Proof of a property of Whitehead product

Question

Let $\alpha\in \pi_n(X)$ and $\beta\in\pi_k(X)$. Let $[\alpha,\beta]\in \pi_{n+k-1}(X)$ be the Whitehead product of $\alpha$ and $\beta$. I am having trouble understanding the following property of the Whitehead product: $$[\alpha,\beta]=(-1)^{nk}[\beta,\alpha]$$ I would appreciate it if someone could give me a proof sketch of this property. I am also grateful for any sharing of the intuition behind this equation.

Answer 1

The attaching map $w_{n,k}$ for the top cell of $S^n\times S^k$ is constructed by writing $D^{n+k}\cong D^n\times D^k$ and noting that

$$S^{n+k-1}\cong \partial D^{n+k}\cong\partial(D^n\times D^k)\cong (\partial D^n\times D^k)\cup (D^n\times\partial D^k)=(S^{n-1}\times D^k)\cup (D^n\times S^{k-1}).$$

This is then mapped to $S^n\vee S^k=(S^n\times\ast)\cup(\ast\times S^k)\subseteq S^n\times S^k$ in the obvious way. This is just the product cell decomposition when each $S^r\cong D^r/S^{r-1}$ is given the CW structure with two cells.

The standard orientation induced from $\mathbb{R}^r$ is fixed on each disc $D^r$, and this in turn induces an orientation on its boundary $S^{r-1}$. Products are oriented in the standard way, and in this way the cofibre of the map described above is $S^n\times S^k$ up to orientation preserving homotopy equivalence.

Now the Whitehead product $[\alpha,\beta]$ is the composition

$$[\alpha,\beta]:S^{n+k-1}\xrightarrow{w_{n,k}} S^n\vee S^k\xrightarrow{(\alpha,\beta)}X$$

and we have a similar definition for $[\beta,\alpha]$ (replacing $w_{n,k}$ with $w_{k,n}$, of course). But notice that we can also write $[\alpha,\beta]$ as

$$[\alpha,\beta]:S^{n+k-1}\xrightarrow{w_{n,k}}S^n\vee S^k\xrightarrow{T}S^k\vee S^n\xrightarrow{(\beta,\alpha)}X$$

where $T$ is the restriction of the interchange

$$T:S^n\times S^k\rightarrow S^k\times S^n.$$

Now $T$ maps the subspace $(S^{n-1}\times D^k)\cup(D^n\times S^{k-1})\subseteq S^n\times S^k$ homeomorphically onto $(S^{k-1}\times D^n)\cup(D^k\times S^{n-1})\subseteq S^k\times S^n$ and thus lifts to a self map $t:S^{n+k-1}\rightarrow S^{n+k-1}$ such that

$$w_{n,k}T\simeq w_{k,n}t$$

Of course self maps of $S^{n+k-1}$ are classified up to homotopy by their (co)homological degree, so to check what map $t$ is it suffices to study its action on cohomology. But in cohomology the map $T$ acts on $H^*(S^k\times S^n)$ as

$$T^*(s_k\times s_n)=(-1)^{nk}s_n\times s_k,$$

where $s_r\in H^rS^r$ denotes the standard generator (recall that we have fixed orientations on all spheres). Since each quotient map $S^r\times S^s\rightarrow S^r\times S^k/S^r\vee S^s\cong S^{r+s}$ induces an isomorphism on $H^{n+k}$, an easy calculation show that $\Sigma t\simeq (-1)^{nk}$. Of course, this means that

$$t\simeq (-1)^{nk}.$$

Putting everything together now gives us

$$[\alpha,\beta]=(\alpha,\beta)w_{n,k}=(\beta,\alpha)Tw_{n,k}\simeq (\beta,\alpha)w_{k,n}t\simeq(-1)^{nk}(\beta,\alpha)w_{k,n}=(-1)^{nk}[\beta,\alpha].$$