Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 !

share|improve this question
    
Of course, if it could be proved using an easy self-contained argument, it would be nice. –  GreginGre Jul 7 '13 at 13:10
add comment

1 Answer

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.

share|improve this answer
    
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 ? –  user36685 Jul 7 '13 at 13:56
    
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. –  Aakumadula Jul 7 '13 at 14:47
    
ok. Thanks for your answers. I'll try to find a counterexample. –  GreginGre Jul 7 '13 at 16:21
    
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... –  GreginGre Jul 7 '13 at 22:14
    
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 $SU(2)$. –  Aakumadula Jul 7 '13 at 23:58
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.