I have a simple question: transitivity of $Spin(7)$ in triples of orthogonal vectors. Let $Spin(7)\subset SO(8)$ act on $\mathbb{R}^8$, and $e_1,e_2,e_3$, $v_1,v_2,v_3$ be two triples of mutually orthogonal vectors.

Is there a (unique?) transformation in $Spin(7)$ that takes $e_i$ to $v_i$, $i=1,2,3$?

If so, which is the easiest way to convince oneself? If not, why? And what about quadruples of vectors?

Thank you very much for you attention and answers.

  • $\begingroup$ If such a transformation always exists, it cannot be unique. Indeed the dimension of the manifold $M$ of triples of mutually orthogonal vectors is 18, while $\dim Spin(7)=28$. If $Spin(7)$ acts transitively on $M$ then the dimension of stabilizer of each triple is $28-18=10$. $\endgroup$ – orbits Apr 16 '16 at 17:30

We know that $\operatorname{Spin}(7)$ acts transitively on unit vectors, with stabilizer $G_2$, so we need only prove that $G_2$ acts transitively on pairs of orthogonal unit vectors. But then apply the same argument, since $G_2$ acts transitively on the unit sphere in $\mathbb{R}^7$, with stabilizer $SU(3)$, which also acts transitively on the unit sphere in $\mathbb{R}^6$. So $\operatorname{Spin}(7)$ acts transitively on triples of mutually orthogonal unit vectors. But the dimension count ensures that it is not unique, apparently, following Semyon Alesker's argument.

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    $\begingroup$ But I think that $\operatorname{Spin}(7)$ has dimension 21. $\endgroup$ – Ben McKay Apr 16 '16 at 18:28
  • $\begingroup$ Yes of course, and the stabiliser of $SU(3)$ acting on $S^5$ is $SU(2)$, which has dimension $3$. In particular, Spin(7) does not act transitively on quadruples. $\endgroup$ – Sebastian Goette Apr 16 '16 at 18:31
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    $\begingroup$ In fact $\operatorname{Spin}(7)$ acts on the orthogonal quadruples preserving the Cayley 4-planes. $\endgroup$ – Ben McKay Apr 16 '16 at 18:34
  • $\begingroup$ @BenMcKay : you are right. The dimension is 21. Sorry for the mistake. $\endgroup$ – orbits Apr 16 '16 at 18:41

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