Contragradient representation of a finite group

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.

-

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.

-
Thanks for your answer. That is clear to me now. Just one thing, when I calculated above the matrix for $\rho^\ast_s$ in the basis $\Lambda_i$, when I apply that to an element of the dual (a row vector) I have to apply it to the left of $\rho^\ast_s$ yes? – user38268 Aug 11 '12 at 3:46
@BenjaLim If by "apply it to the left" you mean right-multiply in terms of matrices, then yes. That is $(Av)^T w=v^TA^T w$ in a nutshell. – anon Aug 11 '12 at 3:48
I understand that $(Av)^T\cdot w = v^TA^T \cdot w$, but how does that relate to just doing $w \rho^\ast_s$ where $w$ is a row vector? – user38268 Aug 11 '12 at 3:50
@BenjaLim You mean $(\rho_s^*w)$ (the contragrediant action is still a left-action)? Applying a dual vector to a vector is the same as matrix multiplication with transposes. So, $x\mapsto v^Tx$ is an example of $v$'s effect on vectors $x$, and $x\mapsto (Av)^Tx=v^TA^Tx=v^T(A^Tx)$ is the effect of $Av$'s on $x$. – anon Aug 11 '12 at 3:54
Sorry when I said $w\rho^\ast_s$ before, I meant that we are multiplying a $1 \times n$ matrix with a $n \times n$ matrix – user38268 Aug 11 '12 at 3:55