# Some Inner Product Calculations

In Nathan Hunter’s thesis, the inner product of two monomials $x^M$ and $x^L$, with $M \leq L$ are given as $$\begin{eqnarray}\langle x^M , x^L \rangle &=& 2 \pi \sum_{n=-NM}^M {M \choose \frac{NM + n}{N+1}} {L \choose \frac{NL + n}{N+1}} \\ && \quad N^{(2n – M – L)/(N+1)} \left( \frac{N M + n}{N+1} + 1 \right)^{-1} \left( \frac{N L + n}{N+1} + 1 \right)^{-1} \frac{s(n+1)}{s – (n+1)},\end{eqnarray}$$ where we take, for instance $${M \choose \frac{NM + n}{N+1}} = 0$$ unless $\frac{NM + n}{N+1}$ is an integer between $0$ and $M$. This latter condition implies that, if we assume $M, L < d,$ then we may in fact take the sum over $n= -N(d-1), \ldots, (d-1)$. Organizing terms in a symmetric way, we see $$\begin{eqnarray} \langle x^M , x^L \rangle &=& 2 \pi \sum_{n=-N(d-1)}^{d-1} \frac{s(n+1)}{s – (n+1)} \\ &\times& \left[{M \choose \frac{NM + n}{N+1}} \frac{N^{(n-M)/(N+1)}}{ \left( \frac{N M + n}{N+1} + 1 \right)} \right] \left[{L \choose \frac{NL + n}{N+1}} \frac{N^{(n-L)/(N+1)}}{ \left( \frac{N L + n}{N+1} + 1 \right)} \right].\end{eqnarray}$$ Combining the appropriate denominator(s) in the summand with the binomial coefficients and simplifying produces $$\langle x^M , x^L \rangle = 2 \pi \sum_{n=-N(d-1)}^{d-1} \frac{s(n+1)}{s – (n+1)} \left[ {M+1 \choose \frac{M-n}{N+1}} \frac{N^{-\frac{M-n}{N+1}}}{M+1}\right] \left[ {L+1 \choose \frac{L-n}{N+1}} \frac{N^{-\frac{L-n}{N+1}}}{L+1}\right].$$

There are $d-1 + N(d-1) + 1 = (N+1)(d-1) + 1$ terms in the sum, though most of them are zero. To see this note that, if $\langle x^M , x^L \rangle$ is to not be identically zero, then we must have $NM + n \equiv NL + n \equiv 0 \bmod (N+1).$ In which case, $N \equiv M \bmod(N+1).$ We will exploit this later, but for now, let us define the Gram matrix $\mathbf X = \left[ \langle x^M, x^L \rangle \right]_{L,M=0}^{d-1},$ the $\left((N+1)(d-1) + 1\right) \times d$ matrix $$\mathbf A = \left[ {m+1 \choose \frac{m-n}{N+1}} \frac{N^{-\frac{m-n}{N+1}}}{m+1}\right]_{m,n}; \quad n=-N(d-1), \ldots, d-1; \quad m=0,\ldots,d-1,$$ and the $\left((N+1)(d-1) + 1\right)$ diagonal matrix $$\mathbf S = \left[ 2 \pi \delta_{n,k} \frac{s(n+1)}{s – (n+1)} \right]_{n,k}; \quad n, k=-N(d-1), \ldots, d-1.$$

Claim: $\mathbf X = \mathbf {A S A}^T.$

Let $\mathcal N = \{-N(d-1), \ldots, d-1; \quad m=0,\ldots,d-1\}$ and let $c$ be a residue class in $\mathbb Z/(N+1) \mathbb Z$. Define $\mathcal N_c = c \cap \mathcal N$. We also define $\mathcal D = \{0, \ldots, d-1\}$ and $\mathcal D_c = c \cap \mathcal D$. Clearly $\mathcal D_c \subseteq \mathcal N_c$.

Then, because $\langle x^M, x^L \rangle$ is zero whenever $M, N \not \in \mathcal D_c$ for some $c$, we have

Claim: For each $c \in \mathbb Z/(N+1) \mathbb Z$ define $$\mathbf A_c = \left[ {m+1 \choose \frac{m-n}{N+1}} \frac{N^{-\frac{m-n}{N+1}}}{m+1}\right]_{m,n}; \quad n \in \mathcal N_c, \quad m \in \mathcal D_c$$ and $$\mathbf S_c = \left[ 2 \pi \delta_{n,k} \frac{s(n+1)}{s – (n+1)} \right]_{n,k} \quad n,k \in \mathcal N_c.$$ Then, $$\det \mathbf X = \prod_{c \in \mathbb Z/(N+1)\mathbb Z} \det \mathbf A_c \mathbf S_c \mathbf A_c^T.$$

To simplify this further, let us enumerate $\mathcal N_c$ and $\mathcal D_c$. Let $N_c = \# \mathcal N_c$ and $D_c = \# \mathcal D_c$. Then, write $\mathcal D_c = \{d_1, \ldots, d_{D_c} \}$ and $\{n_1, \ldots, n_{N_c – D_c}\}$ for the (negative) elements of $\mathcal N_c$ that are not in $\mathcal D_c$. Then we can write $\mathbf A_c$ as the block matrix $\mathbf A_c = [ \mathbf B_c \;\; \mathbf T_c ]$ where $$\mathbf B_c = \left[ {m+1 \choose \frac{m-n}{N+1}} \frac{N^{-\frac{m-n}{N+1}}}{m+1} \right]_{m,n} \quad m \in \mathcal D_c, \quad n \in \mathcal N_c \setminus \mathcal D_c.$$ And, $$\mathbf T_c = \left[{m+1 \choose \frac{m-n}{N+1}} \frac{N^{-\frac{m-n}{N+1}}}{m+1} \right]_{m,n} \quad m,n \in \mathcal D_c.$$ Notice that $\mathbf T_c$ is lower triangular with $1$s on the diagonal.

We then write $\mathbf S_c$ as the block diagonal matrix $$\mathbf S_c = \begin{bmatrix} \mathbf N_c & \boldsymbol{0} \\ \boldsymbol{0} & \mathbf D_c \end{bmatrix}.$$ Note that $\mathbf D_c$ is $D_n \times D_n$ and contains all terms which have poles at positive integers, and $\mathbf N_c$ is $(N_c – D_c)$ square and contains all terms which have poles at negative integers.

Claim: $$\det \mathbf X = \prod_{c \in \mathbb Z/(N+1)\mathbb Z}\det\left[\mathbf B_c \mathbf N_c \mathbf B_c^T + \mathbf T_c \mathbf D_c \mathbf T_c^T \right].$$

We can go a step farther, and factor out $\det( \mathbf T_c \mathbf D_c \mathbf T_c^T )$ so that $$\det\left[\mathbf B_c \mathbf N_c \mathbf B_c^T + \mathbf T_c \mathbf D_c \mathbf T_c^T \right] = \det( \mathbf T_c \mathbf D_c \mathbf T_c^T ) \det(\mathbf I + \mathbf T^{-T}_c \mathbf D^{-1}_c \mathbf T_c^{-1} \mathbf B_c \mathbf N_c \mathbf B_c^T).$$ This is useful because $\mathbf T_c$ is square triangular with $1$s on the diagonal. That is,

Claim: $$\det \mathbf X = \prod_{c \in \mathbb Z/(N+1)\mathbb Z} \det \mathbf D_c \times \prod_{g \in \mathbb Z/(N+1)\mathbb Z} \det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T).$$

Next, notice that $$\prod_{c \in \mathbb Z/(N+1)\mathbb Z} \det \mathbf D_c = \frac{(2 \pi)^d}{d!} \prod_{\ell=1}^d \frac{s}{s – \ell}.$$ This is the moment function for the (good old-fashioned) Mahler measure!

Theorem: $$\det \mathbf X = \frac{(2 \pi)^d}{d!} \prod_{\ell=1}^d \frac{s}{s – \ell} \times \prod_{g \in \mathbb Z/(N+1)\mathbb Z} \det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T).$$

Note that the entries of $\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T$ are rational functions of $s$ with poles lying in $\mathcal N_g \cup \{0\}$ and hence $\det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T)$ is analytic in $\Re(s) > 0$.

Corollary: As a function of $s$, $\det \mathbf X$ is a rational function of $s$ with simple poles in $1 + \mathcal N$.

Proof. The first assertion is clear, because the entries of $\mathbf X$ are rational functions with poles in $\mathcal N$. However, it is not immediately clear that the poles are simple. To see this we use Sylvester’s Identity: $\det(\mathbf I + \mathbf M \mathbf L) = \det(\mathbf I + \mathbf L \mathbf M)$ so long as $\mathbf M \mathbf L$ and $\mathbf L \mathbf M$ are square matrices (though they do not necessarily need to be the same size). From the theorem we see that $$\det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T) = \det(\mathbf I + \mathbf B_g^T \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g ).$$ If $n \in \mathcal N_g \setminus \mathcal D_g$ then write $\mathbf E_n = \left[ \delta_{j,k} + \delta_{j,n} (s – n-2) \right]_{j, k \in \mathcal N_g\setminus \mathcal D_g}$. This is the matrix formed by replacing the diagonal entry in the identity matrix corresponding to $n$ with $(s-n-1)$. Clearly $\det \mathbf E_n = (s – n-1)$. Then $$\rho_n := \lim_{s \rightarrow n+1} (s-n-1) \det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T) = \lim_{s \rightarrow n+1} \det(\mathbf E_n + \mathbf B_g^T \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf E_n).$$ Because the determinant is a continuous function, and because the entries of $\mathbf E_n + \mathbf B_g^T \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf E_n$ are continuous except in $s \in 1 + \mathcal N_g \setminus \mathcal D_g,$ $$\rho_n = \det(\lim_{s \rightarrow n+1} \mathbf E_n + \lim_{s \rightarrow n+1} \mathbf B_g^T \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf E_n).$$ Only the entries of $\mathbf D_g, \mathbf N_g$ and $\mathbf E_n$ depend on $s$. Thus, $$\rho_n = \det(\lim_{s \rightarrow n+1} \mathbf E_n + \mathbf B_g^T \mathbf T^{-T}_g \left(\lim_{s \rightarrow n+1} \mathbf D^{-1}_g\right) \mathbf T_g^{-1} \mathbf B_g \lim_{s \rightarrow n+1} \left(\mathbf N_g \mathbf E_n)\right).$$ The entries of $\mathbf E_n$ and $\mathbf D^{-1}_g$ are analytic functions of $s$, except for the latter at $s = 0$. In particular, $\lim_{s \rightarrow n+1} \mathbf E_n$ and $\lim_{s \rightarrow n} \mathbf D^{-1}_g$ are numeric matrices (in fact with rational entries, times a power of $\pi$ in the latter case). Next we note that $\mathbf N_g \mathbf E_n$ is a diagonal matrix with entries analytic at $s = n+1$, with the exception of the entry corresponding to $n$ itself, which has a removable singularity at $s = n+1$. Thus $\lim_{s \rightarrow n+1} \mathbf N_g \mathbf E_n$ is a numeric matrix, in fact with rational entries times a power of $\pi$. It follows that $\rho_n$ exists, and it is the residue at $s=n+1$ of a simple pole at $\det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T)$. Because, for different cosets $g$, the poles of $\det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T)$ lie in disjoint sets, it follows that $$\prod_{g \in \mathbb Z/(N+1)\mathbb Z} \det(\mathbf I + \mathbf T^{-T}_g \mathbf D^{-1}_g \mathbf T_g^{-1} \mathbf B_g \mathbf N_g \mathbf B_g^T).$$ has simple poles at $s \in 1 + \mathcal N \setminus D$. Because $\frac{(2 \pi)^d}{d!} \prod_{\ell=1}^d \frac{s}{s – \ell}$ has only simple poles as $s \in 1 + \mathcal D,$ it follows that $\det \mathbf X$ is a rational function with only simple poles in $1 + \mathcal N$.