Another way to see that there
There is no higher dimensional Octonion a topological obstruction.
First we have to specify what do we expect from a octonion projective space, because it is the following.
Assume one difficult to prove that something does not exists and let h be if it has not been defined.
We expect that there is a notion of (projective) subspaces of codimension k. Our first assumption is that the cohomology class complement of a hyperplane is the affine octonion space. Then , since cup product agrees with intersection one can use induction (applying the same argument on tranverse subvarieties, you would expect that $h^2$ would be the class of a codimension 2 subspace, $h^3$ of hyperplane) to give a codimension 3 and so cell structure on . $\mathbb{OP}^n$. This already determines its additive cohomology, at leastwhich has a generator in each degree multiple of 8, if your construction is good enough up to talk about subspaces in the usual way8n.I'm not going to be precise because the statement
Our second assumption is that a higher dimensional Octonion projective space does not exist subspace of codimension k is not precise: it depends on which properties do you expect to hold for such a constructionthe intersection of k hyperplanes; this intersection is then automatically transverse. By analogy But cup product agrees with intersection on transverse intersections. So if h is the realsfundamental class of a hyperplane, complex and quaternions and with the 2-dimensional case, which we actually can construct, we also expect cohomology in degree 8k is generated by $h^k$.
We conclude that these classes generate the cohomology .ring of $\mathbb{OP}^n$ must be $\mathbb{Z}[h]/(h^n)$.
Then a construction with Steenrod operations rules out the possibility that a space with such a cohomology can exist. Namely there is no space with cohomology ring $\mathbb{Z}[x]/(x^m)$ \mathbb{Z}[x]/(x^n)$ (with m>3) unless x has degree 2 or 4. For a proof see Hatcher, Algebraic topology, corollary 4.L.10.
