Maximum symmetry metric on Cayley plane $ F_4/{\operatorname{Spin}(9)}$

Question

The maximum symmetry metric on real projective space $ \mathbb{RP}^n $ is the round metric.

The maximum symmetry metric on complex projective space $ \mathbb{CP}^n $ is the Fubini–Study metric; see Maximum symmetry metric on $ \mathbb{C}P^n $.

Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. See for example Ku, Mann, Sicks, and Su - Degree of symmetry of a product manifold.

Two metrics are considered to be equivalent if they are isometric up to a constant multiple.

I'm interested in manifolds $ M $ for which there is a unique up to equivalence metric with isometry group of dimension $ N(M) $.

Is the pushforward of the biinvariant metric of $ F_4 $ onto the Cayley projective plane $$ \mathbb{OP}^2 \cong F_4/{\operatorname{Spin}(9)} $$ a maximum symmetry metric in this sense?

Answer 1

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$A maximum symmetry metric on $M:=\mathbb{O}P^2$ must be equivalent to the one you described.

Here's one way to see it.

Suppose $G$ is the isometry group of some fixed Riemannian metric on $M$ of maximal symmetry. Then $G$ acts effectively on $M$. From Theorem 2.2 of

Chang, Theodore, and Tor Skjelbred. “Lie Group Actions on a Cayley Projective Plane and a Note on Homogeneous Spaces of Prime Euler Characteristic.” American Journal of Mathematics, vol. 98, no. 3, 1976, pp. 655–78. JSTOR, https://doi.org/10.2307/2373811.

it follows that $G$ has rank at most $4$. Now a simple inspection shows that $F_4$ has the largest dimension of all rank $4$ Lie groups. In fact, because we'll need it later, let me note that $\Sp(4)$ and $\Spin(9)$ are both 36 dimensional, and have the second largest dimension among at-most-rank-$4$ groups.

So, since $G$ is maximum symmetry, we conclude that the identity component of $G$ is , up to cover, $F_4$. However, $F_4$ has trivial center, so the identity component of $G$ is $F_4$.

We still need to pin down the metric. To that end, select $p\in M$ and consider the the isotropy group $G_p$. We have $\dim G - \dim G_p = \dim (G\cdot p)\leq \dim M = 16$, so $\dim G_p \geq 36$. As $G_p\subseteq G$ has rank at most $4$, we thus conclude that either $\dim G_p = 36$ (which implies the $G$ action is transitive) or that $G_p = G$, so the $G$-action has a fixed point. The latter case cannot happen: if it does, then then by taking the differential, we obtain a $G$-action on $T_p M$, so obtain a 16-dimensional representation of $G$. However, the smallest non-trivial representation of $F_4$ is $26$-dimensional.

So we now know that $G$ acts transitively, which means a maximum symmetry metric on $M$ is determined once we know it at $p$. Moreover, the induced $G_p$-action on $T_p M$ must be by isometries.

To finish, we note that the transitive actions on $M$ are known: there is an essentially unique such action (see, e.g., Onishchik's Topology of Transitive Transformation Groups, Table 10 on pg. 265). In particular, $G_p = \Spin(9)$, and the embedding of $\Spin(9)$ into $F_4$ is the usual one. Thus, the isotropy action of $\Spin(9)$ is irreducible (in fact, it's transitive on the unit sphere), which implies that there is only one $G_p$-invariant inner product on $T_p M$. Thus, there is, up to scale, a unique maximum symmetry metric on $M$.