At @random123's request, I'm will try to argue that looking for tilting perverse sheaves takes one very close to a setting in which [BBM] work.
Suppose that we have a suitably stratified variety
\begin{equation}
X = \bigsqcup_{\lambda \in \Lambda} X_\lambda
\end{equation}
and we wish to understand $D = D^b_\Lambda(X,k)$, the derived category of $\Lambda$-constructible sheaves of $k$-vector spaces.
Often we try to understand derived categories by finding a tilting generator $T$, i.e. an object $T$ which generates $D$ and has no higher extensions. In this case $D$ is equivalent to a suitable derived category of modules over $End(T)$ ("tilting theory").
In general, describing all tilting complexes in $T$ is hopeless. (In examples I am familiar with, there are more than I can possibly comprehend, and classifying them is hopeless -- it is somewhat analogous to classifying all $t$-structures on $D$.)
However, the theory of highest weight categories (aka quasi-hereditary algebras) leads one to look for tilting complexes $T$ which are perverse and may be written as a direct sum $T = \bigoplus T_\lambda$ in a way that is "compatible with the stratification", more precisely:
- $T_\lambda$ is supported on $\overline{X}_\lambda$.
- The set $\{ T_\mu \; | \; \mu \le \lambda \}$ generates the full subcategory of $D$ generated consisting of complexes supported on $\overline{X}_\lambda$.
Firstly, note that "one $T_\lambda$ per strata" implies that there are no local systems on strata, in other words that $\pi_1(X_\lambda) = \{ 1 \}$. (Or at least that fundamental groups have no finite-dimensional representations.)
Next, it is a nice exercise to show that these assumptions imply that (if one denotes by $j : X_\lambda \hookrightarrow \overline{X}_\lambda$ the inclusion, one has a surjection $Hom^i(T_\lambda, T_\lambda) \to Hom^i(j^*T_\lambda, j^*T_\lambda)$. The latter group is equal to $H^i(X_\lambda, k)$, because I hope I convinced you in the previous paragraph that $j^*T_\lambda$ is a shift of a constant sheaf on $X_\lambda$.
In other words, we are forced to take a stratification with strata which look a lot like they are contractible. Thus we are very close to the setting in which [BBM] work.
(Please see comments following question for more specialized comments.)
Aside: I understand that the title "tilting exercises" refers to the medieval practice of riding towards each other carrying long jousting poles. When reading this beautiful paper, I was mystified as to why they also cite Oliver Twist. Only recently did I understand, but leave this as a tilting exercise for the interested reader.
Best Answer
No, this is not a purely $\ell$-adic phenomenon.
Let $X = \mathbb P^1$, $S$ the stratification with one stratum so sheaves constructible with respect to this stratification are lisse and complexes constructible with respect to this stratification complexes with lisse cohomology.
Let $F = G = \mathbb Q[1]$, clearly constructible with respect to $S$ and perverse.
Then $Hom^2_{D^b_S(X,k) } (F,G) =Hom^2_{D^b(X,k) } (F,G) = H^2(X,\mathbb Q) = \mathbb Q$ but $\operatorname{Perv}(X)$ is just the category of lisse sheaves on $\mathbb P^1$, i.e. the category of vector spaces, and has all higher Ext groups vanish.