$2×2=3+1$ for $\operatorname{GL}_2$ If $V$ is the natural representation for $\operatorname{GL}(2,q)$, then $V⊗V$ appears to decompose into the direct sum of a (strange?) one-dimensional module and a three dimensional module.  I've managed to confuse myself about where $\operatorname{Sym}(V)$ and $Λ(V)$ come into play, and maybe some duals are messed up.  I am also interested in taking those 1 and 3 dimensional modules and dividing them by the determinant.

Can someone explain the 1, 2, and 3 dimensional modules of $\operatorname{GL}(2,q)$ using the natural module, symmetric powers, exterior powers, and duals?

Some specific questions that if the answer is no, might require a nice gentle answer as to sane formulas for both sides of the inequality:


*

*Does $V^*⊗V^*$ decompose into $\operatorname{Sym}(V^*)⊕1$, where is the trivial module?

*Is $W/\det = W^*$, where $W/\det$ is the representation where $g$ acts as $g$ divided by its determinant (taken in $\operatorname{GL}_2$).

*Does it make sense to talk about the determinant after one thinks of $g$ as acting on $W$?  That is, as an $n×n$ matrix, rather than a $2×2$?

*Is the action of $\operatorname{GL}(2,q)$ on polynomials (by change of variable) called $\operatorname{Sym}(V)$ or $\operatorname{Sym}(V^*)$?

*What about the action on wedgies, $Λ(V)$ or $Λ(V^*)$?


If it is easier, I think I only care about $\operatorname{PGL}$ for today, but I wouldn't be surprised if I liked $\operatorname{GL}$ tomorrow.  This confusion is all probably due to sticking to $\operatorname{SL}(2,q)$ where $W =W^*$ and the determinant is 1 so it doesn't really matter when you divide by it.
 A: I have some answers but not others.  Also, I am frightened of finite fields, so I will only talk about representations over $\mathbb{C}$.


*

*The tensor product $V\otimes V\;$ is the direct sum of the alternating square $\Lambda^2(V)$ (which is one-dimensional) and the symmetric square $\text{Sym}_2(V)$ (which is three-dimensional).

*The action of $\mathrm{GL}(2,\mathbb{C})$ on $\Lambda^2(V)$ is multiplication by determinants, i.e. $gx = \det(g)x$ for any real number $x$.  The action on the dual $\Lambda^2(V^*)$ is division by determinants.
Note: By $\Lambda^2(V)$ here I mean the second exterior power of $V$, i.e. linear combinations of wedge products $v \land w$ where $v,w\in V$.  The action is $g(v\land w) = (gv)\land (gw)$, e.g.
$$
\begin{bmatrix}
2 & 0 \\ 0 & 3
\end{bmatrix}(e_1\land e_2) \;=\; (2e_1) \land (3e_2) \;=\; 6(e_1\land e_2).
$$

*The action of $\mathrm{GL}(2,\mathbb{C})$ on $\text{Sym}_2(V)$ can be thought of as the action of $\mathrm{GL}(2,\mathbb{C})$ on the space of quadratic forms on $\mathbb{C}^2$ via  $q \mapsto g q g^T$.  This generalizes to polynomials.

*The action of $\mathrm{GL}(2,\mathbb{C})$ on $\text{Sym}_2(V^*)$ can be thought of as the action of $\mathrm{GL}(2,\mathbb{C})$ on the space of quadratic forms on $\mathbb{C}^2$ via $q \mapsto (g^{-1})^T q g^{-1}$.  This generalizes to polynomials.
Note: After thinking about it, I decided to remove the names "change of basis" and "change of variables" on #3 and #4, since I'm not sure which one you mean when you say "change of variables".
Difference Between #3 and #4:  This is a bit complicated, because there are actually four different ways to change variables with a matrix.  For example, given the matrix $\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$ with variables $x$ and $y$, we can define new variables $u$ and $v$ in any one of the following ways:
A. $\begin{bmatrix}u \\ v\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$
B.  $\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}u \\ v\end{bmatrix}$
C. $\begin{bmatrix}u & v\end{bmatrix} = \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$
D. $\begin{bmatrix}x & y\end{bmatrix} = \begin{bmatrix}u & v\end{bmatrix}\begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}$
This leads to four different actions of matrices on polynomials. Of these four actions, two are left actions (namely those derived from methods A and D), and two are right actions (namely those derived from methods B and C).  For left actions, applying $g$ first and then $h$ is the same as applying $hg$:
$$
h(gp) \;=\; (hg)p.
$$
(For right actions, applying $g$ and then applying $h$ would be the same as applying $gh$).  Since we are only interested in left actions, this leaves us with two ways that a matrix might act on a polynomial.  One of these (method D) gives the action of matrices on $\text{Sym}_2(V)$ while the other (method A) gives the action of matrices on $\text{Sym}_2(V^*)$.
