differential-topologylie-algebraslie-groups
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!!!!!!!!
Left and right translations are not Lie group homomorphisms; they don't even preserve the identity, and the induced map on Lie algebras is obtained by looking at derivatives at the identity. However, conjugation by a fixed element $g \in G$ is, and the induced map on $\mathfrak{g}$ gives a representation $G \to \text{Aut}(\mathfrak{g})$ called the adjoint representation. This is itself a Lie group homomorphism, and differentiating it gives the adjoint representation
$$\mathfrak{g} \ni x \mapsto (y \mapsto [x, y]) \in \text{Der}(\mathfrak{g})$$
of $\mathfrak{g}$ (and this is one way to define the Lie bracket).
Here is a better (more elegant) proof for both questions using Cartan-Von Neumann's theorem on closed subgroups.
Let $H$ be the graph of $\gamma$, then $H$ is closed (and thus a Lie subgroup of $G\times \mathbb{R} \cong \mathbb{R}\times G$, so $H\hookrightarrow \mathbb{R}\times G$ is smooth) The projection from $H$ to $\mathbb{R}$ is a bijective lie group morphism, and therefore its inverse is smooth. Now just have a look at the following diagram
As for the second one, if $\varphi : G\to H$ is continuous, let $\Gamma_\varphi : G\to G\times H$ be the "graph map" which sends $g$ to $\big(g, \varphi(g) \big)$. Note that $\Gamma_\varphi $ is a group homeomorphism from $G$ onto its image $\mathrm{Graph}(\varphi)$ whose inverse is the restriction of the projection $G\times H \to G$. Now, $\mathrm{Graph}(\varphi)$ is clearly closed (therefore a Lie subgroup), and the inverse of $\Gamma_\varphi$ is smooth with constant rank, and thus $\Gamma_\varphi $ is a diffeomorphism. If $q: G\times H \to H$ is the second projection (smooth), you just have to note that $\varphi=q\circ \Gamma_\varphi $ and we are done :)
Best Answer
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 $$d\phi(\mathrm{ad}(X) Y) = \mathrm{ad}(d\phi(X)) d\phi(Y)$$ $$\implies d\phi([X, Y]_{\mathfrak{g}}) = [d\phi(X), d\phi(Y)]_{\mathfrak{h}}.$$