17
$\begingroup$

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:

  • 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 $p$.

  • 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".

  • 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 $K(0) = H\mathbb Q$).

Improve this question
$\endgroup$
8
  • $\begingroup$ The question of $n>1$ at odd primes remains open, doesn't it? At the other question, you assert that Saal ruled it out, but I don't see it. mathoverflow.net/q/410802/4639 $\endgroup$
    – Ben Wieland
    Commented Dec 17, 2021 at 4:02
  • $\begingroup$ @BenWieland Oh wow you're right -- what I had in my head was the fact that $KO_\ast M \neq 0 \Rightarrow KU_\ast M \neq 0$, but this doesn't imply that $KU_\ast M_{(p)} \neq 0$... $\endgroup$
    – Tim Campion
    Commented Dec 17, 2021 at 5:27
  • $\begingroup$ @BenWieland In fact, I don't think we've even established that there do exist closed manifolds $M$ such that $M_{(p)}$ is of type 1 when $p$ is odd -- only that there exist such of type $>0$! $\endgroup$
    – Tim Campion
    Commented Dec 18, 2021 at 19:03
  • $\begingroup$ Although (and I could be wrong about this) -- if $p$ is odd and $M$ is a closed 4-manifold such that $M_{(p)}$ is of type $>0$, then I believe that the AHSS for $KO_{(p)}^\ast(\Sigma M_{(p)})$ collapses, so that $\widetilde{KO}^\ast(M_{(p)}) \neq 0$, so that $\widetilde{KU}^\ast(M_{(p)})\neq 0$, so that $M_{(p)}$ is of type 1. So the question, as you said, is whether $M$ can be chosen such that $M_{(p)}$ has type $\geq 2$... $\endgroup$
    – Tim Campion
    Commented Dec 18, 2021 at 19:13
  • $\begingroup$ (1) A really degenerate example is $\mathbb RP^2$ at an odd prime. It has type $\geqslant 2$, which means that to rule out the existence of type 2 requires actually using the existence of $p$-torsion. (2) I'm pretty sure the lens space example has type 1: Lens spaces themselves have $p$-torsion in $K$-theory; indeed, the $p$-torsion in homology assembles into $p^n$-torsion in $K$. (3) I think that in studying surgery theory, Sullivan proved that every oriented manifold is $L$-oriented and $L[1/2]=KO[1/2]$. I suspect one can use this to rule out high type for unorientable manifolds. $\endgroup$
    – Ben Wieland
    Commented Dec 18, 2021 at 20:19

2 Answers 2

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
8
  • 3
    $\begingroup$ Lens manifolds are orientable though. So this belief seems to clash with one of Tim's observations. $\endgroup$
    – Gregory Arone
    Commented Dec 13, 2021 at 18:28
  • 2
    $\begingroup$ Otherwise said, if $M$ is not oriented then $M$ has a nontrivial element in $KO$-theory, and so in $K(1)$. Oh gosh, this was simple. $\endgroup$
    – Nicholas Kuhn
    Commented Dec 13, 2021 at 18:28
  • $\begingroup$ @GregoryArone Thanks for catching that. I should be more careful with my beliefs in the future. $\endgroup$
    – Saal Hardali
    Commented Dec 13, 2021 at 18:35
  • 1
    $\begingroup$ Turns out the answer for $n=1$ and $p$ odd is also yes. $\endgroup$
    – Tim Campion
    Commented Dec 15, 2021 at 22:00
  • 2
    $\begingroup$ @SaalHardali Your last comment is off: the product of a projective plane and lens space will still have nonzero rational homology (just not in the top dimension). But Tim's comments points to the answer: one needs a little bit more, but not too more, cleverness to do this. $\endgroup$
    – Nicholas Kuhn
    Commented Dec 16, 2021 at 23:15
1
$\begingroup$

Here is my understanding of the situation at odd primes, following Ben Wieland's comment. In all the following, fix an odd prime $p$.

First, note that as in Saal's answer, every closed manifold $M$ with vanishing rational homology is nonorientable, and thus admits a nonvanishing 2-torsion class in $\widetilde{KO}^0(M)$, and thus $\widetilde{KU^\wedge_2}^\ast(M) \neq 0$. Thus $M$ is not literally $p$-local. The remaining question is what the type of the $p$-localization $M_{(p)}$ can be. Ben's comment shows that the type of $M_{(p)}$ can be arbitrary. Let me attempt to expand out Ben's construction.

  • Let $X$ be a finite complex such that $X_{(p)}$ is a type $n$ spectrum.

Embed $X$ in a big sphere $S^N$, and

  • Let $M$ be the boundary of a regular neighborhood of $X$.

Then $M$ is a closed manifold of dimension $N-1$. Assume that $N = 2d+1$ is odd, and

  • Let $M' = M \# \mathbb R\mathbb P^{N-1} = M \# \mathbb R\mathbb P^{2d}$ be the connected sum of $M$ with real projective space.

Claim: $M'_{(p)}$ is of type $n$.

Analysis of $\mathbb R\mathbb P^{2d}$:

Observe that $\widetilde{H}_\ast(\mathbb R \mathbb P^{2d};\mathbb Z_{(p)}) = 0$, so that $\mathbb R \mathbb P^{2d}_{(p)} = 0$. It follows that the attaching map for the top cell $S^{2d-1} \to \mathbb R\mathbb P^{2d} \setminus \ast$ induces an isomorphism in $K(m)_\ast$ for all $m$.

Analysis of $M$:

We have a homotopy pushout

$$(\ast) \quad S^N \simeq X \cup_M \Sigma^N (DX)$$

(Here $DX$ is the Spanier-Whitehead dual of $X$; $\Sigma^N DX$ is modeled by the complement of $X$ in $S^N$; note that $DX_{(p)}$, like $X$ has type $n$.)

The Mayer-Vietoris square for $(\ast)$ shows that for $m < n$, we have $K(m) \wedge M = \Sigma^{N-1} K(m)$ . For $m \geq n$, since $K(m)_\ast S^N$ is 2-dimensional over $K(m)_\ast$ and $K(m)_\ast (X \amalg DX)$ is at least 4-dimensional; it follows that $\widetilde{K(m)}_\ast M \neq 0$ for $m \geq n$. Thus for all $m$, we have $\widetilde{K(m)}_\ast M = \Sigma^{N-1} K(m)_\ast \oplus L(m)$, where $L(m) = 0$ iff $m < n$.

Into $(\ast)$, we may include a homotopy pushout diagram $(S^N\setminus \ast) \simeq (X \setminus \ast) \cup_{M \setminus \ast} ((\Sigma^N (DX)) \setminus \ast)$. The cofiber is a homotopy pushout diagram $S^N = \ast \cup_{S^{N-1}} \ast$. From this, we may conclude that the map $M \to S^{N-1}$ collapsing all but the top cell (which is the double cofiber of the attaching map $S^{N-2} \to M \setminus \ast$) kills exactly $L(m)$ in $\widetilde{K(m)_\ast}$, leaving $\widetilde{K(m)}_\ast (M \setminus \ast) = L(m)$.

Analysis of the connected sum:

Now from the homotopy pushout $M' \simeq (M \setminus \ast) \cup_{S^{N-2}} (\mathbb R \mathbb P^{N-1} \setminus \ast)$, we conclude that $\widetilde{K(m)}_\ast (M') = L(m)$, which vanishes exactly when $m < n$. Thus $M'_{(p)}$ has type $n$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .