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I am interested in computing tensor products of perverse sheaves on (partial) flag varieties. For a specific example - consider the product of the big projective on $\mathbb{P}^1$ with itself (This is the projective cover of the skyscraper sheaf on the 0-dimensional stratum). Does this have a simple description? How can I compute its cohomology?

Any general tips or computational tricks in this context are very welcome. I am especially interested in tilting perverse sheaves such as the one above.

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    $\begingroup$ Maybe interpret tensor product as external tensor product (living on $X \times X$) followed by pullback along the diagonal map... It is not stable in mind, but maybe the external tensor product will again be some projective cover or something describable? then you need to understand the pullback operation, perhaps it has some meaning as a functor (i.e. not just for this specific object) $\endgroup$
    – Sasha
    Commented Jun 16, 2021 at 7:54

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First a general comment: as Sasha alludes to, there are two tensor products of complexes of sheaves. Let $i : X \hookrightarrow X \times X$ denote the diagonal. We have, and given complexes of sheaves $\mathcal{F}$ and $\mathcal{G}$ on $X$ we can form their external tensor product $\mathcal{F} \boxtimes \mathcal{G}$ on $X \times X$. The two tensor products are \begin{gather*} \mathcal{F} \otimes \mathcal{G} := i^*(\mathcal{F} \boxtimes \mathcal{G}) \quad \text{and} \\ \mathcal{F} \otimes^{!} \mathcal{G} := i^!(\mathcal{F} \boxtimes \mathcal{G}). \end{gather*} (I think I learnt this notation from some very nice old notes of Ginzburg.) This description makes it clear that $\mathbb{D}(\mathcal{F} \otimes \mathcal{G}) = \mathbb{D}\mathcal{F} \otimes^{!} \mathbb{D}\mathcal{G}$ etc. In particular, we have control of the stalks of $\mathcal{F} \otimes \mathcal{G}$ in terms of the stalks of $\mathcal{F}$ and $\mathcal{G}$, and costalks of $\mathcal{F} \otimes^{!} \mathcal{G}$ in terms of costalks of $\mathcal{F}$ and $\mathcal{G}$, but for example the costalks of $\mathcal{F} \otimes \mathcal{G}$ are potentially very complicated.

Now to your specific question: consider the big tilting sheaf $T_s$ on $\mathbb{P}^1$. (This is isomorphic to the projective cover you ask about.) Its stalks are $\mathbb{Q}[1]$ on the open locus, and $\mathbb{Q}$ at the point stratum. In particular, the stalks of $T_s \otimes T_s$ are $\mathbb{Q}[1] \otimes \mathbb{Q}[1] = \mathbb{Q}[2]$ on the open locus, and $\mathbb{Q}$ on the point stratum. In particular we have a distinguished triangle \begin{equation} j_!\mathbb{Q}[2] \to T_s^{\otimes 2} \to \mathbb{Q}_{0} \stackrel{[1]}{\to} \end{equation} where $j$ denotes the inclusion of the open stratum, and $0$ denotes the point stratum. Because the appropriate Ext group vanishes, we deduce that \begin{equation} T_s^{\otimes 2} \cong j_!\mathbb{Q}[2] \oplus \mathbb{Q}_{0}. \end{equation} (Already in this example we see the failure of the tensor product to be self-dual.)

What about in general? We know the stalks of tilting sheaves on $G/B$ (see Yun's "weights of mixed tilting sheaves" for a lovely account). They are concentrated in a single degree, with dimension given by Kazhdan-Lusztig polynomials at 1. Hence the stalks of $T_x \otimes T_y$ along the stratum $BzB/B$ are concentrated in degree $-2\ell(z)$ and are computable via a product of KL polynomials at $1$. I am not sure if the tensor product splits as above. But maybe that is your job :)

An aside: if one instead considers tensor products of $IC$ sheaves several small example suggest that the answer is always pure. (This has gotten me excited at least twice in my life, and I know I am not the only one!) However this is not the case in general, as the example of $IC_{st} \otimes IC_{ts}$ in $SL_3/B$ shows. (In this example the tensor product is the constant sheaf on the union of the two closed Schubert curves corresponding to $s$ and $t$.)

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  • $\begingroup$ Thanks! That’s very helpful $\endgroup$
    – Adam Gal
    Commented Jun 17, 2021 at 23:02

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