Simplest example of non-trivial Toda bracket in spaces

Question

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), 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.

Answer 1

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.

Answer 2

The definition you've most likely encountered is the following:

For maps $W\xrightarrow{f} X \xrightarrow{g} Y \xrightarrow{h} Z$, such that adjacent maps compose to $0$, $g$ extends to a map on the cone $\overline{g}: C_f\rightarrow Y$, but then the composition $X\rightarrow C_f \rightarrow Y \rightarrow Z$ is equal to $h\circ g$, so homotopic to zero, so $h\circ\overline{g}$ extends over the cone of $X\rightarrow C_f$, which is homotopy equivalent to the suspension $\Sigma W$, giving you a map $\Sigma W\rightarrow Z$.

There's a dual procedure: Since $h\circ g $ is nullhomotopic, $g$ lifts to the homotopy fiber $F_h$ of $h$. Now $W\rightarrow X \rightarrow F_h \rightarrow Y$ is equal to $g\circ f$, so nullhomotopic. Therefore $f$ lifts over the homotopy fiber of $F_h\rightarrow Y$, which is $\Omega Z$.

The map $W\rightarrow \Omega Z$ constructed this way is adjoint to the map $\Sigma W \rightarrow Z$ given by the other construction, so they are equivalent.

The second method can nicely be used to interpret so-called secondary cohomology operations in terms of Toda brackets:

For a cohomology class $\alpha\in H^n(X; \mathbb{Z}/2)$, suppose $Sq^1 \alpha = 0$. We can represent $\alpha$ by a map $\alpha: X\rightarrow K(\mathbb{Z}/2,n)$. We can also represent $Sq^1$ by maps, namely from $K(\mathbb{Z}/2,n)\rightarrow K(\mathbb{Z}/2,n+1)$ (for any $n$, those are related by applications of the loopspace functor). This gives us maps

$$ X \xrightarrow{\alpha} K(\mathbb{Z}/2,n) \xrightarrow{Sq^1} K(\mathbb{Z}/2, n+1) \xrightarrow{Sq^1} K(\mathbb{Z}/2, n+2) $$ where we can form a Toda bracket since $Sq^1 Sq^1 = 0$ and we assumed $Sq^1 \alpha = 0$.

By using that the Bockstein fits into a fibration sequence $$ K(\mathbb{Z}/2, n) \rightarrow K(\mathbb{Z}/4, n) \rightarrow K(\mathbb{Z}/2, n)\xrightarrow{Sq^1} K(\mathbb{Z}/2, n+1) $$ which is basically the definition of the Bockstein, you can explicitly identify all the spaces and maps in above procedure. Since all of these spaces represent cohomology in various coefficients (and the maps natural homomorphisms between those), you can interpret everything in terms of lifting cohomology classes to other coefficients. In particular, you can compute this. The result $\langle \alpha, Sq^1, Sq^1\rangle$ appears otherwise as second differential in the Bockstein spectral sequence!

There are other examples like this, for reference read the stuff about secondary cohomology operations in Mosher-Tangora's cohomology operations book. They don't explicitly state things in terms of Toda brackets there, but it's a fun exercise to reinterpret this stuff (for example, the celebrated Peterson-Stein formula is just the juggling relation).

Answer 3

Just wanted to expand @Achim Krause 's final comment: Mosher and Tangora has a nice treatment that ramps up from more to less elementary (they do get to statements in terms of Toda brackets eventually). They work stably by considering $S^n$ for $n$ large, and include the following examples involving the (suspensions of) Hopf maps $\eta, \nu, \sigma$:

• Assume $\eta^2 = 0$; then use part of the bracket construction and the relation $Sq^2Sq^2 = 0$ to derive a contradiction.

• Similar strategy to show $\eta^3 = 4 \nu$

• $\eta^2 \in \langle 2, \eta, 2 \rangle$

• $2 \nu \in \langle \eta, 2, \eta \rangle$

• $\nu^2 \in \langle \eta, \nu, \eta \rangle$

• $8 \sigma \in \langle\nu, 8, \nu\rangle$