<s>[INST] 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.
 [/INST] 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.
</s>