Tensor product and semisimplicity of perverse sheaves

Question

Let $X/\mathbb{C}$ be a smooth algebraic variety. Let $D_c^b(X,\bar{\mathbb{Q}}_{\ell})$ be the category defined in 2.2.18, p.74 of "Faisceaux pervers" (by Beilinson, Bernstein and Deligne). Let $\mathrm{Perv}(X)$ be the subcategory of perverse sheaves. If $M,N$ are two simple objects of $\mathrm{Perv}(X)$, then is the derived tensor product $M\otimes^LN\in D_c^b(X,\bar{\mathbb{Q}}_{\ell})$ a direct sum of finitely many degree shifts of semisimple objects of $\mathrm{Perv}(X)$?

Answer 1

No. For example if $X = \mathbb A^1$, $j \colon \mathbb G_m \to \mathbb A^1$ the open immersion, $\mathcal L$ a rank one lisse sheaf on $\mathbb G_m$ with monodromy of order $2$, and $M = N = j_! \mathcal L[1]= j_* \mathcal L[1]$ (here $j_! = j_*$ because the local monodromy of $\mathcal L$ around $0$ is nontrivial so the pushforward has no sections at $0$ as these would have to be local monodromy invariant sections of the original sheaf) which is perverse and simple by 5.2.2(a) on p. 130 of BBD which ensures that $j_* $ of any lisse sheaf on a smooth open subset of a curve (taken in the non-derived sense) is perverse after a shift by $1$.

then $M \otimes N = j_! \overline{\mathbb Q}_\ell[2]$ which is not a sum of shifts of simple perverse sheaves. Instead, we have the short exact sequence of perverse sheaves

$$0 \to i_* \mathbb Q_\ell \to j_!\mathbb Q_\ell[1] \to \mathbb Q_\ell[1] \to 0$$

where $i$ is the inclusion of $0$. To check this short exact sequence, start from the short exact sequence of sheaves $$ 0 \to j_! \mathbb Q_\ell\to \mathbb Q_\ell \to i_* \mathbb Q_\ell \to 0$$ which gives a distinguished triangle in the derived category and then rotate that distinguished triangle, and finally check each member of the rotated distinguished triangle is perverse. (One can also apply general results).

By unique decomposition into simple objects, if $j_!\mathbb Q_\ell[1]$ were a sum of simples, every simple factor of $\mathbb Q_\ell[1]$ would be among them. But the stalk of $\mathbb Q_\ell[1]$ at $0$ is nonzero, so some simple factor of $\mathbb Q_\ell[1]$ has nonzero stalk at $0$ (in fact $\mathbb Q_\ell[1]$ is itself simple with nonzero stalk at $0$), so $j_! \mathbb Q_\ell[1]$ would have nonzero stalk at $0$, contradiction.

Shifts of simples are irrelevant since $j_!\mathbb Q_\ell[1]$ is itself perverse.