Mathematics - 2017-02-28 (page 2 of 5)

Ok, now it makes sense, it's not locally s.c. because the bad point has no s.c. nbhd, but the loops in those nbhds can still be nullhomotoped in the whole cone

Balarka Sen

Yup

BAYMAX

2:13 PM

Any reference for this , looks like mess after googling - any help is great! --- "solution of system of nonlinear ODE equation using R-K method" ?

or any book which has lucid explanation , i have to code it ! time limit given to me is half day ,

Alessandro Codenotti

Nice! That's a funny space, it looks like a mess but as far as homotopy is concerned it couldn't look better

I have a good idea of why cones are contractible, I'll try to write a proper proof

Balarka Sen

While you're at it, prove that suspensions (X times I/X times {0}, X times {1}) are simply connected. They are not in general contractible but I guess you don't have the tools to prove that yet

Alessandro Codenotti

Hm, ok, to prove that they're not contractible in general I guess I need to show that $S^2$ isn't and then notice that it is homeomorphic to the suspension of $S^1$?

Balarka Sen

@Alessandro Yeah. Also, I misspoke. I meant suspensions of path-connected spaces are simply connected (what goes wrong otherwise?)

You don't have the tools to show $S^2$ is not contractible yet.

Yes but I wanted Alessandro to come up with that :)

Alessandro Codenotti

2:42 PM

The intuitive idea is that if there's no path between $x$ and $y$ in the space $X$ then the paths from $x$ to $y$ in $SX$ will need to go through one of the 2 vertices and paths through different vertices won't be homotopic

Balarka Sen

Yeah, eg S(S^0) like Akiva said. But I do wonder if that's literally true. Is the suspension over the topologist's sine curve non simply connected?

I would suspect so but I am not sure.

Alessandro Codenotti

Hm, I'm not sure, the example I had in mind was a disjoint union of circles, connected but not path connected spaces are annoying

Balarka Sen

Ya I'm just trying to come up with stupid examples in algebraic topology

Danu

@BalarkaSen Do you know what the index of a complex $A\to B\to C$ is supposed to mean?

Balarka Sen

No idea. Maybe it means a grading on $A \oplus B \oplus C$?

Danu

2:49 PM

The "obvious" choice is the Fredholm index of $f^*\oplus g$ where $f:A\to B$ and $g:B\to C$

But the only example where I know what both quantities are supposed to be comes out with a sign off...

Balarka Sen

I have no idea.

Danu

Me neither :) Just hoping maybe you'd seen this

Julius

Hi. I have a sequence of real, square, finite matrices $\{A_k\}_{k=1}^{\infty}$ with zeros on the diagonals, and where $A_k$ is $k\times k$. I need a relatively weak condition on $\{A_k\}$ so that $\|(I-A_k)^{-1}\|_{max}\to\infty$ as $k\to\infty$. Any suggestions?

Semiclassical

hi chat

Danu

@MikeMiller So I learned about the half-de Rham complex $0\to\Omega^0(X)\to \Omega^1(X)\to \Omega^2_+(X)\to 0$ where the first map is $d$ and the second $d^+$. I learned that the alternating sum of dimensions of cohomology vector spaces is $b_0-b_1+b_2^+=1/2(\chi+\sigma)$, and that this is $-\operatorname{ind}(d^*\oplus d^+)$. Is that alternating sum what one would call the "index of the complex", or is that supposed to be the actual index of that operator?

Balarka Sen

2:54 PM

@Alessandro Ok, yes, it's true, but my proof involves homology.

$SX$ is simply connected if and only if $X$ is path connected.

pilko

I have a question about methodology. In questions like $\lim\limits_{(x,y)->(0,0)}f(x,y)$ if use the substitution $y=mx$ then I study the limit given the behaviour of $m$. I get correct result, but often get comments like studying along lines is not sufficient (well, I never said m was a constant, but a variable) and anyway never get upvotes while polar substitution for instance get plenty. So do I go against what people learned when using this method ?

Balarka Sen

Do you know the Siefert-van Kampen theorem?

Alessandro Codenotti

I think I have a reasonable example of a space which is connected, not path connected and whose suspension doesn't look simply connected so I'm convinced at least at an intuitive level

Not yet, but we'll cover it in the course

Semiclassical

@pilko Having a limit along lines isn't sufficient to have a limit. See here for an example: math.tamu.edu/~tvogel/gallery/node15.html

Balarka Sen

Interesting, what's your example? You're gonna need that theorem to prove $SX$ is simply connected for path connected $X$ though.

pilko

2:57 PM

@Semi

typical

I just said, m=m(x,y)

Semiclassical

No, you said y=mx.

If you write it like that, people are going to assume m is constant.

pilko

but then I study m->0, m bounded, m-> infty It is clearly not a constant

Semiclassical

Because you've explicitly got $m$ as a variable there.

If somebody tells me they're studying $f(x,y)$ as $(x,y)\to (0,0)$ then I'm naturally going to assume that $m$ is a constant unless otherwise stated.

Balarka Sen

@pilko You never mentioned it's a function of $x$ and $y$ man.

You just said it's a variable. A variable can be anything. It can vary over $\Bbb R$; you still look at lines and not other curves in that case.

pilko

@Balarka I sadi y=mx do you prefer I wrote m=y/x

Balarka Sen

3:00 PM

No I prefer you mention $m$ is a function of $x$ and $y$.

Semiclassical

(In fairness, he never said it wasn't. But, to quote XKCD: Communicating poorly and then acting smug when misunderstood is not cleverness.)

Plus, if you're doing that you've just replaced $y$ with an unknown function of $x,y$

pilko

Will it be better understood if I use another name than m. I don't know u.

Semiclassical

It'd be better if you wrote $y=m(x,y)x$.

Or if that looks bad, "$y=mx$ with $m=m(x,y)$."

pilko

Ok, I'll try in my next answer to such kind of question.

Semiclassical

Keep in mind as well that these are seen as rather low-level problems, so people are not necessarily going to read them in the detail you'd want.

So if you write in such a way that it's easy for them to miss what you're doing, then that's what will happen.

Alessandro Codenotti

3:04 PM

Call $I_n$ the segment joining $(0,0)$ and $(1,1/n)$ in $\Bbb R^2$ and let $X=(\bigcup I_n)\cup([1/2,1]\times\{0\})$, this should be connected but not path connected and I find its suspension not too counterintuitive to think about

Semiclassical

Alternatively, write it as $y=xz$. $z$ is more typically a variable than a parameter, so it forces the reader to stop and check whether it's so.

Balarka Sen

Is that like the infinite broom.

pilko

Thanks for the participation, I now understand where the lack of understanding was.

Semiclassical

(I think it's also the fact that "it's enough to look at lines" is a common enough misconception that people assume that's whats happening unless something forces them to think otherwise.)

Alessandro Codenotti

@BalarkaSen oh, nice, I didn't know it has a name

We saw it in an exercise in class last semester

Balarka Sen

3:07 PM

It's a cool space

Ok, I got an exercise for you. Prove that $S^2$ is simply connected with the tools you have so far.

Hint: I have a sketch of an argument - "take a path in $S^2$, remove a point from it's complement and you have a path in R^2 by stereographic projection. Straightline homotopy there. So any path in $S^2$ can be nullhomotoped"

Semiclassical

@TedShifrin Corollary: Math education is all about the exercises you're given. Math research is all about the exercises you create for yourself.

4

Balarka Sen

But this is a garbage argument. Figure out why and fix it.

BAYMAX

@Semiclassical where you got this corollary >?

Alessandro Codenotti

Uhm, interesting, I'll think about it

Semiclassical

I'm not being entirely serious. But the point is that in education, you typically have some authority who is providing the exercises (be it the teacher or a textbook).

To the extent that math is 'all about the exercises you do', then, being educated in mathematics is 'all about' which exercises you are given.

In research, by contrast, there's not an authority to pick exercises for you. (To the extent that your advisor does so, it's so that you can learn to do it yourself.)

So math research is therefore 'all about' the exercises you create yourself.

Ramanujan

3:16 PM

Hello @BalarkaSen

BAYMAX

ok@Semiclassical

Balarka Sen

Hi

@AlessandroCodenotti I also think the argument generalizes to $S(X)$ if $X$ is a path connected CW complex, but I'll let you verify

Alessandro Codenotti

I will. I need to find out what's wrong with that argument first though!

Balarka Sen

Mhm

Danu

@BalarkaSen More like incomplete than garbage as far as I can tell

Alessandro Codenotti

3:29 PM

No spoilers!

Balarka Sen

@Danu Nah it's garbage man

I don't know why you'd think it's incomplete, but let's not spoil it

Danu

4:10 PM

@MikeMiller Regarding my earlier question, Morgan uses "Euler characteristic" so I guess my question is answered. What remains is: Is "index" a word often used for "Euler characteristic" in this context?

Balarka Sen

I don't know nothing about anything but is this passage relevant?

Little Rookie

Hello all, why cant we simply apply arithmetic on an infinite series the usual way as we do for a finite series?

Danu

Not quite but it's essentially the same thing @BalarkaSen: In this case my operators are acting on the odd cohomology so I get $-1$ times the index equal to the Euler characteristic

Balarka Sen

Ok, I see.

Semiclassical

Because limits of infinite sequences are a lot weirder than finite sequences.

Little Rookie

4:16 PM

In what sense is their behaviour weird?

For instance rearrangement of terms in a series may give different result?

Semiclassical

Sure.

With a finite number of terms, there can be no weirdness. The sum is finite, so changing the order of terms can't modify things.

Any finite sequence always has a last term, and that includes the finite sequence of partial sums corresponding to a sum of finitely many terms.

With an infinite series, though, you're really talking about an infinite sequence of partial sums.

And there's no guarantee that you can make an infinite number of changes to that without changing the result.

Little Rookie

Oh well, i still don't see why spending effort to prove the algebraic limit theorem for series when it is obvious.

BAYMAX

Is Poisson bracket same as Poisson structure ?

Semiclassical

Wikipedia would seem to say yes:

Poisson manifold

A Poisson structure on a smooth manifold M {\displaystyle M} is a Lie bracket { ⋅ , ⋅ } {\displaystyle \{\cdot ,\cdot \}} (called a Poisson bracket in this special case) on the algebra C ∞ ( M ) {\displaystyle {C^{\infty }}(M)} of smooth functions on M {\displaystyle M} , subject to the Leibniz...

BAYMAX

yeah it's written under efinition section @Semiclassical , thanks!

SteamyRoot

4:30 PM

mondaypunday.com/322

Semiclassical

speed of light x 3 and 1/3 * speed of light?

SteamyRoot

You're on the right path

Semiclassical

If there's a pun I'd guess it's based on speed of light = c.

Little Rookie

@Semiclassical is it right that i can factor out a number from a infinite series?
If \sum a_{n} converges, then \sum k\cdot a_{n} = k\cdot \sum a_{n}.

Semiclassical

use $'s around those to put them in math mode. (use the latex in chat link to make it render)

But yes, scalar multiplication is fine.

One way to understand it is to consider the terms of the sum as coefficients of an infinite series (which may or may not converge)

SteamyRoot

4:39 PM

The pun indeed uses "c". The ratio isn't quite 1/3, though.

Random Variable

@DanielFischer An observation that might be useful: In the upper half-plane when $\text{Im} (z)$ is large, we're basically dealing with the function $$\frac{\sqrt{\pi} i e^{3\pi i/4}}{\sqrt{2a}} \frac{e^{({\color{red}{2 \pi - a}})iz}+e^{-iaz}}{z^{3/2}(e^{2 \pi i z}-1)}$$

halirutan

Before I'm going to ask this on main, maybe someone can comment because I'm sure this is a common problem. Let's say I have measured the orientation of cells (eg from an image like this here and the orientations are not wildly distributed but have a main direction. I might end up with an histogram like this

Daniel Fischer

@RandomVariable If that were $e^{(2\pi - a)iz} + e^{iaz}$, then $2\pi$ would be the threshold between a bounded numerator and an exponentially growing one.

halirutan

It's clear that the two peaks belong to the same main orientation, but all statistical descriptions like mean, variance etc will fail for obvious reasons.

How is something like this tackled? Is there some "cyclic statistics"?

(the real problem is that here, I can of course shift the angle so to make a histogram from -pi to pi, but I don't know the main orientation upfront and it don't has to exist at all)

Daniel Fischer

4:55 PM

@Semiclassical It's about $\pi\cdot c$.

Random Variable

5:24 PM

@DanielFischer That was actually my first thought. But I thought it might still be able to explain something. I guess not.

Matti

Why do we have 0 < |x-b| < delta in the epsilon delta definition of limit, but we dont need the 0 < |x-b| part in the definition of continuity?

Little Rookie

Hello all, can i have a hint on how to prove the following statement:
If $\sum a_{n}$ and $\sum b_{n}$ converge absolutely, then $\sum a_{n}+b_{n}$ converges absolutely.

SteamyRoot

Triangle inequality?

Little Rookie

lower and upper bound?

SteamyRoot

Eh?

Not sure what you mean with that (or why you'd want a lower bound)

Little Rookie

5:39 PM

$\sum |a_{n}+b_{n}| \leq \sum |a_{n}| + \sum |b_{n}|$

This?

SteamyRoot

That'll do, yes

Little Rookie

How do i continue from here?

SteamyRoot

$\sum_{n = 1}^N |a_n + b_n| \leq \sum_{n = 1}^N |a_n| + \sum_{n = 1}^N |b_n| \leq \sum_{n = 1}^\infty |a_n| + \sum_{n = 1}^\infty |b_n| $

So what can you say about the sequence $S_N = \sum_{n = 1}^N |a_n + b_n|$ ?

Little Rookie

It is bounded above by some real number?

SteamyRoot

Yup.

Any other properties?

Little Rookie

5:46 PM

I can't think of any =/

SteamyRoot

What can you say about $S_N$ and $S_{N+1}$ ?

Little Rookie

Oh i see

$S_{N}$ in increasing and bounded above. By monotonic convergence theorem, $S_{N}$ is convergent. Hence $\sum |a_{n}+b_{n}|$ is convergent.

Thanks!

SteamyRoot

Indeed :)

Little Rookie

Triangle inequality is everywhere in real analysis!

shredalert

Hello, may I ask a question about textbooks here?

ramanujan_dirac

6:02 PM

Can I take a poll if MSE users prefer tea or coffee?

Little Rookie

I like coffee, but i prefer tea. Healthier choice :)

shredalert

Red tea here.

Little Rookie

The prove for "a Cauchy sequence is convergent" in Stephen Abbott's book is so confusing. :(

Liad

hi

$(X,d)$ is a metric space, my teacher said that $X$ is an open set. i guess it is easy to prove it but im not sure why it is correct. could someone explain?

given $x \in X$ , why there must exists $r \gt 0$ s.t $B_r(x) \subset X$ ?

Alessandro Codenotti

It's a union of open balls, namely $\bigcup\limits_{x\in X} B_r(x)$

Krijn

6:09 PM

@halirutan If you ask this on main, could you send me a link? I'm interested

Liad

@AlessandroCodenotti why? why for each $x\in X$ there is $r >0$ .. ?

Alessandro Codenotti

@Liad how do you define $B_r(x)$?

Liad

$\{y : d(x,y) < r\}$

ramanujan_dirac

Its been a couple of months since I quit coffee and only drink tea, but I miss it so much, as much as I love tea. The good thing about tea is that it comes in a lot of types

Alessandro Codenotti

I'm not quite satisfied by this definition, what's y?

Liad

6:12 PM

$y \in X$

Alessandro Codenotti

Ok, $B_r(x)\subseteq X$ for all $x$ and $r$ by definition

Liad

for all r ?

Alessandro Codenotti

Yep

DHMO

$d(d,A) + \frac f 2 = d(d,B) - \frac f 2$ parce que $d(d,A) + f = d(d,B)$
$d(d,B) - \frac f 2 < d(d,B) - d(d,y)$ parce que $\frac f 2 > d(d,y)$
$d(d,B) - d(d,y) \le d(y,B)$ parce que $d(d,B) \le d(y,B) + d(d,y)$ (triangle equality)

Liad

if $X=[0,1] $ , for $r = 2$ , the ball around x is not in X for all x in X

Alessandro Codenotti

6:14 PM

Sure it is, the ball of radius 2 around 1/2 in $[0,1]$ is $[0,1]$

Because the set of points in $[0,1]$ whose distance from 1/2 is less than 2 is the whole segment

Liad

huh... i had a misunderstanding of the definition :P

thanks!

Alessandro Codenotti

You're welcome

shredalert

I've been studying group theory and am finding modern algebra fascinating. Which books would be suitable for a thorough course in modern algebra if I had to choose between Jacobson's Basic Algebra I&II and Mac lane & Birkhoff's Algebra?

Alessandro Codenotti

What's the ball of radius 1/10 around 3 in $\Bbb N$?

Astyx

Hi chat

Little Rookie

6:18 PM

@ramanujan_dirac Whats the reason for quitting coffee? I still drink coffee occasionally when i feel sleepy in the middle of the night.

@Astyx bonjour

Hi @Astyx

Hello @Astyx

bye chat

Bye @DHMO

6:30 PM

Hi cat

Astyx

Hi @Akiva

Alessandro Codenotti

@AkivaWeinberger Hi DogAteMy

Maks

6:50 PM

Hey, does someone know how matlab gets the column space of a matrix ?

Mahmoud

Hello !

Maks

Hello @Mahmoud

s.harp

maks have you tried a[i,:] ?

(oh the column space, not the columns sorry

Mahmoud

Mathematician Henri Poincaré had an interesting name, which got me thinking, is a point by definition an infinitely small square ? If not, and I was to zoom on it really far, what will I see ?

Maks

Because for example for:
$ \begin{bmatrix}
0 & 2 & 0 & 1\\
1 & 3 & 0 & 1\\
-1 & -1 & 0 & 0 \\
3 & 0 & 3 & 0 \\
2 & 1 & 1 & 0
\end{bmatrix} $

I get that the column space is

\begin{bmatrix}
0 & 2 & 0\\
1 & 3 & 0\\
-1 & -1 & 0 \\
3 & 0 & 3 \\
2 & 1 & 1
\end{bmatrix}

But matlab says its

\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
1 & -1 & 0\\
0 & 0 & 1 \\
-1 & 1 & 1/3
\end{bmatrix}

Which is more useful to me

So I would like to know how did matlab get to that column space

@s.harp no idea?

s.harp

7:00 PM

Nope, if they dropped the last 1/3 then it could have just been Gramm-Schmidt

ramanujan_dirac

@LittleRookie: It disturbs my sleep cycle, and I don't sleep as well, if I drink coffee too late in the day. So, soSo, I don't keep coffee at home or office, cause I get tempted to have the occasional cup.

Balarka Sen

@Alessandro Did you figure out the problem?

Mike Miller

7:22 PM

@Danu Yes. Index and Euler characteristic of an elliptic complex are synonymous. Compare the two-step elliptic complex $0 \to H \to H' \to 0$, where the middle map is an elliptic operator. Then the euler characteristic is clearly the dimension of the kernel minus the dimension of the cokernel.

Alessandro Codenotti

7:33 PM

I didn't think much about it @Balarka, but first of all I need to consider loops rather than paths. Are there continuous and surjective functions $[0,1]\to S^2$? I don't see why not

Balarka Sen

Yes, there are.

Danu

@MikeMiller Thanks for the confirmation: So the index of the complex may be $(-1)$ times the index of the "obvious" operator

Mike Miller

Sure.

Why are you doing ASD stuff?

Danu

ASD? Anti self-dual? I think everything is self-dual in my notes :P

Mike Miller

that's the opposite of what's done in the literature, but sure

i doubt it - you're studying $F_A^+ = 0$, right?

Danu

7:42 PM

Oh, okay... I think it's the same as Morgan's book

Mike Miller

his book on what?

Danu

@MikeMiller I'm studying $F^+_A=\sigma(\Phi,\Phi)$

SW theory

Mike Miller

ah, right, sorry

Danu

Not Donaldson stuff (that's ASD right?)

Mike Miller

i would expect a different complex that you're taking the index of, then...

yes

Danu

7:43 PM

Right

I'm thinking about that complex because the SW theory naturally has a complex associated to it which is like the linearized SW equations and gauge group action, and breaks up this one complex into two decoupled parts, one of which is this half de Rham complex

Alessandro Codenotti

@BalarkaSen ok, so that's a problem, as $S^1$ shows I can't just ignore the surjective loops (my internet is terrible tonight, it keeps going down, I might leave suddenly)

Mike Miller

@Danu Sure.

Balarka Sen

@Alessandro Right. But you can fix it, which is what you should be thinking up next.

No worries, I should probably head to bed too

Alessandro Codenotti

Might be a good idea, I'll think about it and let you know tomorrow what I came up with

Danu

@MikeMiller It's such nice stuff

Mike Miller

7:47 PM

I agree.

Danu

I envy your knowledge of these matters

Mike Miller

Morgan's here right now and I should read a book he wrote before he leaves...

Danu

oh man

even your exposure to awesome mathematicians already makes me jealous :P

Mike Miller

We're running that one conference.

Danu

ah, yes

very nice

Daminark

8:02 PM

Hey everyone!

MickLH

@Daminark ay

Daminark

How's it going?

MickLH

The time has come, to minimize my algebra program, and type $\LaTeX$

Daminark

cringes in Latex

MickLH

It's not that bad, though here I am procrastinating lol

ramanujan_dirac

8:11 PM

@MikeMiller: I have read some of witten's paper on the Euler char/ Witten index, and Seiberg-itten's paper on 4d gauge theory. Is there a good pure math reference for the same?

Akiva Weinberger

$L\!^AT_{\large E}X$

Daminark

$^{\mathbb{LA_TEX}}$

(removed)

Akiva Weinberger

Did you just

MickLH

Idk if you guys give a shit about crypto but I'm calling it how I see it right now. Out with the primes in with the lattices!

Akiva Weinberger

Hashtag hashtag-shapes.

#HashtagShapes

@MickLH Unless you mean a difference type of lattice…?

MickLH

8:19 PM

I actually mean exactly that kind (I think?)

Mike Miller

@ramanujan_dirac I don't know anything about Witten's paper, and probably not the SW paper either. But if you want a mathematical reference for SW theory you can see Morgan's book on it.

I also really like Salamon's big notes.

MickLH

@AkivaWeinberger Most people are researching high dimensional lattices for the purpose, but I've put a lot of time into low dimensional lattices because I believe they can be just as useful for encryption. In fact I'm using exactly a 2D lattice.

The "trick" is that the basis is exponentially large in the security parameter

PVAL-inactive

hI @Mike

Mike Miller

Hi

PVAL-inactive

I'm leaving in 40 minutes for NY

and I'm really sick

Mike Miller

8:23 PM

What's there?

That sucks

PVAL-inactive

a seminar

MickLH

also new york exclusive apple breeds

Mike Miller

you're speaking?

PVAL-inactive

Yah

Mike Miller

cool

I can't find you.

PVAL-inactive

8:26 PM

Binghamton

Mike Miller

got it

Have you told me about this work?

PVAL-inactive

ya

in some sense

Ted Shifrin

Sorry you're sickly, @PVAL. I seem to be, too. Give a good talk regardless! I'm rootin' for ya.

@MikeM: Say hi to John Morgan for me ... Haven't seen him in a while. :)

Mike Miller

OK

Ted Shifrin

I just got an email from Danny R. about a conference at UMass in April. You going to that?

Mike Miller

8:39 PM

Yeah.

Well, applying for funding.

Ted Shifrin

Good. :)

Astyx

Salut @Ted

Ted Shifrin

Salut @Astyx ... @LeGrandDODOM, @Danu, @Hippa et al: Working on my plans for Europe. Looks like I'm going to try to be in Munich and then Paris something like the first week of June.

Astyx

Cool !

Ted Shifrin

Maybe we can arrange an MSE lunch or dinner :)

Astyx

8:43 PM

That would be cool :)

SteamyRoot

:o

Daminark

Hey @Ted!

Ted Shifrin