Many sources give an easy definition of a Toda bracket $\{f,g,h\}$ of appropriate maps $W \to X \to Y \to Z$ in spaces as a subset of the homotopy classes of maps $[\Sigma W, Z]$ (for example, Ravenel's *Complex Cobordism*.) However, the only real usages of Toda brackets that I can find are in the context of heavier lifting, in what appear to be difficult computations of stable homotopy groups of spheres.


It's not difficult to construct maps satisfying the nullhomotopy criteria, but either it's not clear how to work things out or I wind up finding only a trivial bracket.


Some sources give more conceptual accounts (such as [this paper](http://projecteuclid.org/euclid.hha/1139840198)), but again the only examples are quite complicated from my inexperienced point of view. For instance, the only example in the linked paper is an element in $\pi\_{11}(S^6)$ constructed from maps I'm not familiar with.


What's the simplest natural example of a non-trivial Toda bracket in spaces? It would be especially helpful if the example illustrated the intuition or geometry behind Toda brackets, but I understand that may not be how things are.



Simplest example of non-trivial Toda bracket in spaces
I think that the simplest nontrivial case comes from the maps
$$ S^5 \xrightarrow{\Sigma^2\eta} S^4 \xrightarrow{2\iota} S^4 \xrightarrow{\Sigma\eta} S^3 $$
Both two-stage composites are trivial, so we get a Toda bracket which is a subset of $[S^6,S^3]$. This group (or at least the $2$-primary part) is cyclic of order 4, generated by an element called $\nu'$. One way to describe $\nu'$ is to note that $S^3$ can be identified with the group $SU(2)$, so we have a commutator map $S^3\times S^3\to S^3$. This sends $S^3\times 1$ and $1\times S^3$ to the basepoint, so it induces a map from $S^6=S^3\wedge S^3$ to $S^3$; this is $\pm\nu'$. Alternatively, using $SO(3)=S^3/\{\pm 1\}$ we see that $\pi\_3(SO(3))=\mathbb{Z}$, and we can apply the unstable $J$-homomorphism to a generator to get a map $S^6\to S^3$, which is again $\pm\nu'$. Toda showed that the above Toda bracket is $\{\nu',-\nu'\}$ (so the indeterminacy is $\{0,2\nu'\}$).


If you are happy to work stably, then the relevant group is $\pi\_3(S)=\mathbb{Z}/8.\nu$, where $\nu$ is the quaternionic Hopf map, and $\nu'$ becomes $2\nu$, so $\langle\eta,2,\eta\rangle=\{2\nu,-2\nu\}$.


However, if you want to build intuition then then you may well be better off thinking about Massey products; these are closely analogous to Toda brackets, but you can write down examples by pure algebra.


