Does the orthogonal group $O(4,\mathbb{Q})$ have an exceptional outer automorphism analogous to that of its subgroup, the Coxeter/Weyl group $W(F_4)$?


3 Answers 3


$\def\QQ{\mathbb{Q}}\def\Out{\mathrm{Out}}\def\HH{\mathbb{H}}$I think tim penttila's answer may be misleading: $\Out(O(4, \QQ))$ is nontrivial and quite large, and the exceptional automorphism of $W(F_4)$ is induced by a non-inner automorphism of $O(4, \QQ)$. What Connors - The automorphisms of $O^+_4(V)$ in the anisotropic case (the paper tim pentilla cites) shows is that $\Out(O(4, \QQ))$ is generated by the "obvious" automorphisms, but not that $\Out(O(4, \QQ))$ is trivial. So the answer to your question depends on whether you consider an "obvious", but not inner, automorphism of $O(4, \QQ)$ to be exceptional.

Many thanks to tim pentilla for finding this paper!

Connors considers a four dimensional vector space $V$ over a field $k$ (not of characteristic $2$) with an anisotropic quadratic form $q$, and shows that $\Out(O(V,q))$ is generated by two sorts of automorphisms.

(The type that won't interest us) Let $Z$ be the center of $O(V,q)$ and let $\chi : O(V,q) \to Z$ be a character. Then $g \mapsto \chi(g) g$ is an automorphism of $O(V,q)$. In our case, $Z = \pm \mathrm{Id}$ and I think (not sure) that the commutator subgroup of $O(4, \QQ)$ is $SO(4,\QQ)$, so this just gives the automorphism $g \mapsto (\det g) g$.

(The type that will interest us) In the generality which Connors considers, let $g : V \to V$ be a semilinear map which preserves orthogonality. In our case, we can simplify: The automorphism group of the field $\QQ$ is trivial, so semilinear maps are linear; a linear map which preserves orthogonality is one such that there is a scalar $c$ with $q(g(v)) = c q(v)$. Conjugation by $g$ is an automorphism of $O(V,q)$, which Connors calls $\phi_g$.

If $c$ is square in $k$, then conjugation by $g$ is the same as conjugation by $\tfrac{1}{\sqrt{c}} g$, so $\phi_g$ is inner. However, if $c$ is not square, then $\phi_g$ is not inner. Proof: Suppose, to the contrary, that there is some $g'$ in $O(V,q)$ with $ghg^{-1} = g' h (g')^{-1}$ for all $h \in O(V,q)$. Then $g^{-1} g'$ commutes with every $h \in O(V,q)$, so $g^{-1} g'$ is a scalar, say $g=ag'$. Then $c = a^2$.

Now, in the particular case of $\QQ^4$ with the standard quadratic form, there are many automorphisms of this second type. Namely, any positive rational number is of the form $w^2+x^2+y^2+z^2$ for rational $(w,x,y,z)$. Identiying $\QQ^4$ with the quaternions, left-multiplication by $w+xi+yj+zk$ will achieve $c=w^2+x^2+y^2+z^2$. So $\Out(O(4, \QQ))$ contains a copy of $\QQ^+/(\QQ^{\times})^2$.

(The $F_4$-Coxeter group) Let $\HH$ denote the Hurwitz quaternions: Quaternions of the form $w+xi+yj+zk$ with $w$, $x$, $y$, $z$ either all in $\mathbb{Z}$ or all in $\tfrac{1}{2}+\mathbb{Z}$. Let $\HH_k$ be the set of Hurwitz quaternions of norm $k$.

The $F_4$ root system can be represented as $\HH_1 \cup \HH_2$; here $\HH_1$ is the $24$ short roots and $\HH_2$ is the $24$ long roots. Let $g$ be left multiplication by $1+i$. Then $g(\HH_1) = \HH_2$ and $g(\HH_2) = \HH_4 = 2 \HH_1$. So conjugation by $g$ turns reflections over the short roots into reflections over the long roots and vice versa.

The Coxeter group $W(F_4)$ is generated by reflections in the $F_4$-roots. Conjugation by $g$ is its exceptional automorphism.


No. See [Edward A. Connors, The Automorphisms of $O^+_4(V)$ in the Anisotropic Case Proceedings of the American Mathematical Society, Vol. 54, No. 1 (Jan., 1976), pp. 16-18].

  • $\begingroup$ Thanks for the helpful answer! I've left another answer adding some more context and possibly disagreeing, but this is definitely helpful; welcome to MO! $\endgroup$ Sep 13, 2021 at 16:15
  • $\begingroup$ @DavidESpeyer's answer referenced above. $\endgroup$
    – LSpice
    Sep 13, 2021 at 17:53

@DavidESpeyer We don't disagree. As you suspected, I don't consider an automorphism of O(4,q) induced by an element of GO(4,Q) exceptional, even if it induces an exceptional automorphism of the Weyl group. Why would I?

And you're welcome for me finding the paper. Dieudonné's On the automorphisms of the classical groups doesn't cover the anisotropic case. (This was the first place I looked.) It had to be known, and probably some time in the period 1951-1980 (Dieudonné in his 1951 paper Sur les groupes orthogonaux rationnels à trois et quatre variables doesn't know it yet.) He still doesn't seem to know it by 1963 and La géométrie des groupes classiques, 2ieme ed.

Wonenburger, María J. ,The automorphisms of the group of rotations and its projective group corresponding to quadratic forms of any index. Canadian J. Math. 15 (1963), 302–303, did dimension at least 5 for characteristic not 2. It's still not known by Robert Steinberg in his Yale Lectures on Chevalley groups 1967-8, nor by Humphreys in On the automorphisms of infinite Chevalley groups, Canadian J. Math. 21 (1969), 908–911, nor by Borel and Tits, Homomorphismes "abstraits'' de groupes algébriques simples. Ann. of Math. (2) 97 (1973), 499–571. It seemed to me to be known by the time of the work of Zun Xian Li in 1987 [Sci. Sinica Ser. A 30, 1121–1132] and even by Hefendehl in 1980 [ Geom. Dedicata 9, 129–152.]

Having narrowed the time range to 1973-1980, a little more searching found Connors' paper.

  • 1
    $\begingroup$ You can edit your answer if you want to expand on it. You can also comment directly under David's answer if you want to. Note that answers on MO are not necessarily sorted by age, so that many people will not read them in the order they were created. By default, the number of votes an answer gets is the sorting criterion. $\endgroup$ Sep 14, 2021 at 8:58
  • $\begingroup$ Not if you don't have enough reputation you can't- and new users don't have any reputation... $\endgroup$ Sep 14, 2021 at 12:46
  • $\begingroup$ Well you do now. $\endgroup$ Sep 14, 2021 at 13:50

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