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

16 votes
1 answer
2k views

Construction of the Lie functor: left vs. right invariant vector fields on Lie groups and Li...

When constructing the Lie algebra $L(G)$ of a Lie group $G$, one usually uses the identification of the tangent space $T_1 G$ with left invariant vector fields $\mathcal{V}^l(G)$ to construct the Lie …
Christoph Wockel's user avatar
7 votes

Summary of Lie-Algebra integration tactics

The method of integration that I like most is via the theory of central extensions. To start with note that if $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}_n$, then the subgroup $\langle\exp(\math …
Christoph Wockel's user avatar
7 votes
0 answers
941 views

Injectivity of Lie group exponential function

If $G$ is a (finite-dimensional) Lie group, then the exponential function $\exp\colon\mathfrak{g}\to G$ is injective on some identity neighbourhood. If, moreover, $\mathfrak{g}$ is semi-simple and $\m …
Christoph Wockel's user avatar
9 votes
2 answers
467 views

Integral versus real (universal) characteristic classes

I'm pretty confused about the precise relation of the integral and the real cohomology of the classifying space $BG$ of a compact Lie group $G$. The natural map $H^n(BG;\mathbb{Z})\to H^n(BG;\mathbb{R …
Christoph Wockel's user avatar
3 votes

Splitting of a Short-Exact Sequence of Lie Groups

There is a crossed module associated to this situation, which will be trivial if $G \to \pi_0(G)$ splits on the level of gropus. Thus an obstruction for the existence of a splitting is the non-trivial …
Christoph Wockel's user avatar
6 votes

Is a measurable homomorphism on a Lie group smooth?

Perhaps it is interesting to know that for group cocycles the answer is in general no (even up to coboundaries). For instance the extension $\mathbb{Z} \to \mathbb{R} \to S^1$ is described by a 2-cocy …
Christoph Wockel's user avatar