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Oct 7, 2011 at 15:27 comment added pmnev Dear Bruno, I did not mean to question formality of De Rham dga for a surface (I know that it holds due to a general theorem for Kaehler manifolds). I was rather asking whether formality holds in more strict sense, that you can find a dga morphism (not just any $A_\infty$) from cohomology to forms. To this I received a very clear answer from Robert Bryant, see below.
Oct 7, 2011 at 14:50 comment added Bruno V. Dear Pasha, a dga algebra is formal iff there exists an $A__\infty$-quasi-isomorphism between it and its cohomology with induced (stric) algebra structure, not only an $A_\infty$-morphism. (Not this is just a typo, I am sure).
Oct 4, 2011 at 13:11 vote accept pmnev
Oct 4, 2011 at 12:48 answer added Robert Bryant timeline score: 14
Oct 4, 2011 at 0:47 comment added José Figueroa-O'Farrill @pmnev,algori: Thanks for your comments. I understand now what the question actually asks.
Oct 4, 2011 at 0:24 comment added algori pmnev -- I think you are right. Moreover, I think one can get a genuine morphism of algebras at the expense of enlarging the target: namely, if one replaces the de Rham algebra with the cobar of its bar.
Oct 4, 2011 at 0:20 comment added algori Jose -- this just shows that Sullivan's minimal model maps quasi-isomorphically to both algebras.
Oct 4, 2011 at 0:08 comment added pmnev As far as I understand, formality just means that there exists some A_\infty morphism from cohomology to de Rham algebra. Generally it would have polylinear components. My question was whether one can find an A_\infty morphism which has only linear component.
Oct 3, 2011 at 23:50 comment added José Figueroa-O'Farrill Does this not follow from the formality of compact Kähler manifolds, proved in Deligne, Pierre; Griffiths, Phillip A.; Morgan, John W.; Sullivan, Dennis (1975), "Real homotopy theory of Kähler manifolds", Inventiones Mathematicae 29 (3): 245–274, doi:10.1007/BF01389853, MR0382702 ?
Oct 3, 2011 at 14:27 history asked pmnev CC BY-SA 3.0