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

  • 2
    $\begingroup$ 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

1 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.
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.