Multiplicative Brown representability?

Question

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?

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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.

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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.

Answer 1

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.

I was surprised to find that Dan Christensen and I did not discuss this monoidal structure in our 1998 paper on phantom maps and homology theories, as it would fit very naturally there. Probably someone else has published this at some point, but I don't know where.