Sorry for the Clicbait title, but take $A$ a graded $\mathbb R$ commutative (in the graded sense) algebra of finite dimension.

Does there exist a smooth manifold $M$ having $A$ as a DeRham cohomology ring : $H^*(M) \simeq A$ ?

In the negative case, what would be a natural restriction that I missed on the structure of the DeRham ring ?

I was thinking of taking generators for the algebra and taking product of spheres. But i have problems when those generators verfies relations among them.

  • 1
    $\begingroup$ I've thought about this for a little while to no avail (other than some obvious conditions like what $A_0$ must be and that $A$ must arise as $A' \otimes_{\Bbb Z} A$ for some finitely generated $\Bbb Z$-algebra $A'$). It's worth noting that it is sufficient to provide such cohomology rings for finite CW complexes, as every finite CW complex is homotopy equivalent to a small open neighborhood of itself in some large Euclidean space. $\endgroup$
    – user98602
    Jan 19, 2019 at 20:51
  • $\begingroup$ So it seems that you are saying that this is equivalent to the same question replacing manifold by CW complexes and deRham by $\mathbb Z$ cohomology. Didn't noticed your condition on the integrability of the algebra. On the other hand I don't really understand your comment on $A_0$, could you please expand a bit ? On this subject an MO post redirects on an article Andersen and Grodal resolving this problem $\endgroup$ Jan 21, 2019 at 10:08
  • $\begingroup$ .... resolving it by the negative $\endgroup$ Jan 21, 2019 at 10:25
  • $\begingroup$ 0) Real cohomology, not $\Bbb Z$ cohomology. What I was saying above is that our algebra is induced up from some $\Bbb Z$-algebra, but we do not try to realize the $\Bbb Z$-algebra. This was mentioned so we don't use dumb relations like $x^3 = \pi y^2 + z$. 1) You always have a canonical isomorphism of algebras $H^0(X) \cong \Bbb R^{\# \text{components of } X}$. This is a product of copies of $\Bbb R$. The algebra $\Bbb C$ is not. 2) That paper does not answer your question in the negative, because it is not about real cohomology rings. (I am aware of the restrictions given by that paper.) $\endgroup$
    – user98602
    Jan 21, 2019 at 14:47
  • $\begingroup$ In fact that paper seems to give the answer in the positive for rings concentrated in even degrees. $\endgroup$
    – user98602
    Jan 21, 2019 at 15:04


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