Congruence subgroups in arithmetic lattices of $\mathrm{SO}(n,1)$ I am currently reading the paper Deformation Spaces Associated to Hyperbolic Manifolds by Johnson and Millson, Section 7, and the highlighted bit below has been giving me difficulty:

Specifically, how do we define this ideal $\frak{a}$ that allows us to get this group $\Gamma=\Gamma(\frak{a})$? I have tried following up the reference to Geometric construction of cohomology for arithmetic groups by Millson and Raghunathan, but unfortunately despite my best efforts I have really struggled to follow the latter paper and extract anything useful, since I am not only new to hyperbolic geometry, but also completely unfamiliar with number theory.
If someone could concretely describe how to define such an ideal and the group $\Gamma$, I would really appreciate it!
 A: I think the reference to Millson-Raghunathan may be a little misleading. All that is needed is that if $\Gamma \subset G(\mathbb Z)$ is an arithmetic group in a linear algebraic group $G$ defined over $\mathbb Q$, then  there exists an ideal $m\mathbb Z$ of $\mathbb Z$ such that the group $\Gamma (m)$ of matrices in $\Gamma$ congruent to identity modulo $m$, is ${\bf torsion-free}$.  
In your notation, the group $\Phi$ is an arithmetic subgroup of $SO(Q)$ but defined over a number field $K$. By a standard restriction of scalars argument, you may think of $\Phi=SO(Q)(O_K)$ as commensurable to $G(\mathbb Z)$, where $G=R_{K/\mathbb Q}(SO(Q))$ and  where $R$ denotes the restriction of scalars. You may choose the ideal $\mathfrak a$ of $O_K$ to be generated by a rational integer $m$ as in the preceding paragraph.
A: I want to extend the answer of Venkataramana a bit in order clarify how 
you can choose a suitable ideal $\mathfrak{a} \subseteq \mathcal{O}$.
Here is an abstract version of Minkowski's classical theorem.
Let $G$ be an affine group scheme of finite type over the ring of integers $\mathcal{O}$ of some algebraic number field.
If $\mathfrak{a} \subseteq \mathcal{O}$ is an ideal such that for every 
prime number $p$ we have $ p\mathcal{O} \not\subset\mathfrak{a}^{(p-1)}$,
then the principal congruence subgroup 
$$ \Gamma(\mathfrak{a}) = \ker\bigl( G(\mathcal{O}) \to G(\mathcal{O}/\mathfrak{a}) \bigr)$$ is torsion-free (cf. III.2.3 in this thesis).
In particular, here is a concrete example: The ideal
$\mathfrak{a} = m \mathcal{O}$ for any integer $m \geq 3$ always yields a torsion-free principal congruence subgroup.
