I am reading J.J.Duistermaat and J.A.C.Kolk's Lie Groups. I cannot figure out (1.8.6). The textbook said and I quote:
Because $ad$ is a Lie algebra homomorphism: $\mathfrak g\to\mathfrak{gl(g)}$, one has $e^{ad(ad\,X)}\circ ad\,Y=ad(e^{ad\,X}\,Y)$.
I know that a Lie algebra homomorphism is a function that preserve the Lie bracket, but I cannot figure out how to prove the statement (1.8.6).
I think there is a typo there (as in the book has a typo). That $\circ$ should be a $\cdot$. As written, the left hand side lives in a weird place:
$\operatorname{ad}X \in \mathfrak{gl}(\mathfrak{g})$, so $\operatorname{ad}(\operatorname{ad}X) \in \mathfrak{gl}(\mathfrak{gl}(\mathfrak{g}))$ and $e^{\operatorname{ad}(\operatorname{ad}X)}$ is an automorphism on $\mathfrak{gl}(\mathfrak{g})$. It doesn't make sense to compose this with something from $\mathfrak{gl}(\mathfrak{g})$ in general. However it makes lots of sense for it to act on it:
$$ e^{\operatorname{ad}(\operatorname{ad}X)}\cdot \operatorname{ad}Y = \operatorname{ad}Y + [\operatorname{ad}X, \operatorname{ad}Y]+ \frac{1}{2}[\operatorname{ad}X,[\operatorname{ad}X,\operatorname{ad}Y]] + \cdots$$
Then the statement follows from the Jacobi identity. The Jacobi identity can be framed as follows:
$$ [\operatorname{ad}X, \operatorname{ad}Y] = \operatorname{ad}[X,Y]$$
