Timeline for Algebraic theorems with no known algebraic proofs
Current License: CC BY-SA 4.0
34 events
| when toggle format | what | by | license | comment | |
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| Nov 26 at 2:29 | comment | added | Martin Brandenburg | Can people here please stop adding their answers in the comment section and instead post them as answers? Thanks. | |
| Nov 22 at 13:45 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| Nov 20 at 13:31 | answer | added | tkf | timeline score: 14 | |
| Nov 20 at 2:00 | answer | added | Jesse Elliott | timeline score: 5 | |
| Nov 20 at 0:00 | comment | added | Andres Mejia | do proofs from motivic homotopy theory count? | |
| Nov 19 at 20:45 | answer | added | R. van Dobben de Bruyn | timeline score: 16 | |
| Nov 19 at 20:12 | comment | added | R. van Dobben de Bruyn | Ah I see. If you really insist on non-associative algebras, you are probably right that there is no algebraic proof. The associative case is apparently due to Frobenius, for which there are many proofs, including many algebraic ones. | |
| Nov 19 at 18:53 | comment | added | jjcale | Is there a purely algebraic definition for "totally real number field" ? | |
| Nov 19 at 9:33 | answer | added | Adam Epstein | timeline score: 4 | |
| Nov 19 at 8:57 | answer | added | Tobias Fritz | timeline score: 19 | |
| Nov 19 at 7:55 | answer | added | Henri Cohen | timeline score: 7 | |
| Nov 19 at 7:52 | answer | added | Piotr Achinger | timeline score: 39 | |
| Nov 19 at 7:34 | comment | added | Carl-Fredrik Nyberg Brodda | A former example: every subgroup of a surface group is a surface group (if finite index) or free (if infinite index). This was well-known by topological means for a long time (certainly Klein and Fricke knew it) but only given an algebraic proof in the 1970s. | |
| Nov 19 at 7:15 | history | edited | no upstairs |
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| Nov 19 at 1:37 | history | became hot network question | |||
| Nov 18 at 23:29 | comment | added | Sam Hopkins | @R.vanDobbendeBruyn see mathoverflow.net/questions/401542 where some related results are discussed. But if there really are purely algebraic proofs of these theorems of Hopf and Kervaire/Milnor, I would be quite interested to know that (and the Wikipedia page could be updated). | |
| Nov 18 at 21:49 | comment | added | Sam Hopkins | @R.vanDobbendeBruyn but the theorems are about not necessarily associative algebras, right? Am I misunderstanding your comment? | |
| Nov 18 at 20:59 | comment | added | R. van Dobben de Bruyn | I'm going to politely disagree with the example in the post: at least for associative division algebras, their classification is essentially the computation that $\operatorname{Br}(\mathbf R) \cong \mathbf Z/2\mathbf Z$, which is easy to prove using Galois cohomology: $H^2(\mathbf R,\mathbf C^\times) \cong \hat H{}^0(\mathbf R,\mathbf C^\times) = \mathbf R^\times/N(\mathbf C^\times)$. I consider this an algebraic proof. That said, I have no idea if Galois cohomology tells you anything about non-associative algebras. | |
| Nov 18 at 20:53 | comment | added | Benjamin Steinberg | @WillBrian, my point is that if $\beta \mathbb N$ has idempotents so does every other compact right topological semigroup because it is free on one generator in this category | |
| Nov 18 at 20:42 | comment | added | Will Brian | @BenjaminSteinberg: Yes, I agree. I expect that any proof of this theorem needs to make use of the topology on $\beta \mathbb N$, either openly or sneakily. That's what makes it a good example of what the OP is asking for (unless it's dismissed as not "real algebra" and more set theory). | |
| Nov 18 at 20:06 | comment | added | Benjamin Steinberg | @WillBrian, this seems equivalent to the statement that every compact right topological semigroup has an idempotent so I'm not sure how a nontopological proof can be expected | |
| Nov 18 at 19:31 | comment | added | Will Brian | There is an important semigroup operation $+$ defined on the set of ultrafilters on $\mathbb N$. It is an important and nontrivial theorem that there are non-principal ultrafilters $u$ for which $u+u = u$. The statement of this theorem, and all of the relevant definitions, can be expressed without any reference to any topology on the set of ultrafilters. But every proof of the theorem that I know involves topology in a nontrivial way. (I'm putting this as a comment, rather than an answer, because I imagine it might stretch too far what you mean by an "algebraic theorem.") | |
| Nov 18 at 19:23 | comment | added | Benjamin Steinberg | Yes, it should say are free | |
| Nov 18 at 19:17 | history | made wiki | Post Made Community Wiki by Stefan Kohl♦ | ||
| Nov 18 at 18:43 | comment | added | Sam Hopkins | @BenjaminSteinberg are you missing some words in your second comment? The proofs that "torsion-free groups of cohomological dimension 1" what? | |
| Nov 18 at 18:39 | comment | added | Benjamin Steinberg | I believe that the proofs that torsion-free groups of cohomological dimension 1 are at least somewhat topological or geometric, although I’m not sure where you put the dividing line between algebra and combinatorial topology. | |
| Nov 18 at 18:36 | comment | added | Benjamin Steinberg | I don’t believe there are any purely algebraic proofs of Gromov’s theorem that groups of polynomial growth have a nilpotent subgroup of finite index. | |
| Nov 18 at 18:28 | comment | added | J. W. Tanner | Indeed, it has been remarked that the Fundamental Theorem of Algebra is neither fundamental nor a theorem of algebra | |
| Nov 18 at 18:25 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
added 45 characters in body
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| Nov 18 at 18:05 | answer | added | Peter Mueller | timeline score: 32 | |
| Nov 18 at 18:01 | answer | added | J. W. Tanner | timeline score: 7 | |
| Nov 18 at 17:52 | comment | added | Peter Mueller | @J.W.Tanner There are algebraic proofs which only use two facts from analysis: Real polynomials of odd degree have a real root, and each complex number has a complex square root. At some point an algebraic proof has to use the definition or easy properties of real or complex numbers. | |
| Nov 18 at 17:48 | comment | added | J. W. Tanner | Fundamental theorem of algebra? | |
| Nov 18 at 17:36 | history | asked | Sam Hopkins | CC BY-SA 4.0 |