most recent 30 from http://mathoverflow.net 2013-06-19T01:42:30Z http://mathoverflow.net/feeds/question/89033 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89033/differential-forms-with-poles-on-the-diagonal Jethro 2012-02-20T18:50:17Z 2012-02-21T02:30:12Z

<p>This question arises because I'm reading Frenkel and Ben Zvi's book "vertex algebras and algebraic curves" at the moment.</p> <p>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).</p> <p>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.</p> <p>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 $z = w = x$.</p> <p>My question is what is this isomorphism? It looks like some kind of residue analogue, but I'm not sure.</p> <p>Thanks.</p>

http://mathoverflow.net/questions/89033/differential-forms-with-poles-on-the-diagonal/89060#89060 tu_junwu 2012-02-20T23:51:28Z 2012-02-20T23:51:28Z

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

http://mathoverflow.net/questions/89033/differential-forms-with-poles-on-the-diagonal/89066#89066 Kevin Costello 2012-02-21T01:51:46Z 2012-02-21T01:51:46Z

<p>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. </p> <p>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 $X \times X$.</p> <p>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. </p> <p>Does that help?</p>

http://mathoverflow.net/questions/89033/differential-forms-with-poles-on-the-diagonal/89071#89071 tu_junwu 2012-02-21T02:30:12Z 2012-02-21T02:30:12Z

<p>Hi, Keven, good to hear from you here ^_^</p> <p>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 $Res_\Delta f(z,w)dzdw$.</p>