Came by from this link https://en.wikipedia.org/wiki/Complex_conjugate_representation
I am wondering if $\Pi$ a representation, then $\bar{\Pi}$ and dual of $\Pi$ are isomorphic representations.
I can see $\Pi$ and $\bar{\Pi}$ is naturally isomorphic as vector space, hence $\bar{\Pi}$ and the dual is isomorphic as vector space. However, it seems like the natural isomorphism is not a representation isomorphism. How do we deduce this?
Best Answer
Let $G$ be a group, let $\Pi$ be a representation of $G$ on a finite-dimensional $\mathbf{C}$-vector space $V$. We will write simply $gv=\Pi(g)v$ for the action of $G$ on $V$ via $\Pi$. Suppose $V$ is equipped with a non-degenerate Hermitian form $(\cdot,\cdot)$. That is, $(v_1,v_2) \in \mathbf{C}$ is linear in $v_2$, satisfies $(v_2,v_1)=\overline{(v_1,v_2)}$, and for each $v_1 \in V$ there is some $v_2 \in V$ with $(v_1,v_2) \neq 0$. If in addition the form is compatible with the action of $G$ on $V$ in the sense that $$(g v_1,g v_2)=(v_1,v_2) \quad \hbox{for all $g \in G$ and $v_1,v_2 \in V$}$$ then the map $v \mapsto (v,\cdot)$ defines a $G$-equivariant conjugate-linear isomorphism of $V$ onto the dual space $V^*$, or in other words an isomorphism from the conjugate of $\Pi$ to the dual of $\Pi$.
Here is one situation in which such a form is guaranteed to exist: if $G$ is a finite group (or more generally a compact topological group, where sums are replaced by integrals with respect to a Haar measure) then you can start with any positive-definite Hermitian form $(\cdot,\cdot)'$ on $V$ and produce a $G$-invariant positive definite Hermitian form by averaging over $G$, $$(v_1,v_2):= \sum_{g \in G} (g v_1,gv_2)'.$$ Here we use that a sum of positive-definite forms is positive definite and hence non-degenerate.