# The Hilbert symbols of quaternion algebras over a totally real field

Let $$k$$ be a totally real number field. Every quaternion algebra $$B$$ over $$k$$ can be written as $$B = \left(\frac{a,b}{k}\right),$$ for some constants $$a,b \in k^\times$$. My question is, can I always choose these constants so that $$\sigma(a) > 0$$ for all real embeddings $$\sigma \colon k \hookrightarrow \mathbb{R}$$ at which $$B$$ splits?

A priori, one only has $$\sigma(a) > 0$$ or $$\sigma(b) > 0$$ for each of the embeddings $$\sigma \colon k \hookrightarrow \mathbb{R}$$ which split $$B$$. But in many examples one can use the symmetries $$\bigl(\frac{a,\,b}{k}\bigr) = \bigl(\frac{b,\,a}{k}\bigr) = \bigl(\frac{a,\,-ab}{k}\bigr)$$ to achieve the above condition.

First choose $$a$$. You can take any $$a$$ such that $$K = k(\sqrt{a})$$ is a splitting field of $$B$$, so by Grunwald--Wang (or elementary congruences and sign conditions) you can enforce $$\sigma(a)>0$$ at all places where $$B$$ splits.
Now pick any $$b$$ such that $$(\frac{a,b}{k})\cong B$$. Multiplying $$b$$ by any norm from $$K$$ does not change the isomorphism class of resulting algebra. So all you need is to find an element $$c\in K^\times$$ whose norm down to $$k$$ has prescribed signs at the real places that split $$B$$. At those places, $$K/k$$ is split, so it reduces the problem to prescribing the signs of $$\sigma(c)$$ for every real embedding $$\sigma$$ of $$K$$ above the real places of $$k$$ that split $$B$$. This amounts to finding an element of $$K$$ that lies in some open cone in $$K\otimes_\mathbb{Q}\mathbb{R}$$, which is always possible.