Some time ago I was told there's an interesting classical Satake correspondence which I will write as

$$[\mathop{\mathrm{disk}} \Rightarrow G] \,\backslash\, [\mathop{\mathrm{disk}^\times} \Rightarrow G] \,/\, [\mathop{\mathrm{disk}} \Rightarrow G] \,=\, X_*/W \,=\, G^\vee\mbox{-reps}$$

(where $G$ is reductive, $\mathrm{disk}$ could be the spectrum of either $\mathbb Z_p$ or $k[[t]]$ and $\Rightarrow $ denotes algebraic morphism) and its categorified/geometric version (equivariant perverse sheaves on affine grassmanian of $G$ form the tensor category of representations of $G^\vee$).

I think I'm missing the larger context here, though. I don't mean the context of perverse sheaves, geometric Langlands, etc. On the contrary, I feel like I miss any intuition for classical representation theory. Why would statements like this be interesting?

I wasn't able to find anything in wikipedia or nLab.

One thing I know is that the correspondence allows us to construct the Langlands dual group in a natural way. But still, it would be interesting to know if it's a part of larger picture and if there are related results.

Question: is there an intuition for Satake correspondence that would make its statement obvious?


What you have written above isn't classical Satake; it's the generalized Bruhat decomposition. Classical Satake is a much more interesting theorem, which says that the Hecke algebra of $G(\mathcal{K})$ over $G(\mathcal{O})$ (the compactly supported $G(\mathcal{O})$ bi-invariant functions on $G(\mathcal{K})$ with convolution multiplication) is isomorphic to the representation ring of $G^\vee$.

Why is this interesting? Because the Hecke algebra $L^2[G(\mathcal{O})\backslash G(\mathcal{K})/G(\mathcal{O})]$ is the endomorphism algebra of $L^2(G(\mathcal{K})/G(\mathcal{O}))$, so there's a bijection between $W$-orbits on $T^\vee$ and representations of $G(\mathcal{K})$ appearing in the $L^2$ above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.