On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:29:20Z http://mathoverflow.net/feeds/question/55927 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55927/on-the-nose-commutative-cup-product-longrightarrow-characteristic-0 On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$? Charles Rezk 2011-02-18T23:35:43Z 2011-02-19T05:52:46Z <p>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.</p> <blockquote> <p><strong>Theorem.</strong> 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.</p> </blockquote> <p>I'm pretty sure this "Theorem" is true. But I don't have a proof at hand. </p> <p>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.</p> <p><em>Added.</em> 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. </p> http://mathoverflow.net/questions/55927/on-the-nose-commutative-cup-product-longrightarrow-characteristic-0/55929#55929 Answer by SGP for On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$? SGP 2011-02-18T23:54:57Z 2011-02-18T23:54:57Z <p>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 <a href="http://www.digizeitschriften.de/dms/img/?PPN=PPN356556735_0035&amp;DMDID=dmdlog30" rel="nofollow">Cartan's Theories Cohomologiques</a></p> http://mathoverflow.net/questions/55927/on-the-nose-commutative-cup-product-longrightarrow-characteristic-0/55946#55946 Answer by Tyler Lawson for On-the-nose commutative cup product $\Longrightarrow$ characteristic $0$? Tyler Lawson 2011-02-19T02:56:13Z 2011-02-19T05:52:46Z <p>Your theorem needs at least one further assumption. Otherwise, we can let F be the functor on spaces sending $X$ to the commutative DGA <code>$H^*(X;R)$</code>, with zero differential.</p> <p>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.</p> <p>EDIT: Let's suppose you make this assumption. Take the diagram <code>$* \leftarrow \mathbb{CP}^\infty \rightarrow *$</code>, 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 <code>$E_\infty$</code>-algebras. The category of <code>$E_\infty$</code>-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.</p> <p>However, in any characteristic the Steenrod operations are stable operations and an invariant of weak equivalence of <code>$E_\infty$</code>-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.</p> <p>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.</p>