# 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.

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?

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

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