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University lecturer.

Jul
8
comment Connections on a Lie Group
By the canonical connection I mean the canonical connection of the second kind in the sense of Nomizu.
Jul
2
awarded  Curious
Jun
24
comment Decomposition of Lie subspaces
I can check my claim with $S^7=Spin(7)/G_2$.
Jun
24
comment Decomposition of Lie subspaces
For a symmetric space, the canonical connection has zero torsion and so $\frak{m}$$_1=0$. So the result is trivially true, isn't it?
Jun
24
comment Decomposition of Lie subspaces
Let me be a little more specific. Is it possible to view $\frak{m}$$_1$ as the Lie algebra of some subgroup of $G$?
Jun
23
comment 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 $(\frak{m}$,$T_o)$.
Jun
23
asked Decomposition of Lie subspaces
May
21
comment Stiefel manifolds and polar decompositions
Thanks for the suggestions; I'll look into it.
May
20
comment Decomposition of $S^7=Spin(7)/G_2$
@ClaudioGorodski If you view $Spin_7$ as $8\times 8$ real matrices, what do you get for $\text{exp}\> {\frak p}$?
May
20
comment Decomposition of $S^7=Spin(7)/G_2$
@JoséFigueroa-O'Farrill I think the situation is the same. By $SO(3,1)$ I mean the proper orthochronous Lorentz group $SO^+(3,1)$, which is connected.
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
@Deane: An example of what I'm talking about. Take hyperbolic space $H^3=SO(3,1)/SO(3)$. This is a Riemannian symmetric space of noncompact type. Elements of the Lorentz group $SO(3,1)$ can be decomposed into a product of a rotation and a boost. This is by virtue of the fact that we have a Cartan decomposition at the Lie algebra level. The boosts $B({\bf u})$ can be parameterized by elements ${\bf u}\in H^3$ and it can be shown that $B({\bf u})B({\bf v})$=$B({\bf u}+{\bf v})O$ where $O$ is a rotation and ${\bf u}+{\bf v}$ is the relativistic velocity addition.
May
19
awarded  Critic
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
@Andre: We can consider elements of $Spin(7)$ as $7\times 7$ real matrices. What I'm looking for is a type of polar decomposition in terms of $G_2$. This doesn't have to be complicated.
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
@Andre: I'm making no such assumption. I'm not even approaching the problem from the bundle point of view. As I already mentioned, the result I'm using is in Helgason and it uses the exponential map, not bundles. Results of this type are generally not easy; see my earlier post on Stiefel manifolds: <mathoverflow.net/questions/139542/>;. However, in the case of $S^7$, I wonder if the octonionic structure doesn't help.
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
That's a nice description in terms principal bundles as to why there should be a local decomposition. However, without an explicit diffeomorphism $\pi^{-1}U\simeq U\times G_2$, it doesn't tell me how to decompose an element of $Spin(7)$.
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
@Ryan: I'm being sloppy; all these equalities are potentially just holding in a neighborhood of some chosen point of the manifold. They can be global however. I just don't know for $S^7$.
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
@Ryan: You're saying that that isn't true locally in some neighborhood?
May
19
comment Decomposition of $S^7=Spin(7)/G_2$
@Andre: A result in Helgason says that such a decomposition exists for reductive spaces; at least locally. In the case of $S^2$ we have $SO(3)=SO(2)\times S^2$.