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 ?


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 $L=0$.

| cite | improve this answer | |
  • 1
    $\begingroup$ Dear SM, perhaps this will simplify things (?). Since, as you say, it is local, and the local monodromy is diagonalizable, you can reduce to the case where $E$ has rank one with nontrivial monodromy about all components of $D$. Then check $\mathbb{R}j_*E[d] = j_*E[d]= j_!E[d]$ which should yield perversity. Also $j_*E[d]$ should vanish along $D$, which out to imply that this is the minimal extension, i.e. $j_{!*}E[d]$. But perhaps I am overlooking something. $\endgroup$ – Donu Arapura Mar 22 '13 at 15:25
  • 1
    $\begingroup$ Yes, I think you're right, and the fact that the intermediate extension is ${}circ j_*E[d]$ simply follows from $j_!E=j_*E$ (and the definition). $\endgroup$ – user31960 Mar 22 '13 at 17:27
  • 1
    $\begingroup$ Another point of view which might be useful: If $X = X_1 \times X_2$ then $IC(L_1 \boxtimes L_2) = IC(L_1) \boxtimes IC(L_2)$ and so one can reduce to the case of a line. $\endgroup$ – Geordie Williamson Mar 23 '13 at 8:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.