Stratification along which a constructible complex is smooth Let $X$ be a smooth complex algebraic variety. A constructible complex $F$ on $X$ has a singular support $SS(F)\subset T^*X$. Assume you are given a stratification of $X$ such that $SS(F)$ is the union of the conormal bundles to some of these strata. Is there some condition assuring that $F$ is constructible with respect to the given stratification ?
I am more specifically interested with the Bruhat stratification of a reductive group.
 A: I think the answer is yes for the Bruhat stratification, and moreover for the orbit stratification of any space $X$ with an action of a group $G$ with finitely many orbits, such that any irreducible lisse sheaf on an orbit is $G$-equvariant.
The proof is that we may reduce to the case of an irreducible perverse sheaf, since every complex is an interated extension of shifted perverse sheaves, thus if these perverse sheaves are constructible with respect to a given stratification then the complex is as well, and to the singular support of these perverse sheaves is contained in the singular support of the complex.
Then the conormal bundle of the support of the irreducible perverse sheaf is always contained in its singular support, so the support must be (the closure of) one of the strata. On this stratum, the conormal bundle of every irreducible component of the singular locus of the perverse sehaf is contained in its singular support, so the singular locus must be contained in a union of lower-dimensional strata, and hence the sheaf is lisse (and irreducible) on that stratum. Hence because every irreducible lisse sheaf on a Bruhat cell is constant, and therefore $B$-equivariant, the perverse sheaf is a middle extension of a $B$-equivariant lisse sheaf, and thus is $B$-equivariant, and hence is constructible with respect to the stratification into orbits of $B$.
