1
$\begingroup$

I'm looking at a very specific action of the semidirect product (wreath product) $(S_2 \times S_2) \rtimes S_2 = S_2 \wr S_2$ on $\mathbb{Q}^3$. Namely, the generators acts as follows on a basis $x_1, x_2, x_3$ of $\mathbb{Q}^3$ (write $S_2 = \{ 1, \sigma \}$): \begin{align*} ((1,1), \sigma): \hspace{2mm} &x_1 \mapsto x_2, \hspace{2mm} x_2 \mapsto x_1, \hspace{2mm} x_3 \mapsto -x_3 \\ ((\sigma,1), 1): \hspace{2mm} &x_1 \mapsto -x_1, \hspace{2mm} x_2 \mapsto x_2, \hspace{2mm} x_3 \mapsto x_3 \\ ((1, \sigma), 1): \hspace{2mm} &x_1 \mapsto x_1, \hspace{2mm} x_2 \mapsto -x_2, \hspace{2mm} x_3 \mapsto x_3 \end{align*}

Question: How do I find the irreducible subrepresentations of this representation?

What I did so far was to find the conjugacy classes of $(S_2 \times S_2) \rtimes S_2$ and calculate the character of the above representation. But how does this help me?

If we would only look at the direct product rather than the semidirect one, the above action on the basis seems to imply that the representation we're looking at is simply the sum of three times the sign irrep of $S_2$. But I don't understand how to go on in the semidirect case...

Thanks for any help!

$\endgroup$
  • $\begingroup$ The group $S_2\wr S_2$ is isomorphic to the dihedral group $D_4$ of symmetries of a square. That has a two-dimensional irreducible complex representation. It looks like that will be a component of your representation as well. Hmm. Shouldn't be too difficult to calculate the character... $\endgroup$ – Jyrki Lahtonen Dec 9 '16 at 10:56
  • $\begingroup$ @JyrkiLahtonen Thanks for your comment! As I said, I already calculated the character, which is completely straightforward. But how do you know that one of the irreducible subrepresentations is $2$-dimensional (couldn't it be that it consists of three $1$-dimensional irreps)? Forgive my lack of knowledge of representation theory. $\endgroup$ – Tom Bombadil Dec 9 '16 at 13:14
  • $\begingroup$ Because $(S_2\times S_2)$ is abelian, you can also apply the reasoning in the extras added to this answer to get the characters - see the paragraph starting with Edit: $\endgroup$ – Jyrki Lahtonen Dec 9 '16 at 14:03
  • $\begingroup$ Also, I cannot resist remarking that, as we see $S_2\wr S_2$ is a Sylow $2$-subgroup of $S_4$. The construction continues: $(S_2\wr S_2)\wr S_2$ is a Sylow $2$-subgroup of $S_8$ et cetera. Consecutive wreath powers give Sylow $2$-subgroups of $S_{2^k}$. See this thread for other descriptions of those Sylow groups. The wreath product structure is visible there in both answers, but not explicitly! $\endgroup$ – Jyrki Lahtonen Dec 9 '16 at 14:09
1
$\begingroup$

I give an ad hoc argument specific to this case.

By perusing the listed action of the generators we see that the subspaces $$ V_1=\{(x_1,x_2,x_3)\in\Bbb{Q}^3\mid x_3=0\} $$ and $$ V_2=\{(x_1,x_2,x_3)\in\Bbb{Q}^3\mid x_1=x_2=0\} $$ are both stable under the action of $G=S_2\wr S_2$.

Furthermore, if we identify $V_1$ with the $xy$-plane via $(x,y)\leftrightarrow (x,y,0)$, we see that:

  • The element $((1,1),\sigma)$ acts as a reflection w.r.t. the line $x=y$.
  • The element $((\sigma,1),1)$ acts as a reflection w.r.t. the $y$-axis.
  • The element $((1,\sigma),1)$ acts as a reflection w.r.t. the $x$-axis.

Those reflections generate the group $D_4$ of symmetries of a plane square with center at the origin and sides parallel to the coordinate axes. This already implies that the subspace $V_1$ is an irreducible representation of $G$, because this defining representation of $D_4$ is known to be irreducible. Even without this bit of information we can deduce the irreducibility of $V_2$ as follows. A putative non-trivial subrepresentation $W$ will necessarily be $1$-dimensional. But the vectors of a 1-dimensional representation of any group must be eigenvectors for all the elements of the group. Here there are no such vectors, because the eigenvectors of a plane reflection are exactly the line of reflectio (eigenvalue $\lambda=+1$) and the vectors perpendicular to it (eigenvalue $\lambda=-1$). But the line $x=y$ makes a 45 degree angle with the others, so no common eigenvectors exist.

As a 1-dimensional representation $V_2$ is obviously also irreducible.

$\endgroup$
  • $\begingroup$ What a wonderfully clear answer! Thank you! $\endgroup$ – Tom Bombadil Dec 9 '16 at 13:54

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.