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]
http://ncatlab.org/nlab/show/moduli+space+of+connections#FlatConnectionsOverATorus
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?
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 $T$.