# External tensor product of two (perverse) sheaves

Motivation: I was reading through Frenkel's article on geometric Langlands program, and the external tensor product of two perverse sheaves occurred in the definition of the geometric Langlands conjectures. There should be a reference somewhere, but the closest I could find is this research note "Exterior Tensor Product Of Perverse Sheaves" by Lyubashenko, which glossed over the definitions far too quickly for me to grasp the content.

Question

1. Suppose $X$ and $Y$ are sheaves of vector spaces over two spaces $A$ and $B$ respectively (here "space" means topological space - but I'd also like to know how to do it for varieties or schemes, if that's different in a non-trivial way). My question is, how to define the external tensor product of the sheaves $X$ and $Y$, which should be a sheaf over the direct product $A \times B$?.

I've been thinking about it, and the idea I have is this (similar to how to construct the tensor product of two sheaves over the same space): we need to construct for each open set $U$ in $A \times B$, a vector space $F(U)$, and then sheafify this pre-sheaf. If $U$ is an open set of the form $A_1 \times B_1$ where $A_1, B_1$ are open in $A, B$, then this is straightforward: simply take the tensor product of the vector spaces corresponding to $A_1, B_1$ in the sheaves $X$ and $Y$, and the restriction maps on these are fairly clear. What is not clear to me is how to do these when $U$ is not of that form, but a union of some family of sets of the form $A_i \times B_i$. I tried by thinking there should be restriction maps from $F(U)$ to $F(A_i \times B_i)$ for each $i$, but I can't see how to explicitly construct the vector space just from that fact.

1. After doing that, how to go from there to an exterior tensor product of two perverse sheaves $X'$ and $Y'$ over $A$ and $B$, but now where $A$ is a variety but $B$ is an algebraic stack?
-
You should change the title, because perverse sheaves aren't really sheaves, and in general there's no tensor product defined for sheaves. A good rule of thumb is: if the word "sheaf" has no modifier attached, it means a sheaf of sets, because this is the fundamental case of an ordinary sheaf upon which all others are built. I say ordinary here in terms of ordinary category theory. In higher category theory, the theory of sheaves is replaced by the theory of (infinity,1)-sheaves or equivalently, (infinity,1)-stacks, which give us a model for higher grothendieck toposes. – Harry Gindi Jan 20 '10 at 8:09
Harry: It depends on the context. When I say "sheaf" in almost any remotely algebraic geometric context, I usually mean implicitly O-module. Sometimes I even mean coherent O-module. When I'm feeling really naughty, I might even mean locally free O-module. I don't think I'm the only person who does these things. – Kevin H. Lin Jan 20 '10 at 9:13
@Harry: your knowledge is very impressive! – Martin Brandenburg Jan 21 '10 at 6:17
Harry- Perverse sheaves are in constructed in the derived category of all sheaves of vector spaces (with constructible cohomology), not O-modules. Now you look like an idiot. – Ben Webster Jan 22 '10 at 3:00
Sorry I didn't mean to start a silly argument over some trivialty about the spelling of topos/topoi, there's no need to call anyone an idiot. I might be wrong since I'm just a beginner, but I still don't think it's too hard to proceed from tensor products of two sheaves of vector spaces to tensor product of two perverse sheaves - the only thing to check seems that the tensor product of two perverse sheaves is actually a perverse sheaf, which seems straightforward (e.g. it seems enough to prove that the tensor product lies in the heart of the perverse $t$-structure). – Vinoth Jan 23 '10 at 0:49

You don't need to construct $F(U)$ explicitly for $U$ that are not products. Any point $(a,b)$ of $A\times B$ has a basis of neighbourhoods of the form $A_1 \times B_1$, so the presheaf defined on just these open sets is enough data to sheafify and obtain the corresponding sheaf. Since sheafification preserves stalks, you will find that the stalk of $X\boxtimes Y$ at $(a,b)$ is equal to the stalk of $X$ at $a$ tensored with the stalk of $Y$ at $b$.

With this construction in hand we can define exterior tensor product for complexes of sheaves, and then hence for perverse sheaves.

To treat the case where one of $A$ or $B$ is a stack, the main point will be to have a precise definition of what is meant by a perverse sheaf on a stack. Once this is understood, the definition of exterior tensor product will proceed in the same way as for the case of varieties.

-
Silly question: this seems to only define a sheaf on the product topology. You mention the case of varieties: I would expect in that case that you would want a sheaf on the Zariski product A x B. Is this expectation off base? – Tom Church Jan 20 '10 at 7:11
I was thinking of the situation of varieties over the complex numbers, in which case peverse sheaves are sheaves in the complex topology. (I would guess that this is the context in which Frenkel is working.) – Emerton Jan 20 '10 at 14:02
Dear Tom, I should add that in other contexts, one might instead work with perverse sheaves on the etale site (of A x B, say, in the present context), and then one could use the definition Strom Borman gives in his answer, for example. (One would not normally use the Zariski site, for the same reason that there are normally no interesting local systems --- of which perverse sheaves are a generalization --- in the Zariski site.) – Emerton Jan 22 '10 at 3:56

Let $p_A: A \times B \to A$ and $p_B: A \times B \to B$ be the projection maps. The external tensor product of the sheaves $X$ and $Y$ is the normal tensor product of the pullback sheaves $p_A^*X$ and $p_B^*Y$.

(At least that is the formalism for external products for vector bundles.)

-

So we have topological spaces $A,B$ and sheaves $F,G$ on $A,B$ of vector spaces over some fixed field $k$ and want to construct a sheaf $A \otimes_k B$ on the product space $A \times B$. You can write it down explicitly:

Let $W \subseteq A \times B$ be open. Then $(F \otimes_k G)(W)$ consists of those elements $s \in \prod_{(a,b) \in W} F_a \otimes_k G_b$, such that for all $(a,b) \in W$ there are open sets $a \in U \subseteq A, b \in V \subseteq B$ and $t \in F(U) \otimes_k G(V)$ such that $U \times V \subseteq W$ and for all $(c,d) \in U \times V$, we have $t_{c,d} = s_{c,d}$. Here $t \mapsto t_{c,d}$ denotes the canonical map $F(U) \otimes_k G(V) \to F_c \otimes_k G_d$.

Note that this obviously(!) yields a sheaf on $A \times B$. On stalks, there is a canonical map $(F \otimes_k G)_{a,b} \to F_a \otimes_k G_b$; a calculation shows that it is bijective. Remark that this agrees with the definition given by Strom Borman (the same universal property holds). But here you have a description of the sections of $F \otimes_k G$. In particular, you see that if $F$ and $G$ are the sheaves of $\mathbb{K}$-valued continuous functions on $A$ resp. $B$, then $F \otimes_\mathbb{K} G$ is a rather small subsheaf of the continuous functions on $A \times B$.

The whole things makes more sense, when we take $A,B$ to be two $S$-schemes (or more generally, locally ringed spaces). Then we have the fibred product $A \times_S B$ which can be constructed as above (I've written this up here (in german)). Here, the tensor product is the "right" sheaf.

-