If $[X, Y]=0$, then $\operatorname{Ad}_{e^{tX}}Y = Y$ for all $t\in \mathbb{R}$?

Question

Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. Let $X$ and $Y$ be commuting elements of $\mathfrak{g}$, i.e., $[X, Y]=0$. I want to show that $\operatorname{Ad}_{e^{tX}}Y = Y$ for all $t\in \mathbb{R}$. I know that tangent map of $\operatorname{Ad}$ is $\operatorname{ad}$ and $\operatorname{ad}(X)Y = [X, Y]=0$. Can this imply $\operatorname{Ad}_{e^{tX}}Y = Y$ for all $t\in \mathbb{R}$? Any hints or reference are appreciated.

Answer 1

It's a general fact that $$ {\rm Ad}_{\exp (tX) } (Y) = \exp ({\rm ad}_{tX} )(Y) = Y + [tX, Y] + \frac 1 {2!} [tX, [tX, Y]] + \frac 1 {3!} [tX, [tX, [tX, Y]]] + \dots $$ See here. (I've replaced the $X$ in the references with $tX$, but the result obviously still applies.)

And since ${\rm ad}_{tX} (Y) = [tX, Y] = t[X, Y] =0$, we have $\exp ({\rm ad}_{tX})(Y) = Y$, which gives the result you want.