# Leray spectral sequence of the inclusion of an open subvariety

Let $X$ be a smooth variety over a field $k \subset \mathbb{C}$ and $Z$ a smooth subvariety. Let $U=X-Z$. I'm trying to understand what information do the Leray spectral sequences attached to the inclusions

$j: U(\mathbb{C}) \hookrightarrow X(\mathbb{C})$

and

$i: Z(\mathbb{C}) \hookrightarrow X(\mathbb{C})$

provide. More precisely, if $F$ is a sheaf on $U$ one has a spectral sequence

$E_2^{p,q}:=H^p(X, R^q j_\ast F) \Longrightarrow H^{p+q}(U, F)$

Is there a long exact sequence associated to it such that, when $F$ is the constant sheaf $\mathbb{Q}$, gives something like excision for usual cohomology?

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Yup, the sequence you get is the one written here mathoverflow.net/questions/121910/… by another anonymous user. – Donu Arapura Feb 15 '13 at 18:55
@Donu: I cannot see the connection, perhaps you can elaborate this in an answer? – Martin Brandenburg Feb 15 '13 at 19:02
Marthin, see below: – Donu Arapura Feb 15 '13 at 19:36

Here are a few more details: Suppose that $F=\mathbb{Q}_X$. In this case, $$R^qj_*F = \begin{cases} \mathbb{Q}_X & \text{if q=0}\\ \mathbb{Q}_Z & \text{if q= 2 c-1, c=codim(Z)}\\ 0 & \text{otherwise} \end{cases}$$ To see this, first note that the computation is local, so we may replace the pair $(X,Z)$ by $(\mathbb{C}^n, \mathbb{C}^{n-c})$ using the implicit function theorem. After this, it should be easy. With this calculation the $E_2$ term of Leray is concentrated along two lines, and it is not hard to see that it reduces to the above "localization" sequence given in excision in algebraic de Rham cohomology

Notes: Sorry, I had a typo in the formula yesterday. If $Z$ is smooth but not pure dimensional, then $R^qj_*F|_{Z_i}= \mathbb{Q}_{Z_i}$ where $q=2c_i-1$ for each component. The final result will be a bit more cumbersome to state. Some related calculations occur in Deligne's Theorie de Hodge I, II.

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Thanks, Donu! Your answer is very useful. I'm not very familiar with tubular neighbourhoods. Could you give some references? – lerex Feb 15 '13 at 22:16
Another question: what happens if $Z$ is not of pure dimension? Do you have the same $R^qj_\astF=\mathbb{Q}_{Z_i}$ if $q=q2c_i$ for each irreducible component? – lerex Feb 15 '13 at 23:47

For a closed immersion $i : Z \hookrightarrow X$ of schemes the spectral sequence is degenerate and becomes the elementary isomorphism $H^p(X,i_* F) \cong H^p(Z,F)$.

For a morphism $j : U \hookrightarrow X$ of schemes the five term exact sequence associated to the spectral sequence becomes

$0 \to H^1(X,j_* F) \to H^1(U,F) \to \Gamma(X,R^1 j_* F) \to H^2(X,j_* F) \to H^2(U,F).$

In general this cannot be simplified, even if $j$ is an open immersion. But this is useful, for example for the computation of the etale cohomology of $\mu_n$ and $\mathbb{G}_m$ on a curve $X$, where $j$ is the inclusion of the generic point (see Chapter 10 in Tamme's book on etale cohomology).

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