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

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)?

share|improve this question
1  
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

1 Answer 1

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 $\alpha_1\wedge\beta_1 = 0$.

share|improve this answer
    
Very nice argument, thanks a lot! –  pmnev Oct 4 '11 at 13:11

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.