
> 
> UPDATE: Because I was hoping that state the question as concisely as
>  possible, the original post did not include a precise definition of
>  arithmetic 3-manifold only a reference to Maclachlan and Reid's book where it is
>  defined. However, this lead to an ambiguity that caused conflicting
>  answers. With the included definition, Ian Agol's answer is correct.
>  However, given a more restrictive definition of arithmetic 3-manifold, which is 
>  (less standard, but) used in a paper of Gromov and Guth, John Pardon's answer is correct. I
>  apologize for the confusion and appreciate the thoughtful responses of
>  the MO community.
> 
> 
> 


Following a definition was taken from Maclachlan and Reid, ``The Arithmetic of Hyperbolic 3-manifolds'', Chapter 8.2:


Let $k$ be a number field with exactly one complex place and $A$ a quaternion algebra over $k$ which is ramified at all real places. Furthermore assume $A\subset M\_2(\mathbb{C})$ via a complex embedding of $k$. If $\mathcal{O}$ is an order in $A$ and $\mathcal{O}^1$ is the set of elements of $\mathcal{O}$ of determinant 1, then a subgroup $\Gamma$ of $SL(2,\mathbb{C})$ is *arithmetic* if it is commensurable with $\mathcal{O}^1$ and a subgroup $\Gamma \subset PSL(2,\mathbb{C})$ is *arithmetic* if it is commensurable with $P\mathcal{O}^1$.
If we consider the natural action of $PSL(2,\mathbb{C})$ on the upper-half space model of $\mathbb{H}^3$, a finite volume hyperbolic 3-manifold (or 3-orbifold) $Q \cong \mathbb{H}^3/\Gamma$ is *arithmetic* precisely when $\Gamma$ is arithmetic. 


Arithmetic 3-manifolds tend to be atypical many ways. For example, one non-generic property of arithmeticity observed by A.Borel in "[Commensurability classes and volumes of hyperbolic 3-manifolds](http://archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1981_4_8_1/ASNSP_1981_4_8_1_1_0/ASNSP_1981_4_8_1_1_0.pdf)" is the following: for a fixed $V>0$, there are only finitely arithmetic 3-manifolds of volume less than $V$. 


Along those lines, I was wondering about the following: 



> 
> Question: For a fixed Heegaard genus $g$, are there only finitely many arithmetic manifolds of Heegaard genus $<g?$
> 
> 
> 



Heegaard genera of arithmetic 3-manifolds
The answer is no, for trivial reasons: if a hyperbolic 3-manifold fibers over $S^1$ with fiber of genus $g$, then the Heegaard genus is bounded by $2g+1$. However, cyclic covers of this manifold dual to the fiber will retain this property, so there are infinitely many manifolds of bounded Heegaard genus.
One may obtain similar infinite families from semi-fiberings. 


The error in Pardon's answer is that the Gromov-Guth estimate (or earlier estimate of Lackenby) only applies to congruence arithmetic manifolds (at least, Gromov-Guth seem to be missing this hypothesis). 


On the other hand, this is the only way in which there can be an infinite family of arithmetic 3-manifolds of bounded Heegaard genus. The sketch of the argument is given an arithmetic 3-manifold of genus $g$, it will cover a minimal orbifold, which is congruence. Then one applies Lackenby's argument to show that the volume of the orbifold is bounded, and thus by Borel, such manifolds cover finitely many minimal orbifolds. Then one knows that for each such orbifold, all the covers with bounded genus (in fact bounded rank) come from covers of finitely many fiberings or semi-fiberings (see [Biringer-Souto](http://jlms.oxfordjournals.org/content/84/1/227) for (most of) the argument). 


Conjecturally, this argument applies to manifolds of bounded rank, but so far is only known with an extra hypothesis of lower bound on the injectivity radius, which would follows from [Lehmer's conjecture](http://en.wikipedia.org/wiki/Lehmer%27s_conjecture) (see again Biringer-Souto). I've made many attempts at proving this without Lehmer's conjecture, but have failed. The cusped case is ok, since in that case there's a lower bound on the length of simple closed geodesics. 


