In Hall Chap. $3$ Example $3.41$, it is shown that the exponential map is not surjective for $SL_2(\mathbb{C})$, by showing that there exists no matrix $X\in sl_2(\mathbb{C})$ such that $e^X = \begin{pmatrix} -1 & 1\\ 0 & -1\end{pmatrix}$.

Hall also later goes on to show in Chap. $3$ Corollary $3.47$ that for a connected matrix Lie group, every element $A\in G$ can be written in the form $A = e^{X_1}e^{X_2}\cdots e^{X_m}$ for $X_1, X_2, \cdots, X_m\in g$, where $g$ is the Lie algebra of $G$.

Now this is where my confusion arises. According to the Baker-Campbell-Hausdorff (BCH) formula, we can write $e^Xe^Y=e^Z$, where $Z = X + Y + \frac{1}{2}[X,Y]+\cdots$. Now, $sl_2(\mathbb{C})$ contains all traceless matrices, and $[X,Y]$ is traceless for any $X,Y$. So,$X,Y\in sl_2(\mathbb{C})$ should imply that $Z\in sl_2(\mathbb{C})$ too. So, if we iteratiely apply the BCH formula to $A = e^{X_1}e^{X_2}\cdots e^{X_m}$ (which we can, as $SL_2(\mathbb{C})$ is connected), we should end up with $A = e^U$ for $U\in sl_2(\mathbb{C})$, which contradicts what was proved at the start.

So where am I going wrong?


1 Answer 1


The BCH formula doesn't converge for all $X$ and $Y$, so you can't apply it in all cases.


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