# Integrability of certain distribution associated to a connection form on the total space of a principal bundle (Principal Frobenius condition)

Let $$P\to M$$ be a $$G$$-principal bundle where $$P,M$$ are smooth manifolds and $$G$$ is a Lie group with Lie algebra $$\mathfrak{g}$$, whose center is denoted by $$C(\mathfrak{g})$$.

Let $$\omega$$ be the connection form of a connection for our principal bundle.

We define a distribution on the total space $$P$$ as follows: $$(*)\qquad \{v\in T_xP\mid \omega(v)\in C(\mathfrak{g}),\quad x\in P\}$$

This defines a $$G$$-invariant distribution on $$P$$.

Under what algebraic conditions on $$\omega$$, $$(*)$$ is an integrable distribution? What is a precise example of a foliation which can be generated in this way and the Lie algebra $$\mathfrak{g}$$ is not commutative? Is there an example of this situation such that we have a leaf with non-trivial holonomy? On the other extreme, what is an example of a distribution $$(*)$$ which is not integrable?

As a second question, is there a geometric interpretation for the following algebraic condition: $$(\omega \wedge d\omega)(X,Y,Z)\in C(\mathfrak{g}),\quad \forall X,Y,Z\in T_x P,\; x\in P\quad ?$$

For every $$u\in C({\cal g})$$ there exists a fundamental vector $$u^*$$ defined on $$P$$ by $$u^*(x)={d\over{dt}}_{t=0}xexp(tu)$$; if $$u,v\in C({\cal g})$$, $$0=[u,v]^*=[u,^*,v^*]$$. Frobenius theorem implies the distribution
$$(*)\qquad \{v\in T_xP\mid \omega(v)\in C(\mathfrak{g}),\quad x\in P\}$$
• A proof of the Frobenius theorem works like so, let $X_1,..,X_p$ be vector fields in involution, one creates $X'_1,...,X'_p$ from $X_1,..,X_p$ which generates the same distribution such that $[X`_i,X'_j]=0$, then the flow of $X'_i$ define the foliation, so the involution in Frobenius theorem needs to be true for only one set of vector fields in involution. – Tsemo Aristide Jun 29 at 15:54
• So can we conclude from your answer that if we have a G principal bundle such that $\mathfrak{g}$ has trivial center then every arbitrary connection is flat since it is integrable? Is it realy the case? – Ali Taghavi Jun 29 at 19:45