# Let $G,H$ be linear lie groups. Prove that any continuous homomorphism $\Phi:G\to H$ is differentiable.

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.

• 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. – Ahr Jan 30 at 11:26
• Can you tell what $\partial_t$ here? – MathBS Jan 30 at 12:26
• 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 $0$. – Ahr Jan 30 at 12:27
• Does this answer your question? – Moishe Kohan Feb 2 at 0:03