# Perverse sheaves and tensor product

If $X$ is a connected algebraic variety of finite type over $k$ (with $k$ a field of positive characteristic) of dimension $d$, and if $\mathcal{F}$ and $\mathcal{G}$ are perverse sheaves on $X$ so $(\mathcal{F}\otimes\mathcal{G})[-d]$ is also a preverse sheaf?

In the same setting above if we have a unipotent connected group over $k$ acting on $X$ transitively what can we say about the equivariant derived category of $X$ (the category of equivariant complexes in $X$) ?

• One aid for thinking about tensor products. There are actually 2 "tensor products": $\mathcal{F} \otimes \mathcal{G} := i_\Delta^* (\mathcal{F} \boxtimes \mathcal{G})$ and $\mathcal{F} \stackrel{!}{\otimes} \mathcal{G} := i_\Delta^!(\mathcal{F} \boxtimes \mathcal{G})$. (Notation from Ginzburg.) Anyway, if $\mathcal{F}$ and $\mathcal{G}$ are self-dual then we have $\mathbb{D}( \mathcal{F} \otimes \mathcal{G}) = \mathcal{F} \stackrel{!}{\otimes} \mathcal{G}$ which explains why the tensor product fails to be perverse: one version satisfies stalk condition, the other version costalk cond. – Geordie Williamson Jun 13 '16 at 20:24

This is extremely false. Consider the skyscraper sheaf on a smooth point of a positive dimensional variety; this is always perverse (since it is Verdier self-dual). The tensor product of this with itself will be the same sheaf again, so when you shift, you mess up perversity.

• silly question: could you explain why the (derived) tensor product of a skyscraper with itself is the skyscraper again? I know in the coherent world this is super false, but I have no intuition for constructible stuff. – Yosemite Sam Jun 13 '16 at 19:43
• @YosemiteSam Note, I didn't say derived tensor product, and neither did the OP. It's irrelevant for this point, since if there were higher derived functors, that would just be a second reason it's not perverse. I don't believe there are, since I'm taking tensor product of vector spaces, which is exact. It fails for coherent sheaves because there you mean tensor product as $\mathcal{O}_X$ modules, which can fail to be flat. – Ben Webster Jun 13 '16 at 20:17
• thanks. So, if I understand correctly, your reasoning is that if I take a constant sheaf of rings R and have a skyscraper S with stalk a free module, then tensoring with S is exact? (and thus has no higher derived functors) – Yosemite Sam Jun 13 '16 at 20:21
• That's both more special (why restrict to skyscrapers?) and more general (free stalks should be enough) than what I said, but yes. Constant sheaves of rings are quite different from structure sheaves. – Ben Webster Jun 13 '16 at 20:25
• thanks, for some reason I was unnecessarily mystified by constructible things. – Yosemite Sam Jun 13 '16 at 20:28

I don't really know much about perverse sheaves, but the answer for the first general question is "No", I think, even for curves.

Here is a simple example for $\bar{\mathbf{Q}}_{\ell}$-sheaves, which is the context I understand a bit. Take a Kummer sheaf on the multiplicative group over a finite field associated to a non-trivial character $\chi$ (often denoted $\mathcal{L}_{\chi}$), and define a perverse sheaf $F$ on the projective line as the middle-extension of this sheaf to the projective line (suitably normalized by shifting); this is just the extension by zero. Let $G$ be its dual; it is the perverse sheaf associated in the same way to $\mathcal{L}_{\bar{\chi}}$. Then $F\otimes G[-1]$, unless I misunderstand the definitions, is the extension by zero of the lisse sheaf $\mathcal{L}_{1}$ on the multiplicative group, i.e., the extension by zero to the projective line of the trivial rank $1$ sheaf. This is not perverse (because it would be simple perverse, being of rank $1$, and therefore, because of the classification of these and because it is not punctual, it would be the middle extension of its restriction to the multiplicative group, but the latter is the trivial sheaf on the projective line).

• If I understand correctly, the extension by zero of $\mathcal{L}_1$ is perverse; pushforward by affine open inclusions preserves perversity. – Ben Webster May 4 '16 at 15:56
• You're most probably right! I'm barely starting to try to learn these things... – Denis Chaperon de Lauzières May 4 '16 at 16:26
• On reflecting, I am not so sure... Indeed, pushforward from the multiplicative group to the projective line will preserve perverseness, but the extension by zero of the trivial sheaf is not its pushforward (denoting by $j$ the open immersion, extension by zero is $j_!\bar{\mathbf{Q}}_{\ell}$ instead of $j_*\bar{\mathbf{Q}}_{\ell}$; the latter is the trivial sheaf on the projective line). – Denis Chaperon de Lauzières May 8 '16 at 16:26
• These are switched under Verdier duality (well, you have shift both so that they are perverse) so one is perverse if the other one is. Both pushforwards preserve perversity under open affine inclusions. – Ben Webster May 9 '16 at 1:37

I know next to nothing about perverse sheaves, except that they are often used as a geometric construction of highest weight categories from Lie theory. In a highest weight category, you have things like simple objects, standard objects, projectives, injectives, and tiltings. Most of these classes will not be preserved under tensor product. However, the tensor product of tilting modules is a tilting module (in the context of finite-dimensional reps of reductive algebraic groups in positive characteristic). Thus your statement ought to be true if you make it about tilting perverse sheaves, instead of all perverse sheaves, but under an appropriate multiplication for perverse sheaves, such as convolution, and for appropriate X. I don't know in what generality this would hold.

Notes on tilting modules in positive characteristic representation theory: http://www.imsc.res.in/~knr/vdk/mathieu.pdf Geometric survey of tilting theory: http://arxiv.org/pdf/math/0301098.pdf