Suppose we have a principal bundle $p: P \to M$ with principal connection form $\omega: TP \to \mathfrak{g}$.

Given any arbitrary curve $\gamma: I \to P$ in the total space of the principal bundle, where $I=[0,1]$, we can use the nonabelian fundamental theorem of calculus [R. W. Sharpe Thm 7.14] to integrate $\gamma^*\omega: TI \to \mathfrak{g}$ to a map $g: I \to G$ such that $g(0) = e$ and $g^* \omega_G = \gamma^* \omega$, where $\omega_G: TG \to \mathfrak{g}$ is the usual Maurer-Cartan form on $G$.

Now we can define a sort of transport of the starting point $\gamma(0)$ along this curve by \begin{equation}\gamma(0) \mapsto \gamma(t).g(t)^{-1}.\end{equation} Note: I might need to replace $g(t)^{-1}$ by $g(t)$ in the formula above.

On the other hand, we can use the standard definition of parallel transport: take the projection $p \circ \gamma: I \to M$, and lift it again to a horizontal curve $\widetilde{\gamma}: I \to P$ with the same starting point $\widetilde{\gamma}(0)=\gamma(0)$. Then we can parallel transport the starting point $\widetilde{\gamma}(0)=\gamma(0)$ by traveling along the horizontal curve: \begin{equation}\mathrm{Pt}_{p \circ \gamma}(t)\gamma(0)= \widetilde{\gamma}(t). \end{equation}

Do these two notions of transport agree? In other words, do we have an equality $\widetilde{\gamma}(t) = \gamma(t).g(t)^{-1}?$

For anyone not familiar with the nonabelian FTC, I'll give the global version here:

Theorem (nonabelian fundamental theorem of calculus): Suppose we are given a connected manifold $M$ equipped with a $\mathfrak{g}$-valued $1$-form $\omega: TM \to \mathfrak{g}$. Then there exists a smooth map $f: M \to G$ such that $\omega =f^* \omega_G$ if and only if $\omega$ satisfies the Maurer-Cartan equation $d\omega + \frac{1}{2}[w \wedge \omega]=0$ and the monodromy representation $\pi_1(M, x) \to G$ is trivial. Moreover, if these conditions are satisfied then the map $f$, called the integral of $\omega$, is unique up to left translation by an element of $G$.

Corollary: On the interval $I=[0,1]$ any 1-form $\omega: TI \to \mathfrak{g}$ has an integral $f: I \to G$ unique up to choice of $f(0) \in G$.

  • 1
    $\begingroup$ Could you state the nonabelian fundamental theorem of calculus? $\endgroup$
    – Deane Yang
    Feb 1, 2017 at 2:21
  • $\begingroup$ Sure thing; just added it to the end of the question. $\endgroup$
    – ಠ_ಠ
    Feb 1, 2017 at 3:43

1 Answer 1


Yes, they coincide (at least if I understand you correctly).

Let $\check{\gamma}$ be a path in $M$. If we fix a lift $\gamma$ in P, then the horizontal lift $\gamma_\omega$ of $\check{\gamma}$ is necessarily of the form $\gamma_\omega(t) = \gamma(t) \cdot g(t)$ for some curve $g: [0,1] \to G$. Differentiating this relation with respect to $ t $ yields $$ \omega(\dot \gamma_\omega(t)) = \omega(\dot \gamma (t) . g(t)) + \omega(\gamma(t) . \dot g(t)) \\ = \omega(\dot \gamma (t) . g(t)) + \omega(\gamma(t) \cdot g(t) . g(t)^{-1} \dot g(t)) \\ = Ad^{-1}_{g(t)} \omega(\dot \gamma (t)) + g(t)^{-1} \dot g(t).$$ Using the fact that that $\gamma_\omega$ is horizontal, we consequently get $\dot g(t) g^{-1}(t) = - \omega(\dot \gamma (t))$.

A few words about the notation used in the above calculation:

  • The lower dots are just the partial derivatives of the the action. In particular $p . A = \frac{d}{d \epsilon} (p \cdot exp(\epsilon A))$ for $A$ a Lie algebra element and $p \in P$.
  • Equivariance of the connection implies $\omega(X . g) = Ad^{-1}_g \omega(X)$ for every $X \in T_p P$.
  • The connection is the identity on vertical vector, which amounts to $\omega(p. A) = A$ for every $p \in P$ and $A \in \mathfrak{g}$.
  • $\begingroup$ Thanks! I am just confused about what $\gamma(t).\dot{g}(t)$ means? Since $\dot{g}(t) \in T_{g(t)}G$, how is it acting on $\gamma(t) \in P$? $\endgroup$
    – ಠ_ಠ
    Feb 1, 2017 at 12:57
  • 1
    $\begingroup$ Where did $\gamma(t)$ disappear? $\endgroup$ Feb 1, 2017 at 13:04
  • $\begingroup$ Sorry, I was a bit in a hurry. I just updated my answer. Let me know if there is still something that is not clear. $\endgroup$ Feb 1, 2017 at 15:13
  • $\begingroup$ Thanks for the clarification! It all makes sense now. It's interesting that we actually get the right (rather than left) Maurer-Cartan form in the end. $\endgroup$
    – ಠ_ಠ
    Feb 1, 2017 at 23:07

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