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Recall that the Bousfield class of a spectrum $E$, written $\langle E\rangle$, is the class of spectra $X$ such that $X\wedge E$ is not contractible. For example the Bousfield class of any of the spheres $\mathbb{S}^k$ is the class of all noncontractible spectra.

Now take the complex projective space $\mathbb{CP}^n$, choose a basepoint and consider the suspension spectrum $\Sigma^\infty\mathbb{CP}^n$. I think it follows from the thick subcategory theorem of Hopkins-Smith that the Bousfield class $\langle \Sigma^\infty\mathbb{CP}^n\rangle$ is equal to $\langle \mathbb{S}^k\rangle$ (it's a "type 0" finite spectrum). But that theorem seems rather high-powered for the job it's doing here.

So my $\textbf{question}$ is: can the the Bousfield class $\langle \Sigma^\infty\mathbb{CP}^n\rangle$ be computed directly?

For example for $n=1$, $\Sigma^\infty\mathbb{CP}^1\simeq \mathbb{S}^2$ so things are looking good. For $n=2$, since $\Sigma^\infty\mathbb{CP}^2$ is the cone $C(\eta)$ of the Hopf map $\eta:\mathbb{S}^3\rightarrow\mathbb{S}^2$ one can use the facts that

  1. $C(\eta^k)$ can be gotten as the cofiber of (suspensions of) $C(\eta^{<k})$'s and
  2. $\eta^4$ is null

to deduce that if $X\wedge \Sigma^\infty\mathbb{CP}^2$ is contractible then so is $X\wedge (\mathbb{S}\vee \mathbb{S}^5)$, so $X$ is contractibe, and hence $\langle \Sigma^\infty\mathbb{CP}^2\rangle=\langle \mathbb{S}\rangle$.

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    $\begingroup$ Hopkins' and Smith's proof of their thick subcategory theorem uses arguments not so different than yours. Note that you have used the nilpotence of $\eta$ in your little argument for $\mathbb CP^2$; they, of course, have the Nilpotence theorem to play with. $\endgroup$ Apr 28 at 1:57

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Here is an easy argument which sometimes works. I have updated and extended it to incorporate comments from Dylan Wilson and Maxime Ramzi.

For any finite spectrum $X$, we have (co)unit maps $S\xrightarrow{\eta}DX\wedge X\xrightarrow{\epsilon}S$. Given any $f\colon X\to X$ we can form the composite $$ \Lambda(f) = (S \xrightarrow{\eta} DX\wedge X \xrightarrow{1\wedge f} DX\wedge X \xrightarrow{\epsilon} S) \in [S,S] = \mathbb{Z}. $$ This can be computed by the usual Lefschetz formula $$ \Lambda(f) = \sum_{n\in\mathbb{Z}} (-1)^n \text{trace}(f_*\colon H_n(X;\mathbb{Q}) \to H_n(X;\mathbb{Q})) $$ If there exists $f$ with $\Lambda(f)\neq 0\pmod{p}$, then $S$ is $p$-locally a retract of $DX\wedge X$, so the Bousfield classes $\langle X\rangle$ and $\langle S\rangle$ are equal.

Now consider the case where $X=\Sigma^\infty\mathbb{C}P^n$ for some $n>0$. In the unstable category, cellular approximation gives $$ [\mathbb{C}P^n,\mathbb{C}P^n] = [\mathbb{C}P^n,\mathbb{C}P^\infty] = [\mathbb{C}P^n,K(\mathbb{Z},2)] = H^2(\mathbb{C}P^n) = \mathbb{Z}. $$ If we write $f_q$ for the map corresponding to $q\in\mathbb{Z}$, then the effect on the cohomology ring $$ H^*(\mathbb{C}P^n)=\mathbb{Z}[x]/x^{n+1} = \mathbb{Z}\{1,x,\dotsc,x^n\} $$ is given by $f_q^*(x^k)=q^kx^k$. Put $\lambda(n,q)=\Lambda(\Sigma^\infty f_q)$. When calculating this, it is the trace of $f_q^*$ on reduced cohomology that is relevant, giving $$ \lambda(n,q) = q + q^2 + \dotsb + q^n = q(q^n-1)/(q-1). $$

  • If $p\nmid n$ then $\lambda(n,1)\neq 0\pmod{p}$
  • Now suppose that $p\mid n$ but $p-1\nmid n$ (so $p>2$). Let $q$ be a primitive root mod $p$, so in particular $q-1$ is invertible mod $p$, so there is no problem with interpreting the formula $\lambda(n,q)=q(q^n-1)/(q-1)$ modulo $p$. We then find that $\lambda(n,q)\neq 0\pmod{p}$
  • Now suppose that $p(p-1)\mid n$. By considering the cases $q=0\pmod{p}$, $q=1\pmod{p}$ and $q\neq 0,1\pmod{p}$ separately, we find that $\lambda(n,q)=0\pmod{p}$ for all $q$.

In conclusion:

  • If $p(p-1)\not\mid n$ then the above method proves that $\langle\mathbb{C}P^n\rangle=\langle S\rangle$
  • If $p(p-1)\mid n$ then we still know from nilpotence theory that $\langle\mathbb{C}P^n\rangle=\langle S\rangle$, but the above method does not suffice to prove it.
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    $\begingroup$ doesn't the same argument (by considering the Lefschetz trace) prove the general case? (the map $\mathbb{C}P^n \to \mathbb{C}P^n$ with Euler class $p$ times the generator should have trace equal to 1 mod $p$, giving the sphere as a retract of $\mathbb{C}P^n \wedge D\mathbb{C}P^n$.) $\endgroup$ Apr 28 at 15:27
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    $\begingroup$ @DylanWilson see my correction to the answer - the relevant trace here is the trace on the reduced homology, which is zero mod $p$. $\endgroup$ Apr 28 at 16:03
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    $\begingroup$ Isn't it possible to use a map with Euler class $q$ then ? I think in this case the reduced trace is $\sum_{1\leq i\leq n} q^i = q \frac{q^n-1}{q-1}$. This nonzero (hence invertible) mod $p$ if and only if both $q$ and $q^n-1$ are. So we just pick a $q$ not divisible by $p$, and such that $q^n$ is not $1$ mod $p$. If all nonzero $q$'s satisfy $q^n = 1$ mod $p$, then $(p-1)$ divides $n$. So if $(p-1) \nmid n$ or $p\nmid n$, this can be made to work. $\endgroup$ Apr 28 at 16:17
  • $\begingroup$ @MaximeRamzi I think you are right, I went through that analysis previously but miscalculated slightly. So that just leaves the case where $n$ is divisible by $p(p-1)$. $\endgroup$ Apr 28 at 16:55
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    $\begingroup$ @JamesCameron I think that $F\to S$ is not null in the case $X=\mathbb{C}P^2$ with $p=2$. $\endgroup$ Apr 28 at 17:48

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