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

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.

• I can understand why that bugs you, unfortunately I don't have an answer. I also don't know by what means your "standard representation" is "standard". In my eyes your former representation is the "standard" one. Just observe, that also the map $\sigma \mapsto \operatorname{Id}$ is a representation; nowhere in the definition of a group representation it says that it has to be "meaningful" or in any way pleasing to the human eye. May 20, 2020 at 7:59

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.

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.

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.

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 $$S$$.

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.

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 $$\rho_S(g)$$.

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.

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.