Let $X$ be a projective surface and let $x\in X$ be a smooth point. Consider the blow up $Bl_{x}X$ of $X$ in $x$, and let $E$ be the exceptional divisor. Suppose we know that $E$ is the only (-1)-curve in $Bl_{x}X$, so that $\alpha(E) = E$ for any $\alpha\in Aut(Bl_{x}X)$. Then any automorphisms $\alpha\in Aut(Bl_{x}X)$ induces an automorphism of $X$ fixing $x$, let us denote $Aut_{x}(X)\subseteq Aut(X)$ the subgrup of this automorphisms. We get a injective morphism $$f:Aut(Bl_{x}X)\rightarrow Aut_{x}(X).$$ What can one say about the image of this morphism? Are there cases where $f$ is an isomorphism, or on the contrary cases in which the image is trivial? In the simple case $X = \mathbb{P}^{2}$, what happens?
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$\begingroup$ Why is the morphism $f$ injective? I see clearly why an automorphism in $\ker(f)$ is trivial away from $E$, but why is it also the identity on $E$? $\endgroup$– AntoineCommented May 12, 2023 at 19:48
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2 Answers
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Let $\sigma:X\to X$ be an automorphism fixing a point $x\in X$. Then since the inverse image of $x$ is a Cartier divisor in both $Bl_x X$ and $Bl_x \sigma(X)$, by the universal property of blow-ups, we get an automorphism $\tilde{\sigma}$ of the blow-up $Bl_x X$. Since $\tilde{\sigma}$ restricts to $\sigma$ on the complement of the exceptional divisor, it is clear that $\sigma$ is in the image of $f$.
answered Jun 13, 2011 at 13:19
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Another way to explain why $\sigma$ extends to an automorphism of the blow-up is the following. Note that the blowup of a point is the relative Proj of the sheaf of graded algebras $\oplus_{k=0}^\infty I^k$, where $I$ is the ideal of the point $x$. Now if $\sigma$ preserves the point $x$ then it also preserves the ideal $I$, hence induces an automorphism of the above graded algebra, and hence of its Proj.
