Cartan closed subgroup theorem

Put $\mathfrak{g}=T_{e}G$ and define $\mathfrak{h}=\{X\in \mathfrak{g}:\exp (tX)\in H\text{ for every }t\in \mathbb{R}\}\mathrm{.}$ This set contains $0$ and is closed under real scalar multiplication. If $X,Y\in \mathfrak{h}$ and $t\in \mathbb{R}$, then for all sufficiently large positive integers $n$, [F2] gives $\exp (tX\mathrm{/}n)\exp (tY\mathrm{/}n)=\exp Z_n,Z_n=\mathrm{BCH}(tX\mathrm{/}n,tY\mathrm{/}n)\mathrm{.}$ Both factors lie in $H$. The homogeneous expansion and uniform convergence in [F2] give $n{Z}_n\to t(X+Y)$. By [F3], $\exp (n{Z}_n)=(\exp Z_n{)}^n\in H$; closedness of $H$ gives $\exp (t(X+Y))\in H$. Since $t$ was arbitrary, $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a linear subspace.

Choose a complement $\mathfrak{b}$ with $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{b}$ by [F4]. The smooth map $\mathrm{\Psi }:\mathfrak{h}\times \mathfrak{b}⟶G,\mathrm{\Psi }(X,Y)=\exp X\exp Y,$ has differential $(X,Y)\mapsto X+Y$ at $(0,0)$, an isomorphism. By [F4], after shrinking around $(0,0)$ it is a diffeomorphism onto an identity neighborhood.

We claim that some exponential neighborhood $U\subseteq \mathfrak{g}$ satisfies $H\cap \exp U=\exp (U\cap \mathfrak{h})\mathrm{.}$ The inclusion from right to left follows from the definition of $\mathfrak{h}$. If no such neighborhood existed, take a nested sequence of coordinate balls $U_n$ shrinking to $0$, all inside the injectivity domain in [F1] and with $\exp U_n$ inside the image in step 1.2. By [A1], select $h_n\in (H\cap \exp U_n)\setminus \exp (U_n\cap \mathfrak{h})\mathrm{.}$ Write $h_n=\exp X_n\exp Y_n$ using the inverse in step 1.2. Then $(X_n,Y_n)\to (0,0)$, $X_n\in \mathfrak{h}$, and $\exp Y_n\in H$. For all sufficiently large $n$, $Y_n\mathrm{\ne }0$: otherwise injectivity of the common exponential chart would put $h_n$ in $\exp (U_n\cap \mathfrak{h})$.

Fix a norm on $\mathfrak{b}$, put $c_n=\parallel Y_n\parallel$, and discard the finitely many zero terms. The unit vectors $c_n^{-1}{Y}_n$ have a convergent subsequence by [F5]; relabel it so that $c_n^{-1}{Y}_n\to Y\in \mathfrak{b}$ with $\parallel Y\parallel =1$. For arbitrary $t\in \mathbb{R}$, choose the integer $k_n=\lfloor t\mathrm{/}{c}_n\rfloor$. Then $k_n{c}_n\to t$, so $k_n{Y}_n\to tY$. By [F3], $\exp (k_n{Y}_n)=(\exp Y_n{)}^{k_n}\in H\mathrm{.}$ Closedness gives $\exp (tY)\in H$. Since this holds for every $t$, $Y\in \mathfrak{h}\cap \mathfrak{b}=\{0\}$, contradicting $\parallel Y\parallel =1$. The claim in step 2.1 follows.

Choose a linear coordinate isomorphism $E:\mathfrak{g}\to {\mathbb{R}}^m$ carrying $\mathfrak{h}$ to ${\mathbb{R}}^k\times \{0\}$. By step 2.1, the chart $E\circ \log$ on $\exp U$ sends $H\cap \exp U$ to the coordinate slice $E(U)\cap ({\mathbb{R}}^k\times \{0\})$. For each $h\in H$, left translation carries this chart to a slice chart at $h$ because $L_h(H)=H$. Hence [F6] makes $H$ an embedded submanifold with its subspace topology.

Ambient multiplication and inversion preserve $H$. In the slice charts from step 3.2 their restrictions have smooth coordinate representatives, so they make $H$ a Lie group and its inclusion into $G$ a smooth embedded homomorphism. Its tangent space at $e$ is $\mathfrak{h}$, and [F6] confirms that this space is bracket closed.

Any other smooth structure making the same subset $H$ an embedded Lie subgroup has the same subspace topology by definition. In every ambient slice chart from step 3.2, both intrinsic structures use the restriction to the same Euclidean slice, so the identity between them is locally a diffeomorphism and hence globally a diffeomorphism. This proves uniqueness. The cases $H=\{e\}$, $H=G$, dimensions zero and one, and disconnected $H$ are included. A subgroup contains the identity, so the empty case cannot occur; no metric, nondegeneracy, manifold-boundary, or interval-endpoint hypothesis occurs. Choice is used exactly as stated in [A1] and through [F1]–[F3].