For a given spectra $X$ we have $X/p$ defined as the cofiber $X \to^p X$ where the map is basically defined via defining it on the sphere spectrum $\mathbb{S}$! To define $\cdot p$ on the sphere spectrum we just need to define it on $\mathbb{S^1}$ where it's $z \to z^p$ in the complex numbers (and then via suspension we define it for higher spheres).

We thus obtain $\mathbb{S}/p$ and we can consider $$s_p: \mathbb{S}/p \to^p \mathbb{S}/p$$

I'm watching higher algebra on youtube higher algebra- padic spectra where it is claimed that $s_p$ is nullhomotopic for $p \neq 2$, but is not nullhomotopic for $p = 2$, while $s_p * s_p$ should always be nullhomotopic.

I'm trying to verify those claims. Instead of computing the pushout of $\cdot p$ on the sphere spectrum which is $\Sigma^{\infty}(\mathbb{S}^0)$, since $\Sigma^{\infty}$ commutes with colimits we can do the colimit in topological spaces. Because $\mathbb{S}^0$ is scarily degenerate I'm going to use $\mathbb{S}^1$

We can check that the cofiber $\mathbb{S}^1 \to^2 \mathbb{S}^1$ is $\mathbb{R}P^2$, and more generally it's the disk where you glued the boundary $p$ times instead of just twice, we'll denote this space as $\mathbb{D}_p$.

Now let us discuss $\cdot p$ on $\mathbb{D}_p$ and whether it's nullhomotopic. For $p=2$ it certainly is; since $\cdot 2$ vanishes on $\pi_1(\mathbb{D}_2)$ we can lift $\cdot 2 : \mathbb{D}_2 \to \mathbb{D}_2$ to $\cdot 2 : \mathbb{D}_2 \to \mathbb{S^2}$ where it lands on the upper hemisphere and so is contractible. More directly, every $\cdot p$ on $\mathbb{D}_p$ is nullhomotopic via squeezing towards the center of the disk along time.

My proof must be wrong because I was told there is a problem for $p=2$, that there one needs to multiply by $4$. Where was I wrong? What did I miss?

  • 1
    $\begingroup$ I don't think you're in the stable range for your calculation yet. $\endgroup$
    – JHF
    Jun 10 at 18:01
  • $\begingroup$ @JHF but I showed that it's ALREADY nullhomotopic. How can stabiliziation ruin it? Can't I suspension loop space the proof of the nullhomotopy? $\endgroup$
    – user135743
    Jun 11 at 5:45
  • $\begingroup$ I think the point is that multiplication by $p$ map is not defined until you suspend. $\endgroup$
    – JHF
    Jun 13 at 18:07
  • $\begingroup$ @JHF I thought it was defined as I mentioned, how else would you define it after suspenion? $\endgroup$
    – user135743
    Jun 13 at 18:22
  • $\begingroup$ How are you defining the map $\cdot p$ on $\mathbb{D}_p$? $\endgroup$ Jun 13 at 18:28


You must log in to answer this question.

Browse other questions tagged .