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I already posed this question, but my formulation was quite erroneous and unclear so I decided to repost it (which is hopefully not against the rules).

On page 116 in Harris' book "Algebraic Geometry - A First Course" an action of a group $G$ on a projective variety $X \subset \mathbb{P}(V) \cong \mathbb{P}^n$ is defined to be linear if it lifts to the homogeneous coordinate ring $S(X)$ of $X$.

Now my problem is seeing exactly what this is supposed to mean. The term "lift" suggests, at least to me, that the action of G on $S(X)$ somehow restricts to a subset $X' \subset S(X)$ where $X'$ is identified in some way with the original variety.

I tried to construct such an embedding of $X$ using the identification of the underlying (n+1)-dim. K-vector space V with the homogeneous polynomials of degree 1 via $V \cong Sym^1(V) \hookrightarrow \bigoplus_{n=0}^\infty Sym^n(V) \cong K[X_0,...X_n]$

This is, however, bound to fail, and so far I have no idea how to interpret this lift of a group action. Anyways, thanks in advance.

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Any action of $G$ on $S(X)$ induces an action of $G$ on $\mathrm{Proj} S(X) = X$. This can be a meaning of lifting (or consider the affine cone of $X$ which is the affine variety associated to $S(X)$ and you have a canonical map from the affine cone minus the origin to $X$). – user18119 Nov 9 '11 at 18:16
Thank you, do you happen to have a reference on how this action on Proj S(X) = X is induced? I cannot find it in this book. – Paul Nov 9 '11 at 19:41
up vote 2 down vote accepted

I just looked at Harris's book. I think I understand what he means. Let $\hat{X}=\mathrm{Spec}(S(X))\subseteq \mathbb A^{n+1}$ be the affine cone of $X$. There are canonical maps
$$\mathrm{GL}_{n+1, \hat{X}}\to \mathrm{PGL}_{n+1, X}\to \mathrm{Aut}(X)$$ from the group of linear automorphisms of $\mathbb A^{n+1}$ leaving stable $\hat{X}$ to the group of automorphisms of $\mathbb P^n$ leaving stable $X$ and from the latter to the group of automorphisms of $X$. An action of $G$ on $X$ is projective (resp. linear) if $\rho : G\to \mathrm{Aut}(X)$ lifts to $G\to \mathrm{PGL}_{n+1, X}$ (respectively to $G\to \mathrm{GL}_{n+1, \hat{X}}$). The simplest case $X=\mathbb P^n$ is insightful.

Finally, to answer the question in the comments (and this explains the first map in the above displayed formula), a linear automorphism of $K[T_0,\dots, T_n]$ is automatically an automorphism of homogeneous algebras, hence induces an automorphism of the associated projective space. The same holds for $S(X)$ instead of $K[T_0, \dots, T_n]$.

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I don't have the book, but I presume that what Harris means is: the action on $X$ is linear if it can be extended to an action on the projective space which contains $X$. Since the automorphism group of projective space is $PGL$, the action will then be expressable in linear terms, that is, in terms of matrices acting comme on sait.

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In particular, there is no sensible way to «identify $X$ with a subset of $S(X)$...» – Mariano Suárez-Alvarez Nov 9 '11 at 18:58
Thanks so far. I cite: "By a projective action on a variety $X \subset \mathbb{P}^n$ we mean an action of G on $\mathbb{P}^n$ such that $\varphi (g, X) = X$ for all $g \in G$." This seems to be identical to the definition you gave me. According to Harris, a $linear$ action is stricty stronger. Or did I just misunderstand your answer? – Paul Nov 9 '11 at 19:40
While each element of PGL acts linearly on $\mathbb P^n$, the linear tems are not uniquely defined because $PGL=GL/K^*$. This explains the difference with the action of $GL$ on $\mathbb P^n$. – user18119 Nov 11 '11 at 0:40

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