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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.

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Yes.

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.

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