8
$\begingroup$

I've gotten stuck on a slightly random combinatorial question, and I'm doing a bit of a shot in the dark here to see if someone else has thoughts about it. I'm interested in studying a permutation of an odd number of elements $\{1,\dots, 2n+1\}$ which preserves parity, i.e. it sends odd numbers to odd numbers and even numbers to even numbers. What I want to show is that such a permutation has at least as many inversions of pairs with opposite parity as it does of pairs with the same parity.

"Please solve this conjecture" isn't a good MathOverflow question, so let me try to summon a better one:

Does anyone know of any tools for studying the structure of inversion sets of permutations that would be useful for keeping track of parity like this?

I feel like there's some natural approach to this I'm missing. One obvious possibility would be if there was some natural factorization compatible with the length function i.e. a factorization $\sigma=\sigma'\sigma''$ where $\ell(\sigma)=\ell(\sigma')+\ell(\sigma'')$. The problem is that it’s hard to do this while keeping $\sigma'$ and $\sigma''$ parity preserving. For example $(13)(24)$ has no factorization like this.

$\endgroup$
2
  • 5
    $\begingroup$ What about $(0,1,2,3,4,5,6)\mapsto (2,5,0,3,6,1,4)$? It has 9 inversions, 5 with equal parity, 4 with opposite parity. (I understand "inversion" as $(i,j)$ such that $i<j$ and $f(i)>f(j)$ and equal/opposite parity according to whether $j-i$ is even or odd.) $\endgroup$
    – YCor
    Commented Feb 13 at 8:39
  • $\begingroup$ @YCor Shoot, you're right. Well, there goes that proof method (I had hoped to use this as an intermediate step to a result which is probably still true). $\endgroup$
    – Ben Webster
    Commented Feb 14 at 3:11

0

You must log in to answer this question.

Browse other questions tagged .