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i am searching for a concrete as possible description of the (there are two but the are obtained from each other by tensoring with the signature representation) irreducible representation of dimension 5 of the symmetric group $S_5$ on 5 elements.

[Fulton Harris] to my knowledge only computes the character.

Wikipedia says it has something to do with an exceptional transitive embedding of $S_5$ into $S_6$. Would love if someone could tell me a bit more about that.

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The embedding of $S_5$into $S_6$ allows to construct $5$-dimensional representations of $S_6$, not $S_5$. – Andrea Mori Jul 12 '13 at 10:29
What field are you working over? If you work over $\mathbb{C}$ then the irreducible representations of the symmetric group are the Specht modules. – Wintermute Jul 12 '13 at 11:54
It seems to me wikipedia suggests that there is a 5-dim irreducible representation of $S_6$ and that the restricted representation which is a representation of $S_5$ is also irreducible. – mna Jul 12 '13 at 17:00
up vote 4 down vote accepted

You can construct one of these as follows (the other comes, as you observed, by tensoring it with the sign character). Undoubtedly you know that $S_5$ has six Sylow 5-subgroups, each with a normalizer $N$ of order 20. It is easy to convince yourself of the fact that the conjugation action on this set of six elements is doubly transitive. Therefore, by the usual result, this 6-dimensional representation splits into a direct sum of a 1-dimensional trivial representation and a 5-dimensional irreducible one.

This gives, indeed, the exceptional transitive embedding $f$ of $S_5$ in $S_6$. Even more so: by studying the type of elements present in $N$ (and its conjugates), you can deduce a number of things about the cycle structure of elements of $f(S_5)$. Furthermore, $f(S_5)$ obviously has six conjugates in $S_6$. This conjugation action gives rise to the famous non-inner automorphism of $S_6$. Using the bits that you get from the cycle structure of elements in $f(S_5)$ you can deduce that this outer automorphism interchanges the conjugacy classes of $(12)$ and $(12)(34)(56)$ (among other things).

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The symmetry group of the dodecahedron (including reflections) is $S_5$. If you draw the six lines from the center of a face to the opposite face, the group also operates on this $6$-element set, which gives you $S_5\to S_6$. Then consider the operation of $S_6$ on $\mathbb R^6$ by permuting the basis vectors and the invariant space orthogonal to $(1,1,\ldots,1)$

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1… thinks that the full icosahedral symmetry group is not $S_5$. – mna Jul 12 '13 at 17:58

The irreducible $\mathbb{C}$-representations of $S_n$ are very interesting.

The family $\{ S_n\}_{n\geq 1}$ of Symmetric groups is an infinite family of groups whose all irreducible representations over $\mathbb{C}$ are actually matrix representations with "integer entries".

With this remark, we actually find some representations of $S_n$ over $\mathbb{Z}$.

In some answers, it is shown that $S_5$ sits in $S_6$ in an interesting way: is acts on six letters $\{1,2,\cdots,6\}$ $2$-transitively. Therefore, consider the vectors $\{v_1,v_2,\cdots,v_6\}$ which are independent over $\mathbb{C}$. The action of $S_5$ on $6$ letters can be passed to an action of $S_5$ on the vector space of dimension $6$ over $\mathbb{C}$: if $\sigma\in S_5$ takes $i$ to $j$, then the action on vector space is given by $\sigma(v_i)=v_j$ ($1\leq i,j\leq 6$). Consider the subspace spanned by $\{ v_1-v_2, v_2-v_3, \cdots, v_5-v_6\}$. It has dimension $5$. The $2$-transitive action will imply that this subspace is invariant under $S_5$. Now, consider an element $(1\,2)(3\,4)(5\,6)$ of $S_5$ (note: we are considering $S_5$ acting on $6$ symbols explained in other answers.) The action of this element on the subspace is given by

$v_1-v_2\mapsto v_2-v_1=-(v_1-v_2)$

$v_2-v_3\mapsto v_1-v_4=(v_1-v_2)+(v_2-v_3)+(v_3-v_4)$

$v_3-v_4\mapsto v_4-v_3=-(v_3-v_4)$

$v_4-v_5\mapsto v_3-v_6=(v_3-v_4)+(v_4-v_5)+(v_5-v_6)$

$v_5-v_6\mapsto v_6-v_5=-(v_5-v_6)$.

Thus, with respect to the basis $\{ v_1-v_2, v_2-v_3, \cdots, v_5-v_6\}$, the matrix of $(1\,2)(3\,4)(5\,6)$ is given by

$\begin{pmatrix} -1& 1 & & & \\ & 1 & & & \\ & 1 & -1 & 1 & \\ & & & 1 & \\ & & & 1& -1 \\ \end{pmatrix}$.

The unfilled entries are $0$. In this way, we cover the (an) irreducible representation of $S_5$ of dimension $5$, which has all integer entries.

Interesting Remark: As we see that the matrix entries are $0,1$ or $-1$. Therefore, if $F$ is a field of characteristic different from $2$ (to consider $1$ and $-1$ distinct), then the above matrix representation acn actually be considered as a matrix representation over the field $F$; it is irreducible if and only if characteristic of $F$ does not divide $|S_5|$.

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I suppose you could get it by making use of the appropriate Young polytabloids. This might be useful.

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