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 


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*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. 


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*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?

 A: 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. 
A: 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.)
A: 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.
