How to find an isomorphism between two representations of $S3$ $(\Bbb{R}^2,\rho), (W_1(\subset\Bbb{R}^3),\pi)$? How to find an isomorphism between a two representations $(\Bbb{R}^2,\rho), (W_1(\subset\Bbb{R}^3),\pi)$?
For the symmetry group $S_3$, I'm basically asked to show that $\rho$ representation of rotations and reflections of an equilateral triangle in $\Bbb{R}^2$ centered in the origin is isomorphic to the $\pi$ representation of permutations of 3 elements in a set in $W_1=\{(x_1,x_2,x_3)\in \Bbb{R}^3:x_1+x_2+x_3=0\}$. I found $\rho(g)$ to act as a $2\times 2$ matrix that rotates or reflect vertices with respect to this $g$, but the matrix representing $\pi(\sigma)$ could be for instance $$
    \begin{pmatrix}
    0& 0& 1 \\
    1 & 0& 0 \\
    0& 1& 0 \\
    \end{pmatrix}
$$.
Although $\dim W_1=2$, I still don't see how one can find an isomorphism invertible matrix between a subspace of $\Bbb{R}^3$ to $\Bbb{R}^2$. How does $W_1$ change representation, unless one takes $\{e_1=(1,-1,0),(0,1,-1)\}$ for a basis, two dimensional, and yet is still in $\Bbb{R}^3$.
 A: As you have noted, the dimension of $W_1$ is 2, which means that it is isomorphic to $\mathbb R^2$ as a vector space. It remains to find an isomorphism that commutes with the group actions on these two spaces, which would show that the representations are isomorphic.
The rotations of the triangle intuitively correspond to cyclic permutations of the three axes in $\mathbb R^3$ and the reflections of the triangle intuitively correspond to swapping two axes in $\mathbb R^3$. Note that the projection of $(1, 0, 0)^T$ onto $W_1$ is $(2/3, -1/3, -1/3)^T$ and we have similar expressions for the projections of $(0, 1, 0)^T$ and $(0, 0, 1)^T$. If we can map the corners of the triangle in $\mathbb R^2$ to these three projections, then we should be in good shape.
But that isn't too difficult to do. Let's say that the corners of the triangle are at $(1, 0)^T$, $(-1/2, \sqrt{3}/2)^T$ and $(-1/2, -\sqrt{3}/2)^T$, then the map $v \mapsto Av$ with
$$
A = \frac{1}{3}\begin{bmatrix}
2 & 0 \\
-1 & \sqrt{3} \\
-1 & -\sqrt{3}
\end{bmatrix}
$$
does the trick.
Now all that remains is confirming that $A\rho(\sigma) = \pi(\sigma)A$.
