Timeline for Results with short, advanced proofs or long, elementary proofs
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 8, 2023 at 7:40 | comment | added | Lennart Meier | While slightly longer, the proof via Sperner's lemma is of surpreme beauty. And I say this as an algebraic topologists without any particular combinatorial leanings. | |
| Nov 7, 2021 at 23:15 | comment | added | anomaly | @ThibautDemaerel: Of course--- and maybe the equivalence of de Rham and singular cohomology is another example of the sort of thing I was talking about the original post, at least in terms of integration of forms to detect nonzero cohomology classes. | |
| Nov 7, 2021 at 18:59 | comment | added | mlk | @anomaly Students going towards applied analysis will know Brouwer's theorem, as it leads them to the Schauder fix point theorem, which is used a lot to prove existence of solutions for all kinds of problems. But usually they know nothing about cohomology (at least not under the name) and may only have seen differential forms once from a distance. So to them the standard proof often is the one via Stokes theorem, though written explicitly, without differential forms. | |
| Nov 7, 2021 at 18:33 | comment | added | 5th decile | @anomaly That approach and the one via Stokes theorem are closely related. | |
| Nov 7, 2021 at 18:11 | comment | added | anomaly | While Sperner's Theorem certainly works, isn't the standard proof of Brouwer via noting that $\text{id}:H^{n-1}(S^{n-1}) \to H^{n-1}(S^{n-1})$ can't factor through $H^n(D^n) = 0$? I'm considering singular rather than de Rham cohomology above, but I think most students encounter de Rham cohomology and differential forms roughly around the same time. | |
| S Nov 7, 2021 at 16:48 | history | answered | 5th decile | CC BY-SA 4.0 | |
| S Nov 7, 2021 at 16:48 | history | made wiki | Post Made Community Wiki by 5th decile |