Given a manifold endowed with a connection $(M, \nabla)$, I want to see how the curvature tensor appears parallel-transporting a vector around a closed loop. To avoid complications with holonomies, I'd like to do it first working naively with infinitesimal loops (I think this is more appropriate for Physics.SE rather than MSE, since this is how GR books do it).

Figure 3.5

Since this kind of calculation is all over the place and can be found in many GR textbooks, I will just set up the problem here. Let $$P_1\mapsto x^\mu, \qquad P_2\mapsto x^\mu+\delta a^\mu \qquad P_3\mapsto x^\mu+(\delta a^\mu+\delta b^\mu)\qquad P_4\mapsto x^\mu+\delta b^\mu \tag{0}\label{0}$$ where I have called $\delta a^\mu=A^\mu\delta t$ and $\delta b^\mu=B^\mu\delta t$ the infinitesimal coordinate displacements associated to such directions. Consider the vector $V^\sigma$ at $P_1$ and parallel transport it around the loop $P_1 P_2 P_3P_4$. Let's call the outcome of this $V'^\sigma$, now we are concerned with the difference with the original vector and we expect $$\delta V^\sigma=V'^\sigma-V^\sigma=R^\sigma{}_{\rho\mu\nu}V^\rho\delta a^{\mu}\delta b^{\mu}.\tag{1}\label{1}$$

The sketch of the calculation is the following, we compute the variations associated to each parallel transport \begin{align} &P_1P_2:\qquad (\delta V^\sigma)_{P_1P_2}=-\Gamma^\sigma_{\mu\nu}\delta a^\mu V^\nu|_{P_1}\tag{2a}\label{2a} \\ &P_2P_3:\qquad (\delta V^\sigma)_{P_2P_3}=-\Gamma^\sigma_{\mu\nu}\delta b^\mu V^\nu|_{P_2}\tag{2b}\label{2b} \\ &P_3P_4:\qquad (\delta V^\sigma)_{P_3P_4}=\Gamma^\sigma_{\mu\nu}\delta a^\mu V^\nu|_{P_3} \tag{2c}\label{2c}\\ &P_4P_1:\qquad (\delta V^\sigma)_{P_4P_1}=\Gamma^\sigma_{\mu\nu}\delta b^\mu V^\nu|_{P_4} \tag{2d}\label{2d} \end{align} and then add them up to find the total variation (I guess this is what makes it easier with infinitesimal loop as this would be vectors from different tangent spaces in principle, which can't be added). Then, we expand the $RHS$ to express all quantities at point $1$ and this is the crucial point. Many books decide to neglect second order terms, but in practice they only do so for quadratic terms $\delta a\delta a$ or $\delta b\delta b$, keeping the mixed terms $\delta a\delta b$ and since the linear terms eventually cancel, they end up with \eqref{1}. Referring to my equations, the second order quadratic terms appear in \eqref{2c} and \eqref{2d}, as it's easy to understand looking at \eqref{0} Of course, one should either neglect all second order terms or none, so this must be wrong but I don't see why it works. Just to be clear, I'm asking why (conceptually) we are just neglecting some second order terms.


Using the procedure described above, I expand $\Gamma$ and $V$ around $1$ above and using the parallel transport equations for the derivatives of $V$. \eqref{2a} is unchanged, while we have for the others (ignoring terms of order 3 and above):

\begin{align} &(\delta V^\sigma)_{P_1P_2}=-\Gamma^\sigma_{\mu\nu}\delta a^\mu V^\nu|_{P_1}\tag{3a}\label{3a} \\ &(\delta V^\sigma)_{P_2P_3}=-\Gamma^\sigma_{\mu\nu}\delta b^\mu V^\nu|_{P_2}=-\Gamma^\sigma_{\mu\nu}\delta b^\mu V^\nu|_{P_1}-\frac{\partial\Gamma^{\sigma}_{\mu\nu}}{\partial x^{\alpha}}\delta a^{\mu}\delta b^{\alpha} V^{\nu}|_{P_1}+\Gamma^{\sigma}_{\mu\nu}\delta b^{\mu}\Gamma^{\nu}_{\alpha\beta}\delta a^{\alpha}V^{\beta}|_{P_1}\tag{3b}\label{3b} \\ &(\delta V^\sigma)_{P_3P_4}=\Gamma^\sigma_{\mu\nu}\delta a^\mu V^\nu|_{P_3}=\Gamma^\sigma_{\mu\nu}\delta a^\mu V^\nu|_{P_1}+\frac{\partial\Gamma^{\sigma}_{\mu\nu}}{\partial x^{\alpha}}\color{red}{\delta a^{\mu}}(\color{red}{\delta a^{\alpha}}+\delta b^{\alpha}) V^{\nu}|_{P_1}-\Gamma^{\sigma}_{\mu\nu}\color{red}{\delta a^{\mu}}\Gamma^{\nu}_{\alpha\beta}(\color{red}{\delta a^{\alpha}}+\delta b^{\alpha})V^{\beta}|_{P_1} \tag{3c}\label{3c}\\ &(\delta V^\sigma)_{P_4P_1}=\Gamma^\sigma_{\mu\nu}\delta b^\mu V^\nu|_{P_4}=\Gamma^\sigma_{\mu\nu}\delta b^\mu V^\nu|_{P_1}+\frac{\partial\Gamma^{\sigma}_{\mu\nu}}{\partial x^{\alpha}}\color{red}{\delta b^{\mu}\delta b^{\alpha}} V^{\nu}|_{P_1}-\Gamma^{\sigma}_{\mu\nu}\color{red}{\delta b^{\mu}}\Gamma^{\nu}_{\alpha\beta}\color{red}{\delta b^{\alpha}}V^{\beta}|_{P_1}\tag{3d}\label{3d} \end{align} The terms highlighted in red are the quadratic terms that we'd want to vanish to find \eqref{1}, but as you can see they don't when we sum all these equations. I've found this here on PF, adapting to my notation:

[...] if, instead of comparing the result [at $P_3$] of parallel-transport in one direction [$P_1\to P_2\to P_3$] to parallel-transport in the other direction [$P_1\to P_4\to P_3$], you can parallel transport completely around the loop and compare the transported vector with the initial vector. If you do that, you'll get some extra terms that are quadratic in $\delta a$ and $\delta b$ . Also, there's the problem that you don't end up precisely at the same place you started, so you have to throw in another line segment to "close the parallelogram". These two effects actually cancel exactly (if you ignore higher-order terms), but it's a little messy to prove that.

Is this correct? May it be the solution?

  • $\begingroup$ Hi Mr. Feynman. The quadratic terms of the form $\delta a\delta a$ should cancel rather than being neglected. Did you try to work it out? $\endgroup$
    – Qmechanic
    Dec 24, 2023 at 11:03
  • $\begingroup$ Hello @Qmechanic and thank you for your feedback. It helps to know that they should cancel, rather than just be neglected. I have worked it out and they survived, but I will repeat the calculation following the same steps describes above to see be sure. In case I still get this problem, I'll update my question with an appendix including my result. However, the gist of it was that the expansion of $(2c)$ would contain $\partial\Gamma(\delta a)^2$ and the expansion of $(2d)$ would contain a $\partial\Gamma(\delta b)^2$ term and I didn't get any other pure quadratic term. $\endgroup$ Dec 24, 2023 at 11:30
  • $\begingroup$ Ok, I updated my answer with my result, showing that I get the terms in red. Is it possible that I'm forgetting to expand or to include anything? $\endgroup$ Dec 24, 2023 at 14:43
  • $\begingroup$ The terms $\delta a\delta a$ and $\delta b\delta b$ correspond to a "degenerate" parallelogram that looks like a single line segment (since one of the sides is zero) and since parallel transport along a single path is commutative (meaning that it doesn't depend how you move through that path), those terms do not matter. As Qmechanic said, this should be seen explicitly, but this heuristic reasoning shows why. $\endgroup$ Dec 24, 2023 at 16:07
  • 1
    $\begingroup$ If you want to do it rigorously, you’ll need to assume that the Lie bracket of $A,B$ cancel, or else complete the parallelogram with the Lie bracket, this will give another quadratic term. A useful special case are coordinate basis vector fields for which the Lie brackets are zero. $\endgroup$
    – LPZ
    Dec 24, 2023 at 22:08

1 Answer 1


Second order expression for parallel transport.

To start off, let $W\subset\Bbb{R}^m$ be open and let $E=W\times\Bbb{R}^k$ (a trivial vector bundle), and suppose we have smooth functions $\Gamma_i:W\to M_{k\times k}(\Bbb{R})$, whose entries we shall denote $\Gamma_i=\left(\Gamma^{\alpha}_{\,\beta i}\right)$, for $i\in\{1,\dots, m\}$ and $\alpha,\beta\in\{1,\dots, k\}$. Next, fix any smooth curve $x:I\to W$. Then a vector-valued curve $u:I\to \Bbb{R}^k$ is parallel-transported along the base curve $x$ (relative to $\Gamma_1,\dots\Gamma_m$) if the following ODE is satisfied for each $\alpha\in\{1,\dots, k\}$:

\begin{align} \dot{u}^{\alpha}(t)&=-\Gamma^{\alpha}_{\,\beta i}(x(t))\,u^{\beta}(t) \dot{x}^i(t), \end{align} i.e in slightly more index-free notation, \begin{align} \dot{u}(t)&=-\Gamma_i(x(t))\dot{x}^i(t)\,\cdot u(t), \end{align} where the $\cdot$ denotes a matrix product of a $k\times k$ matrix with a column vector in $\Bbb{R}^k$.

Next, carrying out a second-order Taylor expansion of $u(\epsilon)$, we get (denoting $p,v,a$ to mean $x(0),\dot{x}(0),\ddot{x}(0)$ respectively) \begin{align} u(\epsilon)&=u(0)+\epsilon\dot{u}(0)+\frac{\epsilon^2}{2}\ddot{u}(0)+O(\epsilon^3)\\ &=u(0)-\epsilon\Gamma_i(p)\,v^i\,\cdot u(0) +\frac{\epsilon^2}{2}\left[-\frac{\partial\Gamma_i}{\partial x^l}(p)v^lv^i\,\cdot u(0)-\Gamma_i(p)a^i\,\cdot u(0)-\Gamma_i(p)v^i\,\cdot \dot{u}(0)\right]+ O(\epsilon^3)\\ &=u(0)-\epsilon\Gamma_i(p)v^i\,\cdot u(0)+\frac{\epsilon^2}{2}\left[-\frac{\partial\Gamma_i}{\partial x^l}(p)v^lv^i\,\cdot u(0)-\Gamma_i(p)a^i\,\cdot u(0)+\left(v^i\Gamma_i(p)\right)^2\,\cdot u(0)\right]+O(\epsilon^3), \end{align} where we have repeatedly used the parallel-transport ODE to write successively $\dot{u}(0)$ and $\ddot{u}(0)$ in terms of $u(0)$, and also in the last bracketed term, I really mean the square of a matrix: $\left(\sum_{i=1}^mv^i\Gamma_i(p)\right)^2$. Notice how every term now appears with $u(0)$ being multiplied out front. In other words, as an equation relating matrices (or more precisely, endomorphisms of $\Bbb{R}^k$) we get the following for the parallel-transport operator: \begin{align} P_{x,[0,\epsilon]}&=\text{id}-\epsilon \Gamma_i(p)v^i+\frac{\epsilon^2}{2}\left(-\frac{\partial\Gamma_i}{\partial x^l}(p)v^lv^i-\Gamma_i(p)a^i+(\Gamma_i(p)v^i)^2\right)+O(\epsilon^3).\tag{$*$} \end{align} The equation $(*)$ is the key formula for us which gives us a second-order expression for parallel-transport along a curve $x$ and over the interval $[0,\epsilon]$.

Second order expression for parallel-transport along a rectangular loop

We shall now consider a rectangular loop, for small enough $\epsilon,\delta>0$, and any distinct $i,j\in\{1,\dots, m\}$, which starts at point $p\in W$, goes to the right to $p+\epsilon e_i$, then moves up to $p+\epsilon e_i+\delta e_j$ then moves left to $p+\delta e_j$, and finally back down to $p$. In what follows, expressions like $\frac{\partial\Gamma_i}{\partial x^i}(p)$ are NOT summed over.

  • We start off in the bottom leg from $p\to p+p+\epsilon e_i$. Our curve can be explicitly parametrized as $x(t)=p+te_i$ for $t\in[0,\epsilon]$. So, we have $x(0)=p,\dot{x}(0)=e_i$ and $\ddot{x}(0)=0$, so the formula $(*)$ above simply becomes \begin{align} P_{\text{bottom}}&=\text{id}-\epsilon\Gamma_i(p)+\frac{\epsilon^2}{2}\left(\Gamma_i(p)^2-\frac{\partial\Gamma_i}{\partial x^i}(p)\right)+O(\epsilon^3).\tag{1} \end{align}
  • Next, we consider the right leg of the rectangle, parametrized by $x(t)=p+\epsilon e_i+te_j$ for $t\in [0,\delta]$. So, $x(0)=p+\epsilon e_i$, $\dot{x}(0)=e_j,\ddot{x}(0)=0$; plugging this into $(*)$ gives \begin{align} P_{\text{right}}&=\text{id}-\delta\Gamma_j(p+\epsilon e_i)+\frac{\delta^2}{2}\left(\Gamma_j(p+\epsilon e_i)^2-\frac{\partial\Gamma_j}{\partial x^j}(p+\epsilon e_i)\right)+O(\delta^3). \end{align} This formula can now be expanded in powers of $\epsilon$ as well to give \begin{align} P_{\text{right}}&=\text{id}-\delta\Gamma_j(p)-\delta\epsilon\frac{\partial\Gamma_j}{\partial x^i}(p)+\frac{\delta^2}{2}\left(\Gamma_j(p)^2-\frac{\partial\Gamma_j}{\partial x^j}(p)\right)+O_3,\tag{2} \end{align} where $O_3$ stands for all the third order terms $O(\epsilon^3)+O(\epsilon^2\delta)+O(\epsilon\delta^2)+O(\delta^3)$.
  • Next we parametrize the top of the rectangle in the leftward direction: $x(t)=p+\epsilon e_i+\delta e_j-te_i$ for $t\in[0,\epsilon]$. So, we get from $(*)$, and then Taylor expanding the resulting terms (and be careful with extra minus signs since we’re traversing left), \begin{align} P_{\text{top}}&=\text{id}-\epsilon\cdot (-1)\Gamma_i(p+\epsilon e_i+\delta e_j)+\frac{\epsilon^2}{2}\left(\left(-\Gamma_i(p+\epsilon e_i+\delta e_j)\right)^2-\frac{\partial\Gamma_i}{\partial x^i}(p+\epsilon e_i+\delta e_j)\cdot(-1)^2\right)+O(\epsilon^3)\\ &=\text{id}+\epsilon\Gamma_i(p)+\epsilon^2\frac{\partial\Gamma_i}{\partial x^i}(p)+\epsilon\delta\frac{\partial\Gamma_i}{\partial x^j}(p)+ \frac{\epsilon^2}{2}\left(\Gamma_i(p)^2-\frac{\partial\Gamma_i}{\partial x^i}(p)\right)+O_3\\ &=\text{id}+\epsilon\Gamma_i(p)+ \epsilon\delta\frac{\partial\Gamma_i}{\partial x^j}(p)+\frac{\epsilon^2}{2}\left(\Gamma_i(p)^2+ \frac{\partial\Gamma_i}{\partial x^i}(p)\right)+O_3.\tag{3} \end{align}
  • Finally, we consider the left side in the downward direction: $x(t)=p+\delta e_j-te_j$ for $t\in [0,\delta]$. Then, $(*)$ and a Taylor expansion yield \begin{align} P_{\text{left}}&=\text{id}-\delta\cdot (-1)\Gamma_j(p+\delta e_j)+\frac{\delta^2}{2}\left(\left(-\Gamma_j(p+\delta e_j)\right)^2-\frac{\partial\Gamma_j}{\partial x^j}(p+\delta e_j)\cdot (-1)^2\right)+O(\delta^3)\\ &=\text{id}+\delta\Gamma_j(p)+\delta^2\frac{\partial\Gamma_j}{\partial x^j}(p)+\frac{\delta^2}{2}\left(\Gamma_j(p)^2-\frac{\partial\Gamma_j}{\partial x^j}(p)\right)+O_3\\ &=\text{id}+\delta\Gamma_j(p)+\frac{\delta^2}{2}\left(\Gamma_j(p)^2+\frac{\partial\Gamma_j}{\partial x^j}(p)\right)+O_3.\tag{4} \end{align}

Since the parallel-transport along the rectangular loop of size $\epsilon,\delta$ is given by the 4-fold composition of equations $(1),(2),(3),(4)$, we deduce that \begin{align} P_{\text{rectangular loop}}&=P_{\text{left}}\circ P_{\text{top}}\circ P_{\text{right}}\circ P_{\text{bottom}}\\ &=\text{id}-\epsilon\delta\left(\Gamma_i(p)\cdot\Gamma_j(p)-\Gamma_j(p)\cdot\Gamma_i(p)+\frac{\partial\Gamma_j}{\partial x^i}(p)-\frac{\partial\Gamma_i}{\partial x^j}(p)\right)+O_3.\tag{$**$} \end{align} The $\cdot$ here is matrix multiplication (or if you wish, composition of endomorphisms on $\Bbb{R}^k$). Note that it is only in equations (1)-(4) above that we had to do analysis while this final step is simple (albeit a little tedious) algebra: just compose the four expressions, open up all the brackets (being careful that matrix multiplication, or rather composition of endomorphisms is non-commutative), collect like terms, and then simplify. In particular, the coefficients of $\epsilon^2$ and $\delta^2$ vanish. Since this ‘simple’, I will not write out the intermediate algebra. Notice also that in components, the bracketed term is the formnula for the Riemann curvature $R^{\cdot}_{\,\cdot\,ij}=(R^{\alpha}_{\,\beta i j})$.

Putting it all together

Let $(E,\pi, M,,\nabla)$ be a vector bundle with linear connection. Fix the following objects:

  • a point $p\in M$ and two linearly independent tangent vectors $h_p,k_p\in T_pM$ and two different indices $i,j\in\{1,\dots, m\}$.
  • coordinate chart $(W,x=(x^1,\dots, x^m))$ around $p$ such that $\frac{\partial}{\partial x_i}(p)=h$ and $\frac{\partial}{\partial x^j}(p)=k$, over which $E$ is trivializable with vector-bundle chart $(\pi^{-1}(W),(x,u)=(x^1,\dots, x^m,u^1,\dots, u^k))$. Relative to this local trivialization, let $\Gamma_i=(\Gamma^{\alpha}_{\,\beta i})$ be the Christoffel symbols of $\nabla$.
  • Finally, consider the (piecewise $C^{\infty}$) loop $\lambda_{\epsilon,\delta}$ in $M$ based at point $p$ such that its coordinate representation, $x(t)$, is the union of the four curves I explicitly wrote above.

Then, from $(**)$, we get \begin{align} P_{\lambda_{\epsilon,\delta}}&=\text{id}_{E_p}-\epsilon\delta\,\mathbf{R}\left(\frac{\partial}{\partial x^i}(p)\wedge\frac{\partial}{\partial x^j}(p)\right) +O_3(\epsilon,\delta)\\ &=\text{id}_{E_p}-\epsilon\delta\,\mathbf{R}(h_p\wedge k_p)+O_3(\epsilon,\delta),\tag{$***$} \end{align} where in the second equal sign we have used the well-known coordinate formula for the Riemann curvature endomorphism in terms of Christoffel symbols (I’m assuming one has already defined the curvature in terms of the commutator of covariant derivatives… alternatively you can stop at the first equal sign and check that the stuff in brackets actually gives a well-defined expression, or more precisely an $\text{End}(E)$-valued $2$-form on $M$). Notice that this is an equality of endomorphisms on the fiber $E_p$.

Therefore, we see that $(***)$ tells us exactly that when we parallel-transport along a small (finite, not infinitesimal) rectangular loop generated by linearly independent tangent vectors, then the second order effect is precisely the curvature $\mathbf{R}(h_p\wedge k_p)$.


You simply have to start from the parallel-transport ODE (which is a linear ODE) and then use that to figure out how this determines the Taylor expansion of the resulting parallel-transport operator. Regarding your issue about writing out only some but not all second order corrections, I don’t really understand your question (neither do I follow the usual presentations in GR books). In my mind, this is a simple analysis fact: systematically write out Taylor expansions at each stage, and combine them. If you were to do trial and error, you’d see that at zeroth order, you of course get just the identity operator. If you tried to only first-order Taylor expand everything, then you’ll see that at the end, the first order term vanishes. So, you go back and redo everything for second-order, and now you realize you get something non-trivial.

Btw, a very common mistake is to write first order expansions for each leg of the parallel-transport, and then compose the four results, which will of course produce some second order terms, and then claim that is the full second-order expansion. This is obviously incorrect (and you will be missing some terms, or you won’t be able to cancel them out etc).

And finally, here is another answer with more commentary and also the statement of the theorem but no proof (which I was too lazy to provide back then so I did now) but also with a picture to describe the loops and charts.

  • $\begingroup$ Excellent answer. I was sloppy in my question, so thank you for writing a rigorous answer as well. As per GR books, to my understanding they do what I laid out above and you described in the conclusions: they do a first order expansion at each leg and then some second order terms appear (which as you say are not all the story) but only the $\varepsilon^2,\delta^2$ are neglected as higher order terms, while $\delta\varepsilon$ is not. $\endgroup$ Dec 25, 2023 at 8:29
  • $\begingroup$ just a minor remark: Is the choice to work with $W\subset \mathbb{R}^m$ (and I guess the $U$ in the second line is a $W$) and trivial bundles WLOG? I mean, it is the usual argument of considering $M=\mathbb{R}^m$ since manifold are locally euclidean, right? $\endgroup$ Dec 25, 2023 at 9:03
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    $\begingroup$ @Mr.Feynman yes, $U$ should be $W$, that’s a typo. And yes, it is without loss of generality because the definition of a vector bundle (and manifold) give us a coordinate chart and local trivialization as explained in my answer. But I should highlight that in my answer, the local trivialization is important because I keep writing the identity map $\text{id}$ in equations (1)-(4), but this means $\text{id}_{\Bbb{R}^k}$. In other words, if I merely considered a loop based at $p$, I cannot split it up ‘abstractly’ and do a Taylor expansion that would be two different fibers of a vector bundle $\endgroup$
    – peek-a-boo
    Dec 25, 2023 at 9:08
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    $\begingroup$ and so an identity map would be meaningless for different fibers in a vector bundle. We first trivialize, then compute, then put it all together (see $(**)$) and then observe that this is a chart and trivialization-independent result. $\endgroup$
    – peek-a-boo
    Dec 25, 2023 at 9:09
  • $\begingroup$ and finally, it’s been a long time since I’ve looked at the physics book arguments, so I don’t know if they truly disregarded the $\epsilon^2,\delta^2$ terms with the reasoning being ‘they are negligible’ (because that’s complete garbage, and I can’t believe so many textbooks would make such an elementary error). So, instead I think they’re all just quick with the computations showing only a few steps in between and letting the reader work out the algebra to discover for themselves that $\epsilon^2,\delta^2$ terms cancel out. $\endgroup$
    – peek-a-boo
    Dec 25, 2023 at 9:11

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