# Multiplicative Brown representability?

The Brown representability theorem can be convenient way to construct a spectrum. But to get a ring spectrum of even a very unstructured form seems to be harder. There's even currently a statement on the nlab to the effect that any multiplicative cohomology theory is represented by a ring spectrum, but my strong suspicion is that this statement was made in error.

Assumption: Let $$E^\ast: Spaces^{op} \to GrAb$$ be a cohomology theory. To avoid subtleties with the unpointed version of Brown's theorem, let's just assume that $$E^\ast$$ is represented by a spectrum $$E$$. Now when we impose "multiplicative" structure on $$E^\ast$$, the question is whether the "multiplication" is also "representable". I can think of at least two possible ways to make this precise:

Question 1: Suppose that $$E^\ast: Spaces^{op} \to GrAb$$ admits the structure of a lax (symmetric) monoidal functor. Then does there exist a map $$\mu: E \wedge E \to E$$ inducing the (symmetric) monoidal structure of $$E^\ast$$ in the appropriate sense?

I think that's the cleanest form of the question, but I think it would be more classical to ask it in the following form:

Question 2: Suppose that $$E^\ast: Spaces^{op} \to GrAb$$ admits a lift to a functor $$E^\ast: Spaces^{op} \to GrRing$$. Then does there exist a map $$\mu: E \wedge E \to E$$ such that the multiplication $$E^\ast(X) \otimes E^\ast(X) = \pi_{-\ast} F(\Sigma^\infty_+,E) \otimes \pi_{-\ast}F(\Sigma^\infty_+,E) \to \pi_{-\ast}F(\Sigma^\infty_+ X, E) = E^\ast(X)$$ is always given by the formula $$f\otimes g \mapsto \mu_\ast \circ (f \wedge g) \circ \Delta^\ast$$ where ($$\Delta: X \to X \times X$$ is the diagonal)?

Remark: The category "$$GrRing$$" could be interpreted in multiple ways -- if our rings are associative, we could ask for a homotopy associative $$\mu$$; if our rings are graded-commutative we could ask for a homotopy commutative $$\mu$$, etc.

Note: The Yoneda lemma does allow us to give an "$$E_0$$ version" of Brown representability. That is, from the structure of unit elements $$u_X \in E^\ast(X)$$ we can construct a unit map $$u: \mathbb S \to E$$, just by the Yoneda lemma. That's why the question is about the multiplication.

Remark: Actually -- we might even just as well have our cohomology theory be defined as a functor $$E^\ast: Spectra^{op} \to GrAb$$, and the question still seems to be interesting.

• If you haven't seen Dmitri Pavlov's "Structured Brown Representability via Concordance" then I can say with high confidence you'd have fun poking around through it. Mar 22, 2021 at 23:52

I'll write $$h_E$$ for the functor from finite spectra to abelian groups given by $$h_E(X)=\pi_0(E\wedge X)=[DX,E]$$. This is an object of the category $$\mathcal{A}$$ of all additive functors from finite spectra to abelian groups. Given $$A,B,C\in\mathcal{A}$$, a pairing from $$A$$ and $$B$$ to $$C$$ means a natural map $$A(X)\otimes B(Y)\to C(X\wedge Y)$$. By the usual Day construction, there is a symmetric monoidal product $$\boxtimes$$ on $$\mathcal{A}$$ such that $$\mathcal{A}(A\boxtimes B,C)$$ is the set of pairings from $$A$$ and $$B$$ to $$C$$. This preserves colimits in both variables. We can use Spanier-Whitehead duality to identify $$\mathcal{A}$$ with a category of contravariant functors, and then use standard properties of Day convolution to see that $$h_E\boxtimes h_F=h_{E\wedge F}$$ when $$E$$ and $$F$$ are finite. We can then take colimits over finite subspectra to see that $$h_E\boxtimes h_F=h_{E\wedge F}$$ in all cases.
Any of the kinds of input data specified in the question will give rise to a pairing from $$h_E$$ and $$h_E$$ to $$h_E$$, and thus to a map $$h_{E\wedge E}=h_E\boxtimes h_E\to h_E$$. By Brown representability, this will arise from a map $$\mu\colon E\wedge E\to E$$ of spectra, which will be unique mod phantoms. If the original pairing is commutative or associative, then $$\mu$$ will be commutative or associative mod phantoms. Corollary 2.15 of "Morava K-theories and localisation" gives a commonly satisfied criterion under which the group of phantom maps in degree zero is trivial.