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Let $X$ be a Calabi-Yau 3-fold with Picard number one. How can one show that the automorphism group $Aut(X)$ is finite and moreover coincides with the birational automorphism group $Bir(X)$?

It seems this is a well-knwon fact, but I cannot find any reference. Since any automorphism group of $X$ preserves the ample generator of $Pic(X)$, this question should reduce to the projective geometry.

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The group $\mathrm{Aut}(X)$ is finite because, as you point out, it preserves a very ample line bundle; hence it is an algebraic group (a closed subgroup of a projective group). Therefore it suffices to prove that its Lie algebra $H^0(X,T_X)$ is trivial. By Serre duality this is dual to $H^3(X,\Omega ^1_X)=H^{1,3}$, which is conjugate to $H^{3,1}=H^1(X,\mathcal{O}_X)$, which is trivial by definition of Calabi-Yau varieties.

Now since $K_X$ is trivial, any birational automorphism is an isomorphism outside sets of codimension $\geq 2$ (see for instance here, p. 2 for a much more general statement). This implies again that it preserves a very ample line bundle, hence that it is biregular.

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  • $\begingroup$ Thanks for the answer. I have a question about the last statement. How do you know the map is biregular if it preserve the a very ample line bundle? $\endgroup$
    – Truong
    Dec 28, 2013 at 18:07
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    $\begingroup$ Let $u$ be a birational map, $L$ very ample, $u^*L\cong L$. Then $u$ induces a linear automorphism of $V:=H^0(X,L)$; the embedding $X\hookrightarrow \mathbb{P}(V^*)$ defined by $L$ is equivariant w.r.t. $(u,u^*)$. But this means that $u$ is induced by an automorphism of $\mathbb{P}(V^*)$, hence is biregular. $\endgroup$
    – abx
    Dec 28, 2013 at 18:19
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Even though an (very good) answer was already accepted, if you want to know more on automorphism and birational groups of Calabi - Yau 3 folds you can look at papers by Keiji Oguiso.

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  • $\begingroup$ It might help if you were to link to one or two that you think would be helpful. $\endgroup$
    – Todd Trimble ♦
    Dec 28, 2013 at 22:57

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