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

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

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 $$M$$.

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