I am reading the corollary 21.6 in the book "Morse Theory" by John Milnor, but I've encountered a statement for which I have no ideas.

Let $G$ be a simply connected Lie group with a bi-invariant metric and Lie algebra $ \mathfrak{g} $. Then $ \mathfrak{g} $ splits as $ \mathfrak{g'} \oplus \mathfrak{c} $ of Lie algebras, where $ \mathfrak{c} $ is the center of $ \mathfrak{g} $ and $ \mathfrak{g'} $ is the orthogonal complement of $ \mathfrak{c} $. The proof of this is easy.

But then Milnor asserts that $G$ splits as a direct product $ G' \times G'' $ of Lie groups such that the Lie algebras of $G'$ and $G''$ are $ \mathfrak{g'} $ and $ \mathfrak{c} $ respectively. I have no clue of how to prove it, and the reference provided by Milnor is not useful.

Can anyone give some ideas of the proof or some useful references?


The main result you need is:

Suppose $G$ and $H$ are Lie groups with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ and that $G$ is simply connected. Let $f:\mathfrak{g}\rightarrow \mathfrak{h}$ be a Lie algebra homomorphism. Then, there is a unique $F:G\rightarrow H$ for which $d_e F = f$.

Believing this for a second, note that $G'\times G''$ is a simply connected Lie group as is $G$. Applying the above to the identity map between the two Lie algebras, we get a map $F_1:G\rightarrow G'\times G''$ as well as a map $F_2:G'\times G''\rightarrow G$.

The composition $F_2\circ F_1: G\rightarrow G$ satisfies $d_e(F_2\circ F_2) = Id$, but so does $Id:G\rightarrow G$. By uniqueness above, this implies $F_2\circ F_1 = Id$. Similarly, $F_1\circ F_2$ is the identity, so they are both Lie group isomorphisms.

The following proof of the highlighted fact can be found in Wolfgang Ziller's notes Proposition 1.20.

Given $f:\mathfrak{g}\rightarrow \mathfrak{h}$, consider the graph $\mathfrak{k}=\{(x,f(x))\in \mathfrak{g}\oplus\mathfrak{h}: x\in\mathfrak{g}\}$. Since $f$ is a homomorphism, the graph is a subalgebra. Hence, there is a unique connected subgroup $K$ of $G\times H$ with Lie algebra $\mathfrak{k}$. The projection $\pi_1:G\times H\rightarrow G$, when restricted to $K$ is a covering because $d\pi_1$, when restricted to $\mathfrak{k}$ is the identity. Since $G$ is simply connected, $\pi_1|_{K}$ is an isomorphism between $K$ and $G$. Then the map $\pi_2\circ(\pi_1|_{K})^{-1}: G\rightarrow H$ induces $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.