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I'm self-working on two theorems on Lie transformation group from the book Kobayashi transformation group in differential geometry, one is the following

Theorem Let $\mathfrak{S}$ the group of differentiable transformation of manifold $M$ and $\mathcal{S}$ be the set of all vector fields $X\in\mathfrak{X}(M)$ which generates 1-parameter subgroup $\varphi_{t}=\text{exp}(tX)$ of transformation for $M$ such that $\varphi_{t}\in\mathfrak{S}$. The set $\mathcal{S}$ with brackets of vector field define a Lie algebra. Then if $\mathcal{S}$ is finite-dimensional Lie algebra of vector fields on $M$ then $\mathfrak{S}$ is Lie group of transformation and $\mathcal{S}$ its Lie algebra.

the idea of the proof is quite simple, take the Lie algebra $\mathfrak{g}^{*}$ generated by $\mathcal{S}$, if $\mathfrak{g}^{*}$ if finite-dimensional then by Lie third theorem there exist simply connected Lie group $\mathfrak{S}^{*}$ the last can be chosen such that $\mathfrak{S}^{*}\subset \mathfrak{S}$ using local action etc ... , even more it can be shown is normal connected subgroup and open, for now, everything seems normal but then the author claiming that the smooth structure can be transformed to another connected component or simply say to $\mathfrak{S}$ if I'm not wrong, this where I can't get it, unless we accept that the left translation or the right for $g\in\mathfrak{S}$, the mapping $L_{g}:\mathfrak{S}^{*}\longrightarrow g.\mathfrak{S}^{*}$ to be differentiable.

My question is about the idea of how we transfer smooth structure from connected normal subgroup?

There's a paper of Richard S.Plais A global formulation of the lie theory of transformation groups, contains lots of stuff about this, but I couldn't see where the answer precisely.

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Since $\mathcal{G}^*$ is an open subgroup of a topological group $\mathcal{G}$, it is a union of topological components of $\mathcal{G}$. The proof: if some point $g$ lies outside of $\mathcal{G}^*$, then left translation by that point moves $\mathcal{G}^*$ to an open set $g\mathcal{G}^*$. If this set is not disjoint from $\mathcal{G}^*$, say it overlaps in some point $gh$, then since $\mathcal{G}^*$ is a subgroup, and $g\mathcal{G}^*$ contains $gh$, $\mathcal{G}^*$ contains $h$, and so contains $h^{-1}$, and therefore contains $ghh^{-1}=g$, a contradiction. So $g\mathcal{G}^*$ is an open set around $g$ not intersecting $\mathcal{G}^*$, hence $\mathcal{G}^*$ is closed. So closed and open, and so a component. We therefore define a smooth structure on $\mathcal{G}$ by left translation of the smooth structure of $\mathcal{G}^*$. This is the same as right translation of the smooth structure, because the adjoint action of $\mathcal{G}$ on $\mathcal{G}^*$ preserves the smooth structure, i.e. acts by diffeomorphisms. (This last sentence is proven in Kobayashi's book somewhere, if I remember correctly.)

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  • $\begingroup$ correct me if i'm wrong so if any point $g\in mathcal{G}$ taking chart of identity in $mathcal{G}^{*}$, $(U,\phi)$, then $(g.U,\phi(g^{-1}))$ can considered as chart of $g$ in $mathcal{G}$. $\endgroup$ – Bey Alexander Jun 20 at 8:23
  • $\begingroup$ Yes, you left translate the charts around. $\endgroup$ – Ben McKay Jun 20 at 8:29

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