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$V=\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$

$\mathbb{C}$ has standard basis $e_1, e_2$ and V has basis $e_{ijk} := e_i \otimes e_j \otimes e_k $


$\pi$ is a representation of $S_3$ on V which permutes tensor factors of V.

e.g. $\pi$(123)($u \otimes v \otimes w$)$=w \otimes u \otimes v$


I need to compute the character of $\pi$ and decompose $\pi$ irreducible representations.


I know that the character will be $X=Tr$($\pi$(g)) for all $g \in G=S_3$ but how do I compute the matrix that I need to take the trace of? And can I use this character to find the irreducible representations of $\pi$?

Many thanks :)

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  • $\begingroup$ Since you are just permuting the entries, there is another way to think about the character (which is the way combinatorialists think about): it counts the number of fixed points. The way to see this is that if you have a permutation matrix, the only nonzero entries on the diagonal are the fixed points which is what the trace counts. $\endgroup$ – Cameron Williams Mar 25 '16 at 19:14
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Your representation is a case of permutation representation : the elements of $S_3$ permute the elements of your chosen basis.

Precisely, take $X= \{1,2\}^3$, and make $S_3$ act on $X$ in the natural way : $\sigma\cdot (x_1,x_2,x_3) = (x_{\sigma(1)}, x_{\sigma(2)}, x_{\sigma(3)})$. Then your representation is the permutation representation corresponding to this group action.

Now the character of a permutation representation just counts the fixed points of the group action on $X$.

To find the decomposition in irreducible representations, you can just compute the inner product of your character with the (well-known) characters of the irreducible representations of $S_3$.

Note that there are also other informations you can get for free from the fact that it's a permutation representation : for instance, the multiplicity of the trivial representation is the number of orbits.

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  • $\begingroup$ thanks for this. could you sketch out or point me to somewhere I could learn more about going from the fact that the character counts the fixed point of the group action to explicitly calculating it? $\endgroup$ – thinker Apr 4 '16 at 21:00
  • $\begingroup$ If $G$ acts on a set $X$, the the associated representation has a canonical basis $(e_x)_{x\in X}$, and the matrix of $g\in G$ in this basis is a permutation matrix, so all its coefficients are $0$ or $1$. Then the trace of this matrix is the number of $1$ in the diagonal, and there is a $1$ in the diagonal if the corresponding vector is fixed (if you're not convinced, write explicitly the matrices for $S_3$ acting on $\{1,2,3\}$). So the character counts the fixed points. $\endgroup$ – Captain Lama Apr 4 '16 at 21:05

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