# Semicontinuity for complexes

In algebraic geometry, the very useful semicontinuity theorem tells you the following:

Let $X \to Y$ be a projective morphism of schemes, and $F$ a coherent sheaf on $X$ which is flat over $Y$. Then the dimensions of the cohomology on fibers, $h^i(X_y)$ are upper semicontinuous functions of $y$, i.e., the locus where in $Y$ where $h^i(X_y) \ge n$ is Zariski closed for any $n$.

What condition on a complex $F$ with coherent cohomology ensures that $\dim \mathbb{H}^i(X_y, F \otimes^L \mathcal{O}_{X_y})$ is an upper semicontinuous function of $y$?

• Call the morphism $\pi$. I think you're good if $R\pi_*F$ is quasi-isomorphic to a perfect complex. This is true if $F$ is flat over $Y$, and as far as I can tell is the only non-trivial input in this case. Feb 21, 2013 at 0:13
• @Keerthi: If $Y$ is regular then any (bounded) complex with coherent cohomology is (quasi-isomorphic to) a perfect complex.
– naf
Feb 21, 2013 at 3:59
• Keerti: I don't think your claim is true even for derived tensor products; $F$ could be a line bundle in which case the derived tensor product is just the usual restriction to the fibre. For an explicit example, let $Y$ be a smooth surface, $X$ the blowup of a point and $F = \mathcal{O}_X(E)$, where $E$ is the exceptional divisor. For all $y$ not equal to the blown up point, $H^0$ is one dimensional, but it is zero dimensional for the blown up point.
– naf
Feb 22, 2013 at 5:09
• In EGA III, Theoreme 7.7.5, a semi-continuity theorem is proved for complexes of sheaves that are flat over $Y$.
– naf
Feb 22, 2013 at 5:30
• ulrich--That reference seems to answer the question as well as one could hope to. Maybe you should put it down as an answer? Feb 22, 2013 at 12:34