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What is an interesting class of examples of hyperbolic 3-manifolds, each of which satisfies the following conditions?

1. It is compact

2. Its trace field contains a unique imaginary quadratic extension.

3. Its quaternion algebra is isomorphic to one of the form $\Big(\frac{a,b}{K}\Big)$, where $a,b\in K\cap\mathbb{R}$.

Since the word 'interesting' is not well-defined, I'll settle for any examples, but it would be cool if they have some nice combinatorial or geometric characterization.

Then there is a follow-up question (more for algebraic number theorists). How could I replace conditions 2 and 3 with something in terms of the algebra's ramification set? That is, there is a number field $F=K\cap\mathbb{R}$ so that the algebra is isomorphic to $\Big(\frac{a, b}{F(\sqrt{-d})}\Big)$ where $a,b\in F$, and $d\in F^+$. But yet it is still a division algebra (else the manifold is most likely not compact). Can this be (at least for some $F, d$ choices) phrased in terms of the divisors of $a$ and $b$?

I expect one would need to already know the definitions involved to answer this, but I will supply the arithmetic ones below for the sake of readers. Afterall the more people who know this, the more people I have to talk to!


To a hyperbolic 3-manifold $M$ is associated a Kleinian group $\Gamma\cong\Pi_1(M)$ represented in $\mathrm{PSL}_2(\mathbb{C})$. The trace field of $M$ is the field $\mathbb{Q}(\{\mathrm{tr}(\gamma)\mid\gamma\in\Gamma\})$, which I'll denote by $k_0 M$. Using the character variety and algebraic geometry, it follows that this is a number field (a finite extension of $\mathbb{Q}$), and by Mostow rigidity it is a manifold invariant.

Using the same setup, the quaternion algebra of $M$ is defined as $$\big\{\sum_{i=t}^nt_i\gamma_i\mid t_i\in k_0M,\gamma_i\in\Gamma,n\in\mathbb{N}\big\}$$ and is commonly denoted by $A_0M$. A quaternion algebra is a 4-dimensional central-simple algebra, and it can be proven that $A_0M$ is such a thing using the Skolem-Noether theorem. This is a stronger manifold invariant. These algebras (provided the field is not characteristic 2, which doesn't matter here since we're using number fields) necessarily take the following form. If $K$ is the field it is over, then the algebra looks like $$K\oplus Ki\oplus Kj\oplus Kij$$ where $i^2=a, j^2=b, ij=-ji$, with $a,b\in K\setminus\{0\}$. And we denote this by the Hilbert symbol $\Big(\frac{a,b}{K}\Big)$. A property of these algebras is that they are identified up to isomorphism by ramification of their places (field embeddings and prime ideals) over $K$. Quaternion algebras of non-compact manifolds never have ramification over their primes. Quaternion algebras of compact manifolds typically do have ramification of their primes, but there are some strange examples where they don't.

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  • $\begingroup$ Let me mention that I'm writing my thesis right now and have a bunch of cool results for things that satisfy these conditions. Thanks to @BenLinowitz below, the results can be expressed much more neatly. I will post a link in the comments to relevant papers on the archive once they are there (next few months probably). $\endgroup$ – j0equ1nn Oct 23 '15 at 23:32
  • $\begingroup$ I'm glad that my response was useful to you. While it is implicit in my response, I should add that there are infinitely many manifolds satisfying your conditions. In a moment I will add a paragraph making this explicit. $\endgroup$ – user1073 Oct 24 '15 at 1:32
  • $\begingroup$ @BenLinowitz: I noticed what you said about the infinite class of examples, but thanks for including the explanation. It's nice for more people to see how quaternion orders play in to arithmetic groups like this. $\endgroup$ – j0equ1nn Oct 24 '15 at 20:51
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Suppose that $M$ is a compact arithmetic hyperbolic $3$-manifold which is derived from a quaternion algebra. Let $k$ denote the trace field of $M$ and $B=\left(\frac{a,b}{k}\right)$ be the associated quaternion algebra.

It is known (see Maclachlan-Reid, Theorem 8.3.2) that $k$ has a unique complex place. Noting that every proper subfield of a number field with a unique complex place is totally real, we see that the only way for $k$ to contain an imaginary quadratic field is for $k$ to actually be an imaginary quadratic field.

Furthermore, Theorem 8.2.3 of Maclachlan-Reid implies that $M$ will be compact so long as $B\neq \mathrm{M}_2(\textbf{Q}(\sqrt{-d}))$. Combining this with the observation from the previous paragraph, we see that your conditions 1 and 2 will be satisfied if and only if $B$ is a quaternion division algebra which is defined over an imaginary quadratic field.

Now we need to incorporate your third condition, which seems to me to be the most interesting. Suppose that $M$ contains an immersed totally geodesic surface. Then the results of Section 9.5 of Maclachlan-Reid imply that there is an indefinite quaternion algebra $B'$, defined over $\textbf{Q}$, such that $B\cong B'\otimes_\textbf{Q} k$. Because $\left(\frac{a,b}{\textbf{Q}}\right)\otimes_\textbf{Q} k\cong \left(\frac{a,b}{k}\right)$, the Hilbert symbol of $B$ satisfies your condition 3.

Putting all of this together, we see that if $M$ is an arithmetic hyperbolic $3$-manifold which is derived from a quaternion division algebra $B$ defined over an imaginary quadratic field and $M$ contains an immersed totally geodesic surface then $M$ satisfies your three conditions. An arithmetic hyperbolic $3$-manifold contains one totally geodesic surface if and only if it contains infinitely many commensurability classes of totally geodesic surfaces (this also follows from the results of Maclachlan-Reid, Section 9.5), so the aforementioned class of manifolds all contain infinitely many commensurability classes of totally geodesic surfaces. Whether this makes these manifolds interesting...I can't say.

I do not know of a geometric characterization of your third condition, though would like to point out Proposition 5 of Chinburg and Reid's paper Closed hyperbolic $3$-manifolds whose closed geodesics all are simple:

Proposition: Let $M$ be an arithmetic hyperbolic $3$-manifold derived from a quaternion algebra $B=\left(\frac{a,b}{k}\right)$. If $M$ has a non-simple closed geodesic then $a,b$ can be chosen so that $a\in k$ and $b\in k\cap \textbf{R}$.

This proposition shows that if the Hilbert symbol of $B$ cannot be written in a certain form then all closed geodesics of $M$ are simple. (Providing examples of such manifolds was the point of Chinburg and Reid's paper.) It is not known whether this Hilbert symbol obstruction is the only thing which prevents $M$ from having non-simple closed geodesics. If you believe that it is the only obstruction, then you would expect any arithmetic hyperbolic $3$-manifold $M$ satisfying your conditions to have lots of non-simple closed geodesics.

Regarding your request to reinterpret condition 3 in terms of the primes which ramify in $B$, the results of Section 4 of the Chinburg-Reid paper should be very helpful and are meant to provide exactly such an interpretation.

Added: There are infinitely many hyperbolic 3-manifolds which satisfy the OP's three conditions. One such family may be obtained as follows. Let $B$ be a rational quaternion division algebra which is split at the real place of $\bf{Q}$. Let $k$ be an imaginary quadratic field which does not embed into $B$. Then $A:=B\otimes_{\textbf{Q}} k$ is a quaternion division algebra over $k$. (Here I have used the fact that a quadratic field $L$ embeds into $B$ if and only if $B\otimes_\textbf{Q} L \cong \mathrm{M}_2(L)$.) Let $\mathcal O$ be a maximal order of $A$ and $\mathcal{O}^1$ the multiplicative subgroup of $\mathcal{O}^*$ generated by elements of reduced norm $1$. Let $\psi: A\hookrightarrow \mathrm{M}_2(\textbf{C})$ be the map induced by the inclusion $A\hookrightarrow A\otimes_k \textbf{C}\cong \mathrm{M}_2(\textbf{C})$. Finally, let $\Gamma_\mathcal{O}$ denote the image in $\mathrm{PSL}_2(\textbf{C})$ of $\mathcal{O}^1$ under the map $\psi$ composed with the projection $P: \mathrm{SL}_2(\textbf{C})\rightarrow \mathrm{PSL}_2(\textbf{C})$. Then $\Gamma_\mathcal{O}$ is a discrete subgroup of $\mathrm{PSL}_2(\textbf{C})$ of finite covolume which is cocompact and has trace field $k$. Let $\Gamma$ be a finite index subgroup of $\Gamma_\mathcal{O}$ which is torsion-free and $M=\textbf{H}^3/\Gamma$ be the corresponding hyperbolic $3$-manifold. The arguments that I gave in my original response show that $M$ satisfies the three conditions of the OP's question. There are infinitely many commensurability classes of such $M$ because there are infinitely many imaginary quadratic fields $k$ which do not embed into $B$.

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  • $\begingroup$ I had a short conversation with Alan Reid where he told me that I was basically thinking about examples with immersed totally geodesic surfaces. I understood what he was saying for about 10 minutes and then lost it, so thanks so much for this. I wonder now about non-arithmetic ones. I shall also try to look at the Chinburg-Reid paper but I don't have access on that link. I will try emailing Reid.. $\endgroup$ – j0equ1nn Oct 23 '15 at 2:44

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