Geometric Langlands

Idea

Let $G$ be a complex reductive group with maximal torus $T\subset G$.

  1. Let $\Lambda$ and $\Lambda^\vee$ be the lattices of characters and 1-parameter subgroups of $T$,
  2. There are inclusions $\Lambda \subset \mathfrak{t}^\ast$ and $\Lambda^\vee\subset \mathfrak{t}$.
  3. Write the set of roots and coroots as $\Delta \subset \Lambda$ and $\Delta^\vee\subset \Lambda^\vee$.

The root and coroot data can be written as $$ (\Delta\subset\Lambda \subset \mathfrak{t}^\ast, \Delta^\vee\subset\Lambda^\vee\subset \mathfrak{t}). $$

Given $G’$ another complex reductive group with maximal torus $T$ with root-coroot data $$ (\Delta’\subset\Lambda’ \subset \mathfrak{t}’^\ast, \Delta’^\vee\subset\Lambda’^\vee\subset \mathfrak{t}’), $$ assume that there is an isomorphism $\phi:\mathfrak{t}\to \mathfrak{t}’^\ast$ interchanging the root and coroot data, viz. identifying $\Lambda$ with $\Lambda’^\vee$ and $\Delta$ with $\Delta’^\vee$, and that isomorphism $\mathfrak{t}’\to \mathfrak{t}^\ast$ induced by $\phi$ identifies $\Delta’$ with $\Delta^\vee$ and $\Lambda’$ with $\Lambda^\vee$.

Then $G’$ is the Langlands dual group of $G$, denoted by ${}^L G$. Note that ${}^L(^L G) = G$.

Remark. Examples are ${}^L \text{GL}_n \cong \text{GL}_n$, ${}^L\text{SL}_n \cong \text{PSL}_n$, ${}^L\text{SO}_{2n} \cong \text{SO}_{2n}$, ${}^L\text{SO}_{2n+1}\cong \text{Sp}_{2n}$; ${}^L T$ for a complex torus is the dual torus. In general the center $Z(G)$ of $G$ is naturally isomorphic to $\text{Hom}(\pi_{1}({}^L G),\mathbb{C}^\ast)$, the Pontrjagin dual of the fundamental group of ${}^L G$.

Now let $X$ be a smoth projective curve over $\mathbb{C}$ with genus $g \gt 1$ and let $\text{Bun}_G$ be the moduli stack of principal $G$-bundles.

Conjecture(Geometric Langlands) . For each irreducible ${}^L G$-local system $E$ on $X$, there is an irreducible holonomic $\mathcal{D}$-module $\mathcal{F}_E$ on $\text{Bun}_G$, such that $\mathcal{F}_E$ is a Heck eigensheaf with eigenvalue $E$.

Using perverse sheaves the field $\mathbb{C}$ can be replaced by $\mathbb{F}_q$.