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I've received conflicting messages on this point -- on the one hand, I've been told that "forming a natural home for algebraic $K$-theory" was one motivation for the development of motivic homotopy theory. On the other hand, I've been warned about the fact that algebraic $K$-theory isn't always $\mathbb A^1$-local. By "algebraic $K$-theory,", I mean the algebraic $K$-theory of perfect complexes of quasicoherent sheaves (I think -- let me know if I should mean something else).

  • I'm pretty sure that Thomason and Trobaugh show that algebraic $K$-theory always (in the quasicompact, quasiseparated case) satisfies Nisnevich descent.

  • Under certain conditions, algebraic $K$-theory is $\mathbb A^1$-local and has some kind of compatibility with $\mathbb G_m$ which should make it $\mathbb P^1$-local. I think Weibel (already Quillen) calls this "the fundamental theorem of algebraic $K$-theory".

So putting this together, let $S$ be a scheme, and let $SH(S)$ be the stable motivic ($\infty$-)category over $S$.

Questions:

  1. Is algebraic $K$-theory of smooth schemes over $S$ representable as an object of $SH(S)$?

  2. How about if we put some conditions on $S$ -- say it's regular, noetherian, affine, smooth over an algebraically closed field? Heck, what if we specialize to $S = Spec(\mathbb C)$?

  3. Does it make a difference if we redefine $SH(S)$ to be certain sheaves of spectra over the site of smooth schemes affine over $S$ or something like that?

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    $\begingroup$ I would also like to know the answer to this. $\endgroup$ – Yonatan Harpaz Jun 19 '18 at 18:46
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    $\begingroup$ You wrote that Weibel calls it "the fundamental theorem of algebraic K-theory," but that theorem already had that name in Quillen's "Higher algebraic K-theory I." The theorem establishes that the K-theory of a regular ring R agrees with the K-thy of R[t]; in other words, algebraic K-thy is A^1-invariant on regular rings. You need that A^1-invariance to represent K as a motivic spectrum. Nil K-thy measures the failure of K-thy to be A^1-invariant, and nil K-thy can be nonvanishing on nonregular rings; you can think of nil K-grps. as an obstruction to representability of K by a motivic spectrum. $\endgroup$ – user124192 Jun 19 '18 at 21:02
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    $\begingroup$ That stuff is pretty well-documented in the literature. For the definition of nil K-thy (NK) and relationship to A^1-invariance, Grayson "Higher algebraic K-theory II." For an example of nonvanishing NK-grps, try F_p[x]/x^p; see Hesselholt "The tower of K-theory of truncated polynomial algebras". For general structure of NK (it's a Dieudonne module), Grayson's "K-theory of endomorphisms." For whether NK forms a motivic spectrum, the trouble is its own A^1-invariance: see the material on iterated nil K-thy and htpy K-thy in Weibel's K-book in sections IV.11 and IV.12. $\endgroup$ – user124192 Jun 19 '18 at 22:42
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    $\begingroup$ A commutative ring A is called K-regular if K(A[t_1, ... ,t_n]) -> K(A) is an equivalence for all n. Regular rings are K-regular, but there are also nonregular K-regular rings; I see this was something that was asked about in a question you asked in January, linked in the sidebar to this question. See J. Rosenberg's "Algebraic K-theory of operator algebras" for the theorem that every commutative C^*-algebra is K-regular. $\endgroup$ – user124192 Jun 19 '18 at 23:06
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    $\begingroup$ It is not known if algebraic K-theory satisfies Nisnevich descent "always", only on quasi-compact quasi-separated schemes. It was recently proved that K-theory takes infinite sums to infinite products. $\endgroup$ – Marc Hoyois Jun 20 '18 at 5:15
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Let me assume that $S$ is a regular Noetherian scheme (for example a field). Then algebraic K-theory is a motivic spectrum, and in fact it is represented by the $\mathbb{P}^1$-spectrum that is $BGL_\infty\times\mathbb{Z}$ in each level (so it is a $\mathbb{P}^1$-periodic motivic spectrum).

This is theorem 4.3.13 in

Morel, Fabien; Voevodsky, Vladimir, $\bf A^1$-homotopy theory of schemes, Publ. Math., Inst. Hautes Étud. Sci. 90, 45-143 (1999). ZBL0983.14007.

A correct proof of the above result can be found in this survey (thanks to Marc Hoyois for pointing out that the original proof was incorrect).

The proof of the fact that algebraic K-theory is a motivic spectrum goes exactly how you described:

  • It satisfies Nisnevich descent on qcqs schemes by Thomason-Trobaugh's main theorem.
  • It is $\mathbb{A}^1$-invariant on regular Noetherian schemes (which every scheme smooth over $S$ is ) by Quillen's fundamental theorem of K-theory
  • It is $\mathbb{P}^1$-periodic by the projective bundle formula (also in Thomason-Trobaugh)

When $S$ is not regular noetherian the situation is more complicated. You can still represent $K$ by some version of the infinite Grassmannian (proposition 4.3.14 in Morel-Voevodsky), but this object won't be $\mathbb{A}^1$-invariant anymore. What you can consider is the homotopy K-theory presheaf $KH=Sing_* K$. This object is indeed represented by $BGL_\infty\times \mathbb{Z}$. This result has been announced in Voevodsky's ICM address and proven in

Cisinski, Denis-Charles, Descent by blow-ups for homotopy invariant $K$-theory, Ann. Math. (2) 177, No. 2, 425-448 (2013). ZBL1264.19003.

(thanks to Marc Hoyois for the reference to this result)

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  • $\begingroup$ Thanks Denis, this is great! So of the conditions I was willing to assume, regular noetherianness is needed, but affineness and being defined over a field are not. $\endgroup$ – Tim Campion Jun 19 '18 at 21:51
  • $\begingroup$ Whoa thank you but maybe you want to wait a few hours before accepting? Someone might come around that knows what happens in the non regular case. $\endgroup$ – Denis Nardin Jun 19 '18 at 21:53
  • $\begingroup$ I guess I was mostly interested in knowing that generically it is representable, but maybe you're right. $\endgroup$ – Tim Campion Jun 19 '18 at 22:00
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    $\begingroup$ For a general qcqs scheme S, the "same" motivic spectrum represents Weibel's KH. This was announced by Voevodsky in his ICM address and proved by Cisinski: arxiv.org/abs/1003.1487 $\endgroup$ – Marc Hoyois Jun 20 '18 at 5:11
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    $\begingroup$ It's worth pointing out that the proof that $BGL\times\mathbb Z$ represents algebraic K-theory (in the regular case) in Morel-Voevodsky is wrong. The correct proof, which uses $\mathbb A^1$-localization in a crucial way, is given in Morel's book "Théorie homotopique des schémas", or see Section 6 in arxiv.org/abs/1605.00929. $\endgroup$ – Marc Hoyois Jun 20 '18 at 5:23

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