I am currently reading "Algebraic Geometry" by J. Harris, specifically I am trying to learn about algebraic groups, but I am having a few problems. Here is a passage (p. 117) that I struggle with:
- Could I think of the action of GL$_2(K)$ on $V$ as just the normal matrix vector multiplication by choosing a basis for $V$?
- What exactly is Sym$^2(V)$? I have read online that it is a subspace of $V \otimes V$, containing all the symmetric tensors, for example is $e_1 \otimes e_2 + e_2 \otimes e_1$ such a tensor, where $e_1, e_2$ are a basis for $V$. A natural action of GL$_2(K)$ on Sym$^2(V)$ could be $A(x \otimes y) = Ax \otimes Ay$. Is that correct so far? Also, could I for simplicity just think of Sym$^2(V)$ as being the space of symmetric $2 \times 2$ matrices?
- If 1. is true, then one could think of the action of PGL$_2(K)$ on $\mathbb{P}^1$ as $[A][x]=[Ax]$ right? Since invertible matrices map 1-dimensional subspaces to 1-dimensional subspaces.
- How exactly does PGL$_2(K)$ now act on $\mathbb{P}($Sym$^2(V)$$)$? How is the action defined? I can see why $\mathbb{P}($Sym$^2(V)$$)$ is isomorphic to $\mathbb{P}^2$, since they both have the same dimension.
- Why do we need char($K$)$\neq 2$? Do we need it because we divide by $2$ at some point?
Edit: Just to make sure that I understood correctly: since Sym$^2(V)$ has dimension 3, it is isomorphic to $K^3$, so there is some isomorphism $\Phi:K^3 \rightarrow$ Sym$^2(V)$. Now we can define an action of GL$_2(K)$ on $K^3$ by: $A \cdot (a,b,c) = \Phi^{-1}(A(\Phi(a,b,c)))$.
In the same way we can also get an action of PGL$_2(K)$ on $\mathbb{P}^2$. First, we consider the action of PGL$_2(K)$ on $\mathbb{P}($Sym$^2(V)$$)$ by $[A]\cdot [x\otimes y] = [Ax \otimes Ay]$. Then, there is a isomorphism $\Psi : \mathbb{P}^2 \rightarrow \mathbb{P}($Sym$^2(V)$$)$, so we define the group action of PGL$_2(K)$ on $\mathbb{P}^2$ to be: $[A] \cdot [a:b:c] = \Psi^{-1} ([A](\Psi([a:b:c])))$
Is this true so far?