To start, let's work with mod $p$ cohomology $H\mathbb F_p$ where $p$ is a prime. Consider the following three things:

  1. The bigraded abelian group of all unstable cohomology operations, comprising all natural transformations of set-valued functors $H^m(-;\mathbb F_p) \Rightarrow H^n(-;\mathbb F_p)$, i.e. $$Unst(H\mathbb F_p) = (\pi_0 Map(K(\mathbb F_p,m), K(\mathbb F_p,n)))_{m,n \in \mathbb N}$$

  2. The bigraded abelian group of additive cohomology operations, i.e. the subgroup $Unst^{add}(H\mathbb F_p) \subset Unst(H\mathbb F_p)$ comprising natural transformations of abelian group-valued functors $H^m(-;\mathbb F_p) \Rightarrow H^n(-;\mathbb F_p)$.

  3. The bigraded abelian group of all stable cohomology operations $St(H\mathbb F_p) = (\pi_0 Map (\Sigma^m H \mathbb F_p, \Sigma^n H\mathbb F_p))_{m,n \in \mathbb N}$. There is a natural map $St(H\mathbb F_p) \to Unst^{add}$. Of course we have $St(H\mathbb F_p)_{m,n} = \mathcal A^{n-m}$ where $\mathcal A^\ast = \pi_{-\ast} Map(H\mathbb F_p,H \mathbb F_p)$ is the Steenrod algebra. So there is also a natural map $St(H\mathbb F_p)_{0,k} \to \prod_{n-m = k} Unst^{add}(H\mathbb F_p)_{m,n}$.

Now, the standard calculation of $\mathcal A^\ast$ proceeds by calculating $Unst(H\mathbb F_p)$. Looking at the results, I believe the answer to the following questions are affirmative:

Question 1: Is the natural map $St(H\mathbb F_p) \to Unst^{add}(H\mathbb F_p)$ a surjection?

Question 1': Is the natural map $St(H\mathbb F_p)_k \to \prod_{n-m = k} Unst^{add}(H\mathbb F_p)_{m,n}$ an injection?

For instance, when $n$ is not a power of $p$ the $n$th-power operation $(-)^n: H^m(-;\mathbb F_p) \Rightarrow H^{nm}(-;\mathbb F_p)$ is not a stable operation, but this is already explained by the fact that it is not an additive operation.

Assuming I have it right and the answers to Question 1 and 1' are "yes", I have some follow-up questions:

Question 2: Is there a "good reason" for the affirmative answers to Questions 1,1'?

Question 3: Do these facts generalize to an arbitrary spectrum $E$ in place of $H\mathbb F_p$?

I expect the answer to Question 3 is "no" in this generality, but we can ask:

Question 4: Do these facts generalize to an arbitrary sum of Eilenberg-MacLane spectra $E$ in place of $H\mathbb F_p$?

As a small bit of evidence that this might be a general phenomenon, note that if a cohomology operation $\phi: K(A,m) \to K(B,n)$ is additive, then by the Yoneda lemma it corresponds to a map of abelian group objects in the homotopy category. This looks like at least the first step to showing that $\phi$ is a map of infinite loop spaces, and thus lifts to a map of spectra. It would be really convenient if every map between topological abelian groups which is a map of abelian group objects in the homotopy category could be rectified to a map of topological abelian groups, but this is false: it would imply that every additive cohomology operation between sums of Eilenberg-MacLane spectra would be a map of $H\mathbb Z$-modules. Counterexamples are given e.g. by every Steenrod power operation except for $Sq^1$ which coincides with the Bockstein.

  • $\begingroup$ @MaximeRamzi Er... of course you're right, there's something wrong with the way I've set things up. I think your proposal is probably the right fix -- there is a natural map $St \to Unst^{add}$ and the question is whether it's a surjection. $\endgroup$
    – Tim Campion
    Mar 24, 2021 at 16:02
  • $\begingroup$ Sorry I deleted my comment because I got confused for a second.But yeah, the thing you denote $St$ isn't "really" stable cohomology operations, there's a lot of things in the kernel $\endgroup$ Mar 24, 2021 at 16:04
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    $\begingroup$ Aren't the Adams operations an example of additive cohomology operations that are not stable? $\endgroup$ Mar 24, 2021 at 17:18
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    $\begingroup$ 1'. That's true. Again this follows from the description of the cohomology ring. This also holds over integers coefficients. $\endgroup$ Mar 24, 2021 at 19:11
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    $\begingroup$ I'm stupid and have to correct my comment on Q1. Contrary, the answer seems to be "yes" also. I suppose that primitive elements of free commutative Hopf-algebra over F_p are given exactly by p-th powers of (primitive) generators. It is easy to see dualizing everything and consider this thing as divided powers-algebra over F_p. Hence they are actually obtained by applying stable operations to fundamental class. $\endgroup$ Mar 24, 2021 at 19:27

1 Answer 1


I think I've worked out an approach to this which I find enlightening, though technically a bit fiddly.

Claim 1: Let $A$ be an $E_\infty$ space and let $B = K(G,d)$ be an Eilenberg-MacLane space where $G$ is abelian and $d \geq 1$. Then any map of $H$-spaces $f: A \to B$ lifts to an $E_\infty$ map $f: A \to B$.

Corollary 2: Let $A$ be a spectrum and let $B$ be an Eilenberg-MacLane spectrum. Then any additive cohomology operation $A^\ast \Rightarrow B^\ast$ is stable.

Proof: By shifting and taking connective covers if necessary, we may represent $A$ by an $E_\infty$-space and $B$ by an Eilenberg-MacLane space. Then the additivity of the operation means that it is represented by a map of $H$-spaces. By Claim 1, this lifts to a map of $E_\infty$-spaces, i.e. a map of spectra, and so is stable.

(Key) Construction 3: If $B = K(G,d)$ is an Eilenberg-MacLane space, we may model $B$ via an $n+1$-coskeletal simplicial set which has a unique $m$-cell for each $m < n$, a unique $n$-cell for each $g \in G$, a unique $n+1$-cell for each relation in $G$. This simplicial set is a Kan complex. Thus our map $f: A \to B$ may be represented by a map of simplicial sets into this particular model of $B$.

Lemma 4: Model $B$ as in Construction 3. Then any pointed maps into $B$ related by a pointed homotopy are equal.

Proof: Let $\phi,\psi: X \to B$ be maps related by a pointed homotopy $H$. Necessarily $\phi$ and $\psi$ coincide on the $n-1$-skeleton of $X$. By the pointedness of $H$, its components at any cell of dimension $\leq n-1$ are trivial. Therefore the component of $H$ at any cell $x \in X_n$ exhibits $\phi(x),\psi(x)$ as homotopy rel their boundaries. By construction of $B$, this implies that $\phi(x) = \psi(x)$. Then $B$ is suitably coskeletal so that we must in fact have $\phi = \psi$.

Proof of Claim 1: Let $f: A \to B$ be a map of $H$-spaces, represented where $B$ is the simplicial set from Construction 3. Then there is a pointed homotopy between $\mu \circ f^{\times n}$ and $f \circ \mu$ (where $\mu$ ambiguously denotes the multiplication on $A$ or $B$). By Lemma 4, we have $\mu \circ f^{\times n} = f \mu$. Thus the constant homotopy is a $\Sigma_n$-equivariant homotopy between these two maps. Passing to $\Sigma_n$-homotopy fixed points, we obtain a diagram as follows, where the bottom square and the outer rectangle commute up to pointed homotopy:

$$ \require{AMScd} \begin{CD} E\Sigma_n \times_{\Sigma_n} A^n @>E\Sigma_n \times_{\Sigma_n} f^n>> E\Sigma_n \times_{\Sigma_n} B^n\\ @VVV @VVV\\ A @>f>> B \\ @VVV @VVV\\ B\Sigma_n \times A @>id \times f>> B\Sigma_n \times B \end{CD} $$

The map $B \to B\Sigma_n \times B$ is a split monomorphism, so the top square commutes up to pointed homotopy as well. Invoking Lemma 4 again, the top square commutes strictly. Since this is true for every $n$, we have that $f$ is a map of $E_\infty$-spaces as desired.

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    $\begingroup$ Er -- the same proof would seem to show that $f$ is a map of simplicial abelian groups, assuming that $A$ is a simplicial abelian group, which is false. Probably there is something up with the claim that that diagram commutes... $\endgroup$
    – Tim Campion
    Apr 12, 2021 at 0:30
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    $\begingroup$ I think the issue is with homotopies. Eg the standard simplicial model for BG has the properties that you describe. But a homotopy is allowed to be nontrivial on a 1-simplex $p \times \Delta^1$ so long as p is not the basepoint. $\endgroup$ Apr 12, 2021 at 3:36

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