3
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ Please, note a modification in the question. $\endgroup$
    – A. Gupta
    Apr 30 '20 at 5:51
  • $\begingroup$ Yes, we mean exactly the restriction to $\bar B$ ("latter action" is now made explicit in the question). $\endgroup$
    – A. Gupta
    Apr 30 '20 at 6:16
  • 1
    $\begingroup$ The only exception is when $n=4$ in which case a product of two disjoint transposition also works. $\endgroup$ Apr 30 '20 at 17:22
  • $\begingroup$ 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 $-\{e_3 \wedge e_4\}$. $\endgroup$
    – A. Gupta
    May 1 '20 at 7:12
  • $\begingroup$ 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). $\endgroup$
    – A. Gupta
    May 1 '20 at 7:28
2
$\begingroup$

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 $C<S_n$ 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 $C$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.