Timeline for Are Milnor K-groups algebraic groups?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 16, 2020 at 15:29 | comment | added | Eric Peterson | It’s just a guess, and not a very careful one. Even if the quotient group isn’t representable, it could be that the numerator and denominator in the quotient are individually representable—in which case that pair of representing objects gives a Hopf algebroid, and the connected components of the groupoid give the cosets of the quotient. The numerator is definitely representable, and the denominator looks plausibly representable: I think that the set of generators is indeed representable, and then you just need the free group they generate to be available… | |
| S Nov 16, 2020 at 9:05 | history | edited | Ben McKay | CC BY-SA 4.0 |
Corrected grammar and spelling, improved formatting
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| S Nov 16, 2020 at 9:05 | history | suggested | gmvh | CC BY-SA 4.0 |
Corrected grammar and spelling, improved formatting
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| Nov 16, 2020 at 8:40 | review | Suggested edits | |||
| S Nov 16, 2020 at 9:05 | |||||
| Nov 16, 2020 at 3:02 | comment | added | M masa | I got it. Milnor K-group of a field cannot be representable as an algebraic group. I am interested in your suggestion about Hopf algebroid. Why you think so ? | |
| Nov 15, 2020 at 23:42 | comment | added | Eric Peterson | A second thought: Wikipedia says that $K^M_2(\mathbb C)$ is an uncountable uniquely divisible group (so: torsion-free), while $K^M_2(\mathbb R)$ is the sum of such a group and $C_2$. The map induced by inclusion of fields can’t be an inclusion on K-groups, and that prevents K-groups from being representable. | |
| Nov 15, 2020 at 16:31 | comment | added | Eric Peterson | My guess is no, but there may a Hopf algebroid (i.e., a representable functor into groupoids rather than groups) whose $\pi_0$ recovers $K^M_n$. | |
| Nov 15, 2020 at 15:08 | comment | added | LSpice | I think it's just too big to be of finite type for $n > 1$, and finite type is part of the business for me, but I guess you aren't requiring that? (I'm not very familiar with the Hopf-algebra perspective and whether it encodes the same information I'm used to.) | |
| Nov 15, 2020 at 14:57 | history | asked | M masa | CC BY-SA 4.0 |