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Intuitively, cohomology operations are ways to locally compute a cocycle $\alpha\in H^i(X, G)$ from any cocycle $\beta\in H^j(X, H)$. Formally, they are in one-to-one correspondence with homotopy classes of maps $K(G, i)\rightarrow K(H, j)$, which in turn correspond to elements $\theta \in H^j(K(G,i), H)$. Roughly speaking, $\alpha$ can be computed locally from $\beta$ by pulling back a fixed representative of $\theta$ along the map $X\rightarrow K(G, i)$ representing $\alpha$.

The cup product is a way to locally compute a cycle $\alpha\in H^{i+j}(X, G)$ from any two cycles $\beta\in H^i(X, G)$ and $\gamma\in H^j(X, G)$, and could therefore be interpreted as a "2-variable cohomology operation", which leads me to the following questions:

  • Is there a general theory of multi-variable cohomology operations? I've come across "secondary cohomology operations", but that seems to be different from what I'm looking for.
  • Are there $n$-variable cohomology operations besides combinations of the cup product and 1-valued (ordinary) cohomology operations?
  • Elements of $H^i(X,G)\times H^j(X,H)$ are represented by maps $X\rightarrow K(G,i)\times K(H,j)$. So 2-variable cohomology operations $H^i(X,G)\times H^j(X,H)\rightarrow H^k(X, J)$ should be in one-to-one correspondence with elements in $H^k(K(G,i)\times K(H,j), J)$. Applying Kuenneth's formula to the latter seems to be a way towards arguing that the answer to the previous point is "no", but how does this work exactly?
  • Characteristic classes (of the tangent bundle) are ways to compute a cycle just from the local presentation of the manifold. Does it make sense to interpret characteristic classes as 0-variable cohomology operations?
  • Is the cup product with a characteristic class a cohomology operation? It certainly is a way to locally compute a cocycle from another cocycle. It is not a homomorphism of cohomology groups, but that's not a must for a cohomology operation according to some sources. Even though, I'm a bit confused about the last point: As a natural transformation between cohomology functors, it would have to be a group homomorphism, right? Or can one simply consider the cohomology functors to Set instead of AbelianGroup?
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  • $\begingroup$ Yes, guess that's better, renamed now. $\endgroup$
    – Andi Bauer
    Commented Jan 27, 2022 at 14:46
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    $\begingroup$ This is going to depend on the abelian groups of coefficients: the Künneth formula becomes much more complicated when the coefficients are not fields (and in fact is replaced by a spectral sequence...) To answer one of the questions, cohomology operation in general are not required to respect the abelian group structure (an easy example is the cohomology operation $x\mapsto x^n$ for some integer $n$), those that do are called additive cohomology operations. On the other hand, they are always required to be natural in the space. $\endgroup$
    – Denis Nardin
    Commented Jan 27, 2022 at 15:05
  • $\begingroup$ Thinking more about it, I'd now guess that the cup product with a characteristic class is an element of $H^j(BO(n)\times K(G, i), H)$ rather than a cohomology operation, and cohomology operations are not all locally computable functions on cocycles (e.g., on a simplicial complex), but only the ones which do not depend on the tangent bundle (no idea how to formulate this independence for simplicial complexes though). $\endgroup$
    – Andi Bauer
    Commented Jan 27, 2022 at 16:08
  • $\begingroup$ Characteristic classes are not cohomology operations (well, they are in the category of spaces equipped with a vector bundle but that's not what we're talking about). There is no notion of characteristic class for a general topological space or simplicial complex so I'm not sure I understand your intuition here. $\endgroup$
    – Denis Nardin
    Commented Jan 27, 2022 at 16:19
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    $\begingroup$ I might be mistaken, but all that I can see in that paper is that there are formulas that recover the characteristic classes for combinatorial manifolds (or, as they claim, for homology manifolds). However it's not clear to me that those formulas would produce something well-defined for an arbitrary simplicial complex. Indeed I know that can't be true since the only 0-ary cohomology operations are the constants. $\endgroup$
    – Denis Nardin
    Commented Jan 27, 2022 at 16:55

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