# How will horizontal and vertical vector field behave under lie bracket?

With fiber bundle $$\pi:P\to M$$ and an Ehresmann connection on $$P$$, $$V_1$$ and $$V_2$$ are vertical vector fields on $$P$$, $$H_1$$ and $$H_2$$ are horizontal vector fields on $$P$$.

1. Is $$[V_1,V_2]$$ vertical?
2. Is $$[H_1,H_2]$$ horizontal?
3. Is $$[V_1,H_1]$$ zero?

(1) and (3) seems to work if $$V$$ is fundamental vector field and $$P$$ a principal G-bundle.

It is said that lie algebra on $$R^3$$ work like cross product, from this perspective, all will not work. Is there some intuitive method to describe what bracket (more fundamentally, composition of directional derivative) of vector field look like?

• As a comment on your final paragraph, that is certainly not the case. There is no cross product on $\mathbb{R}^n$ unless $n=3$. The standard bracket on $\mathbb{R}^n$ is the abelian one although of course this is not what you are looking for. Jul 11, 2022 at 7:24
• @Callum I'v fixed it.
– jw_
Jul 11, 2022 at 7:51
• Regarding (2): The vertical component of $[H_1, H_2]$ is the curvature $R(H_1, H_2)$, i.e., it vanishes for all $H_1, H_2$ iff the connection is (locally) flat. Jul 11, 2022 at 9:07

The first claim follows from the fact that the vertical subbundle is integrable, since for any point $$q\in M$$, the fibers $$\pi^{-1}(q)$$ are smooth submanifolds of $$P$$.
Counterexamples to the other two claims can be found for instance from contact structures. Consider the Ehresmann connection on $$\mathbb{R}^3\to\mathbb{R}^2$$ given by the standard contact structure in $$\mathbb{R}^3$$, where the horizontal subbundle is $$H = \ker(dz-y dx)$$ and the vertical subbundle is $$V = \ker dx\cap\ker dy$$. The commutator of the horizontal vector fields $$H_1 = \partial_x + y\partial_z$$ and $$H_2=\partial_y$$ is then the vertical vector field $$[H_1,H_2] = -\partial_z$$.
For a counterexample to the last claim, we want an example where the horizontal subbundle is not invariant under vertical flows. So we may for instance perturb the horizontal subbundle of the previous example to $$H= \ker(dz-(y+z)dx)$$. Then for $$V_1=\partial_z$$ and $$H_1=\partial_x+(y+z)\partial_z$$ we obtain a nonzero commutator $$[V_1,H_1] = \partial_z$$.