Morphisms of $\mathbb E_l$-rings between $\mathbb E_k$-rings for $l<k$

Question

Given two commutative rings $A$ and $B$, any map of rings $A\to B$ will automatically preserve the commutative structure. This is to say, the forgetful functor $\operatorname{CRing}\to \operatorname{Ring}$ is fully faithful.

The analogous situation in brave new algebra is richer: we have $\mathbb E_k$-rings for all $1\le k \le \infty$, encompassing the notion of commutativity for up to $k$ factors.

Q: Does a result such as above still hold, i.e. is the forgetful functor $\operatorname{Alg}_{\mathbb E_k}\to \operatorname{Alg}_{\mathbb E_l}$ for some fixed $1\le l < k\le \infty$ fully faithful?

I would not be surprised if the answer turned out to be negative. After all, an $\mathbb E_k$-structure is not a condition but genuine structure instead, so a functor preserving it seems like a stronger condition. If that is the case, I would be very grateful if some examples could be provided of $\mathbb E_k$-ring maps that are not $\mathbb E_{k+1}$-ring maps (or just not $\mathbb E_\infty$-ring maps).

Of course, if the answer is negative, it would seem natural to ask under what conditions on an $\mathbb E_k$-rings $A$ and $B$ does every $\mathbb E_l$-ring map $A\to B$, for some fixed $l < k,$ give rise to an $\mathbb E_k$-ring map.

Answer 1

Let's consider these in characteristic zero, so that we can use differential graded algebras and so that we can model $E_\infty$ things by strictly commutative things and $E_1$ things by associative things.

Let $A$ be the commutative DGA $\Bbb Q[x,y]$ where $x$ and $y$ are in degrees $2n$ and $2m$ respectively, with $dx = dy = 0$. Then the set of maps $A \to B$ in the homotopy category of commutative DGAs can be naturally identified with $H_{2n} B \times H_{2m} B$.

Now let's think of $A$ as an associative DGA; it's no longer semifree, so to compute maps in the homotopy category we first have to find a resolution $\tilde A \to A$. Here's one: the associative $DGA$ given by $$ \tilde A = \Bbb Q\langle x,y,z \rangle $$ where $z$ is in degree $2n+2m=1$ and $dz = xy - yx$. Then $\tilde A$ is semifree, and there is a natural projection map $\tilde A \to A$ that sends $z$ to zero. With some work, one finds that it is a quasi-isomorphism.

As a result, the set of maps $A \to B$ in the homotopy category of $E_1$-algebras are the same as maps $\tilde A \to B$ in the homotopy category of $E_1$-algebras.

Let $B = \Bbb Q[z] / (z^2)$ with $|z| = 2n + 2m + 1$ and $dz = 0$. Then there is only the trivial map $A \to B$ sending $x$ and $y$ to zero, but the projection $\tilde A \to B$ sending $x$, $y$, and $z^2$ to zero is a nontrivial map in the homotopy category. This shows that the forgetful map is not full.

Answer 2

Let $R$ be an $E_\infty$-ring spectrum and write $R\{t\}$ (resp. $R[t]$) for the free $E_\infty$-$R$-algebra (resp. free $E_1$-$R$-algebra) on one generator $t$ (in degree zero). This notation is compatible with your previous question, i.e. $R[t]$ is nothing else than the polynomial $E_\infty$-$R$-algebra on one generator.

The forgetful map $\alpha : Maps_{E_\infty}(R[t], A) \to Maps_{E_1}(R[t], A) \approx \Omega^\infty(A)$ is identified with the canonical map $\varepsilon^* : Maps_{E_\infty}(R[t], A) \to Maps_{E_\infty}(R\{t\}, A) \approx \Omega^\infty(A)$ given by composition with the canonical $E_\infty$-ring homomorphism $\varepsilon : R\{t\} \to R[t]$. The discussion in the previous question implies that whenever $R$ is not a $\mathbf{Q}$-algebra, there exist choices of $A$ such that this map $\varepsilon_R^*$ is not invertible. For those choices, the map $\alpha$ will also fail to be invertible, so in particular there will exist $E_1$-ring homomorphisms $R[t] \to A$ which do not lift to $E_\infty$-ring homomorphisms.