Let $G,H$ be two linear lie groups (i.e. closed subgroup of $GL(n,\Bbb{R})$).
We say a continuous map $\Phi:G\to H$ is differentiable if
- For any differentiable $\gamma:(-\epsilon,\epsilon)\to G$, the map $\Phi\circ \gamma$ is differentiable.
- For any two differentiable maps $\alpha,\beta:(-\epsilon,\epsilon)\to G$ with $\alpha(0)=\beta(0),\alpha'(0)=\beta'(0)$, we have $(\Phi\circ \alpha)'(0)=(\Phi\circ\beta)'(0)$
Now I have a continuous group homomorphism $\Phi:G\to H$, we have to show that $\Phi$ is differentiable with respect to the above definition.
I have proved the case assuming $\gamma:\mathbb{R}\to G$ to be continuous group homomorphism. In that case I have $$\Phi\circ\gamma(t+s)=\Phi(\gamma(t)\gamma(s))=\Phi(\gamma(t))\Phi(\gamma(s))=(\Phi\circ\gamma)(t)(\Phi\circ\gamma)(s)$$ This shows that $\Phi\circ \gamma:\Bbb{R}\to H$ is a group homomorphism which is continuous as both $\gamma,\Phi$ is continuous.
Now I know a result which says
If $\alpha:\mathbb{R}\to G$ is a continuous group homomorphism then there is $A\in M(n,\mathbb{R})$ such that $$\alpha(t)=e^{tA}\ \forall t\in\Bbb{R}$$
Applying this result on $\Phi\circ\gamma$ we have $\Phi\circ\gamma(t)=e^{tA}$ for some $A\in M(n,\Bbb{R})$. Now $t\mapsto e^{tA}$ is differentiable, hence $\Phi\circ\gamma$ is differentiable.
But I'm unable to prove it for any differentiable map $\gamma:(-\epsilon,\epsilon)\to G$.
Can anyone help me in this regard? Thanks for help in advance.
