Notation in Fulton-Harris for dual representation In Fulton-Harris's Representation Theory, on page 4 they write 

This in turn forces us to define the dual representation by
  $$\rho^*(g) = \,\,^t\rho(g)^{-1}:V^* \to V^*.$$

I'm confused by this notation. Looking at other sources, I've learned that they mean $$\rho^*(g)(L)=L\circ \rho(g)^{-1}\quad \text{for } L:V\to \mathbb{C}.$$
But what is that notation they're using? I want to understand it in case it comes up again in the same text.
Also, why is it natural to want $$\langle \rho^*(g)(v^*), \rho(g)(v)\rangle = \langle v^*, v \rangle,$$ as they mention?
 A: If $f \colon E \to F$ is a linear map between vector spaces, ${}^tf$ is one common denotation of the adjoint/dual/transpose map $F^\ast \to E^\ast$ given by $\lambda \mapsto \lambda \circ f$. Other common denotations include $f^\ast,\; f^T,\; f'$ (the latter if the dual space is denoted $E'$).

Also, why is it natural to want $$\langle \rho^*(g)(v^*), \rho(g)(v)\rangle = \langle v^*, v \rangle,$$ as they mention?

I'm not sure. I would reach that identity as a consequence of what I consider a natural desideratum. The dual representation of $\rho$ shall of course be intimately related to $\rho$. We have a natural way to move from $GL(V)$ to $GL(V^\ast)$, transposition. But transposition is contravariant, so $\delta(g) = {}^t\rho(g)$ would not be a representation of $G$, but of $G^{\text{op}}$. So we need another contravariant operation to obtain a representation on $V^\ast$. The obvious contravariant operation on $GL(V^\ast)$ is inversion, so $g \mapsto {}^t\rho(g)^{-1}$ is a representation of $G$ on $V^\ast$ that is intimately related to $\rho$. That makes it an excellent candidate to be called the dual representation of $\rho$.
