Definition of dual representation of a finite group I am trying to understand the dual representation. But I am struggling to understand even the definition.
Let $\rho: G \to GL(V)$ be a representation of a finite group. Then we have the dual space $V^{*}$ that consists of all linear maps from $V$ to $\mathbb{C}$. We then get the dual representation of $G$
$$
\rho^*: G\to GL(V^{*})
$$
My question is: How is this defined?
From here I see  something like
$$\rho^*(g)(f) = f(\rho(g^{-1})) $$
But I don't understand this. If $f\in V^*$, then one can't evaluate $f$ at $\rho(g^{-1})$.
From here I see something like
$$
\rho^*(g) = \rho(g^{-1})^T
$$
Here there is a transpose. But this also doesn't make sense to me because $p^*(g)$ is supposed to take an element of $V^*$ to $V^*$.
I guess I am looking for a definition like
$$
\rho^*(g)(f)(v) = \dots
$$
for $g\in G, f\in V^*, v\in V$.
 A: Yup these are the same, let me explain:
Ok so given any linear map $\phi:V \to W$ we obtain a dual map $\phi^*: W \to V$, this map sends a linear function $f$ on $W$ to the linear function on $v \to f(\phi(v))$ on $V$.  One of the key properties of this is that it is a contravariant functor, which is just a fancy way of saying $(\phi \circ \psi)^* = \psi^* \circ \phi^*$.
That's a great abstract definition, but let's be more concrete.  If we fix bases for $V$ and $W$ then $\phi$ is represented by a matrix $A$. In the corresponding dual bases the map $\phi^*$ is given by $A^T$, and the contravariance mentioned above is just the familiar fact that $(AB)^T = B^TA^T$.
Ok now let's have a group $G$ acting on a vector space $V$, where a group element $g$ acts by a linear map $\phi_g$ and satisfy $\phi_g \circ \phi_h = \phi_{gh}$. If we take the duals of these maps (which again corresponds to transposes) we get maps $\phi_g^*: V^* \to V^*$. 
These satisfy $\phi_g^* \phi_h^* = \phi_{hg}^*$, which note is not the right formula for a representation of $G$ because the order is switched.  Instead this means we have a representation of $G^{op}$ the opposite group of $G$, which has the same elements but multiplication defined by $g \cdot_{G^{op}} h := h \cdot_G g$.
Oh no, what is this mysterious opposite group?  Well it's canonically isomorphic to $G$ by the map $g \to g^{-1}$.  Note that this really using the fact that we have a group and not just a monoid, for those there isn't always a dual representation.
Anyway, now we can compose this canonical isomorphism $G \to G^{op}$ with the dual representation $G^{op} \to GL(V^*)$ to get out representation $G \to GL(V^*)$ sending $g$ to $\phi_{g^{-1}}^*$. If we pick a basis for $V$, the maps $\phi_g$ on $V$ get represented by matrices $A_g$. In the dual basis for $V^*$ the maps $\phi_{g^{-1}}$ are then represented by $A^T_{g^{-1}}$.
A: It is $f\circ\rho(g^{-1})$ the composition, $\rho(g^{-1}:V\rightarrow V$ and $f:V\rightarrow K$, so these maps can be composed.
