Let $V$ be a complex finite-dimensional vector space. Then there always exists an isomorphism $V \simeq V^*$, where $V^*$ is the dual space. The isomorphism can be fixed by choosing a non-degenerate bilinear or sequilinear form on $V$: $$\langle \dot \,, \dot \, \rangle : V\times V \rightarrow \mathbb{C}$$ Now, let $G$ be a Lie group acting on $V$ (label this representation $\rho$). Then the dual representation $\rho^*$ to $\rho$ should preserve the bilinear\sequilinear form defined above: $$ \langle{\rho}^*(g)\varphi, \rho(g)v\rangle = \langle\varphi, v\rangle. $$

Taking a sequilinear form seems to be more natural than a bilinear one (since $V$ is complex).
Do the two form types give different dual representations of $G$?
If yes, then which one is usually referred to as the dual representation?

  • $\begingroup$ The isomorphism $V\cong V^*$ induced by a non-degenerate sesquilinear form is only linear over $\mathbb{R}$, not over $\mathbb{C}$, isn't it? $\endgroup$ – Jeremy Rickard Jul 28 '15 at 13:47
  • $\begingroup$ @JeremyRickard. The isomorphism is $v \rightarrow \langle v, \dot \, \rangle$. Seems that depending on where you put the $v$ in the sequlinear form, it can be either linear or antilinear over $\mathbb{C}$. $\endgroup$ – rcode Jul 28 '15 at 14:09
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    $\begingroup$ If you do it that way, then it doesn't map to $V^*$, as $\langle v,?\rangle$ isn't a linear map $V\to\mathbb{C}$, it's an antilinear map. $\endgroup$ – Jeremy Rickard Jul 28 '15 at 14:28
  • $\begingroup$ You are right, I should have thought of this. $\endgroup$ – rcode Jul 28 '15 at 19:48

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