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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?

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1 Answer 1

up vote 7 down vote accepted

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!

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