Let $f: R\to S$ be a map between two commutative Noetherian rings. Let $G_0(R)=K_0(mod R)$ be the Grothendieck group of finite generated modules over $R$. It means $G_0(R)$ is the quotient of the free abelian group on all isomorphism class of finitely generated modules over $R$ by the subgroup generated by relations coming from short exact sequences.

If $fd_RS<\infty$, one can define a map $f^*:G_0(R)\to G_0(S)$ by: $$f^*([M]) = \sum_{i\geq0} (-1)^i [Tor_R^i(M,S)] $$

(for reference, see Section 7, Chapter 2 of Weibel's book on K-theory.

Now, if $R$ is not regular or $S$ is not a complete intersection in $R$, then having finite flat dimension is somewhat a miraculous condition. So my question is: Can a map $f^*$ be defined in a more general situation than for finite flat dimension maps?

EDIT: Let me elaborate a little bit because of some interesting answers and comments below (especially Clark's answer). The main motivation I have in mind is the case of $R$ being a hypersurface. Then most $R$- modules have infinite resolutions, but it is well known that their resolution is eventually periodic. So, even though they are not homologically finite, the modules can be homologically described with finite data (i.e. finite number of matrices).

The fact above is crucial in many results I know about hypersurfaces. For a random recent example, see here. In particular, in this situation one can define( at least when $S$ is finite $R$-module):

$$f^*([M])= [Tor^{2n}(M,S)] - [Tor^{2n-1}(M,S)]$$

for sufficiently big $n$.

So that's one of the reasons I wonder if there is more systematic map one can define.

  • $\begingroup$ I would also like to know the answer to this question. $\endgroup$ – David E Speyer Dec 18 '09 at 21:44
  • $\begingroup$ Great! I have not had much success with technical questions on MO so far, so may be you will change that (: $\endgroup$ – Hailong Dao Dec 18 '09 at 22:03
  • 1
    $\begingroup$ I'm probably stating the obvious... but I'd say the whole point is that we have two K-groups associated with a ring (one formed by looking at finitely generated modules, and one by looking at finitely generated projective modules), which coincide for regular rings. One of them is covariant functor (with respect to all homomorphisms), one of them is contravariant (with respect to finite homomorphisms). This is very similar to other situations (e.g.: homology pushes forward, cohomology pulls back; functions restrict, distributions have trace...) so the set-up seems natural to me. $\endgroup$ – t3suji Dec 19 '09 at 0:23
  • $\begingroup$ The link you provide saying a recent example is broken, could you please provide a new link ? Thanks $\endgroup$ – sde Nov 3 '19 at 4:09
  • $\begingroup$ @sde: by examining the link, it seems I meant this one: arxiv.org/abs/0912.1629 $\endgroup$ – Hailong Dao Nov 10 '19 at 23:09

Probably not for $G$-theory, unfortunately. That $G_0$ is a contravariant for morphisms of schemes that are globally of finite Tor-dimension is SGA VI, Exp. IV, 2.12. The corresponding statement for the $G$-theory spectra is in Thomason-Trobaugh, 3.14.1.

In effect, given a morphism of schemes $f:X\to Y$, one wishes to show that the induced functor $f^{\star}$ on categories of modules gives rise to a functor between the ∞-categories of cohomologically bounded psuedocoherent complexes of $\mathcal{O}$-modules. Preservation of pseudocoherence is automatic, but the statement that $f^{\star}E$ is cohomologially bounded when $E$ is so is equivalent to the assertion that $f$ is globally of finite Tor-dimension [SGA VI, Exp. III, Pr. 3.3].

This doesn't quite show that there is no way to produce a pullback map on $G$-theory for more general kinds of morphisms, but it does make it clear that it cannot be induced by the functoriality of the ∞-categories of complexes of modules.

  • $\begingroup$ Interesting, thanks! Let me try to digest what you wrote a little bit. $\endgroup$ – Hailong Dao Jan 18 '10 at 1:03
  • $\begingroup$ So what you say is requiring the functor to preserve homologically bounded complexes would force finite flat dimensions. That makes a lot of sense, but is there a deep reasons why we want to only work with cohomologically bounded complexes (other than we don't know what to do otherwise)? $\endgroup$ – Hailong Dao Jan 18 '10 at 1:10
  • $\begingroup$ Exactly. There is a good reason to see that you have to do something like work with cohomologically bounded complexes: it's to avoid the Eilenberg Swindle. If you're allowed to have complexes with no bound on the cohomological dimension (on one side or the other), then you get zero $G$-theory. You could try another condition that would keep the Swindle at bay, however: for instance, you could try putting a different complete t-structure on the derived category of pseudocoherent complexes, and you could demand boundedness with respect to that t-structure. Could be interesting. $\endgroup$ – Clark Barwick Jan 18 '10 at 1:31
  • 1
    $\begingroup$ I meant my edited question above. $\endgroup$ – Hailong Dao Jan 18 '10 at 1:56
  • 2
    $\begingroup$ Oh, interesting. The standard reference on t-structures is MR0751966 (86g:32015) Beĭlinson, A. A.; Bernstein, J.; Deligne, P. Faisceaux pervers. (French) [Perverse sheaves] Analysis and topology on singular spaces, I (Luminy, 1981), 5--171, Astérisque, 100, Soc. Math. France, Paris, 1982. One can also have a look at Jacob Lurie's "Stable Infinity Categories (DAG I)" on his website. $\endgroup$ – Clark Barwick Jan 18 '10 at 15:16

My initial thought was to work with something like the Grothendieck group or modules, but instead use a version of the Grothendieck group of the derived category (my first search on Mathscinet turned up a paper from 1969 in Bucur), where your relations come from the triangles instead of short exact sequences. Then the map could just be derived tensor with S. I am a bit rusty with the derived category so I am not sure if this would respect the relations, however.

  • $\begingroup$ Preserving the relations is the tricky part (I assume you are thinking about the Grothendieck group of the bounded complexes). $\endgroup$ – Hailong Dao Dec 19 '09 at 4:42
  • $\begingroup$ Well, I was thinking about the ones that were bounded only on the left; what I wrote I guess would not even be defined unless one is in this situation. $\endgroup$ – Frank Moore Dec 19 '09 at 16:13
  • 1
    $\begingroup$ As far as I understand, if you take Grothendieck group of the derived category of complexes bounded on one side only, you get zero. $\endgroup$ – t3suji Dec 19 '09 at 22:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.