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I am trying to determine the standard representation of $S_5$. I understand that it will be a map from group elements to $\mathbb{C}^4$. The character table is as follows.

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I understand that the starting point is the permutation representation i.e. the map $\phi: S_5 \to \mathbb{C}^5$. The bases are $f_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$, $f_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$, $f_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$, $f_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}$, $f_5 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$.

Then the standard representation is the map from the group elements to the complement of the one dimensional subspace in $\mathbb{C}^5$. This vector space $V$ is as follows.

$$ V = \{(x_1, x_2, \dots, x_n) | x_1 + x_2 + \ldots + x_n\} $$

What should be my next step if I want to work out the matrices corresponding to the group elements in the standard representation? I understand that their might be more than one choices of basis vectors. What is the most straightforward choices for basis vectors?

UPDATE:

The easiest choice of bases according to @OrangesKid is:

$e_1 = f_1 - f_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$, $e_2 = f_2 - f_3 = \begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \\ 0 \end{pmatrix}$, $e_3 = f_3 - f_4 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \\ 0 \end{pmatrix}$, and $e_4 = f_4 - f_5 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}$.

Omar Shehab
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2 Answers2

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A simple choice of basis vectors is $e_1$, $\ldots$, $e_4$ where $e_i = f_i - f_{i+1}$ and $f_i$ is the standard basis of $\mathbb{C}^5$. It is not an orthonormal basis, but you can determine readily the matrices corresponding to each $\sigma$ and the character of the representation.

orangeskid
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  • What is the intuition behind this choice? – Omar Shehab Dec 03 '15 at 16:08
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    You want vectors in the subspace $x_1+ \cdots + x_5 = 0$. These $e_i$ do form a basis. It works in general for the standard representation of $S_n$ ( again irreducible). It's just a straightforward choice. – orangeskid Dec 03 '15 at 16:28
  • I have worked out the basis vectors. How can I compute the matrix for a group element, say, $(1, 3)$? – Omar Shehab Dec 03 '15 at 18:03
  • @Omar Shehab: Note that a permutation $\sigma$ acts takes the vector $f_i$ to $f_{\sigma(i)}$. So $(1,3)$ swaps the vectors $f_1$, $f_3$ and does not move the other $f_i$s. So $\sigma( e_1) = \sigma( f_1 - f_2) = f_3 - f_2 = - e_2$, – orangeskid Dec 03 '15 at 18:06
  • the comment is not readable. Could you please reformat it? – Omar Shehab Dec 03 '15 at 18:07
  • @OmarShehab: Perhaps more interesting is $\sigma(e_3) = \sigma(f_3 - f_4) = f_1 - f_4 = e_1 + e_2 + e_3$. – orangeskid Dec 03 '15 at 18:09
  • as per my calculation, $$\sigma (e_1) = -e_2$$, $$\sigma (e_2) = -e_1$$, $$\sigma (e_3) = e_1 + e_2 + e_3$$ and $$\sigma (e_4) = e_4$$. So, $$\sigma = (1, 3) \mapsto \begin{pmatrix} 0 & -1 & 0 & 0 \ -1 & 0 & 0 & 0 \ 1 & 1 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix}$$. Am I doing it right? – Omar Shehab Dec 03 '15 at 18:28
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As far as your question in the comments to orangekid's answer, the symmetric group $S_n$ is the same as the reflection group of type $A_{n-1}$ described in the wikipedia article here. The basis described by orangekid is a choice of simple system for the root system associated to $A_{n-1}$. It has a very beautiful geometry.

David Hill
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