# Direct image of the relative dualizing sheaf

Let $f: X \to Y$ be a flat Cohen-Macaulay (CM) morphism with pure relative dimension $n$ between two projective varieties $X$ and $Y$. Then the relative dualizing sheaf $\omega_{X/Y}$ is $Y$-flat. My question is, what extra conditions for the fibers of $f$ do we need (if there are) to ensure that $f_* \omega_{X/Y}$ is locally free? For example, what do we need when $n=1$ or CM is enough? For simplicity, we may assume that general fibers of $f$ are normal (or smooth).

PS: I read from Viehweg's book "Quasi-projective moduli for polarized manifolds" that $f_* \omega_{X/Y}$ is locally free if all fibers are normal with at worst rational singularities. Probably this is a bit too strong for $n=1$.

Thanks!

• For $n=1$ (so ${\rm{R}}^1f_{\ast}$ is right-exact on coherent sheaves), it suffices that $f$ has geometrically connected fibers. Indeed, fibral trace ${\rm{H}}^1(X_y \omega_{X_y/y})\rightarrow k(y)$ is then an isomorphism, so ${\rm{R}}^1f_{\ast}(\omega_{X/Y})$ is locally monogenic by Nakayama. Thus, the base-change compatible trace to $\mathscr{O}_Y$ is a fiberwise isomorphism, so an isomorphism. The formation of ${\rm{R}}^1f_{\ast}(\omega_{X/Y})$ thus commutes with base change to fibers, so $f_{\ast}\omega_{X/Y}$ is a vector bundle compatible with any base change. – nfdc23 Nov 6 '15 at 14:55
• Some statements along these lines are in section 7 of Kollár-Kovács (click here) but they are probably not quite what you want. – Karl Schwede Nov 6 '15 at 15:00
• Correction: for my comment above I should have also assumed the geometric fibers to be reduced (or more generally that $O_{X_y}(X_y) = k(y)$ for each $y \in Y$); that was implicit in knowing the fibral trace is an isomorphism. – nfdc23 Nov 6 '15 at 15:09
• @KarlSchwede Thank you very much for your reference. I will read that. – Tong Nov 6 '15 at 18:04
• @nfdc23 Thank you for your answer. In my case, there could be some multiple fibers. – Tong Nov 6 '15 at 18:06