6-Dimensional Irrep of $S_5$

Question

So I am computing the character table for $S_5$, and the only thing I have yet to understand is how we know the character values for the row relating to the 6-dimensional irrep. The irreps I have are: trivial, sgn, standard, and the product of the standard and sgn. I then computed the dimensions of the remaining irreps using elementary number theory. If I can compute the row for the 6-dimensional irrep, then the rest is not hard by column orthogonality.

The sources I found just sort of state the values or say it is the "exterior square of the standard representation," which we have not discussed in class, so I am not sure if I can use that (not to mention I do not understand it). Is there another way to view this irrep?

Answer 1

It may be easier to view the 6-d rep as a component of the exterior square of the natural (non-irreducible) 5-dimensional representation $V$. At least I think it is easier to figure out the character of $\wedge^2V$. Bear with me for a moment.

Let $x_1,x_2,x_3,x_4,x_5$ be a basis of $V$ with $S_5$ acting by permuting the indices, $\sigma(x_i)=x_{\sigma(i)}$. A basis of $\wedge^2V$ then consists of the wedge products $e_{i,j}:=x_i\wedge x_j, 1\le i<j\le 5$. We follow the usual rules: $x_i\wedge x_j=-x_j\wedge x_i$ and $x_i\wedge x_i=0$, so the ten vectors $e_{ij},1\le i<j\le5$ form a basis of $\wedge^2V$.

The group $S_5$ still acts by permuting the indices, $\sigma(e_{i,j})=e_{\sigma(i),\sigma(j)}$. Let's denote the character of $\wedge^2V$ by $\psi$. I will round up the diagonal entries of the matrices of permutations from all the conjugacy classes of $S_5$. All w.r.t. the above basis. Basically we need to keep an eye on pairs of indices such that $\sigma(\{i,j\})=\{i,j\}$, in some order:

• Obviously $1\cdot e_{i,j}$ for all pairs $(i,j)$, so $\psi(1_{S_5})=10$.
• The 2-cycle $(12)$ fixes $e_{3,4}$, $e_{3,5}$ and $e_{4,5}$, maps $e_{1,2}$ to its negative, and shuffles the rest of them around. So its trace is $\psi((12))=3-1=2.$
• The 3-cycle $(123)$ fixed $e_{4,5}$, but that is the only non-zero diagonal entry in its matrix, so $\psi((123))=1$.
• The product of two disjoint 2-cycles $(12)(34)$ negates both $e_{1,2}$ and $e_{3,4}$ but the rest of diagonal entries are all zero, and $\psi((12)(34))=-2$.
• All the diagonal entries of the matrix representing the 4-cycle $(1234)$ are zero, so $\psi((1234))=0$. No pair of indices is stabilized as a set.
• The permutation $(12)(345)$ negates $e_{1,2}$ but the rest of the diagonal entries are zero, so $\psi((12)(345))=-1$.
• All the diagonal entries of the 5-cycle $(12345)$ are zero, and $\psi((12345))=0$.

Let $\chi_1$ be the 4-dimensional irreducible component of $V$. Given all this (and the census on the sizes of conjugacy classes) it is easy to calculate that $$ \langle\psi,\chi_1\rangle=1\qquad\text{and}\qquad\langle\psi,\psi\rangle=2. $$ This implies that $\psi=\chi_1+\chi_2$ for some previously unknown irreducible character $\chi_2$. The values of $\chi_2$ are easily calculated from the known values of $\psi$ and $\chi_1$.

Calculating the character values of $\wedge^2V$ was elementary combinatorics. The rest followed from basic representation theory.

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You can also view $\wedge^2V$ as the antisymmetric part of the tensor product $V\otimes V$. That is the eigenspace corresponding to the eigenvalue $-1$ of the involution $S:V\otimes V\to V\otimes V$ defined on elementary tensors as $S(x\otimes y)=y\otimes x$.

Answer 2

If you like a concrete presentation, the exterior power of the standard representation is not hard to construct directly. The standard representation is the restriction of the standard $5$-dimensional permutation representation to the subspace with sum of the coordinates zero, or easier the quotient by the $1$-dimensional invariant subspace generated by $(1,1,1,1,1)$. The image of the standard basis in that quotient consist of five vectors $v_1,\ldots,v_5$ in the $4$-dimensional quotient, of which any $4$ form a basis and whose sum is $0$. So choosing $v_1,\ldots,v_4$ as basis, $S_5$ acts by permuting indices, where one rewrites $v_5=-v_1-v_2-v_3-v_4$ whenever it arises.

The elements of second exterior power basically represent the $2$-dimensional subspaces of this $4$-dimensional space, equipped with an oriented area (so that scalar multiplication keeps the subspace, but multiplies the oriented area on it). Concretely any basis of a two-dimensional subspace represents a vector in the exterior square, with two bases of the same subspace representing the same vector whenever the ($2$-dimensional) change of basis has determinant$~1$ (i.e., is oriented area preserving). If $v,w$ are two vectors of the $4$-dimensional (quotient) space, the corresponding element of the exterior square is denoted by $v\wedge w$; one has $w\wedge v=-v\wedge w$, and $v\wedge v=0$. A basis of the $6$-dimensional exterior square of the $4$ dimensional quotient is $\def\B{\mathcal B}\B=[v_1\wedge v_2,~v_1\wedge v_3,~v_1\wedge v_4,~v_2\wedge v_3, ~v_2\wedge v_4,~v_3\wedge v_4]$.

It remains to describe how $S_5$ acts on the exterior square. It acts on each $v\wedge w$ by acting on $v$ and $w$ separately and forming the $\wedge$ of the results; the operation is bilinear in its two arguments, which permits expressing the result in the basis$~\B$. I will give the action of the four adjacent transpositions that generate $S_5$ as matrices on the basis$~\B$; for other permutations it suffices to multiply a corresponding sequence of these matrices. The first three adjacent transposition just act on the indices $i,j$ in $v_i\wedge v_j$; in the case where the two indices get interchanged the result is minus the staring basis vector. For the final adjacent transposition of $4$ and $5$, rewriting $v_5$ is needed, followed by expansion by linearity, and some simplification of resulting wedges (any term $v_i\wedge v_i$ is simply dropped). The matrices are $$ \pmatrix{-1&0&0&0&0&0\cr0&0&0&1&0&0\cr0&0&0&0&1&0\cr0&1&0&0&0&0\cr0&0&1&0&0&0\cr0&0&0&0&0&1} ,\quad \pmatrix{0&1&0&0&0&0\cr1&0&0&0&0&0\cr0&0&1&0&0&0\cr0&0&0&-1&0&0\cr0&0&0&0&0&1\cr0&0&0&0&1&0} ,\quad \pmatrix{1&0&0&0&0&0\cr0&0&1&0&0&0\cr0&1&0&0&0&0\cr0&0&0&0&1&0\cr0&0&0&1&0&0\cr0&0&0&0&0&-1} ,\\ \pmatrix{1&0&-1&0&1&0\cr0&1&-1&0&0&1\cr0&0&-1&0&0&0\cr0&0&0&1&-1&1\cr0&0&0&0&-1&0\cr0&0&0&0&0&-1} $$

As for the character of this representation, just compute the trace of a representative of each conjugacy class. One gets $$ \matrix{ \hbox{type} & (1,1,1,1,1) & (2,1,1,1) & (2,2,1) & (3,1,1) & (3,2) & (4,1) & (5) \cr \hbox{perm} & e & s_1 & s_1s_3 & s_1s_2 & s_1s_2s_4 & s_1s_2s_3 & s_1s_2s_3s_4\cr \hbox{trace}& 6&0&-2&0&0&0&1} $$