Timeline for Maximal geodesic spheres in the "octooctonic projective plane"
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 18, 2021 at 9:00 | comment | added | Sebastian Goette | 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 $S^{64}$. | |
| Mar 16, 2021 at 20:43 | comment | added | John Baez | Okay, I get it now. Yes, I noticed EVI, but I don't think that's what I want here. How do you know "there is no suitable symmetric representation of $S^{64}$? | |
| Mar 15, 2021 at 8:23 | comment | added | Sebastian Goette | 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. | |
| Mar 13, 2021 at 19:41 | comment | added | John Baez | @SebastianGoette - Thanks! What's "a standard form"? And if I want a 64-dimensional totally geodesic sphere, what dimension should $V$ (and/or $\mathfrak{h}$) be? As a $\mathbb{Z}_2$-graded Lie algebra $\mathfrak{e}_8$ is the direct sum of $\mathfrak{spin}(16)$ and its 128-dimensional spinor representation, and I am able, at least in theory, to do calculations with these. | |
| Mar 12, 2021 at 9:35 | comment | added | Sebastian Goette | 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 $\mathfrak h\cap\mathfrak{spin}(16)^\perp=V$. | |
| Mar 12, 2021 at 5:22 | comment | added | John Baez | Oh, that explains the subscript $r$ in the sphere $S_r$ - it's the distance! | |
| Mar 12, 2021 at 5:22 | history | edited | John Baez | CC BY-SA 4.0 |
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| Mar 11, 2021 at 12:50 | comment | added | Claudio Gorodski | 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). | |
| Mar 11, 2021 at 2:11 | comment | added | John Baez | Is there an error in the comment at mathoverflow.net/a/348222 or am I just misinterpreting it? It begins "every geodesic sphere $S_r$, 0<r<π/2, in the octonion (or Cayley) projective plane $\mathrm{Ca}P^2$ is isometric to a canonical deformation of the round metric on $S^{15}$". | |
| Mar 10, 2021 at 1:02 | comment | added | Claudio Gorodski | 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 $OP^2$. | |
| Mar 9, 2021 at 19:19 | history | asked | John Baez | CC BY-SA 4.0 |