*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 $\mathbb{Q}(\mathrm{tr}\Gamma)\neq k\Gamma$?


**Definitions:**


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.


**Properties:**


$\mathbb{Q}(\mathrm{tr}\Gamma)$ 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).



For an arithmetic hyperbolic 3-manifold group, when is its trace field not its invariant trace field?
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](http://www.math.columbia.edu/~neumann/preprints/nrarith.pdf) 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)
    try:
       if kI[0].degree() < k[0].degree():
          print c, "invariant trace field", kI, "trace field:", k, "\n"
          distinct_field_manifolds.append(c)
          if len(distinct_field_manifolds) >= CUTOFF:
              break;
    except:
          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 $\mathbb{Q}(i\sqrt{7})$. 


