Q:

Linear group action on a projective variety

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.

A:

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 $K[T_0, \dots, T_n]$. 

