# Convolution of $\ell$-adic sheaves is commutative if the group is commutative

[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 !

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.