I am a PhD student working in the field of numerical simulation. In several papers, the term "De Rham map" pops up (for instance in the very good thesis by Jérôme Bonelle : https://tel.archives-ouvertes.fr/tel-01116527v2/document ).

I am unsure about the definition of this term. I have come to believe that this term generally means "operation that have continuous objects correspond to discrete ones" i.e. "means of defining the actual values of the degrees of freedom of a discrete object from a continuous object", but I gradually suspect that it might rather mean "result of the integration of a cochain on a differentiable manifold". Of course, the two notions coincide in the litterature I have come accross.

So my question is : What is the definition of the term "De Rham map" ?


  • $\begingroup$ Got lots of hits googling 'De Rham'. Maybe you should investigate and explain why these are not relevant. $\endgroup$
    – BruceET
    Dec 18, 2017 at 17:05
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because numerous Internet pages (incl Wikipedia) seem to deal with the question. Suggestion to google received no response. $\endgroup$
    – BruceET
    Jan 5, 2018 at 15:07
  • $\begingroup$ the suggestion was irrelevant and an intense Google search was obviously already made with no result, hence the question. But you may close the question anyway, I doubt it will receive any quality answer now that a few days have passed. $\endgroup$ Jan 5, 2018 at 15:40

1 Answer 1


In the context of approximating second order boundary value problems we invented the name "de Rham map" with A. Bossavit in paper T. Tarhasaari, L. Kettunen, A. Bossavit: Some realizations of a discrete Hodge operator: A reinterpretation of finite element techniques. IEEE Trans. Magn., Vol 35(3), 1999, pp. 1494-97.

Intuitively, the idea is as follows. Finite element and finite difference kind of methods provide one with approximative solutions of boundary value problems in finite dimensional spaces. To build such approximations one needs a map from the (infinite dimensional) linear spaces of fields to some finite dimensional spaces in which the fields are approximated.

Having a "mesh", that is, a cellular complex, and a field, say p-form $f$, the de Rham map sends $f$ to integrals of $f$ on p-chains (which in turn are formal sums of p-cells of the complex/mesh). Or, alternatively, a simplified view, the de Rham map sends f to an array of integrals of $f$ on the (oriented) p-cells of the complex.

For example, the de Rham map sends the 2-form magnetic flux $b$ to an array of real numbers representing fluxes $\int_c b$ on the 2-cells $c$ (such as triangles) of the complex/mesh. In terms of classical language, if magnetic flux density ${\bf B}$ is considered as a (smooth) vector field, pair $({\bf B}, c)$ is mapped to $\int_c {\bf B}\cdot {\bf n}\, {\rm da}$.

Formally, the definition is something like: Let $F^p$ be the space of differential forms on manifold $\Omega$, $C^p$ the space of $p$-cochains in cellular complex $K$, and $K$ is a cellular tessellation of $\Omega$. Then, map $\mathcal{C}$ from $F^p$ to $C^p$ is the de Rham map, if for all $p$-chains $c\in K$ and smooth $p$-forms $f\in F^p$ map $\mathcal{C}$ satisfies $$\mathcal{C}f: C^p \rightarrow \mathbb{R}, \quad c\mapsto \int\limits_c f\,.$$

  • $\begingroup$ Thank you very much for your answer. It confirms what I thought I had understood. I believe the accepted term to more generally mean "mapping from a continuous fizld to its discrete representatuon " is "reduction operator". The De Rham map being a particular instance of reduction operator. Thank you again. $\endgroup$ Jun 20, 2019 at 15:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .