4
$\begingroup$

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 $\mathbb OP^n$

$\endgroup$
3
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Cayley_plane I don't think there are octonionic projective spaces of higher dimension. $\endgroup$ – Angina Seng Dec 7 '19 at 7:51
  • $\begingroup$ So, Why they aren't exist? $\endgroup$ – C.F.G Dec 7 '19 at 8:14
  • 2
    $\begingroup$ This has nothing to do with symmetric groups. Also quaternion and octonion tags seem more relevant. $\endgroup$ – runway44 Dec 8 '19 at 21:54
10
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Very nice, much better than my answer (+1). $\endgroup$ – Paul Frost Dec 9 '19 at 17:02
  • $\begingroup$ @Paul: I appreciate the nice comment, but I would say it complements your answer, rather than saying it's better. $\endgroup$ – Jason DeVito Dec 9 '19 at 17:37
  • $\begingroup$ @JasonDeVito: Very nice explanation as always. Honestly, I was waiting for your interesting answer!!! $\endgroup$ – C.F.G Dec 10 '19 at 4:44
8
$\begingroup$

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".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.