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
- 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.
- 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?
Best Answer
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.