@Balarka When its well enough approximated by its derivative projecting it onto a linear subspace and taking the straight line homotopy will be an isotopy. (i.e its a graph over some linear subspace (the subspace corresponding to the image of its derivative).
It's important to realize that the proof that Diff(M) acts transitively on points is actually a proof that Diff_0(M) (connected component of the identity) acts transitively on points. Or that proof would be useless at constructing isotopies.
So the topological difficulty of contracting the disk is resolved by using the annulus theorem, which says that the difference of the larger and the smaller disk is topologically a sphere cross interval, so one can contract along the interval factor.
Anyhow, it's totally encyclopedic and unreadable. But I actually had to use a version of Sard's Theorem in my research that I found only in Federer's encyclopedia.
@BalarkaSen Essentially it can be done in a lot of ways all involving some nontrivial business with point set topology. The classical way I think is showing surfaces have linear triangulations and hence smooth structures.
Yeah. Federer did great mathematics, but he is a horrible expositor. I heard him lecture at a summer AMS meeting on geometric measure theory, and I spent the whole time wishing one of my former professors at Berkeley/StonyBrook were giving the talks. Great, great expositor.
I think though theres a lot of work (which I don't understand) in going from Smale's proof to a proof in the topological category. You have to show they have topological handlebody structures I guess.
@TedShifrin Yeah, food service is generally not great, but it has its bright spots. Just doesn't pair well with full-time academic plans in my experience
@MikeMiller I am not sure if it'd be a waste of time; I'd want to spend more time on what I haven't spent more time on. But maybe I'll leave the decision for some other time since it's obviously not an immediate worry
@TedShifrin Here's a more topological approach. 1) This is obviously true for any fixed Riemann surface. One way of seeing this is to write $S^1$ as approximated by an increasing union of finite sets. If it fixes the infinite union of those, it fixes the circle, and is trivial. 2) Show that if the question is true for a surface in $\mathcal M_{g}$, it's also true for nearby surfaces. Now extend this fact to a choice of compactification of the moduli space.
(2) is the hard part, and also doesn't give an effective bound.
I was raised in a very traditional circumstance. Grandparents will still ask my cousins, invariably at least 25 years old, to put the phones away at the table.
What's distressing is how many kids are trying to do more complex arithmetic and have to do basic addition/subtraction on their fingers. The hell with knowing basic multiplication facts.
Still, @Brody, I'm all in favor of the new standards even though everyone bitches. But kids are not learning anything by rote at all. I hate rote, but, seriously ...
Pushing through it only possibly works if you are not alone or else you will fall asleep as soon as you get into some place you can. Even then it just messes you up.
@MikeMiller Speaking of which, did you ever find a reference/find time for finding a reference for the punctured sphere homology thing? Defining the boundary map was troublesome, IIRC
Admittedly there's very few people who write those kind of things and they are usually hard to understand.
I remember one case where I read something he gave an algebraic topological proof of the existence of almost complex structures on symplectic manifolds (you can possibly guess what the proof was) which confused me for a long time because it didn't give me any control in coordinates (the "right" proof certainly does).
In particular to prove say symplectic non-squeezing, you need to show that for any symplectic embedding of a ball $B^2n \into M$ there exists a compatibile almost complex structure which is the standard holomorphic structure on that ball.
@PVAL-inactive You can play the same game actually. By contractibility of stuff all the obstruction theory is automatically zero and you can extend that almost complex structure on the ball to one everywhere.
Yeah that's fair. But a lot of the Fukaya category people frustrate me since they just like to make things more and more algebraic without thinking of any of the other issues.
Nto that my flavor of Floer homology is any safer. Or that I'm not writing something that does the same thing. Dang.
if I have the equation $x \mod 3 = x \mod 5$ how could I rearrange this (if possible) to $x \mod 15 = 0$. Also, the equals sign should obviously be the congruence symbol, but I don't know how to do that in LaTeX.
@MikeMiller, @PVAL-inactive, do you know where to find that information? Of how to find the inverse of $\mod 5$ to get the equation $x \mod 5 = x\mod 3$ to $x\mod 15 = 0$?
