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Can the theorem be proved invariantly, without any reference to charts,frames, basis vectors or coordinates?

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    $\begingroup$ Please state what do you exactly mean by Gauss-Bonnet theorem (there are several formulations with this name). And how do you explain what is a surface "without charts"? $\endgroup$
    – Alexandre Eremenko
    Commented Jul 19, 2019 at 15:15
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    $\begingroup$ The "standard" Chern proof using Cartan's structure equations and transgression of differential forms is coordinate-free. $\endgroup$
    – alvarezpaiva
    Commented Jul 19, 2019 at 17:37
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    $\begingroup$ You can make the Chern proof coordinate free. Also, the Hopf proof is coordinate free, but it's for submanifolds rather than abstract Riemann manifolds. $\endgroup$
    – Ryan Budney
    Commented Jul 19, 2019 at 19:40
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    $\begingroup$ @Riu: yes, the standard textbook proofs use charts and coordinates, but, as alvarezpaiva says, the Chern proof, when using Cartan's structure equations, goes through directly on the frame bundle, without using any charts, or coordinates. $\endgroup$
    – Ben McKay
    Commented Jul 19, 2019 at 21:02
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    $\begingroup$ Related: mathoverflow.net/questions/336303/… $\endgroup$
    – Francois Ziegler
    Commented Jul 19, 2019 at 21:21

2 Answers 2

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Yes. A beautiful conceptual coordinate-free proof is presented by Berwick-Evans in https://arxiv.org/abs/1310.5383. It obtains both sides of the Chern–Gauss–Bonnet theorem as two limits of a partition function associated to a certain (rather simple) 0|2-dimensional supersymmetric sigma-model.

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Chern's famous proof of Gauss-Bonnet in all dimensions is basis-vector and coordinate free (and also intrinsic, i.e. not requiring some embedding in say a euclidean space). Reason for this is the use of differential forms, which have both these properties, which in turn are properties of its underlying algebra, the exterior algebra. The exterior product is basis-free. Do Carmo's book is a good place to learn about this aside from Chern's quite accessible and short paper.

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  • $\begingroup$ Read the description ,I think we have a misunderstanding on what coordinate free is. I asked for a proof without basis vectors and coordinates. $\endgroup$
    – Riu
    Commented Jul 20, 2019 at 16:21
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    $\begingroup$ The very nature of differential forms is that they do not require a choice basis. One can find a projection onto a chosen basis, but this is neither required nor done in Chern's proof. Chern's proof also does not use coordinates in the sense of any substrate space where points in space can be identified by coordinates. Rather it uses Cartan's moving frames to traverse over the surface in question. As all that is needed is derived from the surface itself independent of any embedding, this is an intrinsic proof. If this does not answer your question you need to clarify how it doesn't. $\endgroup$
    – Georg Essl
    Commented Jul 20, 2019 at 16:30
  • $\begingroup$ Oh well then I'm sorry. I also meant no frames .. $\endgroup$
    – Riu
    Commented Jul 20, 2019 at 17:25

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