All Questions
Tagged with lie-groups smooth-manifolds
58 questions
4
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Fixed points of the action of an algebraic group
Hello!
If a compact Lie group $K$ acts smoothly on a smooth manifold $M$, then the set $M^K$ of fixed points under this action is a smooth submanifold of $M$. This is proved for example in ...
6
votes
1
answer
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Does a Trivial Tangent Bundle Induce a Multiplication?
Let $M$ be a connected smooth manifold, and assume that it is parallelisable; that is, its tangent bundle is trivial. Does $M$ admit an H space structure? That is, does there exist a smooth map $\mu:...
3
votes
3
answers
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analytic structure on lie groups
I need a reference for a result I have heard only very vaguely "A lie group (smooth) has a compatible analytic manifold structure".
(Would even appreciate a concise way to refer to the result..)
I ...
1
vote
1
answer
558
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Understanding manifold GL+(3,R)/SO(3) ?
I'm trying to better understand the manifold GL+(3,R)/S0(3) which is diffeomorphic to positive definite symmetric matrices. My motivation is to understand U in F = RU where F in GL+(3,R) = deformation ...
6
votes
2
answers
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Question on transversal slice of Lie group
Assume we have action of Lie group $G$ on a manifold $X$. Fix some orbit $\mathcal{O}$, it is known there exist transversal slice $S$ with respect to this orbit. Fix some point $x$ in $\mathcal{O}$, ...
6
votes
2
answers
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Lie Semigroups?
Why is a Lie group wanted instead of a semigroup, what does the group structure give? References on this would be much appreciated.
I'm currently pondering manifolds and lie groups and their ...
41
votes
3
answers
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What is the classifying space of "G-bundles with connections"
Let $G$ be a (maybe Lie) group, and $M$ a space (perhaps a manifold). Then a principal $G$-bundle over $M$ is a bundle $P \to M$ on which $G$ acts (by fiber-preserving maps), so that each fiber is a $...
12
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3
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Non-Lie Subgroups
A result of Borel and Lichnerowicz states that the holonomy group of a connection on a principal $G$-bundle is a Lie subgroup of $G$ (Cartan had earlier asserted this, but apparently without proof).
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