Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is perhaps somewhat related to this question. Fix a field $k$ of characteristic $p>0$. Suppose that $A$ is an $E_\infty$-algebra over $k$. Then $A$ also has an $A_\infty$-algebra structure, and therefore so does its homology $HA$. Its homology is also graded commutative.

I'm looking for an extended version of graded commutativity, a result like this: if $m_n$ is any of the higher multiplications in the $A_\infty$ structure on $HA$, then for any $1 \leq j \leq n$, $$ x m_n(a_1 \otimes \cdots \otimes a_n) = \pm m_n(a_1 \otimes \cdots \otimes x a_j \otimes \cdots \otimes a_n). $$ This is certainly true if $n=2$ by graded commutativity. What about for larger values?

(I'm tempted to tag any question about $E_\infty$-algebras as "commutative algebra", but I suppose that would be misleading...)

Edit: as Fernando points out in his comment, this is too much to expect in general. The $A_\infty$-algebra structure on $HA$ is not unique, so perhaps the right question is, are there conditions on $x$ and the $a_i$ so that, for some choice of $m_n$, $x m_n(\dots) = \dots$?

Along with Fernando's example, another one to consider is the mod $p$ cohomology of a cyclic group of order $p$, with $p$ odd. If $x$ is the generator of $H^1$ and $y$ is the generator of $H^2$, then $m_p(x^{\otimes p}) = \pm y$, so $x m_p(x^{\otimes p}) = \pm xy \neq 0$ while I think $m_p(x^2 \otimes x^{\otimes p-1}) = 0$, since $x^2=0$.

share|improve this question
2  
Is $x\in HA$ any element? If so then $m_n(a_1\otimes\cdots\otimes a_n)= \pm a_1\cdots a_n\cdot m_n(1\otimes\cdots\otimes 1)=0$ if the $A$-infinity structure is normalized (you can always assume this). –  Fernando Muro Mar 13 '11 at 14:24

1 Answer 1

as Fernando points out in his comment, this is too much to expect in general. The A∞-algebra structure on HA is not unique

There is a unique $A_\infty$-algebra structure on $H(A)$ such that $H(A)$ and $A$ are weakly equivalent (i.e. $A_\infty$-quasi-isomorphic).

About your question, in characteristic 0 you could replace your $E_\infty$-algebra by a weakly equivalent DG commutative algebra... then on its cohomology you would get a $C_\infty$-structure (which is a strictly commutative $A_\infty$-structure).

In positive characteristic this is more difficult. But there is still the possibility to deal with divided power algebras (see e.g. http://math.univ-lille1.fr/~fresse/PartitionHomology.pdf on page 18 for a hint).

I know this is not really an answer to your question. But I hope to can help.

share|improve this answer
    
Damien, the structure is not unique. Perhapes you are mislead by the fact that it is essentially unique, but there are tons precisely because of that. –  Fernando Muro Apr 26 '11 at 12:11
    
By essentieally unique, do you mean that it is unique up to a unique $A_\infty$-isomorphism ? If so, then we agree :-) –  DamienC Apr 26 '11 at 12:52
    
@Damien: Yes, exactly. –  Fernando Muro Apr 26 '11 at 19:59
    
The non-uniqueness is important, because $m_n(a_1 \otimes \dots \otimes a_n)$ may have very different values for two different (but isomorphic) $A_\infty$ structures. Anyway, I will look at the citation you mentioned -- I mainly care about the positive characteristic case. –  John Palmieri Apr 27 '11 at 20:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.