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Boris Rosenfeld claimed that the 128-dimensional compact Riemannian symmetric space on which $\mathrm{E}_8$ acts as isometries could be seen as the "octooctonionic projective plane", $(\mathbb{O} \otimes \mathbb{O})\mathrm{P}^2$. This is the symmetric space called EVIII by Cartan.

This claim has always been hard to understand, since $\mathbb{O} \otimes \mathbb{O}$ is not a division algebra. Nonetheless there seems to be something to it, as witnessed by the magic square.

John Huerta pointed out to me that if $(\mathbb{O} \otimes \mathbb{O})\mathrm{P}^2$ really is something like an octooctonionic projective plane, we could hope it contains a lot of "octooctonionic projective lines". We aren't sure what an "octooctonionic projective line" should be, but one naive guess is that it's $\mathbb{O} \otimes \mathbb{O}$ together with a point at infinity — and thus, a 64-sphere.

Of course there are lots of 64-spheres smoothly embedded in $(\mathbb{O} \otimes \mathbb{O})\mathrm{P}^2$, but here we want 'nice' ones, raising this question:

  • What are the maximal totally geodesic spheres in the 128-dimensional compact Riemannian symmetric space EVIII?

For comparison, note that the octonionic projective plane $\mathbb{O}\mathrm{P}^2$ is the 16-dimensional compact Riemannian symmetric space called FII by Cartan. This contains a lot of totally geodesic 8-spheres, which can rigorously be seen as copies of $\mathbb{O}\mathrm{P}^1$.

I've come across a paper that claims to classify all maximal totally geodesic spheres in compact symmetric spaces:

However, it doesn't seem to discuss the case of EVIII.

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  • $\begingroup$ The existence of a codimension one totally geodesic submanifold in a symmetric space implies that the symmetric space has constant curvature (I think due to Onishchik and later generalized to other homogeneous ambient spaces, I think perhaps naturally reductive homogeneous spaces). So there cannot be 15-dimensional totally geodesic spheres in $OP^2$. $\endgroup$
    – Claudio Gorodski
    Commented Mar 10, 2021 at 1:02
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    $\begingroup$ I think 'geodesic sphere' means 'distance sphere', that is, the locus of points at a given distance from a given point (distance in the metric space structure induced from the Riemannian metric). $\endgroup$
    – Claudio Gorodski
    Commented Mar 11, 2021 at 12:50
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    $\begingroup$ I believe it is possible to answer your question by looking closely at the Lie algebras $\mathfrak e_8\supset\mathfrak{spin}(16)$ and by computing the algebraic version of the curvature tensor, which is $\langle[u,v],[w,x]\rangle$, for $u$, $v$, $w$, $x\in\mathfrak{spin}(16)^\perp$. To be the tangent space of a totally geodesic sphere, the curvature has to have standard form on a subspace $V$ of $\mathfrak{spin}(16)^\perp$. This is a necessary condition. Also necessarily there must exist a subgroup of $E_8$ with Lie algebra $\mathfrak h$ such that $\mathfrak h\cap\mathfrak{spin}(16)^\perp=V$. $\endgroup$
    – Sebastian Goette
    Commented Mar 12, 2021 at 9:35
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    $\begingroup$ Hi @John. I mean, $V$ is the tangent space of one of these spheres at the origin, so we want $\dim V=64$. Standard form means as on a round $S^n$, that is $\langle[u,v],[w,x]\rangle=C(\langle u,w\rangle\langle v,x\rangle-\langle u,x\rangle\langle v,w\rangle)$ for some $C>0$. Also, a totally geodesic submanifold of a Riemannian symmetric space is necessarily itself symmetric. And the bad news is: there is no suitable symmetric representation of $S^{64}$. You can probably embed EVI totally geodesically in EVIII, but that's not what you want. $\endgroup$
    – Sebastian Goette
    Commented Mar 15, 2021 at 8:23
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    $\begingroup$ If there was, then we would have a group of rank at most $8$ that acts transitively on $S^{64}$. This group would therefore be a candidate for a special holonomy group according to Jim Simons' thesis, and hence one of $SO(n)$, $(S)U(n)$ and $Sp(n) (\cdot Sp(1))$, $G_2$ and $Spin(7)$ (which arise as holonomy groups) or $Spin(9)$ or $Sp(n) \cdot S^1$ (which don't). None of these has rank at most 8 and acts transitively on $S^{64}$. $\endgroup$
    – Sebastian Goette
    Commented Mar 18, 2021 at 9:00

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