5
$\begingroup$

I was writing this question, and I came up with an answer, so I thought I would answer it myself:

In considering representations of $S_n$, among others, we have the "sign representation", that is the one-dimensional representation $$ \rho_{\Sigma}:S_n \to \mathbb{C}^{\times}: \tau \mapsto \text{sgn}(\tau). $$

When we are finding irreducible representations of $S_n$, sometimes the argument in books I read proceeds as follows: we find an irreducible representation $\rho$ on $V$. We note that the character of $\rho \otimes \rho_{\Sigma}$ is distinct from that of $\rho$. Then we conclude that $\rho \otimes \rho_{\Sigma}$ must be another irreducible representation.

The character table tells us that $\rho \otimes \rho_{\Sigma}$ is different from $\rho$, but how do we know that it is irreducible?

$\endgroup$

2 Answers 2

19
$\begingroup$

You don't need any character theory to do this. Let $V$ be an irreducible representation of any group $G$ (the group is not necessarily finite and $V$ is not necessarily finite-dimensional) and let $L$ be a $1$-dimensional representation. I claim that $V \otimes L$ is still irreducible. The reason is that tensoring with $L$ is invertible: the natural map $L^{\ast} \otimes L \to 1$ (where $1$ is the trivial representation) is an isomorphism, so

$$(V \otimes L) \otimes L^{\ast} \cong V.$$

Consequently, if $W$ is a proper nonzero submodule of $V \otimes L$, then $W \otimes L^{\ast}$ is a proper nonzero submodule of $V$. More abstractly, tensoring with $L$ is an automorphism of the category of representations of $G$, and automorphisms of categories preserve categorical properties of their objects like irreducibility.

$\endgroup$
7
  • 2
    $\begingroup$ This idea works in many other contexts where a tool like character theory isn't available, e.g. tensoring with an invertible module or an invertible sheaf preserves all categorical properties of modules or sheaves respectively. $\endgroup$ Apr 9, 2014 at 5:26
  • $\begingroup$ Thanks for the answer. I haven't heard of $L^*$; I'm assuming that it's the following map: if $\rho: G \to \mathbb{C}^\times$, then $$\rho^*: G \to \mathbb{C}^\times : g \mapsto \rho(g)^{-1}.$$ Is that right? By the way, I really enjoy reading your answers on Stack; they are very enlightening. $\endgroup$
    – Eric Auld
    Apr 9, 2014 at 11:37
  • $\begingroup$ @Eric: $L^{\ast}$ is the dual representation (en.wikipedia.org/wiki/Dual_representation). That's the correct description for one-dimensional representations, but in general you also need to take the transpose (thinking of the target as being matrices). $\endgroup$ Apr 10, 2014 at 3:55
  • $\begingroup$ I was interested in your opinion on this post, but I'm not sure if my tag for you worked in the other post: math.stackexchange.com/questions/834140/… $\endgroup$
    – Eric Auld
    Jun 14, 2014 at 18:28
  • 1
    $\begingroup$ @alpha123: yes, that's right. In fact we don't need to be working over a field. $\endgroup$ Jun 25, 2021 at 1:35
3
$\begingroup$

Let $\chi$ be the character of $\rho$ and $\chi'$ be the character of $\rho \otimes \rho_{\Sigma}$. Then observe that $\langle \chi, \chi\rangle=\langle \chi', \chi'\rangle$. But $\langle \chi, \chi\rangle=1$ since $\rho$ is irreducible, so $\langle \chi', \chi'\rangle=1$, and $\chi'$ is irreducible as well.

$\endgroup$
3
  • 4
    $\begingroup$ Note that this isn't just true for the sign representation. If $\rho$ is an irreducible representation and $\chi$ a one dimensional one, then $\rho \otimes \chi$ is irreducible too. $\endgroup$
    – ah11950
    Apr 8, 2014 at 16:14
  • $\begingroup$ @ah11950 Why should that be true? I don't see why $\langle \chi, \chi \rangle$ should equal $\langle \chi', \chi'\rangle$ in this case. $\endgroup$
    – Eric Auld
    Apr 8, 2014 at 16:52
  • 1
    $\begingroup$ Do you know that $\chi(g^{-1}) = \overline{\chi(g)}$? If so, when writing out the definition of inner product, it becomes clear; all your terms associated to $\chi$ will just cancel with their conjugates. $\endgroup$
    – ah11950
    Apr 8, 2014 at 16:56

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.