Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 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 $g \cdot [x_1 \dots x_d] = [g \cdot( x_1 \dots x_d)]$.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.