Involutions of the second type in a division algebra

2017-12-03 03:04:05

I'm trying to figure out some details about involutions of division algebra, thought maybe someone here might have a better insight.

Let $k$ be a $p$-adic or number field, and let $K=k[\sqrt{\delta}]$ be a non-trivial extension of degree $2$. For $x\in K$, let $\bar{x}$ denote the conjugate of $x$ under the non-trivial $K/k$ automorphism. Let $D$ be a division algebra of degree $\ell$ with $Z(D)=K$. For simplicity, we shall assume that $\ell$ is prime (or even $\ell=3$ is enough for the moment).

An involution of the second type of $D$ is a $k$-linear anti-automorphism of $\tau:D\to D$ which coincides with $\bar{ }$ on $K$, and is of order two. That is to say that for any $t,s\in D$ and $\alpha,\beta\in $K$

$$(1)\:\tau(\alpha t+\beta s)=\bar{\alpha}\tau(t)+\bar{\beta}\tau(s),\quad(2)\: \tau(st)=\tau(t)\tau(s),\quad\text{and}\quad(2)\:\tau^2(t)=t.$$

Note that if $\tau,\eta$ are two involutions of type 2 of $D$, then $\tau\circ\mu$ is a $K$-automorphism of $D$. It follo

  • This is such an old question that I guess you know this story very well by now. Still I thought it could be useful to record an answer.

    Actually, the strategy outlined by @Jyrki in the comments does work. As he mentions, it is necessary to go through the local theory, and here are the details: let $d$ be a square free integer and consider $\mathbf{Q}[\sqrt{d}]$, and let $D$ be its discriminant, i.e. $D = \begin{cases}d \text{ if } d \text{ is congruent to } 1 \text{ modulo }4\\ 4d \text{ otherwise } \end{cases}$. It has two real places (resp. one complex place) if $d$ is positive (resp. negative). Let us work out the finite places, which are of 3 kinds:

    1) $p$ dividing $D$: then $p$ ramifies, and you have one embedding (up to conjugation) $\nu_p\colon \mathbf{Q}[\sqrt{d}]\to \mathbf{Q}_p[\sqrt{d}]\colon a+b\sqrt{d}\mapsto a+b\sqrt{d}$ (note that $\mathbf{Q}_p[\sqrt{d}]$ is a ramified extension of $\mathbf{Q}_p$).

    2) $p$ not dividing $D$ such that $d$ is not a square root in $\mat

    2017-12-03 05:17:50