Let $V$ be a complex representation of a compact Lie group $G$. Prove that $\overline{V}$ and $V^*$ are isomorphic representations of $G$. Let $G$ a compact Lie group and $V$ a representation of $G$.
I have to proof that $\overline{V}$ $\cong$ $V^*$, where $V^*:=\text{Hom}(V,\mathbb{C})$ and $\overline{V}$=$V$ as a finite vector subspace but it has the operation $(\lambda, v)\mapsto \overline{\lambda} \cdot v$.
Any suggestion? Thanks in advance!
 A: I assume that $V$ is a finite dimensional complex representation of $G$.  First, there exists a $G$-invariant Hermitian form on $V$.  Consider an arbitrary positive definite Hermitian form $\langle\!\langle\bullet,\bullet\rangle\!\rangle:V\times V\to \Bbb C$ on $V$ (the convention is that $\langle\!\langle\bullet,\bullet\rangle\!\rangle$ is linear in the first variable, and anti-linear in the second variable).  Define
$$\langle u,v\rangle =\int_G\langle\!\langle g\cdot u,g\cdot v\rangle\!\rangle \ d\mu(g),$$
where $\mu$ is the normalized Haar measure on $G$ (which exists as $G$ is a compact Lie group).  
In the case where $G$ is a finite group, $\mu$ is simply $\frac{1}{|G|}c$, where $c$ is the counting measure.  Therefore, $\langle\bullet,\bullet\rangle$ is the avarage of $\langle\!\langle \bullet,\bullet\rangle\!\rangle$ over $G$.  Note that $\langle\bullet,\bullet\rangle$ is a $G$-invariant positive definite Hermitian form on $V$.  There exists a $G$-equivariant isomorphism $\varphi:\overline{V}\to V^*$ via the assignment $v\mapsto \langle \bullet,v\rangle$.  
Note that $$\varphi(\lambda\cdot v)=\varphi(\bar{\lambda}v)=\langle \bullet,\bar{\lambda} v\rangle =\lambda\langle\bullet,v\rangle=\lambda\varphi(v)=\lambda\cdot\big(\varphi(v)\big)$$ for all $\lambda\in\Bbb C$ and $v\in \overline{V}$.  Since $\varphi$ is clearly an injective map (noting that $\langle\bullet,\bullet\rangle$ is non-degenerate) and $\dim\overline{V}=\dim V=\dim V^*$, the map $\varphi$ must be an isomorphism of vector spaces.  We now need to show that $\varphi$ is $G$-equivariant.
For $g\in G$, $G$-invariance of $\langle\bullet,\bullet\rangle$ implies that
$$\varphi(g\cdot v)(u)=\langle u,g\cdot v\rangle=\langle g^{-1}\cdot u,v\rangle=\varphi(v)(g^{-1}\cdot u)$$
for all $u\in V$ and $v\in \overline{V}$.  However, the action of $G$ on $V^*$ is given by
$$(g\cdot f)(u)=f(g^{-1}\cdot u).$$
So, with $f=\varphi(v)$, we get that
$$\varphi(g\cdot v)=g\cdot\varphi(v)$$
for all $v\in \overline{V}$.
