I'm currently working with the standard representation of the symmetric group $S_n$. Recall that the standard representation of $S_n$ permutes the basis elements of the $n \times n$ identity matrix and applies it to an element $v \in \mathbb{C}^n$.
I have noted that in $\mathbb{C}^4$, the subspace defined by $K=$ Span$\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$ is $S_4$-invariant. We note that
$$K^\bot = Span \begin{Bmatrix} \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \end{Bmatrix}$$
Now, normally we would write $\pi(12)$ and $\pi(1234)$ as permutations of $4 \times 4$ matrices, but when we restrict $\pi$ to $K^\bot$, I want to treat $K^\bot$ as a subspace of $\mathbb{C}^3$ instead of $\mathbb{C}^4$ and therefore I will need new mappings to generate all of the mappings of $\pi$. How can I do this?
NOTE: The entire reason that I need to do this is because I am seeking to prove that $\pi \big|_{K^\bot}$ is irreducible. Unfortunately, with $K^\bot$ as a subspace of $\mathbb{C}^4$, I obtain $\langle \chi_\pi, \chi_\pi \rangle > 1$, which is clearly an issue. But I'm thinking that if I can view this instead in $3$-space, I'll resolve my issue. I just don't know how to do so.
Linear algebra teaches us to write linear operators with respect to bases as matrices.
Your basis for $K^\perp$ is
$$ a=\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0\end{bmatrix}, \quad b = \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \quad c = \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}. $$
Check how $(12)$ transforms these:
$$ (12)a=-a $$ $$ (12)b = b-a $$ $$ (12) c=c-a $$
which tells us $(12)=\begin{bmatrix} -1 & -1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$