# If a connection on a principal $G$-bundle restricts to an $H$-subbundle, must its holonomy lie in $H$?

Let $P \to M$ be a principal $G$-bundle, equipped with a principal connection $D$.

Let $Q \subset P$ be a principal subbundle with fiber $H$, where $H \leq G$ is a (let's say closed and connected) Lie subgroup.

Question: Suppose that $D$ restricts to a principal connection on $Q$. Does it follow that $\text{Hol}(D)$ is a subgroup of $H$?

As usual, this is not a homework question, but simply me trying to better organize my understanding.

This question asked 11 days ago (which has received little attention) is essentially asking the same thing, but is more optimistic.

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Yes. In fact there as a “smallest” reduction to which $D$ restricts, whose fibre type is $\operatorname{Hol}(D)$, called the holonomy bundle. – canaaerus Feb 9 '14 at 10:35
@canaaerus Cool! Could you make your comment an answer, and perhaps provide either details and/or references? – Jesse Madnick Feb 9 '14 at 10:36

The answer is essentially "yes". More precisely, $\text{Hol}(D)$ is only well-defined up to conjugation in $G$ and the statement is that if $D$ restricts to some $H$ subbundle $Q\subset P$ then $\text{Hol}(D)$ is the conjugacy class of a subgroup of $H$.

This is an immediate consequence of the definitions of the terms involved (the Theorem of Ambrose-Singer is not necessary). Here is a quick sketch of the argument (mostly uncovering definitions).

A connection $D$ on a prinicipal $G$-bundle $\pi:P\to M$ is given by a distribution $\mathcal H\subset TP$, $G$-invariant and transversal to the fibers of $P\to M$. The holonomy of $D$ is defined by picking a base point $p\in P$, then for each closed curve $\gamma:[0,1]\to M$ such that $\gamma(0)=\gamma(1)=\pi(p)$, you first lift it horizontally to the unique curve $\tilde\gamma:[0,1]\to P$ such that $\gamma=\pi\circ\tilde\gamma$, $\tilde\gamma(0)=p$ and ${d\over dt}\tilde\gamma\in\mathcal H$, then its holonomy is defined to be the element $h\in G$ such that $\tilde\gamma(1)=\tilde\gamma(0)h.$ Then $\text{Hol}(D,p)\subset G$ is the set of all holonomies of all such $\gamma$.

If you switch to another base point instead of $p$ then it follows immediately from the $G$-invariance of $\mathcal H$ that $\text{Hol}(D,p)$ gets conjugated, hence $\text{Hol}(D)$ is only well defined up to conjugation.

If $Q\subset P$ is an $H$-subbundle, where $H\subset G$ is a subgroup, then "$D$ restricts to $Q$" means that $\mathcal H$ is tangent to $Q$, i.e. $\mathcal H|_Q\subset TQ$. In particular, if you pick a base point $q\in Q$, all lifted $\tilde\gamma$ remain in $Q$ hence $\text{Hol}(D,q)\subset H$.

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Let $p$ denote the basepoint for the holonomy group. The Ambrose-Singer theorem says that the Lie algebra of the holonomy group of $D$ is the subspace of the Lie algebra of $G$ spanned by elements of the form $\Omega_q(X,Y)$ where $q$ is a point in the holonomy bundle at $p$, $X$ and $Y$ are horizontal tangent vectors at $q$, and $\Omega$ is the curvature of $D$. If $D$ restricts to $Q$ then its curvature lies in the Lie algebra of $H$, so by exponentiation the holonomy group is a subgroup of $H$.

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Ah, so simple, thanks! – Jesse Madnick Feb 9 '14 at 10:40