Diameters of Lens spaces

Question

Fix coprime integers $p,q>0$. The discrete, finite group $\mathbb{Z}/p\mathbb{Z}$ operates on $S^3\subseteq\mathbb{C}^2$ (considered as a Riemannian manifold with the induced metric) via $(k+p\mathbb{Z}).(z_1,z_2)=(e^{2\pi ik/p}z_1,e^{2\pi ikq/p}z_2)$. It is readily checked that this is a free, proper action by isometries, so the quotient $L(p,q)=S^3/(\mathbb{Z}/q\mathbb{Z})$ carries a natural smooth and Riemannian structure, so that the projection $\pi\colon S^3\rightarrow L(p,q)$ becomes a Riemannian covering map. My question is what the diameter $\mathrm{diam}(L(p,q))$ is.

Since $\pi$ is a covering map, any piecewise smooth curve in $L(p,q)$ lifts to a piecewise smooth curve in $S^3$, which moreover has the same length since $\pi$ is a local isometry. It follows that $\mathrm{diam}(L(p,q))\le\mathrm{diam}(S^3)=\pi$. On the other hand, any piecewise smooth curve connecting $\pi(1,0)$ and $\pi(0,1)$ lifts to a piecewise smooth curve connecting $(1,0)$ and a point of the form $(0,e^{2\pi ik/p})$ for some $k\in\mathbb{Z}$. These points are orthogonal, hence all have distance $\pi/2$. It follows that $\mathrm{diam}(L(p,q))\ge\mathrm{dist}(\pi(1,0),\pi(0,1))\ge\pi/2$. In total, we have the potentially crude bounds $\pi/2\le\mathrm{diam}(L(p,q))\le\pi$.

In principle, $\mathrm{dist}([z],[w])=\mathrm{dist}(z,\{w,1.w,\dots,(p-1).w\})$ for $z,w\in S^3$ (the same lifting argument gives $\ge$ and the reverse inequality follows from the completeness of $S^3$) and the geodesics on $S^3$ are well-known, so it doesn't appear impossible to calculate this "by hand", but the formulas seem to get tedious quickly and I'm hoping there's a more conceptual approach.

Answer 1

The diameter of $L(p; q)$ is exactly $\pi/2$. There might be a more direct way to see this but it follows from Morse theory with the Riemannian distance function, see Gromov's "Curvature, diameter and Betti numbers".

Given a Riemannian manifold $M$ and a point $x \in M$, the analogue of a Morse function is the distance $f : M \to \Bbb R$, $f(z) = d(x, z)$. A point $y \neq x $ is said to be a critical point if for every $0 \neq v \in T_y M$, there exists a minimal geodesic $\gamma$ joining $x$ and $y$ such that angle between $v$ and $\gamma$ is at most $\pi/2$. The intuiton is that if $y \neq x$ is a regular point there is a neighborhood $U$ of $y$ in $x$ and a nonzero vector field $X$ on $U$ such that $X(z)$ and a minimizing geodesic $[z, x]$ make an angle of $> \pi/2$, i.e., $\langle X, \gamma' \rangle < 0$ where $\gamma$ are the minimal geodesics emanating from $x$ to points in $U$ i.e., $f$ strictly decreases along the flowlines of $X$. In this situation $X$ is termed a "gradient-like vector field" in Morse theory language.

In this context we get isotopy lemma similar to the one in Morse theory; if $B \subset B' \subset M$ are a pair of concentric geodesic balls around $x$ such that $\overline{B' \setminus B}$ does not contain any critical points of $f$ then obtain an isotopy along the flowlines of the aforementioned gradient-like vector field which shrinks $B'$ to $B$; aka, critical points keep track of change in topology as we fill the Riemannian manifold by larger and larger geodesic balls.

This is apparently a theorem of Groves-Shiohama: $M$ is a (necessarily compact, by Myers) Riemannian manifold of curvature $K \geq 1$ with diameter $d(M) > \pi/2$ then $M$ is homeomorphic to a sphere.

The proof is a short application of the machinery above. Pick $x \in M$, consider $f : M \to \Bbb R$, $f(z) = d(x, z)$ as above. Let $y \in M$ be a point where maximum is achieved; this is in fact a critical point of $f$ as else we can find a gradient-like vector field near this point along the flowlines of which $f$ strictly decrease, but going back along the flowline of $y$ will give a point on which $f$ takes more value, contradiction. I claim $f$ has no other critical points. Indeed, suppose $z \in M$, $z \neq x, y$. Then applying Toponogov comparison theorem to the triangle $[x, y], [x, z], [y, z]$ we obtain that the angle at the hinge at $z$ is obtuse, hence $z$ is regular. We can prove $M \cong S^n$ in an analogous way as in Morse theory; pick geodesic balls $B_1, B_2 \subset M$ around $x$ and $y$ respectively; then $M \setminus B_2$ is isotopic to $B_1$ by flowling along the gradient-like vector field furnished by the above discussion as there are no more critical points in-between. Pasting pairs of balls about the boundary sphere gives a topological sphere, which concludes the theorem.

You have already obtained $L(p; q)$ has diameter at least $\pi/2$, so combined with the above theorem this proves $d(L(p; q)) = \pi/2$.