Let $X$ be a rigid Calabi-Yau threefold. Does $X$ have only finitely many automorphisms?


N.B. A smooth projective threefold $X$ over $\mathbb C$ is a rigid Calabi-Yau variety if $h^i(X,\mathcal O\_X) =0$ for all $i>0$, $K\_X$ is numerically trivial and $\mathrm{H}^1(X,T\_X) =0$ (or equivalently $h^{2,1}(X) = 0$).



Can a rigid CY threefold have infinitely many automorphisms
Yes (to the question in the title; no to the question in the first line). You can find an example in [this paper](http://arxiv.org/abs/1306.1590) by Oguiso and Truong. The variety ``$X$'' should do what you want.


Briefly, let $\omega = (1+\sqrt{3}i)/2$ and let $E$ be the elliptic curve $\mathbb C / (\mathbb Z + \omega \mathbb Z)$. Then $E$ has an automorphism $\tau$ of order $3$ given by multiplication by $\omega$. Let $X$ be the crepant resolution of $(E \times E \times E)/\tau$ with the diagonal action. Then $X$ is a Calabi-Yau threefold, with many automorphisms, induced by the action of $\textrm{SL}\_3(\mathbb Z[\omega])$ on $E \times E \times E$. This $X$ is apparently rigid; there are refs in the paper.


