Complex-qutrit stabilizer in F4

Statement

Let $J=H_3(\mathbb O)$ , let $F_4=\operatorname{Aut}(J)$ , and let $B\subset J$ be a Jordan subalgebra isomorphic to $H_3(\mathbb C)$ . The identity component of its setwise stabilizer is

\operatorname{Stab}_{F_4}(B)^0 \cong\frac{\mathrm{SU}(3)\times\mathrm{SU}(3)}{\mu_3},

where $\mu_3=\{(\zeta I,\zeta I):\zeta^3=1\}$ is the diagonal central subgroup. This is a connected closed subgroup of compact $F_4$ .

The two factors

Using $\mathbb O\cong\mathbb C\oplus\mathbb C^3$ , one obtains a real vector-space decomposition

H_3(\mathbb O)\cong H_3(\mathbb C)\oplus M_3(\mathbb C).

Representatives $(g,h)\in\mathrm{SU}(3)\times\mathrm{SU}(3)$ act by

(X,M)\longmapsto(gXg^\dagger,hMg^\dagger).

The first factor acts by unitary conjugation on the complex qutrit algebra; the second fixes that algebra pointwise and acts on its orthogonal complement. The diagonal $\mu_3$ is precisely the kernel.

Why the identity component matters

The full stabilizer $\operatorname{Stab}_{F_4}(B)$ is disconnected. Besides its unitary component, it has elements whose restriction to $B$ is induced by an antiunitary transformation. Consequently one must not replace $\operatorname{Stab}_{F_4}(B)^0$ by the full stabilizer in intersection theorems without changing the resulting group.

References

John C. Baez and Paul Schwahn, “The Standard Model Gauge Group from the Exceptional Jordan Algebra,” 2026, §3. arXiv:2606.15235.
Ichirô Yokota, Exceptional Lie Groups, 2009, §2.12, Remark 2. arXiv:0902.0431.
Ilka Agricola, Thomas Friedrich, and Jos Höll, “Sp(3) structures on 14-dimensional manifolds,” Journal of Geometry and Physics 69 (2013), 12–30, Appendix A. Article.