Let $B$ be an indefinite quaternion algebra over $\mathbb{Q}$ of discriminant $D$, and $\mathcal{O}_N$ be an Eichler order of level $N$. Is there an element $x\in \mathcal{O}_N$ such that its reduced norm $\mathrm{Nrd}(x)=-1$?

There exists an element $x\in B$ such that $\mathrm{Nrd}(x)=-1$. This follows from the norm theorem. But I don't know if the existence stays to be true if $B$ is replaced with $\mathcal{O}_N$.

The question is interesting in the following way. According to the paper of S. Molina, section 2.2, Shimura curves $X_0(D,N)$ (defined as the upper half plane quotient by the group of norm one elements in $\mathcal{O}_N$) parametrizes abelian surfaces with QM by $\mathcal{O}_N$. Fix an isomorphism $i: B\otimes \mathbb{R} \to \mathrm{Mat}_2(\mathbb{R})$. Given $z$ in the upper half plane, the $\Gamma_0(D,N)$-orbit of $z$ corresponds to the isomorphic class of pairs $(\mathbb{C}^2/\Lambda_z, i)$, where $\Lambda_z$ is the lattice $i(\mathcal{O}_N)\begin{bmatrix} z \\ 1\end{bmatrix}$.

Question: why all abelian surfaces with QM by $\mathcal{O}_N$ arises this way?

There is nothing special in taking $i(\mathcal{O}_N)\begin{bmatrix} z \\ 1\end{bmatrix}$. We can just take any invertible left-ideal $I$ of $\mathcal{O}_N$ and look at the lattice $i(I)\begin{bmatrix} z \\ 1\end{bmatrix}$. Since the class number is one, any such ideal is equal to $\mathcal{O}_Ny$ for some $y\in B$. So after multiplying a scalar, the lattice $i(I)\begin{bmatrix} z \\ 1\end{bmatrix}$ is equal to $i(\mathcal{O}_N)\begin{bmatrix} i(y)z \\ 1\end{bmatrix}$.

Everything works well if $\mathrm{Nrd}(y)>0$. However, in the case $\mathrm{Nrd}(y)<0$, then $i(y)z$ is no longer on the upper half plane. This can be remedied if there is an element $x\in \mathcal{O}_N$ such that $\mathrm{Nrd}{x}=-1$, in which case $i(\mathcal{O}_N)\begin{bmatrix} i(y)z \\ 1\end{bmatrix}$ coincides with $i(\mathcal{O}_N)\begin{bmatrix} i(xy)z \\ 1\end{bmatrix}$ up to a scalar. If such an element does not exists, I am afraid that we might have to throw in the lower half plane as well in defining the Shimura curve, and the quotient by $\Gamma_0(D,N)$ will not be connected.

In fact, in section IX.5 of "Introduction to Algebraic and Abelian Functions" by S. Lang, he explicitly assumed that there is an element in the order whose reduced norm is -1. Is there an theorem that says such an element will always exists in the indefinite case, and his assumption is just for exposition convenience rather than out of necessity?


1 Answer 1


I can think of two ways to answer your question.

The first is to revisit what it is I think you want to conclude, which is that every right fractional O-ideal is (principal and) generated by an element with positive reduced norm. It is in this way that the question most naturally arises when considering double coset spaces giving rise to Shimura curves. I hate to plug my own papers, but see Section 2.2 in http://www.cems.uvm.edu/~voight/articles/classno-ants-fixed-errata.pdf. In this context (and with this notation), what you want to know is that the map $$ B_+ \backslash \widehat{B}^\times / \widehat{\mathcal{O}}^\times \to F_+ \backslash \widehat{F}^\times / \widehat{\mathbb{Z}}_F^\times = \mathrm{Cl}^+(\mathbb{Z}_F) $$ is a bijection; this is a consequence of strong approximation, which can be found in Vigneras "Arithmetique des Algebres de Quaternions", Theoreme III.4.3. Or if you'll forgive their woefully incomplete nature, see Section 4.2 in http://www.cems.uvm.edu/~voight/crmquat/book/quat-modforms-041310.pdf; here I also try to explain the link between the "usual" formulation of strong approximation (in terms of elements of $B^\times$ of reduced norm 1) and representatives of ideal classes.

In the case you are interested in (swapping right for left ideals), $F=\mathbb{Q}$ has trivial narrow class group $\mathrm{Cl}^+(\mathbf{Z})=\{(1)\}$; so every right fractional $\mathcal{O}$-ideal has a totally positive generator.

The second way to answer your question is to prove independently that there is a unit of norm one. This follows from the above reasoning by taking any element of negative norm $\alpha$, considering the right fractional $\mathcal{O}$-ideal $\alpha \mathcal{O}$ generated by it, and then noting that it is generated by a totally positive element $\beta$; then $\beta = \alpha\mu$ and by comparing norms we have $\mathrm{nrd}(\mu) = -1$. But you might also find it to be a helpful exercise to derive this from the "usual" statement of strong approximation, involving the elements of reduced norm 1.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .