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

  1. For any differentiable $\gamma:(-\epsilon,\epsilon)\to G$, the map $\Phi\circ \gamma$ is differentiable.
  2. 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.

  • $\begingroup$ You have for $\gamma$ such a differentiable fonction, $\gamma'(0).\partial_t=X\in \mathfrak{g}$, now you may find a local differentiable fonction $\mu$ of some neighborhood of $0$ in $\mathbb{R}$ such that $\Phi\circ\gamma(t)=\Phi\circ exp(tX+\mu(t)))$, that should give you the result. $\endgroup$ – Ahr Jan 30 at 11:26
  • $\begingroup$ Can you tell what $\partial_t$ here? $\endgroup$ – MathBS Jan 30 at 12:26
  • 1
    $\begingroup$ Here $\partial_t$ is just the "canonical" section trivializing the tangent bundle of $(-\epsilon, \epsilon)$. If you prefer $\gamma$ is a differential map, $\gamma'(0)$ maps the tangent space $T_0(\epsilon, \epsilon)\simeq \mathbb{R}$ to $\mathfrak{g}$, such and $\partial_t$ is simply $1\in \mathbb{R}\simeq T_0(\epsilon, \epsilon)$. To sum up, $\gamma'(0).\partial_t$ is the derivative of $\gamma$ in $0$. $\endgroup$ – Ahr Jan 30 at 12:27
  • $\begingroup$ Does this answer your question? $\endgroup$ – Moishe Kohan Feb 2 at 0:03

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