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Suppose $X$ is a (quasi-projective) variety over a field $k$, and let $\mathbb{P}^n(k)$ be the ambient projective space.

When can one decide that the automorphism group of $X$ is induced by a subgroup of the automorphism group $\mathbf{P G L}_{n + 1}(k)$ of $\mathbb{P}^n(k)$ ?

When $X$ itself is a sub-projective space over $k$, this is of course well known, but when $X$ is e.g. an affine subspace over $k$, this is in general not true (due to the existence of non-linear automorphisms).

Is it true when $X$ is projective ?

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3 Answers 3

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I guess this should be true if and only if $f : X \to X$ fixes a polarization, i.e. there is an ample $A$ such that $f^\ast A \sim A$.

If there is such an $A$, then take the embedding of $X$ into $\mathbb P^n$ by a very ample $mA$. The automorphism of $X$ is then induced by the linear map $f^\ast : \mathbb PH^0(X,mA) \to \mathbb PH^0(X,mA)$. Conversely, if a map comes from a linear map on projective space, then the polarization given by restricting the hyperplane is fixed.

Typically automorphisms won't fix any polarization, so don't come from $\mathbb P^n$. If you actually want to check it for a specific map, the first thing to do work out the induced map on $H^2(X;\mathbb R)$. If there is an invariant ample divisor, its first chern class will be invariant under the pullback. So you may be able to show that this doesn't happen. If there is a fixed class, you need to know (a) whether it is represented by an ample divisor, which may or may not be easy to check depending on what $X$ is and (b) whether the divisor is linearly equivalent (rather than just numerically equivalent) to its pullback, which requires a bit more scrutiny.

As Francesco points out, $K_X$ always pulls back to itself under an automorphism, so if it's either ample or antiample you have what you want.

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  • $\begingroup$ Nice. And this shows exactly what is happening with the elliptic curve example, since for an ample divisor $A$ and a translation $T$ by $P_0$, one has $T^*A\sim A$ if and only if $P_0$ is killed by $\deg(A)$, which is $\ge1$ by the ampleness assumption. $\endgroup$ Dec 8, 2015 at 18:58
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Is it true when X is projective ?

Let $X\subset\mathbb P^2$ be an elliptic curve and let $T:X\to X$ be a translation map $T(P)=P+P_0$ by a non-torsion point $P_0$. Then $T$ is an automorphism, but it is not induced by an element of $\text{PGL}_3$. Maybe a better question is whether, given a projective variety and an automorphism, does there exist some projective embedding for which the automorphism is induced by an element of $\text{PGL}$.

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  • $\begingroup$ That's not true either -- such an automorphism always fixes a polarization on X, given by the restriction of the hyperplane from P^n. But general automorphisms don't fix any polarization; infinite order automorphisms of surfaces are a typical counterexample. The action of an infinite order element of $SL_2(\mathbb Z)$ on $E \times E$ ($E$ an elliptic curve) should give an example. To see that it doesn't fix a polarization, you can just look at the induced map on $H^2$. $\endgroup$
    – user47305
    Dec 8, 2015 at 14:19
  • $\begingroup$ @Joe: why is $T$ not induced by an element of $\mathbf{PGL}_3(k)$ ? $\endgroup$
    – THC
    Dec 8, 2015 at 14:36
  • $\begingroup$ @THC Take $P_0$ to be generic and look at the addition formula. It's pretty clearly non-linear in the coordinates of $P$. (Not a proof, but that's the intuition.) $\endgroup$ Dec 8, 2015 at 15:33
  • $\begingroup$ @Mark I wasn't suggesting that the latter question had an affirmative answer. I was merely pointing out to the OP a more natural question, which I suspected also had a negative answer. $\endgroup$ Dec 8, 2015 at 15:35
  • $\begingroup$ Ah, sorry! I think you are right that non-torsion translations on $E$ are the basic example one should keep in mind, for either form of the question. (In that case I don't think it can be linear wrt to any projective embedding.) $\endgroup$
    – user47305
    Dec 8, 2015 at 16:28
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What you want is true if $X$ is embedded using some integer power of $K_X$, in particular it is true for canonically and anti-canonically embedded varieties.

In fact, any automorphism preserves the canonical class, so it necessarily preserves the hyperplane class under our assumption.

For the same reason, the answer is yes for varieties $X$ such that $\textrm{Pic}(X)= \mathbb{Z}$. In fact, if $L$ is the ample generator of the Picard group of $X$, then the hyperplane class is $H=kL$ for $k \geq 1$ and this must be preserved by all automorphisms, because they clearly preserve $H$.

Then the Noether-Lefschetz theorem implies that what you want is true for the very general smooth surface $X \subset \mathbb{P}^3$, and for all smooth hypersurfaces $X \subset \mathbb{P}^n$ with $n \geq 4$.

For general embeddings $X \hookrightarrow \mathbb{P}^n$ the answer is negative, as shown by the proposed counterexamples.

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  • $\begingroup$ Similarly, if $\operatorname{Pic}(X) \cong \mathbf Z$ and the embedding is linearly normal. $\endgroup$ Dec 8, 2015 at 14:48
  • $\begingroup$ Yes, in fact I just added this, thanks. $\endgroup$ Dec 8, 2015 at 14:55

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