Timeline for Compelling evidence that two basepoints are better than one
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2018 at 12:17 | comment | added | HJRW | I suspect that the proof proposed here can be exactly translated into the nice proof using group actions given by Benjamin Steinberg here: arxiv.org/abs/1006.3833 . At the risk of being polemical, I also suspect that this illustrates exactly why there is no really compelling evidence that "two basepoints are better than one". For most of the examples listed, one can work with group actions instead of groupoids and everything works out fine. This is certainly what I do. | |
| Jan 29, 2012 at 15:39 | comment | added | Ronnie Brown | It is possible to explain the theory of matrices without mentioning vector spaces and linear mappings. That does make it more difficult to explain some aspects of the theory of rank of a matrix, and even more difficult to explain what is discussed under normed vector spaces, topological vector spaces, etc., etc. | |
| Jan 25, 2012 at 7:52 | comment | added | Ronnie Brown | The book Higgins, P.J. Notes on categories and groupoids,Mathematical Studies, Volume 32.Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1-195. gives a generalisation of Grushko's theorem and a proof using groupoids. This generalisation has not been proved by other means as far as I know. Higgins started off with a translation into the algebra of covering morphisms of groupoids of the topological proof. Groupoids allow a nice version of the theory of covering spaces: see the book "Topology and groupoids". | |
| Oct 9, 2010 at 14:16 | comment | added | HJRW | Mariano, I don't think that's entirely fair. The 'usual' purely algebraic proof involves Nielsen reduction (see, for instance, Lyndon and Schupp). Although we think of it as morally similar to the topological proof, there are technicalities that make it a little different. (For instance, the algebraic proof naturally deals with ordered bases, whereas the topological proof spits out a basis as a set.) Greg's proposal does seem like a closer translation into algebra of the topological proof. +1 to Greg. | |
| Oct 5, 2010 at 7:58 | comment | added | Greg Graviton | Fair enough. (I don't know enough about groupoids to judge them). Interestingly, there are counterexamples to the "body of theorems" requirement, though. For instance, there are no non-trivial theorems about functors in general, but they are a really slick way of talking about one ever reappearing pattern. | |
| Oct 4, 2010 at 19:16 | comment | added | Ryan Budney | The point is that group theory has a sufficient body of non-trivial theorems where the terminology becomes a useful conceptual macro. In almost every mathematical paper there are "local definitions" of concepts useful for some stage in a proof. These ideas only earn broad acceptance once people see how convienient the theory is. What I think Daniel is getting at is, "is there such body of theorems for groupoids"? Or are groupoids just a terminology preference of authors -- some choose to use them, some choose to stick to concepts that do not explicitly invoke groupoids, but are equivalent. | |
| Oct 4, 2010 at 15:45 | comment | added | Greg Graviton | Well, you can shuffle cards or perform other permutations without ever mentioning groups, too... Do you have a proof in mind that uses a different construction? I would give the label "groupoid" to anything that looks like the above; more precisely, my criterion for the label "groupoid" is that $h \in H$ is represented as a "composition of edges in a graph". | |
| Oct 4, 2010 at 12:43 | comment | added | Mariano Suárez-Álvarez | The theorem that subgroups of free groups are free can be proved purely algebraically (well, I consider combinatorial group theory as done by manupulating words to be algebra...) without even mentioning groupoids, and I think that the notion does not quite impose itself in the argument. Potatos potatos? | |
| Oct 4, 2010 at 9:24 | history | answered | Greg Graviton | CC BY-SA 2.5 |