# Looking for various proofs there is no faithful representation $\rho:S_4\rightarrow GL(\mathbb{R}^2)$

I stumbled on the question when thinking about a representation $\rho:D_4\rightarrow GL(\mathbb{R}^2)$ as symmetries of the four points $\{(\pm 1, \pm 1)\}$. There didn't seem to be a good geometric picture for how to swap two adjacent vertices without moving the others via linear mapping.

To show $\rho:S_4\rightarrow GL(\mathbb{R}^2)$ is never faithful it is enough to show that the image of the transpositions can not be distinct. It's possible to work this out by calculating all $2\times 2$ matrices of order $2$ and showing there simply aren't enough valid distinct choices for the images.

However, the problem I had in mind came from geometry so I was wondering about other proofs for this problem. Namely:

What are other proofs there does not exist a faithful representation $\rho:S_4\rightarrow GL(\mathbb{R}^2)$?

Here's a proof I like:

Assume (for contradiction) that $$\rho$$ is a faithful representation. We note that $$S_4$$ contains the commuting transpositions $$\tau_1 = (1\;2)$$ and $$\tau_2 = (3\;4)$$. Since neither transposition is in the center of $$S_4$$, neither transposition can be mapped to a multiple of the identity. Because $$\rho(\tau_i)^2 = I$$, we may deduce that each $$\rho(\tau_i)$$ is diagonalizable with eigenvalues $$\pm 1$$. Because the $$\rho(\tau_i)$$ commute, they are simultaneously diagonalizable. Because they are distinct and $$\rho$$ is faithful, they are not mapped to the same matrix.

So, up to an isomorphism of representations (i.e. a change of basis), we have $$\rho(\tau_1) = \pmatrix{1\\&-1} \quad \rho(\tau_2) = \pmatrix{-1\\&1}$$ This means that $$\rho(\tau_1 \tau_2) = \rho(\tau_1) \rho(\tau_2) = -I$$ So, we have $$\rho(\tau_1 \tau_2) \in Z(GL_2)$$. However, $$\tau_1\tau_2 \notin Z(S_4)$$, which since $$\rho$$ is faithful would necessarily imply that $$\rho(\tau_1 \tau_2) \notin Z(GL_2)$$.

So, no faithful representation exists.

Note: It may seem like overkill in this context, but the same proof can be extended to show that there is no faithful representation $$\rho: S_n \to GL_{n/2}$$ for any even $$n \geq 4$$.

A more typical approach is as follows: the table of irreducible characters on $$S_4$$ is given by $$\begin{array}{c|rrrrr} \text{classes:} & 1 & (1\,2) & (1\,2\,3) & (1\,2\,3\,4) & (1\,2)(3\,4)\\ \text{sizes:} & 1&6&8&6&3\\ \hline \chi_1 & 1&1&1&1&1\\ \chi_2 & 1&-1&1&-1&1\\ \chi_3 & 2&0&-1&0&2\\ \chi_4 & 3&1&0&-1&-1\\ \chi_5 & 3&-1&0&1&1 \end{array}$$ (copied from Dummit and Foote). Thus, the character induced by any representation $$\rho:S_4 \to GL(\Bbb R^2)$$ must either have the form $$\chi_i + \chi_j$$ with $$i,j \in \{1,2\}$$ or $$\chi_3$$. Correspondingly, $$\rho$$ must be isomorphic either to the direct sum $$\rho_i \oplus \rho_j$$ with $$i,j \in \{1,2\}$$ or to $$\rho_3$$ (where $$\rho_j$$ is a representation inducing character $$\chi_j$$).

$$\rho_1$$ is the trivial representation and $$\rho_2$$ is the sign representation, so $$\rho_i \oplus \rho_j$$ fails to be faithful for $$i,j \in \{1,2\}$$ (since $$\rho_2$$ is not faithful). Alternatively, since $$\chi_1((1\,2\,3)) = \chi_2((1\,2\,3)) = 1$$ and the degrees of $$\chi_1,\chi_2$$ are $$1$$, it must be that $$\rho_1((1\,2\,3)) = \rho_2((1\,2\,3)) = 1$$. Thus, $$(1,2,3) \in \ker (\rho_1 \oplus \rho_2) = \ker(\rho_1) \cap \ker(\rho_2) \subset \ker(\rho_i) \cap \ker(\rho_j)$$ for $$i,j \in \{1,2\}$$.

It now suffices to note that $$\rho_3$$ is also not faithful. To that end, it suffices to follow our second approach above: note that $$\chi_3((1\,2)(3\,4)) = 2$$ implies that $$\rho_3((1\,2)(3\,4)) = I$$, which is to say that $$(1\,2)(3\,4) \in \ker(\rho_3)$$.

Aside: As this answer notes, the representation $$\rho_3$$ corresponding to $$\chi_3$$ is the composition of the canonical surjection $$S_4 \to S_4/V \cong S_3$$, where $$V$$ is the subgroup generated by the double transpositions, and the standard 2-dimensional representation $$\rho_\mathrm{standard} : S_3 \to \mathrm{GL}(2,\mathbb{C})$$.

• Definitely not overkill; pretty cool actually!
– Eoin
Jul 2, 2015 at 23:52