But characterizing squares via non-squares allows you to talk about negative integers as well by slapping minuses in the right places, so a priori it's merpy
Most professors give out much higher grades in grad courses. Don't know about NYU and your particular course, but most places they give mostly A's and B's, occasional C's or none.
I had colleagues who gave all A's in courses past first-year grad courses. I never did that. Berkeley when I was there was essentially automatic A's for everyone after qualifying-exam level courses.
I learned about that stuff in the context of branched double covers in one of the first papers I ever read in grad school. Atiyah's paper on non-multiplicative signature in a fiber bundle.
If $L \to X$ is a line bundle and $s \in \Gamma(X, L^m)$, $m \geq 1$, then the image of $s$ is an $m$-sheeted cover of $X$ branched along the divisor $s^{-1}(0)$. Does that sound right?
If you want to see my compromise, on my webpage I have the lecture I gave (using LaTeXed slides) at an Math Association meeting when I won a teaching award. But I wanted it to be useful to people to look at afterward, too.
The worst thing is that they told me to spend the first 15 minutes giving a general introduction understandable to anyone in their geometry group and putting things into context.
@Danu Well, if it makes you feel any better - for my public Master's thesis defense, I could not assume the audience knew the definition of a manifold :P
@Semiclassic: He nowhere stipulated that the $z^j$ had anything to do with the symplectic structure. Anthony assumed they were canonical coordinates on the ambient manifold (and then knowing we have a symplectic submanifold it should follow that the first 2k are canonical coordinates on the submanifold). I think the OP wants an arbitrary coordinate system that flattens out the submanifold.
Yes, that's just the canonical immersion theorem ...
So I think it's going to follow from the fact that the submanifold is symplectic. We write down $\omega$ in the $z$-coordinates and get a block matrix form.
@Semiclassic: The OP said that Ratiu/Marsden said it's easy to see that you can choose coordinates so that ... ... in which case I don't see why not use Darboux coordinates. So I'm confuzled.
I should point out that I did slightly edit the OP: it initially read "choose a coordinate $z^i$" rather than "choose coordinates". But that modification was done on the basis of having tracked down the Marsden passage via Google books, and it indeed refers to "choose coordinates"
Quick sanity check; Darboux's theorem only gives us a local chart. But that's all we actually need here, since we just want to confirm that the matrix is invertible at any given point
Oh yeah he's my favorite lecturer here so far. I'll get to do discrete and combo next year so that'll be fun
How many of the lectures you (and this goes to everyone) been to where you got the vibe that the "it is easy to show that..." shtick was abused too much?
