But then I want to do the second one

NOOOOO ...

Why?

My book wants me to do it like that

:|

Your book sucks rotten eggs.

So you use one contour for the first and the opposite contour for the other. But it's stoooopid.

@TedShifrin They say "In utilizing semicircular contours our first inclination is to deal with the complex function $$\frac{cos 3z}{z^2+4}, \:\:\: (1)$$ However with this choice for $f(z)$ we are doomed to failure because the modulus of (1) does not go to zero in either the upper or lower half-plane"

Right, I said something less long-winded above.

Sorry, I didn't get that

@Ted I finished the talk. Now I have time to do some damn work.

Well, I said to be careful with the numerator. If you just blithely use $\cos 3z$ you're screwed.

On the paper PVAL said I've been writing as long as I remember.

@MikeM: If it's not too much trouble to send me a scanned .pdf, I'm interested. But don't waste time on me.

@Lozansky: At any rate, do you have a further question?

@TedShifrin Lots

Ah

I should start charging you and Danu :P

LOL

Probably a smart move

So you are saying I could get away with evaluating the integral for $Re(e^{i3x})$

Or what was the approach?

Yes, that's what I'm saying. Precisely.

Chemistry is tiring.

Well, no. I'm saying to evaluate the integral for $e^{3iz}/(z^2+4)$ by the residue theorem, and take the real part at the end. Check it out.

@Balarka: I love chemistry.

Or I did.

@Ted There's not enough there to send. when I have an actual first draft it's yours.

Oh, OK, @MikeM ... I just meant whatever your lecture notes were going to be. But no biggie.

(I didn't mean your paper with PVAL.)

Oh, the talk. Those are on paper.

I can tell you the story.

That's what I meant. You and PVAL are way over my head.

Hmm, I wonder how I am going to spend the next few minutes awake.

Maybe by doing differential geometry.

We're not writing anything together, for clarity. I think he's just poking at the fact that I have said i'm writing it foreever.

Oh, true.

@Balarka: Better to go to sleep.

@TedShifrin Interesting. I've always thought it was just memorization and no real understanding.

That's not entirely true.

@0celo: That's what most people think math is.

My high school chemistry teacher senior year was totally awesome. And then taking P Chem in college made everything make crystal clear sense. I loved it.

I took QM this year instead of P chem

I hate it

Speaking terrible classes, I have to compute pdfs

byes

@TedShifrin Oh I have the solution of that sup difference quotient problem if you want.

Oh, @0celo, I found some lecture notes that do that strong derivative stuff. I guess the point is that you have to assume differentiable everywhere, $C^1$ (or derivative continuous) only at one point, in order to prove a slightly more general inverse function theorem. I'm still far from convinced that I should care.

@TedShifrin I asked someone about it.

It's apparently very useful in operator theory/functional analysis.

I heard a version of it a long time ago.

Interesting, 0celo, because my operator theory colleagues never once mentioned it to me. Maybe they were in a different corner of the subject.

But thanks for bringing it to my attention.

@TedShifrin It's in Reed & Simon.

(Allegedly)

I wouldn't do it in an undergraduate analysis course, though. I think it's unnecessarily technical.

Fair enough. Still not needed for a first analysis course for undergraduates.

I don't believe in doing the most general versions in first courses.

But my opinions on pedagogy are not universally popular :P

Here.

@TedShifrin We did Stone-Weierstrass in terms of function algebras, Arzela Ascoli on separable spaces, Frechet/Gateaux derivative

According to the prof, if we don't see it here, chances are we'll never see it.

I don't think that's good teaching, in general.

Why?

There's only 4 undergrads in the class

out of 20

Because students learning things for the first time are in no position to appreciate/grok the fanciest versions.

So this is not a first analysis course? What is it?

Well...it's confusing

You're supposed to take it after the 1D real analysis course

and he's teaching it like one has seen all of this in 1D in one form or another

Interesting, @Balarka. So you don't even need to assume the derivative is continuous at a point to get the IFT. You just need everywhere invertible. Fascinating.

Mhm.

I thought I had a counterexample without the continuity at a point, but I don't.

so he spent time on topology, but assumed all the stuff about convergence, sequences, uniform continuity, etc.

0celo should tell his professor to do this more general version :P

He uses the Brouwer fixed point theorem instead of the Banach fixed point theorem to get the proof working for differentiable instead of C^1, apparently.

It's cool. I just wouldn't teach this unless I had students I expected really to be able to apply it (where the usual theorem doesn't apply).

Hmm

I'm fairly sure #2 is wrong

Certainly, for standard manifolds it's irrelephant. Maybe relephant for people doing Lipschitz manifolds instead of $C^1$ manifolds.

Theorem 2, 0celo?

The usual counterexample should work. Where does Tao mention strong diffability?

Yah, I won't read the proof. The statement is of a theoretical interest anyhow.

@0celo7 What "usual counterexample"?

He has a stronger theorem. He doesn't assume continuity of the derivative, but he assumes everywhere invertible.

@BalarkaSen $x/2+x^2\sin(1/x)$ should work.

Nope, the derivative is 0 lots of places.

That's what I was sorting out earlier when I was being quiet.

And if you make it $2x + x^2\sin(1/x)$ then it's invertible.

oh maybe if you're differentiable in a whole open set

I mean, duh, it's filled with critical points.

There's also a stronger version of the usual $C^1 \implies$ differentiable. But it's a theoretical curiosity and I would never put that in my book. I would have put it in as an exercise if I'd known it before writing the book.

@0celo7 The point is not just differentiable, but the derivative being invertible locally too.

@BalarkaSen No, he's saying $Df(x_0)$ is invertible

Which in 1D is equivalent to $f'(x_0)\ne 0$.

Admittedly, Terry's notation is confusingly chosen.

@TedShifrin Ok, so back to what we were talking about.

The class is supposed to be an intro

But all the people in the class have had analysis before

So he changed the material a little

Yeah. I disagree, but I'm only me.

But that doesn't make sense, because we started with Banach spaces on day 1

I am ducking out of meta discussions.

I spent my career being a very aggressive teacher, but I don't necessarily advocate a lot of things that people end up teaching.

Go to sleep, @Balarka, or go do geometry. There are fabulous exercises in the next section.

Yeah that's what I don't understand. Why can't I appreciate Stone-Weierstrass just because we did a crazy general version of it?

We also did Bernstein polynomials btw

Next chapter being the Gauss-Codazzi (grumble), or connections?

Gauss-Codazzi.

:(

Alllright.

I'm giving you cool exercises: 13, 14, 15, 16, 17, 18. 19 is interesting for the examples it gives you, but it involves elliptic integrals and it's only worthwhile if you can use a computer to draw pictures.

Thanks.

Codazzi allows you to prove things like Prop. 3.4, which generalizes to all sorts of beautiful higher-dimensional (and projective) versions. I love it.

Soon I'll ask you to tell me about geodesics on a standard torus :P Next section.

what are we doing?

christoffel symbols

I'm derailing you and getting Balarka back on geometry track :P

/joke

Well, Balarka, eventually you'll read the differential forms version and you'll be in love with 'em. :P

Does he know the essential property of a connection yet? What it does?

NO. We're not there yet. Patience, jackass, patience!

You've already told him, anyhow.

What essential property?

Yeah, but he didn't listen

I am to be blamed for the misunderstanding; it's in the next section and I peeked in.

Then let him get there in due time.

@0celo7 He knows the LC connection of a surface and I want him to define a connection on a bundle.

@Balarka: Eventually you conceded that my getting you to learn multivariable calc carefully was a good idea.

@MikeMiller Ehresmann or Koszul?

Of course, @Ted.

Oh, good grief, leave Balarka alone.

lol

Mike just doesn't want you understanding surfaces in $\Bbb R^3$, @Balarka, because he doesn't know anything.

What?

I'm so confused

@TedShifrin I didn't actually read prop 3.4 carefully. I admittedly don't find it very interesting; should I?

Ok what book is this?

"A First Course in Differential Geometry", it's Ted's notes not a book.

im stuck trying to solve a quadratic, i get the wrong answer but don't know where i went wrong as my answer book doesn't show the steps

any one able to look at my workings to see what i might of got wrong?

don't ask to ask

If two (parameterizations of) surfaces $f_0, f_1: U \to \Bbb R^3$ are isometric, does that mean they are isometric through parameterizations, that is to say, there is a 1-parameter family $f_t : U \to \Bbb R^3$ of immersions, all of which are isometric, and starts and ends at $f_0$ and $f_1$ respectively? Assume $f_0, f_1$ are orientation-preserving parameterizations.

I dare not work in a more general context for immersions between arbitrary Riemannian manifolds because I don't know how the story works there.

this was my calculation steps

Do you not have the quadratic formula?

Ah

When you take the square root you need a +-

There is a very visual one between the helicoid and the catenoid for example.

i was using the complete the square approach

good point on the +- i can't find the symbol in word for it

@Balarka: Perhaps this will entice you more. Suppose $M^n\subset\Bbb R^N$ or $\Bbb P^N$ is a submanifold. If the Gauss map has constant rank $k<n$, then the fibers of the Gauss map are all (subsets of) linear spaces.

@WDUK shouldn't it be 4x^2-7x-3=0?

@Balarka: The helicoid and catenoid are very special, being minimal, so complex analysis enters the picture. (I assume you're looking at that exercise.) It's not true in general, I don't think. But I'll have to ponder that.

@artic good spot!

Yeah, I was looking at that.

that might explain it!

@TedShifrin Ah, ok.

thanks for the feedback ! time to recalculate

@Balarka: Without the constant rank assumption, it fails. I thought that was a fascinating result when I first learned it in the general projective setting.

@BalarkaSen You should get the new Pokemon games in a few weeks

never played em

not a fan of pokemon either

@TedShifrin The next question then will of course be, what's the fundamental group of the space of parametrized surfaces in $\Bbb R^3$? :D

Sounds like nonsense.

But parametrizations are only local, remember. We're cheating.

We're allowing parametrizations by non-open sets for convenience (or we're parametrizing only an open subset of our surface).

If you want the fundamental group of the space of smooth submanifolds of $\Bbb R^3$, I can do that. If you want the fundamental group of the space of smooth embeddings of a given manifold, I can do that too. But "space of parameterized surfaces" doesn't parse.

Do you need compact in any of that, Mike?

@TedShifrin No, but I do need closed (as a subset).

And then the topology might not be what one expects.

@Balarka: BTW, an easier comment on your question is — given two embedded submanifolds that are diffeomorphic, need they be isotopic? That's a bit less structure. Of course the answer is generally no.

Yeah, that's not the right formulation, that was silly. I want to say, which self-isometries of a parametrized surface $f$ in $\Bbb R^3$ are given by a 1-parameter family of isometric parametrized surfaces which starts and ends at $f$?

Certainly you don't have to assume anything, the space will just be really nasty.

In general, there aren't self-isometries, @Balarka.

Unless you start out very symmetric.

Please vote for this MSE feature request:

@TedShifrin There are local ones though? Replace the isometry bit with equal $I$'s.

at all points

This is something in our lives where individual votes have an effective power.

Deciding whether two pieces of surfaces are isometric through some (yet to be found) function is very difficult. It's an overdetermined PDE.

The space of closed smooth surfaces in $\Bbb R^3$, appropriately topologized (so that spheres of increasing radius contained in the right half-space with one point at the origin converges, as the radius goes to infinity, to the $x=0$ plane), is homotopy equivalent to the one-point compactification of the space of affine oriented 2d subspaces of $\Bbb R^3$.

@Ted Do you know what a Dehn twist is?

Yup.

@EnjoysMath: Where do we vote?

For the post

I am assuming that's how a feature or bug gets made

Oh, gotcha.

(like normal post voting)

Done.

I can't believe I never thought of feature requesting this issue before. It's always bugged me!

@MikeMiller Interesting

It's annoying, yes. Worse trying to answer a question on my phone or iPad, though. Almost impossible.

If I consider the standard simple closed curves {a_1,...a_2g} which form the most obvious presentation of the fundamental group and then take the composition of dehn twists around each a_i

I think that means the space should be homotopy equivalent to the one-point compactification of $S^2 \times \Bbb R$, aka a 3-sphere with two points glued together.

Thanks @TedShifrin

d_1,...d_2g

Don't I have at least countably many components, @MikeM?

Do you know how to realize a hyperbolic metric on $\Sigma_2g$ which has d_1...d_2g as an isometry?

Hell no, I don't, @PVAL, but I'm sure Thurstonites do.

Hi guys

Hi @dsillman

How does one actually define the space of closed smooth surfaces in R^3? Is there an easy way to see the explicit topology?

@TedShifrin I've seen proofs of the triviality of the word (d_1...d_2g)^(4g-2) which do something else than showing the map is an isometry for some metric and then using the 4g-2 theorem. I would really like to be able to do it that way

Seems hard to imagine, though, @PVAL. Seems like lengths of the curves along which you do the Dehn twist are increasing.

@Ted Note the topology I put on it. I'm allowing compact things to limit to noncompact things.

Indeed it's path connected since I include the empty set and can just "push everything to the right" more or less.

Well (d_1..d_2g)^4g-2=1, so you'd really expect it to be an isometry with some hyperbolic metric.

I still think that space has countably many components. But can one genus be in the same component as another? Oh, you're going off to infinity and coming back?

Seems like it.

Freaky, I see, @PVAL.

It's times like these that I claim to be a complex algebraic geometer. :P

But I'll renounce that as soon as Danu returns.

Actually, these days I can claim to be stooopid and nothing.

@Ted I bet its an automorphism of an algebraic curve as well.

LOL, most curves have no automorphisms :)

I still don't see how to reduce the genus. I can join a surface to it minus a bunch of punctures, but so what?

Maybe I'm missing something.

Yeah, I don't see it, either, @Balarka.

Oh, push all the handle to infinity

Make it tiny first, I guess.

That's a pretty silly topology, if that's how it works

Right, @Ted.

@TedShifrin Yes well lots of finite groups are groups of automorphisms of algebraic curves.

Well, there's a famous Riemann bound, @PVAL, of course, but the generic curve has none, I think.

How do you show that?

Hmm, probably by keeping track of Weierstrass points or something.

The bound is 2(84g-2) iirc and is usually attributed to Hurwitz.

84(g-1), I think

That number doesn't look quite right.

Balarka has it.

I guess Riemann gave a famous argument for it. I thought I remembered that from G&H.

Anyhow, Balarka, I just came to give you my Gauss map ruled surface story. G'night!

Night.

I really liked the cute little elementary number theory trick to prove it. I learned that as an undergrad. I don't think I understood the proofs of RH or RR at the time.

The more I read formal proofs using quantifiers and logical symbols instead of words, the more math looks like a forgotten language. xD

EEEK! There was an error in my textbook!

It said that if $x\in\mathbb{R}$, then $x^2+1>0$!

It should be $x^2+1\geq 1$

Or $x\in\mathbb{Z}$

@dsillman2000 That happens

@dsillman2000 Well that isn't wrong

How so?

Plot that

It's a parabola that never crosses the x axis

So it's strictly positive

Well true

But it implies that $x^2$ could be in $(-1,0]$

Which it can't be and also be real

or i mean $(-1,0)$

No it doesn't imply that

How so?

It would still be greater than 0

100 > 2 is true

but it doesn't imply 100 = 3

true

but it does mean that $x^2$ CAN be negative

an inequality tells you what your number can't be, in some sense

@dsillman2000 no

$x^2>-1$ is true for any real $x$.

So is $x^2\ge 0$.

Yes

True

Alright fair enough

But it could be simplified more!

Right

The bound can be improved