases \cite{Ha, KR1, KR2}.
One of the important cases is the case where $C=R/I_1 \oplus \dots \oplus R/I_r$ is a
direct sum of cyclic modules. This case was first considered by Kirby-Rees \cite{KR1, KR2} and
they gave an interesting formula for the classical Buchsbaum-Rim multiplicity $e(C)=e^0(C)$ in terms of mixed multiplicities of ideals (see also \cite{Bi} for more direct approach).
\begin{Theorem} {\rm (Kirby-Rees \cite{KR2})}
Let $I_1, \dots , I_r$ be $\fkm$-primary ideals in $R$. Then we have a formula
$$e(R/I_1 \oplus \dots \oplus R/I_r)=\sum_{\stackrel{i_1, \dots , i_r \geq 0}{i_1+\dots +i_r=d}}e_{i_1 \cdots i_r}(I_1, \dots , I_r), $$
where $e_{i_1 \cdots i_r}(I_1, \dots , I_r)$ is the mixed multiplicity of $I_1, \dots , I_r$ of type $(i_1, \dots , i_r)$.
\end{Theorem}
For the other multiplicities $e^j(R/I_1 \oplus \dots \oplus R/I_r)$ where $j=1, \dots , r-1$,
Kirby-Rees \cite{KR2} considered the special case where $I_1 \subset \dots \subset I_r$ and proved the following.
\begin{Theorem} {\rm (Kirby-Rees \cite{KR2})}
Let $I_1, \dots , I_r$ be $\fkm$-primary ideals in $R$. Suppose that $I_1 \subset \dots \subset I_r$. Then for any $j=1, \dots , r-1$,
$$e^j(R/I_1 \oplus \dots \oplus R/I_r)=e(R/I_{j+1} \oplus \dots \oplus R/I_r). $$
In particular, the last positive associated Buchsbaum-Rim
multiplicity
$$e^{r-1}(R/I_1 \oplus \dots \oplus R/I_r)=e(R/I_r)$$
is the Hilbert-Samuel multiplicity of $I_r$.
\end{Theorem}
The purpose of this article is to compute $e^j(R/I_1 \oplus \dots \oplus R/I_r)$ for any $\fkm$-primary
ideals $I_1, \dots , I_r$ in $R$ and give a formula for the last positive associated Buchsbaum-Rim multiplicity $e^{r-1}(R/I_1 \oplus \dots \oplus R/I_r)$ in terms of the ordinary Hilbert-Samuel multiplicity of a sum of ideals.
Here is the main result.
\begin{Theorem}\label{main}
Let $I_1, \dots , I_r$ be arbitrary $\fkm$-primary ideals in $R$. Then we have a formula
$$e^{r-1}(R/I_1 \oplus \dots \oplus R/I_r)=e(R/I_1 + \dots + I_r). $$
In particular, if $I_1, \dots , I_{r-1} \subset I_r$,
$$e^{r-1}(R/I_1 \oplus \dots \oplus R/I_r)=e(R/I_r).$$
\end{Theorem}
This extends the Kirby-Rees formula
for the last positive associated Buchsbaum-Rim multiplicity
and of our previous result \cite{Ha}. Our approach is a direct computation of the Buchsbaum-Rim function
of two variables by using some ideas which is different from the one in \cite{KR2}.
We note that it seems to be difficult to get the general formula by applying their approach \cite{KR2}.
Moreover, our approach indicates the general formula for any other associated Buchsbaum-Rim multiplicities
$e^j(C)$ for $j=1, \dots , r-1$ which we will discuss and present it elsewhere.
The proof of Theorem \ref{main} will be given in section 3.
Section 2 is a preliminary character.
In section 2, we will give a few elementary lemmas that we will use in the proof of Theorem \ref{main}.
Our notation will be also fixed in this section.
Throughout this article, let $(R, \fkm)$ be a Noetherian local ring with the maximal ideal $\fkm$ of dimension $d>0$.
Let $r>0$ be a fixed positive integer and let $[r]=\{1, \dots , r\}$. For a finite set $A$, ${}^{\sharp} A$ denotes the number of elements of $A$.
Vectors are always written in bold-faced letters, e.g., $\boldsymbol i =(i_1, \dots , i_r)$.
We work in the usual multi-index notation. Let $I_1, \dots , I_r$ be ideals in $R$ and
let $t_1, \dots , t_r$ be indeterminates.
Then for a vector $\boldsymbol i =(i_1, \dots , i_r) \in \mathbb Z_{\geq 0}^r$,
we denote $\boldsymbol I^{\boldsymbol i}=I_1^{i_1} \cdots I_r^{i_r}, \boldsymbol t^{\boldsymbol i}=t_1^{i_1}
\cdots t_r^{i_r}$ and $| \boldsymbol i | =i_1+ \dots + i_r$.
For vectors $\boldsymbol a, \boldsymbol b \in \mathbb Z^r$,
$\boldsymbol a \geq \boldsymbol b \stackrel{{\rm def}}{\Leftrightarrow} a_i \geq b_i \ \mbox{for all} \ i=1, \dots , r.$
Let $\boldsymbol 0=(0, \dots , 0)$ be the zero vector in $\mathbb Z_{\geq 0}^r$ and let
$\boldsymbol e=(1, 1, \dots , 1) \in \mathbb Z_{\geq 0}^{r}$.
\section{Preliminaries}
In what follows, let $I_1, \dots , I_r$ be $\fkm$-primary ideals in $R$ and let $C=R/I_1 \oplus \dots \oplus R/I_r$.
In order to compute the associated Buchsbaum-Rim multiplicity of $C$, by taking a minimal free presentation $R^n \stackrel{\varphi}{\to} R^r \to C \to 0$
where the image of $\varphi$ is given by $M:=\Im \varphi=I_1 \oplus \dots \oplus I_r \subset F:=R^r$, we may
assume that $S=R[t_1, \dots , t_r]$ is a polynomial ring and
$R[M]=R[I_1t_1, \dots , I_rt_r]$ is the multi-Rees algebra of $I_1, \dots , I_r$.
Then it is easy to see that for any $p, q \geq 0$, the module $M^pS_q$ can be expressed as
$${\displaystyle M^{p}S_{q}=
\sum_{\substack{| \boldsymbol n | =p+q \\ \boldsymbol n \geq \boldsymbol 0} }
\Bigg( \sum_
{\substack{| \boldsymbol i |=p \\ \boldsymbol 0 \leq \boldsymbol i \leq \boldsymbol n}} \boldsymbol
I^{\boldsymbol i}
\Bigg)
\boldsymbol t^{\boldsymbol n}. }
$$
Here we consider a finite set $H_{p, q}:=\{ \boldsymbol n \in \mathbb Z_{\geq 0}^r \mid |\boldsymbol n |=p+q \}. $
For any $\boldsymbol n \in H_{p, q}$, let
$${\displaystyle J_{p,q}({\boldsymbol n}):=\sum_
{\substack{| \boldsymbol i|=p \\ \boldsymbol 0 \leq \boldsymbol i \leq \boldsymbol n}} \boldsymbol I^
{\boldsymbol i}
}, $$
which is an ideal in $R$.
Then the function $\Lambda(p,q)$ can be described as
$${\displaystyle \Lambda(p, q) = \sum_{\boldsymbol n \in H_{p,q}}
\ell_R(R/J_{p, q}({\boldsymbol n})). }$$
For a subset $\Delta \subset H_{p, q}$, we set
$$\Lambda_{\Delta}(p, q):=\sum_{\boldsymbol n \in \Delta}
\ell_R(R/J_{p, q}({\boldsymbol n})). $$
Here we define special subsets of $H_{p, q}$, which will be often used in the proof of Theorem \ref{main}.
For $p, q>0$ and $k=1, \dots , r$, let
$$\Delta_{p, q}^{(k)}:=\{\boldsymbol n \in H_{p, q} \mid n_1, \dots , n_k>p, n_{k+1}+ \dots + n_r \leq p \}. $$
With this notation, we begin with the following.
\begin{Lemma}\label{lem1}
Let $p, q>0$ and $k=1, \dots , r$.
Then for any $\boldsymbol n \in \Delta_{p, q}^{(k)}$, we have the equality
$$J_{p, q}(\boldsymbol n)=(I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)} \prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j}. $$
\end{Lemma}
\begin{proof}
Let $\boldsymbol n \in \Delta_{p, q}^{(k)}$. Then
\begin{multline*}
J_{p, q}(\boldsymbol n) = \sum_{\substack{| \boldsymbol i|=p \\ \boldsymbol 0 \leq \boldsymbol i \leq \boldsymbol n}} \boldsymbol I^{\boldsymbol i}
\\
\shoveleft{=\sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \cdots \\ 0 \leq i_r \leq n_r}} \Bigg( \sum_{\substack{i_1, \dots , i_k \geq 0 \\ i_1+\dots +i_k=p-(i_{k+1}+ \dots +i_r)}} I_1^{i_1} \cdots I_k^{i_k} \Bigg) I_{k+1}^{i_{k+1}} \cdots I_r^{i_r}
}\\
\shoveleft{=\sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \cdots \\ 0 \leq i_r \leq n_r}} (I_1+\dots +I_k)^{p-(i_{k+1}+\dots +i_r)} I_{k+1}^{i_{k+1}} \cdots I_r^{i_r}
}\\
\shoveleft{= \sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \cdots \\ 0 \leq i_r \leq n_r}} (I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)+(n_{k+1}-i_{k+1})+ \dots + (n_r-i_r)} I_{k+1}^{i_{k+1}} \cdots I_r^{i_r}
}\\
\shoveleft{= (I_1+\cdots +I_k)^{p-(n_{k+1}+\dots +n_r)}\sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \cdots \\ 0 \leq i_r \leq n_r}} (I_1+\dots +I_k)^{(n_{k+1}-i_{k+1})+ \dots + (n_r-i_r)} I_{k+1}^{i_{k+1}} \cdots I_r^{i_r}.}
\end{multline*}
Here one can easily compute the above last sum as
$$\sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \cdots \\ 0 \leq i_r \leq n_r}} (I_1+\dots +I_k)^{(n_{k+1}-i_{k+1})+ \dots + (n_r-i_r)} I_{k+1}^{i_{k+1}} \cdots I_r^{i_r}
=\prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j}. $$
Then we have the desired equality.
\end{proof}
\begin{Lemma}\label{lem2}
Let $p, q>0$ with $q \geq (p+1)r$ and let $k=1, \dots , r$ and $0 \leq m \leq p$. Then
$${}^\sharp \left\{ (n_1, \dots , n_k) \in \mathbb Z_{\geq 0}^k \left| \begin{array}{l}
n_1, \dots , n_k>p, \\
n_1+\dots +n_k=p+q-m
\end{array}
\right.
\right\}={q-(k-1)p-1-m \choose k-1}. $$
\end{Lemma}
\begin{proof}
Let $S:=\{ (n_1, \dots , n_k) \in \mathbb Z_{\geq 0}^k \mid n_1, \dots , n_k>p, n_1+\dots +n_k=p+q-m \}$.
Then the map
$\phi : S \to \{ (n_1, \dots , n_k) \in \mathbb Z_{\geq 0}^k \mid n_1+\dots +n_k=p+q-m-k(p+1) \}$
given by
$\phi(\bsn)=\bsn-(p+1) \boldsymbol e$ is bijective so that the number ${}^\sharp S$
is just ${q-(k-1)p-1-m \choose k-1}$ which is the number of monomials of
degree $p+q-m-k(p+1)$ in $k$ variables.
\end{proof}
By Lemmas \ref{lem1} and \ref{lem2}, we have the explicit form of the function $\Lambda_{\Delta_{p, q}^{(k)}}(p, q)$.
\begin{Proposition}\label{basicfunction}
Let $p, q>0$ with $q \geq (p+1)r$ and let $k=1, \dots , r$. Then
$$\Lambda_{\Delta_{p, q}^{(k)}}(p, q)=\sum_{\stackrel{n_{k+1}, \dots , n_r \geq 0}{n_{k+1}+ \dots +n_r \leq p}}
{q-(k-1)p-1-(n_{k+1}+\dots +n_r) \choose k-1} \ell_R(R/\fka), $$
where $\displaystyle{
\fka:=(I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)} \prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j}.
}$
In particular, we have the inequality
$$\Lambda_{\Delta_{p, q}^{(k)}}(p, q) \leq {q-(k-1)p-1 \choose k-1} \lambda_{L}(p), $$
where $\displaystyle{L=R/(I_1+\dots +I_k) \oplus \bigoplus_{j=k+1}^r R/(I_1+\dots +I_k+I_j)}$.
\end{Proposition}
\begin{proof}
Let $p, q>0$ with $q \geq (p+1)r$ and let $k=1, \dots , r$. Then
\begin{multline*}
\Lambda_{\Delta_{p, q}^{(k)}}(p, q)=\sum_{\bsn \in \Delta_{p, q}^{(k)}} \ell_R(R/J_{p, q}(\bsn))
\\
\shoveleft{= \sum_{\bsn \in \Delta_{p, q}^{(k)}} \ell_R \bigg(
R/(I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)} \prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j}
\bigg) \ \ \ \mbox{by Lemma \ref{lem1}}
}\\
\shoveleft{= \sum_{\substack{n_{k+1}, \dots , n_r \geq 0 \\ n_{k+1}+\dots +n_r \leq p}}
\Bigg[
{}^\sharp \left\{ (n_1, \dots , n_k) \in \mathbb Z_{\geq 0}^k
\left|
\begin{array}{l}
n_1, \dots , n_k>p, \\
n_1+\dots +n_k=p+q-(n_{k+1}+\dots +n_r)
\end{array}
\right.
\right\}
}\\
\times \
\ell_R \bigg( R/(I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)} \prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j} \bigg)
\Bigg]
\\
\shoveleft{= \sum_{\substack{n_{k+1}, \dots , n_r \geq 0 \\ n_{k+1}+\dots +n_r \leq p}}
\Bigg[
{q-(k-1)p-1-(n_{k+1}+\dots +n_r) \choose k-1}
}\\
\times \ \ell_R \bigg( R/(I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)} \prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j} \bigg)
\Bigg] \ \ \ \mbox{by Lemma \ref{lem2}}.
\end{multline*}
This proves the first assertion. For the second one, we first note that the above last term is at most
$$ {q-(k-1)p-1 \choose k-1} \sum_{\substack{n_{k+1}, \dots , n_r \geq 0 \\ n_{k+1}+\dots +n_r \leq p}}
\ell_R \big( R/(I_1+\dots +I_k)^{p-(n_{k+1}+\dots +n_r)} \prod_{j=k+1}^r(I_1+\dots +I_k+I_j)^{n_j} \big). $$
Then, since the above last sum is just the ordinary Buchsbaum-Rim function $\lambda_L(p)$ of
$L:=R/(I_1+\dots +I_k) \oplus \bigoplus_{j=k+1}^r R/(I_1+\dots +I_k+I_j )$ by definition, we get the desired
inequality.
\end{proof}
\begin{Remark}\label{rem}
{\rm
As stated in the above proof, the function $\lambda_L(p)$ in Proposition $\ref{basicfunction}$ is
the ordinary Buchsbaum-Rim function of $L$ where $L$ is a direct sum of $(r-k+1)$ cyclic modules.
Therefore the function $\lambda_L(p)$ is a polynomial function of degree $d+r-k$ for all large enough $p$.
}
\end{Remark}
\section{Proof of Theorem \ref{main}}
We prove Theorem \ref{main}. We work under the same situation and use the same notation as in section 2.
In order to investigate the asymptotic property of the function $\Lambda(p, q)$,
we may assume that
\begin{equation}\label{largepq}
q\geq(p+1)r \gg 0.
\end{equation}
In what follows, we fix integers $p, q$ which satisfy the condition (\ref{largepq}).
Let $H:=H_{p, q}$ and let $J(\boldsymbol n):=J_{p, q}(\boldsymbol n) $ for $\boldsymbol n \in H$.
We note here that for any $\bsn \in H$, there exists $i=1, \dots , r$ such that $n_i > p$
because of the condition (\ref{largepq}).
Then the set $H$ can be divided by $r$-regions as follows:
$$H=\coprod_{k=1}^rH^{(k)}, $$
where $H^{(k)}:=\{ \boldsymbol n \in H \mid {}^{\sharp} \{ i \mid n_i > p \}=k \}. $
Hence the function $\Lambda(p, q)$ can be expressed as follows:
$$\Lambda(p, q)=\sum_{k=1}^r \Lambda_{H^{(k)}} (p, q). $$
Therefore it is enough to compute each function $\Lambda_{H^{(k)}}(p, q)$.
When $k=r$, we can compute the function explicitly as follows.
\begin{Proposition}\label{k=r}
$$\Lambda_{H^{(r)}} (p, q)={q-(r-1)p-1 \choose r-1} \ell_R (R/(I_1+\dots +I_r)^p). $$
\end{Proposition}
\begin{proof}
This follows from Proposition \ref{basicfunction} since $H^{(r)}=\Delta_{p, q}^{(r)}$.
\end{proof}
Thus we can reduce the problem to compute functions $\Lambda_{H^{(k)}}(p, q)$ for $k=1, \dots , r-1$.
Let $1 \leq k \leq r-1$. To compute $\Lambda_{H^{(k)}}(p, q)$,
we divide $H^{(k)}$ into ${r \choose k}$-regions as follows:
$$H^{(k)}=\coprod_{\stackrel{A \subset [r]}{{}^{\sharp}A=r-k}} D_A^{(k)}, $$
where $D_A^{(k)}:=\{ \bsn \in H^{(k)} \mid n_i >p \ \mbox{for} \ i \notin A, n_i \leq p \ \mbox{for} \ i \in A \}$.
Then the function $\Lambda_{H^{(k)}}(p, q)$ can be expressed as follows:
$$\Lambda_{H^{(k)}}(p, q)=\sum_{\stackrel{A \subset [r]}{{}^{\sharp}A=r-k}} \Lambda_{D_A^{(k)}} (p, q). $$
When $k=r-1$, we can also compute the function explicitly and get the inequality as in Proposition \ref{basicfunction}.
Here is the inequality we will use in the proof of Theorem \ref{main}.
\begin{Proposition}\label{k=r-1}
There exists a polynomial $g_{r-1}(X) \in \mathbb Q[X]$ of degree $d+1$ such that
$$\Lambda_{H^{(r-1)}}(p, q) \leq {q-(r-2)p-1 \choose r-2}g_{r-1}(p). $$
\end{Proposition}
\begin{proof}
It is enough to show that for any $j=1, \dots , r$,
$$\Lambda_{D_{\{j\}}^{(r-1)}}(p,q) \leq {q-(r-2)p-1 \choose r-2} \lambda_{L_j}(p), $$
where $L_j=R/ (I_1+ \dots +\hat{I_j}+ \dots +I_{r})\oplus R/(I_1+ \dots +I_{r})$ because the function
$\lambda_{L_j}(p)$ is a polynomial function of degree $d+1$ (see Remark \ref{rem}).
We may only consider the case where $j=r$. Then it follows directly from Proposition \ref{basicfunction}
since $D_{\{r\}}^{(r-1)}=\Delta_{p, q}^{(r-1)}$.
\end{proof}
When $1 \leq k \leq r-2$, we can get the same inequality,
although the situation is not simple as in the case where $k=r-1$.
\begin{Proposition}\label{k<r-1}
For any $1 \leq k \leq r-2$, there exists a polynomial $g_{k}(X) \in \mathbb Q[X]$ of degree $d+r-k$
such that $$\Lambda_{H^{(k)}}(p, q) \leq {q-(k-1)p-1 \choose k-1}g_{k}(p). $$
\end{Proposition}
\begin{proof}
Let $1 \leq k \leq r-2$. To prove the desired inequality, it is enough to show that
for any subset $A \subset [r]$ with ${}^{\sharp} A=r-k$, there exists a polynomial
$h_A(X) \in \mathbb Q[X]$ of degree $d+r-k$ such that
$$\Lambda_{D_A^{(k)}}(p, q) \leq {q-(k-1)p-1 \choose k-1}h_{A}(p). $$
To show this, we may only consider the case where $A=\{k+1, k+2, \dots , r\}$.
We then put $D^{(k)}:=D^{(k)}_{\{k+1, \dots , r\}}$.
To investigate $\Lambda_{D^{(k)}}(p, q)$, we divide $D^{(k)}$ into two-parts:
$$D^{(k)}=E_{-}^{(k)} \coprod E_{+}^{(k)}, $$
where
$$E_{-}^{(k)}:=\{ \bsn \in D^{(k)} \mid n_{k+1}+ \dots + n_r \leq p \}, $$
$$E_{+}^{(k)}:=\{ \bsn \in D^{(k)} \mid n_{k+1}+ \dots + n_r > p \}. $$
With this notation, we have the following two lemmas.
\begin{Lemma}\label{k-}
Let $1 \leq k \leq r-2$. Then
$$\Lambda_{E^{(k)}_{-}}(p, q) \leq {q-(k-1)p-1 \choose k-1} \lambda_L(p)$$
where $\displaystyle{L=R/(I_1+\dots +I_k) \oplus \bigoplus_{j=k+1}^r R/(I_1+\dots +I_k+I_j)}$.
\end{Lemma}
\begin{proof}
This follows from Proposition \ref{basicfunction} since $E_{-}^{(k)}=\Delta_{p, q}^{(k)}$.
\end{proof}
\begin{Lemma}\label{k+}
Let $1 \leq k \leq r-2$. Then
there exists a polynomial $h(X) \in \mathbb Q[X]$ of degree $d+r-k$ such that
$$\Lambda_{E^{(k)}_{+}}(p, q) \leq {q-(k-1)p-1 \choose k-1} h(p). $$
\end{Lemma}
\begin{proof}
Let $1 \leq k \leq r-2$. Then we first note that for any $\bsn \in E_{+}^{(k)}, $
\begin{eqnarray}
J(\bsn)&=&\sum_{\substack{\boldsymbol 0 \leq \boldsymbol i \leq \bsn \\ | \boldsymbol i |=p}} \boldsymbol I^{\boldsymbol i} \notag \\
&=&\sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \dots \\ 0 \leq i_r \leq n_r \\ i_{k+1}+\dots +i_r \leq p}}
\Bigg( \sum_{\substack{i_1, \dots , i_k \geq 0 \\ i_1+ \cdots +i_k=p-(i_{k+1}+ \cdots + i_r) }}
I_1^{i_1} \cdots I_k^{i_k} \Bigg) I_{k+1}^{i_{k+1}} \cdots I_r^{i_r} \notag \\
&=& \sum_{\substack{0 \leq i_{k+1} \leq n_{k+1} \\ \dots \\ 0 \leq i_r \leq n_r \\ i_{k+1}+\dots +i_r \leq p}}
(I_1+ \dots +I_k)^{p-(i_{k+1}+ \dots + i_r)} I_{k+1}^{i_{k+1}} \cdots I_r^{i_r}.
\end{eqnarray}
Here we claim the following.
\medskip
{\bf Claim 1.} There exists an $\fkm$-primary ideal $\fkb$ in $R$ such that
for any $\bsn \in E_{+}^{(k)},$
$$\ell_R(R/J(\bsn)) \leq \ell_R(R/\fkb^p). $$
\medskip
Let $\fkb$ be an $\fkm$-primary ideal in $R$ such that $\fkb \subset I_j$
for any $j=k+1, \dots , r$ (such as $\fkb=I_{k+1} \cdots I_r$).
Let $\bsn \in E_{+}^{(k)}. $ Since $n_{k+1}, \dots , n_r \geq 0$ and $n_{k+1}+\dots +n_r > p$,
there exist integers $a_{k+1}, \dots , a_r \in \mathbb Z$
such that
$$\left\{
\begin{array}{l}
0 \leq a_j \leq n_j \ \mbox{for any} \ j=k+1, \dots , r, \ \mbox{and}\\
a_{k+1}+ \cdots +a_r=p.
\end{array}
\right. $$
Then, by the above expression (2) of $J(\bsn)$,
$J(\bsn)
\supset I_{k+1}^{a_{k+1}} \cdots I_r^{a_r}
\supset \fkb^{a_{k+1}+\cdots + a_r}
= \fkb^p$.
Hence we have $\ell_R(R/J(\bsn)) \leq \ell_R(R/\fkb^p)$.
\medskip
Therefore
$$\Lambda_{E_{+}^{(k)}} (p, q)
= \sum_{\bsn \in E_{+}^{(k)}} \ell_R(R/J(\bsn))
\leq \sum_{\bsn \in E_{+}^{(k)}} \ell_R(R/\fkb^p)
={}^{\sharp} E_{+}^{(k)} \cdot \ell_R(R/\fkb^p).
$$
{\bf Claim 2. } There exists a polynomial $h^{\circ}(X) \in \mathbb Q[X]$ of degree $r-k$ such that
$${}^{\sharp} E_{+}^{(k)} \leq {q-(k-1)p-1 \choose k-1} \cdot h^{\circ}(p). $$
To show this, we divide $E_{+}^{(k)}$ as follows:
\begin{eqnarray*}
E_{+}^{(k)} &=& \{ \bsn \in H^{(k)} \mid n_1, \dots , n_k > p, n_{k+1}, \dots , n_r \leq p, n_{k+1}+ \cdots + n_r > p \} \\
&=& \coprod_{\substack{0 \leq n_{k+1} \leq p \\ \cdots \\ 0 \leq n_r \leq p \\ n_{k+1}+ \cdots + n_r > p}} F(n_{k+1}, \dots , n_r)
\end{eqnarray*}
where
$$F(n_{k+1}, \dots , n_r):=\left\{(n_1, \dots , n_r) \in \mathbb Z_{\geq 0}^r \left|
\begin{array}{l}
n_1, \dots , n_k > p, \\
n_1+ \cdots + n_k = p+q-(n_{k+1}+ \dots + n_r)
\end{array}
\right.
\right\}. $$
Therefore
\begin{eqnarray*}
{}^{\sharp}E_{+}^{(k)}&=&\sum_{\substack{0 \leq n_{k+1} \leq p \\ \cdots \\ 0 \leq n_r \leq p \\ n_{k+1}+ \cdots + n_r > p}} {}^{\sharp} F(n_{k+1}, \dots , n_r) \\
&=& \sum_{\substack{0 \leq n_{k+1} \leq p \\ \cdots \\ 0 \leq n_r \leq p \\ n_{k+1}+ \cdots + n_r > p}}
{q-(k-1)p-1-(n_{k+1}+ \dots + n_r) \choose k-1} \ \ \ \ \ \mbox{by Lemma \ref{lem2}} \\
&\leq & {q-(k-1)p-1 \choose k-1} \cdot
{}^{\sharp} \left\{ (n_{k+1}, \dots, n_r) \in \mathbb Z_{\geq 0}^{r-k} \left|
\begin{array}{l}
n_{k+1}, \dots , n_r \leq p, \\
n_{k+1}+ \cdots +n_r > p
\end{array}
\right.
\right\} \\
&\leq & {q-(k-1)p-1 \choose k-1} \cdot
{}^{\sharp} \{ (n_{k+1}, \dots, n_r) \in \mathbb Z_{\geq 0}^{r-k} \mid p < n_{k+1}+ \cdots +n_r \leq (r-k)p \} \\
&=& {q-(k-1)p-1 \choose k-1} \cdot \left\{ {r-k+(r-k)p-1 \choose r-k}-{r-k+p-1 \choose r-k} \right\}.
\end{eqnarray*}
This proves Claim 2.
\medskip
Consequently, we have that
$$\Lambda_{E_{+}^{(k)}} (p, q)
\leq {}^{\sharp} E_{+}^{(k)} \cdot \ell_R(R/\fkb^p)
\leq {q-(k-1)p-1 \choose k-1} \cdot h^{\circ}(p) \cdot \ell_R(R/\fkb^p).$$
Since $\ell_R(R/\fkb^p)$ is a polynomial in $p$ of degree $d$,
the polynomial $h(X)$ which corresponds to $h(p)= h^{\circ}(p) \cdot \ell_R(R/\fkb^p)$ is our desired one.
\end{proof}
By Lemmas \ref{k-} and \ref{k+},
$$\Lambda_{D^{(k)}}(p,q)
=\Lambda_{E_{-}^{(k)}}(p,q)+\Lambda_{E_{+}^{(k)}}(p,q)
\leq {q-(k-1)p-1 \choose k-1}\Big(\lambda_L(p)+h(p)\Big).
$$
This proves Proposition \ref{k<r-1}. \end{proof}
\medskip
Now let me give a proof of Theorem \ref{main}.
\begin{proof}[Proof of Theorem \ref{main}]
By Propositions \ref{k=r-1} and \ref{k<r-1}, for any $k=1, \dots , r-1$,
there exists a polynomial $g_k(X) \in \mathbb Q[X]$ of degree $d+r-k$ such that
$$\Lambda_{H^{(k)}}(p, q) \leq {q-(k-1)p-1 \choose k-1}g_k(p). $$
Since $\displaystyle{\Lambda(p, q)=\Lambda_{H^{(r)}}(p, q)+\sum_{k=1}^{r-1} \Lambda_{H^{(k)}}(p, q)}, $
we have that by Proposition \ref{k=r},
\begin{eqnarray*}
\Lambda(p, q)-{q-(r-1)p-1 \choose r-1} \ell_R(R/(I_1+ \dots +I_r)^p) &\leq & \sum_{k=1}^{r-1} {q-(k-1)p-1 \choose k-1} g_k(p).
\end{eqnarray*}
Therefore, there exists a polynomial $g(X, Y) \in \mathbb Q[X, Y] $ with $\deg_Y g(X, Y) \leq r-2$ such that
$$\Lambda(p, q)-{q-(r-1)p-1 \choose r-1} \ell_R(R/(I_1+ \dots +I_r)^p) \leq g(p, q). $$
The LHS in the above inequality is a polynomial function of two variables with non-negative integer values so that
the function $\Lambda(p, q)$ can be expressed as
$$\Lambda(p, q)={q-(r-1)p-1 \choose r-1} \ell_R(R/(I_1+ \dots +I_r)^p)+f(p, q) $$
for some $f(X, Y) \in \mathbb Q[X, Y]$ with $\deg_Y f(X, Y) \leq r-2$.
Then, by comparing the coefficients of $p^dq^{r-1}$ in the above equality, we obtain the equality
$$e^{r-1}(R/I_1 \oplus \dots \oplus R/I_r)=e(R/I_1+ \dots +I_r). $$
Then we get the desired formula.
\end{proof}
\section*{Acknowledgments}
The author would like to thank the referee for his/her careful reading and constructive suggestions.
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Hyperoptic has connected its first customers in Edinburgh to its up-to-1Gbps Fibre to the Premises (FTTP) broadband services.
Residents of the fancy Platinum Point development, which looks like a weird inversion of the MI6 building, in Newhaven (pictured) can get an entry-level 20Mbps down and 1Mbps up service or symmetrical 100Mbps and 1Gbps services. As they’re delivered via FTTP, the services are not impacted by your location in the same way that ADSL and Fibre to the Cabinet (FTTC) are.
This prestigious development, where two bed flats can be had for a mere £265,000, aren’t the only places where Hyperoptic plans to set up shop; back in May, the FTTP-only ISP announced plans to bring its wares to more corners of the Scottish capital.
Hyperoptic’s national team manager Tim Huxtable said: “The feedback following our announcement that we were coming to Edinburgh has been phenomenal – there is a huge appetite for our gigabit connectivity, which is over 128 times faster than traditional ADSL speeds.
Read next: Brexit, Schmexit: Hyperoptic to reach 500,000 UK homes - thanks to EU cash
“As a nation we are becoming evermore reliant on fast and reliable access to the Internet; buffering and peak time slowdowns just get more frustrating and it’s well documented that people would now actively avoid moving into a property with slow speeds.”
While Hyperoptic does typically target these kind of developments, it’s also worked with local authorities to bring its services to council estates in Wandsworth and Salford; there’s no word yet if the City of Edinburgh Council plans to follow suit.
Either way, the arrival of Hyperoptic in Edinburgh will put pressure on rival ISPs to catch up; BT, which offers FTTP services to customers connected to the Waverley and Portobello exchanges, recently announced that it’s upping top speeds from 330Mbps to 1Gbps this December.
BT and Virgin Media have also announced plans to reach 2 million customers with FTTP over the next four years, although it’s unclear how many of these (if any) will be based in Edinburgh.
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839 F2d 1206 State of Illinois Hartigan v. Panhandle Eastern Pipe Line Company
839 F.2d 1206
56 USLW 2409, 1987-2 Trade Cases 67,869.
No. 85-2601.
United States Court of Appeals,
Seventh Circuit.
Argued Feb. 26, 1986.
Decided Jan. 22, 1988.
Order Granting Rehearing En Banc April 28, 1988.*
Paul H. LaRue, Chadwell & Kayser, Ltd., Chicago, Ill., for defendant-appellant.
Richard L. Miller, Burke & Smith, Chtd., Chicago, Ill., for plaintiff-appellee.
Before BAUER, Chief Circuit Judge, POSNER, Circuit Judge, and FAIRCHILD, Senior Circuit Judge., 97 S.Ct. 2061, 52 L.Ed.2d 707 (1977).
Plaintiff argues, in substance, that the public utility regulation creates a sufficiently close approximation of a pre-existing cost-plus contract for a fixed quantity so as to fulfill that exception. 431 U.S. at 736, 97 S.Ct. at 2069. The district court agreed and denied Panhandle's motion to dismiss. Permission for this appeal was obtained pursuant to 28 U.S.C. Sec.,2, 97 S.Ct. at 2074.
fuels, and they have resorted to conservation in response to an increased price.
CILCO sold 46,400,000 Mcf of gas in 1982, but its sales decreased by 14.7% in 1983 and 2.3% in 1984. It lost 1,061 customers in 1983 and 693 in 1984. Increased prices thus reduced CILCO's sales and profits as well as being passed on to customers to the extent of sales made. Assuming that the prices charged CILCO by Panhandle included an unlawful overcharge, it had adverse consequences on both CILCO and its customers. Again, if the customers' claims were to be recognized, there must be apportionment of recovery between the direct and indirect purchasers.
As the Illinois Brick Court observed, with respect to the pre-existing cost-plus contract,
[i.
As is evident, plaintiff here cannot really claim that CILCO's relationship with its customers falls literally within the confines of a pre-existing cost-plus contract for a fixed quantity. Plaintiff does, however, assert that the relationship is the "functional equivalent" of a cost-plus contract for a fixed quantity, and relies on In re Beef Industry Antitrust Litigation, 600 F.2d 1148 (5th Cir.1979).
In Beef Industry, defendants were retail chains which allegedly conspired to fix at low levels the prices to be paid packers for beef. The packers allegedly computed the price they paid cattlemen by a formula based on the price the packers were to receive. The plaintiffs were cattlemen, claiming that the unlawful undercharge was passed on to them. The Fifth Circuit held that plaintiffs had alleged the functional equivalent of cost-plus contracts, and thus an exception to the Illinois Brick rule which would otherwise have prevented recovery by an indirect seller.3
In Beef Industry, the court noted that economic forces made the supply of fat cattle "inelastic in the short term." 600 F.2d at 1154. It is not clear, however, that the court relied on this inelasticity as the functional equivalent of the predetermination of volume inherent in a cost-plus contract for a fixed quantity. "Functional equivalence is not lost simply because the proponent of passing-on theory cannot demonstrate that the middleman suffered no loss in volume as the result of raising the price to his customers." 600 F.2d at 1164.
In our view, however, the Illinois Brick Court regarded the predetermination of quantity as an essential element of the exception. Thus there could be a "functional equivalent" of a cost-plus contract for a fixed quantity only where factors such as obligations imposed by law or economic forces or a combination of them made inevitable an exact passing on of price variation applied to a predetermined quantity to the same extent as a contract so providing.4
In Hanover Shoe v. United Shoe Machinery Corp., 392 U.S. 481, 494, 88 S.Ct. 2224, 2232, 20 L.Ed.2d 1231 (1968), in suggesting that "a pre-existing 'cost-plus' contract" would be an exception to the Hanover rule against a pass-on defense, there was no express reference to the fixed quantity element of the cost-plus contract. In Illinois Brick, in describing the same exception, there was such a reference, and an explanation of its importance:
In such a situation, the [direct] purchaser is insulated from any decrease in its sales as a result of attempting to pass on the overcharge, because its customer is committed to buying a fixed quantity regardless of price....
As we have noted, supra [431 U.S.] at 735-736 [97 S.Ct. at 2069-2070], Hanover Shoe itself implicitly discouraged the creation of exceptions to its rule barring pass-on defenses, and we adhere to the narrow scope of exemption indicated by our decision there.
431 U.S. at 736, 745, 97 S.Ct. at 2069, 2074.
It appears in the case before us that Panhandle's alleged antitrust violations would produce actual injury to both the direct and indirect purchasers. The indirect purchasers pay a higher price per therm of gas because Panhandle's price is reflected in the price they pay. CILCO, the direct purchaser, is injured by loss of sales and profit. Where this is so, we conclude that Illinois Brick does not permit the indirect purchasers to sue for their part of the injury. It may be that because of the utility regulation requiring that CILCO's cost per therm be passed through to its customers, the avoidance of multiple recovery against Panhandle will be less complex than in a case where the direct purchaser's price to the indirect purchaser will be the result of less channeled market forces. It may be easier to separate accurately the injury to the indirect purchasers from the injury to CILCO. Illinois Brick did not, however, leave it to the discretion of the lower courts to expand the exceptions to include situations within some range of approximation of the exceptions defined in Illinois Brick. We read Illinois Brick as requiring that in this, as well as more complex cases, it is only the direct purchaser who may bring a treble damage action.
The order appealed from is REVERSED, with directions to dismiss the complaint as to all claims by or on behalf of indirect purchasers.
POSNER, Circuit Judge, concurring and dissenting.
Central Illinois Light Company, a retail distributor of natural gas, bought natural gas from the defendant, an interstate gas pipeline company, at prices allegedly inflated because of violations of the antitrust laws by the defendant, and resold the gas to its customers. The question--and it is a difficult one--is whether any of those customers can sue the pipeline company on the theory that CILCO passed on the entire cost of the overcharge to them in the form of higher rates.
Hanover Shoe, Inc. v. United Shoe Machinery Corp., 392 U.S. 481, 88 S.Ct. 2224, 20 L.Ed.2d 1231 (1968), held that it is not a defense to a damages action that a buyer forced to pay a higher price because of the seller's antitrust violation passed on the cost of the violation to his own customers by raising his prices to them--unless the buyer had a cost-plus contract with them. Illinois Brick Co. v. Illinois, 431 U.S. 720, 97 S.Ct. 2061, 52 L.Ed.2d 707 (1977), announced a corollary to Hanover Shoe: the "indirect purchaser" (that is, the customer of the buyer, who is the "direct purchaser") cannot sue to recover the part of the overcharge that the buyer passed on to him. The Court again recognized an exception for the cost-plus contract, noting that it insulates the direct purchaser from "any decrease in its sales as a result of attempting to pass on the overcharge, because its customer is committed to buying a fixed quantity regardless of price." Id. at 736, 97 S.Ct. at 2069. Fastening on the words "fixed quantity," my brethren hold that the cost-plus exception is never available when the indirect purchasers are free to vary the quantity they buy from the direct purchaser.
This is not the first court to confine the exception so. See Mid-West Paper Products Co. v. Continental Group, Inc., 596 F.2d 573, 577 n. 9, 580 (3d Cir.1979); In re Midwest Milk Monopolization Litigation, 730 F.2d 528, 533 (8th Cir.1984); Lefrak v. Arabian American Oil Co., 487 F.Supp. 808, 819 (E.D.N.Y.1980); cf. Arizona v. Shamrock Foods Co., 729 F.2d 1208, 1212 n. 2 (9th Cir.1984). But the previous cases involved privately negotiated cost-plus contracts rather than cost-plus contractual provisions required by public utility regulation, which may make a difference as we shall see. Nor is it clear from the Supreme Court's opinion in Illinois Brick whether the reference to fixed quantity was intended to state an independent requirement of the cost-plus exception or merely to describe a normal contract situation, where the buyer's obligation is to buy the quantity specified in the contract. Not all contracts have this feature, but there is no indication that the Court meant to distinguish between fixed-quantity and variable-quantity contracts. Not only is the reference to fixed quantity dictum, because the case did not involve a fixed-quantity contract; the Court's entire discussion of cost-plus contracts is dictum, because the case did not involve a cost-plus contract but merely an argument (which the Court rejected, see 431 U.S. at 744, 97 S.Ct. at 2073) that a buyer's practice of rule-of-thumb cost-plus pricing should be enough to allow his customers to sue.
Although several district courts have rejected an exception for cost-plus regulation, see Go-Tane Service Stations, Inc. v. Ashland Oil, Inc., 508 F.Supp. 200, 204 (N.D.Ill.1981); City of Cleveland v. Cleveland Electric, Illuminating Co., 538 F.Supp. 1320 (N.D.Ohio 1980); U.S. Oil Co. v. Koch Refining Co., 518 F.Supp. 957 (E.D.Wis.1981), the cases are factually distinguishable; and in the case whose facts are most like those of the present case the district court held that indirect purchasers could sue, because public utility regulation had created "a straight cost passthrough." In re New Mexico Natural Gas Antitrust Litigation, 1982-1 Trade Cases p 64,685, at p. 73,722 (D.N.Mex.1982). Cf. County of Oakland v. City of Detroit, 628 F.Supp. 610, 613 (E.D.Mich.1986); Illinois v. Borg, Inc., 548 F.Supp. 972, 975-76 (N.D.Ill.1982). (An additional wrinkle in that case, however, was that the direct purchasers were in cahoots with the defendants; this was an independent ground for allowing the indirect purchasers to sue.) It is possible to allow indirect purchasers to sue in a case such as the present one without embracing the ill-defined "functional equivalent" approach of In re Beef Industry Antitrust Litigation, 600 F.2d 1148 (5th Cir.1979); see also, e.g., Gulf Oil Corp. v. Dyke, 734 F.2d 797, 809 (T.E.C.A.1984), effectively criticized in In re Midwest Milk Monopolization Litigation, 529 F.Supp. 1326, 1337-38 (W.D.Mo.1982), aff'd, 730 F.2d 528 (8th Cir.1984); Comment, A Legal and Economic Analysis of the Cost-Plus Contract Exception in Hanover Shoe and Illinois Brick, 47 U.Chi.L.Rev. 743, 756-70 (1980); cf. Abbott Dairies Division v. Butz, 584 F.2d 12, 16-17 (3d Cir.1978), and found to be inapplicable to conditions in the beef industry itself in In re Beef Industry Antitrust Litigation, 710 F.2d 216, 219-20 (5th Cir.1983). However, for reasons to be explained, I would not allow all the indirect purchasers in this case to sue--just the residential consumers.
To determine whether (or to what extent) this case is within the rule of Illinois Brick, we must consider the reasons for confining the right to sue to the direct purchaser; for it is the reasons behind a rule that determine its scope. First, the direct purchaser is closer to the violation, hence more likely to discover it. We therefore want to make sure that he has a powerful incentive to bring the violator to book, and we do this by holding out to him the prospect of recovering the entire damages for the violation if he wins the suit. Second, it is difficult to apportion damages between direct and indirect purchasers by the methods of litigation. A direct purchaser who finds himself paying a higher price for inputs would love to pass on all of the additional cost to his customers in the form of a higher price, but he can't, because a price that much higher will so reduce the demand for his product that his profits will fall unacceptably. The optimal adjustment to the increased cost of the input will be a price increase smaller than the increase in input cost. But this means that the increased cost will be divided between the two tiers, the direct and indirect purchasers--but in what proportions will often be hard to determine. An additional complication is that the higher input price may induce the direct purchaser to use more of an alternative input, and this substitution will affect the proportion of the initial overcharge that the direct purchaser can recoup.
Where the direct purchaser has a cost-plus contract with his customers that requires them to buy a fixed quantity, the reasons for confining the right to seek damages to the direct purchaser cease to be fully persuasive. There is no longer a problem of apportionment, because the whole of any price increase will have been passed on to the customers. Yet the other reason for confining the right to seek damages to the direct purchaser survives: he has better information about the violation. And, despite the cost-plus nature of the contract, he has everything to gain from suing. He will not have to share any of the damages that he recovers with his customers unless the contract contains a clause (or a court is persuaded to adopt an imaginative conception of unjust enrichment) that entitles them to any rebate which he might receive, probably years later, on an input used in performing his side of the bargain.
In the present case, where cost-plus pricing is the product of public utility regulation rather than a purely private contract, the reasons balance out slightly differently, but the case for applying the cost-plus exception of Illinois Brick is no weaker once the balance is restruck. The public utility has less to gain from suit than the direct purchaser in the purely private contract case has because the public utility commission may force the utility to pass on to the consumers any and all damages that the utility recovers. If so, the utility will have no incentive to sue; and it is no surprise that CILCO did not sue (and apparently the statute of limitations has now run). And although the amount of gas purchased by the utility's customers is not fixed in their contract with the utility, there is no problem of apportionment with respect to the only class of customers that I would allow to sue, the residential customers.
To see this, notice first that the retail distribution of natural gas through a grid of pipes to the customers' homes or places of business is a classic natural monopoly, Omega Satellite Products Co. v. City of Indianapolis, 694 F.2d 119, 123, 126 (7th Cir.1982); so, in the absence of regulation, a gas utility would charge a monopoly price. Although the efficacy of public utility regulation has been questioned (see, e.g., Stigler & Friedland, What Can Regulators Regulate? The Case of Electricity, 5 J.Law & Econ. 1 (1962); Moore, The Effectiveness of Regulation of Electric Utility Rates, 36 So.Econ.J. 365 (1970)), the facts of this case suggest, and the parties seem to agree, that residential natural-gas rates in Illinois are lower because of regulation than they would be in an unregulated market, implying that CILCO has unused monopoly power in that market. The situation in the industrial market is different. Some industrial consumers of natural gas have good alternatives, and as to them CILCO apparently had no unused monopoly power that would enable it to shift the whole of the cost increase to them. Instead CILCO sought and obtained regulatory permission to reduce its profit margin on sales to these customers, thereby offsetting in part the higher rates enabled by the automatic pass-through provision.
The residential consumers have no good alternatives to natural gas, and as to them CILCO's profit-maximizing course of action was, it appears, to allow its rates to rise by the exact amount of the increase in its gas costs. Rate regulation evidently had succeeded in keeping CILCO's rates to a level where the demand for its gas was inelastic. In this region of a firm's demand curve, an increase in price will by definition increase the firm's revenues, because the price increase will not be offset by an equal or greater proportional decline in quantity demanded--that is what it means to say that demand is inelastic. And since the increase in price will reduce the firm's costs (by causing demand for its product to fall, since consumers will buy less at the higher price) at the same time that it causes the firm's revenues to rise, the firm's profits must increase. CILCO had therefore (with a qualification to be considered shortly) every incentive to raise its price by the full amount allowed by the regulatory commission--that is, by the full amount of the gas overcharge. This implies that the entire overcharge was passed on to the utility's residential consumers on all sales made to these consumers.
Put more simply, if because of regulation a utility's rates are below what it would like to charge, it will raise those rates by the full amount allowed by the regulatory commission unless such an increase would carry the utility above its optimal level. It is unclear from the record filed in this court how large the allowance was; conceivably the commission allowed CILCO to double its rates--yet even so, if regulation had forced CILCO to charge half or less of its preferred price, CILCO would pass through the entire cost increase. No one suggests that CILCO in fact absorbed any of the increase, so far as sales to its residential customers were concerned.
The mechanics of the pass-through provision are important. The "Uniform Purchase of Gas Adjustment Clause" that the Illinois commission required CILCO to include in its contracts not only entitled but directed CILCO, if it paid Panhandle Eastern Pipe Line Company an extra penny per million cubic feet of gas, to add exactly one penny to each customer's bill for every Mcf of gas sold to that customer. So if all of its customers had continued buying the same amount of gas, CILCO would have suffered no loss on account of the overcharge. To the extent that the utility lost residential sales because it was charging a higher price--and no doubt it lost some sales, because demand is never perfectly inelastic--the loss was a loss to the utility, was not passed on to its customers (at least in any sufficiently direct way to escape the rule of Illinois Brick ), and hence is not an allowable factor in computing the customers' damages.
To illustrate the distinction between the two types of loss, suppose that the price to CILCO's residential customers before the overcharge to CILCO by Panhandle Eastern Pipe Line Company was $1 per Mcf, the overcharge was 10cents per Mcf, and therefore by operation of the Uniform Purchase of Gas Adjustment Clause the retail price rose to $1.10 because CILCO made no offsetting reduction in another component of the price, as it did with its industrial customers. Suppose further that at the new, higher price CILCO sold only 950,000 Mcf, whereas at the old price it had sold one million Mcf; and suppose that nothing plausibly accounts for the decrease in sales except the price increase, which induced consumers to use less gas. Cf. Illinois Power Co. v. Commissioner, 792 F.2d 683, 687-88 (7th Cir.1986). The loss to each consumer would be the number of Mcf he bought at the new price times 10cents; the loss to CILCO would be its lost profits on the sales it did not make.
Clearly, then, there is no problem of apportionment in the suit by the residential customers. Those customers are not seeking damages for gas they didn't buy, and the damages for the gas they did buy can simply be read off from their gas bills. (Well, almost: at the beginning of each year the utility estimates the amount of rate increase necessary to cover any increase in its gas bill, and there is an adjustment at the end of the year based on actual experience.) The only problem would come if CILCO had tried to sue for its lost sales, for then there would be more than one set of plaintiffs. But each set would be suing in respect of different sales--not, as in a Hanover Shoe or Illinois Brick case, the same sales. There would be no haggling over how much of the overcharge on each sale had been borne ultimately by the direct purchaser and how much by the indirect purchaser. And the proof of CILCO's lost sales would be straightforward--at least as straightforward as is possible in an antitrust case. It would involve comparing the utility's sales before and after the increase in gas prices, correcting for other factors, besides the increase, that might have affected those sales. Such correction is not always easy but is a conventional aspect of calculating damages in antitrust cases; it has to be done in every case where the plaintiff claims to have lost sales because of the defendant's unlawful conduct and the defendant argues that the loss was due partly or entirely to other factors. More important, it is a computational problem that has nothing to do with the problem that concerned the Supreme Court in Hanover and Illinois Brick. To repeat, the Court was concerned with the situation where two purchasers of the same thing--the initial purchaser and the purchaser from the initial purchaser--are or could be complaining that both had been hurt, and the problem is to apportion the loss between them. Here only the residential consumers could complain about a loss from the overcharge on the gas they bought, while only CILCO could complain about a loss caused by the overcharge on gas that the residential consumers did not buy.
Finally, unless the indirect purchasers are allowed to sue, the antitrust violation is likely to go unremedied, because the direct purchaser has even less incentive to sue than in a contractual cost-plus setting. CILCO might have an incentive to sue in respect of its lost sales (a distinct item of damages, as we have seen), but this would depend on how many sales it lost and on whether it would have to pass through any damages to its customers. In fact the incentive was not enough to induce it to sue.
I am not suggesting that the rule of Illinois Brick has no application to cases where the direct purchaser is subject to rate regulation. Although cost-plus is the spirit of rate regulation, the flesh is weak and often therefore the utility has considerable flexibility in pricing, much like an unregulated firm. Indeed, to the extent that the utility operates free from effective rate regulation--either because it faces competition that depresses its rates below the regulated level (apparently CILCO's situation with its industrial customers), or because the regulators are unable to prevent it from charging monopoly prices to its captive customers--its situation is identical to that of an unregulated seller. But in the case of CILCO's residential customers, regulation apparently succeeded in forcing the utility to operate deep in the inelastic region of its demand curve, with the result that 100 percent passing on of any cost increase was the optimal strategy for the utility to follow.
With the rapid and unanticipated increases in fuel prices during the 1970s, utilities pressed for and obtained the right to include automatic fuel pass-through provisions in their contracts with customers, provisions that would allow the utility to pass on every dollar in higher prices that it paid for gas or other fuels to its customers without going through the time-consuming process of obtaining regulatory authorization to raise rates. Such provisions are also found in unregulated contracts and should be treated the same there when the contract requires the buyer to take either a fixed quantity or his requirements, since a buyer cannot reduce his purchases under a requirements contract merely because he is dissatisfied with the terms of the contract as they have worked themselves., 130 F.2d 471, 473-74 (3d Cir.1942); White & Summers, Handbook of the Law Under the Uniform Commercial Code 126 (2d ed. 1980). (This example shows, by the way, why a rigid requirement of fixed quantity would be a senseless limitation on the cost-plus exception of Hanover Shoe and Illinois Brick: a buyer under a requirements contract does not have discretion as to the amount to take under the contract.) If on the other hand the buyer has complete flexibility as to how much to buy, a cost-plus provision is ineffectual; the buyer can always condition an agreement to buy a specific amount on the seller's agreeing to modify the contract by reducing the price. This is another reason for supposing that an agreement to take a fixed quantity, or, what is equivalent for these purposes, an agreement to take one's requirements, is implicit in the cost-plus exception rather than being an independent requirement for invoking it. The unexhausted monopoly power of a regulated utility takes the place of a fixed-quantity or requirements provision. The utility can force the whole of the cost increase through to its residential customers without sacrificing any profits, and did so.
I would not allow the industrial customers to sue, however. By cutting its profit margin to them, CILCO raised its price by less than the increase in gas cost; and while the apportionment of that increase between the utility and its industrial customers is easy to make--precisely because the utility was required to get approval for reducing its profit margin--I would be reluctant to complicate the administration of the Illinois Brick rule by trying to distinguish between difficult and easy apportionment cases. And the Court seemed unwilling to listen to such arguments. However, for every cubic foot of gas bought by a residential customer, we know that the whole overcharge was passed on to the customers, in accordance with the fuel pass-through provision.
It might seem an unimportant detail whether a buyer reacts to an overcharge by raising its price by less than the overcharge (as CILCO did with its industrial customers), thus losing fewer customers, or by raising its price by the full overcharge and thereby losing more customers than it would if it swallowed part of the overcharge. But in the second case the problem of apportioning losses on the same sales does not arise. It might also seem impermissible under Illinois Brick to inquire into the amount of passing on but the Court made an explicit exception for cases where there is a cost-plus contract. There is such a contract here--the automatic fuel pass-through provision--and although it is not a contract for a fixed quantity or (what I contend is equivalent for purposes of the exception) the buyer's requirements, the existence of public utility regulation is an adequate substitute in the circumstances.
We can never be absolutely certain that regulation has resulted in a 100 percent pass through; for all we know, CILCO would have sought a rate increase but for the gas overcharge, and by forbearing to do so in effect absorbed part of the overcharge. But by the same token, the seller under a fixed-quantity cost-plus contract might forbear to insist on a 100 percent pass through in order to curry favor with the buyer for the sake of future deals. No counterfactuals are certain, but the doubts here are too small to warrant us in insisting that this potentially serious antitrust violation, which may have imposed on consumers of natural gas aggregate damages of almost $50 million, go unremedied, as apparently it will.
The suit by the residential customers is within the scope of the cost-plus exception to the rule of Illinois Brick, and I would therefore affirm the denial by the district court of the defendant's motion to dismiss the complaint insofar as the complaint seeks damages on behalf of CILCO's residential customers.
ORDER
On consideration of the petition for rehearing and suggestion for rehearing en banc filed by counsel for plaintiff-appellee in the above-entitled cause, a vote of the active members of the Court was requested. A majority of the judges in regular active service above named** voted to GRANT the petition and suggestion for rehearing en banc. Accordingly.
IT IS ORDERED that the aforesaid petition for rehearing and suggestion for rehearing en banc be, and the same are hereby, GRANTED.
IT IS FURTHER ORDERED that the judgment and opinion entered in this case on January 22, 1988 be, and are hereby, VACATED. This case will be reheard en banc on Thursday, May 26, 1988 at 930 AM.
Under the Purchase Gas Adjustment (PGA) process, CILCO estimates the purchases, costs and sales in a future period, computes the cost for that period and adjusts its rates prospectively to reflect the estimate. At the end of each year, the estimated and actual costs are reconciled. In 1983, CILCO overcollected more than $3 million, and in 1984 more than $1 million. Overcollections are refunded through adjustments in the following year. A customer who terminates service before the refund does not receive it. A customer who used less gas in the refund period than he used in the overcollection period does not receive a full refund
On remand, it developed that plaintiff could not, in fact, demonstrate the habitual use by the packers of a predetermined formula, as alleged, preventing the packers from absorption of some of the undercharge. Judgment was granted to defendants. 542 F.Supp. 1122 (N.D.Tex.1982), affirmed 710 F.2d 216 (5th Cir.1983)
Plaintiff has cited a case involving the regulated distribution of natural gas and a PGA mechanism similar to the one before us. New Mexico Natural Gas, CCH 1982-1 Trade Cases p 64,685, p. 73,714 (D.N.M.1982) . Treble damage actions were brought by and on behalf of consumers who purchased from a defendant distributor, who purchased from defendant producers. Producer defendants sought summary judgment based on Illinois Brick. The New Mexico court denied the motion, finding an exception because two policy considerations underlying Illinois Brick were not significant in the case before it. One of the policy considerations was the difficulty in tracing the pass-on of an overcharge. The New Mexico court decided that the PGA clauses obviated this concern. The second policy consideration was the risk of multiple liability if indirect as well as direct purchasers could sue. The New Mexico court found this concern was not significant in the case before it. The court relied on the fact, which it acknowledged was unusual, that the distributor was a defendant and alleged to be a partner with defendant producers in the price-fixing conspiracy, and would be precluded from suing its co-conspirators. The New Mexico court did not address the question whether the demand for gas is elastic and the distributor (direct purchaser) may have been injured by loss of sales volume. The New Mexico Gas case is distinguishable
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Item#:
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1:12 Scale Ceiling light with red shades. Non-Electric
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In this contract:
- We, us or our means Stalbridge Building Supplies Limited registered in England and Wales with Company Number 08916300 whose registered office is at Stalbridge Building Supplies Ltd, Manchester House, High Street, Stalbridge, DT10 2LL. VAT No 995 9139 50; and
- You or your means the person using our site to buy goods from us.
If you don't understand any of this contract and want to talk to us about it, please contact us by:
- Telephone 01963 363372 calls will be answered, Monday to Friday 7:30am to 5pm.
1 Introduction
1.1 If you buy goods on our site you agree to be legally bound by this contract.
1.2 This contract is only available in English. No other languages will apply to this contract.
2 Ordering goods from us
2.1 When you place your order at the end of the online checkout process (e.g. when you click on the place order and pay), we will acknowledge it by email “(Order Acknowledgement”). This Order Acknowledgement does not, however, mean that your order has been accepted..
2.3 Unless we contact you to confirm that we are unable to accept your order, our acceptance of your order will occur automatically on the date we deliver the goods to you.
2.4 All orders are subject to stock availability. If we are unable to supply any of the goods that you have ordered we will inform you as soon as possible. In the event that a product that is out of stock is part of an order and we cannot contact you, we will send what goods we have in stock, refunding the purchase price of any unavailable product where payment has already been made.
2.5 Where you place an order for an age restricted good such as knives, blades, solvents and axes you confirm that you are over the age of 18 and that, where applicable, delivery will be accepted by a person over 18. We reserve the right to cancel your order if we reasonably believe you do not meet the age requirements.
3 Returns
3.1 You have the right to cancel this contract within 28 days without giving any reason.
3.2 The cancellation period will expire after 28 days from the date of delivery to you or your collection of the goods from a trade counter (as applicable).
3.3 To exercise the right to cancel, you must inform us of your decision to cancel this contract by a clear statement (e.g. a letter sent by post, fax or email) or by returning the goods to the store.
3.4 To meet the cancellation deadline, it is sufficient for you to send your communication concerning your exercise of the right to cancel before the cancellation period has expired.
3.5 The right to cancel the contract set out above is subject to the following exclusions:
3.5.1 goods which are cut, made to measure or mixed to your requirements, or otherwise customised or made to your specifications will not be exchanged or refunded unless they are faulty or incorrectly delivered;
3.5.2 goods which are liable to deteriorate or expire rapidly, including perishable goods such as gypsum or cement based products, will not be exchanged or refunded unless they are faulty or incorrectly delivered;
3.5.3 your right of cancellation does not apply to goods which are not suitable for return due to health or hygiene reasons, if you have opened the goods packaging after delivery or collection; and
3.5.4 your right of cancellation does not apply to goods which become mixed inseparably with other items after delivery or collection (which may be the case where the goods are installed).
3.6 If you cancel this contract, we will reimburse to you all payments received from you, including the costs of delivery (except for the supplementary costs arising if you chose a type of delivery other than the least expensive type of standard delivery offered by us).
3.7 We may make a deduction from the reimbursement for loss in value of any goods supplied, if the loss is the result of unnecessary handling by you.
3.8 We will make the reimbursement without undue delay, and not later than:
3.8.1 14 days after the day we received back from you any goods supplied; or
3.8.2 (if earlier) 14 days after the day you provide evidence that you have returned the goods; or
3.8.3 if there were no goods supplied, 14 days after the day on which we are informed about your decision to cancel this contract.
3.9 We will make the reimbursement using the same means of payment as you used for the initial transaction, unless you have expressly agreed otherwise; in any event, you will not incur any fees as a result of the reimbursement.
3.10 If the goods have been delivered to you:;
3.10.2 you will have to bear the direct cost of returning the goods;
3.10.3 you are only liable for any diminished value of the goods resulting from the handling other than what is necessary to establish the nature, characteristics and functioning of the goods.
4 Delivery
4.1 There are three types of goods that you can order from us, which, depending on your location, will affect delivery (see specific sections below):
4.1.1 'Standard' goods are goods that do not fall into our 'Bulk' categories;
4.1.2 'Bulk' goods;
4.1.3 Direct Delivery
4.2 Items can only be delivered within our normal trading area, and this will be clearly marked on the website. Our normal delivery area will be as described to you at the time you place your order or as shown on our website. We reserve the right not to deliver outside this area.
4.3 Direct deliveries will be fulfilled by the supplier. In most instances the supplier or their courier service will contact you directly regarding a delivery day and time. Delivery may take 7-10 days depending on the goods. If your order also contains items not stated as direct delivery, then these items will be delivered separately.
4.4 The estimated date and time window for delivery of the goods will be discussed on the telephone once we have received your order.
4.5 If something happens which:
4.5.1 is outside of our control; and
4.5.2 affects the estimated date of delivery,
we will let you have a revised estimated date for delivery of the goods.
4.6 Delivery of the goods will take place when we deliver them to the address that you gave to us.
4.7 Unless you and we agree otherwise, if we cannot deliver your goods within 30 days, we will:
4.7.1 let you know;
4.7.2 cancel your order; and
4.7.3 give you a refund.
4.8 If nobody is available to take delivery, please contact us using the contact details at the top of this page.
4.9 You are responsible for the goods when delivery has taken place. In other words, the risk in the goods passes to you when you take possession of the goods.
4.10 We may deliver your goods in instalments.
4.11 Unless otherwise agreed in writing, our delivery price includes the cost of delivery on weekdays during our normal working hours of 7.30am and 5.00pm. An additional charge may be made if we agree to your request to deliver outside normal working hours.
4.12 If you keep our delivery vehicle waiting for an unreasonable time or the delivery driver is obliged to return without completing delivery, or if due to the nature of the goods we have to provide additional staff to unload the goods, a reasonable additional charge will be made that reflects the extra services provided.
4.13 We pride ourselves on a reliable delivery service and always do our utmost to deliver on the date or time frame specified. However, please note it acts as a guideline only and cannot be guaranteed, although rare, factors beyond our control can sometimes cause delay and we will endeavour to contact you as soon as we are able to in these cases.
5 Click & Collect
5.1 The “Click & Collect” service is available on all goods marked as available for branch collection on our site.
5.2 If the goods are in stock at the store, we will aim to have the goods ready for collection within two hours after we receive your order and process your payment.
5.3 Any goods not in stock on the day of ordering we will let you know when they become available. It is possible for you to collect the goods which are available in store and collect the remaining goods when they are available.
5.4 Goods are available for collection between 7.30am – 5.00pm.
5.5 Collection within two hours applies to orders that have been placed during trading hours of the branch. Orders made after 4.00pm will be available for collection in branch from 7.30am the following working day.
5.6 Collection times are approximate.
5.7 You will receive an email order confirmation once your order has been placed notifying you of the order number. A second email will be sent to confirm your order is ready for collection.
5.8 “Click & Collect” orders can only be collected from branches between 7.30am – 5.00pm.
5.9 If you have not collected your order within 48 hours of receiving confirmation of your order being ready to collect you will be contacted by a member of branch staff. Orders not collected within 5 working days will have the goods re-allocated and you will be contacted to arrange a full refund.
5.10 When collecting your order we need to see a copy of the Order Acknowledgement (printed or electronic) and a valid form of ID, for example driving licence, passport, debit card or credit card. Please note that orders will be not released to courier companies or third parties who are unable to provide the correct information needed to verify them.
5.11 If you would like your order to be collected by someone else on your behalf they will need to bring photographic ID and the order confirmation email (printed or electronic).
6 Payment
6.1 We accept Visa and MasterCard credit cards and Maestro, Delta, Visa, Electron and Solo debit cards. We do not accept 'prepaid' credit cards or American Express..
6.3 All payments by credit card or debit card need to be authorised by the relevant card issuer.
6.4 If your payment is not received by us and you have already received the goods, you:
6.4.1 must pay for such goods within 14 ‘cooling off’ period under clauses 4.
6.7 The price of the goods:
6.7.1 is in pounds sterling (£)(GBP);
6.7.2 includes VAT at the applicable rate.
7 Nature of the goods
7.1 The Consumer Rights Act 2015 gives you certain legal rights (also known as ‘statutory rights’), for example, the goods:
7.1.1 are of satisfactory quality;
7.1.2 are fit for purpose;
7.1.3 match the description, sample or model; and
7.1.4 are installed properly (if we install any goods).
7.2 We must provide you with goods that comply with your legal rights.
7.3 The packaging of the goods may be different from that shown on the site.
7.4 Any goods sold:
7.4.1 at discount prices;
7.4.2 as remnants; or
7.4.3 as substandard;
will be identified and sold as such. Please check that they are of a satisfactory quality for their intended use.
7.5 We will take reasonable steps to display as accurately as possible the colours, appearance and other detailing of our goods goods (or its packaging). The goods (and their packaging) may vary slightly from those images.
7.6 Natural goods may show some colour variation. To the extent permitted by law, we accept no liability associated with such variations.
7.7 Any information on the website and in our catalogues and other media regarding sizing, weights, capacities, specifications, dimensions and measurements of goods is included as a guide only. If you are concerned about the precise size, weight, capacity, specifications, dimensions or measurements of any goods you require, we recommend that you contact us prior to placing an order and/or purchasing goods.
7.8 If your order includes goods that are made according to measurements you provide us, please ensure these measurements are correct and accurate as your right of return as set out in clause 4 will not be available unless the goods are faulty or incorrectly delivered.
7.9 The prices displayed on the website may be different prices for the same goods purchased in the stores. These prices may be higher or lower.
7.10.
7.11 If a good’s correct price is higher than the price stated on our site, we will let you know as soon as reasonably possible. We will then offer you the option of reconfirming your order at the correct price or cancelling it in exchange for a full refund.
7.12 All prices are subject to change without prior notice.
7.13 re-credit you for any sum that has been paid by you or debited from your credit card or debit card for the goods.
7.14 Whilst we try to maintain continuity of supply in relation to our product lines, we reserve the right to discontinue any goods at any time and we shall be under no obligation to supply you with discontinued goods in the future. If you have already placed your order, we will notify you as soon as reasonably possible that the goods are not available and offer you an alternative product if one is available or a full refund.
7.15 In the case of certain goods, variations may arise in the finish of those goods where they originate from different factory batches. We shall not be liable for any loss caused by such variation where such variation does not diminish the quality of the goods and where goods bought for a specific job or purpose are not purchased at the same time.
7.16.
8 Manufacturer Warranties
Some of our goods are sold with a manufacturer’s warranty or guarantee. The goods description will provide details of any warranty or guarantee. Any compliant, query or claim under the manufacturer’s guarantee must be made direct to the manufacturer. These rights are in addition to your statutory rights.
9 Faulty goods
9.1 Your legal rights under the Consumer Rights Act 2015 (also known as ‘statutory rights’), are set out below.
Summary of your key legal rights
This is a summary of your key legal rights. These are subject to certain exceptions. For detailed information please visit the Citizens Advice website or call 03454 04 05 06.
For goods. For more details, please visit the website on the ‘Your Europe’ portal: .
12.4 If you want to take court proceedings, the relevant courts of England and Wales will have exclusive jurisdiction in relation to this contract.
12.5 The laws of England and Wales will apply to this contract.
13 Third party rights
| 103,735
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TITLE: Proving combinatorical identity (total number of multichoose on n items)
QUESTION [1 upvotes]: How do I prove this identity
$$\sum_{k=0}^{n}{\binom{m+k-1}{k}}=\binom{m+n}{n}$$
Can anybody also shed some light on intuition (may be a double counting proof) about how right hand side of the identity turns out to be the total number of multichoose possible over n items.
REPLY [0 votes]: Consider the set $[m+n]=\{1,2\cdots m+n\}$. If you take a size $n$ subset $S$ from this $[m+n]$, it will contain the first $n-k$ integers for some $k$ where $0\leq k\leq n$, but not the $n-k+1$th element.
If I know that a size $n$ subset contains the first $n-k$ elements, then I can only choose the last $k$ elements. Furthermore, I have to choose from $m+n-(n-k)=m+k$ elements, since the first $n-k$ have already been included. This is slightly off from what's on the left hand side. This is why I also required that the subset have the first $n-k$ elements, but not the $n-k+1$th, as I also have to exclude that element from my choices when I fill in the rest of the set. This means I can fill in the rest of the set ${m+k-1\choose k}$ ways.
This added condition is important, since it negates double counting- I look for the largest $n-k$ such that my set contains those elements, so there's only one such $k$ for each set. The left side of that identity counts those subsets for all possible $k$, which gives our identity.
| 59,038
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100% Soy wax tealight candles. We use a 10% scent to wax ratio for the perfect smell.
Size: Tealight
Color: Random (16 Different Colors)!
8 oz Black & Gold Tin
100% Soy wax candle made to order in a black and gold tin with lid. We use a 10% scent to wax ratio for the perfect smell.
Size: 8 oz
6 oz Glass Jar 1 w/ 3D Printed Lid
Each jar comes with a random colored 3D printed lid showing scent name
7 oz Glass Jar 2 w/ 3D Printed Lid
Size: 7 oz
8 oz Glass Jar 3 w/ 3D Printed Lid
| 150,929
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\begin{document}
\title[ A weighted isoperimetric inequality in a wedge ]{ A weighted
isoperimetric inequality in a wedge }
\author{ F. Brock$^1$ - F. Chiacchio$^2$ - A. Mercaldo$^2$}
\thanks{}
\date{}
\begin{abstract}
\noindent Let $c, k_1 , \ldots , k_N $ be non-negative numbers, and define a
measure $\mu $ in the wedge $W:= \{x\in \mathbb{R} ^N :\, x_i >0 , i=1,
\ldots ,N\} $ by $d\mu = e^{c|x|^2 } x_1 ^{k_1 }\cdots x_N ^{k_N } \, dx $.
It is shown that among all measurable subsets of $W$ with fixed $\mu$
-measure, the intersection of $W$ with a ball centered at the origin renders
the weighted perimeter relative to $W$ a minimum.
\bigskip
\textsl{Key words:} relative isoperimetric inequalities, Polya-Szeg\"o
principle, degenerate elliptic equations.
\textsl{2000 Mathematics Subject Classification:} 26D20, 35J70, 46E35
\end{abstract}
\maketitle
\setcounter{footnote}{1} \footnotetext{
Leipzig University, Department of Mathematics, Augustusplatz, 04109 Leipzig,
Germany, e-mail: brock@math.uni-leipzig.de}
\setcounter{footnote}{2} \footnotetext{
Dipartimento di Matematica e Applicazioni \textquotedblleft R.
Caccioppoli\textquotedblright , Universit\`{a} degli Studi di Napoli
``Fe\-derico II", Complesso Monte S. Angelo, via Cintia, 80126 Napoli,
Italy, e-mails: francesco.chiacchio@unina.it, mercaldo@unina.it}
\section{Introduction}
Let a measure $\nu $ be defined by $d\nu =\phi (x)\, dx$, where $\phi $ is a
positive Borel measurable function defined on a subset $\Omega $ of
$\mathbb{R} ^N $.
If $M$ is Lebesgue measurable set with $M\subset \Omega $, we define
the $\nu $-measure of $M$
\begin{equation} \label{intronu}
\nu (M)=\int_{M}d\nu =\int_{M}\phi(x)\, dx
\end{equation}
and the $\nu $-perimeter of $M$ relative to $\Omega$
\begin{equation*}
P_{\nu }(M,\Omega )=\sup \left\{ \int_{M}\mbox{div }(\mathbf{v}(x)\phi(x))\,
dx:\, \mathbf{v}\in C_0 ^1(\Omega ,\mathbb{R}^{N}),\,|\mathbf{v}|\leq 1\
\right\}\, .
\end{equation*}
Note that if $M$ is a smooth set, then
\begin{equation*}
P_{\nu }(M,\Omega )=\int_{\partial M\cap \Omega }\phi(x)\, d{\mathcal{H}}
_{N-1}(x).
\end{equation*}
The isoperimetric problem reads as
\begin{equation} \label{Imu}
I_{\nu , \Omega }(m) :=\inf \{P_{\nu }(M,\Omega ):\, M\subset \Omega ,\,\nu
(M)=m\},\quad m>0.
\end{equation}
One says that $M$ is an isoperimetric set if $\nu (M)=m$ and $I_{\nu ,
\Omega }(m)= P_{\nu }(M,\Omega )$.
In this paper we consider the case that $\Omega$ is the wedge $W$ in
$\mathbb{R}^N$, where
\begin{equation} \label{wedge1}
W:= \{ x \in \mathbb{R}^N :\, x_i >0 , \, i=1 , \ldots ,N \},
\end{equation}
and we determine functions $\phi$ having $B_{R}\cap W $ as an
isoperimetric set. \newline
Here and throughout the paper, $B_R$ and $B_R(x)$ denote the ball of radius
$R$ centered at zero and at $x$, respectively. \newline
In a recent paper \cite{BCM2}, the authors studied the case that $\Omega $ is the half-space
\begin{equation*}
H:= \{ x\in \mathbb{R}^N :\, x_N >0 \},
\end{equation*}
and, among other things, the following two results
were proved. \\[0.2cm]
\textbf{Theorem A:} \textsl{(see \cite{BCM2}, Theorem 1.1) \newline
Let
\begin{equation*}
\varphi (x) = x_N ^k \mbox{ exp } \{ c |x|^2 \} , \quad x\in H,
\end{equation*}
where $c,k\geq 0 $. Then $I_{\nu ,H } ( m) = P_{\nu } ( B_R \cap H, H) $ for
every $R>0$ such that $m= \nu (B_R \cap H)$.} \\[0.2cm]
\textbf{Theorem B:} \textsl{(see \cite{BCM2}, Theorem 2.1 and Lemma 2.1)
\newline
Let $\phi \in C^2 (W)$, and assume $\phi $ is in separated form,
\begin{equation*}
\phi (x) = \prod _{i=1 } ^N \phi _i (x_i ) ,
\end{equation*}
where $\phi _i \in C^2 ((0, \infty )) \cap C ([0, \infty )) $, $\phi _i (x_i
) >0 $ if $x_i >0 $, ($i=1, \ldots ,N$). Further, suppose that $I_{\nu ,W }
(m) = P_{\nu } (B_R \cap W, W)$ for every $R>0 $ such that $m= \nu (B_R \cap
W) $. Then
\begin{equation} \label{weight}
\phi (x) = a \mbox{ exp } \{ c|x|^2 \} \prod _{i=1} ^N x_i ^{k_i } , \quad
x\in W,
\end{equation}
where $a>0$, $k_i \geq 0$, ($i=1, \ldots , N$), and $c\in \mathbb{R}$.}
\\[0.2cm]
Theorems A and B are imbedded in a wide bibliography related to the
isoperimetric problems for \textsl{manifolds with density } (see, for
instance,
\cite{BBMP, Bo, Borell, BCM, BMP, CMV, CJQW, DDNT, MS, Mo, Mo2, RCBM, S}).
The following result accomplishes Theorem B, and its proof will be given in
Section 2: \\[0.2cm]
\textbf{Theorem 1 :}
\textsl{Let $c, k_i $ be nonnegative numbers, ($i= 1, \ldots ,N$), and let a
measure $\mu $ on $W$ be defined by
\begin{equation} \label{mumeasure}
d\mu := \mbox{ exp } \{ c|x|^2 \} \prod_{i=1 } ^N x_i ^{k_i } \, dx .
\end{equation}
Then $I_{\mu ,W } (m) = P_{\mu } (B_R \cap W, W)$ for every $R>0 $ such that
$m= \mu (B_R \cap W) $. } \\[0.2cm]
It is customary to write isoperimetric inequalities as a
relation between the perimeter and the measure of a set. \newline
Set
\begin{eqnarray*}
\mathbf{k } & := & (k_1 , \ldots , k_N ), \\
|\mathbf{k}| & := & \sum _{i=1} ^N k_i , \\
h(r) & := & e^{cr ^ 2 } r^{N-1 +|\mathbf{k} | } , \\
H(r) & := & \int_0 ^r e^{ct ^ 2 } t^{N-1 +|\mathbf{k} | } \, dt , \quad
(r\geq 0), \quad \mbox{and } \\
\kappa & := & \int_{{\mathbb{S}} ^{N-1} \cap W } x_1 ^{k_1 } \cdots x_N
^{k_N } \, d{\mathcal{H}} _{N-1} (x).
\end{eqnarray*}
With these notations Theorem 1 can be reformulated as
\begin{equation} \label{pmu1}
P_{\mu } (M, W) \geq \kappa h\left( H^{-1} ( \mu (M)/\kappa ) \right ) ,
\end{equation}
with equality if $M= B_R \cap W$, for some $R>0 $. Note, in the special case
$c=0$, (\ref{pmu1}) reads
\begin{equation} \label{pmu2}
P_{\mu } (M, W) \geq \kappa ^{1/(N+|\mathbf{k}|)} \left( ( N+|\mathbf{k}|)
\mu (M) \right) ^{(N-1+|\mathbf{k}|)/(N+|\mathbf{k}|)} .
\end{equation}
Theorem 1 has numerous applications, and they will be analysed in a
forthcoming paper \cite{BCM4}:\newline
The fact that sets $B_R \cap W $, ($R>0$), are isoperimetric for the
weighted measure $\mu $ imply a Polya-Szeg\"{o} - type inequality, comparing
the weighted Sobolev norms of a given function and its weighted
rearrangement (compare with \cite{T}, p. 125). In turn, this allows to find the
best constants in some Sobolev inequalities for functions defined in the
wedge $W$. \newline
Furthermore, Theorem 1 gives rise to sharp comparison result for weighted
elliptic problems in subsets of $W$
(compare with \cite{T0}, \cite{T2}, and \cite{BBMP}).
\section{Proof of the main result}
In this section we prove Theorem 1. A crucial role is played by the
following \\[0.2cm]
\textbf{Lemma 1: } \textsl{Let $k>0$, $l>0$, and define a function $\sigma
\in C^{\infty } ((0, \pi /2 )) \cap C([0, \pi /2 ] )$ implicitly by
\begin{equation} \label{st}
\int_0 ^{\theta } \sin ^k t \cos ^l t \, dt = c_1 \int _0 ^{\sigma (\theta )
} \sin ^{k+l } s \, ds , \quad \theta \in [0, \pi /2 ],
\end{equation}
where
\begin{equation} \label{c1}
c_1 := \frac{ \int_0 ^{\pi /2} \sin ^k t \cos ^l t \, dt }{ \int _0 ^{\pi }
\sin ^{k+l } s \, ds } .
\end{equation}
Then }
\begin{equation} \label{s'>1}
\sigma ^{\prime } (\theta ) \geq 1, \quad \theta \in (0, \pi /2 ).
\end{equation}
\textsl{Proof: \/} Equation (\ref{st}) implies
\begin{equation} \label{eqn_s'}
\sin ^k \theta \cos ^l \theta = c_1 \sigma ^{\prime } (\theta ) \sin ^{k+l }
\sigma (\theta ) , \quad \theta \in (0, \pi /2) ,
\end{equation}
$\sigma (0)=0$, and $\sigma (\pi /2 )= \pi $. Set
\begin{equation*}
f(\theta ):= \sin ^k \theta \cos ^l \theta -c_1 \sin ^{k+l} \sigma (\theta )
, \quad \theta \in [0, \pi /2 ] ,
\end{equation*}
and note that $f\in C^{\infty } ((0, \pi /2 ))\cap C([0, \pi /2 ]) $, with
$f(0)=f(\pi /2 )=0$. Then (\ref{s'>1}) holds iff
\begin{equation} \label{f>0}
f(\theta ) \geq 0, \quad \theta \in (0, \pi /2) .
\end{equation}
Assume
(\ref{f>0}) was not true. Then there exists $\theta _0 \in (0, \pi /2)$ with
\begin{equation} \label{def_t0}
f(\theta _0 )<0, \ f^{\prime }(\theta _0 ) =0, \ f^{\prime \prime }
(\theta_0 )\geq 0.
\end{equation}
This, in turn, also implies that
\begin{equation} \label{s'<1}
\sigma ^{\prime }(\theta _0 ) <1 .
\end{equation}
By (\ref{eqn_s'}) we have
\begin{equation} \label{f'simple}
f^{\prime } (\theta ) = c_1 \sigma ^{\prime }(\theta ) \sin ^{k+l } \sigma
(\theta ) \left( k\cot \theta - l\tan \theta - (k+l) \cot \sigma (\theta )
\right) , \quad \theta \in (0, \pi /2).
\end{equation}
The second condition in (\ref{def_t0}) and (\ref{f'simple}) give
\begin{equation} \label{cond1}
k\cot \theta _0 -l \tan \theta _0 - (k+l ) \cot \sigma (\theta _0 ) =0.
\end{equation}
Then, differentiating (\ref{f'simple}), the third condition in (\ref{def_t0}
), and (\ref{cond1}) give
\begin{equation*}
f^{\prime \prime } (\theta _0 )= c_1 \sigma ^{\prime } (\theta _0 ) \sin
^{k+l } \sigma (\theta _0 ) \left( -\frac{k}{\sin ^2 \theta _0 } - \frac{l}
{\cos ^2 \theta _0 } + \frac{(k+l) \sigma ^{\prime } (\theta _0 ) }{\sin ^2
\sigma (\theta _0 ) } \right) \geq 0,
\end{equation*}
which implies
\begin{equation} \label{cond3}
1 > \sigma ^{\prime }(\theta _0 ) \geq \left( \frac{k}{\sin ^2 \theta _0 } +
\frac{l}{\cos ^2 \theta _0 } \right) \frac{\sin ^2 \sigma (\theta _0 ) }{k+l}
.
\end{equation}
On the other hand, (\ref{cond1}) yields
\begin{equation*}
\frac{(k \cot \theta _0 - l \tan \theta _0 )^2 }{ (k+l )^2 } = \frac{1-\sin
^2 \sigma (\theta _0 )}{\sin ^2 \sigma (\theta _0 )} ,
\end{equation*}
that is,
\begin{equation} \label{cond4}
\sin ^2 \sigma (\theta _0 ) = \frac{ (k+l)^2 }{ (k+l)^2 + (k \cot \theta _0
- l \tan \theta _0 )^2 }.
\end{equation}
Plugging (\ref{cond4}) into (\ref{cond3}), we find
\begin{equation} \label{cond5}
1 > \sigma ^{\prime } (\theta _0 ) \geq \frac{ (k+l ) \left( \frac{k}{\sin
^2 \theta _0 } + \frac{l}{\cos ^2 \theta t_0 } \right) }{ (k+l)^2 + (k\cot
\theta _0 -l \tan \theta _0 ) ^2 }.
\end{equation}
This implies
\begin{equation} \label{cond6}
(k+l )^2 + (k\cot \theta _0 - l \tan \theta _0 ) ^2 > (k+l) \left( \frac{k}{
\sin ^2 \theta _0 } + \frac{l}{\cos ^2 \theta _0 } \right) .
\end{equation}
Setting $\sin ^2 \theta _0 =: z\in (0,1)$, this yields
\begin{equation*}
(k+l )^2 + k^2 \frac{1-z}{z} + l^2 \frac{z}{1-z} -2kl > (k+l) \left( \frac{k
}{z} +\frac{l}{1-z} \right) ,
\end{equation*}
that is,
\begin{equation*}
\frac{kl}{z } + \frac{kl}{1-z} <0,
\end{equation*}
which is impossible.
Hence (\ref{f>0}) follows, and Lemma 1 is proved.
$\hfill \Box $ \\[0.2cm]
Now we prove Theorem 1 in two steps. We firstly face, using Lemma above, the bidimensional case and then the result is achieved in its full generality via an induction argument over the dimension $N$.
\noindent \textsl{Proof of Theorem 1. Step 1: The case $N=2$. }\newline
We write $(x,y)$ for points in $\mathbb{R}^2 $, and $W := \{ (x,y)\in
\mathbb{R}^2 :\, x>0, y>0\}$. Introduce a measure on $W $ by
\begin{equation*}
d\mu = \mbox{exp}\, \{c(x^2 +y^2 ) \} x^l y^k \, dx\, dy,
\end{equation*}
where $c,l$ and $k$ are nonnegative constants. \newline
Let $M$ be a smooth set contained in $W $, and choose $R>0 $ such that $\mu
(B_R \cap W ) = \mu (M)$. Then
\begin{equation} \label{decomp1}
\partial M \cap W = \bigcup_{k=1} ^m C_k ,
\end{equation}
where the $C_k $'s are mutually non-intersecting, smooth curves, and
\begin{equation} \label{perimeter1}
P_{\mu } (M, W ) = \sum_{k=1} ^m \int _{C_k } \mbox{exp}\, \{ c(x^2 + y^2
)\} x^l y^k \, ds , \quad (ds: \ \mbox{ Euclidean arc length differential }).
\end{equation}
Let $C$ be one of the curves in the decomposition (\ref{decomp1}), with
parametrization
\begin{equation*}
C:= \{ (x(t),y(t)): \, t\in [a,b]\} ,
\end{equation*}
where $x,y\in C^1 ( [a,b])$ and $a,b\in \mathbb{R} $, $a<b$. Then
\begin{equation} \label{perimeter2}
\int _{C } \mbox{exp}\, \{ c(x^2 + y^2 )\} x^l y^k \, ds = \int_a ^b
\mbox{exp}\, \{ c(x^2 (t)+y^2 (t) ) \} x^l (t) y^k (t) \sqrt{(x^{\prime }
(t))^2 + (y^{\prime } (t) )^2 } \, dt .
\end{equation}
Let $p: W\to (0, +\infty )\times (0, \pi /2) $ be the polar coordinates
mapping given by
\begin{equation*}
p (r\cos \theta , r\sin \theta ) := (r, \theta ), \quad (r> 0, \, 0<\theta <
\pi /2 ),
\end{equation*}
and write $(r(t), \theta (t)):= p (x(t), y(t)) $, ($a\leq t\leq b$). Then
also
\begin{equation} \label{perimeter3}
\int _{C } \mbox{exp}\, \{ c(x^2 + y^2 )\} x^l y^k \, ds = \int_a ^b e^{cr^2
(t) } r^{k+l } (t) \sqrt{ r^2 (t) (\theta ^{\prime } (t))^2 + (r^{\prime }
(t) )^2 } \, \sin ^k \theta (t) \cos ^l \theta (t) \, dt .
\end{equation}
(Here and in the following, "$\, ^{\prime } \, $" denotes differentiation
w.r.t. $t$.) \newline
Next, let $U: (0, +\infty )\times (0, \pi /2 ) \to (0, +\infty ) \times (0,
\pi )$ be the mapping given by
\begin{equation*}
U(r, \theta ) := (r, \sigma (\theta )) , \quad (r>0, \, 0<\theta <\pi /2 ),
\end{equation*}
where $\sigma $ is the function defined by (\ref{st}) and (\ref{c1}).
Finally, let $\mathbb{R} ^2 _+ $ be the upper half-plane $\{ (u , v )\in
\mathbb{R} ^2 :\, v >0 \} $, and define a diffeomorphism from $W$ onto
$\mathbb{R} ^2 _+ $ by
\begin{equation*}
T:= p^{-1} \circ U\circ p.
\end{equation*}
Writing $(u, v )$ for points in $\mathbb{R} ^2 _+ $, and
$(x,y)\equiv (r \cos \theta , r \sin \theta )$ for points in $W$, we have
\begin{equation*}
(u,v) \equiv T(x,y) = (r\cos \sigma (\theta ), r \sin \sigma (\theta ) ) .
\end{equation*}
Introduce a measure $\widetilde{\mu } $ on $\mathbb{R} ^2 _+ $ by
\begin{equation*}
d\widetilde{\mu} := \mbox{exp}\, \{ c(u ^2 + v ^2 ) \} v ^{k+l} \, du \, dv
.
\end{equation*}
Using the notation $(u(t),v(t)) := T(x(t),y(t))$, and since $p (u(t), v(t))
= (r(t), \sigma (\theta ((t)))$, ($a\leq t\leq b$), we find, similarly as
above,
\begin{equation} \label{perimeter4}
P_{\widetilde{\mu} } (T(M), \mathbb{R} ^2 _+ ) = \sum _{k=1} ^m
\int_{T(C_k )} \mbox{exp}\, \{ c(u ^2 + v ^2 ) \} v ^{k+l} \, ds ,
\end{equation}
and
\begin{eqnarray} \label{perimeter5}
& & \int_{T(C )} \mbox{exp}\, \{ c(u ^2 + v ^2 ) \} v ^{k+l} \, ds \\
& = & \int_a ^b \mbox{exp}\, \{ c(u ^2 (t) + v ^2 (t) ) \} v ^{k+l} (t)
\sqrt{ (u^{\prime } (t))^2 + (v^{\prime } (t)) ^2 } \, dt \notag \\
& = & \int_a ^b e^{cr^2 (t) } r^{k+l} (t) \sqrt{ r^2 (t) \left( \frac{d\sigma
}{d\theta} \right) ^2 (\theta ^{\prime } (t))^2 + (r ^{\prime } (t)) ^2 } \,
\sin ^{k+l } \sigma (\theta (t)) \, dt . \notag
\end{eqnarray}
By Lemma 1 we have $\frac{d\sigma }{d \theta } \geq 1$, which implies,
together with (\ref{perimeter5}), (\ref{st}) and (\ref{c1}),
\begin{eqnarray} \label{perimeter6}
& & \int_{T(C )} \mbox{exp}\, \{ c(u ^2 + v ^2 ) \} v ^{k+l} \, ds \\
& \leq & \int_a ^b e^{cr^2 (t) } r^{k+l} (t) \sqrt{ r^2 (t) (\theta ^{\prime
} (t))^2 + (r ^{\prime } (t)) ^2 } \frac{d\sigma }{d\theta } \, \sin ^{k+l}
\sigma (\theta (t) ) \, dt \notag \\
& = & \frac{1}{c_1 } \int_a ^b e^{cr^2 (t) } r^{k+l} (t) \sqrt{ r^2 (t)
(\theta ^{\prime } (t))^2 + (r ^{\prime } (t)) ^2 } \, \sin ^k \theta (t)
\cos ^l \theta (t) \, dt \notag \\
& = & \frac{1}{c_1 } \int _C \exp \, \{ c( x^2 + y^2 ) \} x^l y^k \, ds ,
\notag
\end{eqnarray}
with equality if $r(t)=$ const. for $t\in [a,b]$. In view of
(\ref{perimeter1}), (\ref{perimeter3}), (\ref{perimeter4}) and (\ref{perimeter6})
we conclude that
\begin{equation} \label{perimeter7}
c_1 P_{\widetilde{\mu} }( T(M), \mathbb{R} ^2 _+ ) \leq P_{\mu } (M, W) ,
\end{equation}
with equality if $M= B_R \cap W$. \newline
Finally, using (\ref{st}), an elementary calculation shows that
\begin{eqnarray} \label{measure1}
\mu (M) & = & \int\limits _M \! \! \int \mbox{exp}\, \{ c(x^2 + y^2 )\} x^l
y^k \, dx\, dy \\
& = & \int\limits _{ p(M)} \! \! \! \! \int e^{cr^2 } r^{k+l+1 } \sin ^k
\theta \cos ^l \theta \, dr\, d\theta \notag \\
& = & c_1 \int\limits _{ p(M)} \! \! \! \! \int e^{cr^2 } r^{k+l+1 } \sin
^{k+l} \sigma (\theta ) \frac{d\sigma }{d\theta } \, dr\, d\theta \notag \\
& = & c_1 \int\limits _{p(T(M))} \! \! \! \! \int e ^{cr^2 } r^{k+l+1 } \sin
^{k+l} \sigma \, dr\, d\sigma \notag \\
& = & c_1 \widetilde{\mu} (T(M)). \notag
\end{eqnarray}
Now Theorem A, for $N=2$, tells us that
$P_{\widetilde{\mu} } (T(M), \mathbb{R} ^2 _+ ) \geq P_{\widetilde{\mu} } (T(B_R \cap W ), \mathbb{R} ^2 _+ )$.
Together with (\ref{perimeter7}) and (\ref{measure1}) this proves the
assertion for smooth sets. \newline
If $M$ is a measurable subset of $W$ with finite $\mu$-perimeter, then, by
the very properties of the weighted perimeter, there exists a sequence of
smooth sets $\{ M_n \} $, ($M_n \subset W$, $n\in \mathbb{N}$), such that
$\lim_{n\to \infty } \mu (M_n \Delta M ) =0$, and $\lim_{n\to \infty } P_{\mu
} (M_n ) = P_{\mu } (M)$. Hence the assertion for $M$ follows by
approximation. $\hfill \Box $ \\[0.2cm]
\textsl{Step 2: The general case. } \newline
We proceed by induction over the dimension $N$. We write $y= (x ^{\prime
},x_N , x_{N+1 } ) $ for points in $\mathbb{R}^{N+1} $, where $x^{\prime
}\in \mathbb{R}^{N} $, and $x_{N+1 } \in \mathbb{R} $, and
\begin{equation*}
W_{N+1} := \{ y= (x^{\prime }, x_N , x_{N+1} ) \in \mathbb{R} ^{N+1} : \,
x_i >0 , \, i=1,\ldots ,N+1 \} .
\end{equation*}
Assume that the assertion holds true for some $N\in \mathbb{N} $,
($N\geq 2 $).
\newline
Let a measure $\mu $ on $W_{N+1} $ be given by
\begin{equation*}
d \mu = \mbox{ exp } \{ c (|x^{\prime}|^2 + x_N ^2 + x_{N+1} ^2 ) \} \,
\prod_{i=1 } ^{N+1} x_i ^{k_i }\, dy .
\end{equation*}
We define two measures $\nu _1 $ and $\nu _2 $ by
\begin{eqnarray*}
d \nu _1 & = & \mbox{ exp } \{ c |x^{\prime}|^2 \}\, \prod_{i=1 } ^{N-1} x_i
^{k_i } \, dx^{\prime }, \\
d \nu _2 & = & x_{N} ^{k_N } x_{N+1} ^{k_{N+1}} \mbox{ exp } \{ c ( x_N ^2 +
x_{N+1} ^2 ) \} \, dx_N dx_{N+1 } ,
\end{eqnarray*}
and note that $d\mu = d\nu _1 d\nu _2 $. \newline
Let $M $ be a subset of $W_{N+1}$ having finite and positive $\mu $-measure.
\newline
We define $2$-dimensional slices
\begin{equation*}
M(x^{\prime }) := \{ (x_N , x_{N+1 } ):\, (x^{\prime }, x_N , x_{N+1} ) \in
M\} , \quad (x^{\prime }\in \mathbb{R}^{N-1} ).
\end{equation*}
Let $M^{\prime }:= \{ x^{\prime }\in \mathbb{R}^{N-1} :\, 0< \mu _2
(M(x^{\prime })) \} $, and note that $\nu _2 (M(x^{\prime })) < +\infty $
for a.e. $x^{\prime }\in M^{\prime }$. For all those $x^{\prime }$, let
$Q(x^{\prime })$
be the quarter disc in $\mathbb{R} ^2 _+ $ centered at
$(0,0)$ with $\nu_2 (M(x^{\prime })) = \nu _2 (Q(x^{\prime }))$. (For
convenience, we put $Q(x^{\prime })= \emptyset $ for all $x^{\prime }\in
M^{\prime }$ with $\nu _2 (M(x^{\prime })) =+\infty $.) Let
\begin{equation*}
W_2 := \{ (x_N , x_{N+1} ):\, x_N >0, \, x_{N+1 } >0 \} .
\end{equation*}
Since Theorem 1 holds in the two-dimensional case, we have that
\begin{equation} \label{isopnu2}
P_{\nu_2 } (Q(x^{\prime }), W_2 ) \leq P_{\nu _2 } (M(x^{\prime }), W_2 )
\quad \mbox{for a.e. } \ x^{\prime }\in M^{\prime }.
\end{equation}
Let
\begin{equation*}
Q:= \{ y=(x^{\prime }, x_N , x_{N+1} ) :\, (x_N , x_{N+1} )\in Q(x^{\prime
}) ,\, x^{\prime }\in M^{\prime }\} .
\end{equation*}
The product structure of the measure $\mu $ tells us that \newline
\textbf{(i)} $\mu (M)= \mu (Q)$, and \newline
\textbf{(ii)} the isoperimetric property for slices, (\ref{isopnu2}),
carries over to $M$, that is,
\begin{equation} \label{isopnu}
P_{\mu } (Q, W_{N+1 }) \leq P_{\mu } (M, W_{N+1} ) ,
\end{equation}
(see for instance Theorem 4.2 of \cite{BBMP}). \newline
Note again, the slice $Q (x^{\prime }) = \{ ( x_N , x_{N+1 } ) :\,
(x^{\prime }, x_N , x_{N+1 } ) \in Q \} $ is a quarter disc $\{ (r\cos
\theta , r \sin \theta ):\, 0 < r < R(x^{\prime }) ,\, \theta \in (0, \pi/2
) \} $, with $0<R(x^{\prime }) <+\infty $, ( $x^{\prime }\in M^{\prime }$).
Set
\begin{equation*}
K:= \{ (x^{\prime }, r) : \, 0< r < R(x^{\prime }), \, x^{\prime }\in
M^{\prime }\} ,
\end{equation*}
and introduce a measure $\alpha $ on
\begin{equation*}
W_N := \{ (x^{\prime },r ) :\, x_i >0 , \, i=1,\ldots ,N-1, \, r>0 \}
\end{equation*}
by
\begin{equation*}
d\alpha := a r^{k_N+k_{N+1}+1 } \mbox{ exp } \{ c(|x^{\prime}|^2+ r^2 )\} \,
dx^{\prime }dr,
\end{equation*}
where
\begin{equation*}
a := \int_0 ^{\pi / 2} \mbox{cos} ^{k_N} \theta \,\,\mbox{sin} ^{k_{N+1}
}\theta \, d\theta.
\end{equation*}
An elementary calculation then shows that
\begin{equation*}
\mu (Q) = \alpha (K),
\end{equation*}
and
\begin{equation*}
P_{\mu } (Q, W_{N+1} ) = P_{\alpha } (K, W_N ).
\end{equation*}
Let $B_R $ denote the open ball in $\mathbb{R}^N $ centered at the origin,
with radius $R$, and choose $R>0 $ such that
\begin{equation*}
\alpha (B_R \cap W_N ) = \alpha ( K).
\end{equation*}
By the induction assumption it follows that
\begin{equation} \label{isopalpha}
P_{\alpha } (B_R \cap W_N , W_N ) \leq P_{\alpha } (K, W_N ).
\end{equation}
Finally, let $M^{\bigstar } $ defined by
\begin{equation*}
M^{\bigstar } := \{ y= (x^{\prime }, x_N , x_{N+1 } ):\, |x^{\prime}|^2+ x_N
^2 + x_{N+1 } ^2 < R^2 , \, x_i >0 ,\, i=1,..,N+1 \} .
\end{equation*}
Then
\begin{equation*}
\mu (M^{\bigstar }) = \mu (M)
\end{equation*}
and
\begin{equation*}
P_{ \mu } (M^{\bigstar} , W_{N+1} ) = P_{\alpha } (B_R \cap W_N , W_N ).
\end{equation*}
The equalities above, together with (\ref{isopalpha}) and (\ref{isopnu}), yield
\begin{equation*}
\label{isopN+1} P_{\mu } (M^{\bigstar }, W_{N+1} ) \leq P_{\mu } (M, W_{N+1}
),
\end{equation*}
that is, the isoperimetric property holds for $N+1 $ in place of $N$
dimensions. The Theorem is proved. $\hfill \Box $
| 10,404
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We understand that the student's who wall through the doors of Ursuline College, come from all walks of life, and pursue their own unique faith journey. Moreover, it is our goal to support their spiritual journey by providing frequent opportunities to encounter God, and to allow that relationship to flourish. While we continue to practice physical distancing, we can still come together as a communion of disciples growing in the way of Jesus Christ. Please click on the links below to find ways we can live out our faith this year. Virtual Events And Liturgies ► Prayer Service Fall 2020: Caring for our Common Home ► Science and Faith: A Conversation With Br. Guy Consolmagno, November 9th, 1:00 PM ► Remembrance Day Service: November 11th, 10:45 AM Chaplaincy Google Classroom/ Google Meet ► Here, you will find morning prayers and the google meet link for live liturgical services and prayers during the year. The class code is: zydeolh UCC Virtual Chapel ► Here, you will find multimedia resources to explore the wonder of God and our faith. Chaplaincy Team Sign Up ► Here, you will find opportunities to help lead our school community with prayer, leadership, and service. School Chaplaincy is carried out in a collaborative and cooperative manner, in order to promote the spiritual and human development of all members in our Catholic School Community. If you feel called to help with the faith life of our school, please make an appointment by contacting me with the information found below. Furthermore, I offer an open, friendly, and nonjudgmental environment for you to share your thoughts and concerns. I look forward to meeting you! Mr. Dula, Chaplaincy Leader 519-351-2987, ext. 25421 thomas.dula@sccdsb.net
| 182,379
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| 380,819
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TITLE: Algebraic Elements are Integral, if their Minimal Polynomial is.
QUESTION [3 upvotes]: In an upcoming exercise class in commutative algebra I would like to discuss how to detect, whether an algebraic element $\alpha$ over $\Bbb Q$ is integral over $\Bbb Z$. The claim is that it is precisely the case if $\operatorname{Mipo}_{\Bbb Q}(\alpha) \in \Bbb Z[X]$. This is indeed true, see for example this question.
The usual proof of this fact goes along the lines of showing that the minimal polynomial has integral coefficients, so since $\Bbb Z$ is integrally closed the minimal polynomial lies in $\Bbb Z[X]$. However, I think I was told another proof some while ago. I tried to remember it, but it seems to circumvent the integral closedness of $\Bbb Z$, which confuses me. So I would really appreciate a second opinion.
Let $\alpha$ be algebraic over $\Bbb Q$. Then $\alpha$ is integral over $\Bbb Z$ if and only if $\operatorname{Mipo}_\Bbb{Q}(\alpha) \in \Bbb Z[X]$.
One direction is immediate since minimal polynomials are assumed to be monic. So let $\alpha$ be integral over $\Bbb Z$. This means that $\Bbb Z[\alpha]$ is a finitely generated $\Bbb Z$-module. Since $\Bbb Z[\alpha]$ is torsion-free and $\Bbb Z$ is a principal ideal domain, $\Bbb Z[\alpha]$ is a free $\Bbb Z$-module of rank $k \in \Bbb N$. Tensoring with $\Bbb Q$ shows that $\Bbb Z[\alpha] \otimes \Bbb Q \cong \Bbb Q[\alpha] = \Bbb Q(\alpha)$ is a free $\Bbb Q$-module of rank $k$. By IBN $k = \deg \operatorname{Mipo}_\Bbb{Q}(\alpha)$.
Now multiplication with $\alpha$ gives a $\Bbb Z$-linear endomorphism $\cdot\alpha:\Bbb Z[\alpha] \rightarrow \Bbb Z[\alpha]$. It gives rise to a characteristic polynomial $\operatorname{CP}(\cdot\alpha) \in \Bbb Z[X]$, which is a monic polynomial of degree $k$. Cayley-Hamilton implies that evaluating it at $\cdot\alpha$ yields the zero-endomorphism, ie. that $\operatorname{CP}(\cdot\alpha)(\cdot\alpha) = 0$. In particular $\operatorname{CP}(\cdot\alpha)(\cdot\alpha)(1) = \operatorname{CP}(\cdot\alpha)(\alpha) = 0$.
We have found a monic polynomial in $\Bbb Z[X]\subseteq \Bbb Q[X]$, which annihilates $\alpha$ and has the same degree as $\operatorname{Mipo}_\Bbb{Q}(\alpha)$, so both polynomials have to coincide.
Does this proof fail, and if so where?
Thank you for your help.
REPLY [3 votes]: Your proof is correct, but it uses the fact that $\mathbb{Z}$ is a PID, a property stronger than integral closedness.
If $R$ is not integrally closed, there exists an $\alpha \in Q(R)-R$ that is integral over $R$. Its minimum polynomial over $Q(R)$, however, is $X-\alpha$, which is not in $R[X]$. (@ChrisH: $R[\alpha]$ is finitely generated as an $R$-module, but not free.)
Edit:
To see that $R[\alpha]$ is not free over $R$, consider Chris' example $R = k[x,y]/(x^2-y^3)$, and take $\alpha = x/y$. Then $\alpha^2 = y$, so $\alpha$ is integral over $R$. But the minimum polynomial over $Q(R)$ is $X-\alpha$ $\notin$ $R[X]$. Assuming $S$ = $R[\alpha]$ is free, it is certainly flat over $R$, so that $(I\cap J)S$ = $IS \cap JS$ for every pair of ideals $I,J$ of $R$ (cf. Matsumura, Commutative Ring Theory, Th.7.4.ii). Taking $I=xR$ and $J=yR$, we have $IS$ = $xR$ + $y^2R$ and $JS$ = $xR$ + $yR$, with intersection $xR+y^2R$. But $I \cap J$ = $x^2R$ + $xyR$, which is already an ideal of $S$, but does not contain the element $x$, or $y^2$, for that matter. Contradiction.
End of insert.
When $R$ is an integrally closed domain, $K = Q(R)$ its quotient field, $L$ a field extension of $K$, and $\alpha \in L$ an element that is integral over $R$, the minimum polynomial $f$ of $\alpha$ over $K$ is in $R[X]$. For $f$ factorizes as $\prod_{1 \leq i \leq n}(X-\alpha_i)$ in $E[X]$, where $E \supseteq K(\alpha)$ is a splitting field for $f$ over $K$, $n = \text{deg}(f)$, and $\alpha = \alpha_1$. Let $g \in R[X]$ be a monic polynomial such that $g(\alpha) = 0$. Then $f$ divides $g$ in $K[X]$. So every $\alpha_i$ is a root of $g$, and therefore integral over $R$. The coefficients of $f$ are (elementary symmetric) polynomials in the $\alpha_i$ (with coefficients in $\mathbb{Z}$), hence they are also integral over $R$. But the coefficients are in $K$, and because $R$ is integrally closed in $K$, they must be in $R$.
Compare Corollary 2.6.a in Fröhlich & Taylor, Algebraic Number Theory. There, the additional assumption is made that $\alpha$ is separable over $K$, which is not necessary for the result at hand.
| 33,696
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\section{Ramification Point}\label{section:ramification-point}
\subsection{Introduction}
In section \ref{subsection:deck-group-generators}, we concluded that the deck transformation group $\Aut(f)$ is generated by $(n-1)$ \textit{parabolic} transformations from $\Aut(\mathbb{H})$. On the other hand, we will discover properties of such subgroups of $\mbox{GL}_2(\mathbb{R})$. Notice that $\gamma(\tau)=(\frac{\gamma}{\det{\gamma}})(\tau)$, so we will only consider $\mbox{SL}_2(\mathbb{R})$ from now. Consider $\mbox{SL}_2(\mathbb{R})$ with the normal $\mathbb{R}^4$ topology. The general matrix multiplication
$$\mbox{SL}_2(\mathbb{R})\times\mbox{SL}_2(\mathbb{R})\rightarrow\mbox{SL}_2(\mathbb{R}),\quad (\gamma_1,\gamma_2)\mapsto \gamma_1\gamma_2$$
and inversion
$$\mbox{SL}_2(\mathbb{R})\rightarrow\mbox{SL}_2(\mathbb{R}),\quad \gamma\mapsto\gamma^{-1}$$
are continuous maps.
\begin{definition}
We say that $\mbox{SL}_2(\mathbb{R})$ acts on $\mathbb{H}$ continuously if the following map
\begin{equation}
\mbox{SL}_2(\mathbb{R})\times\mathbb{H}\rightarrow \mathbb{H},\qquad (\gamma,\tau)\mapsto\gamma(\tau)\notag
\end{equation}
is continuous.
\end{definition}
In general, if $G$ is a subgroup of $\mbox{SL}_2(\mathbb{R})$. An orbit of a point $\tau\in\mathbb{H}$ is the set of images of $\tau$ under actions in $G$, i.e.,
$$\mbox{Orbit}_G(\tau)=\{g(\tau)\,|\,g\in G\}.$$
The stabilizer of a point $\tau\in\mathbb{H}$ are the elements in $G$ that fixes $\tau$, i.e.,
$$\mbox{Stab}_G(\tau)=\{g\,|\,g(\tau)=\tau, g\in G\}.$$
We say that two points $\tau,\tau'\in\mathbb{H}$ are equivalent under the action of $G$ if $\tau'\in\mbox{Orbit}_G(\tau)\tau$, i.e.,
$$\tau\sim\tau'\qquad\mbox{if and only if}\qquad\tau'=g(\tau),\,\mbox{for some } g\in G.$$
The set $\mbox{Orbit}_G(\tau)$ is the set of points that are equivalent to $\tau$ under the action of $G$.
\begin{definition}
If $G$ is a discrete subgroup of $\mbox{SL}_2(\mathbb{R})$, define the quotient space $\mathbb{H}/G$ to be the space of equivalent classes of $\mathbb{H}$ under the action of $G$, i.e.,
$$\mathbb{H}/G=\mathbb{H}/\sim,$$
equipped the induced topology through the quotient map.
\end{definition}
Let $\mbox{SL}_2(\mathbb{R})/G$ to be the left cosets of $G$ in $\mbox{SL}_2(\mathbb{R})$, i.e.,
$$\mbox{SL}_2(\mathbb{R})/G=\bigcup_{h\in H}hG,$$
where $H$ is the left coset representation of $G$ in $\mbox{SL}_2(\mathbb{R})$. The topology on $\mbox{SL}_2(\mathbb{R})/G$ is induced from the left group action.
\begin{definition}\label{def:elliptic-point}
A point $P\in\mathbb{H}$ is \textit{elliptic} of a discrete subgroup $G\subset\mbox{SL}_2(\mathbb{R})$ if it is invariant under the action of an \textit{elliptic} element in $G$.
\end{definition}
The goal for this section is to show that a point $P\in\mathbb{H}/G$ is \textit{elliptic} if and only if the point $P\in\mathbb{H}/G$ is a ramification point.
\subsection{Discrete Group, Quotient Space and Ramification Point}
All the conclusions can be found in \cite[p. 2-20, Chapter 1]{ShimuraIntroToArith}. For the completion of this article, I still state them.
\begin{lemma}\label{lemma:fact1-SO2-homeomorphism}
$\mbox{SO}_2(\mathbb{R})$ is compact in $\mbox{SL}_2(\mathbb{R})$. Furthermore, the map
\begin{equation}
\tilde{\alpha}:\mbox{SL}_2(\mathbb{R})/\mbox{SO}_2(\mathbb{R})\rightarrow\mathbb{H},\qquad [\gamma]\mapsto \tilde{\alpha}([\gamma])=\gamma(i)
\end{equation}
is a homeomorphism.
\end{lemma}
\begin{proof}
See \cite[p. 2, Theorem 1.1]{ShimuraIntroToArith} or \cite[p. 25, Proposition 2.1 (d)]{JSMilne}
\end{proof}
\begin{comment}
\begin{proof}
To show the compactness, define a map
$$\begin{array}{cccc}
\varphi: & S^1 & \rightarrow & \mbox{SL}_2(\mathbb{R}),\\
& e^{i\theta} & \mapsto & \varphi(e^{i\theta})=\left(\begin{array}{cc}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{array}\right).
\end{array}$$
Notice that $\varphi$ is continuous and $S^1$ is compact, so its image $\varphi(S^1)=\mbox{SO}_2(\mathbb{R})$ is also compact.
To show $\tilde{\alpha}$ is a homeomorphism, it is equivalent to show that it is bijective and continuous and open. We define a map $\alpha$ on $\mbox{SL}_2(\mathbb{R})$ as the following,
$$\alpha:\mbox{SL}_2(\mathbb{R})\rightarrow\mathbb{H},\qquad \gamma\mapsto \alpha(\gamma)=\gamma(i).$$
Since $\mbox{SL}_2(\mathbb{R})$ acts transitively on $\mathbb{H}$, i.e., $\mbox{Orbit}_{\mbox{SL}_2(\mathbb{R})}(\tau)=\mathbb{H}$ for any $\tau\in\mathbb{H}$, so the map $\alpha$ is onto, which implies that $\tilde{\alpha}$ is surjective. For any $\gamma_1,\gamma_2\in\mbox{SL}_2(\mathbb{R})$, the condition $\tilde{\alpha}([\gamma_1])=\tilde{\alpha}([\gamma_2])$ holds if and only if the equation $\gamma_1(i)=\gamma_2(i)$ holds. If $\gamma_1(i)=\gamma_2(i)$ is true, then we have equality $$\gamma_2^{-1}(\gamma_1(i))=(\gamma_2^{-1}\gamma_1)(i)=i,$$
which implies the following relation
$$(\gamma_2^{-1}\gamma_1)\in\mbox{Stab}_{\mbox{SL}_1(\mathbb{R})}(i)=\mbox{SO}_2(\mathbb{R}),\quad\mbox{equivalently}\quad[\gamma_1]=[\gamma_2],$$
thus $\alpha$ is injective. Therefore $\tilde{\alpha}$ is a bijection. The continuity of $\tilde{\alpha}$ follows from the continuity of $\alpha$.\\
Now we only need to show it is open. Observe that if $x+yi\in\mathbb{H}$ such that $\alpha(\gamma)=x+yi$ for some $\gamma=\left(\begin{array}{cc}a & b\\
c & d
\end{array}\right)\in\mbox{SL}_2(\mathbb{R})$, then we have the following conclusion
$$ad-bc=1,\quad\frac{ai+b}{ci+d}=x+yi,\qquad\mbox{implies}\qquad (x^2+y^2)d-bx-ay=0,$$
since $y\neq 0, c=\frac{dx-b}{y}$. Suppose $\Omega\subseteq\mbox{SL}_2(\mathbb{R})/\mbox{SO}_2(\mathbb{R})$ is open, we will show that any point $x_0+y_0i\in\tilde{\alpha}(\Omega)$ is an interior point. Notice that $x_0+y_0i$ is in the image $\tilde{\alpha}(\Omega)$, so there exists an element $\gamma_0=\left(\begin{array}{cc}
a_0 & b_0\\
c_0 & d_0
\end{array}\right)\in\mbox{SL}_2(\mathbb{R})$ such that $[\gamma_0]\in\Omega$ and $\gamma_0(i)=x_0+y_0i$, we have the following equation
$$(x_0^2+y_0^2)d_0-b_0x_0-a_0y_0=0.$$
Define a function
$$\begin{array}{cccl}
s: & \mathbb{R}^4\times\mathbb{R} & \rightarrow & \mathbb{R},\\
& (x,y,a,b,d) & \mapsto & (x^2+y^2)d-bx-ay.
\end{array}$$
Notice that $\partial s/\partial d=x^2+y^2>0$ since $y>0$. By implicit function theorem, there is a continuous differentiable function $g(x,y,a,b)=d$ on a neighborhood $\Omega_0$ of $(x_0,y_0,a_0,b_0)$ in $\mathbb{R}^4$ with $s(x,y,a,b,d)=0$. Notice that $d=g(x_0,y_0,a,b_0)=\frac{y_0}{x_0^2+y_0^2}a+\frac{x_0}{x_0^2+y_0^2}b_0$ is linearly dependent on $a$, thus there is $\{x_0\}\times\{y_0\}\times(a_0-\varepsilon,a_0+\varepsilon)\times\{b_0\}\subseteq\Omega_0$ for some $\varepsilon>0$ such that $(d_0-\varepsilon',d_0+\varepsilon')=U_{d_0}\subseteq g(\Omega_0)$ for some $\varepsilon'>0$. Recall the fact that $\Omega\subseteq\mbox{SL}_2(\mathbb{R})/\mbox{SO}_2(\mathbb{R})$ is open, without loss of generality, we assume $\{a_0\}\times\{b_0\}\times\{c_0\}\times U_{d_0}\subseteq\Omega$. Therefore the solution set $g^{-1}(U_{d_0})\subseteq\mathbb{R}^4$ is open, so there is an neighborhood $U_{x_0}\times U_{y_0}\times\{a_0\}\times\{b_0\}\subseteq g^{-1}(U_{d_0})$ such that $U_{x_0}\times U_{y_0}$ is an open neighborhood of $(x_0,y_0)$, thus it is an interior point. We can conclude that $\alpha$ is a homeomorphism.
\end{proof}
\end{comment}
\begin{lemma}\label{lemma:fact2-discrete-compact-finite}
Assume $\Gamma$ is a discrete subgroup of $\mbox{SL}_2(\mathbb{R})$, if $V_1,V_2$ are any compact subsets in $\mathbb{H}$, then the set $\{\gamma\in\Gamma|\gamma(V_1)\cap V_2\neq\emptyset\}$ is finite.
\end{lemma}
\begin{comment}
\begin{proof}
Assume $V_1\subseteq\mathbb{H}$ is compact, we will show that $\alpha^{-1}(V_1)$ is also compact in $\mbox{SL}_2(\mathbb{R})$, where $\alpha$ is the same map defined in lemma \ref{lemma:fact1-SO2-homeomorphism}. Since $\mbox{SL}_2(\mathbb{R})$ has the normal topology in $\mathbb{R}^4$, we can consider the collection of balls $B_i=\{(a,b,c,d)\,|\,|(a,b,c,d)|<i\}\subseteq\mbox{SL}_2(\mathbb{R})$ which forms an open cover of $\mathbb{R}^4$, we have the following conclusion
$$\mbox{SL}_2(\mathbb{R})\subseteq\cup_{i=1}^{\infty}B_i\quad\mbox{implies}\quad V_1\subseteq\cup_{i=1}^{\infty}\alpha(B_i).$$
Thus there exists a finite number $m\in\mathbb{Z}$ such that $V_1\subseteq\alpha(B_m)$ since $V_1$ is compact. Therefore the image of the closure $\overline{B_m}$ of $B_m$ covers $V_1$, i.e., $V_1\subseteq\alpha(\overline{B_m})\subseteq\mathbb{H}$. Consider the composition with the inverse of $\alpha$,
$$\alpha^{-1}(V_1)\subseteq\alpha^{-1}(\alpha(\overline{B_m})),$$
we conclude that $\alpha^{-1}(V_1)$ is closed because $\alpha$ is continuous and $V_1$ is compact so it is closed. Consider the right hand side $\alpha^{-1}(\alpha(\overline{B_m}))=\tilde{\alpha}^{-1}(\alpha(\overline{B_m}))\times\mbox{SO}_2(\mathbb{R})$, $\alpha(\overline{B_m})$ is compact since $\overline{B_m}$ is compact, and $\tilde{\alpha}^{-1}(\alpha(\overline{B_m}))$ is compact since $\tilde{\alpha}$ is a homeomorphism. Therefore the set $\tilde{\alpha}^{-1}(\alpha(\overline{B_m}))\times\mbox{SO}_2(\mathbb{R})$ is compact since $\mbox{SO}_2(\mathbb{R})$ is compact. The pre-image $\alpha^{-1}(V_1)$ of $V_1$ is a closed subset in a compact subset, so it is also compact. Similarly, $\alpha^{-1}(V_2)$ is also compact. The fact that the intersection of a discrete set with a compact set $\Gamma\cap(\alpha^{-1}(V_1)\cup\alpha^{-1}(V_2))$ is finite, so there is only finitely $\gamma$ such that $\gamma(V_1)\cap V_2\neq\emptyset$.
\end{proof}
\end{comment}
\begin{proof}
See \cite[p. 3, Proposition 1.6]{ShimuraIntroToArith} or \cite[p. 26, Proposition 2.4]{JSMilne}.
\end{proof}
\begin{prop}\label{prop:discreteneighborhood}
If $\Gamma\in\mbox{SL}_2(\mathbb{R})$ is a discrete subgroup, then for any $\tau\in\mathbb{H}$, there is a neighborhood $U$ of $\tau$ such that if $\gamma\in\Gamma$ and $U\cap\gamma(U)\neq\emptyset$, then $\gamma(\tau)=\tau$.
\end{prop}
\begin{proof}
See \cite[p. 3, Proposition 1.7]{ShimuraIntroToArith} or \cite[p. 27, Proposition 2.5 (b)]{JSMilne}
\end{proof}
\begin{comment}
\begin{proof}
For any $\tau\in\mathbb{H}$, let $V$ be a compact neighborhood of $\tau$ in $\mathbb{H}$. By lemma \ref{lemma:fact2-discrete-compact-finite}, we assume there is a finite subset $\{\gamma_1,\ldots,\gamma_l\}\subseteq\Gamma$ such that $\gamma(V)\cap V\neq\emptyset$. Suppose $\{\gamma_1,\ldots,\gamma_s\}$ are the ones that fixes $\tau$, i.e., $\gamma_i(\tau)=\tau$ for $1\le i\le s$. And for any element $\gamma_i$ which is from the rest of the set $\{\gamma_{s+1},\ldots,\gamma_l\}$ does not fix $\tau$, i.e., $\gamma_i(\tau)\neq\tau(s+1\le i\le l)$. Take some disjoint neighborhoods $U_i^{(1)}$ of $\tau$ and $U_i^{(2)}$ of $\gamma_i(\tau)$ in $V$, consider the intersection
$$U=\cap_{i=s+1}^l (U_i^1\cap\gamma_i^{-1}(U_i^2)),$$
then $U$ is the neighborhood we are looking for.
\end{proof}
\end{comment}
Next, we are able to show that if $\Gamma$ is a discrete subgroup of $\mbox{SL}_2(\mathbb{R})$, a point $[\tau]=[\tau_0]\in\mathbb{H}/\Gamma$ is ramified if and only if following condition
\begin{equation}\label{eq:elliptic-point-equivalent-condition}
\mbox{Stab}_{\Gamma}(\tau)-\{I\}\neq \emptyset
\end{equation}
holds, for any $\tau\in[\tau_0]\subset\mathbb{H}$. Recall Definition \ref{def:elliptic-point}, a point $P$ satisfying condition \eqref{eq:elliptic-point-equivalent-condition} if and only if $P$ is an \textit{elliptic} point. The following proposition can be concluded from \cite[p. 8, Proposition 1.18]{ShimuraIntroToArith}, we will present proof here since it is not obvious.
\begin{prop}\label{proposition-5.7}
Let $\Gamma$ be a discrete subgroup of $\mbox{SL}_2(\mathbb{R})$. Then on the quotient space $\mathbb{H}/\Gamma$, a point $\tau\in\mathbb{H}$ is ramified if and only if it is an \textit{elliptic} point of $\Gamma$.
\end{prop}
\begin{proof}
If $\tau\in\mathbb{H}$ is not an elliptic point, then by Proposition \ref{prop:discreteneighborhood}, there is a neighborhood $U$ of $\tau$ such that $\gamma(U)\cap U=\emptyset$, for any $\gamma\in\Gamma$. It implies that $\tau$ is not ramified. If $\tau\in\mathbb{H}$ is an elliptic point of $\Gamma$, assume there is an elliptic element $\gamma\in\Gamma$ such that $\gamma(\tau)=\tau$. Let $\sigma\in\mbox{SL}_2(\mathbb{R})$ such that $\sigma(i)=\tau$, then the composition $\sigma^{-1}\circ\gamma\circ\sigma\in\mbox{Stab}(i)=\mbox{SO}_2(\mathbb{R})$, so $<\sigma^{-1}\circ\gamma\circ\sigma>=\sigma^{-1}\circ<\gamma>\circ\sigma$ is a subgroup of the conjugate $\sigma^{-1}\Gamma\sigma$, which is still a discrete subgroup of $\mbox{SL}_2(\mathbb{R})$. Also we have the following correspondence relation
\begin{equation*}
<\gamma>\cong \sigma^{-1}<\gamma>\sigma=\mbox{SO}_2(\mathbb{R})\bigcap\sigma^{-1}\Gamma\sigma,
\end{equation*}
the right hand side is an intersection of a compact set and a discrete set, so it has to be finite. Therefore if $m=\#\mbox{Stab}_{\Gamma}(\tau)$, $\tau$ is a ramification point of index $m$.
\end{proof}
| 52,150
|
TITLE: Action of a group on set of morphisms
QUESTION [1 upvotes]: In the theory of cohomology of groups, I came across some problem. It might be silly question, but I get puzzled around it long time.
Let $G$ be a group and $A$ abd $B$ be abelian group with action of $G$ on each of them.
Then $A$ and $B$ are $G$-modules (or $\mathbb{Z}[G]$ modules).
Consider the set $\mbox{Hom}(A,B)$ of homomorphisms of abelian group $A$ into $B$. Given action of $G$ on this set by (given $x\in G$ and $a\in A$)
$$f^x(a):=(f(a^{x^{-1}}))^x \hskip1cm (*).$$
My question is simple, I don't know where is my fault of understanding.
Can't we consider the action by $$f^x(a):=(f(a))^x\hskip1cm (**)$$ or $$f^x(a):=f(a^x) \hskip1cm (***)?$$ In these cases, the action of $G$ is essentially considered only on one module. But What is the reason to consider the action of $G$ on $\mbox{Hom}(A,B)$ as in ( * ) in the Cohomology of groups? Also, why inverse comes in (*)?
In the classroom, I was writing in notebook the action of $G$ as in $(***)$, but as soon as the speaker wrote $(*)$, I puzzled with above question.
REPLY [1 votes]: A nice thing about this action is that the fixed points $\mbox{Hom}(A,B)^G$ are precisely the $G$-linear homomorphisms $\operatorname{Hom}_G(A,B)$.
Also note that $f^x(a):=f(a^x)$ is a right action not a left action, because under this definition $(f^x)^y(a)=f(a^{yx})=f^{yx}(a)$, that's why you include the inverse.
| 194,932
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Art Week 2019
25th October 2019
Every year the children take part in a ‘Finstall Art Week’, working with some professional artists, as well as taking part in a wide range of art activities throughout the week. This year the children worked with paint, felt, clay and other mediums, as well as working with dance and yoga instructors. Here are a few pictures from the week’s activities.
| 408,472
|
\begin{document}
\title[Distances between pairs of vertices and vertical profile]
{Distances between pairs of vertices and vertical profile
in conditioned Galton--Watson trees}
\date{December 17, 2008}
\author{Luc Devroye}
\address{School of Computer Science, McGill University, 3480
University Street, Montreal, Canada H3A 2K6}
\email{lucdevroye@gmail.com}
\urladdrx{http://cg.scs.carleton.ca/~luc/}
\author{Svante Janson}
\address{Department of Mathematics, Uppsala University, PO Box 480,
SE-751~06 Uppsala, Sweden}
\email{svante.janson@math.uu.se}
\urladdrx{http://www.math.uu.se/~svante/}
\keywords{Random Galton--Watson tree, paths of given length, vertical
profile, probabilistic analysis of algorithms, branching process.
}
\subjclass[2000]{60C05; 05C05}
\thanks{\emph{CR Categories.} 3.74, 5.25, 5.5.}
\thanks{The first author's research was sponsored by NSERC Grant A3456 and FQRNT Grant 90-ER-0291}
\begin{abstract}
We consider a \cGWt{} and
prove an estimate of the number of pairs of vertices
with a given distance, or, equivalently,
the number of paths of a given length.
We give two proofs of this result, one probabilistic and the other
using generating functions and singularity analysis.
Moreover,
the second proof yields a more general estimate for generating
functions, which is used to prove a conjecture by
Bousquet--M\'elou and Janson \cite{SJ185},
saying that the vertical
profile of a randomly labelled \cGWt{} converges in distribution,
after suitable normalization,
to the density of ISE (Integrated Superbrownian Excursion).
\end{abstract}
\maketitle
\section{Introduction and results}\label{S:intro}
Let $\tn$ be a \cGWt, \ie, the random rooted tree $\cT$ obtained as the
family tree of a \GWp{} with some given offspring distribution $\xi$,
conditioned on the number of vertices $|\cT|=n$. We will always assume
that
\begin{align}\label{Axi}
\E\xi=1 \qquad \text{and} \qquad
0<\gss\=\Var\xi<\infty
\end{align}
In other words, the \GWp{} is critical and with finite variance, and
$\P(\xi=1)<1$.
(Note that this entails $0<\P(\xi=0)<1$.)
It is well-known that this assumption is without essential loss of
generality, and that the resulting random trees are essentially the
same as the simply generated families of trees introduced by Meir and
Moon \cite{MM}.
The importance of this construction
lies in that
many combinatorially interesting random trees are of this type,
for example
random plane (= ordered) trees,
random unordered labelled trees (Cayley trees),
random binary trees,
and (more generally) random $d$-ary trees.
For further examples see \eg{}
Aldous \cite{AldousII} and Devroye \cite{Devroye}.
We consider
only $n$ such that $\tn$ exists, \ie,
such that $\P(|\cT|=n)>0$.
The \emph{span} of $\xi$ is defined to be the largest integer $d$ such
that $\xi\in d\bbZ$ a.s. If the span of $\xi$ is $d$, then $\tn$ exists
only for $n\equiv 1 \pmod d$, and it exists for all large such $n$.
We consider in this paper two types of properties of $\tn$ that turn
out to have proofs using a common argument.
First, for an arbitrary rooted tree $\tau$,
let $P_k(\tau)$, $k\ge1$, be the number of (unordered) pairs of
vertices \set{v,w} in $\tau$ such that the distance $d(v,w)=k$;
equivalently, $P_k(\tau)$ is the number of paths of length $k$ in $\tau$.
Our first result is an estimate, uniform in all $k$ and $n$,
of the expectation of this number $P_k(\tn)$ for a conditioned
Galton--Watson tree $\tn$.
We let in this paper $C_1,C_2$ and $c_1,c_2$ denote various positive constants
that may depend on (the distribution of) $\xi$, and sometimes later
$\eta$ introduced below, but not on $n$, $k$ and other variables
unless explicitly stated.
Recall that we tacitly assume \eqref{Axi}.
\begin{theorem}
\label{T1}
There exists a constant $\CC$
such that for all
$k\ge1$ and $n\ge1$,
$\E P_k(\tn)\le \CCx nk$.
\end{theorem}
One way to interpret this result is that the expected number of
vertices of distance $k$ from a randomly chosen vertex in $\tn$ is
$O(k)$. In other words, if $\tnx$ is $\tn$ randomly rerooted, and
$\wz_k(\tau)$ is the number of vertices of distance $k$ from the root in
a rooted tree $\tau$, then the following holds.
\begin{corollary}
\label{C1}
$\E \wz_k(\tnx)=O(k)$, uniformly in all $k\ge1$ and $n\ge1$.
\end{corollary}
This can be compared to \cite[Theorem 1.13]{SJ167}, which shows that
\begin{equation}
\label{wk}
\E \wz_k(\tn)=O(k),
\end{equation}
again uniformly in $k$ and $n$.
Note that in the special case when $\tn$ is a random (unordered)
labelled tree, $\tnx$ has the same distribution, so \refC{C1} reduces
to \eqref{wk}. However, in general, a randomly rerooted \cGWt{} is not
a \cGWt.
\begin{remark}
The emphasis is on uniformity in both $k$ and $n$. If we, on the
contrary, fix $k$ and consider limits as \ntoo, we have
$\E \wz_k(\tn)\to1+k\gss$, see Meir and Moon \cite{MM} and Janson
\cite{SJ167,SJ188}. It is shown in \cite{SJ188} that the sequence
$\E \wz_k(\tn)$ is not always monotone in $n$.
\end{remark}
We give a probabilistic proof of \refT{T1}, and thus of \refC{C1} too,
in \refS{Spf1}.
We also give a second proof by first proving
a corresponding estimate for the generating function.
(We present two different proofs, since we find both
methods interesting, and both methods yield
as intermediary steps in the proofs
other results that we find interesting.)
Let $\fn(z)$ be the generating function defined by
\begin{equation*}
\fn(z)
\=\sum_{k=1}^\infty \E P_k(\tn)\,z^k
.
\end{equation*}
We will use standard singularity analysis, see \eg{}
Flajolet and Sedgewick \cite{FS},
and define the domain, for $0<\gb<\pi/2$ and $\gd>0$,
\begin{equation*}
\gdbd\=\set{z\in\bbC:|z|<1+\gd,\, z\neq1,\, |\arg(z-1)|>\pi/2-\gb}.
\end{equation*}
Note that $|\arg(z-1)|>\pi/2-\gb$ is equivalent to
$|\arg(1-z)|<\pi/2+\gb$.
\begin{theorem}
\label{Tgen1}
For every $\xi$ there exist positive constants $\CC$, $\gb$, $\gd$ such
that for all $n\ge1$, $\fn$ extends to an analytic function in $\gdbd$
with
\begin{equation}
\label{fn}
|\fn(z)| \le \CCx n|1-z|\qww,
\qquad z\in\gdbd.
\end{equation}
\end{theorem}
By standard singularity analysis (\ie, estimate of the Taylor
coefficients of $f_n(z)$ using Cauchy's formula and a suitable contour
in $\gdbd$), \eqref{fn} implies $\E P_k(T_n)=O(n k)$,
see Flajolet and Sedgewick \cite{FS},
Theorem VI.3 and (for the uniformity in $n$) Lemma IX.2 (applied
to the family \set{f_n(z)/n}).
Hence, \refT{T1} follows from \refT{Tgen1}.
For each pair of
vertices $v,w$ in a rooted tree,
the path from $v$ to $w$ consists of two (possibly empty) parts, one
going from $v$ towards the root, ending at the last common ancestor
$\vw$ of $v$ and $w$, and another part going from $\vw$ to $w$ in the
direction away from the root.
We will also prove extensions of the results above for $\tn$, where we
consider separately the lengths of these two parts.
Define
the corresponding bivariate generating function
(now considering ordered pairs $v,w$)
\begin{equation}
\label{hn}
\hn(x,y)
\=
\E\sumvwtn x^{d(v,\vw)} y^{d(w,\vw)}.
\end{equation}
\begin{theorem}
\label{Tgen2}
For every $\xi$ there exist positive constants $\CC\CCdef\CCgenii$,
$\gb$, $\gd$ such
that for all $n\ge1$,
\begin{equation*}
|\hn(x,y)| \le \CCx n|1-x|\qw|1-y|\qw,
\qquad x,y\in\gdbd.
\end{equation*}
\end{theorem}
Note that, by \eqref{hn} and \eqref{fn}
\begin{equation*}
\hn(z,z)=\E \sumvwtn z^{d(v,w)} = n+2\fn(z).
\end{equation*}
Hence \refT{Tgen1} follows from \refT{Tgen2}.
We prove \refT{Tgen2} in \refS{Sgen2}.
If we define $\yylm(\tau)\=\#\set{(v,w)\in \tau:d(v,\vw)=\ell,\,d(w,\vw)=m}$,
then
singularity analysis as above (but twice) shows that \refT{Tgen2}
implies the following.
(Since $P_k=\frac12\sum_{\ell=0}^k \yyxx{\ell}{k-\ell}$, this too
implies \refT{T1}.)
\begin{theorem}
\label{T11}
There exists a constant $\CC$
such that for all
$\ell,m\ge0$ and $n\ge1$,
$\E \yylm(\tn)\le \CCx n$.
\end{theorem}
One motivation for these results is that they (more precisely,
\refT{Tgen2}) are used to prove the second type of result in this
paper.
For this, we assume that we are given a further random variable
$\eta$. Given a rooted tree $\tau$,
we take an independent copy $\etae$ of $\eta$ for
every edge $e\in \tau$.
We give each vertex $v$ the label $\lv$ obtained by summing $\etae$
for all $e$ in the path from the root $\oo$ to $v$. (Thus,
$\lx{\oo}=0$.)
We assume that
\begin{align}\label{Aeta}
\E\eta=0 \qquad \text{and} \qquad 0<\gssy\=\Var\eta<\infty.
\end{align}
We further assume that $\eta$ is integer valued and with
span $1$; thus
all labels are integers, and all integers are possible labels.
We let $X(j;\tau)$ be the number of vertices in $\tau$ with
label $j$; the sequence $(X(j;\tau))_{j=-\infty}^{\infty}$
is the \emph{vertical profile} of the labelled tree.
For the random tree $\tn$, we assume that the variables $\etae$ are
independent of $\tn$. The vertical profile $X(j;\tn)$ then is a random function
defined for $j\in\bbZ$; we write $\xn(j)\=X(j;\tn)$ and extend the
domain of $\xn$ to $\bbR$ by linear interpolation between the integer points.
Our next theorem says that this function $\xn$, suitable
normalized, converges in distribution in the space
$\cor$ of continuous functions on
$\bbR$ that tend to 0 at $\pm\infty$; we equip $\cor$ with the usual
uniform topology defined by the supremum norm.
Let, further, $\fise$ denote the density of the random measure ISE
introduced by Aldous \cite{AldousISE}; $\fise$ is a random continuous
function
with (random) compact support, see
Bousquet--M\'elou and Janson \cite[Theorem 2.1]{SJ185}.
\begin{theorem}\label{T2}
With the assumptions above,
including \eqref{Axi} and \eqref{Aeta},
let $\gam\=\gsy\qw\gs\qq$.
Then, as \ntoo,
\begin{equation}\label{t2a}
\frac1n \gam\qw n\qqqq \xn\bigpar{\gam\qw n\qqqq x}
\dto \fise(x),
\end{equation}
in the space $\cor$ with the usual uniform topology.
Equivalently,
\begin{equation}\label{t2b}
n\qwwi \xn\bigpar{n\qqqq x}
\dto \gam\fise(\gam x).
\end{equation}
\end{theorem}
Note that the random functions on the left and \rhs{s} of \eqref{t2a}
and \eqref{t2b}
are density functions, \ie,
non-negative functions with integral 1.
\begin{corollary}\label{C2}
If \ntoo{} and $j_n/n\qqqq \to x$, where
$-\infty<x<\infty$,
then
$n\qwwi X(j_n;\tn)\dto \gam \fise(\gam x)$.
\end{corollary}
The limit law is characterized in \cite{BM} by a formula for its
Laplace transform.
\refT{T2} was conjectured in \cite{SJ185}, and proved there in two
special cases, \viz{} when $\xi$ has the Geometric distribution
$\Ge(1/2)$ and thus $\tn$ is a
random ordered tree, and $\eta$ is uniformly distributed on either
\set{-1,1} or \set{-1,0,1}.
Moreover, it was shown there \cite[Remark 3.7]{SJ185} that the proof
given in \cite{SJ185} applies generally under the assumptions above,
provided the following estimate holds.
\begin{lemma}
\label{L0}
Under the assumptions above,
there exists a constant $\CC$ such that for all $n\ge1$ and
$t\in[-\pi,\pi]$,
\begin{equation}
\label{l2}
\E\biggabs{\frac1n\sum_j X(j;\tn)e^{\ii jt}}^2
\le \frac{\CCx}{1+nt^4}.
\end{equation}
\end{lemma}
We prove \refL{L0}, and thus \refT{T2}, in \refS{Sl2},
assuming \refT{Tgen2}.
Finally, we prove \refT{Tgen2}, using singularity analysis again, in
\refS{Sgen2}, which completes the proof of all other results.
\begin{ack}
This research was mainly done during a workshop at Bellairs Research
Institute in Barbados, March 2006,
and completed during a visit of SJ to
Centre de recherches math\'ematiques, Universit\'e de
Montr\'eal, October 2008.
\end{ack}
\section{First proof of \refT{T1}} \label{Spf1}
In a rooted tree $\tau$, let
$Q_k(\tau)$, $k\ge1$, denote the number of (unordered) pairs of vertices
at path distance $k$ from each other such that the path
between them visits the root,
and let $Q'_k(\tau)$ be the number of such pairs where
the root cannot be one of the two vertices in the pair;
thus $Q_k(\tau)=Q'_k(\tau) + Z_k(\tau)$.
Then, in the Galton--Watson tree $\cT$, if $\xi$ is the number of
children of the root, and the subtrees rooted at these children are
denoted $\cT_1,\dots,\cT_\xi$,
\begin{equation}
\label{qkt}
Q'_k(\cT) = \sum_{(r,s): 1 \le r < s \le \xi} \sum_{j=0}^{k-2} Z_j (\cT_r)
Z_{k-2-j} (\cT_s)
\end{equation}
and thus, since we assume $\cT$ to be critical, \ie, $\E\xi=1$, so
$\E Z_k(\cT)=1$ for every $k$,
\begin{equation}
\label{qk}
\E Q_k(\cT)=\E Z_k(\cT)+\E Q'_k(\cT) = 1+ \E\frac{\xi(\xi-1)}2 (k-1)
=1+(k-1)\frac{\gss}2.
\end{equation}
Let $\htn$ denote the random subtree of $T_n$
rooted at a uniformly selected random vertex.
(Note the difference from $\tnx$ in \refC{C1}; in $\tnx$ we keep
all $n$ vertices, but in $\htn$ we keep only the vertices below the
new root.)
Then, clearly,
$$
\EXP \{ P_{k}(T_n) \} = n \EXP \{ Q_k (\hT_n) \}.
$$
Consequently, \refT{T1} is equivalent to:
\begin{theorem}
\label{TP1}
There exists a constant $\CC$
such that for all
$k\ge1$ and $n\ge1$,
$\E Q_k(\htn)\le \CCx k$.
\end{theorem}
In order to prove this, we will need a related, but different,
estimate for the \cGWt{} $T_n$.
\begin{theorem}
\label{TQ}
There exists a constant $\CC\CCdef\CCTQ$
such that for all
$k\ge1$ and $n\ge1$,
$\E Q_k(\tn)\le \CCx k\sqrt n$.
\end{theorem}
It is easy to see $\E Q_k(\tn)\ge \cc n^{3/2}$ when $k\sim\sqrt n$,
so the estimate in \refT{TQ} then is of the right order; in particular, the
estimate in \refT{TP1} for $\htn$ does \emph{not} hold for $\tn$.
To prove these theorems we use a few more or less standard estimates.
\begin{lemma}
\label{L2}
Assume, as above, \eqref{Axi},
and let $d$ be the span of $\xi$.
Let $S_n\=\sum_{i=1}^n\xi_i$, where $\xi_i$ are independent copies of $\xi$.
Then, for $n\equiv 1\pmod d$,
\begin{equation}\label{tail}
\P(|\cT|=n)=\frac1n\P\xpar{S_n=n-1}
\sim \frac{d}{\sigma \sqrt{2 \pi}\, n^{3/2}}
\quad\text{as }\ntoo.
\end{equation}
More generally, let $W_\ell\=\sum_{i=1}^\ell|\cT_i|$ be the size of the
union of $\ell$ independent copies of $\cT$, or equivalently, the
total progeny of a Galton--Watson process started with $\ell$
individuals,
with offspring distribution $\xi$.
Then, for all $\ell\ge1$ and
$n\ge1$,
\begin{equation}\label{l2b}
\P(W_\ell=n)=\frac{\ell}n\P(S_{n}=n-\ell)
\le \CC \ell n\qcw\exp(-\cc\ell^2/n).
\CCdef\CClb \ccdef\cclb
\end{equation}
In particular,
\begin{equation}\label{l2c}
\P(W_\ell=n)\le \CC n\qw.
\end{equation}
\end{lemma}
\begin{proof}
The identity in \eqref{l2b} is well-known, see \eg, Dwass
\cite{Dwass}, Kolchin \cite[Lemma 2.1.3, p.~105]{Kolchin} and
\citet{Pitman:enum}.
The identity in \eqref{tail} is the special case $\ell=1$, and
the well-known tail estimate in \eqref{tail} then follows by the local
central limit theorem, see, \eg, Kolchin \cite[Lemma 2.1.4, p.~105]{Kolchin}.
Similarly, the inequality in \eqref{l2b} follows by the
estimate
$\P(S_n=n-\ell)\le \CClb n\qqw \exp(-\cclb\ell^2/n)$
from \cite[Lemma 2.1]{SJ167}.
The inequality $e^{-x}\le x\qqw$ yields \eqref{l2c}.
\end{proof}
\begin{lemma}
\label{L1a}
For every $r>0$ there is a constant $\CC(r)\CCdef\CCr$ such that
for all $k\ge0$ and $n\ge1$, $\E Z_k(T_n)^r\le \CCr(r) n^{r/2}$.
\end{lemma}
\begin{proof}
For any rooted tree $\tauq$, let $\tauq\hk$ be the tree pruned at height
$k$, i.e., the subtree consisting of all vertices of distance at most $k$
from the root.
Let $\tau$ be a given rooted tree of height $k$, and let
$m\=Z_k(\tau)$, the number of leaves at maximal depth.
Note that if $\tau= \tauq\hk$ for some tree $\tauq$, then $|\tauq|=n$
if and only if $\tauq$ has $n-|\tau|$ vertices at greater depth than $k$,
and thus $N\=n-|\tau|+m$ vertices at depth $k$ or greater.
Hence, with $W_m$ as in \refL{L2} and using \eqref{tail} and
\eqref{l2b}, for any $r>0$ and assuming $N>0$ (otherwise the
probability is 0),
\begin{equation*}
\begin{split}
\P\bigpar{T_n\hk=\tau}
&=
\frac{\P\bigpar{\cT\hk=\tau,\,|\cT|=n}}{\P(|\cT|=n)}
=\frac{\P(\cT\hk=\tau)\P(W_m=N)}{\P(|\cT|=n)}
\\&
\le \CC n\qc\P(\cT\hk=\tau)m N\qcw e^{-\cclb m^2/N}
\\&
\le \CC(r) n\qc\P(\cT\hk=\tau)m N\qcw (m^2/N)^{-r/2}
\\&
= \CCx(r) n\qc N^{r/2-3/2} m^{1-r}\P(\cT\hk=\tau).
\end{split}
\end{equation*}
If $r\ge3$, this yields, since $N\le n$, the estimate
\begin{equation*}
\P(T_n\hk=\tau)\le \CCx(r) n^{r/2} m^{1-r}\P(\cT\hk=\tau),
\end{equation*}
and
summing over all $\tau$ of height $k$ with $Z_k(\tau)=m$ we obtain
\begin{equation*}
\P(Z_k(T_n)=m)\le \CCx(r) n^{r/2} m^{1-r}\P(Z_k(\cT)=m).
\end{equation*}
Consequently,
\begin{equation*}
\begin{split}
\E Z_k(T_n)^r &=\summi m^r \P(Z_k(T_n)=m)
\\&
\le \CCx(r) n^{r/2} \summi m\P(Z_k(\cT)=m)
\\&
= \CCx(r) n^{r/2} \E Z_k(\cT)
= \CCx(r) n^{r/2}.
\end{split}
\end{equation*}
This proves the result for $r\ge3$, and the result for $0<r<3$ follows
by Lyapounov's (or H\"older's) inequality.
\end{proof}
\begin{lemma}
\label{L1b}
For all $k\ge1$ and $n\ge1$, $\E Z_k(T_n)\le \CC(k\bmin\sqrt n)
\CCdef\CCLi$.
Equivalently,
for all $k\ge0$ and $n\ge1$, $\E Z_k(T_n)\le \CC((k+1)\bmin\sqrt n)
\CCdef\CCLia$.
\end{lemma}
\begin{proof}
The estimate $\E Z_k(T_n)\le \CC k$ is \eqref{wk}, which is proved in
\cite[Theorem 1.13]{SJ167}.
The estimate $\E Z_k(T_n)\le \CC\sqrt n$ is proved in \refL{L1a}.
\end{proof}
\begin{remark}
The estimate $\E Z_k(T_n)\le \CC\sqrt n$ was proved by
Drmota and Gittenberger \cite{DrmotaG:width}, assuming that $\xi$
has an exponential moment;
in fact, they then
prove the stronger bound
$\E Z_k(T_n)\le \CC\sqrt n \exp(-\cc k/\sqrt n)$.
The bound in \refL{L1b} can be further improved to
$\E Z_k(T_n)\le \CC k \exp(-\cc k^2/ n)$, but we do not know a reference for
this estimate. (Details may appear elsewhere.)
\end{remark}
\begin{remark}
Note that \refL{L1a} yields an estimate $O(n^{r/2})$ of the $r$th moment of
$Z_k(T_n)$ for an arbitrary $r$ assuming only a second moment of $\xi$.
This is in contrast to the estimate \eqref{wk}, where the
corresponding estimate $\E Z_k(T_n)^r = O(k^r)$ is valid
(for integer $r\ge1$ at least)
if $\xi$ has a finite $r+1$:th moment, but not otherwise (not even
for a fixed $k\ge2$);
one direction is by Theorem 1.13 in \cite{SJ167}, and the converse
follows from the discussion after Lemma 2.1 in \cite{SJ167}.
\end{remark}
\begin{proof}[Proof of \refT{TQ}]
We have
$Q_k (T_n)=Q'_k (T_n)+Z_k (T_n)$, and $\E Z_k (T_n)\le \CCLi k$ by
\refL{L1b},
so it suffices to show that $\E Q'_k (T_n)\le \CC k\sqrt n$.
We use \eqref{qkt}, condition on $|\cT|=n$ and take
expectations. Using the symmetry and recalling that
$\cT_1,\dots,\cT_\xi$ are independent and
$(\cT_i\mid |\cT_i|=n_i)\eqd(\cT\mid |\cT|=n_i)\eqd T_{n_i}$ for any
$n_i$,
we obtain,
with $p_\ell\=\P(\xi=\ell)$ and $q_m\=\P(|\cT|=m)$,
\begin{equation*}
\begin{split}
\EXP \left\{ Q'_k (T_n) \right\}
&=
\frac{ \EXP \left\{ \IND{\xi \ge 2, |\cT|=n } \sum_{1 \le r<s \le \xi}
\sum_{j=0}^{k-2} Z_{j} (\cT_r) Z_{k-2-j} (\cT_s) \right\}}
{ \PROB \{ |\cT| = n \} } \\
&=
\frac{ \EXP \left\{ \IND{|\cT|=n }\binom{\xi}2 \sum_{j=0}^{k-2} Z_{j}
(\cT_1) Z_{k-2-j} (\cT_2) \right\}}
{\PROB \{ |\cT| = n \} } \\
&=
q_n\qw \sum_{\ell=2}^\infty p_\ell \binom{\ell}2
\sum_{n_1,n_2\ge1}
q_{n_1} q_{n_2}
\P\lrpar{\sum_{i=3}^\ell|\cT_i|=n-1-n_1-n_2}
\\&\qquad\qquad\times
\sum_{j=0}^{k-2}
\EXP \left\{ Z_{j}(T_{n_1}) \right\}
\EXP \left\{ Z_{k-2-j}(T_{n_2}) \right\}.
\end{split}
\end{equation*}
We begin with the inner sum over $j$, $\sa$ say.
By symmetry, we consider only $n_1\le n_2$, and then we obtain from
\refL{L1b} the estimates
$\E Z_{k-2-j}(T_{n_2})\le \CCLia((k-1-j)\bmin n_2\qq)\le \CCLia(k\bmin n_2\qq)$
and
\begin{equation*}
\sum_{j=0}^{k-2} \EXP \left\{ Z_{j}(T_{n_1}) \right\}
\le
\begin{cases}
\sum_{j=0}^{k-2} \CCLia(j+1)
\le
\CCLia k^2
,
\\
\E\sum_{j=0}^{\infty} Z_{j}(T_{n_1}) =n_1.
\end{cases}
\end{equation*}
Hence
\begin{equation}\label{sa}
\sa \le \CC(k^2\bmin n_1)(k\bmin n_2\qq).
\end{equation}
Let $m\=n_1+n_2$ and sum over $n_1,n_2$ with a given sum $m$. We have
by \eqref{sa} and \eqref{tail},
\begin{equation}\label{sbm}
\begin{split}
\sbm&:= \sum_{n_1+n_2=m} q_{n_1}q_{n_2}\sa
\\&
\le
2\sum_{n_1=1}^{m/2} q_{n_1}q_{m-n_1}\CCx(k^2\bmin n_1)(k\bmin (m-n_1)\qq)
\\
&\le
\CC\sum_{n_1=1}^{m/2} n_1\qcw (m-n_1)\qcw(k^2\bmin n_1)
(k\bmin (m-n_1)\qq)
\\
&\le
\CC\frac{k\bmin m\qq}{m\qc} \sum_{n_1=1}^{m/2} \frac{k^2\bmin n_1}{n_1\qc}
\\
&\le
\CC\frac{k\bmin m\qq}{m\qc} (k\bmin m\qq)
=
\CCx\frac{k^2\bmin m}{m\qc}.
\end{split}
\end{equation}
We define further
\begin{equation}\label{scl}
\scl\=\sum_{m=2}^{n-1} \sbm \P\Bigpar{\sum_{i=3}^\ell|\cT_i|=n-1-m};
\end{equation}
thus
\begin{equation}\label{qg}
\E Q_k'(T_n) =
q_n\qw \sum_{\ell=2}^\infty p_\ell \binom{\ell}2 \scl
\le
\CC n\qc \sum_{\ell=2}^\infty p_\ell {\ell}^2 \scl.
\end{equation}
We will show that $\scl\le \CC k/n$, uniformly in $\ell\ge2$, and the result
follows by \eqref{qg}, recalling that $\sum_\ell p_\ell\ell^2=\E\xi^2<\infty$.
(The proof can be simplified in the case $\E\xi^3<\infty$, when it
suffices to show that $\scl\le \CC k\ell/n$.)
First, if $\ell=2$, the only non-zero term in \eqref{scl} is for
$m=n-1$, which yields, by \eqref{sbm},
\begin{equation*}
\scx2=\sbx
\le
\CC\frac{k^2\bmin n}{n\qc}
\le
\CCx\frac{k\sqrt n}{n\qc}.
\end{equation*}
For $\ell>2$, we split the sum in \eqref{scl} into two parts, with
$m\le n/2$ and $m>n/2$. We have, by \eqref{sbm},
\begin{equation*}
\begin{split}
\sum_{m=n/2}^{n-1}
\sbm \P\Bigpar{\sum_{i=3}^\ell|\cT_i|=n-1-m}
&\le
\CC \frac{k^2\bmin n}{n\qc} \P\Bigpar{\sum_{i=3}^\ell|\cT_i|\le n/2}
\\
\le
\CCx \frac{k^2\bmin n}{n\qc}
\le \CCx \frac k n.
\end{split}
\end{equation*}
Similarly, using
\eqref{sbm} and \eqref{l2c} (with $\ell$ replaced by $\ell-2$),
\begin{equation*}
\begin{split}
\sum_{m=1}^{n/2}
\sbm \P\Bigpar{\sum_{i=3}^\ell|\cT_i|=n-1-m}
&\le
\CC \sum_{m=1}^{n/2} \frac{k^2\bmin m}{m\qc}\cdot \frac 1n
\\
&\le
\frac \CCx n \sum_{m=1}^{\infty} \frac{k^2\bmin m}{m\qc}
\le \CC \frac k n.
\end{split}
\end{equation*}
Thus $\scl\le \CC k/n$, and the theorem follows by \eqref{qg}.
\end{proof}
\begin{proof}[Proof of Theorems \ref{TP1} and \refT{T1}]
Aldous \cite{Aldous:fringe} has studied the behavior of a random subtree
$\htn$ in a conditional Galton--Watson tree $\tn$.
In particular, he has the following identity
\cite[p.~242]{Aldous:fringe}, for any fixed ordered tree $\tau$ of
order at most $n$ (provided that the probabilities
in the denominators are nonzero):
$$
\frac{\PROB \{ \hT_n = \tau \} }{ \PROB \{ \cT = \tau \} }
=
\frac{(n-|\tau|+1) \PROB \{ |\cT| = n - |\tau| +1 \} }
{ n \PROB \{ |\cT| = n \} }
\cdot
\frac{\gamma }{ p_0},
$$
where $\gamma$ is the expected proportion of leaves in $\hT_{n-|\tau|+1}$
and $p_0 = \PROB \{ \xi = 0 \}$.
We will simply bound $\gamma \le 1$, but it is well-known that as
$n-|\tau|+1 \to \infty$, $\gamma \to p_0$,
see \eg{} Kolchin \cite[Theorem 2.3.1, p.~113]{Kolchin}.
Thus, using the well-known tail estimate \eqref{tail},
for all (permitted) $n$ and $\tau$
\begin{equation*}
\begin{split}
\frac{\PROB \{ \hT_n = \tau \} }{ \PROB \{ \cT = \tau \} }
&\le \CC
\frac{(n-|\tau|+1) \PROB \{ |\cT| = n - |\tau| +1 \} }
{ n \PROB\{|\cT| = n \} }
\le \CC\CCdef\CCb \sqrt\frac{ n }{ n-|\tau|+1} .
\end{split}
\end{equation*}
Hence,
\begin{equation*}
\begin{split}
\EXP \{ Q_k (\hT_n) \}
&= \sum_{\tau} Q_k (\tau) \PROB \{ \hT_n = \tau \} \\
&\le \CCb\sum_{\tau}
Q_k (\tau) \sqrt\frac{ n }{ n-|\tau|+1} \PROB \{ \cT = \tau \}\\
&=
\CCb\E \lrpar{Q_k (\cT) \sqrt\frac{ n }{ n-|\cT|+1}\,} \\
&\le \CC \EXP \{ Q_k (\cT) \} + \CCb \sum_{n \ge \ell > n/2} \sqrt\frac{ n }{
n-\ell+1} \, \EXP \left\{ \IND{|\cT|=\ell} Q_k (\cT) \right\}.
\end{split}
\end{equation*}
We have $\E\xcpar{Q_k(\cT)}\le\CC k\CCdef\CCk$ by \eqref{qk}, and,
using \eqref{tail} and \refT{TQ},
\begin{equation*}
\EXP \left\{ \IND{|\cT|=\ell} Q_k (\cT) \right\}
=\P\xcpar{|\cT|=\ell} \EXP \left\{ Q_k (T_\ell) \right\}
\le \CC \ell\qcw k\ell\qq
=\CCx k/\ell.
\end{equation*}
Consequently,
\begin{equation*}
\begin{split}
\EXP \{ Q_k (\hT_n) \}
&\le \CCk k+\CC \sum_{n \ge \ell > n/2} \frac{k }{ \ell} \,
\sqrt\frac{ n }{ n-\ell+1} \\
&\le \CCk k+\CC \frac kn \sum_{j=1}^n \sqrt\frac{n}{ j} \\
&\le \CC k. \\
\end{split}
\end{equation*}
This proves \refT{TP1}, which by the argument at the beginning of the
section yields \refT{T1}.
\end{proof}
\section{Proof of \refL{L0} and \refT{T2}} \label{Sl2}
Denote the \lhs{} of \eqref{l2} by $\psint$.
Since
$
\sum_j X(j;\tn)e^{\ii jt} = \sumvtn e^{\ii t L_v}
$,
we have
\begin{equation}
\label{b1}
\psint
=n\qww
\E\,\biggabs{\sumvtn e^{\ii t L_v}}^2
=n\qww
\E\sumvwtn e^{\ii t (L_v-L_w)}.
\end{equation}
Condition on $T_n$ and consider two vertices $v$ and $w$ in $\tn$. If
$\vw$ is the last common ancestor of $v$ and $w$, then $L_v-L_{\vw}$
and $L_w-L_{\vw}$ are (conditionally, given $\tn$) independent sums of
$d(v,\vw)$ and $d(w,\vw)$ copies of $\eta$,
respectively. Consequently, letting $\gfeta(t)\=\E e^{\ii t\eta}$ be the
characteristic function of $\eta$,
\begin{equation*}
\begin{split}
\E \bigpar{e^{\ii t (L_v-L_w)}\mid\tn}
&=
\E \bigpar{e^{\ii t (L_v-L_{\vw})}\mid\tn}
\E \bigpar{e^{-\ii t (L_w-L_{\vw})}\mid\tn}
\\&
=\gfeta(t)^{d(v,\vw)}\,\overline{\gfeta(t)}^{d(w,\vw)}.
\end{split}
\end{equation*}
Hence, by \eqref{b1} and \eqref{hn},
\begin{equation*}
\psint
= n\qww \E\sumvwtn\gfeta(t)^{d(v,\vw)}\overline{\gfeta(t)}^{d(w,\vw)}
=n\qww \hn\bigpar{\gfeta(t),\overline{\gfeta(t)}}
\end{equation*}
and \refT{Tgen2} yields
\begin{equation}
\label{b2}
\psint \le \CCgenii n\qw|1-\gfeta(t)|\qww.
\end{equation}
Since $\E \eta=0$ and $\E \eta^2=\gssy<\infty$,
we have
$\gfeta(t)=\exp\bigpar{-\tfrac12\gssy t^2+o(t^2)}$
for small $|t|$; moreover, since $\eta$ has span 1, $\gfeta(t)\neq1$ for
$0<|t|\le\pi$. It follows
that $\psi(t)\=(1-\gfeta(t))/t^2$ is a continuous non-zero function on
$[-\pi,\pi]$ (with $\psi(0)\=\frac12\gssy$); hence, by compactness,
$|\psi(t)|\ge \cc$ for some $\cc>0$, and thus
\begin{equation*}
|1-\gfeta(t)|\ge \ccx t^2,
\qquad |t|\le\pi.
\end{equation*}
It now follows from \eqref{b2}, and the obvious fact that
$\psint\le1$, that
\begin{equation*}
(1+nt^4)\psint
\le
1+nt^4\psint
\le
1+ \CCgenii \frac{t^4}{|1-\gfeta(t)|^2}
\le \CC.
\end{equation*}
This proves \refL{L0}, which as remarked in \refS{S:intro} implies
\refT{T2} by \cite[Remark 3.7]{SJ185}.
\section{Proof of \refT{Tgen2}} \label{Sgen2}
We use some further generating functions.
Recall that $\cT$ is the (unconditioned) \GWt{}
with offspring distribution $\xi$, and define
\begin{align*}
\Phi(z) &\= \E z^\xi, \\
F(z) &\= \E z^{|\cT|}, \\
G(z,x) &\= \E \Bigpar{z^{|\cT|}\sumvt x^{d(v,\oo)}}, \\
H(z,x,y) &\= \E \Bigpar{z^{|\cT|}\sumvwt x^{d(v,\vw)}y^{d(w,\vw)}}
=\sum_{n=1}^\infty \P(|\cT|=n)\hn(x,y)z^n.
\end{align*}
These functions are defined and analytic at least for $|z|,|x|,|y|<1$.
Let us condition on the degree $\doo$ of the root of $\cT$, recalling
that $\doo\eqd\xi$.
If $\doo=\ell$, then $\cT$ has $\ell$ subtrees $\cT_1,\dots,\cT_\ell$
at the root $\oo$, and conditioned on $\doo=\ell$, these are independent
and with the same distribution as $\cT$;
we denote their roots (the neighbours of
$\oo$), by $\oo_1,\dots,\oo_\ell$.
Assume $\doo=\ell$, and let $|z|,|x|,|y|<1$.
First, $|\cT|=1+\sumil|\cT_i|$ and thus
$z^{|\cT|}=z\prodil z^{|\cT_i|}$. Taking the expectation, we obtain, as
is well-known, first
\begin{equation*}
\E\bigpar{z^{|\cT|}\mid\doo=\ell}
=
z\E \prodil z^{|\cT_i|}
= z F(z)^\ell,
\end{equation*}
and then
\begin{equation*}
F(z)=\E\bigpar{z^{|\cT|}}
= z \suml \P(\xi=\ell)F(z)^\ell
=z\phifz.
\end{equation*}
Similarly, separating the cases
$v\in\cT_i$, $i=1,\dots,\ell$, and $v=\oo$,
\begin{equation*}
\sumvt x^{d(v,\oo)}
=
\sumil\sum_{v\in\cT_i} x^{d(v,\oo_i)+1} +1.
\end{equation*}
Hence,
\begin{align*}
\E\bigpar{z^{|\cT|}\sumvt x^{d(v,\oo)}\mid\doo=\ell}
&=
\E\sumil z z^{|\cT_i|}\sum_{v\in\cT_i} x^{d(v,\oo_i)+1}
\prod_{j\neq i} z^{|\cT_j|}
+\E \Bigpar{z\prodil z^{|\cT_i|}}
\\
&= \ell zx G(z,x)F(z)^{\ell-1}+ z F(z)^\ell
\end{align*}
and
\begin{equation*}
G(z,x)=\sum_{\ell=0}^\infty\P(\xi=\ell)\ell zxG(z,x)F(z)^{\ell-1}+F(z)
=zx\Phi'(F(z))G(z,x)+F(z)
\end{equation*}
which gives
\begin{equation}\label{yg}
G(z,x)=\frac{F(z)}{1-zx\Phi'(F(z))}.
\end{equation}
Similarly,
\begin{align*}
\E\Bigpar{z^{|\cT|}&\sumvwt x^{d(v,\vw)}y^{d(w,\vw)}\mid\doo=\ell}
=
\\
&\E\sumil z z^{|\cT_i|}\sum_{v,w\in\cT_i} x^{d(v,\vw)}y^{d(w,\vw)}
\prod_{j\neq i} z^{|\cT_j|}
\\&
\quad+
\E\sum_{i\neq j} z z^{|\cT_i|}\sum_{v\in\cT_i} x^{d(v,\oo_i)+1}
z^{|\cT_j|}\sum_{w\in\cT_j} y^{d(w,\oo_j)+1}
\prod_{k\neq i,j} z^{|\cT_k|}
\\&
\quad+
\E\sumil z z^{|\cT_i|}\sum_{v\in\cT_i} x^{d(v,\oo_i)+1}
\prod_{k\neq i} z^{|\cT_k|}
\\&
\quad+
\E\sumjl z z^{|\cT_j|}\sum_{w\in\cT_j} y^{d(w,\oo_j)+1}
\prod_{k\neq j} z^{|\cT_k|}
+\E \Bigpar{z\prodil z^{|\cT_i|}}
\end{align*}
leading to
\begin{multline*}
H(z,x,y)
=
z\Phi'(F(z))H(z,x,y)
+
zxy\Phi''(F(z))G(z,x)G(z,y)\\
+
zx\Phi'(F(z))G(z,x)
+
zy\Phi'(F(z))G(z,y)
+F(z)
\end{multline*}
which gives
\begin{multline}
\label{yh}
H(z,x,y)
=\\
\frac{zxy\Phi''(F(z))G(z,x)G(z,y)
+
z\Phi'(F(z))\bigpar{xG(z,x)+ yG(z,y)}
+F(z)}
{1-z\Phi'(F(z))}
\end{multline}
Assume now for simplicity that $\xi$ has span 1. (The case when the
span is $d>1$ is treated similarly with the standard modification that
we have to give special treatment to neighbourhoods of the $d$:th unit
roots.)
Then, by \cite[Lemma A.2]{SJ167},
for some $\gd>0$ and $\gb\le\pi/4$, $F$ extends to an analytic function
in $\gdbd$ with $|F(z)|<1$ for $z\in\gdbd$ and
\begin{equation}
\label{a5}
F(z)=1-\sqrt2\gs\qw\sqrt{1-z}+\oqq,
\qquad
\text{as $z\to1$ with } z\in\gdbd.
\end{equation}
We will prove the following companion results.
\begin{lemma}\label{Lx}
If\/ $\xi$ has span $1$, then there exists $\gb,\gd>0$ such that
$F$ extends to an analytic function in $\gdbd$ and, for some
$\cc\ccdef\cclxa,\cc\ccdef\ccj>0$,
if $x,z\in\gdbd$, then
\begin{align}\label{sofie}
|1-z\Phi'(F(z))|&\ge \cclxa|1-z|\qq,
\\
|1-xz\Phi'(F(z))|&\ge \ccj|1-x|.\label{julie}
\end{align}
Consequently, $G(z,x)$ and $H(z,x,y)$ extend to analytic functions
of $x,y,z\in\gdbd$, and, for
all $x,y,z\in\gdbd$,
\begin{align}
|G(z,x)| &\le \CC|1-x|\qw, \label{erika}
\\
|H(z,x,y)| &\le \CC|1-z|\qqw|1-x|\qw|1-y|\qw. \label{magnus}
\end{align}
\end{lemma}
Standard singularity analysis
\cite[Lemma IX.2]{FS}
applied to \eqref{magnus} yields
\begin{align*}
|\P(|\cT|=n)\hn(x,y)|
&
\le \CC n \qqw|1-x|\qw|1-y|\qw,
\qquad
x,y\in\gdbd,
\end{align*}
which proves \refT{Tgen2} because, as is well known, a singularity
analysis of \eqref{a5} yields
\begin{align*}
\P(|\cT|=n)\sim \cc n^{-3/2}.
\end{align*}
It thus remains only to prove \refL{Lx}.
\begin{proof}[Proof of \refL{Lx}]
Since $\E\xi^2<\infty$, $\Phi'$ and $\Phi''$ extend to continuous
functions on the closed unit disc with $\Phi'(1)=\E\xi=1$ and
$\Phi''(1)=\E\xi(\xi-1)=\gss$. Hence, \eqref{a5} yields, for $z\in\gdbd$,
\begin{equation*}
\begin{split}
\Phi'(F(z))&=\Phi'(1)+\Phi''(1)\bigpar{F(z)-1}+o(|F(z)-1|)
\\&
=1-\sqrt2\gs\sqrt{1-z}+\oqq
\end{split}
\end{equation*}
and
\begin{equation}
\label{david}
z\Phi'(F(z))
=\Phi'(F(z))+O(|z-1|)
=1-\sqrt2\gs\sqrt{1-z}+\oqq.
\end{equation}
Let $\be\=\set{z:|z-1|<\eps}$, and take $\gb<\pi/4$.
Since $z\in\ogdbdx$ entails $|\arg(1-z)|\le\pi/2+\gb$ and thus
$|\arg\sqrt{1-z}|\le\pi/4+\gb/2$, it follows from \eqref{david} that,
for some small $\eps>0$,
if $z\in\overline{\gdbde}$ with $z\neq1$,
then
\eqref{sofie} holds,
$\bigl|{z\Phi'(F(z))-1}\bigr|=O(\eps\qq)$,
\begin{equation}
\label{emma}
\begin{split}
\bigl|\arg\bigpar{z\Phi'(F(z))-1}\bigr|
&
>|\arg(-\sqrt{1-z})|-\gb/2
\\&
\ge
\pi-(\pi/4+\gb/2)-\gb/2
=3\pi/4-\gb,
\end{split}
\end{equation}
and consequently, since $3\pi/4-\gb>\pi/2$,
if $\eps$ is small enough,
\begin{equation}
\label{samuel}
|z\Phi'(F(z))|<1.
\end{equation}
Similarly, if $x\in\gdbd$, then $|\arg(1-x)|<\pi/2+\gb$ and
\begin{equation*}
x\qw=\bigpar{1-(1-x)}\qw = 1+(1-x)+o(|1-x|),
\qquad x\to1,
\end{equation*}
so if $\eps>0$ is small enough,
then, for $x\in\gdbd\cap\be$,
\begin{equation}\label{jesper}
|\arg(x\qw-1)|<\pi/2+2\gb.
\end{equation}
If we choose $\gb\le\pi/16$, it follows from \eqref{emma} and
\eqref{jesper} that the triangle with vertices in $1$, $x\qw$ and
$z\Phi'(F(z))$ has an angle at least $\pi/4-3\gb\ge\pi/16$ at 1,
and thus by elementary
trigonometry (the sine theorem),
\begin{equation*}
|x\qw- z\Phi'(F(z))| \ge \cc |x\qw-1|,
\end{equation*}
and so \eqref{julie} holds, when $z,x\in\gdbd\cap\be$,
provided $\gb,\gd,\eps$ are small enough.
It remains to treat the case when $x$ or $z$ does not belong to $\be$,
\ie, $|x-1|\ge\eps$ or $|z-1|\ge\eps$. We do this by compactness
arguments.
First, let
\begin{align*}
A&\=\set{z\Phi'(F(z)):z\in\gdbd\cap\be}
\\
\br&\=\set{x\qw:x\in\overline{\gdbr}\setminus\be,\, |x|\ge1/2}.
\end{align*}
Then $B\=\bigcap_{\rho>0}\br
\subset\set{\zeta:|\zeta|\ge1}\setminus\set1$,
and it follows from \eqref{samuel} that
$\ba\cap B=\emptyset$.
Since $\ba$ and all $\br$ are compact, it follows that
$\ba\cap\br=\emptyset$ for some $\rho>0$, and thus, if
$z\in\gdbd\cap\be$ and $x\in\gdbr\setminus\be$ with $|x|\ge1/2$, then
$|x\qw-z\Phi'(F(z))|\ge \cc$ for some $\ccx>0$, which implies \eqref{julie}
for such $z$ and $x$.
Moreover, if $z\in\gdbde$ and $|x|<1/2$, \eqref{samuel} shows that
$|1-xz\Phi'(F(z))|\ge 1-|x|\ge1/2$, so \eqref{julie} then holds if
$\ccj\le1/3$.
Finally, if $z\in\gdbd$, then
$|F(z)|<1$ \cite[Lemma A.2]{SJ167} as stated above,
and thus $|\Phi'(F(z))|<1$.
If $0<\gb_1<\gb$ and $0<\gd_1<\gd$, then
$\overline{\gdbdi}\subset\gdbd\cup\set1$, and thus by compactness
\begin{equation*}
\ce\=\sup\bigset{|\Phi'(F(z))|:z\in\overline{\gdbdi}\setminus\be}<1.
\end{equation*}
Consequently, if $\gd_2\le\gd_1$ is small enough and $x,z\in\gdbdii$ with
$|z-1|\ge\eps$, then
\begin{equation*}
|xz\Phi'(F(z))|\le(1+\gd_2)^2\ce<1.
\end{equation*}
Hence \eqref{julie} holds in this case too for some $\ccj>0$, and
similarly \eqref{sofie} holds for $z\in\gdbdii\setminus\be$.
This completes the proof of \eqref{sofie} and \eqref{julie}, for some new
$\gb,\gd>0$
(\viz, $\gb_1$ and $\min(\gd_2,\rho)$).
$G(z,x)$ now can be defined for all $x,z\in\gdbd$ by \eqref{yg},
and \eqref{erika}
holds by \eqref{julie}.
Similarly,
$H(z,x,y)$ can be defined for all $x,y,z\in\gdbd$ by \eqref{yh},
and \eqref{magnus} holds by \eqref{sofie}, \eqref{erika}, and the fact
that $\Phi'$ and $\Phi''$ are bounded on the unit disc. (Recall that
$|F(z)|<1$ for $z\in\gdbd$.)
This completes the proof of \refL{Lx}, and thus of \refT{Tgen2} and of
all results in this paper.
\end{proof}
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\newcommand\TAMS{\emph{Trans. \AMS} }
\newcommand\AnnMS{\emph{Ann. Math. Statist.} }
\newcommand\AnnAP{\emph{Ann. Appl. Probab.} }
\newcommand\AnnPr{\emph{Ann. Probab.} }
\newcommand\CPC{\emph{Combin. Probab. Comput.} }
\newcommand\JMAA{\emph{J. Math. Anal. Appl.} }
\newcommand\RSA{\emph{Random Struct. Alg.} }
\newcommand\ZW{\emph{Z. Wahrsch. Verw. Gebiete} }
\newcommand\DMTCS{\jour{Discr. Math. Theor. Comput. Sci.} }
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\newcommand\arxiv[1]{\webcite{arXiv:#1}}
\def\nobibitem#1\par{}
| 58,939
|
TITLE: On Buchstab et al's "forgotten" sieve and the Goldbach conjecture for certain integers
QUESTION [4 upvotes]: There is a somewhat forgotten sieve-theoretic approach to the Goldbach conjecture, due to Buchstab et al, see e.g. pp.247-248 of R.D. James.
On p.247, James defines some function $F$ such that for any fixed $a \in \mathbb{N}$ and even $x \geq 6$:
$F(x ; 2, a, 1) = F(x; 2)$ with $a=1$ is the number of positive integers $n \leq x$ such that $n \equiv a\pmod{2}$. Thus $F(x; 2) = x/2$.
$F(x; 2, √x, a)=F(x; 2, √x)$ with $a=1$ is the number of odd positive integers $n<x$ (without double counting $n$ and $x-n$), such that each of $n$ and $x-n$ is either a prime or equal to 1. Thus if it could be shown that $F(x; 2, √x) \geq 2$, it would follow that there exists at least one representation $x= n+(x-n)$ whereby each of $n$ and $x-n$ is either a prime or equal to 1. Thus if $x-1$ is composite, it would suffice to show that $F(x; 2, √x) = F(x; 2, √x) \geq 1$.
On the bottom of p.248, James states that
$$
F(x; 2, √x) = F(x; 2) - 2\sum_{r=1}^{k} F(x; 2p_r, p_{r-1}) = x/2 - 2\sum_{r=1}^{k} F(x; 2p_{r}, p_{r-1})
,$$
where $p_i$ denotes the $i$-th odd prime $\leq √x$. T. Kubalalika, in his preprint [2], lets $6 \leq x \equiv 2\pmod{4}$ where $x-1$ is composite. Now suppose that $x$ is a counterexample to Goldbach, so that $F(x; 2, √x)=0$. Putting this into the above equality gives
$$
x/2 = 2\sum_{r=1}^{k} F(x; 2p_r, p_{r-1}),
$$
contradicting the fact that $x/2$ is odd. One therefore deduces that if $x\equiv 2\pmod{4}$ and $x-1$ is composite, then $x$ is a sum of two primes.
My question is, given the strength of Buchstab et al's sieve (as evidenced by how easily it leads to a proof of the above result), are there any modern improvements to it, such that it could possibly lead to even more powerful results ? A quick Google search seems to suggest that the sieve became forgotten as soon as the Hardy-Littlewood circle method lead to Vinogradov's 3-primes theorem.
References
[1] R. D. James, "Recent progress in the Goldbach problem" Bulletin of the American Mathematical Society 55, 246-260 (1949), MR0028893, Zbl 0034.02301.
[2] T. Kubalalika, "On the binary Goldbach conjecture for certain even integers", figSHARE preprint.
REPLY [13 votes]: The result (aka Buchstab's identity) you mentioned is not forgotten. In modern sieve theory texts such as Halberstam & Richert's Sieve Methods and Friedlander & Iwaniec's Opera de Cribro, the identity is written as
$$
S(\mathcal A,z)=S(\mathcal A,w)-\sum_{w\le p<z}S(\mathcal A_p,p)
$$
where $S(\mathcal A,z)$ counts the integers in $\mathcal A$ that are free of prime divisors $<z$.
One of its generalization (Kuhn's weighted sieve) is used to prove Chen's theorem. When $N$ is a positive even integer, $\mathcal A=\{N-p:p\le N\}$, we have
$$
\begin{aligned}
r_{1,2}(N)&=\#\{p\le N:n-p\text{ prime or product of two primes}\} \\
&>S(\mathcal A,N^{1/10})-
\frac12\sum_{N^{1/10}\le p<N^{1/3}}S(\mathcal A_p,N^{1/10})-\frac\Omega2+O(N^{9/10})
\end{aligned}\tag1
$$
in which
$$
\Omega=\#\{p\le N:N-p=p_1p_2p_3,N^{1/10}\le p_1<N^{1/3}\le p_2<(N/p_1)^{1/2}\}.
$$
By evaluating the right hand side using Jurkat-Richert's theorem and Selberg's sieve, Chen found out that for large $N$ there is
$$
r_{1,2}(N)>{0.67N\over\log^2N}\prod_{2<p|N}{p-1\over p-2}\prod_{p>2}\left(1-{1\over(p-1)^2}\right).
$$
| 71,428
|
TITLE: How to prove that any finite extension field for this field would be cyclic
QUESTION [1 upvotes]: This question was asked in mid term exam of Field Theory quite earlier this year and I was unable to completely solve and so I am looking for help here.
Let $K$ be a field, $\bar K$ an algebraic closure of $K$, and $\sigma \in \operatorname{Aut}_K(\bar{K})$. Let
$F = \{u \in \bar{K} : \sigma(u)=u\}$.
Then $F$ is a field and every finite dimensional extension of $F$ is cyclic.
I have proved $F$ to be a field but unable to prove that any finite dimensional extension would be cyclic. I took the extension to be $K = F(u_1,\dots,u_k)$ but I am unable to get any ideas why it should be cyclic and I would require some help.
REPLY [1 votes]: OK. Just to add a few details.
(1) If $F$ has characteristic $p$, and $x\in F$, pick a $p$-th root $y$ of $x$ in $\overline{F}$, then $y^p = x = \sigma(x) = \sigma(y^p) = \sigma(y)^p\Rightarrow (y-\sigma(y))^p = 0\Rightarrow y = \sigma(y)\Rightarrow y\in F$, therefore every element of $F$ has a $p$-th root in it. If $F$ has characteristic $0$, it's already perfect. Either way, $F$ is perfect, and any extension of $F$ is separable.
(2) For a finite extension $E$ of $F$, pick its normal closure $\hat E$ of $E$, so that $\sigma|_{\hat E}\in G:=\text{Gal}(\hat E/F)$ is well-defined. By $E/F$ is separable, $\hat E/F$ is Galois, so we can apply the fundamental theorem of Galois theory. Let $H=\langle \sigma|_{\hat E}\rangle\le G$ be the subgroup generated by $\sigma|_{\hat E}$. By Galois correspondence $$[G:H]=[(\hat E)^H:F]$$ where $(\hat E)^H:=\{x\in \hat E: \tau(x)=x, \forall \tau\in H\}$. If $x\in (\hat E)^H$, then $x$ is fixed by $\sigma|_{\hat E}$ hence $\sigma$, therefore by the definition of $F$, $x\in F$. This shows that $(\hat E)^H=F$. Therefore $[G:H]=1$, $G=H$ is a cyclic group.
(3) As $G$ is cyclic in particular abelian, all its subgroups are normal. In particular, $\{\tau\in G: \tau(x)=x, \forall x\in E\}$ is normal. Thus as part of the fundamental theorem, $E/F$ is already Galois, hence $\hat E=E$, and it's been shown that $G=\text{Gal}(E/F)$ is cyclic.
| 180,854
|
TITLE: Equations that make no sense (explaining the vector bundle isomorphism)
QUESTION [0 upvotes]: I'm stuck at two places in these lecture notes:
1) Consider on pp 40:
What does the last equation mean? This makes no sense to me. $\varphi_i$, $i=1,2$ can't denote the $i$th coordinate function, since that takes 2 arguments, not one. What does it the denote? And what is the $\cdot$? The scalar product?!
2) Consider on pp. 53
($T^r_s(E)$ are the $\binom{r}{s}$-tensors on the vector space $E$, as defined previously in the notes - see the link.)
I cannot understand the equation $e(\varphi^* (\beta)) = \varphi (e) (\beta)$, or why $\varphi_0^1$ is in some sense isomorphic to $\varphi$.
Here $\varphi: E \rightarrow F$ for finite-dim. real vector spaces $E,F$ and $\varphi^*$ is the adjoint, $e \in E^{**}$ and $\beta \in F^*$ (this is all explained in the lines above this example).
Actually the equation makes no sense in my opinion, since $\varphi(e)$ is an element of $F$, so it makes no sense to apply $\beta$ here.
REPLY [2 votes]: As to your first question, the map $\varphi$ has two components $$(u, f)\longmapsto \varphi_1(u)\quad \textrm{and}\quad (u, f)\longmapsto \varphi_2(u)\cdot f,$$
where $\varphi_1:U\longrightarrow F$ and $\varphi_2:U\longrightarrow \mathsf{Hom}(F, F^\prime)$ where $\mathsf{Hom}(F, F^\prime)$ is the vector space of linear maps from $F$ to $F^\prime$. This applies to the definition of local vector bundle homomorphism.
If you change "homomorphism" by "isomorphism" you must change $\mathsf{Hom}(F, F^\prime)$ by $\mathsf{Iso}(F, F^\prime)$ where $\mathsf{Iso}(F, F^\prime)$ is the vector space of linear isomorphism from $F$ to $F^\prime$.
As to your second question, when he writes $\varphi_2(u)\cdot f$ it means you are evaluating the linear map $\varphi_2(u):F\longrightarrow F^\prime$ in the point $f\in F$.
As to your last question, I believe he is defining $\varphi_0^1:E\longrightarrow F$ via duality. There are natural pairings $$\langle \cdot , \cdot\rangle_E:E\times E^*\longrightarrow \mathbb R$$ and $$\langle \cdot, \cdot\rangle_F: F\times F^*\longrightarrow \mathbb R$$ and he defines $\varphi_0^1:E\longrightarrow F$ as the unique map such that $$\langle \varphi_0^1(e), \beta\rangle_F=\langle e, \varphi^*(\beta)\rangle_E=\langle \varphi(e), \beta\rangle$$ for every $e\in E$ and $\beta\in F^*$.
| 154,986
|
DISCOVER STAFFING is currently working with a client in Doraville seeking an Electronic Repair Technician. Must be an expert in electronics with at least two years experience. Will be troubleshooting, soldering, and reading schematics. Must understand electronics down to the component level. $14-$15/hour depending on experience. Company will review resumes, conduct interviews and perform a soldering test for consideration.
Please send your resume to laura@discoverstaffing.com for our review. Candidates must be local to the Doraville area and have reliable transportation.
| 30,632
|
By SCOTT TAYLOR
It would not be a stretch to say that the Rozin family is Canadian now.
Readers of The Jewish Post & News might recall that three years ago, Bernie Bellan introduced you to Roie Rozin, his wife Pnina and their three kids, who had decided to move to Canada because middle child Guy had a special skill.
Guy Rozin was an Israeli hockey player and the family’s story is one of courage, faith and sacrifice, all so a then-13-year-old could play the game he loved in the country where hockey is king.
And it all started with a young man named David Levin, the first Israeli hockey player to move to Canada and find some level of success. Levin left Israel in 2014, moved to Canada and found success with the Sudbury Wolves of the Ontario Hockey League. In fact, he wore an A for the Wolves this season and had 42 points in 43 games.
One hot day in 2015, Levin’s father, who just happened to be Guy’s coach, suggested to Roie and Pnina that their then 12-year-old son should follow in David’s footsteps, move to Canada and pursue his hockey career at the highest level possible. It was a crazy suggestion, but for a father who was once a professional soccer player, it was not as crazy as one might think.
“At first we joked about it,” said Roie as he, Pnina, Guy and youngest son Etay spoke at the Tim Horton’s in Charleswood (How Canadian is that?). “But it wasn’t long before we were serious. We decided to leave everything behind. We had good jobs. I was a warehouse manager and Pnina was a nurse. We had good salaries and a good life, but we decided to come to Winnipeg so that Guy could play hockey in Canada.”
Yep, it sounds crazy. Completely nuts, in fact. Since arriving in Canada almost four years ago, Roie has been working as a cook in a local restaurant while Pnina, who is still trying to get her Canadian certification as a nurse, has been working as a healthcare aide. They are not making the kind of money they did in Israel and they aren’t working in the professional fields that made them successful, but to do something grand, something sweeping in life, sacrifices often have to be made.
“It was clear to both of us that if Guy was going to have a chance to become a hockey player, he would have to do it in Canada,” Roie said. “So, we decided to move. We chose Winnipeg because we had an advocate, Abe Anhang. We didn’t speak English so we needed help just to get jobs and set up our lives.”
With that, the Rozins – Roie, Pnina and their three kids, daughter Roni, now 18, Guy, now 16 and Etay, now 12 – applied for Permanent Residence under the Nominee Program (because Pnina is a registered nurse) and they moved from the small city of Kfar Yonah, Israel, to Winnipeg in July of 2016.
“I thought it might be hard finding a team and having a place to play, but it wasn’t,” said Guy. “I felt I was good enough.”
Guy arrived in Winnipeg and hit the ice flying. He was first taken under the wing of Monarchs’ City Minor Bantam AAA coach Jeff Sveinson and then City Bantams AAA head coach John Fehr. He has now won three straight AAA championships with the Monarchs – Minor Bantam, Bantam and Midget. Last month, he helped the AAA Midget Monarchs beat the Sharks to win the 2019 title.
Despite breaking his wrist early in the season, he was a force down the stretch and in the playoffs. He finished the regular season with six goals and 15 points in 16 games.
To top off his season, Rozin flew to Bulgaria to help Israel finish 3-1-1 and in second place at the IIHF U-18 Division III World Championship. He scored the winning goal 40 seconds into overtime in a 4-3 win over New Zealand and also scored the eventual winning goal in a big 5-2 win over Mexico. He finished the tournament with three goals and an assist in five games.
“Bulgaria was very good,” said Roie, who accompanied Guy to the World Championship. “We lost to Bulgaria in overtime and finished second so Israel didn’t advance to Division II, but it was a very good tournament and Guy played well.
“He played on the second line and helped the players he was playing with. Their level of play was very different from Guy’s. But he was excited to see his old friends from Israel and I think that motivated him to play as well as he did. We had an American coach and he liked Guy.”
Guy’s immediate goal is to make the Provincial Midget AAA Wild next season. If not, he’d be fine playing another season with the Monarchs.
“My plan is to get him into Focus Fitness with AJ Zeglen and to work out with Dave Cameron at the Iceplex,” Roie said. “He is going to the Waywaysee-cappo Junior camp and he will attend the (Winnipeg) Blues camp and the Selkirk Steelers camp. If he doesn’t make the Wild or a junior team we will be happy to play with the Monarchs again.
“We know he has to get bigger and stronger. He should have a growth spurt yet but he must get stronger. That’s why I have hired AJ as his personal trainer.”
While the kids appear to be adapting to their new home in Winnipeg and their new school, Gray Academy, with relative ease, it hasn’t been as easy for mom and dad. However, sometimes, you do things in life to give your children a better opportunity, and for both Roie and Pnina there have been no regrets.
“It’s been fun coming to Winnipeg,” said Guy. “I enjoy it, I’ve been successful and I like playing here.”
When asked how he’s adapted to Winnipeg’s winters, Roie said he had no problem with the cold and the snow.
“I love it,” he said laughing. “I love the cold and snow. It’s hockey weath
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A report written by By Simon Bahceli and published on the Cyprus mail today says that with Cyprus and Turkey sliding headlong towards a clash over hydrocarbon resources in the eastern Mediterranean, Turkish Cypriots – whose rights Turkey says it is protecting by obstructing Greek Cypriot exploration – appear oblivious to the possible benefits to their community if the exploration and subsequent extraction went ahead.
So far, claims the article, the north’s leaders have said little, repeating only the mantra that the Greek Cypriots’ “unilateral” exploration and drilling would irreparably damage already-faltering reunification talks. It would also be in breach of the constitutional bicommunality of the Cyprus Republic, they say. Discussion of the benefits are few and far between, while nearly all of the talking, supposedly on behalf of the Turkish Cypriots, has been done by Ankara. Rational discussion of the issue in the north has been effectively drowned out.
Turkish Cypriot mistrust of the Greek Cypriot leadership is an unfortunate reality, says the article. And adds that the reason for this is because too many simply do not believe pledges from the south that they will share the offshore natural resources with “all Cypriots”.
The article says that it is not only Greek Cypriots who feel the Turkish Cypriot community, and Turkey too for that matter, should wise up to the massive boon the island may be about to receive if exploratory drillings go ahead in Cyprus’ Aphrodite field deep below the eastern Mediterranean this autumn. Regional editor of the Middle East Economic Survey (MEES) Gary Lakes believes the financial and environmental benefits could be massive for all on the island. Furthermore, he envisages significant benefits for Turkey too – but only if rationality was somehow, miraculously allowed to prevail.
As Lakes points out, Noble Energy, the Texas-based company that plans to carry out an assessment of deposits in the Aphrodite field next month, has estimated that around 280 billion cubic metres of natural gas lie in wait below the seabed. If these estimates are proven accurate, Cyprus will make billions of euros in revenue, and will be awash with cheap and cleanly-produced electricity. Lakes says Cyprus’ predicted annual demand for natural gas in 15 years will probably be around 1 billion cubic metres per year. The Liquid Natural Gas (LNG) plant Noble Energy has proposed building in Cyprus would handle around 15 billion cubic metres per year. This means plenty to sell on to lucrative markets in fuel-hungry Europe.
While Lakes says much depends on whether the Aphrodite’s deposits are as big as predicted, and of course the market price, some have predicted that Cyprus could earn up to €10 billion annually from the gas. If this were the case, the Turkish Cypriot community’s share of the earnings (if the Greek Cypriots keep their promise) would outstrip many times over the aid Ankara provides. Moreover, these earnings would be rightfully earned, as opposed to handouts that have created dependence and have to a great extent stripped the community of its self-esteem.
There are other equally important benefits, such as the fact that burning natural gas to make electricity will greatly reduce carbon emissions and improve the general air quality around power stations. Turkish Cypriots know too well the damage their two dated generators do to the areas they occupy.
And it is not only the Cypriots that could stand to gain from the deposits, says Lakes. Turkey could benefit by transporting Cypriot natural gas through its Nabucco pipeline in Anatolia directly into western Europe. According to Lakes and other experts, this would be by far the most cost effective way of transporting the gas.
“If the situation were different, then it could be piped through Turkey. It’s a shame that the political situation prevents it,” Lakes says, concluding: “There is too much politics in the mix”
With the first gas not due to arrive till at least 2014, Greek Cypriot head of Energy Services Solon Kassinis is calling on Turkish Cypriots to, rather than paying lip-service to Turkey’s threats, “join forces with Greek Cypriots” to make the project work.
“If they don’t choose to work with us on this, they will lose a lot,” he predicts.
Kassinis insists he has done much to try and convince Turkish Cypriots that Cyprus’ hydrocarbon deposits are the property of all Cypriots, and that they should shun Turkey’s efforts to politicise the situation. He adds that plans for an LNG terminal include a blueprint for an infrastructure that would take the gas to all parts of the island, “including the north”.
“We’re not going to sell it or take it away from the Turkish Cypriots or the Maronites, or any other resident of Cyprus,” he told the Sunday Mail. He also called for patience and a political climb-down, at least until it became clear what deposits are there. “At the moment we are just looking to see what is there, and will only be able to tell you what we have after we verify the reserves,” he said.
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