bio | website | |
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location | Los Angeles | |
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visits | member for | 3 years, 7 months |
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University lecturer.
Nov 17 |
comment |
Gaussian Curvature of Exponentiated 2-Planes
Anton: you expect what is true? |
Nov 17 |
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Gaussian Curvature of Exponentiated 2-Planes
Okay, thanks Anton. It would be nice to know what happens in the case of compact symmetric spaces. |
Nov 17 |
accepted | Gaussian Curvature of Exponentiated 2-Planes |
Nov 17 |
awarded | Popular Question |
Nov 15 |
comment |
Gaussian Curvature of Exponentiated 2-Planes
Anton, what if I add the condition that the manifold $M$ is compact? I'm specifically interested in the case of Riemannian symmetric spaces of compact type. |
Nov 15 |
revised |
Gaussian Curvature of Exponentiated 2-Planes
deleted 2 characters in body |
Nov 15 |
asked | Gaussian Curvature of Exponentiated 2-Planes |
Sep 24 |
awarded | Autobiographer |
Sep 23 |
comment |
Tori in Compact Riemannian Symmetric Spaces
@IgorBelegradek: Thanks for the reference. |
Sep 22 |
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Tori in Compact Riemannian Symmetric Spaces
I guessed that's what you meant but just wanted to make sure. Thanks again for the nice answer. |
Sep 22 |
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Tori in Compact Riemannian Symmetric Spaces
Thanks for the answer Robert but there's something puzzling me. The flats you describe above need not be tori, right? For example, if we have a symmetric space of noncompact type then the exponential map is a diffeomorphism and so the space can't contain a torus. Have I misunderstood something? |
Sep 22 |
accepted | Tori in Compact Riemannian Symmetric Spaces |
Sep 21 |
asked | Tori in Compact Riemannian Symmetric Spaces |
Jul 2 |
awarded | Curious |
Jun 24 |
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Decomposition of Lie subspaces
I can check my claim with $S^7=Spin(7)/G_2$. |
Jun 24 |
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Decomposition of Lie subspaces
For a symmetric space, the canonical connection has zero torsion and so $\frak{m}$$_1=0$. So the result is trivially true, isn't it? |
Jun 24 |
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Decomposition of Lie subspaces
Let me be a little more specific. Is it possible to view $\frak{m}$$_1$ as the Lie algebra of some subgroup of $G$? |
Jun 23 |
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Decomposition of Lie subspaces
I think this answers my question, Jose. Thanks. However, I was wondering whether there were any results of this nature in the literature. The Lie group result is well-known and I would have thought that it was generalized to reductive spaces. |
Jun 23 |
accepted | Decomposition of Lie subspaces |
Jun 23 |
comment |
Decomposition of Lie subspaces
@JoséFigueroa-O'Farrill What I meant was $T_o(\frak{m}_1,\frak{m}_1)\ne 0$. Also, in the decomposition, $\frak{m}_0$ is a subspace and $\frak{m}_1$ is a semi-simple subalgebra of $(\frak{m}$,$T_o)$. |