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I have added a section Relation to Sullivan models with the statement that the co-binary part of the Sullivan differential equals the $\mathbb{Q}$-linear dual of the Whitehead product.
There is a hidden factor of 2 in that statement, which I would like to understand better. But I’ll give that statement now it’s own entry, and then ask my question there…
finally added the original references
Then I made explicit the subtlety with $[\phi,\phi]_{Wh}$ (here).
Finally I reorganized a little, starting to put the previous material into appropriate subsections.
Thanks for catching this. Both fixed now.
[ The missing re-ordering was a silly typo, but the difference in sign came from thinking L-infinity algebras as opposed to dg-Lie algebras, via the pertinent isomorphism (equation (3) in arXiv:hep-th/9209099). ]
Hi Noah, once we fix our models of the four spaces involved in the pushout diagram into which the attaching map fits, and fix once and for all how to include spheres into discs (this fixing is part of the definition of a CW-complex, as is the fixing of models of $S^{n_1 + n_2 -1}$ and $D^{n_1 + n_2}$), then the attaching map is unique (on the nose), by the universal property of a product together with the fact that the map $S^{n_1} \vee S^{n_2} \rightarrow S^{n_1} \times S^{n_2}$ is a monomorphism. Thus, once we have fixed our definition of a CW-complex, the attaching map is unique on the nose up to a choice of model for $S^{n_1} \vee S^{n_2}$, and our choice of the latter is irrelevant since it is part of a map which we are ultimately interested in considering up to homotopy.
We should add an explanation of this kind to the page!
When it comes to signs, the decomposition into $n_1$ and $n_2$,. i.e. the choice to work with $S^{n_1} \times S^{n_2}$ and not $S^{n_2} \times S^{n_1}$ is part of the input to the construction of a Whitehead product, i.e. the two possibilities are to be regarded as different.
Not sure if this answers your question?
Hi Richard, thanks for the explanation. I still don’t think I completely understand.
First, I agree that you’ve given a validly specified construction of “a” Whitehead product.
However, there are other constructions of a Whitehead product where it’s not completely clear if they agree with this one. For example, if you look at Section 8 of Brunerie there’s a construction of a Whitehead product that uses iterated suspensions. In particular, if you translate into the language of CW complexes and unpack the induction, it’s using a CW complex for the 2-sphere where you have two 2-cells (northern and southern hemispheres), two 1-cells (eastern and western equators), and two 0-cells. In this particular case I think it’s probably not hard to fix an identification between this sphere and the standard sphere and then check whether Brunerie’s Whitehead product is yours (this is exactly the exercise that lead me to ask this question). It’s pretty clear that in this case there’s an arbitrary choice of convention (whether you prefer the northern or southern hemisphere) which gives an overall sign. But my point is that this is another Whitehead product and it’s not clear that either deserves to be called “the” Whitehead product.
And importantly, neither Wikiedpia nor the nLab seems to think that you need to work with a specific CW-description. I’d much rather have a model-free description of the Whitehead product. But at first glance it seems like the situation is quite bad, because to identify each sphere with the standard sphere you have to pick a sign and you have to do this independently in each dimension! So it’s a torsor for $(Z/2Z)^\infty$. But my suspicion is that most of those are “poorly behaved” (e.g. in terms of properties of the resulting Whitehead product), and that one is supposed to impose some kind of additional condition with respect to suspension. At that point one can hope that there are exactly two Whitehead products.
[Removed, will come back later when in less haste!].
Hi Noah, an elaboration on my removed comment! To construct the Whitehead product, we have to construct an attaching map of the required kind, in particular fitting into a pushout diagram of the required kind, in any way you like (using any models you like). Once we have done that, we can see that any other choices for the pushout diagram give, up to a choice of model for (i.e. up to automorphisms of) the source and target, an attaching map equal to the one we first constructed, for the reasons I gave: the universal property of a product and of a monomorphism. Finally, observe that the models for the source and target of the attaching map are irrelevant since we are ultimately working up to homotopy.
In particular, it doesn’t matter what CW-structures or more generally what models we choose: any construction fits up to isomorphism into a pushout diagram of the required kind (in particular such that the right vertical map is a monomorphism), and then it must be equal to the first one we constructed up to models of the source and target, which are, as observed, irrelevant.
I don’t follow the step “Finally, observe that the models for the source and target of the attaching map are irrelevant since we are ultimately working up to homotopy.” My apology if I’m being dumb. Spheres have nontrivial automorphisms even up to homotopy.
Good to discuss it, I often find treatments of this kind of thing unsatisfactory!
If one is working with homotopy groups on the nose, then the two maps $\phi_1$ and $\phi_2$ require a choice of model of the wedge sum, and the choice of model of the target homotopy group requires a choice of model of $S^{n_1 + n_2 - 1}$.
The point is really that in practise mathematicians almost never work up to equality, they work up to isomorphism or weaker. In particular, we never really fix a model up to equality of a homotopy group, we allow it to be replaced silently up to isomorphism. That is what is at stake here: when I wrote that the choices of models are irrelevant, I meant that since it only really makes sense to work up to isomorphism anyway when it comes to homotopy groups, we cannot worry about choices of those models.
I do think though that we can add something to the page to help clarify things. One thing that is really critical, for instance, is that the definition of an attaching map must require that the right vertical map is a monomorphism.
We seem to still be entirely speaking past each other. I have no problem with working up to isomorphism. I still think there’s a $2^\infty$ component space of Whitehead products unless you impose some additional compatibility, and that even with this additional compatibility there’s still two Whitehead products. I can’t even sort out whether your comment is disagreeing with my claim in that direction or not.
Yes, I am disagreeing with you, and am agreeing with the nLab page that the ’the’ is justified. I was trying to point out that your arguments are relying on specifics of the models of the spaces involved, which cannot be relevant when working up to isomorphism.
In #15, you seemed to follow my explanation until its final step. In #16, I explained this final step. Again: a purely category theoretic argument shows that the attaching map is unique up automorphisms of its source and target. If you wish to consider homotopy groups up to isomorphism, you have to identify the attaching maps obtained by automorphisms of the source, since if you replace that source sphere by one isomorphic to it, you get the same homotopy group up to isomorphism in which the Whitehead product lives.
In other words: if you are working with homotopy groups up to isomorphism, you have to identify Whitehead products obtained under those isomorphisms to have a well-defined notion.
The same considerations apply to the target of the attaching map, namely the wedge sum.
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