# 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!

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

## 1 Answer

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 $$v\in \overline{V}$$.