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This question is about the state of current knowledge regarding Voevodsky's algebraic cobordism of a point $\mathrm{MGL}^{*,*}(\mathrm{Spec}\,k)$. That the geometric diagonal $\mathrm{MGL}^{2*,*}(\mathrm{Spec}\,k)$ is isomorphic to the Lazard ring $\mathbb{L}$ if $char(k)=0$ has been shown by Levine using the Hopkins-Morel isomorphism. For positive characteristic fields, Hoyois showed the same isomorphism holds after inverting the characteristic exponential. It is also well-known that $\mathrm{MGL}^{n,n}(\mathrm{Spec}\,k)$ is identified with the Milnor $K$-theory of a field.

Spitzweck, in Algebraic Cobordism in mixed characteristic, section 7, computed some of the homotopy groups $\pi_{p,q}\mathrm{MGL}_S$ of the algebraic cobordism spectrum in the stable homotopy category $SH(S)$ over $S$, $S$ being the spectrum of a Dedekind domain of mixed characteristic. For instance, he described the cases $(p,q)=(2n+1,n)$, $(n+1,n)$. My question is:

  1. Are similar descriptions already known in the case $S=\mathrm{Spec}\,k$?
  2. For $S=\mathrm{Spec}\,k$, is it known that $\mathrm{MGL}^{2n+i,n}(\mathrm{Spec}\,k)$ vanishes for $i\geq 1$.
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  • $\begingroup$ Are you only interested in the characteristic zero case? Some of the results you recalled are only known in that case. $\endgroup$ – Marc Hoyois Mar 22 '16 at 18:26
  • $\begingroup$ Even though characteristic zero is what I am mainly interested in, I am curious about positive characteristics too. These questions can be asked after inverting the characteristic exponential, as in your paper referred above. I am sorry I should have mentioned this in the question. $\endgroup$ – user86186 Mar 23 '16 at 4:26
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The answer to both question is yes, provided you invert $p$ in characteristic $p$ (though conjecturally this is not necessary). In fact, as far as I can see, all of Spitzweck's computations apply to fields as well. He only assumes mixed characteristic because some of the arguments are more difficult in that case (e.g. Corollary 5.9). So for example $\mathrm{MGL}^{2n-1,n}(\operatorname{Spec} k) = k^\times \otimes L_{1-n}$ and $\mathrm{MGL}^{n-1,n}(\operatorname{Spec} k)$ is an extension of $H^{n-1,n}(\operatorname{Spec} k)$ by a quotient of $K_{n+1}^M(k)$.

For the vanishing see also this question: The vanishing of $MGL^{2n+i,n}(X)$; do spectra of smooth projective varieties generate $SH_{l}$?

It's worth noting that the vanishing of $\mathrm{MGL}^{p,q}(\operatorname{Spec}k)$ for $p>q$ and the computation $\mathrm{MGL}^{n,n}(\operatorname{Spec}k)=K_n^M(k)$ hold without inverting the characteristic (they come from Morel's results about the sphere spectrum).

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  • $\begingroup$ Thank you for the answer. I noticed that some of my questions are already answered in the thread you mentioned. I am, however, a bit confused regarding the vanishing of $\mathrm{MGL}^{p,q}(\mathrm{Spec}\,k)$ for $p>q$ as $\mathrm{MGL}^{2n,n}(\mathrm{Spec}\,k)$ or $\mathrm{MGL}^{2n-1,n}(\mathrm{Spec}\,k)$ are not zero. $\endgroup$ – user86186 Mar 23 '16 at 13:41
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    $\begingroup$ They are zero if $n$ is positive! $\endgroup$ – Marc Hoyois Mar 23 '16 at 13:43
  • $\begingroup$ Are the results on $\mathrm{MGL}^{2n-1,n}(\mathrm{Spec}\,k)$ and $\mathrm{MGL}^{n-1,n}(\mathrm{Spec}\,k)$ that you mention written down somewhere? Or they are deduced from Spitzweck's results? $\endgroup$ – user86186 Mar 23 '16 at 13:45
  • $\begingroup$ Ah! ok! I was assuming in my mind that $\mathrm{MGL}^{p,q}(\mathrm{Spec}\,k)$ vanishes for $q<0$ like motivic cohomology. Thanks for clarifying. $\endgroup$ – user86186 Mar 23 '16 at 13:47
  • $\begingroup$ I have only quoted 7.6 and 7.7 in Spitzweck. $\endgroup$ – Marc Hoyois Mar 23 '16 at 13:48

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