# Central simple quaternion algebra: why is the matrix for $\rho(v)$ antidiagonal?

Let $F$ be a field of characteristic $0$. Let $D$ be a central, simple quaternion division algebra over $F$. Let $x \in D$, not in $F$. Then $K = F[x]$ is a field of degree two over $F$, and $D$ is a two dimensional vector space over $K$. Let $y \in D$, not in $K$, so that $1,y$ is a basis for $D$ as a left $K$-module.

For $b \in D$, the $K$-linear map $D \rightarrow D$ given by $y \mapsto yb$ defines an injection of $F$-algebras:

$$D \rightarrow \textrm{End}_{K}(D)^{\textrm{op}}$$

and the choice of basis $1, y$ gives an isomorphism of $K$-algebras (hence of $F$-algebras) $\textrm{End}_{K}(D)^{\textrm{op}} \cong \textrm{Mat}_2(K)$. Let $\rho: D \rightarrow \textrm{Mat}_2(K)$. To describe $\rho$ explicitly, if $b \in D$, and we write

$$b = \alpha + \beta y, yb = \gamma + \delta y$$

then

$$\rho(b) = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$$

Now, I'm trying to understand the argument in Hilbert Modular Forms and Iwasawa Theory by Hida:

I understand how we change the basis $1, y$ to a basis $1, v$ (and replace $\rho$ by a corresponding map of the same name) such that $v$ is an eigenvector for $\rho(x)$ with eigenvalue $x^{\tau}$, i.e. $vx = x^{\tau}v$. It follows that $va = a^{\tau}v$ for all $a \in K$, and then $\rho(v)\rho(a) = \rho(a^{\tau}) \rho(v)$.

I don't understand how we can conclude that

$$\rho(v) = \begin{pmatrix} 0 & \alpha \\ \beta & 0 \end{pmatrix}$$

I see how we can immediately get $\alpha = 1$ by choice of $v$. But it seems to me like we should have $v = 0 \cdot 1 + 1 \cdot v$ and $v^2 = \gamma + \delta v$ and therefore

$$\rho(v) = \begin{pmatrix} 0 & 1 \\ \beta & \delta \end{pmatrix}$$

How do we know that $\delta = 0$?

As usual, I figure it out right after I post the question. Write $v^2 = \gamma + \delta v$ for $\gamma, \delta \in K$. We have
$$v^2x = \gamma x + \delta v x = \gamma x + \delta x^{\tau} v$$
$$v^2x = vvx = v x^{\tau}v = xvv = x \gamma + x \delta v = \gamma x + \delta x v$$
and therefore $\delta x^{\tau} = \delta x$. Since $x \neq x^{\tau}$, we get $\delta = 0$.