MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello everybody. I would first like to apologise for the basic question; I'm not expert on Lie Theory. Can someone please help me with the following questions

Let $\mathrm{G}=\mathrm{K} \exp(\mathfrak{p})$ be a Real reductive group (as in Knapp's Lie Groups: Beyond an Introduction) and let $\mathfrak{a}$ be a maximal Lie subalgebra of $\mathfrak{g}=\operatorname{Lie}(\mathrm{G})$ such that $\mathfrak{a} \subseteq \mathfrak{p}$. Since ${[\mathfrak{p},\mathfrak{p}]} \subset \mathfrak{k}$ ($\mathfrak{k}=\operatorname{Lie}(\mathrm{K})$), the Lie algebra $\mathfrak{a}$ is commutative. Let $A:=\exp(\mathfrak{a})$ denote the corresponding commutative Lie subgroup of $\mathrm{G}$.

My questions are:

  1. Is $\mathrm{A}$ a closed subgroup of $\mathrm{G}$?
  2. Is $\mathrm{A}$ a reductive subgroup of $\mathrm{G}$?
  3. In otherwise, Can someone give some examples to show that 1. and 2. are not always true?

Any answer or comment will be greatly appreciated. Thanks.

share|cite|improve this question
Answers are: 1. yes, 2. yes. A is a maximal split torus (or rather the connected component of its real points). – Anton Nov 29 '11 at 20:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.