# Constructing a space with prescribed cohomology ring

The most general way I can formulate my question is the following:

Question 1: Given a Gorenstein quotient ring $S$ of a polynomial ring over a field $K$, can one construct a (topological) space $X$ such that the (even degree part of the) singular cohomology ring of $X$ with coefficients in $K$ is isomorphic to $S$?

Edit: as mentioned in the comments, the answer depends on the grading of the variables. I would be most interested in a uniform grading (i.e all variables have the same degree d, without restriction on the value of d).

For the specific cases I have in mind, $S$ is a quotient of a polynomial ring over $\mathbb{C}$ by an ideal generated by monomials and binomials. Below is an example of such a ring $S$ that I would like to realize as the cohomology ring of some space:

$$S = \mathbb{C}[x_1,\ldots,x_7]/(x_7^2, x_3x_7-x_4x_7, x_2x_7-x_5x_7, x_1x_7-x_6x_7, x_6^2, x_3x_6-x_5x_6, x_2x_6-x_4x_6, x_1x_6-x_6x_7, x_5^2, x_3x_5-x_5x_6, x_2x_5-x_5x_7, x_1x_5-x_4x_5, x_4^2, x_3x_4-x_4x_7, x_2x_4-x_4x_6, x_1x_4-x_4x_5, x_3^2, x_1x_3-x_2x_3, x_2^2, x_1x_2-x_2x_3, x_1^2).$$

The ring $S$ above is Gorenstein and all rings that I am considering for my question are also Gorenstein (i.e satisfy Poincare duality).

Edit 2: I will put this here rather than leaving the additional question in the comments because it goes deeper into what I was really interested in.

Question 2: Since the answer to Question 1 seems to be affirmative, can the space X be chosen to be "nice" (e.g. compact manifold) under the assumption that $S$ (the cohomology ring of $X$) satisfies Poincare duality?

• I wonder if you want to specify the grading of $x_i$'s. It is important as the ring $\mathbb{Z}/2[x]/x^2$ can be realized as space if $|x| \in \lbrace0,1,2,4,8\rbrace$. Otherwise they dont. This is a consequence of 'the Hopf invariant one problem'. – Prasit Aug 12 '14 at 15:19
• @Prasit: I am really puzzled by your comment. Would not this ring with $|x|=d$ be the cohomology ring of $S^d$? [The question you ask the OP totally makes sense though.] – Vladimir Dotsenko Aug 12 '14 at 15:57
• To the OP: this ring you suggest as an example is a very nice one! How does it occur in your work? – Vladimir Dotsenko Aug 12 '14 at 16:00
• Oh I made some mistakes above comment. I meant $\mathbb{Z}/2[x]/x^3$. That is there is $RP^2$, $CP^2$, $HP^2$ and $OP^2$ but the sequence stops here. – Prasit Aug 12 '14 at 16:16
• @Prasit: thank you for your important remark concerning grading. I have edited the question to clarify this. – Alexandra Seceleanu Aug 12 '14 at 21:17

I did not understand exactly your question. What is important here, is the ring $k$. If $k=\mathbf{Z}$ then the problem you are asking for is hard, if $k$ is any commutative ring then the problem is very hard. If $k$ is a field of characteristic 0 or $\mathbf{F}_{p}$ then there is always a solution. When the characteristic is 0, the problem is solved by Sullivan (under finiteness condition + simply connected+ condition on generalized Steenrod 0-operation cf comments). I will try to explain the situation over $\mathbf{F}_{p}$. Detnote by $\mathsf{E}_{\overline{\mathbf{F}}_{p}}$ the model category of $E_{\infty}$- differential graded $\overline{\mathbf{F}}_{p}$-algebras (in positive degree). Mandell's theorem says the following Quillen adjunction : $$C^{\ast}(-,\overline{\mathbf{F}}_{p}):\mathsf{sSet}^{op} \longrightarrow \mathsf{E}_{\overline{\mathbf{F}}_{p}}: Map_{\mathsf{E}_{\overline{\mathbf{F}}_{p}}}(-,\overline{\mathbf{F}}_{p})$$ which induces an equivalence between $\infty$-subcategories (I will not go to the details here). Suppose $S$ is graded commutative $\mathbf{F}_{p}$-algebra, such that in each degree $S_{n}$ is finite dimensional $\mathbf{F}_{p}$-module, and $S_{0}=\mathbf{F}_{p}$ and $S_{1}=0$. In particular $S\otimes\overline{\mathbf{F}}_{p}$ is an object of $\mathsf{E}_{\overline{\mathbf{F}}_{p}}$. The space you are looking for is given by the derived $\mathbb{R}Map_{\mathsf{E}_{\overline{\mathbf{F}}_{p}}}(S\otimes\overline{\mathbf{F}}_{p} ,\overline{\mathbf{F}}_{p})\simeq \mathbb{R}Map_{\mathsf{E}_{\mathbf{F}_{p}}}(S ,\overline{\mathbf{F}}_{p}):=X$. Applying Mendell's Theorem, we obtain that $$C^{\ast}(X,\overline{\mathbf{F}}_{p})\simeq S\otimes\overline{\mathbf{F}}_{p}.$$ By definition, the cohomology of $X$ with coefficients in $\overline{\mathbf{F}}_{p}$ is $S\otimes \overline{\mathbf{F}}_{p}$ since $S$ is formal by definition. Then you conclude for $\mathbf{F}_{p}$ coefficients.

PS: the functor $C^{\ast}(-,k)$ is the functor of cochain complex.

• For me K is always field. In my specific example $K=\mathbb{C}$. – Alexandra Seceleanu Aug 12 '14 at 21:24
• Then, Sullivan theorem is enough! the only thing you need to check is that $S_{0}=k$ and $S_{1}=0$ + finiteness condition (which is true in you case if I'm not wrong). Then there is always space $X$ such that $H^{\ast}(X,k)\simeq S$ – Ilias A. Aug 12 '14 at 21:28
• You mean if we can choose $X$ to be a compact manifold (since $S$ verifies the Poincare duality? I don't know a reference for that, but I'm pretty sure there is some papers on the subject. May be someone more qualified will give an answer to your comment. – Ilias A. Aug 12 '14 at 21:38
• @Fedotov Could you elaborate on how you go from $\overline{\mathbf F}_p$ coefficients to $\mathbf F_p$ coefficients? – Dan Petersen Aug 12 '14 at 21:38
• @Fedotov This answer is wrong. It is not true that any $\mathbf F_p$-algebra occurs as the mod p cohomology of a space. Look at the introduction to the paper that Matthias Wendt linked to, and the references there. – Dan Petersen Aug 16 '14 at 5:30

As Matthias Wendt says, the general problem is hard. However, if $K$ has characteristic zero, the problem is completely solved by Sullivan's approach to rational homotopy theory: such a topological space always exists. See his paper "Infinitesimal computations in topology". Briefly you should take your algebra $S$, consider it as a commutative differential graded algebra and choose a cofibrant replacement, i.e. a Sullivan model. Then plug this model into the "spatial realization" functor to get a topological space whose Sullivan cochains are quasi-isomorphic to the algebra you started with. In particular the cohomology ring of this space is $S$.

This would work more generally also if $S$ had elements of odd degree - but for simplicity I think you should assume that it vanishes in degree one.

As mentioned in Prasit's comment, the grading needs to be fixed. In general, the question of realizing graded rings as cohomology rings is a fairly difficult one.

In the special case where $S$ is a polynomial ring, this is called Steenrod's problem. This was resolved only very recently, see the paper "The Steenrod problem of realizing polynomial cohomology rings" by Andersen and Grodal, link to arXiv version, link to published version.

If the field $K$ of coefficients is a finite field, you also have an action of the Steenrod algebra on the cohomology of a space. Depending on your problem, you might also want to fix the Steenrod module structure on the cohomology ring you are trying to realize. In that case, there is an extensive literature on realizability of Steenrod modules, see e.g.

• N.J. Kuhn. On Topologically Realizing Modules Over the Steenrod Algebra. Annals of Mathematics (2), Vol. 141, No. 2 (1995), pp. 321-347

and references in there. (The Hopf invariant one problem mentioned in Prasit's comment and the Steenrod problem are mentioned in the introduction.)

To answer your second question, you can apply the Sullivan-Barge realization theorem. Here is a description and references. The theorem is usually stated over $\mathbb{Q}$, but I see no reason why it shouldn't apply also over $\mathbb{C}$.

In the easiest case, if the formal dimension $n$ of your ring $S$ is not a multiple of $4$, then yes, it can be realized as the complex cohomology ring of some closed $n$-manifold.