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We define the Whitehead product as follows:

Given basepoint-preserving maps $f:S^k \to X$ and $g: S^l \to X$, let $[f,g]:S^{k+l-1} \to X$ be the composition of $$ S^{k+l-1} \to S^k \vee S^l \overset{f \vee g}{\to} X$$

That is the composition with the attaching map of the $(k+l)$-cell to the wedge $S^k \vee S^l$.

It is then stated that in the case $k=l=1$, this is the commutator of the fundamental group $\pi_1(X)$.

For general $k,n$, I have tried to explicitly write out what the Whitehead product maps would look like but I am not getting it.

Let us say we are attaching a $(k+l)$-cell to $S^k \vee S^l$ we can view the Whitehead product $[f,g]$ as the following,

$$ \partial(D^k \times D^l) \to S^k \vee S^l \overset{f\vee g}{\to} X$$ $$ \partial D^k \times D^l \cup D^k \times\partial D^l \to S^k \vee S^l \overset{f \vee g}{\to} X $$

$$S^{k-1} \times D^l \cup D^k \times S^{l-1} \to S^k \vee S^l \overset{f \vee g}{\to} X $$

where the attachment is as follws $S^{k-1} \hookrightarrow S^k$ and $S^{l-1} \hookrightarrow S^l$ via the inclusion maps and $D^l \to S^l$, $D^k \to S^k$ via quotient maps.

Even with the description I am still not sure how to define $f \vee g$ on the image of the four distinct maps from the attachment. I looked at Seifert Van-Kampen as recommended in a comment but I still am not seeing how I should define the composition map defining the Whitehead product.

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  • $\begingroup$ Write down explicitly all the maps. $\endgroup$ Mar 28, 2016 at 21:36
  • $\begingroup$ That's what I am not sure how to do. I know that we attach a $2$-cell to $aba^{-1}b^{-1}$ if we let $a,b$ be generators of the two copies of $S^1$. However, I really don't know how I apply $f \vee g$ to this. $\endgroup$
    – user7090
    Mar 28, 2016 at 21:42
  • $\begingroup$ The best advice I (probably anyone) can give is to think harder. You need to figure out the induced map $\pi_1(S^1 \vee S^1) \to \pi_1(X)$. Think about van Kampen's theorem. $\endgroup$
    – user98602
    Apr 4, 2016 at 19:14

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The Whitehead product is more easily thought about cubically. So in your formulae replace $D$ by $I$, where $I= [0,1]$ is the unit interval. Basic formulae are that

$$I^m \times I^n \cong I^{m+n}, \quad \partial (I^m \times I^n) = (\partial I^m \times I^n)\cup (I^m \times \partial I^n), $$where $\partial$ denotes the boundary. So if $a: \partial I^m \to X, b: \partial I^n \to X$ then we get $[a,b]: \partial(I^m \times I^n) \to X $ as $(r,s) \mapsto a(r) $ or $b(s)$ according to which part of the decomposition $(r,s)$ belongs to.

This view is useful also for proving results on say $[a,b+c]$, as composing cubes is clear.

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