# Can the Bousfield class of projective space be computed directly?

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) $$C(\eta^{'s and
2. $$\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 $$\langle \Sigma^\infty\mathbb{CP}^2\rangle=\langle \mathbb{S}\rangle$$.

• 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. Apr 28 at 1:57

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.
• 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 $\mathbb{C}P^n \wedge D\mathbb{C}P^n$.) Apr 28 at 15:27
• @DylanWilson see my correction to the answer - the relevant trace here is the trace on the reduced homology, which is zero mod $p$. Apr 28 at 16:03
• 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. Apr 28 at 16:17
• @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 $p(p-1)$. Apr 28 at 16:55
• @JamesCameron I think that $F\to S$ is not null in the case $X=\mathbb{C}P^2$ with $p=2$. Apr 28 at 17:48