Differential forms with poles on the diagonal

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.

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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?

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 Thanks! Yes that helps a lot. – Jethro Feb 22 2012 at 18:45

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 $Res_\Delta f(z,w)dzdw$.

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

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