What is the Cayley projective plane? - MathOverflow

What is the Cayley projective plane?

2009-10-22T19:09:35Z

One can build a projective plane from R^n, C^n and H^n and is then tempted to do the same for octonions. This leads to the construction of a projective plane known as OP^2, the Cayley projective plane.

What are the references for the properties of the Cayley projective plane ? In particular, I would like to know its (co)homology and homotopy groups.

Also, what geometric intuition works when working with this object? Does the intuition from real projective space transfer well or does the non-associativity make a large difference? For example, I would like to know why one could have known that there is no OP^3.

Answer by Ilya Nikokoshev for What is the Cayley projective plane?

2009-10-22T19:11:53Z

See, e.g., Baez: http://math.ucr.edu/home/baez/octonions/node8.html or even better http://math.ucr.edu/home/baez/week173.html

Answer by Hugh Thomas for What is the Cayley projective plane?

2009-10-22T20:21:52Z

OP^2 is a minuscule (and cominuscule) flag variety G/P associated to root system E_6. One can do Schubert calculus there. There's a description of the cohomology ring in terms of jeu de taquin (like the classical description for Grassmannians); you just have to replace the kx(n-k) rectange which you use for Grassmannians by a rather funny-looking partially ordered set. See arXiv:math/0701215.

I have no idea about your other questions, though.

Answer by Mark Hovey for What is the Cayley projective plane?

2009-11-07T15:16:56Z

As I recall, the Cayley projective plane is painful to build, but it is a 2-cell complex, with an 8-cell and a 16-cell. The cohomology is Z[x]/(x^3) where x has degree 8, as you would expect. Its homotopy is unapproachable, because it is just two spheres stuck together, so you would pretty much have to know the homotopy groups of the spheres to know it. The attaching map of the 16-cell is a map of Hopf invariant one, from S^15 to S^8, the last such element.

I think the real reason that the Cayley projective plane exists is because any subalgebra of the octonions that is generated by 2 elements is associative. That is just enough associativity to construct the projective plane, but not enough to construct projective 3-space. And this is why you should not expect there to be a projective plane for the sedonions (the 16-dimensional algebra that is to the octonions what the octonions are to the quaternions), because every time you do the doubling construction you lose more, and in particular it is no longer true that every subalgebra of the sedonions that is generated by 2 elements is associative.
Mark

Answer by John Stillwell for What is the Cayley projective plane?

2009-11-07T18:22:11Z

The geometric reason that there is no octonion projective 3-space is that the Desargues theorem holds in any projective space, and the Desargues theorem implies that the coordinate algebra is associative. (Essentially due to Hilbert, Grundlagen der Geometrie, 1899.)

Incidentally, the term "Cayley plane" is a misnomer, due to misidentification of the octonions with Cayley. John Graves discovered the octonions in December 1843, and they were rediscovered by Cayley a couple of years later. The octonion projective plane was first constructed by Ruth Moufang in the 1930s.

Answer by Jason DeVito for What is the Cayley projective plane?

2009-11-21T01:46:40Z

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 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.

Answer by Gil Kalai for What is the Cayley projective plane?

2009-11-24T10:03:40Z

There is some curious combinatorial aspects of the real- complex- Quaternionic- and Cayley- projective planes. A theorem of Brehm and Kuhnel asserts that a 2m-dimensional manifold which is not a sphere cannot be triangulated with less than 3m+3 vertices. Equality imposes strict topological restrictions on the manifold (similar to known restrictions for "tight embeddings.") There are known triangulations of RP^2 with 6 vertices, of RC^2 with 9 vertices of RH^2 (probably) with 15 vertices and finding a 27 vertex triangulations of RO^2 is open. See U. Brehm and W. Kühnel, 15-vertex triangulations of an 8-manifold, Math. Annalen 294, 167-193 (1992) and references cited there. As far as I remember, all the 4 projective planes have important tight smooth embeddings ("Veronese type"). Here are 2 links to a paper and web page.

Answer by Andrea Ferretti for What is the Cayley projective plane?

2009-11-24T10:23:14Z

There is a topological obstruction.

First we have to specify what do we expect from a octonion projective space, because it is difficult to prove that something does not exists 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 complement of a hyperplane is the affine octonion space. Then one can use induction (applying the same argument on the hyperplane) to give a cell structure on $\mathbb{OP}^n$. This already determines its additive cohomology, which has a generator in each degree multiple of 8, up to 8n.

Our second assumption is that a subspace of codimension k is the intersection of k hyperplanes; this intersection is then automatically transverse. But cup product agrees with intersection on transverse intersections. So if h is the fundamental class of a hyperplane, the cohomology in degree 8k is generated by $h^k$.

We conclude that 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 $\mathbb{Z}[x]/(x^n)$ (with m>3) unless x has degree 2 or 4. For a proof see Hatcher, Algebraic topology, corollary 4.L.10.

Answer by Benoît Kloeckner for What is the Cayley projective plane?

2010-05-10T09:09:28Z

It is easy to construct the octonionic projective line in almost the usual way, thanks to $2$-associativity: one take the quotient of $\mathbb{O}^2$ by the relation $(x,y)\sim (1,yx^{-1})$ which turns out to be equivalent to the relation $(x,y)\sim(xy^{-1},1)$ when both $x,y$ are non-zero.

The octonionic plane cannot be constructed in the same way, because one has to deal with three coordinates. But one can simply consider the quotient of the subset $\mathbb{O}^3_\bullet$ of $\mathbb{O}^3$ consisting of triples having at least one real coordinate. Then the three equivalence relations $(1,y,z)\sim(y^{-1},1,zy^{-1})$, $(x,1,z)\sim(xz^{-1},z^{-1},1)$ and $(x,y,1)\sim(1,yx^{-1},x^{-1})$, extended $\mathbb{R}$-linearly. From this the realization of the plane by three $\mathbb{R}^{16}$ charts is straightforward.

I think this construction is discussed in a book by Salzmann and other authors.

Answer by Artie Prendergast-Smith for What is the Cayley projective plane?

2010-06-04T07:09:52Z

One description of the Cayley plane that doesn't seem to have been mentioned yet is as the homogeneous space F₄ / Spin(9), mentioned in this paper. (More precisely, F₄ here means the simply-connected compact Lie group of type F₄.)

Answer by Carl McTague for What is the Cayley projective plane?

2010-06-04T14:52:53Z

A quick note about the homotopy groups of the Cayley plane. Mimura computed some of them. Specifically for i=8,9,10,...,23 he computed that $\pi_i\mathbf{CaP}^2$ equals Z, Z/2, Z/2, Z/24, 0, 0, Z/2, Z/120, (Z/2)$^{\oplus3}$, (Z/2)$^{\oplus4}$, Z/24$\oplus$ Z/2, Z/504$\oplus$ Z/2, 0, Z/6, Z/4, Z$\oplus$ Z/120$\oplus$ (Z/2)$^{\oplus2}$, respectively. See Theorem 7.2 of his 1967 paper The homotopy groups of Lie groups of low rank:
http://www.ams.org/mathscinet-getitem?mr=206958
http://projecteuclid.org/euclid.kjm/1250524375

Answer by Abhinav Kumar for What is the Cayley projective plane?

2011-02-20T15:10:50Z

There's a very pretty construction using the exceptional Jordan algebra of dimension 27, described in the book "On Quaternions and Octonions" by Conway and Smith, as well as in Baez's article on the Octonions (see http://math.ucr.edu/home/baez/octonions/node12.html). To summarize, you take Hermitian 3 by 3 matrices over the Octonions, with Jordan product $A \circ B = (AB + BA)/2$. That gives you the exceptional Jordan algebra. If you further restrict to matrices which are of unit trace and idempotent, you get $\mathbb{OP}^2$. You say point $P$ lies on line $L$ if $P \circ L = 0$.

As some of the answers above have pointed out, the usual construction of higher dimensional projective spaces doesn't work because the Desargues theorem holds automatically in these (see Courant and Robbins, "What is Mathematics", pg. 171, for a nice illustration and quick proof), and would imply that $\mathbb{O}$ is associative, which it isn't. Note that the plane $\mathbb{OP}^2$ is non-Desarguian.