If we regard the comparison isomorphism of cohomologies , we are given an isomorphism of Langlands dual group into itself; i wondered if it had any interest whatsoever, i cannot figure myself.

If we regard the comparison isomorphism of cohomologies , we are given an isomorphism of Langlands dual group into itself; i wondered if it had any interest whatsoever, i cannot figure myself.

for example if X is a smooth sheme over $\textbf{C}$, we have the classical comparison isomorphism between de Rham and singular cohomology:$\phi: H_{dr}(X/\textbf{C})\equiv H^{*}(X(\textbf{C},\textbf{Q})\otimes \textbf{C}$.

If $F=k(X)$ and G is a connected reductive group on F, then Satake equivalence gives us a monoidal isomorphism between the categories of G-equivariant perverse sheaves denoted by $Perv_{H}^{G}(X,coeff)$ ( for a cohomology H on a ring coeff) and the category of representations of Langlands dual group $G^{L}$ But $\phi$ induces a morphism between the category of perverse sheaves $Perv_{dr}^{G}(X, \textbf{C})$ and $Perv_{B}^{G}(X, \textbf{C})$ with respect to these cohomologies, so it should a monoidal endofunctor of $\textrm{Rep}(G^L)$

probably it is very trivial, but i wondered if Satake equivalence of categories of perverse sheaves worked only for perverse sheaves with respect to de Rham cohomology, and in particular, roughly if there is a way to "see" comparison maps of cohomologies through it; i will try to precise maybe

If we regard the comparison isomorphism of cohomologies , we are given an isomorphism of Langlands dual group into itself; i wondered if it had any interest whatsoever, i cannot figure myself.