Let $\mathcal X \rightarrow S$ be a flat family of projective varieties over a discrete valuation ring $S$ such that the generic fibre $\mathcal X_{\eta}$ (say) is smooth projective variety and the special fibre $\mathcal X_0$ (say) is a normal crossing divisor in $\mathcal X$.

Question: Does there exist a vector bundle $\mathcal E$ over $\mathcal X$ such that $\mathcal E|_{_{\mathcal X_{\eta}}}\cong \Omega^1_{_{\mathcal X_{\eta}}}$ and $\mathcal E|_{_{\mathcal X_0}}\cong \Omega^1_{_{\mathcal X_0}}(\mathrm{Log} D)$, where $D$ is the singular locus of $\mathcal X_0$?

It will be very helpful if anyone can explain this construction or provide a reference. Thank you..


1 Answer 1


First of all, it's unclear what you mean by $\Omega^1_{X_0}(\log D)$ since $X_0$ is singular.

Second, if you make up such a definition then most probably such a vector bundle will not exist. Note that the Euler characteristics $\chi(X_\eta, \Omega^1_\eta)$ and $\chi(X_0, \Omega^1_{X_0}(\log D)$ have to agree. I suggest checking this for an elliptic curve degenerating to a node.

Finally, what people usually consider in your situation is the vector bundle $\Omega^1_{X/S}(\log X_0)$, so differentials on $X$ with log poles along the components of $X_0$, divided by the pull-back $t^{-1}dt$ where $t$ is a uniformizer of $S$. This is indeed locally free and restricts to $\Omega^1_{X_\eta}$.

  • 2
    $\begingroup$ X_0 has normal crossing singularity, we can take the normlisation of X_0, and consider the Log cotangent bundle with logarithmic poles along the preimage of the singular locus of X_0. Then the log cotangent bundle of X_0 is the subsheaf of the log cotangent bundle above whose elements have opposite residues along the pullback divisors..this is mentioned in the famous paper of Mumford and Deligne on the irreducibility of moduli of curves.. $\endgroup$
    – user149914
    Feb 20, 2020 at 8:12
  • $\begingroup$ If we consider the elliptic curve case then it is just the relative dualising sheaf. Therefore It is actually $\Omega^1_{X_0} (Log D)$. This supports my question. $\endgroup$
    – user149914
    Mar 15, 2020 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.