Let $X$ and $Y$ be smooth projective varieties over $\mathbb C$ which have the isomorphic universal covers (in the analytic category).

It's simple to see that $\dim X = \dim Y$.

Is the Kodaira dimension of $X$ equal to the Kodaira dimension of $Y$?

Is $X$ of general type if and only if $Y$ is of general type?

More vaguely: which invariants of $X$ and $Y$ coincide? (What can one reconstruct from the universal cover? Which cohomology groups?)


If the fundamental group is finite, the universal covering has the same cohomologies with values in a field. Indeed there is a useful topological result: assume that $A$ is an $n$-divisible Abelian group (i.e. for any $a \in A$ there exists $a' \in A$ s.t. $a = n \cdot a'$). Than for any $n$-branched covering $\tilde{X} \to X$ one has $H_i(\tilde{X}, A) = H_i(X, A)$ (by the way I'm not pretty sure about the generalisation of this fact for sheaves).

If your variety has non-trivial $b_1$, after passing to the universal cover the first cohomology group obviously disappear and so the top cohomology group does (since the universal cover is not compact). But all the others remain the same. In holomorphic category, I think, the Hodge numbers are preserved as well, but perhaps one must be more careful at this point.

Nevertheless, there is an argument which looks a bit unnatural here, though it definitely works. For a locally trivial fibration the canonical sheaf of the total space is an extension of canonical sheaf of the base by the canonical sheaf of the fibre. Here the fibre is zero-dimensional, so it has trivial canonical sheaf and the Kodaira dimension doesn't change.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.