# 6-Dimensional Irrep of $S_5$

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?

• It's well-worth learning about exterior powers. Also, this rep is the Specht module associated to the partition $3\,1^2$, but I'll bet you haven't seen this in class either. Anyway, I'd recommend the book by Fulton and Harris if you really want to learn about these. – Lord Shark the Unknown Feb 28 at 7:12
• It might be easier to treat the (two) $5$d representations first and then use the fact that the every irrep $\phi$ occurs $\dim \phi$ times in the regular representation together with what you know know about the character of the regular representation. – Travis Feb 28 at 7:21

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 $$e_{i,j}:=x_i\wedge x_j, 1\le i. We follow the usual rules: $$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 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 $$\sigma(\{,j\})=\{i,j\}$$, 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.

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 $$S(x\otimes y)=y\otimes x$$.

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 $$\matrix{ \hbox{type} & (1,1,1,1,1) & (2,1,1,1) & (2,2,1) & (3,1,1) & (3,2) & (4,1) & (5) \cr \hbox{perm} & e & s_1 & s_1s_3 & s_1s_2 & s_1s_2s_4 & s_1s_2s_3 & s_1s_2s_3s_4\cr \hbox{trace}& 6&0&-2&0&0&0&1}$$