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Let $M$ be a differential manifold. Let $\mathfrak{X}(M)$ be the set of smooth vector field over $M$. Let $g$ be a finite Lie subalgebra of $\mathfrak{X}(M)$. Let $G$ be a connected, simply connected Lie group whose Lie algebra is $g$.

Suppose $g$ has the property that $\forall X\in g$, $X$ is complete. We can thus define the transformation $X_t\in\mathrm{Diff}(M)$. The group generated by all such transformations is a subgroup of $\mathrm{Diff}(M)$, denoted $H$.

If $\dim g=1$, $\exp(tX)\rightarrow X_t$ defines a homomorphism $G\rightarrow H$.

In general, how do we prove that there exists such a homomorphism?

(in Kobayashi, page 13, it is stated that the group $G$ acts locally on $M$, which, I believe, is equivalent to the existence of such a homomorphism. The homomorphism is also used in the proof of lemma 1 on the same page).

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    $\begingroup$ It's not clear to me that $G$ and $H$ are isomorphic, even when $\dim.g=1$. For example, if $M=S^1$ and $g$ is the span of the vector field $X(p)= ip$, then I think $G=\mathbb{R}$ while $H=S^1$. $\endgroup$ – Jason DeVito Feb 24 '20 at 12:43
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    $\begingroup$ Can you use the fact that if two Lie groups have the same Lie algebra then their universal covers are isomorphic? $\endgroup$ – Jason DeVito Feb 24 '20 at 14:28
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    $\begingroup$ Let $G, H$ be Lie groups and $\mathfrak{g, h}$ their Lie algebras. If $\Phi: G\to H$ is a Lie group homomorphism it is clear that $d\Phi:\mathfrak{g}\to\mathfrak h$ is a Lie-algebra homomorphism. This cannot always be turned around, but if $G$ is simply connected then it can: For any Lie-algebra homomorphism $\phi: \mathfrak g\to \mathfrak h$ there is a Lie-group homomorphism $\Phi:G\to H$ integrating this if $G$ is simply connected. For your case your group $G$ is simply connected and $\mathfrak g = \mathfrak h$, so you take as Lie-algebra homomorphism the identity. $\endgroup$ – s.harp Feb 24 '20 at 14:43
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    $\begingroup$ I see, I was assuming $H$ was already a finite dimensional Lie group. If I was going to prove the existence of the homomorphism $G\rightarrow H$, I'd probably follow Kobayashi's proof that $H$ really is a Lie subgroup, then use s.harp's argument. I do not think that Kobayashi's local action is equivalent to such a homomorphism (a priori) since the local action is only defined on a neighborhood of $e\in G$. For $U$ in Koboyashi's book, one can work as follows: there is an open neighborhood of $0\in T_e G$ for which $\exp|_V:V\rightarrow G$ is a diffeo on its image. Call the image $U$. $\endgroup$ – Jason DeVito Feb 24 '20 at 15:31
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    $\begingroup$ What you need is called the subgroup-subalgebra correspondence. If $h\subset p$ is a sub-algebra then there is a unique connected Lie subgroup (not necessarily closed) $H\subset P$ having $h$ as its Lie algebra. However this is a finite-dimensional statement, and uniqueness at least might fail if $h$ is infinite dimensional. For finite $h$ it seems reasonable that this statement is true even if $P$ is infinite dimensional, take a look at this write-up I found by googling for example, maybe the proof doesn't need finite dimensionality of $P$. math.ucla.edu/~vsv/liegroups2007/10.pdf $\endgroup$ – s.harp Feb 24 '20 at 17:38
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Here I give a proof idea:

Let $N=G\times M$ and $g_\times$ be the vector field defined by $(g,m)\mapsto(X_{\mathfrak{X}(G)}(g),X_{\mathfrak{X}(M)}(m))$. Then by Frobenius theorem there exists a unique submanifold $\mathcal{L}_m$ passing by $(e,m)$ integrating $g_\times$. We prove that $\mathcal{L}_m\simeq G$ by the first projection, and there exists $\mathrm{Diff}(\mathcal{L}_m)\rightarrow\mathrm{Diff}(M)$ a group homomorphism induced by the second projection (which concerns picking a pre-image in $\mathcal{L}_m$ the choice of which which leaves the overall operation invariant). And finally we observe that the composite morphism:

$$G\rightarrow\mathrm{Diff}(G)\rightarrow\mathrm{Diff}(\mathcal{L}_m)\rightarrow\mathrm{Diff}(M)$$

sends $\exp(tX)$ to $X_t$.

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