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Suppose we have a polynomial, is it always the Poincare polynomial of some manifold? I guess the answer is no, but don't know any example. Even more, if we have a ring, is it the cohomology ring of some manifold. We can consider the real cofficient case. For example, $\mathbb{C}[x]/(x^3=1)$ is the cohomology ring of $\mathbb{CP}^2$. Is there any manifold has cohomology $\mathbb{C}[x]/(-x^3=1)$?

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Manifolds satisfy poincare duality so this limits the poincare polynomial. Im not sure if this is the only limitation. –  Eric O. Korman Apr 14 '12 at 2:19
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If the cohomology ring of a closed manifold is of the form $k[x]/(x^n)$ then $n$ can only be $1$, $2$, $4$ or $8$. That gives examples of rings which do not occur. –  Mariano Suárez-Alvarez Apr 14 '12 at 3:20
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Do you mean $x^3 = 0$ in the cohomology ring of $\mathbb{C}P^2$? @Mariano: I think you mean the degree of $x$, there are no limitations on $n$ ($k = \mathbb{Z}/2\mathbb{Z}$ and $\mathbb{R}P^n$). Further if the degree of $x$ is $8$, then $n$ is at most 3. –  Jason DeVito Apr 14 '12 at 3:22
    
Yeah: that's what I meant :) –  Mariano Suárez-Alvarez Apr 14 '12 at 3:24
    
If you allow the manifold to have boundary, there's no limitations on the Poincare polynomial. –  Ryan Budney Apr 14 '12 at 5:08

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