It's well known that for Noetherian separated regular schemes the canonical map $$K(X) \longrightarrow G(X)$$ (Quillen uses $K'$ instead of $G$, though) is a weak equivalence.

This statement is usually called Poincaré duality.

One can also define $K$-theory with compact support (for sufficiently nice schemes $X$) by choosing a compactification $X \hookrightarrow \overline{X}$ and setting $K_c(X)$ as the homotopy kernel of $K (X) \rightarrow K (\overline{X}\setminus X)$. I have no idea on whether such $K_c$ and $K$ enjoy some kind of Poincaré duality.

When I hear something like Poincaré duality I expect some kind of cap product map with some fundamental virtual class $$H^{\bullet} \longrightarrow H_{d -\bullet}^{BM}$$ or, dually, $$H_c^{\bullet} \longrightarrow H_{d - \bullet}$$. Of course, there's a cap product $$K(X) \wedge G(X) \longrightarrow G(X)$$ induced by tensor product which when restricted to tensoring with $\mathscr{O}_X$ gives the Poincaré duality.

However, I'm not satisfied with such analogy. I, hence, ask the following.

1) Is there any sense in which $G$ is a $K$-theory with compact support? Or maybe it's even the opposite: $K$ is a $G$-theory with compact support?

2) If yes, is there any relation between $K_c (X)$ and $G(X)$?

3) If no, is there any kind of duality between $K (X)$ and $K_c (X)$?

4) If I'm actually sounding silly since in ordinary Poincaré duality both sides of the isomorphism are always simultaneously of the same kind (compact or not compact, for instance, $H_{\bullet}^{BM}$ is somehow non compact as $H^{\bullet}$), how can I see the duality as some isomorphism from a cohomology to a homology? In other words, why $K(X)$ should be a cohomology theory and $G(X)$ a homology theory?

5) If one uses some Atiyah-Hirzebruch spectral sequence for $G$-theory, would it be the case that the graded pieces of the $\gamma$ filtration define a motivic cohomology with compact support up to torsion?

6) What about 5 for $K_c$ instead of $G$? What about $G_c$?

7) After applying the Atiyah-Hirzebruch sequence to all the possibilities ($K$, $K_c$, $G$, $G_c$) what sort of Poincaré duality one acquires?

Thanks in advance.


I've added new questions in order to correct my lack of attention to concordance of the "kind" (compact or noncompact) of the domain and codomain in the duality.


Given the comments below by Marc Hoyois and Gasterbiter, $K_c (X)$ should be defined as the homotopy colimit over $r$ of $K (\overline{X}, r (\overline{X}\setminus X))$, where the prefix $r$ denotes the infinitesimal thickening of order $r$ (following the notation of https://arxiv.org/abs/1211.1813).

Also, as noted below, $G$ should behave as a Borel-Moore homology. The analogy, therefore, is that

$$K(X) \longrightarrow G(X)$$ is the analogous of the first duality expressed above (cohomology-BM homology), whereas $$K_c(X) \longrightarrow G_c(X)$$ should correspond to the second duality (the compact version), where $G_c (X) := G (\overline{X}, \overline{X}\setminus X)$ (Btw, how do I state these dualities using the six functor formalism instead of using this "underline $c$"?).

Therefore, only the last questions remain. I will restate them here.

1) If one applies he Atiyah-Hirzebruch spectral sequence to $K$, $K_c$, $G$ and $G_c$, then what will be the graded pieces of the $\gamma$-filtration up to torsion (take $X$ as general as possible)? Or even better, in the level of spectra, what kind of decomposition one acquires?

For instance, in the case of smooth $X$, $K(X) \wedge \mathbb{S}_{\mathbb{Q}} \cong \bigvee_i H \mathbb{Q} \wedge (\mathbb{P}^1)^{\wedge i}$ (I have no idea what happens when $X$ is not smooth, though).

2) Does one recover some kind of Poincaré duality from the graded pieces mentioned in 1?

  • 5
    $\begingroup$ G-theory is like Borel-Moore homology. The simplest reason is its functoriality: it is covariant for proper maps and contravariant for etale maps, like BM homology. Your $K_c$ is not well-defined (now that we know K-theory satisfies pro-cdh descent, we can define $K_c(X)$, but this involves taking the limit over all nilpotent thickenings of $\bar X-X$). $\endgroup$ – Marc Hoyois Nov 9 '18 at 15:34
  • 1
    $\begingroup$ @MarcHoyois Thanks for the comment! It's somehow surprising my $K_c$ is ill defined. What exactly will fail? If I recall correctly, Gillet seems to define the $K$-theory of a pair in the analogous way when proving higher GRR, but maybe I'm overseeing something... By the way, do you know some reference where such $K_c (X)$ that you mentioned is studied? $\endgroup$ – user40276 Nov 9 '18 at 21:04
  • 2
    $\begingroup$ @user40276 check out section 4.1 in arxiv.org/pdf/1211.1813.pdf $\endgroup$ – user123627 Nov 10 '18 at 6:25
  • 1
    $\begingroup$ @Gasterbiter Thanks for the comment. That definition still weird to me, though. I guess its the just the best that one can do when something fails to be smooth (be it $\overline{X}$ or $\overline{X}\setminus X$). In any case, I suppose that by dévissage the naive definition is safe enough for $G$. $\endgroup$ – user40276 Nov 13 '18 at 18:53
  • $\begingroup$ How do you define $G_c$? $G$-theory does not have pullbacks along closed immersions. $\endgroup$ – Marc Hoyois Nov 16 '18 at 3:00

To my knowledge, one can only make this analogy fully consistent with Weibel's homotopy invariant $K$-theory $KH$ and $G$-theory (although the proofs of what I claim below rely heavily on our understanding of classical algebraic $K$-theory). Then, using the canonical map $K(X)\to KH(X)$, the pairing relating $KH$ and $G$ induce the pairing relating $K$ and $G$ which fits in the folkloric description of Poincaré duality relating $K$ and $G$.

Indeed, classical Poincaré duality is a particular instance of Grothendieck-Verdier duality (i.e. Grothendieck duality for ordinary sheaves of abelian groups). Indeed, if $D(X)$ denotes the derived category of sheaves on a (nice locally compact) space $X$, then we have, for any continuous map $f:X\to Y$ a pullback functor $f^*:D(Y)\to D(X)$ which has a right adjoint $f_*: D(X)\to D(Y)$, and there is push-forward functor $f_!:D(X)\to D(Y)$, the right derived functor of the direct image with compact support functor, which has a right adjoint $f^!: D(Y)\to D(X)$. There is natural map $f_!\to f_*$ which is invertible for $f$ proper, and so on.

Now, if $X$ is a space and $a:X\to \{pt\}$ denotes the canonical maps to the point,

  • the cohomology of $X$ with coefficients in $\mathbf Z$ is $a_* a^*(\mathbf Z)$
  • the cohomology with compact support is $a_! a^*(\mathbf Z)$
  • Borel-Moore homology is $a_* a^!(\mathbf Z)$
  • Homology is $a_! a^!(\mathbf Z)$

For nice enough spaces (e.g. algebraic varieties), these are perfect complexes of abelian groups (whence dualizable objects in the derived category of abelian groups), and taking the dual in the derived category exchanges $*$ and $!$. In particular, the dual of homology $a_! a^!(\mathbf Z)$ is cohomology $a_* a^*(\mathbf Z)$, while the dual of homology with compact support $a_! a^*(\mathbf Z)$ is Borel-Moore homology $a_* a^!(\mathbf Z)$. Poincaré duality consists in identifying, when $X$ is smooth complex orientable of dimension $d$, $a_! a^* (\mathbf Z)$ and $a_! a^!(\mathbf Z)(-d)[-2d]$ (where $A(-n)=A\otimes H^2(\mathbf P^1(\mathbf C),\mathbf Z)^{\otimes n}$).

Using Morel-Voevodsky's motivic stable homotopy category $SH$, we can extend this to schemes: to simplify, I will restrict to schemes of finite type over a field $k$. Then, given a a commutative motivic ring spectrum $E$ in $SH(k)$, we may define $D(X)$ as the category of $E$-modules in $SH(X)$ (to be precise, of $a^*(E)$-modules in the $(\infty,1)$-category $SH(X)$, where $a:X\to\mathrm{Spec}\, k$ denotes the structural map). And we have most of the features above, replacing $\mathbf Z$ by $E$ (e.g. we have cohomology $a_* a^*(E)$ and so forth). In the case where $E=KGL$ is the object which represents Weibel's $KH$ in $SH$, this gives a context in which Grothendieck's six operations apply.

Here, Poincaré duality identifies $a_! a^* (KGL)$ and $a_! a^!(KGL)(-d)[-2d]$ (for $X$ smooth of dimension $d$). Dually, it corresponds to

$$a_* a^!(KGL)\simeq a_* a^*(KGL)(d)[2d]$$

but Bott-periodicity also says that $KGL\simeq KGL(d)[2d]$.

Furthermore, for possibly singular $X$, one can check that the global sections of $a_* a^!(KGL)$ really give back $G$-theory:

$$\Gamma(\mathrm{Spec} \,k,a_* a^!(KGL))=G(X)$$ (this is essentially a reformulation of $K$-theoretic Poincaré duality as formulated in the question above, of Quillen's localization theorem for $G$-theory and of homotopy invariance for $G$-theory). And we have:

$$\Gamma(\mathrm{Spec} \,k,a_* a^*(KGL))=KH(X)$$

We could define homotopy invariant $K$-theory with compact support: $$KH_c(X):=\Gamma(\mathrm{Spec} \,k,a_! a^*(KGL))$$ and $KH$-homology as $\Gamma(\mathrm{Spec} \,k,a_! a^!(KGL))$.

As for the coniveau spectral sequence (a.k.a the motivic Atiyah-Hirzebruch spectral sequence), applied to $G$-theory, Marc Levine has showed that the $E_2$-term will be motivic cohomology defined through Bloch's cycle complexes (also for $X$ singular): this is why the cohomology defined through Bloch cycle complex should not be called "motivic cohomology" but rather "motivic Borel-Moore homology".

Rationally, $KGL$ is naturally a $H\mathbf Q$-algebra, where $H\mathbf Q$ denotes the $\mathbf Q$-linear motivic Eilenberg-MacLane spectrum. In fact, $KGL$ becomes the free Bott-periodic $H\mathbf Q$-algebra. Furthermore, the category of $H\mathbf Q$-modules in $SH(X)$ is then equivalent to $DM(X,\mathbf Q)$ (the category of motivic sheaves over $X$), and the change of scalars functor from $H\mathbf Q$-modules to $KGL$-modules commutes with the six operations (at least if we restric to compact objects). In particular, one recovers Poincaré duality in $KGL$-modules from the one in motives, but with a "Todd-twist": the classical formulations of Poincaré duality as above involve Thom isomorphisms, which themselves rely on a choice of an orientation. Poincaré duality on homotopy $K$-theory as described above corresponds to the orientation of $KGL$ defined by the multiplicative formal group law, while the one coming from seeing rationalized $KGL$ as a $H\mathbf Q$-algebra corresponds to the additive formal group law. Relating the two through an explicit isomorphism is exactly the purpose of Grothendieck-Riemann-Roch theorems.

  • $\begingroup$ Thanks for the wonderful answer! That clarified a lot of things. I, however, feel tempted to ask for more. 1) At which point everything breakdown in the non-homotopy invariant version (i.e., working inside Nisnevich sheaves with transfer without the $\mathbf{A}^1$-localisation)? 2) How general can one take $a : X \rightarrow S$ in order to carry the above constructions? 3) What would be the analogous decomposition $KH (X)_{\mathbf{Q}} \cong \bigvee_{i \in \mathbf{Z}} H\mathbf {Q} \wedge (\mathbf{P}^1)^{\wedge i}$ for $KH$-homology, $G$-theory and $KH_c$-theory?... $\endgroup$ – user40276 Nov 21 '18 at 2:58
  • $\begingroup$ ... (I suppose that one just has to apply the usual decomposition of $KGL$ by the eingenspaces of the Adams operations, but I'm unsure about the exact result)? $\endgroup$ – user40276 Nov 21 '18 at 2:58
  • 1
    $\begingroup$ @user40276 Since G-theory is homotopy invariant, any cohomology theory whose associated Borel-Moore homology would be G-theory must be homotopy invariant for regular schemes. Furthermore, the formalism of Grothendieck six operations implies descent by blow ups. Therefore, assuming resolution of singularities, we see that there is no hope to get a full picture as above for non-homotopy invariant K-theory. $\endgroup$ – Denis-Charles Cisinski Nov 22 '18 at 20:51
  • 1
    $\begingroup$ One can extend this picture replacing Spec(k) by any regular scheme S. One can even work over General noetherian schemes, but this would be at the price of having several different flavors, either in the direction of absolute purity, or replacing G-theory by more relative versions of K-theory. Your last question 3) is answered by the last paragraph of my answer! $\endgroup$ – Denis-Charles Cisinski Nov 22 '18 at 21:04

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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