# Immersions of the hyperbolic plane

Is it possible to isometrically immerse the hyperbolic plane into a compact Riemannian manifold as a totally geodesic submanifold? Any nice examples?

Edit: Although I did not originally say so, I was looking for injective immersions or at least for immersions that do not factor through a covering onto a compact surface. Thank you for your answers and comments, they've been very helpful.

• I like Ian's answer, and I see that you have accepted it, but I wonder whether you wanted to specify any further properties of your immersion. For example, there is the trivial example of an immersion into a compact manifold as a totally geodesic submanifold by simply regarding the hyperbolic plane as the simply-connected cover of a compact Riemann surface $S$ of genus $g\ge 2$. Isn't that the simplest example satisfying your stated criteria? (If you want the image to be a proper submanifold, simply take the cross product of this example with any compact Riemannian manifold.) Aug 18, 2018 at 11:12
• Ian guessed I was looking for something more like an injective immersion rather than something you would get from a covering map. I should have specified that in the OP. Aug 18, 2018 at 11:28
• To add to Bryant's comment: generically the image of a plane is dense (Ratner, Shah) for compact hyperbolic 3-manifolds and McMullen Mohammedi Oh for non compact. Aug 20, 2018 at 23:18

Yes, it immerses isometrically into certain solvmanifolds. Take an Anosov map of $T^2$, such as $\left[\begin{array}{cc}2 & 1 \\1 & 1\end{array}\right]$. The mapping torus admits a locally homogeneous metric modeled on the 3-dimensional unimodular solvable Lie group. The matrix has two eigenspaces with eigenvalues $\frac{3\pm\sqrt{5}}{2}$, and the suspensions of lines on the torus parallel to these eigenspaces give immersed manifolds modeled on $\mathbb{H}^2$. If the eigenspace line contains a periodic point of the Anosov map, the mapping torus of it will be an annulus. But otherwise it will be an immersed injective totally geodesic hyperbolic plane, which I think is what you're asking for.

• Sorry how could it become an annulus? What will correspond to its boundary? Aug 18, 2018 at 7:14
• @მამუკაჯიბლაძე : I mean an open annulus, or infinite cylinder. Geometrically it will be a quotient of the hyperbolic plane by a translation. Aug 18, 2018 at 13:53

Here is a general construction. Take a non-trivial representation of $H=\text{SL}_2(\mathbb{R})$ into a semisimple Lie group $G$, take $K<G$ a maximal compact subgroup and take $\Gamma<G$ an irreducible cocompact lattice. Endow $X=G/K$ with the standard symmetric space structure and consider the image of $H$ in $X$ which is a totally geodesic hyperbolic space. Its image in $\Gamma\backslash X$ will be a totally geodesic immersion of a hyperbolic plane into a compact Riemannian manifold.

Further, if $H$ is not a factor of $G$, up to Baire generically conjugating $\Gamma$ in $G$, we can get that the image of $H$ will be non-compact and if $X$ is of dimension $\geq 5$ (e.g $G=\text{SO}(5,1)$ or $G=\text{SL}_3(\mathbb{R})$) we can get that the immersion is injective (thanks to Ian Agol for correcting an inaccuracy here in a previous version of my answer).

• I suppose X should have dimension at least 5. Aug 19, 2018 at 2:52
• @Ian, no, $X$ could be of any dimension $n>2$ by taking $G=\text{SO}(n,1)$. In fact, the case $n=3$ could be achieved in a way very closed to your example, replacing $T^2$ with a higher-genus surface and the Anosov map with a pseudo-Anosov one. Aug 19, 2018 at 6:06
• The question was asking for an embedding, not immersion (I interpreted this as injective immersion). The pseudo-Anosov case will give QI embedded planes in the hyperbolic metric, or isometrically embedded ones in the singular solve metric. I don’t think that the singular metric can be perturbed to give isometrically embedded hyperbolic planes. Aug 19, 2018 at 14:02
• In fact, I think one can show that the hyperbolic plane cannot isometrically and injectively immerse in a closed hyperbolic 3-manifold. Aug 20, 2018 at 5:24
• if you don’t mod out by K on the right, then you can get hyperbolic planes immersed. For example, if $G=SL_2R$, then we can identify $\Gamma\backslash G$ with the unit tangent bundle of a hyperbolic surface, and hyperbolic planes to stable or unstable leaves of the Anosov flow, which is analogous to my example of an Anosov flow. But we are taking the orbit by an upper triangular group (which is the hyperbolic plane) instead of $SL_2R$. Aug 20, 2018 at 14:58

This is an attempt to visualize the answer by Ian Agol. I am not sure it is correct. If it is, it must be a fundamental domain for the group action from that answer, and the surfaces - totally geodesic images of various projective plane embeddings.

Definitely the simplest examples are the covering maps from the upperf half space to a compact hyperbolic surface that Bryant mentioned, although these are not injective. A slight variation is to consider the main diagonal embedding $\imath: \mathbb{H}^2 \to \mathbb{H}^2 \times \mathbb{H}^2$ and then use different projections on each factor. For example, the first can be the orbit projection $\pi : \mathbb{H} \to \mathbb{H}^2/G$ where $G \subset PSL(2, \mathbb{R})$ contains no translations in the real axis and the second could be $\pi \circ T$ where $T$ is any of such translations. Then $(\pi, \pi \circ T) \circ \imath$ is one to one.

On the other hand, every compact hyperbolic 3-manifold has plenty of totally geodesic immersed $\mathbb{H}^2$ by just projecting totally geodesics $\mathbb{H}^2 \subset \mathbb{H}^3$, as Bryant pointed it is not clear whether the projection could be made injective.

This should be a comment but I am not able to comment yet :)

• I don't think the 'generic' such hyperbolic plane will be injected into the quotient compact hyperbolic $3$-manifold. More likely, the image will intersect itself nontrivially in a family of disjoint geodesic curves. (For comparison, just consider the case of a geodesic line in a compact Riemann surface of genus $2$. The 'generic' such line will intersect itself (countably) many times. Aug 19, 2018 at 10:18
• Right, I was naive, thanks. The euclidean intuition of an irrational line in the torus is definetly not good here Aug 19, 2018 at 10:26