I've read many times that moving coframes where a convenient tool for computations in Riemannian Geometry, especially on surfaces, but never really used it. Lately to get a better feel of the method, I've tried to reprove classical results on surfaces using moving coframes and the Cartan structure equations :

Cartan's equations for surfaces : if $(\eta^1,\eta^2)$ is a orthonormal coframe on $(\Sigma^2,g)$, then $d\eta^1=\omega\wedge\eta^2$ and $d\eta^2=-\omega\wedge\eta^1$ for a unique $1$-form $\omega$, which moreover satisfies $d\omega=-K_g\eta^1\wedge\eta^2$.

What is really nice is that in these three equations we define at the same time the Levi-Civita connection (through $\omega$ which is the connection $1$-form) and the curvature, without ever actually computing covariant derivative.

I have been amazed at how some result get much easier in this framework, like the fact that $K_g=0$ implies local isometry to the euclidean metric, and also at how curvature computation become more straight forward.

Moreover a lot of the objects of interest in geometric analysis can be defined solely in terms of $1$-forms, the Hodge star $\ast$, exterior differential of forms and wedge product :

  • $g(\nabla u,\nabla v)=\ast(du\wedge \ast dv)$.
  • $\Delta u=\ast\ d\ast d u$.

As a geometric analyst, my favorite equation is probably Bochner's formula, which in the case of surfaces reads as : $$\tfrac{1}{2}\Delta |\nabla u|^2=\langle\nabla u,\nabla\Delta u\rangle+|\mathrm{Hess}\, u|^2+K_g|\nabla u|^2.$$ The usual proof of the Bochner formula involves commutations of covariant derivatives. My question is :

Question : Is it possible to prove the Bochner formula for surfaces relying on Cartan's structure equations instead of commutation of covariant derivative ?

It basically boils down to two subquestions :

  • How does one compute $d\ast d\left(\ast(du\wedge \ast du)\right)$ ?
  • Can one express $|\mathrm{Hess}\, u|^2$ in terms of the Hodge star, the connection form and exterior derivatives ?

1 Answer 1


The straightforward way to do this is to note that we have, for some $u_1$ and $u_2$, $$ \mathrm{d}u = u_1\,\eta^1 + u_2\,\eta^2 $$ Taking the exterior derivative of this and using Cartan's Lemma then shows that, for some $u_{11}$, $u_{12}$, and $u_{22}$, $$ \begin{aligned} \mathrm{d}u_1 &= \phantom{-}u_2\,\omega + u_{11}\,\eta^1 + u_{12}\,\eta^2\\ \mathrm{d}u_2 &= -u_1\,\omega + u_{12}\,\eta^1 + u_{22}\,\eta^2 \end{aligned} $$ Then taking exterior derivatives of these equations and using Cartan's Lemma yields $$ \begin{aligned} \mathrm{d}u_{11} &= & 2u_{12}\,&\omega + u_{111}\,\eta^1 + u_{112}\,\eta^2\\ \mathrm{d}u_{12} &= &(u_{22}-u_{11})\,&\omega + u_{121}\,\eta^1 + u_{122}\,\eta^2\\ \mathrm{d}u_{22} &= &-2u_{12}\,&\omega + u_{221}\,\eta^1 + u_{222}\,\eta^2\\ \end{aligned} $$ for some $u_{ijk}$ satisfying $u_{112}-u_{121} = -K\,u_2$ and $u_{122}-u_{221} = K\,u_1$.

Now just plug these formulae into the left hand side of Bochner's formula and compute.

Comment: I understand that the OP would like to have some kind of exterior algebra 'magic' that just makes the calculation a consequence of $\mathrm{d}^2=0$, but that's an unreasonable request, it seems to me. The strength of the method of calculating by moving coframes is that it does show that all the local identities that you can prove are consequences of $\mathrm{d}^2=0$ and exterior algebra, though you may have to apply this principle multiple times if the identity you want involves higher derivatives.

For example, in this particular case, the general structure equations for an orthonormal coframe (i.e., $g = {\eta_1}^2 + \cdots + {\eta_n}^2$) are that there are unique $1$-forms $\eta_{ij}=-\eta_{ji}$ satisfying $$ \begin{aligned} \mathrm{d}\eta_i &= -\eta_{ij}\wedge\eta_j\\ \mathrm{d}\eta_{ij} &= -\eta_{ik}\wedge\eta_{kj} + \tfrac12 R_{ijkl}\,\eta_k\wedge\eta_l \end{aligned}\tag1 $$ where $R_{ijkl}=-R_{ijlk}$ and all terms are summed on repeated indices.

Now, if $u$ is a function in the domain of $\eta$, we have $$ \mathrm{d}u = u_i\,\eta_i\tag2 $$ for some functions $u_i$. Applying the exterior derivative and then Cartan's Lemma to (2) yields $$ \mathrm{d}u_i = -u_j\,\eta_{ij} + u_{ij}\,\eta_j\tag3 $$ for some functions $u_{ij}=u_{ji}$. Applying the exterior derivative and then Cartan's Lemma to (3) yields $$ \mathrm{d}u_{ij} = -u_{ik}\,\eta_{kj} -u_{jk}\,\eta_{ki}+ u_{ijk}\,\eta_k\tag4 $$ where $u_{ilk}-u_{ikl} = R_{ijkl}\,u_j$ (and, of course, $u_{ijk}=u_{jik}$).

Now note that $|\nabla u|^2 = u_iu_i$ and then plug in the above formulae for derivatives to get $$ \tfrac12\Delta|\nabla u|^2 = u_{ij}u_{ij} + u_iu_{ijj}. $$ Since, by the above formulae, $u_{ijj}=u_{jij} = u_{jji} + R_{jlji}\,u_l$, this yields $$ \tfrac12\Delta|\nabla u|^2 = u_iu_{jji} + u_{ij}u_{ij} + R_{jlji}\,u_iu_l\,, $$ and the rest is just interpreting the three terms, where, in general dimension, the final term involving curvature becomes $$ \mathrm{Ric}_g(\nabla u,\nabla u). $$

Whether you choose to use moving coframes or not is a matter of taste, of course, since you can equally well work with the definition of curvature in terms of the Lie brackets of the dual vector fields. (Of course, the $u_i$, $u_{ij}$ and $u_{ijk}$ are just the result of 'covariantly differentiating' $u$ with respect to these vector fields.)

  • $\begingroup$ I was kind of hoping for something less heavy on index book keeping, but I'll give it a try. Thanks. $\endgroup$ Commented Nov 13, 2016 at 17:22
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    $\begingroup$ I have to say that I think this calculation might be easier to do in an arbitrary dimension rather than specifically in dimension $2$. Or at least write $\omega$ as $\omega_{21}$. Also, although the notation gets cumbersome, I often prefer to write $u_i = \partial_i u$ ($\partial$ because the connection is not used), $u_{ij} = \nabla_j\partial_iu$, etc. This helps me keep track of what everything means. Or do this at first but switch back to Robert's notation after you have a clear idea of what everything is. $\endgroup$
    – Deane Yang
    Commented Nov 13, 2016 at 20:35
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    $\begingroup$ @DeaneYang: I agree that the invariant nature of the calculation is clearer in arbitrary dimension, but that wasn't what was asked. Also, I have to say that I dislike using the notation $\partial_i u$ to mean the derivative of $u$ in the direction of the $i$-th vector field in the dual orthonormal frame field (which is what you are doing). I find that this is too confusing to students, who often forget the meaning and regard it as a derivative with respect to a coordinate chart. Moreover, $u_i$ is simpler to write than $\partial_iu$. $\endgroup$ Commented Nov 13, 2016 at 21:27
  • $\begingroup$ Robert, I think you have a good point about $\partial_iu$. It's a convention I use for myself, but after all these years I find myself still tweaking the notation I use for calculations. $\endgroup$
    – Deane Yang
    Commented Nov 13, 2016 at 23:22

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