What do cohomology operations have to do with the non-existence of commutative cochains over $\mathbb{Z}$?

Question

Let $X$ be a topological space. In elementary algebraic topology, the cup product $\phi \cup \psi$ of cochains $\phi \in H^p(X), \psi \in H^q(X)$ is defined on a chain $\sigma \in C_{p+q}(X)$ by $(\phi \circ \psi)(\sigma) = \phi(_p\sigma)\psi(\sigma_q)$ where $p_\sigma$ and $\sigma_q$ denote the restriction of $\sigma$ to the front $p$-face and the back $q$-face, respectively. (More generally, any diagonal approximation $C_{\ast}(X) \to C_{\ast}(X) \otimes C_{\ast}(X)$ could be used; this is the Alexander-Whitney one.) The cup product defined by the Alexander-Whitney diagonal approximation as above is associative for cochains but skew-commutative only up to homotopy (this results from the fact that the two diagonal approximations $C_{\ast}(X) \to C_{\ast}(X) \otimes C_{\ast}(X)$ given by Alexander-Whitney and its "flip" (with the signs introduced to make it a chain map) agree only up to chain homotopy.

The commutative cochain problem attempts to fix this: that is, to find a graded-commutative differential graded algebra $C_1^*(X)$ associated functorially to $X$ (which may be restricted to be a simplicial complex) which is chain-equivalent to the usual (noncommutative) algebra $C^{\ast}(X)$ of singular cochains.

In Rational homotopy theory and differential forms, Griffiths and Morgan mention briefly that there is no way to make the cup-product skew-commutative on cochains (that is, to solve the commutative cochain problem) with $\mathbb{Z}$-coefficients, and that this is because of the existence of cohomology operations. It is also asserted that these cohomology operations don't exist over $\mathbb{Q}$ (presumably excluding the power operations). Could someone explain what this means?

Answer 1

Via the Dold-Kan correspondence, the category of cosimplicial abelian groups is equivalent to the category of nonpositively graded chain complexes of abelian groups (using homological grading conventions). Both of these categories are symmetric monoidal: chain complexes via the usual tensor product of chain complexes, and cosimplicial abelian groups via the "pointwise" tensor product. But the Dold-Kan equivalence is not a symmetric monoidal functor.

However, you can make it lax monoidal in either direction. The Alexander-Whitney construction makes the functor (cosimplicial abelian groups -> cochain complexes) into a lax monoidal functor, so that for every cosimplicial ring, the associated chain complex has the structure of a differential graded algebra. However, it is not lax symmetric monoidal, so the differential graded algebra you obtain is generally not commutative even if you started with something commutative.

There is another construction (the "shuffle product") which makes the inverse functor (cochain complexes -> cosimplicial abelian groups) into a lax symmetric monoidal functor. In particular, it carries commutative algebras to commutative algebras. So every commutative differential graded algebra (concentrated in nonpositive degrees) determines a cosimplicial commutative ring.

One way of phrasing the phenomenon you are asking about is as follows: not every cosimplicial commutative ring arises in this way, even up to homotopy equivalence. For example, if $A$ is a cosimplicial ${\mathbb F}_2$-algebra, then the cohomology groups of $A$ come equipped with some additional structures (Steenrod operations). Most of these operations automatically vanish in the case where $A$ is obtained from a commutative differential graded algebra.

If $R$ is a commutative ring and $X$ is a topological space, you can obtain a cosimplicial commutative ring by associating to each degree $n$ the ring of $R$-valued cochains on $X$ (the ring structure is given by ``pointwise'' multiplication). These examples generally don't arise from commutative differential graded algebras unless $R$ is of characteristic zero. For example when $R = {\mathbb F}_2$, the $R$-cohomology of $X$ is acted on by Steenrod operations, and this action is generally nontrivial (and useful to know about).

Answer 2

One such example comes from Steenrod squares. Let $C^{*}(X)$ be the singular cochain algebra of a topological space $X$ with coefficients mod 2. The cup product $\smile$ is not commutative, but there are operations $\smile_i:C^{*}(X)\otimes C^{*}(X)\to C^{*}(X)$ of degree $-i$ such that $\smile_0=\smile$ and $\smile_i$ is $(-1)^i$ commutative up to homotopy $\smile_{i+1}$. For instance, $$a\smile b-(-1)^{|a||b|}b\smile a=d(a\smile_1 b)+da\smile_1 b+a\smile_1 db.$$

If a cohomology class $[a]\in H^n(X,\mathbf{Z}/2)$ is represented by a cycle $a$, then one way to define $Sq^i([a])$ is to set $Sq^i([a])=[a\smile_{n-i} a]$. More details (and eventually the correct signs) can be found somewhere in the Topology part of Markl, Schneider, Stasheff, Operads ... (which I don't have at hand).

Answer 3

One further answer to this is in Mandell's paper "Cochain Multiplications". We can view your hypothetical commutative cochains $C_1^*(X)$ as a functor landing in $E_\infty$ algebras over $\mathbb Z$. In that case, if the functor satisfies a reasonable list of axioms then it follows that it is naturally equivalent to the usual cochain functor. It follows from that that it has non-trivial Steenrod operations on $\mathbb Z/p$ cohomology, and that contradicts commutativity, as the Steenrod operations on a strictly commutative algebra are almost all zero.

Answer 4

A nice example is the following. Let $G=\mathbb{Z}/p$ and look at the DGA $C^*$ of singular cochains on $BG$ with coefficients in $\mathbb{F}_p$. Let us assume $p>2$.

Its cohomology is an exterior algebra in a generator $x$ in degree $1$ tensored with a polynomial algebra of a generator $y$ in degree $2$.

It cannot be formal, since $y$ is a Massey power of $x$. Thus it also cannot be quasiisomorphic to a graded commutative DGA. If there was such a DGA $D^*$, then we can define a quasiisomorphism $H^*(D^*)\rightarrow D^*$ by sending the cohomology classes $x$ and $y$ to any representing cocycles. Thus $C^*$ would be formal.