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

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 $$ and operations of arity divisible by $$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?

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.

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).
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.