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I am reading Representation Theory by Fulton and Harris on my own to get an introduction to the topic. I have a question about notation which really boils down to a question about the adjoint and transpose of a transformation.

Let $V$ be a complex vector space and $\rho:G\to GL(V)$ a representation of the group $G.$ The book then motivates the definition of $\rho^{\ast}: G\to V^{\ast}$ by saying we want the $\rho^{\ast}$ to preserve the pairing of $V^{\ast}$ and $V,$ i.e. we want the following:

$$ (1) \qquad \langle \rho^{\ast}(g)(v^{\ast}), \rho(g)(v) \rangle = \langle v^{\ast}, v \rangle$$

Then they claim that the only way to do this is by defining $$\rho^{\ast}(g) := {}^t\rho(g^{-1})$$

My question quite simply is this: Does the superscript $t$ stand for the transpose or the adjoint (i.e. conjugate transpose)?

If it stands for the transpose then,

$$\langle \rho^{\ast}(g)(v^{\ast}), \rho(g)(v) \rangle = \langle v^{\ast}(\rho(g^{-1})), \rho(g)v\rangle$$

And by evaluating the above pairing directly we get (1) which is what we wanted. If however the superscript $t$ stands for the adjoint and the dual pairing behaves like an inner product then we would get,

$$\langle \rho^{\ast}(g)(v^{\ast}), \rho(g)(v) \rangle = \langle v^{\ast}, \rho(g^{-1})\rho(g) v\rangle$$

and again we get (1). Am I misinterpreting something? The book claims there is only one way to define $\rho^{\ast}$ to preserve the dual pairing but it seems like using either the transpose or the adjoint works.

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You're both right but you have to be careful with what is well-defined. If $V$ is a complex vector space and $A: V \to V$ then there is a transpose map $A^t : V^* \to V^*$. If you choose a basis for $V$ and corresponding dual basis for $V^*$, then the matrix form $A^t$ is just the usual transpose of the matrix form for $A$.

However, the conjugate transpose does not make invariant sense for $V$ or $V^*$. Instead the conjugate transpose is related to the vector space $\bar V^*$ of conjugate-linear maps $V \to \mathbb C$: given $A : V\to V$ we get a map $\bar A^t : \bar V^* \to \bar V^*$ by $$ \langle \bar A^t \phi, v \rangle = \langle \phi, Av\rangle, \phi \in \bar V^*, v \in V. $$ What you showed is that a rep on $V$ also induces a rep on $\bar V^*$ (in a similar manner to the induced rep on $V^*$).

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Thanks! That makes sense, I have never see the superscripted t written on the left before for transpose which sparked the original question in my mind. I thought it could be some nonstandard way of indicating the conjugate transpose. Thanks again! – mike vaiana Mar 9 '13 at 3:26
Yea I've seen that notation before but I think it is rare. Maybe it's purpose is to make it cleaner to write transpose inverse (which you use when defining dual reps). – Eric O. Korman Mar 9 '13 at 3:30

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