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Is there a seven-dimensional geometric object whose full group of symmetries is isomorphic to the compact Lie group $G_2$, or does the same problem as with the special orthogonal groups occur?

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    $\begingroup$ Isn't $G_2$ transitive on the unit sphere? Thus any invariant object is a union of spheres and hence has symmetry group $O_7$. $\endgroup$
    – Will Sawin
    Nov 5, 2021 at 18:59
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    $\begingroup$ Is the space of imaginary octonions considered "geometric"? $\endgroup$ Nov 6, 2021 at 4:46

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The OP doesn't say what is meant by a 'geometric object', so it's hard to give a definitive answer. However, if one assumes that the geometric object is a smooth manifold $M^7$ and that the action is smooth, then there are a few things one can say:

First, the $\mathrm{G}_2$-stabilizer of any point $p\in M^7$ has to be a closed subgroup $H_p\subset \mathrm{G}_2$ which is therefore a compact Lie group of rank at most $2$ and dimension at least $7$. The rank $1$ closed connected subgroups of $\mathrm{G}_2$ are either of dimension $1$ or $3$, so they are ruled out, so $H_p$ must be of rank $2$ and hence must contain a maximal torus of $\mathrm{G}_2$. Looking at the root diagram, one sees that the only possibilities for such subgroups of dimension at least $7$ are conjugates of $\mathrm{SU}(3)$ (which is a maximal proper subgroup) and $\mathrm{G}_2$ itself. Thus, the orbits of $\mathrm{G}_2$ on $M^7$ must be either $6$-spheres or points.

Obviously, there must be at least one $6$-sphere orbit, otherwise the action will be trivial. We may as well discard the orbits that are points (if there are any), and we may as well restrict to a connected component of what remains, in which case $M^7$ is diffeomorphic to $\mathbb{R}\times S^6$, with the standard $\mathrm{G}_2$-action on $S^6$. Let $t:\mathbb{R}\times S^6\to\mathbb{R}$ be projection on the first factor.

Now, $\mathrm{G_2}$ preserves a nondegenerate $2$-form $\omega$ on $S^6$ (and a compatible Riemannian metric $g$, which has constant sectional curvature $+1$). In fact, the $G_2$-invariant differential forms on $S^6$ are generated by $1$, $\omega$, $\mathrm{d}\omega$ (which is nowhere vanishing), and $\ast_g\mathrm{d}\omega$. Meanwhile, the group of diffeomorphisms of $S^6$ that preserve $\omega$ is exactly $\mathrm{G}_2$.

Thus, for example, the group of diffeomorphisms of $\mathbb{R}\times S^6$ that preserve the $2$-form $t\,\omega$ is exactly $\mathrm{G}_2$. Thus, if you count the manifold $\mathbb{R}\times S^6$ endowed with the $2$-form $t\,\omega$ as a geometric object, you have a $7$-dimensional geometric object whose symmetry group is $\mathrm{G}_2$.

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  • $\begingroup$ @DanielSeibald: If there's something about my answer that doesn't satisfy you, could you let me know? If you are satisfied with the answer, you should accept it, so that it doesn't keep coming up as 'open'. $\endgroup$ Nov 11, 2021 at 10:18

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