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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: $$(\omega \wedge d\omega)(X,Y,Z)\in C(\mathfrak{g}),\quad \forall X,Y,Z\in T_x P,\; x\in P\quad ?$$

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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

$$(*)\qquad \{v\in T_xP\mid \omega(v)\in C(\mathfrak{g}),\quad x\in P\}$$

is integrable.

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  • $\begingroup$ Fundamental vector fields are vertical vector fields. But to check Frobenius condition we need consider arbitrary vector fields in our distribution. Right? $\endgroup$ – Ali Taghavi Jun 29 at 15:44
  • $\begingroup$ 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. $\endgroup$ – Tsemo Aristide Jun 29 at 15:54
  • $\begingroup$ 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? $\endgroup$ – Ali Taghavi Jun 29 at 19:45
  • $\begingroup$ My apology for asking this question again: According to your answer, does the conclusion in my previous comment, true? $\endgroup$ – Ali Taghavi Jul 3 at 6:32

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