The EHP sequence consists of maps $S^n\stackrel{E}{\to} \Omega S^{n+1} \stackrel{H}{\to} \Omega S^{2n+1}$. This is a homotopy fibration sequence if $n$ is odd. For even $n$ it is a homopy fibration sequence after localizing at $2$. Applying $\Omega^n$, one obtains the sequence $\Omega^n S^n{\to} \Omega^{n+1} S^{n+1} {\to} \Omega^{n+1} S^{2n+1}$

By J homomorphism I just mean the map $O(n)\to \Omega^n S^n$ given by one-point compactification.

It is mentioned on page 30 of this book that at least if one localizes at $2$, then the following diagram commutes (up to homotopy?) (EDIT: as John Klein points out, the right square only commutes after being looped once, and this is what is claimed in the linked book of Ravenel). Here the right vertical map is the suspension. $$ \begin{array}{cccccc} O(n) &\to & O(n+1)& \to & S^n \\ \downarrow & & \downarrow & & \downarrow \\ \Omega^n S^n &{\to}& \Omega^{n+1} S^{n+1} &{\to} &\Omega^{n+1} S^{2n+1} \end{array} $$

Question 1: Is there a reference where it is proved that the right square commutes? Is it necessary to localize at $2$ for it?

This diagram implies that the induced homomorhism on relative homotopy groups $\pi_k(O(n+1),O(n))\to\pi_k( \Omega^{n+1} S^{n+1}, \Omega^{n} S^{n})$ is an isomorphism roughly for $k < 2n$, after localizing at $2$. In fact, there is a good reason to believe that this is an isomorphism without any localization. If true, this must be well-known.

Question 2: What is a good reference to the asserion that this map of relative homotopy groups is an isomorphism in the metastable range $k\le 2n-\epsilon$?

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    $\begingroup$ The closely related fact that $G_n/O_n\to G/O$ is approximately $2n$-connected is stated on page 124 of Walls "Surgery on Compact Manifolds", and he gives as reference Haefliger, A., Differentiable embeddings of $S^n$ in $S^{n+q}$ for $q>2$. Ann. of Math. 83 (1966), 402–436. $\endgroup$ – Tom Goodwillie Aug 19 '16 at 17:57
  • $\begingroup$ I want to understand this. I've been curious about it for years. Your question is prompting me to think about it. I'll let you know if I think of anything sensible to say. $\endgroup$ – Tom Goodwillie Aug 19 '16 at 21:36
  • $\begingroup$ I went through Haefliger's paper. While there clearly are similar looking statements there, I had trouble locating exactly the one I am looking for. If someone can give me chapter and verse, I will be grateful. $\endgroup$ – Gregory Arone Aug 20 '16 at 15:45
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    $\begingroup$ @GregoryArone : I just checked your link to Ravenel's book (it's the second edition). On page 30, he only claims that the diagram commutes after looping once. He doesn't assert that the diagram itself commutes. On page 56 of the first edition he doesn't say anything at all about whether the diagram commutes––it's inferred. Obviously, he corrected his misstatement in edition 2. $\endgroup$ – John Klein Sep 6 '16 at 2:23
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    $\begingroup$ @JohnKlein You are right. Not sure how I could miss it. I edited the question. $\endgroup$ – Gregory Arone Sep 6 '16 at 17:39

Added 9/7/16: I just got access to the paper:

James, I. M. On the iterated suspension. Quart. J. Math., Oxford Ser. (2) 5, (1954). 1–10

which is an explicit reference to Greg's questions on the level of homotopy groups. See section 2.


${}^\dagger$Added 9/2/16: I realize now that what I wrote below doesn't answer Question (1). It supplies an argument for Question (2).


The answer is yes.$^\dagger$

Let $G_n$ be the topological monoid of self equivalences of $S^{n-1}$, not necessarily preserving the basepoint.

Let $F_n$ be the topological monoid of self equivalences of $S^n$, preserving the basepoint.

Unreduced suspension defines a homomorphism $G_n \to F_n$.

Forgetting the basepoint defines a homomorphism $F_n \to G_{n+1}$.

In effect, you are asking whether the map $$ O(n+1)/O(n) = S^n = G_{n+1}/F_n \to F_{n+1}/F_n $$ is about $(2n)$-connected.

Since the diagram $$ \require{AMScd} \begin{CD} G_{n+1} @>>> G_{n+1}/F_n \\ @VVV @VVV \\ F_{n+1} @>>> F_{n+1}/F_n \end{CD} $$ is $\infty$-cartesian it suffices to show that the map $G_{n+1} \to F_{n+1}$ is about $(2n)$-connected.

Think of $G_{n+1}$ as a set of components of the based mapping space $F_\ast(S^{n}_+,S^{n})$ and think of $F_{n+1}$ as given similarly by components of $\Omega^{n+1}S^{n+1} = F_\ast(S^{n+1},S^{n+1}) \cong F_\ast(S^{n},\Omega\Sigma S^{n+1})$. Then we have an $\infty$-cartesian square $$ \begin{CD} G_{n+1} @>>> F_{n+1} \\ @V\cap VV @VV\cap V\\ F_\ast(S^{n}_+,S^{n}) @>>E^{\sharp}> \Omega^{n+1}S^{n+1}\, . \end{CD} $$ Consequently, we need to understand the homotopy fibers of the lower horizontal map, labeled $E^\sharp$, taken at the identity $1$.

Consider the commutative diagram $$ \begin{CD} \text{"?"} @>>> F_\ast(S^{n}_+,S^{n}) @>E^\sharp>> \Omega^{n+1}S^{n+1} \\ @VVV @VV E V @| \\ \Omega \Sigma S^{n} @> i>> F_\ast(S^{n}_+,\Omega \Sigma S^{n}) @>R>> \Omega^{n+1}S^{n+1} \\ @V (a) VV @VV HV @VVV \\ F_\ast(S^{n}_+,\Omega \Sigma S^{2n}) @= F_\ast(S^{n}_+,\Omega \Sigma S^{2n}) @>>> \ast \end{CD} $$ where the horizontal maps are fiber sequences, with the homotopy fibers of the first two rows taken at $1$. We need to identify "?" in the $2n$ range (more precisely, we need to show that it is about $(2n)$-connected.

The middle column is a fiber sequence in degrees up to approximately $2n$ (to get an actual fiber sequence, we only need to replace the first term by $F_\ast(S^n_+,\text{fib}(H))$; this works since the map $S^n \to \text{fib}(H)$ is about $(3n)$-connected). The map $E^\sharp $ is given by reduced suspension $E$ restricted to $S^{n+1} \subset \Sigma(S^n_+)$ (the map $R$ is defined by restriction).

The fiber of $R$ is given by $\Omega \Sigma S^n$ by use of the cofiber sequence of based spaces $S^{n+1}\to \Sigma (S^n_+) \to S^1$. However, we have to remember that the inclusion of the fiber picks up the identity map of $S^{n+1}$ by means of the adjunction. The map $H$ is the James-Hopf invariant.

The map $(a)$ is defined by the composition.

Discussion. Note that both the source and target of $(a)$ have the homotopy type of $S^n$ in degrees $< 2n$ approximately. At first glance one might think this map is trivial, as it would be if we had taken fibers at the trivial map rather than at 1. But since to define it we've implicitly used an equivalence $\Sigma (S^n_+) \simeq S^{n+1} \vee S^1$, which doesn't desuspend, this will make the map (a) nontrivial.

The desired connectivity statement about "?" is a direct consequence of the following claim.

Claim: The map (a) has degree one on $\pi_n$ (which is the same as its degree on $H_n$).

Here's a sketch of the argument proving the claim.

Consider the composition $$ \begin{CD} S^n @>j>> \Omega\Sigma S^n @>i >> F_\ast(S^n_+,\Omega\Sigma S^n) = F(S^n,\Omega\Sigma S^n) \end{CD} $$ where $j$ is adjoint to the identity map. The map $j$ represents a generator of $\pi_n$ of $\Omega\Sigma S^n$.

The adjoint of the composite $i\circ j$ is a map $$ g: S^n \times S^n\to \Omega \Sigma S^n\, . $$ By the definition of $i$, the map $g$ is given by loop multiplication of the following two maps: $(x,y) \mapsto (t\mapsto x \wedge t)$ and $(x,y) \mapsto (t\mapsto y \wedge t)$.

Let $J(X)$ be the James construction on a (nice) based space. Then $J(X) \to \Omega\Sigma X$ is a weak equivalence, where the map is induced from the inclusion $X\to \Omega \Sigma X$ using the multiplicative structure on $\Omega \Sigma X$ (strictly speaking, we should replace loops by Moore loops to get the map). Explicitly, the map sends a word $(x_1,\dots, x_j)$ to the product $\prod_i \gamma_i$ where $\gamma_i(t) := t\wedge x_i$. Let $J_2(X) = X \cup (X \times X)$ be filtration two of the James construction.

Then the map $g$ factors up to homotopy through the map $$ h: S^n \times S^n\to J_2(S^n) $$ given by $h(x,y) = (x,y)$.

The James-Hopf invariant $H: J(X) \to J(X\wedge X)$ sits in a homotopy commuting diagram $$ \begin{CD} J_2(X) @> q >> J_2(X)/X = X\wedge X \\ @V\cap VV @VV\cap V \\ J(X) @>>H > J(X\wedge X) \end{CD} $$ where $q$ is the quotient map. If $X$ is $r$-connected then the vertical maps are at least $(3r)$-connected.

It follows that the degree of the map $(a)$ on $\pi_n$ is identified with the degree of the composition $$ \begin{CD} S^n \times S^n @>g >> J_2(S^n) @>q>> J_2(S^n)/S^n = S^n \wedge S^n \end{CD} $$ in homology in dimension $2n$. But this is just the evident quotient map which has degree $+1$.

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    $\begingroup$ According to Bill Richter: "The statement [Question 1]" is false, and I proved that the right square does not commute. Dan Kahn showed me that James had a simple proof that the right square homotopy commutes after a loop, so I didn't publish my proof that I was so proud of." $\endgroup$ – John Klein Sep 2 '16 at 11:47

In my comment to my first answer, I noted that I didn't address Greg's question (1). This answer aims to address that question. I am indebted to Bill Richter for explaining the following argument to me.

We are trying to understand whether or not the square $$ \require{AMScd} \begin{CD} O(n+1) @>>> S^n \\ @VVV @VVV \\ F_{n+1} @>> H > \Omega^{n+1} S^{2n+1} \end{CD} $$ commutes. I mentioned in my comment that James proved it commutes after a loop. I will sketch a proof of a related result which is both more and less general than the one of James.

Let $\lambda = \Sigma H$ denote the Boardman and Steer Hopf invariant. This is a natural transformation of based function spaces $$ F_\ast(A,B) \to F_\ast(\Sigma^2 A,\Sigma B\wedge \Sigma B) $$ given by suspending the James-Hopf invariant.

Let's consider the related problem of whether or not the diagram $$ \begin{CD} G_{n+1} @> \pi >> S^n \\ @VVV @VV E V \\ F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2} \end{CD} $$ commutes. If it does then Greg's question (1) will commute if we replace $H$ by $\lambda$, since the map $O(n+1) \to S^n$ factors through $G_{n+1}$.

Claim: The last diagram above homotopy commutes after taking a single loop.

Reformulate the claim as follows: let $B = \Sigma A$ be a suspension and let $B\to G_{n+1}$ be any map. Then the claim is equivalent to the assertion that the diagram $$ \begin{CD} B @> \pi >> S^n \\ @VVV @VV E V \\ F_{n+1} @>>\lambda > \Omega^{n+2} S^{2n+2} \end{CD} $$ commutes (by taking $B = \Sigma \Omega G_{n+1}$).

To verify the reformulation of the claim, we will use the Boardman and Steer Cartan formula. Let $X = S^n$, then the action of $G_{n+1}$ on $X$ induces a map $$ F: B \times X \to X\, . $$ The Hopf construction of $F$ is a map $h_F: \Sigma B\wedge X \to \Sigma X$ whose adjoint is the left vertical (inclusion) map of the square. Thus we need to compute $\lambda(h_F): \Sigma B \wedge \Sigma X \to \Sigma X \wedge \Sigma X$.

Take the suspension of this map $\Sigma F: \Sigma (B \times X) \to \Sigma X$. Then we have a factorization up to homotopy $$ \begin{CD} \Sigma (B\times X) @>\Sigma F >> \Sigma X \\ @V p_1 + p_2 + p_{12} VV @| \\ \Sigma B \vee \Sigma X \vee \Sigma B\wedge X @>>(f,g,h_F) > \Sigma X \end{CD} $$ where $p_1$ is the projection onto $\Sigma B$, $p_2$ the projection onto $\Sigma X$ and $p_{12}$ is the quotient map onto $\Sigma B \wedge X$. The map $f$ is the suspension of the restriction of $F$ to $B \times \ast$, the map $g$ is the suspension of the restriction of $F$ to $\ast \times X$.

Since $\Sigma F$ is a suspension, its Hopf invariant $\lambda(\Sigma F)$ is trivial. So, if we take the Hopf invariant of the composite $$ \Sigma F = (f,g,h_F) \circ (p_1 + p_2 + p_{12}) $$ and use the composition formula and the Cartan formula (in Boardman and Steer), we obtain, after some rewriting, the formula $$ \lambda(h_F) = (f \wedge g) \circ \Sigma p_{12} \qquad (*) $$ Implicit in this calculation, which I've omitted, is the fact that the reduced diagonal maps of both $B$ and $X$ are trivial since these spaces are suspensions. I refer the reader to Boardman and Steer's paper for the details.

Returning to our original notation, note that the map $f$ is the suspension of composition $$ \begin{CD} B @>>> G_{n+1} @>\pi >> S^n \, , \end{CD} $$ and the map $g$ is the identity map of $S^{n+1}$. The map $\lambda(h_F)$ is adjoint to the composite $$ \begin{CD} B\to G_{n+1} \to F_{n+1} @>\lambda >> \Omega^{n+2} S^{2n+2}\, \end{CD} $$ which is one of the composites of the reformulated claim.

The map $(f \wedge g) \circ \Sigma p_{12}$ is the composite $$ \begin{CD} \Sigma^2 (B \times S^n) @>>> \Sigma^2 (G_{n+1} \times S^n) @>q >> \Sigma^2 G_{n+1} \wedge S^n @>\Sigma \pi \wedge 1 >> \Sigma S^n \wedge \Sigma S^n \end{CD} $$ where $q$ is the quotient map. By definition, the last composition is adjoint to the composition $$ \begin{CD} B @>>> G_{n+1} @> \pi >> S^n @> E >> \Omega^{n+2} S^{2n+2} \end{CD} $$ which is the other composite of the reformulated claim. Thus the claim follows from equation $(\ast)$.

Comments: (1). The above argument shows that the diagram of Greg's question (1) commutes after looping. The equation $(\ast)$ isn't generally valid when $B$ (or $X$) isn't a suspension.

For general $B$, there is a correction term in the formula involving the reduced diagonal $B\to B \wedge B$. This gives evidence that Greg's diagram will fail to commute before taking loops.

A potential counterexample to the diagram commuting is to take $n = 2$ and $B =\Bbb RP^3 = SO(3) \subset O(3)$, since the diagonal $\Bbb RP^3 \to \Bbb RP^3 \wedge \Bbb RP^3$ is stably essential (it's detected by the the cup product).

In fact, Bill Richter (unpublished) shows that if $n > 2$ then Greg's diagram fails to homotopy commute (Richter also explicitly identifies the deviation from the diagram commuting).

(2). The reader might ask why we've worked with the Boardman-Steer Hopf invariant rather than the one of James. The answer is that James' invariant doesn't satisfy a Cartan formula––––there are "partial" Cartan formulae, which were known to Barratt, but this is lost knowledge...

(3). The Boardman and Steer paper I alluded to is:

Boardman, J. M.; Steer, B. On Hopf invariants. Comment. Math. Helv. 42 1967 180–221.

See also:

Boardman, J. M.; Steer, B. Axioms for Hopf invariants. Bull. Amer. Math. Soc. 72 1966 992–994.

(4) The result of James about the diagram looping after one suspension isn't stated by James. In effect, James proves the (reformulated) claim in the special case when $B = S^p$ is a sphere (so Greg's diagram will commute on homotopy groups). The result is given by Corollary 15.9 of the paper:

James, I. M. On the suspension triad. Ann. of Math. (2) 63 (1956), 191–247.

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