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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 $e_4 = f_4 - f_5 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}$.

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  • $\begingroup$ I have added the standard bases. $\endgroup$ – Omar Shehab Dec 3 '15 at 17:54
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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.

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  • $\begingroup$ What is the intuition behind this choice? $\endgroup$ – Omar Shehab Dec 3 '15 at 16:08
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    $\begingroup$ 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. $\endgroup$ – orangeskid Dec 3 '15 at 16:28
  • $\begingroup$ I have worked out the basis vectors. How can I compute the matrix for a group element, say, $(1, 3)$? $\endgroup$ – Omar Shehab Dec 3 '15 at 18:03
  • $\begingroup$ @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 $\sigma( e_1) = \sigma( f_1 - f_2) = f_3 - f_2 = - e_2$, $\endgroup$ – orangeskid Dec 3 '15 at 18:06
  • $\begingroup$ the comment is not readable. Could you please reformat it? $\endgroup$ – Omar Shehab Dec 3 '15 at 18:07
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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.

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