# $E_{\infty}$ algebra in characteristic zero

I asked this question on MSE(https://math.stackexchange.com/q/1579026/239218). Let $A^{\bullet}$ be a cosimplicial commutative algebra over a field $\Bbbk$. Denote with $N(A)^{\bullet}$ the conormalized Moore complex. Since $A^{\bullet}$ is equipped with a product, the Alexander–Whitney map induces an associative product $$\Delta : N(A)^{\bullet}\otimes N(A)^{\bullet}\to N(A)^{\bullet}.$$ which is commutative in cohomology.

If $A$ is good enough, the product above may be interpreted as a piece of some complicated operads $P$ (Eilenberg–Zilber operad, or Barratt–Eccles operad). In particular they are a cofibrant replacement of the operad $\mathsf{Comm}$, and they induce an $E_{\infty}$ structure on $N(A)^{\bullet}$ (see cochain on a simplicial set). It seems to me that these two operads are defined in order to find an $E_{\infty}$ structure on $N(A)^{\bullet}$, when the characteristic of the field is positive (so the hard case).

My question is: what happens for characteristic zero? Are there some simpler $E_{\infty}$ structure on $N(A)^{\bullet}$?

Here a suggestion: $(N(A)^{\bullet}, \Delta)$ is a differential associative algebra, where the product is graded commutative in cohomology. Is that an $E_{\infty}$ algebra?

• In characteristic zero, $C_\infty$ is a (much simpler) cofibrant replacement for the commutative operad. You get the first part of your $C_\infty$ structure by symmetrizing the product (then higher parts of the structure come from symmetrizing homotopies and higher homotopies for associativity). Roughly, to do this to "level $n$" involves dividing by $n!$ so this is only possible in characteristic zero. You can also view this as being a consequence of every $S_n$ representation being projective over $\mathbb{Q}$. – Gabriel C. Drummond-Cole Dec 17 '15 at 9:52
• Ok, I see, by symmetrizing the homotopies we get a $C_{\infty}$ structure (hence an $A_{\infty}$ structure) on $N(A)^{\bullet}$. Let's call this structure $m_{\bullet}$. It sems to me that $(N(A)^{\bullet}, m_{\bullet})$ and $(N(A)^{\bullet}, d, \Delta)$ are quasi isomorphic as $A_{\infty}$ algebras, is it correct? – Cepu Dec 17 '15 at 11:54
• Yes, that's the idea (provided you normalize your symmetrization, of course). – Gabriel C. Drummond-Cole Dec 17 '15 at 12:39
• Thanks, it seems to me a general argument: given an associative algebra $(B,d, \mu)$ as initial data (plus some conditions...) , then there is a $C_{\infty}$ algebra structure $m_{\bullet}$ on $B$ such that $(B,d, \mu)$ and $(B,m_{\bullet})$ are quasi isomorphic as $A_{\infty}$ algebras. Do you know some references? – Cepu Dec 18 '15 at 10:56
• Depends on what you mean by "is"...In the sense that there exist higher homotopies that extend $(N(A), d, \Delta)$ to a full $E_\infty$ structure, yes. By itself though, $(N(A), d, \Delta)$ is not enough data to be $E_\infty$. – Justin Young Dec 18 '15 at 14:11

In characteristic zero, Comm is already "cofibrant enough" so you don't need $E_\infty$ or even $C_\infty$. In particular, Comm is $\Sigma$-cofibrant, which means it's cofibrant in the model category of symmetric sequences (equivalently, it has lifts with respect to trivial fibrations, but the lifts are maps of symmetric sequences, not necessarily of operads). One of the recurring themes when you study the homotopy theory of operads is that $\Sigma$-cofibrant is good enough for everything you want to do.

For instance, suppose you want to have a good homotopy theory on the algebras over your operad, here CDGAs and $E_\infty$-algebras. For chain complexes in characteristic zero, all operads are $\Sigma$-cofibrant and we have transferred model structures on algebras over any operad. The next question is: if we have a weak equivalence of operads does it induce a Quillen equivalence of algebras? This is called rectification, and the answer is yes for characteristic zero. A reference is my paper "Model Structures on Commutative Monoids in General Model Categories", Theorem 4.6 of the submitted version, which is longer than the arxiv version. I'll update the arxiv after I finish making the improvements suggested by the referee.

The point is that in order to have rectification between Comm and $E_\infty$ what you really need is a property we're familiar with from the study of spectra. There, it's the statement that for cofibrant $X$, the natural map $X\wedge_{\Sigma_n} (E\Sigma_n)_+ \to X/\Sigma_n$ is a weak equivalence. In general, it's the statement that $X\otimes_{\Sigma_n} Q_{\Sigma_n}S \to X/\Sigma_n$ is a weak equivalence, where $Q_{\Sigma_n}$ is the cofibrant replacement of the unit in the projective model structure on $M^{\Sigma_n}$, i.e. objects of $M$ with a $\Sigma_n$ action. I named this the rectification axiom and proved it implies rectification occurs. I then check it holds for Ch(k) in characteristic zero, in positive model structures on spectra, and NOT for spaces (via an explicit counterexample). My thinking on this actually started on MathOverflow with this thread.

• This answer is great if you are doing something where you can work in the homotopy category, but is this helpful if you really want to work with a particular algebra (like the original poster) or if you want a model where a weak equivalence is represented by a single map and not a zig-zag? – Gabriel C. Drummond-Cole Dec 19 '15 at 7:28