2

Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{R}$ or $\mathbb{C}$. Then there exists a connected and simply connected Lie group $G$ whose Lie algebra is $\mathfrak{g}$. The question is, is there any general way to find $Z(G)$, the center of $G$, from $\mathfrak{g}$? Thanks in advance.

Seewoo Lee
  • 15,670
  • 1
    It's been a long time, but I believe the weight lattice modded out by the root lattice gives you the center. Perhaps this only works for semi-simple things... – Jason DeVito - on hiatus Sep 20 '18 at 18:54
  • Note that $\mathrm{SL}_2$ and $\mathrm{PGL}_2$ both have (up to isomorphism of Lie algebras) $\mathfrak{sl}_2$ as their Lie algebra. But the first group has a centre, and the second does not. – Joppy Sep 20 '18 at 23:54
  • @Joppy Why $\mathrm{PGL}{2}$ has a Lie algebra $\mathfrak{sl}{2}$? Isn't it $\mathfrak{gl}_{2}$? – Seewoo Lee Sep 21 '18 at 02:37
  • 1
    $\mathrm{PGL}_2$ is three-dimensional as a Lie group, so its Lie algebra must also be three-dimensional. – Joppy Sep 21 '18 at 02:41
  • 1
    @Joppy I see...but there should be a unique connected and simply connected Lie group that corresponds to a given finite-dimensional Lie algebra over $\mathbb{R}$ or $\mathbb{C}$. – Seewoo Lee Sep 21 '18 at 05:14
  • This answer by Marc v. Leeuwen provides detailed insight into this: https://math.stackexchange.com/a/1755110/96384 I tried to work through an example in my answer to this: https://math.stackexchange.com/q/2596716/96384 – Torsten Schoeneberg Sep 26 '18 at 20:47

0 Answers0