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I've found a paper of Spanier's (Higher Order Operations) where he uses the theory of "carriers" to study $n$-th order operations. The set-up is rather general; for example a particular case defines the Massey triple product. Spanier also talks about Toda's 'toral construction', which as far as I can tell is Toda's name for what (everybody else now) calls a Toda bracket.

I am particular interetsed in the results of Section 6 of this paper. Here Spanier sets up an operation associated to a sequence of spaces $$ A_0 \xrightarrow{a_1} A_1 \xrightarrow{a_2} A_2 \to \cdots \xrightarrow{a_n}A_n $$

In particular he defines an operation $\langle a_1,a_2,\ldots,a_n \rangle$ corresponding to a subset of $[\Sigma^{n-2}A_0,A_n]$, which gives me hope that this is a Toda bracket.

The result I'm particularly interested in is Theorem 6.3: A sequence,as above, is split if there is a commutative diagram of the form $$ \begin{array}{cccccccccc} &&&& B_2 &&&&B_{n-1} \\ &&&\nearrow_{b_2} & \downarrow_{c_2} &&&& \downarrow_{c_{n-1}}\\ A_0 & \xrightarrow{a_1} & A_1 & \xrightarrow{a_2} & A_2 & \xrightarrow{a_3} & \cdots & \xrightarrow{} & A_{n-1} \xrightarrow{a_n} & A_n \end{array} $$ such that $b_2a_1,b_3c_2,\cdots,a_nc_{n-1}$ are all null.

Spanier proves that such a splitting exists if and only if the operation$\langle a_0,a_1,\ldots,a_{n-1} \rangle$ is defined and vanishes. (Note that an $n$-th order is defined if and only if all lower order operations are defined and each contain the zero element. An operation vanishes if it is defined and contains the zero element.)

One thing that worries me slightly about the proof is that Spanier uses equalities $$ \langle a_1,\cdots,a_{n-2},g'b_{n-1} \rangle = g'_\# \langle a_1,\cdots,a_{n-2},b_{n-1} \rangle $$

If I'm not mistaken the shuffling/juggling lemma for Toda brackets is that $$ w \circ \langle x,y,z \rangle \subset \langle x,y,wz \rangle $$ (and I'm guessing something similar for higher brackets), so I'm not so sure that these are just Toda brackets.

To summarise my questions:

1.) Is the operation constructed a Toda bracket? (I guess in the sense of Kochman if there is some ambiguity as to what a higher Toda bracket is)

2.) If not, do the obstructions to splitting sequences as above lie in the genuine Toda brackets? I'll be working with sequences of spectra, rather than spaces if it makes a difference

I would also appreciate if there are any obvious references that I'm missing on this, as the paper by Spanier is the only place I've seen this discussed.

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For the case of spectra, possibly the appendix of Brooke shipley's "An Algebraic model for rational $S^1$-equivariant homotopy theory" may be helpful- see appendix A, particularly proposition A5. Paper is on her webpage. –  Mike-Doherty Sep 6 '13 at 9:13
@Mike-Doherty: Thanks, I'll have to read it more closely, but that looks very promising! –  Drew Heard Sep 6 '13 at 13:31

1 Answer 1

up vote 3 down vote accepted

Just to close this off - thanks to Mike-Doherty it appears that the answer is yes (and in fact for spaces this goes back to the paper "The decomposition of stable homotopy" by Joel Cohen.)

Using the terminology of Shipley's paper, given a sequence of maps $$ A_{n-1} \xrightarrow{f_{n-1}} A_{n-2} \xrightarrow{f_{n-2}} \cdots \xrightarrow{f_1} A_0 $$ then there is an object $X \in \{ f_1,\ldots,f_{n-1} \}$ if and only if $0 \in \langle f_1,\ldots,f_{n-1} \rangle$. The statement that $X \in \{ f_1,\ldots,f_{n-1} \}$ basically corresponds to the existence of the following commutative diagram:

$$ \begin{array}{cccccccccc} &&&& \Sigma^{-n+1}F_{n-1}X &&&& \Sigma^{-1} F_1X \\ &&&\nearrow & \downarrow &&&& \downarrow\\ \Sigma^{-n} X & \to & A_{n-1} & \xrightarrow{f_{n-1}} & A_2 & \xrightarrow{f_{n-2}} & \cdots & \xrightarrow{} & A_{1} \xrightarrow{f_1} & A_0 \end{array} $$

It is easy to see that the 3-fold bracket is just the usual Toda bracket. In addition the bracket constructed by Kochman is contained in the bracket constructed by Cohen.

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Dear @Drew: If you want to close off this question, you may want to accept this answer. –  Ricardo Andrade Sep 24 '13 at 1:03
@RicardoAndrade...and done! –  Drew Heard Sep 24 '13 at 1:55

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