I need to find a decomposition of $\mathbb{C}S_3$ in the following way:


with $e_i=|G|^{-1}\sum\limits_{g\in G}\chi_i(\text{id})\chi_i(g^{-1})g$ and $\chi_1,\chi_2,\chi_3$ being the irreduceable characters.

$\chi_1 $ being the trivial character, $\chi_2$ being the singum function and $\chi_3: \text{id}\mapsto 2,\, [(12)]\mapsto 0, \, [(123)]\mapsto -1$

I calculated:

$e_1=1/6\,( \,\text{id}+(1 2) +(13)+(23)+(123)+(132)\,)$

$e_2=1/6\,( \,\text{id}-(1 2) -(13)-(23)+(123)+(132)\,)$

$e_3=1/6\,( \,4\text{id}-2(123)-2(132)\,)$

But from here I dont know how to proceed.. could someone explain to me how I can use my $e_i$ to get the decomposition?

thanks :)

  • $\begingroup$ Are you wanting to calculate bases of each direct summand? The key fact is that each $e_i$ is a projection ($e_i^2=e_i$) and they are orthogonal ($e_ie_j=0$ if $i\neq j$). A reasonable way to proceed is to choose a basis for $\mathbb{C}S_3$ and compute the matrix for each $e_i$. Calculating the column space of the matrix for $e_i$ gives coordinates for a basis for $\mathbb{C}S_3e_i$. You should get six basis vectors total in the end, since it's a direct sum decomposition. One for the first, one for the second, and four for the third. $\endgroup$ Jul 26, 2021 at 20:48
  • $\begingroup$ A trick to speed things up is that it's easy to find bases for the first two subspaces, and then you might be able to intuit what the third has to be, since it must be an invariant subspace that doesn't nontrivially intersect the sum of the first two. (Or in other words, if you calculate the invariant inner product, you can take the orthogonal complement of the sum of the first two spaces.) $\endgroup$ Jul 26, 2021 at 20:52
  • $\begingroup$ Yes, exactly i want to calculate a basis for each summand. Thank you, so I might use the Basis $B=(g:g\in S_3)$? And then calculate $ge_1,\,ge_2,\,ge_3$ for all $g \in S_3$? is this correct? $\endgroup$ Jul 26, 2021 at 21:36
  • $\begingroup$ That's a natural basis to use, and that's what you need to calculate for the method. $\endgroup$ Jul 26, 2021 at 22:11
  • $\begingroup$ so $\mathbb{C}S_3e_1=\mathbb{C}e_1$ for example, what does that tell me about my basis vector? $\endgroup$ Jul 27, 2021 at 7:32

1 Answer 1


My solution:

Following the explanation in the comments we get:

Choosing the canonical basis $\mathcal{B}=S_3$

$$\mathbb{C}S_3e_1=\mathbb{C}e_1\Longrightarrow \mathbb{C}(1,1,1,1,1,1)^T$$

$$\mathbb{C}S_3e_2=\mathbb{C}e_2\Longrightarrow \mathbb{C}(1,-1,-1,-1,1,1)^T$$

$\mathbb{C}S_3e_3 $ is the orthogonal complement of the span two other vectors combined.So:

$$\mathbb{C}S_3e_3=\text{Span}((1,0,0,0,0,-1)^T,\,(1,0,0,0,-1,0)^T),\,(0,1,-1,0,0,0)^T, (0,1,0,-1,0,0)^T)$$


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