Let $\mathfrak{g}$ be a Lie algebra of Lie group $G$. Consider two sets $S_1$ and $S_2$ consisting of elements from $\mathfrak{g}$. Now suppose that $[S_1, S_1]= S_1$, which will mean that elements in set $S_1$ form a Lie subalgebra, while in general $[S_2, S_2] \not\subset S_2 \text{ and }[S_2, S_2] \not \subset S_1$ . Also suppose that $[S_1, S_2]= \mathfrak{g}$. In other words, elements from both sets form the entire Lie algebra.

Now, consider the following $z=log(e^x e^y)$ where $ x\in S_1$ and $y \in S_2$. I would like to show that in general $ z\not \in S_1 $ and $z \not \in S_2$. It seems to me that the proof should formalize the following intuition:

  1. Quite clearly, $z$ will in general not be an element of $S_1$ since BCH will involve commutators with elements from $S_2$.

  2. Also, for any $x \in S_1$ and $y \in S_2$, $z$ will be in a subalgebra generated by $x$ and $y$; since the commutators between elements in $S_1$ and $S_2$ together form the entire Lie algebra there will be a some $x \in S_1 $ and $y \in S_2$ such that $z \not \in S_1$ and $z \not \in S_2$.

  3. If it was possible that for all $x$ and $y$ , $z \in S_1 $ or $z \in S_2$, then (I think) the $\log$ would not be an isomorphism between sets in the neighborhood of the $0$ and sets in the neighborhood of $I$.

I want to set it up by contradiction, but I don't know how to deal with the sets rather than just the existence of a specific elements $x$ and $y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.