I am reading Serre's Linear Representations of Finite Groups and in an exercise in there he asks to show if $\rho$ is a representation of a finite group on $\textrm{GL}(V)$ with $V$ a finite dimensional complex vector space that there exists a unique linear representation $\rho^\ast:G \to \textrm{GL}(V^\ast)$ such that
$$(\rho^{\ast}_sf)(\rho_sx) = f(x) $$
for all $s \in G, x \in V$, $f \in V^\ast$. Now I believe I have shown existence as follows: Let $\rho^\ast_sf$ be the map from $G$ to $\textrm{GL}(V^\ast)$ defined by saying that given any $f \in V^\ast$, $x \in V$ we have that
$$(\rho^\ast_sf)(x) = f\left(\rho_{s^{-1}}(x)\right).$$
Now we see that $\rho_s^\ast$ is a linear map for each $s$ and we can write down a matrix for $\rho^\ast_s$ in the standard basis $\{\Lambda_i\}$ of $V^\ast$. Here I am taking $\{e_i\}$ as my standard basis for $V$, so as usual we have that
$\Lambda_i(e_j) = \delta_{ij}$.
I have found that the matrix of $\rho_s^\ast$ in the basis I stated above is the matrix $(\rho_{s^{-1}})^T$. Now once I have this, it is easy to see that $\rho^\ast_s$ defined a homomorphism from $G$ to $GL(V^\ast)$ and that
$$\textrm{Tr}(\rho^\ast_s) = \overline{\textrm{Tr}(\rho_s)}$$
where the bar indicates complex conjugation.
The problem I am having now is in showing uniqueness. I can't seem to get uniqueness from the property that such a representation must satisfy. What I have tried is this. Suppose we have another homomorphism $\sigma: G \to GL(V^\ast)$. Then we must have that
$$(\sigma_sf)(\rho_sx) = fx$$
for all $f \in V^\ast$, $s \in G$ and $x \in V$. Then by definition of an element in $\textrm{End}(V^\ast)$ we must have that $ f(\sigma_s\rho_s(x)) = f(x)$. Since this must hold for all $f \in V^\ast$ and $x \in V$, I get that
$$\sigma_s\rho_s = \textrm{Id}.$$
Now don't I want $\sigma = (\rho_{s}^{-1})^T$ from here? I can't seem to get it from the expression above. How can I get uniqueness?
Thanks.
Best Answer
If $u,v$ are maps with the same domain and codomain, and $u(x)=v(x)$ for every $x$ in the domain, then $u=v$ as maps. This applies to the symmetries of a function space just the same:
$$\begin{array}{c l} & \forall f\in V^*, & \forall x\in V: & (\rho_s^*f)(\rho_sx)=f(x)=(\sigma_sf)(\rho_sx) \\ \iff & \forall f\in V^*, & \forall v\in V: & (\rho_s^*f)(v)=(\sigma_sf)(v) \\ \iff & \forall f\in V^*: & & \rho_s^*f=\sigma_sf \\ \iff & & &\rho_s^*=\sigma_s. \end{array}$$
Note the substitution $u=\rho_sx$ ($\rho_sV=V$ because $\rho_s$ is always invertible). Applies for every $s\in G$.
The moral: recognize reparametrizations of universal quantification, and recognize that universally quantified equalities in a space pass up higher to equalities in the relevant function space.