Closed subgroup theorem

Remarks

The proof first shows that $\mathfrak{h}$ is a linear subspace, using the limits $\exp(X + Y) = \lim_n (\exp(X/n)\exp(Y/n))^n$ and closedness of $H$. Then one chooses a complement $\mathfrak{m}$ of $\mathfrak{h}$ in $\mathfrak{g}$ and shows that $(Y, X) \mapsto \exp(Y)\exp(X)$ on $\mathfrak{m} \times \mathfrak{h}$ is a chart near $e$ in which $H$ is the slice $Y = 0$. The key point is that $H$ has no elements of the form $\exp(Y)$ with small nonzero $Y \in \mathfrak{m}$; otherwise a limit argument gives a nonzero element of $\mathfrak{h} \cap \mathfrak{m}$. Example: $SO(n) = \set{A \in GL_n(\R) : A^T A = I, \det A = 1}$ is closed, so it is a Lie group; its Lie algebra is the space of skew-symmetric matrices. The closedness hypothesis is necessary: the image of $t \mapsto (e^{it}, e^{i\sqrt2 t})$ in the torus $S^1 \times S^1$ is a subgroup that is dense and not closed, and it is not an embedded submanifold.

Going deeper

For a student

The theorem is often stated for matrix groups and attributed to von Neumann in that case and to Élie Cartan in general. A typical use: the stabiliser $H$ of a point under a smooth action is closed, hence a Lie subgroup. Then the quotient $G / H$ is a manifold, a homogeneous space. The formula for $\mathfrak{h}$ is the practical way to compute Lie algebras: for $SL_n(\R)$ it gives the matrices of trace $0$, because $\det e^{tX} = e^{t \tr X}$. It also gives Cartan's theorem that a continuous homomorphism of Lie groups is smooth, by applying it to the graph.