Edit: In my original post I failed to require the group to be a manifold group. The answer below from @BenLinowitz works in that case. I am really interested though in when the group is torsion-free, so have edited this to reflect that.

For a non-elementary finite-covolume Kleinian group $\Gamma$, let $\mathbb{Q}(\mathrm{tr}\Gamma)$ and $k\Gamma$ be the trace field and the invariant trace field of $\Gamma$, in the sense of Maclachlan-Reid and Neumann-Reid.

If $\Gamma$ is derived from a quaternion algebra, then $\mathbb{Q}(\mathrm{tr}\Gamma)=k\Gamma$. What I want to know is, what is an example of a case where $\Gamma$ is arithmetic and torsion-free (but not derived from a quaternion algebra, of course), and $\mathbb{Q}(\mathrm{tr}\Gamma)\neq k\Gamma$?


The trace field of $\Gamma$ is $\mathbb{Q}(\mathrm{tr}\Gamma):=\mathbb{Q}\big(\{\mathrm{tr}\gamma\mid\gamma\in\Gamma\}\big)$.

The invariant trace field of $\Gamma$ is $k\Gamma:=\mathbb{Q}(\mathrm{tr}\Gamma^{(2)})$, where $\Gamma^{(2)}:=\langle\gamma^2\mid\gamma\in\Gamma\rangle$ (the group generated by squares in $\Gamma$).

$\Gamma$ is derived from a quaternion algebra if there exists a quaternion algebra $B$ that has a unique complex place, and is ramified at all real places, and an order $\mathcal{O}\subset B$ so that $\Gamma$ is a finite-index subgroup of $\mathrm{P}\mathcal{O}^1:=\{x\in\mathcal{O}\mid\mathrm{nrd}(x)=1\}/\{\pm1\}$.

$\Gamma$ is arithmetic if it is commensurable to a group derived from a quaternion algebra.


$\mathbb{Q}(\mathrm{tr}\Gamma)$ and $k\Gamma$ are number fields.

If $\Gamma$ and $\Gamma'$ are commensurable (up to conjugation), then $k\Gamma=k\Gamma'$.

The definition given of arithmeticity is equivalent to $\Gamma$ being Kleinian and and an arithmetic group in the sense of Borel (the ramification conditions guarantee discreteness).

  • $\begingroup$ If $\Gamma \subset SL_2({\mathbb C})$ is a Kleinian subgroup, will you please define what the trace field and the invariant trace fields are? One can then think about your question (since I have not seen Maclachlan Reid) . $\endgroup$ Mar 13, 2016 at 4:36
  • $\begingroup$ @Venkataramana I've added those definitions. I'm going to put some more info there too, just wanted to post this in case you're still looking at it right now. $\endgroup$
    – j0equ1nn
    Mar 13, 2016 at 4:58

3 Answers 3


For cusped 3-manifolds that are link complements in $\mathbb{Z}/2\mathbb{Z}$ homology spheres, the trace field and invariant trace field are equal (see Neumann and Reid Arithmetic of Hyperbolic Manifolds Corollary 2.3), so it makes sense to look at manifolds with non-peripheral $\mathbb{Z}/2\mathbb{Z}$ homology or closed manifolds.

In fact, $m007(3,1)$ aka Vol3 (the third smallest volume manifold in the Hodgson-Weeks' census of closed orientable manifolds) is an example of a arithmetic hyperbolic manifold of this form. Its trace field is degree 4 ($\mathbb{Q}(\sqrt{1 + i \sqrt{3}})$) and its invariant trace field is $\mathbb{Q}(i\sqrt{3})$. In some sense, this is the 'first' such example, but there are many more of this form in the Hodgson-Weeks' census and the early manifolds in this census have a high probability of being arithmetic (compare to the Chapter 13.4 of Maclachlan and Reid's "The Arithmetic of Hyperbolic 3-Manifolds", with the caveat that the surgery descriptions of closed manifolds are non-unique), so if you wanted more examples of this form the following code will help:

sage: import snappy
sage: CC = snappy.OrientableClosedCensus()
sage: distinct_field_manifolds = []
sage: CUTOFF = 10
sage: for c in CC:  
    citfg = c.invariant_trace_field_gens()
    ctfg = c.trace_field_gens()
    kI = citfg.find_field(degree=30,prec=200)
    k = ctfg.find_field(degree=30, prec=200)
       if kI[0].degree() < k[0].degree():
          print c, "invariant trace field", kI, "trace field:", k, "\n"
          if len(distinct_field_manifolds) >= CUTOFF:
          print "field computation failed for c=",c, "kI= ", kI, "k=",k

(Note there are a lot of manifolds in the census and computations in exact arithmetic are expensive so the code cuts out after the first 10, of course that can be adjusted by CUTOFF.)

UPDATE: There are cusped arithmetic manifolds with this property as well. For example, 'm009' which has homology $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, trace field $\mathbb{Q}(\sqrt{\frac{5}{2} - \frac{i \sqrt{7}}{2}})$ and invariant trace field $\mathbb{Q}(i\sqrt{7})$.

  • $\begingroup$ This is a great answer and thanks for supplying the code! I'm relatively new to Sage, which is part of why I'm in need of help finding examples such as this. The case of non-arithmetic cusped manifolds is turning out to be far more difficult. See for instance my question over here mathoverflow.net/questions/233477/… @IanAgol give me the right nudge to answer the compact case but the non-compact case is still elusive. $\endgroup$
    – j0equ1nn
    Mar 18, 2016 at 1:13
  • $\begingroup$ You might want to look at m037 which is non-arithmetic and has trace field smaller than cusp field. $\endgroup$ Mar 18, 2016 at 1:34

Reid gives an example in his paper A note on trace fields of Kleinian groups. This example appears as Example 3.3.1 in Maclachlan-Reid. Before getting into the details, the "big picture" is that the invariant trace field was developed when people realized that you could have commensurable Kleinian groups with different trace fields. As commensurable groups have the same invariant trace field, this will give you the example you are after.

Consider the subgroup $\Gamma=\langle g,h\rangle$ of $\mathrm{PSL}_2(\mathbb C)$ where
$$g=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \qquad h=\begin{pmatrix} 1 & 0 \\ -\omega & 1 \end{pmatrix}$$

for $\omega=\frac{-1+\sqrt{-3}}{2}$. All of the matrix entries of $\Gamma$ lie in the ring of integers $\mathbb Z[\omega]$ of $\mathbb Q(\sqrt{-3})$, and in fact $\Gamma$ is a subgroup of the Bianchi group $\mathrm{PSL}_2(\mathbb Z[\omega])$ of index $12$. Therefore $\Gamma$ is discrete and $\mathbb Q(\mathrm{tr}\Gamma)=k\Gamma=\mathbb Q(\sqrt{-3})$.

Let $x=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$ so that the image $Px$ of $x$ in $\mathrm{PSL}_2(\mathbb C)$ normalizes $\Gamma$ and has square equal to the identity. The group $\Gamma'=\langle \Gamma, Px\rangle$ is therefore a subgroup of the normalizer of $\Gamma$ which contains $\Gamma$ with index $2$. In particular $\Gamma$ and $\Gamma'$ are directly commensurable, hence $$k\Gamma'=k\Gamma=\mathbb Q(\sqrt{-3}).$$

Notice however that $\Gamma'$ contains the element $xhg=\begin{pmatrix} i & i \\ i\omega & -i+i\omega \end{pmatrix}$. It follows that the trace field of $\Gamma'$ contains both $\omega$ and $i$ and is in fact equal to $\mathbb Q(i,\omega)$.

Thus $\Gamma'$ has trace field $\mathbb Q(i,\sqrt{-3})$ and invariant trace field $\mathbb Q(\sqrt{-3})$.

  • $\begingroup$ Great, this answers my question but I notice that $x$ is elliptic. Is there an example where both groups are torsion-free? I neglected to put this in the question but I was wondering about groups of manifolds $\endgroup$
    – j0equ1nn
    Mar 13, 2016 at 18:31
  • $\begingroup$ Presumably one can come up with an example in which both of the groups are torsion-free, though I haven't given the matter much thought. I'll let you know if I come up with anything. $\endgroup$
    – user1073
    Mar 13, 2016 at 20:25
  • $\begingroup$ Ben, I know you answered the question I posted and then I changed it. But I'm afraid nobody else is going to contribute as long as I've accepted an answer. I hope you don't mind my un-accepting the answer for this reason. $\endgroup$
    – j0equ1nn
    Mar 14, 2016 at 5:43

Here is a Fuchsian construction that I think works: fix a prime number $p$, and let $$ \Gamma(p) = \left\{ g \in \mathrm{PSL}_2(\mathbb Z) :\: g = 1 \pmod{p}\right\} $$ and let $\gamma \in \mathrm{PSL}_2(\mathbb R)$ be any hyperbolic that normalizes $\Gamma(p)$ and the square of which lies in $\mathrm{PSL}_2(\mathbb Z)$. Then the group $\langle \gamma, \Gamma(p) \rangle$ is discrete and you can use its LERF property to get an arithmetic torsion-free group containing $\gamma$.

So we have to find $p$ such that there is a $\gamma$ with non-integral trace. For this take $p$ such that there is a unit $u \in \mathbb Z[\sqrt p]^\times$ of norm $-1$. Then you can take $$ \gamma = \begin{pmatrix} (u+u^{-1})/2 & \sqrt p(u - u^{-1}) \\ (u - u^{-1})/\sqrt p & (u+u^{-1})/2 \end{pmatrix}. $$

If you want something in dimension three you can of course embed this into a supergroup of a Bianchi group and use LERF to get somrthing that is torsion-free.


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