Can the Bousfield class of projective space be computed directly?

Question

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 $\langle \Sigma^\infty\mathbb{CP}^2\rangle=\langle \mathbb{S}\rangle$.

Answer 1

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.