# Rational curves on the image of the pluricanonical maps

Let $$X$$ be a compact complex manifold with canonical bundle $$K_X$$. Assume the Kodaira dimension $$\kappa(X)$$ is positive (but not maximal, i.e., $$X$$ is not of general type). Let $$\varphi_m : X \dashrightarrow Y_m \subset \mathbb{P}^{N_m}$$ denote the $$m$$th pluricanonical map given by the sections of the $$m$$th tensor power $$K_M^{\otimes m}$$.

Suppose $$X$$ does not contain any rational curves. Does $$Y_m$$ contain any rational curves?

In more detail, if $$K_X$$ is semi-ample, the base of the Iitaka map $$\varphi : X \to X_{\text{can}}$$ given by the linear system $$|K_X^{\otimes \ell} |$$ for $$\ell>0$$ sufficiently large has ample canonical bundle (I've seen this stated, but am not certain about $$K_{X_{\text{can}}}$$ being ample, for general $$X$$ with $$K_X$$ semi-ample). In any case, if $$K_{X_{\text{can}}}$$ is ample, then Mori's newness result says that $$X_{\text{can}}$$ has no rational curves.

I'm wondering if this type of phenomenon occurs for the pluricanonical maps, in general, eventually. That is,

Suppose $$X$$ does not contain any rational curves. For $$m>0$$ sufficiently large, do the base spaces $$Y_m$$ of the pluricanonical maps have rational curves?

• For sure one can construct examples of pairs $(X,m)$ for which the answer to your yes, but there are also examples where $Y_m$ does not contain any rational curves. What kind of answer are you looking for?
– Pop
Oct 1 at 21:54
• @Pop Thank for your comment. I have added some (hopefully not additionally confusing) remarks on the type of example/result I'm looking for. Oct 1 at 22:23

Not only could $$Y_m$$ contain a rational curve for all $$m$$, $$Y_m$$ could be a rational curve for all $$m$$.

Take $$C$$ a hyperelliptic curve, $$E$$ an elliptic curve, $$\tau$$ the hyperelliptic involution on $$C$$, $$\sigma$$ the translation on $$E$$ by a point of order $$2$$.

Let $$X = (E \times C)/ (\sigma \times \tau)$$.

Since $$\sigma$$ has no fixed points, $$\sigma \times \tau$$ has no fixed points, so $$K_X$$ pulls back to $$K_E\otimes K_C= K_C$$, so $$H^0 (X, K_X^{\otimes m})$$ is a subspace of $$H^0(E \times C, K_C^{\otimes m})= H^0(C, K_C^{\otimes m}),$$ specifically, the $$(\sigma \times \tau)$$-invariant part.

The action of $$\sigma$$ is trivial so this is just the $$\tau$$-invariant part. For $$g$$ the genus of $$C$$, $$K_C$$ is the pullback of $$\mathcal O_{\mathbb P^1}(g-1)$$ along $$\mathbb P^1 = C/\tau$$, so the $$\tau$$-invariant part of $$H^0(C, K_C^{\otimes m})$$ is simply $$H^0(\mathbb P^1, \mathcal O_{\mathbb P^1} ( m(g-1)))$$ and thus $$Y_m = \mathbb P^1$$.

• Thank you very much, this is perfect and precisely what I should have in mind. Do you happen to know an example where $X$ is hyperbolic, i.e., does not have rational or elliptic curves? Oct 2 at 1:34
• @AmorFati Probably one can do something similar by replacing $E$ by a higher-dimensional variety that doesn't contain elliptic curves. But does one usually consider a higher-dimensional abelian variety to be hyperbolic? Oct 2 at 12:34
• Thank you for this comment. Perhaps more precisely, I should request that $X$ is hyperbolic in the sense of Brody or (equivalently) Kobayashi. Then Abelian varieties would not be hyperbolic. Oct 2 at 21:03