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There is a saying in wiki of Maximal torus:

"Let $G$ be a compact, connected Lie group and let ${\displaystyle {\mathfrak {g}}}$ be the Lie algebra of G. A maximal torus in G is a maximal abelian subgroup, but the converse need not hold."

Because union of two different Abelian subgroup is not a group in general, the maximal Abelian subgroup depends on which Abelian subgroup $A'$ you start with. $A$ is maximal Abelian subgroup if you can not find a bigger Abelian group containing $A$.

My first question is : Whether all maximal Abelian subgroup of a connected Lie group $G$ has the same dimension? If $G$ is compact and connected, does the answer will change?

Because closed subspace of a compact space is compact. Every compact Abelian Lie group is isomorphic to $T^n$. So the wiki's agrument is possilbe only when a maximal abelian subgroup is not closed or not connected.

My second question is : what's the concrete example of wiki's argument?

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    $\begingroup$ The closure of an abelian subgroup in a Hausdorff topological group is again an abelian subgroup. So a maximal abelian subgroup is always closed. Thus the problem is about connectedness, a connected compact Lie group could contain disconnected maximal abelian subgroups. $\endgroup$ Commented Dec 5, 2016 at 20:46
  • $\begingroup$ @DanielFischer Thanks. So can you give me an example of maximal abelian subgroup of compact lie group is not connected? $\endgroup$
    – 346699
    Commented Dec 5, 2016 at 20:50
  • $\begingroup$ Not without some trial-and-error search. I'll wait whether somebody knows an example off-hand first. $\endgroup$ Commented Dec 5, 2016 at 20:55
  • $\begingroup$ @DanielFischer Another question: Should the dimension of this disconnected maximal Abelian subgroup be smaller than the maximal torus ? If your argument is right, the connected component of this disconnected subgroup should be a torus. Its dimension should be smaller than maximal torus. $\endgroup$
    – 346699
    Commented Dec 5, 2016 at 20:57
  • $\begingroup$ Yes, it would have to be. If the dimension were that of a maximal torus, the subgroup would contain a maximal torus. $\endgroup$ Commented Dec 5, 2016 at 20:59

1 Answer 1

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Let's answer the second question first: Since in a Hausdorff topological group the closure of an abelian subgroup is again an abelian subgroup, a maximal abelian subgroup of a Hausdorff topological group is always closed, and in the case of Lie groups, hence an embedded Lie group. Since a connected compact abelian Lie group (of finite dimension) is a torus, a maximal abelian subgroup that is not a maximal torus must be disconnected.

Take $G = SO_3(\mathbb{R})$. This is a compact connected Lie Group of dimension $3$, and its maximal tori are embedded copies of $SO_2(\mathbb{R})$. In $G$, the two matrices

$$A = \begin{pmatrix} - 1& 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\qquad\text{and}\qquad B =\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$

generate a Klein $4$-group $K = \{ I, A, B, AB\}$. This is a maximal abelian subgroup of $SO_3(\mathbb{R})$ which is not a maximal torus.

For a $3\times 3$-matrix commutes with $A$ and with $B$ if and only if it is diagonal, and the diagonal matrices in $SO_3(\mathbb{R})$ are precisely $I, A, B, AB$, thus there is no abelian subgroup of $SO_3(\mathbb{R})$ properly containing $K$, i.e. $K$ is maximal abelian.

Since $K$ has dimension $0$, and maximal tori in $SO_3(\mathbb{R})$ have dimension $1$, not all maximal abelian subgroups have the same dimension. In fact, a maximal abelian subgroup which is not a maximal torus must necessarily have lower dimension than the maximal tori. The connected component of the identity in a compact abelian subgroup is a torus, and if it had the same dimension as maximal tori, it would also be a maximal torus.

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  • $\begingroup$ "maximal abelian subgroup which is not a maximal torus must necessarily have lower dimension than the maximal tori" -- You mean for a compact Lie group, right? $\endgroup$ Commented Sep 23, 2019 at 15:33
  • $\begingroup$ Yes. If we drop the compactness, $G \times \mathbb{R}^n$ for a compact Lie Group $G$ and $n > \operatorname{dim} G$ gives an obvious counterexample. And we also need connectedness, for if $M$ is a maximal torus of $G$, then $M \times \{1,-1\}$ is a maximal abelian subgroup of $G \times \{1,-1\}$ of the same dimension that is not a maximal torus. $\endgroup$ Commented Sep 30, 2019 at 12:00
  • $\begingroup$ @DanielFischer I made an edit to your post - I hope you don't mind. (Long no time no see, by the way. best wishes!) $\endgroup$
    – peter a g
    Commented Jan 17, 2023 at 13:49
  • $\begingroup$ I know it's been a long time you posted this answer, but can you elaborate on the phrase "a maximal abelian subgroup which is not a maximal torus must necessarily have lower dimension."? I understand that this requires that the larger group $G$ be compact and connected, but why does it hold in this case? $\endgroup$ Commented Jan 17, 2023 at 17:54
  • $\begingroup$ I have an answer (from Bröcker and tom Diecks' Representations of Compact Lie Groups, Section IV Theorem 2.3(2)). Namely, if $T\subseteq G$ is a maximal torus and $g\in G$ commutes with $T$, then $g^k\in T$ for some finite $k$ (else the closure of the group generated by $T$ and $g$ has larger dimension than $T$ and is abelian, contradicting maximality of $T$. Now it follows that the group $\langlte T,g\rangle$ has an element $z$ generating a dense subset: we already know there is an $x\in T$ generating a dense subset of $T$, so use $z=yg$ where $y^{k} = x$. (continued) $\endgroup$ Commented Jan 17, 2023 at 19:27

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