Here is the character table of S3:

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Consider $V=\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$ with basis $e_{ijk} := e_i \otimes e_j \otimes e_k $

Let $\pi$ be the representation of $S_3$ on V which permutes tensor factors of V.

e.g. $\pi$(12)($u \otimes v \otimes w$)$=v \otimes u \otimes w$

$\pi$(123)($u \otimes v \otimes w$)$=w \otimes u \otimes v$

Compute the character of $\pi$ and decompose $\pi$ into irreducible representations

$\chi(g)=tr(g\pi)$ but I do not know what matrix we will take the trace of. I believe it will be $3 \times 3$ with entries reflecting how $\pi$ permutes elements, but am not sure of its exact construction

Also how can we use the information from a character to decompose $\pi$ into irreps?

Many thanks


As you've already written $(e_{i,j,k})_{(i,j,k)\in \{0, 1\}^3}$ is a basis of $V$. We are interested in the matrices representing the linear maps

$$\pi(g): V \rightarrow V, \ \pi(g) (e_{a_1,a_2,a_3})=(e_{a_{g(1)}, a_{g(2)}, a_{g(3)}}),$$

where $g\in S_3$. Note that $V$ is 8-dimensional and therefore the matrix representing $\pi(g)$ will be a $8\times 8$ (permutation) matrix. The trace of such a matrix is just the number of basis elements, which remain fixed under the permutation, i.e.

$$\Gamma_4(g) = tr(\pi(g)) = \vert \{ (a_1, a_2, a_3) \in \{0,1 \}^3 : a_{g(1)}=a_1, a_{g(2)}=a_2, a_{g(3)}=a_3 \} \vert.$$

Therefore, $\Gamma_4(e)=8$, $\Gamma_4((12))=4$, $\Gamma_4((123))=2$.

For the decomposition of the representation you can use the decomposition of the character. The row orthogonality is thelling you that you can write

$$ \Gamma_4 = b_1 \Gamma_1 + b_2 \Gamma_2 + b_3 \Gamma_3 $$


$$ b_j = \frac{1}{\vert S_3 \vert} \sum_{g\in S_3} \Gamma_4(g) \overline{\Gamma_j(g)}.$$

Let $V_1, V_2, V_3$ representatives of the isomorphism class of irreducible representations of $S_3$ then we have the decomposition

$$ V\simeq V_1^{\bigoplus b_1} \oplus V_2^{\bigoplus b_2} \oplus V_3^{\bigoplus b_3}$$

where we mean isomorphic as representations.


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