Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I am having a difficult time showing that if $\phi: G \rightarrow H$ is a Lie group homomorphism, then $d\phi: \mathfrak{g} \rightarrow \mathfrak{h}$ satisfies the property that for any $X, Y \in \mathfrak{g},$ we have that $d\phi([X, Y]_\mathfrak{g}) = [d\phi(X), d\phi(Y)]_\mathfrak{h}$.
Any help is appreciated!!!!!!!!
Let $x \in G$. Since $\phi$ is a Lie group homomorphism, we have that $$\phi(xyx^{-1}) = \phi(x) \phi(y) \phi(x)^{-1} \tag{$\ast$}$$ for all $y \in G$. Differentiating $(\ast)$ with respect to $y$ at $y = 1$ in the direction of $Y \in \mathfrak{g}$ gives us $$d\phi(\mathrm{Ad}(x) Y) = \mathrm{Ad}(\phi(x)) d\phi(Y). \tag{$\ast\ast$}$$ Differentiating $(\ast\ast)$ with respect to $x$ at $x = 1$ in the direction of $X \in \mathfrak{g}$, we obtain $$d\phi(\mathrm{ad}(X) Y) = \mathrm{ad}(d\phi(X)) d\phi(Y)$$ $$\implies d\phi([X, Y]_{\mathfrak{g}}) = [d\phi(X), d\phi(Y)]_{\mathfrak{h}}.$$
