In my previous post on the major index, I mentioned the statistics $\DeclareMathOperator{\inv}{inv} \DeclareMathOperator{\maj}{maj} \DeclareMathOperator{\code}{code} \inv$ and $\maj$ on permutations, and the fact that they were equidistributed. I have since learned of a more enlightening way of proving this result, due to Carlitz.

Let’s start by recalling the definitions of $\inv$ and $\maj$. An **inversion** of a permutation of the numbers $1,2,\ldots,n$ is a pair of numbers which appear ``out of order’’ in the sequence, that is, with the larger number appearing first. For instance, for $n=4$, the permutation $3,1,4,2$ has three inversions: $(3,1)$, $(3,2)$, and $(4,2)$. We write \[\inv(3142)=3.\]

For the major index, define a **descent** of a permutation to be an index $i$ for which the $i$th number is greater than the $(i+1)$st. The **major index** is defined to be the sum of the descents. For instance, the permutation $3,1,4,2$ has two descents, in positions $1$ and $3$, so the major index is $4$. We write \[\maj(3142)=4.\]

My colleague David Harden recently pointed me to Molien’s theorem, a neat little fact about the invariant polynomials under the action by a finite group. It turns out that this has a nice interpretation in the case of the symmetric group $S_n$ that brings in some nice combinatorial and group-theoretic arguments.

The general version of Molien’s theorem can be stated thus: Suppose we have a finite subgroup $G$ of the general linear group $GL_n(\mathbb{C})$. Then $G$ acts on the polynomial ring $R=\mathbb{C}[x_1,\ldots,x_n]$ in the natural way, that is, by replacing the column vector of variables $(x_1,\ldots,x_n)$ with their image under left matrix multiplication by $G$.

Let $R^G$ be the invariant space under this action. Then $R$ is graded by degree; that is, for each nonnegative integer $k$, the space $R_k^G$ of $G$-invariant homogeneous polynomials of degree $k$ are a finite dimensional subspace of $R^G$, and $R^G$ is the direct sum of these homogeneous components. What is the size (dimension) of these homogeneous components?

In my last post, we saw that the number of positive up-down walks (going either up by $1$ or down by $1$ at each step) of length $2n$ starting from $(0,0)$ is $\binom{2n}{n}$. Furthermore, the number of these that return to height $0$ at the end is the $n$th Catalan number $\frac{1}{n+1}\binom{2n}{n}$. It is natural to ask, then: how many positive walks of length $2n$ end at height $h$ for a given $h$?

This question was brought up at the dinner table last week by Carlos Shine, who leads the Brazilian IMO team and was also one of my coworkers at the IdeaMath summer camp last week in San Jose. In classic mathematician style, we asked the waiter for a pen and began scribbling on the nearest napkin.

The first thing we noticed was that the ending height is always even, since we change the parity by $1$ at each of the $2n$ steps. So let the ending height be $2k$, and define $f(n,k)$ to be the number of positive up-down walks of length $2n$ ending at height $2k$. We can simply list out the possibilities for the first few values of $n$ to make the following table of values of $f(n,k)$.

It is rather well-known that the number of up-down walks that start at $0$ and stay nonnegative until ending at $0$ after $2n$ steps is the $n$th Catalan number, $C_n=\frac{1}{n+1}\binom{2n}{n}$. Such paths, in which we either go up by $1$ or down by $1$ at each step and never fall below the $x$-axis, are called *Dyck paths*. An example for $n=5$ is shown below.

What is not so well-known is that:

The number of up-down walks of length $2n$ that stay nonnegative, but do not necessarily return to $0$ at the end, is simply $\binom{2n}{n}$.

This fact came up in my research a few days ago, and I figured that this had to be a trivial theorem that was easy to prove. Several hours of failed attempts later, I found myself googling “positive walks central binomial coefficients” and came across a wonderful recent paper of M. van Leeuwen.

Classical Euclidean geometry is one of the oldest branches of mathematics, and in my opinion still as beautiful as ever. Here is a cute little geometric gemstone that was introduced to me recently by a fellow staff member at MOSP.

Consider a triangle $ABC$ drawn in the plane. Let $M$ be the midpoint of $BC$, and let $O$ be the circumcenter of triangle $ABC$, that is, the center of the circle $\Omega$ passing through $A$, $B$, and $C$.

Now let $\omega$ the circumcircle of triangle $BOC$, and let $D$ be the point on $\omega$ outside of $ABC$ for which $MD$ is perpendicular to $BC$. Then it turns out that the line $AD$ can be obtained by reflecting the median $AM$ across the angle bisector of angle $A$! This reflection of $AM$ is called a *symmedian* of the triangle.

How can we go about proving this? Well, it suffices to show that angle $BAD$ is congruent to angle $CAM$. One way to go about showing angle measures are equal is by finding similar triangles. Triangles $AMC$ and $ABD$ look promising; let’s start by calculating angle $ABD$. This is equal to angle $B$ plus angle $CBD$, which is an inscribed angle on circle $\omega$. By the inscribed angle theorem (there is a beautiful interactive demonstration at that link!) angle $CBD$ is equal to angle $DOC$, which is half the arc $BC$ on circle $\Omega$. Thus angle $DOC$, and hence angle $CBD$, is equal to angle $A$ of the triangle.