# Question about the precise statement of Leray spectral sequences and a simple example

On Bott's paper "Homogeneous vector bundles" there is the following statement of Leray spectral sequence:

Let $X$ and $Y$ be paracompact and locally compact spaces and $f : Y \to X$ be a proper map. Let $\mathscr F$ be a sheaf over $Y$ with cohomology of $Y$ with coefficients in $\mathscr F$ denoted by $H^*(Y;\mathscr F)$. Let $\mathscr H^q_V = H^q(f^{-1}(V); \mathscr F|_{ f^{-1}(V) })$ be the associated Leray pre-sheaf. Consider the sheaf $H^q$ whose fiber at each point $x \in X$ is given by the inductive limit of $H^q_V$ with $V \ni x$.

Theorem: Then there exist a spectral sequence $E_r$, with $$E^{p,q}_2 = H^p(X; \mathscr H^q),$$ whose final term is associated to $H^*(Y; \mathscr F)$.

(*) In particular, if $\mathscr H^q$ is 0 except for $q = k$, the spectral sequence is trivial and therefore $$H^p(Y; \mathcal F) = H^{p-k}(X; \mathscr H^k).$$

I have two questions:

1) Is it not true that $\mathscr H^0$ is always different from zero? If this is true, then the condition (*) does not make sense to me.

2) To understand this sequence I am doing some computations on the following example: Let $X = Y = S^1 = \{z = e^{i\theta} : \theta \in \mathbb{R}\}$ and let $f : S^1 \to S^1$ be given by $f(z) = z^2.$ and let $\mathscr F$ be the constant sheaf $\mathbb{C}$. In this case, for small $V \subset S^1$, we have that $f^{-1}(V)$ is a disjoint union of two open segments and so $\mathscr H^0_V = H^0(f^{-1}(V); \mathscr F|_{ f^{-1}(V) }) = \mathbb C^2$ and $\mathscr H^1_V = 0$. So we are in the condition (*), so $$H^p(S^1; \mathbb C) = H^p(S^1; \mathbb C^2).$$ Where am I making a mistake here?

• You are making a number of mistakes. $\mathcal{H}^0$ may very well be zero; in 2), $f_*(\mathcal{F})$ is not a constant sheaf. MSE would be a better fit for these elementary questions. – abx May 16 '17 at 5:14
• To elaborate a little on abx's comments. It may be the case that $\mathscr{H}^0$ is zero, because $\mathscr{F}(f^{-1}(U))$ might always vanish even though $\mathscr{F}$ is nonzero. You might have in mind the case where $\mathscr{F}$ is a constant sheaf: this does not happen then if $Y$ nonempty. Second, as abx says, in your example $\mathscr{H}^0$ is not a constant sheaf because there is nontrivial monodromy when you travel around $S^1$: it swaps the two factors of $\Bbb C$. (For the record, I wouldn't regard elementary questions on sheaf cohomology from nonspecialists as "beneath" this site.) – Tyler Lawson May 16 '17 at 5:24
• I would say that almost any question on sheaf cohomology is better suited to MO than MSE. – Neil Strickland May 16 '17 at 22:13