On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$? - MathOverflow

On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$?

2011-02-18T23:35:43Z

I was discussing with a student today the nature of the non-commutativity of cup-product at the level of cochains. In trying to explain what happens, I came up with the following statement.

Theorem. Fix a commutative ring $R$. Suppose $F:\mathrm{Top}^{\mathrm{op}}\to \mathrm{cDGA}_R$ is a contravariant functor from spaces to commutative DGAs over $R$, such that $X\mapsto H^*(F(X))$ is ordinary cohomology with coefficents in $R$. Then $R$ contains $\mathbb{Q}$ as a subring.

I'm pretty sure this "Theorem" is true. But I don't have a proof at hand.

The question is: does anyone have a proof, or know of one in the literature? I'm particularly interested in seeing a proof which is relatively "elementary", in the sense of not requiring much more heavy machinery than is needed in order to make the statement.

Added. Tyler points out in his answer that this can't be true as stated. We should add a hypothesis, such as: F takes homotopy pushouts of spaces to homotopy pullbacks of chain complexes.

Answer by SGP for On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$?

2011-02-18T23:54:57Z

It seems to be known at least to Thom, Sullivan, Swan, Cartan and Miller (1960's). I do not have a precise reference, but do see Cartan's Theories Cohomologiques

Answer by Tyler Lawson for On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$?

2011-02-19T02:56:13Z

Your theorem needs at least one further assumption. Otherwise, we can let F be the functor on spaces sending $X$ to the commutative DGA $H^*(X;R)$ , with zero differential.

I'm not sure what statement to add. The thing I initially wanted to write is that your functor should take homotopy pushouts to homotopy pullbacks. However, commutative DGAs aren't closed under homotopy pullback.

EDIT: Let's suppose you make this assumption. Take the diagram $* \leftarrow \mathbb{CP}^\infty \rightarrow *$ , form the homotopy pushout $\Sigma \mathbb{CP}^\infty$, and apply your functor $F$ to the associated square pushout diagram. You get a commutative square of commutative differential graded algebras, which is in particular a commutative square of $E_\infty$ -algebras. The category of $E_\infty$ -algebras has homotopy pullbacks, and so you can construct by this method a weak equivalence $F(\Sigma \mathbb{CP}^\infty) \rightarrow P$ of $E_\infty$-algebras, where $P$ is the homotopy pullback.

However, in any characteristic the Steenrod operations are stable operations and an invariant of weak equivalence of $E_\infty$ -algebras. If $pR = 0$, the element $x \in H^2(F(\mathbb{CP}^\infty))$ supports nonzero power operations (namely, powers) at any prime, and so the associated element $$\sigma x \in H^3(P) \cong H^3(F(\Sigma \mathbb{CP}^\infty))$$ would also support nonzero power operations. But for commutative DGAs those operations are automatically zero.

Ideally one could make this for more general $R$ where $p$ is not invertible using a secondary operation and $B\mathbb{Z}/p$ instead, but the kids were up early and I'm too tired to figure out how right now.