Mathematics - 2016-03-30 (page 1 of 3)

Akiva Weinberger

00:00

Looking up coarse spaces now

*structure

Mike Miller

I do know a good place to learn about those. John Roe has a book on Coarse geometry which should be mostly accessible.

Balarka Sen

Speaking of Gromov-ish stuff, have a look at Gromov's new GAFA paper. "Singularities, Expanders, and Topology of Maps". Unreadable, but at least it can be extracted that he's doing completely crazy stuff. Interesting, but crazy.

Mike Miller

I'm alright.

Balarka Sen

:P

He calls it quantitative algebraic topology. Studying "size" of homotopies. (e.g., a simple question would be this: if you have a nullhomotopic loop on a manifold, how small can you make the disk you nullhomotope it along?)

Mike Miller

There's plenty of insane mathematics I actually need to understand.

I'll do that first.

Balarka Sen

00:11

On a brighter note at least you have readable references.

Mike Miller

Dubious.

rorty

Is there anyone here that primarily works in statistics or a related field?

Mike Miller

Clarinetist, I think. But he's not here right now.

I guess to read John Roe's book you want a passing familiarity with C* algebras. But this should not be hard to pick up if you know what a Banach space is and are willing to spend a little bit of time reading an intro to the subject.

Akiva Weinberger

Paging @Clarinetist

Mike Miller

I'm browsing this coarse geometry book and I like it. This could be a good talk idea, I'm putting it in reserve.

Semiclassical

00:35

evening chat

Mary Star

Is the order of the group $S_3\times\mathbb{Z}_2$ the product of the order of $S_3$ and $\mathbb{Z}_2$, so $6\cdot 2=12$ ?

Akiva Weinberger

@MaryStar I believe so

Mary Star

Ah ok... Thanks!! :-) @AkivaWeinberger

Balarka Sen

I caught a cold. Darn it.

Bungo

@MaryStar $|G \times H| = |G||H|$ for any finite groups

Mike Miller

00:43

@BalarkaSen seems like at any given time you're halfway dead

Balarka Sen

fighting death has become a hobby

no, that doesn't sound right. too cool. I meant being dead has become a hobby.

remind me: are all ideals of the localization $A_\wp$ of the form $I_\wp$?

where $I$ is an ideal of $A$.

Akiva Weinberger

@MikeMiller "Mostly accessible?"

Balarka Sen

Nevermind, it's true.

Pick an ideal $J$ of $A_\wp$. Elements of that are of the form $a/b$, $b \notin \wp$. These contain stuff of the form $a/1$, the collection of which can be naturally identified with elements of $A$. These form an ideal - call it $I$ - of $A$. Then $J$ is clearly $I_\wp$.

Mike Miller

01:04

@AkivaWeinberger: Modulo learning about $C^*$ algebras.

Balarka Sen

Actually what I wrote down about was not a complete prove: I think I just proved $I_\wp$ is contained in $J$. But who cares, the other inclusion should not be too messed up.

*proof.

Akiva Weinberger

So do course structures basically formalize the idea of boundedness?

In the same way that topologies formalize continuity and uniform structures formalize uniform continuity

Kari

I feel like I just walked in on some wicked generalisations.

Uniform structures and course structures. :-0

Mike Miller

Not quite, since something that only formalizes boundedness should probably have $\Bbb Z^2$ and $\Bbb Z$ equivalent, yeah? But coarse structures also can tell you about the dimension, the number of ends ($\Bbb N$ and $\Bbb Z$), say.

Akiva Weinberger

Ah

Mike Miller

01:10

Somehow it's about uniformity without worrying about continuity.

Akiva Weinberger

I'm not sure how to think about these

Mike Miller

It's just the sort of object that cares about large-scale behavior, not local behavior.

Balarka Sen

Today's topic of discussion is not my cup of tea. silently crawls away

Mike Miller

You're no longer interested in quasi-isometries? :P

Balarka Sen

Sure I am. But I don't understand this coarse space business.

So, "whereof one cannot speak, thereof one must be silent".

Akiva Weinberger

01:14

So, by saying that $\Bbb Z$ and $\Bbb Z^2$ are not coarsely equivalent, we're saying that any bijection between the two breaks something in a big way

Mike Miller

One never understands something they've spent no time on.

@AkivaWeinberger I don't remember but I don't think coarse equivalences have to be bijections. But I last thought about this years ago.

Akiva Weinberger

Oh, right, they don't. $\Bbb R^n\cong\Bbb Z^n$

Basically, we have two maps $f:\Bbb R^n\to\Bbb Z^n$ and $g:\Bbb Z^n\to\Bbb R^n$, that let us go back and forth between them.

But going from $\Bbb R^n$ to $\Bbb Z^n$ and back (the map $gf$) doesn't break anything in terms of large-scale stuff

Everything is still roughly where it started

And, $fg$ is probably going to be the identity in this example so it doesn't break things too badly either

Wait, no, that doesn't work.

Kari

May I ask what you mean by 'breaking things', @AkivaWeinberger?

Mike Miller

I think that sequence of (non)equalities $\Bbb R^n \cong \Bbb Z^n \not\cong \Bbb Z^m$, $m \neq n$, tells us a lot about what coarse geometry "is".

Akiva Weinberger

I'm not entirely sure. Besides, assuming the identity doesn't "break things", this gives us $\Bbb Z^2\cong\Bbb Z$, so I'm wrong

Actually, @Mike, are you sure that $\Bbb Z^2\ncong\Bbb Z$?

Mike Miller

01:23

yes

this would be a pretty crappy notion if not

Balarka Sen

Z^2 is not quasi-isometric to Z under standard metrics, so if this coarse equivalence is worth it's money they shouldn't be equivalent in this notion either, because what I have gathered so far is that it's a generalization to non-metric objects.

Mike Miller

ams.org/notices/200606/whatis-roe.pdf

Akiva Weinberger

You know what, I think I forgot that $f$ and $g$ need to be "coarse"

Got it.

Mike Miller

I find the discussion of index theory inspiring, but you might not :)

Akiva Weinberger

@Kari In technical terms, $f$ and $g$ have to be coarse maps, and $fg$ and $gf$ have to be close to the identity.

Kari

01:33

Wicked. Thanks, @Akiva! :-)

Balarka Sen

I forgot how one defines tangent space of a projective variety.

Take a chart, cut along, you have an affine variety, and then tangent space?

Akiva Weinberger

I forgot how one defines projective variety.

AKA never learned

Balarka Sen

Zero set of a bunch homogeneous polynomials inside $\Bbb{CP}^n$.

Semiclassical

there's a relation between coarse spaces and index theory? i may have to look at that

though given your mention of Banach being part of that, i probably shouldn't be surprised

Mike Miller

@Semiclassical It's at the end of the above link.

Semiclassical

01:36

hmm

Mike Miller

This is not so much related to Banach spaces. The point is that you can't actually take the index of an elliptic operator on a noncompact manifold.

So they say "there's an index group, depending on the coarse type of the Riemannian manifold, that the index of an elliptic operator actually lives in."

Semiclassical

hmm. a bit too abstract for me to get more than just that slogan out of it

probably this is one of those spots where i wish i knew k-theory

Akiva Weinberger

Why do people name theories after letters?

Semiclassical

for better or worse, i'd need something relatively concrete as an example

Mike Miller

I'm not sure if it's K-theoretic. I guess it probably is.

Akiva Weinberger

01:41

I thought a pretty big part of algebra was that it doesn't matter which letters we use

Semiclassical

well, it mentions K-theory in that survey paper

Mike Miller

I'm not sure if it's K-theoretic. I guess it probably is.

Semiclassical

" Nevertheless, one can define an “index group” (actually the K-theory of a certain C∗-algebra), which only depends on the coarse structure and which allows the index of D to be well-defined as an element of this group. "

Balarka Sen

too many deja vu's today

Mike Miller

@AkivaWeinberger: Because it's about the functor K.

Balarka Sen

01:44

I wonder why they named it K in the first place.

Kyle S.

Is anyone remotely familiar with algebraic varieties?

Semiclassical

from wikipedia: "[K-theory] can be said to begin with Alexander Grothendieck (1957), who used it to formulate his Grothendieck–Riemann–Roch theorem. It takes its name from the German Klasse, meaning "class.""

Mike Miller

@Semiclassical: You win!

Semiclassical

that's from the K-theory page

Mike Miller

(re: K-theory of C* algebras)

Kyle S.

01:46

I'm trying to wrap my head around a statement one of my advisors said, and it's just not making sense.

Semiclassical

ah

Balarka Sen

@KyleS. I know the basics, but not guaranteeing that I would be able to help.

You can ask the question. Someone will help if not I.

Kyle S.

I'm only really dipping into the basics here, I think.

Mike Miller

My impression is John Roe spent a lot of time thinking about index theory first - I think the relation between C* algebras and index theory originally comes from Connes work on foliated spaces - and somehow people realized this was also the thing to do on noncompact manifolds.

Kyle S.

I'm actually dealing with algebraic manifolds, I'm pretty sure.

Balarka Sen

01:47

Well, I won't know if I would be able to help or not until you ask the question :)

Mike Miller

You should just ask, not ask to ask.

Kyle S.

:P

Give me a minute to organize the question and try to get all the background info needed in there

Balarka Sen

@Semiclassical Ah, OK.

Kool.

Kyle S.

We have several systems of (algebraic) equations and are considering the varieties defined by them (ie: the zero sets of those polynomials). Varieties can have singular points, but away from them they look like manifolds. Since the equations generating the variety are algebraic, we have that they are smooth away from singularities.

damn

better

anyways..

We have several different systems of equations defining multiple varieties, all of which end up being submanifolds of a larger space. When looking at intersections between these submanifolds (away from singularities), I want to ensure that we can always make a perturbation such that their intersection is transverse

Which you can do with smooth submanifolds, but my advisor claims that we can't do that here

"You are correct that we are dealing with algebraic manifolds. However if we were to perturb one of them it would change the conditions that we are using."

(conditions being conditions on the larger space we've defined that actually ensure the varieties won't have singular points)

Balarka Sen

Transversality in manifold category is different from transversality in algebraic category. That being said, I am not sure what you are trying to do.

Give two smooth varieties, you want to make them intersect in a smooth manifold?

The question is not quite clear to me.

Kyle S.

01:54

I'm not sure if it's clear to me either. Sort of hoping that trying to explain it to someone else makes it come together a bit...

I think I'm ultimately dealing with two submanifolds of a larger manifold, which I've ensured are smooth nearly everywhere, and want to show that their intersection is transverse. The fact that they come from varieties might not matter that much

Balarka Sen

@KyleS. In the manifold category you make two smooth manifold transverse to each other so that the intersection is a smooth manifold. In the algebraic category this always happens, in the sense that if you have a a variety $X$ in $\Bbb P^n$ and a hypersurface $H$, they will always intersect in a variety of dimension exactly less than $1$ whenever $X$ is not contained in $H$.

@KyleS. OK, so you're asking if given two smooth varieties I can always make their intersection transverse in the manifold sense.

Mike Miller

but not necessarily in a transverse way... he wants to perturb the defining equation for the varieties so that they're trandverse at non-singular points

Balarka Sen

I doubt this can always be done.

Kyle S.

@BalarkaSen That's what my advisor is claiming, but I just wasn't sure why.

The above quote was all he gave me, heh.

Semiclassical

simplest way to prove it would be to give an example, though i'm not in a position myself to do that

Balarka Sen

01:59

I don't have an example off the top of my head, but it's true say that every smooth manifold is cut out by algebraic equations in R^n (Nash's theorem). But if you could perturb to make them transverse every smooth manifold could be transversely cut out by equations - which is not true recalling a discussion @MikeMiller and I was having a few days ago.

I don't know, really. This is just a heuristic.

Akiva Weinberger

So, there's this thing called nonstandard topology, where we define concepts in the space $X$ in terms of something called $^*X$, which is roughly $X$ plus infinitesimals.

Mike Miller

I do not parse your claim.

Kyle S.

If it matters at all my larger space is simply $\mathbb{R}^n$, and each of the submanifolds have codimension 1.

Balarka Sen

Probably because my claim is something silly. Of course making transverse could change the topology of the intersection.

Akiva Weinberger

Specifically, we can define topology in terms of a function $\mu:X\to P(^*X)$, where $\mu(x)$ is roughly the set of things infinitely close to $x$.

I think uniform structures corresponds to the slightly larger function $\mu:{}^*X\to P({}^*X)$.

Mike Miller

02:02

oh yeah I definitely do not care about this.

Akiva Weinberger

I'm not sure if there's a similar thing for coarse structures

Mike Miller

do you really need the blank space? $^*X$

oic the point.

Akiva Weinberger

$^*\!X$

More annoying to type

Well, technically you can define $\mu(x)$ for $x\in{}^*\!X$ with topological spaces, but I think they end up being too small.

Balarka Sen

@KyleS. The reason I am objecting is the following. If $M$ and $N$ are the algebraic submanifolds in some ambient manifold, $M$ and $N$ can be made transverse by a small homotopy. But what's the guarantee that intermediates of this homotopy will actually be algebraic varieties?

This seems a bit too much to want, at least to me.

Akiva Weinberger

Perhaps coarse structures correspond to a function that gives you things infinitely far away from $x$?

Kyle S.

02:07

@BalarkaSen Hmmm.

Mike Miller

@BalarkaSen: I think it sounds reasonable. I think his advisor's point is that the intersections after perturbing will be different algebraic varieties, and that's bad for their situation, probably.

Kyle S.

All good thoughts. Gives me some more to think on. @MikeMiller yes, different varieties after perturbation would definitely be bad.

Balarka Sen

By intermediates being algebraic variety there I actually mean intermediates being all algebraic varieties isomorphic to $M$.

Mike Miller

Yeah, that's what I would worry about. I think that's probably a problem.

@BalarkaSen: Even then, I'm not worried about that. I'm worried about the intersection changing.

Kyle S.

It seems weird that perturbations would change into a new variety, but I'm also not at all familiar with varieties so..

Balarka Sen

02:12

I am not sure if I buy that. If you have two osculating circles, you perturb to make them intersect each other in two points, you change the topology of the intersection, no? Why's it a trouble here?

Mike Miller

He said he didn't want that to happen.

Kyle S.

Funny you mention circles, because the overall space is representing configurations of circles in the plane.

Balarka Sen

@MikeMiller But then this cannot be even done in the manifold category? How can you make two osculating circles in the plane transverse to each other, while remaining the intersection a point?

Mike Miller

I don't care enough about this to argue more, especially because what's desired is unclear.

Of course you can't.

messages swapped order

Balarka Sen

Right. I think the question is pretty vague right now and I believe my interpretation is probably what Kyle is looking for. But of course he has to decide that.

Kyle S.

02:17

@BalarkaSen @MikeMiller Thanks for your time, and letting me try to formulate it into a coherent question. It'll help me tomorrow when trying to understand what my advisor is trying to say.

Balarka Sen

About my question: I think this cannot be done. I don't have an example but I think it's an interesting question.

OK, I have to get some sleep. G'night.

Akiva Weinberger

Your biological clock makes no sense to me, @Balarka

Mike Miller

@Akiva Don't try

PS you should learn about coarse spaces etc because I want to convince someone to do noncommutative geometey and tell me what the deal is

and I think those are a gateway drug

Balarka Sen

@MikeMiller I rethought: I think you're right that that's not an issue. Bad things like x-axis and damping sine curves cannot happen, i.e., "nontrasversality at infinity" cannot occur.

Akiva Weinberger

The hell is noncommutative geometry? EDIT: looked it up

@BalarkaSen Isn't it past your arbitrary, self-appointed bedtime?

Balarka Sen

02:30

I am less skeptic of the truth of my interpretation now. Oh well.

@AkivaWeinberger Yep.

Something something spectrum of a C^* algebra something.

I forgot, don't want to remember. I am really off to bed, g'night.

Mike Miller

@Akiva It's studying operator algevras using methods of topology and geometry; or conversely studying topology and geometey using methods of operator algebras.

Baum, Connes, Roe are the names I think of immediately.

Akiva Weinberger

I know none of these people

Wait, didn't Baum write The Wizard of Oz?

Big Endian

I am working on a difficult problem with polynomials and I'm wondering if someone might like to help me with it. Is there anyone online who would take some time to hear out my task?

Jessy Cat

Hello! I am in need of some help:

0

Q: Show composition of harmonic function and analytic function is harmonic without calculating 2nd derivatives

This is not a duplicate. I realize similar questions to this have been asked, but what I am asking is slightly different: I need to prove the following, arguing by complex differentiability only, and NOT by calculating $\displaystyle \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\par...

Eric Stucky

"Just ask; don't ask to ask"

11

Jessy Cat

02:41

@Akiva, he did! :)

Dunno if that's the same dude. But, hey if Lewis Carroll can do it...

Big Endian

Okay, I will summarize the problem then. I have an expression in the form AD^2 + BND + CN^2 where both N and D are second order polynomials. I want to factor this expression into the product of two second order polynomials. All coefficients are real.

Additionally, my polynomials N and D have a few properties which may be useful

N is the product of two first order polynomials in the form a+x where |x|<1 and D the product of two first order polynomials of the form 1+ax, where again |x|<1

sorry, I mean |a|<1

is this clear enough?

I believe that the problem is tractable

Mike Miller

@Akiva Anyway it's cool stuff but I'm pretty divorced from it!

I would like to hear more :)

Akiva Weinberger

Who got custody of the kids?

…it's past my bedtime, too

Jessy Cat

Meh. Nobody looks at my questions. Unless 10:48 Eastern US time is considered "after hours" on stack exchange...

Mike Miller

Connes.

Akiva Weinberger

02:49

@JessyCat Apparently 9am Eastern is the best time for posting things on the internet.

Like, in general.

Jessy Cat

@Akiva, fascinating. Spock eyebrow raise

Akiva Weinberger

(Assuming it's something in English)

Also, there's a chance that I'm completely wrong

Big Endian

Either of you have any clues on my problem?

Jessy Cat

Might just be nobody likes me.

Akiva Weinberger

@JessyCat Perhaps

Jessy Cat

02:51

That's probably it then.

Akiva Weinberger

Maybe the name "Jessy" triggers SE-users' PTSD

Or "Cat"

Jessy Cat

Why? Was there some kind of traumatic SE tragedy in the past involving someone named Jessy and a cat?

Akiva Weinberger

Maybe

Jessy Cat

How 'bout that.

Akiva Weinberger

I have no idea but it can't be ruled out

Big Endian

02:52

To be fair I looked at your question and simply didn't understand it

Jessy Cat

Definitely not the simplest explaination.

@BigEndian, asked in a weird way?

Big Endian

I'm not much of a math person

That's all

Jessy Cat

O

Akiva Weinberger

I don't know much about harmonic functions either

Big Endian

I'm in electrical engineering, computer science, use math a lot but nothing like that

Jessy Cat

02:54

Well, Imma let it sit for a while. Both on SE and in my brain. Let things percolate.

Akiva Weinberger

I know they exist and they have something to do with partial derivatives and complex analysis but that's basically it

Jessy Cat

I've found I do my best work earlier in the day.

Akiva Weinberger

It's still past my bedtime…

Jessy Cat

Lemme see...who knows stuff?

Akiva Weinberger

…has been for days.

Jessy Cat

02:55

@robjohn @Dr.MV you guys know stuff

0

Q: Show composition of harmonic function and analytic function is harmonic without calculating 2nd derivatives

This is not a duplicate. I realize similar questions to this have been asked, but what I am asking is slightly different: I need to prove the following, arguing by complex differentiability only, and NOT by calculating $\displaystyle \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\par...

complex-analysis complex-numbers harmonic-functions analytic-functions

Akiva Weinberger

G'night

Jessy Cat

G'night.

Akiva Weinberger

Like @TedShifrin says, You all misbehave without me.

Jessy Cat

Don't let the evil SE cats bite

SE Cat, Jessy Cat. Rhymes ;)

Akiva Weinberger

Not as if you haven't been misbehaving already

Big Endian

02:56

Jessy, scratch my back on this polynomial problem thing?

Ther

I'll give you a virtual cookie!

Jessy Cat

@BigEndian, virtual catnip.

Honestly, though it looks awful.

Big Endian

agreed, virtual catnip

Jessy Cat

Much too awful for my poor little sleepy cat brain right now.

Big Endian

trust me it is awful

Jessy Cat

But yeah, it's doable. I think

But I'm not an algebraist, so...

Big Endian

02:58

well maybe you want to work on it tomorrow

Jessy Cat

Possibly.

Big Endian

I'll seriously do some work for you in return

if you need any code written

Python, C, MATLAB, C#, VB.NET...

Jessy Cat

If you know anybody who's awesome with complex variables, send them my way.

Sweet, but unlikely I will.

Need it that is.

Big Endian

well, I'll check in the chat tomorrow if you feel like solving this

Jessy Cat

anyway, I'm gonna go fight in the alley.

Nite.

Big Endian

02:59

g'night

Akiva Weinberger

03:13

There's a Long Line but no Long Spiral

The limit ordinals mess it up

rorty

Hi all, do you think it's unreasonable to work through a introductory analysis text in a month? specifically Abbott's text

Mike Miller

03:31

@rorty: What's your math background?

Big Endian

I wrote up my question onto the math SE

0

Q: Factoring binary quadratic form in two second order polynomials

I have a binary quadratic form in $N$ and $D$, $AD^2 + BND + CN^2$, where $A$, $B$, and $C$ are real coefficients and $N$ and $D$ are both second order polynomials of $x$ with real roots $\lvert r \rvert <1$. I want to factor this expression into the product of two second order polynomials of $x$...

polynomials factoring quadratic-forms

user19405892

hi

do all divergent sums go to either -$\infty$ or $\infty$?

Mike Miller

No. Exercise: Think of one that doesn't.

user19405892

03:48

I guess $\sum_{n=1}^{\infty}(-1)^n$ is an example

Mike Miller

Yup.

user19405892

How would you prove that $\displaystyle \sum_{n=1}^{\infty} \sin \left(\frac{1}{n}+n\pi\right)$ converges?

misheekoh

Hello, if anybody is familiar with Linear Programming / Optimization and familiar with the Complementary slackness condition, mind helping math.stackexchange.com/questions/1719752/…?

user147690

@user19405892 Did you think about it?

rorty

@MikeMiller In undergrad I took only a few proofy math courses (linear algebra, discrete math, intro probability theory, etc) but was a poor math student (not very mature). I am now trying to build a solid undergrad math foundation through self study. I just finished Pinter's abstract algebra book and did 90% of the exercises (with a lot of help from math.se), which took me 4 months (including weeks where i would be too lazy / distracted).

I am wondering how long it would take to do the Analysis text just because I have 4 months of free time left and am trying to allocate my time properly (I want to do other things too, math and otherwise)

Mike Miller

04:16

@rorty: Go for it.

r9m

@RandomVariable You around?

@user19405892 $\displaystyle \sum_{n=1}^{\infty} \sin \left(\frac{1}{n}+n\pi\right) = \sum_{n=1}^{\infty} (-1)^{n} \sin \left(\frac{1}{n}\right)$ and $\sin \dfrac{1}{n} \to 0$ as $n \to \infty$ (converges by alternating series test).

crocket

04:50

0

Q: How do I derive a proper natural deduction proof for $\{(\phi\leftrightarrow(\psi\leftrightarrow\psi))\}\vdash\phi$?

I tried to derive a natural deduction proof for the sequent as below, but it feels wrong. Below is the latex code. How should I prove this correctly? \documentclass[oneside,12pt]{article} \usepackage[a4paper]{geometry} \usepackage{microtype} \usepackage[T1]{fontenc} \usepackage{enumitem} \us...

logic natural-deduction sequent-calculus

Never mind the above question. I solved it by myself.

crocket

05:45

Please come have a look at

0

Q: The textbook's way of deriving a natural deduction proof of $\vdash((\phi\leftrightarrow\psi)\leftrightarrow\phi)$ feels wrong.

The problem is "Show that if we have a derivation $D$ of $\psi$ with no undischarged assumptions, then we can use it to construct, for any statement $\phi$, a derivation of $((\phi\leftrightarrow\psi)\leftrightarrow\phi)$ with no undischarged assumptions." Here is my solution. Below is the t...

logic natural-deduction sequent-calculus

Evinda

07:15

@robjohn Hello!!!

@robjohn I want to show using the fourier transform that the fundamental solution of $\frac{\partial{E}}{\partial{t}}-a^2 \Delta{E}=\delta(t,x), x \in \mathbb{R}^n$, is given by $E(t,x)=\frac{H(t)}{(2 a \sqrt{\pi t})^n} e^{-\frac{|x|^2}{4a^2 t}}$.

$H$ is the Heaviside function.

We have:

$\frac{1}{(2 \pi)^n} \int_{\mathbb{R}^n} \widehat{\phi(\xi)} e^{i x \xi} d \xi=\phi(x)=\frac{\partial{E}}{\partial{t}}-a^2 \Delta E=\left( \frac{\partial}{\partial t}-a^2 \Delta \right)E=\left( \frac{\partial}{\partial t}-a^2 \Delta \right) \frac{1}{(2 \pi)^n} \int_{\mathbb{R}^n} \hat{E}(\xi) e^{ix \xi…

Evinda

07:36

@robjohn Do you have an idea?

robjohn

@Evinda how are you defining $\delta(t,x)$

skill patrol

Hi @robjohn how are you?

robjohn

@skillpatrol pretty good. how are you doing?

skill patrol

Not too bad thanks @robjohn

Algebra 2015

07:51

hello :)

How to determine the characteristic polynomial of the corresponding matrix.

\begin{vmatrix}1+(n-1)x&x&x&\ldots&x\\0&1-x&0&\ldots&0\\0&0&1-x&\ldots&0\\..&..&..&‌..&..\\0&0&0&\ldots&1-x\end{vmatrix}=\left(1+(n-1)x\right)(1-x)^{n-1}

Evinda

@robjohn It's the Dirac delta function.

Algebra 2015

@evinda

are u here ?

?

@skill patrol

are u here?

How to determine the characteristic polynomial of the corresponding matrix.
\begin{vmatrix}1+(n-1)x&x&x&\ldots&x\\0&1-x&0&\ldots&0\\0&0&1-x&\ldots&0\\..&..&..&‌..&..\\0&0&0&\ldots&1-x\end{vmatrix}=\left(1+(n-1)x\right)(1-x)^{n-1}

robjohn

08:12

@Evinda is it just $\delta(t)\delta(x)$?

Evinda

@robjohn It's not defined... Do we define it usually like that?

robjohn

@Algebra2015 the determinant of an upper triangular matrix is the product of the diagonal elements.

Algebra 2015

ok. but i really cant get the final record

of the polynom

(1+(n−1)x)(1−x)n−1

from this: (1+(n−1)x)(1−x)n−1

to the solution for any natural number n

?

:(

What are zeroes ?

robjohn

08:30

@Algebra2015 do you want the determinant of that matrix, or the characteristic polynomial of that matrix for a given $x$?

Algebra 2015

he characteristic polynomial of that matrix, depending on the natural number n

Mary Star

@Bungo Ah ok... Thank you!! :-)

Algebra 2015

?

robjohn

@Algebra2015 The characteristic polynomial would be $(1-\lambda+(n-1)x)(1-\lambda-x)^{n-1}$

Algebra 2015

ok. which are zeros?

robjohn

08:41

@Algebra2015 what do you mean?

Algebra 2015

need to find zeroes

the rate also

robjohn

@Algebra2015 Is what you really want the eigenvalues? why not ask that instead?

@Algebra2015 rate of what?

Algebra 2015

ok

or degree of the polynom

what is the degree and the leading coefficient. I can do it w= 5ith n

but not with n

roots

for n

and then the final form

user147690

@Sodre Hi

Algebra 2015

08:56

@robjohn

so, how to write down for n ?

user147690

09:35

Hey @TobiasKildetoft have you read any of Introduction to Quantum groups by Lusztig?

Tobias Kildetoft

@AlexClark No, not any of Lusztig (though bits of several others)

I generally don't find Lusztig to be a very good expositor

user147690

Is the algebra $\bf{f}$ standard enough in the literature that you know what I mean?

Tobias Kildetoft

@AlexClark Assuming you mean the generator corresponding to a negative root, sure

user147690

Oh sorry, I meant the $\Bbb Q(q)$ quotient algebra

Tobias Kildetoft

@AlexClark Not sure what you mean

user147690

09:41

Okay so Lusztig defines $'\bf{f}$ as the free associative $\Bbb Q(q)$-algebra with $1$ with the generators $\theta_i, i\in I$ where $(I,\cdot)$ is the Cartan datum

user147690

And then he constructs some radical of a form and shows it's a two sided ideal and quotients to give $\bf{f}$

user147690

Does any of this relate to what you do? I was just going to ask a question about $\bf{f}$ for $q=1$

Tobias Kildetoft

@AlexClark No, that all seems completely alien (this is typical for anything written by Lusztig)

user147690

Haha sure, hopefully I can work this thingy out

Tobias Kildetoft

It may help to figure out what these things are called in other sources and checking those out too

Jantzen is pretty good as far as I know

user147690

09:45

Mcnamara said $\bf{f}$ could be treated as $U(\frak{n}_+)$ and I was trying to check how they differed

user147690

Thanks, I'll check it out

Tobias Kildetoft

What does the Cartan datum contain?

user147690

Fancy way of saying the finite indexing set $I$ and the symmetric bilinear form on the free abelian group $\Bbb Z[I]$ with that $i\cdot i \in \{2,4,\cdots\}$, $2\frac{i\cdot j}{i\cdot i}$ yada

user147690

Oh in my case you mean?

user147690

For $n_+$?

Tobias Kildetoft

09:50

@AlexClark I meant how many of the roots are "in" $I$ (i.e. all of them, the positive ones or just the simple ones)

user147690

@TobiasKildetoft I am looking at two papers, for the non-quantised paper, it is simple roots, for the quantised one I believe it is positive roots

user147690

Also what was the Jantzens source, a textbook or papers?

Tobias Kildetoft

@AlexClark Ok, that makes sense if it should give something related to $n_+$

Jantzen's is a texbook. Lectures on quantum groups or something like that

user147690

Thanks, now I see it

Tobias Kildetoft

(btw, it is usually a good bet that any notation in a paper by Lusztig is non-standard and made up randomly on the spot. Or at least it often feels that way, even when he writes about topics with long-established notational traditions)

user147690

09:55

@TobiasKildetoft Wow haha, that's definitely a good heads up

user147690

It felt a little painful

user147690

He was changing between $v_i, \mathpzc{v}_i, v, \mathbold{v}$ over and over again in the same paragraph

user147690

apparently I don't know the script he is using haha

user147690

Embarrassing for me, it's the letter \nu I've just realised :P $v_i,\nu_i, v, \nu$

Tobias Kildetoft

10:15

@AlexClark bold is \mathbf

no idea what the pzc should have been

Clarinetist

11:34

@Akiva Here. @rorty, yes, I work in statistics. Do you have questions?

Balarka Sen

12:27

Hi @Danu, @Akiva.

Danu

Hi

In Forster's book, he uses Leray coverings in his practical calculations of sheaf cohomology groups

The way this works strongly reminds me of SvK proofs

Taking two simply connected subsets and then you just have to look at the intersection

Balarka Sen

I don't even know what sheaf cohomology is. You should teach me at some point of time.

Danu

Does this analogy make sense? Can it be made precise?

Don't let me teach you anything

Balarka Sen

Why not?

Danu

Because you're way ahead of me

Balarka Sen

12:34

That's false. I have just studied more algebraic topology than you, but that doesn't mean I am "ahead" or anything. E.g., you know sheaf cohomology: I don't :D

Akiva Weinberger

@BalarkaSen Remember a while back I was talking about the sides of a triangular prism, with both sets of edges identified cyclically? I'm 99% sure that that's embeddable in $\Bbb R^3$, but I can't visualize it. And that's all I'll say about that.

user147690

How do you choose your topics to study btw @BalarkaSen? Just wondering if you'll get to Lie algebras some time :P

Balarka Sen

@AkivaWeinberger Well the fundamental group of that is $\langle a, b : a^3 = b^3 \rangle$, no? The abelianization of that has torsion I think, so it's not possible.

Alexander duality is a thing can be embedded in $\Bbb R^3$ only if $H_1$ has no torsion.

Yup, $H_1$ of your thing has 3-torsion.

Alexander duality *says.

@AlexClark I don't know. I just start off, and if someone forced me to study something I protest for a week and then start learning. That's about how I learnt everything I learnt.

user147690

@BalarkaSen How do they force you?

user147690

Convince you it is worth learning?

Balarka Sen

12:49

Yes, that's what my real life profs do. Or otherwise, if it's a virtual person like Mike or Ted, ignore me. The latter's more effective.

skill patrol

you have profs in high school?

user147690

@skillpatrol How do you not know this about Balarka after all these years?

user147690

@BalarkaSen Well I don't yet know anything that would motivate you to learn lie algebras :P

Balarka Sen

All the better!

user147690

And I don't know enough math for it to work ignoring you

user147690

12:53

So @BalarkaSen, how's about dat Humphrey's intro to lie algebras, prettttty slick textbook wouldn't you say?

Balarka Sen

No, you do, but it's just that being ignored by Mike or Ted is more humiliating and depressing than someone else. Not entirely for not being able to badger them during that ignore period.

@AlexClark I haven't seen it. Should I?

user147690

@BalarkaSen When you have time :P. I think it's a pretty nice textbook, but again, I have no idea in regard to your mathematical development why you should learn lie algebras

skill patrol

@AlexClark unless he goes to some fancy exclusive private school

user147690

Although lie algebras -> enveloping algebras -> quantised enveloping algebras -> braid theory :P

Balarka Sen

@skill Not high school professors. Some actual mathematicians help me learn the math I learn. We discussed this before, you might have forgotten.

skill patrol

12:57

Right.

Balarka Sen

@AlexClark My taste so far lies in topology/geometry(which I don't know much about), so if you can tell me something interesting about lie algebras which relate to topology/geometry I'd probably be excited.

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