The Steenrod algebra $\mathcal{A}^*$ is the algebra of cohomology operations for the spectrum $\mathrm{H}\mathbb{F}_2$, i.e. $\mathcal{A}^*\simeq (\mathrm{H}\mathbb{F}_2)^*(\mathrm{H}\mathbb{F}_2)$. The dual Steenrod algebra $\mathcal{A}^\vee _*$ can be interpreted as the coalgebra of homology cooperations, i.e. $\mathcal{A}^\vee _\ast\simeq (\mathrm{H}\mathbb{F}_2) _*(\mathrm{H}\mathbb{F}_2)$ and it coacts on mod 2 homology groups of cell complexes. This is a consequence of the perfect pairing between mod 2 homology and cohomology. Unfortunately, this does happens because $\mathbb{F}_2$ is a field, which means that we may in trouble if change our coefficients to something else. The situation may get even worse with generalized cohomology theories.

Now let us say we are given a generalized cohomology theory $E$. The ring $E^\ast E$ acts on cohomology groups of any space $X$: we have map $E^*E\otimes_{E^\ast} E^\ast(X)\rightarrow E^\ast(X)$, which is natural in $X$. The coaction may be a little tricky: it is not true that $E_\ast E$ coacts on $E_*(X)$. However, upon requiring that $E_\ast E$ is a flat $E_*$-module, we get a coaction as well. So let us say that we are in this situation. In addition to this we have a pairing $E^*E\otimes_{E_*}E_\ast E\rightarrow E_\ast$.

My questions is: do these pairings work well together or there are some tricky things that has to account for? In particular, $E_\ast E$ has distinct left and right $E_*$-module structures, and I fear this may come into play.

  • 3
    $\begingroup$ the distinct left and right module structures definitely come in to play. They show up whenever you want to computations, look at a cobar resolution that would give you the ANSS. You can also think about it from a categorical viewpoint. The Hopf Algebroid you have is a cogroupoid object, one unit represents the source of a morphism, the other represents the target of a morphism. I believe Ravenel talks about this in one of his appendices to the "Green Book". $\endgroup$ Aug 9, 2012 at 0:45
  • $\begingroup$ Yeah, I started suspecting something like this, when I was looking at $\mathrm{K}$ rationally. One can understand $K_\ast K\otimes \mathbb{Q}$ easily, so I wanted to understand $(K\wedge \mathbb{Q})^\ast(K\wedge \mathbb{Q})$ by just dualizing. This even ended up having elements that behaved much like the Adams operations. However, I could not recover the action of these operations on coefficient ring. I assume I made a mistake not taking into account these difference between left and right module structures. $\endgroup$
    – nerses
    Aug 9, 2012 at 13:21

1 Answer 1


Propositions 17.10 and 17.11 of Switzer's book seem to be examples of things working nicely.

For those who don't have the book to hand: let $E$ be a commutative ring spectrum such that $E_\ast(E)$ is flat as a right module over $E_\ast=E_\ast(*)$, and let $X$ be any spectrum.

Let $\psi_X\colon E_\ast(X)\to E_\ast(E)\otimes_{E_\ast} E_\ast(X)$ be the coaction map, and let $c\colon E_\ast(E)\to E_\ast(E)$ be the conjugation map (which switches the left and right counits).

Then for any elements $a\in E^\ast(E)$, $y \in E^\ast(X)$ and $u\in E_\ast (X)$ with $\psi_X(u) = \sum_i e_i\otimes u_i$, we have $$ a\cdot u = \sum_i \langle a,ce_i\rangle u_i $$ and $$ \langle ay,u\rangle = \sum_i (-1)^{|y||e_i|}\langle a,e_i\langle y,u_i\rangle \rangle, $$ where $\langle \cdot , \cdot \rangle\colon E^\ast(X)\otimes_{E_\ast} E_\ast(X)\to E_\ast$ is the Kronecker pairing.

  • $\begingroup$ Thanks, Mark, this was pretty useful. The important thing for me was the fact that $\langle y, u_i\rangle$ acts on the right of $e_i$ in the second equation. It seems to give me the right results. $\endgroup$
    – nerses
    Aug 11, 2012 at 0:13

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