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2 votes

How to derive this Lie group multiplication formula?

First note that $$\frac{d}{dt}\Big|_{t=0}\exp^{tu} = u$$ by definition. Now note that $\exp^{tu}\exp^{s\nu}\exp^{-tu}$ is a product of three elements of the $G$, and we can't assume this product ...
quarague's user avatar
  • 6,133
2 votes

Non-simply connected Lie groups with nontrivial Lie algebra cohomology.

Whitehead's lemma gives $H^2(\mathfrak{g}, \mathbb{R}) = 0$ for every real semisimple Lie algebra $\mathfrak{g}$, so we can take any Lie group $G$ which is not simply connected but whose Lie algebra ...
Qiaochu Yuan's user avatar
1 vote

Applying an $SO(3)$ geodesic onto a unit vector results in a circle on a sphere

For the converse, there is a quick geometric construction. Consider the plane containing the circle and any vector $\omega=[\omega_x,\omega_y,\omega_z]^T$ normal to the plane. Pick any $b_0$ on the ...
pwensing's user avatar
1 vote

"Linear independency" of Lie Brackets

Yes, indeed they always will be whatever basis you choose in the case of $\mathfrak{so}(3)$ although this is a more in depth result and not true for every Lie algebra. Obviously the structure ...
Callum's user avatar
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