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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.

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up vote 2 down vote accepted

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 $g \cdot [x_1 \dots x_d] = [g \cdot( x_1 \dots x_d)]$.

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Thanks! So the expression that the space of homogeneous polynomials "is naturally identified with" $Sym^d(V^*)$ essentially means that the space of homogeneous polynomials is DEFINED as $Sym^d(V^*)$? Otherwise, I don't see how this identification is natural if I have to choose a basis. – Paul Oct 28 '11 at 18:38
Yes, that is the right way to think about it. Keep in mind that there is no such thing as a homogeneous polynomial function on projective space. Homogeneous polynomials are really global sections of certain line bundles on projective space, which you'll learn as you go further. Homogeneous polynomials on $\mathbb{P}V$ are naturally identified with $Sym^d(V^*) = H^0(\mathbb{P}V, \mathcal{O}(d))$, the right hand side being (one of several) notation for the global sections of the aforementioned line bundle. Similarly, $k[x_0,x_1,\dots,x_n]_d = H^0(\mathbb{P}^n, \mathcal{O}(d))$. – Michael Joyce Oct 28 '11 at 20:02

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