Let $X := \mathbb{P}^1$, $S\subset X$ a finite set of points, $U := X - S$, and $j : U\rightarrow X$ the inclusion.

Let $F$ be a complex local system on $U$ of rank $r$, and let $F_0$ be a typical fiber, so $F_0$ is a complex vector space. On p14 of Katz's book Rigid local systems, he says that $$\chi(X,j_*F) = r\cdot\chi(U,\mathbb{C}) + \sum_{s\in S}\text{dim}_{\mathbb{C}}F_0^{I(s)}$$ where $I(s)\cong\mathbb{Z}$ denotes the local monodromy group at $s$ (the fundamental group of a punctured neighborhood of $s$).

Why is this true? This is a very naive question, and I'm clearly just missing some basic points about the cohomology of local systems. I'm hoping someone here can help fill me in.

In the simple case $S = \{s\}$, let $D$ be a small neighborhood of $s$, $D^* := D - s$, then Mayer Vietoris would give: $$\chi(X,j_*F) = \chi(U,F) + \chi(D,j_*F) - \chi(D^*,F)$$ Since $h^0(D,j_*F) = h^0(D^*,F) = F_0^{I(s)}$, we have $$\chi(X,j_*F) = \chi(U,F) - h^1(D,j_*F) + h^1(D^*,F) + h^2(D,j_*F) - h^2(D^*,F)$$

  1. I think we should have $\chi(U,F) = r\cdot \chi(U,\mathbb{C})$ (why?).
  2. Assuming that local system cohomology is homotopy invariant, we should have $h^2(D^*,F) = 0$.
  3. By comparison with group cohomology, we have $h^1(D^*,F) = r$.

Thus the desired result would follow in this simple case as long as $\dim_{\mathbb{C}}F_0^{I(s)} = h^2(D,j_*F) - h^1(D,j_*F)$. Is this true?


1 Answer 1


The answer to your question at the end is negative. In fact, $h^2(D, j_*F)= h^2(D, j_* F)=0$. In fact, the cohomology of a sufficiently small disc around a point in any complex variety, with coefficients in any fixed constructible sheaf, vanishes in all positive degrees.

This is because taking global sections on a small disc around a point is the same as taking the stalk at that point and is an exact functor.

The error is in your step 3. In fact group cohomology shows $ h^1(D*, F)= h^0(D^*,F) = \dim_{\mathbb C} F_0^{I_S}$, since the group cohomology $H^1(\mathbb Z, M)$ is the coinvariants of $M$ and thus has the same dimension as the invariants of $M$ for finite-dimensional $M$.

  • $\begingroup$ Thank you so much! That's exactly what I was looking for. $\endgroup$ Sep 22 at 13:14

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