Let $S$ be a normal surface over an algebraically closed field $k$ and let $s$ be a point of $S$. Let ${\mathcal F}$ be a reflexive sheaf on $S$ of generic rank $n$ . Consider the (derived) fiber of ${\mathcal F}$ at $s$. Are the dimensions of its cohomologies a priori bounded (for fixed $n$)?

Can I ask a dumb question, what's a derived fiber in this context? – Karl Schwede Apr 15 '11 at 12:03

Derived fiber means the derived functor of the restriction functor to the point... – Alexander Braverman Apr 15 '11 at 12:13

1The derived fiber will have $H^0$ and $H^{1}$ where $H^0$ is the usual fiber (taking the derived fiber is the same as looking at $Tor$ with the skyscraper sheaf at $s$). I will be happy to have a bound just on $H^0$. – Alexander Braverman Apr 15 '11 at 12:50

Thanks. By the way, do you have a particular reflexive sheaf in mind? For example the reflexification of a sheaf of differentials, or the pushforward of some other sheaf via a finite map? – Karl Schwede Apr 15 '11 at 18:37

NO, I really need an arbitrary reflexive sheaf – Alexander Braverman Apr 15 '11 at 23:54
$\def\cO{\mathcal{O}}$ I can give a positive answer for $H^0$ in the case that $S = \mathrm{Spec} A$ for a graded ring $A$ and $\mathcal{F}$ is the sheaf associated to a graded $A$module.
The proof will consist of a number of elementary reductions, followed by a result of Atiyah, ad then some more elementary arguments.
Since $A$ is normal, we know that $\mathrm{Proj} \ A$ is a smooth curve $X$. Let $g$ be the genus of $X$; let $\cO(1)$ be the line bundle on $X$ coming from the graded ring $A$; let $k$ be the degree of $\cO(1)$. These are all constants.
For any vector bundle $E$ on $X$, the section module is $$\Gamma(E) := \bigoplus_{n=\infty}^{\infty} H^0(X, E \otimes \cO(n)).$$ Section modules are reflexive and every graded reflexive $A$module is the section module of a vector bundle. The rank of $\Gamma(E)$ is the rank of the vector bundle $E$, which we will call $r$.
Step 1: We may assume that $E$ is not a direct sum of lower rank vector bundles. This is because $\Gamma$ and $\otimes_A k$ both commute with direct sum so, if we can bound the irreducible case, we just have to add together our bounds for each partition of $r$.
Let $d$ be the degree of $E$.
Step 2: We may assume that $0 \leq d < kr$. This is because tensoring with $\cO(1)$ shifts $d$ by $kr$ and gives the same section module up to a grading shift, so we get the zero fiber after this tensoring operation.
Step 3: Here is where we pull out the early parts of Atiyah's "Vector Bundles over an Elliptic Curve". As explained in section I.4 of Atiyah's paper, $E$ has a filtration $0=E_0 \subset E_1 \subset E_2 \subset \cdots E_r=E$ such that $E_i/E_{i1}$ is a line bundle $L_i$ and the degrees of the $L_i$ obey $$\deg L_i  \deg L_1 \leq C_1(g,r)$$ for a constant depending only on $g$ and $r$ (Atiyah's lemmas 4 and 6). Since $\sum \deg L_i = d$, we can rewrite this as $$\deg L_i  d/r  \leq C_2(g,r).$$ (This uses that $E$ is not a nontrivial direct sum, as this is a standing assumption for Atiyah.) Since $0 \leq d \leq kr$, we can rewrite this as $$\deg L_i \leq C_3(g,r,k).$$
So there are only finitely many sequences $(\deg L_1, \cdots, \deg L_r)$. We will prove a bound for each one.
Step 4: We prove the following result by induction on $r$: Continue to fix the genus $g$ of $X$. Let $(d_1, \ldots, d_r)$ be a given sequence of integers. Then there are constants $P$ and $Q$, and a sequence of integers $(C_P, C_{P+1}, \ldots, C_Q)$, such that, for any vector bundle $E$ on $X$ with a filtration of degree sequence $(d_i)$, the zero fiber of $\Gamma(E)$ is supported in grades between $P$ and $Q$, and the part in grade $i$ has dimension at most $C_i$.
The base case, $r=0$, is vacuous.
So, let $E$ lie in a short exact sequence $0 \to L \to E \to E' \to 0$, where we have bounds of the assumed form for $E'$ and we know the degree of $L$. We will, of course, be using the short exact sequences $$0 \to H^0(L \otimes \cO(n)) \to H^0(E \otimes \cO(n)) \to H^0(E' \otimes \cO(n)) \to H^1(L \otimes \cO(n))$$ for various values of $n$.
When $n$ is negative enough, meaning less than $\deg L$ and less than $P'$, then the first and third term vanish, so the second one does as well and we get no contribution in those grades.
When $n$ is positive enough, meaning larger than $2g\deg L$, the fourth term vanishes. So, for $n$ slightly larger than that, we have a complex with exact rows: $$\begin{matrix} 0 & \to & H^0(L \otimes \cO(n1)) \otimes H^0(\cO(1)) &\to& H^0(E \otimes \cO(n1)) \otimes H^0(\cO(1)) &\to& H^0(E' \otimes \cO(n1)) \otimes H^0(\cO(1)) &\to& 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & H^0(L \otimes \cO(n)) &\to& H^0(E \otimes \cO(n)) &\to& H^0(E' \otimes \cO(n1)) &\to& 0 \\ \end{matrix}$$
For large enough $n$, we inductively know that the third column is surjective; for $n$ larger than $\deg L + 2g$, the first column is (by the standard RiemmanRoch argument). So, by the snake lemma, we get that the middle column is surjective for large enough $n$, and there is no contribution to the zero fiber in that degree.
We are left bounding the size of the zero fiber in the intermediate degrees. We can just bound the size of $\Gamma(E)$ itself. From RiemmanRoch, $\dim H^0(X, E \otimes \cO(n)) \leq (d+rk)  r(g1)$, and we are done.

Thank you. In fact, the case when $S$ is a cone over a curve is exactly what interests me and when ${\mathcal F}$ comes from a graded $A$module I think that I've had a slightly different argument. However, I really need it for arbitrary ${\mathcal F}$... – Alexander Braverman Apr 16 '11 at 11:55