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It's my understanding that the octonians aren't associative enough to have a projective plane in the usual sense. That is, you'd want to define $\mathbb{O}P^{2}$ as the collection of Cayley lines in $\mathbb{R}^{16}$. However, "Cayley lines" in $\mathbb{R}^{16}$ doesn't make sense due to the lack of associativity.

However, equivalent (for $\mathbb{K}P^{2}$ with $\mathbb{K} \in${$\mathbb{R}, \mathbb{C}, \mathbb{H}$} ) formulations of due still work. Set $k = dim_{\mathbb{R}} \mathbb{K}$.

For example, topologically, $\mathbb{K}P^{2}$ is obtained by attaching a $2k$ ball to the sphere $S^{k}$ via the $k$-dimensional Hopf map (meaning, where the base sphere is $k$ dimensional). The same is true for $\mathbb{K} = \mathbb{O}$, topologically. The (black box) fact that only fibration with fiber $S^{7}$ and total space a sphere is the fibration $S^{7}\rightarrow S^{15}\rightarrow S^{8}$ leads to the fact that there is no higher $\mathbb{O}P^n$.

From this description, it's not too hard (using the same techniques which work on $\mathbb{C}P^{2}$ and $\mathbb{H}P^{2}$), to show that $H^{*}(\mathbb{O}P^{2}, \mathbb{Z}) = \mathbb{Z}[x]/x^{3}$ with $|x| = 8$.

As another example of an equivalent formulation, one can start with a $2k$-dimensional ball in $\mathbb{R}^{2k}$ and quotient out the boundary by the $k$ dimensional Hopf map. One can put a particular radial metric on the ball and check that it's well defined and smooth under the quotienting (I forget exactly what the metric is). This construction yields $\mathbb{K}P^2$ with the Fubini-Study metric.

This construction also works when $\mathbb{K} = \mathbb{O}$. This construction is nice because it shows $\mathbb{O}P^2$ has a Fubini-Study metric so that curvatures lie between 1 and 4 and the cut locus relative to a point is an $S^8$. In other words, this construction shows the geometry is very similar to that of $\mathbb{K}P^2$ for the division algebras $\mathbb{K}$ over $\mathbb{R}$. One can also use this description (with some hard work, or so I'm told), to show that $\mathbb{O}P^2$ is isometric to the homogeneous space $F_{4}/Spin(9)$ with normal homogeneous metric.

1

It's my understanding that the octonians aren't associative enough to have a projective plane in the usual sense. That is, you'd want to define $\mathbb{O}P^{2}$ as the collection of Cayley lines in $\mathbb{R}^{16}$. However, "Cayley lines" in $\mathbb{R}^{16}$ doesn't make sense due to the lack of associativity.

However, equivalent (for $\mathbb{K}P^{2}$ with $\mathbb{K} \in${$\mathbb{R}, \mathbb{C}, \mathbb{H}$} ) formulations of due still work. Set $k = dim_{\mathbb{R}} \mathbb{K}$.

For example, topologically, $\mathbb{K}P^{2}$ is obtained by attaching a $2k$ ball to the sphere $S^{k}$ via the $k$-dimensional Hopf map (meaning, where the base sphere is $k$ dimensional). The same is true for $\mathbb{K} = \mathbb{O}$, topologically. The (black box) fact that only fibration with fiber $S^{7}$ and total space a sphere is the fibration $S^{7}\rightarrow S^{15}\rightarrow S^{8}$ leads to the fact that there is no higher $\mathbb{O}P^n$.

From this description, it's not too hard (using the same techniques which work on $\mathbb{C}P^{2}$ and $\mathbb{H}P^{2}$), to show that $H^{*}(\mathbb{O}P^{2}, \mathbb{Z}) = \mathbb{Z}[x]/x^{3}$ with $|x| = 8$.

As another example of an equivalent formulation, one can start with a $2k$-dimensional ball in $\mathbb{R}^{2k}$ and quotient out the boundary by the $k$ dimensional Hopf map. One can put a particular radial metric on the ball and check that it's well defined and smooth under the quotienting (I forget exactly what the metric is). This construction yields $\mathbb{K}P^2$ with the Fubini-Study metric.

This construction also works when $\mathbb{K} = \mathbb{O}$. This construction is nice because it shows $\mathbb{O}P^2$ has a Fubini-Study metric so that curvatures lie between 1 and 4 and the cut locus relative to a point is an $S^8$. In other words, this construction shows the geometry is very similar to that of $\mathbb{K}P^2$ for the division algebras $\mathbb{K}$ over $\mathbb{R}$. One can also use this description (with some hard work, or so I'm told), to show that $\mathbb{O}P^2$ is isometric to the homogeneous space $F_{4}/Spin(9)$ with normal homogeneous metric.
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