Timeline for Why is there a duality between spaces and commutative algebras?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 20, 2015 at 23:13 | comment | added | Tim Campion | Here's why a category of commutative algebras is co-distributive: Suppose $(C,\otimes)$ is a symmetric monoidal category with biproducts $\oplus$ such that $\otimes$ distributes over $\oplus$ (guaranteed by closedness or by $\otimes$ respecting the unique enrichment of $C$ in commutative monoids). Then in the category $\mathrm{Comm}(C)$ of commutative monoid objects in $C$, $\otimes$ becomes the binary coproduct while $\oplus$ becomes the binary product, so binary coproduct distributes over binary product. Co-extensivity seems a bit more complicated | |
| Sep 20, 2015 at 15:17 | vote | accept | Yonatan Harpaz | ||
| Sep 11, 2015 at 19:41 | comment | added | Qiaochu Yuan | It seems I should also mention the notion of an extensive category (ncatlab.org/nlab/show/extensive+category) somewhere in this discussion; this is a more conceptual condition satisfied by categories of spaces which in particular implies distributivity, but e.g. unlike the condition of being cartesian closed, it holds for $\text{Top}$. In any case, one might ask whether / why opposites of categories of commutative algebra-like objects tend to be extensive. | |
| Sep 11, 2015 at 16:03 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Sep 11, 2015 at 10:31 | comment | added | Yonatan Harpaz | The case of pro-$p$-finite spaces is a bit illusive. It looks like a right Kan extension case, but for this to hold the "base" category must include all the $p$-finite Eilenberg-MacLane spaces. In order for the opposite of this category to map into a ring of commutative-algebra objects one needs to realize various Steenrod powers. Curiously enough, Steenrod powers can actually be defined intrinsically for every $\mathbb{E}_\infty$-algebra over $\mathbb{F}_p$ using cohomology of the symmetric group. | |
| Sep 11, 2015 at 10:00 | comment | added | Yonatan Harpaz | Right Kan extension is what's going on in the example of Boolean algebras. Left Kan extension (in the $\infty$-categorical sense), is a bit what's going on in the example of rational spaces. I'm still not completely sure how to accommodate for the case of $\mathrm{C}^*$-algebras and the case of pro-$p$-finite spaces. | |
| Sep 11, 2015 at 9:57 | comment | added | Yonatan Harpaz | Thanks! I really like this answer. One can even expend a bit on point 4 as follows. As said, in geometric categories products tend to distribute over coproducts. On the other hand, in commutative-algebraic categories, coproducts tend to distribute over products. Since $\mathrm{FinSet}$ is free in that sense, we get that $\mathrm{FinSet}$ should map to any reasonable geometric category $C$, and $\mathrm{FinSet}^{op}$ should map to any reasonable commutative-algebraic category $D$. One may then try to extend these maps into a map $C\to D^{op}$, either by left or right Kan extension. -> | |
| Sep 10, 2015 at 23:05 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |