Compact hyperbolic 3-manifolds with prescribed quaternion algebra, quaternion parameters as ramification condition

Question

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!

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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.

Answer 1

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 $B$.