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

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


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

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

enter image description here

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

$$v^2x = \gamma x + \delta v x = \gamma x + \delta x^{\tau} v$$

and on the other hand,

$$v^2x = vvx = v x^{\tau}v = xvv = x \gamma + x \delta v = \gamma x + \delta x v$$

and therefore $\delta x^{\tau} = \delta x$. Since $x \neq x^{\tau}$, we get $\delta = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.