0

I am taking a course in Lie algebra and due to the fact that I am struggling with the concepts I decided to try to solve most of the exercises of a problem sheet, here I came across with this problem and thought it was easy at first but now I'm stuck:

a) Find the centers of $SU_{n}$, $SO_{n}(\mathbb{R})$ and $SL_{n}(\mathbb{C})$
b) Find all connected Lie groups with the tangent Lie algebra $\mathfrak{su}_{n}(\mathbb{C})$, $\mathfrak{sl}_{n}(\mathbb{C})$, where $n$ is prime.

I solved the first part without any problems, but in the second half of the problem, I had no idea how to continue.
So, for the center of $SU_{n}$ i got $n\mathbb{Z}$.
For the center of $SO_{n}(\mathbb{R})$ I got $\{\pm I\}$ when $n\geq3$ and itself when $n=2$.
And for the center of $SL_{n}(\mathbb{C})$ I got $\{\lambda I:\lambda^{n}=1\}$. Now, how should I proceed with the second part of the problem? Any help would be really appreciated.

AdrinMI49
  • 706
  • 1
    Over which field are the Lie groups and Lie algebras in part b)? Have you seen some posts here, e.g., this one? Start with $\mathfrak{sl}(2)$. – Dietrich Burde Oct 24 '23 at 08:44
  • I would say they are over any field, as the problem doesn't state any. Thank you very much, I'll see if that answer helps. Besides I was talking with a friend in the class and she told me that if I find a discrete subgroup that could help me find all the connected groups. – AdrinMI49 Oct 24 '23 at 09:50
  • All fields? So don't forget to consider the field of $p$-adic numbers $\Bbb Q_p$, and $p$-adic Lie groups, see here, for example. – Dietrich Burde Oct 24 '23 at 13:40
  • Following your first question, I asked today my profesor for clarification and he told me he forgot to include the field, which is ℂ, Ill edit the post, and I think I got the answer, so I will try to respond myself haha – AdrinMI49 Oct 24 '23 at 16:10
  • When you say that the center is $\mathbb{Z}$, which subgroup isomorphic to $\mathbb{Z}$ do you mean? The center of $\operatorname{SU}(n)$ is actually much larger than $\mathbb{Z}$ anyway. It is a torus, the product of $n-1$ circles. – Joshua Tilley Oct 28 '23 at 01:39

1 Answers1

0

As I understood the answer was really simple, but I may be wrong.
We know that the center of a group is isomorphic to $\mathbb{Z}^{n}$ and $\mathbb{Z}^{n}$ is a discrete space, so we can characterize all connected Lie groups with the following quotients: $$\mathfrak{su}_{n}(\mathbb{C})/z_{1},$$ and $$\mathfrak{sl}_{n}(\mathbb{C})/z_{2},$$ where $z_{1}$ is the center of $SU_{n}(\mathbb{C})$ and $z_{2}$ is the center of $SL_{n}(\mathbb{C})$

AdrinMI49
  • 706