I was wondering how much is known about the integral motivic cohomology groups of $\mathrm{Spec}\, k$, $H^{n,p}_{\mathrm{mot}}(\mathrm{Spec}\, k,\mathbb{Z})$. One knows that $H^{n,n}_{\mathrm{mot}}(\mathrm{Spec}\, k,\mathbb{Z})\simeq K^M_n(k)$, the Milnor $K$-theory of the field $k$. Also, because of the isomorphism to higher Chow groups $H^{n,p}_{\mathrm{mot}}(X,\mathbb{Z})\simeq CH^p(X,2p-n)$ when $k$ is perfect, we get $H^{2p,p}_{\mathrm{mot}}(\mathrm{Spec}\, k,\mathbb{Z})\simeq CH^p(\mathrm{Spec}\, k)=0$ if $p\neq 0$ and $\mathbb{Z}$ otherwise. This also implies $H^{n,p}_{\mathrm{mot}}(\mathrm{Spec}\, k,\mathbb{Z})$ vanishes for $n>p$.

For which $(n,p)$ is $H^{n,p}_{\mathrm{mot}}(\mathrm{Spec}\, k,\mathbb{Z})$ known to vanish? In other words, is there any $n\neq p$ for which the group is non-trivial?

  • 8
    $\begingroup$ Not much is known, the vanishing result mentioned in the question is essentially all (unless you ask for specific fields). For finite fields, a lot more groups are known to vanish. By Borel's computations we know that $H^1(\operatorname{Spec} k,\mathbb{Z}(n))$ is nontrivial for number fields with a complex place. $\endgroup$ – Matthias Wendt Mar 15 '16 at 14:25
  • 3
    $\begingroup$ If the higher Chow groups were to vanish for all $n\neq p$, then the spectral sequence from motivic cohomology to $K$-theory would degenerate immediately, in which case Quillen $K$-theory and Milnor $K$-theory would be identical, which they are frequently not (for example for finite fields). So whenever $K^M\neq K^Q$, there must be other non-vanishing motivic cohomology groups. $\endgroup$ – Steven Landsburg Jul 11 '16 at 7:11
  • 1
    $\begingroup$ We have $H^{n-d,n}(Spec(k), \mathbb{Z}/p) = K_{n-d}^M(k)/p$ for $d \ge 0$, which also implies (via the Bockstein sequence) that there must be many non-vanishing integral groups. $\endgroup$ – Tom Bachmann Jul 11 '16 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.