Finite order elements in the integral symplectic group

TL;DR

This paper investigates the orders of elements in the integral symplectic group, proving that the set of possible orders is bounded for each dimension and demonstrating that the number and maximum order grow at least exponentially with the dimension.

Contribution

It establishes the boundedness of element orders in symplectic groups and analyzes the exponential growth of their count and maximum order as the dimension increases.

Findings

Set of element orders is bounded for each g

Number of possible orders grows at least exponentially with g

Maximum order of elements also grows at least exponentially

Abstract

For $g\in \mathbb{N}$, let $G=\Sp(2g,\mathbb{Z})$ be the integral symplectic group and $S(g)$ be the set of all positive integers which can occur as the order of an element in $G$. In this paper, we show that $S(g)$ is a bounded subset of $\mathbb{R}$ for all positive integers $g$. We also study the growth of the functions $f(g)=|S(g)|$, and $h(g)=max\{m\in \mathbb{N}\mid m\in S(g)\}$ and show that they have at least exponential growth.

Equations

$$A^{\top }JA=J$$

$$S(g)=\{m\in \mathbb{N}\mid \exists A\ne 1\in G\text{ with }o(A)=m\}.$$

$$p_1+p_2+\cdots +p_n<\frac{1}{2}n{p}_n.$$

$$\prod_{p\leq x}\bigg{(}1-\frac{1}{p}\bigg{)}>\frac{e^{-\gamma}}{\log x}\bigg{(}1-\frac{0.2}{(\log x)^{2}}\bigg{)}.$$

$$m_a$$

$$r_a$$

$$m_{a}(\alpha_{1},\alpha_{2},\ldots\alpha_{n})=\frac{q_{1}^{\alpha_{1}}\big{(}1-\frac{1}{q_{1}}\big{)}q_{2}^{\alpha_{2}}\big{(}1-\frac{1}{q_{2}}\big{)}\ldots q_{n}^{\alpha_{n}}\big{(}1-\frac{1}{q_{n}}\big{)}}{\displaystyle\prod_{i=1}^{n}\bigg{(}1-\frac{1}{q_{i}}\bigg{)}}.$$

$$M_{a}=\frac{\bigg{(}q_{1}^{\alpha_{1}}\big{(}1-\frac{1}{q_{1}}\big{)}\bigg{)}^{n}}{{\displaystyle\prod_{i=1}^{n}\bigg{(}1-\frac{1}{q_{i}}\bigg{)}}}\leq\frac{\bigg{(}\frac{2g+1}{n}\bigg{)}^{n}}{\displaystyle\prod_{i=1}^{k}\bigg{(}1-\frac{1}{p_{i}}\bigg{)}}.$$

$$m$$

$$\displaystyle\leq\frac{\max\limits_{1\leq a\leq 2^{k}-1}\bigg{(}\frac{2g+1}{n}\bigg{)}^{n}}{\displaystyle\prod_{i=1}^{k}\bigg{(}1-\frac{1}{p_{i}}\bigg{)}}$$

$$\displaystyle\leq\frac{e^{\frac{2g+1}{e}}}{\displaystyle\prod_{i=1}^{k}\bigg{(}1-\frac{1}{p_{i}}\bigg{)}}$$

$$m\leq 3(5/4)^{\pi(2g+1)-2}e^{\frac{2g+1}{e}}\leq 3e^{\big{(}\frac{2g+1}{e}+g-1\big{)}}\leq 3e^{3g}.$$

$$\prod_{i=1}^{k}\bigg{(}1-\frac{1}{p_{i}}\bigg{)}>\frac{1}{2}\frac{e^{-\gamma}}{\log(2g+1)}.$$

$$m\leq\frac{e^{\frac{2g+1}{e}}}{\displaystyle\prod_{i=1}^{k}\bigg{(}1-\frac{1}{p_{i}}\bigg{)}}\leq 2e^{\gamma}\log(2g+1)e^{\frac{2g+1}{e}}.$$

$$\sum _{p\le x}p<\frac{1}{2}x\pi (x)$$

$$\sum _{p\le x}p=\sum _{p\le p_n}p<\frac{1}{2}n{p}_n\le \frac{1}{2}\pi (x)x.$$

$$\pi (\sqrt{g\log (g)})$$

$$\displaystyle\leq\frac{\sqrt{g\log g}}{\log(\sqrt{g\log g})}\bigg{(}1+\frac{3}{2\log 23}\bigg{)}$$

$$=\frac{3\sqrt{g\log (g)}}{\log (g\log (g))}.$$

$$\beta <\sum _{p\le x}p$$

$$<\frac{3}{2}\frac{g\log (g)}{\log (g\log (g))}=\frac{3}{2}g.$$

$$\frac{\sqrt{g\log (g)}}{\log (g\log (g))}<\pi (\sqrt{g\log (g)}).$$

$$f(g)\ge 2^{\pi (\sqrt{g\log (g)})}>2^{\frac{\sqrt{g\log (g)}}{\log (g\log (g))}}>2^{\frac{1}{2}\sqrt{\frac{g}{\log (g)}}}>e^{\frac{1}{4}\sqrt{\frac{g}{\log (g)}}}.$$

Full text

Finite order elements in the integral symplectic group

Kumar Balasubramanian

Kumar Balasubramanian

Department of Mathematics

IISER Bhopal

Bhopal, Madhya Pradesh 462066, India

bkumar@iiserb.ac.in

, 

M. Ram Murty

M. Ram Murty

Department of Mathematics and Statistics

Queen s University

Kingston, Ontario K7L 3N6, Canada

murty@mast.queensu.ca

 and 

Karam Deo Shankhadhar

Karam Deo Shankhadhar

Department of Mathematics

IISER Bhopal

Bhopal, Madhya Pradesh 462066, India

karamdeo@iiserb.ac.in

Research of Kumar Balasubramanian was supported by DST-SERB Grant: YSS/2014/000806.

Research of M. Ram Murty was partially supported by an NSERC Discovery grant.

Abstract

For $g\in\mathbb{N}$, let $G=\operatorname{Sp}(2g,\mathbb{Z})$ be the integral symplectic group and $S(g)$ be the set of all positive integers which can occur as the order of an element in $G$. In this paper, we show that $S(g)$ is a bounded subset of $\mathbb{R}$ for all positive integers $g$. We also study the growth of the functions $f(g)=|S(g)|$, and $h(g)=max\{m\in\mathbb{N}\mid m\in S(g)\}$ and show that they have at least exponential growth.

1. Introduction

Given a group $G$ and a positive integer $m\in\mathbb{N}$, it is natural to ask if there exists $k\neq 1\in G$ such that $o(k)=m$, where $o(k)$ denotes the order of the element $k\in G$. In this paper, we make some observations about the collection of positive integers which can occur as orders of elements in $G=\operatorname{Sp}(2g,\mathbb{Z})$. Before we proceed further we set up some notation and briefly mention the problems studied in this paper.

Let $G=\operatorname{Sp}(2g,\mathbb{Z})$ be the group of all $2g\times 2g$ matrices with integral entries satisfying

[TABLE]

where $A^{\top}$ is the transpose of the matrix $A$ and $J=\begin{pmatrix}0_{g}&I_{g}\\ -I_{g}&0_{g}\end{pmatrix}$.

Throughout we write $m=p_{1}^{\alpha_{1}}\dots p_{k}^{\alpha_{k}}$, where $p_{i}$ is a prime and $\alpha_{i}>0$ for all $i\in\{1,2,\ldots,k\}$. We also assume that the primes $p_{i}$ are such that $p_{i}<p_{i+1}$ for $1\leq i<k$. We write $\pi(x)$ for the number of primes less than or equal to $x$. Also for $A\in G$ we let $o(A)$ denote the order of $A$. We let $\phi$ denote the Euler’s phi function. It is a well known fact that the function $\phi$ is multiplicative, i.e., $\phi(mn)=\phi(m)\phi(n)$ if $m,n$ are relatively prime and satisfies $\phi(p^{\alpha})=p^{\alpha}(1-\frac{1}{p})$ for all primes $p$ and positive integer $\alpha\in\mathbb{N}$ (see [2] for a proof). Let

[TABLE]

In this paper we show that $S(g)$ is always a bounded subset of $\mathbb{R}$ for all positive integers $g$. Once we know that $S(g)$ is a bounded set, it makes sense to consider the functions $f(g)=|S(g)|$, where $|S(g)|$ is the cardinality of $S(g)$ and $\displaystyle h(g)=max\{m\mid m\in S(g)\}$, i.e., $h(g)$ is the maximal possible (finite) order in $G=\operatorname{Sp}(2g,\mathbb{Z})$. We show that the functions $f$ and $h$ have at least exponential growth.

The above problem derives its motivation from analogous problems from the theory of mapping class groups of a surface of genus $g$. We know that given a surface $S_{g}$ of genus $g$, there is a surjective homomorphism $\psi:\operatorname{Mod}(S_{g})\rightarrow\operatorname{Sp}(2g,\mathbb{Z})$, where $\operatorname{Mod}(S_{g})$ is the mapping class group of $S_{g}$. It is a well known fact that for $f\in\operatorname{Mod}(S_{g})$ ($f\neq 1$) of finite order, we have $\psi(f)\neq 1$. Let $\tilde{S}(g)=\{m\in\mathbb{N}\mid\exists f\neq 1\in\operatorname{Mod}(S_{g})\textrm{ with }o(f)=m\}$. The set $\tilde{S}(g)$ is a finite set and it makes sense to consider the functions $\tilde{f}(g)=|\tilde{S}(g)|$ and $\tilde{h}(g)=max\{m\in\mathbb{N}\mid m\in\tilde{S}(g)\}$. It is a well known fact that both these functions $\tilde{f}$ and $\tilde{h}$ are bounded above by $4g+2$. We refer the reader to [5] for an excellent introduction to the mapping class group and the proofs of some of these facts.

2. Some results we need

In this section we mention a few results that we need in order to prove the main results in this paper.

Proposition 2.1** (Bürgisser).**

*Let $\displaystyle m=p_{1}^{\alpha_{1}}\dots p_{k}^{\alpha_{k}}$, where the primes $p_{i}$ satisfy $p_{i}<p_{i+1}$ for $1\leq i<k$ and where $\alpha_{i}\geq 1$ for $1\leq i\leq k$. There exists a matrix $A\in\operatorname{Sp}(2g,\mathbb{Z})$ of order $m$ if and only if

1. a)

$\displaystyle\sum_{i=2}^{k}\phi(p_{i}^{\alpha_{i}})\leq 2g$, if $m\equiv 2(\bmod\,4)$. 2. b)

$\displaystyle\sum_{i=1}^{k}\phi(p_{i}^{\alpha_{i}})\leq 2g$, if $m\not\equiv 2(\bmod\,4)$.

Proof.

See corollary 2 in [1] for a proof. ∎

Proposition 2.2** (Dusart).**

*Let $p_{1},p_{2},\dots,p_{n}$ be the first $n$ primes. For $n\geq 9$, we have

[TABLE]

Proof.

See theorem 1.14 in [3] for a proof. ∎

Proposition 2.3** (Dusart).**

For $x\geq 2$, \pi(x)\leq\frac{x}{\log x}\bigg{(}1+\frac{1.2762}{\log x}\bigg{)}. For $x\geq 599$, \pi(x)\geq\frac{x}{\log x}\bigg{(}1+\frac{1}{\log x}\bigg{)}.

Proof.

See theorem 6.9 in [4] for a proof. ∎

Proposition 2.4** (Dusart).**

*For $x\geq 2973$,

[TABLE]

*where $\gamma$ is the Euler’s constant.

Proof.

See theorem 6.12 in [4] for a proof. ∎

Proposition 2.5** (Rosser).**

For $x\geq 55$, we have $\pi(x)>\frac{x}{\log x+2}$.

Proof.

See theorem 29 in [6] for a proof. ∎

3. Main Results

In this section we prove the main results of this paper. To be more precise, we prove the following.

1. a)

$S(g)$ is a bounded subset of $\mathbb{R}$. 2. b)

$f(g)=|S(g)|$ has at least exponential growth. 3. c)

$h(g)=max\{m\mid m\in S(g)\}$ has at least exponential growth.

3.0.1. Boundedness of $S(g)$

For each $g\in\mathbb{N}$, let $S(g)=\{m\in\mathbb{N}\mid\exists\,A\neq 1\in G\textrm{ with }o(A)=m\}$. In this section we show that $S(g)$ is a bounded subset of $\mathbb{R}$.

Let $m=p_{1}^{\alpha_{1}}\ldots p_{k}^{\alpha_{k}}\in S(g)$. Suppose $p_{i}>2g+1$ for some $i\in\{1,2,\ldots,k\}$. This would imply that $\phi(p_{i}^{\alpha_{i}})=p_{i}^{\alpha_{i}-1}(p_{i}-1)>2g$, which contradicts proposition 2.1. It follows that all primes in the factorization of $m$ should be $\leq 2g+1$ and hence $k\leq g+1$.

Theorem 3.1**.**

For $g\in\mathbb{N}$, $S(g)$ is a bounded subset of $\mathbb{R}$.

Proof.

For $g\in\mathbb{N}$, fix $k=\pi(2g+1)$ and $P=\{p_{1},p_{2},\dots,p_{k}\}$ be the set of first $k$ primes arranged in increasing order. The prime factorization of any $m\in S(g)$ involves primes only from the set $P$. The total number of non-empty subsets of $P$ is $2^{k}-1$. Let us denote the collection of these subsets of $P$ as $\{P_{1},P_{2},\ldots P_{2^{k}-1}\}$. For $1\leq a\leq 2^{k}-1$, let $P_{a}$ denote the subset $\{q_{1},q_{2},\dots,q_{n}\}$ of $P$, where $n=n(P_{a})$ is the number of primes in the subset $P_{a}$. For a fixed $a$ (and hence fixed $P_{a}$), define

[TABLE]

where $\alpha_{i}>0$. The key idea of the proof is to maximize the function $m_{a}$ considered as a function of the real variables $(\alpha_{1},\alpha_{2},\ldots\alpha_{n})$ with respect to the inequality constraint $r_{a}\leq 2g+1$. We let $M_{a}$ denote this maximum. Using the Lagrange multiplier method we see that the function $m_{a}$ attains the maximum $M_{a}$ precisely when $q_{i}^{\alpha_{i}}(1-\frac{1}{q_{i}})=q_{j}^{\alpha_{j}}(1-\frac{1}{q_{j}})$ for all $1\leq i,j\leq n$. Under the above condition, the constraint $r_{a}\leq 2g+1$ gives us $q_{i}^{\alpha_{i}}(1-\frac{1}{q_{i}})\leq\frac{2g+1}{n}$, for any $1\leq i\leq n$. Now

[TABLE]

From this it follows that for $1\leq a\leq 2^{k}-1$,

[TABLE]

Therefore, for $m\in S(g)$, we have

[TABLE]

In the above computation, we have used the fact that for $x>0$, \bigg{(}\frac{2g+1}{x}\bigg{)}^{x} attains the maximum when $x=(2g+1)/e$.

Observing that \displaystyle\prod_{i=1}^{k}\bigg{(}1-\frac{1}{p_{i}}\bigg{)}\geq\frac{1}{2}\frac{2}{3}\bigg{(}\frac{4}{5}\bigg{)}^{\pi(2g+1)-2}, we have

[TABLE]

∎

Corollary 3.2**.**

For $g\in\mathbb{N}$, $f(g)\leq h(g)\leq 3e^{3g}$.

Proof.

For $m\in S(g)$, we have $m\leq 3e^{3g}$. The result follows. ∎

Remark 3.3*.*

Upper bound for $S(g)$ for $g\geq 1486$: The bound obtained in theorem 3.1 is an absolute upper bound for $S(g)$. For $g\geq 1486$ , we can improve the above upper bound as follows: Using proposition 2.4, we get

[TABLE]

Therefore it follows that for $m\in S(g)$, we have

[TABLE]

3.0.2. Growth of $f(g)$ and $h(g)$

In the previous section, we computed an upper bound for the functions $f(g)$ and $h(g)$. In this section we show that $f(g)$ and $h(g)$ have at least exponential growth.

Lemma 3.4**.**

For $x\geq 23$, we have

[TABLE]

where the sum is over all primes $p\leq x$.

Proof.

Let $n$ be such that $p_{n}\leq x<p_{n+1}$, where $p_{n}$ denotes the $n^{th}$ prime number. It follows from proposition 2.2, that for $x\geq 23$, we have

[TABLE]

∎

Before we proceed further, we set up some notation which we need in the following results.

Let $K(\geq e)\in\mathbb{N}$ be such that for $\sqrt{K\log K}\geq 23$.

Lemma 3.5**.**

For $g\geq K$, $\pi(\sqrt{g\log(g)})<\frac{3\sqrt{g\log(g)}}{\log(g\log(g))}.$

Proof.

For $y>1$, we have \pi(y)<\frac{y}{\log(y)}\bigg{(}1+\frac{3}{2\log(y)}\bigg{)} (see proposition 2.3). Using this estimate we get,

[TABLE]

∎

Lemma 3.6**.**

Let $x=\sqrt{g\log(g)}$ and $\displaystyle m=m(g)=\prod_{p\leq x}p$. Then for $g\geq K$, we have $m\in S(g)$.

Proof.

By proposition 2.1, it is enough to show that $\beta=\displaystyle\sum_{2\neq p\leq x}(p-1)\leq 2g$. Using lemma 3.4 and lemma 3.5 , we have

[TABLE]

∎

For $g\geq K$, let $A(g)=\{p\in\mathbb{N}\mid p\leq\sqrt{g\log(g)}\}$ and $m=m(g)$ be as above. If $d$ is any divisor of $m$, then it is easy to see that $d\in S(g)$. Also it is clear that the divisors $d$ of $m$ are in bijection with the number of subsets of $A(g)$. Since any divisor $d$ of $m$ is an element in $S(g)$ and the number of divisors correspond bijectively with subsets of $A(g)$, it follows that $f(g)=|S(g)|\geq 2^{\pi(\sqrt{g\log(g)})}$ (since number of subsets of $A(g)=2^{\pi(\sqrt{g\log(g)})}$).

We will now show that $|S(g)|>e^{\frac{1}{4}\sqrt{\frac{g}{\log(g)}}}$ from which it follows that the function $f(g)=|S(g)|$ has at least exponential growth.

Theorem 3.7**.**

Let $L\in\mathbb{N}$ such that $\sqrt{L\log L}\geq 55$. Then $f(g)=|S(g)|>e^{\frac{1}{4}\sqrt{\frac{g}{\log(g)}}}$ for all $g\geq L$.

Proof.

From proposition 2.5, we have for all $g\geq L$,

[TABLE]

From this it follows that for all $g\geq L$, we have

[TABLE]

∎

Corollary 3.8**.**

Let $L\in\mathbb{N}$ be as in the above theorem. Then $h(g)>e^{\frac{1}{4}\sqrt{\frac{g}{\log(g)}}}$ for all $g\geq L$.

Proof.

Since $h(g)\geq f(g)$, the result follows. ∎

Remark 3.9*.*

For $g\log g\geq(599)^{2}$, we can improve the above lower bound $e^{\frac{1}{4}\sqrt{\frac{g}{\log g}}}$ to $e^{\sqrt{\frac{g}{4\log g}}}$ by using proposition 2.3.