2
$\begingroup$

I am working out of Representation Theory, A First Course, Fulton and Harris. In particular I am in Part 1, Section 1.1 where all groups are taken to be finite and all vector spaces are finite dimensional $\mathbb{C}$- vector spaces.

They define the (..a?) representation on $\mathrm{Hom}(V,W)$ by identifying $\mathrm{Hom}(V,W) = V^{*} \otimes W$, since they have already shown how the dual of a representation and the tensor product of representations can be made into a representation. I am familiar with the identification mentioned above (as vector spaces), but I would like to verify the induced action (as a representation) on $\mathrm{Hom}(V,W)$. In order to do so I first need to understand the representation on $V^{*}$. This leads to my first question.

Given a $G$-representation $V$, with homomorphism $\rho$, the text defines the dual representation (verbatim, page 4) as

$$\rho^{*}(g) = \, ^{t}\rho(g^{-1})\colon V^{*} \to V^{*}.$$

First of all, is there a reason the transpose is written on the left? and further if $\mathrm{dim}(V) = n$, then if I am interpreting the above correctly $ \, ^{t}\rho(g^{-1})$ is (can be identified with) an $n$ x $n$ invertible matrix. Given $\lambda \in V^{*}$, $\lambda \colon V \to \mathbb{C}$, $\lambda$ can be identified with a 1 x $n$ matrix. Is the action of $g\lambda$ computed by identifying $\lambda$ with its transpose which would appear as a vector in $V$, then computing the product $^{t}\rho(g^{-1})$$\lambda$ which outputs a new vector in $V$, which can be transposed to be seen as a new map $V \to \mathbb{C}$? Hopefully my question here is clear. I will reformulate if it is not.

Now, assuming I do understand the representation on $V^{*}$ and the representation on a tensor product of representations, in the case of $V^{*} \otimes W$ we would have $$\begin{align*}g(\lambda \otimes w) &= g\lambda \otimes gw\\ &= \rho_{1}^{*}(g) \lambda \otimes \rho_2(g)w\\ &= \, ^{t}\rho_{1}(g^{-1})\lambda \otimes \rho_2(g)w.\\ \end{align*}$$

The final line should justify why we define the action of $g$ on $\phi \in \mathrm{Hom}(V,W)$ as $g\phi = g\phi(g^{-1}v)$?. This definition should be such that the appropriate diagram commutes. Going one direction through the diagram I get that $\lambda \otimes w$ maps to the linear map from $V$ to $W$ given by $\lambda(v)w$ then after being acted on by $g$ becomes the map that takes $v$ to $g(\lambda(g^{-1}v)w$, I do not see how this is the same as if I traveled the other direction in the diagram.

$\endgroup$
  • $\begingroup$ I have continued trying to figure this out since positing and it seems I misinterpreted the definition of the dual representation, it it is actually pre composing the linear function with $g^{-1}$. This seems to make things work out, but I will leave the question up because someone may still have an illuminating answer :) $\endgroup$ – Prince M Apr 21 '18 at 22:41
2
$\begingroup$

I like the following way of thinking about the representation of $V^*$:

suppose $\lambda:V \to \Bbb{C}$ is a linear functional and $g \in G$. Then $\rho^*(g)(\lambda)=\lambda \circ \rho(g^{-1})$. Also, if we are suppressing $\rho$ and $\rho^*$ and if $v \in V$, then $(g\lambda)(v)=\lambda(g^{-1}v)$. This is the same as what is written in the book and a little easier to think about.

I do not know why they wrote the $^t$ on the left. If you are thinking in terms of matricies, then $[\rho^*(g)\lambda]=[\lambda][\rho(g^{-1})]$

In this case, we get that: $$\begin{align*}g(\lambda \otimes w) &= g\lambda \otimes gw\\ &= \lambda \circ g^{-1}\otimes gw\\ \end{align*}$$

which would correspond to the map which takes $v$ to $\lambda(g^{-1}v)gw$. Call the map corresponding to $\lambda \otimes w$, $\phi$. then $g\phi(g^{-1}v)=g(\lambda(g^{-1}v)w)=\lambda(g^{-1}v)gw$ (since $\lambda(g^{-1}v)$ is a scaler) which is what we expected. Note that not every map in $\text{Hom}(V,W)$ corresponds to a pure tensor $\lambda \otimes w$. But they do correspond to linear combinations of such tensors (you may be familiar with this though).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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