TITLE: Fulton and Harris: Exercise 1.3 in section 1.1
QUESTION [6 upvotes]: This is exercise 1.3 on page 5 of Fulton and Harris Representation Theory: A First Course.
Exercise: Let $G$ be a finite group, let $V$ be an $n$-dimensional $\mathbb C$-vector space and let $\rho: G\to \mathrm{GL}(V)$ be a representation of $G$. Suppose that for all $g$ in $G$, the determinant of $\rho(g)$ is $1$ (i.e., $\rho: G\to \mathrm{SL}(V)$). Show that the spaces $\land^kV$ and $\land^{n-k}(V^*)$ are isomorphic as representations of $G$. Hint: The hypotheses ensure that $\land^nV$ is trivial, and the bilinear map $\land^kV\otimes \land^{n-k}V\to \land^nV = \mathbb C$ is a perfect pairing, i.e., it makes each space the dual of the other, c.f. Sect. B.3.
I have some ideas but no solution. Since $V$ is $n$-dimensional, $\land^n V$ is $1$-dimensional, so I suppose we can identify it with $\mathbb C$. Each $g$ in $G$ acts on $\mathbb C$ by a nonzero scalar, but I don't know how to use $\det \rho(g) = 1$ to show that this scalar must be $1$. Maybe I can identify $\rho(g)$ with its vector of columns in $V^{\oplus n}$ and use the fact that $\det$ is multilinear alternating?
The map $\land^kV\otimes \land^{n-k}V\to \land^nV$ looks like evaluation. If I could figure out the appropriate coevaluation $\mathbb C\to \land^{n-k}V\otimes\land^kV$, I could show that $\land^kV$ and $\land^{n-k}V$ are dual and thus isomorphic as vector spaces (since they are finite-dim). In that case $\land^{n-k}(V^*)\cong (\land^{n-k}V)^*\cong \land^kV$, but I would still have to show that this isomorphism respects $G$.

REPLY [6 votes]: (1) First, you have an isomorphism $\bigwedge^nV\otimes\left(\bigwedge^k V\right)^*\cong\bigwedge^{n-k}V$ which is given by $(w_1\wedge\dots\wedge w_n)\otimes \omega\mapsto\omega(w_1\wedge\dots\wedge w_k)\cdot w_{k+1}\wedge\dots\wedge w_n$ which is an element of $\bigwedge^{n-k}V$.
(2) Second, $G$ acts on $\bigwedge^nV$ trivially since the determinants are all $1$, so this is a copy of the trivial representation $\mathbb C$.
(3) So you go back to point 1 and what you have is actually $G$-linear (just check it, basically from the definition of how $G$ acts on exterior products) and so you get a sequence of $G$-isomorphisms $$\left(\bigwedge^kV\right)^*=\left(\bigwedge^kV\right)^*\otimes\mathbb C\cong\left(\bigwedge^kV\right)^*\otimes\bigwedge^nV\cong\bigwedge^{n-k}V.$$
You can then use a remark preceding the exercise to finish.