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Question: Let $p$ be an odd prime. Does there exist a closed manifold $M$ with $\widetilde H^\ast(M; \mathbb Q) = 0$ but $\widetilde H^\ast(M; \mathbb F_p) \neq 0$?

When $p = 2$, an example is given by $\mathbb R \mathbb P^2$.

After discussion, this question turns out to be equivalent to this other question. That is, over at that question Saal Hardali explained that if $M$ is a closed manifold, then the chromatic type of $M$ at a prime $p$ is either 0 or 1. Both possibilities are realized at $p=2$; the question is whether chromatic type 1 is realized at odd primes. Chromatic type 0 just means having nonvanishing rational (co)homology. So the question is whether there exists a closed manifold $M$ which is rationally contractible but whose $p$-localization is nontrivial for an odd $p$. This turned out to be mistaken, thanks to Ben Wieland for pointing this out at the other question.

Side Question: When $p=2$, what are some other examples of $M$ with $\widetilde H^\ast(M;\mathbb Q) = 0$ but $\widetilde H^\ast(M;\mathbb F_2) \neq 0$ besides $M = \mathbb R \mathbb P^{2n}$ and products thereof?

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    $\begingroup$ They are manifolds, but rationally are not a point (they are orientable) $\endgroup$
    – Thomas Rot
    Commented Dec 15, 2021 at 16:24
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    $\begingroup$ @ThomasRot Non-orientable odd-dimensional closed manifolds cannot be rational homology balls, because every odd-dimensional closed manifold has zero Euler characteristic (use Poincare duality on the oriented double cover.) So your second approach sounds more promising. $\endgroup$
    – mme
    Commented Dec 15, 2021 at 16:30
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    $\begingroup$ What if you take $\mathbb{RP}^{2n} \# M$, where $M$ is an (orientable) rational homology sphere? $\endgroup$
    – Marco Golla
    Commented Dec 15, 2021 at 17:43
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    $\begingroup$ @TimCampion: Marco Golla's xample is an answer to the main question I think. $\endgroup$
    – Thomas Rot
    Commented Dec 15, 2021 at 18:11
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    $\begingroup$ I'll try to write up an answer later, but it boils down to Mayer–Vietoris: connected summing with a rational homology sphere doesn't do anything to the rational homology, but it keeps all the p-torsion if there was any to begin with. $\endgroup$
    – Marco Golla
    Commented Dec 15, 2021 at 18:14

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As mme noted in the comments, such examples cannot exist in odd dimensions, for Euler characteristic reasons. They can't exist in dimension 2 either, by classification. I claim that in all other dimensions $2n > 2$ we have (plenty of) examples.

Let $N$ be a rational homology $2n$-sphere, that is a $2n$-manifold with $H_*(N; \mathbb{Q}) = H_*(S^{2n}; \mathbb{Q})$. For every prime $p$ there exists a rational homology $2n$-sphere with $\dim_{\mathbb{F}_p} H_*(N;\mathbb{F}_p) > 2$. For instance, you can take a spun lens space (any spun rational homology $2n-1$-sphere would do).

Now, the integral homology of $M = \mathbb{RP}^{2n} \# N$ splits as a direct sum of that of the two summands in all dimensions strictly between 0 and 2n, and it vanishes in dimension 2n (because $M$ is non-orientable) and it is $\mathbb{Z}$ in dimension 0 (because $M$ is connected). (This is Exercise 6 in Section 3.3 of Hatcher's Algebraic topology.) That is, $M$ is a rational homology ball, and its homology has as much $p$-torsion as that of $N$.

For the side question, if we choose $N$ to have no 2-torsion in its homology (e.g. spinning an odd lens space should do the trick), this gives plenty of examples.

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    $\begingroup$ In case it is helpful for Tim, let me remind the definition of "spinning an n-manifold": if $M$ is closed of dimension $n$, puncture it by deleting the interior of a fixed embedding of the smooth n-ball: this is a compact manifold $P(M) = M \setminus D^n$ with parameterized boundary $S^{n-1}$ . Now consider $P(M) \times S^1$, which has boundary equipped with a diffeomorphism to $S^{n-1} \times S^1$. Fill in this boundary by gluing in $S^{n-1} \times D^2$. Said another way, take $M \times S^1$ and do surgery along $p \times S^1$. The homology is obtained by a Mayer-Vietoris computation. $\endgroup$
    – mme
    Commented Dec 15, 2021 at 19:44
  • $\begingroup$ Try picturing this for M = S^1 and you'll see that this gives the picture of the 2-sphere by taking an arc from north pole to south pole and spinning it 360 degrees. $\endgroup$
    – mme
    Commented Dec 15, 2021 at 19:51
  • $\begingroup$ Thanks! I must admit I am currently remedially reminding myself what the homology of a nonorientable manifold looks like! $\endgroup$
    – Tim Campion
    Commented Dec 15, 2021 at 22:01

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