Timeline for Why do we use the diagonal for diagonal approximations ?
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13 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2012 at 13:11 | comment | added | Ralph | @Mariano: I think there is an associative product on the cohm. of Thompson's group $F$ that is induced by a non-diagonal hom. $F \times F \to F$: math.cornell.edu/~kbrown/papers/homology.pdf (2.2). But I don't know if it has a unit. | |
| Dec 19, 2012 at 20:34 | comment | added | Mariano Suárez-Álvarez | @Theo, as you say, the diagonal is the unique comultiplication that turns the group into a counital coassociative coalgebra in the appropriate category —but that is sort of irrelevant here, where what one wants is to turn the cohomology of the group into an associative algebra, and one could possibly have the latter without the former! I honestly do not know if, given a group, the diagonal is the unique map from which one can get an associative product in cohomology, but I am pretty sure that it is the only natural way to do it (natural in the sense of category theory, of course) | |
| Dec 17, 2012 at 0:29 | vote | accept | tj_ | ||
| Dec 16, 2012 at 5:10 | comment | added | Theo Johnson-Freyd | @Mariano: The diagonal map is unique in the following sense. In any cartesian category (i.e. category with finite products, in which the symmetric monoidal structure is chosen to be categorical product), on any object the diagonal map is the unique comultiplication making that object into a counital coassociative coalgebra. So the answer to @TJ's question is that the other maps do not give a unital cup product, as can be seen explicitly from @John's answer, in which "cup product with $\beta$" is sometimes the zero map. | |
| Dec 16, 2012 at 2:52 | comment | added | Mariano Suárez-Álvarez | It is interesting to note that Cartan and Eilenberg, in the chapter on products of their book, consider cup products induced by arbitrary "diagonal" maps. They have to assume coassociativity to get associativity of the product and so on, of course (there is more leeway, anyways, as one needs only a diagonal map on resolutions of $\mathbb Z$, and those can be taken non-coassociative and worse :-) ) | |
| Dec 16, 2012 at 2:29 | comment | added | Mariano Suárez-Álvarez | Naturality of the cup product is «a big thing», and that comes straight from the naturality of the diagonal map. I'll bet that there are not a lot of other natural maps into the product! | |
| Dec 16, 2012 at 1:57 | comment | added | Dylan Wilson | @TJ: Sorry, meant to say it quotients out the reduced cohomology of one of the factors. This is what is now explained below in John's answer. Also, both John and Hiro give much better answers to (1). | |
| Dec 16, 2012 at 1:03 | comment | added | Hiro Lee Tanaka | Nice observations, TJ. (1) The diagonal map is a gift. It makes any set (any space) into a coalgebra. So any symm. monoidal contravariant functor from sets (spaces) gives you an algebra for every set (space). This is why, morally, cohomology of a space has the structure of an algebra. So when you have a functor like "cochains on a group," it's natural to examine the diagonal, motivated (for me) by this standard fact from topology, to yield an algebraic structure like the cup product. | |
| Dec 16, 2012 at 0:58 | answer | added | John Palmieri | timeline score: 2 | |
| Dec 16, 2012 at 0:19 | comment | added | tj_ | I don't understand what it means that the map I'll get quotients out one of the factors in the tensor product ? | |
| Dec 15, 2012 at 23:09 | comment | added | Dylan Wilson | "...use others than..." should be "use others then" | |
| Dec 15, 2012 at 23:09 | comment | added | Dylan Wilson | Answer to 1: Because it always exists. If you use the others than... Answer to 2: You just get the map $H^*G \otimes H^*G \rightarrow H^*G$ that quotients out one of the factors in the tensor product. | |
| Dec 15, 2012 at 22:54 | history | asked | tj_ | CC BY-SA 3.0 |