# Is every additive cohomology operation stable?

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.

• @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. – Tim Campion Mar 24 at 16:02
• 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 – Maxime Ramzi Mar 24 at 16:04
• Aren't the Adams operations an example of additive cohomology operations that are not stable? – Connor Malin Mar 24 at 17:18
• 1'. That's true. Again this follows from the description of the cohomology ring. This also holds over integers coefficients. – Bad English Mar 24 at 19:11
• 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. – Bad English Mar 24 at 19:27

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.

• 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... – Tim Campion Apr 12 at 0:30
• 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. – Tyler Lawson Apr 12 at 3:36