Theorem (Yamabe): Let $G$ be a Lie group, and let $H$ be an arc-wise connected subgroup of $G$.Then $H$ is a Lie subgroup of $G$.

I am reading this theory form an appendix in a book called (bilinear control systems, Elliott) hence there is no enough details about proofs.. But for this theorem there was a discussion instead of a proof, which I need help to understand initially for the matrix Lie groups case.

Here is the discussion: Let $G$ be a (real) matrix Lie group and $H$ be an arc-wise connected subgroup of $G$. Let $\mathfrak{h}$ be the set of $n \times n$ matrices $L$ with property that for any neighborhood $U$ of the identity matrix in $G$, there exists an arc $\alpha: [0,1] \rightarrow H$ s.t $\alpha(0)=I,\alpha(t)\in e^{tL}U , 0\leq t\leq 1$. First show that if $X\in \mathfrak{h}$ then so does $\lambda X$, $\lambda \in\mathbb{R}$. Use (Lie product theorem) to show that for sufficiently small $X$ and $Y$, $X+Y \in \mathfrak{h}$. Therefore $\mathfrak{h}$ is a matrix Lie algebra. Use the Brouwer fixed point theorem to show that $H$ is the connected matrix Lie subgroup of $G$ that corresponds to $\mathfrak{h}$.

You can use another notations and symbols if you wish. Thanks in advance :-)


I found a proof on page 354 of Structure and Geometry of Lie Groups (Hilgert and Neeb) with a free google preview. It uses the Brouwer fixed point theorem, and $A(H)$ is the sub algebra you describe.


Since $H$ is a subgroup of $G$, all that need to be verified is that $H$ is a (differentiable) manifold. But as $G$ is a manifold, any (arcwise) connected component of $G$ is also a manifold.

  • 4
    $\begingroup$ How do we know that $H$ is a whole connected component of $G$? $\endgroup$ – sds Jun 5 '13 at 20:27

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