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Is there any example of a compact manifold $M$ of dimension $n>10000$ such that

  1. $M$ admits an embedding into $\mathbb R^{n+2}$,

  2. $M$ is hyperbolic; i.e., it admits a Riemannian metric with curvature $\equiv -1$?

More generally, are there some criteria sufficient for existence of embedding of compact manifolds $M^n$ of large dimension in $\mathbb R^{n+2}$?

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As long as it embeds with a trivial normal bundle (which can be arranged by some restrictions on $M$), the manifold's fundamental class would be dual to a bounding $n+1$-manifold in $\mathbb R^{n+2}$. – Ryan Budney Oct 3 '12 at 16:22
Dima: If immersion is enough for you, then every closed hyperbolic n-manifold has a finite cover which admits an immersion in ${\mathbb R}^{n+1}$ (Sullivan+Hirsch+Smale). On the other hand, we do not have any "examples" of hyperbolic n-manifolds for large n, where "example" means a description in terms of triangulation or handle decomposition. They are all described by algebraic means. We do not even have a single hyperbolic n-manifold or orbifold with known 1-st homology group (say, for $n\ge 10^4$). – Misha Oct 3 '12 at 16:25
Also, see for a survey of the embedding theory. Unfortunately, all sufficient conditions for embeddings require high connectivity that hyperbolic manifolds will not have. – Misha Oct 3 '12 at 17:09
I'm just running through the standard argument that co-dimension two knots have Seifert surfaces, but adapted to your case. If $M$ embeds in $\mathbb R^{n+2}$, the complement of the embedding has $H^1 \simeq \mathbb Z$. Represent that cohomology class as a map from the complement to the circle, and take the fiber of a regular value. Provided the normal bundle to the embedded manifold $M$ is trivial we can ensure this is a Seifert surface / bounding manifold for $M$. You could ensure the normal bundle is trivial by making $M$ orientable and $H^2 M$ trivial. – Ryan Budney Oct 3 '12 at 18:16
@Ryan: This is just to note that there are no known examples of hyperbolic $n$-manifolds (for large $n$) with vanishing 2-nd Betti number (or any Betti number for this matter). For all what we know, they all may have nonzero 2nd Betti number. – Misha Oct 3 '12 at 21:06

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