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16 votes
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Geodesic preserving diffeomorphisms of constant curvature spaces

For $\mathbb{R}^n$: the fundamental theorem of projective geometry (proof: https://www3.nd.edu/~andyp/notes/FunThmProjGeom.pdf) says that the bijections of $\mathbb{R}^n$ taking lines to lines are ...
Ben McKay's user avatar
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15 votes
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Are the symmetric spaces $\operatorname{SU}(n)/{\operatorname{SO}(n)}$ always nontrivial in the bordism rings for $n>2$?

There is a fibration $SU(n) \overset p\to SU(n)/SO(n) \overset j\to BSO(n)$, where the $j$ is the classifying map of $p$, viewed as (the projection of) a principal $SO(n)$-bundle. The Stiefel–Whitney ...
jdc's user avatar
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15 votes
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Is there a contractible hyperbolic 3-orbifold of finite volume?

Yes. For example, let $M$ be the figure-eight knot complement. So $M$ is a hyperbolic manifold with volume a bit more than 2. The manifold $M$ has a two-fold symmetry $\tau$ that fixes, pointwise, a ...
Sam Nead's user avatar
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12 votes
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Is SO(2n+1)/U(n) a symmetric space?

Let me complement Claudio's answer. There is indeed a definition of symmetric space which works for any Riemannian manifold $M$: For any point $p\in M$ there is an involutive isometry $\iota_p$ of $M$ ...
Friedrich Knop's user avatar
12 votes

Is SO(2n+1)/U(n) a symmetric space?

I think you are right to say that both homogeneous spaces are diffeomorphic as smooth manifolds. Therefore if one of them is a symmetric space, then one can transfer this structure to the other one. ...
Claudio Gorodski's user avatar
12 votes
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Definition of locally symmetric space of reductive groups

There is a very natural, intrinsic definition of a "symmetric space", as a manifold (Riemannian or Hermitian) with an extra symmetry of a certain prescribed type. It is then a theorem, not a ...
David Loeffler's user avatar
11 votes
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Covolumes of unit groups of division algebras

First, you need to avoid definite quaternion algebras over $\mathbb{Q}$: in this case, the unit groups are finite, so the index cannot grow with $N$. With that out of the way, your algebra $D$ ...
John Voight's user avatar
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10 votes

Almgren's regularity Theorem ; a simple example?

As I mentioned in my comment, a pair of orthogonal $2$-discs in $\mathbb{R}^4$ is known to be minimizing among all orientable $2$-currents with the same boundary by the technique of calibrations: One ...
Robert Bryant's user avatar
9 votes
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Explicit fundamental domain for the action of $\operatorname{O}(n,1)(\mathbb{Z})$ on $\operatorname{O}(n,1)(\mathbb{R})$

For information about these groups up through dimension 17, see: Vinberg, È. B. The groups of units of certain quadratic forms. (Russian) Mat. Sb. (N.S.) 87(129) (1972), 18–36. English translation [...
Toffee's user avatar
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9 votes
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Relationship between fans and root data

(1) A (connected) reductive group $G$ over an algebraically closed field $k$ is described by a combinatorial object called the based root datum ${\rm BRD}(G)$. (2) A spherical homogeneous space $Y=...
Mikhail Borovoi's user avatar
9 votes
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A different notion of a decomposable symmetric tensor (besides Veronese)

Your $\vee$ is essentially multiplication of polynomials. The variety of tensors $x_1 \vee \dotsb \vee x_m$ corresponds to polynomials that factor as products of linear factors. Points of the (...
Zach Teitler's user avatar
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8 votes
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Compact quaternionic Kahler manifolds of negative curvature: examples

Any (Riemannian) symmetric space admits a cocompact lattice. This is due to A. Borel, Compact Clifford-Klein forms of symmetric spaces, Topology 2, 1963, pp.111-122. The quaternionic hyperbolic space ...
Igor Belegradek's user avatar
8 votes
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Can all hermitian symmetric spaces be realised as coadjoint orbits?

This is true. One can use a few facts from Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces to show that, indeed, $K = \mathrm{Stab}_G(Z)$. Since $M=G/K$ is an irreducible Hermitian ...
Robert Bryant's user avatar
7 votes

Geodesic sphere in the octonion projective plane

Yes; every geodesic sphere $S_r$, $0<r<\pi/2$, in the octonion (or Cayley) projective plane $Ca P^2$ is isometric to a canonical deformation of the round metric on $S^{15}$ with respect to the ...
Renato G. Bettiol's user avatar
7 votes
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The automorphism group of a symplectic symmetric space

The affine group of $(M,\nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of ...
Robert Bryant's user avatar
7 votes

Applications of Jordan algebras

Jordan algebras were originally introduced by Pascual Jordan in a hope to generalize the orthodox formulation of quantum mechanics, but this program was not successful as far as the generalization of ...
Zurab Silagadze's user avatar
7 votes

Jacobi fields on non-symmetric spaces

A particularly simple non-homogeneous example in which one can explicitly integrate the Jacobi equations is the complete metric on $\mathbb{R}^2$ given by $$ g = (x^2{+}y^2{+}2)\bigl(\mathrm{d}x^2+\...
Robert Bryant's user avatar
7 votes

Spherical roots, restricted roots, and the dual group of a symmetric variety

There is a small survey published in an Oberwolfach report by Bart Van Steirteghem comparing the different normalizations of spherical roots in more detail. The standard normalization is obtained by ...
Friedrich Knop's user avatar
6 votes
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Relationship between Multiplicative Ergodic Theorems

The equivalence of "sublinear shadowing" property and the Oseledec theorem (or, rather, of what is called "Lyapunov regularity") is due to Kaimanovich MR0947327 (89m:22006), and is explained there in ...
R W's user avatar
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6 votes

When can the metric be reconstructed (up to scaling) from knowing the conjugate points?

I guess the most famous cases are: (1) Riemannian tori without conjugate points are flat (Hopf, Burago and Ivanov) and (2) Auf-Wiedersehen metrics on the sphere (Green, Berger) are round (i.e. your ...
alvarezpaiva's user avatar
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6 votes
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Holonomy groups of compact Riemannian symmetric spaces

At the request of the OP I put my comment as an answer: in general, the holonomy group and the isotropy group have the same identity component (this is a theorem of E. Cartan). So if you assume that $...
abx's user avatar
  • 37.3k
6 votes

Bounded non-symmetric domains covering a compact manifold

A silly generalization would be a direct product of Koadaira's surface and, say, a Riemann surface. A better construction is below. I will use Griffiths, Phillip A., Complex-analytic properties of ...
Misha's user avatar
  • 31k
6 votes
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Invariants for the isotropy representation of a Riemannian symmetric space

One reference is in Helgason's 1984 book Groups and Geometric Analysis. The result you want appears there as Corollary 5.12. The notation he uses is $X=G/K$ is a symmetric space where $G$ is ...
Robert Bryant's user avatar
6 votes

Is there a contractible hyperbolic 3-orbifold of finite volume?

As pointed out by Moishe Kohan in the comments below, the following doesn't answer the question as asked, because my group $\Gamma$ is not contained in $SO(3,1)$. Anyway, here is an easy description ...
IJL's user avatar
  • 3,431
6 votes
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Buildings as generalizations of symmetric spaces

(Wanted to post as a comment but didn't have enough reputation) I have a partial answer only for your first question. Let $F$ be a non-Archimedean field. For p-adic(or $\pi-$adic) symmetric space $\...
sriram's user avatar
  • 131
6 votes

References for $K$-orbits in $G/B$

A good reference for this is "On Rationality Properties of Involutions of Reductive Groups" by Helminck and Wang (Advances in Math. 1993). The decomposition of $K\backslash G/B$ is studied ...
Paul Broussous's user avatar
5 votes
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Irreducible Symmetric Pairs

It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. Be aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-...
Friedrich Knop's user avatar
5 votes
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The uniqueness of a $K$-fixed vector in a spinor representation

There is no contradiction. To use your physical language, the fully filled state is not $K$-fixed as a vector. Rather, the group ${\rm U}(n)$ acts on it by the determinantal character. The subtle ...
Victor Protsak's user avatar
5 votes
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Help with definition of Liouville measure

The construction doesn’t really simplify on symmetric spaces. On $TM\cong T^*M$ (using the metric) consider the canonical 1-form $\alpha=“\langle p,dq\rangle”$ and symplectic form $d\alpha$ and ...
Francois Ziegler's user avatar
5 votes
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Cut locus for simply connected manifolds

The equality $D_p = C_p$ holds for compact symmetric spaces (CROSSes) of rank $1$, but not in general for higher rank symmetric spaces. A rank $1$ CROSS is isometric to a round sphere or projective ...
Jason DeVito - on hiatus's user avatar

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