4
votes
1answer
287 views

What does the Jacquet-Langlands correspondence say about quaternion algebras of class number one?

If F is a totally real number field of degree n, and A is a definite quaternion algebra over F, I understand (not really) the Jacquet Langlands correspondence to construct a modular form in n ...
8
votes
2answers
382 views

Which quaternion algebras have class number one?

Over Q, the definite quaternion algebras with a unique conjugacy class of maximal orders, i.e. "with class number one", are those with discriminant 2,3,5,7, and 13. Three questions: What is a ...
3
votes
2answers
312 views

How do you find maximal orders in quaternion algebras?

Let A be the 4-dimensional algebra over Q with basis 1,i,j,k, and multiplication table $$ i^2 = -1 \quad j^2 = -11 \quad k^2 = -11 \quad ij = k \quad jk = 11i \quad ki = j $$ So, A is the unique ...
3
votes
1answer
208 views

Question about the definition of the genus 0 curves in Gross' paper “Heights and the Special values of L-series”

Let $N \in \mathbb{Z}$ be a prime number, and let $B = \left( \dfrac{a, b}{\mathbb{Q}} \right)$ be the unique quaternion algebra over $\mathbb{Q}$ ramified at $N$ and at $\infty$. Then, in section 3 ...
5
votes
3answers
841 views

Octonions and the dance of the seven veils

Let us take the octonions as having all integer coefficients and the multiplication table at BAEZ We have a standard conjugation operator with $\bar{1} = 1$ and $\bar{e_i}= - e_i,$ extend by ...
18
votes
1answer
1k views

The Quaternion Moat Problem

"One cannot walk to infinity on the real line if one uses steps of bounded length and steps on the prime numbers. This is simply a restatement of the classic result that there are arbitrarily large ...
4
votes
5answers
1k views

Why is every quadratic subfield of a Galois extension of the rationals with the quaternions as Galois group real?

Suppose that L is a field extension of the rationals with Galois group the quaternions Q={1,-1,i,-i,j,-j,k,-k}. Furthermore assume that L contains a quadratic subfield K. I have learned from this link ...