# Why does the tensor product of an irreducible representation with the sign representation yield another irreducible representation?

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?

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.

• 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. Apr 9, 2014 at 5:26
• 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. Apr 9, 2014 at 11:37
• @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). Apr 10, 2014 at 3:55
• 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/… Jun 14, 2014 at 18:28
• @alpha123: yes, that's right. In fact we don't need to be working over a field. Jun 25, 2021 at 1:35

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.

• 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. Apr 8, 2014 at 16:14
• @ah11950 Why should that be true? I don't see why $\langle \chi, \chi \rangle$ should equal $\langle \chi', \chi'\rangle$ in this case. Apr 8, 2014 at 16:52
• 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. Apr 8, 2014 at 16:56