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Wille Liu
Wille Liu
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For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type over(*Edit. locally of finite type) over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

Edit Yes, a priori $\mathrm{Aut}(X^n)$ may be discrete and infinite. Sorry for that.

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type (*Edit. locally of finite type) over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

Edit Yes, a priori $\mathrm{Aut}(X^n)$ may be discrete and infinite. Sorry for that.

added 86 characters in body
Source Link
Wille Liu
Wille Liu
  • 1.1k
  • 5
  • 17

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

Source Link
Wille Liu
Wille Liu
  • 1.1k
  • 5
  • 17

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.