Skip to main content
MathOverflow

Return to Answer

Added \operatorname to Aut, Bir and Proj.
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $\operatorname{Aut}(X)=\operatorname{Bir}(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong\operatorname{Proj}(R)=X_{can}$$$$X\cong\operatorname{Proj}(R)=X_\text{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $\operatorname{Aut}(X)=\operatorname{Bir}(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong\operatorname{Proj}(R)=X_{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $\operatorname{Aut}(X)=\operatorname{Bir}(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong\operatorname{Proj}(R)=X_\text{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

Added \operatorname to Aut, Bir and Proj.
Source Link
edit approved Jul 14, 2022 at 22:01
Samuel Adrian Antz
edit approved Jul 14, 2022 at 22:01
Samuel Adrian Antz
  • 345
  • 1
  • 3
  • 12

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $Aut(X) = Bir(X)$$\operatorname{Aut}(X)=\operatorname{Bir}(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong Proj(R)=X_{can}$$$$X\cong\operatorname{Proj}(R)=X_{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $Aut(X) = Bir(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong Proj(R)=X_{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $\operatorname{Aut}(X)=\operatorname{Bir}(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong\operatorname{Proj}(R)=X_{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

Source Link
Puzzled
Puzzled
  • 9k
  • 1
  • 38
  • 65

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $Aut(X) = Bir(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong Proj(R)=X_{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.