# Effective actions by non-commutative groups have non-commuting fundamental vector fields?

I have a bit of a contradiction in my brain and I was hoping once again that excellent Mathoverflow community could help me out :)

Let $\rho_g$ be the action associated to a non-abelian Lie Group $G$ on a manifold $M$ and take $\eta,\nu \in \mathfrak{g}$ such that $[\eta,\nu]=\zeta$. It seems that in order for the group action to be effective, the fundamental vector fields $X_{\eta}, X_{\nu}$ associated to $\eta,\nu$ do not commute as given the usual Lie algebra homomorphism $A:\mathfrak{g}\rightarrow \mathfrak{X}(M)$ we have that $[X_{\eta}, X_{\nu}]=A([\eta, \nu])=A(\zeta)\neq 0$ if the action is effective.

On the other hand, if such an action is Hamiltonian with respect to a symplectic form $\omega$ on $M$ we know that the orbits are isotropic submanfolds and so $\omega(X_{\eta}, X_{\nu})=0$. However this implies that the Hamiltonian vector fields $X_{\eta}, X_{\nu}$ commute... Where have I gone wrong?

Edit: I'm fine to assume that $G$ is compact, if this makes everything easier

• The orbits are not isotropic. Rotation of a sphere is Hamiltonian. – Ben McKay Jul 6 '18 at 14:22
• Dear @AliTaghavi, I'm really only interested in the action of compact groups so I'm fine assuming that the orbits are submanifolds - I have added this to the q. Thanks – R Mary Jul 6 '18 at 15:40
• In Ben's example, the symplectic manifold is the sphere, and the SO(3)-orbit is the entire sphere (i.e. the SO(3)-action is transitive). You seem to be confusing this with the group orbit of rotations about one axis, which are one-dimensional. As he said, in general the group orbits are not isotropic. – user17945 Jul 6 '18 at 23:55
• @AliTaghavi: the isometry group action of the real projective plane (in its constant curvature metric) is not a Hamiltonian action, because the real projective plane has no symplectic form. The action of translations on the flat torus is not Hamiltonian, because on a compact manifold, any function has critical points, so any Hamiltonian vector field has zeros. Similar problems arise for a cylinder: the action is not Hamiltonian, because you can't get the ruling lines to be the level sets of a function. I think hyperbolic plane has Hamiltonian isometry group action. – Ben McKay Jul 7 '18 at 11:24
• @AliTaghavi The orbits of any Lie group action are immersed submanifolds. – Ben McKay Jul 7 '18 at 11:24