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For $\mathbb{R} P^{n}$ we have,

$$ H^*( \mathbb{R} P^2 ; \mathbb{Z}_2) \cong \mathbb{Z}_2 [ \alpha ] / (\alpha^3)$$

where $|\alpha|=1$.

Let us assume there exists $X$ such that,

$$ H^*( X ; \mathbb{Z}_2) \cong \mathbb{Z}_2 [ \beta ] / (\beta^3) $$

but $| \beta |=2$.

Then the rings have the form, $a_0 + a_1 \alpha + a_2 \alpha^2$ and $b_0 + b_1 \beta + b_2 \beta^2$.

So as polynomials these rings look the same except for the condition given on the variables $\alpha$ and $\beta$.

How does the dimension of $\alpha$ and $\beta$ come into play when viewing the polynomial rings?

I think, naively, I would assume these space are isomorphic and that dimension doesn't add any strong information about the structure of the spaces.

David G. Stork
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Yuugi
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  • Whether $\alpha$ lives in even or odd degree affects whether $\alpha$ commutes or anticommutes with itself; in the latter case, $\alpha^2$ is $2$-torsion. There are also more sophisticated things you can say involving Steenrod operations. (Incidentally, the $X$ you want to exist does exist; you can take $\mathbb{CP}^2$.) – Qiaochu Yuan Dec 19 '15 at 00:15

1 Answers1

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Of course it says a lot about the structure of the spaces.

If $X, Y$ are homotopy equivalent topological spaces, $H^*(X)$ is isomorphic to $H^*(Y)$ not just as rings, but as graded rings.

Thus, if $H^*(X) \cong \Bbb Z[\beta]/(\beta^3)$ such that $|\beta| = 2$, $X$ cannot be homeomorphic to $\Bbb{RP}^2$, because $H^*(X)$ has no element of degree $1$, whereas $H^*(\Bbb{RP}^2)$ does - hence, even though $H^*(X)$ and $H^*(\Bbb{RP}^2)$ are isomorphic as rings, they are not isomorphic as graded rings. On a different note, as Qiaochu mentioned, such an $X$ would be $\Bbb {CP}^2$.

Balarka Sen
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