Does every complex orientable $E_\infty$-ring admit an $E_\infty$ complex orientation?

Question

A ring spectrum $E$ is complex oriented if it is equipped with a ring map $MU\rightarrow E$. It is complex orientable if such a ring map exists.

An $E_\infty$-ring $E$ is $E_\infty$-complex oriented if it is equipped with an $E_\infty$-ring map $MU\rightarrow E$. It is $E_\infty$-orientable if such a map exists.

The topic of $E_\infty$-orientations has been well studied, for example here Ando gives a necessary condition for a complex orientation to lift to an $E_\infty$ one, and here Hopkins and Lawson produce an obstruction theory for that same lifting problem.

However, as far as I can tell both references provide techniques that are really designed to solve the problem of finding an $E_\infty$ orientation with a $\textit{prescribed}$ underlying (plain) orientation (as opposed to finding any $E_\infty$-orientation whatsoever).

So my question is: If an $E_\infty$ ring spectrum admits a complex orientation, does it admit an $E_\infty$ complex orientation?

Answer 1

The answer is no. I guess I owe Fernando Muro 10 dollars.

Let $\mathbb{S}\rightarrow\Sigma^{-2}\mathbb{CP}^\infty$ be the inclusion of the bottom cell, and let $f:F\rightarrow\mathbb{S}$ be the fiber. Then a complex orientation of a ring spectrum $R$ is a nullhomotopy of the composite $F\rightarrow\mathbb{S}\xrightarrow{\text{unit}} R$. Let $MX$ be the pushout in the category of $E_\infty$ algebras of the span $\mathbb{S}\leftarrow\text{Free}_{E_\infty}(F)\rightarrow\mathbb{S}$ where the left leg is the map induced by the zero map $F\rightarrow \mathbb{S}$ and the second is the map induced by $f$. $MX$ is the initial complex oriented $E_\infty$-ring: the space of maps $MX\rightarrow R$ is equivalent to the space of nullhomotopies above (=complex orientations).

Let $H=H\mathbb{F}_2$. Then $H\otimes MX$ is the pushout of $H\leftarrow H\otimes\text{Free}_{E_\infty}(F)\rightarrow H$ in the category of $E_\infty$-$H$-algebras. But note that $H\otimes f$ is null (because $H$ is complex oriented). So $H\otimes MX$ is equivalent, as an $E_\infty$-$H$-algebra, to $H\otimes\text{Free}_{E_\infty}(\Sigma F)$. In particular, $H_*MX$ is equivalent, as a module over the dyler lashof algebra, to the free (graded) commutative ring generated by the free Dyler-Lashof module generated by $H_*\Sigma F\simeq\mathbb{F}_2\{b_1,b_2,..\}$, $|b_i|=2i$.

Now $MX$ is evidently (canonically) complex oriented, so it receives a ring map $MU\rightarrow MX$. On homology that induces the map $H_*MU\simeq \mathbb{F}_2[a_1,a_2,...]\rightarrow H_*MX$ sending $a_i$ to $b_i$.

Suppose $\phi: MU\rightarrow MX$ is an $E_\infty$-ring map. Then for degree reasons, its effect on homology must send $a_1$ to some multiple of $b_1$. But $Q^6(a_1)=a_1^4$ while $Q^6(b_1)$ is another free generator of $H_*MX$. So $a_1$ must be sent to zero. Let $MX\rightarrow MU$ be the $E_\infty$-map corresponding to any complex orientation of $MU$. Then on the one hand the composite $MU\xrightarrow{\phi}MX\rightarrow MU$ also kills $a_1$ in homology, but on the other hand it is a ring map $MU\rightarrow MU$, all of which are equivalences.