What is 'formal' ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:53:38Z http://mathoverflow.net/feeds/question/13005 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13005/what-is-formal What is 'formal' ? Xiao Xinli 2010-01-26T03:02:53Z 2010-02-01T08:41:00Z <p>The key step in Kontsevich's proof of deformation quantization of Poisson manifolds is the so-called formality theorem where 'a formal complex' means that it admits a certain condition. I wonder why it is called 'formal'. I only found the definition of Sullivan in Wikipedia: 'formal manifold is one whose real homotopy type is a formal consequence of its real cohomology ring'. But still I am confused because most of articles I found contain the same sentence only and I cannot understand the meaning of 'formal consequence'. Does anyone know the history of this concept?</p> http://mathoverflow.net/questions/13005/what-is-formal/13019#13019 Answer by Kevin Lin for What is 'formal' ? Kevin Lin 2010-01-26T06:44:35Z 2010-02-01T08:23:31Z <p>I would guess that the terminology goes back to the work of Sullivan and Quillen on rational homotopy theory. You should probably also look at the paper of <a href="http://www.springerlink.com/content/m48544t785221635/" rel="nofollow">Deligne-Griffiths-Morgan-Sullivan</a> on the real homotopy theory of Kähler manifolds. Actually, I think that at least some familiarity with the DGMS paper is an important prerequisite for understanding many of Kontsevich's papers.</p> <p>I am not totally sure, but I believe that the definitions are as follows:</p> <ul> <li><p>A differential graded algebra $(A,d)$ is called formal if it is quasi-isomorphic (in general, if we work in the category of dg algebras and not, say, the category of A-infinity algebras, we need a "zig-zag" of quasi-isomorphisms) to $H^\ast(A,d)$ considered as a dg algebra with zero differential.</p></li> <li><p>A space X is called formal (over the rationals resp. the reals) if its cochain dg algebra $C^\ast(X)$ (with rational resp. real coefficients) with the standard differential is a formal dg algebra.</p></li> </ul> <p>One of the things I'm not sure about is whether in the definition we should require $H^\ast(A,d)$ to be commutative; but for spaces this is not an issue since $H^\ast(X)$ is always (graded-)commutative.</p> <p>The DGMS paper proves that if X is a compact Kähler manifold, then the de Rham dg algebra consisting of (real, $C^\infty$) differential forms on X with the standard de Rham differential is a formal dg algebra.</p> <p>The phrase "the real (resp. rational) homotopy type of X is a formal consequence of the real (resp. rational) cohomology ring of X", which appears in e.g. the DGMS paper, simply means that the real (resp. rational) homotopy theory of X is determined by (and is probably explicitly and algorithmically computable from?) the cohomology ring of X. In other words, if X and Y are formal (over the rationals resp. the reals) and have isomorphic (rational resp. real) cohomology rings, then their respective (rational resp. real) homotopy theories are the same (and are explicitly computable, if we know the cohomology ring(s)?). For example, the ranks of their homotopy groups will be equal.</p> <p>Actually I am not totally sure whether what I said in the last paragraph is true. I think it's true when X and Y are simply connected. I'm not sure about what happens more generally.</p> <p>In the context of rational homotopy theory, I think the term "formal" is fine, for the reasons I've explained above. Perhaps in the more general context of dg algebras, the use of the term "formal" makes less sense. However, I think that it is still reasonable, for the following reasons. Let me use the more "modern" language of A-infinity algebras. In general, it is not true that a dg algebra $(A,d)$ is quasi-isomorphic to $H^\ast(A,d)$ considered as a dg algebra with zero differential. However, it is a "standard" fact (Kontsevich-Soibelman call this the "homological perturbation lemma" (for example, it's buried somewhere in <a href="http://arxiv.org/abs/math/0011041" rel="nofollow">this paper</a>), and you can find it in the operads literature as the "transfer theorem") that you can put an A-infinity structure on $H^\ast(A,d)$ which makes $A$ and $H^\ast(A,d)$ quasi-isomorphic as A-infinity algebras. The A-infinity structure manifests itself as a series of $n$-ary products satisfying various compatibilities. Intuitively at least, these $n$-ary products should be thought of as being analogous to Massey products in topology. So $H^\ast(A,d)$ with this A-infinity structure does carry some "homotopy theoretic" information. In this language then, a dg algebra $(A,d)$ is formal if it is quasi-isomorphic, as an A-infinity algebra, to $H^\ast(A,d)$ with all higher products zero. In other words, all of the "Massey products" vanish*, and thus the only remaining "homotopy theoretic" information is that coming from the ordinary ring structure on $H^\ast(A,d)$.</p> <p><hr /></p> <p>*Don Stanley notes correctly that vanishing of Massey products is weaker than formality. However, I believe that triviality of the A-infinity structure is equivalent to formality. In the language of the DGMS paper, which does not use the A-infinity language, they say that formality is equivalent to the vanishing of Massey products "in a uniform way". I believe this uniform vanishing is the same as triviality of A-infinity structure. From the paper:</p> <blockquote> <p>... a minimal model is a formal consequence of its cohomology ring if, and only if, all the higher order products vanish in a uniform way.</p> </blockquote> <p>and also</p> <blockquote> <p>[Choosing a quasi-isomorphism from a minimal dg algebra to its cohomology] is a way of saying that <em>one may make uniform choices so that the forms representing all Massey products and higher order Massey products are exact</em>. This is stronger than requiring each individual Massey product or higher order Massey product to vanish. The latter means that, given one such product, choices may be made to make the form representing it exact, and there may be no way to do this uniformly.</p> </blockquote> <p>(Sorry for the proliferation of parentheses, and sorry for my lack of certainty on all of this, I have not thought about this in a while. People should definitely correct me if I'm wrong on any of this.)</p> http://mathoverflow.net/questions/13005/what-is-formal/13028#13028 Answer by Agusti Roig for What is 'formal' ? Agusti Roig 2010-01-26T09:17:44Z 2010-01-26T09:17:44Z <p>Maybe you could take a look at</p> <p>Y. Félix, J. Oprea, D. Tanré; Algebraic models in Geometry, Oxford Graduate Text in Math. 17 (2008)</p> <p>where they talk about formality in the context of rational homotopy theory, RHT, (for instance, in sections 2.7 and 3.1.4). Also the more classical, but excellent little book</p> <p>D. Lehmann; Théorie homotopique des formes différentielles, Astérisque 45</p> <p>is worth reading (section V.9).</p> <p>As for formality in the context of operads, allow me a little self-promotion :-) :</p> <p>F. Guillén, V. Navarro, P. Pascual, Agustí Roig, Moduli spaces and formal operads; Duke Math. J. 129, 2 (2005).</p> <p>In this work, we translate some classical results concerning formality in RHT to chain operads. For instance, the Deligne-Griffiths-Morgan-Sullivan theorem about formality of Kähler manifolds, formality's independence of the ground field... And extend them also to modular operads.</p> http://mathoverflow.net/questions/13005/what-is-formal/13045#13045 Answer by Don Stanley for What is 'formal' ? Don Stanley 2010-01-26T14:57:56Z 2010-01-26T15:30:32Z <p>Formal can mean slightly different things in different contexts.</p> <p>A commutative differential graded algebra (CDGA) is formal if it is quasi-isomorphic to it's homology. This is stronger than having all the higher Massey products equal to 0 (I think there are such examples in the Halperin-Stasheff paper). </p> <p>To a space you can associate a CDGA (via Sullivan's $A_{pl}$ functor) which is basically the deRham complex when the space is a manifold. In nice cases this functor induces an equivalence from the rational homotopy category to the homotopy category of CDGA. Quasi-isormorphic CDGA correspond to (rationally) homotopy equivalent spaces. You can also tensor with the reals to get real CDGA. </p> <p>If A is a CDGA which is quasi-isomorphic to $A_{pl}(X)$ for a space $X$ then A is often called a model of X. A space is formal if $A_{pl}$ of it is formal. So a formal space is modeled by its cohomology. In that sense its rational homotopy type is a formal consequence of its cohomology.</p> <p>I think you have to be slightly careful with using $C^*$. This functor lands in differential graded algebra which are not commutative, so possibly the notion of formality could be different. In particular if you consider two CDGA there may be more strings of quasi-isomorphisms between them as DGAs then as CDGAs. I believe it is unknown if two CDGA that are quasi-isomorphic as DGA have to be quasi-isomorphic as CDGA. </p> http://mathoverflow.net/questions/13005/what-is-formal/13063#13063 Answer by Agusti Roig for What is 'formal' ? Agusti Roig 2010-01-26T19:11:21Z 2010-02-01T08:41:00Z <p>Paraphrasing Groucho Marx: if you don't like my first answer..., well I have another one. :-)</p> <p>Here it is: let $X$ be a simply connected differentiable manifold.</p> <p>Rational homotopy theory tells us that the <em>rational homotopy type</em> of $X$ (that is, its homotopy type modulo torsion) is contained in its <em>minimal model</em>, <code>$M_X$</code>, which is a <em>commutative</em> differential graded (cdg) algebra.</p> <p>By definition, this means that you have a quasi-isomorphism (<em>quis</em>, a morphism of cdg algebras inducing an isomorphism in cohomology)</p> <p>$$M_X \longrightarrow \Omega^*(X) \ .$$</p> <p>Here, $\Omega^* (X)$ is the algebra of differential forms of $X$ and the <em>minimality</em> of $M_X$ means that, in a certain, but precise, sense, it is the smallest cdg algebra for which such a quis exists.</p> <p>The fact that $M_X$ <em>contains</em> the rational homotopy type of $X$ implies, for instance, that you can obtain the ranks of the homotopy groups of $X$ from it:</p> <blockquote> <p>rank $\pi_n(X) =$ number of degree n generators (as an algebra) of $M_X$, for $n \geq 2$.</p> </blockquote> <p>Nice, isn't it? :-)</p> <p>The problem is that the algebra $\Omega^*(X)$ is, in general, not computable, so you can not obtain from it the minimal model $M_X$. And here is where formality comes to help you.</p> <p>Almost by definition, $X$ is a <em>formal</em> space if there exists two quis</p> <p>$$\Omega^*(X) \longleftarrow M_X \longrightarrow H^*(X;\mathbb{Q})<br />$$</p> <p>Hence, if $X$ is formal you can compute its minimal model $M_X$, and hence its rational homotopy type, directly from the cohomology algebra <code>$H^*(X; \mathbb{Q})$</code>, which is nicer (smaller, more computable) than $\Omega^*(X)$.</p> <p>And the final point is that there are plenty of examples of spaces which are known to be formal.</p> <p>(Final remark: Actually, you'd have to put <code>$A_{PL}^*(X;\mathbb{Q})$</code> instead of $\Omega^*(X)$ to work over the rationals, but this you can find it explained in the references we have provided for you.)</p>