Mathematics - 2015-12-15 (page 4 of 7)

bolbteppa

10:01

Oh I was just referring to the general area he was asking about, e.g. Peter-Weyl or Weights etc... I've never seen property T tbh and now I see why, it's just a generalization of the graph theory notion of an 'expansion constant' if you know what that is? terrytao.wordpress.com/2011/12/06/… This makes sense to arise in group representation theory considering the relationship of groups to graphs, but I don't even know graph theory sadly :p

Balarka Sen

thanks for the Tao post.

bolbteppa

I would imagine that Smale comment is true, I've heard similar stories about the guy who motivated most of Perelman's proof, they all seem to be on some Hawaiian/cool beach

Balarka Sen

Perelman? That seems doubtful.

bolbteppa

Hamilton

Balarka Sen

oh, I see, I thought you were claiming Perelman has been on a beach.

bolbteppa

10:05

haha

Balarka Sen

yeah, Hamilton can be trusted

Vrouvrou

10:27

Hello, i have this example: let $(E,\theta)$ be a topological space such that $card(\theta)<\infty$ Then E is compact

$E$ must be a Hausdorff space no ?

@robjohn have you an idea please ?

Tobias Kildetoft

@Vrouvrou Why would it have to be Hausdorff?

Vrouvrou

Compact space= Hausdorff space+ from any open cover we can extract a finite cover

@TobiasKildetoft

Tobias Kildetoft

@Vrouvrou No, that is the wrong definition of compact, though some people use it

Vrouvrou

@TobiasKildetoft and what is the right definition please

Tobias Kildetoft

@Vrouvrou the same without the Hausdorff part

user174558

10:41

@TobiasKildetoft I would not call that the wrong definition. It is just another definition.

user174558

@Vrouvrou Have you tried doing this on your own? What are your thoughts on this?

Tobias Kildetoft

@JasperLoy Sure, but I decided to take a stronger stance on the issue :)

user174558

Strangely, Bourbaki uses the stronger definition as well.

Vrouvrou

but in my book there is the first definition and in the exercise is stated without hausdorff space

Tobias Kildetoft

@Vrouvrou That is very strange then

@JasperLoy and then they use quasicompact for the weaker version?

user174558

10:45

@TobiasKildetoft I don't know about that.

user174558

@Vrouvrou Definition is just definition. We can call a cat a dog. Also, this question is really simple and you really should attempt it on your own.

Tobias Kildetoft

@JasperLoy When his book defines compact to include Hausdorff, then the problem becomes impossible

user174558

@TobiasKildetoft Then that realisation is also trivial, and as a math student he should know that.

user174558

Maybe he has already put me on ignore, but for his own good, I will say this again. You really need to think more before asking, since you seem to be doing advanced math but asking very basic questions.

Vrouvrou

my problem is in hausdorff space

iwriteonbananas

10:48

Hey @BalarkaSen

Vrouvrou

not in finding a finite cover @JasperLoy

Paradox 101

Can anyone let me know what the criteria for checking for existence and uniqueness to solutions for nonlinear differential equations is?

user174558

@Vrouvrou Anyway, now you know that there are different definitions of compact in the literature. Also, books can have mistakes, just like all of us.

Huy

11:41

@Paradox101: maybe you want Picard Lindelöf?

0celo7

12:03

Has anyone heard of the distinction "pure manifold" before? If yes, do you know a counterexample?

Arthur Fischer

12:40

@0celo7 From Wikipedia, a pure manifold is one in which each point has a nbhd homeomorphic to the same Euclidean space. The disjoint union of a sphere and a line would be a manifold (each point has a nbhd homeomorphic to some Euclidean space), but is not pure. (Some authors require that manifolds are pure, and so omit this adjective.)

0celo7

@ArthurFischer Yes, I meant not disconnected.

Lembik

hi.. could someone help me with a really simple math notation question please. $\forall x,y \in \{0,1\}^n$ s.t. $x\ne y$, $Mx \neq My
\iff \neg \exists v \neq 0 \in \{-1,0,1\}$ s.t. $Mv = 0$.

0celo7

Argh.

@ArthurFischer I came up with $S^1\sqcup \mathbb{R}^2$ instantly.

Lembik

what's the right way to write the restriction that $x \ne y$ here?

Huy

@Lembik: why is that not the right way?

Lembik

12:45

@Huy it seems clumsy to me

I just want to say that $x$ and $y$ are binary vectors not identical

Huy

and what else would $x \neq y$ mean

Lembik

and the righthand side doesn't look good to me either

as it looks a little like $0\in\{-1,0,1\}$ which is not what I mean

Huy

if $v \neq 0$ then why include $0$ in that set? I think you meant to raise it by the $n$th power as well

Lembik

@Huy I did!

Arthur Fischer

@0celo7 I'm pretty sure a connected manfold (without boundary) must be pure. (The collection of all points with nbhd homeo to $n$ would be open for each $n$, so we get a cover by disjoint open sets.)

Lembik

12:48

$\forall x,y \in \{0,1\}^n$ s.t. $x\ne y$, $Mx \neq My
\iff \neg \exists v \neq 0 \in \{-1,0,1\}^n$ s.t. $Mv = 0$

@Huy do you think it is ok as it is?

Huy

if you want to be very very correct, you could write something like $$(\forall x \in \{0,1\}^n \, \forall y \in \{0, 1\}^n: x \neq y \implies Mx \neq My) \iff (\not \exists v \in \{-1,0,1\}^n: v \neq 0 \implies Mv = 0)$$ or something like this

Lembik

oh that looks better.. I think

0celo7

@ArthurFischer I'm not sure I follow. What guarantees that the $n$ is the same for each open set?

Huy

what happens on the intersection

Lembik

@Huy how do you mean?

Huy

12:53

not to you

Lembik

oh :)

0celo7

@Huy Indeed...

Lembik

I was hoping you were still produced a better version

Arthur Fischer

@0celo7 For each $n$ let $U_n$ to be the family of all points with a(n open) nbhd homeom to $\mathbb{R}^n$. (Equivalently $U_n$ is the union of all open subsets homeo to $\mathbb{R}^n$.) So the $U_n$ cover your space, and are pairwise disjoint open sets.

Huy

@Lembik: I'd be fine with your version unless it's some kind of intro to logic course where everything must me perfectly rigorous

Arthur Fischer

12:54

If a connected space is the union of a pairwise disjoint family of open sets, then all but one of these sets is empty.

0celo7

@ArthurFischer Ok

Lembik

@Huy thanks.. that's very helpful

Huy

@Lembik: out of curiousity, why is $v$ allowed to contain $-1$ as entry

Lembik

@Huy v is the difference between two x, y vectors

$x- y = v$

Huy

ok

that should be included though, I don't think the statement is true if it's not included

Lembik

13:08

@Huy are you sure?

oh.. wait.. there is a bug I think

$\neg \exists x \ne y \in \{0,1\}^n$, $Mx = My
\iff \neg \exists v \neq 0 \in \{-1,0,1\}^n$ s.t. $Mv = 0$

@Huy that's right isn't it? I just subtracted $My$ from both sides

r9m

@robjohn I see you are wigging a moustache :P

robjohn

@r9m None of the hats fit, so I've made do

r9m

@robjohn the frown has extended beyond the mean square :P How much meaner are you planning it get? Isn't it Christmas? :P

robjohn

@r9m Bah! Humbug! ;-)

r9m

@robjohn haha!! :)))

Lembik

13:20

hi @robjohn

I noticed math.stackexchange.com/questions/1573728/… and thought.. only robjohn can solve it :)

robjohn

@Lembik good morning

Lembik

good afternoon!

it's clear that for fixed $A$ the entropy is at most $n$

but does that mean it is at most $n$ for random $A$? I am thoroughly confused :)

robjohn

@Lembik How are you defining the entropy?

Lembik

@robjohn I assume it's just shannon entropy

as she says that the entropy of $x$ is $n$

binary Shannon entropy to be more precise

@robjohn does that make sense?

L33ter

13:37

how do you guys put those weird shapes around your name ?

Huy

you're a weird shape

wtf

Lembik

13:59

@Huy L33ter is referring to the icons that some people have I assume

Agawa001

@L33ter answer question, get some hats

Michael

14:36

How would you prove that $5^n=9^n+2$ . For what values of n is this thingy is divisible by 4.

Mike Miller

@Huy: THe Maslov index shows up in symplectic geometry, J-holomorphic curves stuff. I don't know much about it other than that it's important.

OFFSHARING

@Michael For all $n$ it seems. Kindergarten question? Use that $9^n=(4+5)^n$ and then

$$5^n+3=(4+1)^n+3$$

Q.E.D.

Mike Miller

First class sounds fun. I don't know much rep theory.

OFFSHARING

14:53

@Michael It's wiser and simpler to start with $$9^n=(8+1)^n=4M+1$$

then, we get $$5^n +3+4M$$

or

$$(4+1)^n+3+4M$$

or

$$4N+4$$

That's all.

My apologies that I solve number theory problems with no pratice for a long period of time and doing only integrals, series and limits in the last years.

I think I might even put down some of this type but far harder, at the Olympiad level, maybe, not sure. I slayed in the very past enough crazy hard questions of this type.

However, the present looks different to me now, other interests in mathematics.

Back to my work.

Mike Miller

15:25

Hi @iwriteonbananas.

iwriteonbananas

Hey Mike

We're doing vector bundles in algebraic topology now. Next week we'll do transversality etc., yay

Mike Miller

Great!

iwriteonbananas

I was about to ask you for some help with the following:

$H_*(\Bbb RP^\infty; \Bbb Z_2)$ is a module over the ring $H^*(\Bbb RP^\infty; \Bbb Z_2)$ via the cap product. Is this a free module?

My guess is yes and I've been trying to prove it, but to no success.

Mike Miller

oops

@iwriteonbananas: Can't possibly be true. Take $\alpha^n \in H^n(\Bbb{RP}^\infty)$ and, say, $a \in H_1(\Bbb{RP}^\infty)$. Then $\alpha^n a = 0$.

DanielSank

How does one approach an expression like $\int dx \delta(f(x,y))$?

I understand how to handle expressions like $\int dx \, \delta(f(x))$; you Taylor expand the $f$ and wind up with a factor of $1/|df/dx|$.

But, I'm having trouble understanding how to integrate over only one variable as in the first expression.

iwriteonbananas

15:32

@MikeMiller Ahh, good point.

Mike Miller

@iwriteonbananas: $H_*$ is the limit of $\Bbb Z_2[x]/x^{n+1}$. I guess this is analagous to $\lim \Bbb Z/2^n$, which is a Prufer group. I've never thought about this before, so I don't know if there's any actual meaning to this analogy or what.

iwriteonbananas

What does $\lim \Bbb Z/2^n$ mean?

Mike Miller

There's an injective map from each to the next, yeah? I just mean the normal direct limit of groups.

iwriteonbananas

Oh, I see

Mike Miller

I'm not sure that what I said there is worth spending time on.

Because I'm not sure if I said anything of value.

iwriteonbananas

15:37

Ok, I'll just move on then

Mike Miller

15:50

lol "In a later paper, we use these to derive results of bihermitian manifolds"

how utterly precise! results! I love results!

iwriteonbananas

hahaha

bihermitian manifolds are a great source of results.

Mike Miller

it's like a cop movie. "Sure, bihermitian manifolds may be nonstandard in their methods, but damnit, they get results!"

Semiclassical

@MikeMiller where i've encountered the Maslov index is in the context of Bohr-Sommerfeld quantization in quantum mechanics

iwriteonbananas

lool

Mike Miller

that means nothing to me :)

Semiclassical

16:01

snerk

asymptotic analysis of a differential equation. but one of the fruits of that is that you can approximate the location of eigenvalues by requiring that a certain integral on phase space (i.e. on a symplectic manifold) takes on specific values

Mike Miller

gromov-witteny stuff? that's roughly what I was saying above

funny how different our language is for the same sort of thing

Semiclassical

so you get equations like $\oint p\,dq = h(n+\nu)$ where $p,q$ are the position and conjugate momentum, $n$ is a nonnegative integer, and $\nu$ is the appropriate maslov index

for a one-dimensional well with 'soft edges', you have $\nu=1/2$ and so you get 'half-integer' values. simplest example is for a quantum harmonic oscillator, for which that's exact

Balarka Sen

hi @iwriteonbananas'

Semiclassical

you can also end up with a soft edge and a hard edge (usually in the context of radial equations) and then the index is $3/4$

alas, i don't know gromov-witten stuff :/

iwriteonbananas

@BalarkaSen Hey Balarka

Semiclassical

16:07

in quantum mechanics, it's stuff under the rubric of the WKB approximation

Balarka Sen

upvote/downvote!

@MikeMiller Any new hats?

Mike Miller

Essentially all the rest of the hats come from consistently posting a lot of good questions and answers on the site. So...

Balarka Sen

Yeah, well, I can't find any good questions.

Semiclassical

oh, mike. a question for you

does the phrase 'riemann-hilbert problem' mean anything to you?

Mike Miller

Nope

Is that a diffy q thing

Semiclassical

16:12

more of a complex analysis thing

a boundary value problem in complex analysis, really

i've also seen the phrase 'riemann-hilbert correspondence', but that's baffling to me

Mike Miller

The number of differential equations I can say something interesting about I can count on two hands

iwriteonbananas

I'm trying to show that the tautological line bundle $$E=\{([x_0,\cdots, x_n],v):[x]\in \Bbb RP^n \text{ and } v\in \operatorname{span}(x_0,\cdots x_n)\}$$ over $\Bbb RP^n$ is not trivial.

Mike Miller

what tools you got

iwriteonbananas

I'm getting that all sections are trivial, is that true?

Mike Miller

no

and obviously not: pick a chart (over which it automatically trivializes). pick a bump function. let that be your section.

all sections have a zero, though

iwriteonbananas

16:14

Hmm ok

Semiclassical

what i have in mind when I say RH problem is this part of the Wiki page: en.wikipedia.org/wiki/…

Mike Miller

it seems interesting, and if you're asking about an interesting differential equation, odds are I know nothing :)

Semiclassical

whereas the Riemann-Hilbert correspondence is a lot more abstract, the linkage being the connection to monodromy

Riemann–Hilbert correspondence

In mathematics, the Riemann–Hilbert correspondence is a generalization of Hilbert's twenty-first problem to higher dimensions. The original setting was for the Riemann sphere, where it was about the existence of regular differential equations with prescribed monodromy groups. First the Riemann sphere may be replaced by an arbitrary Riemann surface and then, in higher dimensions, Riemann surfaces are replaced by complex manifolds of dimension > 1. There is a correspondence between certain systems of partial differential equations (linear and having very special properties for their solutions) and...

Balarka Sen

how does $xy = 1$ look like in $\Bbb C^2$?

Semiclassical

it was a shot in the dark, anyways

Mike Miller

16:18

@Semiclassical: but that looks like something I can parse

i might be too cold to parse it right now

Semiclassical

i wondered if you might find that more readable. i find it pretty obscure

Mike Miller

haha

Semiclassical

i understand differential equations a lot more than i understand bundles :P

Mike Miller

I think maybe you should think of a holomorphic vector bundle as a sort of patched-together system of PDEs

(a section of the bundle being holomorphic is a certain special PDE)

Semiclassical

that sounds similar to the Gauss-Manin stuff i knew at one point

Mike Miller

16:23

the first part of this page btw is the only part I understand; I'm into connections, not so much D-modules

I think D-modules are the attempt to salvage differential equations in a more general algebraic geometric context

Semiclassical

yeah, i dunno about d-modules

i have a sneaking suspicion that one of the main language barriers we have is the word 'instanton', since we use it in rather different ways

Mike Miller

I'm turning into the new instanton guy, I think, because the last instanton guy left.

Semiclassical

hah

Mike Miller

So I guess I should one day understand what you mean by instanton.

Semiclassical

an instanton to me is a certain kind of solution to a schrodinger equation

i suspect that what you do is more towards what an instanton is in field theory

Balarka Sen

16:27

@Semiclassical that seems like a understandable definition.

Mike Miller

An instanton to me is a solution to the ASD equations on a 4-manifold, which I think is the same thing as a solution to the Yang-Mills equations on a 4-manifold

Semiclassical

if so, i'm probably making it sound easier than it is :P

Mike Miller

I meant to check this at some point but never did

Semiclassical

ASD = anti self dual?

i do quantum mechanics rather than QFT on a regular basis, so the definition for me would be simpler

Mike Miller

yeah

Semiclassical

16:30

this gives a nice introduction on the level of physics for instantons: tpi.umn.edu/shifman/lectures4students/ABC_of_Instantons.pdf

not helpful without background, though

Mike Miller

wait: are the YM equations just that the curvature is harmonic? because that's not enough - self-dual curvature forms are also harmonic

Semiclassical

shrug

amusingly, the second author for that just finished giving a lecture down the hallway about 10 minutes ago (he's a prof here)

and he was talking about instantons :P

and he does refer to YM theories in there

Mike Miller

who can blame him? instantons are great

Semiclassical

heh

if i was going to try to describe the difference between your definition and mine, i'd probably take a page from this bit on nLab re: instantons: ncatlab.org/nlab/show/Yang-Mills+instanton

where near the end it gives as a slogan “a Yang-Mills instanton describes the tunneling between two Chern-Simons theory vacua”.

whereas in regular quantum mechanics, it'd just be "an instanton describes the tunneling between two classical vacuum states"

iwriteonbananas

@MikeMiller Meh, I can't figure out why a section must have a zero.

Semiclassical

16:43

they state that in terms of gradient flows between flat connections, so more your neck of the woods

iwriteonbananas

So a section is of the form $[x]\mapsto ([x], \sigma([x]))$ for some cts map $\sigma:\Bbb RP^n\to \Bbb R^{n+1}$

with $\sigma([x])$ in the span of $x$

dorothy

hi all.. I asked this recently but it is getting little love.. math.stackexchange.com/questions/1573728/… Is it unclear?

Semiclassical

it may help to include the definition of Shannon entropy in the question itself, in addition to including the link

the more self-contained a problem is, the more willing someone is to spend some time looking at it

dorothy

@Semiclassical ok. I was not sure if someone unfamiliar with the definition would want to work on it

Semiclassical

perhaps not, but it's not wise to exclude part of your audience

Mike Miller

16:50

@iwriteonbananas: Start with $n=1$. See what happens.

Semiclassical

i might even go so far as to explicitly express the quantity you're trying to maximize. just because someone doesn't know entropy doesn't mean they might now know something about approximations regarding large matrices

Huy

Just FYI, Papers is 50% off. Great app to organize PDFs and references.

iwriteonbananas

@MikeMiller Ok, I can kind of see it but I'm still not able to make it rigorous

Mike Miller

17:09

Can you name it make it rigorous for $n+1$?

iwriteonbananas

Sorry what?

Mike Miller

$n=1$

Phonetypin

Semiclassical

2011

iwriteonbananas

@MikeMiller It seems to me like continuity would break if $\sigma$ were nonzero

Mike Miller

@iwriteonbananas: I just showed you how to construct a nonzero $\sigma$. Let's try again. In this case, you may as well think of $\Bbb{RP}^1$ as a quotient of the interval $I=[0,\pi]$ by identifying the boundary points. Then the demand is that the map $f: I \to \Bbb R^2$ have $f(0)=f(\pi)$ and $f(\theta)$ in the line of angle $\theta$. Equivalently (why?), it's a function $g: I \to \Bbb R$ with $g(0) = -g(1)$.

Err, $f'$ not the best notation. Editing.

Well, that's easy to do. Take $g(x) = \cos(x)$.

In the planar setting, this corresponds to $f(\theta) = (\cos(\theta),\theta)$, where I'm using polar coordinates (and allowing $r$ to be negative).

But the intermediate value theorem says that $g$ must be zero at some point.

The construction I mentioned before is taking $g$ to be a bump function that's 0 near the endpoints.

By the way, your question just inspired me to post this question. Not 'inspired' as in 'you will solve your question with Borsuk-Ulam', but 'inspired' as in 'I was thinking about your question when I got into the shower and then thinking about that question by the time I got out'. It is not possible for me, or anyone, to explain the showerthoughts that led me from one to the other.

OFFSHARING

17:26

$$\huge{\text{Another absolutely amazing breakthrough!!!}}$$

TIME TO CELEBRATE!!!

Clarinetist

Is anyone here familiar with the concept of a generalized inverse?

of a matrix with real entries?

iwriteonbananas

@MikeMiller Thanks for the help. you meant to write $g(0)=-g(\pi)$, right?

Mike Miller

Oh, yeah. Thanks for the correction.

I think the first two sentences up there sounded perhaps short with you. Sorry, I didn't intend that.

Vrouvrou

someone have an idea : math.stackexchange.com/questions/1577084/…

please

iwriteonbananas

No worries

Paradox 101

17:37

How do you convert an nth order nonlinear differential equation into a system of $n$ first order differential equations?

Balarka Sen

meh, guess I am giving up on hats after all.

need to get math done.

Mike Miller

@BalarkaSen: I just posted a question you might like. I have absolutely no idea how to fo it.

Balarka Sen

good question.

thanks for the edit, that's exactly the interpretation i had in mind.

good to know i am still awake

Mike Miller

17:52

I really quite firmly believe that if it's true for $X = [0,1]$, it's true for all (of a reasonable class of) $X$.

Balarka Sen

I don't have a reason to believe it's true.

Mike Miller

'cuz'

Balarka Sen

Even for [0, 1]

so you're essentially asking if two $S^n \to \Bbb R^n$ are homotopic, we can choose a homotopy such that the Z/2-equivariant point vary continuously.

Mike Miller

I'm not choosing the homotopy. The homotopy is fixed. I'm asking if there's a path of equivariant points.

iwriteonbananas

Gaaah, I still can't generalize to arbitrary $n$ T__T. I need a break.

Mike Miller

17:59

@iwriteonbananas: $\Bbb{RP}^n$ is a quotient of $D^n$. Maybe try looking at it via analogy to $n=1$ via that.

Balarka Sen

ok, fair enough. you want to choose an eqv pt.

iwriteonbananas

I'm guessing I need to use the Brouwer FPT at some point

Balarka Sen

not a htpy.

Mike Miller

I want to choose a path of equivariant points. One point for each $t$.

iwriteonbananas

That's what I was trying to do, viewing $\Bbb RP^n$ as $D^n/S^0$

Balarka Sen

18:00

yes, that is what I meant. ok.

Mike Miller

@iwriteonbananas: I would think about it in terms of the explicit equivalence relation instead of $S^0$.

Balarka Sen

let's think for $n= 1$. I want a crazy sort of homotopy.

I bet this is false unless your homotopies are good (say, smooth).

iwriteonbananas

@MikeMiller Yeah, ok. Is it true that we're gonna apply Brouwer at some point?

Mike Miller

When you can actually cause trouble for me, I'll be worried.

@iwriteonbananas Seems that way to me.

iwriteonbananas

wait

Mike Miller

18:06

What is $x$?

iwriteonbananas

Ugh, I really need a break...

lol

Balarka Sen

Take one.

Mike Miller

You could also, I suppose, think of this as maps $f: S^n \to \Bbb R$ with $f(x)=-f(-x)$. Your section is given by $xf(x)$.

And now I think I've given away the problem.

iwriteonbananas

@MikeMiller I'm not sure what you did there, but I'm gonna drink a tea before I embarrass myself anymore

Mike Miller

:)

Hi @Carl. I see your point, but then further I'm not sure I agree: if someone downvoted every user and then left a comment every time about why he downvoted them, but nonetheless downvoted every post that user made...

(I'm not going to reply on that meta thread. I don't want to get dragged into any further fights.)

Willem D'Haeseleer

18:21

Hey Folks, is there anyway I can put this in a single equation, so I can solve it ?
A + B = 10
A + C = 20
B + C = 24

Mike Miller

(A-3)^2+(B-7)^2+(C-17)^2=0

Balarka Sen

So, you're looking for a path $\sigma$ in $S^1 \times I$ such that $\sigma$ and $-\sigma$ (the path obtained from taking antipodal points at each intermediate circle) map to the same path by the htpy $F : S^1 \times I \to \Bbb R$. just looking for different interpretations of the same thing to approach this.

Mike Miller

@BalarkaSen: Right. I'm not sure if that will help.

Willem D'Haeseleer

@MikeMiller But now you already solved the values in your equation...

Mike Miller

Well, that's the only way I could think to put it in a single equation :)

You need to do some algebra. Take the second equation and subtract the first to get C-B = 10. Now add this to the last equation to get 2C=34.

Noam

18:25

$$\lim_{x \to \infty} \left(\left(\sqrt{9x+1}-\sqrt{3x}-\sqrt[3]{2x}\right)\left(\sqrt{2x+1}\right)\right)$$

Somebody has an idea how to calculate the limit of this ?

Balarka Sen

we can ask this in a more restricted setting, where the initial maps $f_0, f_1 : S^1 \to \Bbb R$ the homotopy is between are just the constant maps. then we're asking the same question for a map $S^2 \to\Bbb R$.

i doubt this is true.

Mike Miller

Well, come up with a counterexample!

evinda

Hey @DanielFischer
Or don't we deduce from the lemmas that $\langle x, A^2 x \rangle \leq \sup \{ |\langle x, Ax \rangle|: ||x||_2=1\}^2$ ?

Carl Mummert

@MIke, I agree with not commenting too much more on the thread. I had thought about it before, but with other comments I decided not to. Those seem to be deleted, so I left one, which is my self-imposed limit there. I think that leaving comments every time might start to look like harassment. And downvoting has a visible effect on someone's rep score. But just voting to close, if nobody else votes that way, doesn't have much real effect on anything.

@WillemD'Haeseleer: solve equations 2 and 3 for A and B, then plug into equation 1 to get an equation in terms of C only

Mike Miller

@CarlMummert: Sure, I understand your point. My response to Bill was looking at it from my own perspective: if someone voted to close every differential geometry question on the site, I'd get mad.

Would probably tell a mod. Would they do anything about it? Who knows.

Balarka Sen

18:29

@MikeMiller I mean, lots of maps $f : S^2 \to \Bbb R$ such that there does not exist a path $\sigma : [0, 1] \to S^2$ between the two poles s.t. $f(\sigma(t)) = f(-\sigma(t))$, yeah?

Mike Miller

Write one down.

Balarka Sen

let me ponder for a while.

Mike Miller

My spoiling is no longer valid. But I think you will have an exceptional amount of trouble doing what you claim.

Hippalectryon

@MikeMiller :o so you found how to get that monkey hat ? Was it the meta downvotes ?

Balarka Sen

The point is that $f$ can be horrendous, because no smoothness assumption is made. I think I can do this, I'll ponder a bit.

Clarinetist

18:32

For a matrix $X \in M_{N \times k}(\mathbb{R})$, define $P_{X} = X(X^{T}X)^{-1}X^{T}$. Let $\mathbf{1} \in \mathbb{R}^N$ be a column vector. How do I show that $\text{tr}(P_{X}-P_{\mathbf{1}}) = k-1$?

Ted Shifrin

Salut, M le Méchant. Hi @Clarinet

Clarinetist

Afternoon @Ted. I bought the Mechanics book. Looking forward to reading it after my final

Ted Shifrin

Oh cool, @Clarinet. Remember to work exercises :)

Clarinetist

@Ted Will do. Any wisdom for the question I have right there?

Mike Miller

@BalarkaSen: I've gotten annoyed with the pseudo-philosophical comments. When you can give me some math, I'll pay attention.

Balarka Sen

18:34

well, I'm thinking!

Ted Shifrin

When you say to let $\mathbf 1$ be a column vector, you mean the vector of all ones?

Clarinetist

@TedShifrin Yes

Ted Shifrin

You also need to assume that $X$ has rank $k$ or this makes no sense.

Mike Miller

@TedShifrin: We are (were) talking about this question I had in the shower today.

Ted Shifrin

So use linearity of trace and figure out the trace of the projection matrix.

Clarinetist

18:35

@TedShifrin Yep, that's assumed

Oh boy

Hmm

How strange. So the rank of $P_{X}$ is equal to the rank of $X$? That's news to me

Ted Shifrin

That is a natural questiion, @MikeM. It's similar to the sorts of questions I've had Balarka trying to work on. You need an implicit function theorem set up to study how the fixed-point (or antipodal) set varies, in general.

You don't mean rank, @Clarinet. You mean trace.

Clarinetist

Oops, eys

*yes

Where the heck would this be in my stats book...

Ted Shifrin

Yup. Think about eigenvalues?

What are eigenvalues of a projection map?

Clarinetist

Of course, we haven't even talked about eigenvalues in this class. But I have a proof that says that a symmetric matrix $A$ has trace equal to the sum of the eigenvalues, and $P_{X}$ is of course symmetric

Ted Shifrin

BTW, @MikeM @Balarka, that nincompoop removed his garbage answer on that Galois theory question.

Balarka Sen

18:39

Good to hear.

Ted Shifrin

Any matrix has its trace = sum of eigenvalues, @Clarinet.

I have gotten a ton of points on MSE for giving a one-line explanation. You can find it.

Hippalectryon

@TedShifrin o/

JeSuis

@TedShifrin hi

Clarinetist

Sigh, I really need to brush up on some of this linear algebra stuff. It's nuts how stats classes don't even touch this stuff

JeSuis

@Hippalectryon salut

Agawa001

18:40

@Hippalectryon \0

Hippalectryon

@JeSuis @Agawa001 :D

Ted Shifrin

salut, @JeSuis

You'll get to it all in my book eventually, @Clarinet.

Mike Miller

@CarlMummert: Looks like they sucked you back in :)

Ted Shifrin

OK, I just dropped in to say hi. Im headed off to the charter high school in a few.

Clarinetist

Eigenvalues are either $0$ or $1$. Proof was much more straightforward than I was expecting

Agawa001

18:41

i invite u to share with me a giant pie lied on the peak of the mountain in my picture @Hippalectryon

Clarinetist

Thanks @Ted, I should be able to figure it out from here

JeSuis

Tu fais des probas Hippa ?

Huy

"The topics will be roughly about definitions of quantum groups, representations, and two main applications: solutions to Yang--Baxter equations, and constructions of quantum knot invariants (Jones polynomials, HOMFLYPT, etc)." anyone know what this might be about?

Ted Shifrin

I hope so, @Clarinet :)

hi/bye @Huy, @Agawa

Huy

hi @Ted

Hippalectryon

18:42

@Agawa001 Sure :D

Mike Miller

@Huy: I know the words.

Hippalectryon

@JeSuis J'en ai fait un peu

Agawa001

comment ça va mr @Ted

Huy

@MikeMiller and a bit more than that?

PVAL

@MikeMiller Do you know any proof of Borsuk-Ulam that could be useful for this, or just the standard non-constructive homological one? Googling gives some short classical papers giving "constructive proofs" of Borsuk-Ulam. i.e. this one ams.org/journals/proc/1979-073-01/S0002-9939-1979-0512075-9/…

Mike Miller

18:42

Not much. The construction I know of the Jones polynomial is via skein relations.

@PVAL: I only personally know non-constructive ones (though I learned a non-constructed one via differential geometry at some point).

@Huy My impression is that it's about fancy, modern-day representation theory. There is a group of people who like to apply representation theory to knots.

JeSuis

je prends une v.a.d intégrable et une fonction à valeurs réelles j'aimerais exprimerl'espérance $E[X]$

Huy

@MikeMiller: and is that useful or is it just being fancy?

Mike Miller

Useful to what?

Huy

idk

anything

except maybe knots

Mike Miller

I mean, one can ask that question about a lot of math. People study representation theory because they think it's interesting. All I know is knots.

I am dubious about its value in knot theory.

Huy

18:47

idk, usually there is some greater motivation behind it except "this seems interesting", at least that was my impression

maybe not a very obvious and direct application, but something "behind" it

doesn't have to be non-mathematical either

Mike Miller

There probably is. I know very, very little representation theory. Not the classical stuff, much less this.

JeSuis

Je dois donc monter que la variable $Y:=f(X)$ est intégrable. J'écris $\sum_{y\in f(E)}\vert y\vert p_Y(y)$ et là je ne vois pas trop quoi faire. @Hippalectryon

Huy

ok, thx

PVAL

Here is one "opinion" on the value of the Jones' polynomial in topology (and topology in mathematics) math.rutgers.edu/~zeilberg/Opinion1.html

Mike Miller

lol

Hippalectryon

18:50

@JeSuis Tu veux dire $E(f(X))$ ?

JeSuis

@Hippalectryon oui

Mike Miller

I can tell you that a number of people will be very excited when computing the Seiberg-Witten invariants is made combinatorial.

"Of course these assumptions are highly idealized, since there are infinitely many tautologies, but we can assume that the logician is immortal." A frightening assumption indeed.

JeSuis

j'ai aussi (je pense) $\{Y=y\}=\cup_{x\in E; f(x)=y} \{X=x\}$

Clarinetist

@TedShifrin Thank you for the hint. Very helpful. $$\text{tr}(P_{X}-P_{\mathbf{1}}) = \text{tr}(P_{X}) - \text{tr}(P_{\mathbf{1}}) = r(X) - r(\mathbf{1}) = k - 1$$

Agawa001

something bugs here if $c$ is $0$ then how comes lim 0/0 = 1 ?

JeSuis

18:59

j'ai donc à calculer la somme sur les y dans f(E) des $\vert y\vert 1_{y}(f(x))$, et ça je dirais intuitivement que c'est $\vert f(x)\vert$ car cette somme ne vaut qu'un quand $y=f(x)$. Right ? @Hippalectryon

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