Polynomial Forms on Dual Projective Spaces

Question

Let $E$ be a finite dimensional complex vector space. Let $\mathbb{P}(E)$ be the projective space of lines through the origin of $E$. Fulton, in his book "Young Tableaux", then defines the $\textit{dual projective space}$ to be the collection of all hyperplanes in $E$, $\mathbb{P}^*(E)$. He mentions that the motivation for this is that then $E$ then becomes the space of linear forms on $\mathbb{P}^{*}(E)$, and, likewise, $\mathsf{Sym}_{\bullet}(E)$ is the algebra of polynomial forms on $\mathbb{P}^{*}(E)$.

My question is, how exactly does this work? That is, how is an element $v \in E$ a linear form on $\mathbb{P}^{*}(E)$?

Answer 1

First, an hyperplane corresponds to a linear form up to scaling that is to an element of $\Bbb P(E^*)$. Then, by definition a linear form on $\Bbb P(E)$ is exactly a linear form on $E$, i.e an element of $E^*$.

Maybe you are surprised because this does not define a function on $\Bbb P(E)$ since it's only defined up to scaling, so in particular the value of a linear form at a point is not defined ! This is because in general, there are no maps $f : \Bbb P^n \to \Bbb P^1$ which are algebraic and non-constant, let alone a map $g : \Bbb P^n \to \Bbb A^1 = k$.

So instead of having functions, you will have sections of line bundle. This is what happens here, "linear forms" are sections of the line bundle $\mathcal O_{\Bbb P^n}(1)$, but not "functions" in the usual sense.

This is almost as nice, for example consider the 3 sections $x^2,xy,y^2$ of the bundle $\mathcal O_{\Bbb P^1}(2)$, we can use them to get a nice map $\varphi : \Bbb P^1 \to \Bbb P^2, [x_0 : x_1] \to [x_0^2:x_0x_1:x_1^2]$

Answer 2

The most natural way to view $\mathbb{P}^*(E)$ is as equivalence classes of surjections from $E$ to a line, where $[f: E \twoheadrightarrow L] = [g: E \twoheadrightarrow L']$ if and only if there exists some isomorphism $\varphi: L \xrightarrow{\sim} L'$ such that $\varphi f = g$. The equivalence class $[f: E \twoheadrightarrow L]$ represents the hyperplane $\ker f$. (Check that this works!)

So, the points of $\mathbb{P}^*(E)$ are (equivalence classes of) surjections from $E$ to a line. A fixed $v \in E$ determines a "function" $\mathbb{P}^*(E)$, in the sense that we can say for a point $[f: E \twoheadrightarrow L]$ whether $f(v)$ is zero or not. Also, given two functions $v, w \in E$, we can determine over the point $[f: E \twoheadrightarrow L]$ the value of the ratio $f(v) / f(w) \in k$, which is actually a number (provided $w$ does not vanish at $[f]$). This is exactly how projective coordinates are supposed to work.