# How many distinct homeomorphism classes of lens spaces are there with a fixed p?

This question is about the topological classification of lens spaces. Fix $p$ a positive integer, not necessarily a prime. From Brody, The topological classification of lens spaces, Annals of Math. (2) 71 pp 163-184, the lens spaces $L(p,q_1)$ and $L(p,q_2)$ are homeomorphic if and only if either $q_1\equiv \pm q_2$ (mod $p$) or $q_1q_2\equiv \pm 1$ (mod $p$). Now the question is, for fixed $p$, how many homeomorphism classes of lens spaces are there with this fixed $p$. In particular, is there a formula $f(p)$ that gives the number of homemorphism classes as a function of $p$.

Example: When $p=7$, then $q$ can be chosen from the set $\{1,2,3,4,5,6\}$ (since any such $q$ is coprime to 7) and we find that $L(7,1)\approx L(7,6)$ (where $\approx$ denotes homeomorphism) and $L(7,2)\approx L(7,3)\approx L(7,4)\approx L(7,5)$ so that there are two distinct homeomorphism classes of lens spaces with $p=7$. Thus $f(7)=2$.

Example: When $p=8$, then $q$ can be chosen from the set $\{1,3,5,7\}$ (since $p$ and $q$ need to be coprime) and then $L(8,1)\approx L(8,7)$ and $L(8,3)\approx L(8,5)$ so in this case there are again two homeomorphism classes of lens spaces with $p=8$. Thus $f(8)=2$.

Remark: For any $p$, the permissible $q$'s can be found and then Brody's formula can be applied to each such pair of $q$'s to determine which are homeomorphic to each other. Thus $f(p)$ can be computed by hand for any $p$.

My question is, is there a formula for $f(p)$?

• This should be a straightforward computation using Burnside's lemma (you're counting orbits under the action of the finite group generated by multiplying by $\pm 1$ and taking inverses), more appropriate for math.SE than here. Feb 2, 2014 at 20:28

This is an interesting question which can be solved by applying Burnside's lemma (as suggested by Qiaochu Yuan in a comment) and some knowledge of the group of units modulo $n$. The relevant facts about the groups of units mostly derive from the Chinese remainder theorem, quadratic reciprocity, and the knowledge that the group of units modulo a power of an odd prime is cyclic. This is all rather elementary group theory and number theory, but I found it to be a very instructional exercise. I have written the details below, partly for my own benefit.\newcommand{\totient}{\varphi}$$\newcommand{\congruent}{\equiv}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\ZZmod}{\ZZ/ #1 \ZZ}$$\newcommand{\set}{\lbrace #1 \rbrace}$$\newcommand{\units}{U(#1)}$$\newcommand{\size}{\mathopen{}\mathclose{\left\lvert #1 \right\rvert}} I will first describe the final answer. Let n > 2 be an integer. The number \psi(n) of distinct homeomorphism classes of 3-dimensional lens spaces whose fundamental group has order n is$$ \psi(n) = \frac14 \bigl( \totient(n) + 2^{l(n) + \epsilon(n)} \bigr) $$where \totient is Euler's totient function. Moreover, \epsilon(n) is 1 if -1 is a quadratic residue modulo n, and zero otherwise. As will be shown below, \epsilon(n) = 1 if and only if 4 does not divide n and each of the odd prime divisors of n is congruent to 1 modulo 4. Finally, describing l requires the prime factorization of n:$$ n = {p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k} $$where the numbers p_i are distinct primes, and the \alpha_i are positive integers. Assume further that p_1 is the smallest prime divisor of n. Then l(n) is given by$$ l(n) = \begin{cases} k & \text{if }\ p_1 \neq 2 \\ k + \min\set{\alpha_1 - 2 , 1} & \text{if }\ p_1 = 2 \end{cases} $$Notation. To maintain a clear distinction between rings and groups, \ZZmod{n} will denote the ring of integers modulo n, where n is a positive integer. On the other hand, C_n will denote a cyclic group with n elements. The group of units in the ring \ZZmod{n} is denoted by \units{\ZZmod{n}}. I will generally use multiplicative notation for the groups appearing in this answer. Applying Burnside's lemma. We follow here the suggestion of Qiaochu Yuan in a comment below the question. Fix an integer n > 2. Consider the group G = C_2 \times C_2, where C_2 = \set{-1,1} with generator -1. Make G act on \units{\ZZmod{n}} in the following way: the generator (-1,1) of the first copy of C_2 acts by taking a number to its negative (the additive inverse in \ZZmod{n}), and the generator (1,-1) of the second copy of C_2 takes a number to its multiplicative inverse. From Brody's classification mentioned in the question, the set of homeomorphism classes of lens spaces with fundamental group of order n is in bijection with the set of orbits for the action of G on \units{\ZZmod{n}}. Let \psi(n) denote the size of the quotient set \units{\ZZmod{n}}/G. The so-called Burnside's lemma implies that$$ 4\, \psi(n) = \size{G} \cdot \psi(n) = \sum_{g\in G} \size{\units{\ZZmod{n}}^g} \tag{1a} $$where \units{\ZZmod{n}}^g denotes the set of fixed points of the action of g on \units{\ZZmod{n}}, and \size{\bullet} returns the size of a set. The fixed point sets corresponding to the first copy of C_2 are quickly determined. The unit (1,1)\in G fixes everything in \units{\ZZmod{n}} and thus contributes$$ \size{\units{\ZZmod{n}}^{(1,1)}} = \size{\units{\ZZmod{n}}} = \totient(n) $$The fixed points for the generator (-1,1)\in G of the first copy of C_2 are the units x \in \units{\ZZmod{n}} such that 2x = 0 in \ZZmod{n}. The existence of such an invertible x entails that 2 \congruent 0 \mod n. Since n>2,$$ \size{\units{\ZZmod{n}}^{(-1,1)}} = 0 $$Square roots modulo n. The fixed points for the remaining elements of G are related to the number of square roots modulo n. For each a\in\units{\ZZmod{n}}, define the set of square roots of a:$$ R(n,a) = \set{ x\in\units{\ZZmod{n}} \mid x^2 = a } $$In particular, the set R(n,a) is non-empty if and only if a is a quadratic residue modulo n. Now, the fixed point set for the elements (1,-1) and (-1,-1) in G consist precisely of the units which square to 1 and -1, respectively:$$ \begin{align*} \units{\ZZmod{n}}^{(1,-1)} &= R(n,1) \\ \units{\ZZmod{n}}^{(-1,-1)} &= R(n,-1) \end{align*} $$Replacing everything back into (1a):$$ 4\, \psi(n) = \totient(n) + \size{R(n,1)} + \size{R(n,-1)} \tag{1b} $$More generally, for any abelian group A and an element a\in A, define the set:$$ R(A,a) = \set{ x \in A \mid x^2 = a } $$Then the sets of square roots modulo n are an instance of this general definition: R(n,a) = R\bigl(\units{\ZZmod{n}},a\bigr). Consider the squaring homomorphism \sigma : A\to A on an abelian group A which sends an element x \in A to its square x^2 — we are using multiplicative terminology here. Then R(A,a) is the fibre of \sigma over a \in A: R(A,a) = \sigma^{-1}(a). In particular, if 1 \in A denotes the identity element of A, R(A,1) is the kernel of the squaring homomorphism, and is therefore an abelian group. The fibres of a group homomorphism are always torsors over the kernel, so R(A,a) is a torsor over the group R(A,1) for any a\in A. Hence, \size{R(A,a)} = \size{R(A,1)} whenever R(A,a) is non-empty. We conclude from equation (1b):$$ 4\, \psi(n) = \totient(n) + \size{R(n,1)}(1 + \epsilon(n)) \tag{1c} $$where \epsilon(n) = 1 if R(n,-1)\neq\emptyset and \epsilon(n) = 0 if R(n,-1)=\emptyset. In other words, \epsilon(n) = 1 if and only if -1 is a quadratic residue modulo n. Square roots and the Chinese remainder theorem. As before, set the prime factorization of n to n = {p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k}, where the numbers p_i are distinct primes, and the \alpha_i are positive integers. According to the Chinese remainder theorem, there is an isomorphism of rings$$ \ZZmod{n} = (\ZZmod{{p_1}^{\alpha_1}}) \times \cdots \times (\ZZmod{{p_k}^{\alpha_k}}) $$determined by the canonical projections from \ZZmod{n} onto each of the factors in the target. Consequently, we deduce an isomorphism between the groups of units:$$ \units{\ZZmod{n}} = \units{\ZZmod{{p_1}^{\alpha_1}}} \times \cdots \times \units{\ZZmod{{p_k}^{\alpha_k}}} $$Since R\bigl(A\times B,(a,b)\bigr) = R(A,a) \times R(B,b), taking the set of square roots of a\in\units{\ZZmod{n}} on each side of the previous expression produces:$$ R(n,a) = R({p_1}^{\alpha_1},a) \times \cdots \times R({p_k}^{\alpha_k},a) \tag{2} $$In each term above, a really stands for the projection of a in the corresponding group. Determining when -1 is a quadratic residue. Recall that a is a quadratic residue modulo m if and only if R(m,a) \neq \emptyset. As a consequence of the isomorphism (2), a \in \units{\ZZmod{n}} is a quadratic residue modulo n if and only if a is a quadratic residue modulo {p_i}^{\alpha_i} for every i\in\set{1,\ldots,k}. The following claim reduces the case of a prime power modulus to the case of a prime modulus. Proposition: Let p be an odd prime, and \alpha a positive integer. A unit a \in \units{\ZZmod{p^\alpha}} is a quadratic residue modulo p^\alpha if and only a is a quadratic residue modulo p. Proof: Consider the projection map \pi : \ZZmod{p^\alpha} \to \ZZmod{p}, which is a ring map. Then \pi^{-1}\bigl(\units{\ZZmod{p}}\bigr) = \units{\ZZmod{p^\alpha}}: an integer m becomes invertible in \ZZmod{p^\alpha} if and only if it is coprime to p^\alpha, which happens precisely when m is not divisible by p and thus invertible in \ZZmod{p}. Hence, we obtain a short exact sequence of groups:$$ 1 \To \pi^{-1}(1) \To \units{\ZZmod{p^\alpha}} \To \units{\ZZmod{p}} \To 1 \tag{ES} $$The group \pi^{-1}(1) = 1 + p(\ZZmod{p^\alpha}) has order p^{\alpha-1}, which is odd. Therefore, the squaring homomorphism on \pi^{-1}(1) is necessarily a self-bijection, and so every element in \pi^{-1}(1) is a square. Chasing around the preceding short exact sequence, it follows that when a \in \units{\ZZmod{p^\alpha}} is a quadratic residue modulo p, it is also a quadratic residue modulo p^\alpha. ■ The first supplement to the law of quadratic reciprocity now implies the following result. Corollary: If p is an odd prime, and \alpha is a positive integer, then -1 is a quadratic residue modulo p^\alpha if and only p \congruent 1 \mod 4. It remains to deal with the case of powers of the prime two. Observe that -1 is a quadratic residue modulo 2. However, -1 is not a quadratic residue modulo 4, and thus cannot be a quadratic residue modulo 2^\alpha, for any \alpha > 1. The following proposition collects what we have learned in this section. Proposition: Given an integer n > 1, -1 is a quadratic residue modulo n — i.e. \epsilon(n) = 1 — if and only if: • each odd prime divisor of n is congruent with 1 modulo 4, • and 4 does not divide n. Counting square roots of unity. To finish the proof, we need to find the number of square roots of the unit in the ring \ZZmod{n}, that is, the size of R(n,1). By (2), this is reduced to calculating the size of R({p_i}^{\alpha_i},1) for each i. We will use the fact that R(C_m,1) \cong C_1 (where 1 \in C_m is the identity element) is trivial if m is odd, and R(C_m,1) \cong C_2 if m is even. Claim: For p an odd prime, and \alpha a positive integer, R(p^\alpha,1) \cong C_2. Proof: This follows from the fact that \units{\ZZmod{p^\alpha}} is a cyclic group of even order equal to p^{\alpha-1}(p-1) (stated on the relevant wikipedia page). Alternatively, it may be deduced from two results stated in the proof of the first proposition from the previous section: the short exact sequence (ES), and the observation that the squaring homomorphism is a bijection on the kernel of that exact sequence. ■ Claim: For \alpha > 2, R(2^\alpha,1) \cong C_2 \times C_2. Moreover, R(4,1) \cong C_2 and R(2,1) \cong C_1. In particular, \size{R(2^\alpha,1)} = 2^{\min\set{\alpha-1,2}} for each positive integer \alpha. Proof: This is a consequence of the following facts (see the wikipedia page): • \units{\ZZmod{2}} \cong C_1, so that R(2,1) \cong C_1; • \units{\ZZmod{4}} \cong C_2, so that R(4,1) \cong C_2 • for \alpha > 2, \units{\ZZmod{2^\alpha}} \cong C_2 \times C_{2^{\alpha-2}}, so that R(2^\alpha,1) \cong C_2 \times C_2. For reference, section 4.5 of Helmut Hasse's book Number theory proves the results, identifying the groups \units{\ZZmod{p^\alpha}}, which were used in this proof and in the previous one. ■ Using the isomorphism (2), these two claims prove that R(n,1) is the product of a certain number of cyclic groups of order 2. We calculate:$$ \size{R(n,1)} = \prod_{i=1}^k \size{R({p_i}^{\alpha_i},a)} = 2^{l(n)} $$Here, l(n) = l({p_1}^{\alpha_1} \cdots {p_k}^{\alpha_k}) is as defined at the beginning:$$ l(n) = \begin{cases} k & \text{if }\ p_1 \neq 2 \\ k + \min\set{\alpha_1 - 2 , 1} & \text{if }\ p_1 = 2 \end{cases} $$where we assume that p_1 is the smallest prime number which divides n. Finally, equation (1c) implies$$ 4\, \psi(n) = \totient(n) + 2^{l(n)}(1+\epsilon(n)) = \totient(n) + 2^{l(n) + \epsilon(n)}$since$\epsilon(n)$only takes the values$0$and$1\$.