6
$\begingroup$

Take the familiar Universal Coefficient Theorem for ordinary homology with $\mathbb{Z}$-coefficients and ordinary cohomology with coefficients in some abelian group $A$:$$0\rightarrow \text{Ext}_\mathbb{Z}^1(H_{i-1}(X;\mathbb{Z}),A)\rightarrow H^i(X;A)\rightarrow \text{Hom}_\mathbb{Z}(H_i(X;\mathbb{Z}),A)\rightarrow 0$$This exact sequence can be rewritten in the language of topological spectra:$$0\rightarrow\text{ext}_{H\mathbb{Z}}^1([\Sigma^{i-1}S,H\mathbb{Z}\wedge X],HA)\rightarrow [X,\Sigma^iHA]\rightarrow [[\Sigma^iS,H\mathbb{Z}\wedge X],HA]\rightarrow 0$$where $[-,-]$ denotes the internal hom $\text{hom}_{H\mathbb{Z}}(-,-)$ of $H\mathbb{Z}$-module spectra. (I am confused about some basic points [see my question], but I presume that $\text{ext}$ makes sense in this context.)

How can one generalize this sequence to generalized (co)homology theories? One guess is to replace $H\mathbb{Z}$ with some commutative ring spectrum $E$ and $HA$ with an $E$-module spectrum $F$. I imagine this guess is far too ambitious and a more conservative theorem is the correct generalization.

$\endgroup$
2
  • $\begingroup$ See for example Adams' Lectures on Generalized Cohomology, in Volume 1 of his collected works. $\endgroup$
    – Peter May
    Sep 5, 2015 at 21:11
  • $\begingroup$ @AlexTurzillo combine the classical UCSS with the Atiyah-Hirzebruch spectral sequence. $\endgroup$ Sep 5, 2015 at 21:33

1 Answer 1

11
$\begingroup$

There is more than one possible generalization. The most common is the universal coefficient spectral sequence. Given a (homotopy) commutative ring spectrum $E$ and a spectrum $X$, there is under certain conditions a spectral sequence $$ Ext^{p,q}_{E_*}(E_*(X), E_*) \Rightarrow E^{q-p}(X) $$ This is true for example if

  • $E$ is an $A_\infty$-ring spectrum (e.g. if $E =ko, ku, KO, KU, TMF, MO, MSO, MSpin \dots$) [EKMM, IV.4]
  • $E$ is even and Landweber exact (e.g. if $E = MU, E(n), E_n, \dots$) [Adams' lectures on generalized cohomology, which Peter alluded to, and Devinatz: Morava Modules and Brown-Comenetz Duality, Prop 1.3, and the discussion thereafter]

Sometimes, this spectral sequence is not extremely useful though. For example, take $E = KO$. In general, $KO_*X$ might have infinite cohomological dimension over $KO_*$ so that the spectral sequence might be difficult to control. In this case another perspective is more useful: Anderson duality.

Consider the functor $E \mapsto Hom(\pi_*E, \mathbb{Q}/\mathbb{Z})$ from the homotopy category of spectra to graded abelian groups. As $\mathbb{Q}/\mathbb{Z}$ is injective, this is a cohomological functor, by Brown representability represented by a spectrum $I_{\mathbb{Q}/\mathbb{Z}}$. There is an evident map $H\mathbb{Q} \to I_{\mathbb{Q}/\mathbb{Z}}$ whose fiber we denote by $I$. For a spectrum $E$, we define its Anderson dual $IE$. to be the function spectrum $F(E, I)$. It is easy to show that we get for every spectrum $X$ a short exact sequence $$ 0 \to Ext^1_{\mathbb{Z}}(E_{k-1}X, \mathbb{Z}) \to (IE)^kX \to Hom_{\mathbb{Z}}(E_kX, \mathbb{Z}) \to 0.$$ That means that there exists always a short exact sequence computing from $E$-homology the $IE$-cohomology. This is, of course, only useful if we can identify $IE$. Luckily, this has been done for a few spectra:

  • $IH\mathbb{Z} \simeq H\mathbb{Z}$
  • $IKU \simeq KU$
  • $IKO \simeq \Sigma^4 KO$ [see e.g. Heard, Stojanoska]
  • $ITmf \simeq \Sigma^{21}Tmf$ (at least at primes $>2$) [see Stojanoska]

For example, for $KO$ this means that we get a universal coefficient sequence $$ 0 \to Ext^1_{\mathbb{Z}}(KO_{k-1}X, \mathbb{Z}) \to KO^{k+4}X \to Hom_{\mathbb{Z}}(KO_kX, \mathbb{Z}) \to 0.$$

$\endgroup$
3

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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