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4 votes
1 answer
241 views

Let $G \subset \mathrm{GL}(n)$ be a matrix Lie group. Does there exist an affine connection under which the matrix and manifold exponential coincide?

Let $G\subset \operatorname{GL}(n)$ be a matrix Lie group. I am curious about curves $\gamma(t) = g \exp(tv)$, where $g \in G$, $v \in \mathfrak{g}$, and $\exp(.)$ is the matrix exponential. If we ...
Spencer Kraisler's user avatar
7 votes
1 answer
387 views

Torsion-free Cartan connections

Let $M$ a differentiable manifold and $H\subset G$ a Lie group with a closed subgroup such that $G/H$ is connected. A $H\subset G$-Cartan connection on $M$ can be defined by A principal $G$-bundle on ...
jpdm's user avatar
  • 141
3 votes
0 answers
155 views

Maximally symmetric affine manifold

As a physicist who knows (something) about General Relativity, I'm accustomed to the term "maximally symmetric space" being an $n$-dimensional manifold with $\frac{n(n+1)}{2}$ Killing vectors. A ...
Dox's user avatar
  • 690
3 votes
1 answer
456 views

Left invariant connections on a Lie group

The exponential map associated with the (-) - connection on a Lie group is generally not surjective. This is because, for this connection, the one-parameter subgroups and geodesics coincide. If we ...
Oliver Jones's user avatar
  • 1,378
5 votes
2 answers
662 views

Most natural connection on Lie group: comparison of different pictures

Let $G$ be a Lie group (not necessarily compact). One can equip $G$ with the left invariant metric (or right invariant but in general there is no biinvariant metric in the noncompact case). Once the ...
truebaran's user avatar
  • 9,330