4
$\begingroup$

I am considering the following set up:Let $G$ be a finite group,let $Rep(G)$ denote the category of finite dimensional representations over $\mathbb{C}$. Let $V,W$ be representations of $G$ in $Rep(G)$. One can define a bilinear form on $Rep(G)$ or inner product in $K_0(Rep(G))$ (in Teleman's notes) as $dim_\mathbb{C}Hom(V,W)^G$ which is $G$ invariant of $Hom(V,W)$.Then there is a homomorphism:$\chi$:$K_0(Rep(G))\rightarrow Fun(G,\mathbb{C})$,such that $\chi(V,\rho)=Tr(\rho(g))$.Notice that the inner product on $Fun(G,\mathbb{C})$ is $\frac{1}{|G|}\Sigma_{g\in G}\bar{\chi(W,g)}\chi(V,g)$.I would like to consider the geometrization of this set up as following:

I identify the category of $Rep(G)$ with category of quasi coherent sheaves on quotient stack $[spec(\mathbb{C})/G]$ and equip with a bilinear form on $Qcoh([spec(\mathbb{C})/G])$ as the Euler characteristic: $\chi(V,W):=\Sigma_{i}(-1)^i dimExt^i(V,W)$,$V,W$ are representation of $G$ which can be regarded as quasi coherent sheaves. Since the category $Rep(G)$ is semisimple,the higher extension are vanishing,hence $\chi(V,W)$ is just $dimHom_{Qcoh}(V,W)=h^0([spec(\mathbb{C})/G],V\otimes W^*)\cong dimHom(V,W)^G$

My question is what is the inner product on the right handside? How can I geometrize the $\frac{1}{|G|}\Sigma_{g\in G}\bar{\chi(W,g)}\chi(V,g)$.Notice that if we consider the case of infinite group $G$,say Lie group. The summation over $g\in G$ should become the intergration over $G$. I have the guess that the inner product on the right hand side should be Hizebruch-Riemann-Roch for computing $\chi(V,W)$ which is $\int_{[spec(\mathbb{C})/G].}ch(V)ch(W)^*Td([spec(\mathbb{C})/G])$,it looks like similar to the formular $\frac{1}{|G|}\Sigma_{g\in G}\bar{\chi(W,g)}\chi(V,g)$

But I dont know how to make it very precise and how to prove or disprove my guessing. I found a paper online http://faculty.missouri.edu/~edidind/Papers/rrforDMstacks.pdf

It talks about the Grothendieck Riemann Roch for quotient stack. I am trying to understand his work and probably solve my question.

Thanks for your help. Any comments and suggestions are welcomed

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.