If $f$ is a mod $p$ Katz modular form of weight $k$ for $k \geq 2$, there is a classical modular form $g$ such that reduction of $g$ is $f$.

Is there a similar property holds for Hilbert modular forms of higher weights? Namely, let $f$ be a mod $p$ Hilbert modular form of parallel weight $k$ for some big $k$ (in the sense of Katz). Then, does there exist a classical Hilbert modular form $g$ of same weight that gives $f$ by reduction? If so, is there any bound for $k$?


1 Answer 1


Let $\overline{\mathcal{M}}(N,\mathcal{O})$ be the minimal compactification of the Hilbert modular scheme over $\mathbb{Z}_p$ for a totally real field $F$ and $\omega$ be the determinant of the co-normal sheaf ($p\nmid N$). It is known that $\omega$ is ample. We have an exact sequence of sheaf such that the first map is multiplying by $p$.

$0\rightarrow \omega^{k} \rightarrow \omega^{k} \rightarrow \omega^{k}/p\rightarrow 0$

Since $\overline{\mathcal{M}}(N,\mathcal{O})$ is projective an $\omega$ is ample, then for a large $k$, $H^{1}(\overline{\mathcal{M}}(N,\mathcal{O}), \omega^{k})$ is trivial. Moreover, the support of $\omega^{k}/p$ is the special fibre of $\overline{\mathcal{M}}(N,\mathcal{O})$, Hence by applying the global section to the above exact sequence, we find that $M_k(N,\mathbb{Z}_p) \otimes \mathbb{F}_p$ surjects to $M_k(N,\mathbb{F}_p)$.

You can jump on the weight by multiplying by the Hasse modular form.

In the case where we have a modular curve $X_1(N)$, we can use the same argument combined with Riemann-Roch to show that $H^1(X_1(N),\omega^{k})$ is trivial for a large $k$.

  • $\begingroup$ Is it known explicitly for which $k,p$ this is true? Also is there a similar result for forms with character? I have recently posted a question here: mathoverflow.net/q/306397/21698 $\endgroup$
    – fretty
    Jul 23, 2018 at 8:21

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