I admit I am not a differential geometer (a probabilist actually). However recently I get interested and I would like to have more intuitions and insight of what is the Riemann curvature.

This is the way I see it so far (please correct me if I am wrong):

  • We start from a connection $\nabla$.
  • This defines a parallel transport along paths $\gamma$ that is a linear application on the vector bundle.
  • If $\gamma$ is a loop, this application may be different to the identity.
  • The Riemann curvature is different from the identity when $\gamma$ is a very small loop (at first order).

This kind of definition seems very similar to the one of the rotational (as presented in physics classes).

  • We start from a vector field (one form) $u$,

  • If $\gamma$ is loop, $I=\oint_{\gamma}u\cdot d\gamma$ can be different to $0$.

  • The rotation $\operatorname{rot}(u)$ is this value $I$ when $\gamma$ is a very small loop (at first order).

and we have the wonderful Stokes Theorem which for the $\operatorname{rot}$ follows very naturally from this definition (we glue small loops together to get a big loop) $$\oint_\gamma u \cdot d\gamma = \iint_\mathcal{S}\operatorname{rot}(u)\cdot d\sigma $$ with $\mathcal{S}$ a surface delimited by $\gamma$.

So here is my question: Does there exist an equivalent for the Riemann curvature? That is: can one calculate the parallel transport of $\nabla$ along a loop $\gamma$ from the Riemann curvature on $\mathcal{S}$?

  • $\begingroup$ Consider a connection on a vector bundle over a 1-dimensional manifold: the circle. You can easily find all such connections up to isomorphism, and see that the parallel transport does not have to be trivial, but the curvature must be zero. $\endgroup$
    – Ben McKay
    Aug 7, 2020 at 14:08
  • $\begingroup$ Is this something like what you’re looking for? math.nyu.edu/~yangd/papers/holonomy.pdf $\endgroup$
    – Deane Yang
    Aug 7, 2020 at 18:28

1 Answer 1


In the case that the Riemannian manifold $M$ in question has dimension $2$ and is oriented and $\gamma([0,1])\subset M$ is the piecewise-$C^1$ oriented boundary of a compact domain $S\subset M$, we have the famous Gauss-Bonnet Theorem, which asserts that the holonomy around $\gamma$ is equal to rotation by the angle $$ \theta = \int_S K\,dA. $$ Thus, yes, in this case, the holonomy around $\gamma$ can be 'computed' from the Riemann curvature tensor.

This may seem a little unsatisfying because the parallel transport around $\gamma$ is defined using only information in an open neighborhood of $\gamma$ (and, of course, one can get away with less than that), but the above formula uses information that could, a priori, come from very far away from the image $\gamma([0,1])$. However, simple examples show that, even for surfaces, one can have a closed curve with arbitrary holonomy for which the metric is flat on a neighborhood of the curve. Thus, the holonomy cannot be computed purely locally from the Riemann curvature tensor.

(By the way, Ben's cautionary remark is based on a $\gamma$ that is not the boundary of any surface, so there couldn't be a formula of the kind you are seeking that would address his 'counterexample'.)

Once one goes to higher dimensions, even for Riemannian $3$-manifolds$(M,g)$, there is no formula known that would start with the Riemann curvature tensor and construct 'something' that could be integrated over every oriented compact surface $S$ with connected boundary $\partial S$ so as to yield the element of $\mathrm{SO}(T_{p}M)$ that is the holonomy of the oriented curve $\partial S$ at the boundary point $p\in \partial S$.

Actual nonexistence would be hard to prove without making some assumptions about what form the 'something' constructed out of the Riemann curvature tensor might take. However, if you make reasonable assumptions, you can rule things out.

For example, it is not hard to show that there is no universal $2$-form on $(M^3,g)$ constructed polynomially out of the Riemann curvature tensor (even when one is allowed to use $g$ and $\nabla$ as well) that has the property that its integral over any oriented compact surface $S\subset M$ with circular boundary $\partial S$ gives even the angle of the rotation of the Riemannian holonomy around $\partial S$. (Note that this angle in $[0,\pi]$ does not depend on the point $p\in\partial S$ that one choses as the 'initial point'.)

  • $\begingroup$ Thank you a lot for your answer! I still have two simple question to make it clear : 1- if the Riemann curvature is 0 on $S$ then the parallele transport of $\nabla$ on the loop $\partial S$ is the identity. 2-If $\nabla_1$ and $\nabla_2$ are two connections with the same Riemann curvature tensor on $S$, does the two parallele transport of $\nabla_1$ and $\nabla_2$ on $\partial S$ are the same? $\endgroup$
    – RaphaelB4
    Aug 31, 2020 at 12:03
  • 1
    $\begingroup$ @RaphaelB4: Yes, if $S$ is a compact, oriented surface with $\partial S$ being a circle and with a smooth Riemannian metric $g$ whose Riemannian curvature is zero, then the parallel transport of $\nabla$ around the entire (oriented) loop $\partial S$ is the identity. Meanwhile, if $g_1$ and $g_2$ are two metrics on $S$ with vanishing Riemannian curvature, then the parallel transports of $\nabla_1$ and $\nabla_2$ along a proper subinterval $I\subset \partial S$ from one end of $I$ to the other need not be the same. $\endgroup$ Aug 31, 2020 at 14:04
  • $\begingroup$ I think the interesting case is $g_1$ and $g_2$ with non vanishing Riemannian curvature $R_1 = R_2$ and $I=\partial S$ the entire loop. Can we still say something here? $\endgroup$
    – RaphaelB4
    Aug 31, 2020 at 15:42
  • 1
    $\begingroup$ @RaphaelB4: It wasn't clear what you meant. Remember that the $2$-form integrand $K\,dA$ is constructed from the Riemann curvature tensor, the metric $g$ and the orientation of the surface by an algebraic formula. As long as $K_1\,dA_1 = K_2\,dA_2$, then the holonomies of the two metrics around the boundary circle $\partial S$ will be the same. Meanwhile, note that, if one is, as is standard, regarding the Riemann curvature tensor as a tensor of type $(1,3)$, then one can easily have $R_1 = R_2\not=0$ without having $g_1=g_2$. $\endgroup$ Aug 31, 2020 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.