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Akatsuki
Akatsuki
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This question is moved from my StackExchange.

Assume the base field is algebraically closed. Let $X\subset \mathbb P^n$ be a fixed smooth hypersurface of degree $d$. For any hyperplane $H\subset \mathbb P^n$, we denote $X_H$ to be the intersection of $X$ with $H$. I believe the following statement is true:

If the automorphism group of $X$ is trivial${\rm Aut}(X)=\{id\}$, then there exists at least one $H$ such that the automorphism group of $X_H$ is also trivialsmooth and ${\rm Aut}(X_H)=\{id\}$.

For simplicity, we exclude the case of cubic curves and quartic surfaces to ensure every automorphism is linear. It would be rather easy to show that if $X$ has a non-trivial automorphism, then at least one of the $X_H$ will inherit a non-trivial automorphism. For the statement we want, I guess we need to use the automorphism groups of such hypersurfaces are discrete, and the spirit of the proof might be, assuming every $X_H$ admits a non-trivial automorphism, then they somehow rise to an automorphism of $X$. But I do not know how to make this precise. Could anyone give some hint or reference on this?

Also, I guess if we moreover assume the base field is of characteristic zero and $d>2$, then we can even drop out the condition that ${\rm Aut}(X)=\{id\}$. However, I have no clue how to prove it.

Thanks in advance.

This question is moved from my StackExchange.

Assume the base field is algebraically closed. Let $X\subset \mathbb P^n$ be a fixed smooth hypersurface of degree $d$. For any hyperplane $H\subset \mathbb P^n$, we denote $X_H$ to be the intersection of $X$ with $H$. I believe the following statement is true:

If the automorphism group of $X$ is trivial, then there exists at least one $H$ such that the automorphism group of $X_H$ is also trivial.

For simplicity, we exclude the case of cubic curves and quartic surfaces to ensure every automorphism is linear. It would be rather easy to show that if $X$ has a non-trivial automorphism, then at least one of the $X_H$ will inherit a non-trivial automorphism. For the statement we want, I guess we need to use the automorphism groups of such hypersurfaces are discrete, and the spirit of the proof might be, assuming every $X_H$ admits a non-trivial automorphism, then they somehow rise to an automorphism of $X$. But I do not know how to make this precise. Could anyone give some hint or reference on this?

Also, I guess if we moreover assume the base field is of characteristic zero and $d>2$, then we can even drop out the condition that ${\rm Aut}(X)=\{id\}$. However, I have no clue how to prove it.

Thanks in advance.

This question is moved from my StackExchange.

Assume the base field is algebraically closed. Let $X\subset \mathbb P^n$ be a fixed smooth hypersurface of degree $d$. For any hyperplane $H\subset \mathbb P^n$, we denote $X_H$ to be the intersection of $X$ with $H$. I believe the following statement is true:

If ${\rm Aut}(X)=\{id\}$, then there exists at least one $H$ such that $X_H$ is smooth and ${\rm Aut}(X_H)=\{id\}$.

For simplicity, we exclude the case of cubic curves and quartic surfaces to ensure every automorphism is linear. It would be rather easy to show that if $X$ has a non-trivial automorphism, then at least one of the $X_H$ will inherit a non-trivial automorphism. For the statement we want, I guess we need to use the automorphism groups of such hypersurfaces are discrete, and the spirit of the proof might be, assuming every $X_H$ admits a non-trivial automorphism, then they somehow rise to an automorphism of $X$. But I do not know how to make this precise. Could anyone give some hint or reference on this?

Also, I guess if we moreover assume the base field is of characteristic zero and $d>2$, then we can even drop out the condition that ${\rm Aut}(X)=\{id\}$. However, I have no clue how to prove it.

Thanks in advance.

Source Link
Akatsuki
Akatsuki
  • 131
  • 6

Automorphism group of a hypersurface and its sections

This question is moved from my StackExchange.

Assume the base field is algebraically closed. Let $X\subset \mathbb P^n$ be a fixed smooth hypersurface of degree $d$. For any hyperplane $H\subset \mathbb P^n$, we denote $X_H$ to be the intersection of $X$ with $H$. I believe the following statement is true:

If the automorphism group of $X$ is trivial, then there exists at least one $H$ such that the automorphism group of $X_H$ is also trivial.

For simplicity, we exclude the case of cubic curves and quartic surfaces to ensure every automorphism is linear. It would be rather easy to show that if $X$ has a non-trivial automorphism, then at least one of the $X_H$ will inherit a non-trivial automorphism. For the statement we want, I guess we need to use the automorphism groups of such hypersurfaces are discrete, and the spirit of the proof might be, assuming every $X_H$ admits a non-trivial automorphism, then they somehow rise to an automorphism of $X$. But I do not know how to make this precise. Could anyone give some hint or reference on this?

Also, I guess if we moreover assume the base field is of characteristic zero and $d>2$, then we can even drop out the condition that ${\rm Aut}(X)=\{id\}$. However, I have no clue how to prove it.

Thanks in advance.