Need help understanding matrix representations of the symmetric group $S_3$.

Question

I have the following map for a representations of $S_3$:

$$e \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad (1\; 2) \mapsto \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad (1\; 3)\mapsto \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$$

$$(2\; 3)\mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad (1\; 2\; 3) \mapsto \begin{pmatrix} 0& 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \quad (1\; 3\; 2)\mapsto \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$$

We can check that any $\sigma \in S_3$ and its image under the map represents the same permutation. For example, consider multiplying the matrix associated with $(2\; 3)$ with the column vector $[a\; b\; c]$:

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} a\\c\\b \end{pmatrix}$$

The second and third element in the column vector are interchanged, with the first element remaining fixed. This is the kind of behavior I expected any representation of $S_3$ will be exhibit. However, when I consider the "standard representation" of $S_3$, given as: $$e \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad (1\; 2) \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad (1\; 3)\mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & -1 & 0 \end{pmatrix}$$

$$(2\; 3)\mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & \\ 0 & 1 & -1 \end{pmatrix}, \quad (1\; 2\; 3) \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix}, \quad (1\; 3\; 2)\mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 1 \\ 0& -1 & 0 \end{pmatrix}$$

I don't know how to interpret the result I get from looking at it the way I was looking above, using matrix multiplication. For instance,

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} a\\b\\c\end{pmatrix}= \begin{pmatrix} a\\-b+c\\c \end{pmatrix},$$ from which I have no idea what to infer.

I realize I'm not explaining myself very well, but I hope someone can maybe take a stab at answering my question anyway. Thank you very much for your help.

Answer 1

Your two representations are equivalent, the permutation representation, they just use different bases to represent linear operators as different matrices. The first uses the standard basis.

The permutation representation of $S_3$ is reducible; the span of $(1,1,1)$ is an invariant subspace, as is its complement comprised of $(x,y,z)$ satisfying $x+y+z=0$. Since decomposition into irreducible representations is the primary focus of representation theory, the 2D subspace is what's actually called the "standard representation". In general, the permutation representation of $S_n$ is a direct sum of a 1D trivial subrepresentation and the standard representation of dimension $n-1$, just like for $S_3$.

In the second set of matrices you present, the basis $\{(1,1,1),(1,-1,0),(0,1,-1)\}$ is used instead. To calculate the $2\times2$ part of the matrix for say $(12)$, write out

$$ (12)\left(\color{red}{a}\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\color{blue}{b}\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix}\right)=a\begin{bmatrix}-1\\ \phantom{+}1 \\ \phantom{+}0\end{bmatrix}+b\begin{bmatrix}\phantom{+}1\\ \phantom{+}0\\-1\end{bmatrix} $$

$$ = -a\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+b\left(\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix}\right)=\color{green}{(-a+b)}\begin{bmatrix}\phantom{+}1\\-1\\ \phantom{+}0\end{bmatrix}+\color{purple}{b}\begin{bmatrix}\phantom{+}0 \\ \phantom{+}1\\-1\end{bmatrix} $$

which matches

$$ \begin{bmatrix} -1 & 1 \\ \phantom{+}0 & 1 \end{bmatrix} \begin{bmatrix} \color{red}{a} \\ \color{blue}{b} \end{bmatrix} = \begin{bmatrix} \color{green}{-a+b} \\ \phantom{+}\color{purple}{b} \end{bmatrix}. $$

(Apologies to the color-blind.)

These $2\times2$ matrices also represent the possible permutations of $\{0,1,\infty\}$ in the Riemann sphere using Mobius transformations.

Answer 2

I don't believe that I fully understand your question, so I apologise if this is not a complete answer to your question, but I hope it will serve of some explanation.

The first representation of $S_3$ that you have constructed is an example of what is called a permutation representation. The idea is as follows.

Suppose that $G$ is a group and $S$ is a non-empty set equipped with a (left) $G$-action.

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If you are unfamiliar with group actions, it is a function $a : G \times S \to S$ such that $a(e,s) = s$ for all $s \in S$, and $a(g_1,a(g_2,s)) = a(g_1g_2,s)$ for all $g_1,g_2 \in G$ and $s \in S$. We would typically write $g \cdot s$ as shorthand for $a(g,s)$ if there is no ambiguity.

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Then from this group action we can form a representation for $G$ which is called a permutation representation. Suppose that $\mathsf{K}$ is a field, and for now we shall assume that $S$ is a finite set of size $m > 0$ say. Then let $V_S$ be an $m$-dimensional vector space with a basis of $m$-elements indexed by the elements of $S$.

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If this doesn't make sense to you, another way of thinking about this is as follows. Let $\mathcal{B}$ be any $\mathsf{K}$-basis for $V_S$, and order the elements in some way (doesn't matter which way) so we can write $\mathcal{B} = \left\{ v_i | 1 \leq i \leq m \right\}$. Then order the elements of $S$ in any way (again, doesn't matter), so that we can write $S = \left\{ s_i | 1 \leq i \leq m \right\}$. Then we can index $\mathcal{B}$ by the elements of $S$ by $v_{s_{i}} = v_i$. But since the ordering doesn't matter we can just write $\mathcal{B} = \mathcal{B}_S = \left\{ v_s | s \in S \right\}$ with no worry.

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Then we can define a representation of $G$ on $V_S$ by

$$ \rho_S : G \to \operatorname{Aut}_{\mathsf{K}}(V_S) \ : \ \rho_S(g)(v_s) = v_{g \cdot s}. $$

Now notice that since $g \cdot s \in S$ for all $g \in G$, $s \in S$, $\rho_S(g)$ maps to set $\mathcal{B}_S$ to itself, and so each $\rho_S(g)$ is a well-defined $\mathsf{K}$-linear automorphism of $V_S$, and so $\rho_S$ is a well-defined function. But notice that $\rho_S(e) = \operatorname{Id}_V$, and

$$ (\rho_S(g_1) \circ \rho_S(g_2))(v_s) = \rho_S(g_1)(v_{a(g_2,s)}) = v_{a(g_1,a(g_2,s))} = v_{a(g_1g_2,s)} = \rho_S(g_1g_2)(v_s), $$

and so $\rho_S$ is a well-defined group homomorphism, in other words $\rho_S$ is a representation of $G$. Now the reason that $\rho_S$ is called a permutation representation of $G$ is precisely because it maps a basis, i.e $\mathcal{B}_S$, of $V_S$ to itself. That means that the matrix representatives of each $\rho_S(g)$ with respect to the basis $\mathcal{B}_S$ are permutation matrices, or in other words have precisely $1$ non-zero entry in each row and column, with each non-zero entry being a $1$.

Now notice that the group action is recoverable from the representation since $a(g,s)$ is the index of the image of $v_s$ under $\rho_S(g)$.

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Now why is this relevant? Well thus far we have considered the most general situation of a group acting on a set. If we impose extra conditions on this action, the permutation representation can have different properties. The condition that we are concerned about here is called faithfulness. This means that for all $g_1 \neq g_2 \in G$, there exists $s \in S$ such that $g_1 \cdot s \neq g_2 \cdot s$. Morally, the group action allows $S$ to "separate" the elements of $G$. In this special case, the representation $\rho_S$ becomes an injective group homomorphism, and so you can "see" the group itself via $\rho_S$ since $\rho_S$ is an "embedding" of $G$ in the linear automorphisms of $V_S$.

This is precisely what is going on with your first representation. The set $S$ here is $\left\{1,2,3 \right\}$ and $S_3$ acts on this set precisely how you expect it would. This is a faithful group action, and so our representation is an embedding of $S_3$ in $\operatorname{GL}_3$ as the permutation matrices as you showed.

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However importantly not all representations of groups are permutation representations, and not all representations of a group are faithful. Take the trivial representation $G \to \mathsf{K}^{*}$ such that $g \mapsto 1$ for all $g \in G$. Certainly not faithful unless $G$ is the trivial group.

The take home message is that a faithful group representation fully characterises your group, and that a faithful permutation representation makes this as explicit as possible. But not all representations are obviously permutation representations, or faithful.