Let $G$ be a compact, connected Lie group. Let $x, y \in G$ be an arbitrary pair of commuting elements. Is there necessarily a torus $T \leq G$ containing $x$ and $y$?

Apparently not:

Commutativity and Maximal Tori in Connected, Compact Lie Groups

The issue arises due to discrete abelian subgroups. Consider the $SO_3(\mathbb{R})$ example from the link. Thinking geometrically, a maximal torus can be described as the circle subgroup of rotations fixing a particular direction in $\mathbb{R}^3$. Consider two orthogonal directions. Then rotation by $\pi$ radians about these two axes will commute, but they certainly do not live in a common maximal torus.

My concern comes from trying to compute the following character variety on a torus:

$\chi_G(\Sigma_1)=\{\rho: \pi_1(\Sigma_1) \to G \ \vert \rho \ \text{is a group homomorphism}\}/\text{conjugation by} \ G$

where $\Sigma_1 \cong S^1 \times S^1$ is a torus. Since $\pi_1(\Sigma_1)$ is free abelian of rank 2, it follows that such a homomorphism is determined by a choice of commuting elements $x, y \in G$. What should I make of the following argument, particularly the part characterizing a flat connection on $S^1 \times S^1$? [Edit: The link has been updated to reflect this issue]


It seems that there are oddball homomorphisms not fitting into this general setup for some compact groups. I would appreciate more examples (in other compact, connected gauge groups) of commuting elements that fail to live in a common maximal torus.

Of course, $SO_3(\mathbb{R})$ is not simply connected, but the gauge groups I am working with all are. I don't know enough about discrete abelian subgroups of compact Lie groups. Can this issue be avoided by further assuming that the gauge group is simply connected?

  • $\begingroup$ Perhaps you are right that this would fit at MO. Once your bounty has ended, if it remains answerless, ping us and we'll migrate it over. $\endgroup$ – davidlowryduda Oct 28 '14 at 2:39
  • $\begingroup$ For simply-connected groups, this phenomenon does not occur: Is this what you wanted to know? $\endgroup$ – Moishe Kohan Oct 28 '14 at 19:53

If $G$ is simply-connected, then this phenomenon cannot occur: This follows from Jim Humphreys' answer to this mathoverflow question. He works with algebraic groups over algebraically closed field, but, in you can you should just take the complexification $G^c$ of your compact Lie group. The reason is that tori in $G$ are intersections of tori in $G^c$ with $G$ and, conversely, each torus in $G^c$ is conjugate to a torus $T^c$ which intersects $G$ along a subtorus $T$, whose complexification is $T$.

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  • $\begingroup$ I'm chewing on your answer now. I must admit that I was hoping to see a proof involving real geometry and Lie theory or algebraic topology, but I understand that complexification might be the way to go. I'll let the bounty run a bit longer in case someone has a different perspective they want to share. Am I correct in thinking that there is something not quite right about the argument in the nLab link in the question? $\endgroup$ – James Staff Oct 30 '14 at 15:59
  • $\begingroup$ @JamesStaff: Yes, they are plain wrong on this and SO(3) is a counter example. $\endgroup$ – Moishe Kohan Oct 30 '14 at 20:40

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