Mathematics - 2020-06-04 (page 2 of 2)

user462942

5:01 PM

@BalarkaSen Yeah ...

user462942

@TobiasKildetoft Would you recommend someone to go into representation theory?

Thorgott

all is pointless in the end

so having a point is pointless in and of itself

hence being pointless is pointless too

ergo no need to worry

Balarka Sen

I think in an ideal world scientific work should be something people do on the side of their actual job, which would consist of something less fake and runs the society (I believe this is the point of teaching - academicians have to teach because it runs the society, but then again teaching is not as valued as it should be, a pity).

Tobias Kildetoft

If they find it interesting, sure. But just like all branches of math, it is not for everyone

user462942

@TobiasKildetoft I see

user462942

5:10 PM

@Thorgott Yeah :)

user462942

@BalarkaSen I see ...

user462942

I'm supposed to start my PhD this fall, in 3 short months. But the area is Applied Math, and I'm considering not going and instead applying for other PhD programs that will let me work in pure math.

Ted Shifrin

5:55 PM

With regard to the earlier discussion, I can't imagine myself having been happy in any occupation other than teaching/academia, other than possibly owning a restaurant and being a semi-serious chef, but I would not have lasted in that endeavor more than a few months.

Semiclassical

The tricky thing is entering academia/teaching nowadays

just such a different career context

Ted Shifrin

Yeah, who knows how long it'll be before we return to the "intimacy" of in-person teaching, office hours, etc. My experience with my Multivariable Math videos makes me realize that I could easily entertain a flipped classroom if I had the means of recording lectures (with live students). Still not my ideal, but I could see it working.

Semiclassical

sure. i meant even before that, though

more in terms of job market etc

Ted Shifrin

Yeah, it's gotten so much more money/grant driven than even 15 years ago.

skillpatrol

6:10 PM

Just found this:

Ted Shifrin

There have been cartoons of the emperor kneeling on (and killing) the statue of liberty.

skillpatrol

#DeleteFacebookNow

Ted Shifrin

#DeleteTheWorld

Semiclassical

startribune.com/sack-cartoon-i-can-t-breathe/570815132

that was the Minneapolis Star Tribune's editorial cartoonist

skillpatrol

Also, I guess Facebook has been banning people for calling out white supremacists?
Zuckerberg is such a slimeball

Guarding the emperor.

Ted Shifrin

6:18 PM

A few of his high-placed staff have tendered their resignations.

Hawk

so i have a confession. after going through all that yesterday it still isn't clear to me why we needed to do all of that work with that improper integral

Ted Shifrin

@Hawk: It's up to your professor/course what you "need" to do. I was trying to justify doing a calculus polar coordinates integral to "justify" existence of the double improper integral over the plane.

But ask your professor what is expected.

Hawk

its not for a course

Ted Shifrin

What is it for?

Hawk

i just found this randomly on some exam

skillpatrol

6:22 PM

@Semiclassical now they're saying Floyd tested positive for Covid?

Ted Shifrin

Well, if you're going through the trouble of giving a rigorous definition of the improper integral, I expect justification is necessary. If it's on a standard engineering calculus exam, then of course not.

Hawk

i literally found it on some phd exam

Ted Shifrin

Yes, @skill, I heard that on PBS last night.

If it's on a PhD exam, @Hawk, then rigorous justification (such as I was trying to do) is required, for sure.

Thorgott

if it's on a PhD exam, it's a Lebesgue integral

Hawk

so i think you were telling me what must be justified is that the improper integral over disk (polar) is equivalent to the improper integral over rectangles

no they actually asked for a Rieman one lol

Thorgott

6:24 PM

yikes, why would they do that

Semiclassical

@skillpatrol yeah

Hawk

i looked through like 3 books trying to find the definition and the definition was right in wikipedia

its an old exam i think

skillpatrol

Where did you find the exam? @Hawk

Hawk

some phd exam, i can find it if i go through my web browsing history

Thorgott

there're essentially no reasons to do Riemann integration other than didactical ones

Hawk

6:26 PM

honestly most phd exams that want to tes trieman integrals ask about the 1-dim case and want you to tell them why the rational sometimes and irrational sometimes is not integrable

*appears to be the trend

Thorgott

that sounds like a surprisingly easy question for a PhD exam

Hawk

apparently stumping me

skillpatrol

Is it an entrance exam to a Ph.D. program?

Hawk

i think it is (they all are)

Ted Shifrin

Even in my first-year rigorous multivariable exam, I expect rigorous justification when Fubini is used, etc.

skillpatrol

6:31 PM

May I ask why? @TedShifrin

Ted Shifrin

@Hawk: This sounds more like an undergraduate exam than a PhD exam. PhD exams typically are on graduate level coursework.

Some schools give an exam to incoming PhD students that is on undergraduate material (e.g., Berkeley does this). But then they want algebra, analysis, linear algebra, with justifications.

Hawk

"https://www.mun.ca/math/graduate/grad-compexam/CE_-Analysis-_2008.pdf "i found it.

Ted Shifrin

@skill: What's the point of a proof-based course if you don't ask for justifications? Of course, I expect them to do the computations as well. And the justifications are a few points out of 10 or 15.

skillpatrol

I see.

Ted Shifrin

The link doesn't work, @Hawk.

I mean math students need to learn that they need to indicate that they've checked hypotheses (even for the mean value theorem or fundamental theorem of calculus). Otherwise, it's just an engineering course.

Hawk

6:35 PM

one sec let me just upload it on some file dumping site,.

file.io/i0odUBke

skillpatrol

right, makes sense

Hawk

if that doesn't work i m just gonna post the screenshot

skillpatrol

Doesn't work

404

Hawk

imgur.com/a/pq2Ew6K exam has four pages it tests real anal, lesbegue theory, functional analysis and complex analysis.

skillpatrol

Got it thanks :-)

Ted Shifrin

6:45 PM

It's real analysis, so, yes, justification is the point. Everyone can write down the polar coordinates integral. That's part of it, of course.

Without the first part (define the improper integral), I would suspect that no student would think to justify it; but once that definition is there, you have to use it.

Hawk

i know i m going back and forth with this, but why isn't change of variable enough? why do we have to show integral_disk\leq integral_sq, \leq intagral_disk?

Thorgott

Do you know a version of the change of variables theorem for improper Riemann integrals?

Hawk

7:03 PM

i've always thought there was one version of it. idk what analysis book might discuss it

Thorgott

I'm actually not sure whether there is one

Hawk

what is the one you are thinking about?

Thorgott

I'm not thinking about one, I don't know whether there is one

that said, the function in the exam is non-negative, so change of variables will work out just fine in this case

but I wouldn't know how to justify that when only limiting myself to the theory of Riemann integration

Hawk

yeah but ted say there still something to prove.

Thorgott

yes, you have to justify a change of variables if you want to do one

Hawk

7:16 PM

why wouldn't the change of variable hold if we switch it up to improper integral?

Thorgott

what do you mean

Hawk

i mean how would chang of variable possibly change the value of integral?

Thorgott

that's not an argument

Ted Shifrin

I have said this five times. I'll say it one more time. The natural way of doing an improper integral in polar coordinates corresponds to integrating over disks of increasing radii. That is not your definition of the double improper integral in part a. Thus, you must prove that they are equivalent.

user462942

8:27 PM

@TedShifrin I'm considering giving up my PhD slot ...

user462942

That's scary to do, I met the dept. folks for almost a week, just before the pandemic, and I felt so at home with them. But, I think the subject matter isn't something that I feel I can be do and at the same time distinguish myself from the many others.

user462942

Everything else about it is amazing -- the funding, the seminars, the people.

user462942

I don't want to do it for a year and transfer either; that'll be bad karma coming back to bite me later.

skillpatrol

Have you seen this? @Joanna

Thorgott

8:47 PM

is this seriously a 100 reasons list only going up to 98?

skillpatrol

yup

Balarka Sen

Need to finish point set assignment

Grumble

skillpatrol

how many questions?

Balarka Sen

5

Thorgott

another sleepless night for Balarka

Balarka Sen

8:52 PM

It's just hard to write shit you know

Munkres has a godawful definition of perfect normality

With the right definition everything is a tautology

I don't want to write a textbook explaining why two definitions are the same

It just is man just read Engelking

Thats what I will write to my TA

Thorgott

did you know that a local homeomorphism from a pathwise connected Hausdorff space to a simply connected Hausdorff space is a global homeomorphism iff proper

Balarka Sen

Yeah that's because simply connected spaces dont have nontrivial covering spaces

proper local homeomorphism is a covering map

Thorgott

nice

Balarka Sen

This is how you prove the global inverse function theorem, which states that a proper map $f : \Bbb R^n \to \Bbb R^n$ with $Df$ invertible everywhere is a diffeomorphism

I remember we had this conversation a while back so maybe worth bringing it up

Thorgott

ah, makes sense, that statement + local inverse function theorem

Balarka Sen

9:00 PM

yeah

Do you ever plan to look at the fold map stuff

Thorgott

hmm, what's a connected, but not path-connected manifold?

yes, I will get to that eventually

Balarka Sen

None, because manifolds are locally R^n, locally path connected + connected = globally path connected

Strictly_increasing

Could someone please help me answer to the first two questions? The third one is already solved. It's about some passages of proof os Strong Law of Large Numbers

https://math.stackexchange.com/questions/3705396/some-doubts-about-the-proof-of-strong-law-of-large-numbers

Thorgott

but I have to give a seminar talk in a bit over a month, so I'm currently preparing for that

Balarka Sen

Ah OK no worries then

Thorgott

9:06 PM

oh, I hadn't considered that implication before, let's see: local path-connectedness implies the path-connected components are open, but they form a partition, so there's only one path-connected components by connectedness, hence the manifold is path-connected

makes sense

so this immediately generalizes to smooth proper maps from any manifold into a simply connected manifold

any connected*

Balarka Sen

Ya

That proper local diffeomorphisms are covering maps follows from Ehresmann fibration theorem that you proved a while ago by the way

An F-bundle with F a 0-dimensional manifold is a covering map!

Here's a fun exercise: List as many simply connected compact manifolds as possible. Make novel examples by complicating the topology as much as possible

Bonus points if you can keep the dimension low

Thorgott

that's an interesting perspective, but surely there's an honest point-set proof of that fact

Balarka Sen

for homeomorphisms, yeah

Thorgott

uh, empty set for all dimensions, singletons are the only 0-dimensional ones, there are no 1-dimensional ones, the only 2-dimensional one should be a sphere

spheres work in all finite dimensions $\ge2$

also all products of such manifolds are of that type again

Balarka Sen

yup. so you have products of spheres of dimension > 1

thats a good start

nothing topologically novel yet!

Thorgott

9:21 PM

certainly not

but I have to consider 3-manifolds and up now

Balarka Sen

mhm

of course, you're also faced with the immediate question of why your list is not already overcounting; eg why is $S^2 \times S^2 \times S^2$ not homeomorphic to $S^3 \times S^3$? but these are technical problems whose answers are "obviously not" but hard to justify

Ted Shifrin

hard to justify?

you mean with just point-set? yup, impossible.

Balarka Sen

yeah, with the tools Thorgott has

Ted Shifrin

well, a little bit of deRham cohomology won't be bad

Thorgott

just compute homotopy groups
(I have no clue how to compute homotopy groups)

user462942

9:31 PM

@skillpatrol Nope, thanks for it :) I read some of it just now ...

user462942

@Thorgott Are you a professor?

user462942

(I know you mentioned you don't write papers.)

Balarka Sen

For my specific example above you can get away with Euler characteristic; for any finite simplicial complex $X$, define $\chi(X) = V - E + F - \cdots$. This happens to be a topological invariant (not so obvious but common knowledge). $\chi(S^n) = 1 + (-1)^n$ (you can either do the count on the boundary of the $(n+1)$-simplex, a simplicial complex homeomorphic to $S^n$, or just say "it has one vertex and one $n$-dimensional face"). Moreover, $\chi(X \times Y) = \chi(X) \cdot \chi(Y)$ - a simple counting exercise.

But this breaks down for, say, $S^3 \times S^5$ and $S^3 \times S^3 \times S^2$. So shrug

In general, as Ted mentioned, you compute (co)homology groups.

Thorgott

9:53 PM

Is there a super direct argument for $S^1\times S^1$ and $S^2$? Best I can do is one being simply connected, the other not

I'll only learn deRham cohomology in a couple weeks, but won't that depend on the differentiable structure?

@Joanna nah, I'm a mere undergrad

Ted Shifrin

@Thorgott: Yes, but there's a major theorem that says that what you're computing is the same as the (topological) cohomology, which depends just on the continuous structure.

Balarka Sen

@Thorgott Simple connectivity is indeed the easiest argument

Again, of course, you can do an "Euler characteristic philosophy". Suppose you're trying to prove $S^1 \times S^1$ is not diffeomorphic to $S^2$ instead (in 2D and 3D, asking if two manifolds are homeomorphic and asking if they are diffeomorphic happen to be the same questions). Then you simply note that one has a nowhere vanishing vector field, the other does not!

In fact, let's phrase this completely elementarily. Suppose there is a homeomorphism $S^1 \times S^1 \cong S^2$. Let $f$ be a homeomorphism of $S^1 \times S^1$ which rotates the torus by a small angle. This is (1) homotopic to the identity (2) has no fixed points

Thus, we also have a map $f : S^2 \to S^2$ which is (1) homotopic to the identity (2) has no fixed point.

I claim there is no such map

Ok, I guess this isn't entirely elementary. You can show by brute force that a map $f : S^2 \to S^2$ with no fixed points is homotopic to the antipodal map $S^2 \to S^2$, $x \mapsto -x$. But it's not elementary to see the antipodal map is not homotopic to the identity map.

Thorgott

10:13 PM

in which direction do you rotate the torus?

guess that doesn't really matter

Balarka Sen

yeah

Thorgott

wait, the antipodal map is not homotopic to the identity?

Balarka Sen

Nope! For odd dimensional spheres, yes, but not even dimensional spheres

Thorgott

I thought that's what sphere eversion is about

Balarka Sen

Ah yeah that's exactly why sphere eversion is surprising. You can move through R^3 to break that

Thorgott

10:16 PM

oh right, sphere eversion takes place in R^3, that's what I was missing

hmm, are real projective spaces simply connected

Balarka Sen

I nearly forgot why sphere eversion was counterintuitive to the commonplace mathematician

Thorgott

yes in 0 and 1 dimension

I feel it breaks in dim 2

Balarka Sen

yah

well not in dimension 1, RP^1 = S^1

but thats a bad dimension anyway

Thorgott

oops, right

RP^2 is a sphere quotient, so the identification must fuck up the simple connectedness

Balarka Sen

indeed

Thorgott

10:23 PM

in fact, it must be equatorial

by "it" i mean the issue

Balarka Sen

spot on

Thorgott

well, I can only (essentially) think of one loop on the equator

so I walk from one point on the equator to its antipodal point, which is a loop after identification

but this feels contractible

I'm so bad at trying to picture this in my head

shiddingatspeedoflight

11:18 PM

Hello. My question got kinda buried so I thought I would try my chance and ask here. It's pretty basic, it's about finding the range of an unknown from an inequation with 2 unknowns. math.stackexchange.com/questions/3702404/… Thanks for all you guys do!

Thorgott

I don't understand where your inequality is coming from

Semiclassical

to be normalized, you need a/3+b/2=1, so take b=2-2a/3.

then that gives ax^2+(2-2a/3)x, which should be nonnegative on [0,1]. since x>=0, that amounts to ax+(2-2a/3)>=0. that's equivalent to the stated inequation

so there is a reason for it. presumably it's left out because it's only relevant to how the inequation arises, not the inequation itself

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