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?

  • 1
    $\begingroup$ 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? $\endgroup$
    – Pop
    Commented Oct 1, 2022 at 21:54
  • $\begingroup$ @Pop Thank for your comment. I have added some (hopefully not additionally confusing) remarks on the type of example/result I'm looking for. $\endgroup$
    – AmorFati
    Commented Oct 1, 2022 at 22:23

1 Answer 1


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 $Y_m = \mathbb P^1$.

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    $\begingroup$ 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? $\endgroup$
    – AmorFati
    Commented Oct 2, 2022 at 1:34
  • $\begingroup$ @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? $\endgroup$
    – Will Sawin
    Commented Oct 2, 2022 at 12:34
  • $\begingroup$ 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. $\endgroup$
    – AmorFati
    Commented Oct 2, 2022 at 21:03

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