It is a well known fact that the geometric meaning of a linear connection's curvature can be realized as the measure of a change in a fiber element as it is parallel transported along a closed loop.

As far as I am aware, the formula goes as $$ F(X,Y)\sigma=-\lim_{t,s\rightarrow 0}\frac{1}{ts}\left(P^X_tP^Y_sP^X_{-t}P^Y_{-s}\sigma-\sigma\right), $$ where $P^X_t$ is parallel transport along $X$'s integral curves for time $t$, and I here assume that $[X,Y]=0$.

Unfortunately, none of my references contain a proof of this statement, at least not one that is useful to me now.

I can prove this statement using coordinate-based methods that often involve "expanding to first order" and other rather handwave-y methods (Weinberg, Wald etc. though Wald's procedure is actually fairly interesting and rigorous, it is not what I am looking for), but that's not what I am looking for.

I have also seen proofs of this statement (Lee's Manifolds and Differential Geometry) which involves starting with vectors $u$ and $v$ at points and extending them to vector fields in special ways, so that they are parallel along some curves, which simplifies things greatly. This is once again not what I am looking for, because it already presupposes that we know that $F$ is tensorial and that $F(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$.

In essence, what I am looking for is a derivation that $$ -\lim_{t,s\rightarrow 0}\frac{1}{ts}\left(P^X_tP^Y_sP^X_{-t}P^Y_{-s}\sigma-\sigma\right)=\nabla_X\nabla_Y\sigma-\nabla_Y\nabla_X\sigma $$ for commuting vector fields $X,Y$, without supposing we already know what curvature is (I need this for didactic reasons, for possible use in a general relativity lecture).

I have tried to do this on my own, but I am really terrible at formally manipulating these "flow" type of objects like $P^X_t$, and so far my attempts failed.

I would appreciate any reference, textbooks, lecture notes, articles, papers, which prove this formula (with the added caveat that I outlined in bold above). If it does so without assuming $X$ and $Y$ commute, it would especially be stellar, but I don't need that.

  • 1
    $\begingroup$ You might be able to adapt the calculation here: deaneyang.com/papers/holonomy.pdf to get what you want. $\endgroup$
    – Deane Yang
    Jun 15, 2017 at 13:07
  • $\begingroup$ Also lemma 3.5 in "Smooth functors vs Differential Forms" arxiv.org/abs/0802.0663 $\endgroup$ Jun 15, 2017 at 21:20
  • $\begingroup$ Are you sure the formula you report is correct? I would have thought that, in the limit expression for the curvature of $\nabla$, $P_t^X$ were the parallel transport along the geodesic (with respect to $\nabla$) along the direction $X\in T_p M$ at time $t$, not the flow of a vector field extending $X$ locally. Or maybe it's the same? $\endgroup$
    – Qfwfq
    Jun 15, 2017 at 22:24
  • $\begingroup$ @Qfwfq I want to know this too. my idea is that it doesn't matter the curve as long as the parallel transport map of the connection is the one used. but I'm not sure. $\endgroup$ Jul 27, 2021 at 18:55

1 Answer 1


A simple way is as follows.

Given a point $p$, the vector field $(P_{-t}^X\sigma)(p)$ is obtained from $\sigma(\gamma(s))$ by backward parallel transport along the curve $\gamma(t):=\exp(tX)(p)$, up to time $0$. Denoting this backward parallel transport (from time $t$ to time $0$) by $Q_t^\gamma$, we have $$ \frac{d}{dt}(P_{-t}^X\sigma)(p)=\frac{d}{dt}(Q_t^\gamma \sigma(t))=Q_t^\gamma(D_t\sigma(t))=(P_{-t}^X)(\nabla_X\sigma)(p), $$ where $\sigma(t):=\sigma(\gamma(t))$ is a vector field along $\gamma$ and $D_t$ is the covariant derivative along $\gamma$. So $$P_{-t}^X\sigma=\sigma+\int_0^t P_{-t'}^X(\nabla_X\sigma)\,dt'=\sigma+t\nabla_X\sigma+O(t^2). $$ Hence, $$\begin{align}&P_t^XP_s^YP_{-t}^XP_{-s}^Y\sigma=P_t^XP_s^YP_{-t}^X(\sigma+s\nabla_Y\sigma)+O(s^2)\\ &=P_t^XP_s^Y(\sigma+s\nabla_Y\sigma+t\nabla_X\sigma+ts\nabla_X\nabla_Y\sigma)+O(s^2)+O(t^2)\\ &=P_t^X(\sigma+t\nabla_X\sigma+ts\nabla_X\nabla_Y-ts\nabla_Y\nabla_X\sigma)+O(s^2)+O(t^2)\\ &=\sigma+ts\nabla_X\nabla_Y-ts\nabla_Y\nabla_X\sigma+O(s^2)+O(t^2)\end{align}$$ (of course the parallel transport applied to the error terms has the same order of magnitude, since it is obtained by solving a first-order ODE).

Now you just have to observe that the second order expansion of the smooth function $f(t,s):=(P_t^XP_s^YP_{-t}^XP_{-s}^Y\sigma)(p)$ has the form $\sigma(p)+tsV$ for some $V\in T_pM$: the coefficients of $s$, $t$, $s^2$ and $t^2$ all vanish since $f(t,0)\equiv\sigma(p)$ and $f(0,s)\equiv\sigma(p)$.

  • $\begingroup$ This IMO is the derivation of RCT that makes the most sense $\endgroup$ Jul 26, 2021 at 16:53
  • $\begingroup$ Before seeing this I asked about an approximation of the parallel transport along the flow of a vector field. The result looks similar but I'm not so sure about all the steps. Can you take a look if you're available? mathoverflow.net/questions/398269/… $\endgroup$ Jul 26, 2021 at 17:02

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