In Fulton and Harris' Representation Theory, right at the beginning when they introduce representations, they note
The dual $V^{\ast} = \mbox{Hom}(V,{\mathbb C})$ of $V$ is also a representation, though not in the most obvious way: we want the rwo representations of $G$ to respect the natural pairing (denoted $\langle\hspace{.1in}, \hspace{.1in}\rangle$) between $V^{\ast}$ and $V$, so that if $\rho:G\rightarrow \mbox{GL}(V)$ is a representation and $\rho^{\ast}:G\rightarrow \mbox{GL}(V^{\ast})$ is the dual, we should have $\langle\rho^{\ast}(g)(v^{\ast}), \rho(g)(v)\rangle = \langle v^{\ast},v\rangle$ for all $g\in G$, $v\in V$, and $v^{\ast}\in V^{\ast}$. This in turn forces us to define the dual representation by $\rho^{\ast}(g) = ^{t}\rho(g^{-1}):V^{\ast}\rightarrow V^{\ast}$ for all $g\in G$.
I have a few questions about this.
- What is this natural pairing they are referring to? Is it that we can make our basis such that $e^{\ast}_{i}(e_{j}) = \delta_{ij}$?
- (This may be answered by the question above) What is the equality between these two relationships implying?
- What is this notation in the definition of the dual representation — is this the transpose of the image of the inverse of $g$? Where is this coming from?
Best Answer
The natural bilinear pairing between $V^*$ and $V$ is the pairing $\langle \mathbf{f},v\rangle = \mathbf{f}(v)$ for each $\mathbf{f}\in V^*$ and each $v\in V$.
The equality means we want the representation of $V$ and that of $V^*$ to "respect" the relationship between $V^*$ and $V$. Let $v_1,\ldots,v_n$ be a basis for $V$, and let $v_1^*,\ldots,v_n^*$ be the dual basis. If $\rho\colon G\to \mathrm{GL}(V)$ be a representation and we want $\rho^*$ to satisfy the desired property, then we have $$\langle \rho^*(g)(v_i^*),\rho(g)(v_j)\rangle = \langle v_i^*,v_j\rangle = \delta_{ij}$$
That means that $\rho^*(g)(v_1^*),\ldots,\rho^*(g)(v_n^*)$ must be the dual basis to $\rho(g)(v_1),\ldots,\rho(g)(v_n)$.
But because a map $V\to W$ induces a map in dual spaces going the other way, $W^*\to V^*$, the way to achieve this is to map $V^*\to V^*$ by the map induced by $\rho(g^{-1})$, rather than the map induced by $\rho(g)$. And the map induced by $\rho(g^{-1})$ has matrix given by the conjugate transpose of the matrix given by $\rho(g^{-1})$.