In this [interview by Eric Weinstein to Roger Penrose, Timestamp 1:24:05.](https://youtu.be/mg93Dm-vYc8?t=5045), what result is the host talking about?


Transcription of the relevant part:



> 
> "If you have two sets of symmetries, known as Lie groups, that act transitively on the same sphere in usual position, then either their intersection acts transitively on that sphere, or the dimension of that sphere is $15$. And I believe the intersection of the groups looks like the electro-strong group. So it's very close to the... particle spectrum of theoretical physics... pulled out of nowhere just by talking about sphere transitive group actions"
> 
> 
> 


Edit: it seems like the host is trying to recall a particular result. Given how bizarre and peculiar the result seems to be (in line with dimension $4$ being special for differentiable structures on Euclidean spaces, or dimension $7$ in the case of exotic spheres), I would like to know if it's a real thing and, in case it's real, what's the exact statement.


In particular,


1. I don't care if the exact quoted statement is true or false;
2. I only want to know if there's a result that sounds very similar to that one and is actually true and, if you're aware of such a result, what's its exact statement.



A result on Lie group actions on 15-dimensional spheres?
My guess is that Weinstein was thinking of this fact, but didn't get it out correctly:


For every $n\not=15$, there is a compact Lie group $H\_n\subseteq\mathrm{SO}(n{+}1)$ that acts transitively on the $n$-sphere such that any Lie group $G$ that acts transitively and effectively on the $n$-sphere contains a subgroup $G'$ that acts transitively on the $n$-sphere and is conjugate to $H\_n$ in $\mathrm{Diff}(S^n)$.


There are two non-isomorphic subgroups, $\mathrm{Spin}(9)$ and $\mathrm{Sp}(4)$ of $\mathrm{SO}(16)$, both of dimension $36$, that act transitively on $S^{15}$ such that any Lie group $G$ that acts transitively on $S^{15}$ contains a subgroup $G'$ that is conjugate to (exactly) one of these two subgroups in $\mathrm{Diff}(S^{15})$.


Note:


$\bullet$ For $m\not=0,3$, $H\_{2m}\simeq \mathrm{SO}(2m{+}1)$,
while $H\_0\simeq\mathrm{O}(1)$ and $H\_6 \simeq \mathrm{G}\_2$,


$\bullet$ for $m\not=0$, $H\_{4m+1}\simeq \mathrm{SU}(2m{+}1)$ while $H\_1\simeq\mathrm{SO}(2)$, and


$\bullet$ for $m\not=4$, $H\_{4m-1}\simeq\mathrm{Sp}(m)$.


This follows from Borel's classification of the Lie groups acting transitively on spheres.


**N.B.:** The phrase 'and effectively' in the above statement is needed to rule out the following kinds of (ineffective) actions: First, $\mathbb{Z}$ has a transitive action on $S^0 = \{-1,1\}\subset\mathbb{R}$ but has no subgroup isomorphic to $\mathrm{O}(1)\simeq \mathbb{Z}\_2$. Second, the simply-connected cover of $H\_1=\mathrm{SO}(2)$ is isomorphic to $\mathbb{R}$, and it acts transitively on $S^1$ without containing a subgroup isomorphic to $H\_1 = \mathrm{SO}(2)$. Third, for $m\not=0,3$, $H\_{2m}\simeq\mathrm{SO}(2m{+}1)$ has a nontrivial double cover $\mathrm{Spin}(2m{+}1)$ that acts transitively on $S^{2m}$ but does not contain a subgroup isomorphic to $H\_{2m}$.


