# action of symmetric group on the second exterior power

Let $$e_i \wedge e_j \ (i < j)$$ be a basis for the $$\mathbb Z$$-module $$\wedge^2 \Gamma$$, where $$\Gamma = \mathbb Z^n$$. Clearly $$S_n$$ acts on the module $$\wedge^2 \Gamma$$ via $$\pi(e_i \wedge e_j) = e_{\pi(i)} \wedge e_{\pi(j)} \ \ \ \forall \pi \in S_n.$$ By restriction this induces an action on the subset $$\bar B = \{ \epsilon e_i \wedge e_j \ (i < j), \ \epsilon \in \{-1, 1\} \}$$.

Which (non-trivial) cyclic subgroups of $$S_n$$ have maximal number of orbits in this action on $$\bar B$$. The answer seems to be the subgroups generated by transpositions $$\pi = (ij)$$. But can there be other permutations $$\pi$$ that are not transpositions but with the same number of orbits?

• Please, note a modification in the question. Apr 30 '20 at 5:51
• Yes, we mean exactly the restriction to $\bar B$ ("latter action" is now made explicit in the question). Apr 30 '20 at 6:16
• The only exception is when $n=4$ in which case a product of two disjoint transposition also works. Apr 30 '20 at 17:22
• Please note: For $n = 4$ there are $7$ orbits for the group $\langle (12) \rangle$, namely, $\{\pm e_1\wedge e_2\}$, $\{e_1 \wedge e_3, e_2 \wedge e_3 \}$, $\{-e_1 \wedge e_3, -e_2 \wedge e_3 \}$, $\{e_1 \wedge e_4, e_2 \wedge e_4 \}$, $\{-e_1 \wedge e_4, -e_2 \wedge e_4 \}$, $\{e_3 \wedge e_4 \}$ and $-\{e_3 \wedge e_4\}$. May 1 '20 at 7:12
• For the action of $\langle (12)(34) \rangle$ there are six orbits only: $\{e_1 \wedge e_2, -e_1 \wedge e_2 \}$, $\{e_3 \wedge e_4, -e_3 \wedge e_4\}$, $\{e_1 \wedge e_3, e_2 \wedge e_4 \}$, $\{-e_1 \wedge e_3, -e_2 \wedge e_4 \}$, $\{e_1 \wedge e_4, e_2 \wedge e_3 \}$ and $\{-e_1 \wedge e_4, -e_2 \wedge e_3 \}$. The reason behind this is that $\langle (12) \rangle$ fixes more points in $\bar B$ (on average). May 1 '20 at 7:28

By the Lemma that is not Burnside's, the number of orbits is the average number of fixed points. An element fixes $$\epsilon e_i \wedge e_j$$ iff it fixes both $$i,j$$ (because if it swaps them it reverses the sign, and otherwise it won't even preserve the span $$\mathbb{Z} e_i \wedge e_j$$. It follows that $$\sigma \in S_n$$ will have $$2\binom{\#\mathrm{Fix}(\sigma)}{2}$$ fixed points on $$\bar{B}$$ because we need to choose pairs $$(i,j)$$ in its fixed point set.
Now a transposition has the largest number of fixed points of any non-identity element. Accordingly let $$G < S_n$$ be any non-trivial subgroup and let $$C be the subgroup generated by a transposition. Then we have
$$\# \bar{B}/G = \frac{1}{\# G} \sum_{\sigma\in G} 2\binom{\#\mathrm{Fix}(\sigma)}{2} \leq \frac{1}{\# G} n(n-1) + \left(1-\frac{1}{\# G}\right)(n-2)(n-3) = (n-2)(n-3) + \frac{1}{\# G} (4n-6) \leq (n-2)(n-3) + \frac{1}{\# C} (4n-6) = \# \bar{B}/C \,.$$
It follows that the number of orbits of $$C$$ is maximal, with equality iff $$G$$ is conjugate to $$C$$.