Four-cocycles in commutative semigroup cohomology

Abstract

The symmetric cohomology group of a commutative semigroup is defined in dimension 4 and is shown to equal its triple cohomology group.

Appendices

Proof of Proposition 1.3

Proposition 1.3

If $c \in C_1$, then $c = \Theta _1 s$ for some $s \in Z^4$ if and only if it has properties

Proof

Every 4-cocycle $s \in Z^4$ has property (P1), since $[Q|P] = [P|Q]$, and properties (Z1a), (Z1b), and (Z1c) (applied to s, as in Lemma 1.2). Then $c = \Theta _1 s$ inherits (Z1a), (Z1b), (Z1c), and (P1) from s, in the form above.

Conversely, let $c \in C_1$ enjoy properties (P1), (Z1a), (Z1b), and (Z1c). Inspired by (Y), define:

for all $m \geqq 3$ and $P_1\,,\, \dots \,,\, P_{m} \in F_2$, so that $s\,(P_1\,,\, \dots \,,\, P_{m}) \in G_{\pi \,(P_1 \cdots P_{m})}$. (The variable in s is an ordered sequence, since $s\,(P_1\,,\, \dots \,,\, P_{m})$ may depend on the order of $P_1\,,\, \dots \,,\, P_{m}$.)

If $m = 2$, then (def 1c) reads

$$\begin{aligned} s\,(P,Q) ~&=~ \gamma c\,(P) ~+~ \gamma c\,(Q) ~+~ c\,(P,Q) \\&~~~~-~ c\,([\zeta P] \,,\, [\zeta Q]) ~+~ c\,([\eta P] \,,\, [\eta Q]) ~=~ c\,(P,Q)\,, \end{aligned}$$

by (Z1b); thus (def 1c) also holds when $m = 2$. Sums that are devoid of terms will be regarded as equal to 0. This applies to sums $\sum _{j=2}^{m}$ if $m = 1$; then (def 1c) also holds if $m = 1$.

Property (Y) also holds for s, in the form

$$\begin{aligned} s\,(P_1 \,,\, \dots \,,\,P_m) ~&=~ {\sum \limits _{j=1}^{m}} \gamma s\,(P_j) ~+~ {\sum \limits _{j=2}^{m}} \,\gamma s\,(P_1 \cdots P_{j-1} \,,\, P_j) \\&~~~~-~ {\sum \limits _{j=2}^{m}} \,\gamma s\,([\zeta P_1 \,;\, \dots \,;\,\zeta P_{j-1}] \,,\, [\zeta P_j]) \\&~~~~+~ {\sum \limits _{j=2}^{m}} \,\gamma s\,([\eta P_1 \,;\, \dots \,;\,\eta P_{j-1}] \,,\, [\eta P_j]) \end{aligned}$$

(Ya)

Lemma A.1

s has property (Xa):

$$\begin{aligned}&s\,(P_1 \,,\, \dots \,,\, P_m) \\&\quad =~ \gamma s\,(P_1 \,,\, \dots \,,\, P_{m-1}) ~+~ \gamma s\,(P_m) ~+~ s\,(P_1 \cdots P_{m-1} \,,\, P_m) \\&\quad ~~~-~ s\,([\zeta P_1 \,;\, \dots \,;\,\zeta P_{m-1}] \,,\, [\zeta P_m]) ~+~ s\,([\eta P_1 \,;\, \dots \,;\,\eta P_{m-1}] \,,\, [\eta P_m]). \end{aligned}$$

(Xa)

Proof

The terms of (Ya) that do not contain $P_m$ add up to $\gamma s\,(P_1\,,\, \dots \,,\, P_{m-1})$, by (def 1c). $\square$

Lemma A.2

s has property

$$\begin{aligned} s\,(P_1\,,\, \dots \,,\, P_{m}) ~=~ s\,(P_{\sigma 1} \,,\, \dots \,,\,P_{\sigma m}) \hbox { for every permutation }\sigma \hbox { of }1,2,\dots m. \end{aligned}$$

(P)

Proof

Property (P) holds when $m \leqq 2$, by (P1), and is proved by induction on m when $m \geqq 3$. Since $\sigma$ is a product of transpositions $(i~~i+1)$ it suffices to prove (P) when $\sigma = (i~~i+1)$. Then (P) holds if $i+1 < m$, by (Xa) and the induction hypothesis. With $\sigma = (m-1~~m)$, the equality to prove is, by (def 1c):

$$\begin{aligned}&s\,(P_1\,,\, \dots \,,\, P_{m})\\&\quad =~ {\sum \limits _{i=1}^{m-2}} \gamma c\,(P_i) ~+~ \gamma c\,(P_{m-1}) ~+~ \gamma c\,(P_m)&\hbox {(line 1)} \\&\qquad +~{\sum \limits _{i=2}^{m-2}} \gamma c\,(P_1 \cdots P_{i-1} \,,\, P_i) \\&\qquad \qquad +~ \gamma c\,(P_1 \cdots P_{m-2} \,,\, P_{m-1}) ~+~ c\, (P_1 \cdots P_{m-1} \,,\, P_m)\\&\qquad -~ {\sum \limits _{i=2}^{m-2}} \gamma c\,([\zeta P_1 \,;\, \dots \,;\,\zeta P_{i-1}] \,,\, [\zeta P_i])&\hbox {(line 4)} \\&\qquad \qquad -~ \gamma c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}] \,,\, [\zeta P_{m-1}]) ~-~ c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] \,,\, [\zeta P_m]) \\&\qquad +~ {\sum \limits _{i=2}^{m-2}} \gamma c\,([\eta P_1 \,;\, \dots \,;\,\eta P_{i-1}] \,,\, [\eta P_i])&\hbox {(line 6)} \\&\qquad \qquad +~ \gamma c\,([\eta P_1\,;\, \dots \,;\, \eta P_{m-2}] \,,\, [\eta P_{m-1}]) ~+~ c\,([\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] \,,\, [\eta P_m]) \\&\quad {\mathop {=}\limits ^{?}}~ s\,(P_1\,,\, \dots \,,\, P_{m-2} \,,\, P_m \,,\, P_{m-1}) \\&\quad =~ {\sum \limits _{i=1}^{m-2}} \gamma c\,(P_i) ~+~ \gamma c\,(P_m) ~+~ \gamma c\,(P_{m-1})&\hbox {(line 8)} \\&\qquad +~{\sum \limits _{i=2}^{m-2}} \gamma c\,(P_1 \cdots P_{i-1} \,,\, P_i) \\&\qquad \qquad +~ \gamma c\,(P_1 \cdots P_{m-2} \,,\, P_m) ~+~ c\, (P_1 \cdots P_{m-2} \, P_m \,,\, P_{m-1})&\hbox {(line A)} \\&\qquad -~ {\sum \limits _{i=2}^{m-2}} \gamma c\,([\zeta P_1 \,;\, \dots \,;\,\zeta P_{i-1}] \,,\, [\zeta P_i]) \\&\qquad \qquad -~ \gamma c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}] \,,\, [\zeta P_m]) \\&\qquad \qquad -~ c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2} \,;\, \zeta P_m] \,,\, [\zeta P_{m-1}])&\hbox {(line D)} \\&\qquad +~ {\sum \limits _{i=2}^{m-2}} \gamma c\,([\eta P_1 \,;\, \dots \,;\,\eta P_{i-1}] \,,\, [\eta P_i]) \\&\qquad \qquad +~ \gamma c\,([\eta P_1\,;\, \dots \,;\, \eta P_{m-2}] \,,\, [\eta P_m]) \\&\qquad \qquad +~ c\,([\eta P_1\,;\, \dots \,;\, \eta P_{m-2} \,;\, \eta P_m] \,,\, [\eta P_{m-1}]).&\hbox {(line G)} \end{aligned}$$

(1)

The lines of (1) are numbered 1, ..., 9, A, B, ..., G (in base 17?). The sums ${\sum \limits _{i=2}^{m-2}}$ are devoid of terms if $m = 3$, and are regarded as 0.

Equality (1) is proved as follows. By (Z1a),

If $P = P_{m-1}$, $Q = P_1 \cdots P_{m-2}$, and $R = P_m$, then $PQ = P_1 \cdots P_{m-1}$ and (Z1c) yields:

$$\begin{aligned} c\,&(P_1 \cdots P_{m-1} \,,\, P_m) ~+~ \gamma c\,(P_{m-1} \,,\, P_1 \cdots P_{m-2}) ~+~ \gamma c\,(P_m) \\&\quad -~ c\,([\zeta P_{m-1} \,;\, \zeta (P_1 \cdots P_{m-2})] \,,\, [\zeta P_m])&\hbox {(line 2)} \\&\quad +~ c\,([\eta P_{m-1} \,;\, \eta (P_1 \cdots P_{m-2})] \,,\, [\eta P_m]) \\&=~ c\,(P_{m-1} \,,\, P_1 \cdots P_{m-2} \, P_m) ~+~ \gamma c\,(P_{m-1}) ~+~ \gamma c\,(P_1 \cdots P_{m-2} \,,\, P_m) \\&\quad -~ c\,([\zeta P_{m-1}] \,,\, [\zeta (P_1 \cdots P_{m-2}) \,;\, \zeta P_m])&\hbox {(line 5)} \\&\quad +~ c\,([\eta P_{m-1}] \,,\, [\eta (P_1 \cdots P_{m-2}) \,;\, \eta P_m]). \end{aligned}$$

(4)

Next, let $P = [\zeta P_m]$, $Q = [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}]$, and $R = [\zeta P_{m-1}]$, so that

$$\begin{aligned} PQ ~=~ [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2} \,;\, \zeta P_m] ~~\hbox {and}~~QR ~=~ [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}]\,. \end{aligned}$$

By (ii) in Lemma 0.1, $\zeta P = \zeta P_m$, $\zeta R = \zeta P_{m-1}$, $\eta P = [\pi P_m]$, $\eta R = [\pi P_{m-1}]$,

$$\begin{aligned} \zeta Q ~=~ \zeta \,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}] ~&=~ (\zeta P_1) \cdots (\zeta P_{m-2}) ~=~ \zeta \,(P_1 \cdots P_{m-2}); ~~\hbox {and}~~\\ \eta Q ~=~ \eta \,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}] ~&=~ [\pi \zeta P_1\,,\, \dots \,,\, \pi \zeta P_{m-2}] ~=~ [\pi P_1\,,\, \dots \,,\, \pi P_{m-2}]. \end{aligned}$$

Hence (Z1c) yields:

$$\begin{aligned}&c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2} \,;\, \zeta P_m] \,,\, [\zeta P_{m-1}]) \\&\qquad +~ \gamma c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}] \,,\, [\zeta P_m]) ~+~ \gamma c\,([\zeta P_{m-1}])&\hbox {(line 2)}\\&\qquad -~ c\,([\zeta P_{m-1}] \,,\, [\zeta (P_1 \cdots P_{m-2}) \,;\, \zeta P_m]) \\&\qquad +~ c\,([\pi P_{m-1}] \,,\, \left[ [\pi P_1\,,\, \dots \,,\, \pi P_{m-2}] \,;\, [\pi P_m]]\right] )\,,&\hbox {(line 4)} \\&\quad =~ c\,([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] \,,\, [\zeta P_m]) ~+~ \gamma c\,([\zeta P_m]) \\&\qquad +~ \gamma c\,([\zeta P_{m-1}] \,,\, [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-2}]) \\&\qquad -~ c\,([\zeta P_{m-1} \,;\, \zeta (P_1 \cdots P_{m-2})] \,,\, [\zeta P_m])&\hbox {(line 7)} \\&\qquad +~ c\,(\left[ [\pi P_{m-1}] \,;\, [\pi P_1\,,\, \dots \,,\, \pi P_{m-2}]\right] \,,\, [\pi P_m]). \end{aligned}$$

(5)

Finally, let $P = [\eta P_{m-1}]$, $Q = [\eta P_1\,;\, \dots \,;\, \eta P_{m-2}]$, and $R = [\eta P_m]$. As above,

$$\begin{aligned} PQ ~=~ [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] ~~\hbox {and}~~QR ~=~ [\eta P_1\,;\, \dots \,;\, \eta P_{m-2} \,;\, \eta P_m]. \end{aligned}$$

By (ii) in Lemma 0.1, $\zeta P = \eta P_{m-1}$, $\zeta R = \eta P_m$, $\eta P = [\pi P_{m-1}]$, $\eta R = [\pi P_m]$, and

$$\begin{aligned} \zeta Q ~=~ \zeta \,[\eta P_1\,;\, \dots \,;\, \eta P_{m-2}] ~&=~ (\eta P_1) \cdots (\eta P_{m-2}) ~=~ \eta \,(P_1 \cdots P_{m-2})\,,~~\hbox {and}~~\\ \eta Q ~=~ \eta \,[\eta P_1\,;\, \dots \,;\, \eta P_{m-2}] ~&=~ [\pi \eta P_1\,,\, \dots \,,\, \pi \eta P_{m-2}] ~=~ [\pi P_1\,,\, \dots \,,\, \pi P_{m-2}]. \end{aligned}$$

Hence (Z1c) yields:

$$\begin{aligned}&c\,([\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] \,,\, [\eta P_m]) \\&\qquad +~ \gamma c\,([\eta P_{m-1}] \,,\, [\eta P_1\,;\, \dots \,;\, \eta P_{m-2}]) ~+~ \gamma c\,([\eta P_m])&\hbox {(line 2)} \\&\qquad -~ c\,([\eta P_{m-1} \,;\, \eta (P_1 \cdots P_{m-2})] \,,\, [\eta P_m]) \\&\qquad +~ c\,(\left[ [\pi P_{m-1}] \,;\, [\pi P_1\,,\, \dots \,,\, \pi P_{m-2}]\right] \,,\, [\pi P_m])&\hbox {(line 4)} \\&\quad =~ c\,([\eta P_{m-1}] \,,\, [\eta P_1\,;\, \dots \,;\, \eta P_{m-2} \,;\, \eta P_m]) \\&\qquad +~ \gamma c\,([\eta P_{m-1}]) ~+~ \gamma c\,([\eta P_1\,;\, \dots \,;\, \eta P_{m-2}] \,,\, [\eta P_m]) \\&\qquad -~ c\,([\eta P_{m-1}] \,,\, [\eta (P_1 \cdots P_{m-2}) \,;\, \eta P_m])&\hbox {(line 7)} \\&\qquad +~ c\,([\pi P_{m-1}] \,,\, \left[ [\pi P_1\,,\, \dots \,,\, \pi P_{m-2}] \,;\, [\pi P_m]\right] ). \end{aligned}$$

(6)

Adding equalities (2), (3), (4), (5), and (6) yields (1) as follows. In (1), terms

111 and 181,	112 and 183,	113 and 182,
121 and 191,	141 and 1B1,	161 and 1E1

cancel each other. Terms

211 and 552,	212 and 413,	213 and 622,	311 and 442,	312 and 661,
313 and 522,	421 and 571,	431 and 631,	451 and 531,	461 and 671,
541 and 681,	581 and 641

cancel within (2), (3), (4), (5), (6). The remaining terms of (2), (3), (4), (5), and (6) match the remaining terms of (1):

131 = 412,	132 = 411,	151 = 561,	152 = 551,	171 = 621,	172 = 611,
1A1 = 443,	1A2 = 441,	1C1 = 521,	1D1 = 511,	1F1 = 662,	1G1 = 651.

Thus (1) follows from (2), (3), (4), (5), and (6), and (1) holds. $\square$

By (P), $s\,(P_1\,,\, \dots \,,\, P_{m})$ depends only on $[P_1\,|\, \dots \,|\, P_{m}]$ and may be denoted by $s\,[P_1\,|\, \dots \,|\, P_{m}]$; and we may regard s as a 4-cochain $s \in C^4$.

Lemma A.3

s has property (X): for all $X \in F_3$ and $Q \in F_2$,

$$\begin{aligned} s\,(X\,[Q]) = \gamma s\,(X) + \gamma s\,[Q] + s\,[\varphi X \,|\, Q] - s\,\left[ \chi X \,|\, [\zeta Q]\right] + s\,\left[ [\psi X \,|\, [\eta Q]\right] \,. \end{aligned}$$

(X)

Proof

Now that s has property (P), property (X) follows from (Xa) in Lemma A.1. $\square$

Lemma A.4

s has properties (Z1e) and (Z1x): for all $P,Q \in F_2$ and $X \in F_3$,

Proof

With $P = [A;B]$, so that $\zeta P = AB$ and $\eta P = [\pi A \,,\, \pi B] = [a,b]$, property (Z1a) yields

$$\begin{aligned} s\,[A;B] ~-~ s\,[AB] ~+~ s\,[a,b] ~=~ 0\,, \end{aligned}$$

for all $A,B \in F_1$. With $p = \pi P = \pi \zeta P = \pi \eta P$ and $q = \pi Q$, this yields

$$\begin{aligned}&s\,[\zeta P \,;\, \zeta Q] ~-~ s\,[(\zeta P)(\zeta Q)] ~+~ s\,[p,q] ~=~ 0,&\hbox {(line 1)} \\&s\,[\eta P \,;\, \eta Q] ~-~ s\,[(\eta P)(\eta Q)] ~+~ s\,[p,q] ~=~ 0, \hbox { whereas}&\hbox {(line 2)} \\&s\,[PQ] ~-~ s\,[\zeta (PQ)] ~+~ s\,[\eta (PQ)] ~=~ 0, \text{ by } \text{(Z1a). }&\hbox {(line 3)} \end{aligned}$$

Since $\zeta$ and $\eta$ are homomorphisms, adding line 3 to line 2 minus line 1 yields (Z1e).

Property (Z1x) follows from (Z1e). With $X = [P_1\,|\, \dots \,|\, P_{m}]$, (Z1x) reads:

$$\begin{aligned} s\,[P_1 \cdots P_{m}] ~-~ s\,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m}] ~+~ s\,[\eta P_1\,;\, \dots \,;\, \eta P_{m}] ~{\mathop {=}\limits ^{?}}~ 0. \end{aligned}$$

If $m = 1$, then (Z1x) reduces to (Z1a). Let $m \geqq 2$, and let $Y = [P_1\,|\, \dots \,|\, P_{m-1}]$, so that $X = Y\,[P_m]$. Applying (Z1e): $s\,[PQ] = s\,[\zeta P \,;\, \zeta Q] - s\,[\eta P \,;\, \eta Q]$ to

$$\begin{aligned} P_1 \cdots P_{m} ~&=~ (P_1 \cdots P_{m-1})(P_m)\,, \\ [\zeta P_1\,;\, \dots \,;\, \zeta P_{m}] ~&=~ [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}][\zeta P_m],, ~~\hbox {and}~~\\ [\eta P_1\,;\, \dots \,;\, \eta P_{m}] ~&=~ [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}][\eta P_m] \end{aligned}$$

yields:

$$\begin{aligned}&s\,[P_1 \cdots P_{m}] ~-~ s\,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m}] ~+~ s\,[\eta P_1\,;\, \dots \,;\, \eta P_{m}] \\&=~ s\,[\zeta (P_1 \cdots P_{m-1}) \,;\, \zeta P_m] ~-~ s\,[\eta (P_1 \cdots P_{m-1}) \,;\, \eta P_m] \\&\quad -~ s\,\left[ \zeta [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] \,;\, \zeta [\zeta P_m]\right] ~+~ s\,\left[ \eta [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] \,;\, \eta [\zeta P_m]\right] \\&\quad +~ s\,\left[ \zeta [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] \,;\, \zeta [\eta P_m]\right] ~-~ s\,\left[ \eta [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] \,;\, \eta [\eta P_m]\right] ~=~ 0, \end{aligned}$$

since

$$\begin{aligned} \zeta [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] ~&=~ (\zeta P_1) \cdots (\zeta P_{m-1})\,, \\ \zeta [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] ~&=~ (\eta P_1) \cdots (\eta P_{m-1})\,, \\ \eta [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] ~&=~ [p_1\,,\, \dots \,,\, p_{m-1}] ~=~ \eta [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}]\,, \end{aligned}$$

and $\eta \,[\zeta P_m] = [p_m] = \eta \,[\eta P_m]$. $\square$

Lemma A.5

s has property (Z): for all $X_1\,,\, \dots \,,\, X_{n} \in F_3$,

$$\begin{aligned}&s\,(X_1 \cdots X_n) \\&{\mathop {=}\limits ^{?}}~ s\,[\varphi X_1 \,|\, \dots \,|\,\varphi X_n] ~-~ s\,[\chi X_1 \,|\, \dots \,|\,\chi X_n] \\&~~~~+~ s\,[\psi X_1 \,|\, \dots \,|\,\psi X_n] + {\sum \limits _{i=1}^{n}} \gamma s(X_i) \end{aligned}$$

(Z)

Proof

Property (Z) is proved by induction on n. If $n = 1$, then (Z) becomes property (Z1x) in Lemma A.4. Now let $n > 1$. Let

$$\begin{aligned} T = X_1 \cdots X_{n-1}\,,~~&U = [\varphi X_1 \,|\, \dots \,|\,\varphi X_{n-1}]\,, \\ V = [\chi X_1 \,|\, \dots \,|\,\chi X_{n-1}]\,,~~&W = [\psi X_1 \,|\, \dots \,|\,\psi X_{n-1}]\,. \end{aligned}$$

The induction hypothesis is

$$\begin{aligned} s\,(T) ~=~ s\,(U) ~-~ s\,(V) ~+~ s\,(W) ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i)\,. \end{aligned}$$

$\square$

Lemma A.6

$\varphi T = \varphi U$; $\chi T = \varphi V$; $\psi T = \varphi W$; $\chi U = \chi V$; $\psi U = \chi W$; and $\psi V = \psi W$.

Proof

This follows from $\varepsilon ^n_i\,\varepsilon ^{n+1}_{j+1} = \varepsilon ^n_j\,\varepsilon ^{n+1}_i$ in Lemma 0.1. If $\mathbb {X}= [X_1\,,\, \dots \,,\, X_{n-1}]$, then $T = \varepsilon ^3_3\, \mathbb {X}$, $U = \varepsilon ^3_2\, \mathbb {X}$, $V = \varepsilon ^3_1\, \mathbb {X}$, and $W = \varepsilon ^3_0\, \mathbb {X}$, whereas $\varphi = \varepsilon ^2_2$, $\chi = \varepsilon ^2_1$, and $\psi = \varepsilon ^2_0$. Hence

$$\begin{aligned} \varphi T ~=~ \varepsilon ^2_2\, \varepsilon ^3_3\, \mathbb {X}~=~ \varepsilon ^2_2\, \varepsilon ^3_2\, \mathbb {X}~=~ \varphi U, ~~~&\chi T ~=~ \varepsilon ^2_1\, \varepsilon ^3_3\, \mathbb {X}~=~ \varepsilon ^2_2\, \varepsilon ^3_1\, \mathbb {X}~=~ \varphi V, \\ \psi T ~=~ \varepsilon ^2_0\, \varepsilon ^3_3\, \mathbb {X}~=~ \varepsilon ^2_2\, \varepsilon ^3_0\, \mathbb {X}~=~ \varphi W, ~~~&\chi U ~=~ \varepsilon ^2_1\, \varepsilon ^3_2\, \mathbb {X}~=~ \varepsilon ^2_1\, \varepsilon ^3_1\, \mathbb {X}~=~ \chi V, \\ \psi U ~=~ \varepsilon ^2_0\, \varepsilon ^3_2\, \mathbb {X}~=~ \varepsilon ^2_1\, \varepsilon ^3_0\, \mathbb {X}~=~ \chi W, ~~~&\psi V ~=~ \varepsilon ^2_0\, \varepsilon ^3_1\, \mathbb {X}~=~ \varepsilon ^2_0\, \varepsilon ^3_0\, \mathbb {X}~=~ \psi W. \end{aligned}$$

$\square$

Resuming the proof of (Z), if in the above $X_n = [Q]$ has length 1, then (Z) reads

$$\begin{aligned}&s\,(X_1 \cdots X_{n-1}\, [Q]) \\&\quad {\mathop {=}\limits ^{?}}~ s\,\left[ \varphi X_1\,|\, \dots \,|\, \varphi X_{n-1} \,|\, \varphi \,[Q]\right] ~-~ s\,\left[ \chi X_1\,|\, \dots \,|\, \chi X_{n-1} \,|\, \chi \,[Q]\right] \\&\qquad +~ s\,\left[ \psi X_1\,|\, \dots \,|\, \psi X_{n-1} \,|\, \psi \,[Q]\right] ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,[Q]; \end{aligned}$$

equivalently

$$\begin{aligned} s\,(T\,[Q]) ~&{\mathop {=}\limits ^{?}}~ s\,\left( U\,[Q]\right) ~-~ s\,\left( V\,\left[ [\zeta Q]\right] \right) ~+~ s\,\left( W\,\left[ [\eta Q]\right] \right) \\&~~~~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,[Q]. \end{aligned}$$

(Za)

Using (X) in its equivalent form puts (Za) in the form

$$\begin{aligned}&\gamma s\,(T) ~+~\gamma s\,[Q] ~+~ s\,[\varphi T \,|\, Q] ~-~ s\,\left[ \chi T \,|\, [\zeta Q]\right] ~+~ s\,\left[ \psi T \,|\, [\eta Q]\right]&\hbox {(line 1)} \\&\quad {\mathop {=}\limits ^{?}}~ \gamma s\,(U) ~+~ \gamma s\,[Q] ~+~ s\,[\varphi U \,|\, Q] ~-~ s\,\left[ \chi U \,|\, [\zeta Q]\right] ~+~ s\,\left[ \psi U \,|\, [\eta Q]\right] \\&\qquad -~ \gamma s\,(V) ~-~ \gamma s\,\left[ [\zeta Q]\right] ~-~ s\,\left( \varphi V \,|\, \left[ \zeta \, [\zeta Q]\right] \right) \\&\qquad +~ s\,\left( \chi V \,|\, \left[ \zeta \, [\zeta Q]\right] \right) ~-~ s\,\left( \psi V \,|\, \left[ \eta \, [\zeta Q]\right] \right)&\hbox {(line 4)}\\&\qquad +~ \gamma s\,(W) ~+~ \gamma s\,\left[ [\eta Q]\right] ~+~ s\,\left( \varphi W \,|\, \left[ \zeta \, [\eta Q]\right] \right) \\&\qquad -~ s\,\left( \chi W \,|\, \left[ \zeta \, [\eta Q]\right] \right) ~+~ s\,\left( \psi W \,|\, \left[ \eta \, [\eta Q]\right] \right) \\&\qquad +~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,[Q].&\hbox {(line 7)} \end{aligned}$$

In this equality, terms

12 and 22,	13 and 23,	14 and 33,	15 and 53,
24 and 41,	25 and 61,	42 and 62

cancel each other, by (ii) in Lemma 0.1 and by Lemma A.6. Terms 72, 32, and 52 add up to 0, by (Z1a): $s\,[Q] - s\,[\zeta Q] + s\,[\eta Q] = 0$. The induction hypothesis cancels term 11 and the remaining terms 21, 31, 51, and 71. Hence (Za) holds; and (Z) holds when $X_n$ has length 1.

When $X_n$ has length at least 2, another small calculation proves (Z) by induction on the length of $X_n$. Let $X_n = Y\,[Q]$, where $Y \in F_3$ and $Q \in F_2$, so that

$$\begin{aligned} \varphi X_n&~=~ (\varphi Y)(\varphi \,[Q]) ~=~ (\varphi Y)\,Q, \\ \chi X_n&~=~ (\chi Y)(\chi \,[Q]) ~=~ (\chi Y)\,[\zeta Q]\,, ~~\hbox {and}~~\\ \psi X_n&~=~ (\psi Y)(\psi \,[Q]) ~=~ (\psi Y)\,[\eta Q]. \end{aligned}$$

The induction hypothesis is

$$\begin{aligned} s\,(X_1 \cdots X_{n-1}\, Y) ~=~&s\,[\varphi X_1 \,|\, \dots \,|\,\varphi X_{n-1} \,|\, \varphi Y] ~-~ s\,[\chi X_1 \,|\, \dots \,|\,\chi X_{n-1} \,|\, \chi Y] \\&+~ s\,[\psi X_1 \,|\, \dots \,|\,\psi X_{n-1} \,|\, \psi Y] ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,(Y); \end{aligned}$$

equivalently, with $T,\, U,\, V,\, W$ as before,

$$\begin{aligned} s\,(TY) ~=~ s\,(U\, [\varphi Y]) ~-~ s\,(V\, [\chi Y]) ~+~ s\,(W\, [\psi Y]) ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,(Y). \end{aligned}$$

Equality (Z):

$$\begin{aligned}&s\,(X_1 \cdots X_{n-1}\, Y\,[Q]) \\ ~{\mathop {=}\limits ^{?}}~&s\, (\varphi X_1 \,|\, \dots \,|\,\varphi X_{n-1} \,|\, (\varphi Y)\,Q) ~-~ s\, (\chi X_1 \,|\, \dots \,|\,\chi X_{n-1} \,|\, (\chi Y)\,[\zeta Q]) \\&\quad +~ s\, (\psi X_1 \,|\, \dots \,|\,\psi X_{n-1} \,|\, (\psi Y)\,[\eta Q]) ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,(Y\,[Q]) \end{aligned}$$

becomes

$$\begin{aligned} s\, (TY\,[Q]) ~\mathop {=}\limits ^{?}&~ s\, (U\, [(\varphi Y)\,Q]) ~-~ s\,\left( V\, \left[ (\chi Y)[\zeta Q]\right] \right) \\&~~+~ s\,\left( W\, \left[ (\psi Y)[\eta Q]\right] \right) ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,(Y\,[Q]). \end{aligned}$$

(Z)

By the induction hypothesis and (X),

$$\begin{aligned} s\,(TY)&=~ s\,(U\, [\varphi Y]) ~-~ s\,(V\, [\chi Y]) ~+~ s\,(W\, [\psi Y]) ~+~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,(Y)\\&=~ \gamma s\,(U) ~+~ \gamma s\,[\varphi Y] ~+~ s\, [\varphi U \,|\, \varphi Y] \\&\quad \quad -~ s\,\left[ \chi U \,|\, [\zeta \varphi Y]\right] ~+~ s\,\left[ \psi U \,|\, [\eta \varphi Y]\right] \\&\quad -~ \gamma s\,(V) ~-~ \gamma s\,[\chi Y] ~-~ s\, [\varphi V \,|\, \chi Y] \\&\quad \quad +~ s\,\left[ \chi V \,|\, [\zeta \chi Y]\right] ~+~ s\,\left[ \psi V \,|\, [\eta \chi Y]\right] \\&\quad +~ \gamma s\,(W) ~+~ \gamma s\,[\psi Y] ~+~ s\, [\varphi W \,|\, \psi Y] \\&\quad \quad -~ s\,\left[ \chi W \,|\, [\zeta \psi Y]\right] ~+~ s\,\left[ \psi W \,|\, [\eta \psi Y]\right] \\&\quad +~ {\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i) ~+~ \gamma s\,(Y)\,, \end{aligned}$$

so that (X) and (Z1c) yield

$$\begin{aligned}&s\, (TY\,[Q])\\&\quad = \gamma s\,(TY) + \gamma s\,[Q] + s\,[\varphi (TY),\,Q] - s\,\left[ \chi (TY),\,[\zeta Q]\right] \\&\qquad \qquad + s\,\left[ \psi (TY),\,[\eta Q]\right] \,, ~~\text{ by } \text{(X) }\\&\quad = \gamma s\,(U) + \gamma s\,[\varphi Y] + \gamma s\,[\varphi U \,|\, \varphi Y]\\&\qquad \quad \qquad - \gamma s\,\left[ \chi U \,|\, \zeta \varphi Y\right] + \gamma s\,\left[ \psi U \,|\, [\eta \varphi Y]\right]&\hbox {(line 2)} \\&\qquad \qquad - \gamma s\,(V) - \gamma s\,[\chi Y] - \gamma s\,[\varphi V \,|\, \chi Y] \\&\qquad \quad \qquad + \gamma s\,\left[ \chi V \,|\, \zeta \chi Y\right] -\gamma s\,\left[ \psi V \,|\, [\eta \chi Y]\right] \\&\qquad \qquad + \gamma s\,(W) + \gamma s\,[\psi Y] + \gamma s\,[\varphi W \,|\, \psi Y]&\hbox {(line 5)} \\&\qquad \qquad \quad - \gamma s\,\left[ \chi W \,|\, \zeta \psi Y\right] + \gamma s\,\left[ \psi W \,|\, [\eta \psi Y]\right] \\&\qquad \qquad + \sum \limits _{i=1}^{n-1} \gamma s\,(X_i) + \gamma s\,(Y)\\&\qquad \qquad \qquad \quad + \gamma s\,[Q]&\hbox {(line 8)} \\&\qquad \qquad \qquad \quad + s\, [\varphi T \,|\, (\varphi Y)\,Q] + \gamma s\,[\varphi T] + \gamma s\,[\varphi Y \,|\, Q]\\&\qquad \qquad \qquad \quad - s\, \left[ [\zeta \varphi T] \,|\, [\zeta \varphi Y \,;\, \zeta Q]\right] + s\, \left[ [\eta \varphi T] \,|\, [\eta \varphi Y \,;\, \eta Q]\right]&\hbox {(line A)}\\&\qquad \qquad \qquad \quad - \gamma s\,[\varphi T \,|\, \varphi Y] - \gamma s\,[Q]\\&\qquad \qquad \qquad \quad + s\, \left[ [\zeta \varphi T \,;\, \zeta \varphi Y] \,|\, [\zeta Q]\right] - s\, \left[ [\eta \varphi T \,;\, \eta \varphi Y] \,|\, [\eta Q]\right] \\&\qquad \qquad \qquad \quad - \gamma s\,\left[ \chi T \,|\, [(\chi Y) \,;\, [\zeta Q]]\right] - \gamma s\,[\chi T] - \gamma s\,\left[ \chi Y \,|\, [\zeta Q]\right]&\hbox {(line D)} \\&\qquad \qquad \qquad \quad + s\, \left[ [\zeta \chi T] \,|\, [\zeta \chi Y \,;\, \zeta [\zeta Q]]\right] - s\, [[\eta \chi T] \,|\, \eta \chi Y \,;\, \eta [\zeta Q]] \\&\qquad \qquad \qquad \quad + \gamma s\,[\chi T \,|\, \chi Y] + \gamma s\,\left[ [\zeta Q]\right] \\&\qquad \qquad \qquad \quad - s\, \left[ [\zeta \chi T \,;\, \zeta \chi Y] \,|\, \zeta [\zeta Q]]\right] + s\, \left[ [\eta \chi T \,;\, \eta \chi Y] \,|\, [\eta [\zeta Q]]\right]&\hbox {(line G)} \\&\qquad \qquad \qquad \quad + s\, \left[ \psi T \,|\, [(\psi Y) \,;\, [\eta Q]]\right] + \gamma s\,[\psi T] + \gamma s\,\left[ \psi Y \,|\, [\eta Q]\right] \\&\qquad \qquad \qquad \quad - s\, \left[ [\zeta \psi T] \,|\, [\zeta \psi Y \,;\, \zeta [\eta Q]]\right] + s\, \left[ [\eta \psi T] \,|\, [\eta \psi Y \,;\, \eta [\eta Q]]\right] \\&\qquad \qquad \qquad \quad - \gamma s\,\,[\psi T \,|\, \psi Y] - \gamma s\,\,\left[ [\eta Q]\right]&\hbox {(line J)} \\&\qquad \qquad \qquad \quad + s\, \left[ [\zeta \psi T \,;\, \zeta \psi Y] \,|\, [\zeta [\eta Q]]\right] - s\, \left[ [\eta \psi T \,;\, \eta \psi Y] \,|\, [\eta [\eta Q]]\right] . \end{aligned}$$

(7)

By (X),

$$\begin{aligned}&~ s\, (U\, [(\varphi Y)\,Q]) ~-~ s\,\left( V\, \left[ (\chi Y)[\zeta Q]\right] \right) \\&\qquad +~ s\,\left( W\, \left[ (\psi Y)[\eta Q]\right] \right) ~+~{\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i)~+~\gamma s\,(Y\,[Q])\\&\quad =~ \gamma s\,(U) ~+~ \gamma s\,[(\varphi Y)\,Q] ~+~ s\,[\varphi U \,|\, (\varphi Y)\,Q] \\&\qquad \qquad -~ s\, \left[ \chi U \,|\, [\zeta [(\varphi Y)\,Q]]\right] ~+~ s\, \left[ \psi U \,|\, [\eta [(\varphi Y)\,Q]]\right]&\hbox {(line 2)} \\&\qquad -~ \gamma s\,(V) ~-~ \gamma s\,\left[ (\chi Y)\,[\zeta Q]\right] ~-~ s\,\left[ \varphi V \,|\, (\chi Y)\,[\zeta Q]\right] \\&\qquad \qquad +~ s\, \left[ \chi V \,|\, \left[ \zeta [(\chi Y)\,[\zeta Q]]\right] \right] ~-~ s\, \left[ \psi V \,|\, \left[ \eta [(\chi Y)\,[\zeta Q]]\right] \right]&\hbox {(line 4)} \\&\qquad +~ \gamma s\,(W) ~+~ \gamma s\,\left[ (\psi Y)\,[\eta Q]\right] ~+~ s\,\left[ \varphi W \,|\, (\psi Y)\,[\eta Q]\right] \\&\qquad \qquad -~ s\, \left[ \chi W \,|\, \left[ \zeta [(\psi Y)\,[\eta Q]]\right] \right] ~+~ s\, \left[ \psi W \,|\, \left[ \eta [(\psi Y)\,[\eta Q]]\right] \right] \\&\qquad \qquad +~{\sum \limits _{i=1}^{n-1}} \gamma s\,(X_i)&\hbox {(line 7)} \\&\qquad \qquad +~ \gamma s\,(Y) ~+~ \gamma s\,[Q] ~+~ \gamma s\,[\varphi Y \,|\, Q] ~-~ \gamma s\,\left[ \chi Y \,|\, [\zeta Q]\right] \\&\quad \quad \quad \quad ~+~ \gamma s\,\left[ \psi Y \,|\, [\eta Q]\right] \,; \end{aligned}$$

(8)

and (Z1a), applied to $T,\, Y,\, Q$, and $Y\,[Q]$, yields

$$\begin{aligned} \gamma s\,[\varphi T] ~-~ \gamma s\,[\chi T] ~+~ \gamma s\,[\psi T] ~&=~ 0, \\ \gamma s\,[\varphi Y] ~-~ \gamma s\,[\chi Y] ~+~ \gamma s\,[\psi Y] ~&=~ 0, \\ \gamma s\,[Q] ~-~ \gamma s\,\left[ [\zeta Q]\right] ~+~ \gamma s\,\left[ [\eta Q]\right] ~&=~ 0, \\ \gamma s\,\left[ (\varphi Y)\,Q\right] ~-~ \gamma s\,\left[ (\chi Y)[\zeta Q]\right] ~+~ \gamma s\,\left[ (\psi Y)[\eta Q]\right] ~&=~ 0, \end{aligned}$$

(9)

since $\varphi \,(Y\,[Q]) = (\varphi Y)\, Q$, $\chi \,(Y\,[Q]) = (\chi Y)[\zeta Q]$, and $\psi \,(Y\,[Q]) = (\psi Y)[\eta Q]$.

The left hand side of (Z) is (7) and its right hand side is (8). The lines of (7) are numbered 1,...,9,A,B,...,K. Terms 712, 732, and 752, terms 792, 7D2, and 7H2, terms 7B2, 7F2, and 7J2, and terms 812, 832, and 852, add up to 0, by (9). In (7) and (8), terms

713 and 7B1,	721 and 741,	722 and 761,	733 and 7F1,	742 and 762,
753 and 7J1,	7A1 and 7E1,	7A2 and 7I1,	7C1 and 7G1,	7C2 and 7K1,
7E2 and 7I2,	7G2 and 7K2,	821 and 841,	822 and 861,	842 and 862

cancel in pairs, by Lemmas A.6 and 0.1. The remaining terms match in pairs:

711 = 811,	731 = 831,	751 = 851,	771 = 871,	772 = 881,	781 = 882,
791 = 813,	793 = 883,	7D1 = 833,	7D3 = 884,	7H1 = 853,	7H3 = 885,

by the same Lemmas. Thus (7) = (8), and (Z) holds. $\square$

It follows from Lemma A.5 that s is a 4-cocycle. Then (def 1a) and (def 1b) show that $\Theta _1 s = c$. This completes the proof of Proposition 1.3. $\square$

Proof of Proposition 1.4

Proposition 1.4

If $c \in C_1$, then $c = \Theta _1 s$ for some 4-coboundary s if and only if there exists a 3-cochain $u \in C^3 (S,G)$ such that, for all $P,Q \in F_2$:

Proof

It was shown in Sect. 1 that every 4-coboundary s has properties (B1a) and (B1b) for some $u \in C^3$. If $c = \Theta _1 s$, then c inherits these properties from s.

Conversely, let $c \in C_1$ satisfy (B1a) and (B1b) for some $u \in C^3$. Define s as in “Appendix A”:

for all $m \geqq 3$ and $P_1\,,\, \dots \,,\, P_{m} \in F_2$. Then $s\,(Q,P) = c\,(Q,P) = c\,(P,Q) = s\,(P,Q)$, so that $s\,(P,Q)$ depends only on [P|Q] and may be denoted by $s\,[P|Q]$.

Lemma B.1

s has properties (X) and (Y), in the form

for all $m \geqq 2$ and $P_1,\, \dots ,\, P_m \in F_2$.

Proof

Property (Ya) follows from (def 1c). Property (Xa) follows from (Ya): as in “Appendix A”, the terms of (Ya) that do not contain $P_m$ add up to $\gamma s\,[P_1\,,\, \dots \,,\, P_{m-1}]$. $\square$

The proof of Proposition 1.4 resumes with the proof that

$$\begin{aligned} s\,(P_1\,,\, \dots \,,\, P_{m}) ~&=~ u\,(P_1\cdots P_m) ~-~ u\,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m}] \\&~~~~+~ u\,[\eta P_1\,;\, \dots \,;\, \eta P_{m}] ~-~ {\sum \limits _{j=1}^{m}} \gamma u(P_j) \end{aligned}$$

(Ba)

for all $P_1,\, \dots ,\, P_m \in F_2$. First,

$$\begin{aligned} s\, (P) ~&=~ c\, (P) ~=~ u\,[\eta P] ~-~ u\,[\zeta P] ~~\hbox {and}~~\\ s\,(P,Q) ~&=~ c\, (P,Q) \\&=~ u\,(PQ) ~-~ u\,[\zeta P \,;\, \zeta Q] ~+~ u\,[\eta P \,;\, \eta Q] ~-~ \gamma u\,(P) ~-~ \gamma u\,(Q) \end{aligned}$$

for all $P,\, Q \in F_2$, by (B1a) and (B1b). Hence (Ba) holds for s whenever $m \leqq 2$. When $m \geqq 3$, (B) is proved by induction on m. By (Xa), the induction hypothesis, and (ii) in Lemma 0.1:

$$\begin{aligned}&s\, (P_1\,,\, \dots \,,\, P_{m}) \\&\quad =~ \gamma s\,(P_1\,,\, \dots \,,\, P_{m-1}) ~+~ \gamma s\,[P_m] ~+~ s\,[P_1 \cdots P_{m-1} \,|\, P_m] \\&\qquad -~ s\, \left[ [\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] \,|\, [\zeta P_m]\right] ~+~ s\, \left[ [\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] \,|\, [\eta P_m]\right] \\&\quad =~ \gamma u\,(P_1 \cdots P_{m-1}) ~-~ \gamma u\,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}]&\hbox {(line 1)}\\&\qquad \quad +~ \gamma u\,[\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] ~-~ {\sum \limits _{j=1}^{m-1}} \gamma u\,(P_j) \\&\qquad +~ \gamma u\,[\eta P_m] ~-~ \gamma u\,[\zeta P_m]&\hbox {(line 3)} \\&\qquad +~ u\,(P_1 \cdots P_{m}) ~-~ u\,[\zeta (P_1 \cdots P_{m-1}) \,;\, \zeta (P_m)]\\&\qquad \quad +~ u\,[\eta (P_1 \cdots P_{m-1}) \,;\, \eta (P_m)] ~-~ \gamma u\,(P_1 \cdots P_{m-1}) ~-~ \gamma u\,(P_m)&\hbox {(line 5)} \\&\qquad -~ u\, ([\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}][\zeta P_m]) ~+~ u\,[(\zeta P_1) \cdots (\zeta P_{m-1}) \,;\, \zeta P_m]\\&\qquad \quad -~ u\,[\pi P_1\,,\, \dots \,,\, \pi P_{m-1}] \,;\, [\pi P_m]) ~+~ \gamma u\,[\zeta P_1\,;\, \dots \,;\, \zeta P_{m-1}] ~+~ \gamma u\,[\zeta P_m]\\&\qquad +~ u\, ([\eta P_1\,;\, \dots \,;\, \eta P_{m-1}][\eta P_m]) ~-~ u\,[(\eta P_1) \cdots (\eta P_{m-1}) \,;\, \eta P_m]&\hbox {(line 8)} \\&\qquad \quad +~ u\,[\pi P_1\,,\, \dots \,,\, \pi P_{m-1}] \,;\, [\pi P_m]) ~-~ \gamma u\,[\eta P_1\,;\, \dots \,;\, \eta P_{m-1}] ~-~ \gamma u\,[\eta P_m]\,. \end{aligned}$$

In this equality, the following terms cancel each other:

11 and 52,	12 and 72,	21 and 92,	31 and 93,	32 and 73,	71 and 91,
42 and 62,	51 and 82	(since $\zeta$ and $\eta$ are homomorphisms).

The remaining terms, 41, 61, 81, 22, and 53 yield (Ba).

By (Ba), $s\,(P_1\,,\, \dots \,,\, P_{m})= s\,(P_{\sigma 1} \,,\, \dots \,,\,P_{\sigma m})$ for every permutation $\sigma$ of 1, 2, $\dots ,\, m$. Hence $s\,(P_1\,,\, \dots \,,\, P_{m})$ depends only on $[P_1\,|\, \dots \,|\, P_{m} ]$ and s can be regarded as a 4-cochain. Then (B) holds and s is a 4-coboundary. $\square$

Proofs of Lemma 2.2 and Proposition 2.3

Lemma 2.2

Every $s \in Z_1$ has properties

Proof

First, (P2) holds, since $[B|A] = [A|B]$.

If $P = \left[ [a]\right]$, then $\zeta P = [a]$, $\eta P = [a]$, and (Z1a): $s\,(P) - s\,([\zeta P]) + s\,(\eta P]) = 0$ yields (Z2a):

$$\begin{aligned} 0 ~=~ s\,(P) ~-~ s\,(\zeta P) ~+~ s\,(\eta P) ~=~ s\,[a] ~-~ s\,[a] ~+~ s\,[a]\,. \end{aligned}$$

If $P = [A]$ and $Q = [B]$, then $\zeta P = A$, $\zeta Q = B$ and $\eta P = [a]$, $\eta Q = [b]$, and (Z1b): $s\,([\zeta P], [\zeta Q]) - s\,([\eta P], [\eta Q]) = \gamma s\,(P) + \gamma s\,(Q)$ yields (Z2b).

If $P = [A]$ and $R = [C]$, then (Z1c) becomes (Z2c):

$$\begin{aligned}&s\,(PQ|R) ~+~ \gamma s\,(P|Q) ~+~ \gamma s\,(R) \\&\qquad -~ s\,([\zeta P \,;\, \zeta Q] \,|\, [\zeta R]) ~+~ s\,([\eta P \,;\, \eta Q] \,|\, [\eta R]) \\&\quad =~ s\,(P|QR) ~+~ \gamma s\,(P) ~+~ \gamma s\,(Q|R) \\&\qquad -~ s\,([\zeta P] \,|\, [\zeta Q \,;\, \zeta R]) ~+~ s\,([\eta P] \,|\, [\eta Q \,;\, \eta R]) \end{aligned}$$

To prove (Z2d), apply (Z1c) three times: to [A], [B], [C; D]:

$$\begin{aligned}&s\,(A;B \,|\, C;D) ~+~ \gamma s\,(A|B) ~+~ \gamma s\,(C;D) \\&\qquad -~ s\,(A;B \,|\, CD) ~+~ s\,([a];[b] \,|\, [c,d])\\&\quad =~ s\,(B;C;D \,|\, A) ~+~ \gamma s\,(A) ~+~ \gamma s\,(C;D \,|\, B) \\&\qquad -~ s\,(B;CD \,|\, A) ~+~ s\,([b];[c,d] \,|\, [a]); \end{aligned}$$

(a)

to [D], [B; C], [A]:

$$\begin{aligned}&s\,(B;C;D \,|\, A) ~+~ \gamma s\,(B;C \,|\, D) ~+~ \gamma s\,(A) \\&\qquad -~ s\,(BC;D \,|\, A) ~+~ s\,([b,c];[d] \,|\, [a])\\&\quad =~ s\,(A;B;C \,|\, D) ~+~ \gamma s\,(D) ~+~ \gamma s\,(B;C \,|\, A) \\&\qquad -~ s\,(A;BC \,|\, D) ~+~ s\,([a];[b,c] \,|\, [d]); \end{aligned}$$

(b)

and to [A; B], [C], [D]:

$$\begin{aligned}&s\,(A;B;C \,|\, D) ~+~ \gamma s\,(A;B \,|\, C) ~+~ \gamma s\,(D) \\&\qquad -~ s\,(AB;C \,|\, D) ~+~ s\,([a,b];[c] \,|\, [d])\\&\quad =~ s\,(A;B \,|\, C;D) ~+~ \gamma s\,(A;B) ~+~ \gamma s\,(C|D) \\&\qquad -~ s\,(C;D \,|\, AB) ~+~ s\,([c];[d] \,|\, [a,b]). \end{aligned}$$

(c)

Finally, (Z1a): $s\,(P) = s\,(\zeta P) - s\,(\eta P)$ yields:

$$\begin{aligned} \gamma s\,[C;D] ~&=~ \gamma s\,(CD) ~-~ \gamma s\,[c,d] ~~\hbox {and}~~\\ \gamma s\,[A;B] ~&=~ \gamma s\,(AB) ~-~ \gamma s\,[a,b]\,. \end{aligned}$$

(d)

When equalities (a), (b), (c), (d) are added, terms common to both sides cancel out:

a11 and c31,	a13 and d11,	a31 and b11,	a32 and b13,
b31 and c11,	b32 and c13,	c32 and d31.

The remaining terms are the terms of (Z2d): b12, c21, b41, a21, c12; a12, c22, b42; a22, d12, d13; b33, a41, b21, c41, a33; c33, a42, b22; c42, d32, and d33. This proves (Z2d). $\square$

Proposition 2.3

When $c \in C_2$, then $c = \Theta _2 s$ for some $s \in Z_1$ if and only if it has properties (P2), (Z2a), (Z2b), (Z2c), and (Z2d).

Proof

By Lemma 2.2, every $s \in Z_1$ has these properties, which are then inherited by $\Theta _2 s$.

Conversely, let $c \in C_2$ have properties (P2), (Z2a), (Z2b), (Z2c), and (Z2d). Inspired by (Z1a) and (Y2) in Lemma 2.1, define s by

for all $P \in F_2$, $m \geqq 1$, and $B_1\,,\, \dots \,,\, B_{m} \in F_1$. Note that $s\, (P \,|\, B_1\,,\, \dots \,,\, B_{m})$ may depend on the order of $B_1\,,\, \dots \,,\, B_{m}$; that it does not is Corollary C.5 below.

If $n=1$, then (def 2a) reads $s\,(A_n) = c\,(A_n)$. If $m=1$, then (def 2b) reduces to $s\,(P \,|\, B_m) = c\, (P \,|\, B_m)$. Thus $s\,(A) = c\,(A)$, and $s\,(P|A) = c\,(P|A)$, for all $A \in F_1$ and $P \in F_2$. Hence:

Lemma C.1

s has properties (P2), (Z1a), (Z2a), (Z2b), (Z2c), and (Z2d). $\square$

Moreover, $c = \Theta _2 s$ will follow from $s \in Z_1$. The latter follows from three lemmas.

Lemma C.2

s has properties (Z1b) and

$$\begin{aligned}&s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m}) \\&\quad =~ \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m-1}) ~+~ \gamma s\,(B_m) ~-~ \gamma s\,(P) \\&\qquad +~ s\,(P\,[B_1\,;\, \dots \,;\, B_{m-1}] \,|\, B_m) ~-~ \gamma s\,([B_1\,;\, \dots \,;\, B_{m-1}] \,|\, B_m) \\&\qquad -~ s\,(\zeta P \,;\, B_1 \cdots B_{m-1} \,|\, B_m) ~+~ s\,(\eta P \,;\, [b_1\,,\, \dots \,,\, b_{m-1}] \,|\, [b_m]) \\&\qquad +~ s\,(B_1 \cdots B_{m-1} \,;\, B_m \,|\, \zeta P) ~-~ s\,([b_1\,,\, \dots \,,\, b_{m-1}] \,;\, [b_m] \,|\, \eta P)\,, \end{aligned}$$

(X2)

for all $m \geqq 2$, $P \in F_2$, and $B_1\,,\, \dots \,,\, B_{m} \in F_1$; equivalently,

$$\begin{aligned}&s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m} \,,\, C) \\&\quad =~ \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m}) ~+~ \gamma s\,(C) ~-~ \gamma s\,(P) ~+~ s\,(PQ \,|\, C) ~-~ \gamma s\,(Q \,|\, C) \\&\qquad -~ s\,(\zeta P \,;\, \zeta Q \,|\, C) ~+~ s\,(\eta P \,;\, \eta Q \,|\, [c]) \\&\qquad +~ s\,(\zeta Q \,;\, C \,|\, \zeta P) ~-~ s\,(\eta Q \,;\, [c] \,|\, \eta P)\,, \end{aligned}$$

(X2a)

for all $P,\, Q = [B_1\,;\, \dots \,;\, B_{m}] \in F_2$ and $C \in F_1$.

Proof

If $m \geqq 2$, the terms of (def 2b) that do not contain $B_m$ add up to $\gamma s\,(P \,|\, B_1 \,,\,$$\cdots \,,\, B_{m-1})$; adding the remaining terms yields (X2).

Property (Z1b) follows from (Z2b) and (Z1a):

$$\begin{aligned}&s\,(\zeta P \,|\, \zeta Q) - s\,(\eta P \,|\, \eta Q) \\&\quad = \gamma s\,(\zeta P) + \gamma s\,(\zeta Q) + s\,[p|q] - \gamma s\,(\eta P) - \gamma s\,(\eta Q) - s\,[p|q] \\&\quad = \gamma s\,(P) + \gamma s\,(Q)\,. \end{aligned}$$

$\square$

Lemma C.3

For all $P,\, Q = [B_1\,;\, \dots \,;\, B_{n}],\, R = [C_1\,;\, \dots \,;\, C_{m}] \in F_2$,

$$\begin{aligned}&s\,(PQ \,|\, C_1\,,\, \dots \,,\, C_{m}) + \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n}) + \gamma s\,(R) \\&\quad - s\,(\zeta P \,;\, \zeta Q \,|\, \zeta R) + s\,(\eta P \,;\, \eta Q \,|\, \eta R) \\&= s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n} \,,\, C_1\,,\, \dots \,,\, C_{m}) + \gamma s\,(P) + \gamma s\,(Q \,|\, C_1\,,\, \dots \,,\, C_{m}) \\&\quad - s\,(\zeta Q \,;\, \zeta R \,|\, \zeta P) + s\,(\eta Q \,;\, \eta R \,|\, \eta P)\,. \end{aligned}$$

(Z1ca)

Proof

If $R = [C]$ has length 1, then (X2a) yields

$$\begin{aligned}&s\, (P \,|\, B_1\,,\, \dots \,,\, B_{n} \,,\, C) \\&\quad = \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n}) + \gamma s\,(C) - \gamma s\,(P) + s\,(PQ \,|\, C) - \gamma s\,(Q \,|\, C)\\&\qquad - s\,(\zeta P \,;\, \zeta Q \,|\, C) + s\,(\eta P \,;\, \eta Q \,|\, [c]) + s\,(\zeta Q \,;\, C \,|\, \zeta P) - s\,(\eta Q \,;\, [c] \,|\, \eta P)\,. \end{aligned}$$

Hence

$$\begin{aligned}&s\, (P \,|\, B_1\,,\, \dots \,,\, B_{n} \,,\, C) + \gamma s\,(P) + \gamma s\,(Q \,|\, C)\\&\qquad - s\,(\zeta Q \,;\, C \,|\, \zeta P) + s\,(\eta Q \,;\, [c] \,|\, \eta P) \\&\quad = \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n}) + \gamma s\,(C) + s\,(PQ \,|\, C)\\&\qquad - s\,(\zeta P \,;\, \zeta Q \,|\, C) + s\,(\eta P \,;\, \eta Q \,|\, [c])\,, \end{aligned}$$

and (Z1ca) holds when $|R| = 1$.

If R has length $m \geqq 2$, then (Z1ca) is proved by induction on m. The equality

$$\begin{aligned}&s\,(PQ \,|\, C_1\,,\, \dots \,,\, C_{m} \,,\, C) + \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n}) + \gamma s\,(R\,[C]) \\&\qquad - s\,(\zeta P \,;\, \zeta Q \,|\, (\zeta R)\, C) + s\,(\eta P \,;\, \eta Q \,|\, (\eta R)[c]) \\&\quad {\mathop {=}\limits ^{?}} s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n} \,,\, C_1\,,\, \dots \,,\, C_{m} \,,\, C) + \gamma s\,(P) + \gamma s\,(Q \,|\, C_1\,,\, \dots \,,\, C_{m} \,,\, C)\\&\qquad - s\,(\zeta Q \,;\, (\zeta R)\, C \,|\, \zeta P) + s\,(\eta Q \,;\, (\eta R)[c] \,|\, \eta P) \end{aligned}$$

(Z1ca)

follows from the induction hypothesis:

$$\begin{aligned}&\gamma s\,(PQ \,|\, C_1\,,\, \dots \,,\, C_{m}) + \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n}) + \gamma s\,(R) \\&\qquad - \gamma s\,(\zeta P \,;\, \zeta Q \,|\, \zeta R) + \gamma s\,(\eta P \,;\, \eta Q \,|\, \eta R) \\&\quad = \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{n} \,,\, C_1\,,\, \dots \,,\, C_{m}) + \gamma s\,(P) + \gamma s\,(Q \,|\, C_1\,,\, \dots \,,\, C_{m}) \\&\qquad - \gamma s\,(\zeta Q \,;\, \zeta R \,|\, \zeta P) + \gamma s\,(\eta Q \,;\, \eta R \,|\, \eta P)\,; \end{aligned}$$

(1)

from (X2a):

from (Z2d), applied to $\zeta P,\, \zeta Q,\, \zeta R,\, C$:

$$\begin{aligned}&\gamma s\,(\zeta Q \,;\, \zeta R \,|\, C) - s\,(\zeta (PQ) \,;\, \zeta R \,|\, C) + s\,(\zeta P \,;\, \zeta \,(QR) \,|\, C) \\&\qquad - s\,(\zeta P \,;\, \zeta Q \,|\, (\zeta R)\,C) + \gamma s\,(\zeta P \,;\, \zeta Q \,|\, \zeta R)&\hbox {(line 2)} \\&\qquad + \gamma s\,(\zeta P \,|\, \zeta Q) + s\,([p,q];[r] \,|\, [c]) - s\,([p] \,;\, [q,r] \,|\, [c]) \\&\qquad + s\,([p];[q] \,|\, [r,c]) + \gamma s\,\left( (\zeta R)\,C\right) - \gamma s\,[r,c] \\ =&\gamma s\,(\zeta Q \,;\, \zeta R \,|\, \zeta P) - s\,(\zeta Q \,;\, (\zeta R)\,C \,|\, \zeta P) + s\,(\zeta (QR) \,;\, C \,|\, \zeta P) \\&\qquad - s\,\left( \zeta R \,;\, C \,|\, \zeta (PQ)\right) + \gamma s\,(\zeta R \,;\, C \,|\, \zeta Q)&\hbox {(line 6)} \\&\qquad + \gamma s\,(\zeta R \,|\, C) + s\,([q];[r,c] \,|\, [p]) - s\,([q,r];[c] \,|\, [p]) \\&\qquad + s\,([r] \,;\, [c] \,|\, [p,q]) + \gamma s\,\left( \zeta (PQ)\right) - \gamma s\,[p,q]\,; \end{aligned}$$

(5)

from (Z2d), applied to $[c],\, \eta R,\, \eta Q,\, \eta P$:

$$\begin{aligned}&\gamma s\,(\eta Q \,;\, \eta R \,|\, \eta P) - s\,(\eta Q \,;\, (\eta R)[c] \,|\, \eta P) + s\,(\eta (QR) \,;\, [c] \,|\, \eta P) \\&\qquad - s\,\left( \eta R \,;\, [c] \,|\, \eta (PQ)\right) + \gamma s\,(\eta R \,;\, [c] \,|\, \eta Q)&\hbox {(line 2)} \\&\qquad + \gamma s\,(\eta R \,|\, [c]) + s\,([q];[r,c] \,|\, [p]) - s\,([q,r];[c] \,|\, [p]) \\&\qquad + s\,([r] \,;\, [c] \,|\, [p,q]) + \gamma s\,\left( \eta (PQ)\right) - \gamma s\,[p,q] \\&\quad =\gamma s\,(\eta Q \,;\, \eta R \,|\, [c]) - s\,(\eta (PQ) \,;\, \eta R \,|\, [c]) + s\,(\eta P \,;\, \eta \,(QR) \,|\, [c]) \\&\qquad - s\,(\eta P \,;\, \eta Q \,|\, (\eta R)[c]) + \gamma s\,(\eta P \,;\, \eta Q \,|\, \eta R)&\hbox {(line 6)} \\&\qquad + \gamma s\,(\eta P \,|\, \eta Q) + s\,([p,q];[r] \,|\, [c]) - s\,([p] \,;\, [q,r] \,|\, [c]) \\&\qquad + s\,([p];[q] \,|\, [r,c]) + \gamma s\,\left( (\eta R)\,[c]\right) - \gamma s\,[r,c]\,; \end{aligned}$$

(6)

finally, from (Z1a) applied to PQ and $R\,[C]$, and from (Z1b) applied to P, Q and to $R,\,[C]$:

$$\begin{aligned} \gamma s\,(PQ)&= \gamma s\,\left( \zeta (PQ)\right) - \gamma s\,\left( \eta (PQ)\right) \,, \\ \gamma s\,(R\,[C])&= \gamma s\,\left( (\zeta R)\, C\right) - \gamma s\,((\eta R)[c])\,,\\ \gamma s\,(\zeta P \,|\, \zeta Q) - \gamma s\,(\eta P \,|\, \eta Q)&= \gamma s\,(P) + \gamma s\,(Q)\,, \\ \gamma s\,(R) + \gamma s\,(C)&= \gamma s\,(\zeta R \,|\, C) - \gamma s\,(\eta R \,|\, [c])\,, \end{aligned}$$

(7)

thanks to Lemmas C.1 and C.2. When equalities (1) through (7) are added, terms

111 and 221,	113 and 741,	121 and 522,	122 and 662,	131 and 311,
133 and 411,	141 and 551,	142 and 611,	222 and 312,	223 and 711,
231 and 321,	232 and 422,	241 and 512,	242 and 652,	251 and 561,
252 and 621,	313 and 733,	322 and 421,	331 and 513,	332 and 653,
341 and 553,	342 and 613,	412 and 742,	413 and 734,	431 and 511,
432 and 651,	441 and 562,	442 and 622,	531 and 731,	532 and 672,
533 and 673,	541 and 681,	542 and 722,	543 and 683,	571 and 743,
572 and 632,	573 and 633,	581 and 641,	582 and 712,	583 and 643,
631 and 744,	642 and 713,	671 and 732,	682 and 723

cancel in pairs. The remaining terms, 211, 112, 721; 521, 661; 351, 132, 451; 552, and 612, constitute (Z1ca). Thus (Z1ca) follows from (1) through (7). $\square$

Lemma C.4

For all $P = [A_1\,;\, \dots \,;\, A_{n}]$ and $Q = [B_1\,;\, \dots \,;\, B_{m}] \in F_2$,

$$\begin{aligned} s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m}) = s\,(Q \,|\, A_1\,,\, \dots \,,\, A_{n}) \end{aligned}$$

(P1a)

Proof

The following particular case of (P1a):

$$\begin{aligned} s\,(B \,|\, A_1\,,\, \dots \,,\, A_{n}) = s\,([A_1\,;\, \dots \,;\, A_{n}] \,|\, B) \end{aligned}$$

(P1b)

holds when $n = 1$, by (P2), and is proved by induction on n if $n \geqq 2$. With $P = [A_1\,;\, \dots \,;\, A_{n}]$, (X2a) and the induction hypothesis yield:

$$\begin{aligned}&s\,(B \,|\, A_1\,,\, \dots \,,\, A_{n} \,,\, C) \\&\quad = \gamma s\,(B \,|\, A_1\,,\, \dots \,,\, A_{n}) + \gamma s\,(C) - \gamma s\,(B) + s\,([B]\,P \,|\, C) - \gamma s\,(P \,|\, C) \\&\qquad - s\,(B \,;\, \zeta P \,|\, C) + s\,([b] \,;\, \eta P \,|\, [c]) + s\,(\zeta P \,;\, C \,|\, B) - s\,(\eta P \,;\, [c] \,|\, [b]) \\&\quad = \gamma s\,(P \,|\, B) + \gamma s\,(C) - \gamma s\,(B) + s\,([B]\,P \,|\, C) - \gamma s\,(P \,|\, C) \\&\qquad - s\,(B \,;\, \zeta P \,|\, C) + s\,([b] \,;\, \eta P \,|\, [c]) + s\,(\zeta P \,;\, C \,|\, B) - s\,(\eta P \,;\, [c] \,|\, [b])\,. \end{aligned}$$

Applying (Z2c) to C, P, B then yields

$$\begin{aligned}&s\,(P\,[C] \,|\, B) \\&= s\,([B]\,P \,|\, C) + \gamma s\,(C) + \gamma s\,(P \,|\, B) - s\,(B \,;\, \zeta P \,|\, C) + s\,([b] \,;\, \eta P \,|\, [c]) \\&\quad - \gamma s\,(P \,|\, C) - \gamma s\,(B) + s\,(C \,;\, \zeta P \,|\, B) - s\,([c] \,;\, \eta P \,|\, [b]) \\&= s\,(B \,|\, A_1\,,\, \dots \,,\, A_{n} \,,\, C)\,. \end{aligned}$$

Thus (P1a) holds if $m = 1$.

If $m \geqq 2$ the proof is by induction on m. By (X2a) and the induction hypothesis,

$$\begin{aligned}&s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m} \,,\, C) \\&\quad = \gamma s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m}) + \gamma s\,(C) - \gamma s\,(P) + s\,(PQ \,|\, C) - \gamma s\,(Q \,|\, C) \\&\qquad - s\,(\zeta P \,;\, \zeta Q \,|\, C) + s\,(\eta P \,;\, \eta Q \,|\, [c]) + s\,(\zeta Q \,;\, C \,|\, \zeta P) - s\,(\eta Q \,;\, [c] \,|\, \eta P)\\&\quad = \gamma s\,(Q \,|\, A_1\,,\, \dots \,,\, A_{n}) + \gamma s\,(C) - \gamma s\,(P) + s\,(PQ \,|\, C) - \gamma s\,(Q \,|\, C) \\&\qquad - s\,(\zeta P \,;\, \zeta Q \,|\, C) + s\,(\eta P \,;\, \eta Q \,|\, [c]) + s\,(\zeta Q \,;\, C \,|\, \zeta P) - s\,(\eta Q \,;\, [c] \,|\, \eta P)\,. \end{aligned}$$

Applying (Z1ca) to C, Q, P then yields

$$\begin{aligned}&s\,(Q\,[C] \,|\, A_1\,,\, \dots \,,\, A_{n}) \\&\quad = s\,(C \,|\, B_1\,,\, \dots \,,\, B_{m} \,,\, A_1\,,\, \dots \,,\, A_{n}) + \gamma s\,(C) + \gamma s\,(Q \,|\, A_1\,,\, \dots \,,\, A_{n}) \\&\qquad - s\,(\zeta Q \,;\, \zeta P \,|\, C) + s\,(\eta Q \,;\, \eta P \,|\, [c]) \\&\qquad - \gamma s\,(C \,|\, B_1\,,\, \dots \,,\, B_{m}) - \gamma s\,(P) + s\,(C \,;\, \zeta Q \,|\, \zeta P) - s\,([c] \,;\, \eta Q \,|\, \eta P)\,. \end{aligned}$$

By (P1b),

$$\begin{aligned}&s\,(C \,|\, B_1\,,\, \dots \,,\, B_{m} \,,\, A_1\,,\, \dots \,,\, A_{n}) = s\,(PQ \,|\, C) ~~\hbox {and}~~\\&\quad \gamma s\,(C \,|\, B_1\,,\, \dots \,,\, B_{m}) = \gamma s\,(Q \,|\, C)\,; \end{aligned}$$

hence $s\,(Q\,[C] \,|\, A_1\,,\, \dots \,,\, A_{n}) = s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m} \,,\, C)$. $\square$

Corollary C.5

$s\, (P \,|\, B_{\sigma 1} \,,\, \dots \,,\,B_{\sigma m}) = s\, (P \,|\, B_1\,,\, \dots \,,\, B_{m})$ for every permutation $\sigma$ of $1 \,,\, \dots \,,\,m$ and every $P \in F_2$ and $B_1\,,\, \dots \,,\, B_{m} \in F_1$.

Proof

With $P = [A_1\,;\, \dots \,;\, A_{k}]$, Lemma C.4 yields

$$\begin{aligned}&s\, (P \,|\, B_{\sigma 1} \,,\, \dots \,,\,B_{\sigma m}) = s\, ([B_{\sigma 1} \,;\, \dots \,;\,B_{\sigma m}] \,|\, A_1\,,\, \dots \,,\, A_{k}) \\&= s\, ([B_1\,;\, \dots \,;\, B_{m}] \,|\, A_1\,,\, \dots \,,\, A_{k}) = s\, (P \,|\, B_1\,,\, \dots \,,\, B_{m}) \,. \end{aligned}$$

$\square$

By Corollary C.5, $s\,(P|Q)$ is well defined by: $s\,(P \,|\, Q) = s\,(P \,|\, B_1\,,\, \dots \,,\, B_{m})$, for all P and $Q = [B_1\,;\, \dots \,;\, B_{m}] \in F_2$. Hence s has property (P1) and $s \in C_1$. Then Lemmas C.2 and C.3 show that $s \in Z_1$, so that $c = \Theta _2 s$. This completes the proof of Proposition 2.3. $\square$

Proof of Proposition 3.2

Proposition 3.2

If $c \in C_3$, then $c = \Theta _3 s$ for some $s \in Z_2$ if and only if it has properties

Proof

Every $s \in Z_2$ has properties (P2), (Z2a), (Z2b), and (Z2d), by definition. With $Q = [B]$, property (Z2c):

$$\begin{aligned}&s\,([A]\,Q \,|\, C) + \gamma s\,(Q \,|\, A) + \gamma s\,(C) - s\,(A \,;\, \zeta Q \,|\, C) + s\,([a] \,;\, \eta Q \,|\, [c]) \\&= s\,([C]\,Q \,|\, A) + \gamma s\,(Q \,|\, C) + \gamma s\,(A) - s\,(C \,;\, \zeta Q \,|\, A) + s\,([c] \,;\, \eta Q \,|\, [a]) \end{aligned}$$

yields (Z3c):

$$\begin{aligned}&s\,(A \,;\, B \,|\, C) + \gamma s\,(B \,|\, A) + \gamma s\,(C) - s\,(A \,;\, B \,|\, C) + s\,([a] \,;\, [b] \,|\, [c]) \\&= s\,(C \,;\, B \,|\, A) + \gamma s\,(C \,|\, B) + \gamma s\,(A) - s\,(C \,;\, B \,|\, A) + s\,([c] \,;\, [b] \,|\, [a])\,. \end{aligned}$$

To prove (Z3e), apply (Z2c) to $C,\, [B \,;\, E],\, A$, then to $A,\, [C \,;\, E],\, B$, and then to $B,\, [A \,;\, E],\, C$:

When equalities (1), (2), and (3) are added, terms

111 and 231,	113 and 233,	131 and 311,	133 and 313,
211 and 331,	213 and 333

cancel each other. The remaining terms, 141, 121, 241; 221, 341, 321; 112, 232, 212; 332, 312, 132; 142, 122, 242; 222, 342, and 322, constitute (Z3e). Hence (Z3e) holds for s.

If $c = \Theta _3 s$, then c inherits properties (P2), (Z2a), (Z2b), (Z3c), (Z2d), and (Z3e) from s.

Conversely, assume that $c \in C_3$ has properties (P2), (Z2a), (Z2b), (Z3c), (Z2d), and (Z3e). Inspired by (Z2c), define s recursively by:

$$\begin{aligned}&s\, (A) = c\, (A)\,, \\&s\, (P \,|\, B) = c\, (P \,|\, B) ~~\hbox {if}~~|P| \leqq 2 ~~{\hbox {or if}}~~B \leqq P, \\&s\, (P \,|\, B) \\&\quad = c\, ([B]\,Q \,|\, M) + \gamma s\,(Q \,|\, B) + \gamma c\,(M) - \gamma s\,(Q \,|\, M) - \gamma c\,(B) \\&\qquad - c\,(B \,;\, \zeta Q \,|\, M) + c\, ([b] \,;\, \eta Q \,|\, [m]) + c\,(M \,;\, \zeta Q \,|\, B) - c\, ([m] \,;\, \eta Q \,|\, [b])\\&\qquad ~~\hbox {if}~~|P| \geqq 3, B \geqq M = \mathrm{min}\,P, ~~\hbox {and}~~P = [M]\,Q, \end{aligned}$$

(def 3)

for all $A,B \in F_1$ and $P \in F_2$. The two definitions of $s\, (P \,|\, B)$ agree when $|P| \geqq 3$ and $B = M$.

Now $s \in C_2$ and s inherits properties (P2), (Z2a), (Z2b), (Z2d), (Z3c), and (Z3e) from c. It remains to show that s also has property (Z2c), so that $s \in Z_2$.

If $|Q| = 1$, then (Z2c) reduces to (Z3c) and holds. If $|Q| \geqq 2$, then (Z2c) is proved by induction on |Q|. Let $M = \mathrm{min}\,Q$. There are three cases to consider.

Case 1: $A \leqq C, M$. Then $A \leqq [C]\,Q$ and $A = \mathrm{min}\,([A]\,Q)$. By (def 3), $s\, ([C]\,Q \,|\, A) = c\, ([C]\,Q \,|\, A)$ and (Z2c) holds:

$$\begin{aligned} s\, ([A]\,Q \,|\, C)&= c\, ([C]\,Q \,|\, A) + \gamma s\,(Q \,|\, C) + \gamma c\,(A) - \gamma s\,(Q \,|\, A) - \gamma c\,(C)\\&\quad - c\,(C \,;\, \zeta Q \,|\, A) + c\, ([c] \,;\, \eta Q \,|\, [a]) \\&\quad + c\,(A \,;\, \zeta Q \,|\, C) - c\, ([a] \,;\, \eta Q \,|\, [c]) \\&= s\, ([C]\,Q \,|\, A) + \gamma s\,(Q \,|\, C) + \gamma s\,(A) - \gamma s\,(Q \,|\, A) - \gamma s\,(C)\\&\quad - s\,(C \,;\, \zeta Q \,|\, A) + s\, ([c] \,;\, \eta Q \,|\, [a]) \\&\quad + s\,(A \,;\, \zeta Q \,|\, C) - s\, ([a] \,;\, \eta Q \,|\, [c])\,; \end{aligned}$$

Case 2: $C \leqq A, M$: follows from Case 1 by exchanging A and C.

Case 3: $M < A, C$. Then $M = \mathrm{min}\,([A]\,Q) = \mathrm{min}\,([C]\,Q)$. Let $Q = [M]\,R$. By (def 3),

$$\begin{aligned} s\, ([A]\,Q \,|\, C)&= s\, ([M][A]\,R \,|\, C) \\&\quad = c\, ([C][A]\,R \,|\, M) + \gamma s\,([A]\,R \,|\, C) + \gamma c\,(M) \\&\qquad - \gamma c\,([A]\,R \,|\, M) - \gamma c\,(C)\\&\qquad - c\,(C \,;\, A\,(\zeta R) \,|\, M) + c\, ([c] \,;\, [a](\eta R) \,|\, [m]) \\&\qquad + c\,(M \,;\, A\,(\zeta R) \,|\, C) - c\, ([m] \,;\, [a](\eta R) \,|\, [c])\,,\\ \gamma s\,(Q \,|\, A)&= \gamma s\,([M]\,R \,|\, A) \\&\quad = \gamma c\,([A]\,R \,|\, M) + \gamma s\,(R \,|\, A) + \gamma c\,(M) \\&\quad - \gamma c\,(R \,|\, M) - \gamma c\,(A)\\&\qquad - \gamma c\,(A \,;\, \zeta R \,|\, M) + \gamma c\,([a] \,;\, \eta R \,|\, [m]) \\&\qquad + \gamma c\,(M \,;\, \zeta R \,|\, A) - \gamma c\,([m] \,;\, \eta R \,|\, [a])\,,\\ s\, ([C]\,Q \,|\, A)&= s\, ([M][C]\,R \,|\, A) \\&\quad = c\, ([A][C]\,R \,|\, M) + \gamma s\,([C]\,R \,|\, A) + \gamma c\,(M) \\&\qquad - \gamma c\,([C]\,R \,|\, M) - \gamma c\,(A)\\&\qquad - c\,(A \,;\, C\,(\zeta R) \,|\, M) + c\, ([a] \,;\, [c](\eta R) \,|\, [m]) \\&\qquad + c\,(M \,;\, C\,(\zeta R) \,|\, A) - c\, ([m] \,;\, [c](\eta R) \,|\, [a])\,,~~\hbox {and}~~\\ \gamma s\,(Q \,|\, C)&= \gamma s\,([M]\,R \,|\, C) \\&= \gamma c\,([C]\,R \,|\, M) + \gamma s\,(R \,|\, C) + \gamma c\,(M) \\&\qquad - \gamma c\,(R \,|\, M) - \gamma c\,(C)\\&\qquad - \gamma c\,(C \,;\, \zeta R \,|\, M) + \gamma c\,([c] \,;\, \eta R \,|\, [m]) c\\&\qquad + \gamma c\,(M \,;\, \zeta R \,|\, C) - \gamma c\,([m] \,;\, \eta R \,|\, [c])\,. \end{aligned}$$

Hence (Z2c) becomes:

$$\begin{aligned}&c\, ([C][A]\,R \,|\, M) + \gamma s\,([A]\,R \,|\, C) + \gamma c\,(M) - \gamma c\,([A]\,R \,|\, M) - \gamma c\,(C)\\&\qquad - c\,(C \,;\, A\,(\zeta R) \,|\, M) + c\, ([c] \,;\, [a](\eta R) \,|\, [m])&\hbox {(line 2)} \\&\qquad + c\,(M \,;\, A\,(\zeta R) \,|\, C) - c\, ([m] \,;\, [a](\eta R) \,|\, [c]) \\&+ \gamma c\,([A]\,R \,|\, M) + \gamma s\,(R \,|\, A) + \gamma c\,(M) - \gamma c\,(R \,|\, M) - \gamma c\,(A) \\&\qquad - \gamma c\,(A \,;\, \zeta R \,|\, M) + \gamma c\,([a] \,;\, \eta R \,|\, [m]) \\&\qquad + \gamma c\,(M \,;\, \zeta R \,|\, A) - \gamma c\,([m] \,;\, \eta R \,|\, [a])&\hbox {(line 6)} \\&\qquad + \gamma c\,(C) - c\, (A \,;\, M\,(\zeta R) \,|\, C) + c\, ([a] \,;\, [m](\eta R) \,|\, [c]) \\&{\mathop {=}\limits ^{?}} c\, ([A][C]\,R \,|\, M) + \gamma s\,([C]\,R \,|\, A) + \gamma c\,(M) - \gamma c\,([C]\,R \,|\, M) - \gamma c\,(A) \\&\qquad - c\,(A \,;\, C\,(\zeta R) \,|\, M) + c\, ([a] \,;\, [c](\eta R) \,|\, [m]) \\&\qquad + c\,(M \,;\, C\,(\zeta R) \,|\, A) - c\, ([m] \,;\, [c](\eta R) \,|\, [a])&\hbox {(line A)} \\&\qquad + \gamma c\,([C]\,R \,|\, M) + \gamma s\,(R \,|\, C) + \gamma c\,(M) - \gamma c\,(R \,|\, M) - \gamma c\,(C) \\&\qquad - \gamma c\,(C \,;\, \zeta R \,|\, M) + \gamma c\,([c] \,;\, \eta R \,|\, [m])&\hbox {(line C)} \\&\qquad + \gamma c\,(M \,;\, \zeta R \,|\, C) - \gamma c\,([m] \,;\, \eta R \,|\, [c]) \\&\qquad + \gamma c\,(A) - c\, (C \,;\, M\,(\zeta R) \,|\, A) + c\, ([c] \,;\, [m](\eta R) \,|\, [a]).&\hbox {(line E)} \end{aligned}$$

(4)

To prove (4), use the induction hypothesis:

$$\begin{aligned}&\gamma s\,([A]R \,|\, C) + \gamma s\,(R \,|\, A) + \gamma s\,(C) \\&\qquad - \gamma s\,(A \,;\, \zeta R \,|\, C) + \gamma s\,([a] \,;\, \eta R \,|\, [c]) \\&\quad = \gamma s\,([C]R \,|\, A) + \gamma s\,(R \,|\, C) + \gamma s\,(A) \\&\qquad - \gamma s\,(C \,;\, \zeta R \,|\, A) + \gamma s\,([c] \,;\, \eta R \,|\, [a])\,; \end{aligned}$$

(5)

apply (Z3e) to $A, C, M, \zeta R$:

$$\begin{aligned}&c\, (A \,;\, C\,(\zeta R) \,|\, M) - c\, (M \,;\, C\,(\zeta R) \,|\, A) + c\, (C \,;\, M\,(\zeta R) \,|\, A) \\&\quad \quad - c\, (A \,;\, M\,(\zeta R) \,|\, C) + c\, (M \,;\, A\,(\zeta R) \,|\, C) - c\, (C \,;\, A\,(\zeta R) \,|\, M) \\&\quad + \gamma c\,(C \,;\, \zeta R \,|\, M) - \gamma c\,(M \,;\, \zeta R \,|\, C) + \gamma c\,(M \,;\, \zeta R \,|\, A) \\&\quad \quad - \gamma c\,(A \,;\, \zeta R \,|\, M) + \gamma c\,(A \,;\, \zeta R \,|\, C) - \gamma c\,(C \,;\, \zeta R \,|\, A) \\&\quad - c\, ([a] \,;\, [c,r] \,|\, [m]) + c\, ([m] \,;\, [c,r] \,|\, [a]) - c\, ([c] \,;\, [m,r] \,|\, [a]) \\&\quad \quad + c\, ([a] \,;\, [m,r] \,|\, [c]) - c\, ([m] \,;\, [a,r] \,|\, [c]) + c\, ([c] \,;\, [a,r] \,|\, [m]) \\&\quad = 0\,; \end{aligned}$$

(6)

and apply (Z3e) to $[a], [c], [m], \eta R$, with sides reversed:

$$\begin{aligned} 0&= c\, ([a] \,;\, [c](\eta R) \,|\, [m]) - c\, ([m] \,;\, [c](\eta R) \,|\, [a]) + c\, ([c] \,;\, [m](\eta R) \,|\, [a]) \\&\quad \quad - c\, ([a] \,;\, [m](\eta R) \,|\, [c]) + c\, ([m] \,;\, [a](\eta R) \,|\, [c]) - c\, ([c] \,;\, [a](\eta R) \,|\, [m]) \\&\quad + \gamma c\,([c] \,;\, \eta R \,|\, [m]) - \gamma c\,([m] \,;\, \eta R \,|\, [c]) + \gamma c\,([m] \,;\, \eta R \,|\, [a]) \\&\quad \quad - \gamma c\,([a] \,;\, \eta R \,|\, [m]) + \gamma c\,([a] \,;\, \eta R \,|\, [c]) - \gamma c\,([c] \,;\, \eta R \,|\, [a]) \\&\quad - c\, ([a] \,;\, [c,r] \,|\, [m]) + c\, ([m] \,;\, [c,r] \,|\, [a]) - c\, ([c] \,;\, [m,r] \,|\, [a]) \\&\quad \quad + c\, ([a] \,;\, [m,r] \,|\, [c]) - c\, ([m] \,;\, [a,r] \,|\, [c]) + c\, ([c] \,;\, [a,r] \,|\, [m])\,. \end{aligned}$$

(7)

In (4), terms	411 and 481,	413 and 483,	414 and 441,	415 and 4B5,
	443 and 4B3,	444 and 4B4,	445 and 485,	484 and 4B1

cancel each other. When equalities (5), (6), and (7) are added, terms:

521 and 642,

522 and 742,

541 and 643,

542 and 743

also cancel each other, and lines 5 and 6 of (6) cancel lines 5 and 6 of (7). The remaining terms match the remaining terms of (4):

412 = 511,	421 = 623,	422 = 723,	431 = 622,	432 = 722,	442 = 512,
451 = 641,	452 = 741,	461 = 633,	462 = 733,	471 = 513,	472 = 621,
473 = 721,	482 = 531,	491 = 611,	492 = 711,	4A1 = 612,	4A2 = 712,
4B2 = 532,	4C1 = 631,	4C2 = 731,	4D1 = 632,	4D2 = 732,	4E1 = 533,
4E2 = 613,	4E3 = 713.

Thus (4) follows from (5), (6), and (7); this proves (Z2c). $\square$

Proof of Proposition 3.3

Proposition 3.3

If $c \in C_3$, then $c = \Theta _3 s$ for some $s \in B_2$ if and only if it has properties

for some 3-cochain $u \in C^3 (S,G)$.

Proof

These conditions are necessary, since every $s \in B_2$ has properties (B2a) and

$$\begin{aligned} s\,(P|A) ~&=~ u\,(P\,[A]) - u\,[\zeta P \,;\, A] + u\,\left[ \eta P \,;\, [a]\right] - \gamma u\,(P) - \gamma u\,[A]\,, \\&~~~~~{\hbox {for all}}~~A \in F_1 ~~\hbox {and}~~P \in F_2 \end{aligned}$$

(B2b)

for some cochain $u \in C^3$.

Conversely, let $c \in C_3$ satisfy (B2a) and (B3b). Define $s \in C_2$ recursively, as in “Appendix D”:

$$\begin{aligned} s\, (A) ~&=~ c\, (A)\,, \\ s\, (P \,|\, B) ~&=~ c\, (P \,|\, B) ~~\hbox {if}~~|P| \leqq 2 ~~{\hbox {or if}}~~B \leqq P,\\ s\, (P \,|\, B) ~&=~ c\, ([B]\,Q \,|\, M) ~+~ \gamma s\,(Q \,|\, B) ~+~ \gamma c\,(M) ~-~ \gamma s\,(Q \,|\, M) ~-~ \gamma c\,(B) \\&\quad -~ c\,(B \,;\, \zeta Q \,|\, M) ~+~ c\, ([b] \,;\, \eta Q \,|\, [m])\\&\quad ~+~ c\,(M \,;\, \zeta Q \,|\, B) ~-~ c\, ([m] \,;\, \eta Q \,|\, [b])\\&\qquad ~~\hbox {if}~~|P| \geqq 3, ~~B \geqq M = \mathrm{min}\,P, ~~\hbox {and}~~P = [M]\,Q, \end{aligned}$$

(def 3)

for all $A,B \in F_1$ and $P \in F_2$. Then s inherits (B2a) and (B3b) from c. The proof that s also has property (B2b) is by induction on |P|. If $|P| \leqq 2$, or if $A \leqq P$, then (B2b) follows from (B3b). Assume that $|P| \geqq 3$ and that $A > M = \mathrm{min}\,P$. Let $P = [M]\,Q$. By (def 3),

$$\begin{aligned} s\, (P \,|\, A)&= c\, ([A]\,Q \,|\, M) + \gamma s\,(Q \,|\, A) + \gamma c\,(M) - \gamma s\,(Q \,|\, M) - \gamma c\,(A)\\&\quad - c\,(A \,;\, \zeta Q \,|\, M) + c\, ([a] \,;\, \eta Q \,|\, [m]) \\&\quad + c\,(M \,;\, \zeta Q \,|\, A) - c\, ([m] \,;\, \eta Q \,|\, [a])\,. \end{aligned}$$

The induction hypothesis is:

$$\begin{aligned} \gamma s\,(Q \,|\, A) = \gamma u\,(Q\,[A]) - \gamma u\,[\zeta Q \,;\, A] + \gamma u\,\left[ \eta Q \,;\, [a]\right] - \gamma u\,(Q) - \gamma u\,[A]\,. \end{aligned}$$

Since $M \leqq [A]\,Q$ and $|[A \,;\, \zeta Q]| = |[a] \,;\, \eta Q| = |[M \,;\, \zeta Q]| = |[m] \,;\, \eta Q| = 2$, (B3b) yields:

$$\begin{aligned} c\, ([A]\,Q \,|\, M)&= u\,([A]\,Q\,[M]) - u\,[A\,(\zeta Q) \,;\, M] \\&\quad + u\,\left[ [a](\eta Q) \,;\, [m]\right] - \gamma u\,([A]\,Q) - \gamma u\,[M]\,, \\ \gamma c\,(Q \,|\, M)&= u\,(Q\,[M]) - u\,[\zeta Q \,;\, M] + u\,\left[ \eta Q \,;\, [m]\right] - \gamma u\,(Q) - \gamma u\,[M]\,,\\ c\,(A \,;\, \zeta Q \,|\, M)&= u\,[A \,;\, \zeta Q \,;\, M] - u\,[A\,(\zeta Q) \,;\, M]\\&\quad + u\,\left[ [a,q] \,;\, [m]\right] - \gamma u\,[A \,;\, \zeta Q] - \gamma u\,[M]\,,\\ c\, ([a] \,;\, \eta Q \,|\, [m])&= u\, ([a] \,;\, \eta Q \,;\, [m]) - u\, ([a](\eta Q) \,;\, [m])\\&\quad + u\, ([a,q] \,;\, [m]) - \gamma u\,([a] \,;\, \eta Q) - \gamma u\,[m]\,, \\ c\,(M \,;\, \zeta Q \,|\, A)&= u\,[A \,;\, \zeta Q \,;\, A] - u\,[M\,(\zeta Q) \,;\, A]\\&\quad + u\,\left[ [m,q] \,;\, [a]\right] - \gamma u\,[M \,;\, \zeta Q] - \gamma u\,[A]\,, ~~\hbox {and}~~\\ c\, ([m] \,;\, \eta Q \,|\, [a])&= u\, ([m] \,;\, \eta Q \,;\, [a]) - u\, ([m](\eta Q) \,;\, [a])\\&\quad + u\, ([m,q] \,;\, [a]) - \gamma u\,([m] \,;\, \eta Q) - \gamma u\,[a]\,; \end{aligned}$$

and (B2a) yields

$$\begin{aligned} \gamma c\,(A)&= \gamma u\,[a] - \gamma u\,[A]\,, ~~\hbox {and}~~\\ \gamma c\,(M)&= \gamma u\,[m] - \gamma u\,[M]. \end{aligned}$$

Hence

$$\begin{aligned}&s\, (P \,|\, A)&\hbox {(line 1)} \\&\quad = u\,([A]\,Q\,[M]) - u\,[A\,(\zeta Q) \,;\, M] \\&\qquad + u\,\left[ [a](\eta Q) \,;\, [m]\right] - \gamma u\,([A]\,Q) - \gamma u\,[M] \\&\qquad + \gamma u\,(Q\,[A]) - \gamma u\,[\zeta Q \,;\, A] + \gamma u\,\left[ \eta Q \,;\, [a]\right] - \gamma u\,(Q) - \gamma u\,[A]&\hbox {(line 4)} \\&\qquad + \gamma u\,[m] - \gamma u\,[M] \\&\qquad - \gamma u\,(Q\,[M]) + \gamma u\,[\zeta Q \,;\, M] - \gamma u\,\left[ \eta Q \,;\, [m]\right] + \gamma u\,(Q) + \gamma u\,[M] \\&\qquad - \gamma u\,[a] + \gamma u\,[A]&\hbox {(line 7)} \\&\qquad - u\,[A \,;\, \zeta Q \,;\, M] + u\,[A\,(\zeta Q) \,;\, M] \\&\qquad - u\,\left[ [a,q] \,;\, [m]\right] + \gamma u\,[A \,;\, \zeta Q] + \gamma u\,[M] \\&\qquad + u\, ([a] \,;\, \eta Q \,;\, [m]) - u\, ([a](\eta Q) \,;\, [m])&\hbox {(line A)} \\&\qquad + u\, ([a,q] \,;\, [m]) - \gamma u\,([a] \,;\, \eta Q) - \gamma u\,[m] \\&\qquad + u\,[A \,;\, \zeta Q \,;\, A] - u\,[M\,(\zeta Q) \,;\, A]&\hbox {(line C)} \\&\qquad + u\,\left[ [m,q] \,;\, [a]\right] - \gamma u\,[M \,;\, \zeta Q] - \gamma u\,[A] \\&\qquad - u\, ([m] \,;\, \eta Q \,;\, [a]) + u\, ([m](\eta Q) \,;\, [a]) \\&\qquad - u\, ([m,q] \,;\, [a]) + \gamma u\,([m] \,;\, \eta Q) + \gamma u\,[a]\,.&\hbox {(line F)} \end{aligned}$$

In this equality, terms

22 and 82,	31 and A2,	32 and 41,	33 and 52,	42 and 92,	43 and B2,
44 and 64,	51 and B3,	62 and D2,	63 and F2,	65 and 93,	71 and F3,
72 and D3,	81 and C1,	91 and B1,	A1 and E1,	D1 and F1

cancel each other. The remaining terms: 11, 21, C2, E2, 61, and 45, constitute (B2b):

$$\begin{aligned} s\,(P|A) ~=~ u\,(P\,[A]) ~-~ u\,[\zeta P \,;\, A] ~+~ u\,\left[ \eta P \,;\, [a]\right] ~-~ \gamma u\,(P) ~-~ \gamma u\,[A]\,. \end{aligned}$$

$\square$

Proof of Lemma 6.3

Lemma 6.3

$v\,(B,A) = v\,(A,B)$, for all $A,B \in F_1$.

Recall (Lemma 6.2) that, given $t \in Z_5$, $v\,(A,B)$ is well-defined by

$$\begin{aligned} v\,(A,B) ~&=~ 0 ~~\hbox {if}~~|AB| \leqq 4, \\ v\,(A,[b]) ~=~ v\,([b],A) ~&=~ {\left\{ \begin{array}{ll} 0 &{}\hbox {if }b \leqq A, \\ \gamma v\,(A',[b]) - t\,([m],A',[b]) &{}\hbox {if }|A| \geqq 4\hbox { and }b \geqq m, \\ \end{array}\right. } \\ v\,(A,B) ~&=~ \gamma v\,(A',B) ~+~ v\,([m],A'B) ~-~ t\,([m],A',B) ~+~ L(A,B)\\&\quad ~~\hbox {if}~~|A|,|B| \geqq 2 ~~\hbox {and}~~|AB| \geqq 5, \end{aligned}$$

where $m = \mathrm{min}\,A$ and L(A, B) is well-defined if $|A| \geqq 2$ by Lemma 6.1:

$$\begin{aligned} L(A,[b]) ~&=~ {\left\{ \begin{array}{ll} 0 &{}\hbox {if }|A| \geqq 4\hbox { or if }b \leqq m; \\ t\,([m],A',[b]) &{}\hbox {if }|A| \leqq 3\hbox { and }b \geqq m; \\ \end{array}\right. }\\ L(A,B) ~&=~ {\left\{ \begin{array}{ll} t\,([m],A',B) &{}\hbox {if }|A| = |B| = 2 \\ \gamma L(A,B') ~+~ t &{}([m,a'],[b'],[n]) ~+~ t\,([a',b'],[n],[m]) \\ &{}\hbox {if }n = \mathrm{min}\,B, |B| \geqq 2,\hbox { and }|AB| \geqq 5. \\ \end{array}\right. } \end{aligned}$$

Lemma F.1

If $|B| = 1$, $|A| \geqq 2$, and $m = \mathrm{min}\,A$, then

$$\begin{aligned} v\,(A,B) ~=~ \gamma v\,(A',B) ~+~ v\,([m],A'B) ~-~ t\,([m],A',B) ~+~ L(A,B)\,; \end{aligned}$$

if $|A| = 1$, $|B| \geqq 2$, and $n = \mathrm{min}\,B$, then

$$\begin{aligned} v\,(A,B) ~=~ \gamma v\,(A,B') ~+~ v\,(AB',[n]) ~+~ t\,(A,B',[n]) ~+~ L(B,A)\,. \end{aligned}$$

Proof

Let $|A| \geqq 2$, $m = \mathrm{min}\,A$, and $B = [b]$. If $b \leqq m$, then $v\,(A,[b]) = 0$ and $b \leqq A'$, so that $v\,(A',[b]) = 0$, $\mathrm{min}\,(A'B) = b \leqq m$, $(A'B)' = A'$; $L(A,B) = 0$,

$$\begin{aligned}&v\,([m],A'B) ~=~ \gamma v\,(A',[m]) ~-~ t\,([b],A',[m]) ~=~ -~ t\,([b],A',[m])\,, ~~\hbox {and}~~\\&\gamma v\,(A',B) ~+~ v\,([m],A'B) ~-~ t\,([m],A',B) ~+~ L(A,B) \\&\quad =~ -~ t\,([b],A',[m]) ~-~ t\,([m],A',B) ~=~ 0 ~=~ v\, (A,B) \,, ~~{\hbox { by (S3b)}}. \end{aligned}$$

If $|A| \leqq 3$ and $b \geqq m$, then $v\,(A,[b]) = 0$, $v\,(A',[b]) = 0$, $m \leqq A'[b]$, and $v\,([m],A'[b]) = 0$; $L(A,B) = t\,([m],A',B)$, and

$$\begin{aligned}&\gamma v\,(A',B) ~+~ v\,([m],A'B) ~-~ t\,([m],A',B) ~+~ L(A,B) \\&\quad =~ -~ t\,([b],A',[m]) ~-~ t\,([m],A',B) ~=~ 0 ~=~ v\, (A,B) \,, ~~{\hbox { by (S3b)}}. \end{aligned}$$

If $|A| \geqq 4$ and $b \geqq m$, then $v\,(A,[b]) = \gamma v\,(A',[b]) - t\,([m],A',[b])$, $m \leqq A'[b]$, $v\,([m],A'B) = 0$, $L(A,B) = 0$, and

$$\begin{aligned}&\gamma v\,(A',B) ~+~ v\,([m],A'B) ~-~ t\,([m],A',B) ~+~ L(A,B) \\&\quad =~ \gamma v\,(A',[b]) ~-~ t\,([m],A',[b]) ~=~ v\,(A,B) \,. \end{aligned}$$

This proves the first equality. Since $v\,(B,A) = v\,(A,B)$ whenever $|A| = 1$ or $|B| = 1$, exchanging A and B yields the second equality.

Lemma 6.3 is proved by induction on |AB|, using Lemma F.1 and property (Z5d) of $t \in Z_5$:

$$\begin{aligned}&\gamma t\,(B,C,D) ~-~ t\,(AB,C,D) ~+~ t\,(A,BC,D) ~-~ t\,(A,B,CD) ~+~ \gamma t\,(A,B,C) \\&\quad +~ t\,([a,b],[c],[d]) ~-~ t\,([a],[b,c],[d]) ~+~ t\,([a],[b],[c,d]) ~=~ 0. \end{aligned}$$

The equality $v\,(B,A) = v\,(A,B)$ already holds if $|A| = 1$, if $|B| = 1$, and if $|AB| \leqq 4$. Now let $|A|,|B| \geqq 2$ and $|AB| \geqq 5$. With $m = \mathrm{min}\,A$ and $n = \mathrm{min}\,B$, the equality $v\,(A,B) = v\,(B,A)$ reads:

$$\begin{aligned}&\gamma v\,(A',B) ~+~ v\,([m], A'B) ~-~ t\,([m],A',B) ~+~ L(A,B) \\&\quad {\mathop {=}\limits ^{?}}~ \gamma v\,(B',A) ~+~ v\,([n], B'A) ~-~ t\,([n],B',A) ~+~ L(B,A)\,. \end{aligned}$$

(1)

CASE 1: $|A| = 3$ and $|B| = 2$. Then $B = [b',n]$, $v\,(A',B) = 0$, $v\,(B',A) = 0$, $v\,(A'B',[n]) = 0$, and $v\,([m], A'B') = 0$.

SUBCASE 1a: $m \leqq n$. Then $m \leqq A'B$, $v\,([m], A'B) = 0$, $L(A,B') = t\,([m],A',B')$, and

$$\begin{aligned} L(A,B) ~=~ \gamma t\,([m],A',B') ~+~ t\,([m,a'],[b'],[n]) ~+~ t\,([a',b'],[n],[m])\,; \end{aligned}$$

$\mathrm{min}\,(AB') = m$, $(AB')' = A'B'$, and

$$\begin{aligned} v\,(AB', [n]) ~=~ \gamma v\,(A'B', [n]) ~-~ t\,([m], A'B', [n]) ~=~ -~ t\,([m], A'B', [n)\,; \end{aligned}$$

$L(B,A') = t\,([n],B',A')$ and

$$\begin{aligned} L(B,A) ~=~ \gamma t\,([n],B',A') ~+~ t\,([n,b'],[a'],[m]) ~+~ t\,([b',a'],[m],[n])\,. \end{aligned}$$

Hence (1) becomes:

$$\begin{aligned}&0 ~+~ 0 ~-~ t\,([m],A',B) \\&~~~+~ \gamma t\,([m],A',B') ~+~ t\,([m,a'],[b'],[n]) ~+~ t\,([a',b'],[n],[m]) \\ {\mathop {=}\limits ^{?}}~&0 ~-~ t\,([m],\, A'B',\, [n]) ~-~ t\,([n],B',A) \\&~~~+~ \gamma t\,([n],B',A') ~+~ t\,([n,b'],[a'],[m]) ~+~ t\,([b',a'],[m],[n])\,. \end{aligned}$$

(1a)

To prove (1a), use property (Z5d):

$$\begin{aligned}&\gamma t\,(B,C,D) ~-~ t\,(AB,C,D) ~+~ t\,(A,BC,D) ~-~ t\,(A,B,CD) ~+~ \gamma t\,(A,B,C) \\&\quad +~ t\,([a,b],[c],[d]) ~-~ t\,([a],[b,c],[d]) ~+~ t\,([a],[b],[c,d]) ~=~ 0. \end{aligned}$$

Applied to $[m],\, A',\, B',\, [n]$, (Z5d) yields:

$$\begin{aligned}&\gamma t\,(A',B',[n]) ~-~ t\,([m]A',B',[n]) ~+~ t\,([m],A'B',[n]) \\&\quad -~ t\,([m],A',B'[n]) ~+~ \gamma t\,([m],A',B') \\&\quad +~ t\,([m,a'],[b'],[n]) ~-~ t\,([m],[a',b'],[n]) ~+~ t\,([m],[a'],[b',n]) ~=~ 0; \end{aligned}$$

equivalently, by (S3b) and (S3d),

$$\begin{aligned}&\gamma t\,(A',B',[n]) ~-~ t\,([m]A',B',[n]) ~+~ t\,([m],A'B',[n]) \\&\qquad -~ t\,([m],A',B'[n]) ~+~ \gamma t\,([m],A',B') \\&\quad =~ -~ t\,([m,a'],[b'],[n]) ~+~ t\,([a',b'],[m],[n]) \\&\qquad -~ t\,([a',b'],[n],[m]) ~+~ t\,([n,b'],[a'],[m])\,. \end{aligned}$$

(2)

Note that (2) holds whenever $|A| \geqq 2$ and $|B| \geqq 2$. The terms of (1a) match the terms of (2):

111 = 221,	121 = 222,	122 = 231,	123 = 241,	131 = 213,
132 = 212,	141 = 211,	142 = 242,	143 = 232,

by (S3b) (‘0’ terms have been omitted when counting terms). Thus (1) holds.

SUBCASE 1b: $n \leqq m$ (and $|A| = 3$, $|B| = 2$, so that $B = [b',n]$, $v\,(A',B) = 0$, $v\,(B',A) = 0$, $v\,(A'B',[n]) = 0$, and $v\,(A'B',[m]) = 0$). Then $\mathrm{min}\,(A'B) = n$, $(A'B)' = A'B'$, $L(A,B') = t\,([m],A',B')$,

$$\begin{aligned} v\,([m],A'B) ~&=~ \gamma v\,(A'B',[m]) ~-~ t\,([n],A'B',[m]) ~=~ t\,([m],A'B',[n]) ~~\hbox {and}~~\\ L(A,B) ~&=~ \gamma t\,([m],A',B') ~+~ t\,([m,a'],[b'],[n]) ~+~ t\,([a',b'],[n],[m])\,; \end{aligned}$$

$n \leqq B'A$, $v\,([n],B'A) = 0$, $L(B,A') = t\,([n],B',A')$, and

$$\begin{aligned} L(B,A) ~=~ \gamma t\,([n],B',A') ~+~ t\,([n,b'],[a'],[m]) ~+~ t\,([b',a'],[m],[n])\,. \end{aligned}$$

Hence (1) becomes

$$\begin{aligned}&0 ~+~ t\,([m],A'B',[n]) ~-~ t\,([m],A',B) \\&\qquad +~ \gamma t\,([m],A',B') ~+~ t\,([m,a'],[b'],[n]) ~+~ t\,([a',b'],[n],[m]) \\&\quad {\mathop {=}\limits ^{?}}~ 0 ~+~ 0 ~-~ t\,([n],B',A) \\&\qquad +~ \gamma t\,([n],B',A') ~+~ t\,([n,b'],[a'],[m]) ~+~ t\,([b',a'],[m],[n])\,. \end{aligned}$$

(1b)

Equality (1b) differs from (1a) only in that $t\,([m],A'B',[n])$ has migrated from the right hand side of (1a) to the left hand side of (1b). Hence (1b) is equivalent to (1a) and therefore follows from (2); and again (1) holds.

CASE 2: $|A| = 2$ and $|B| = 3$: follows from Case 1 by exchanging A and B.

CASE 3: $|A| \geqq 2$, $|B| \geqq 2$, and $|AB| \geqq 6$. Then (1) is proved by induction on |AB|, using Lemma F.1. Exchanging A and B if necessary one may assume that $m \leqq n$. Then $m \leqq A'B$, $v\,([m],A'B) = 0$, $v\,([m],A'B') = 0$, $\mathrm{min}\,(AB') = m$, and $(AB')' = A'B'$.

By the induction hypothesis, or by Lemma F.1 if $|A'| = 1$ or if $|B'| = 1$,

$$\begin{aligned} v\,(A',B) ~&=~ v\,(B,A') \\&=~ \gamma v\,(B',A') ~+~ v\,([n],A'B') ~-~ t\,([n],B',A') ~+~ L(B,A')\,, \\ v\,(B',A) ~&=~ v\,(A,B') \\&=~ \gamma v\,(A',B') ~+~ v\,([m],A'B') ~-~ t\,([m],A',B') ~+~ L(A,B') \\&=~ \gamma v\,(A',B') ~-~ t\,([m],A',B') ~+~ L(A,B')\,, ~~\hbox {and}~~\\ v\,(AB',[n]) ~&=~ \gamma v\,(A'B',[n]) ~-~ t\,([m],A'B',[n])\,. \end{aligned}$$

Hence (1) becomes

$$\begin{aligned}&\gamma v\,(B',A') ~+~ \gamma v\,([n],A'B') ~-~ \gamma t\,([n],B',A') ~+~ \gamma L(B,A') \\&\qquad +~ 0 ~-~ t\,([m],A',B) ~+~ L(A,B) \\&\quad {\mathop {=}\limits ^{?}}~ \gamma v\,(A',B') ~-~ \gamma t\,([m],A',B') ~+~ \gamma L(A,B') \\&\qquad +~ \gamma v\,(A'B',[n]) ~-~ t\,([m],A'B',[n]) \\&\qquad -~ t\,([n],B',A) ~+~ L(B,A)\,. \end{aligned}$$

(1c)

As above, (Z5d) yields

$$\begin{aligned}&\gamma t\,(A',B',[n]) ~-~ t\,([m]A',B',[n]) ~+~ t\,([m],A'B',[n]) \\&\qquad -~ t\,([m],A',B'[n]) ~+~ \gamma t\,([m],A',B') \\&\quad = -~ t\,([m,a'],[b'],[n]) ~+~ t\,([a',b'],[m],[n]) \\&\qquad -~ t\,([a',b'],[n],[m]) ~+~ t\,([n,b'],[a'],[m])\,. \end{aligned}$$

(2)

and, by Lemma 6.1,

$$\begin{aligned}&\left( L(A,B) ~-~ \gamma L(A,B')\right) - \left( L(B,A) ~-~ \gamma L(B,A')\right) \\&\quad =~ -~ (\delta t)([m],A',B',[n])&\hbox {(line 2)} \\&\quad =~ t\,([m,a'],[b'],[n]) ~-~ t\,([m],[a',b'],[n]) ~+~ t\,([m],[a'],[b',n])\,.&\hbox {(line 3)} \end{aligned}$$

(3)

Terms 111 and 131,  112 and 141 cancel each other in (1c), by (S3b). When (2) and (3) are added, lines 3 and 4 of (2) cancel line 3 of (3), by (S3d) and (S3b). The remaining terms: 211, 314; 221, 311; 222, 312; 213; 212, and 313 match the remaining terms of (1c). Therefore (1) holds. $\square$

Proof of Lemma 6.4

Lemma 6.4 computes the symmetric 3-cochain $r = t - \delta v\in Z_5$, given $t \in Z_5$. First, recall that

$$\begin{aligned} (\delta v)(A,B,C) ~=~ \gamma v\,(B,C) ~-~ v\,(AB,C) ~+~ v\,(A,BC) ~-~ \gamma v\,(A,B)\,. \end{aligned}$$

Moreover, t, $\delta v$, and $r = t - \delta v$ are symmetric 3-cochains and have properties (S3b) and (S3d).

If $|ABC| \leqq 4$, then $(\delta v)(A,B,C) = 0$, since $v\,(A,B) = 0$ whenever $|AB| \leqq 4$; hence $r\,(A,B,C) = t\,(A,B,C)$. Now assume that $|ABC| \geqq 5$. Let $m = \mathrm{min}\,A$, $n = \mathrm{min}\,B$, and $p = \mathrm{min}\,C$.

CASE 1: $m \leqq n,p$ and $|A| = 1$. Then $\mathrm{min}\,(AB) = m$, $(AB)' = B$, $m \leqq BC$, $v\,([m],BC) = 0$,

$$\begin{aligned} v\,(AB,C) ~=~ \gamma v\,(B,C) ~-~ t\,(A,B,C) ~+~ L(AB,C)\,, \end{aligned}$$

$v\,(A,BC) = 0$, $v\,(A,B) = 0$, $(\delta v)(A,B,C) = t\,(A,B,C) - L(AB,C)$, and

$$\begin{aligned} r\,(A,B,C) ~=~ t\,(A,B,C) ~-~ (dv)(A,B,C) ~=~ L(AB,C)\,. \end{aligned}$$

CASE 2: $n \leqq m,p$ and $|B| = 1$. By (S3d) and Case 1,

$$\begin{aligned} r\,(A,B,C) ~=~ r\,(B,A,C) ~-~ r\,(B,C,A) ~=~ L(AB,C) ~-~ L(BC,A)\,. \end{aligned}$$

CASE 3: $p \leqq m,n$ and $|C| = 1$. By (S3b) and Case 1,

$$\begin{aligned} r\,(A,B,C) ~=~ -~ r\,(C,B,A) ~=~ -~ L(BC,A)\,. \end{aligned}$$

CASE 4: $m \leqq n,p$ and $|A| \geqq 2$. Then $m \leqq A'BC$, $v\,([m],A'BC) = 0$, $v\,([m],A'B) = 0$, $\mathrm{min}\,(AB) = m$, $(AB)' = A'B$, and

$$\begin{aligned} v\,(AB,C) ~&=~ \gamma v\,(A'B,C) ~-~ t\,([m],A'B,C) ~+~ L(AB,C) \,, \\ v\,(A,BC) ~&=~ \gamma v\,(A',BC) ~-~ t\,([m],A',BC) ~+~ L(A,BC) \,, \\ \gamma v\,(A,B) ~&=~ \gamma v\,(A',B) ~-~ \gamma t\,([m],A',B) ~+~ \gamma L(A,B)\,, \end{aligned}$$

by the definition of v, or by Lemma F.1 if $|C| = 1$, or if $|B| = 1$, so that

$$\begin{aligned} r\,(A,B,C) ~&=~ t\,(A,B,C) ~-~ (dv)(A,B,C) \\&=~ t\,(A,B,C) ~-~ \gamma v\,(B,C) \\&\quad +~ \gamma v\,(A'B,C) ~-~ t\,([m],A'B,C) ~+~ L(AB,C) \\&\quad -~ \gamma v\,(A',BC) ~+~ t\,([m],A',BC) ~-~ L(A,BC) \\&\quad +~ \gamma v\,(A',B) ~-~ \gamma t\,([m],A',B) ~+~ \gamma L(A,B)\,. \end{aligned}$$

On the other hand, $(\delta t)([m],[a'],[b],[c]) = (\delta t)([m],A',B,C)$, by (Z5d), and

$$\begin{aligned}&\gamma r\,(A',B,C) ~-~ (\delta t)([m],[a'],[b],[c]) \\&\qquad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) \\&\quad =~ \gamma t\,(A',B,C) ~-~ \gamma v\,(B,C) ~+~ \gamma v\,(A'B,C) \\&\qquad \quad -~ \gamma v\,(A',BC) ~+~ \gamma v\,(A',B) \\&\qquad -~ \gamma t\,(A',B,C) ~+~ t\,([m]A',B,C) ~-~ t\,([m],A'B,C) \\&\qquad \quad +~ t\,([m],A',BC) ~-~ \gamma t\,([m],A',B) \\&\qquad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B)\,. \end{aligned}$$

In this last sum, terms $\gamma t\,(A',B,C)$ cancel each other; the remaining terms yield

$$\begin{aligned} r\,(A,B,C) ~&=~ \gamma r\,(A',B,C) ~-~ (\delta t)([m],[a'],[b],[c]) \\&\quad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B)\,. \end{aligned}$$

CASE 5: $n \leqq m,p$ and $|B| \geqq 2$. By (S3d) and Case 4,

$$\begin{aligned} r\,(A,B,C) ~=~&r\,(B,A,C) ~-~ r\,(B,C,A) \\ =~&\gamma r\,(B',A,C) ~-~ (\delta t)([n],[b'],[a],[c]) \\&\quad +~ L(AB,C) ~-~ L(B,AC) ~+~ \gamma L(B,A) \\&-~ \gamma r\,(B',C,A) ~+~ (\delta t)([n],[b'],[c],[a]) \\&\quad -~ L(BC,A) ~+~ L(B,CA) ~-~ \gamma L(B,C) \\ =&~\gamma r\,(A,B',C) ~-~ (\delta t)([n],[b'],[a],[c]) ~+~ (\delta t)([n],[b'],[c],[a]) \\&+~ L(AB,C) ~+~ \gamma L(B,A) \\&-~ L(BC,A) ~+~ L(B,CA) ~-~ \gamma L(B,C)\,; \end{aligned}$$

and (Z5c), (S3b), (S3d) yield

$$\begin{aligned}&-~ (\delta t)([n],[b'],[a],[c]) ~+~ (\delta t)([n],[b'],[c],[a]) \\&\quad =-~ t\,([n,b'],[a],[c]) ~+~ t\,([n],[b',a],[c]) ~-~ t\,([n],[b'],[a,c]) \\&\qquad +~ t\,([n,b'],[c],[a]) ~-~ t\,([n],[b',c],[a]) ~+~ t\,([n],[b'],[c,a]) \\&\quad =t\,([a],[b',c],[n]) ~-~ t\,([n],[b',a],[c]) ~+~ t\,([c],[b',n],[a])\,. \end{aligned}$$

CASE 6: $p \leqq m,n$ and $|C| \geqq 2$. By Case 4 and (S3b),

$$\begin{aligned} r\,(A,B,C) ~&=~ -~ r\,(C,B,A) \\&=~ -~ \gamma r\,(C',B,A) ~+~ (\delta t)([p],[c'],[b],[a]) \\&\quad -~ L(CB,A) ~+~ L(C,BA) ~-~ \gamma L(C,B) \\&=~ \gamma r\,(A,B,C') ~-~ (\delta t)([a],[b],[c'],[p]) \\&\quad -~ L(BC,A) ~+~ L(C,AB) ~-~ \gamma L(C,B)\,, \end{aligned}$$

since

$$\begin{aligned}&(\delta t)([p],[c'],[b],[a]) \\&\quad =~ -~ t\,([p,c'],[b],[a]) ~+~ t\,([p],[c',b],[a]) ~-~ t\,([p],[c'],[b,a]) \\&\quad = t\,([a,b],[c'],[p]) ~-~ t\,([a],[b,c'],[p]) ~+~ t\,([a],[b],[c',p]) \\&\quad =~ -~ (\delta t)([a],[b],[c'],[p])\,. \end{aligned}$$

$\square$

Proofs of Lemmas 6.6 and 6.7

Lemma 6.6 proves properties

of

$$\begin{aligned} g\,(a,b,c,d) ~=~ -~ f(a,b,c,d) ~+~ f(b,c,a,d) ~-~ f(b,c,d,a) ~+~ f(c,d,b,a) \end{aligned}$$

when $f \in C_6$ has property (P6): $f\,(y,x,z,t) = f\,(x,y,z,t)$.

Proof

Property (G0) follows from (G1). Properties (Ga), (Gb), (Gc), (Gd), (G1) and (G2) follow from (P6):

$$\begin{aligned} g\,(a,b,b,a)&~=~ -~ f(a,b,b,a) ~+~ f(b,b,a,a) ~-~ f(b,b,a,a) ~+~ f(b,a,b,a) \\&=~ 0, \\ g\,(d,c,b,a)&~=~ -~ f(d,c,b,a) ~+~ f(c,b,d,a) ~-~ f(c,b,a,d) ~+~ f(b,a,c,d) \\&=~ -~ g\,(a,b,c,d)\,, \\ g\,(a,b,c,d)&~-~ g\,(b,c,d,a) ~+~ g\,(c,d,a,b) ~-~ g\,(d,a,b,c) \\&=~ -~ f(a,b,c,d) ~+~ f(b,c,a,d) ~-~ f(b,c,d,a) ~+~ f(c,d,b,a) \\&\quad +~ f(b,c,d,a) ~-~ f(c,d,b,a) ~+~ f(c,d,a,b) ~-~ f(d,a,c,b) \\&\quad -~ f(c,d,a,b) ~+~ f(d,a,c,b) ~-~ f(d,a,b,c) ~+~ f(a,b,d,c) \\&\quad +~ f(d,a,b,c) ~-~ f(a,b,d,c) ~+~ f(a,b,c,d) ~-~ f(b,c,a,d) ~=~ 0, \\ g\,(a,b,c,d)&-~ g\,(b,a,c,d) ~+~ g\,(b,c,a,d) ~-~ g\,(b,c,d,a) \\&=~ -~ f(a,b,c,d) ~+~ f(b,c,a,d) ~-~ f(b,c,d,a) ~+~ f(c,d,b,a) \\&\quad +~ f(b,a,c,d) ~-~ f(a,c,b,d) ~+~ f(a,c,d,b) ~-~ f(c,d,a,b) \\&\quad -~ f(b,c,a,d) ~+~ f(c,a,b,d) ~-~ f(c,a,d,b) ~+~ f(a,d,c,b) \\&\quad +~ f(b,c,d,a) ~-~ f(c,d,b,a) ~+~ f(c,d,a,b) ~-~ f(d,a,b,c) ~=~ 0, \\ g\,(a,b,c,d)&~-~ g\,(a,b,d,c) \\&\quad =~ -~ f(a,b,c,d) ~+~ f(b,c,a,d) ~-~ f(b,c,d,a) ~+~ f(c,d,b,a) \\&\qquad +~ f(a,b,d,c) ~-~ f(b,d,a,c) ~+~ f(b,d,c,a) ~-~ f(d,c,b,a) \\&\quad =~ -~ f(c,b,d,a) ~+~ f(b,d,c,a) ~-~ f(b,d,a,c) ~+~ f(d,a,b,c) \\&\qquad +~ f(c,b,a,d) ~-~ f(b,a,c,d) ~+~ f(b,a,d,c) ~-~ f(a,d,b,c) \\&\quad =~ g\,(c,b,d,a) ~-~ g\,(c,b,a,d)\,, \\ g\,(e,a,b,c)&~-~ g\,(e,a,c,b) ~+~ g\,(e,b,c,a) \\&\qquad -~ g\,(e,b,a,c) ~+~ g\,(e,c,a,b) ~-~ g\,(e,c,b,a) \\&\quad =~-~ f(e,a,b,c) ~+~ f(a,b,e,c) ~-~ f(a,b,c,e) ~+~ f(b,c,a,e) \\&\qquad +~ f(e,a,c,b) ~-~ f(a,c,e,b) ~+~ f(a,c,b,e) ~-~ f(c,b,a,e) \\&\qquad -~ f(e,b,c,a) ~+~ f(b,c,e,a) ~-~ f(b,c,a,e) ~+~ f(c,a,b,e) \\&\qquad +~ f(e,b,a,c) ~-~ f(b,a,e,c) ~+~ f(b,a,c,e) ~-~ f(a,c,b,e) \\&\qquad -~ f(e,c,a,b) ~+~ f(c,a,e,b) ~-~ f(c,a,b,e) ~+~ f(a,b,c,e) \\&\qquad +~ f(e,c,b,a) ~-~ f(c,b,e,a) ~+~ f(c,b,a,e) ~-~ f(b,a,c,e) \\&\quad =~ -~ f(a,e,b,c) ~+~ f(e,b,a,c) ~-~ f(e,b,c,a) ~+~ f(b,c,e,a) \\&\qquad +~ f(a,e,c,b) ~-~ f(e,c,a,b) ~+~ f(e,c,b,a) ~-~ f(c,b,e,a) \\&\quad =~ g\,(a,e,b,c) ~-~ g\,(a,e,c,b)\,. \end{aligned}$$

$\square$

Lemma 6.7

Let t be a symmetric 3-cochain on $F_1$, and let $f = \Phi t$. If t has property (Z5c): $t\,([a],[b],[c]) ~=~ 0$, then, for all $a,b,c,d \in S$,

If t also has property (Z5d):

$$\begin{aligned}&\gamma t\,(B,C,D) ~-~ t\,(AB,C,D) ~+~ t\,(A,BC,D) ~-~ t\,(A,B,CD) ~+~ \gamma t\,(A,B,C) \\&\quad +~ t\,([a,b],[c],[d]) ~-~ t\,([a],[b,c],[d]) ~+~ t\,([a],[b],[c,d]) ~=~ 0, \end{aligned}$$

then, for all $a,b,c,d,e \in S$,

$$\begin{aligned}&\gamma g\,(b,c,d,e) ~-~ g\,(ab,c,d,e) ~+~ g\,(a,bc,d,e) \\&\quad -~ g\,(a,b,cd,e) ~+~ g\,(a,b,c,de) ~-~ \gamma g\,(a,b,c,d) = 0. \end{aligned}$$

(Z6)

In particular, (G3), (G4), and (Z6) hold if $t \in Z_5$.

Proof

Property (G3) follows from properties (S3b), (S3d), and (Z5c) of t:

$$\begin{aligned}&(\delta t)([a],[b],[c],[d])\\&\quad =~ -~ t\,([a,b],[c],[d]) ~+~ t\,([a],[b,c],[d]) ~-~ t\,([a],[b],[c,d]) \\&\quad =~ -~ t\,([a,b],[c],[d]) ~+~t\,([b,c],[a],[d]) \\&\qquad -~t\,([b,c],[d],[a]) ~+~t\,([c,d],[a],[b]) \\&\quad =~ -~ f\,(a,b,c,d) ~+~ f\,(b,c,a,d) ~-~ f\,(b,c,d,a) ~+~ f\,(c,d,a,b) \\&\quad =~ g\,(a,b,c,d)\,. \end{aligned}$$

Property (G4) also follows from (G3) and (S3b), (S3d):

$$\begin{aligned}&g\,(a,b,c,d) ~-~ g\,(a,b,d,c) \\&\quad =~ -~t\,([a,b],[c],[d]) ~+~ t\,([a],[b,c],[d]) ~-~ t\,([a],[b],[c,d]) \\&\qquad +~t\,([a,b],[d],[c]) ~-~ t\,([a],[b,d],[c])) ~+~ t\,([a],[b],[d,c]) \\&\quad =~ t\,([a],[b,c],[d]) ~+~ t\,([d],[a,b],[c]) ~+~ t\,([c],[b,d],[a])\,. \end{aligned}$$

By (Z3c) and (Z5d), $(\delta t)([a],[b],[c],[d]) = (\delta t)(A,B,C,D)$; in particular

$$\begin{aligned} (\delta t)([ab],[c],[d],[e]) ~&=~ (\delta t)([a,b],[c],[d],[e])\,, \\ (\delta t)([a],[bc],[d],[e]) ~&=~ (\delta t)([a],[b,c],[d],[e])\,, \\ (\delta t)([a],[b],[cd],[e]) ~&=~ (\delta t)([a],[b],[c,d],[e])\,, \\ (\delta t)([a],[b],[c],[de]) ~&=~ (\delta t)([a],[b],[c],[d,e])\,. \end{aligned}$$

Property (Z6) follows from these equalities and (G3):

When the right hand sides, (1) to (6), are added, all terms cancel in pairs:

111 and 311,	112 and 411,	113 and 511,	211 and 312,
212 and 412,	221 and 512,	222 and 611,	313 and 413,
321 and 521,	322 and 612,	421 and 522,	422 and 613.

Hence the left hand sides add up to 0. $\square$

Proof of Lemma 6.10

Lemma 6.10

If $f \in C_6$ has properties (P6) and (Z6), then $r\,(A,B,C) \in G_{abc}$ is well-defined, for all $A,B,C \in F_1$, by:

$$\begin{aligned}&{\mathrm{(0)}~{\textit{if}}~~} |ABC| \leqq 4, ~~\hbox {then}~~r\,(A,B,C) ~=~ t\,(A,B,C)\,; \\&{\mathrm{(1)}~{\textit{if}}~~} |A| = 1 ~~\hbox {and}~~m \leqq n,p, ~~\hbox {then}~~r\,(A,B,C) ~=~ L(AB,C)\,; \\&{\mathrm{(2)}~{\textit{if}}~~} |B| = 1 ~~\hbox {and}~~n \leqq m,p, ~~\hbox {then}~~r\,(A,B,C) ~=~ L(AB,C) - L(BC,A)\,; \\&{\mathrm{(3)}~{\textit{if}}~~} |C| = 1 ~~\hbox {and}~~p \leqq m,n, ~~\hbox {then}~~r\,(A,B,C) ~=~ - L(BC,A)\,; \\&{\mathrm{(4)}~{\textit{if}}~~} |A| \geqq 2 ~~\hbox {and}~~m \leqq n,p, ~~\hbox {then}~~\\&\quad r\,(A,B,C) ~=~ \gamma r\,(A',B,C) ~-~ g\,(m,a',b,c) \\&\qquad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B)\,; \\&{\mathrm{(5)}~{\textit{if}}~~} |B| \geqq 2 ~~\hbox {and}~~n \leqq m,p, ~~\hbox {then}~~\\&\quad r\,(A,B,C) ~=~ \gamma r\,(A,B',C) ~-~ g\,(n,b',a,c) ~+~ g\,(n,b',c,a) \\&\qquad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C)\,; \\&{\mathrm{(6)}~{\textit{if}}~~} |C| \geqq 2 ~~\hbox {and}~~p \leqq m,n, ~~\hbox {then}~~\\&\quad r\,(A,B,C) ~=~ \gamma r\,(A,B,C') ~-~ g\,(a,b,c',p) \\&\qquad -~ L(BC,A) ~+~ L(C,AB) ~-~ \gamma L(C,B)\,, \end{aligned}$$

where $m = \mathrm{min}\,A$, $n = \mathrm{min}\,B$, $p = \mathrm{min}\,C$, and t, g, L are as in Lemmas 6.8 and 6.9.

Proof

By Lemma 6.8, property (Z5c): $t\,([x],[y],[z]) = 0$, and properties (S3a), (S3b), (S3c), and (S3d) apply to $t\,(A,B,C)$, as long as $|ABC| \leqq 4$. By Lemmas 6.6 and 6.7, g has properties (Ga), (Gb), (Gc), (Gd), (G0), (G1), (G2), (G3), and (G4).

Lemma I.1

If $|ABC| = 3$, then $r\,(A,B,C) = 0$, $r\,(A,B,C)$ is well-defined by Lemma 6.10, and r has properties (S3b) and (S3d).

Proof

By definition, $L(A,B) = 0$ if $|A| = 2$ and $|B| = 1$, since either $b \leqq m$, or $b \geqq m$ and $L(A,B) = t\,([m],A',[b]) = 0$ by (Z5c). If now $|ABC| = 3$, then $|A| = |B| = |C| = 1$, $L(AB,C) = 0$, $L(BC,A) = 0$, and cases (0), (1), (2), and (3) all yield $r\,(A,B,C) = 0$. $\square$

The proof that $r\,(A,B,C)$ is well-defined by Lemma 6.10, and has properties (S3b) and (S3d), is by induction on |ABC|. By Lemma I.1, these properties hold if $|ABC| = 3$. The induction hypothesis is that $r\,(A,B,C)$ is well-defined by Lemma 6.10, and has properties (S3b) and (S3d), whenever |ABC| has a lower value. The next Lemmas all assume this induction hypothesis.

Lemma I.2

Let $r_k\,(A,B,C)$ denote the value of $r\,(A,B,C)$ obtained from case (k).

• 1. If $|A| = 1$ and $m \leqq n,p$, then $r_3\,(C,B,A) ~=~ -~ r_1\,(A,B,C)$

  $~~\hbox {and}~~r_2\,(B,A,C) ~-~ r_3\,(B,C,A) ~=~ r_1\,(A,B,C)$.

• 2. If $|B| = 1$ and $n \leqq m,p$, then $r_2\,(C,B,A) ~=~ -~ r_2\,(A,B,C)$

  $~~\hbox {and}~~r_1\,(B,A,C) ~-~ r_1\,(B,C,A) ~=~ r_2\,(A,B,C)$.

• 3. If $|C| = 1$ and $p \leqq m,n$, then $r_1\,(C,B,A) ~=~ -~ r_3\,(A,B,C)$

  $~~\hbox {and}~~r_3\,(B,A,C) ~-~ r_2\,(B,C,A) ~=~ r_3\,(A,B,C)$.

• 4. If $|A| \geqq 2$ and $m \leqq n,p$, then $r_6\,(C,B,A) ~=~ -~ r_4\,(A,B,C)$

  $~~\hbox {and}~~r_5\,(B,A,C) ~-~ r_6\,(B,C,A) ~=~ r_4\,(A,B,C)$.

• 5. If $|B| \geqq 2$ and $n \leqq m,p$, then $r_5\,(C,B,A) ~=~ -~ r_5\,(A,B,C)$

  $~~\hbox {and}~~r_4\,(B,A,C) ~-~ r_4\,(B,C,A) ~=~ r_5\,(A,B,C)$.

• 6. If $|C| \geqq 2$ and $p \leqq m,n$, then $r_4\,(C,B,A) ~=~ -~ r_6\,(A,B,C)$

  $~~\hbox {and}~~r_6\,(B,A,C) ~-~ r_5\,(B,C,A) ~=~ r_6\,(A,B,C)$.

Proof

1. If $|A| = 1$ and $m \leqq n,p$, then (A, B, C) is in case (1), (C, B, A) and (B, C, A) are in case (3), (B, A, C) is in case (2), and

$$\begin{aligned} r_3\,(C,B,A) ~&=~ -~ L(AB,C) ~=~ -~ r_1\,(A,B,C) ~~\hbox {and}~~\\ r_2\,(B,A,C) ~-~ r_3\,(B,C,A) ~&=~ L(AB,C) - L(AC,B) + L(AC,B) \\&=~ L(AB,C) ~=~ r_1\,(A,B,C)\,. \end{aligned}$$

3. If $|C| = 1$ and $p \leqq m,n$, then exchanging A and C in part 1 yields $r_3\,(A,B,C) = -~r_1\,(C,B,A)$ and $r_2\,(B,C,A) - r_3\,(B,A,C) = r_1\,(C,B,A) = -~r_3\,(A,B,C)$.

2. If $|B| = 1$ and $n \leqq m,p$, then (A, B, C) and (C, B, A) are in case (2), (B, C, A) and (B, A, C) are in case (1), and

$$\begin{aligned} r_2\,(C,B,A) ~=~ L(BC,A) ~&-~ L(AB,C) ~=~ -~ r_2\,(A,B,C) ~~\hbox {and}~~\\ r_1\,(B,A,C) ~-~ r_1\,(B,C,A) ~=~&L(AB,C) ~-~ L(BC,A) ~=~ r_2\,(A,B,C)\,. \end{aligned}$$

4. If $|A| \geqq 2$ and $m \leqq n,p$, then (A, B, C) is in case (4) and (C, B, A) is in case (6). By the induction hypothesis, $r\,(C,B,A')$ and $r\,(A',B,C)$ are well-defined and $r\,(C,B,A') = -~r\,(A',B,C)$. Then property (Gb) of g yields

$$\begin{aligned} r_6\,(C,B,A) ~&=~ \gamma r\,(C,B,A') ~-~ g\,(c,b,a',m) \\&\quad -~ L(AB,C) ~+~ L(A,BC) ~-~ \gamma L(A,B) ~=~ -~ r_4\,(A,B,C)\,. \end{aligned}$$

5. If $|B| \geqq 2$ and $n \leqq m,p$, then (A, B, C) and (C, B, A) are in case (5). By the induction hypothesis, $r\,(C,B,A')$ and $r\,(A',B,C)$ are well-defined and $r\,(C,B,A') = -~r\,(A',B,C)$. Hence

$$\begin{aligned} r_5\,(C,B,A) ~&=~ \gamma r\,(C,B',A) ~-~ g\,(n,b',c,a) ~+~ g\,(n,b',a,c) \\&~~~~~~~+~ L(CB,A) ~+~ \gamma L(B,C) ~-~ L(BA,C) ~-~ \gamma L(B,A) \\&=~ -~ r_5\,(A,B,C)\,. \end{aligned}$$

Similarly, (B, A, C) and (B, C, A) are in case (4), $r\,(B',A,C)$ and $r\,(B',C,A)$ are well-defined, and $r\,(B',A,C) - r\,(B',C,A) = r\,(A,B',C)$. Hence

$$\begin{aligned}&r_4\,(B,A,C) ~-~ r_4\,(B,C,A)\\&=~ \gamma r\,(B',A,C) ~-~ g\,(n,b',a,c) ~+~ L(BA,C) ~-~ L(B,AC) ~+~ \gamma L(B,A) \\&~~-~ \gamma r\,(B',C,A) ~+~ g\,(n,b',c,a) ~-~ L(BC,A) ~+~ L(B,CA) ~-~ \gamma L(B,C) \\&=~ r_5\,(A,B,C)\,. \end{aligned}$$

6. If $|C| \geqq 2$ and $p \leqq m,n$, then (A, B, C) is in case (6) and (C, B, A) is in case (4). Exchanging A and C in part 4 yields $r_6\,(A,B,C) = -~r_4\,(C,B,A)$.

Also, (B, A, C) is in case (6), (B, C, A) are in case (5). By the above, $r_6\,(B,A,C) = -~r_4\,(C,A,B)$. By part 5, $r_5\,(B,C,A) = r_4\,(C,B,A) - r_4\,(C,A,B)$. Hence

$$\begin{aligned} r_6\,(B,A,C) ~-~ r_5\,(B,C,A) ~=~ -~ r_4\,(C,B,A) ~=~ r_6\,(A,B,C)\,. \end{aligned}$$

4. If now $|A| \geqq 2$ and $m \leqq n,p$, then exchanging A and C in the above yields

$$\begin{aligned} r_6\,(B,C,A) ~-~ r_5\,(B,A,C) ~=~ -~ r_4\,(A,B,C)\,. \end{aligned}$$

$\square$

Lemma I.3

If $|ABC| \leqq 4$, then, in cases (1) through (6), $r_k\,(A,B,C) = t\,(A,B,C)$.

Proof

If $|ABC| = 3$, then (A, B, C) is in case (i) = (1), (2), or (3), and $r_k\,(A,B,C) = 0 = t\,(A,B,C)$, by Lemma I.1 and (Z5c). Now let $|ABC| = 4$.

If $|A| = 1$ and $a \leqq B,C$, then $\mathrm{min}\,(AB) = a$, $(AB)' = B$, and $L(AB,C) = t\,([a],B,[c])$ if $|C| = 1$, $t\,([a],B,C)$ if $|C| = 2$. In either case, $r_1\,(A,B,C) = L(AB,C) = t\,(A,B,C)$.

If $|B| = 1$ and $b \leqq A,C$, then, by the above, $r_1\,(B,A,C) = t\,(B,A,C)$ and $r_1\,(B,C,A) = t\,(B,C,A)$. By Lemma I.2 and property (S3d) of t,

$$\begin{aligned} r_2\,(A,B,C) ~&=~ r_1\,(B,A,C) ~-~ r_1\,(B,C,A) \\&=~ t\,(B,A,C) ~-~ t\,(B,C,A) ~=~ t\,(A,B,C)\,. \end{aligned}$$

If $|C| = 1$ and $c \leqq A,B$, then, similarly, $r_1\,(C,B,A) = t\,(C,B,A)$. By Lemma I.2 and (S3b),

$$\begin{aligned} r_3\,(A,B,C) ~=~ -~ r_1\,(C,B,A) ~=~ -~ t\,(C,B,A) ~=~ t\,(A,B,C)\,. \end{aligned}$$

If $|A| = 2$ and $m = \mathrm{min}\,A \leqq B,C$, then $|B| = |C| = 1$, $r\,(A',B,C)$ is well-defined and $r\,(A',B,C) = 0$, $\mathrm{min}\,(AB) = m$, $(AB)' = A'B = [a',b]$, $L(A,B) = t\,([m],[a'],[b]) = 0$ by (Z5c),

$$\begin{aligned} L(AB,C) ~&=~ t\,([m],[a',b],[c])\,, ~~L(A,BC) = t\,([m],[a'],[b,c])\,, \\ g\,(m,a',b,c) ~&=~ -~ t\,([m,a'],[b],[c]) ~+~ t\,([m],[a',b],[c])\\ {}&~-~ t\,([m],[a'],[b,c])\,, \end{aligned}$$

by (G3), and

$$\begin{aligned} r_4\, (A,B,C) ~&=~ \gamma r\,(A',B,C) ~-~ g\,(m,a',b,c) \\&\quad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) \\&=~ t\,([m,a'],[b],[c]) ~-~ t\,([m],[a',b],[c]) ~+~ t\,([m],[a'],[b,c]) \\&\quad +~ t\,([m],[a',b],[c]) ~-~ t\,([m],[a'],[b,c]) \\&=~ t\,(A,B,C)\,. \end{aligned}$$

If $|C| = 2$ and $\mathrm{min}\,C \leqq A,B$, then, by Lemma I.2, (S3b), and the above,

$$\begin{aligned} r_6\,(A,B,C) ~=~ -~ r_4\,(C,B,A) ~=~ -~ t\,(C,B,A) ~=~ t\,(A,B,C)\,. \end{aligned}$$

If $|B| = 2$ and $\mathrm{min}\,B \leqq A,C$, then, by Lemma I.2, (S3d), and the above,

$$\begin{aligned} r_5\,(A,B,C) ~&=~ r_4\,(B,A,C) ~-~ r_4\,(B,C,A) \\&=~ t\,(B,A,C) ~-~ t\,(B,C,A) ~=~ t\,(A,B,C)\,. \end{aligned}$$

$\square$

Lemma I.4

Cases (1) through (6) are compatible with each other.

Proof

As before, the induction hypothesis is that cases (1) through (6) in the definition of $r\,(A,B,C)$ are compatible with each other, and that $r\,(A,B,C)$ has properties (S3b) and (S3d), whenever |ABC| has a lower value. By Lemmas I.1 and I.3, $r\,(A,B,C)$ has these properties if $|ABC| \leqq 4$. Now assume that $|ABC| \geqq 5$.

Cases (1) and (4), (2) and (5), (3) and (6) are mutually exclusive, leaving 12 overlaps to consider.

CASE 1-2: $m \leqq n,p$, $|A| = 1$, $n \leqq m,p$, and $|B| = 1$. Then $|C| \geqq 3$ (since $|ABC| \geqq 5$), $L(BC,A) = 0$, and

$$\begin{aligned} r_1\,(A,B,C) ~=~ L(AB,C) ~=~ L(AB,C) ~-~ L(BC,A) ~=~ r_2\,(A,B,C)\,. \end{aligned}$$

CASE 1-3: $m \leqq n,p$, $|A| = 1$, $p \leqq m,n$, and $|C| = 1$. Then $|B| \geqq 3$ and

$$\begin{aligned} r_1\,(A,B,C) ~=~ L(AB,C) ~=~ 0 ~=~ -~ L(BC,A) ~=~ r_3\,(A,B,C)\,. \end{aligned}$$

CASE 2-3: $n \leqq m,p$, $|B| = 1$, $p \leqq m,n$, and $|C| = 1$. By Lemma I.2 and Case 1-2,

$$\begin{aligned} r_2\,(A,B,C) ~=~ -~ r_2\, (C,B,A) ~=~ -~ r_1\,(C,B,A) ~=~ r_3\,(A,B,C)\,. \end{aligned}$$

CASE 2-4: $m \leqq n,p$, $|A| \geqq 2$, $n \leqq m,p$, and $|B| = 1$. Then $L(A,B) = 0$, since $b = m$, $\mathrm{min}\,(BC) = b$, $(BC)' = C$, and

$$\begin{aligned}&r_2\,(A,B,C) ~=~ L(AB,C) ~-~ L(BC,A) \\&\quad =~ L(AB,C) ~-~ \gamma L(BC,A') ~-~ t\,([b,c],[a'],[m]) ~-~ t\,([c,a'],[m],[b]) \,, \end{aligned}$$

since $|ABC| \geqq 5$. On the other hand,

$$\begin{aligned} L(A,BC) ~&=~ \gamma L(A,C) ~+~ t\,([m,a'],[c],[b]) ~+~ t\,([a',c],[b],[m])\,, \\ g\,(m,a',b,c) ~&=~ -~t\,([m,a'],[b],[c]) ~+~ t\,([a',b],[m],[c]) \\&\quad -~ t\,([a',b],[c],[m]) ~+~ t\,([b,c],[a'],[m])\,, \end{aligned}$$

by (G3); $|B| = 1$, $n \leqq a',p$, $r\,(A',B,C) = r_2\,(A',B,C)$ by the induction hypothesis, and

$$\begin{aligned} r_4\,(A,B,C) ~&=~ \gamma r_2\, (A',B,C) ~-~ g\,(m,a',b,c) \\&\quad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) \\&=~ \gamma L(A'B,C) ~-~ \gamma L(BC,A') \\&\quad +~ t\,([m,a'],[b],[c]) ~-~ t\,([a',b],[m],[c]) \\&\qquad +~ t\,([a',b],[c],[m]) ~-~ t\,([b,c],[a'],[m]) \\&\quad +~ L(AB,C) ~-~ \gamma L(A,C) \\&\qquad -~ t\,([m,a'],[c],[b]) ~-~ t\,([a',c],[b],[m])\,. \end{aligned}$$

In the above, $t\,([m,a'],[b],[c]) - t\,([a',b],[m],[c]) = 0$ since $b = m$, and $\gamma L(A'B,C)- \gamma L(A,C) = 0$ since $A'B = A$. The remaining terms show that $r_4\,(A,B,C) = r_2\,(A,B,C)$.

CASE 3-4: $m \leqq n,p$, $|A| \geqq 2$, $p \leqq m,n$, and $|C| = 1$. Then $r_3\,(A,B,C) = - L(BC,A)$.

On the other hand, $|AB| \geqq 4$ and $L(AB,C) = 0$; $\mathrm{min}\,(BC) = p$, $(BC)' = B$, and

$$\begin{aligned} L(A,BC) ~&=~ \gamma L(A,B) ~+~ t\,([m,a'],[b],[c]) ~+~ t\,([a',b],[c],[m])\,; \\ L(BC,A) ~&=~ \gamma L(BC,A') ~+~ t\,([c,b],[a'],[m]) ~+~ t\,(b,a'],[m],[c])\,; \\ g\,(m,a',b,c) ~&=~ -~ t\,([m,a'],[b][c]) ~+~ t\,([a',b],[m],[c]) \\&\quad -~ t\,([a',b],[c],[m]) ~+~ t\,([b,c],[a'],[m])\,, \end{aligned}$$

by (G3), as in case 2-4; $r\,(A',B,C) = r_3\,(A',B,C) = - L(BC,A')$, by the induction hypothesis; and

$$\begin{aligned} r_4\,(A,B,C) ~&=~ \gamma r_3\,(A',B,C) ~-~ g\,(m,a',b,c) \\&\quad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) \\&=~ -~ \gamma L(BC,A') ~+~ t\,([m,a'],[b][c]) ~-~ t\,([a',b],[m],[c]) \\&\quad +~ t\,([a',b],[c],[m]) ~-~ t\,([b,c],[a'],[m]) \\&\quad -~ \gamma L(A,B) ~-~ t\,([m,a'],[b],[c]) \\&\quad -~ t\,([a',b],[c],[m]) ~+~ \gamma L(A,B) \\&=~ -~ \gamma L(BC,A') ~-~ t\,([a',b],[m],[c]) ~-~ t\,([b,c],[a'],[m]) \\&=~ -~ L(BC,A) ~=~ r_3\,(A,B,C) \,. \end{aligned}$$

CASE 1-6: $m \leqq n,p$, $|A| = 1$, $p \leqq m,n$, and $|C| \geqq 2$. By Lemma I.2 and case 3-4,

$$\begin{aligned} r_1\,(A,B,C) ~=~ -~ r_3\,(C,B,A) ~=~ -~ r_4\,(C,B,A) ~=~ r_6\,(A,B,C)\,. \end{aligned}$$

CASE 2-6: $n \leqq m,p$, $|B| = 1$, $p \leqq m,n$, and $|C| \geqq 2$. By Lemma I.2 and case 2-4,

$$\begin{aligned} r_2\,(A,B,C) ~=~ -~ r_2\,(C,B,A) ~=~ -~ r_4\,(C,B,A) ~=~ r_6\, (A,B,C)\,. \end{aligned}$$

CASE 1-5: $m \leqq n,p$, $|A| = 1$, $n \leqq m,p$, and $|B| \geqq 2$. By Lemma I.2 and cases 2-4, 3-4,

$$\begin{aligned} r_5\, (A,B,C) ~&=~ r_4\,(B,A,C) ~-~ r_4\,(B,C,A) \\&=~ r_2\,(B,A,C) ~-~ r_3\,(B,C,A) ~=~ r_1\,(A,B,C)\,. \end{aligned}$$

CASE 3-5: $n \leqq m,p$, $|B| \geqq 2$, $p \leqq m,n$, and $|C| = 1$. By Lemma I.2 and case 1-5,

$$\begin{aligned} r_3\,(A,B,C) ~=~ -~ r_1\,(C,B,A) ~=~ -~ r_5\,(C,B,A) ~=~ r_5\,(A,B,C)\,. \end{aligned}$$

CASE 5-6: $n \leqq m,p$, $|B| \geqq 2$, $p \leqq m,n$, and $|C| \geqq 2$. By Lemma I.2 and case 4-5,

$$\begin{aligned} r_6\,(A,B,C) ~=~ -~ r_4\,(C,B,A) ~=~ -~ r_5\,(C,B,A) ~=~ r_5\,(A,B,C)\,. \end{aligned}$$

The last two cases, 4-5 and 4-6, constitute Lemmas I.5 and I.6.

Lemma I.5

(CASE 4-5) If $m \leqq n,p$, $|A| \geqq 2$, $n \leqq m,p$, and $|B| \geqq 2$, then $r_4\,(A,B,C) = r_5\,(A,B,C)$.

Proof

First assume that $|ABC| = 5$, so that $|A| = |B| = 2$ and $|C| = 1$. Then $r\,(A',B,C) = t\,([a'],[b',n],[c])$ by Lemma I.3, $\mathrm{min}\,(A'B) = n \leqq c$, $\mathrm{min}\,(BC) = n$, $(BC)' = [b',c]$,

$$\begin{aligned} L(AB,C) ~&=~ 0\,, ~L(A'B,C) = t\,([n],[a',b'],[c])\,, \\ L(A,B'C) ~&=~ t\,([m],[a'],[b',c])\,, ~L(A,B) = t\,([m],[a'],B)\,, \\ L(A,BC) ~&{=}~ \gamma L(A,B'C) ~+~ t\,([m,a'],[b'c],[n]) ~{+}~ t\,([a',b'c],[n],[m])\,, ~~\hbox {and}~~\\ r_4\,(A,B,C) ~&=~ \gamma r\,(A',B,C) ~-~g\,(m,a',b,c) \\&\quad +~L(AB,C) ~-~L(A,BC) ~+~\gamma L(A,B) \\&=~ \gamma t\,([a'],[b',n],[c]) ~-~ g\,(m,a',b'n,c) \\&\quad -~ \gamma t\,([m],[a'],[b',c]) \\&\quad -~ t\,([m,a'],[b'c],[n]) ~-~ t\,([a',b'c],[n],[m]) \\&\quad -~ \gamma t\,([b',n],[a'],[m]) \,, ~~{\hbox { by (S3b)}}. \end{aligned}$$

On the other hand, $r\,(A,B',C) = t\,([m,a'],[b'],[c])$ by Lemma I.3,

$$\begin{aligned} L(B,C) ~&=~ 0\,, ~L(B,A) ~=~ t\,([n],[b'],A) ~=~ t\,([n],[b'],[m,a'])\,, \\ L(BC,A') ~&=~ t\,([n],[b',c],[a'])\,, ~L(AB,C) ~=~ 0\,, \\ L(BC,A) ~&=~ \gamma L(BC,A') ~+~ t\,([n,b'c],[a'],[m]) ~+~ t\,([b'c,a'],[m],[n]) \\ g,(a,b',n,c) ~&-~ g\,(a,b',c,n) ~=~ g\,(n,b',c,a) ~-~ g\,(n,b',a,c) ~~{\hbox { by (G1)}}\,, ~~\hbox {and}~~\\ r_5\,(A,B,C) ~&=~ \gamma r\,(A,B',C) ~-~ g\,(n,b',a,c) ~+~ g\,(n,b',c,a) \\&\quad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C) \\&=~ \gamma t\,([m,a'],[b'],[c]) ~-~ g\,(a,b',c,n) ~+~ g,(a,b',n,c) \\&\quad +~ \gamma t\,([n],[b'],[m,a']) ~-~ \gamma t\,([n],[b',c],[a']) \\&\quad -~ t\,([n,b'c],[a'],[m]) ~-~ t\,([b'c,a'],[m],[n])\,. \end{aligned}$$

The equality $r_4\,(A,B,C) = r_5\,(A,B,C)$ now reads:

$$\begin{aligned}&\gamma t\,([a'],[b',n],[c]) ~-~ g\,(m,a',b'n,c) \\&\qquad -~ \gamma t\,([m],[a'],[b',c]) ~-~ t\,([m,a'],[b'c],[n])&\hbox {(line 2)} \\&\qquad -~ t\,([a',b'c],[n],[m]) ~-~ \gamma t\,([b',n],[a'],[m]) \\&{\mathop {=}\limits ^{?}}~ \gamma t\,([m,a'],[b'],[c]) ~-~ g\,(ma',b',c,n) ~+~ g\,(ma',b',n,c) \\&\qquad +~ \gamma t\,([n],[b'],[a',m]) ~-~ \gamma t\,([n],[b',c],[a'])&\hbox {(line 5)} \\&\qquad -~ t\,([n,b'c],[a'],[m]) ~-~ t\,([b'c,a'],[m],[n])\,. \end{aligned}$$

(1)

To prove (1), use (G3):

$$\begin{aligned}&\gamma g\,(m,a',b',n) \\&\quad =~ -~ \gamma t\,([m,a'],[b'],[n]) ~+~ \gamma t\,([m],[a',b'],[n])~-~ \gamma t\,([m],[a'],[b',n])\,,\\ -~&\gamma g\,(m,a',b',c)&\hbox {(line 3)} \\&\quad =~ \gamma t\,([m,a'],[b'],[c]) ~-~ \gamma t\,([m],[a',b'],[c]) ~+~ \gamma t\,([m],[a'],[b',c])\,, \\ -~&g\, (m,a',b'c,n)&\hbox {(line 5)} \\&\quad =~ t\,([m,a'],[b'c],[n]) ~-~ t\,([m],[a',b'c],[n]) ~+~ t\,([m],[a'],[b'c,n]) \end{aligned}$$

(2)

apply (Z6) to $m, a', b', n, c$ and to $m\, a', b', c, n$:

$$\begin{aligned}&\gamma g\,(a',b',n,c) ~-~ g\, (ma',b',n,c) ~+~ g\,(m,a'b',n,c) \\&\quad -~ g\,(m,a',b'n,c) ~+~ g\,(m,a',b',nc) ~-~ \gamma g\,(m,a',b',n) ~=~ 0, \\ 0 ~&=~ \gamma g\,(a',b',c,n) ~-~ g\, (ma',b',c,n) ~+~ g\,(m,a'b',c,n) \\&\quad -~ g\,(m,a',b'c,n) ~+~ g\,(m,a',b',cn) ~-~ \gamma g\,(m,a',b',c)\,; \end{aligned}$$

(3)

and use (G0) and (G4):

$$\begin{aligned}&g\, (m,a'b',c,n) ~-~ g\, (m,a'b',n,c) ~=~ 0, \hbox { since }m = n, \\&\gamma g\,(a',b',c,n) ~-~ \gamma g\,(a',b',n,c) \\&=~ \gamma t\,([a'],[b',c],[n]) ~+~ \gamma t\,([n],[b',a'],[c]) ~+~ \gamma t\,([c],[b',n],[a'])\,. \end{aligned}$$

(4)

Terms 131 and 162 cancel each other, since $m = n$, and terms 222, 262 vanish, by (S3a). When the equalities in (2), (3), and (4) are added, terms

211 and 323,	231 and 343,	242 and 432,	251 and 341,	311 and 412,
313 and 412,	322 and 342,	331 and 421,	333 and 411

cancel in pairs, some by (S3b). The remaining terms match the remaining terms of (1):

111 = 413,	112 = 321,	121 = 243,	122 = 261,
133 = 223,	141 = 241,	142 = 332,	143 = 312,
151 = 221,	152 = 431,	161 = 263,

by (S3b). Thus (1) follows from (2), (3), and (4).

The general case is proved by induction on |ABC|. Assume $|ABC| \geqq 6$ (still with $|A|,|B| \geqq 2$ and $m = n \leqq p$). The induction hypothesis yields

$$\begin{aligned} r_4\,(A,B,C) ~&=~ \gamma r\,(A',B,C) ~-~ g\,(m,a',b,c) \\&\quad +~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) ~~\hbox {and}~~\\ \gamma r_5\, (A',B,C) ~&=~ \gamma r\,(A',B',C) ~-~ \gamma g\,(n,b',a',c) ~+~ \gamma g\,(n,b',c,a') \\&\quad +~ \gamma L(A'B,C) ~+~ \gamma L(B,A') ~-~ \gamma L(BC,A') ~-~ \gamma L(B,C) \\&=~ \gamma r\,(A',B',C) ~-~ \gamma g\,(a',b',c,n) ~+~ \gamma g\,(a',b',n,c) \\&\quad +~ \gamma L(A'B,C) ~+~ \gamma L(B,A') ~-~ \gamma L(BC,A') ~-~ \gamma L(B,C)\,, \end{aligned}$$

by (G1), as above, whereas

$$\begin{aligned} r_5\, (A,B,C) ~&=~ \gamma r\,(A,B',C) ~-~ g\,(n,b',a,c) ~+~ g\,(n,b',c,a) \\&\quad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C) \\&=~ \gamma r\,(A,B',C) ~-~ g\,(a,b',c,n) ~+~ g\,(a,b',n,c) ~~{\hbox { by (G1)}} \\&\quad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C)\,, ~~\hbox {and}~~\\ \gamma r_4\, (A,B',C) ~&=~ \gamma r\,(A',B',C) ~-~ \gamma g\,(m,a',b',c) \\&\quad +~ \gamma L(AB',C) ~-~ \gamma L(A,B'C) ~+~ \gamma L(A,B')\,. \end{aligned}$$

The equality $r_4\,(A,B,C) = r_5\,(A,B,C)$ now reads:

$$\begin{aligned}&\gamma r\,(A',B',C) ~-~ \gamma g\,(a',b',c,n) ~+~ \gamma g\,(a',b',n,c) \\&\quad +~ \gamma L(A'B,C) ~+~ \gamma L(B,A') ~-~ \gamma L(BC,A') ~-~ \gamma L(B,C)&\hbox {(line 2)} \\&\quad -~ g\,(m,a',b,c) ~+~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) \\&{\mathop {=}\limits ^{?}} \gamma r\,(A',B',C) ~-~ \gamma g\,(m,a',b',c) \\&\quad +~ \gamma L(AB',C) ~-~ \gamma L(A,B'C) ~+~ \gamma L(A,B')&\hbox {(line 5)} \\&\quad -~ g\,(a,b',c,n) ~+~ g\,(a,b',n,c) \\&\quad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C)\,. \end{aligned}$$

(5)

To prove (5), apply Lemma 6.9: since $|A| \geqq 2$, $|B| \geqq 2$, and $m = n$,

$$\begin{aligned} \left( (L(B,A) ~-~ \gamma L(B,A')\right) ~-~ \left( L(A,B) ~-~ \gamma L(A,B')\right) ~=~ g\,(m,a',b',n)\,. \end{aligned}$$

(6)

Since $n \leqq p$ makes $n = \mathrm{min}\,(BC)$ and $(BC)' = B'C$, Lemma 6.9 also yields

$$\begin{aligned} \left( L(BC,A) ~{-}~ \gamma L(BC,A')\right) ~-~ \left( L(A,BC) ~{-}~ \gamma L(A,B'C)\right) ~{=}~ g\,(m,a',b'c,n)\,. \end{aligned}$$

(7)

Next, (Z6), applied to $m, a', b', n, c$ and to $m, a', b', c, n$, yields:

$$\begin{aligned}&\gamma g\,(a',b',n,c) ~-~ g\, (ma',b',n,c) ~+~ g\,(m,a'b',n,c) \\&\qquad -~ g\,(m,a',b'n,c) ~+~ g\,(m,a',b',nc) ~-~ \gamma g\,(m,a',b',n) ~=~ 0, \\&0 ~=~ \gamma g\,(a',b',c,n) ~-~ g\, (ma',b',c,n) ~+~ g\,(m,a'b',c,n) \\&\qquad -~ g\,(m,a',b'c,n) ~+~ g\,(m,a',b',cn) ~-~ \gamma g\,(m,a',b',c)\,; \end{aligned}$$

(8)

and (G0) yields, since $m = n$:

$$\begin{aligned} g\, (m,a'b',c,n) ~-~ g\, (m,a'b',n,c) ~=~ 0. \end{aligned}$$

(9)

In (5), terms

511 and 541,    521 and 551 (since $A'B = AB'$),    524 and 574,    532 and 571

cancel each other. When the equalities in (6), (7), (8), and (9) are added, terms

611 and 823,    715 and 841,    813 and 912,    822 and 842,    833 and 911

cancel in pairs. The remaining terms match the remaining terms of (5):

512 = 831,	513 = 811,	522 = 613,	523 = 712,	531 = 821,
533 = 713,	534 = 614,	542 = 843,	552 = 714,	553 = 615,
561 = 832,	562 = 812,	572 = 612,	573 = 711.

Thus (5) follows from (6), (7), (8), and (9). $\square$

Lemma I.6

(CASE 4-6) If $m \leqq n,p$, $|A| \geqq 2$, $p \leqq m,n$, and $|C| \geqq 2$, then $r_4\,(A,B,C) = r_6\,(A,B,C)$.

Proof

First assume that $|ABC| = 5$, so that $|A| = |C| = 2$ and $|B| = 1$. Then $r\,(A',B,C) = t\,([a'],[b],[c',p])$, by Lemma I.3;

$$\begin{aligned} \mathrm{min}\,(AB)&= m, ~(AB)' = [a',b]\,, ~~c' \geqq m, ~L(AB,C') ~=~ t\,([m],[a',b],[c']); \\ \mathrm{min}\,(BC)&= p, ~(BC)' = [b,c']\,, ~L(A,BC') ~=~ t\,([m],[a'],[b,c']); \\ L(AB,C) ~&=~ \gamma L(AB,C') ~+~ t\,([m,a'b],[c'],[p]) ~+~ t\,([a'b,c'],[p],[m]) \\&\qquad =~ \gamma t\,([m],[a',b],[c']) ~+~ t\,([m,a'b],[c'],[p]) ~+~ t\,([a'b,c'],[p],[m]) \,; \\ L(A,BC) ~&=~ \gamma L(A,BC') ~+~ t\,([m,a'],[bc'],[p]) ~+~ t\,([a',bc'],[p],[m]) \\&\qquad =~ \gamma t\,([m],[a'],[b,c']) ~+~ t\,([m,a'],[bc'],[p]) ~+~ t\,([a',bc'],[p],[m]) \,; \end{aligned}$$

$b \geqq m$, and $L(A,B) = 0$. Hence

$$\begin{aligned}&r_4\,(A,B,C) \\&\quad =~ \gamma r\,(A',B,C) ~-~g\,(m,a',b,c) ~+~L(AB,C) ~-~L(A,BC) ~+~\gamma L(A,B) \\&\quad =~ \gamma t\,([a'],[b],[c',p]) ~-~g\,(m,a',b,c) \\&\qquad +~ \gamma t\,([m],[a',b],[c']) ~+~ t\,([m,a'b],[c'],[p]) ~+~ t\,([a'b,c'],[p],[m]) \\&\qquad -~ \gamma t\,([m],[a'],[b,c']) ~-~ t\,([m,a'],[bc'],[p]) ~-~ t\,([a',bc'],[p],[m])\,. \end{aligned}$$

On the other hand, $r\,(A,B,C') = t\,([m,a'],[b],[c'])$, by Lemma I.3; $p = \mathrm{min}\,(BC)$, $a' \geqq p$, $(BC)' = [b,c']$,

$$\begin{aligned} L(BC,A') ~&=~ t\,([p],[b,c'],[a'])\,, ~L(C,(AB)') ~=~ t\,([p],[c'],[a',b])\,, \\ L(BC,A) ~&=~ \gamma L(BC,A') ~+~ t\,([p,bc'],[a'],[m]) ~+~ t\,([bc',a'],[m],[p]) \\&\qquad =~ \gamma t\,([p],[b,c'],[a']) ~+~ t\,([p,bc'],[a'],[m]) ~+~ t\,([bc',a'],[m],[p])\,, \\ L(C,AB) ~&=~ \gamma L(C,(AB)') ~+~ t\,([p,c'],[a'b],[m]) ~+~ t\,([c',a'b],[m],[p]) \\&\qquad =~ \gamma t\,([p],[c'],[a',b]) ~+~ t\,([p,c'],[a'b],[m]) ~+~ t\,([c',a'b],[m],[p])\,; \end{aligned}$$

$b \geqq p$, and $L(C,B) = 0$. Hence

$$\begin{aligned}&r_6\,(A,B,C) \\&\quad =~ \gamma r\,(A,B,C') ~-~g\,(a,b,c',p) ~-~L(BC,A) ~+~L(C,AB) ~-~\gamma L(C,B) \\&\quad =~ \gamma t\,([m,a'],[b],[c']) ~-~ g\,(a,b,c',p) \\&\qquad -~ \gamma t\,([p],[b,c'],[a']) ~-~ t\,([p,bc'],[a'],[m]) ~-~ t\,([bc',a'],[m],[p]) \\&\qquad +~ \gamma t\,([p],[c'],[a',b]) ~+~ t\,([p,c'],[a'b],[m]) ~+~ t\,([c',a'b],[m],[p])\,. \end{aligned}$$

The equality $r_4\,(A,B,C) = r_6\,(A,B,C)$ now reads:

$$\begin{aligned}&\gamma t\,([a'],[b],[c',p]) ~-~ g\,(m,a',b,c) \\&\quad +~ \gamma t\,([m],[a',b],[c']) ~+~ t\,([m,a'b],[c'],[p]) ~+~ t\,([a'b,c'],[p],[m]) \\&\quad -~ \gamma t\,([m],[a'],[b,c']) ~-~ t\,([m,a'],[bc'],[p]) ~-~ t\,([a',bc'],[p],[m]) \\&\quad {\mathop {=}\limits ^{?}}~\gamma t\,([m,a'],[b],[c']) ~-~ g\,(a,b,c',p) \\&\quad -~ \gamma t\,([p],[b,c'],[a']) ~-~ t\,([p,bc'],[a'],[m]) ~-~ t\,([bc',a'],[m],[p]) \\&\quad +~ \gamma t\,([p],[c'],[a',b]) ~+~ t\,([p,c'],[a'b],[m]) ~+~ t\,([c',a'b],[m],[p])\,. \end{aligned}$$

(1)

To prove (1), use (G3):

$$\begin{aligned} -~ \gamma g\,(m,a',b,c') ~&=~ \gamma t\,([m,a'],[b],[c']) \\&\quad -~ \gamma t\,([m],[a',b],[c']) ~+~ \gamma t\,([m],[a'],[b,c'])\,, \\ -~ \gamma g\,(p,c',b,a') ~&=~ \gamma t\,([p,c'],[b],[a']) \\&\quad -~ \gamma t\,([p],[c',b],[a']) ~+~ \gamma t\,([p],[c'],[b,a'])\,, \\ g\,(p,bc',a',m) ~&=~ -~ t\,([p,bc'],[a'],[m]) ~-~ t\,([p],[bc'],[a',m])\,, \\ -~ g\,(p,c',a'b,m) ~&=~ t\,([p,c'],[a'b],[m]) ~+~ t\,([p],[c'],[a'b,m])\,, \end{aligned}$$

(2)

since $p = m$ and $t\,([p],[bc',a'],[m]) = t\,([p],[c',a'b],[m]) = 0$ by (S3a); and use (Z6):

$$\begin{aligned} 0 ~&=~ \gamma g\,(a',b,c',p) ~-~ g\,(ma',b,c',p) ~+~ g\,(m,a'b,c',p) \\&-~ g\,(m,a',bc',p) ~+~ g\,(m,a',b,c'p) ~-~ \gamma g\,(m,a',b,c') \,. \end{aligned}$$

(3)

In (1), terms 123 and 163,   133 and 153 cancel each other, since $m = p$. When the equalities in (2) and (3) are added, terms

211 and 323,    231 and 311,    251 and 321,    261 and 313

cancel each other, by (S3b). The remaining terms match the remaining terms of (1):

111 = 222,	112 = 322,	121 = 221,	122 = 263,	131 = 222,	132 = 253,
141 = 212,	142 = 312,	151 = 241,	152 = 252,	161 = 242,	162 = 262.

Thus (1) follows from (2) and (3).

The general case is proved by induction on |ABC|. Assume $|ABC| \geqq 6$ (still with $|A|,|C| \geqq 2$ and $m = p \leqq n$). By the induction hypothesis,

$$\begin{aligned}&r_4\,(A,B,C) \\&\quad =~ \gamma r_6\,(A',B,C) ~-~ g\,(m,a',b,c) ~+~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B)\\&\quad =~ \gamma r\,(A',B,C') ~-~ \gamma g\,(a',b',c',p) -~ \gamma L(BC,A') ~+~ \gamma L(C,A'B) ~-~ \gamma L(C,B) \\&\qquad -~ g\,(m,a',b,c) ~+~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B) \,. \end{aligned}$$

Similarly,

$$\begin{aligned}&r_6\,(A,B,C) \\&\quad =~ \gamma r_4\,(A,B,C') ~-~ g\,(a,b,c',p) ~-~ L(BC,A) ~+~ L(C,AB) ~-~ \gamma L(C,B)\\&\quad =~ \gamma r\,(A',B,C') ~-~ \gamma g\,(m,a',b,c') ~+~ \gamma L(AB,C') ~-~ \gamma L(A,BC') ~+~ \gamma L(A,B) \\&\qquad -~ g\,(a,b,c',p) ~-~ L(BC,A) ~+~ L(C,AB) ~-~ \gamma L(C,B)\,. \end{aligned}$$

The equality $r_4\,(A,B,C) = r_6\,(A,B,C)$ now reads:

$$\begin{aligned}&\gamma r\,(A',B,C') ~-~ \gamma g\,(a',b',c',p) ~-~ \gamma L(BC,A') ~+~ \gamma L(C,A'B) ~-~ \gamma L(C,B) \\&\qquad -~ g\,(m,a',b,c) ~+~ L(AB,C) ~-~ L(A,BC) ~+~ \gamma L(A,B)\\&\quad {\mathop {=}\limits ^{?}}~ \gamma r\,(A',B,C') ~-~ \gamma g\,(m,a',b,c') ~+~ \gamma L(AB,C') ~-~ \gamma L(A,BC') ~+~ \gamma L(A,B) \\&\qquad -~ g\,(a,b,c',p) ~-~ L(BC,A) ~+~ L(C,AB) ~-~ \gamma L(C,B)\,. \end{aligned}$$

(4)

Since $p = \mathrm{min}\,(BC)$, $(BC)' = BC'$, $m = \mathrm{min}\,(AB)$, and $(AB)' = A'B$, Lemma 6.9 yields:

$$\begin{aligned}&\left( L(BC,A) - \gamma L(BC,A')\right) - \left( L(A,BC) - \gamma L(A,BC')\right) ~=~ g\,(m,a',bc',n)\,, \\&g\,(m,a'b,c',p) ~=~\left( L(C,AB) - \gamma L(C,A'B)\right) - \left( L(AB,C) - \gamma L(AB,C')\right) \,. \end{aligned}$$

(5)

Then (Z6) yields:

$$\begin{aligned}&-~\gamma g\,(a',b',c',p) ~+~ g\, (ma',b,c',p) ~-~ g\, (m,a'b,c',p) \\&~~~~~+~ g\, (m,a',bc',p) ~-~ g\, (m,a',b,c'p) ~+~ \gamma g\,(m,a',b,c') ~=~ 0. \end{aligned}$$

(6)

In (4), terms    411 and 431,    415 and 444,    424 and 435

cancel in pairs. When the equalities in (5) and (6) are added, terms

515 and 621 (since $n = p$),    521 and 613

also cancel each other. The remaining terms match the remaining terms of (4):

412 = 611,	413 = 512,	414 = 523,	421 = 622,	422 = 524,	423 = 513,
432 = 623,	433 = 525,	434 = 514,	441 = 612,	442 = 511,	443 = 522.

Thus (4) follows from (5) and (6). This also completes the proof of Lemma 6.10. $\square$

Proof of Lemma 6.11

Lemma 6.11

If $f \in C_6$ has properties (P6) and (Z6), then r, as defined in Lemma 6.10, is a symmetric 3-cochain with properties (Z5c), (Z5d), and (Z5e).

The proof is divided into several lemmas.

Lemma J.1

r has properties (S3a), (S3b), (S3c), and (S3d). Hence $r \in SC^3(F_1,G)$.

Proof

Since r is well-defined by Lemma 6.10, property (S3d) follows from Lemma I.2. Then (S3a), (S3b), and (S3c) hold, by Lemma 5.1.

Lemma J.2

r has property (Z5c).

Proof

By definition, $r_0\,([a],[b],[c]) = t\,([a],[b],[c]) = 0$. $\square$

Lemma J.3

r has property (Z5d): $(\delta r)(A,B,C,D) = (\delta r)([a],[b],[c],[d])$.

Proof

By Lemma I.3 and by (G3),

$$\begin{aligned} (\delta r)([a],[b],[c],[d]) ~=~ (\delta t)([a],[b],[c],[d]) ~=~ g\,(a,b,c,d)\,. \end{aligned}$$

Hence (Z5d) reads

$$\begin{aligned}&\gamma r\,(B,C,D) ~-~ r\,(AB,C,D) ~+~ r\,(A,BC,D) \\&~~-~ r\,(A,B,CD) ~+~ \gamma r\,(A,B,C) ~=~ g\,(a,b,c,d)\,, \end{aligned}$$

for all $A,B,C,D \in F_1$. This holds if $|ABCD| = 4$, and is proved by induction on |ABCD| if $|ABCD| \geqq 5$, distinguishing several cases that depend on which of $m = \mathrm{min}\,A$, $n = \mathrm{min}\,B$, $p = \mathrm{min}\,C$, $q = \mathrm{min}\,D$ is smallest. Since $(\delta r)(D,C,B,A) = - (\delta r)(A,B,C,D)$ and $g\,(d,c,b,a) = - g\,(a,b,c,d)$, one may assume that m or n is smallest.

CASE 1: $m \leqq n,p,q$ and $|A| = 1$. Then $m = a = \mathrm{min}\,(AB)$, $(AB)' = B$,

$$\begin{aligned}&r_4\,(AB,C,D) ~=~ \gamma r\,(B,C,D) ~-~ g\,(a,b,c,d) \\&\quad +~ L(ABC,D) ~-~ L(AB,CD) ~+~ \gamma L(AB,C)\,,\\&r_1\,(A,BC,D) ~=~ L(ABC,D)\,, \\&r_1\,(A,B,CD) ~=~ L(AB,CD)\,, \\&r_1\,(A,B,C) ~=~ L(AB,C)\,, ~~\hbox {and}~~\\&\gamma r\,(B,C,D) ~-~ r\,(AB,C,D) ~+~ r\,(A,BC,D) \\&\quad -~ r\,(A,B,CD) ~+~ \gamma r\,(A,B,C) ~=~ g\,(a,b,c,d)\,. \end{aligned}$$

CASE 2: $m \leqq n,p,q$ and $|A| = 2$. Then $m = \mathrm{min}\,(AB)$, $(AB)' = A'B$,

$$\begin{aligned} r_4\,(AB,C,D)&~=~ \gamma r\,(A'B,C,D) ~-~g\,(m,a'b,c,d) \\&\quad +~L(ABC,D) ~-~L(AB,CD) ~+~\gamma L(AB,C)\,,\\ r_4\,(A,BC,D)&~=~ \gamma r\,(A',BC,D) ~-~g\,(m,a',bc,d) \\&\quad +~L(ABC,D) ~-~L(A,BCD) ~+~\gamma L(A,BC)\,,\\ r_4\,(A,B,CD)&~=~ \gamma r\,(A',B,CD) ~-~g\,(m,a',b,cd) \\&\quad +~L(AB,CD) ~-~L(A,BCD) ~+~\gamma L(A,B)\,, ~~\hbox {and}~~\\ r_4\,(A,B,C)&~=~ \gamma r\,(A',B,C) ~-~ \gamma g\,(m,a',b,c) \\&\quad +~ \gamma L(AB,C) ~-~ \gamma L(A,BC) ~+~ \gamma L(A,B)\,. \end{aligned}$$

By the induction hypothesis,

$$\begin{aligned}&\gamma r\,(B,C,D) ~-~ r\,(AB,C,D) ~+~ r\,(A,BC,D) \\&\qquad -~ r\,(A,B,CD) ~+~ \gamma r\,(A,B,C) \\&\quad =~ \gamma r\,(B,C,D) ~-~ \gamma r\,(A'B,C,D) ~+~ \gamma r\,(A',BC,D) \\&\qquad \quad -~ \gamma r\,(A',B,CD) ~+~ \gamma r\,(A',B,C) \\&\qquad +~ g\,(m,a'b,c,d) ~-~ g\,(m,a',bc,d) ~+~ g\,(m,a',b,cd) ~-~ \gamma g\,(m,a',b,c) \\&\qquad -~ L(ABC,D) ~+~ L(AB,CD) ~-~ \gamma L(AB,C) \\&\qquad +~ L(ABC,D) ~-~ L(A,BCD) ~+~ \gamma L(A,BC) \\&\qquad -~ L(AB,CD) ~+~ L(A,BCD) ~-~ \gamma L(A,B) \\&\qquad +~ \gamma L(AB,C) ~-~ \gamma L(A,BC) ~+~ \gamma L(A,B) \\&\quad =~ \gamma g\,(a',b,c,d) \\&\qquad +~g\,(m,a'b,c,d) ~-~ g\,(m,a',bc,d)) ~+~ g\,(m,a',b,cd) ~-~ \gamma g\,(m,a',b,c) \\&\quad =~ g\,(ma',b,c,d)\,, ~~{\hbox { by (Z6)}}\,. \end{aligned}$$

CASE 3: $n \leqq m,p,q$ and $|B| = 1$. Then $r_1\,(B,C,D) = L(BC,D)$, $\mathrm{min}\,(AB) = n = b = \mathrm{min}\,(BC)$, $(AB)' = A$, $(BC)' = C$,

$$\begin{aligned} r_4\,(AB,C,D) ~&=~ \gamma r\,(A,C,D) ~-~g\,(b,a,c,d) \\&\quad +~ L(ABC,D) ~-~ L(AB,CD) ~+~ \gamma L(AB,C)\,,\\ r_5\,(A,BC,D) ~&=~ \gamma r\,(A,C,D) ~-~ g\,(b,c,a,d) ~+~ g\,(b,c,d,a) \\&\quad +~ L(ABC,D) ~+~ \gamma L(BC,A) ~-~ L(BCD,A) ~-~ \gamma L(BC,D)\,, \\ r_2\,(A,B,CD) ~&=~ L(AB,CD) ~-~ L(BCD,A)\,, \\ r_2\,(A,B,C) ~&=~ L(AB,C) ~-~ L(BC,A)\,, \end{aligned}$$

and (G1), (Gb) yield

$$\begin{aligned} g\,(b,c,d,a) ~-~ g\,(b,c,a,d) ~&=~ g\,(d,c,a,b) ~-~ g\,(d,c,b,a) \\&=~ g\,(a,b,c,d) ~-~ g\,(b,a,c,d)\,. \end{aligned}$$

(G5)

Hence

$$\begin{aligned}&\gamma r\,(B,C,D) ~-~ r\,(AB,C,D) ~+~ r\,(A,BC,D) ~-~ r\,(A,B,CD) ~+~ \gamma r\,(A,B,C) \\&\quad =~ \gamma L(BC,D) \\&\qquad -~ \gamma r\,(A,C,D) ~+~ g\,(b,a,c,d) ~-~ L(ABC,D) ~+~ L(AB,CD) ~-~ \gamma L(AB,C) \\&\qquad +~ \gamma r\,(A,C,D) ~-~ g\,(b,c,a,d) ~+~ g\,(b,c,d,a) \\&\qquad \quad +~ L(ABC,D) ~+~ \gamma L(BC,A) ~-~ L(BCD,A) ~-~ \gamma L(BC,D) \\&\qquad -~ L(AB,CD) ~+~ L(BCD,A) ~+~ \gamma L(AB,C) ~-~ \gamma L(BC,A) \\&\quad =~ g\,(a,b,c,d)\,, ~~{\hbox { by (G5)}}\,. \end{aligned}$$

CASE 4: $n \leqq m,p,q$ and $|B| \geqq 2$. Then $\mathrm{min}\,(AB) = n = \mathrm{min}\,(BC)$, $(AB') = AB'$, $(BC)' = B'C$,

$$\begin{aligned} r_4\,(B,C,D) ~&=~ \gamma r\,(B',C,D) ~-~ g\,(n,b',c,d) \\&\quad +~ L(BC,D) ~-~ L(B,CD) ~+~ \gamma L(B,C)\,, \\ r_4\,(AB,C,D) ~&=~ \gamma r\,(AB',C,D) ~-~ g\,(n,ab',c,d) \\&\quad +~ L(ABC,D) ~-~ L(AB,CD) ~+~ \gamma L(AB,C)\,, \\ r_5\,(A,BC,D) ~&=~ \gamma r\,(A,B'C,D) ~-~ g\,(n,b'c,a,d) ~+~ g\,(n,b'c,d,a) \\&\quad +~ L(ABC,D) + \gamma L(BC,A) - L(BCD,A) - \gamma L(BC,D)\,, \\ r_5\,(A,B,CD) ~&=~ \gamma r\,(A,B',CD) ~-~ g\,(n,b',a,cd) ~+~ g\,(n,b',cd,a) \\&\quad +~ L(AB,CD) + \gamma L(B,A) - L(BCD,A) - \gamma L(B,CD)\,, \\ r_5\,(A,B,C) ~&=~ \gamma r\,(A,B',C) ~-~ g\,(n,b',a,c) ~+~ g\,(n,b',c,a) \\&\quad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C)\,. \end{aligned}$$

By the induction hypothesis and (G5):

$$\begin{aligned}&\gamma r\,(B,C,D) ~-~ r\,(AB,C,D) ~+~ r\,(A,BC,D) ~-~ r\,(A,B,CD) ~+~ \gamma r\,(A,B,C) \\&\quad =~ \gamma r\,(B',C,D) ~-~ \gamma r\,(AB',C,D) ~+~ \gamma r\,(A,B'C,D) \\&\qquad \quad -~ \gamma r\,(A,B',CD) ~+~ \gamma r\,(A,B',C) \\&\qquad -~ \gamma g\,(n,b',c,d) ~+~ g\,(n,ab',c,d) ~-~ g\,(n,b'c,a,d) ~+~ g\,(n,b'c,d,a) \\&\qquad +~ g\,(n,b',a,cd) ~-~ g\,(n,b',cd,a) ~-~ \gamma g\,(n,b',a,c) ~+~ \gamma g\,(n,b',c,a) \\&\qquad +~ \gamma L(BC,D) ~-~ \gamma L(B,CD) ~+~ \gamma L(B,C) \\&\qquad -~ L(ABC,D) ~+~ L(AB,CD) ~-~ \gamma L(AB,C) \\&\qquad +~ L(ABC,D) ~+~ \gamma L(BC,A) ~-~ L(BCD,A) ~-~ \gamma L(BC,D) \\&\qquad -~ L(AB,CD) ~-~ \gamma L(B,A) ~+~ L(BCD,A) ~+~ \gamma L(B,CD) \\&\qquad +~ L(AB,C) ~+~ \gamma L(B,A) ~-~ L(BC,A) ~-~ \gamma L(B,C) \\&\quad =~ \gamma g\,(a,b',c,d) \\&\qquad -~ \gamma g\,(n,b',c,d) ~+~ g\,(n,ab',c,d) ~+~ g\,(a,n,b'c,d) ~-~ g\,(n,a,b'c,d) \\&\qquad -~ g\,(a,n,b',cd) ~+~ g\,(n,a,b',cd) ~+~ \gamma g\,(a,n,b',c) ~-~ \gamma g\,(n,a,b',c)\,. \end{aligned}$$

The equality to be proved:

$$\begin{aligned}&\gamma g\,(a,b',c,d) ~-~ \gamma g\,(n,b',c,d) ~+~ g\,(n,ab',c,d) \\&\quad \quad +~ g\,(a,n,b'c,d) ~-~ g\,(n,a,b'c,d) \\&\quad -~ g\,(a,n,b',cd) ~+~ g\,(n,a,b',cd) ~+~ \gamma g\,(a,n,b',c) \\&\quad \quad -~ \gamma g\,(n,a,b',c) ~-~ g\,(a,nb',c,d) ~{\mathop {=}\limits ^{?}}~ 0 \end{aligned}$$

(1)

now follows by adding two equalities provided by (Z6):

$$\begin{aligned}&\gamma g\,(a,b',c,d) ~-~ g\,(na,b',c,d) ~+~ g\,(n,ab',c,d) \\&\quad -~ g\,(n,a,b'c,d) ~+~ g\,(n,a,b',cd) ~-~ \gamma g\,(n,a,b',c) ~=~ 0, \\ -~&\gamma g\,(n,b',c,d) ~+~ g\,(an,b',c,d) ~-~ g\,(a,nb',c,d) \\&\quad +~ g\,(a,n,b'c,d) ~-~ g\,(a,n,b',cd) ~+~ \gamma g\,(a,n,b',c) ~=~ 0. \end{aligned}$$

(2)

In the sum, terms 212 and 232 cancel each other, and the remaining terms match the terms of (1):

111 = 211,	112 = 231,	113 = 213,	121 = 241,	122 = 221,
131 = 242,	132 = 222,	133 = 243,	141 = 223,	142 = 233.

Thus (1) follows from (2). $\square$

Lemma J.4

r has property (Z5e):

$$\begin{aligned}&r\,(A,BE,C) ~+~ r\,(B,CE,A) ~+~ r\,(C,AE,B) \\&\qquad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A) ~+~ \gamma r\,(A,E,B) \\&\quad =~ r\,([a],[b,e],[c]) ~+~ r\,([b],[c,e],[a]) ~+~ r\,([c],[a,e],[b])\,. \end{aligned}$$

Proof

By Lemma I.3 and by (G4),

$$\begin{aligned}&r\,([a],[b,e],[c]) + r\,([b],[c,e],[a]) + r\,([c],[a,e],[b]) \\&\quad =~ t\,([a],[b,e],[c]) + t\,([b],[c,e],[a]) + t\,([c],[a,e],[b]) \\&\quad =~ g\,(a,e,b,c) - g\,(a,e,c,b)\,. \end{aligned}$$

Hence (Z5e) is equivalent to

$$\begin{aligned}&r\,(A,BE,C) ~+~ r\,(B,CE,A) ~+~ r\,(C,AE,B) \\&\qquad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A) ~+~ \gamma r\,(A,E,B) \\&\quad =~ g\,(a,e,b,c) ~-~ g\,(a,e,c,b)\,. \end{aligned}$$

(Z5ea)

This property is proved by induction on |ABCE|.

If $|ABCE| = 4$, then (Z5ea) follows from (Z5c) and (G4).

When $|ABCE| \geqq 5$, the proof is divided among several cases, according to which of $m = \mathrm{min}\,A$, $n = \mathrm{min}\,B$, $p = \mathrm{min}\,C$, $q = \mathrm{min}\,E$ is smallest. Since circular permutations of A, B, C leave (Z5e) unchanged one may assume that m or q is smallest.

CASE 1: $m \leqq n,p,q$, $|A| = 1$. Then $\mathrm{min}\,(AE) = m = a$, $(AE)' = E$,

$$\begin{aligned} r_1\,(A,BE,C) ~=~&L(ABE,C)\,, ~r_3\,(B,CE,A) ~=~ - L(ACE,B)\,, \\ r_5\,(C,AE,B) ~=~&\gamma r\,(C,E,B) ~-~ g\,(a,e,c,b) ~+~ g\,(a,e,b,c) \\&~~+~ L(ACE,B) + \gamma L(AE,C) - L(ABE,C) - \gamma L(AE,B)\,, \end{aligned}$$

$r_3\,(C,E,A) = - L(AE,C)$, $r_1\,(A,E,B) = L(AE,B)$, and

$$\begin{aligned}&r\,(A,BE,C) ~+~ r\,(B,CE,A) ~+~ r\,(C,AE,B) \\&\qquad \quad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A) ~+~ \gamma r\,(A,E,B) \\&\qquad =~ L(ABE,C) ~-~ L(ACE,B) \\&\qquad +~ \gamma r\,(C,E,B) ~-~ g\,(a,e,c,b) ~+~ g\,(a,e,b,c) \\&\qquad \quad +~ L(ACE,B) ~+~ \gamma L(AE,C) ~-~ L(ABE,C) ~-~ \gamma L(AE,B) \,, \\&\qquad +~ \gamma r\,(B,E,C) ~-~ \gamma L(AE,C) ~+~ \gamma L(AE,B) \\&\quad =~ g\,(a,e,b,c) ~-~ g\,(a,e,c,b)\,,\hbox { by property (S3b) of }r. \end{aligned}$$

CASE 2: $m \leqq n,p,q$, $|A| \geqq 2$. Then $m = \mathrm{min}\,(AE)$, $(AE)' = A'E$,

$$\begin{aligned} r_4\,(A,BE,C) ~=~&\gamma r\,(A',BE,C) ~-~g\,(m,a',be,c) \\&~~+~L(ABE,C) ~-~L(A,BCE) ~+~\gamma L(A,BE)\,,\\ r_6\,(B,CE,A) ~=~&\gamma r\,(B,CE,A') ~-~g\,(b,ce,a',m) \\&~~-~L(ACE,B) ~+~ L(A,BCE) ~-~ \gamma L(A,CE)\,,\\ r_5\,(C,AE,B) ~=~&\gamma r\,(C,A'E,B) ~-~ g\,(m,a'e,c,b) ~+~ g\,(m,a'e,b,c) \\&~~+~ L(ACE,B) ~+~ \gamma L(AE,C) \\&~~-~ L(ABE,C) ~-~ \gamma L(AE,B) \,, \\ r_6\,(C,E,A) ~=~&\gamma r\,(C,E,A') ~-~ g\,(c,e,a',m) \\&~~-~ L(AE,C) ~+~ L(A,CE) ~-~ \gamma L(A,E)\,, ~~\hbox {and}~~\\ r_4\,(A,E,B) ~=~&\gamma r\,(A',E,B) ~-~ g\,(m,a',e,b) \\&~~+~ L(AE,B) ~-~ L(A,BE) ~+~ \gamma L(A,E) \,. \end{aligned}$$

Hence

$$\begin{aligned}&r\,(A,BE,C) ~+~ r\,(B,CE,A) ~+~ r\,(C,AE,B) \\&\qquad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A) ~+~ \gamma r\,(A,E,B) \\&\quad =~ \gamma r\,(A',BE,C) ~+~ \gamma r\,(B,CE,A') ~+~ \gamma r\,(C,A'E,B) \\&\qquad \quad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A') ~+~ \gamma r\,(A',E,B) \\&\qquad -~ g\,(m,a',be,c) ~-~ g\,(b,ce,a',m) ~-~ g\,(m,a'e,c,b) \\&\qquad \quad +~ g\,(m,a'e,b,c) ~-~ \gamma g\,(c,e,a',m) ~-~ \gamma g\,(m,a',e,b) \\&\qquad +~ L(ABE,C) ~-~ L(A,BCE) ~+~ \gamma L(A,BE) \\&\qquad \quad -~ L(ACE,B) ~+~ L(A,BCE) ~-~ \gamma L(A,CE) \\&\qquad +~ L(ACE,B) ~+~ \gamma L(AE,C) ~-~ L(ABE,C) ~-~ \gamma L(AE,B) \\&\qquad -~ \gamma L(AE,C) ~+~ \gamma L(A,CE) ~-~ \gamma L(A,E) \\&\qquad \quad +~ \gamma L(AE,B) ~-~ \gamma L(A,BE) ~+~ \gamma L(A,E) \\&\quad =~ \gamma g\,(a',e,b,c) ~-~ \gamma g\,(a',e,c,b) \\&\qquad -~ g\,(m,a',be,c) ~-~ g\,(b,ce,a',m) ~-~ g\,(m,a'e,c,b) \\&\qquad +~ g\,(m,a'e,b,c) ~-~ \gamma g\,(c,e,a',m) ~-~ \gamma g\,(m,a',e,b)\,, \end{aligned}$$

by the induction hypothesis, and (Z5ea) reads:

$$\begin{aligned}&\gamma g\,(a',e,b,c) ~-~ \gamma g\,(a',e,c,b) \\&\qquad -~ g\,(m,a',be,c) ~-~ g\,(b,ce,a',m) ~-~ g\,(m,a'e,c,b) \\&\qquad +~ g\,(m,a'e,b,c) ~-~ \gamma g\,(c,e,a',m) ~-~ \gamma g\,(m,a',e,b)\,, \\&{\mathop {=}\limits ^{?}}~ g\,(a,e,b,c) ~-~ g\,(a,e,c,b)\,. \end{aligned}$$

(1)

By (Z6),

$$\begin{aligned}&\gamma g\,(a',e,b,c) ~-~ g\,(ma',e,b,c) ~+~ g\,(m,a'e,b,c) \\&\quad -~ g\,(m,a',eb,c) ~+~ g\,(m,a',e,bc) ~-~ \gamma g\,(m,a',e,b) ~=~ 0, \\&-~\gamma g\,(a',e,c,b) ~+~ g\,(ma',e,c,b) ~-~ g\,(m,a'e,c,b) \\&\quad +~ g\,(m,a',ec,b) ~-~ g\,(m,a',e,bc) ~+~ \gamma g\,(m,a',e,c) ~=~ 0. \end{aligned}$$

(2)

When the equalities in (2) are added, terms 222 and 242 cancel each other; the remaining terms match the terms of (1), by (Gb):

111 = 211,	112 = 231,	121 = 221,	122 = 241,	123 = 233,
131 = 213,	132 = 243,	133 = 223,	141 = 212,	142 = 232.

Thus (1) follows from (2).

CASE 3: $q \leqq m,n,p$, $|E| = 1$. Then $q = e = \mathrm{min}\,(AE) = \mathrm{min}\,(BE) = \mathrm{min}\,(CE)$, $(AE)' = A$, $(BE)' = B$, $(CE)' = C$,

$$\begin{aligned}&r_5\,(A,BE,C) ~=~ \gamma r\,(A,B,C) ~-~ g\,(e,b,a,c) ~+~ g\,(e,b,c,a) \\&\qquad +~ L(ABE,C) ~+~ \gamma L(BE,A) ~-~ L(BCE,A) ~-~ \gamma L(BE,C)\,, \\&r_5\,(B,CE,A) ~=~ \gamma r\,(B,C,A) ~-~ g\,(e,c,b,a) ~+~ g\,(e,c,a,b) \\&\qquad +~ L(BCE,A) ~+~ \gamma L(CE,B) ~-~ L(ACE,B) ~-~ \gamma L(CE,A)\,, \\&r_5\,(C,AE,B) ~=~ \gamma r\,(C,A,B) ~-~ g\,(e,a,c,b) ~+~ g\,(e,a,b,c) \\&\qquad +~ L(ACE,B) ~+~ \gamma L(AE,C) ~-~ L(ABE,C) ~-~ \gamma L(AE,B)\,, \\&r_2\,(B,E,C) ~=~ L(BE,C) ~-~ L(CE,B)\,, \\&r_2\,(C,E,A) ~=~ L(CE,A) ~-~ L(AE,C)\,, \\&r_2\,(A,E,B) ~=~ L(AE,B) ~-~ L(BE,A)\,, \end{aligned}$$

and

$$\begin{aligned}&r\,(A,BE,C) ~+~ r\,(B,CE,A) ~+~ r\,(C,AE,B) \\&\qquad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A) ~+~ \gamma r\,(A,E,B) \\&\quad =~ \gamma r\,(A,B,C) ~+~ \gamma r\,(B,C,A) ~+~ \gamma r\,(C,A,B) \\&\qquad \quad -~ g\,(e,b,a,c) ~+~ g\,(e,b,c,a) ~-~ g\,(e,c,b,a) \\&\qquad \quad +~ g\,(e,c,a,b) ~-~ g\,(e,a,c,b) ~+~ g\,(e,a,b,c) \\&\qquad +~ L(ABE,C) ~+~ \gamma L(BE,A) ~-~ L(BCE,A) ~-~ \gamma L(BE,C) \\&\qquad +~ L(BCE,A) ~+~ \gamma L(CE,B) ~-~ L(ACE,B) ~-~ \gamma L(CE,A)\,, \\&\qquad +~ L(ACE,B) ~+~ \gamma L(AE,C) ~-~ L(ABE,C) ~-~ \gamma L(AE,B)\,, \\&\qquad +~ \gamma L(BE,C) ~-~ \gamma L(CE,B) ~+~ \gamma L(CE,A) \\&\qquad \quad -~ \gamma L(AE,C) ~+~ \gamma L(AE,B) ~-~ L(BE,A) \\&\quad =~ g\,(a,e,b,c) ~-~ g\,(a,e,c,b)\,, \end{aligned}$$

by (S3b) and (G2).

CASE 4: $q \leqq m,n,p$, $|E| \geqq 2$. Then $q = \mathrm{min}\,(AE) = \mathrm{min}\,(BE) = \mathrm{min}\,(CE)$, $(AE)' = AE'$, $(BE)' = BE'$, $(CE)' = CE'$,

$$\begin{aligned}&r_5\,(A,BE,C) ~=~ \gamma r\,(A,BE',C) ~-~ g\,(q,be',a,c) ~+~ g\,(q,be',c,a) \\&\quad +~ L(ABE,C) ~+~ \gamma L(BE,A) ~-~ L(BCE,A) ~-~ \gamma L(BE,C)\,, \\&r_5\,(B,CE,A) ~=~ \gamma r\,(B,CE',A) ~-~ g\,(q,ce',b,a) ~+~ g\,(q,ce',a,b) \\&\quad +~ L(BCE,A) ~+~ \gamma L(CE,B) ~-~ L(ACE,B) ~-~ \gamma L(CE,A)\,, \\&r_5\,(C,AE,B) ~=~ \gamma r\,(C,AE',B) ~-~ g\,(q,ae',c,b) ~+~ g\,(q,ae',b,c) \\&\quad +~ L(ACE,B) ~+~ \gamma L(AE,C) ~-~ L(ABE,C) ~-~ \gamma L(AE,B)\,, \\&r_5\,(B,E,C) ~=~ \gamma r\,(B,E',C) ~-~ g\,(q,e',b,c) ~+~ g\,(q,e',c,d) \\&\quad +~ L(BE,C) ~+~ \gamma L(E,B) ~-~ L(CE,B) ~-~ \gamma L(C,E)\,, \\&r_5\,(C,E,A) ~=~ \gamma r\,(C,E',A) ~-~ g\,(q,e',c,a) ~+~ g\,(q,e',a,c) \\&\quad +~ L(CE,A) ~+~ \gamma L(E,C) ~-~ L(AE,C) ~-~ \gamma L(A,E)\,, \\&r_5\,(A,E,B) ~=~ \gamma r\,(A,E',B) ~-~ g\,(q,e',a,b) ~+~ g\,(q,e',b,a) \\&\quad +~ L(AE,B) ~+~ \gamma L(E,A) ~-~ L(BE,A) ~-~ \gamma L(B,E)\,, \end{aligned}$$

and

$$\begin{aligned}&r\,(A,BE,C) ~+~ r\,(B,CE,A) ~+~ r\,(C,AE,B) \\&\qquad +~ \gamma r\,(B,E,C) ~+~ \gamma r\,(C,E,A) ~+~ \gamma r\,(A,E,B) \\&\quad =~ \gamma r\,(A,BE',C) ~+~ \gamma r\,(B,CE',A) ~+~ \gamma r\,(C,AE',B) \\&\qquad \quad +~ \gamma r\,(B,E',C) ~+~ \gamma r\,(C,E',A) ~+~ \gamma r\,(A,E',B)\\&\qquad -~ g\,(q,be',a,c) ~+~ g\,(q,be',c,a) ~-~ g\,(q,ce',b,a) \\&\qquad \quad +~ g\,(q,ce',a,b) ~-~ g\,(q,ae',c,b) ~+~ g\,(q,ae',b,c)\\&\qquad -~ \gamma g\,(q,e',b,c) ~+~ \gamma g\,(q,e',c,d) ~-~ \gamma g\,(q,e',c,a) \\&\qquad \quad +~ \gamma g\,(q,e',a,c) ~-~ \gamma g\,(q,e',a,b) ~+~ \gamma g\,(q,e',b,a) \\&\qquad +~ L(ABE,C) ~+~ \gamma L(BE,A) ~-~ L(BCE,A) ~-~ \gamma L(BE,C) \\&\qquad +~ L(BCE,A) ~+~ \gamma L(CE,B) ~-~ L(ACE,B) ~-~ \gamma L(CE,A) \\&\qquad +~ L(ACE,B) ~+~ \gamma L(AE,C) ~-~ L(ABE,C) ~-~ \gamma L(AE,B) \\&\qquad +~ \gamma L(BE,C) ~+~ \gamma L(E,B) ~-~ \gamma L(CE,B) ~-~ \gamma L(C,E) \\&\qquad +~ \gamma L(CE,A) ~+~ \gamma L(E,C) ~-~ \gamma L(AE,C) ~-~ \gamma L(A,E) \\&\qquad +~ \gamma L(AE,B) ~+~ \gamma L(E,A) ~-~ \gamma L(BE,A) ~-~ \gamma L(B,E) \\&\quad =~ \gamma g\,(a,e',b,c) ~-~ \gamma g\,(a,e',c,b) \\&\qquad -~ g\,(q,be',a,c) ~+~ g\,(q,be',c,a) ~-~ g\,(q,ce',b,a) \\&\qquad +~ g\,(q,ce',a,b) ~-~ g\,(q,ae',c,b) ~+~ g\,(q,ae',b,c) \\&\qquad -~ \gamma g\,(q,e',b,c) ~+~ \gamma g\,(q,e',c,b) ~-~ \gamma g\,(q,e',c,a) \\&\qquad +~ \gamma g\,(q,e',a,c) ~-~ \gamma g\,(q,e',a,b) ~+~ \gamma g\,(q,e',b,a)\,, \end{aligned}$$

by the induction hypothesis, and (Z5ea) reads:

$$\begin{aligned}&\gamma g\,(a,e',b,c) ~-~ \gamma g\,(a,e',c,b) \\&\quad -~ g\,(q,be',a,c) ~+~ g\,(q,be',c,a) ~-~ g\,(q,ce',b,a)&\hbox {(line 2)} \\&\quad +~ g\,(q,ce',a,b) ~-~ g\,(q,ae',c,b) ~+~ g\,(q,ae',b,c) \\&\quad -~ \gamma g\,(q,e',b,c) ~+~ \gamma g\,(q,e',c,b) ~-~ \gamma g\,(q,e',c,a) \\&\quad +~ \gamma g\,(q,e',a,c) ~-~ \gamma g\,(q,e',a,b) ~+~ \gamma g\,(q,e',b,a)&\hbox {(line 5)} \\&\quad {\mathop {=}\limits ^{?}} g\,(a,e,b,c) ~-~ g\,(a,e,c,b)\,. \end{aligned}$$

(3)

By (G2),

$$\begin{aligned}&\gamma g\,(a,e',b,c) - \gamma g\,(a,e',c,b) \\&\quad = -~ \gamma g\,(e',b,a,c) + \gamma g\,(e',b,c,a) - \gamma g\,(e',c,b,a) \\&\qquad + \gamma g\,(e',c,a,b) - \gamma g\,(e',a,c,b) + \gamma g\,(e',a,b,c)\,, \\ \end{aligned}$$

(4)

$$\begin{aligned}&-~ g\,(e,b,a,c) + g\,(e,b,c,a) - g\,(e,c,b,a) \\&\qquad + g\,(e,c,a,b) - g\,(e,a,c,b) + g\,(e,a,b,c) \\&\quad = g\,(a,e,b,c) - g\,(a,e,c,b)\,. \end{aligned}$$

(5)

By (Z6),

$$\begin{aligned}&\gamma g\,(e',a,b,c) ~-~ g\,(qe',a,b,c) ~+~ g\,(q,e'a,b,c) \\&\quad -~ g\,(q,e',ab,c) ~+~ g\,(q,e',a,bc) ~-~ \gamma g\,(q,e',a,b) = 0,&\hbox {(line 2)} \\&\gamma g\,(e',b,c,a) ~-~ g\,(qe',b,c,a) ~+~ g\,(q,e'b,c,a) \\&\quad -~ g\,(q,e',bc,a) ~+~ g\,(q,e',b,ca) ~-~ \gamma g\,(q,e',b,c) = 0, \\&\gamma g\,(e',c,a,b) ~-~ g\,(qe',c,a,b) ~+~ g\,(q,e'c,a,b)&\hbox {(line 5)} \\&\quad -~ g\,(q,e',ca,b) ~+~ g\,(q,e',c,ab) ~-~ \gamma g\,(q,e',c,a) = 0, \\&-\gamma g\,(e',c,b,a) ~+~ g\,(qe',c,b,a) ~-~ g\,(q,e'c,b,a) \\&\quad +~ g\,(q,e',cb,a) ~-~ g\,(q,e',c,ba) ~+~ \gamma g\,(q,e',c,b) = 0,&\hbox {(line 8)} \\&-~ \gamma g\,(e',a,c,b) ~+~ g\,(qe',a,c,b) ~-~ g\,(q,e'a,c,b) \\&\quad +~ g\,(q,e',ac,b) ~-~ g\,(q,e',a,cb) ~+~ \gamma g\,(q,e',a,c) = 0,&\hbox {(line A)} \\&-~ \gamma g\,(e',b,a,c) ~+~ g\,(qe',b,a,c) ~-~ g\,(q,e'b,a,c) \\&\quad +~ g\,(q,e',ba,c) ~-~ g\,(q,e',b,ac) ~+~ \gamma g\,(q,e',b,a) = 0.&\hbox {(line C)} \end{aligned}$$

(6)

When the equalities in (4) and (5) are added, terms

421 and 5B1,	422 and 531,	423 and 571,	431 and 551,	432 and 591,
433 and 511,	441 and 5B2,	442 and 532,	443 and 572,	451 and 552,
452 and 5A2,	453 and 512,	521 and 5C1,	522 and 5A2,	541 and 581,
542 and 5C2,	561 and 591,	562 and 582

cancel in pairs. The remaining terms of (4) and (5) match the terms of (3):

311 = 411,	312 = 412,	321 = 5B3,	322 = 533,	323 = 573,	331 = 553,
332 = 593,	333 = 513,	341 = 543,	342 = 583,	343 = 563,	351 = 5A3,
352 = 523,	353 = 5C3,	361 = 4B1,	362 = 4B2.

Thus (3) follows from (4) and (5), which proves (Z5ea). $\square$

Proof of Proposition 7.2

Proposition 7.2

If $s \in S Z^4 (S,G)$, then $s = \Gamma \!f$ for some $f \in Z_6$.

Proof

Given $s \in S Z^4$, define $f \in C_6$ as follows: for every $x,y,z,w \in S$,

Case A: if $x = y = z = w$, then $f\,(x,y,z,t) = 0$.

Case B: if $y = z = w \ne x$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y, \\ 0 &{}\text {if }y < x.\end{array}\right. }$

Case C: if $x = z = w \ne y$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(y,x,z,w) &{}\text {if }y< x, \\ 0 &{}\text {if }x < y. \\ \end{array}\right. }$

Case D: if $x = y = w \ne z$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< z, \\ 0 &{}\text {if }z < x. \\ \end{array}\right. }$

Case E: if $x = y = z \ne w$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< w, \\ 0 &{}\text {if }w < x. \\ \end{array}\right. }$

Case F: if $x = y \ne z = w$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< z, \\ 0 &{}\text {if }z < x. \\ \end{array}\right. }$

Case G: if $x = z \ne y = w$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y, \\ s\,(y,x,z,w) &{}\text {if }y < x. \\ \end{array}\right. }$

Case H: if $x = w \ne y = z$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y, \\ s\,(y,x,z,w) &{}\text {if }y < x.\\ \end{array}\right. }$

Case I: if $x = y \ne z \ne w \ne y$, then

$f\,(x,y,z,w) = {{\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< z,w, \\ 0 &{}\text {if }z< x\hbox { or if }w < x.\\ \end{array}\right. }}$

Case J: if $x = z \ne y \ne w \ne z$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y< w, \\ s\,(y,x,z,w) &{}\text {if }y< x,w \\ 0 &{}\text {if }w < y.\\ \end{array}\right. }$

Case K: if $x = w \ne y \ne z \ne w$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y< z, \\ s\,(y,x,z,w) &{}\text {if }y< x,z \\ 0 &{}\text {if }z < y.\\ \end{array}\right. }$

Case M: if $y = z \ne x \ne w \ne y$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y,w, \\ s\,(y,x,z,w) &{}\text {if }y< x< w, \\ 0 &{}\text {if }w < x.\\ \end{array}\right. }$

Case N: if $y = w \ne x \ne z \ne w$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y,z, \\ s\,(y,x,z,w) &{}\text {if }y< x< z, \\ 0 &{}\text {if }z < x.\\ \end{array}\right. }$

Case P: if $z = w \ne x \ne y \ne w$, then $f\,(x,y,z,w) = {\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y,z, \\ s\,(y,x,z,w) &{}\text {if }y< x,z, \\ 0 &{}\text {if }z < x,y.\\ \end{array}\right. }$

Case Q: if x, y, z, w are all distinct, then $f\,(x,y,z,w) ={\left\{ \begin{array}{ll} s\,(x,y,z,w) &{}\text {if }x< y,z,w, \\ s\,(y,x,z,w) &{}\text {if }y< x,z,w, \\ 0 &{}\text {if }z< x,y\hbox { or if }w < x,y.\\ \end{array}\right. }$

These 15 cases are mutually exclusive. As before, $f_X (x,y,z,w)$ denotes the value of $f\,(x,y,z,w)$ in Case X.

That $f\,(y,x,z,w) = f\,(x,y,z,w)$ is clear in Cases A, D, E, F, I, P, and Q. Exchanging x and y also exchanges Cases B and C, G and H, J and M, and K and N: indeed, if $x = z = w \ne y$, then

$$\begin{aligned} f_B\,(y,x,z,w) ~=~ {\left\{ \begin{array}{ll} s\,(y,x,z,w) &{}\hbox {if }y< x \\ 0 &{}\hbox {if }x < y \\ \end{array}\right. } ~~=~ f_C\,(x,y,z,w)\,; \end{aligned}$$

if $x = w \ne y = z$, then

$$\begin{aligned} f_G\,(y,x,z,w) ~=~{\left\{ \begin{array}{ll} s\,(y,x,z,w) &{}\hbox {if }y< x \\ s\,(x,y,z,w) &{}\hbox {if }x < y \\ \end{array}\right. } ~~=~ f_H\,(x,y,z,w)\,; \end{aligned}$$

if $y = z \ne x \ne w \ne y$, then

$$\begin{aligned} f_J\,(y,x,z,w) ~=~ {\left\{ \begin{array}{ll} s\,(y,x,z,w) &{}\hbox {if }y< x< w \\ s\,(x,y,z,w) &{}\hbox {if }x< y,w \\ 0 &{}\hbox {if }w < x\\ \end{array}\right. } ~~=~ f_M\,(x,y,z,w)\,; \end{aligned}$$

if $y = w \ne x \ne z \ne w$, then

$$\begin{aligned} f_K\,(y,x,z,w) ~=~ {\left\{ \begin{array}{ll} s\,(y,x,z,w) &{}\hbox {if }y< x< z \\ s\,(x,y,z,w) &{}\hbox {if }x< y,z \\ 0 &{}\hbox {if }z < x \\ \end{array}\right. } ~~=~ f_N\,(x,y,z,w)\,. \end{aligned}$$

Thus f has property (P6).

Proving property (1):

$$\begin{aligned} f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) ~=~ s\,(a,b,c,d) \end{aligned}$$

(1)

will also establish that f has property (Z6), so that $f \in Z_6$, and that $\Gamma \!f = s$. The proof is divided into 55 different cases, based on the pattern of equalities and inequalities between a, b, c, and d. It helps that both $\Gamma \!f$ and s have property (S4b), so that (1) remains true when the sequence a, b, c, d is reversed.

Case a: $a = b = c = d$. Then $f_A\, (a,b,c,d) = f_A\, (b,c,a,d) = f_A\, (b,c,d,a) = f_A\, (c,d,b,a) = 0 = s\,(a,b,c,d)$, and (1) holds.

Case b1: $a < b = c = d$. Then $f_B\, (a,b,c,d) = s\,(a,b,c,d)$, $f_D\, (b,c,a,d) = 0$, $f_E\, (b,c,d,a) = 0 = f_E\, (c,d,b,a)$, and (1) holds.

Case b2: $a > b = c = d$. Then $f_B\, (a,b,c,d) = 0$, $f_D\, (b,c,a,d) = s\,(b,c,a,d)$, $f_E\, (b,c,d,a) = s\,(b,c,d,a)$, $f_E\, (c,d,b,a) = s\,(c,d,b,a)$, and (1) follows from (S4b) and (S4h):

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ -~ s\,(b,b,a,b) ~+~ s\,(b,b,b,a) ~-~ s\,(b,b,b,a)\\ {}&\quad ~=~ s\,(b,a,b,b) ~=~ s\,(a,b,b,b)\,. \end{aligned}$$

Case c1: $b < a = c = d$. Then $f_C\, (a,b,c,d) = s\,(b,a,c,d)$, $f_B\,(b,c,a,d) = s\,(b,c,a,d)$, $f_B\,(b,c,d,a)= s\,(b,c,d,a)$, $f_D\,(c,d,b,a) = 0$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(b,a,a,a) ~-~ s\,(b,a,a,a) ~+~ s\,(b,a,a,a) ~=~ s\,(a,b,a,a)\,, ~~{\hbox { by (S4h).}} \end{aligned}$$

Case c2: $b > a = c = d$. Then $f_C\, (a,b,c,d) = 0$, $f_B\,(b,c,a,d) = 0 = f_B\,(b,c,d,a)$, $f_D\,(c,d,b,a) = s\,(c,d,b,a) = s\,(a,a,b,a) = - s\,(a,b,a,a)$ by (S4b), and (1) holds.

Cases d1: $c < a = b = d$,  d2: $c > a = b = d$,  e1: $d < a = b = c$, and e2: $d > a = b = c$, follow from Cases c1, c2, b1, b2, applied to d, c, b, a.

Case f1: $a = b < c = d$. Then $f_F\, (a,b,c,d) = s\,(a,b,c,d)$, $f_G\,(b,c,a,d) = s\,(b,c,a,d)$, $f_H\,(b,c,d,a)= s\,(b,c,d,a)$, $f_F\,(c,d,b,a) = 0$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(a,a,c,c) ~-~ s\,(a,c,a,c) ~+~ s\,(a,c,c,a) ~=~ s\,(a,b,c,d)\,, \\ {}&\quad ~~{\hbox { by (S4a) and (S4e).}} \end{aligned}$$

Case f2: $a = b > c = d$: follows from Case f1, applied to d, c, b, a.

Case g1: $a = c < b = d$. Then $f_G\, (a,b,c,d) = s\, (a,b,c,d)$, $f_H\,(b,c,a,d) = s\,(c,b,a,d)$, $f_G\,(b,c,d,a)= s\,(c,b,d,a)$, $f_H\,(c,d,b,a) = s\,(c,d,b,a)$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\, (a,b,a,b) ~-~ s\,(a,b,a,b) ~+~ s\,(a,b,b,a) ~-~ s\,(a,b,b,a) \\&\quad =~ 0 ~=~ s\,(a,b,c,d)\,, ~~{\hbox { by (S4e).}} \end{aligned}$$

Case g2: $a = c > b = d$: follows from Case g1, applied to d, c, b, a.

Case h1: $a = d < b = c$. Then $f_H\, (a,b,c,d) = s\,(a,b,c,d)$, $f_F\,(b,c,a,d) = 0$, $f_F\,(b,c,d,a) = 0$, $f_G\,(c,d,b,a) = s\,(d,c,b,a) = s\,(a,b,b,a) = 0$ by (S4a), and (1) holds.

Case h2: $a = d > b = c$. Then $f_H\, (a,b,c,d) = s\,(b,a,c,d)$, $f_F\,(b,c,a,d) = s\,(b,c,a,d)$, $f_F\,(b,c,d,a)= s\,(b,c,d,a)$, $f_G\,(c,d,b,a) = s\,(c,d,b,a)$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(b,a,b,a) ~-~ s\,(b,b,a,a) ~+~ s\,(b,b,a,a) ~-~ s\,(b,a,b,a) \\&\quad =~ 0 ~=~ s\,(a,b,c,d)\,, ~~{\hbox { by (S4a).}} \end{aligned}$$

Cases i1: $a = b< c < d$ and i2: $a = b< d < c$. Then $f_I\, (a,b,c,d) = s\,(a,b,c,d)$, $f_J\,(b,c,a,d) = s\,(b,c,a,d)$, $f_K\,(b,c,d,a)= s\,(b,c,d,a)$, $f_P\,(c,d,b,a) = 0$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(a,a,c,d) ~-~ s\,(a,c,a,d) ~+~ s\,(a,c,d,a) ~=~ s\,(a,b,c,d)\,, ~~{\hbox { by (S4d).}} \end{aligned}$$

Case i3: $c< a = b < d$. Then $f_I\, (a,b,c,d) = 0$, $f_J\,(b,c,a,d) = s\,(c,b,a,d)$, $f_K\,(b,c,d,a) = s\,(c,b,d,a)$, $f_P\,(c,d,b,a) = s\,(c,d,b,a)$, and (1) follows from (S4d) and (S4b):

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ - s\,(c,a,a,d) + s\,(c,a,d,a) - s\,(c,d,a,a) \\ {}&\quad ~=~ - s\,(d,c,a,a) ~=~ s\,(a,b,c,d)\,. \end{aligned}$$

Case i4: $d< a = b < c$. Then $f_I\, (a,b,c,d) = 0$, $f_J\,(b,c,a,d) = 0$, $f_K\,(b,c,d,a) = 0$, $f_P\,(c,d,b,a) = s\,(d,c,b,a) = - s\,(a,b,c,d)$ by (S4b), and (1) holds.

Case i5: $c< d < a = b$. Then $f_I\, (a,b,c,d) = 0$, $f_J\,(b,c,a,d) = s\,(c,b,a,d)$, $f_K\,(b,c,d,a) = s\,(c,b,d,a)$, $f_P\,(c,d,b,a) = s\,(c,d,b,a)$, and (1) holds as in Case i3.

Case i6: $d< c < a = b$. Then $f_I\, (a,b,c,d) = 0$, $f_J\,(b,c,a,d) = 0$, $f_K\,(b,c,d,a) = 0$, $f_P\,(c,d,b,a) = s\,(d,c,b,a) = - s\,(a,b,c,d)$ by (S4b), and (1) holds.

Case j1: $a = c< b < d$. Then $f_J\, (a,b,c,d) = s\,(a,b,c,d)$, $f_M\,(b,c,a,d) = s\,(c,b,a,d)$, $f_N\,(b,c,d,a)= s\,(c,b,d,a)$, $f_P\,(c,d,b,a) = 0$ since $b < d$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(a,b,a,d) ~-~ s\,(a,b,a,d) ~+~ s\,(a,b,d,a) ~=~ s\,(a,b,a,d)\,, ~~{\hbox { by (S4f).}} \end{aligned}$$

Case j2: $a = c< d < b$. Then $f_J\, (a,b,c,d) = 0$, $f_M\,(b,c,a,d) = 0$, $f_N\,(b,c,d,a) = 0$, and

$$\begin{aligned}&f_K\,(c,d,b,a) ~=~ s\,(c,d,b,a) ~=~ s\,(a,d,b,a) ~=~ -~ s\,(a,b,d,a) \\&\quad =~ -~ s\,(a,b,a,d)\,,~~{\hbox { by (S4f).}} \end{aligned}$$

Case j3: $b< a = c < d$. Then $f_J\,(a,b,c,d) = s\,(b,a,c,d)$, $f_M\,(b,c,a,d) = s\,(b,c,a,d)$, $f_N\,(b,c,d,a)= s\,(b,c,d,a)$, $f_K\,(c,d,b,a) = 0$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(b,a,a,d) ~-~ s\,(b,a,a,d) ~+~ s\,(b,a,d,a) ~=~ s\,(a,b,a,d)\,, ~~{\hbox { by (S4g).}} \end{aligned}$$

Case j4: $d< a = c < b$. Then $f_J\,(a,b,c,d) = 0$, $f_M\,(b,c,a,d) = 0$, $f_N\,(b,c,d,a) = 0$, $f_K\,(c,d,b,a) = s\,(d,c,b,a) = - s\,(a,b,c,d)$ by (S4b), and (1) holds.

Case j5: $b< d < a = c$. Then $f_J\,(a,b,c,d) = s\,(b,a,d,c)$, $f_M\,(b,c,a,d) = s\,(b,c,a,d)$, $f_N\,(b,c,d,a)= s\,(b,c,d,a)$, $f_K\,(c,d,b,a) = 0$, and (1) holds as in case j3.

Case j6: $d< b < a = c$. Then $f_J\,(a,b,c,d) = 0$, $f_M\,(b,c,a,d) = 0$, $f_N\,(b,c,d,a) = 0$, $f_K\,(c,d,b,a) = s\,(d,c,b,a) = - s\,(a,b,c,d)$ by (S4b), and (1) holds.

Case k1: $a = d< b < c$. Then $f_K\,(a,b,c,d) = s\,(a,b,c,d)$, $f_P\,(b,c,a,d) = 0$, $f_P\,(b,c,d,a) = 0$, $f_N\,(c,d,b,a) = 0$, and (1) holds.

Case k2: $a = d< c < b$: follows from case k1, applied to d, c, b, a.

Case k3: $b< a = d < c$. Then $f_K\,(a,b,c,d) = s\,(b,a,c,d)$, $f_P\,(b,c,a,d) = s\,(b,c,a,d)$, $f_P\,(b,c,d,a)= s\,(b,c,d,a)$, $f_N\,(c,d,b,a) = 0$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(b,a,c,a) ~-~ s\,(b,c,a,a) ~+~ s\,(b,c,a,a) ~=~ s\,(a,b,c,a)\,, ~~{\hbox { by (S4g).}} \end{aligned}$$

Case k4: $c< a = d < b$: follows from case k3, applied to d, c, b, a.

Case k5: $b< c < a = d$. Then $f_K\,(a,b,c,d) = s\,(b,a,c,d)$, $f_P\,(b,c,a,d) = s\,(b,c,a,d)$, $f_P\,(b,c,d,a)= s\,(b,c,d,a)$, $f_N\,(c,d,b,a) = 0$, and (1) holds as in case k3.

Case k6: $c< b < a = d$: follows from case k5, applied to d, c, b, a.

Case m1: $b = c< a < d$. Then $f_M\,(a,b,c,d) = s\,(b,a,c,d)$, $f_I\,(b,c,a,d) = s\,(b,c,a,d)$, $f_I\,(b,c,d,a) = s\,(b,c,d,a)$, $f_J\,(c,d,b,a) = 0$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&=~ s\,(b,a,c,d) ~-~ s\,(b,c,a,d) ~+~ s\,(b,c,d,a) ~=~ s\,(a,b,c,d), ~~{\hbox { by (S4d).}} \end{aligned}$$

Case m2: $b = c< d < a$: follows from case m1, applied to d, c, b, a.

Case m3: $a< b = c < d$. Then $f_M\,(a,b,c,d) = s\,(a,b,c,d)$, $f_I\,(b,c,a,d) = 0$, $f_I\,(b,c,d,a) = 0$, $f_J\,(c,d,b,a) = 0$, and (1) holds.

Case m4: $d< b = c < a$: follows from case m3, applied to d, c, b, a.

Case m5: $a< d < b = c$. Then $f_M\,(a,b,c,d) = s\,(a,b,c,d)$, $f_I\,(b,c,a,d) = 0$, $f_I\,(b,c,d,a) = 0$, $f_J\,(c,d,b,a) = 0$, and (1) holds.

Case m6: $d< a < b = c$: follows from case m5, applied to d, c, b, a.

Cases n1: $b = d< a < c$, n2: $b = d< c < a$, n3: $a< b = d < c$, n4: $c< b = d < a$, n5: $a< c < b = d$, and n6: $c< a < b = d$ follow from cases j2, j1, j4, j3, j6, and j5, applied to d, c, b, a.

Cases p1: $c = d< a < b$, p2: $c = d< b < a$, p3: $a< c = d < b$, p4: $b< c = d < a$, p5: $a< b < c = d$, and p6: $b< a < c = d$ follow from cases i2, i1, i4, i3, i6, and i5, applied to d, c, b, a.

Case qa: $a < b,c,d$ and $b \ne c \ne d \ne b$. Then $f_Q\,(a,b,c,d) = s\,(a,b,c,d)$, whereas $f_Q\,(b,c,a,d) = 0$, $f_Q\,(b,c,d,a) = 0$, and $f_Q\,(c,d,b,a) = 0$, since $a < b,c,d$, and (1) holds.

Case qb: $b < a,c,d$ and $a \ne c \ne d \ne a$. Then $f_Q\,(a,b,c,d) = s\,(b,a,c,d)$, $f_Q\,(b,c,a,d) = s\,(b,c,a,d)$, $f_Q\,(b,c,d,a) = s\,(b,c,d,a)$, $f_Q\,(c,d,b,a) = 0$, since $b < a,c,d$, and

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ s\,(b,a,c,d) ~-~ s\,(b,c,a,d) ~+~ s\,(b,c,d,a) ~=~ s\,(a,b,c,d)\,, ~~{\hbox { by (S4d).}} \end{aligned}$$

Cases qc: $c < a,b,d$ and $a \ne b \ne d \ne a$, and qd: $d < a,b,c$ and $a \ne b \ne c \ne a$, follow from cases qb and qa, applied to d, c, b, a. $\square$

Proof of Proposition 7.3

Proposition 7.3

If $f \in Z_6$, then $f \in B_6$ if and only if $\Gamma \!f \in S B^4 (S,G)$.

Proof

Recall that $B_6 \subseteq Z_6$ is the abelian group of all $f \in C_6$ for which there exists a cochain $u \in C^3 (S,G)$ such that

for all $A \in F_1$ and $a,b,c,d \in S$. Property (B4a) implies $u\,[x,y] = u\,[xy]$ for all $x,y \in S$.

Let $s = \Gamma \!f$. If the above holds, then

$$\begin{aligned} s\,&(a,b,c,d) \\&\quad =~ f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ \gamma u\,([c];[d]) ~-~ u\,([a,b,c];[d]) ~+~ u\,([a,b];[c,d]) ~-~ \gamma u\,([a,b];[c])\\&\quad \qquad \qquad +~ \gamma u\,[a,b] ~-~ \gamma u\,[d] ~+~ u\, ([ab,c];[d]) ~-~ u\,([ab];[c,d]) \\&\qquad \quad -~ \gamma u\,([a];[d]) ~+~ u\,([a,b,c];[d]) ~-~ u\,([b,c];[a,d]) ~+~ \gamma u\,([b,c];[a])\\&\quad \qquad \qquad -~ \gamma u\,[b,c] ~+~ \gamma u\,[d] ~-~ u\, ([a,bc];[d]) ~+~ u\,([bc];[a,d]) \\&\quad \qquad +~ \gamma u\,([d];[a]) ~-~ u\,([d,b,c];[a]) ~+~ u\,([b,c];[d,a]) ~-~ \gamma u\,([b,c];[d])\\&\quad \qquad \qquad +~ \gamma u\,[b,c] ~-~ \gamma u\,[a] ~+~ u\, ([bc,d];[a]) ~-~ u\,([bc];[d,a]) \\&\quad \qquad -~ \gamma u\,([b];[a]) ~+~ u\,([b,c,d];[a]) ~-~ u\,([c,d];[b,a]) ~+~ \gamma u\,([c,d];[b])\\&\quad \qquad \qquad -~ \gamma u\,[c,d] ~+~ \gamma u\,[a] ~-~ u\, ([cd,b];[a]) ~+~ u\,([cd];[b,a]) \\&\quad =~ \gamma u\,[c] ~+~ \gamma u\,[d] ~-~ \gamma u\,([a,b];[c]) \\&\qquad \quad \quad \quad +~ \gamma u\,[ab] ~+~ u\, ([ab,c];[d]) ~-~ u\,([ab];[c,d])&\hbox {(line 2)} \\&\quad \quad \quad -~ \gamma u\,[a] ~-~ \gamma u\,[d] ~+~ \gamma u\,([b,c];[a])\\&\qquad \quad \quad \quad -~ \gamma u\,[bc] ~-~ u\, ([a,bc];[d]) ~+~ u\,([bc];[a,d]) \\&\quad \quad \quad +~ \gamma u\,[d] ~+~ \gamma u\,[a] ~-~ \gamma u\,([b,c];[d]) \\&\qquad \quad \quad \quad +~ \gamma u\,[bc] ~+~ u\, ([bc,d];[a]) ~-~ u\,([bc];[d,a])&\hbox {(line 6)} \\&\quad \quad \quad -~ \gamma u\,[b] ~-~ \gamma u\,[a] ~+~ \gamma u\,([c,d];[b]) \\&\qquad \quad \quad \quad -~ \gamma u\,[cd] ~-~ u\, ([cd,b];[a]) ~+~ u\,([cd];[b,a])\,, \end{aligned}$$

(1)

by (B4b) and (B4a). Let

$$\begin{aligned} v\,(x,y,z) ~=~ u\,([x];[y,z]) ~-~ u\,([x,y];[z]) ~+~ \gamma u\,[z] ~-~ \gamma u\,[x]\,. \end{aligned}$$

Then $v \in SC^3 (S,G)$: if $x = z$, then $v\,(x,y,z) = 0$; also, $v\,(z,y,x) = -~v\,(x,y,z)$; and

$$\begin{aligned}&v\,(x,y,z) ~+~ v\,(y,z,x) ~+~ v\,(z,x,y) \\&\quad =~ u\,([x];[y,z]) ~-~ u\,([x,y];[z]) ~+~ \gamma u\,[z] ~-~ \gamma u\,[x] \\&\qquad +~ u\,([y];[z,x]) ~-~ u\,([y,z];[x]) ~+~ \gamma u\,[x] ~-~ \gamma u\,[y] \\&\qquad +~ u\,([z];[x,y]) ~-~ u\,([z,x];[y]) ~+~ \gamma u\,[y] ~-~ \gamma u\,[z] ~=~ 0 \,; \end{aligned}$$

thus (S3a), (S3b), and (S3c) hold. Moreover,

$$\begin{aligned}&\gamma v\,(b,c,d) ~-~ v\,(ab,c,d) ~+~ v\,(a,bc,d) ~-~ v\,(a,b,cd) ~+~ \gamma v\,(a,b,c) \\&\quad =~ \gamma u\,([b];[c,d]) ~-~ \gamma u\,([b,c];[d]) ~+~ \gamma u\,[d] ~-~ \gamma u\,[b] \\&\qquad -~ u\,([ab];[c,d]) ~+~ u\,([ab,c];[d]) ~-~ \gamma u\,[d] ~+~ \gamma u\,[ab] \\&\qquad +~ u\,([a];[bc,d]) ~-~ u\,([a,bc];[d]) ~+~ \gamma u\,[d] ~-~ \gamma u\,[a] \\&\qquad -~ u\,([a];[b,cd]) ~+~ u\,([a,b];[cd]) ~-~ \gamma u\,[cd] ~+~ \gamma u\,[a] \\&\qquad +~ \gamma u\,([a];[b,c]) ~-~ \gamma u\,([a,b];[c]) ~+~ \gamma u\,[c] ~-~ \gamma u\,[a]\,. \end{aligned}$$

(2)

In the right hand side of (1), terms 141 and 161,   143 and 163 cancel each other. The remaining terms match the remaining terms of the right hand side of (2):

111 = 253,	112 = 213,	113 = 251,	121 = 224,	122 = 222,
123 = 221,	131 = 234,	132 = 223,	133 = 252,	142 = 232,
151 = 233,	152 = 244,	153 = 212,	162 = 231,	171 = 214,
172 = 254,	173 = 211,	181 = 243,	182 = 241,	183 = 242.

Hence $s = \delta v\in S B^4 (S,G)$ and $\Gamma \!f = s \in S B^4 (S,G)$.

Conversely, assume that $f \in Z_6$ and $\Gamma \!f = \delta v$ for some $v \in S C^3 (S,G)$. Then

$$\begin{aligned}&f\,(a,b,c,d) ~-~ f\,(b,c,a,d) ~+~ f\,(b,c,d,a) ~-~ f\,(c,d,b,a) \\&\quad =~ \gamma v\,(b,c,d) ~-~ v\,(ab,c,d) ~+~ v\,(a,bc,d) ~-~ v\,(a,b,cd) ~+~ \gamma v\,(a,b,c)\,, \end{aligned}$$

(3)

for all $a,b,c,d \in S$. The proof constructs $u \in C^3$ such that $f = \delta u$. This is done in several stages.

Lemma L.1

For all $A \in F_1$ and $a,b,c \in S$, let

$$\begin{aligned} u\,(A) ~&=~ 0 \text{ for } \text{ all } A \in F_1\,, \\ u\,([a];[b]) ~&=~ 0 \text{ for } \text{ all } a,b \in S\,, \\ u\,([a];[b,c]) ~=~ u\,([b,c];[a]) ~&= {\left\{ \begin{array}{ll} v\,(a,b,c) &{}\hbox {if }c \leqq a,b, \\ v\,(a,c,b) &{}\hbox {if }b \leqq a,c, \\ 0 &{}\hbox {if }a \leqq b,c. \\ \end{array}\right. } \end{aligned}$$

Then $u\,([a];[b,c])$ is well-defined and (B4a), (B4b) hold. Moreover

$$\begin{aligned} u\,([a];[b,c]) ~-~ u\,([a,b];[c]) ~=~ v\,(a,b,c)\,. \end{aligned}$$

Proof

The three cases in the definition of $u\,(a;b,c)$ overlap but are consistent with each other: if $b = c \leqq a$, then $v\,(a,b,c) = v\,(a,c,b)$; if $b = a \leqq c$, then $v\,(a,c,b) = 0$ by (S3a); if $a = c \leqq b$, then $v\,(a,b,c) = 0$. Moreover, if $a \leqq b,c$, then

$$\begin{aligned} u\,(a;b,c) ~-~ u\,(a,b;c) ~=~ -~ u\,(c;a,b) ~=~ -~ v\,(c,b,a) ~=~ v\,(a,b,c)\,, \end{aligned}$$

by (S3b); if $b \leqq a,c$, then

$$\begin{aligned} u\,(a;b,c) ~-~ u\,(a,b;c) ~&=~ v\,(a,c,b) ~-~ v\,(c,a,b) \\&=~ -~ v\,(b,c,a) ~+~ v\,(b,a,c) ~=~ v\,(a,b,c)\,, \end{aligned}$$

by (S3b) and (S3d); if $c \leqq a,b$, then

$$\begin{aligned} u\,(a;b,c) ~-~ u\,(a,b;c) ~=~ v\,(a,b,c) ~-~ 0. \end{aligned}$$

$\square$

The hypothesis is;

$$\begin{aligned}&f\,(x,y,z,t) ~-~ f\,(y,z,x,t) ~+~ f\,(y,z,t,x) ~-~ f\,(z,t,y,x) \\&\quad -~ \gamma v\,(x,y,z) ~-~ \gamma v\,(y,z,t) \\&\quad +~ v\,(xy,z,t) ~-~ v\,(x,yz,t) ~+~ v\,(x,y,zt) ~=~ 0\,; \end{aligned}$$

equivalently, since v has properties (S3b) and (S3d) and $v\,(a,b,c) = u\,(a;b,c) - u\,(a,b;c)$:

$$\begin{aligned}&f\,(x,y,z,t) ~-~ f\,(y,z,x,t) ~+~ f\,(y,z,t,x) ~-~ f\,(z,t,y,x) \\&\quad -~ \gamma u\,(y,z;x) ~+~ \gamma u\,(x,y;z) ~+~ v\,(xy,z,t) ~-~ v\,(yz,x,t) \\&\quad +~ v\,(yz,t,x) ~-~ v\,(zt,y,x) ~+~ \gamma u\,(y,z;t) ~-~ \gamma u\,(z,t;y) ~=~ 0\,. \end{aligned}$$

(4)

This is used to establish other equalities:

Lemma L.2

Given $a,b,c,d \in S$, let

$$\begin{aligned} K(a,b) ~&=~ f\,(c,d,a,b) ~+~ v\,(cd,a,b) ~+~ \gamma u\,(c,d;a)\,, \\ K(b,c) ~&=~ f\,(d,a,b,c) ~+~ v\,(da,b,c) ~+~ \gamma u\,(d,a;b)\,, \\ K(c,d) ~&=~ f\,(a,b,c,d) ~+~ v\,(ab,c,d) ~+~ \gamma u\,(a,b;c)\,, \\ K(d,a) ~&=~ f\,(b,c,d,a) ~+~ v\,(bc,d,a) ~+~ \gamma u\,(b,c;d)\,, \\ K(a,c) ~&=~ f\,(b,d,a,c) ~+~ v\,(bd,a,c) ~+~ \gamma u\,(b,d;a)\,, \\ K(b,d) ~&=~ f\,(c,a,b,d) ~+~ v\,(ca,b,d) ~+~ \gamma u\,(c,a;b)\,, \\ K(c,a) ~&=~ f\,(d,b,c,a) ~+~ v\,(db,c,a) ~+~ \gamma u\,(d,b;c)\,, \\ K(d,b) ~&=~ f\,(a,c,d,b) ~+~ v\,(ac,d,b) ~+~ \gamma u\,(a,c;d)\,, \\ K(a,d) ~&=~ f\,(b,c,a,d) ~+~ v\,(bc,a,d) ~+~ \gamma u\,(b,c;a)\,, \\ K(b,a) ~&=~ f\,(c,d,b,a) ~+~ v\,(cd,b,a) ~+~ \gamma u\,(c,d;b)\,, \\ K(c,b) ~&=~ f\,(d,a,c,b) ~+~ v\,(da,c,b) ~+~ \gamma u\,(d,a;c)\,, ~~\hbox {and}~~\\ K(d,c) ~&=~ f\,(a,b,d,c) ~+~ v\,(ab,d,c) ~+~ \gamma u\,(a,b;d)\,. \end{aligned}$$

Then

$$\begin{aligned} K(c,a) - K(b,a) ~&=~ K(d,b) - K(a,b) \\&=~ K(a,c) - K(d,c) ~=~ K(b,d) - K(c,d)~~\hbox {and}~~\\ K(d,a) - K(b,a) ~&=~ K(c,b) - K(a,b) \\&=~ K(b,c) - K(d,c) ~=~ K(a,d) - K(c,d)\,. \end{aligned}$$

Proof

First note that circular permutations of a, b, c, d change K(a, b) to K(b, c), K(c, d), K(d, a), and similarly for K(a, c) and for K(a, d).

By definition,

$$\begin{aligned}&\left( K(c,a) - K(b,a)\right) - \left( K(d,b) - K(a,b)\right) \\&\quad =~ f\,(d,b,c,a) ~+~ v\,(db,c,a) ~+~ \gamma u\,(d,b;c) \\&\qquad -~ f\,(c,d,b,a) ~-~ v\,(cd,b,a) ~-~ \gamma u\,(c,d;b) \\&\qquad -~ f\,(a,c,d,b) ~-~ v\,(ac,d,b) ~-~ \gamma u\,(a,c;d) \\&\qquad +~ f\,(c,d,a,b) ~+~ v\,(cd,a,b) ~+~ \gamma u\,(c,d;a)\,, \end{aligned}$$

whereas (4), applied to $x,y,z,t = b,d,c,a$, yields

$$\begin{aligned}&f\,(b,d,c,a) ~-~ f\,(d,c,b,a) ~+~ f\,(d,c,a,b) ~-~ f\,(c,a,d,b) \\&\quad -~ \gamma u\,(d,c;b) ~+~ \gamma u\,(b,d;c) ~+~ v\,(bd,c,a) ~-~ v\,(dc,b,a) \\&\quad +~ v\,(dc,a,b) ~-~ v\,(ca,d,b) ~+~ \gamma u\,(d,c;a) ~-~ \gamma u\,(c,a;d) ~=~ 0\,. \end{aligned}$$

Therefore $K(c,a) - K(b,a) = K(d,b) - K(a,b)$. Permuting a, b, c, d circularly twice then yield $K(a,c) - K(d,c) ~=~ K(b,d) - K(c,d)$. Similarly,

$$\begin{aligned}&\left( K(b,d) - K(c,d)\right) - \left( K(d,b) - K(a,b)\right) \\&\quad =~ f\,(c,a,b,d) ~+~ v\,(ca,b,d) ~+~ \gamma u\,(c,a;b) \\&\qquad -~ f\,(a,b,c,d) ~-~ v\,(ab,c,d) ~-~ \gamma u\,(a,b;c) \\&\qquad -~ f\,(a,c,d,b) ~-~ v\,(ac,d,b) ~-~ \gamma u\,(a,c;d) \\&\qquad +~ f\,(c,d,a,b) ~+~ v\,(cd,a,b) ~+~ \gamma u\,(c,d;a)\,, \end{aligned}$$

whereas (4), applied to $x,y,z,t = d,c,a,b$, yields

$$\begin{aligned}&f\,(d,c,a,b) ~-~ f\,(c,a,d,b) ~+~ f\,(c,a,b,d) ~-~ f\,(a,b,c,d) \\&\quad -~ \gamma u\,(c,a;d) ~+~ \gamma u\,(d,c;a) ~+~ v\,(dc,a,b) ~-~ v\,(ca,d,b) \\&\quad +~ v\,(ca,b,d) ~-~ v\,(ab,c,d) ~+~ \gamma u\,(c,a;b) ~-~ \gamma u\,(a,b;c) ~=~ 0\,. \end{aligned}$$

Therefore $K(b,d) - K(c,d) ~=~ K(d,b) - K(a,b)$. This proves the first three equalities in the statement.

Next,

$$\begin{aligned}&\left( K(d,a) - K(b,a)\right) - \left( K(c,b) - K(a,b)\right) \\&\quad =~ f\,(b,c,d,a) ~+~ v\,(bc,d,a) ~+~ \gamma u\,(b,c;d) \\&\qquad -~ f\,(c,d,b,a) ~-~ v\,(cd,b,a) ~-~ \gamma u\,(c,d;b) \\&\qquad -~ f\,(d,a,c,b) ~-~ v\,(da,c,b) ~-~ \gamma u\,(d,a;c) \\&\qquad +~ f\,(c,d,a,b) ~+~ v\,(cd,a,b) ~+~ \gamma u\,(c,d;a)\,, \end{aligned}$$

whereas (4), applied to $x,y,z,t = b,c,d,a$, yields

$$\begin{aligned}&f\,(b,c,d,a) ~-~ f\,(c,d,b,a) ~+~ f\,(c,d,a,b) ~-~ f\,(d,a,c,b) \\&\quad -~ \gamma u\,(c,d;b) ~+~ \gamma u\,(b,c;d) ~+~ v\,(bc,d,a) ~-~ v\,(cd,b,a) \\&\quad +~ v\,(cd,a,b) ~-~ v\,(da,c,b) ~+~ \gamma u\,(c,d;a) ~-~ \gamma u\,(d,a;c) ~=~ 0\,. \end{aligned}$$

Therefore $K(d,a) - K(b,a) ~=~ K(c,b) - K(a,b)$. Permuting a, b, c, d circularly then yields $K(a,b) - K(c,b) ~=~ K(d,c) - K(b,c)$ and $K(b,c) - K(d,c) ~=~ K(a,d) - K(c,d)$. This proves the last three equalities in the statement. $\square$

Lemma L.3

For all $a,b,c,d \in S$, $u\,([a,b];[c,d]))$ is well-defined by:

$$\begin{aligned} \mathrm{(1)}&~~\hbox {if}~~a,b \leqq c,d ~~{\hbox {or if}}~~c,d \leqq a,b, ~~\hbox {then}~~\\&u\,([a,b];[c,d]) ~=~ 0, \\ \mathrm{(2)}&~~\hbox {if}~~a,c \leqq b,d ~~{\hbox {or if}}~~b,d \leqq a,c, ~~\hbox {then}~~\\&u\,([a,b];[c,d]) ~=~ K(b,a) - K(c,a) ~=~ K(a,b) - K(d,b) \\&~~~=~ K(d,c) - K(a,c) ~=~ K(c,d) - K(b,d)\,, \\ \mathrm{(3)}&~~\hbox {if}~~a,d \leqq b,c ~~{\hbox {or if}}~~b,c \leqq a,d, ~~\hbox {then}~~\\&u\,([a,b];[c,d]) ~=~ K(b,a) - K(d,a) ~=~ K(a,b) - K(c,b) \\&~~~=~ K(d,c) - K(b,c) ~=~ K(c,d) - K(a,d)\,, \end{aligned}$$

where K is as defined in Lemma L.2.

Proof

Define $p\,(a,b,c,d)$ as in the statement:

$$\begin{aligned} \mathrm{(1)}&~~\hbox {if}~~a,b \leqq c,d ~~{\hbox {or if}}~~c,d \leqq a,b, ~~\hbox {then}~~\\&p\,(a,b,c,d) ~=~ 0, \\ \text{(2) }&~~\hbox {if}~~a,c \leqq b,d ~~{\hbox {or if}}~~b,d \leqq a,c, ~~\hbox {then}~~\\&p\,(a,b,c,d) ~=~ K(b,a) - K(c,a) ~=~ K(a,b) - K(d,b) \\&~~~=~ K(d,c) - K(a,c) ~=~ Kcb,d) - K(b,d)\,, \\ \mathrm{(3)}&~~\hbox {if}~~a,d \leqq b,c ~~{\hbox {or if}}~~b,c \leqq a,d, ~~\hbox {then}~~\\&p\,(a,b,c,d) ~=~ K(b,a) - K(d,a) ~=~ K(a,b) - K(c,b) \\&~~~=~ K(d,c) - K(b,c) ~=~ K(c,d) - K(a,d)\,, \end{aligned}$$

where the equalities follow from Lemma L.2. Let $p_k\,(a,b,c,d)$ denote the value of $p\,(a,b,c,d)$ in case (k). Note that

$p_2\,(a,b,c,d) = 0$ if $b = c$ or if $a = d$;

$p_3\,(a,b,c,d) = 0$ if $a = c$ or if $b = d$;

$p_2\,(a,b,c,d) = p_3\,(a,b,c,d)$ if $a = b$ or if $c = d$.

Cases (1), (2), and (3) in the definition of p overlap as follows.

(1) and (2):

• If $a,b \leqq c,d$ and $a,c \leqq b,d$, then $b {=} c$ and $p_2\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

• If $a,b \leqq c,d$ and $b,d \leqq a,c$, then $a {=} d$ and $p_2\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

• If $c,d \leqq a,b$ and $a,c \leqq b,d$, then $a {=} d$ and $p_2\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

• If $c,d \leqq a,b$ and $b,d \leqq a,c$, then $b {=} c$ and $p_2\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

(1) and (3):

• If $a,b \leqq c,d$ and $a,d \leqq b,c$, then $b {=} d$ and $p_3\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

• If $a,b \leqq c,d$ and $b,c \leqq a,d$, then $a {=} c$ and $p_3\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

• If $c,d \leqq a,b$ and $a,d \leqq b,c$, then $a {=} c$ and $p_3\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

• If $c,d \leqq a,b$ and $b,c \leqq a,d$, then $a {=} c$ and $p_3\,(a,b,c,d) ~=~ 0 ~=~ p_1\,(a,b,c,d)$.

(2) and (3):

• If $a,c \leqq b,d$ and $a,d \leqq b,c$, then $c = d$ and $p_2\,(a,b,c,d) ~=~ p_3\,(a,b,c,d)$.

• If $a,c \leqq b,d$ and $b,c \leqq a,d$, then $a = b$ and $p_2\,(a,b,c,d) ~=~ p_3\,(a,b,c,d)$.

• If $b,d \leqq a,c$ and $a,d \leqq b,c$, then $a = b$ and $p_2\,(a,b,c,d) ~=~ p_3\,(a,b,c,d)$.

• If $b,d \leqq a,c$ and $b,c \leqq a,d$, then $c = d$ and $p_2\,(a,b,c,d) ~=~ p_3\,(a,b,c,d)$.

Thus $p\,(a,b,c,d)$ is well-defined.

The proof that $p\,(b,a,c,d) = p\,(a,b,c,d)$ is divided into several cases.

If $p\,(a,b,c,d)$ is in case (1) (if $a,b \leqq c,d$ or if $c,d \leqq a,b$), then so is $p\,(b,a,c,d)$ and $p_1\,(b,a,c,d) = 0 = p_1\,(a,b,c,d)$.

If $p\,(a,b,c,d)$ is in case (2) (if $a,c \leqq b,d$ or if $b,d \leqq a,c$), then $p\,(b,a,c,d)$ is in case (3) and $p_3\,(b,a,c,d) ~=~ K(a,b) - K(d,b) ~=~ p_2\,(a,b,c,d)$.

If $p\,(a,b,c,d)$ is in case (3) (if $a,d \leqq b,c$ or if $b,c \leqq a,d$), then $p\,(b,a,c,d)$ is in case (2) and $p_2\,(b,a,c,d) ~=~ K(a,b) - K(c,b) ~=~ p_3\,(a,b,c,d)$.

In each case, $p\,(b,a,c,d) = p\,(a,b,c,d)$.

The proof that $p\,(c,d,a,b) = p\,(a,b,c,d)$ is similar. If $p\,(a,b,c,d)$ is in case (1) (if $a,b \leqq c,d$ or if $c,d \leqq a,b$), then so is $p\,(c,d,a,b)$ and $p_1\,(c,d,a,b) = 0 = p\,(a,b,c,d)$.

If $p\,(a,b,c,d)$ is in case (2) (if $a,c \leqq b,d$ or if $b,d \leqq a,c$), then so is $p\,(c,d,a,b)$ and $p_2\,(c,d,a,b) ~=~ K(a,c) - K(d,c) ~=~ p_2\,(a,b,c,d)$.

If $p\,(a,b,c,d)$ is in case (3) (if $a,d \leqq b,c$ or if $b,c \leqq a,d$), then so is $p\,(c,d,a,b)$ and $p_3\,(c,d,a,b) ~=~ K(b,c) - K(d,c) ~=~ p_3\,(a,b,c,d)$.

In each case, $p\,(c,d,a,b) = p\,(a,b,c,d)$.

Finally,

$$\begin{aligned} p\,(a,b,d,c) ~=~ p\,(d,c,a,b) ~=~ p\,(c,d,a,b) ~=~ p\,(a,b,c,d)\,. \end{aligned}$$

By the above, $p\,(a,b,c,d)$ depends only upon [a, b], [c, d], and $\left[ [a,b] \,;\, [c,d]\right]$, and $u\,([a,b] \,;\, [c,d])$ is well-defined by $u\,([a,b] \,;\, [c,d]) = p\,(a,b,c,d)$, as in the statement. $\square$

Lemma L.4

For all $a,b,c,d \in F_1$, $u\,(a,b,c \,;\, d)$ is well-defined by

$$\begin{aligned} u\,(a,b,c \,;\, d) ~=~ u\,(a,b \,;\, c,d) ~-~ K(c,d) ~=~ u\,(a,c \,;\, b,d) ~-~ K(b,d)\,. \end{aligned}$$

Proof

In case (2), $u\,(a,b \,;\, c,d) ~=~ K(c,d) - K(b,d)$. If now $u\,(a,b \,;\, c,d)$ is in case (1) (if $a,b \leqq c,d$ or if $c,d \leqq a,b$), then $u\,(a,c \,;\, b,d)$ and $u\,(a,d \,;\, b,c)$ are in case (2),

$$\begin{aligned}&u\,(a,b \,;\, c,d) ~-~ K(c,d) ~=~ 0 ~-~ K(c,d)\,, \\&u\,(a,c \,;\, b,d) ~=~ K(b,d) ~-~ K(c,d)\,, ~~\hbox {and}~~\\&u\,(a,b \,;\, c,d) ~-~ K(c,d) ~=~ u\,(a,c \,;\, b,d) ~-~ K(b,d)\,. \end{aligned}$$

If $u\,(a,b \,;\, c,d)$ is in case (2) (if $a,c \leqq b,d$ or if $b,d \leqq a,c$), then $u\,(a,c \,;\, b,d)$ is in case (1),

$$\begin{aligned}&u\,(a,b \,;\, c,d) ~=~ K(c,d) ~-~ K(b,d)\,, \\&u\,(a,c \,;\, b,d) ~-~ K(b,d) ~=~ 0 ~-~ K(b,d)\,, ~~\hbox {and}~~\\&u\,(a,b \,;\, c,d) ~-~ K(c,d) ~=~ u\,(a,c \,;\, b,d) ~-~ K(b,d)\,. \end{aligned}$$

If $u\,(a,b \,;\, c,d)$ is in case (3) (if $a,d \leqq b,c$ or if $b,c \leqq a,d$), then so is $u\,(a,c \,;\, b,d)$ and

$$\begin{aligned}&u\,(a,b \,;\, c,d) ~=~ K(c,d) ~-~ K(a,d)\,, \\&u\,(a,c \,;\, b,d) ~=~ K(b,d) ~-~ K(a,d)\,, ~~\hbox {and}~~\\&u\,(a,b \,;\, c,d) ~-~ K(c,d) ~=~ u\,(a,c \,;\, b,d) ~-~ K(b,d)\,. \end{aligned}$$

The last equality thus holds in all cases.

By the above, $q\,(a,b,c,d)$ is well-defined for all $a,b,c,d \in F_1$ by:

$$\begin{aligned} q\,(a,b,c,d) ~=~ u\,(a,b \,;\, c,d) ~-~ K(c,d) ~=~ u\,(a,c \,;\, b,d) ~-~ K(b,d)\,. \end{aligned}$$

Consequently, $q\,(b,a,c,d) = q\,(a,b,c,d)$ and $q\,(c,b,a,d) = q\,(a,b,c,d)$. This in turn implies $q\,(a,c,b,d) = q\,(b,c,a,d) = q\,(c,b,a,d) = q\,(a,b,c,d)$. Therefore $q\,(a,b,c,d)$ depends only upon [a, b, c] and d, and $u\,[a,b,c \,;\, d]$ is well-defined by: $u\,[a,b,c \,;\, d] = q\,(a,b,c,d)$. $\square$

The definition of u, and the proof of Proposition 7.3, can now be completed as follows. When $P \in F_2$, $u\,(P)$ is already defined if $|P| = 1$, or if $P = [A;B]$ and $|AB| \leqq 4$. Let $u\,(P) = 0$ if $P = [A;B]$ and $|AB| \geqq 5$, or if $|P| \geqq 3$. Then $u \in C^3 (S,G)$ and u has properties (B4a) and (B4b). Furthermore, (B6) holds:

$$\begin{aligned} f\,(a,b,c,d) ~&=~ \gamma u\,(c;d) ~-~ u\,(a,b,c \,;\, d) ~+~ u\,(a,b \,;\, c,d) ~-~ \gamma u\,(a,b \,;\, c)\\&~~~~~+~ \gamma u\,[a,b] ~-~ \gamma u\,[d] ~+~ u\, (ab,c \,;\, d) ~-~ u\,(ab \,;\, c,d)\,, \end{aligned}$$

since $\gamma u\,(c;d) = \gamma u\,[a,b] = \gamma u\,[d] = 0$ and $u\,(a,b,c \,;\, d)$ is defined as

$$\begin{aligned} u\,(a,b,c \,;\, d) ~&=~ u\,(a,b \,;\, c,d) ~-~ K(c,d) \\&=~ u\,(a,b \,;\, c,d) ~-~ f\,(a,b,c,d) ~-~ v\,(ab,c,d) ~-~ \gamma u\,(a,b \,;\, c)\,. \end{aligned}$$

Thus $f \in B_6$. $\square$