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In Fulton and Harris' Representation Theory, right at the beginning when they introduce representations, they note

The dual $V^{\ast} = \mbox{Hom}(V,{\mathbb C})$ of $V$ is also a representation, though not in the most obvious way: we want the rwo representations of $G$ to respect the natural pairing (denoted $\langle\hspace{.1in}, \hspace{.1in}\rangle$) between $V^{\ast}$ and $V$, so that if $\rho:G\rightarrow \mbox{GL}(V)$ is a representation and $\rho^{\ast}:G\rightarrow \mbox{GL}(V^{\ast})$ is the dual, we should have $\langle\rho^{\ast}(g)(v^{\ast}), \rho(g)(v)\rangle = \langle v^{\ast},v\rangle$ for all $g\in G$, $v\in V$, and $v^{\ast}\in V^{\ast}$. This in turn forces us to define the dual representation by $\rho^{\ast}(g) = ^{t}\rho(g^{-1}):V^{\ast}\rightarrow V^{\ast}$ for all $g\in G$.

I have a few questions about this.

  1. What is this natural pairing they are referring to? Is it that we can make our basis such that $e^{\ast}_{i}(e_{j}) = \delta_{ij}$?
  2. (This may be answered by the question above) What is the equality between these two relationships implying?
  3. What is this notation in the definition of the dual representation -- is this the transpose of the image of the inverse of $g$? Where is this coming from?
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I am reading the same book and just got stuck there :) –  haemhweg Jun 17 '13 at 11:27

1 Answer 1

up vote 5 down vote accepted

The natural bilinear pairing between $V^*$ and $V$ is the pairing $\langle \mathbf{f},v\rangle = \mathbf{f}(v)$ for each $\mathbf{f}\in V^*$ and each $v\in V$.

The equality means we want the representation of $V$ and that of $V^*$ to "respect" the relationship between $V^*$ and $V$. Let $v_1,\ldots,v_n$ be a basis for $V$, and let $v_1^*,\ldots,v_n^*$ be the dual basis. If $\rho\colon G\to \mathrm{GL}(V)$ be a representation and we want $\rho^*$ to satisfy the desired property, then we have $$\langle \rho^*(g)(v_i^*),\rho(g)(v_j)\rangle = \langle v_i^*,v_j\rangle = \delta_{ij}$$

That means that $\rho^*(g)(v_1^*),\ldots,\rho^*(g)(v_n^*)$ must be the dual basis to $\rho(g)(v_1),\ldots,\rho(g)(v_n)$.

But because a map $V\to W$ induces a map in dual spaces going the other way, $W^*\to V^*$, the way to achieve this is to map $V^*\to V^*$ by the map induced by $\rho(g^{-1})$, rather than the map induced by $\rho(g)$. And the map induced by $\rho(g^{-1})$ has matrix given by the conjugate transpose of the matrix given by $\rho(g^{-1})$.

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I've wondered, how does one actually show that the property holds? $\langle \rho^*(g)(v^*),\rho(g)(v)\rangle=\langle {}^t\rho(g^{-1})(v^*),\rho(g)(v) \rangle = \langle {}^t\rho(g)^{-1}(v^*),\rho(g)(v) \rangle =?$ I've always just understood the transpose and inverse for it to actually be a representation (since transpose is order reversing) –  Juan S Dec 13 '11 at 1:43
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For a matrix $A$, $A^*$ is defined to be the (unique) matrix such that $\langle A^*v,w\rangle = \langle v,Aw\rangle$ for all $v$ and $w$. One then proves, by considering well-chosen vectors $v$ and $w$, that $A^*$ is the conjugate transpose. Now you can apply that to get $\langle {}^t\rho(g^{-1})(v^*),\rho(g)(v)\rangle = \langle v^*,\rho(g^{-1})\rho(g)(v)\rangle = \langle v^*,v\rangle$, which is the desired equality. –  Arturo Magidin Dec 13 '11 at 3:36
    
Nice, thanks as always! –  Juan S Dec 13 '11 at 3:38

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