# Are compact Lie algebras necessarily compact as a set of matrices?

I'm reading through a paper and came across something confusing; my limited experience with Lie theory is a bit of a hindrance:

The author starts with a compact set of matrices (in the usual Euclidean topology) $\{A_i\}\subset \mathbb{R}^{n \times n}$ and generates a Lie algebra with Levi decomposition $\mathfrak{g} = \mathfrak{s}\oplus\mathfrak{r}$. Under the assumption that $\mathfrak{s}$ is a compact Lie algebra, I'm curious about the compactness of the set $\{e^{-s}r_i e^{s}\,:\,s\in \mathfrak{s}, A_i = s_i + r_i\}$. Clearly the set of $r_i$ is compact (in the usual Euclidean topology), but it seems to me that as a set, $\mathfrak{s}$ could possibly be non-compact (in Euclidean sense) even though it's declared a compact Lie algebra (since in my reading of compact Lie algebras, I haven't come across any of the familiar "closed and bounded" type assertions).

Any help would be greatly appreciated, thanks!

• Here is a simple example: take $\mathbb{R}$ as a trivial one dimensional lie algebra (so Lie Bracket is zero ) then the map $e^{2\pi. i x}: \mathbb{R} \rightarrow S^{1}$ makes $S^{1}$ the unit circle into a compact lie goup. – DBS Apr 1 '15 at 20:13
• DBS: I like that example, thanks. So, being the Lie algebra of a compact Lie group, would $\mathbb{R}$ be considered a compact Lie algebra? – Ironbeard Apr 1 '15 at 20:25
• I wouldn't say that. A Lie group gives rise to a Lie algebra where as going from a Lie algebra to a Lie group is little ambiguous. For example you could have taken the usual exponential lmap $e^{x}: \mathbb{R} \rightarrow \mathbb{R}_{+}$ and arrive at the real non-compact lie group $\mathbb{R}_{+}$. Technically mine is not an example because we are dealing with "tori" but this is not an isolated phenomenon and leads up to compact real forms etc etc. – DBS Apr 1 '15 at 22:56
• It is best to think that a compact Lie algebra is the lie algebra of a compact lie group. But the same Lie algebra structure can also give rise to a non-compact group so it is not an intrinsic property of the lie algebra itself. A Lie algebra is a vector space (over reals/complex) so it is never compact. However if the group that it gives rise to is compact then we have an action of a compact group of a vector space (by the adjoint action) and hence nice things can be said about this action (see wikipedia). – DBS Apr 1 '15 at 23:01
• Much clearer now, thanks for taking the time! – Ironbeard Apr 2 '15 at 20:15

## 2 Answers

Compact Lie groups are topological groups whose topology is compact. A compact Lie algebra is the Lie algebra of a compact Lie group, therefore the name "compact". Lie algebras are vector spaces, and all finite-dimensional Lie algebras are linear, i.e., subalgebra of the Lie algebra of matrices. For Lie algebras "compact" just means that they are reductive with negative-definite Killing form.

• Thanks for the response. If I understand you correctly, then assuming $\mathfrak{s}$ is a compact Lie algebra, it's Lie group (the set $\{e^{st}\, : \, s\in \mathfrak{s}, t\geq 0\}$ with matrix multiplication) is compact (in the usual sense), thus the set in question (\{e^{-s}r_i e^{s}\, : \, s\in \mathfrak{s}, A_i = s_i + r_i \}$) would be compact (in the usual sense). Does this seem right to you? – Ironbeard Apr 1 '15 at 20:17 Lie algebras are vector spaces, in this case subspaces of the vector space of matrices. Over the real and complex numbers, vector spaces are never compact unless they are trivial. • Of course! Thanks for pointing that out in an easy to understand manner. Then it follows that the set in question could not be compact in$\mathbb{R}^{n \times n}$unless$\mathfrak{s}$is trivial. Edits: English – Ironbeard Apr 1 '15 at 20:07 • Actually, in light of Burde's answer, I suppose it's still possible for the set in question to be compact in the usual topology if the set$\{e^s\,:\,s\in \mathfrak{s}\}\$ is. – Ironbeard Apr 1 '15 at 20:20
• @Ironbeard the exponential is often compact. – Matt Samuel Apr 1 '15 at 20:25