Closed-subgroup theorem

Every closed subgroup of a finite-dimensional Lie group is an embedded submanifold and a Lie subgroup. Closedness alone forces the compatible smooth structure; smoothness need not be assumed in advance. The proof constructs the Lie algebra of a closed subgroup from one-parameter subgroups, then uses local product coordinates for a closed subgroup.

Choose a vector subspace $\mathfrak{m}$ complementary to the Lie algebra of a closed subgroup $H$. The differential of $F$ at $(0,0)$ is $(U,V)\mapsto U+V$, so the inverse function theorem makes $F$ a local diffeomorphism. Sufficiently small nonzero $U\in \mathfrak{m}$ cannot satisfy $\exp U\in H$: otherwise limit directions of a closed subgroup give a nonzero element of $\mathfrak{m}\cap \mathfrak{h}$. Consequently $H$ is locally exactly the coordinate slice $U=0$, proving that it is an embedded submanifold.

For a closed subgroup $H$ of a Lie group, this set is a real Lie subalgebra. Scalar closure is immediate; the Lie product formula in a Lie group and closedness of $H$ give addition. It is a closed vector subspace, also directly by continuity of the Exponential map of a Lie group. Conjugating its one-parameter subgroups by elements of $H$ preserves it. Differentiating the Adjoint representation of a Lie group along $\exp (sX)$ gives $[X,Y]\in \mathfrak{h}$ for $X,Y\in \mathfrak{h}$. After the closed-subgroup theorem establishes smoothness, $\mathfrak{h}$ is exactly the tangent space $T_{e}H$.

Let $H$ be a closed subgroup of a Lie group $G$, and let nonzero vectors $X_j$ in the ambient Lie algebra $\mathfrak{g}$ tend to zero with $\exp X_j\in H$ and $X_j/\Vert X_j\Vert \to X$. For any real $t$, choose integers $m_j$ with $m_j\Vert X_j\Vert \to t$. Then $(\exp X_j{)}^{m_j}=\exp (m_j{X}_j)\to \exp (tX)$. Closedness gives $\exp (tX)\in H$. Compactness of the unit sphere supplies a limit direction from every sequence approaching the identity through nonidentity subgroup elements.