Let $M^n$ be a Riemannian manifold with a connection $\nabla$. For a fixed $x\in M$, we define $H:=\text{Hol}_x(\nabla)$ and identify $T_xM\equiv \mathbb{R}^n$. The Holonomy Principle states that:

there is a bijective correspondence between $H$-invariant tensors on $\mathbb{R}^n$ and parallel tensor fields on $M$.

I wasn't able to find any proof I could understand, mainly because of the tensor/bundle dialect, which is still very obscure to me.

I wonder if there is an elementary proof for a particular case, which would shed light on the general idea. For example, if the tensors are $1$-forms, I would have to prove that:

if $\omega_0\in (\mathbb{R}^n)^*$ is $H$-invariant, then there exists $\omega\in\Gamma(TM^*)$ with $\nabla \omega=0$ and $\omega_x=\omega_0$. Conversely, if $\nabla\omega=0$, then $\omega_x$ is $H$-invariant.

Any suggestions?

  • $\begingroup$ You've misstated the holonomy principle. It should say there's a bijective correspondence between $\boldsymbol{H}$-invariant tensors on $\mathbb R^n$ and parallel tensor fields on $M$. $\endgroup$ – Jack Lee Jun 16 at 18:13
  • $\begingroup$ @JackLee, I've just corrected it, thank you $\endgroup$ – rmdmc89 Jun 16 at 19:31

Constructing $\omega$ from $\omega_0$ is straightforward. We'll need $M$ to be connected though.

For any piecewise smooth path $\gamma : [0,1]\to M$, starting at $x$ and ending at $y$, the parallel transport map $P_\gamma : T_xM\to T_yM$ is an isomorphism of the tangent spaces at $x$ and $y$. Then for any other path $\gamma'$ from $x$ to $y$, if $v\in T_x$, we can produce both the vector $P_\gamma v$ and $P_{\gamma'}v$ in $T_yM$. In order to get a single well defined vector in $T_yM$, these had better be equal.

The condition that $$P_\gamma v = P_{\gamma'}v$$ is equivalent to $$P_{\gamma'}^{-1}P_{\gamma}v=v,$$ but $$P_{\gamma'}^{-1} = P_{\bar{\gamma}'},$$ where $\bar{\gamma}'$ denotes the reverse path of $\gamma$. Thus $P_{\gamma'}^{-1}P_\gamma v= P_{\bar{\gamma}'\cdot\gamma}v$, where $\cdot$ denotes concatenation of paths. But $P_{\bar{\gamma}'\cdot\gamma}\in H$. Thus we can uniquely parallel transport a vector $v\in T_xM$ to any other point $y$ in $M$ as long as $v$ is invariant under $H$. An identical argument shows that this is true for any other sort of tensor as well.

To be precise. If $v\in T_xM$ is a vector invariant under the action of $H$, then we can extend it to a parallel vector field on $M$, $V$ by defining $V_y = P_\gamma v$, where $\gamma$ is any piecewise smooth path from $x$ to $y$. The same is true for any $H$-invariant tensor at $x$.

For the converse, we'll also stick to vector fields, since everything generalizes identically to tensor fields.

Conversely, suppose we are given a parallel vector field $V$. Let $\gamma$ be a loop at $x$. Since $V$ is parallel, $V|_{\gamma}$ is a parallel vector field on $\gamma$. However $V_x=V_{\gamma(1)}=V_{\gamma(0)}=P_{\gamma}V_{\gamma(0)} = P_{\gamma}V_x$ by definition of the parallel transport map. Thus $V_x$ must be invariant under $P_\gamma$ for any loop $\gamma$ based at $x$. In other words $V_x$ is invariant under the holonomy action.


To prove this for $1$-forms, the following lemma will be helpful, which basically says that parallel transport of $1$-forms agrees with parallel transport of tangent vectors.

Let $\gamma$ be a piecewise smooth curve in $M$. Parallel transport along $\gamma$ induces an isomorphism $P_\gamma : T_xM\to T_yM$. This produces an isomorphism $P_\gamma^* : T^*_xM\to T^*_yM$ defined by $P_\gamma^* \omega_x = \omega_x \circ P_\gamma^{-1}$. Then $P_\gamma^* \omega_x = \omega_y$ if and only if there is a 1-form field $\omega$ along $\gamma$ such that $\nabla \omega=0$ and $\omega_0 = \omega_x$ and $\omega_1 = \omega_y$.


Suppose we begin with $\omega_x \in T_xM$. Fairly sure the same argument that produces a unique parallel vector field along $\gamma$ extending any tangent vector at the start also applies here producing a 1-form field $\omega$ along $\gamma$ with $\omega_0=\omega_x$ and $\nabla \omega=0$.

It suffices then to prove that $\omega_1 = P_\gamma^*\omega_x$.

Let $u_x\in T_xM$ be arbitrary. Let $u$ be the parallel vector field along $\gamma$. Recall that by definition, $$ \nabla\omega(u)_t = \frac{d}{dt}\omega(u) - \omega(\nabla_u)_t, $$ but $\nabla u = 0$ and $\nabla\omega = 0$, so $\frac{d}{dt}\omega(u)=0$. Thus $\omega(u)$ is constant.

In particular, $\omega_y(u_y) = \omega_x(u_x)$, but $u_y = P_\gamma u_x$. Thus $\omega_y\circ P_\gamma = \omega_x$, since $u_x$ was arbitrary. This rearranges to $\omega_y = P_\gamma^* \omega_x$, as desired.

The proof for 1-forms

The lemma tells us that it doesn't matter that $H$ is defined by parallel transport of tangent vectors, its action on covectors is compatible with parallel transport of covectors.

Now given an $H$-invariant $\omega_x\in T^*_xM$, we can extend it to $\omega$ on all of $M$ by $\omega_y = P_\gamma^* \omega_x$ for any path $\gamma $ from $x$ to $y$.

Conversely, given $\omega$ with $\nabla \omega = 0$, we have that $P_\gamma^* \omega_x = \omega_x$ for all loops $\gamma$ based at $x$ by the lemma.

  • $\begingroup$ In this case $V|_{\gamma}$ is the pullback of $V$, right? $\endgroup$ – rmdmc89 Jun 14 at 19:06
  • 1
    $\begingroup$ @rmdmc89 Yes, that's correct. You might also think of it as $V\circ \gamma$. $\endgroup$ – jgon Jun 14 at 19:07
  • $\begingroup$ I'm trying to adapt it for the case of $1$-forms. If $\nabla\omega=0$, then I must prove $(P_\gamma\omega_x)=\omega_x$. If I'm getting the $H$-action right, for $u\in T_xM$, we would have $(P_\gamma\omega_x)(u)=\omega_x(P_\gamma u)$, right? How do I know $\omega_x(P_\gamma u)=\omega_x(u)$? $\endgroup$ – rmdmc89 Jun 14 at 20:00
  • $\begingroup$ I know I must use $\nabla\omega=0$, but I don't know how $\endgroup$ – rmdmc89 Jun 14 at 20:44
  • $\begingroup$ @rmdmc89 I've edited with a bit of an explanation on why the same argument works for covectors. Also, I've given the correct action in the edit, but notice that the action of the holonomy group on covectors is not $P_\gamma \omega_x = \omega_x \circ P_\gamma$, but instead $P_\gamma \omega_x = \omega_x \circ P_\gamma^{-1}$. Also, to keep things clear, I called the action of $P_\gamma$ on $\omega_x$ $P_\gamma^*\omega_x$, but it's just the action of the holonomy group. $\endgroup$ – jgon Jun 14 at 20:49

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