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?



Formula for the Euler characteristic of a local system on $\mathbb{P}^1$
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 $M$.


