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 $](https://mathoverflow.net/questions/433847/maximum-symmetry-metric-on-mathbbcpn).


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](https://www.ams.org/journals/tran/1969-146-00/S0002-9947-1969-0250340-1).


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?



Maximum symmetry metric on Cayley plane $ F_4/{\operatorname{Spin}(9)}$
$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$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](https://mathoverflow.net/questions/369887/non-isomorphic-complex-lie-groups-with-the-same-exceptional-lie-algebra-for-ma?_gl=1*1t5iech*_ga*Mjc3NDg0Mzc5LjE1NjgyMzcyOTc.*_ga_S812YQPLT2*MTY3NDM0NzE5NS40NDMuMS4xNjc0MzUxMzkyLjAuMC4w "Non-isomorphic complex Lie groups with the same exceptional Lie algebra for g2,f4,e6,e7,e8?"), 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 $M$.


