9
$\begingroup$

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 follows easily (by the Skolem-Noether theorem) that there exists some $\gamma\in D$ such that $\tau(t)=\gamma^{-1}\mu(t)\gamma$ for all $t\in D$.

In the case where $D$ is a quaternion algerbra over $K$ (i.e. a division algebra of degree $2$), one can construct a non-trivial involution of the second type on $D$ in the following way:

  • Since the order of quaternion algebras in the Brauer group of a field is two, it follows that for any field $L$ and quaternion algebra $L$ has a non-trivial $L$-involution (i.e. an $L$-linear anti-automrophism of the algebra). This holds since the fact that $L$ has order two in the Brauer group is equivalent to $L$ being isomorphic to $L^{op}$, the opposite algebra, and hence existence of a non trivial map $L\to L^{op}$, which is the same thing as an anti-automorphism.
  • Let $\mathbf{d}$ be a quaternion algebra over $k$, and let $\tau':\mathbf{d}\to\mathbf{d}$ be the non-trivial $k$ involution.
  • One shows that $D\cong \mathbf{d}\otimes_K K$ as $K$-algebras, and that the map defined on generators by $\tau(t\otimes \alpha)=\tau(t)\otimes \bar{\alpha}$ is a field automorphism.

The question is- what happens for higher degrees?

In the book "The Book of Involutions", Knus presents an argument for the existence of a non-trivial involution of the second type on $D$. Namely, such an involution exists if and only if the norm $N_{K/k}(D)$ is a split $F$ algebra (see $\S 3$ of the book for the definition of the norm algebra, I will add it here if someone here irequests it).

My problem with Knus's proof is that it in not constructive, in the sense that it presents the reader with a bijection between the set of 2nd type involutions of $D$, and some specific set of left-sided ideals in $N_{K/k}(D)$, but shows that such ideals exists if the splitting condition holds. But it is terribly unclear to me, how to go back and construct such an involution once you've shown that it exists.

So, after all this long introduction- here is my question

Question: Does anybody here know of an example of a division algebra of degree 3 (or higher) over a $p$-adic or global field, which has a non-trivial and explicitly presented involution of the second type?

I would be very thankful for any reference or example that anyone can offer.

Thank you.

$\endgroup$
  • $\begingroup$ I'm not sure I got all the pieces fitting together, yet. But, thinking in terms of this example of index $\ell=3$ division algebra over $\Bbb{Q}(\sqrt{-3})$: wouldn't applying the automorphism $\zeta_9\mapsto\zeta_9^2$ to all the entries work there? $\endgroup$ – Jyrki Lahtonen Jul 18 '14 at 21:27
  • $\begingroup$ Scratch the previous comment. That is an automorphism - without ANTI. $\endgroup$ – Jyrki Lahtonen Jul 18 '14 at 21:38
  • $\begingroup$ Thanks for the comment. The answer, apparently, is that they sometimes exists and sometimes not- if your base field is local then there aren't any involutions of the second type. In a global setting one can construct an example, but only when the degree of the algebra is 2. In a more abstract setting- I don't know... $\endgroup$ – kneidell Jul 18 '14 at 21:43
  • $\begingroup$ I don't know either :-) My guess would be that we need the Hasse invariants to somehow be compatible with the action of the non-trivial $k$-automorphism of $K$. But that is just a guess. I will try this tomorrow - too late here to think about it now. $\endgroup$ – Jyrki Lahtonen Jul 18 '14 at 21:47
  • 2
    $\begingroup$ I would expect this type of an involution to exist in the following (global) example case. Let $k=\Bbb{Q}$, $K=k[i]$. Assume that $D$ is of index $\ell$ and that it's non-trivially Hasse invariants are at primes $2+i$ and $2-i$, and they are, respectively $1/\ell+\Bbb{Z}$ and $-1/\ell+\Bbb{Z}$. Then $D$ is anti-isomorphic with its inverse $D^{opp}$ in $Br(K)$ with Hasse-invariants that are negatives of those of $D$. But here it might(?) happen that conjugation of $K$ extends to a $k$-isomorphism between $D$ and $D^{opp}$, $\endgroup$ – Jyrki Lahtonen Jul 19 '14 at 10:26
2
$\begingroup$

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 $\mathbf{Q}_p$: then $p$ is called inert and you have one embedding (up to conjugation) $\nu_p\colon \mathbf{Q}[\sqrt{d}]\to \mathbf{Q}_p[\sqrt{d}]\colon a+b\sqrt{d}\mapsto a+b\sqrt{d}$

3) $p$ such that $d$ is a square root in $\mathbf{Q}_p$: then $p$ is called split and you have two non-equivalent embeddings $\nu_{\pm p}\colon \mathbf{Q}[\sqrt{d}]\to \mathbf{Q}_p\colon a+b\sqrt{d}\mapsto a\pm b\sqrt{d}$.

To fix ideas and notations (and to still go along Jyrki's comment), let's say $d=-1$. By Hasse theorem, there exists a division algebra D over $\mathbf{Q}[i]$ with local Hasse invariant $$([0],[0],[0],[\frac{1}{l},\frac{-1}{l}],[0],[0],[0,0],[0,0]\dots),$$ for $l$ any integer $\geq 1$. Explanation on the notation: the first bracket corresponds to the complex place, and the following brackets correspond to the primes in increasing order, recalling that some primes (the split ones) contain two places.

Let $\tau \colon \mathbf{Q}[i]\to \mathbf{Q}[i]\colon a+bi\mapsto a-bi $ be the nontrivial Galois automorphism. Let $D^{\tau}$ be the conjugate of $D$ (i.e. as a ring, $D^{\tau} = D$, but the $\mathbf{Q}[i]$-algebra structure is twisted by $\tau$). By construction, $D^{\tau}\cong D^{\text{op}} $ (just look at the local invariants!). So, that's your example! (and in fact, it's also the one described by Jyrki in its comment).

It is also satisfactory to know that this example is in fact the generic one when working over a number field. Indeed, as you note in your question, a central simple algebra possesses an involution of the second kind if and only if the corestricted algebra splits. And over a number field, this is the case if and only if all split primes contain local invariant that sum up to $0$ (like $\frac{1}{l}$ and $\frac{-1}{l}$ in the above example). See the paper "On the restriction and corestriction of algebras over number fields" by Kleinert.

To end up, let me make an unrelated observation: for $A$ a division algebra over $\mathbf{Q}[\sqrt{d}]$ and having an involution $\sigma$ of the second kind, you can form the algebraic group SU$(A,\sigma )$ and one could ask what does this group becomes over finite places? The answer is simple: for all non-split primes $p$ and corresponding place $v$, it is the algebraic group SU$(A_v,\sigma_v )$, while for $v$ belonging to a split place, it is the group SL$_1(A_v)$ (note that the involution does not extend to the completion in those cases, thus addressing the concern of @Kimball).

$\endgroup$

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.