How we can construct 4d-dimensional connected closed oriented manifold whose has property $H^{0}=\mathbb{Q},\quad H^{2d}=\mathbb{Q},\quad H^{4d}=\mathbb{Q}.$ Few well known examples are $\mathbb{C}\mathbb{P}^{2}$ for $d=1$, $\mathbb{H}\mathbb{P}^{2}$ for $d=2$ and $\mathbb{O}\mathbb{P}^{2}$ for $d=4.$ If anybody has an idea or references for $d>4$ or other examples for $d=1,2,4,$ please share with me. Also, for $d=3.$
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I have some vague recollection that this is actually impossible for $d \ge 4$, at least integrally. Not sure about rationally. – Qiaochu Yuan Oct 21 '17 at 07:54
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1The Cayley plane is 16 dimensional, so I think you mean that it corresponds to $d=4$. – Tyrone Oct 21 '17 at 16:02
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sorry, I will correct my question. – Michael jordan Oct 22 '17 at 05:41
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2See e.g. Hatcher. Algebraic topology, section 4L (cor. 4L.10 etc). – Grigory M Oct 22 '17 at 11:42
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Over integer see paper of J. Eells, Jr. & N.H. Kuiper, Manifolds which are like projective planes.
King Khan
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Theorem A of Smooth manifolds with prescribed rational cohomology ring by Fowler and Su shows that the only dimensions in which a smooth such manifold can exist is $n = 2^a$ or $8(2^a + 2^b)$ where $a \neq b$. Note, while the authors are concerned with simply connected such manifolds, the conclusion does not require any hypothesis on the fundamental group.
In the earlier paper Rational analogues of projective planes, Su showed that there are infinitely many homeomorphism types of such manifolds in dimension $32$.
In a recent paper, On dimensions supporting a rational projective plane, Kennard and Su show that there are examples in dimensions $128$ and $256$ (Theorem A). Moreover, there are infinitely many dimensions of the form $2^a$ and infinitely many of the form $8(2^a + 2^b)$ with $a \neq b$ for which there is no such manifold which is simply connected.
Note, all known examples have dimension of the form $2^a$.
Michael Albanese
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