$\pi_{2p^kn - 1}(P^{2n+1}(p^r))$ contains a $\mathbb{Z}_{p^{r+1}}$ summand

Question

I am reading Neisendorfer's paper Samelson products and exponents of homotopy groups and related papers. I am stuck on theorem 14.1 on page 21, which says that there exists a $\mathbb{Z}_{p^{r+1}}$ summand in $\pi_{2p^kn - 1}(P^{2n+1}(p^r))$ for any $k \geq 1$ where $p > 3$ is a prime number and $P^{2n+1}(p^r) = D^{2n+1} \cup_{p^r} S^{2n}$.

The overall idea of the proof is to show that there exists an element in position $2p^kn - 2$ in the mod $p$ homotopy Bockstein spectral sequence for $\Omega P^{2n+1}(p^r)$ such that survives-until and then dies-in the $(r + 1)^\text{th}$ page. The first step is to find such an element and to show that it survives.

The method consists in comparing the mod $p$ homotopy Bockstein ss $E_\pi^*$ with the mod $p$ homology Bockstein ss $E_H^*$ via the mod $p$ Hurewicz map $\phi$. Both the spectral sequences are differential graded Lie algebras via, respectively, the Samelson product and the commutator of the Pontryagin product; moreover the Hurewicz map is a morphism of differential graded Lie algebras.

The mod $p$ homology Bockstein ss is simple. To see this, first notice that $H^*(P^{2n}(p^r); \mathbb{Z}) \cong \mathbb{Z}_{p^r}\langle \alpha \rangle$ where $\deg \alpha = 2n$, hence $H_*(P^{2n}(p^r); \mathbb{Z}) \cong \mathbb{Z}_{p^r}\langle\beta\rangle$ where $\deg \beta = 2n - 1$ and finally $H_*(P^{2n}(p^r); \mathbb{Z}_p) \cong \mathbb{Z}_p\langle u, v \rangle$ where $\deg u = 2n - 1$ and $\deg v = 2n$. Then we have $E^1_H \cong H_*(\Omega P^{2n+1}(p^r); \mathbb{Z}_p) \cong H_*(\Omega\Sigma P^{2n}(p^r); \mathbb{Z}_p) \cong \mathbb{Z}_p^\text{Alg} \langle u, v\rangle$ by the Bott-Samelson theorem. Moreover $\beta^s_H = 0$ for $s < r$ and $\beta^r_Hv = u$, so $E^1_H = \cdots = E^r_H$ and $E^{r+1}_H = \cdots = E^\infty_H \cong \mathbb{Z}_p$.

The mod $p$ homotopy Bockstein ss is not that simple, but surely we have at least two elements $\mu, \nu$ respectively in degrees $2n - 1$ and $2n$ such that $\phi\mu = u$ and $\phi\nu = v$ for the mod $p$ Hurewicz theorem. Moreover $\beta^s_\pi = 0$ for $s < r$ and $\beta^r_\pi v = u$, so $E^1_\pi = \cdots = E^r_\pi$.

Now the Lie algebra structures come to play. In any graded differential Lie algebra $(L, d)$ over a field of characteristic $p > 2$ any element $x \in L$ of degree $2n$ gives rise to some cycles $\sigma_k(x), \tau_k(x)$ in degrees $2p^kn - 2$ and $2p^kn - 1$ for any $k \in \mathbb{N}$. Moreover, in the universal enveloping algebra of $L$ holds $dx^{p^k} = \tau_k(x)$.

Specializing to our Bockstein sss, this gives us the cycles $\sigma_k(\nu)$, $\tau_k(\nu)$ in $E_\pi^*$ and $\sigma_k(v), \tau_k(v)$ in $E_H^*$. Moreover, since they are constructed via the Lie algebra structures, we have that $\phi\sigma_k(\nu) = \sigma_k(v)$ and $\phi\tau_k(\nu) = \tau_k(v)$. The claim is that $\sigma_k(\nu)$ survives until $E^{r+1}_\pi$ for any $k$ and $\tau_k(\nu)$ survives until $E^{r+1}_\pi$ for any $k > 1$.

How can I see the survival of $\sigma_k(\nu)$? Probably I am missing something stupid, but at the moment I cannot see it. Thanks in advance to anyone.

Answer 1

Firstly, let me thank Gustavo Granja, who independently contacted Joe Neisendorfer in order to make me have an answer. Secondly, let me thank Joe Neisendorfer for his time and answers.

Briefly, the answer to my question is in proposition 9.6.2 (page 296) in Neisendorfer's Algebraic methods in unstable homotopy theory. First notice that the mod $p$ Hurewicz map $\phi^r: E^r_\pi \to E^r_H$ has image in $PE^r_H$, the module of primitive elements of $E^r_H$. Hence it factors as $\phi^r: E^r_\pi \to PE^r_H \hookrightarrow E^r_H$ and we have $\phi^{r+1}: E^{r+1}_\pi \to HPE^r_H \to E^{r+1}_H$. It is true that $\phi^{r+1}[\sigma_k(\nu)] = [\sigma_k(v)]$ is trivial when regarded in $E^{r+1}_H$ but it is no more trivial when regarded in $HPE^r_H$. Hence $[\sigma_k(\nu)]$ is nontrivial as well, so $\sigma_k(\nu)$ survives until $E^{r+1}_\pi$.

The nontriviality of $[\sigma_k(v)]$ in $HPE^r_H$ has nothing to do with topology, but it is a purely algebraic fact. In fact it always holds $0 \neq [\sigma_k(v)] \in HP(dv, v)$ and this applies here because $HPE^r_H = HPT(dv, v) = HPUL(dv, v) = HP(dv, v)$.

I post here Neisendorfer's complete answers, edited accordingly with the notations used here. Any error is due to my transcription.

First answer

The secret to the question is to look at the fibration sequence $\Omega F^{2n+1}(p^r) \to \Omega P^{2n+1}(p^r) \to \Omega S^{2n+1}$. In mod p homology it looks like $T(\text{ad}^k(v)(u))_{k\geq 0} \to T(v,u) \to T(v)$, three tensor algebras in a row, the middle one on two generators is acyclic with the $r^\text{th}$ Bockstein differential, and the the left one is infinitely generated and decidedly not acyclic.

The important thing is that there are no powers $v^{p^k}$ on the left. They are in the middle and on the right. In the middle tensor algebra they make the Bockstein spectral sequence acyclic. Nothing can be seen after that. But the Bockstein spectral sequence on the left has all the information and it is visible.

The homology Bockstein spectral sequence does not become trivial on the left. In fact, $E^{r+1}$ there is a big polynomial tensor exterior algebra, that is, it is $\bigotimes E(\sigma_k(v)) \otimes P(\tau_k(v))$ for all $k \geq 1$.

Further analysis shows that the left has no torsion of order greater than $p^{r+1}$ and thus $\beta^{r+1} \tau_k(v) = \sigma_k(v)$ for all $k$. This is nontrivial. It needs a brilliant argument with chains. I can say "brilliant" since Moore thought of this and I don't think anyone else could have. Maybe Eilenberg could have.

The Hurewicz homomorphism shows that this is true in homotopy also. Of course, the elements $\text{ad}^k(\nu)(\mu)$ are relative Samelson products in homotopy. Full details are in the paper "Torsion in homotopy groups".

So the main idea is that the middle homology Bockstein spectral sequence vanishes since it is an acyclic tensor algebra (bad!) but the left one survives to the next stage since the $p^\text{th}$ powers are gone (good!). You can see what is happening there, but not in the middle! This was Moore's brilliant idea. It is the heart of the Cohen-Moore-Neisendorfer paper.

Second answer

Consider the mod $p$ Bockstein spectral sequences of $\Omega P^{2n+1}(p^r)$ and later those of $\Omega F^{2n+1}(p^r)$.

The Hurewicz map of both of these mod $p$ Bockstein spectral sequences $\phi: E^r_\pi \to E^r_H$ factors through the primitives $E^r_\pi \to PE^r_H \to E^r_H$.

Factoring through the primitives gives us a stronger representation than we would get by going all the way to homology. Taking Bockstein homology we get $E^{r+1}_\pi \to HPE^r_H \to E^{r+1}_H$ where $\sigma_k(\nu)$ has a nontrivial image in the middle for both spaces and $\tau_k(\nu)$ has a nontrivial image in the middle for the space $\Omega F^{2n+1}(p^r)$. Hence they are nontrivial on the left. The homology of the primitives gives us a nontrivial representation of mod $p$ homotopy at the $(r+1)^\text{th}$ stage.

Looking more closely at $\Omega F^{2n+1}(p^r)$, we can see that $\beta^{r+1} \tau_k(v) = \sigma_k(v)$ in mod $p$ homology. That is, this mod p homology Bockstein spectral sequence has two nontrivial differentials. In other words it has no torsion of order bigger than $p^{r+1}$ but it does have that. This is where we see the precise higher order torsion via the Hurewicz representation of the mod $p$ homotopy. Up to the kernel of the mod $p$ Hurewicz map $\beta^{r+1} \tau_k(\nu) = \sigma_k(\nu)$ in homotopy.

This is the way an integral class of order $p^{r+1}$ is represented in the Bockstein spectral sequence. Thus we see that $\sigma_k(\nu)$ represents integral torsion of order exactly $p^{r+1}$ and in that dimension.

For further details see section 9.6 of the book "Algebraic methods in unstable homotopy theory".