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Let $f \colon X \to Y$ be a proper morphism of (Noetherian) schemes, $\mathcal{F} \in \mathop{Coh}(X)$. Let $i_Z \colon Z \hookrightarrow Y$ be a closed subscheme and take the inverse image $W := X \times_Y Z \overset{i_W}{\hookrightarrow} X$.

If we assume that $\dim X_y \leq k - 1$ for all $y \in Y \setminus Z$, then the sheaf $R^k f_{*}(\mathcal{F})$ is concentrated on $Z$ by flat base change for the open inclusion $Y \setminus Z \to Y$.

Is it true in this case that $$R^k f_{*}(\mathcal{F}) \cong i_{Z*} R^k (f|_W)_{*}(\mathcal{F}|_{W})$$ where $\mathcal{F}|_{W} = i_W^{*}(\mathcal{F})$?

I was not able to derive this from the standard results about base change, nor to find any counterexamples.

Special case: What I actually need is a rather special case of the former. Namely in the case I am insterested in:

  • all objects are varieties over an algebraically closed field
  • $X$, $Z$ and $W$ are smooth
  • $\mathcal{F}$ is a line bundle
  • $k = 1$
  • the map $X \setminus W \to Y \setminus Z$ is an isomorphism.

It interesting to know something about the general case, though.

share|cite|improve this question
In general case, the conditions seem too weak. Why would the direct image $R^kf_*F$ be supported by $Z$ in the scheme-theoretic sense? (Note that you are not assuming anything about the scheme structure of $Z$.) For an easy counterexample, suppose $Y$ is non-reduced, and $Z$ is its underlying reduced subscheme. Now let $X$ be the product of $Y$ with a proper $k$-dimensional scheme, and let $F$ be a sheaf on $X$ whose top direct image is say a line bundle. – t3suji Nov 19 '10 at 15:48
In fact, even in your particular case, it seems that the statement does not hold. Let $Y$ be a plane, $X$ be its blow-up at a point $Z$, and let $F$ be the line bundle corresponding to a high power of the exceptional divisor on $X$. Then the right-hand side and the left-hand side have different length. Right? – t3suji Nov 19 '10 at 15:57
@t3suji: I don't understand your second example. It seems to me that in this case the two sides agree by the standard cohomology and base change results, since there is no second cohomology on any fiber. – Andrea Ferretti Nov 19 '10 at 16:20
As for your first example, I actually do not want to assume anything on $Z$. The interesting case for me is when $Z$ is reduced, and if I'm not wrong if the result holds in this case, it also holds on the non-reduced case. But it would be interesting to assume that $Y$ is reduced. – Andrea Ferretti Nov 19 '10 at 16:23
Dear Andrea: The standard base change results don't apply since there's no flatness around the interesting part of the map. – BCnrd Nov 19 '10 at 16:38
up vote 4 down vote accepted

Here are more details for the second example from comments. Set $Y={\mathbb A}^2$, $Z=0$ (the origin), $X=$ blow-up of $Y$ and $Z$, so that $W$ is the exceptional divisor. Take $F_n=O_X(nW)$. Since the self-intersection of $W$ is $-1$, we see that $F_n/F_{n-1}=F_n|_W\simeq O_W(-n)$ (more properly, the direct image of this sheaf to $X$). Looking at the short exact sequence $$0\to F_{n-1}\to F_n\to F_n/F_{n-1}\to 0\qquad(n>0),$$ we see that $R^0f_*(F_n)=R^0f_*(F_{n-1})$, while $R^1f_*(F_n)$ is an extension $$0\to R^1f_*(F_{n-1})\to R^1f_*(F_n)\to R^1f_*(F_n/F_{n-1})\to 0.$$ The right term of this exact sequence is $i_{Z*}R^1f_*(F_n|_W)$, so your condition fails as long as $R^1f_*(F_{n-1})\ne0$. But this happens for $n>2$ (again, follows from the same sequence).

share|cite|improve this answer
I guess you mean $F_n = O_X(n W)$ (but then I don't understand your parenthetic remark "more properly...") – Andrea Ferretti Nov 19 '10 at 18:30
Thanks, fixed now. The remark is to distinguish between a sheaf on a closed subscheme and its direct image to the whole scheme (which is in a sense the same sheaf...) – t3suji Nov 19 '10 at 19:48

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