Let $B$ a smooth projective connected variety over $\mathbf C$.

Suppose that $K_B$ is torsion. Then, clearly, the Kodaira dimension of $B$ is zero.

Does the converse hold? That is, suppose that $B$ is of Kodaira dimension zero. Does it follow that $K_B$ is torsion?

This is true when $\dim B\leq 2$, but I don't know whether this is true when $\dim B>2$.

If not true when $\dim B>2$, what non-trivial properties can we show $K_B$ to have if $X$ is of Kodaira dimension zero?


First, note that what you want isn't true even in dimension 2, unless you specify that $B$ is minimal. Otherwise one can take a $K3$ surface say and blow up a point: the resulting surface certainly still has $\kappa=0$, but its canonical bundle is $E$, the exceptional divisor, which isn't torsion.

So we should restrict to the case where $B$ is minimal, in other words, that $K_B$ is nef. Then the abundance conjecture (which is a theorem in dimension $\leq 3$) says that $K_B$ is semi-ample, in other words some multiple $mK_B$ is basepoint-free and gives a morphism to a variety of dimension $\kappa(B)$. In this case that implies that $mK_B$ is pulled back from some line bundle on a point, hence is trivial, so $K_B$ is indeed torsion.

On the other hand, if we knew your statement for all minimal $B$ of Kodaira dimension 0, then we would know the abundance conjecture in that case. But that is still very much open.

Update: I just realised the last sentence is completely false. Kawamata proved the abundance conjecture for minimal varieties with $\kappa=0$ in 1985!


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.