# Semisimple Lie algebras with invariant inner-products give compact Lie groups [duplicate]

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Question. Let $\mathfrak{g}$ be a real semisimple Lie algebra admitting an invariant inner-product. Is every connected Lie group with Lie algebra $\mathfrak{g}$ compact?

I know that the converse is true: If $G$ is a compact connected Lie group, then the Haar measure may be used to give an invariant inner-product on $\mathrm{Lie}(G)$. Also, semisimplicity is necessary since $\mathrm{Lie}(\mathbb{R})=\mathbb{R}$ trivially admits an invariant inner-product.

## marked as duplicate by Dietrich Burde, Leucippus, Claude Leibovici, naslundx, WatsonSep 13 '16 at 9:12

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## 2 Answers

Yes, that's true: If $\langle-,-\rangle$ is an invariant inner product on ${\mathfrak g}$, then $\text{Aut}({\mathfrak g})^{\circ}\subset\text{O}({\mathfrak g},\langle -,-\rangle)$ is at least some compact Lie group with Lie algebra ${\mathfrak g}$. Next, the fundamental group of a compact semisimple Lie group is finite (see e.g. https://mathoverflow.net/questions/95637/connected-compact-semisimple-lie-group-finite-fundamental-group), and applying this to $\text{Aut}({\mathfrak g})^{\circ}$ shows that the simply connected Lie group with Lie algebra ${\mathfrak g}$ - call it $\widetilde{G}$ - is compact, too. Finally, an arbitrary connected Lie group $G$ with Lie algebra ${\mathfrak g}$ is a quotient of $\widetilde{G}$, hence compact.

I think this is true, if by "inner product" you mean either positive or negative definite symmetric real bilinear form.

If $\mathfrak{g}$ admits an invariant inner product $<,>$, then in particular the $Ad$ representation of $G$ on $\mathfrak{g}$ must preserve it, meaning that $Ad:G\to O(\mathfrak{g},<,>)$. Of course $O(\mathfrak{g},<,>)$ is a compact group, and since $G$ is semisimple, the kernel of $Ad$ (i.e. $Z(G)$, the center), is a discrete subgroup of $G$.

Hence we have that $G/Z(G)$ is compact. Now we use what is sometimes known as Weyl's theorem: if $H$ is a compact semisimple Lie group, then $\pi_1(H)$ is finite. If we know this, then $G\to G/Z(G)$ must be a finite cover, and since $G/Z(G)$ is compact, $G$ will be compact too.

To prove Weyl's theorem let $H$ be a compact semisimple Lie group, $\mathfrak{h}=Lie(H)$. We use that we have an isomorphism $H_{dr}^1(H,\mathbb{R})\cong H^1(\mathfrak{h},\mathbb{R})$ between the 1st de Rham cohomology of $H$ and the first Lie algebra cohomology of the trivial $\mathfrak{h}$-module $\mathbb{R}$. Since $\mathfrak{h}$ is semisimple, $H^1(\mathfrak{h},\mathbb{R})=0$, so $H_1(H,\mathbb{Z})$ must be finite (since it finitely generated since $H$ is compact), and so $\pi_1(H)\cong H_1(H,\mathbb{Z})$ is finite, using that $\pi_1(H)$ is abelian ($H$ is a topological group).