Skip to main content
5 of 5
Removed residues of the first draft; exanded explanation for the AHSS
Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

I will work stably: everything in sight will be a spectrum.


It is well known and classical that cohomology operations correspond to map of spectra: that is if $E,F$ are two spectra, a natural transformation $E^*X→F^{*+n}X$ correspond by Yoneda to a map of spectra $E→\Sigma^nF$.

On the other hand if you have a map of spectra $E→\Sigma^nF$, you also get a homology operation $E_*X→F_{*-n}X$. This is because homology can be written as $$ E_*X = \pi_*(E\wedge X),\qquad F_{*-n}X=\pi_*(\Sigma^nF\wedge X)\,.$$ So understanding cohomology operations will give you also a supply of homology operations. This is not a dualization as you said, it is just a way to get a homology operation out of a map of spectra: no duality required. Moreover we can prove that all homology operations are of this form.

In fact any such operation is determined by its restriction to finite spectra (since every spectrum is a filtered homotopy colimit of finite spectra), and using the canonical equivalence $\mathrm{Sp}_{fin}\cong\mathrm{Sp}_{fin}^{op}$, such a natural transformation is the same thing as a natural transformation $$E^{-*}X=E_*DX→F_{*+n}DX=F^{-*-n}X$$ when $X$ varies among finite spectra and $D$ is the Spanier-Whitehead dual. But we know that these operations are induced by maps $E→\Sigma^nF$.

Basically, when you restrict to finite spectra homology and cohomology are pretty much the same thing: you just need a Spanier-Whitehead dual to pass from one to the other.

We can say more about the second differential in the AHSS: if we look at the construction of the AHSS that uses the Postnikov tower for the spectrum $E$ instead than the one using the cellular filtration for $X$, we immediately see that the differential $d_2$ is the map of spectra given by the k-invariants of $E$. That is, we can look at the AHSS as the spectral sequence associated to the exact couple

$$\require{AMScd} \begin{CD} \dots @>>> \pi_*(P_{n+1}E\wedge X) @>>> \pi_*(P_nE\wedge X) @>>> \pi_*(P_{n-1}E\wedge X) @>>> \cdots\\ {} @VVV {} @VVV {} @VVV {}\\ \dots @. \pi_*(H\pi_{n+1}E \wedge X)=H_*(X;\pi_{n+1}E) @. \pi_*(H\pi_nE\wedge X)=H_*(X;\pi_nE) @. \pi_*(H\pi_{n-1}E\wedge X)=H_*(X;\pi_{n-1}E) @.\cdots \end{CD}$$

That is the $d_2:H_*(X;\pi_nE)\to H_{*-1}(X;\pi_{n+1}E)$ is induced by the map of spectra $$H\pi_nE\to \Sigma P_{n+1}E\to \Sigma H\pi_{n+1}E\,.$$ So we can leverage our understanding of the cohomology operations to get at the $d_2$ of the AHSS.


Now let us tackle homology cooperations. These are the homotopy classes of $E\wedge E$. I really do not have too much to say about them: they show up because they are important for the Adams spectral sequence (in fact they basically run the show there). They are also the "dual" of the cohomology operations. If $E$ is an $E_\infty$-ring spectrum we can use the universal coefficient spectral sequence to go $$\mathrm{Ext}_{E_*}(\pi_*(E\wedge E),E_*) \Rightarrow \pi_*F_E(E\wedge E,E)=\pi_*F(E,E)$$ where $F_E$ and $F$ are the internal hom in $E$-modules and spectra respectively. For example when $E=Hk$ for $k$ a field (the main case people treat I think) this spectral sequence degenerates telling you that the $k$-linear dual of $\pi_*(Hk\wedge Hk)$ is the ring of cohomology operations. As you can see these have no direct relation to homology operations and they arise in a different setting altogether.


Finally, the Hurewicz homomorphism. This is, by definition, the map on homology theories defined by the map $h:\mathbb{S}\to H\mathbb{Z}$ sending 1 to 1. In fact, if $f:\Sigma^n\mathbb{S}\to X$ is a class in $\pi_nX$ we are sending it to the image of the fundamental homology class of $\Sigma^n\mathbb{S}$. But the fundamental homology class is just the map $$\Sigma^nh=(h\wedge 1):\Sigma^n\mathbb{S}\to H\mathbb{Z}\wedge\Sigma^n\mathbb{S}\,.$$ So we are sending $f$ to $(1_{H\mathbb{Z}}\wedge f)\circ (h \wedge 1) = (h\wedge f)$, that is we are doing the homology operation represented by $h$.

Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103