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Let $M$ be a complete Riemannian manifold of finite volume whose sectional curvatures $\kappa$ satisfy $a \leq \kappa \leq 0$ for some $a \leq 0$. Let $\tilde{M}$ be the universal cover of $M$. The space $\tilde{M}$ has a visual boundary $\tilde{M}(\infty)$ (homeomorphic to a sphere). Let $X \subset \tilde{M}(\infty)$ be the set of endpoints of axes of hyperbolic elements of $\pi_1(M)$.

Question : Must $X$ be dense in $\tilde{M}(\infty)$? I assume that the answer is no, but I'm having trouble coming up with counterexamples.

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  • $\begingroup$ This will be true for surfaces (n=2). $\endgroup$ – Ian Agol Apr 24 '13 at 10:38
  • $\begingroup$ Also true for locally-symmetric spaces. The references to check are Gromov-Ballmann-Schroeder and Ballmann's lectures on spaces of nonpositive curvature. $\endgroup$ – Misha Apr 24 '13 at 11:15
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    $\begingroup$ If $M$ has rank one, then the answer is yes, see Theorem III.3.4 in Ballmann's DMV book. $\endgroup$ – Igor Belegradek Apr 24 '13 at 13:28
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    $\begingroup$ I do not know enough of the locally symmetric case to see why the answer is yes (as Misha says), but if so, then it seems you can combine the above to conclude that the answer is always yes, i.e., use rank rigidity to decompose the universal cover as a product of locally symmetric or rank one factors. Fix a point at infinity $z$, pick a tangent vector $v$ in its direction, project it to factors, and approximate the projections by axis of hyperbolic elements, which commute and stabilize the product of axes, which is a flat. Arguing in that flat find a hyperbolic element with axis ending at $z$. $\endgroup$ – Igor Belegradek Apr 24 '13 at 14:14
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    $\begingroup$ @Igor Belegradek : That sounds like a reasonable approach to me, but I also do not know enough about the locally symmetric case. I guess I need to do some more reading. Thanks! $\endgroup$ – Sven Apr 24 '13 at 14:31
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Here is the proof in the case of locally symmetric spaces; you will have to combine it with "rank rigidity theorem" and Igor's comments to get a complete proof.

Let $G$ be a semisimple Lie group, $K\subset G$ a maximal compact subgroup, $B\subset G$ a Borel subgroup; $X=G/K$ is the symmetric space of $G$. Let $r$ denote the real rank of $G$, i.e., the rank of $X$. Let $\Gamma\subset G$ be a lattice. Consider the action of $\Gamma$ on $G/B$, the complete flag manifold of $G$, also known as Furstenberg boundary of $X$. (Geometrically, one thinks of elements of $G/B$ as Weyl chambers at infinity in the Tits boundary of $X$.) The action of $\Gamma$ on $G/B$ is known to be minimal, i.e., every orbit is dense. You should be able to find a proof in Mostow's book on strong rigidity. The other ingredient you need is a theorem of Prasad and Raghunathan (Annals of Math., 1972) that $\Gamma$ contains a semisimple subgroup $\Lambda$ isomorphic to ${\mathbb Z}^r$: Such subgroup stabilizes (unique) maximal flat $F$ in $X$ and acts on $F$ cocompactly. It is elementary that end-points of axes of elements of $\Lambda$ are dense in the sphere $S^{r-1}$, the geometric boundary of $F$ (rational directions are dense in the unit sphere).

Now, we can put all this together: Given a point $\xi$ in the geometric boundary of $X$, let $\sigma$ be a Weyl chamber at infinity containing $\xi$. Let $\tau$ be a chamber at infinity stabilized by the abelian subgroup $\Lambda\subset \Gamma$ as above. Let $\gamma_n\in \Gamma$ be a sequence such that $\lim_n \gamma_n(\tau)=\sigma$. Since ends of axes of elements $\lambda$ of $\Lambda$ are dense in $\tau$, their images under the sequence $\gamma_n$ will accumulate to $\xi$ as well. These images are ideal points of axes of conjugates $\gamma_n \lambda \gamma_n^{-1}$. qed

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