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?
> 
> 
> 



Rational curves on the image of the pluricanonical maps
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 $Y\_m = \mathbb P^1$.


