Fibers of pushforward of a bundle when the fiber dimension is not constant

I could not decide if I should post this question in MO or Mathstackexchange, so feel free to downvote it if you think it does not belong here. I will delete my post and post it in MathSE in that case.

What I am wondering about is the following: Given a projective map $\pi:X\to B$, where $B$ is integral and $\mathscr{F}$ a coherent sheaf on $X$, which is flat over $Y$. Grauert's theorem tells us that there is an identification $$R^i\pi_*(\mathscr{F})\otimes k(b)\cong H^i(X_b,\mathscr{F}_b),$$ provided that the dimension of $H^i(X_b,\mathscr{F}_b)$ is the same for all $b\in B$. I want to know what happens if we relax the last condition.

To this end, let us consider the following example: Let $C$ be a genus $g$ curve and $\mathscr{L}$ a Poincare line bundle on $C\times Pic^{2g-2}(C)$ and consider ${\pi_2}_*\mathscr{L}$ on $Pic^{2g-2}(C)$. Now we know that over the open set $U$ of nonspecial bundles in $Pic^{2g-2}(C)$ ,we have (by Grauert's theorem) the identification that $${\pi_2}_*\mathscr{L}\otimes k(L)\cong H^0(C,L)$$ for every $L\in U$. What happens in this case over special bundles? For instance, what is $${\pi_2}_*\mathscr{L}\otimes k(K_C)$$ concretely and how far is it from $H^0(C,K_C)$? ($K_C$ being of course the canonical bundle of $C$)

• The dimension of $\pi_{2*}\mathcal{L}\otimes k(K_C)$ is $\binom{g}{2}$ and its image in $H^0(K_C)$ has dimension $g-1$. – Mohan Dec 16 '16 at 19:22
• @Mohan: I think your description of the image is not correct, see my answer. – Sasha Dec 16 '16 at 20:25

This is an explanation and a correction of Mohan's comment. Let us use the base change isomorphism (of derived functors) $$Lj^*R\pi_{2*}\mathcal{L} \cong R\Gamma(C,Li^*\mathcal{L}),$$ where $j$ is the embedding of the point $[K_C]$ and $i$ is the embedding of the fiber $C$ over this point. Since $\mathcal{L}$ is locally free, the right hand side is just $H^\bullet(C,K_C)$. It follows that $$L_0j^*R^1\pi_{2*}\mathcal{L} \cong H^1(C,K_C) \cong k$$ (the base field). With a bit more work one can check that $R^1\pi_{2*}\mathcal{L} \cong \mathcal{O}_{[K_C]}$. Now note that $$L_ti^*\mathcal{O}_{[K_C]} \cong k^{g \choose t}.$$ Canceling out the contribution $k^g$ of $L_1R^1$ from $H^0(C,K_C)$, we see that the contribution of $L_0R^0$ should be zero, hence $L_0R^0$ should cancel out in the spectral sequence with $L_2R^1$. This shows that its fiber is $k^{g \choose 2}$. But this also shows that the map from $L_0R^0$ to $H^0(C,K_C)$ is zero (this is the correction for the comment).

• You are right. I messed up my Koszul complex. – Mohan Dec 16 '16 at 20:58

For a family of curves you can indeed get some mileage even if the dimension of cohomology on the fibers is not constant.

First of all, by "cohomology and base change" you always have that if $d=\dim (X/B)$, then the natural map $$R^d\pi_*(\mathscr{F})\otimes k(b)\to H^d(X_b,\mathscr{F}_b),$$ is always surjective and hence an isomorphism.

You also know that $R^d\pi_*(\mathscr{F})$ is locally free (nearby) if and only if the same natural map for $d-1$ is surjective in a neighbourhood.

For a family of curves $d=1$, so you already covered all the cases.

Then as a next step you could do at least two things:

• Look at the proof of "cohomology and base change" and see if you can figure out the "connecting maps".
• Write down Grothendieck-Riemann-Roch for this map and compare it with the above information. (Mohan may have done this to compute those values).