@TedShifrin: Let me tell you the story after I've figured it out for section. I'm a little confused about it right now.

Also, I'm striking this from the lesson plan, because they would have to be able to tell me the cohomology of $S^n$....

I give up. I'll just answer whatever questions they have.

Sure thing, @MikeM. I wasn't pressuring you. I should have seen this, but it doesn't ring a bell.

You could easily show them Mayer-Vietoris for deRham, but I suspect that comes next semester?

Right.

I think that's one of the coolest applications of partitions of unity.

They have so few tools right now. The first 2/3 of this quarter is setting up the basic definitions of manifolds.

Right.

But you could do some interesting transversality applications.

I think they really should spend the first part of the course just doing the topology from G&P... I think some of the basic stuff is just very boring to a beginner.

I can go back to that now and look at the fundamentals they're doing and give reasons it's interesting, but not so when I was learning it...

I am very fond of the diff top generalization of the argument principle.

The fun applications I've done are Brouwer & Jordan.

G&P sort of sketch that in exercises, but I do it in class, because it's so neat and so important.

Reidemeister's theorem is fun!!!

@TedShifrin Well-defined as in it doesn't send $g$ to multiple things?

The latter in the form of: if $\text{Hom}(\pi_1(M),\Bbb Z/2) = 0$, then every hypersurface separates. the converse is true too but they don't have the tools to prove it.

If $f\colon M\to \Bbb R^n$, $M$ a compact oriented $n$-manifold with boundary, and $0$ is a regular value (and nowhere $0$ on the boundary), then the degree of $f/|f|$ counts the preimages of $0$ with signs.

Ah, I gave you the question Lawson gave me, I think. Prove that a compact hypersurface in a simply connected manifold is always orientable.

That stuff in G&P was really cool indeed @MikeMiller---thanks again for the recommendation.

Sure, that follows from tubular neighborhoods and intersection numbers.

@Danu: Basically, yes. You defined it in terms of $\phi(gH)$. But you could have $gH=g'H$.

There

Other options: an introduction to Morse theory and a sketch of the main theorem (handlebody decomposition), or possibly, since they know now that I can write any noncompact manifold as $M_1 \subset M_2 \subset \dots$ where $M_i$ are all compact, I can give some elementary theorems about noncompact manifolds.

They all support nonvanishing vector fields eg.

@MikeM: Lawson's proof never used tubular neighborhoods, just posited an orientation-reversing loop.

There's a proof Georges gave on this site that's pretty nice Ted of that for $\Bbb R^n$. I think it extends for simply-connected things.

You're ignoring my argument principle thing. It's used in the Hopf degree theorem proof, but I think it is amazingly cool.

@TedShifrin: I just hadn't read it yet, I'm not ignoring it.

LOL

Nice result.

Pontryagin-Thom gets thrown at students like this a lot doesnt it?

what does like this mean?

georges has a proof somewhere of a much more exciting theorem (IMO): Closed hypersurfaces in $\Bbb R^n$ are globally level sets.

It was thrown at me in my oral geometry/topology qual at Berkeley in 1975, @PVAL, but I didn't know it. So Kirby led me through it :)

First year diff top students.

I'm not going to do that even though I like PT.

It's too sophisticated for where they are now.

Right

Hmm, I'm not sure I know how to prove that level set result, without using orientability.

He had some sheafy proof. I also need orientability... and tubes.

Well, in some sense, it's a Cech-cohomology statement.

@TedShifrin But if $gH=g'H$ then $g\phi(eH)=g'\phi(eH)$---I guess I don't see what you're trying to say.

So that could be sheafified.

I'm saying that you should use $G$-equivariance clearly in your argument, @Danu, to say why $\bar\phi$ is well-defined. But there isn't much to it.

@MikeM: It should just be a partition of unity argument, once you can choose all the gradients pointing compatibly "outward."

@TedShifrin How do you specialize this to simply-connected manifolds?

You mean generalize? :)

It's very obviously false if you don't assume simply-connected.

Right. Projective space will do.

So you're saying I should say the following explicitly: Suppose $g\in \hat\phi^{-1}(aK)$ and $g\in \hat\phi^{-1}(bK)$ for $a\neq b$. That's impossible because it implies $g\phi(eH)=g\alpha K$ to be equal to both $aK$ and $bK$, but $\phi$ is well-defined by assumption so this cannot happen.

Or am I still not understanding you?

Oh, you are giving the argument for $\Bbb R^n$ I guess.

Hmm, now I'm confused, @Danu. If you set $\hat\phi(g) = \phi(gH)$, this is well-defined. Why are we using equivariance?

No, @PVAL, I'm not giving any argument.

fine

But if you have an orientation-reversing loop, push it off the submanifold smoothly to make a smooth loop in the ambient manifold.

Spent the last hour grappling with notation. I'm pretty sure this is how I spend most of my time.

We all do that, @MikeM.

Sometimes poor notation makes it worse.

@Danu: Very simply, $\bar\phi = \phi\circ\pi$. So what's the big deal?

Because $\phi(gH)=g\phi(eH)$ and hence every $g$ is nicely associated to some element of $G/K$... I'm still not seeing your problem :P

But you don't need to say that.

Oh, I see. You're claiming that equivariance makes continuity evident. Got it.

You win. I was dumb.

LOL, I'm confused by your reactions :P

Well, as I just wrote up there, $\bar\phi$ is automatically defined by composition, given that $\phi$ exists. But you hadn't explained clearly to me that continuity of $\bar\phi$ would be evident from that equivariance formula.

Partly, I was distracted by three conversations at once.

@TedShifrin: An author was writing a formula in coordinates but did not make it clear they were working in coordinates, so I was trying to figure out how some of their terms were defined.

They turned out not to make sense globally.

Ah @MikeM.

@PVAL: I forgot to ping you with the argument 8 bars up.

@TedS Oh right I wrote up a proof of this recently... (should work for $H_1(M,Z_2)=0$) math.stackexchange.com/questions/1486105/… @MikeM You probably don't want to present that one though...

Oh, never mind.

The global form makes the desired result way more obvious so I don't really know why they decided to work locally.

The proof I wrote is not elementary by any stretch.

But the intersection theory proof is easy, @PVAL, @MikeM. The only thing to justify is smoothingly joining up the two ends of the curve.

They know basic smoothing results, so nothing to justify. I love that proof.

I thought that was so cool when Lawson gave me that problem in coffee hour one day.

Is the argument not somewhere in GP?

Nope. Nowhere.

Huh. I definitely learned it during qual prep.

Matic assigned it when she taught G&P, because she was assigning some of my homework problems. She tried helping the students with all sorts of handlebody arguments to do it. The students came to me :P

The people who came to my discussion section learned it from me.

Handlebodies?? What for?

You might have learned it because Jacob assigned it, because I assigned it to him :P

She had some sort of argument that was way off the deep end. I don't know what it was.

What I should have emphasized to them was to pay attention to how it works in the case of $\Bbb R^2$.

I still dont understand the argument.

@PVAL: We're talking about separations, I think.

Nope.

Oh. I was.

Simple connectivity says that pushed-off curve bounds. So the boundary theorem says the intersection number with the hypersurface is 0. But it is clearly 1.

Oh, i see what you're proving.

I love that argument.

Looking at the comments on the right, I wish there was a popcorn.SE.

@PVAL: Are we ok?

Ya got it.

:)

My goal for today is to get past page 2 of this paper.

Though now I feel like I am using the fact that an orientable vector bundle restricted to $S^1$ is trivial.

@PVAL: That's how I tend to think about it too.

Well, an orientable line bundle on anything is equivalent to its being trivial. :)

That's easy to prove by hand though.

Am I really using that?

I am confuzled.

I fiddled with the AC controls in my office because I noticed them for the first time. Now it's permanently 60 degrees.

That's way too cold.

You will be sick.

You misunderstand. The AC is now stuck there.

I don't consider this desirable.

:(

@TedShifrin No I take it back, the one-dimensional case is enough. For some reason I was thinking about the entire tangent bundle restricted to your loop.

Ah, you can't readjust? You may need to tell someone in the math office to call for help.

I did, they're fixing it tomorrow. I learned my lesson.

No, just the normal bundle, @PVAL.

Ya I see.

Ah @MikeM. We weren't allowed to fiddle, Mike, because thermostats in one office affected a fourth of the offices on the floor.

Hahaha

Does any of you have interest in "physics-related geometry" topics such as e.g. spin geometry, gauge theory, etc?

Mike and PVAL know a lot of that.

I once knew a little.

That's my job, until I get fired.

So melodrama goes up to make up for the lack of drama (for a change)? :)

I come from a physics background and (perhaps as a consequence) am interested in these things; I'd like to get some serious mathematical background on the relevant topics.

I know a littlle about the topology of 4-manifolds that admit physics related structures (mainly symplectic and Stein).

Hence, I've been studying basic differential geometry and am now learning some topology

That probably requires some serious work.

I never thought Stein manifolds had anything remotely to do with physics, @PVAL.

I've been wondering about the algebra component.

Or at least, it has for me.

I've never heard Stein manifolds either, in the context of physics :P

@TedShifrin Well they are somewhat nice symplectic manifolds.

They're complex manifolds that embed holomorphically in $\Bbb C^n$, @Danu.

@MikeMiller Of course; This is a plan with timescale on the order of 2-5 years.

Well, any complex manifold is a symplectic manifold, @PVAL :D

@TedShifrin Hello!

Most of the algebra I've ever seen in this context is algebraic topology. Eg, one wants to compute the topology of the gauge group and its classifying space.

heya @Anthony :)

I'm just wondering whether you could tell me something about recommended reading material.

@TedShifrin Nope, that's not true.

@MikeMiller Yo

@MikeMiller Right. I'm now taking a 1-year course on alg. top.

Oh, Kähler, of course, @PVAL.

@TedShifrin I got completely destroyed by the math GRE, I think I made a bubbling mistake because the score I got is a low lower than I think I could have gotten just from mistakes...

The good news is that I think most of the programs I'm gonna apply to are CS

Whoa, @Anthony. I warned you you should have allowed yourself two takes.

I don't know what you want to learn so I can't give very good advice.

But the programs you're interested in may not need the math GRE because they're cross-disciplinary.

@Anthony: Hey.

@TedShifrin Yeah, I missed to the deadline for the first.

And yeah, that too.

hi

hi Karim

Also, did you hear Ribet got the AMS presidency?

I haven't heard how all the kids back in GA did on the GRE, although I've asked, @Anthony.

:D

@TedShifrin I don't understand the following statement in the book. If R is not commutative, however, the set {ras | r,s \in R} is not necessarily the two sided ideal generated by a, since it need not to be closed under additiion.

yes, he posted it on Facebook, as did other math people. I already congratulated him.

@MikeMiller I'm not very clear on it either---yet.

@TedShifrin You're also gonna want to put a properly in front of that holomorphicly.

Yes, @PVAL, I knew that. I was trying to give the idea to poor Danu.

@MikeMiller How's life?

how is it not closed ?

here is my understand ted

However, I think reading at least one book on algebra would probably be a good idea, no?

@L33ter why would it be closed?

Well, Karim: How do you add $ras+r'as'$?

@Anthony The GRE's are lame :\

isn't it all finite sums

hi @anon

of elements of ras

hi

?

no, all finite sums is the ideal generated.

You gave just a set of elements.

@Danu Yeah, but also I probably should have done better :P

no, that would be $\{\sum_i r_i a s_i\}$ or $\langle ras:r,s\in R\rangle$, not $\{ras:r,s\in R\}$

oh I thought here they meant the finite sums of those specific elements of the set

Set bracket notation is set bracket notation, Karim!

Anyway I should probably get back to work. I'll be back on later everyone~ hasta

yeah

See ya, @Anthony. Keep me posted.

I skip some words sometimes when reading I need to be more careful

nods

I was thinking of reading Atiyah & MacDonald's book (I've already read a more introductory book---though I didn't pay all that attention to the details)

Do you think that'd be overkill @MikeMiller?

@Danu: The algebra shows up when you start studying the Floer homology groups, when some amount of commutative and homological algebra is desirable. I really and honestly have never needed to know much algebra for the 4-manifold stuff.

That is a great book, but you don't really need commutative algebra for math physics, unless you plan to do algebraic algebraic geometry.

Homological algebra is more basic.

I might be needing some algebraic geometry too, if I ever end up getting interested in strings.

Well, true. But I would urge you to try to learn Griffiths/Harris, based on geometry and complex analysis, rather than all the abstract algebra stuff ... starting off, anyhow.

If you have the time, then certainly it can't hurt, but I don't think it's that important.

But for instance, trying to read the book by Lawson & Michelsohn on spin geometry, I got the feeling that I needed more algebra.

Where?

@TedShifrin As in Principles of AG?

I mean, my advice to everyone is basically "Push forward, and fall back if you need to."

yes, @Danu.

@MikeMiller I'm very hesitant about such an approach :P

Do you know basic undergraduate algebra well, like Mike Artin's book?

The stuff in LM I had trouble with was representation theory of Lie groups. I don't know that very well.

Some people don't do well with that, @MikeM, although I don't disagree in principle.

But you won't learn that in Atiyah MacDonald.

Nope.

@TedShifrin Not well, no. "Cursory knowledge" (I've read a book but didn't do many exercises)

You need basic groups, rings, modules. Probably not Galois theory.

Some will disagree with me, but I don't believe you learn much math unless you actually work exercises.

Oh, I'm assuming you know what Ted said as a baseline...

@TedShifrin I agree

And, of course, solid, solid, solid linear algebra.

I know enough to understand what A&M are talking about---at least in the first few pages I've read.

First few pages are minimally tough

@TedShifrin The bane of my existence :P

All of math is linear algebra so that's a sticking point

Yes, linear algebra and multivariable analysis are the usual weak spots for an amazing number of people.

I know... If only my undergraduate degree had offered anything beyond trivialities.

LOL ...

@TedShifrin I learned all the linear algebra and multivariable analysis I know in the past year. :D

I don't think we got very far past the definition of eigenvectors/values. I've brushed up a bit since then but definitely not sufficient.

Um, stop fibbing, @MikeM.

@Danu I imagine most of the people who actually do Spin geometry know a lot about Lie groups. I had to mentor a physics student who wanted to learn about Lie groups (The Lorentz group in particular) and handed him Stillwell's Naive Lie Theory. I dont know if this will help you do much research, but it is a start. If one wanted to do research in Spin geometry or gauge theory one should really learn material from almost all of the first year graduate courses in math.

I do intend on eventually improving on this, but there are so many other things to learn :\

Curtis's Matrix Groups is a nice, concrete introduction, too, @MikeM, @PVAL, @Danu.

I think the best differential geometry book to learn for the sake of gauge theory is Taubes'.

Still need linear algebra and multivariable analysis :P

I'm not disagreeing.

@TedShifrin: I have a holomorphic bundle. What is the canonical unitary connection on it?

Ever heard of Greub, Halperin & Vanstone - Connections, curvature & cohomology? :P

It seems like a mighty tome to me :D

I haven't.

When I say differential geometry the book assumes you know, say, Lee's worth of smooth manifolds.

Do you know if any ever calculated this one (here)? $$\int_0^1 \frac{\log (x) \log ^3(1-x)}{x+1} \, dx$$ @robjohn @r9m

GHV looks like a great book to never finish.

@Chris'ssistheartist I don't think I have

@MikeMiller I've had two courses on diff geo---one covered something corresponding to "selected topics from about half of Lee's book"

Good enough then.

@robjohn I think I have a way beyond mind-blowing.

and one was called Riemannian geometry but was just Chern-Weil theory with last 2 weeks "trivial applications in Riem. geom"

Hey, that sounds like a great course.

It was amazing :D

"Just" chern-Weil theory. Get outta here.

There's a ton of stuff in GHV. I don't know it well.

...but we didn't do any Riemannian geometry :P

I think we spent 2 hours on geodesics

Well, just go read Kobayashi Nomizu.

I usually did Chern-Weil in my second semester of diff/Riem geometry.