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If $SO(N) $ is the connected subgroup of $O(N)$ that contains the identity, is it meaningful to discuss generators of $O(N)$? Can we represent elements of $O(N)$ as the exponential of some quantity in the Lie Algebra of $SO(N)$ or otherwise? Or does it only make sense to talk about the generators of $SO(N)$?

This is coming from a particle-physics perspective, so please excuse my lack of knowledge on the topic.

Eweler
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1 Answers1

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Here are two general facts that should help you:

  1. A connected Lie group is generated by any neighbourhood of the identity.
  2. The exponential map of a compact connected Lie group is surjective.

Here $\mathrm{SO}(n)$ is a compact connected Lie group, so both 1 and 2 apply. In particular, every element $A\in\mathrm{SO}(n)$ can be written as $$A=e^{X}$$ where $$X\in\mathfrak{so}(n)=\{Y\in \mathrm{M}(n,\mathbb{R}):Y+Y^\top=0\}.$$ We cannot generate $\mathrm{O}(n)$ by the exponentials of elements in $\mathfrak{so}(n)$ since by point 2, $\exp(\mathfrak{so}(n))=\mathrm{SO}(n)$ which is a subgroup of $\mathrm{O}(n)$.

However, we have $J:=\operatorname{diag}(-1,1,\ldots,1)\in\mathrm{O}(n)$, and it is not difficult to show that $\mathrm{O}(n)$ is generated by $\mathrm{SO}(n)$ and $J$.

Spenser
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