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Consider the usual cell structure on $\mathbb C \mathbb P^\infty$. The skeleta are the $\mathbb C \mathbb P^n$’s, and there is one cell in each even degree. So we have cofiber sequences $S^{2n+1} \to \mathbb C \mathbb P^n \to \mathbb C \mathbb P^{n+1}$.

Each of these attaching maps $S^{2n+1} \to \mathbb C \mathbb P^n$ is at least stably nilpotent. This follows easily from the nilpotence theorem: $MU_\ast(\mathbb C \mathbb P^n) \to MU_\ast(\mathbb C \mathbb P^{n+1})$ is injective because $MU$ is complex-oriented, so the attaching map is zero on $MU_\ast$; by the nilpotence theorem this implies that the attaching map is stably nilpotent (some suspension of some smash power of the map is zero).

Question 1: Are the attaching maps $S^{2n+1} \to \mathbb C \mathbb P^n$ unstably nilpotent?

Question 2: Is there some way to see that the attaching maps $S^{2n+1} \to \mathbb C \mathbb P^n$ are stably nilpotent without invoking the nilpotence theorem?

Question 3: How many smash powers do we need to make $S^{2n+1} \to \mathbb C \mathbb P^n$ become stably null? How many suspensions?

Notes:

  • When $n = 1$, the attaching map $S^3 \to \mathbb C\mathbb P^1$ is the Hopf fibration $\eta$. That $\eta$ is stably nilpotent follows from Nishida nilpotence because it’s a positive-degree class in the stable homotopy groups of the sphere.

  • I’m actually most interested in Question 2. The other questions are just there because they are more precise questions whose answer would seem to require thinking past the nilpotence theorem :)

  • I should probably observe somewhere here that the map in question $S^{2n+1} \to \mathbb C \mathbb P^n$ is the quotient map usually used to define $\mathbb C \mathbb P^n$. It’s a fiber bundle with fiber $S^1$. I’m not sure if these facts are useful...

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    $\begingroup$ Suppose $f: S^k \to X$ is a map, and factor $f^{\wedge(n+m)} = (f^{\wedge n} \wedge id^{\wedge m}) \circ (\Sigma^{nk} f^{\wedge m})$. If $f$ is stably smash-nilpotent, then you can pick $m$ sufficiently large so that $f^{\wedge m}$ is stably trivial, and then if $k > 0$ you can find $n$ sufficiently large so that $\Sigma^{nk} f^{\wedge m}$ is in the stable range. So stably smash-nilpotent does imply unstably smash-nilpotent, but possibly with a different exponent. $\endgroup$ Commented May 23 at 12:21
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    $\begingroup$ (In fact you can always take $f^{\wedge (m+3)}$, and if $X$ is connected you can take $f^{\wedge (m+1)}$.) $\endgroup$ Commented May 23 at 12:31

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