3
$\begingroup$

[This is a duplicate of this question on Stackexchange]

I am trying to figure out how to prove a very basic statement about convolution of $\ell$-adic/perverse sheaves in Katz's "Rigid local systems" (section 2.5.3, (1) ). The fact is fairly obvious and I guess that the proof is purely formal, but I'm not sure about it since my knowledge of the schemes formalism is quite limited.

Settings: $G/k$ is a (smooth, separated) group scheme over a field $k$, of pure relative dimension $d$. Denote $\mu: G\times_k G\rightarrow G$ the multiplication map and $e: k\rightarrow G$ the identity section. For $K$, $L\in D^b_c(G,\overline{\mathbb{Q}_l})$ (the "derived category" of $\ell$-adic sheaves over $G$, $l\neq \text{char}(p)$) define the product $K\times L\in D^b_c(G\times_k G,\overline{\mathbb{Q}_l})$ as $$K\times L:=pr_1^{*}K\otimes^{\mathbf{L}}pr_2^{*}L$$ with $pr_1$, $pr_2$ the canonical projections $G\times_k G\rightarrow G$, and $\otimes^{\mathbf{L}}$ the derived tensor product, which I'll just denote by $\otimes$ in the following.

Now define their $\star_{*}$ convolution as $$K\star_{*}L:=R\mu_{*}(K\times L)\in D^b_c(G,\overline{\mathbb{Q}_l})$$

The claim is that, if $G$ is commutative, then the $\star_{*}$ convolution is commutative.

What I did:

We want to show that if $K$, $L\in D^b_c(G,\overline{\mathbb{Q}_l})$, then $R\mu_{*}(K\times L)=R\mu_{*}(L\times K)$, i.e., as $\otimes$ is commutative, that $R\mu_{*}(pr_1^* K\otimes pr_2^*L)=R\mu_{*}(pr_2^* K\otimes pr_1^*L)$

  • First, I think that $$(pr_2,pr_1)^* (pr_2^*K\otimes pr_1^*L)\simeq pr_1^*K\otimes pr_2^*L$$ i.e., $(pr_2,pr_1)^*(L\times K)=K\times L$. This sounds reasonable that, if we switch both factors in $G\times_k G$, then we should replace $K\times L$ by $L\times K$. But I'm not sure this is obvious.

  • Moreover, we have the following commutative diagram (as $G$ is commutative) \begin{array}{ccc} G\times G &\xrightarrow{(pr_2, pr_1)}& G\times G\\ |& &| \\ \mu & &\mu \\ \downarrow & &\downarrow\\ G &\xrightarrow{id}&G \end{array}

and I think it's also cartesian. I would like to use some kind of "proper base change" to say that $R\mu_*(pr_2,pr_1)^*=R\mu_*$, and to apply $R\mu_*$ to my (claimed) equality $(pr_2,pr_1)^*(L\times K)=K\times L$ to conclude.

But the problem is that I'm not sure that I can do it directly with my diagram, as $\mu$ is maybe not proper.

  • Another option could just be to use the commutativity of my diagram to write $R\mu_*R(pr_2,pr_1)_*=R\mu_*$, and invoke that $R(pr_2,pr_1)_*=(pr_2,pr_1)^*$ by proper base change applied to the following diagram \begin{array}{ccc} G\times G &\xrightarrow{(pr_2, pr_1)}& G\times G\\ |& &| \\ id & &(pr_2,pr_1) \\ \downarrow & &\downarrow\\ G\times G &\xrightarrow{id}&G\times G \end{array}

Is there anything right in what I wrote ? Is there some easier proof ? Thanks for your help !

$\endgroup$
4
$\begingroup$

The thing that makes everything easy here is that the horizontal maps in your diagram are isomorphisms. For instance, every commutative square where the horizontal maps are isomorphisms is Cartesian.

The fact that the horziontal maps are isomorphisms also means the base change maps are isomorphisms. You can think of this as following from functoriality, but it's really more primitive - the isomorphism induces an isomorphism on the categories of etale sheaves, and we can take corresponding injective resolutions to get isomorphic etale cohomology groups.

For the identity, the key fact is that $(pr_2,pr_1)^* ( pr_2^* K \otimes pr_1^* L) = (pr_2,pr_1)^* pr_2^* K \otimes (pr_2,pr_1)^* pr_1^* L$ and then use functoriality of the pullback. This identity $f^* (A \otimes B) = f^* A \otimes f^*B$ is again "obvious" for an isomorphism $f$ but is known for general maps.

$\endgroup$
  • $\begingroup$ Thanks Will for this answer ! Fun to see that Mathoverflow is in fact a Zürich thing ;-) $\endgroup$ – Yoël Nov 4 '17 at 10:20
  • $\begingroup$ @Yoël As is sheaf convolution, I guess. Well, both of them I first learned how to use before I Came to Zuerich. $\endgroup$ – Will Sawin Nov 4 '17 at 14:04

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.