Let $p$ be a prime and $n \in \mathbb N$. Does there exist a closed manifold which is of type $n$ after $p$-localization?
When $n= 0$ the answer is yes. When $p = 2$ and $n = 1$ we can take $\mathbb R \mathbb P ^2$.
Other than that, I'm not sure.
Notes:
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Recall that a finite CW complex $X$ is said to be of type $n$ if $\widetilde{K(n)}_\ast X \neq 0$ but $\widetilde{K(m)}_\ast X = 0$ for $m < n$, where $K(n)$ is the $n$th Morava $K$-theory at the prime $p$.
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Recall the thick subcategory theorem tells us that for every $n$ (and every $p$) there exists a finite CW complex which is of type $n$ after $p$-localization, and conversely that every finite CW complex is of type $n$ for some $n$ after $p$-localization. The question is whether "finite CW complex" can be upgraded to "closed manifold".
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When $n \geq 1$, a closed manifold of type $n$ can't be orientable (since if it's orientable, then by Poincare duality its reduced rational homology is nonvanishing, and $K(0) = H\mathbb Q$).
Best Answer
After discussing this with Tim we came up with the following answer:
The first steifel whiteny class $\omega_1$ of $M$ can be written as the following composition:
$$M \to BO(n) \to BO \to BAut(\mathbb{S}) \to BAut(\mathbb{Z}) \simeq B\mathbb{Z}/2$$
But if $M$ is of type $\ge 2$ then $[M,BO]\simeq [\Sigma^\infty M, bo] \simeq 0$ since $bo$ is of height $\le 1$. So $M$ must be orientable in cotradiction with the third point.
Conclusion: All closed smooth manifolds are of type $\le 1$.
Oh and I believe that at odd primes, type $1$ complexes can be realized by Lens manifolds.Here I was uncareful. This is wrong as it conflicts with the Tim's third point as was pointed out by Gregory Arone in the comments.