Let $k$ be a field, $K$ a finite extension of $k$, and $K_{n}^{M}(K)$ the $n$-th Milnor K-group of $K$, that is, $$ K_{n}^{M}(K)=K^{\times}\otimes_{\mathbb{Z}}\cdots\otimes_{\mathbb{Z}} K^{\times}/I, $$ where $I$ is the subgroup generated by $\{a_{1}\otimes\cdots\otimes a_{n}\mathrel{\vert}a_{i}+a_{j}=1 \text{ for some }i\neq j \}$.

Question: Is there a Hopf algebra $A$ over $k$ such that ${\rm Hom}_{k}(A,K)=K_{n}^{M}(K)$? In other words, are Milnor K-groups affine algebraic groups?

Of course when $n=1$, we can take $A=k[t^{\pm}]$. So I am interested in the case of $n\geq 2$. Do you have any positive or negative answers about this?

  • $\begingroup$ 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.) $\endgroup$
    – LSpice
    Nov 15, 2020 at 15:08
  • 2
    $\begingroup$ My guess is no, but there may a Hopf algebroid (i.e., a representable functor into groupoids rather than groups) whose $\pi_0$ recovers $K^M_n$. $\endgroup$ Nov 15, 2020 at 16:31
  • 4
    $\begingroup$ 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. $\endgroup$ Nov 15, 2020 at 23:42
  • $\begingroup$ 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 ? $\endgroup$
    – M masa
    Nov 16, 2020 at 3:02
  • $\begingroup$ 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… $\endgroup$ Nov 16, 2020 at 15:29


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