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Let $F$ be a totally real number field, $R$ is a quaternion algebra over $F$ ramified in all infinite places of $F$. Let $\mathcal{O}\subset R$ be an order. By assumption on $R$ its group of units $\mathcal{O}^{\times}/\mathcal{O}^{\times}_F$ is a finite(possibly non-commutative) group.

How one can compute it?

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  • $\begingroup$ Am I missing something? If $F=\mathbb Q$ and $R=M_2(\mathbb Q)$ then $\mathcal O=M_2(\mathbb Z)$ is an order whose group of units is infinite. $\endgroup$
    – user1073
    Commented Sep 26, 2016 at 13:52
  • $\begingroup$ @Ben Linowitz sure, I changed unramified to ramified, thank you! $\endgroup$
    – SashaP
    Commented Sep 26, 2016 at 14:53
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    $\begingroup$ Actually, the condition that you need to impose is that $B$ be a definite quaternion algebra (i.e., ramified at all infinite places of $F$). Otherwise the subgroup $\mathcal O^1$ of the unit group which consists of elements with reduced norm $1$ will embed into a product of $\mathrm{SL}_2(\mathbb R)$'s and $\mathrm{SL}_2(\mathbb C)$'s as a cocompact discrete arithmetic subgroup with finite covolume. In the case that $\mathcal O$ is a definite quaternion order, $\mathcal O^*$ will be finite and can be computed by Magma: math.lsu.edu/doc/magma/html/text1025.htm $\endgroup$
    – user1073
    Commented Sep 26, 2016 at 15:55
  • $\begingroup$ In the case that $\mathcal{O}$ is a maximal order, there's a brief discussion of this problem in section 4.5 of arxiv.org/abs/1006.4381 $\endgroup$
    – Henri Johnston
    Commented Sep 26, 2016 at 16:04
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    $\begingroup$ One more correction: The unit group of $\mathcal{O}_F$ will embed into $\mathcal{O}^{\ast}$ so, unless $F = \mathbb{Q}$, this will always be infinite. You might have wanted to compute $\mathcal{O}^{\ast}/\mathcal{O}^{\ast}_F$? That's what the Magma command Ben Linowitz lists does. $\endgroup$
    – David E Speyer
    Commented Sep 26, 2016 at 18:12

1 Answer 1

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By the ramification hypothesis, the reduced norm $\mathrm{nrd}$ gives a positive definite quadratic form $\mathrm{Tr}\circ \mathrm{nrd} : \mathcal{O}\to \mathbb{Q}$. On the other hand, the reduced norm induces a morphism $f : \mathcal{O}^\times/\mathcal{O}_F^\times \to \mathcal{O}_F^\times/(\mathcal{O}_F^\times)^2$. The latter group is finite, and you can compute it by standard algorithms for units in number fields. For each $u\in \mathcal{O}_F^\times$ in a set of representatives modulo squares, the fiber $f^{-1}(u)$ is finite and admits representatives in the finite set $\{x\in\mathcal{O} \mid \mathrm{Tr}(\mathrm{nrd}(x)) = \mathrm{Tr}(u)\}$ which you can compute by lattice enumeration (Fincke-Pohst algorithm).

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