Suppose $X$ and $Y$ are finite, simply connected, based CW-complexes and $m,n\geq 2$. If $a\in \pi_m(X)$ and $b\in \pi_n(Y)$, then one can regard these as elements of the homotopy groups of $X\vee Y$. If $a$ and $b$ are non-trivial, must the corresponding Whitehead product $[a,b]\in \pi_{m+n-1}(X\vee Y)$ be non-trivial?

I understand some cases using homotopy excision and the smash product but not this general situation.


1 Answer 1


Suppose $\pi_m(X)\cong \mathbb Z/p$ and $\pi_n(Y)\cong \mathbb Z/q$, where $p$ and $q$ are distinct primes. Then $[a, b]=0$, by bilinearity of the Whitehead product.

On the other hand, if $a$ and $b$ are non-zero in $\pi_*\otimes \mathbb Q$, then I suspect that you can use rational homotopy theory to prove that $[a, b]\ne 0$.


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