10
$\begingroup$

When reading about homotopy algebras (e.g. $L_\infty$-algebras, $A_\infty$-algebras), an $\infty$-morphism $f$ is called an $\infty$-quasi-isomorphism if $f_1$ is a quasi-isomorphism.

Recall/Example ($A_\infty$-algebras):

An $A_\infty$-morphism between two $A_\infty$-algebras $(A,\mathfrak{m})$ and $(A', \mathfrak{m}')$ (here $\mathfrak{m}$ and $\mathfrak{m}'$ are the structure maps) is a collection $\{f_k\}_{k\geq1}:(A,\mathfrak{m}) \rightarrow (A',\mathfrak{m}') $ of degree zero (degree preserving) multilinear maps \begin{equation*} f_k: A^{\otimes k}\rightarrow A', \hspace{1cm}k\geq 1 \end{equation*} that satisfy the following relation for $n\geq1$: \begin{equation*} \sum_{k+l=n+1}\sum^k_{i=0} (-1)^{a_1+\dots+a_n}f_k(a_1, \dots, a_i, m_l(a_{i+1}, \dots, a_{i+l}), a_{i+l+1}, \dots, a_n). \end{equation*} \begin{equation*} =\sum_{\substack{1\leq k_1\leq \dots \leq k_j \\ k_1+\cdots+k_j= n}} m'_j(f_{k_1}(a_1, \dots, a_{k_1}), f_{k_2}(a_{k_1+1}, \dots, a_{k_1+k_2}), \dots, f_{k_j}(a_{k_{j-1}+1}, \dots, a_n)) \end{equation*}

Furthermore, we call such morphisms $A_\infty$-quasi-isomorphisms if $f_1$ induces isomorphism in cohomology.

Q1: Why do we normally omit higher arity maps when talking about quasi-isomorphisms?

Q2: Would it be possible to have a weak equivalence that only appears in higher arity maps?

Q3: In case we only care about $f_1$, wouldn't that imply an equivalence at the level of homotopy categories between, for example, Ho(DGLA) and Ho(L$_\infty$), as all the higher arity maps between $A$ and $B$ in L$_\infty$ with the same $f_1$ give isomorphism in Ho(L$_\infty$)?

$\endgroup$
4
  • $\begingroup$ I'm very far from an expert of $A_\infty$-algebras, but I assume that this is in order for a map to be an equivalence iff the induced map in $Ho(A_\infty-alg)$ is an isomorphism $\endgroup$ Commented Apr 12, 2016 at 15:00
  • $\begingroup$ Well, the thing is that when computing the cohomology of a homotopy algebra, e.g. Hochschild cohomology of an $A_\infty$-algebra, we built the Hochschild complex with differentials in terms of the whole $A_\infty$ structure, using higher arity maps. Intuitively, I would think that to construct a weak equivalence in the category A$_\infty$, I would need higher arity maps to detect again what happens in cohomology. (but seems like it's not) --I can understand that if a map $f$ induces an iso in Ho(A$_\infty$), $f_1$ should be a quasi-iso; but I can't see why it should hold the other way around. $\endgroup$ Commented Apr 12, 2016 at 15:33
  • 2
    $\begingroup$ The answer to this more specific question is, basically, "because the spectral sequences starting from the cohomology along the differential induced by $m_1$ converge to the cohomology of the total differential, and a quasi-isomorphism on the $E_1$ pages implies a quasi-isomorphism on the limit pages". See my more general answer to Q1 below for a further elaboration. $\endgroup$ Commented Apr 12, 2016 at 20:21
  • $\begingroup$ That makes sense, thanks! I am in fact more interested in your answer below; but it is nice to see these methods appearing here. $\endgroup$ Commented Apr 12, 2016 at 23:36

5 Answers 5

12
$\begingroup$

Q1: The conventional theory of homotopy algebras is built on the premise that the lower-degree operations dominate over the higher-degree ones, in some sense. A discussion of this can be found in the introduction to my preprint "Weakly curved $\mathrm A_\infty$-algebras over a topological local ring", http://arxiv.org/abs/1202.2697. This does not answer your question fully, but explains the underlying ideology to some extent.

Q2: The theory of curved DG-algebras/curved DG-coalgebras and the co/contraderived categories of CDG-modules/comodules/contramodules over them is built on the premise that the (co)multiplication dominates over the differential (and the differential dominates over the curvature). So the higher-degree operations dominate over the lower-degree ones in these "theories of the second kind". On CDG-coalgebras or DG-coalgebras, there is even a model structure with such weak equivalences (though a precise definition is more complicated and maybe does not accord to what you describe in the question). See my memoir "Two kinds of derived categories, ...", http://arxiv.org/abs/0905.2621.

Q3: It is indeed true that the homotopy category of DG Lie algebras is equivalent to the homotopy category of $\mathrm L_\infty$-algebras, though perhaps for reasons more complicated than described in the question. Similarly, the homotopy category of associative DG-algebras is equivalent to the homotopy category of $\mathrm A_\infty$-algebras.

Basically, with any $\mathrm A_\infty$-algebra one can naturally associate a much bigger DG-algebra quasi-isomorphic to it; while for any DG-algebra one can, if one wishes, construct a (generally speaking) much smaller $\mathrm A_\infty$-algebra quasi-isomorphic to it (in addition to the obvious option of viewing a DG-algebra as an $\mathrm A_\infty$-algebra with vanishing higher operations).

But of course, one cannot just forget the higher operations of an $\mathrm A_\infty$-algebra and obtain a quasi-isomorphic DG-algebra, firstly because the multiplication in an $\mathrm A_\infty$-algebra need not be associative, and secondly, even if it is, the identity map cannot be extended to an $\mathrm A_\infty$-morphism (in most cases).

$\endgroup$
2
  • $\begingroup$ thank you for your answers! I appreciate you sharing your papers. I did not know about curved $A_\infty$-algebras, but I will try to read more about them --I imagine that deformation theory in the setting of curved $A_\infty$-algebras would be difficult (because of gauge/homotopy equivalence?). $\endgroup$ Commented Apr 12, 2016 at 23:22
  • $\begingroup$ In particular, in the first paper you linked, at 0.24 of the introduction, you talk a bit about deformation theory. From what I read, is it right to think that there are statements like: 'given a weak equivalence between two curved algebras A and B (weak equivalence in the sense that you described for Q2), the respective deformation functors are isomorphic' ? $\endgroup$ Commented Apr 12, 2016 at 23:28
10
$\begingroup$

My favorite explanatory analogy for the first question, along the line of Leonid's answer, is that a power series in one variable with no constant term, $a_1x +a_2x^2 +\cdots$ has an inverse under composition if and only if $a_1$ has an inverse in the ground ring.

$\endgroup$
7
$\begingroup$

Here is a down to earth explanation, by way of analogy.

A group homomorphism $\varphi:G\to H$ is an isomorphism of groups iff it is a bijection of sets. Why do we not need to say anything about the group operations coinciding? The reason is simply because the fact that $\varphi$ is a group homomorphism already implies this.

There is an essentially endless list of similar situations, and quasi-isomorphisms of $A_\infty$-algebras is one of them. Of course, the fact that quasi-isomorphisms of $A_\infty$-algebras (over a field!) are invertible up to homotopy is by no means automatic, but it is straightforward to check.

$\endgroup$
4
$\begingroup$

$\text{}$Hi Marcel, I don't believe you're confused about this anymore, but I stumbled upon this question and wanted to add a remark that I think is missing from the other answers.

I claim that before understanding what a quasi-isomorphism of $A_\infty$-algebras is, one should understand what an isomorphism of $A_\infty$-algebras is.

But there is absolutely no ambiguity in what it means for a morphism in a category to be an isomorphism. An isomorphism always means a morphism with a two-sided inverse. And one can easily prove the following result:

Proposition: If $f \colon A \to B$ is a morphism of $A_\infty$-algebras, given by components $f_n \colon A^{\otimes n} \to B$, then $f$ is an isomorphism in the category of $A_\infty$-algebras if and only if $f_1$ is an isomorphism in the category of chain complexes.

Given this result it is very natural that quasi-isomorphisms are defined the way they are.

$\endgroup$
2
$\begingroup$

I will try to answer Q1.

One simple reason why $A_\infty$-algebras appear is that usual DG-algebras are not homotopy stable, in the sense that if $A$ is a DG-algebra and $V$ is a complex which is a deformation retract of $A$ (in the usual topological sense, using the notion of homotopy for complexes), then $V$ does not inherit a DG-algebra structure, but well, an $A_\infty$-algebra structure. In fact, it suffices, that there is a map of complex $V\to A$ that is right homotopy invertible, see this paper. One usually likes to do this when $V$ is quasi-isomorphic to $A$ via some map $V\to A$ (for example, $V=H(A)$), and one can produce a retraction for example when the ground ring is a field.

Now one problem which $A_\infty$-algebras fix is that of localising the category of DG-algebras at quasi-isomorphisms. If $A$ is an $A_\infty$-algebra, one can consider its bar construction $BA$, which is a honest DG-coalgebra, and take $\mathsf{DASH}$ the category with objects the $A_\infty$-algebras and hom-sets the maps DG-coalgebra maps between bar constructions. Every DG-algebra is an $A_\infty$-algebra, and one can think of the category $\mathsf{DA}$ of DG-algebras sitting inside $\mathsf{DASH}$ as a (non-full, and that's the jist of it all) subcategory via the obvious inclusion. There is a usual notion of homotopy between maps of DG-coalgebras which yields the homotopy category $\mathsf{DASH}/\sim$, and it turns out the map $\mathsf{DA} \to \mathsf{DASH}$ gives an equivalence of categories $\mathsf{DA}[\text{Qis}^{-1}] \to \mathsf{DASH}/\sim$. This is explained, and I suppose first proven in this paper by Munkholm. Thus, $A_\infty$-algebras "fix" the problem of quasi-isomorphism of complexes by replacing perhaps a big complex by a small one, usually its homology, at the cost of giving you an ugly differential.

To get back to Q1, note that in the category $\mathsf{DC}$ of conilpotent DG-coalgebras, we also have a notion of quasi-isomorphism. Thus, say that a map $f:A\to A'$ of $A_\infty$-algebras is a weak equivalence if its image in $\mathsf{DASH}$, i.e. $Bf : BA\to BA'$, is a quasi-isomorphism. It turns out that, because $BA$ and $BA'$ are quasi-free (i.e. free as coalgebras, with some funny differential), weak equivalences are homotopy equivalences (there is a model category on $\mathsf{DC}$ where all objects are cofibrant and the fibrant objects are the quasi-free coalgebras, as proven here), and in fact one can show that $Bf$ is a quasi-isomorphism if and only it is an homotopy equivalence, if and only if $f_1$ is a quasi-isomorphism. This means that, so far as quasi-isomorphism (equivalently, homotopy) of $A_\infty$-algebras goes, the only relevant map is the one of length $1$. You can find a proof of this here. As mentioned in the comments, the implication "$f_1$ a quasi-isomorphism then $Bf$ a quasi-isomorphism" is a simple spectral sequence argument, but the reverse implication is a bit more involved, and is proven in loc. cit. with comparison theorems (Zeeman/Moore) of spectral sequences and constructions (Cartan).

$\endgroup$
1
  • $\begingroup$ (Usually the new differential has nice combinatorial interpretations, for example, but from a computational point of view, it is still a bit ugly to work with!) $\endgroup$
    – Pedro
    Commented Jan 7, 2018 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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