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Let $k$ be a number field. Let $B$ be a central division $k$-algebra.

Let us consider an isomorphism $\varphi: B\otimes_k \mathbb{C}\overset{\sim}{\to} M_n(\mathbb{C})$.

Let $\Lambda$ be a subring of $B$ which is also a free $\mathcal{O}_k$-module. Is $\varphi(\Lambda\otimes 1)\cap U_n(\mathbb{C})$ finite ?

In fact, the question boils down to the following one: let $R$ be a subring of $M_n(\mathbb{C})$ , such that $R=\mathcal{O}_kM_1\oplus\cdots\oplus \mathcal{O}_k M_r$, where $M_1,\ldots,M_r$ are $\mathbb{C}$-linearly independent. Is $R\cap U_n(\mathbb{C})$ finite ?

Unfortunately, $R$ is not necessarily closed and discrete, so if it is true, some other kind of arguments have to be used. I think the answer is "yes", and maybe it is due to the fact that $U_n(\mathbb{C})$ is a compact Lie group, but I would need someone to confirm.

Thanks in advance !

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  • $\begingroup$ Of course, if it could be proved using an easy self-contained argument, it would be nice. $\endgroup$
    – GreginGre
    Commented Jul 7, 2013 at 13:10

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It is not true in general. Let $k$ be a real quadratic extension and let $B$ be a quaternionic central division algebra over $k$ such that in one of the embeddings of $k$ in $\mathbb R$, the division algebra becomes (i.e. after tensoring with the archimedean completion) a matrix algebra and in the other it becomes Hamiltonian quaternions. Such division algebras exist in profusion.

Let $\Lambda$ be an order in the division algebra. The order will contain units of norm one infinite order. However, the intersection of the group of units of norm one in $\Lambda$ will lie in $SU(2)$ in the "other" archimedean embedding, and is infinite.

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  • $\begingroup$ Thanks for the quick reply. I will precise the context a bit more then. In fact, I have a unitary involution $\tau$ on $B$, such that $(B,\tau)\otimes_k \mathbb{C}\simeq (M_n(\mathbb{C}),*),$ where $*$ is the transconjugate. In other words, $\varphi(\tau(b)\otimes 1)=\varphi(b\otimes 1)^*$. In particular, the set $U(B,\tau)$ of elements $b$ of $B$ satisfying $\tau(b)b=1$ is identified to a subgroup of $U_n(\mathbb{C})$. The new question is : is $U(B,\tau) \cap \Lambda^\times$ finite ? $\endgroup$
    – user36685
    Commented Jul 7, 2013 at 13:56
  • $\begingroup$ No, even then this will not be true. You can get a totally real field $k_0$ of large degree (and a suitable quadratic $k/k_0$). ; the condition you are imposing affects only one completion, leaving large room for manouvering in the other archimedean embedding of $k$s. $\endgroup$
    – Venkataramana
    Commented Jul 7, 2013 at 14:47
  • $\begingroup$ ok. Thanks for your answers. I'll try to find a counterexample. $\endgroup$
    – GreginGre
    Commented Jul 7, 2013 at 16:21
  • $\begingroup$ Mmmh, one last question. Would it still possible to have a counterexample if the involution $\tau$ restricts on $k$ to complex conjugation ? In particular, $k/\mathbb{Q}$ would have no real embeddings... $\endgroup$
    – GreginGre
    Commented Jul 7, 2013 at 22:14
  • $\begingroup$ No. Here is the construction; let $k_0$ be a totally real cubic extension of $\mathbb Q$. Let $k/k_0$ be a quadratic extension, which becomes complex over one real place of $k_0$ and becomes a product of two reals over the other real embeddings of $k_0$. Let $D$ be a quaternionic central division algebra over $k$ with an involution of the second kind (restricted to $k$, it is the galois auto of $k/k_0$). Then $\Lambda $ is an order on $D$. You can take the group of norm one elts in $\Lambda$. This projects to a lattice in the other real embeddings of $k_0$ but is a DENSE subgroup of $SU(2)$. $\endgroup$
    – Venkataramana
    Commented Jul 7, 2013 at 23:58

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