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.