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?
 A: 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.


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