Signed binomial matrices of small order

Wildon's Weblog 2020-12-28

Let $B(n)$ denote the $n \times n$ Pascal’s Triangle matrix defined by $B(n)_{xy} = \binom{x}{y}$ , where, as in the motivating paper, we number rows and columns of matrices from 0. Let $J(n)_{xy} = [x+y=n-1]$ be the involution reversing the order of rows (or columns) of a matrix, and let

$D^\pm(n) = \mathrm{Diag}(1,-1,\ldots, (-1)^{n-1}).$

Since $n$ is fixed throughout, we shall write $B$ , $J$ and $D^\pm$ for readability in proofs.

Claim 1. For any $a \in \mathbb{N}$ , and any $n \ge 2$ , the matrix $D^\pm(n)B_a(n)$ is an involution.

Proof. Since $n\ge 2$ , the matrix is not the identity. Since $\bigl( D^\pm B_a \bigr)_{xy} = (-1)^x \binom{x+a}{y+a}$ we have

$\begin{aligned} \bigl( D^\pm &B_a D^\pm B_a \bigr)_{xz} = (-1)^x \sum_y (-1)^y \binom{x+a}{y+a} \binom{y+a}{z+a} \\ &= (-1)^x \sum_y \binom{x+a}{z+a}\binom{x-z}{y-z}(-1)^y \\ &= (-1)^{x+z} \binom{x+a}{z+a} \sum_r \binom{x-z}{r}(-1)^r \\ &= (-1)^{x+z} \binom{x+a}{z+a} [x=z] (-1)^{x+z} \\ &= [x=z] \end{aligned}$

as required. $\quad\Box$

Claim 2. For any $n \ge 2$ , the matrix $D^\pm(n)J(n)B(n)$ has order $3$ if $n$ is odd and order $6$ if $n$ is even. $\quad\Box$

Proof. Since $\bigl( D^\pm J B \bigr)_{xy} = (-1)^x \binom{n-1-x}{y}$ we have

$\begin{aligned} \bigl( D^\pm J B D^\pm J B D^\pm J B \bigr)_{xw} &= (-1)^x \sum_y \sum_z (-1)^{y+z} \binom{n-1-x}{y} \binom{n-1-y}{z} \binom{n-1-z}{w} \\ &= (-1)^{x+n-1} \sum_z \Bigl( \sum_y \binom{n-1-x}{y} \binom{-z-1}{n-1-y-z} \Bigr) \binom{n-1-z}{w} \\ &= (-1)^{x+n-1} \sum_z \binom{n-2-x-z}{n-1-z}\binom{n-1-z}{w} \\ &= (-1)^{x} \sum_z (-1)^{z}\binom{x}{n-1-z}\binom{n-1-z}{w} \\ &= (-1)^{x+n-1} \sum_r (-1)^r \binom{x}{r} \binom{r}{w} \\&= (-1)^{x+n-1} (-1)^x [x=w] = (-1)^{n-1}. \end{aligned}$

It is easily seen that the matrix is not a scalar multiple of the identity, therefore it has the claimed order. $\quad\Box$

Claim 3. We have $B(n)^{-1} J(n) B(n) = D^\pm(n) J(n) B(n) J(n)$ and $B(n) J(n) B(n)^{-1} = J(n) B(n) J(n) D^\pm(n)$ .

Proof. The alternating sum used at the end of both the previous two claims can be used to show that $B(n)^{-1}_{xy} = (-1)^{x+y} \binom{x}{y}$ . Hence

$\begin{aligned} \bigl(B(n)J(n)B(n)^{-1} \bigr)_{xz} &= \sum_{y} (-1)^{x+y} \binom{x}{y} \binom{n-1-y}{z} \\ &= (-1)^x\sum_y (-1)^y \binom{x}{y} (-1)^{n-1-y-z} \binom{-z-1}{n-1-y-z} \\ &= (-1)^{x+n-1-z} \binom{x-z-1}{n-1-z} \\ &= (-1)^{x+n-1-z} \binom{n-1-x}{n-1-z} (-1)^{n-1-z} \\ &= (-1)^x \binom{n-1-x}{n-1-z} \\ &= \bigl( D^\pm(n)J(n)B(n)J(n) \bigr)_{xz}\end{aligned}$

as required. The second identity is proved very similarly. $\quad\Box$

Let $X^\circ$ denote the half-turn rotation of an $n \times n$ matrix $X$ , as defined by $X^\circ = J(n)XJ(n)$ . By Claim 3, $B(n)D^\pm(n)B(n)^\circ D^\pm = BD^\pmJBJD^\pm$ is conjugate to $JD^\pm BD^\pm JB$ and so to $(-1)^{n-1} (D^\pm JB)(D^\pm JB)$ . Hence this matrix has order $3$ when $n$ is odd and order $6$ when $n$ is even. We state without proof some further identities along these lines. Let $B_a(n)$ be the shifted Pascal’s Triangle matrix defined by $B_a(n)_{xy} = \binom{x+a}{y+a}$ .

Claim 4. The matrix $B(n)D^\pm(n)B_n(n)D^\pm(n)$ has order $3$ . If $n$ is even then $B(n)D^\pm(n)B_1(n)D^\pm(n)$ has order $12$ . If $n$ is odd then $B_1(n)D^\pm(n)B_1(n)D^\pm(n)$ has order $3$.

The second case seems particularly remarkable. There are some obstructions (related to root systems) to integer matrices having finite order. These identities were discovered as a spin-off of a somewhat involved construction with linear algebra; I have no idea how to motivate them in any other way. For instance, how looking at

$B(6)D^\pm(6)B_1(6)D^\pm(6) = \left( \begin{matrix} 1 & -6 & 15 & -20 & 15 & -6 \\ 1 & -5 & 10 & -10 & 5 & -1 \\ 1 & -4 & 6 & -4 & 1 & 0 \\ 1 & -3 & 3 & -1 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 & 0 \\ 1 & -1 & 0 & 0 & 0 & 0 \end{matrix} \right)$

would one guess that it has order $12$ ?