Real, Complex, Quaternionic and Octonionic Projective spaces Do Octonionic Projective spaces exist or defined similar to $\Bbb RP^n$, $\Bbb CP^n$, $\Bbb HP^n$? If so, are they symmetric spaces?
I am asking this question because I've never seen Octonionic Projective spaces as examples of the Compact Rank One Symmetric Spaces or the manifolds of positive sectional curvature except its 2-dimensional case i.e. Cayley plane $\Bbb OP^2$ (or $\Bbb{Ca}P^2$). What can be say about $\Bbb OP^n$ (or $\Bbb{Ca}P^n$)?

Update: Please give a proof or a reference for non-existence of Octonionic Projective spaces for $n\geq 3$.
Related question: Quaternion Projective Space $\mathbb HP^n$ and Octonionic Projective Space $\mathbb OP^n$
 A: The projective spaces $\mathbb K P^n$ are quotient spaces of $\mathbb K^{n+1} \setminus \{0\}$ under the equivalence relation $v \sim \lambda v$ for $\lambda \in \mathbb K \setminus \{0\}$. Now see Hatcher's Algebraic Topology p. 222:

Associativity of quaternion multiplication is needed for the identification $v ∼ λv$ to be an equivalence relation, so the definition does not extend to octonionic projective spaces, though there is an octonionic projective plane $\mathbb OP^2$ defined in Example 4.47.

Perhaps also the following is useful:
Dray, Tevian, and Corinne A. Manogue. The geometry of the octonions. World Scientific, 2015.
Also have a look at this, especially at Chapter 1.3 "Why Octonions are Bad".
A: Paul is right (and I upvoted him!) that the usual definition of projective spaces requires multiplication to be associative.
However, this seems to be counter to the fact that $\mathbb{O}P^1$ and $\mathbb{O}P^2$ exist.  I have heard the explanation that "the octionions are associative enough" to support the existence of these two spaces, but honestly, I've never understood it.  Instead, I understand projective spaces more geometrically and topologically.
Geometrically:  Each projective space $\mathbb{K}P^n$ with $\mathbb{K}\in\{\mathbb{R}, \mathbb{C}, \mathbb{H}\}$ has a canonical Riemannian metric, called the Fubini-Study metric.
With regards to this metric, the manifolds are homogeneous (better, they are symmetric spaces), with sectional curvatures lying between $1$ and $4$.   In addition, for any $p\in \mathbb{K}P^n$, the cut locus is a copy of $\mathbb{K}P^{n-1}$ and the preimage of a point in $\mathbb{K}P^{n-1}$ in the unit sphere in $T_p \mathbb{K}P^n$ gives the expected Hopf fibration.
One can easily generalize this to what $\mathbb{O}P^n$ should be.  It should be a symmetric space with curvature between $1$ and $4$, and with the cut locus working as expected.  Of course, when $n=1$, $\mathbb{O}P^1\cong S^8$ fits kind of trivially.  When $n=2$, there is such an example:  $F_4/Spin(9)$ with normal homogeneous metric meets all the criteria.
However, we've classified symmetric spaces, and we've classified spaces with curvature between $1$ and $4$.  From either of these classifications, there is no $\mathbb{O}P^n$ for $n > 2$.
Topologically:  The cohomology ring of $\mathbb{K}P^n$ for $\mathbb{K}\in \{\mathbb{C},\mathbb{H}\}$ is a truncated polynomial algebra:  $H^\ast(\mathbb{K}P^n;\mathbb{Z})\cong \mathbb{Z}[\alpha]/\alpha^{n+1}$ where $|\alpha| = \dim_{\mathbb{R}} \mathbb{K}$.  One could reasonably define a "topological projective space" to be a manifold having a truncated polynomial ring as it's cohomology ring.  This is not done, because there are essentially no other examples.
In Hatcher's Algebraic topology book, Theorem 4L.9 gives that $|\alpha|$ must be a power of $2$.  Then, Corollary 4L.10 proves that there is no $\mathbb{O}P^n$ for $n\geq 3$.
Further, the ring $\mathbb{Z}[\alpha]/\alpha^3$ for $|\alpha| = 2^k$, $k\geq 4$ cannot arise by Adam's solution to the Hopf invariant one problem.
