Mathematics - 2016-12-22 (page 1 of 6)

12:00 AM

But the argument from before does not work. What do you get when you take germs at infinity? The space of germs of homeomorphisms of R^n that preserve 0 and are the diffeomorphism away from 0. Which is, what?

Balarka Sen

Diffeomorphisms of the disk fixing boundary, which can of course be complicated in high dimensions.

Mike Miller

No, that's not what it is. I have no idea what it is.

Balarka Sen

Um, oh, right.

Mike Miller

The fiber sequence is diff(D^n, partial) -> Diff(R^n) -> "DTOP(n)" which is what I'll call that... thing.

And the first term is Diff(S^n)/O(n).

Balarka Sen

Yeah, I was thinking of the fiber

Mike Miller

12:04 AM

O(n+1)

If you want to figure out what DTOP is I'd be happy

Null

$A\cup \emptyset=A$ right?

Yeah

or if akiva wants to

What?

I pass to Akiva

12:07 AM

I'm going to watch the finale of season 5 of Sabrina the teenage witch

Akiva Weinberger

Uh, sure, what's DTOP?

or TOP? I saw you mentioning it earlier

@MikeMiller Spoiler alert, she rescues Matt Damon

Balarka Sen

I'm going to read thermodynamics

or sleep, both seems appealing

Kaj Hansen

and yet mutually exclusive

such a shame

Akiva Weinberger

Read the first exercise and then sleep on it

Balarka Sen

who does exercises in physics?? not me

Proven

12:16 AM

Question, what does $(\mathbb{Z}/n\mathbb{Z})^{\times}$ mean notation-wise? I know that $\mathbb{Z}/n\mathbb{Z}$ are the integers modulo $n$, but I'm not sure what the $\times$ notation is supposed to denote

Balarka Sen

"you have three pulleys inside a roller-coaster ride which are in turning falling from outer space with 981 cm/s acceleration, each of them carrying a spring of variable length. suppose they are connected parallely. find the spring force constant"

and all kinds of extremely unrealistic situations

as a bonus, after you do the math, you find that's not at all what the problem meant. the roller-coaster ride was moving in 981 cm/s^2 acceleration, not the whole system... great

Mike Miller

@Akiva The space of germs of homeomorphism of R^n preserving 0 such that it's a diffeomorphisk away from 0

@Balarka Do you not like the spectral sequences?

Akiva Weinberger

@Proven The ones with inverses. $(\Bbb Z/6\Bbb Z)^\times=\{\bar1,\bar5\}$, for example

because $\bar0,\bar2,\bar3,\bar4$ all don't have inverses

@Proven $(\Bbb Z/n\Bbb Z)^\times$ is a group under multiplication (but not addition)

Perturbative

@BalarkaSen, You were talking about Manifolds earlier, surely you would be more interested in stuff like General Relativity than Hooke's Law, no?

Akiva Weinberger

@MikeMiller Don't even know what germs are

Balarka Sen

12:26 AM

@MikeMiller I do, but I can't get much done this week

Have lots to read

Mike Miller

@Akiva What are they teaching you in health?

Balarka Sen

@Perturbative Nah, I like manifolds for their own sake, not because of general relativity.

Mike Miller

@Balarka Ok. I was going to say I can offer other things to work on if you don't. But if you do you do. I thought you were done with school for a while?

Balarka Sen

School's off but there's still a bunch of schoolwork. I can hear what you suggest though, spectral sequences was just for a refresher in algebraic topology and I am not going to read much more than Serre spectral sequences anyway

Abcd

Does Kink of the Graph have to follow the Scale?

Zach Hauk

12:32 AM

Hi

hi chat

Hi @ZachHauk

Hey @Abcd

@BalarkaSen Foliations are near and dear to my heart, and you have the background. I don't know that much about them, so at a point we'd be caught up and learning at the same time.

Semiclassical

@BalarkaSen "who's got two thumbs and does physics exercises? this guy!"

Null

12:33 AM

when working with matrices, can we shift whole lines up and down?

Balarka Sen

:P

Zach Hauk

@Null what?

Null

and, can we shift columns left and right?

Abcd

Does anybody know the answer the answer to my question?

Zach Hauk

@Null you can... but it will change the matrix...

Null

12:34 AM

@ZachHauk to get row echolon form

Zach Hauk

oh yeah thats fine

Semiclassical

@balarka got into a debate about a physics picture lately, and with the prof for the course I was TAing for.

Mike Miller

@AkivaWeinberger Looks like you were right that it's metrizable, by the way. Oops.

Semiclassical

he thought a certain approach in the book didn't make sense, and I'm quite convinced it does.

Balarka Sen

@MikeMiller Great idea. Earlier the differential geometry I started studying with you was too hard for me because it takes a lot of work for me to think in terms of symbols, so I hope there's more topology element in this.

Abcd

12:35 AM

Does Kink of the Graph have to follow the Scale?

Akiva Weinberger

@MikeMiller Looking at Brian's solution? I think it's subtly different

Semiclassical

The stuff here, basically: phet.colorado.edu/en/simulation/legacy/radiating-charge

Perturbative

@Null When you say "shift whole lines up and down", I think you're referring to the elementary row operations

Mike Miller

@AkivaWeinberger Maybe so

Semiclassical

namely, he doesn't think that the picture it generates for the electric field of a moving charge is right.

Mike Miller

12:36 AM

@BalarkaSen Yes. You get over the symbols in differential geometry eventually; I did. (I hate symbol-pushing exercises.) But you only get over that by pushing symbols until you understand the symbols.

Null

@Perturbative yep, but i don't think the same can be made with columns, as i tested with a very small matrix

Semiclassical

(I link that both for sake of specificity -and- because that applet is fun to play around with.)

Balarka Sen

@Semiclassical Neat!

Mike Miller

Connections make sense to me, so do their curvatures, etc. But I spent days staring at stuff in books when I took my first course with my advisor.

@BalarkaSen I can throw some things at you to start with. Do you know what a foliation is?

Akiva Weinberger

@MikeMiller In the same way that the wedge of countably many circles is subtely different from Hawaiian rings (or a noncompact version thereof)

Abcd

12:37 AM

BYE!

Semiclassical

Connections bother me because I feel like I should know what they are from physics, and I don't.

Abcd

nobody is ready to answer here!

Mike Miller

@AkivaWeinberger Ah, sure

@Semiclassical Aren't they just gauge fields?

Semiclassical

Not sure what you mean by that.

You mean, like a local gauge change?

Mike Miller

I don't know what I mean by it either. I'm playing pattern recognition.

Balarka Sen

12:39 AM

@MikeMiller Fair enough, I agree. It took me a lot of work even to push through Ted's stuff (which I am reasonably comfortable with now). Nope, I don't.

Semiclassical

I think you're right, in any case. As an example

Mike Miller

Maybe I want to say it's a field, the curvature is the field strength tensor, and a local gauge change changes the field by $dg$ (roughly).

Semiclassical

In the Schrodinger equation, say, I've typically got $-\frac{1}{2m}\frac{d^2}{dx^2}\psi+V\psi=E\psi$

Mike Miller

@BalarkaSen I need a few minutes

Balarka Sen

sure

@Semiclassical I made the test charge do some violent gymnastics and now I'm sick

Semiclassical

12:44 AM

If for some reason I make the redefinition $\psi(x)\to e^{i\lambda(x)}\psi(x)$, I can recover something that looks mostly like the Schrodinger equation: $-\frac{1}{2m}\left(\frac{d}{dx}+i\lambda\right)^2\psi(x)+V(x)\psi(x)=E\psi(x)$

(not confident I've done the algebra right on that. working somewhat from memory)

Mike Miller

@BalarkaSen A foliation is a decomposition of a smooth manifold into smooth submanifolds (not necessarily embedded) so that locally, the partition looks like the partition of $\Bbb R^n$ into parallel $\Bbb R^k$s.

Note that in this decomposition, there's no reason to believe that one of the smooth submanifolds we started with corresponds to a single $\Bbb R^k$ - it might even correspond to eg $\Bbb Q \times \Bbb R^{n-1}$

but each path component in the chart is one of the $\Bbb R^k$s

The standard example is the decomposition of $T^2$ into irrational lines of the same slope.

Balarka Sen

Ah, alright.

Totally makes sense

Mike Miller

So another way of thinking about this. When we do this, it gives us a rank $k$ subbundle of $TM$ (a "distribution"), agreed?

Semiclassical

Ignore my physics rambling for now, I don't have it right atm

Balarka Sen

@MikeMiller By taking the tangent space of the submanifolds we are decomposing into at each point?

Semiclassical

12:50 AM

It amounts to saying that one augments the standard spatial derivative $\frac{d}{dx}$ to $D=\frac{d}{dx}+\lambda(x)$

Which does sound like a connection-ish thing.

I don't remember the details, though.

Mike Miller

@BalarkaSen Correct.

Balarka Sen

Do these immersed (?) submanifolds need to be homeomorphic?

Mike Miller

Nah. Rarely are.

Balarka Sen

Alrighty. The mental picture of parallelism is a bit messed then but go ahead.

Mike Miller

For instance, take $[0,1] \times \Bbb R$, the left side and the right side are submanifolds, and the submanifolds in the middle look like parabolas pointing downwards, stacked on top of each other. Does this picture make sense so far?

Balarka Sen

12:56 AM

Ohh yeah you posted an image of something like that a long while ago. Reeb foliation IIRC

Yep, makes sense

Mike Miller

Now quotient by $\Bbb Z$ vertically.

Akiva Weinberger

Where's a @TedShifrin problem when you need one

Mike Miller

Then you've got two circles and a bunch of lines in your foliation of $S^1 \times [0,1]$. If you want this on a closed manifold, double that to get a foliation of the torus with two circles and a bunch of lines.

Balarka Sen

Done.

Agreed

Mike Miller

Does the parallelism make sense now?

Balarka Sen

12:58 AM

Yeah, much better.

So I suppose the fibers of the Hopf fibration also gives you a foliation? Actually no there are degenerate cases at the two "ends". Wrong dimensions.

Mike Miller

No, that's exactly right!

Balarka Sen

Oh, I was think of the "foliation" by torii of increasing diameter, which is not really a foliation (well, it's a foliation of the complement of the Hopf link). Yeah, with circles that gives one.

Mike Miller

Now thinking of this distribution, let's just think about distributions in general. We say a distribution is integrable if locally it's equivalent to, well, standard thing on $\Bbb R^n$. Equivalently, that at every point, there's a submanifold $M$ passing through that point and everywhere tangent to the distribution.

Let's suppose it's codimension 1, so that locally it's the kernel of some nowhere zero 1-form $\lambda$. OK?

Balarka Sen

A'right, so every distribution coming from foliations is integrable (but probably not vice versa). Ok.

Mike Miller

No, vice versa is true. Integrable literally means "Gives a foliation".

Zach Hauk

1:10 AM

@AkivaWeinberger can you give me a Ted problem?

Mike Miller

The actual theorem (Frobenius) is that $\lambda \wedge d\lambda = 0$ is equivalent to integrable.

Balarka Sen

Ah, right, given a point $p$, just draw the subamanifold passing through it tangent to the distribution and then take their union for all $p$'s (some of them might coincide but w/e). It should just be that simple.

@MikeMiller Hmm.

Akiva Weinberger

@ZachHauk Unfortunately I don't happen to be Ted.

Zach Hauk

@AkivaWeinberger one that he gave you?

Mike Miller

I wouldn't ask you to prove that. You can find it in any standard smooth manifolds book.

Balarka Sen

1:13 AM

Sure, I can believe it.

Akiva Weinberger

@ZachHauk Do you know the formal definition of curvature of a plane curve?

Zach Hauk

@AkivaWeinberger unfortunately no

Mike Miller

The 1D version is that vector fields have flowlines.

Akiva Weinberger

Alternatively I think there was a sequence of analysis questions he gave me a while ago

Zach Hauk

i haven't studied much analysis but go ahead

Balarka Sen

1:15 AM

Right, got it

Akiva Weinberger

Dec 4 at 23:25, by Ted Shifrin

Suppose $\lim\limits_{n\to\infty}n\left(f(1/n)-f(0)\right) = 1$ and $f$ is continuous at $0$. Can you conclude that the right-hand derivative $f'_+(0)=1$?

@ZachHauk

Mike Miller

@BalarkaSen I'm not super sure where to start with this, I haven't given a foliations lecture in quite some time. There are many things one can study. One can study the set of closed leaves; the dynamics of where leaves accumulate; whether there are any closed foliated subsets other than individual leaves or the whole thing... Whether manifolds can support certain kinds of foliations, or foliations at all.

Balarka Sen

Do any $n$-manifold support a foliation by $k$-dimensional submanifolds, $k < n$?

Mike Miller

No. Set $k = 1$. You're asking for the existence of a nonzero vector field.

For $k = n-1$, taking normal spaces you're asking for a line field as well. It's due to Thurston that you support a codimension 1 foliation iff you have $\chi(M) = 0$.

Balarka Sen

Yeah, should have figured that.

Adeek

1:21 AM

0

Q: intuition regarding monomorphic functions

Instead of talking about the categorical way of defining monomorphism. I wanted to see if I have the correct intuition regarding monomorphic function. A function $f : A \rightarrow B$ is said to be monomorphic if for all sets $Z$ and for all functions such that $\alpha_1,\alpha_2 : Z \rightarrow ...

category-theory

Mike Miller

I can throw a question at you I liked working out (but I suspect at this point you're not going to be able to find the full answer to). Suppose $M$ is foliated by circles. Is the foliation the fibers of some fiber bundle?

Or I can say things if you have questions. Basically I suggest picking up Foliations I, Candel and Conlon.

Easy to find a pdf, very accessible.

Balarka Sen

Great, thanks for everything. I'm going to get it downloaded.

Mike Miller

I never did the exercises so if you work through those it'll give me an excuse t.

Balarka Sen

Also thinking about your problem - I'll see if I can get anything.

Hah

Adeek

how do you guys get those hats ?

Mike Miller

1:23 AM

I just got it. I never looked up why.

Balarka Sen

with hard work and dedication

Adeek

lol

GFauxPas

I got mine by biting into and swallowing a piece of a user with 1 rep

Balarka Sen

crap, bad download. damn ya libgen

Adeek

@BalarkaSen I have 100 mbs internet at my home :D

I could download all the books that I want

hehe

Balarka Sen

1:30 AM

I have a decent internet today so I think it's libgen who's to blame this time

Adeek

I have to download some book about quadratic forms when I go home..

I will judge your statement when I go home :P

Balarka Sen

Ok, got a better copy downloaded. The previous file was probably corrupted.

Adeek

I see

Ramanujan

@MikeMiller you got that hat for using search functionality for 3 consecutive days

Mike Miller

By the way, @BalarkaSen, note that in the example I gave all bundles have trivial total space, so you can pick a nontrivial bundle and the trivial bundle and you can't do any tricks about automorphisms of the base.

Null

1:42 AM

we got now a test exam from our prof. By the looks it's way easier than I imagined. Altho I have to learn the topics better, I feel relieve.

Balarka Sen

@MikeMiller True.

PVAL-inactive

@MikeMiller Is this book supposed to better than Calegari's book?

Do they actually prove all of Novikov's results?

Sophie

I have a set of 6 linear equations in 4 variables, how do I know if there's a solution?

Balarka Sen

I'm a bit sleep-deprived now but you mean that Milnor's theorem says more strongly that the map [X, O] --> [X, TOP] is zero instead of just not being injective, yep? So every bundle (of appropriate rank) has total space homeomorphic to the trivial one.

DHMO

@Sophie use wolframalpha

Balarka Sen

1:51 AM

Just to check.

Mike Miller

@PVAL-inactive I think they do. But that's in the sequel.

I never looked at Calegari's but CC is very friendly.

Sophie

I'm trying to find out if $\{a+b\sqrt{2+\sqrt{2}}+c\sqrt{2-\sqrt{2}}+d\sqrt{2}|a,b,c,d\in\mathbb{Q}\}$ is a field

PVAL-inactive

I can't find the sequel easily

DHMO

@Sophie well that's easy, just multiply (a,b,c,d) by (e,f,g,h) and see if it is still there

Mike Miller

It's just Foliations II. I have a pdf probably from libgen.

Sophie

1:55 AM

@DHMO doesn't guarantee the existence of inverses

PVAL-inactive

@Mike I found it, I was having trouble on libgen

DHMO

@Sophie as I said, just use wolframalpha

Balarka Sen

libgen's being bad today i guess

Null

@Sophie why not check the axioms?

DHMO

@Sophie Is it even closed under multiplication? I am having trouble expressing $\sqrt{2+\sqrt2}\sqrt2$ in terms of (a,b,c,d)

Sophie

2:00 AM

I am checking the axioms, but I'm stupid, that's why I was trying to find inverses before noticing it isn't closed under multiplication

PVAL-inactive

@MikeMiller I've been rather frustrated looking for modern expositions of Novikov's work. Calegari seems to give up on some of the harder ones (e.g. the consequence in my answer to your older question: a closed leaf of a taut foliation is $\pi_1$-injective).

Mike Miller

I believe that's in there.

Null

@DHMO i think noone bests you at axioms hehe

DHMO

@Null lol thanks

Mike Miller

@PVAL-inactive He does.

PVAL-inactive

2:05 AM

@MikeMiller Ya part II seems like it has all of that stuff.

As well as expositions of Gabai's sutured manifold hiearchy.

Mike Miller

I never got around to reading it. I want to but don't have time.

PVAL-inactive

I might have to read this.

Mike Miller

Expositions but hardly proofs.

Do you need to do it soon? If you wait I can join.

I need to hurry up on writing.

PVAL-inactive

@MikeMiller I'm on writing break.

@MikeMiller It claims proofs of Gabai's main results.

Mike Miller

Wow.

PVAL-inactive

2:08 AM

I can't tell if theres big black boxes they are using

Mike Miller

I want to be done by March.

Adeek

1

A: intuition regarding monomorphic functions

Injectivity is good as a first intuition but you want to keep in mind that morphisms in a general category need not be functions. So I don't know if there is a "non-algebraic way" to think about them. You should really think of monomorphisms as morphisms that can be "left-canceled". In particul...

Cool question

PVAL-inactive

@Mike I lied, "the details of this construction are too complex to be treated here".

Mike Miller

That's more reasonable. I'd like to read his papers eventually.

PVAL-inactive

They're beyond dense.

You should read his summary of the construction.

Mike Miller

2:13 AM

I was next to him in Houston. I should have just asked him.

PVAL-inactive

@MikeMiller My only interaction with him, was handing him his packet at a conference in UT. I remember that he insisted that he was given a receipt for the $20 he had to put down for the dinner at that conference.

I think I have been at a few conference with a nametag that he never picked up.

Mike Miller

I'm not gonna leave that message.

Balarka Sen

er

GFauxPas

why do people still use $*$ for ordinary multiplication on math.se in contexts where convolution would make sense

DHMO

@GFauxPas because math.se is math underflow?

Kaj Hansen

2:21 AM

heh, I could answer that I think

GFauxPas

what does that mean

Kaj Hansen

And I haven't taken algebraic geo

DHMO

Balarka Sen

also strange to see schemes being mentioned below in the comments

GFauxPas

oh this is like math.se for hard questions?

DHMO

2:22 AM

ya

GFauxPas

yeah I don't understand these things overflow is talking about XD

oh this is for research

Null

@KajHansen hi, how is it?

GFauxPas

i'll keep in in mind when I start writing papers

Kaj Hansen

Not too bad @Null

Balarka Sen

@KajHansen you should do algebraic geometry, algebraist.

DHMO

2:23 AM

@GFauxPas it does contain some gold though

see this and this and this

Kaj Hansen

I'd write the sequel to Ted's book: Geometry: The Algebraic Approach

Balarka Sen

I don't think Ted would take it in a good way

But seriously, you should get into that stuff

Kaj Hansen

I've joked about that for years now, lol. He's heard it in person

I have book(s) I could use

Balarka Sen

I should have known

Kaj Hansen

I guess I have enough background to get started

Balarka Sen

2:26 AM

You totally do!

I can recommend you something if you want

GFauxPas

thanks DHMO i'll check out th ebig lists

Ramanujan

Stackexchenge was created on 2010?

Terry Tao, Los Angeles

45.6k 15 202 290

nt.number-theory big-list soft-question ca.analysis-and-odes analytic-number-theory co.combinatorics fourier-analysis

Kaj Hansen

Sounds about right

Null

@KajHansen how would this work? Proof Theory: An Empirical Approach

Ramanujan

He is member from 7 years and 2 months

Kaj Hansen

2:34 AM

heh, sounds like a constructivist text

Null

@Ramanujan that's only 1 year give or take. Classic example of dataset noise.

DHMO

2:51 AM

217

A: Examples of common false beliefs in mathematics

The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space. In a somewhat related spirit: the boundary of a subset of (say) Euclidean space has empty interior, and furthermore has Lebesgue measure zero. (This false belief is closely relat...

What is "bullet-ridden squares"?

Balarka Sen

@MikeMiller Isn't exercise 1.1.3 in C-C precisely the Ehresmann's theorem?? "If $\partial M = \emptyset = \partial B$ and $B$ is connected prove that a submersion $f : M \to B$ with compact level sets is a fiber bundle" O_o

Mike Miller

What did they give you before then?

Semiclassical

@DHMO Should be bullet-riddled

Balarka Sen

Intuitive understanding of the foliation, definition of a fiber bundle, example of a foliation which is not a fiber bundle.

DHMO

@Semiclassical thanks

Semiclassical

2:55 AM

He includes it as an exercise in his notes on measure theory here: terrytao.wordpress.com/2010/09/04/…

Balarka Sen

It's the first exercise in the book

Mike Miller

lol

Balarka Sen

oh noes

DHMO

$[0..1]^2\setminus\Bbb Q^2$

Mike Miller

Whressman isn't that hard anyway.

Semiclassical

2:56 AM

Namely: "Show that the bullet-riddled square $ [0,1]^2 \setminus {\bf Q}^2$, and set of bullets $[0,1]^2 \cap {\bf Q}^2$, both have inner Jordan measure zero and outer Jordan measure one. In particular, both sets are not Jordan measurable."

Balarka Sen

Alright. I've never tried to prove it; I want to ponder on it for a while. Maybe I'll sleep on it.

Semiclassical

(I have no idea why that's true, btw. I'm just using my google-fu)

GFauxPas

Isn't the latter just the unit square?

Balarka Sen

G'night

DHMO

@GFauxPas I don't think so

By the way, should we include this definition of $e$ in proofwiki?

13 hours ago, by Akiva Weinberger