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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 $\textrm{Rep}(G^L)$

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  • $\begingroup$ The result by Mirkovic and Vilonen has nothing to do with de Rham cohomology. You may take coefficients in any ring. $\endgroup$
    – S. Carnahan
    Commented Feb 20, 2012 at 11:32
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    $\begingroup$ Which comparison isomorphism? You need to include more detail in your question if you want people to understand it. Please read the "how to ask" page, and follow the advice that is written there. $\endgroup$
    – S. Carnahan
    Commented Feb 21, 2012 at 11:53
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    $\begingroup$ It seems to me that you are giving an isomorphism of two fiber functors on the same category (a Betti and a de Rham one), hence an identification of two Tannakian groups (the "Betti" and "de Rham" Langlands dual groups)?? $\endgroup$ Commented Feb 22, 2012 at 21:20

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