TITLE: Homogeneous polynomials on a vector space $V$, $\operatorname{Sym}^d(V^*)$ and naturality
QUESTION [6 upvotes]: My three questions all relate to Harris' "Algebraic Geometry - A First Course".
Unless mentioned otherwise, $V$ is an n-dimensional vector space over an arbitrary field $k$.
1) At first, I'd like to make sure that I have the right motivation for this construction. According to my humble understanding, the main advantage of $\operatorname{Sym}^d(V^*)$ is having a framework to talk about homogeneous polynomials of a fixed degree on V and being able to evaluate them without the need to choose a basis.
More precisely, there exists an evaluation map $\operatorname{Sym}^d(V^*) \times V \rightarrow k$ via $(\sum\limits_i a_{i_1}a_{i_2}\cdots a_{i_d}\hat{v_{i_1}}\cdot \hat{v_{i_2}} \cdots \hat{v_{i_d}}, v) \mapsto \sum\limits_i a_{i_1}a_{i_2}\cdots a_{i_d}\hat{v_{i_1}}(v) \hat{v_{i_2}}(v) \cdots \hat{v_{i_d}}(v)$
Is that correct?
2) On page 4, it is written that homogeneous polynomials on $\mathbb{P}V$ can be naturally identified with the vector space $\operatorname{Sym}^d(V^*)$. I was able to find the obvious isomorphism from $\operatorname{Sym}^d(V^*)$ to $k[X_0, X_1, \ldots, X_n]_d$ after choosing a basis for $V^*$, but I don't see how such an identification can be made naturally.
3) Let $V$ be two-dimensional, $char\ k \neq 2$ In chapter 10, 10.8 we deal with the action of $PGL_2(k)$ on $\mathbb{P}^2$. Now $PGL_2(k)$ obviously acts on $\mathbb{P}^1$ and Harris mentions that this naturally induces an action on $\mathbb{P}(\operatorname{Sym}^2(V^*))$. However, I fail to realize what this action is supposed to look like, let alone how it is obtained naturally (I DO see how $PGL_2(k)$ acts on $\mathbb{P}(\operatorname{Sym}^2(V))\cong \mathbb{P^2}$, though).
On a side note, where exactly comes the assumption on the characteristic of $k$ into play? When I verified that the set of squares $v\cdot v$ is isomorphic to the image of the quadratic Veronese, I had to divide by two, but I suspect that this is not the only reason to make this assumption.
Thanks in advance for any help. I tried to keep this as brief as possible, but to no avail. Sorry for that.

REPLY [3 votes]: 1) Yep.
2) $Sym^d(V^*)$ is the abstract definition of homogeneous polynomial on $\mathbb{P}V$, and any identification $Sym^d(V^*) \cong k[x_0,x_1,\dots,x_n]_d$ corresponds to choose a basis $x_0, x_1, \dots, x_n$ of $V^*$, which amounts to identifying $\mathbb{P}V$ with $\mathbb{P}^n$ using the dual basis.
3) $PGL(V)$ acts on $V$ (so $PGL_2(k)$ acts on $\mathbb{P}^1$ when we identify $\mathbb{P}^1$ with $\mathbb{P}V$ by choosing a basis).  Then $PGL(V)$ acts on $V^*$ by the rule $(g \cdot x)(v) = x(g^{-1} \cdot v)$.  The inverse in the formula is needed to make it into an action.  Then you get an induced action on $Sym^d(V^*)$ by $g \cdot (x_1 \dots x_d) = (g \cdot x_1) \dots (g \cdot x_d)$.  And then the action on $\mathbb{P}Sym^d(V^*)$ is $g \cdot [x_1 \dots x_d] = [g \cdot( x_1 \dots x_d)]$.