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user1234
user1234
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Not sure this is too interesting, but I however think that the case when $X \rightarrow \text{Spec}(\mathbb{Z})$ is proper and smooth(!) should follow from a result proven independently by Fontaine and Abrashkin that implies that the Hodge-numbers are zero for $i+q \leq 3$$i+j \leq 3$ and $i \neq j$.

Not sure this is too interesting, but I however think that the case when $X \rightarrow \text{Spec}(\mathbb{Z})$ is proper and smooth(!) should follow from a result proven independently by Fontaine and Abrashkin that implies that the Hodge-numbers are zero for $i+q \leq 3$ and $i \neq j$.

Not sure this is too interesting, but I however think that the case when $X \rightarrow \text{Spec}(\mathbb{Z})$ is proper and smooth(!) should follow from a result proven independently by Fontaine and Abrashkin that implies that the Hodge-numbers are zero for $i+j \leq 3$ and $i \neq j$.

Source Link
user1234
user1234
  • 11
  • 2

Not sure this is too interesting, but I however think that the case when $X \rightarrow \text{Spec}(\mathbb{Z})$ is proper and smooth(!) should follow from a result proven independently by Fontaine and Abrashkin that implies that the Hodge-numbers are zero for $i+q \leq 3$ and $i \neq j$.