I realize there are a few different ways of going about proving this, depending on one's background, but there's a particular number theoretic aspect that I am just blanking on, and can't seem to find the info on. (BTW I'm aware of the "well known" result that noncompact arithmetic Kleinian groups are commensurable to Bianchi groups, though cannot seem to find a proof addressing the part bugging me -- so please do not just reference that as the reason).

Let's set up the conventional notation. Let $\Gamma$ be the Kleinian group. Since $\Gamma$ is arithmetic, we have a quaternion algebra $\mathcal{A}$ over a number field $k$, where $k$ has a unique complex place $\rho$, and $\mathcal{A}$ ramifies at all real places of $k$. And we have an order $\mathcal{O}\subset\mathcal{A}$, such that $\Gamma$ is commensurable to $P\rho(\mathcal{O}^1)\subset \text{PSL}_2(\mathbb{C})$.

Let $k_0\Gamma$ and $A_0\Gamma$ be, respectively, the trace field and quaternion algebra of $\Gamma$. Omitting some details, I'm able to show that $k=k_0\Gamma$ and $\mathcal{A}\cong A_0\Gamma$. We also know that, since $\Gamma$ is noncocompact, $\mathbb{H}^3/\Gamma$ has cusps, which contribute parabolic elements to $\Gamma$, and if $p\in\Gamma$ is parabolic, then $p-1\in A_0\Gamma$ is not invertible, hence $A_0\Gamma$ can't be a division algebra. Using some more theory which I'll omit here (but I'm comfortable with that part), then we can get that $A_0\Gamma\cong\mathsf{M}_2(k_0\Gamma)$.

Now here's the number theory part I'm stuck on (which I am kind of embarrassed about, but not enough to not post!)

Suppose that $[k:\mathbb{Q}]>2$. Apparently, this implies that $\mathcal{A}$ would have some ramification. Why? Once this is established, we can use the fact that a split $\mathcal{A}$ never has finite ramification, giving a contradition. Then we'd get that $[k:\mathbb{Q}]\leq2$, but since $k\not\subset\mathbb{R}$, then $[k:\mathbb{Q}]=2$ and we are done.

But why would number field of degree greater than 2 necessarily force the quaternion algebra to ramify? I looked at some examples and observed this is true, but can't seem to understand it as a general property.

Thanks in advance for any help and/or patience.

  • $\begingroup$ I'm referring mainly to the proof of Theorem 8.2.3 of The Arithmetic of Hyperbolic 3-Manifolds by Maclachlan & Reid, on page 259. $\endgroup$
    – j0equ1nn
    Commented Jul 15, 2014 at 3:22
  • $\begingroup$ Incidentally, I think it would be a good topic to study quaternion algebras associated to representations of 3-manifolds. In fact, there should be a tautological one defined over the function field of the character variety. $\endgroup$
    – Ian Agol
    Commented Jul 16, 2014 at 5:50
  • $\begingroup$ Yeah, that's the direction I'm headed for my dissertation. I'm in the process of deciding something in particular to focus on in that vein. $\endgroup$
    – j0equ1nn
    Commented Jul 17, 2014 at 10:11

1 Answer 1


If $\mathcal{A}$ ramifies at a real place, then the group of units cannot contain parabolic elements, because the corresponding Galois conjugate lies in $SU(2)$ (the unit group of the real quaternions). However, a non-cocompact lattice contains unipotent elements (due to the Margulis Lemma, but really the Godement compactness criterion for arithmetic lattices, conjectured by Godement but proved by Borel-Harish-Chandra).

  • $\begingroup$ Okay, so to put it another way, $\mathcal{A}$ must ramify at all its real places, since it is chosen according to the definition of $\Gamma$ being arithmetic. Then as you pointed out, the image of a parabolic can't lie in $\mathcal{H}^1\cong SU(2)$, so it must be the case that $\mathcal{A}$ doesn't have any real places, giving the result. Now that you've made it so clear, that second comment about noncocompact lattices is actually more interesting. $\endgroup$
    – j0equ1nn
    Commented Jul 15, 2014 at 4:58

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