Let $X$ be a smooth complex projective variety and $D$ a normal crossing divisor. Assume that you are given a local system $V$ of complex vector spaces on $X-D$ having **finite monodromy**. Consider the intersection complex $IC\_X(E)$ on $E$. Is it true that $IC\_X(E)$ is a sheaf and not just a complex of sheaves? If this is the case, I guess the only reasonable possibility is 


$IC\_X(E)=j\_\ast E$ 


(maybe up to shift) where $j: X-D \hookrightarrow X$. 
Can anybody provide a prove of this statement ?  

Thanks!



is this intersection complex a sheaf? 
The answer is yes. (I suppose that $V=E$ in your statement.) This might be too complicated, and it's messy, but it's the first that came to mind. (The idea is really quite simple.)


I will write $j\_\*$ for the derived functor and ${}^\circ j\_\*$ for its ordinary $H^0$.


1/ First suppose that your local system $E$ is trivial. Then the statement you want is well-known, and a reference is lemma 4.3.2 of Astérisque 100.


2/ General case. Write $K=E[d]$, where $d=\dim X$ and $K'={}^\circ j\_\* E[d]$. So we're trying to prove that $j\_{!\*}K=K'$. 


The problem is local, so you can assume that $D$ is defined by a global equation $t\_1\dots t\_r=0$. Let $X'=X[t\_i^{1/N}]$ (ie the subscheme of $X\times\mathbb{A}^r$ defined by the equations $T\_i^N=t\_i$, where $T\_1,\dots,T\_r$ are the coordinates on $\mathbb{A}^r$). Let $\pi:X'\rightarrow X$ be the projection, $U'$ be the inverse image of $U:=X-D$ in $X'$ and $j':U'\rightarrow X'$ the inclusion. Then $\pi$ is finite, and its restriction to $U'$ is étale. As the monodromy of $E$ is finite, if we take $N$ big enough, then $\pi^\* E$ is trivial; write $E'$ for its obvious extension to $X'$, i.e., ${}^\circ j'\_\*\pi^\*E$. Then $j'\_{!\*}\pi^\*K=E'[d]$ by 1/. Note that $j\_{!\*}\pi\_\*\pi^\*K$ is a direct summand of $\pi\_\*j'\_{!\*}\pi^\*K$ (this is a very particular case of the decomposition theorem).


Using the trace map, we see that $E$ (resp. $K$) is a direct summand of $\pi\_\*\pi^\* E$ (resp. $\pi\_\*\pi^\*K$). So $j\_{!\*}K$ is a direct summand of $\pi\_\* E'[d]$ and ${}^\circ j\_\* E$ is a direct summand of $\pi\_\* E'$ (note that $\pi\_\*$ is exact in the ordinary sense and in the perverse sense). In particular, $j\_{!\*}K[-d]$ is an ordinary sheaf and ${}^\circ j\_\*E[d]$ is a perverse sheaf.


Using that $j\_{!\*}K[-d]$ is an ordinary sheaf, we get that the canonical morphism $j\_{!\*}K\rightarrow j\_\*K$ factors through a morphism $j\_{!\*}K\rightarrow {}^\circ j\_\* E[d]$, and this morphism is the identity on $U$ so it has to be injective. By the decomposition theorem again, ${}^\circ j\_\*E[d]=j\_{!\*}K\oplus L$, where $L$ is a perverse sheaf supported on $D$ and is such that $L[-d]$ is an ordinary sheaf. Let $i:D\rightarrow X$ be the inclusion. Then $i^!L[-d]$ is still an ordinary sheaf. But it is a direct factor of $i^! {}^\circ j\_\* E$, and, applying $i^!$ to the exact triangle ${}^\circ j\_\* E\rightarrow j\_\* E\rightarrow \tau\_{\geq 1}j\_\* E$ (and using that $i^! j\_\* E=0$), we get $i^! {}^\circ j\_\* E=i^!\tau\_{\geq 1}j\_\* E[-1]$. This is concentrated in (ordinary) degree $\geq 2$, hence $L=0$.


