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Lie Groups are Groups that are additionally smooth manifolds such that the multiplication and the inverse maps are smooth.

78 votes
7 answers
8k views

Example of a manifold which is not a homogeneous space of any Lie group

Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc. What is an example of a connected smooth …
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19 votes
Accepted

Commutativity of the fundamental group of any Lie Group

As Vahid says, it is true for any topological group. Here is a proof. I'm sure there are nicer, more conceptual ones out there, but here goes. Let $G$ be your topological group. Take two loops $\s …
MTS's user avatar
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15 votes

Matrix expression for elements of $SO(3)$

There is a good way to derive the sort of thing you're looking for: use the double cover $SU(2) \to SO(3)$. $SU(2)$ is diffeomorphic to the 3-sphere $S^3 \subseteq \mathbb{C}^2$ $$ SU(2) = \left\{ \b …
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8 votes
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Criterion for nilradical of a maximal parabolic subalgebra to be abelian?

Denote by $\mathfrak{l}$ the Levi factor of the parabolic, so that $\mathfrak{p} = \mathfrak{l} \oplus \mathfrak{n}$, and note that this is a splitting as $\mathfrak{l}$-modules. Also denote by $\mat …
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3 votes
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How can one find generators of basic differential forms on homogeneous spaces?

This actually can be done in much greater generality. Let $G$ be a compact group and $K \subseteq G$ a closed subgroup. Then for any finite-dimensional representation $(V,\pi)$ of $K$ you can form the …
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2 votes

How can I tell whether a manifold is homogeneous?

Not really an answer, but too long to be a comment. I have thought about this question before, without much progress. It seems difficult. Even just to recognize if a manifold has the structure of a …
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2 votes

analytic structure on lie groups

I don't know the original reference, but you can find a proof of the theorem about real-analytic structures on Lie groups in Chapter 1 of Knapp's book "Lie Groups Beyond an Introduction." The proof u …
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1 vote

Explicit Coquasi-Triangular Quantised Coordinate Algebra of a Complex Semi-Simple Lie Group?

The coefficients of $R$ are essentially the coefficients of the braiding of the vector representation of $U_q(\mathfrak{g})$. So, more or less, you are asking for a general formula in terms of Cartan …
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