# Zeta function of $\Z^d$

### Contents

Zeta functions have been associated with groups since 1988—when Grundewald, Segal and Smith introduced the idea. The definition is simple: let $G$ be a group, and let $s$ be a (complex) variable. For each positive integer $n$, let $a_n(G)$ be the number of subgroups of $G$ with index $n$. The zeta function associated with $G$ is the Dirichlet series $\zeta_G(s) = \sum_{n = 1}^{\infty} \dfrac{a_n(G)}{n^s} .$

We will compute the zeta function associated with $\Z^d$ by building an explicit bijection between subgroups of a prime-power index and a complete flag in a finite vector space. But before that journey, let’s do a warm up with $d=1$.

## A warm up: $G = \Z$

If $G = \Z$, then $a_n(G) = 1$ for all $n\geqslant 1$. That is, the unique subgroup of index $n$ in $\Z$ is $n\Z$. Therefore, the zeta function associated with $\mathbb{Z}$ is the Riemann zeta function: $\zeta_{\Z}(s) = 1 + \frac{1}{2^{-s}} + \frac{1}{3^{-s}} + \frac{1}{4^{-s}} + \cdots .$

This sum can be written as an infinite product: $\sum_{n=1}^\infty \dfrac{1}{n^{-s}} = \prod_{p\text{ prime}} \dfrac{1}{1 - p^{-s}},$

which follows from the fact that every positive integer $n$ can be written as a unique product of (powers of) prime numbers. Equivalently, the quotient $\Z/n\Z$ splits uniquely into a direct product of its prime power subgroups (i.e. Sylow subgroups). We can view each of the infinite factors $\dfrac{1}{1 - p^{-s}}$

as a zeta function for a different group. If we consider the abelian group $\Z_p$, equal to the group of $p$-adic integers, then its zeta function is the geometric series: $\zeta_{\Z_p}(s) = \dfrac{1}{1 - p^{-s}} = 1 + \dfrac{1}{p^{-s}} + \dfrac{1}{p^{-2s}} + \dfrac{1}{p^{-3s}} + \cdots .$

## $G = \Z^d$

For essentially the same reason as in the case for $d=1$, the zeta function for $\Z^d$ factors into a infinite product of zeta functions for $\Z_p^d$, for all primes $p$: $\zeta_{\Z^d} (s) = \prod_{p\text{ prime}} \zeta_{\Z_p^d} (s).$

Therefore, we can focus on $\Z_p^d$ for a fixed prime $p$. We provide a combinatorial proof to the following theorem identifying collections of subgroups of $\Z_p^{d}$ with Schubert cells in a finite vector space.

Theorem 1. For $d\geqslant 0$ and all primes $p$, $\zeta_{\Z_p^{d+1}}(s) = \prod_{k=0}^{d}\dfrac{1}{1 - p^{k-s}}.$

The theorem is proved with two steps, but first we establish some notation, following Fulton.

### Schubert cells

Let $\mathcal{G}(d, n)(\F_p)$ be the Grassmannian of $d$-dimensional subspaces of the finite vector space $V = \F_p^n$. We write $\binom{n}{d}_p = \#\mathcal{G}(d, n)(\F_p)$, which are known as the Gaussian binomial coefficients. We denote by

$F_{\bullet}^{(n)} : 0 = F_0 \subset F_1 \subset \cdots \subset F_n = V,$ where $\dim(F_i)=i$, a complete flag in $V$.

Let $\mathcal{P}(d, n)$ be the set of (possibly empty) partitions with at most $d$ parts each of size at most $n$. For $\lambda = (\lambda_1,\dots, \lambda_d)\in\mathcal{P}(d, n)$, let $|\lambda|=\lambda_1+\cdots +\lambda_d$, written this way, we assume $\lambda_i\geqslant \lambda_{i+1}$. For $\lambda \in \mathcal{P}(d, n)$ and a complete flag $F_{\bullet}^{(n+d)}$, the associated Schubert cell is $\Omega_{\lambda}^{\circ} (F_{\bullet}^{(n+d)}) =\left\{U\in\mathcal{G}(d, n+d) ~\middle|~ \begin{array}{c}i \in [d]_0,\ n+i-\lambda_i\leqslant j\leqslant n+i-\lambda_{i+1}, \cr \dim(U\cap F_j)=i \end{array} \right\}$ where the condition $i=0$ is understood to be $U\cap F_j=0$ for $j=n-\lambda_1$. We fix a complete flag $F_{\bullet}^{(n+d)}$, so we drop the flag from the notation of Schubert cells. Then $\dim(\Omega_\lambda^\circ)=dn-|\lambda|$.

#### Schubert cell example

Let $n = 3$ and $d=4$, and set $\lambda = (3, 2, 2, 1)$. The Schubert cell $\Omega_{\lambda}^{\circ}$ is a set of $4$-dimensional subspaces of $\F_p^7$. We can write these subspaces as $4\times 7$ matrices over $\F_p$, where the row span of such a matrix yields a $4$-dimensional subspace. The set $\Omega_{\lambda}^{\circ}$ can, therefore, be identified with the set of matrices of the form: $\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & * & 1 & 0 & 0 & 0 & 0 \cr 0 & * & 0 & 1 & 0 & 0 & 0 \cr 0 & * & 0 & 0 & * & 1 & 0 \end{pmatrix},$

where we place a $1$ in $(i, n+i-\lambda_i)$ entry. The entries with $*$ may be replaced with arbitrary elements from $\F_p$, yielding a subspace. We see there are $p^4$ such subspaces, where $4 = dn - \vert\lambda\vert$.

### Step 1: the bridge

Since $\mathcal{G}(d, n+d)(\F_p)$ is the disjoint union of Schubert cells, $\begin{equation} \binom{n+d}{d}_p = \sum_{\lambda \in\mathcal{P}(d, n)} \# \Omega_{\lambda}^{\circ} = \sum_{\lambda \in\mathcal{P}(d, n)} p^{dn-|\lambda|} = \sum_{\lambda\in\mathcal{P}(d, n)} p^{|\lambda|}. \end{equation}$

Related to the set of partitions $\mathcal{P}(d, n)$ is the set $\mathcal{S}(d,n)= \left\{ (m_1,\dots, m_{d+1})\in \N_0^{d+1}~\middle|~ n = m_1 + \cdots + m_{d+1}\right\}.$

The next lemma is a bridge between our bijective proof between subgroups of $\Z_p^d$ and subspaces of a finite vector space, whose cornerstone is a bijection between $\mathcal{P}(d, n)$ and $\mathcal{S}(d, n)$. In fact, let’s define that function (soon to be bijection) now. Let $\Pi : \mathcal{S}(d, n) \to \mathcal{P}(d, n)$ be given by $(m_1,\dots, m_{d+1}) \longmapsto (m_{i+1} + \cdots + m_{d+1})_{i=1}^d.$

For the case $d=0$, this just maps the element $(n)$ to the empty partition.

$\$1.5em] Lemma 2. For all d\geqslant 0 and n\geqslant 0, the function \Pi is a bijection. Proof. We define another function \Sigma : \mathcal{P}(d, n) \to \mathcal{S}(d, n) via \[ (\lambda_1,\dots, \lambda_d)\longmapsto (\lambda_{k-1}-\lambda_k)_{k=1}^{d+1},$ where$\lambda_0=n$and$\lambda_{d+1}=0. Now we consider the composition \begin{aligned} (\Sigma\Pi)(m_1,\dots, m_{d+1}) &= \Sigma\left((m_{i+1} + \cdots + m_{d+1})_{i=1}^d\right) = (m_1,\dots, m_{d+1}). \end{aligned} Here, we are using the fact thatm_1 = n-(m_2 + \cdots + m_{d+1})$. Furthermore, we have $(\Pi\Sigma)(\lambda_1, \dots, \lambda_d) = \Pi\left( (\lambda_{k-1}-\lambda_k)_{k=1}^{d+1} \right) = (\lambda_1, \dots, \lambda_d).$ Therefore,$\Sigma = \Pi^{-1}$, so$\Pi$is a bijection.$\square$We will use the map$\Pi$to rewrite the zeta function for$\Z_p$in terms of Gaussian binomial coefficients. Lemma 3. For all$d\geqslant 0$and primes$p$, $\begin{equation} \prod_{k=0}^{d}\dfrac{1}{1 - p^{k-s}} = \sum_{n\geqslant 0}p^{-ns}\binom{n+d}{d}_p . \end{equation}$ Proof. For$m = (m_1,\dots, m_{d+1})\in \mathcal{S}(d,n)$, $\begin{equation} \sum_{k=0}^{d}m_{k+1}(k-s) = \sum_{k=1}^{d+1}\left(\sum_{j=k}^{d+1}m_{j+1}\right) - s\sum_{k=1}^{d+1}m_k = |\Pi(m)| - ns. \end{equation}$ The$p^{-ns}coefficient on the left in Equation (2) is \begin{aligned} \sum_{m\in \mathcal{S}(d,n)}~\prod_{k=0}^{d} p^{m_{k+1}k} &= \sum_{m\in \mathcal{S}(d,n)} p^{|\Pi(m)|} \\[1.5em] &= \sum_{\lambda \in\mathcal{P}(d, n)} p^{|\lambda|} \\[1.5em] &= \binom{n+d}{d}_p, \end{aligned} where the first equality follows from Equation (3), the second equality from Lemma 1, and the last equality from Equation (1).\square$### Step 2: an explicit map The last step is to prove the final lemma. Let$\Gamma = \GL_{d+1}(\Z_p)$. For a fixed minimal generating set of$\Z_p^{d+1}$, finite index subgroups$H\leqslant \Z_p^{d+1}$correspond to$\Gamma$-cosets of lower triangular matrices $M_H\in\GL_{d+1}(\Q_p)\cap\mathrm{Mat}_{d+1}(\Z_p),$ where$M_H$is some chosen representative. It follows that$[\Z_p^{d+1} : H] = \det(M_H)$.$\$1.5em] Lemma 4. The number of subgroups of \Z_p^{d+1} of index p^n is \binom{n+d}{d}_p. Proof. Let H\leqslant \Z_p^{d+1} have index p^n, and let M_H=(M_{ij}) be a lower-triangular matrix whose rows generate H. Without loss of generality, we assume the diagonal entries of M_H have the form M_{ii} = p^{m_i} for some m=(m_1,\dots, m_{d+1})\in\N_0^{d+1} and 0\leqslant M_{ij}< M_{jj} = p^{m_j} for all i>j. There is exactly one such matrix in the coset \Gamma M_H. Since the index of H is p^n, we have m\in\mathcal{S}(d, n). Let \mathcal{H}_m be the collection of subgroups of H\leqslant\Z_p^{d+1} whose diagonal entries of M_H are (p^{m_1},\dots, p^{m_{d+1}}). Using the bijection \Pi : \mathcal{S}(d, n)\rightarrow \mathcal{P}(d, n) from Lemma 1, we will show that there is a bijection between subgroups and subspaces: \[ \mathcal{H}_m \longleftrightarrow \Omega_{\Pi(m)}^\circ.$

Once we have done this, we will have the following chain of bijections: \begin{aligned} \left\{H \leqslant \Z_p^{d+1} ~\middle|~ [\Z_p^{d+1} : H] = p^n\right\} &\longleftrightarrow \bigsqcup_{m\in\mathcal{S}(d, n)} \mathcal{H}_m \cr &\longleftrightarrow \bigsqcup_{m\in\mathcal{S}(d, n)} \Omega_{\Pi(m)}^\circ \cr &\longleftrightarrow \bigsqcup_{\lambda\in\mathcal{P}(d, n)} \Omega_{\lambda}^\circ \cr &\longleftrightarrow \mathcal{G}(d, n+d)(\F_p), \end{aligned}

which would prove the lemma.

Now let’s construct the bijection between $\mathcal{H}_m$ and $\Omega_{\Pi(m)}^\circ$. Because $H\in\mathcal{H}_m$ corresponds to $\Gamma M_H$ and $M_H$ is lower triangular, $H$ corresponds to the coset $\mathrm{LT}_{d+1}(\Z_p) M_H$, where $\mathrm{LT}_{d+1}(\Z_p)$ is the group of lower triangular matrices. Since $M_{jj} = p^{m_j}$, it follows that for all $i>j$ there exists integers $a_{i, k}^{(j)}\in [p-1]_0$ for each $k\in [m_j-1]_0$ such that $M_{ij} = a_{i,0}^{(j)} + a_{i,1}^{(j)}p + \cdots + a_{i,m_j-1}^{(j)}p^{m_j-1} + O(p^{m_j}).$

Furthermore, this array of integers $a_{i,k}^{(j)}$ uniquely determines the $\mathrm{LT}_{d+1}(\Z_p)$-coset.

Suppose $(e_1,\dots, e_{n+d})$ is the basis determined by the complete flag $F_\bullet^{(n+d)}$. Let $\lambda = \Pi(m)$. Define a subspace $U_H = \left\langle e_{n+i-\lambda_i} + \sum_{j=1}^i \sum_{k=0}^{m_j-1} a_{i-1,k}^{(j)} \cdot e_{n+j-\lambda_{j-1}+k} ~\middle|~ i\in [d]\right\rangle,$

which by construction is contained in $\Omega_\lambda^\circ$. Since the $a_{i,k}^{(j)}$ uniquely determine the $\mathrm{LT}_{d+1}(\Z_p)$-coset, it follows that $\Omega_\lambda^\circ$ is in bijection with $\mathcal{H}_m$. $\square$

### Mixing the ingredients (Proof of Theorem 1)

Now we can use Lemmas 3 and 4 to prove Theorem 1.

Proof of Theorem 1. By Lemma 3, we have $\prod_{k=0}^{d-1}\dfrac{1}{1 - p^{k-s}} = \sum_{n\geqslant 0}p^{-ns}\binom{n+d}{d}_p ,$

and by Lemma 4, this can be rewritten as $\sum_{n\geqslant 0}p^{-ns}\binom{n+d}{d}_p = \sum_{n\geqslant 0}a_n(\Z_p) p^{-ns} = \zeta_{\Z^{d}_p}(s).$

It follows that $\prod_{k=0}^{d-1}\dfrac{1}{1 - p^{k-s}} = \zeta_{\Z^{d}_p} ,$

which proves the theorem. $\square$

Now we consider two specific examples, and construct a subspace from a subgroup and vice versa.

## Example 1

Let $p=7$, $d=3$, and $n=6$. Let $H\leqslant\Z_7^{3+1}$ with corresponding matrix $M_H = \begin{pmatrix} 7^2 & . & . & . \cr 33 & 1 & . & . \cr 42 & 0 & 7^3 & . \cr 1 & 0 & 321 & 7 \end{pmatrix},$

so $H$ has index $7^6$. The $m\in \mathcal{S}(3, 6)$ corresponding to $H$ is $m=(2, 0, 3, 1)$, so $\lambda = \Pi(m) = (4, 4, 1)$. Thus, the subspace $U_H$ in $\F_7^9$ is given by the row span of the following matrix: $\begin{pmatrix} 5 & 4 & 1 & . & . & . & . & . & . \cr 0 & 6 & 0 & 1 & . & . & . & . & . \cr 1 & 0 & 0 & 0 & 6 & 3 & 6 & 1 & . \end{pmatrix} .$

## Example 2

Let $p = 5$, $d = 4$ and $n = 12$. Let $U\leqslant \F_5^{12+4}$ be the subspace spanned by the rows of the matrix $\begin{pmatrix} 3 & 1 & . & . & . & . & . & . & . & . & . & . & . & . & . & . \cr 2 & 0 & 0 & 1 & . & . & . & . & . & . & . & . & . & . & . & . \cr 0 & 0 & 2 & 0 & 1 & 4 & 3 & 1 & 1 & . & . & . & . & . & . & . \cr 1 & 0 & 4 & 0 & 0 & 3 & 3 & 2 & 0 & 2 & 0 & 2 & 4 & 1 & . & . \end{pmatrix} .$

Then $\lambda = (11, 10, 6, 2)$ and $m = \Pi^{-1}(\lambda)=(1, 1, 4, 4, 2)$, so the corresponding subgroup $H$ in $\Z_5^5$ of index $5^{12}$ is generated by the rows of the matrix $\begin{pmatrix} 5 & . & . & . & . \cr 3 & 5 & . & . & . \cr 2 & 0 & 5^4 & . & . \cr 0 & 2 & 221 & 5^4 & . \cr 1 & 4 & 340 & 552 & 5^2 \end{pmatrix} .$

1. Grunewald, F. J., Segal, D., Smith, G. C. Subgroups of finite index in nilpotent groups. Invent. Math. 93 (1988), no. 1, 185–223.
2. Fulton, W. Young tableaux: with applications to representation theory and geometry. No. 35. Cambridge University Press, 1997.