Let the symmetric group $S_5$ act by permutation on the set $X:=\{S \subset \{1,2,3,4,5\}:|S|=2\}$ and denote by $V$ the associated complex $S_5$-representation, with $\chi:S_5 \rightarrow \mathbb{C}$ its character. I'm supposed to compute $\chi(\sigma)$ for $e, (2,3), (1,2,3), (1,2,3,4), (1,2,3,4,5), (1,2)(4,5), (1,5)(2,3,4)$. I know the character is the trace of the matrix that each element of $S_5$ is mapped onto, and the standard presentation of $S_5$ in terms of transpositions, however I'm young in this field, so I can't really see what the are actual matrices that the transpositions $(1,2), (2,3), (3,4), (4,5)$ are mapped to are in this case. Another difficulty is that I can't quite see how to represent the elements of $X$ as elements of a vector space. Could it be the $5$-dimensional complex vectors with only $2$ entries equal to $1$ and all other entries vanishing? Thanks in advance!
Characters of complex $S_5$-representation
charactersrepresentation-theorysymmetric-groups
Related Solutions
In general if the symmetric group $S_n$ is acting on $\mathcal P (X),$ where $X = \{1,2, \cdots ,n\},$ then the orbits under this action are precisely of the form $\mathcal P_{d} (X),$ for $0 \leq d \leq n,$ where $$\mathcal P_{d} (X) : = \left \{A \subseteq X\ |\ \text {Card}\ (A) = d \right \}.$$
I think I have it now:
Firstly, let's take an element $(a, b) = (\{a_1, a_2, .., a_l \}, \{b_1, b_2, .., b_k \}) \in x_l \times x_k$. A permutation $\sigma$ acts on this as $\sigma (a, b) = (\{\sigma (a_1), .., \sigma (a_l) \}, \{\sigma (b_1), .., \sigma (b_k) \})$.
$a_i$ and $b_i$ take values out of $\{ 1, 2, .., n \}$. For $i \neq j$, $a_i \neq a_j$ and $b_i \neq b_j$. However, $a_i = b_j$ is allowed. With this in mind, define the 'overlap' $r = | a \cap b |$, and notice that $0 \leq r \leq l$. Therefore, $r$ takes on $l+1$ values.
It turns out that $r$ indexes each orbit. To see this, note that each $(a, b)$ has $l + k - r$ unique numbers. We can find a permutation taking us from $(a, b)$ to $(a', b')$ with $r = r'$ by specifying $l + k - r$ transpositions.
If $r \neq r'$, then (without loss of generality) take $r' > r$. A map taking us from $(a, b)$ to $(a', b')$ now has a domain of cardinality $l + k - r$ and co-domain of cardinality $l + k - r'$. The domain is larger than the co-domain, so the map is non-invertible and thus cannot be a group action.
As a concrete example, we can take $n=6, l=2, k=3$:
- $(a,b) = (\{1,2\}, \{1,4,3\})$ with $r=1$
- $(a’,b’) = (\{1,3\}, \{1,2,3\})$ with $r’=2$
There's four possible maps (all non-invertible), two of which are:
- $1 \rightarrow 1, 2 \rightarrow 3, 3 \rightarrow 3, 4 \rightarrow 2$
- $1 \rightarrow 3, 2 \rightarrow 1, 3 \rightarrow 1, 4 \rightarrow 2$
Finally, the inner product is the number of orbits, which is the number of values r takes, which is $l + 1$.
Best Answer
The set $X$ has ten elements, and $S_5$ is acting by permutation on those elements. So think of a ten-dimensional vector space, with axes labeled by the elements of $X$. Every element of $S_5$ then turns into a 10-by-10 permutation matrix. The character of an element is going to be the trace of the matrix. As it turns out, this only depends on the conjugacy class of the element (since $g' = h^{-1}gh$ implies $\phi(g') = \phi(h)^{-1} \phi(g) \phi(h)$ implies $Tr(g') = Tr(g)$, where $\phi$ is the matrix representation).
If $g=e$, what is the corresponding 10-by-10 matrix $\phi(g)$? If $g = (2,3)$, what is the corresponding matrix? Try labelling rows and columns by elements of $X$ and see what happens.
Hint: Under what circumstances does the matrix have a 1 on the diagonal? Under what circumstances does it have a 0?