# Number of inversions

• Nov 14th 2010, 12:53 AM
Number of inversions
What happens to the number of inversions in a permutation group if we reflect the matrix across its antidiagonal?

(I suppose that Leibniz formula for determinants should be used or maybe showing by induction?)

• Nov 14th 2010, 07:39 AM
Drexel28
Quote:

What happens to the number of inversions in a permutation group if we reflect the matrix across its antidiagonal?

(I suppose that Leibniz formula for determinants should be used or maybe showing by induction?)

A little more context would be helpful.
• Nov 14th 2010, 08:21 AM
For instance for n=2
$\displaystyle $\left[ {\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} } \right]$$
So the number of inversions in the two groups:
in $\displaystyle a_{11}a_{12}$ it is 0
in $\displaystyle a_{12}a_{21}$ it is 1
(That's why the value of its matrix is $\displaystyle a_{11}a_{12}-a_{12}a_{21}$)
If we reflect it across its antidagonal:
$\displaystyle $\left[ {\begin{array}{cc} a_{22} & a_{12} \\ a_{21} & a_{11} \\ \end{array} } \right]$$
And the number of inversions are the same. For n=3 it is a little more complicated. So I don't really know what happens with the number of inversions if n>2. I would be glad if you could help me.
• Nov 14th 2010, 09:20 AM
Drexel28
Quote:

For instance for n=2
$\displaystyle $\left[ {\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} } \right]$$
So the number of inversions in the two groups:
in $\displaystyle a_{11}a_{12}$ it is 0
in $\displaystyle a_{12}a_{21}$ it is 1
(That's why the value of its matrix is $\displaystyle a_{11}a_{12}-a_{12}a_{21}$)
If we reflect it across its antidagonal:
$\displaystyle $\left[ {\begin{array}{cc} a_{22} & a_{12} \\ a_{21} & a_{11} \\ \end{array} } \right]$$
And the number of inversions are the same. For n=3 it is a little more complicated. So I don't really know what happens with the number of inversions if n>2. I would be glad if you could help me.

I know what you meant, sort of. This is pretty much what I thought. You're taking the idea of an inversion in a permutation $\displaystyle \sigma\in S_n$ to be two values $\displaystyle x,y\in[n]$ such that $\displaystyle x<y$ but $\displaystyle \sigma(x)>\sigma(y)$. But could you explain this a little more? Combinatorially inversions can be seen as connecting the values of your matrix with lines and an inversion is an incidence of intersecting lines.

So, could you give a precise definition and not an example? It's clear if you interpret combinatorially (as above) that the number of inversions is invariant under such a flip.
• Nov 14th 2010, 11:08 AM
• Nov 14th 2010, 11:15 AM
Drexel28
Quote:

I'm sorry, you've explained to me what I knew an inversion was for a permutation. My question was, how exactly do you want to interpret this with a matrix?
• Nov 14th 2010, 12:04 PM