Hodge numbers in a family - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:51:01Z http://mathoverflow.net/feeds/question/88798 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88798/hodge-numbers-in-a-family Hodge numbers in a family Akhil Mathew 2012-02-18T02:30:58Z 2012-02-18T20:07:24Z <p>Let $X \to Y$ be a smooth projective morphism of smooth varieties over $\mathbb{C}$. Then the fibers $X_y, y \in Y$ have locally constant Hodge numbers $H^q(X_y, \Omega^p_{X_y})$. Namely, one can argue that the Hodge numbers are upper semicontinuous because they represent the kernels of the $\overline{\partial}$-Laplacian, and when you wiggle an operator, the kernel only jumps. However, their sum gives the sum of the Betti numbers of the fibers (since each $X_y$ is smooth and projective), which is locally constant by Ehresmann's fibration theorem. </p> <p>In what generality is this claim about Hodge numbers true? For instance, will upper semicontinuity still be true (I would like to apply the semicontinuity theorem but don't know how the $\Omega^p_{X_y}$ fit into a flat family) if the morphism is, say, only proper? Is any of this true in characteristic $p$ where we don't have Hodge theory? </p> http://mathoverflow.net/questions/88798/hodge-numbers-in-a-family/88803#88803 Answer by Sándor Kovács for Hodge numbers in a family Sándor Kovács 2012-02-18T03:04:50Z 2012-02-18T20:07:24Z <p>Without flatness you have little chance for this to even get off the ground: Let $f:X\to Y$ be the blow up of a smooth (closed) point of $Y$. Then all fibers except one consist of a single point, while the special fiber is a $\mathbb P^n$. That will have non-zero Hodge numbers that the others can't even dream about. </p> <p>So, assume $f$ is flat, then if $X$ is smooth, then $\Omega_X$ is still locally free and so it is flat over $Y$ and hence you get that $\dim H^q(X_y, (\Omega^p_X)_y)$ is semi-continuous. This works in any characteristic and it does not have anything to do with Hodge theory. The shortcoming of this is that you're actually not getting $\dim H^q(X_y, \Omega^p_{X_y})$ even in the smooth case, because for that you would need $\Omega_{X/Y}$, but that's not flat and in some sense not the right object to consider.</p> <p>If $Y$ is also smooth, then $\Omega_Y$ is locally free and if $f$ is dominant, then you still have the short exact sequence $$0\to f^*\Omega_Y\to \Omega_X\to \Omega_{X/Y}\to 0$$ The trouble is that the sheaf on the right is not locally free, so when you take exterior products, then it gets kind of tricky. One possibility is to construct complexes that behave very similarly to $\Omega_{X/Y}^p$ with respect to $\Omega_X^p$ and $\Omega_Y^p$. This is done in <a href="http://books.google.com/books?id=p6YrfFeS4s0C&amp;pg=PA89&amp;lpg=PA89&amp;dq=Relative+de+Rham+complex+for+non-smooth+morphisms&amp;source=bl&amp;ots=Uv5S_X1AcQ&amp;sig=nDB5JEyg_pQ5iWJxNPxwPeSlw0c&amp;hl=en&amp;sa=X&amp;ei=TgJAT5nyPLHciQL71rG2AQ&amp;ved=0CCEQ6AEwAA#v=onepage&amp;q=Relative%2520de%2520Rham%2520complex%2520for%2520non-smooth%2520morphisms&amp;f=false" rel="nofollow">this</a> paper. The primary goal of the paper is not what you want and I am not sure that you actually get semi-continuity, but you can at least check out the construction. A general version of that construction is in <a href="http://books.google.com/books?id=l_CM8szufz0C&amp;lpg=PA57&amp;ots=2gpe1fsF4p&amp;dq=spectral%2520sequences%2520associated%2520to%2520morphisms&amp;lr&amp;pg=PA57#v=onepage&amp;q=spectral%2520sequences%2520associated%2520to%2520morphisms&amp;f=false" rel="nofollow">this</a> paper. (Sorry, neither of them are on arXiv).</p> <p>Also, for singular fibers <code>$\Omega_{X_y}$</code> is not the "right" thing to look at. One can look at the objects that come from the Deligne-Du Bois complex and do Hodge theory for singular varieties. Then the restriction becomes a little tricky, because those are objects in a derived category and restriction is not exact, so you need to do something else. Completion along the fiber gives the right thing, but I don't know if there is a semi-continuity theorem using completion instead of restriction. That might be an interesting question to contemplate. This is related to the paper I linked above. See the references in that for more details.</p>