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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!

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    $\begingroup$ $V$ is a finite-dimensional, I assume. The isomorphism $\bar{V}\cong V^*$ does not hold if $V$ is infinite-dimensional. $\endgroup$ – Batominovski Nov 4 '18 at 8:13
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    $\begingroup$ Exactly the same as in finite group. $\endgroup$ – user10354138 Nov 4 '18 at 8:16
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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 $v\in \overline{V}$.

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