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I'm trying to find all non-isomorphic 2 dimensional representations of the symmetric group, $S_3$. This is all my own work, so I shan't be surprised if it's horrendously wrong.

I understand that two representations cannot be isomorphic if they have different kernels and images, so I'm thinking that one "good" way to find a set of potential representations would be to look at representations of various "faithfulnesses".

Since $\ker(\rho) = \{1\} \iff \rho \text{ is injective}$, I'll look at $\rho: S_3 \rightarrow \mathbb{R^2}$ with varying kernels.

  1. Firstly, a 2D representation, $\rho_1$ is simply $S_3$ acting as $D_3$ (the dihedral group with 6 elements) on $\mathbb{R^2}$, this has $\ker(\rho_1) = \{1\}$.
  2. Look at $\rho_2$, which is just a representation with a non trivial kernel. E.g. $\rho_2((12)) = Id$, this would mean $\rho_2((123)) = \rho_2((13)) \rho_2((12)) = \rho_2((13)) $, but these are cycles of differing order, so the only way to preserve the group structure is if $\rho_2((123)) = \rho_2((13)) = Id$. Of course this then means that $\rho_2$ is the trivial representation
  3. If $\rho_3$ has a non-trivial kernel but, say, $(123) \in \ker(\rho_3)$. This then implies $(132) \in \ker(\rho_3)$, we're left to define $\rho_3$ on the two cycles s.t. they still have order 2 and multiply to give the identity. I think this means they all must be equal and one of three matrices:\begin{pmatrix}-1&0\\0 & 1\end{pmatrix} Or, \begin{pmatrix}1&0\\0 & -1\end{pmatrix} Or, \begin{pmatrix}-1&0\\0 & -1\end{pmatrix}
  4. Should the above be true, then it is clear that there is no longer enough room for a still "smaller" representation, therefore there are 4 different representations total.

Questions

  • Is the above working correct?

  • I'm unsure as to whether I can claim that: Any two faithful representations of the same finite group, $G$, on the same vector space, $V$, are $G$-isomorphic?

  • Can I generate even more representations by considering a different 2 dimensional vector space, which is not $\mathbb{R^2}$? If not, how can this be proven?


Thoughts

  1. My thinking is that "Any two faithful representations of the same finite group, $G$, on the same vector space, $V$, are $G$-isomorphic" is true:

    We'd just need that $T: (\rho_a, V) \rightarrow (\rho_b, V)$ has $$T(\rho_a(g)v) = \rho_b(g)T(v)$$ But there is a bijection between the set of $\{\rho_a(g) : g \in G \}$ and $\{\rho_b(g) : g \in G \}$ and $G$ being finite means that after relabelling they should be exactly the same representation, with $T$, a trivial intertwiner.

  2. I'm also not sure if there is anything similar if the representations aren't faithful , I have tacitly assumed a similar result for 2. and 3. to guarantee that these are all the representations

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  • $\begingroup$ The notation $\rho: S_3\to\Bbb R^2$ is wrong. The representation is a homomorphism $\rho:S_3\to GL_2(\Bbb R)$. $\endgroup$
    – anon
    Oct 31, 2019 at 21:06

1 Answer 1

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Some comments:

(1) If two representations of a group on a vector space are isomorphic, then their images are conjugate subgroups of the general linear group, but can be distinct subgroups.

(2) Kernels are normal subgroups. So if a group homomorphism $\rho: S_3\to\mathrm{GL}_2(\mathbb{R})$ has a kernel containing an element like $(12)$, then it contains all of its conjugates, which are also transpositions, and transpositions generate all of $S_3$ so the kernel is all of $S_3$ and the representation is trivial.

(This generalizes your argument.)

(3) The only (proper nontrivial) normal subgroup of $S_3$ is the alternating subgroup $A_3$, which happens to be the cyclic group $C_3$ generated by the $3$-cycle $(123)$. Then $A=\rho((12))$ must have order $2$, as you note. However, there are infinitely many matrices of order $2$. For instance, reflections across a line are order $2$, and there are infinitely many lines. In fact, every such $A$ acts like a reflection across a line in some coordinates.

Factor $A^2v=v$ as $(A+I)(A-I)v=0$ or $(A-I)(A+I)v=0$ for all $v$, hence there exist nonzero eigenvectors $x$ and $y$ (in the range of $A+I$ and $A-I$) such that $Ax=x$ and $Ay=-y$. Thus, through a change of coordinates, $A$ is conjugate to the diagonal matrix $\mathrm{diag}(1,-1)$. So all the matrices $A$ of order $2$ are conjugate, and the corresponding representations are all equivalent.

This representation is a direct sum of the trivial representation and the sign representation.

(4) A more advanced fact in representation theory states the sum of the (complex) irreps' squared dimensions is the size of the group squared. Here, we have two $1$D reps (trivial and sign) and the $2$D rep (dihedral), and since $1^2+1^2+2^2=3!^2$ (where $|S_3|=3!$), there can't be any more reps. (We're talking about real irreps, but in this case the sum of the squared dimensions is $\le |G|^2$, so the logic still works.)

(5) Two faithful representations of a finite group on a vector space are not necessarily isomorphic. Even if their images are the same subgroup (not just conjugate), they aren't necessarily isomorphic. A way to produce counterexamples is using outer automorphisms. If $G$ is a finite group, $\rho:G\to GL(V)$ a representation, and there is an outer automorphism $\alpha:G\to G$, then $\rho\circ\alpha$ is inequivalent to $\rho$, but has the same image.

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