The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)

It is not difficult to show that this map has a framed manifold description via the Pontryagin construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds in $S^{2n-1}$ (with trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction.)

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{2n-2} \amalg S^{2n-2} \amalg S^{2n-2} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?

  • 1
    $\begingroup$ I imagine Dev Sinha can answer this one fairly quickly. If not, in a few days I'll have a little more time. $\endgroup$ Mar 18, 2016 at 5:39
  • 3
    $\begingroup$ Side comment: I think it would be lovely for someone to work out a proof of the Hilton-Milnor theorem, all in the language of framed bordism and higher linking numbers. $\endgroup$ Mar 18, 2016 at 6:33
  • 1
    $\begingroup$ Shouldn't the link be three copies of $S^{2n-2}$? $\endgroup$ Mar 18, 2016 at 9:40
  • $\begingroup$ Oscar: you're correct. I'll fix it. $\endgroup$
    – John Klein
    Mar 18, 2016 at 10:03
  • $\begingroup$ Dev says he'll write a response over the weekend. $\endgroup$ Mar 18, 2016 at 21:11

1 Answer 1


By Pontryagin's Theorem, you are asking for the preimages of any chosen points in each wedge factor of $S^p \vee S^q \vee S^r$ for the iterated Whitehead product map $p : S^{p+q+r-2} \to S^p \vee S^q \vee S^r$. This can be obtained by chasing through the factorization of this map as $$ S^{p+q+r-2} \to S^{p+q-1} \vee S^r \to S^p \vee S^q \vee S^r.$$

Let's call the three points we want to take preimage of $0_p$, $0_q$ and $0_r$. Then as John said, in the standard model for this product map the preimage of $0_p$ in $S^{p+q-1}$ is a copy of $S^{q-1}$, which we can view as $0 \times S^{q-1} \subset \partial(D^p \times D^q) \subset D^p \times D^q$. Then the preimage of $0_p$ in $S^{p+q+r-2}$ will be the preimage of this $S^{q-1} \subset S^{p+q+1} \vee S^r$, which will be homeomorphic to $S^{q-1} \times S^{r-1}$. Note that this isn't what John anticipated; if he is looking for a (framed) $S^{q+r-2}$, then my answer falls short of that, and in fact I'm not sure if there whether or not there is a model for the Whitehead product where this preimage is spherical. Note as well that I haven't given an embedding of this in $S^{p+q+r-2}$, but that is "straightforward" once one identifies $S^{p+q+r-2}$ as a codimension one subspace of $\partial(D^p \times D^q \times D^r)$. (That is, I'm being lazy.)

Similarly, the preimage of $0_q$ is a copy of $S^{p-1} \times S^{r-1}$. Finally, the preimage of $0_r$ requires just one step and is a $S^{p+q-2}$. Again, embeddings can be "explicitly" worked out, though one runs into repeated use of the homeomorphism $\partial(D^n \times D^m) \cong S^{n+m-1}$, which clouds things a bit.

The reason I've thought of this is the following question: in what sense are these $S^{p-1} \times S^{r-1}$, $S^{q-1} \times S^{r-1}$ and $S^{p+q+r-2}$ linked, and in particular what topological invariants are there of this linking? Here's an answer: Call these three sub manifolds $K, M$ and $N$. Because they are disjoint and disjoint from $\infty \in S^{p+q+r-2}$, taking one point on each defines a map $K \times M \times N \to {\rm Conf}_3({\mathbb R}^{p+q+r-2})$. Taking the image of the fundamental class defines a homology class in the configuration space, which in this example is non-trivial, and in general is a complete framed cobordism invariant. Choosing some cohomology class to evaluate on this gives a numerical linking invariant, and one such would "count over crossings" of $K$ over $L$ over $M$, etc.

This is all worked out in my paper with Ben Walter on Hopf invariants, where we show that such linking invariants "with correction terms" faithfully measure all rational homotopy groups of simply connected spaces.

  • $\begingroup$ I suppose this is the paper? arxiv.org/abs/0809.5084 $\endgroup$
    – j.c.
    Mar 21, 2016 at 22:07
  • $\begingroup$ Yes, that's it. $\endgroup$
    – Dev Sinha
    Mar 21, 2016 at 22:09
  • $\begingroup$ Dev, It seems to me that it might be possible to do embedded framed surgery to turn your $S^{p-1} \times S^{r-1} $ into a framed sphere. That would give a spherical link. Does that make sense? $\endgroup$
    – John Klein
    Mar 21, 2016 at 22:13
  • $\begingroup$ Yes, that makes sense. As you know, a single surgery should do the trick in this case. $\endgroup$
    – Dev Sinha
    Mar 22, 2016 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.