Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question arises because I'm reading Frenkel and Ben Zvi's book "vertex algebras and algebraic curves" at the moment.

Let $X$ be a curve, $\Delta \subset X \times X$ the usual diagonal embedding, and $\mathcal{O}$ the sheaf of functions on $X$. A way to define the sheaf $\Omega$ of differential $1$-forms on $X$ is as $$ \Omega = \frac{\mathcal{O} \boxtimes \mathcal{O}(-\Delta)}{\mathcal{O} \boxtimes \mathcal{O}(-2\Delta)}|_\Delta $$ (where $\mathcal{F}(-\Delta)$ means sections of $\mathcal{F}$ on the diagonal of order $-1$, etc).

I'm pretty sure I understand this; it's a reformulation of the usual definition of $\Omega$ in terms of germs of functions vanishing at a point modulo functions vanishing to second order.

Frenkel and Ben Zvi go on to say that there's an isomorphism $$ \mathcal{O} \cong \frac{\Omega \boxtimes \Omega(2\Delta)}{\Omega \boxtimes \Omega(\Delta)}|_\Delta, $$ i.e., given a thing of the form $f(z, w) dz dw$ with an order 2 pole at $z=w$, we can produce a naturally defined function $g(x)$ which we should think of as living on the diagonal $z = w = x$.

My question is what is this isomorphism? It looks like some kind of residue analogue, but I'm not sure.

Thanks.

share|improve this question
add comment

3 Answers

up vote 4 down vote accepted

The first point to observe is that question is equivalent to showing that the line bundle $$\Omega \boxtimes \Omega(2 \triangle) \mid_{\triangle}$$ is canonically trivial.

Indeed, given any line bundle $L$ on $X \times X$, we have an exact sequence of sheaves $$ L(-\triangle) \to L \to \triangle_\ast (L \mid_{\triangle}) $$ on $X \times X$.

But, $$\left(\Omega \boxtimes \Omega \right)\mid_{\triangle} = \Omega^{\otimes 2}$$ and, as you already know, $$ \mathcal{O}(\triangle) \mid_{\triangle} = \Omega^{-1} $$ so it's clear.

Does that help?

share|improve this answer
    
Thanks! Yes that helps a lot. –  Jethro Feb 22 '12 at 18:45
add comment

Hi, Keven, good to hear from you here ^_^

To confirm the previous two answers, Kevin's suggestion is that we write $f(z,w)dzdw$ for $f(z,w)$ with a pole of order two on the diagonal as $h(z)dz\otimes g(w)dw\otimes \frac{1}{z-w}\otimes \frac{1}{z-w}\in (\Omega\boxtimes\Omega) \otimes \mathcal{O}(\Delta)\otimes \mathcal{O}(\Delta)$. Pulling-back to $X$, and applying the canonical pairing between $\Omega$ and $T$ yields $Res_\Delta f(z,w)dzdw$.

share|improve this answer
add comment

Maybe $f(z,w)dzdw\mapsto Res_{\Delta} [(z-w)fdzdw]$?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.