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I have some question about a special type of hypersurfaces in manifolds. Let $M$ be a compact Riemannian manifold of nonpositive sectional curvature with convex boundary. We call two totally-geodesic immersions $\phi_1: M_1 \rightarrow M$ and $\phi_2: M_2 \rightarrow M$ of closed manifolds $M_1, M_2$ of nonpositive curvature in $M$ parallel, if there is a totally-geodesic embedding $\phi: N \times [a_1, a_2 ] \rightarrow \tilde{M}$ into the universal covering $\tilde{M}$ and covering maps $p_i: N \times \{a_i\} \rightarrow M_i$, which fulfill $\phi_i \circ p_i = \pi \circ \phi \vert_{N \times a_i} $ where $\pi: \tilde{M} \rightarrow M$ denotes the covering projection.

Now I have some questions: Let $F \subset \tilde{M}$ be a closed flat submanifold and denote by $\Gamma(F)$ the subgroup of the group $\Gamma$ of deck translations that fixes $F$. We call $F$ a $\Gamma$-flat if $\Gamma(F)$ acts cocompactly on $F$. Then the following seems to be true, but I don't know why and would be interested in a proof:

  1. If $M$ is not flat, then the set of all flats parallel to a $\Gamma$-flat $F$ in $\tilde{M}$ is isometric to $F \times [a_1, a_2]$ for some compact intervall $[a_1, a_2]$.

  2. If we assume that $[a_1, a_2 ]= [-a, a ]$ then we call the hypersurface $F \times 0$ in $F \times [-a, a ]$ central. Let $F'$ denote the immersed hypersurface in $M$ that is covered by $F$. Prove that every hypersurface parallel to $F'$ is homotopic to $F'$.

I would be very happy if someone could answer these questions!

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Correct TeX errors. –  Misha May 24 '12 at 19:16

1 Answer 1

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First of all, in order for Claim 1 to be true you need to assume that $F$ has codimension 1 in $\tilde{M}$. Second, by "set of all flats" you presumably mean "union of all flats". Now, let's prove Claim 1: Take the union $U$ of all "parallel strips" of the form $F\times [0, a]\subset X$, where $F$ is identified with $F\times 0$; $X=\tilde{M}$. Since the property of being parallel is transitive, $U$ can be exhausted by nested subsets $U_i, i\in {\mathbb N}$, each of which is a "parallel strip." By Arcela-Ascoli theorem, $U$ is closed in $X$ and is isometric to $I\times F$, where $I\subset {\mathbb R}$ is either a closed interval or a ray or $I={\mathbb R}$. The latter is impossible since you are assuming that $X$ is not flat. Suppose that $I$ is a ray. Then pick a sequence of points $x_i\in U$ whose distances $R_i$ to the boundary of $U$ diverge to infinity. Thus, since $M$ is compact, for sufficiently large $i$ the ball $B(x_i, R_i)$ covers the entire manifold $M$ under the projection $X\to M$. Since $U$ is flat, it follows that $M$ is flat as well, contradicting your assumption. This leaves us with $U\cong F\times [a_1,a_2]$ as required.

Claim 2 is false as stated, consider, for, instance, the flat Klein bottle $K=M$ and $F'$ the orientation-reversing simple closed geodesic in $K$. Then there will be orientation-preserving simple geodesic loops $F''\subset K$ which are "parallel" to $F'$ with your definition. However, clearly, $F''$ are not homotopic to $F'$. To get the correct statement, assume that parallel hypersurfaces $F', F''$ are cooriented in $M$ (i.e., have trivial normal bundles). Then stabilizers $\Gamma_{F'}, \Gamma_{F''}$ of the universal covers $\tilde{F}', \tilde{F}''$ of $F'$ and $F''$ in $\Gamma$ are the same and, thus, the hypersurfaces $F', F''$ are homotopic. To see that the stabilizers are the same, note that both stabilizers $\Gamma_{F'}, \Gamma_{F''}$ preserve the parallel set $U$ of $\tilde{F}', \tilde{F}''$. Furthermore, the entire stabilizer $\Gamma_U$ of $U$ in $\Gamma$ contains index 2 subgroup $\Gamma^+_U$ which fixes $a_1, a_2$ as above. Hence, $\Gamma^+_U$ fixes the entire segment $[a_1, a_2]$ and, thus, preserves both $\tilde{F}', \tilde{F}''$. By our assumption that both $F', F''$ are cooriented, thus, both groups $\Gamma_{F'}, \Gamma_{F''}$ are contained in $\Gamma_U$. Hence, they are equal to $\Gamma_U$.

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