Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question
I am reading the same book and just got stuck there :) –  haemhweg Jun 17 '13 at 11:27
add comment

1 Answer

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 $\rho(g^{-1})$.

share|improve this answer
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
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.