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Is it possible to embed de Rham cohomology of a two-dimensional closed surface of genus $g\geq 2$ into the differential graded algebra of differential forms (with de Rham differential and wedge product) on the surface as a differential graded subalgebra (in a way that is compatible with canonical projection from closed forms to cohomology, associating to a closed form its cohomology class)?

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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 ? – José Figueroa-O'Farrill Oct 3 '11 at 23:50
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. – pmnev Oct 4 '11 at 0:08
Jose -- this just shows that Sullivan's minimal model maps quasi-isomorphically to both algebras. – algori Oct 4 '11 at 0:20
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. – algori Oct 4 '11 at 0:24
@pmnev,algori: Thanks for your comments. I understand now what the question actually asks. – José Figueroa-O'Farrill Oct 4 '11 at 0:47
up vote 14 down vote accepted

The answer is no. Suppose that $\alpha_1,\ldots,\alpha_g,\beta_1,\ldots,\beta_g$ were closed $1$-forms on $M$ such that their cohomology classes were a basis of $H^1(M)$ and they satisfied $\alpha_i\wedge\alpha_j = \beta_i\wedge\beta_j = 0$ while $\alpha_i\wedge\beta_j = \delta_{ij}\ \gamma$ where $\gamma$ is a single $2$-form whose cohomology class is nonzero. Then $\gamma$ cannot vanish identically.

Let $U\subset M$ be the open set on which $\gamma$ is nonzero. Then none of the $\alpha_i$ or $\beta_i$ can vanish on $U$. Since $\alpha_1\wedge\alpha_2 = \beta_1\wedge\alpha_2=0$ while $\alpha_2\not=0$ on $U$, it follows that $\alpha_1$ and $\beta_1$ are multiples of $\alpha_2$ on $U$. But this implies that $\alpha_1$ and $\beta_1$ must be linearly dependent on $U$ as well, which implies that $\alpha_1\wedge\beta_1 = 0$.

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Very nice argument, thanks a lot! – pmnev Oct 4 '11 at 13:11

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