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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

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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 FINITE ORDER ELEMENTS IN THE INTEGRAL SYMPLECTIC GROUP KUMAR BALASUBRAMANIAN, M. RAM MURTY, AND KARAM DEO SHANKHADHAR Abstract For g 2 N, let G = Sp(2g; 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 R for all positive integers g. We also study the growth of the functions f(g) = jS(g)j, and h(g) = maxfm 2 N j m 2 S(g)g and show that they have at least exponential growth. 1. Introduction Given a group G and a positive integer m 2 N, it is natural to ask if there exists k 2 G such that o(k) = m, where o(k) denotes the order of the element k. In this paper, we make some observations about the collection of positive integers which can occur as orders of elements in G = Sp(2g; Z). Before we proceed further we set up some notations and briefly mention the questions studied in this paper. Let G = Sp(2g; Z) be the group of all 2g × 2g matrices with integral entries satisfying > A JA = J 0 I where A> is the transpose of the matrix A and J = g g . −Ig 0g α1 αk Throughout we write m = p1 : : : pk , where pi is a prime and αi > 0 for all i 2 f1; 2; : : : ; kg. We also assume that the primes pi are such that pi < pi+1 for 1 ≤ i < k. We write π(x) for the number of primes less than or equal to x. We let ' denote the Euler's phi function. It is a well known fact that the function ' is multiplicative, i.e., '(mn) = '(m)'(n) if m; n are α α 1 relatively prime and satisfies '(p ) = p (1 − p ) for all primes p and positive integer α 2 N. Let S(g) = fm 2 N j 9 A 6= 1 2 G with o(A) = mg: 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. 1 In this paper we show that S(g) is a bounded subset of R for all positive integers g. The bound depends on g. Once we know that S(g) is a bounded set, it makes sense to consider the functions f(g) = jS(g)j, where jS(g)j is the cardinality of S(g) and h(g) = maxfm j m 2 S(g)g, i.e., h(g) is the maximal possible (finite) order of an element in G = Sp(2g; Z). We show that the functions f and h have at least exponential growth. The above question derives its motivation from analogous questions from the theory of mapping class groups of a surface of genus g (see section 2.1 in [4] for the definition). We know that given a closed oriented surface Sg of genus g, there is a surjective homomorphism : Mod(Sg) ! Sp(2g; Z), where Mod(Sg) is the mapping class group of Sg (see theorem 6.4 in [4]). It is a well known fact that for f 2 Mod(Sg)(f 6= 1) of finite order, we have ~ (f) 6= 1. Let S(g) = fm 2 N j 9f 6= 1 2 Mod(Sg) with o(f) = mg. The set S~(g) is a finite set and it makes sense to consider the functions f~(g) = jS~(g)j and h~(g) = maxfm 2 N j m 2 S~(g)g. It is a well known fact that both these functions f~ and h~ are bounded above by 4g + 2 (see corollary 7.6 in [4]). 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. α1 αk Proposition 2.1 (B¨urgisser). Let m = p1 : : : pk , where the primes pi sat- isfy pi < pi+1 for 1 ≤ i < k and where αi ≥ 1 for 1 ≤ i ≤ k. There exists a matrix A 2 Sp(2g; Z) of order m if and only if Xk αi a) '(pi ) ≤ 2g, if m ≡ 2(mod 4). i=2 Xk αi b) '(pi ) ≤ 2g, if m 6≡ 2(mod 4). i=1 Proof. See corollary 2 in [1] for a proof. Proposition 2.2 (Dusart). Let p1; p2; : : : ; pn be the first n primes. For n ≥ 9, we have 1 p + p + ··· + p < np : 1 2 n 2 n Proof. See theorem 1.14 in [2] for a proof. 2 x 1:2762 Proposition 2.3 (Dusart). For x > 1, π(x) ≤ log x 1 + log x . For x 1 x ≥ 599, π(x) ≥ log x 1 + log x . Proof. See theorem 6.9 in [3] for a proof. Proposition 2.4 (Dusart). For x ≥ 2973, Y 1 e−γ 0:2 1 − > 1 − : p log x (log x)2 p≤x where γ is the Euler's constant. Proof. See theorem 6.12 in [3] for a proof. x Proposition 2.5 (Rosser). For x ≥ 55, we have π(x) > log x+2 . Proof. See theorem 29 in [5] for a proof. 3. Main Results In this section we prove the main results of this paper. To be more pre- cise, we prove the following. a) S(g) is a bounded subset of R. b) f(g) = jS(g)j has at least exponential growth. c) h(g) = maxfm j m 2 S(g)g has at least exponential growth. 3.0.1. Boundedness of S(g). In this subsection we show that S(g) is a bounded subset of R. α1 αk Let m = p1 : : : pk 2 S(g). Suppose pi > 2g+1 for some i 2 f1; 2; : : : ; kg. αi αi−1 This would imply that '(pi ) = pi (pi −1) > 2g, which contradicts propo- sition 2.1. It follows that all primes in the factorization of m should be ≤ 2g + 1 and hence k ≤ g + 1. Theorem 3.1. For g 2 N, S(g) is a bounded subset of R. Proof. For g 2 N, fix k = π(2g + 1) and P = fp1; p2; : : : ; pkg be the set of first k primes arranged in increasing order. The prime factorization of any m 2 S(g) involves primes only from the set P . The total number of non- empty subsets of P is 2k − 1. Let us denote the collection of these subsets k of P as fP1;P2;:::P2k−1g. For 1 ≤ a ≤ 2 − 1, let Pa denote the subset fq1; q2; : : : ; qng of P , where n = n(Pa) is the number of primes in the subset 3 Pa. For a fixed a (and hence fixed Pa), define α1 α2 αn ma = ma(α1; : : : ; αn) = q1 q2 : : : qn ; Xn 1 r = r (α ; : : : ; α ) = qαi 1 − ; a a 1 n i q i=1 i where αi > 0. The key idea of the proof is to maximize the function ma considered as a function of the real variables (α1; α2; : : : αn) with respect to the inequality constraint ra ≤ 2g + 1. We let Ma denote this maximum. Using the Lagrange multiplier method we see that the function ma attains the maximum M precisely when qαi (1− 1 ) = qαj (1− 1 ) for all 1 ≤ i; j ≤ n. a i qi j qj Under the above condition, the constraint r ≤ 2g + 1 gives us qαi (1 − 1 ) ≤ a i qi 2g+1 n , for any 1 ≤ i ≤ n. Now qα1 1 − 1 qα2 1 − 1 : : : qαn 1 − 1 1 q1 2 q2 n qn ma(α1; α2; : : : αn) = : Yn 1 1 − q i=1 i From this it follows that for 1 ≤ a ≤ 2k − 1, n n qα1 1 − 1 2g+1 1 q1 n Ma = Yn ≤ : 1 Yk 1 1 − 1 − qi p i=1 i=1 i Therefore, for m 2 S(g), we have m ≤ max Ma 1≤a≤2k−1 n 2g+1 max n 1≤a≤2k−1 ≤ Yk 1 1 − p i=1 i 2g+1 e e ≤ Yk 1 1 − p i=1 i 4 x 2g+1 In the above computation, we have used the fact that for x > 0, x attains the maximum when x = (2g + 1)=e. Yk 1 1 2 4 π(2g+1)−2 Observing that 1 − ≥ , we have p 2 3 5 i=1 i 2g+1 π(2g+1)−2 2g+1 +g−1 3g m ≤ 3(5=4) e e ≤ 3e e ≤ 3e : 3g Corollary 3.2. For g 2 N, f(g) ≤ h(g) ≤ 3e . 3g Proof. For m 2 S(g), we have m ≤ 3e . The result follows. Remark 3.3. Upper bound for S(g) for g ≥ 1486: The bound obtained in theorem 3.1 is an absolute upper bound for S(g). For g ≥ 1486 , we can improve the above upper bound as follows: Using proposition 2.4, we get Yk 1 1 e−γ 1 − > : p 2 log(2g + 1) i=1 i Therefore it follows that for m 2 S(g), we have 2g+1 e e γ 2g+1 m ≤ ≤ 2e log(2g + 1)e e : Yk 1 1 − p i=1 i 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.
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