Conceptual (operadic?) reason for the generalized EHP fiber sequence $J_{q-1}(S^{2n}) \to J S^{2n} \to JS^{2nq}$?

Question

Let $q$ be a prime and $q=p^r$ a power. Then there is a $p$-local fiber sequence from the $q-1$st stage of the James construction on $S^{2n}$, to $J(S^{2n}) = \Omega \Sigma S^{2n}$, to $J(S^{2nq}) = \Omega \Sigma S^{2nq}$. Here the first map is the natural inclusion map for the James filtration, and the second map is the James-Hopf map, adjoint (under $\Sigma \dashv \Omega$) to the projection coming from the Snaith splitting $\Sigma J(X) = \Sigma \vee_{q\geq 1} X^{\wedge q}$ (with $X = S^{2n}$). The EHP sequence arises when $p=q=2$.

What I'd like to understand is the fact that this is a fiber sequence. The proof I'm familiar with (from lectures 3-5 of notes by Akhil Mathew on a course by Mike Hopkins) can be seen by computing what it does on $\mathbb F_p$ homology and considering the Serre spectral sequence. The homology picture is rather suggestive:

• Recall that the James construction is the free $E_1$-space on an $E_0$-space (i.e. a pointed space), and the Snaith splitting tells us that correspondingly $H_\ast(J(X))$ is the free associative algebra on the augmented vector space $H_\ast(X)$: $H_\ast(J(X)) = T \tilde H_\ast(X)$.

• Moreover, the James filtration corresponds to the filtration of the tensor algebra by tensor rank: $H_\ast(J_k(X)) = T_{\leq k} \tilde H_\ast(X)$.

• The James-Hopf map is rather complicated, but one works out using the comultiplication (and computing some mod $p$ multinomials) that $p$-locally, and when $X = S^{2n}$, it looks additively like the obvious projection $H_\ast(J(S^{2n})) = \mathbb F_p[x_{2n}] \to \mathbb F_p[y_{2nq}] = H_\ast(J(S^{2nq}))$.

So the fiber sequence boils down to the additive decomposition $\mathbb F_p[x_{2n}] \cong (\mathbb F_p[x_{2n}] / x_{2n}^q) \otimes \mathbb F_p[x_{2n}^q]$. This closely mirrors the universal properties of $J_{q-1} S^{2n}$ and $J(S^{2nq})$. So in some sense,

The fiber sequence $J_{q-1} S^{2n} \to J S^{2n} \to J S^{2nq}$ says something about decomposing operations in the $E_1$ operad into operations of arity $<q$ and operations of arity divisible by $q$.

But what exactly it says, I'm not sure. So I suppose my question is:

Questions:

1. Is there a conceptual explanation for the $p$-local fiber sequence $J_{q-1} S^{2n} \to J(S^{2n}) \to J(S^{2nq})$?

2. In particular, is there such an explanation which either circumvents or else better explains the seeming contingency that the James-Hopf map is a $H_\ast(-,\mathbb F_p)$-surjection?

3. As an alternative desideratum, is there such an explanation which follows from operadic considerations?

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To flesh out this operadic perspective a bit more, here's a description of a "James-Hopf sequence"in a more general setting: one might formalize (3) above as asking under what conditions the following sequence

$$J^O_{q-1}(X) \to J^O(X) \to J^O(O(q)_+ \wedge_{\Sigma_q} X^{\wedge q})$$

is a fiber sequence.

Claim: ("Destabilization of the stable Snaith splitting"): Let $O$ be an operad with $O(0) = \ast$ and admitting a map from the $A_2$ operad (the latter condition means that every $O$-space is an $H$-space). Let $J^O$ be the free functor from $E_0$-spaces to $O$-spaces. Then the natural map

$$J^O(\vee_{n \geq 1} O(n)_+ \wedge_{\Sigma_n} X^{\wedge n}) \xrightarrow{J^O\varphi} (J^O)^2(X)$$

is an equivalence for every connected space $X$.

Proof: Use the stable Snaith splitting, the fact that $J^O$ preserves stable equivalences, and the fact that a stable equivalence between connected $H$-spaces is an equivalence.

Corollary: ("Operadic James-Hopf map") Let $O$ and $X$ be as above. Then for any $q \geq 1$, there is a "James Hopf" map

$$ J^O(X) \xrightarrow{\eta_{J^O(X)}} (J^O)^2(X) \overset{(J^O\varphi)^{-1}}{\simeq} J^O(\vee_{n \geq 1} O(n)_+ \wedge_{\Sigma_n} X^{\wedge n}) \xrightarrow{J^O(\pi_q)} J^O(O(q)_+ \wedge_{\Sigma_q} X^{\wedge q}) $$

which kills the subspace $J^O_{q-1}(X) \subseteq J^O(X)$. Here the filtration $\dots \subseteq J^O_k(X) \subseteq \dots J^O(X)$ is defined by arity of operations in $O$ as in the James filtration, and $\eta_Y: Y \to J^O(Y)$ is the unit map.

This is analogous to the usual James-Hopf map; note that the "adjointing over" that occurs with the unit map is analogous to the $\Sigma \dashv \Omega$ adjointing that occurs in the usual James-Hopf map.

Answer 1

In my view, the fact of that this is a fibration sequence is something to be cherished, and I wouldn't think that it generalizes without complication.

Regarding your comments that the James-Hopf map is complicated, here is the point: the defining property of the $q$th James-Hopf invariant $j_q: J(X) \rightarrow J(X^{\wedge q})$ is that it is a natural extension to all of $J(X)$ of the composite $J_q(X) \rightarrow J_q(X)/J_{q-1}(X) = X^{\wedge q} \hookrightarrow J(X^{\wedge q})$. Thus, of course, it is trivial when restricted to $J_{q-1}(X)$. It is also clear what how it behaves on homology when restricted to $J_q(X)$, and when $X$ is an even sphere, it then becomes easy to compute in all degrees (using cohomology, as you mentioned).

It was the discovery of a young Ioan James in the mid 1950's that such an extension existed - I'd recommend looking up the paper. He writes down a formula on the point set level that has the properties I just listed. His formula is not canonical, but I think any other explicit formula would be rather similar.

By the way, it is wrong to describe the splitting of $\Sigma J(X)$ as the `Snaith splitting'. It is due to James: indeed, it is quite formal from the properties of $j_q$ that I just wrote down. (Milnor wrote some notes about this, advertised by Adams in his student's guide to algebraic topology.) Snaith was trying to generalize this known result to $\Omega^n \Sigma^n X$ for $n \geq 2$, and discovered that an infinite number of suspensions is needed for the splitting.