ABELIAN SUBGROUPS OF THE FUNDAMENTAL GROUP OF A SPACE WITH NO CONJUGATE POINTS

A locally simply connected length space $X$ with universal cover $\pi:\hat{X}\to X$ has no conjugate points if any two points in $\hat{X}$ can be joined by a unique geodesic. Let $X$ be a compact and locally simply connected length space with no conjugate points and finite Hausdorff dimension $d$. In the Riemannian case, it has been believed for some time that Abelian subgroups of $\pi_{1}(X)$ must be finitely generated; for example, this is stated in [2], although the argument there contains a gap. It will be shown here that each Abelian subgroup is isomorphic to $\mathbb{Z}^{k}$ for some $0\leq k\leq d$.

It was proved by Yau [11] in the case of nonpositive curvature, and O’Sullivan [10] for no focal points, that every solvable subgroup of the fundamental group is a Bieberbach group. Croke–Schroeder [2] mapped out a way to generalize this to spaces with no conjugate points: If a torsion-free solvable group has the property that its Abelian subgroups are all finitely generated and straight, then it must be a Bieberbach group. Lebedeva [7] showed that finitely generated Abelian subgroups of the fundamental group of a compact and locally simply connected length space with no conjugate points must be straight. Combining this with Theorem 1 completes the argument set out by Croke–Schroeder.

Since the exponential map at each point of its universal cover is a diffeomorphism, a Riemannian manifold with no conjugate points must be aspherical. It’s worth pointing out that this condition isn’t enough to guarantee the conclusion of Theorem 1, as Mess [9, 8] showed that for each $n\geq 4$ there exists a compact manifold with universal cover $\mathbb{R}^{n}$ whose fundamental group contains a divisible Abelian subgroup, which cannot be finitely generated.

The second section contains a short proof of Theorem 1. The third section gives a different proof in the Riemannian setting, based on a property of Riemannian norms satisfied by the asymptotic norm of each Abelian subgroup of the fundamental group.

Fix $\hat{p}\in\hat{X}$ and a basepoint $p=\pi(\hat{p})$ for $\pi_{1}(X)$. Overloading notation, each $\gamma\in\pi_{1}(X)$ will be identified with the corresponding deck transformation of $\hat{X}$. Let $\Gamma$ be an Abelian subgroup of $\pi_{1}(X)$, in which the group operation is written additively, and suppose $\sigma_{1},\ldots,\sigma_{k}\in\Gamma$ are linearly independent. Denote by $G$ the subgroup generated by the $\sigma_{i}$. The following are proved in [7]: On $\pi_{1}(X)$, the function

is positively homogeneous over $\mathbb{Z}$. It is bounded below on $\pi_{1}(X)\setminus\{e\}$ by $\mathrm{sys}(X)$, the length of the shortest nontrivial geodesic loop in $X$, so $\pi_{1}(X)$ is torsion free. Its restriction to $\Gamma$ satisfies the triangle inequality, and, with respect to the isomorphism $G\cong\mathbb{Z}^{k}$ that takes each $\sigma_{i}$ to the $i$-th standard basis vector, $|\cdot|_{\infty}$ extends to a norm $\|\cdot\|_{\infty}$ on $\mathbb{R}^{k}$.

Denote by $\|\cdot\|$ the Euclidean norm on $\mathbb{R}^{k}$. From the identifications $G(\hat{p})\cong G\cong\mathbb{Z}^{k}$, $G(\hat{p})$ inherits the coordinate functions $\rho_{1},\ldots,\rho_{k}$ on $\mathbb{Z}^{k}$. Since $\|\cdot\|_{\infty}$ is a norm on $\mathbb{R}^{k}$, there exists $C>0$ such that

for all $u\in\mathbb{R}^{k}$. The number $C$ is a Lipschitz constant for the $\rho_{i}$ on $G(\hat{p})$, and, as in the proof of Kirszbraun’s theorem [5], the functions

are Lipschitz extensions of the $\rho_{i}$ to $\hat{N}$. Each $f_{i}$ is $(G,\mathbb{Z})$-equivariant, in the sense that $f_{i}(\gamma(\hat{x}))-f_{i}(\hat{x})\in\mathbb{Z}$ for all $\hat{x}\in\hat{N}$ and all $\gamma\in G$.

The map $f=(f_{1},\ldots,f_{k}):\hat{N}\to\mathbb{R}^{k}$ is Lipschitz, and $f(\gamma(\hat{x}))-f(\hat{x})\in\mathbb{Z}^{k}$ for all $\hat{x}\in\hat{N}$ and all $\gamma\in G$. By construction, $f(G(\hat{p}))=\mathbb{Z}^{k}$. Since $G$ is Abelian, there exists a map $\phi:\mathbb{T}^{k}\to X$ such that $\phi_{*}(\pi_{1}(\mathbb{T}^{k}))\cong G$. Lift $\phi$ to a map $\hat{\phi}:\mathbb{R}^{k}\to\hat{N}$. The composition $f\circ\hat{\phi}:\mathbb{R}^{k}\to\mathbb{R}^{k}$ descends to a map $\mathbb{T}^{k}\to\mathbb{T}^{k}$ with surjective induced homomorphism, so by degree theory it must be surjective. Thus $f$ is surjective. Since a Lipschitz map cannot increase Hausdorff dimension, $k\leq d$.

It follows that $\Gamma$ has rank at most $d$. If it has rank zero, then the result is trivial. Without loss of generality, suppose it has rank $k>0$. For any $\gamma\in\Gamma$, there exist $n,a_{1},\ldots,a_{k}\in\mathbb{Z}$ such that $n\gamma=\sum_{i=1}^{k}a_{i}\sigma_{i}$. It is well known that the function $F:\Gamma\to\mathbb{Q}^{k}$ defined by $F(e)=(0,\ldots,0)$ and

for $\gamma\neq e$ is a well-defined and injective homomorphism, so $F$ is an isomorphism onto its image $\Gamma_{0}$. This map satisfies

for any $\gamma\neq e$. For any distinct $q_{0},q_{1}\in\Gamma_{0}$, there exist distinct $\gamma_{0},\gamma_{1}\in\Gamma$ such that $F(\gamma_{i})=q_{i}$ for each $i$. For $c=1/C$, one has that

Thus $\Gamma_{0}$ is a discrete subgroup of $\mathbb{R}^{k}$, and, consequently, $\Gamma\cong\mathbb{Z}^{k}$.

For simplicity, it will be assumed in this section that $X$ is a smooth $d$-dimensional Riemannian manifold, although what follows holds when $X$ is $C^{r}$ for some $r$ depending on $d$. As before, let $G$ be an Abelian subgroup of $\pi_{1}(X)$ generated by linearly independent $\gamma_{1},\ldots,\gamma_{k}$. The key step in the proof of Theorem 1 is the construction of a $(G,\mathbb{Z}^{k})$-equivariant map $f:\hat{X}\to\mathbb{R}^{k}$ such that $f(G(\hat{p}))=\mathbb{Z}^{k}$. When $X$ is Riemannian, another such map may be constructed using a nondegenerate collection of Busemann functions.

An important theorem of Ivanov–Kapovitch [4] states that, whenever $\alpha_{1},\alpha_{2}\in\pi_{1}(X)$ commute, the change in the Busemann functions of axes of $\alpha_{2}$ under the action of $\alpha_{1}$ is constant on $\hat{X}$. This was previously proved by Croke–Schroeder [2] for analytic $X$. Thus one may define a function $B:G\times G\to\mathbb{R}$ by setting $B(\alpha_{1},\alpha_{2})$ equal to that change.

Because $B(\alpha,\alpha)=|\alpha|_{\infty}^{2}$ for all $\alpha\in G$, one might hope to show that $B$ extends to an inner product and, consequently, that $\|\cdot\|_{\infty}$ is Riemannian. In fact, $B$ satisfies a number of the properties of an inner product: It is linear over $\mathbb{Z}$ in the first slot (see Corollary 4.2 of [4]), $B(\alpha_{1},n\alpha_{2})=nB(\alpha_{1},\alpha_{2})$ for all $n\in\mathbb{Z}$, and it satisfies a version of the Cauchy–Schwarz inequality,

with equality if and only if $\alpha_{1}$ and $\alpha_{2}$ are rationally related. It follows that $B$ extends to an inner product if and only if it is symmetric, but it’s far from clear that symmetry holds in general (cf. [1]). Regardless, $B$ also resembles an inner product in the following way.

The proof of Lemma 3 is by induction. When $m=1$, the conclusion holds with $\alpha_{1}=\gamma_{1}$. Suppose the conclusion holds for some $1\leq m<k$. If the conclusion fails when $\alpha_{m+1}=\gamma_{m+1}$, then there exists a nonzero $c=(c_{1},\ldots,c_{m+1})$ in the null space of the $(m+1)\times(m+1)$ matrix $[B(\alpha_{j},\alpha_{i})]$. The following lemma then completes the inductive step.

When $m=2$ in Lemma 3, inequality (1) implies that one may take $\alpha_{1}=\gamma_{1}$ and $\alpha_{2}=\gamma_{2}$. When $X$ has no focal points, one may, by the flat torus theorem, take $\alpha_{i}=\gamma_{i}$ for all $i$. However, in the general case for $m\geq 3$, there is no apparent local structure that forces the Busemann functions of the axes of the $\gamma_{i}$ to have linearly independent gradients, and it is not clear that the conclusion of Lemma 3 holds with $\alpha_{i}=\gamma_{i}$ for all $i$.