记些杂七杂八的感想.
本系列完全不定时更新.
From now on, we let $\Sigma$ be a surface.
Firstly, we aimed to calculate $\operatorname{Loc}(Pic\Sigma)$. Here we identify $\operatorname{Pic}\Sigma$ be the space of holomorphic line bundles on $\Sigma$. Then we have a short exact sequence:
\[0\to \operatorname{Jac}\Sigma\to \operatorname{Pic}\Sigma\overset{\operatorname{deg}}{\to}\mathbb{Z}\to 0.\]$\operatorname{Jac}\Sigma$ the degree $0$ line bundles. By the theory of Riemann Surface, one can topologically identify:
\[\operatorname{Jac}\Sigma=\mathbb{C}^g/\mathbb{Z}^{2g}.\]One can identify $\pi_1(\operatorname{Jac})=H^1(\Sigma,\mathbb{Z})$. Then we have abelian category equivalence
\[\begin{align*} \operatorname{Loc}(\operatorname{Pic}\Sigma)&=\mathbf{Vect}^{\mathbb{Z}\text{-gr}}\otimes \operatorname{Loc}(\operatorname{Jac}\Sigma)\\ &=\mathbf{Rep}_{\mathbb{Z}\text{-gr}}(H^1(\Sigma,\mathbb{Z}))\\ &=\mathbf{Vect}^{\mathbb{Z}\text{-gr}}\otimes(\mathbb{C}H^1(\Sigma,\mathbb{Z}),\ast)-\mathbf{mod}. \end{align*}\]Secondly, we shall consider the algebraic geometry of the character scheme $\operatorname{Loc}_{\mathbb{C}^\times}$. To begin with, we can write the character scheme by
\[\begin{align*} \operatorname{Loc}_{\mathbb{C}^\times}(\Sigma)&=\operatorname{Hom}(\pi_1(\Sigma,\mathbb{Z}),\mathbb{C}^\times)/\mathbb{C}^\times\\&=\operatorname{Hom}(H_1(\Sigma,\mathbb{Z}),\mathbb{C}^\times)/\mathbb{C}^\times\\ &=T^\vee/\mathbb{C}^\times.\end{align*}\]Then, by the theory of algebraic geometry, there are
\[\mathbf{QC}(\operatorname{Loc}_{\mathbb{C}^\times}\Sigma)=(\mathbb{C}[T^\vee],\cdot)\text{-}\mathbf{mod}\otimes \mathbf{Rep}(\mathbb{C}^\times).\]With $(\mathbb{C}[T^\vee],\cdot)$ denote the global functions on $T^\vee$. And we can identify $\mathbf{Rep}(\mathbb{C}^\times)$ as $\mathbf{Vect}^{\mathbb{Z}\text{-gr}}$.
But by Fourier transform, we can identify
\[(\mathbb{C}[T^\vee],\cdot)\simeq (\mathbb{C}H^1(\Sigma,\mathbb{Z}),\ast).\]Hence we finally attains
\[\operatorname{Loc}\operatorname{Pic}(\Sigma)\simeq \mathbf{QC}(\operatorname{Loc}_{\mathbb{C}^\times}(\Sigma)).\]从 Ben-Zvi 的一篇课程笔记: Between electric-magnetic duality and the Langlands program 抄过来的.
这个实际上是最最简单的 Betti Langlands Correspondence: $G=\operatorname{GL}_1$ 的 case.
水点轻松的先.
但是这也不轻松, 原文有一些 typo. Check 一下还有些麻烦的.