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Let $X$ be a normal affine variety of dimension at least two over $\mathbb{C}$ and let $U\subset X$ be a dense open. Assume that $\mathrm{codim}(X\setminus U) \geq 2$.

I think Hartog's lemma implies that every automorphism of $U$ extends to an automorphism of $X$. Is that true?

Do we have a surjective homomorphism $\mathrm{Aut}(X) \to \mathrm{Aut}(U)$?

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    $\begingroup$ I cannot think of any natural homomorphism $\mathrm{Aut}(X)\to\mathrm{Aut}(U)$ (except the trivial one!). $\endgroup$ Feb 22, 2018 at 8:31

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I think you have an injective homomorphism the "other way around".

As you suspect, Hartog's Lemma implies that $Hom(U,X) = Hom(X,X)$. Now, $Aut(U)$ naturally injects into $Hom(U,X)$. Also, $Aut(X)$ naturally injects into $Hom(U,X)$. The image of $Aut(U)$ in $Hom(U,X)$ lands inside the image of $Aut(X)\to Hom(U,X)$. Thus, in this way, we see that

$Aut(U)$ is a subgroup of $Aut(X)$.

As abx's example shows: this really requires $X$ to be affine. (Otherwise, you just get an injective map $Aut(U) \to Hom(U,X)$.)

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    $\begingroup$ One step I thought was unclear: To check that the one image is contained in the other image, take both an element of $Aut(U)$ and its inverse from $Aut(U)$ to $Hom(U,U)$ to $Hom(U,X)$ to $Hom(X,X)$. Their composition in both directions is the identity on $U$, hence the identity on $X$, so in fact the extended map is invertible and lies in $Aut(X)$. $\endgroup$
    – Will Sawin
    Feb 21, 2018 at 17:25
  • $\begingroup$ Dear @WillSawin , thank you for your comment. Yes, you are right. I should have clarified that. $\endgroup$ Feb 21, 2018 at 17:32
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No, this is false. Here is a somewhat sophisticated example; one can probably do simpler.

Let $S\subset \mathbb{P}^3$ be a smooth quartic surface, which contains precisely one line $L$. The Hilbert scheme $X:=\operatorname{Hilb}^2(S) $ is a smooth fourfold, and $\operatorname{Hilb}^2(L) $ embeds naturally as a surface $Y\subset X$; put $U:=X \smallsetminus Y$. Let $Z\in U$ be a length 2 subscheme of $S$; then $Z$ lies on a unique line $\ell_Z≠L$ in $\mathbb{P}^3$, which intersects $S$ along a length 4 subscheme. Associating to $Z$ the residual subscheme of $Z$ in $\ell_Z\cap S$ defines an involution of $U$, which does not extend to $X$.

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  • $\begingroup$ Nice example. But isn't $X$ projective (and not affine)? $\endgroup$ Feb 21, 2018 at 16:20
  • $\begingroup$ Oops! Right, sorry, I read too fast. $\endgroup$
    – abx
    Feb 21, 2018 at 16:33
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The answer to your second question is no. Let $X$ be a three-dimensional algebraic torus, then the automorphism group of $X$ is the extension of $GL_3(\mathbb Z)$ by $X$ (acting by translations. Let $U$ be the complement of three points in general positionin a one-dimensional subtorus. Then the automorphism group of $U$ is the subgroup of $GL_3(\mathbb Z)$ fixing a vector. If there were a surjective homomorphism $Aut(X) \to Aut(U)$, the image of the torus would be a normal abelian subgroup, hence when sent to $GL_2(\mathbb Z)$ it would lie in $\pm 1$, but there is no surjective homomorphism $GL_3(\mathbb Z) \to PGL_2(\mathbb Z)$.

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  • $\begingroup$ Why do you write $Aut(X) \to Aut(\mathbb{Z})$? $\endgroup$
    – Xiboto
    Feb 21, 2018 at 20:22
  • $\begingroup$ @Xiboto a typo. $\endgroup$
    – Will Sawin
    Feb 21, 2018 at 22:10

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