—¦

G (XG2 , E) in AT

(XG2 , Y ) — (XG2 , E)

y y

±

±—±

G (XT, E)

(XT, Y ) — (XT, E) in A

—¦

The following elements match up under the vertical isomorphisms:

G

(a, y 0 s) 1y s

±

±—±

G sT · ψ

(a, ξs)

ξs and sT · ψ correspond to the compositions of s with 0 and 1 under adjointness (cf.

description of the nonhomogeneous coboundary operator in §2). Thus, nonhomogeneously,

G Y in A is (E, ψ)˜, then a is determined by the relation

if a: XT

G E.

a —¦ ξs = sT · ψ

(5) in the set of maps XT

Of course, (5) could be used to de¬ne a directly, since the two maps

ξ llllS X ‚‚‚‚ s

‚‚‚

l

llll ‚‚‚

A

llS E

XT ‚‚‚

lll

‚‚‚ l

‚ lll ψ

sT ‚A

ET

28 JONATHAN MOCK BECK

G X. The argument we sketched shows that

are equal after composition with p: E

this de¬nition of a is consistent with the earlier, homogeneous, de¬nition, in the not

necessarily tripleable case. Since a is a 1-cocycle, we have the coboundary zero relation

GY.

(aT · θ) —¦(ξT · a) = Xµ · a in the group of maps XT 2

G Z 1 (X, Y ) on maps is as follows (with Z 1 treated

The e¬ect of ˜: PO(X, Y )

G (E , ψ ), then the corresponding coboundary b =

nonhomogeneously). If f : (E, ψ)

G a is given by the formula

(f )˜: a

G E in A,

(6) b —¦ s = sf as maps X

where s and s are the sections of E and E .

Now we shall de¬ne the inverse functor

Z 1 (X, Y )

PO(X, Y ) o (relative to U T ).

˜’1

G Y is any 1-cocycle, let a˜’1 be the principal T-algebra given by X — Y

If a: XT

GX

as an object in A, with left Y -operation (in A) on the second factor, πX : X — Y

G X — Y as section, and the composition

as projection, (X, 1): X

ξ—θa

G XT — Y T GX —Y

π

(X — Y )T

as T-structure. Here π is induced by the projections, π = (πX T, πY T ), and ξ — θa is an

abbreviation which we shall use consistently for the map (t0 , t1 )(ξ—θa) = (t0 ξ, t1 θ —¦ t0 a) for

G XT , t1 : A G Y T in A. Diagrammatically, ξ —θa is the composition

any maps t0 : A

(ξ,a)—θ X—(π2 ,π1 )

GX —Y —Y GX —Y —Y GX —Y

X—mult.

XT — Y T

G (X, ξ) is indeed

One can verify easily that the projection πX : (X — Y, π(ξ — θa))

a T-algebra map, although the section (X, 1) is only a map in A (unless a = 1, or a

coboundary, of course, as follows from the theorem; if a = 1, then π(ξ — θa) is the T-

structure on the product algebra (X, ξ) — (Y, θ) ). The rest of the veri¬cation that a˜’1

is a (Y, θ)-principal T-algebra will be left till the end.

G a is a map of 1-cocycles, we set b˜’1 = X — Y —¦ b

If b: a

—¦b

a˜’1 = (X — Y, π(ξ — θa)) G (X — Y, π(ξ — θa )) = a ˜’1

X—Y

G X,

where X — Y —¦ b is given by the formula (t0 , t1 )(X — Y —¦ b) = (t0 , t1 —¦ t0 b) for t0 : A

G Y in A. We will sketch later a proof that this is a map of principal algebras.

t1 : A

We can now show that

G ˜˜’1 , ˜’1 ˜ = Z 1 (X, Y ) .

PO(X, Y )

p

G (X, ξ) be a (Y, θ)-principal algebra with section s: X G E. Then we

Let (E, ψ)

G X — Y in A,

know that E

(p,σ)

GX —Y GE

„

E

29

TRIPLES, ALGEBRAS AND COHOMOLOGY

˜˜’1 , all we have to do is show

where σ —¦ ps = E, (x, y)„ = y —¦ xs. To prove that PO

that the A-isomorphism

(p,σ)

G (X — Y, π(ξ — θa))

(E, ψ)

is a T-algebra map. Consider the diagram

(p,σ)T

G (X — Y )T

ET

ET π

(pT,σT )

G XT — Y T

ET

ψ ξ—θa

GX —Y

E (p,σ)

Since ψp = pT · ξ ( p is a T-algebra map) and

(pT, σT )(ξ — θa) = (pT · ξ, (σT · θ) —¦(pT · a))

G Y . We let them both

we have to prove that ψσ = (σT · θ) —¦(pT · a) as maps ET

GE :

operate on ψps: ET

(σT · θ) —¦(pT · a) —¦ ψps = (σT · θ) —¦(pT · a) —¦(pT · ξs)

= (σT · θ) —¦ pT · (a —¦ ξs)

= (σT · θ) —¦(pT · sT · ψ)

= (σT · θ) —¦((ps)T · ψ)

= (σ —¦ ps)T · ψ

=ψ ,

using Lemma 4, and

ψσ —¦ ψps = ψ(σ —¦ ps)

=ψ

Veri¬cation of f ˜˜’1 = f : E G E will be left out; it also uses Lemma 4.

To prove ˜’1 ˜ = Z 1 (X, Y ), let a be a 1-cocycle. Then a˜’1 ˜ is determined by the

equation

(X, 1)T · π(ξ — θa) = a˜’1 ˜ —¦ ξ(X, 1) ,

(X, 1) being the section in a˜’1 . Computing, we have

(X, 1)T · π(ξ — θa) = (XT, 1T )(ξ — θa)

= (ξ, (1T · θ) —¦ a)

= (ξ, a)

30 JONATHAN MOCK BECK

G Y , the operation ( )T · θ preserving the group

because 1T · θ is the neutral map X

law (adaptation of Lemma 2 to the non-abelian case), whereas

ξ(X, 1) = (ξ, 1XT ),

by naturality of neutral maps. Manifestly, a˜’1 ˜ = a.

It remains, ¬nally, to prove that the values of ˜’1 are principal T-algebras and maps

thereof. We begin by showing that the proposed T-structure π(ξ — θa) on a˜’1 is asso-

ciative. This is done by computing both sides of the diagram

µ

G (X — Y )T

(X — Y )T T

π

πT

(XT — Y T )T XT — Y T

(ξ—θa)T ξ—θa

GX —Y

(X — Y )T π(ξ—θa)

We present the result for the left side in the table which follows. Each entry on the

right is the accumulated composition of the maps down to that stage.

(X — Y )T T

πT

(XT — Y T )T (πX T, πY T )T

(ξ—θa)T

(X — Y )T (πX T · ξ, (πY T · θ) —¦(πX T · a))T

π

XT — Y T (πX T T · ξT, ((πY T · θ) —¦(πX T · a))T )

ξ—θa

X —Y (πX T T · ξT · ξ, ((πY T · θ) —¦(πX T · a))T · θ —¦(πX T T · ξT · a))

31

TRIPLES, ALGEBRAS AND COHOMOLOGY

which is (πX T T ·ξT ·ξ, (πY T T ·θT ·θ) —¦(πX T T ·aT ·θ) —¦(πX T T ·ξT ·a)) by the multiplicative

property of ( )T · θ. The right side of the diagram is, similarly3 ,

(X — Y )T T

µ

(X — Y )T

π

XT — Y T (πX T T · Xµ, πY T T · Y µ)

ξ—θa

X —Y (πX T T · Xµ · ξ, (πY T T · Y µ · θ) —¦(πX T T · Xµ · a))

Since ξ and θ are associative, and a satis¬es the 1-cocycle identity, both sides of the

diagram give the same map. Thus the structure π(ξ — θa) is associative.

To prove the structure is unitary, consider the diagram

(X—Y )·

G (X — Y )T

X — Yu

uu

uu

uu

uu

uu

X·—Y · uuu

π

uu

u7

XT — Y T

ξ—θa

X —Y

The composition ·π(ξ — θa) = (X· · ξ, (Y · · θ) —¦(X· · a)). Since ξ and θ are unitary,

G Y is the trivial map 1. (In a sense, the cocycle

it su¬ces to prove that X· · a: X

a is already “normalized”.) This follows from the following peculiar computation. Let

G XT T denote either of X· · XT ·, X· · X·T . Then

X··: X

X·· · Xµ = X· ,

X·· · aT · θ = X· · XT · · aT · θ

= X· · a · Y · · θ

= X· · a ,

X·· · ξT = X· · X·T · ξT

= X·.