Which lens spaces admit degree -1 maps?

Question

Let $p$ and $q$ be relatively prime and let $\mathbb{Z}/p$ act on $S^3 \subset \mathbb{C}^2$ via $1 \times (z_1,z_2) = (e^{\frac{2\pi i }{p}} z_1, e^{\frac{2 \pi i q}{p}} z_2)$. The quotient of $S^3$ by this action is the lens space $L(p,q)$. My question is: for which $p$ and $q$ does $L(p,q)$ admit a map $f : L(p,q) \to L(p,q)$ of degree $-1$?

Answer 1

Let $m, q_0, q_1, \dots, q_n$ be positive integers with $(m, q_i) = 1$ for all $i$. We then define the lens space $L(m; q_0, q_1, \dots, q_n)$ to be the quotient of $S^{2n+1} \subset \mathbb{C}^{n+1}$ by the $\mathbb{Z}_m$ action generated by the map $(z_0, z_1, \dots, z_n) \mapsto (e^{2\pi i q_0/m}z_0, e^{2\pi i q_1/m}z_1, \dots, e^{2\pi i q_n/m}z_n)$.

Some facts about lens spaces:

• If $q_0', q_1', \dots, q_n'$ is another set of integers coprime to $m$ such that $q_i' \equiv q_i \bmod m$, then $L(m; q_0, q_1, \dots, q_n) = L(m; q_0', q_1', \dots, q_n')$.

• The lens space $L(m; q_0, q_1, \dots, q_n)$ has dimension $2n + 1$.

• The fundamental group of $L(m; q_0, q_1, \dots, q_n)$ is $\mathbb{Z}_m$.

• If $(m, r) = 1$, then $L(m; rq_0, rq_1, \dots, rq_n)$ is homeomorphic to $L(m; q_0, q_1, \dots, q_n)$.

As $q_0$ is coprime to $m$, there is an integer $r$ such that $rq_1 \equiv 1 \bmod m$, so we can always take lens spaces to be of the form $L(m; 1, q_1, \dots, q_n)$. For three-dimensional lens spaces (i.e. when $n = 1$), one usually denotes $L(m; 1, q)$ by $L(m; q)$. In as similar vein, in his paper Mappings of Manifolds and Notion of Degree, Olum uses the notation $L(m; q_1, \dots, q_n)$ in the following theorem (theorem V in the paper) to denote the $(2n + 1)$-dimensional lens space $L(m; 1, q_1, \dots, q_n)$.

Theorem: For each integer $k$, $0 \leq k < m'$, satisfying $km \equiv 0 \bmod m'$, there is a homomorphism $\theta_k : \Gamma \to \Gamma'$ determined by $\theta_k(\gamma) = \gamma'^k$, and this exhausts the possible homomorphisms. The degrees of the mappings $L(m; q_1, \dots, q_n) \to L(m'; q_1', \dots, q_n')$ inducing $\theta_k$ are all of the integers congruent to

$$k\left(\frac{km}{m'}\right)^nq_1'\dots q_n'q_1^{-1}\dots q_n^{-1}$$

modulo $m'$.

Here $\Gamma$ and $\Gamma'$ are the fundamental groups of $L(m; q_1, \dots, q_n)$ and $L(m'; q_1', \dots, q_n')$ respectively, with generators $\gamma$ and $\gamma'$. When $m' = m$ and $q_i' = q_i$, we have the following corollary:

Corollary: The lens space $L(m; q_1, \dots, q_n)$ admits a self-map of degree $d$ if and only if $d \equiv k^{n+1} \bmod m$ for some $k$.

In particular, a $(2n + 1)$-dimensional lens space with fundamental group $\mathbb{Z}_m$ admits a degree $-1$ map if and only if $-1 \equiv k^{n+1} \bmod m$ for some $k$. This is always the case for $n$ even (just take $k = -1$). When $n = 1$, which is the case you asked about, we see that $L(m; q)$ admits a degree $-1$ map if and only if $-1$ is a quadratic residue modulo $m$.

So your question is equivalent to the following question: for which $m$ is $-1$ a quadratic residue modulo $m$?

First note that $-1$ is a quadratic residue modulo $2$ because $-1 \equiv 1^2 \bmod 2$. On the other hand, if $-1 \equiv k^2 \bmod 2^l$ then $k$ must be odd so $k^2 \equiv 1 \bmod 4$ and hence $k^2 - 1$ is divisible by $4$, but then $k^2 + 1$ is not divisible by $4$ nor any higher power of $2$, so $-1$ is not a quadratic residue modulo $2^l$ for $l > 1$.

If $p$ is an odd prime, then $-1$ is a quadratic residue modulo $p^j$ if and only if $-1$ is a quadratic residue modulo $p$ if and only if $p \equiv 1 \bmod 4$.

Finally, $-1$ is a quadratic residue modulo $m$ if and only if it is a quadratic residue modulo $p^i$ for every prime power $p^j$ which divides $m$. Therefore $-1$ is a quadratic residue modulo $m$ if and only if $m$ has at most one factor of two and every other prime factor is of the form $4a + 1$. Such numbers are listed here: OEIS A008784.

In conclusion, $L(p; q)$ admits a degree $-1$ map if and only if $p$ is of the above form. In particular, if $p$ is prime, then $p = 2$ or $p \equiv 1 \bmod 4$.

Suppose now that $n$ is odd and $n > 1$. Note that if $-1 \equiv k^{n+1} \bmod m$, then $-1 \equiv (k')^2 \bmod m$ where $k' = k^{(n+1)/2}$, so $-1$ is a quadratic residue modulo $m$ and hence $m$ has to be of the above form. However, this necessary condition is not sufficient. For example, $-1$ is a quadratic residue modulo $5$, but it is not a quartic residue modulo $5$ so lens spaces of the form $L(5; q_1, q_2, q_3)$ do not admit maps of degree $-1$.