You do know that you get convergent subsequences?

Suppose towards a contradiction that nobody could help you...

Yep, it's the next part that confuses me.

That's because you have a typo in your notes.

Several typos.

There was a summer seminar (group work) in my university, using Hirsch's book. Eventually, they spent all of time on that chapter of function spaces and weak/strong topology.

$\lim f(x_{n_k}) = \lim f(y_{n_k}) = f(u)$.

@Frank: That's the most technical and, to a beginner, least interesting stuff. :)

Yep, I can't see how we concluded the limit on the line underneath, @Ted. Where does the infinity come from in $\lim |f(x_{n_k})-f(y_{n_k})|$?

It's wrong, @Khallil. That was the first typo I noted.

What should it be?

And all the $x_k$ should be $x_{n_k}$.

Should it not be $0$?

Of course it should.

(Yep, I've noted the subscript inconsistencies, thank you!)

Either your teacher messed up or you copied stuff wrong.

That was the page that the lecturer uploaded which is why I was unsure, @Ted. :-)

That's what you did or just you copied from your professor?

Wow. None of you caught the mistakes in class before he uploaded?

Sorry, I'm not being too clear. The professor scanned their hand written notes and uploaded them online for all to view, @FrankScience.

(They were from last year's lectures, @Ted!)

@Khallil Okay, you can safely forget it and prove on your own.

(I hadn't been present which is why I was catching up on the online notes!)

I was always upset when I made a stupid mistake in class (when they were video-ing the lectures) and a student would come up after class to correct it. I would like to have had it corrected for the video at the time. Sigh.

That'll learn you to skip classes, Khallil.

I always endeavour to do that during the class, @Ted. :-)

Blergh, I can't see it. I have to wait for tomorrow morning :(

I was really ill for the lecture, @Ted! I wouldn't skip without reason, honest! :-)

What's the simplest function you learn about as a child, @Balarka, other than linear functions?

Understood, @Khallil. I apologize.

Parabola?

Bingo.

@Khallil You should learn to prove things on your own, not barely reading lecture notes.

I understand, @Frank!

Man, everyone's a tough taskmaster here! :) I look like a softie.

Ironically, I understood piecewise functions before I ever learned about $\mathbb{R}$, I think it effected the way I learned everything else after that.

There is a very interesting post: mathoverflow.net/a/13149

On the topic of soft questions actually... Does anyone know of a process which can evaluate $$f(x_1, x_2, \cdots, x_n) = (\operatorname{rank}(x_1),\,\operatorname{rank}(x_2),\,\cdots,\,\operatorname{ra‌​nk}(x_n))$$ on a time complexity of $\mathcal{O}(n)$ when $f : \mathbb{R}^n \rightarrow \mathbb{N}^n$

I don't think I have ever taken afterwards readable notes in a class. If I didn't know something, I'd prove it myself or look for a readable explanation.

So you took notes, but they were not readable? :P

@PVAL, I was very different from you as a student. I did sometimes rework bits and pieces of my notes, but I held on to them until just a few months ago, when I trashed everything from college and grad school when I retired.

Also, if anyone has suggestions on a better way to notate that, I'd love that too. That repeating the rank function felt very inelegant when I just want to say "element-wise rank of a vector"

I've noticed in my century of teaching that students vary a lot in their learning styles. Some hate lectures and read books. Some love lectures and pick it up well from interacting in class. And plenty are in the middle.

Well, in what you wrote, $x$ is fixed, @evinda.

@TedShifrin Is your example about transverse intersection of submanifolds? Just to clarify.

You can't have $\|x-y_n\|\to 0$ for different $x$'s, can you?

@Balarka: It's supposed to be a counterexample of the same stability phenomenon.

The parabola is meant to be the image of the map. You haven't yet picked an appropriate non-compact submanifold to which it's transverse.

@TedShifrin I have written a good amount of notes, but they were notes that I spent time preparing. I never developed the skill to simultaneously understand new material while copying down things accurately and readably.

Ok, that's what I wanted to know. I was unsure whether your map was an embedding. Gotcha.

Right, @PVAL, I understand. And I've had numbers of students like you, too. I'm just saying that I was not like that, and I've had students who aren't.

@TedShifrin Why not?

@PVAL Agree. When I'm copying, I cannot understand anything. My brain isn't multithreaded.

@evinda: Because a sequence must have a unique limit.

Me neither.

@TedShifrin Oh yes, right...

I can only note down the statement of theorems, and (maybe) a sketch of the proof. I cannot copy the whole proof.

I would agree that, in general, students tend to copy down too much, @Frank. One needn't write every word, although it takes lots of experience to know what to leave out.

OK, back later.

See you in a bit :-)

And if I were a teacher, I would leave most of proofs as assignments for students, with hints, of course, since most of them were nontrivial.

May I ask for a hint in proving cty implying uniform cty over a closed bounded interval, @FrankScience?

I feel like the proof is via contrapositive if I'm to suppose that $f$ not being uniformly continuous implies that $f$ isn't continuous.

Is there a big distinction between contrapositive and contradiction?

Heck, I keep thinking about making things transverse at $\infty$. I can't visualize this.

Whereas contradiction entails something else, @Jasper?

(Nice to see you again after not being on here for a while, btw!)

A proof written as contradiction but de facto a contrapositive would be a bad organization of proof.

@Balarka: What happens if you start with $\Bbb R$ as your submanifold? Doesn't work, does it?

@DanielFischer I think that I got it now... Thank you very much!!!

No, I'd think not. If my parabola is transverse to R, my visualization says every perturbation would have a small bit in which it's transverse.

@TedShifrin I thank you too for your answer!!!

Taking notes isn't easy. Two years ago, Miles Reid came to our university, teaching undergraduate algebraic geometry.

I failed to follow the last part.

You know Miles Reid, @FrankScience?

He's at my university! :-)

Whoa, @FrankScience. I am a fan of Miles Reid's comm. alg. book.

But, @Balarka, it's clearly not transverse to the $x$-axis. Right?

So many beautiful pictures!

The last part was on Riemann-Roch theorem.

@TedShifrin I thought you were starting off with a disjoint parabola. No, it's not.

No, he proved.

So how could you change things to force transversality (not changing the map)?

But I failed to follow.

And he mentioned Grothendieck-Serre duality.

I cannot remember details.

My favorite geometric proof of Riemann-Roch (not using sheaf cohomology) is in Griffiths/Harris and Griffiths' Introduction to Algebraic Curves.

Hey @TedShifrin

You again, Karim? :)

the problem I asked earlier is actually direct application of Dirichlet's test

He was talking about algebraic version.

Yes, that works, Karim.

yeah :D

Yeah, very sheafical, @Frank.

@TedShifrin Depends on how you want me to change things. I can of course perturb x-axis at the origin to make it disjoint from the parabola, but I believe that's not what you're asking for.

@TedShifrin Is that one really differing from the sheaf-theoretic proof?

lol at sheafical

cool prof will also be very amused of this proof

Absolutely, @Frank. It's interpreting Riemann-Roch in terms of the geometry of the canonical curve.

@Balarka: Yes, I told you originally to change the submanifold so it wasn't all of $\Bbb R$.

I attented some K-theory talk he was talking about Riemann Roch theorem alot

anyway that talk was above my head

And the K-theory & physics talk I attended didn't even mention K-theory. What a bummer.

Oh wait @TedShifrin

Damn it, this is awesome.

R - {0}

And what's the homotopy, @Balarka?

Slide the parabola downward by $t$.

That won't give a counterexample, will it?

Uhh.

Yes, figured. Wait, I can do this. Hmm.

Series? You mean Caroline Series?

Ah, I know her.

@TedShifrin I skimmed Griffith & Harris. I remembered that it was essentially doing linear algebra on residue or something.

She collaborated with prof, in fact.

I cannot remember the details, but there should be some Serre duality involved.

That's one of the proofs in there, @Frank.

differentials of first kind, second kind, that language.

Oh, I see, the proof I had in mind was on p. 248-249, but they do use residues there, too. One doesn't need to say it that way, but it's cool.

I remember there's a proof on Donaldson's book.

But essentially same as sheaf-theoretic proof, just expressing the same idea in language of PDE. I cannot remember the details.

@TedShifrin I cannot make this rigorous, but the homotopy should "straighten" the parabola at the origin. That is, straighten the two horns horns of the parabola.

@Balarka: Be a simpleton. Don't be so difficult.

Ah.

Gotcha.

Just translate the parabola horizontally.

Whew.

One problem with the way you've jumped into such abstract math is that you overthink simple things.

This is a cool example, but I prefer my transversal at $\infty$ example :)

What a pity! Up until now I haven't started to read Hartshorne.

(Because I found it)

@TedShifrin I am working on it!

I pushed you on this, @Balarka, because I want you to learn to look for simple examples.

But, yes, you should prefer things you yourself find. I heartily approve.

Of course, I have no idea what you mean about transverse at infinity ...

Yeah, I appreciate you being patient with me.

Thank you.

I like this example.

Arg!!! God grant me the willpower to overcome the temptation! Third result in Google results for "anton linear algebra" is a free pdf! I'm poor and I could save $50! Must! Not! Look!

Anton's a crummy book, anyhow.

You get what you pay for.

@TedShifrin I mean, as you push the damped sine curve above, you're pushing the nontransversality from infinity, because your minimas are increasing.

So the original curve $f_0$ is actually not transeverse to $\Bbb R$ at infinity.

That's why it works.

Learn to spell TRANSVERSE :P

If you compactify everything in $\Bbb P^2$, yes, @Balarka.

Boo, I keep adding an e.

It's only that cheap because it's the 10th edition. 11th hardcover is $148.92 from Amazon Prime.

It's still a crummy book, regardless, @idontunderstand.

lol

Artin is the best linear algebra come abstract algebra book I know.

I don't like those notes that much, either, @Jasper. But you and I rarely agree.

Maybe the linear algebra book of Halmos is worthy.

Hoffman-Kunze has slightly harder exercises and a little more content.

@JasperLoy Thank you. Now, if only I could have a @Jasper to correct my transverse mistakes.

@mikeonly: I'll get to you in a bit.

Aluffi is all categories.

@idontunderstand is talking about linear algebra at a very much more plug-and-chug level, guys. Get off your high perches.

And so I believe is anything which is authored by MacLane

I purchased by perch, why should I get off it?

DF Is nice

@BalarkaSen

That pun didn't pan out.

DF is nice for everything

I learned alot of algebra from that book

@L33ter DF's theory is horrible.

But exercises are good, I agree.

@mikeonly, no, the proof is given later, using the limit definition for an arbitrary $v$.

I like that version better @JasperLoy

careful, @Balarka, we can deep fry your perch.

I can define alot = A lot

@TedShifrin

so I am show for any n hat sin(nz) is bounded

by some M

where M is real

@TedShifrin So corollary is just what you get after you've proven the main statement?

in order to use dirchlet test

@JasperLoy Ambiguous, there're a lot of ways you can do that.

@TedShifrin Sorry for my ignorance, now it makes sense for me. Thanks.

@mikeonly: The corollary is being deduced as a special case (corollary means it follows from the proposition). But the general proof is independent of it.

I think in the book I was stupid and did the same proof two times. In the lecture, I didn't do that.

@Jasper @Balarka: perch is a type of fish.

$|sin(nz)| = |\frac{1}{2i}(e^{inz} - e^{-inz})| \leq 1/2 |e^{inx + -ny}| + 1/2|e^{-inx + ny}|$

ok ?

Not helpful, @Jasper.

He ran away from his perch.

nvm

well we know this will be actually less than 1

because of the domain

I don't think that's right at all.

by assumption we have our domain

is the following

-1 < im(z) < 1

Great, I am going to dream of transverse submanifolds today.

so $|e^{y}| < 1$

@BalarkaSen There is an excellent book on transversality, etc.

There is?

How are you applying Dirichlet, Karim?

well I am thinking $a_n = e^{-n}$ and $b_n = sin(nz)$

@BalarkaSen J-P. Serre, Lie Algebras and Lie Groups

where $a_n$ is the real part of the Dirchlet thing and $b_n$ is the imaginary part.

That talks about transversality?

What is the statement of Dirichlet's Theorem?

Speaking of, I need to get my hands on Serre's Tree someday.

OK, so now, what are $a_n$ and $b_n$?

so $a_n = e^{-n}$ and $b_n = sin(nz)$

OK, I agree. So you need to look at the partial sums $\sum_{n=1}^N \sin(nz)$ and see why there's a uniform bound.

alright

Applying Abel-Dirichlet criterion to show the analyticity?

yeas

I remember it from some analysis book that I was reading

That's the standard approach to understand convergence of $\sum \dfrac{\sin n}n$. This is a bit thornier.

But that's nothing to do with analyticity.

@TedShifrin |sin(nz)| < 1/2|e^{-n}| + 1/2|e^n|

You'll never get a uniform bound on partial sums that way, Karim.

He's trying to prove uniform convergence on compacts, @Frank, I think.

yes

correct @TedShifrin

what would you suggest @TedShifrin ?

I don't have a suggestion at the moment. It's not my homework :)

I'm not so familiar with complex analysis. There is a theorem of uniform convergence and normal family, but, well, I think absolute convergence is enough.

Nothing with normal families here.

Otherwise how do you deduce analyticity from uniform convergence?

You need to estimate derivatives.

Here's a hint to try, Karim. $e^{inz} = (e^{iz})^n$.

No you don't, @Frank. One proves once and for all that if $f_n\to f$ uniformly on compacts and $f_n$ are analytic, then $f$ is.

Morera proves that, or Cauchy integral formula proves it ...

Good night, everyone.

Night, @mikeonly. Have fun!

you know @TedShifrin I have a problem next semester I won't do it again like I start out very strong at the start of the semester and then as the semester ends I start doing assignments the day before and such.

I dunno why I do that

Math is too hard to do that.

yes

@TedShifrin Okay.

@Frank: May I ask where you're a student? (You can of course decline to answer. :) )

@TedShifrin Previously in China but now in France.

I don't know whether there exists a sequence of analytic functions $f_n$

Ah, très intéressant.

we have a very good student here doing masters in topology

from china

such that $\sum_nf_n$ uniformly converges in some open set but $\sum\lvert f_n\rvert$ diverges on some generic set.

In fact, I haven't graduated from my university in China so now the status is somewhat special.

Try $f_n(z)=e^{inz}$, in fact, @Frank, on an appropriate domain.

M, N be smooth submanifolds of W. I want to prove that I can isotope M through embeddings to intersect N transversely. Would I not need some kind of tubular neighborhood of M to do that? Then perturb M inside that tubular neighborhood. I haven't thought it out.

Wait til you study chapter 2 of G&P, @Balarka. Damn, stop getting so far ahead.

lol

ok.

I just wondered whether you could do it. I guess this is genericity.

There was a nice exercise that if I have a Morse function on $M$, I can always homotope it to get a Morse function on $M$ which takes value $i$ on the index $i$ critical points.

This is an enhanced version of the exercise in G&P which said if $M$ has a Morse function, I can homotope it to get a Morse function taking distinct values on the critical points.

BTW, @Balarka, I do not like G&P's treatment of Morse functions. I gave my classes the definition that $\text{grad}\, f$ is transverse to the zero section of the tangent bundle. (Or $df$ is transverse to the zero section of the cotangent bundle.) Then all sorts of stuff falls out from standard transversality theorems.

Ultimately, you should understand why that definition agrees with the "standard" one.

I think I wrote up a proof for my prior lemma. Would you mind having a look, @Ted or @Frank?

Hmm, in fact, for the second exercise, I think you can get an arbitrarily close Morse function which takes distinct values on the critical points. Homotopy is weaker.

@TedShifrin It's the same for $f_n(z)=\exp(nz)$, but if $\lim_{n\to\infty}f_n(z)=0$, we must have $\Re z<0$ or something.

@TedShifrin I don't know what a zero section of the tangent bundle means :) By grad f, you mean $\nabla f$?

@TedShifrin Maybe you misunderstood my statement. The generic set on which it doesn't absolutely converge is contained in the prior open set, namely there's an open set on which $f_n$ converges uniformly but absolutely diverges generically.

I find Morse functions somewhat natural, but not that I did many exercises. I find it more natural than transversality.

I meant the definition for submanifolds of $\Bbb R^n$ (or, more generally, Riemannian manifolds).

Well, @Frank, what I was thinking of was $e^{in\theta}$ for $\theta\in\Bbb R$. Then absolutely you get total divergence, but on an appropriate set, you get uniform convergence, I believe.

But what I was asking is about analytic functions.

It should be so generic on the whole domain.

@Khallil: The proof your prof posted was correct except for typos. So you should be able to criticize it yourself.

For maps between Euclidean spaces, one writes it as $\nabla$, and for manifolds one writes it as $\text{grad}$? How outrageous.

Not just on circles, say.

Sure, @Frank, so take $z$ in a neighborhood of the real axis.

But then you cannot get the uniform convergence.

Sure I can, @Frank; it's a geometric series.

@Balarka: In Riemannian geometry, $\nabla$ means a connection, so the notation must be kept distinct. This is a generalization of the gradient of functions in the Euclidean setting.

ok I did it

oh, right, ok.

@TedShifrin No, since the geometric ratio is with modulus no less than $1$, impossible to converge.

so you can show that $| \Sigma^N sin(nz)| < N$

OK, @Frank, I confess I need a Dirichlet type thing in there to make it converge. I can't continue these conversations with 3 things going on. I'm not smart enough any more :)

Not to worry, I read it as $\neg B \implies \neg A$, @Jasper. :-)

@TedShifrin Thanks for the comment, but I heavily doubt it's accessible. It seems to me that oscillating stuff usually happens at the boundary. If I have time, I'll think it over.

Yeah, @Frank, I need to stop and think. When there's peace.

Thank you, @Ted. I'm pretty comfortable with it now. :-)

OK

I need to get to sleep now. Before I go, can I ask something?

You just did.

haha fair enough. that was old, not sure why I fell for it.

I used to do this to my students over and over ...

Anyhow?

ok, I'm going to ask it. So, there's this fact which says if $M$ is a compact closed manifold admitting a Morse function with 2 critical points, then $M$ is homotpy equivalent to $S^n$. How is this true? Since $M$ is compact, the Morse function is bound to have maxima and minima, so by enumeration, there is no other critical points. The only ones have index 0 and n. So you get, locally, a disk on the bottom and a disk on the top using Morse lemma.

But how can I prove that the complement of the union of those two is $S^{n-1} \times I$?

Because there are no other handles.

Basically, you need to prove that if there are no critical points, you get a trivial bundle over the interval.

So you're invoking that $f^{-1}([0, c])$ and $f^{-1}([0, c'])$ are homotopy equivalent if there is no critical point between $c$ and $c'$, but I don't know why it's true.

You use flow of the gradient to prove it.

Ah?

G'night, @Balarka. More to learn in the next year.

Darn.

I was trying to get you to tell me the proof of that. :(

Oh well.