# Holonomy group of $\mathbb{O}P^1$

What is the holonomy group of the 1-dimensional octonionic projective space ?

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$OP^1 = S^8$, as can be found by an elementary search, eg here: math.ucr.edu/home/baez/octonions/node9.html. This should answer your question. –  David Roberts Jul 8 '11 at 3:14

Following David Roberts' comment, and using the fact that the holonomy of the round sphere $S^n$ is $SO(n)$, you get $SO(8)$ as the answer to your question.

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Hi Andre - maybe I was being a little bit mean. But I found the answer on the wikipedia page on holonomy, and is I presume a standard fact in geometry. –  David Roberts Jul 8 '11 at 12:04

Of course, it depends on the Riemannian metric you put on $\mathbb O \mathbb P^1 \cong S^8$. For the round metric you do indeed get $\mathrm{SO}(8)$ holonomy. A priori, one could imagine other "natural" metrics with reduced holonomy. However, we know by the Berger classification that if the metric was not locally symmetric, since the dimension is $8$, the only possibilities for the holonomy would imply the existence of a non-trivial parallel $2$-form or $4$-form, which is impossible in this case because of the topology of $S^8$.

I suppose that one could find some way to write $S^8 = G/H$ as a Riemannian homogenous space, in which case the holonomy would be $H$, but I am not the right person to ask about this. The only way I know how to do this is to take $G = \mathrm{SO}(9)$ and $H = \mathrm{SO}(8)$, which gives us back the round $S^8$ again.

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If by 'holonomy' you mean the holonomy of a Riemannian metric on $S^n$, then the holonomy is always $SO(n)$, no matter what the metric. This is true for all $n$. Now, for some $n$, there can exist a $G$-structure on $S^n$ where $G$ is a proper subgroup of $SO(n)$, and there can be a $G$-compatible connection on such a $G$-structure that has holonomy a subgroup of $G$, but such a connection cannot be torsion-free. –  Robert Bryant Jul 8 '11 at 15:39
Hi Robert. Indeed, I meant the holonomy of a Riemannian metric. I suspected that it would always be $\mathrm{SO}(n)$. Is there an obvious reason why? –  Spiro Karigiannis Jul 8 '11 at 16:24
@Spiro: Using some machinery, here's an argument: The holonomy can't act reducibly, since $S^n$ is not a non-trivial product. The holonomy also can't be one of the proper subgroups of $SO(n)$ on Berger's list, since each of those would have a nonzero parallel $p$-form with $0<p<n$, which would be nonzero in cohomology (proof: wedge with the Hodge dual), but $H^p(S^n)=0$. By Berger, the metric would have to be a symmetric metric, but the classification of Riemannian symmetric spaces shows that the only ones diffeomorphic to $S^n$ are the round $S^n$s, and they have holonomy $SO(n)$. –  Robert Bryant Jul 18 '11 at 19:29
Thanks, Robert. I had the first two steps. I'm just not that familiar with the classification of Riemannian symmetric spaces. It's been on my "must learn" list for many many years. –  Spiro Karigiannis Jul 18 '11 at 21:14