Timeline for "Burnside ring" of the natural numbers and algebraic K-theory
Current License: CC BY-SA 4.0
23 events
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| Apr 8, 2022 at 15:30 | history | became hot network question | |||
| Apr 8, 2022 at 13:36 | answer | added | Benjamin Steinberg | timeline score: 6 | |
| Apr 8, 2022 at 11:53 | comment | added | John Klein | @YCor, yes, that was a mere oversight. | |
| Apr 8, 2022 at 11:49 | history | edited | John Klein | CC BY-SA 4.0 |
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| Apr 8, 2022 at 9:47 | comment | added | YCor | After your edit: you probably need finiteness of the $G$-sets (or assume they're finitely generated $G$-sets i.e. existence of a finite subset $F$ such that $X=GF$). Otherwise taking any $X$, and defining $Y$ as the disjoint union of infinitely many copies of $X$, one gets $X=0$ in the "Grothendieck group". For instance, the Grothendieck group of the monoid of countable $G$-sets with disjoint union is zero. | |
| Apr 8, 2022 at 9:45 | history | edited | YCor | CC BY-SA 4.0 |
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| Apr 8, 2022 at 9:38 | answer | added | Maxime Ramzi | timeline score: 8 | |
| Apr 8, 2022 at 3:01 | comment | added | Benjamin Steinberg | I suspect with your definition N is different than Z | |
| Apr 8, 2022 at 1:58 | comment | added | John Klein | @A.S. sure! [email protected] | |
| Apr 8, 2022 at 1:09 | comment | added | Benjamin Steinberg | Im not sure if the paper I linked is using exactly the same definition. It sounds like they extend the category and then do some homotopy. My guess is with your definition the case of natural numbers is more complex | |
| Apr 7, 2022 at 23:43 | comment | added | user164898 | It's not hard to work out generators and relations for $A(\mathbb{Z})$. You can use a "shearing isomorphism" to express how to decompose a Cartesian product of cyclic finite $\mathbb{Z}$-sets as a disjoint union of cyclic finite $\mathbb{Z}$-sets, and this does almost all the work. I suspect the calculation has been done independently many times over. I'll email you a written account if you think it would be helpful. | |
| Apr 7, 2022 at 22:57 | history | edited | John Klein | CC BY-SA 4.0 |
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| Apr 7, 2022 at 22:51 | history | edited | John Klein | CC BY-SA 4.0 |
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| Apr 7, 2022 at 22:50 | comment | added | John Klein | @BenjaminSteinberg The Dress-Siebeneicher paper apparently gives the statement that $A(\Bbb Z)$ can be viewed as being isomorphic to a dense subring of the ring of universal Witt vectors. | |
| Apr 7, 2022 at 22:43 | comment | added | John Klein | @BenjaminSteinberg the paper you linked to shows that $A(\Bbb N)$ is isomorphic to $A(\Bbb Z)$ (provided that these authors are using the same definition as the one here. The latter has been studied. See for example math.uni-bielefeld.de/~sieben/lothar/… | |
| Apr 7, 2022 at 22:38 | comment | added | Benjamin Steinberg | This was the definition I mentioned in my first comment for monoids. | |
| Apr 7, 2022 at 22:38 | history | edited | John Klein | CC BY-SA 4.0 |
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| Apr 7, 2022 at 22:36 | comment | added | John Klein | @BenjaminSteinberg thanks. I guess I will change the definition and use the one mentioned by Ycor | |
| Apr 7, 2022 at 22:18 | comment | added | YCor | You can make a definition without using transitive $G$-sets. Namely, consider the set $W$ of all isomorphism class of finite $G$-sets. Then take the abelian group generated by elements of $W$ with relators corresponding to $X+Y=Z$ if $Z$ is the disjoint union of $X$ and $Y$. This also inherits a ring structure induced by cartesian product. | |
| Apr 7, 2022 at 21:41 | comment | added | Benjamin Steinberg | You do have that each M-set deconposes as a disjoint union of indecomposable ones but these are not transitive and even a finite monoid can have even infinitely many indecomposable actions. | |
| Apr 7, 2022 at 21:20 | comment | added | Benjamin Steinberg | link.springer.com/article/10.1007/s10485-016-9477-4 may be relevant but there are other papers too | |
| Apr 7, 2022 at 21:02 | comment | added | Benjamin Steinberg | Your definition of a transitive monoid is nonstandard. Cyclic is more usual. It is not true that all actions decompose into a disjoint union of cyclic ones. And I don't think the direct product of two cyclic ones does. People have looked at the semiring with all finite M-sets and disjoint union and direct product | |
| Apr 7, 2022 at 20:40 | history | asked | John Klein | CC BY-SA 4.0 |