Download A non-weakly amenable augementation ideal

A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
B. E. JOHNSON AND M. C. WHITE
Abstract. We show that the augmentation ideal I of L1 (SL(2, R)) is not
weakly amenable and that its unitization I # is weakly amenable.
1. Introduction and Notation
Let G be a locally compact group and X a locally compact topological space.
We will assume that X is a topological G space, that is, there is a continuous map
(g, x, h) ∈ G × X × G 7→ gxh ∈ X
with exe = x (x ∈ X) and
g(hx) = (gh)x
g(xh) = (gx)h
x(gh) = (xg)h
(g, h ∈ G, x ∈ X)
where gx and xg denote gxe and exg. The space C0 (X) is the Banach G bimodule
with gf h(x) = f (hxg), (g, h ∈ G, x ∈ X, f ∈ C0 (X)) and its dual Mb (X), the
space of bounded measures on X, is a dual G bimodule if we put (gµh, f ) = (µ, hf g),
(g, h ∈ G, f ∈ C0 (X), µ ∈ Mb (X)). We shall assume that there is a measure
λX ∈ Mb (X) such that for all φ ∈ L1 (λX ), which we consider as the space of
measures absolutely continuous with respect to λX rather than a space of functions,
the map (g, h) 7→ gφh is continuous from G × G into L1 (λX ). We will usually write
L1 (X) for L1 (λX ) and L∞ (X) for L∞ (λX ).
R
WeR make L1 (X) into an L1 (G) bimodule by defining aφ = G gφ da(g) and
φa = G φg da(g) (a ∈ L1 (G), φ ∈ L1 (X)). L∞ (X) is the dual module.
In [8], the first author showed that the group algebra L1 (G) is always weakly
amenable. Another treatment of this result was published by Despić and Ghahramani [2]. These proofs can be used to show that H1 (L1 (G), L∞ (X)) = 0. In [4,
Problem 1] Grønbæk and Lau show that the augmentation ideal is weakly amenable
in a number of cases, most notably when G is discrete, and they raise the question
of whether the augmentation ideal is always weak amenability. This is equivalent
to whether we always have H1 (L1 (G), L∞ (G)/C1) = 0. We shall show that this
cohomology group is non-zero for G = SL(2, R), so the augmentation ideal for this
group is not weakly amenable. In contrast we will also show that it is zero for
G = SL(2, C). Our results involve considering H1 (L1 (G), L∞ (X)/C1), but, as we
Date: Newcastle, June 26, 2003.
1991 Mathematics Subject Classification. Primary ????
16A61.
Key words and phrases. Cohomology, Banach algebra.
1
46H20, 46J40; Secondary 43A20,
2
B. E. JOHNSON AND M. C. WHITE
are primarily concerned with the counterexample just mentioned, we will not be
working in anything like the generality we have described above. Many results will
be for the case in which G and X are discrete. Some measure theoretic complications occur in the continuous case and we get our counterexample because we are
able to mimic the discrete results in some rather simple continuous situations. For
the continuous examples which we will consider in detail, X will be a simple subset
of a Euclidean space on which G acts by diffeomorphisms of an elementary nature
and λX will be Lebesgue measure.
Our results involve relative cohomology. This was described by Lykova [9], but
we need to extend her results to the case in which we have an amenable subalgebra
of the multiplier algebra of A rather than of the actual algebra. This amenable
subalgebra arises from an amenable subgroup of G. The extension is fairly immediate and is given in Section 2. In Section 3 we give results for discrete G and X,
though where the extension to the continuous case is immediate this is given too.
Section 4 contains the results about H1 (L1 (G), L∞ (G)/C1) for G = SL(2, R) and
SL(2, C) mentioned above. In Section 5 the relationship of these results with the
weak amenability of the augmentation ideal is discussed. In that section we also
show that unitization of every augmentation ideal is weakly amenable, and hence
are able to show that in particular the unitization of the augmentation ideal of
G = SL(2, R) is weakly amenable, while the ideal itself is not. Finally in Section 6
we give some results concerning the multipliers of the augmentation ideal.
2. Relative cohomology
Let A be a Banach algebra, X an A bimodule and B ⊆ A a subalgebra.
n
(A, X), the Hochschild cohomology relative to B is defined using B adjusted
HB
cochains – that is continuous n-linear maps T from A × · · · × A → X with
(∗) bT (a1 , . . . , an ) = T (ba1 , . . . , an )
T (a1 , . . . , aj b, aj+1 , . . . , an ) = T (a1 , . . . , aj , baj+1 , . . . , an )
T (a1 , . . . , an b) = T (a1 , . . . , an )b
where b ∈ B, a1 , . . . , an ∈ A and (1 ≤ j < n).
Let A be a Banach algebra with a bounded approximate identity {eα }, and X
an essential A bimodule, B ⊆ ∆(A) – the double multiplier algebra of A.
As is well known [6] X becomes a ∆(A) bimodule with module operations defined
by
M x = lim(M eα )x
α
xM = lim x(eα M )
α
(x ∈ X), (M ∈ ∆(A)).
Clearly the dual X ∗ then also becomes a ∆(A) bimodule. We should perhaps recall
here, although we shall not use it, that under these circumstances [7, Proposition
19] we have Hn (A, X ∗ ) = Hn (∆(A), X ∗ ).
n
We can now define a cohomology HB
(A, X) using B adjusted cochains – i.e.
n-linear maps satisfying the equations (∗) above.
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
3
Theorem 2.1. Let A be a Banach algebra with a bounded approximate identity. If
B is amenable and X ∗ is a dual A bimodule then
n
HB
(A, X ∗ ) = Hn (A, X ∗ )
Proof. We consider the following resolution of A
ˆ B A ← A⊗
ˆ B A⊗
ˆ B A ← A⊗
ˆ B A⊗
ˆ B A⊗
ˆ BA
A ← A⊗
This is a weakly admissible flat resolution of A and so we can use it to compute
the cohomology of dual modules. In particular the cohomology of Hn (A, X ∗ ) is the
cohomology of the complex
∗
∗
∗
ˆ B A⊗
ˆ A⊗A
ˆ B A⊗
ˆ B A⊗
ˆ A⊗A
ˆ B A⊗
ˆ B A⊗
ˆ B A⊗
ˆ A⊗A
(A⊗
ˆ op X) → (A⊗
ˆ op X) → (A⊗
ˆ op X) .
n
But this is also the definition of the cohomology HB
(A, X) and hence they are
equal.
3. H1 (G, `∞ (X)/C1) and relative cohomology
In this section we will consider a discrete group G acting as a group of bijections of a discrete space X. The bounded group cohomology of G is essentially the same as the `1 (G) algebra cohomology and we will express our results in
terms of the former. We will write Hn (G, X) for Hn (`1 (G), `∞ (X)) and HSn (G, X)
for H`n1 (S) (`1 (G), `∞ (X)) The long exact sequence of cohomology for `∞ (X) and
the submodule C1 of constant functions has a connecting homomorphisms γX :
H1 (G, `∞ (X)/C1) → H2 (G, C). As H1 (G, `∞ (X)) = 0, γX is injective and we will
describe H1 (G, `∞ (X)) by describing the range of γX .
For any subgroup S of G there is a map ηS : HS2 (G, C) → H2 (G, C) arising from
the inclusion CS2 (G, C) ⊆ C 2 (G, C).
Proposition 3.1. Let T ∈ ZS2 (G, C) with k[ηS (T )]k < 1, then k[T ]k < 25. Hence
the map ηS is injective.
Proof. As k[ηS (T )]k < 1 there exists F ∈ `∞ (G) such that kT − δF k < 1. Then
we have |δF (sg, h) − δF (g, h)| < 2 (s ∈ S, g, h ∈ G). So
|(F (g) − F (sg)) − (F (gh) − F (sgh))| < 2
Define (s) ∈ C by (s) = F (1) − F (s). We have for all s, t ∈ S,
|((s) + (t) − (st)| = |((F (1) − F (s)) + (F (1) − F (t))) − (F (1) − F (st))|
≤ |F (1) − F (s) − F (t) + F (st)| < 2.
If kk > 2, pick s ∈ S be such that |(s)| ≥ kk −1, then (s2 ) − 2(s) < 2 and so
2 (s ) ≥ 2(kk) − 2 > kk .
This contradiction shows that kk ≤ 2. Hence |F (1) − F (s)| < 2 and so we have
|F (g) − F (gs)| < 4. Similarly |F (g) − F (sg)| < 4. Now we define
FS (g) = sup Re F (sgt) + i sup Im F (sgt)
s,t∈S
s,t∈S
so that FS is S adjusted and kFS − F k < 8. Hence kT − δFS k ≤ kηS (T ) − η(δFS )k ≤
kηS (T ) − η(δF )k + kη(δF ) − η(δFS )k ≤ 1 + 3 kF − FS k < 25.
This shows that if α ∈ ZS2 (G, C)∩B 2 (G, C) then α ∈ BS2 (G, C) so ηS is injective.
Using the map ηS , we will consider HS2 (G, C) as a subset of H2 (G, C).
4
B. E. JOHNSON AND M. C. WHITE
Proposition 3.2. Let S be a subgroup of G and let θ ∈ `∞ (G × G) be such that
(i)
θ(g, hs) = θ(g, h)
(ii)
For all g, h ∈ G
(g, h ∈ G, s ∈ S)
θ(h, k) − θ(gh, k) + θ(g, hk)
is a constant function of k ∈ G.
Then there is φ ∈ `∞ (G) such that putting
θ00 (g, h) = θ(g, h) − θ(g, e) − (φ(h) − φ(gh) + φ(g))
00
then θ ∈
ZS2 (G, C)
(g, h ∈ G)
00
and θ (g, s) = 0 (g ∈ G, s ∈ S).
Condition (i) implies that θ̃(g, hS) = θ(g, h) defines θ̃ ∈ C 1 (G, `∞ (G/S)). If we
denote the quotient map `∞ (G/S) → `∞ (G/S)/C1 by Q, then (ii) implies that
Qθ̃ ∈ Z 1 (G, `∞ (G/S)/C1). Denoting the expression in (ii) by α(g, h) then one
can check that α ∈ Z 2 (G, C) and [α] = γQθ̃ where [α] is the coset in H2 (G, C)
containing α. The conclusion implies that [θ0 ] = [α] and so the range of γX is in
HS2 (G, C).
Proof. Put θ0 (g, h) = θ(g, h) − θ(g, e)
Then for all g, h, k ∈ G
(g, h ∈ G) so that θ0 (g, e) = 0
(g ∈ G).
θ0 (h, k) − θ0 (gh, k) + θ0 (g, hk) = α(g, h) − (θ(h, e) − θ(gh, e) + θ(g, e))
so the left hand side is independent of k and is equal to its value at k = e; that is
θ0 (h, k) − θ0 (gh, k) + θ0 (g, hk) = θ0 (g, h).
Hence θ0 ∈ Z 2 (G, C) and [θ0 ] = [α]. We also have θ0 (g, hs) = θ0 (g, h) (g, h ∈ G, s ∈
S).
Define a bounded function Φ from S into `∞ (G/S) by Φ(s)(hS) = θ0 (s, h) (s ∈
S, h ∈ G). Then Φ(s)(S) = 0. If we consider G/S as a left S module then `∞ (G/S)
becomes a right S module under the action
(φs)(hS) = φ(shS).
The cocycle condition for θ0 implies that Φ is a bounded crossed homomorphism
and so is principal; that is there is φ ∈ `∞ (G/S), which we consider as a function
on G constant on the cosets of S. Under the action of S on G/S, S itself is a one
point orbit, so replacing φ by the function which is 0 on S and equal to φ on the
rest of G/S if necessary, we can assume that φ is 0 on S.
We need to show that θ00 has the required properties. As θ00 differs from θ0 ∈
2
Z (G, C) by a coboundary, θ00 ∈ Z 2 (G, C). Replacing h by hs in θ00 (g, h) does
not alter it because the same is true of θ(g, h) and θ(h). We have φ(h) − φ(sh) =
Φ(s)(h) = θ0 (s, h) (s ∈ S, h ∈ G), so θ00 (s, h) = 0 because φ(s) = 0. The identity
δθ00 (g, s, h) = 0 then gives θ00 (gs, h) = θ00 (g, sh) and the identity δθ0 (s, g, h) = 0
gives θ00 (g, h) = θ00 (sg, h) (g, h ∈ G, s ∈ S). Thus θ00 ∈ ZS2 (G, C). We have
θ00 (g, e) = 0 because θ0 (g, e) = 0 and φ(e) = 0 (g ∈ G) and so θ00 (g, s) = 0
(g ∈ G, s ∈ S).
Proposition 3.3. Let S be a subgroup of G, l ∈ G and put S 0 = lSl−1 . Then
HS2 (G, C) = HS2 0 (G, C) are the same subset of H2 (G, C).
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
5
Proof. Let θ ∈ ZS2 (G, C) and define ψ ∈ `∞ (G × G) by ψ(g, h) = θ(g, hl). Then for
s0 ∈ S 0 , ψ(g, hs0 ) = θ(g, hs0 l) = θ(g, hls) = θ(g, hl) = ψ(g, h) where s = l−1 s0 l ∈
S. Also ψ(h, k) − ψ(gh, k) + ψ(g, hk) = θ(g, h) so ψ satisfies the conditions of
Proposition 3.2 with S 0 in place of S. This ψ 00 ∈ ZS2 0 (G, C) and [ψ 00 ] = [θ] so
[θ] ∈ HS2 0 (G, C). Interchanging S and S 0 , the result follows.
For x ∈ X denote {g : gx = x} by Zx .
Proposition 3.4. Let the group
of bijections of a set X. Then
T
T G act2 as a group
2
∼
∼
(G, C) where T ⊆ X
H
(G,
C)
Im
γ
H1 (G, `∞ (X)/C1) ∼
= z∈T HZ
=
X =
Zx
x∈X
t
contains at least one element from each G orbit.
Proof. The first equality follows because γX is injective and the last follows from
Proposition 3.1. It is thus enough to show that the last set is Im γX when T contains
2
exactly one element of each G orbit. Suppose α ∈ Z 2 (G, C) has [α] ∈ HZ
(G, C)
t
2
∞
for all t ∈ T . Then for each t ∈ T there is αt ∈ ZZt (G, C) and βt ∈ ` (G) with
α = αt + δβt . Define θ ∈ `∞ (G, X) by
θ(g, ht) = αt (g, h) + βt (g)
(g, h ∈ G, t ∈ T ).
Note that Proposition 3.1 guarantees that θ is bounded. Then for x = kt
θ(h, x) − θ(gh, x) + θ(g, hx)
=αt (h, k) − αt (gh, k) + αt (g, hk)
+ βt (h) − βt (gh) + βt (g)
=αt (g, h) + δβt (g, h) = α(g, h)
Writing (θ̃(g))(x) = θ(g, x) so that θ̃ ∈ C 1 (G, `∞ (X)) and Q for the quotient
map `∞ (X) → `∞ (X)/C1 we have, from the above, Qθ̃ ∈ Z 1 (G, `∞ (X)/C1) and
γX [Qθ̃] = [α]. Thus [α] ∈ Im γX .
Conversely if [α] ∈ Im γX there is θ ∈ `∞ (G × X) with Qθ̃ ∈ B 1 (G, `∞ (X)/C1)
and γX [Qθ̃] = [α]. Put
α0 (g, h) = θ(h, x) − θ(gh, x) + θ(g, hx)
(g, h ∈ G, x ∈ X).
Then α0 is independent of x because θ̃(h) − θ̃(gh) + θ̃(g)h is a constant function.
Moreover the construction of γX gives γQθ̃ = [α0 ] so α0 ∈ [α]. Let t ∈ T and
define ψ ∈ `∞ (G × G) by ψ(g, h) = ψ(g, ht) (g, h ∈ G). Then ψ(g, hs) = ψ(g, h)
for s ∈ Zt and ψ(h, k) − ψ(gh, k) + ψ(g, hk) = α0 (g, h). By Proposition 3.2 with
ψ and α0 in place of θ and α we get ψ 00 ∈ ZZ2 t (G, C) and [ψ 00 ] = [α0 ] = [α] so
2
[α] ∈ HZ
(G, C).
t
4. H1 (L1 (G), L∞ (X)/C1) and relative cohomology
In this section we will extend the results of Section 3 to some continuous situations. First of all, we extend Proposition 3.1. For any closed subgroup S of G
2
1
2
1
there is a map ηS : HL
1 (S) (L (G), C) → H (L (G), C) arising from the inclusion
2
1
2
1
CL1 (S) (L (G), C) → C (L (G), C).
Proposition 4.1. Let T ∈ ZS2 (G, C) with k[ηS (T )]k < 1, then k[T ]k < 25. In
particular the map ηS is injective.
6
B. E. JOHNSON AND M. C. WHITE
Proof. We prove only the second. The general case follows as in Proposition 3.1.Let
φ ∈ C 1 (L1 (G), C) be such that δφ ∈ CL2 1 (S) (L1 (G), C). Rearranging the equation
(δφ)(µa, b) = λ(µ)(δφ)(a, b) (µ ∈ L1 (S), a, b ∈ L1 (G) and writing λ(µ) = µ(G) )
we get
(λ(µ)φ − φµ)(ab − λ(b)a) = 0.
If a0 ∈ I0 (G) then there is a ∈ L1 (G), b ∈ I0 (G) with a0 = ab = ab − λ(b)a so
(λ(µ)φ − φµ)(a0 ) = 0 for all a0 ∈ I0 (G). Thus λ(µ)φ − φµ is a multiple of λ, c(µ)λ
say. It is straightforward to check that c is a continuous linear functional on L1 (S).
We have
(λ(µ)c(ν) − c(µν))λ = λ(µ)(λ(ν)φ − φν) − (λ(µν)φ − φµν)
= (−λ(µ)φ + φµ)ν = −c(µ)λν
= −c(µ)λ(ν)λ
(µ, ν ∈ L1 (S)).
Thus c(µν) = λ(µ)c(ν) + c(µ)λ(ν). If µ is a probability measure we have kck ≥
|cn | = n |c(µ)| so letting n → ∞ we get c(µ) = 0 and so c = 0. Thus φ ∈
CL1 1 (S) (L1 (G), C). Thus if α ∈ ZL2 1 (S) (L1 (G), C) is such that ηS [α] = 0 so α = δφ
2
1
as above, then α ∈ BL
1 (S) (L (G), C) and [α] = [0].
Using ηS we will consider HS2 (L1 (G), C) as a subgroup of H2 (L1 (G), C).
We will assume that X is locally compact and that it has a G-action: that is
there is a continuous map (g, x) 7→ gx from G × X → X with g(hx) = (gh)x and
ex = x (g, h ∈ G, x ∈ X). This gives rise to a right Banach M (G)-module structure
on C0 (G) defined by
Z
φµ(x) =
φ(gx) dµ(g)
(φ ∈ C0 (X), µ ∈ M (G), x ∈ X)
G
and the corresponding dual action on M (X) = C0 (X)∗ .
We assume that M (X) contains a closed submodule which we denote L1 (X)
with the properties
(1) If µ ∈ L1 (X) and ν is absolutely continuous with respect to µ then ν ∈
L1 (X).
(2) If µ ∈ L1 (X) then the map g 7→ gµ is norm continuous.
We denote the dual of L1 (X) by L∞ (X). Usually L1 (X) will be the L1 space
of a single measure and L∞ (X) the corresponding L∞ space. We are interested
in H1 (L1 (G), L∞ (X)/C1). As in the discrete case, there is a connecting homomorphism γX : H1 (L1 (G), L∞ (X)/C1) → H2 (L1 (G), C) which is injective because
H1 (L1 (G), L∞ (X)) = 0.
Of course; if H2 (L1 (G), C) = 0 then we always have H1 (L1 (G), L∞ (X)/C1) = 0.
The main difficulty in extending the results for the discrete case beyond this is there
seems no reason to suppose, in general, that the stability subgroups Zx behave in
a regular fashion. We shall restrict attention to the case in which they do.
Definition 4.2. We say that X satisfies hypothesis H if there is an index I (usually
finite or countably infinite) and for each i ∈ I:
(1) a closed subgroup Si of G;
(2) a locally compact space Yi and a strictly positive measure νi on Yi ;
(3) a G homeomorphism ξi from G/Si × Yi onto an open subset Xi of X where
G acts in the usual way on the coset space G/Si and trivially on Yi ;
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
7
(4) the Xi are mutually disjoint and the complement of their union is of measure zero for all the L1 (X) measures;
ˆ 1 (νi ) onto the elements of L1 (X)
(5) the homeomorphisms ξi carry L1 (G/Si )⊗L
1
concentrated on Xi where L (G/Si ) is the image of L1 (G) under the quotient map from G to G/Si .
Example 4.3. For G = SL(2, R) and X = SL(2, R) where G acts on X by
conjugation we take I = {1, 2}
X1
X2
S1
S2
Y1
Y2
ν1 , ν2
=
=
=
=
=
=
=
{g : g ∈ G, G has distinct real eigenvalues}
{g : g ∈ G, G has distinct non-real eigenvalues}
Diagonal matrices in G
Rotation matrices in G
(−1, 0) ∪ (0, 1)
{z : z ∈ C, |z| = 1, Im z > 0}
Lebesgue measure
Each element x1 of X1 has eigenvalues λ, λ−1 for just one λ ∈ Y1 so there is
g ∈ G with gag −1 = x1 where a is the diagonal matrix with entries λ, λ−1 . Any
two such elements g lie in the same coset of S1 and we take ξ1 to be the inverse of
the map x1 7→ (gS1 , λ).
Each element x2 of X2 has eigenvalues λ, λ−1 for just one λ ∈ Y2 so there is
g ∈ G with gbg −1 = x2 where b is the rotation with eigenvalues λ, λ−1 and a
similar analysis applies.
To proceed we need to elucidate the connection between the elements F of L∞ (G)
with F ν = λ(ν)F (ν ∈ L1 (S)) and L∞ functions constant on the cosets of S. We
will consider a more general situation which we shall need later. We suppose X
to be a locally compact space on which G acts continuously as a group of homeomorphisms. To simplify the measure theory we will assume that G and X are
metrisable and σ-compact. We suppose there is a bounded measure µX on X such
that for all ν ∈ L1 (µX ), gν ∈ L1 (µX ) and g 7→ gν is norm continuous. If F is a
bounded Borel function on X then (g, x) 7→ F (gx) is a bounded Borel function on
G × X and we can apply Fubini’s Theorem to the integral
Z
F (gx) d(νG (g) × νX (x))
G×X
1
1
where νGR ∈ L (G), νX ∈ L (µX ). Because F (gx) is a bounded Borel function of g,
F0 (x) = G F (gx) dνG (g) defines a bounded Borel function of x with
Z
Z Z
F0 (x) dνX (x) =
F (gx) dνX (x) dνG (g)
X
X
ZG
=
(F, gνX ) dνG (g)
G
= (F, νG νX )
= (F νG , νX )
8
B. E. JOHNSON AND M. C. WHITE
so F νG = F0 almost everywhere on X. This shows that if F is constant of the
G-orbits in X then F νG = λ(νG )F for all νG ∈ L1 (G) and we will show that the
second condition implies that F is equal almost everywhere to a function constant
on the G-orbits.
So suppose now that F ∈ L∞ (X) has F νG = λ(νG )F for all νG ∈ L1 (G). There
is a bounded Borel function equal to F almost everywhere so we can assume that F
is a Borel function.RLet F0 be as above for some probability measure νG on G. For
g 0 ∈ G, F0 (g 0 x) = X F (gx) d(νG g 0 )(x) so that, as g 0 7→ νG g 0 is norm continuous,
F0 (g 0 x) = F0 g 0 (x) is continuous in g 0 for all x ∈ X. However,
F0 g 0 = F νG g 0 = λ(νG g 0 )F = λ(νG )F = F νG = F0
almost everywhere in X for each g 0 in G. Consider the function F 0 (g, x) = F0 (gx)−
F0 (x) on G × X. It is a Borel function which, for each g in G is zero for almost all
x in X.
Thus it is zero almost everywhere on G × X and so for almost all x it is zero for
almost all g. As it is continuous in g,
E = {x : x ∈ X, F 0 (g, x) = 0 for all g in G}
= {x : x ∈ X, F 0 (g, x) = 0 for almost all g in G}
Z
= {x : x ∈ X, |F 0 (g, x)| dλ(x) = 0}.
The first expression shows that the set is G invariant and Fubini’s theorem and the
last expression show that it is a Borel set of measure zero. This redefining F0 to
be zero on E we have a bounded Borel function on X which is constant on the G
orbits and equal almost everywhere to F .
Theorem 4.4. Suppose that G is a metrisable and σ-compact and that X satisfies
hypothesis H. Then
\
Im γX ⊇
HS2 i (L1 (G), C).
i∈I
Proof. Let α ∈ Z 2 (L1 (G), C) with [α] ∈ HS2 i (L1 (G), C) for all i ∈ I. Then for
each i ∈ I there is αi ∈ ZS2i (L1 (G), C) and βi ∈ L∞ (G) with α = αi + δβi . Define
θ ∈ L∞ (G × X) by
θ(g, xi ) = αi (g, ξ −1 (xi )1 ) + βi (g)
(g ∈ G, xi ∈ Xi )
where for z ∈ G/Si × Yi , z1 is its projection onto G/Si and we consider αi ∈
L∞ (G × G) as a function of G × G/Si because αi (a, bν) = λ(ν)αi (a, b) (a, b ∈
L( G), ν ∈ L1 (Si )). We then have, for g, h ∈ G and xi ∈ Xi
θ(h, xi ) − θ(gh, xi ) + θ(g, hxi ) = αi (h, k) − αi (gh, k) + αi (g, hk)
+ βi (h) − βi (gh) + βi (g)
= αi (g, h) + δβi (g, h) = α(g, h)
almost everywhere on G × G × X where ξi−1 (xi )1 = kSi . For a ∈ L1 (G) deR
fine θ̃(a)(b) = G×X θ(g, x)a(g) dλ(g) db(x) (a ∈ L1 (G), b ∈ L1 (X)). Then θ̃ ∈
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
9
C 1 (L1 (G), L∞ (X)). If Q is the quotient map L∞ (X) → L∞ (X)/C1 then the equation above gives
λ(a)θ(a0 , b) − θ(aa0 , b) + θ(a, a0 b) = α(a, a0 )λ(b) a, a0 ∈ L1 (G), b ∈ L1 (X)
where θ and α now denote the bilinear functionals on the L1 spaces corresponding
to these L∞ functions. Thus
λ(a)θ̃(a0 ) − θ̃(aa0 ) + θ̃(a)a0 = α(a, a0 )1
so that Qθ̃ ∈ Z 1 (L1 (G), L∞ (X)/C1) and the definition of γX shows that γX [Qθ̃] =
[α].
Corollary 4.5. If the groups Si are amenable then
H1 (L1 (G), L∞ (X)/C1) = H2 (L1 (G), C).
Proof. Amenability of Si implies that HS2 i (L1 (G), C) = H2 (L1 (G), C) so by the
Theorem
\
H2 (L1 (G), C) =
HS2 i (L1 (G), C) ⊆ Im γ ⊆ H2 (L1 (G), C)
i∈I
and γ is surjective.
2
1
Theorem 4.6. Let G = SL(2, R). Then H (L (G), C) is one dimensional.
Proof. Let X = R ∪ ∞ be the real projective line. The group G acts on X by
gx = (g11 x + g12 )/(g21 x + g22 ). Let H be the subgroup {h : h ∈ G, h∞ = ∞}.
Then H consists of the matrices with g21 = 0 and so is solvable and hence amenable.
2
Thus H2 (L1 (H), C) = HH
(L1 (G), C) and we calculate the latter group.
Consider the action of H × H on G given by (h1 , h2 )g = h1 gh−1
2 . There are two
orbits,
H
and
G
\
H.
To
see
this
note
that
H
is
obviously
an
orbit
and the action of
ia
b
h=
on R is x 7→ a2 x + ab (a, b ∈ R, a 6= 0) so H acts transitively on
0 a−1
R Thus if g, g 0 ∈ G \ H then x = g∞ ∈ R and x0 = g 0 ∞ ∈ R so taking h ∈ H with
hx0 = x we have g −1 hg 0 ∞ = ∞ so h0 = g −1 hg 0 ∈ H and g 0 = h−1 gh0 . Thus H × H
acts transitively on G \ H. The 1 cochains are thus elements F of L∞ (G) with
F ν = λ(ν)F for all ν in L1 (H × H). Such a function is equal almost everywhere to
a function constant on the orbits and, since H is of measure zero, the cochains are
2
contant anmost everywhere. This BH
(L1 (G), C) is one dimensional and consists of
the constant functions on G × G.
The 2 cochains are certain L∞ functions on G × G. For such a function write
θ̃(g1 , g2 ) = θ(g1−1 , g2 ). Then H × H × H acts on G × G by (h1 , h2 , h3 )(g1 , g2 ) =
−1
(h1 g1 h−1
2 , h1 g2 h3 ) and the cochains are functions θ such that] θ̃ν = λ(ν)θ̃ for
1
all ν in L (H × H × H). We have seen that such functions are equal almost
everywhere to a function constant on the orbits of H × H × H on G × G. The
quotient map G × G → G/H × G/H is a map of H × H × H modules where the
action is (h1 , h2 , h3 )(x, x0 ) = (h1 x, h1x0 ) and the inverse image of each point lies
in a single orbit. This the orbits in G × G are the inverse images of the orbits in
G/H × G/H. These orbits are {(∞, ∞)}, {∞} × R, R × {∞}, {(x, x) : x ∈ R},
L = {(x, x0 ) : x, x0 ∈ R, x < x0 } and U = {(x, x0 ) : x, x0 ∈ R, x > x0 }. This follwos
because H is transitive on R and if (x, x0 ) and (y, y 0 ) are two pairs of distinct
elements of R, then there is h ∈ H with hx = y, hx0 = y 0 if and only if x > x0 and
y > y 0 or x < x0 and y < y 0 . All these orbits, except L and U have zero measure so
10
B. E. JOHNSON AND M. C. WHITE
2
CH
(L1 (G), C) consists of functions which are constant on these sets and so is two
dimensional.
Using stereographic projection between X and the circle T we can say whether
two ordered 3 element subsets of X havethe same cyclic order. If (x, y, z) is such
a set and g ∈ G then (gx, gy, gz) has the same cyclic order as (x, y, z) because we
have
a
1
ax + b
= − 2
cx + d
c
c x + cd
where the maps x 7→ c2 x + cd, x 7→ ac + x and x → − x1 all preserve cyclic order.
2
Let θ ∈ CH
(L1 (G), C). We need to prove that for almost all (g, h, k) in G × G × G
we have
θ̃(h−1 , k) − θ̃(h−1 g −1 , k) = θ̃(g −1 , h) − θ̃(g −1 , hk).
It is enough to show this when ∞, h−1 ∞, k∞, h−1 g −1 ∞ are distinct. Suppose that
θ̃(g, h) = ξ for (g∞, h∞) ∈ L, θ̃(g, h) = η for (g∞, h∞) ∈ U . The left hand side of
the equation depends on;y on the cyclic order of the points ∞, h−1 , k∞, h−1 g −1 ∞
which is the same as the cyclic order of h∞, ∞, hk∞ g −1 ∞ which determines the
right hand side. By checking the six possible cases we see that the equation holds.
2
2
2
Thus ZH
is 2 dimensional, BH
is one dimensional so HH
is one dimensional.
Corollary 4.7. Let G = SL(2, R). Then H1 (L1 (G), L∞ (G)/C1) is one dimensional
Proof. By [7, page 32], this cohomology group is the same as H1 (L( G), L∞ (X)/C1)
where X = G and G acts on X by conjugation. We have seen in Example 4.3 that X
satisfies hypothesis H where te groups S1 ans S2 are abelian. Thus by Corollary 4.5,
H1 (L1 (G), L∞ (X)/C1) = H2 (L1 (G), C) and we have just shown that this latter
space is one dimensional.
Theorem 4.8. Let G = SL(2, C) then H1 (L1 (G), L∞ (G)/C1) = {0}.
Proof. The proof is similar to Theorem 4.6 with C in place of R throughout. The
essential difference is that the two sets L and U are replaced by the single set
2
{(z.z 0 ) : z, z 0 ∈ C, z 6= z 0 } on which H acts transitively. Thus ZH
(L1 (G), C) is one
1
2
dimensional and so is the same as BH (L (G), C).
Corollary 4.9. Let G = SL(2, C). Then H1 (L1 (G), L∞ (G)/C1) = {0}.
Proof. This follows from Theorem 4.8 because the connecting map is injective. 5. Weak amenability
The following result is essentially from Grønbæk and Lau [4, Theorem 2.6].
Theorem 5.1. Let A be a Banach algebra with a bounded approximate identity
and let I be a codimension 1 closed two-sided ideal.
Hn (A, X ∗ ) = Hn (I, X ∗ )
for each neo-unital Banach A bimodule. The isomorphism being induced by restricting n-cochains from A to I.
We are now in a position to prove the main results referred to in the abstract.
The first, that the augmentation ideal of SL(2, R) is not weakly amenable is just
a reformulation of Corollary 4.7. The latter is a little longer.
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
11
Theorem 5.2. Let G = SL(2, R) and I0 (G) be the augmentation ideal of G. Then
H1 (I0 (G), I0 (G)∗ ) is one dimensional. In particular I0 (G) is not weakly amenable.
Proof. Theorem 5.1 tells us that
H1 (L1 (G), L∞ (G)/C1) = H1 (I0 (G), L∞ (G/C1).
It is clear that I0 (G)∗ = L∞ (G)/C1 and so
H1 (L1 (G), L∞ (G)/C1) = H1 (I0 (G), I0 (G)∗ ).
It remains only to note that Corollary 4.7 tells us these are all one dimensional. As it
is one dimensional it is certainly not zero and so I0 (G) is not weakly amenable. Theorem 5.2 very nearly answers [3, Question 20] of Grønbæk. He asks whether
a Banach algebra must be weakly amenable if its unitization is. In the theorem,
we have a ideal which is not weakly amenable, I0 (SL(2, R)), as a codimension one
ideal in the weakly amenable algebra L1 (SL(2, R)). Sadly L1 (G) does not have a
unit, unless G is discrete, and in the discrete case I0 (G) is always weakly amenable,
as is shown in [4].
However, I0 (SL(2, R)) is indeed a counter-example, even though its unitization,
I0 (G)# , is not L1 (SL(2, R)), both algebras share a number of homological properties. In particular they are both weakly amenable.
Lemma 5.3. Let A be a Banach algebra with a bounded approximate identity {eα }
and χ a non-trivial character on A, with kernel I. Then
(1) Every derivation D : A → A0 restricts to give a derivation D : I → I 0 such
that there exists a function ψ : I → C satisfying D(i)(j) + D(j)(i) = ψ(ij)
for all i, j ∈ I.
(2) Every derivation D : I → I 0 for which there exists such a function ψ extends
to a derivation D̃ : A → A0 .
Proof. So we are not continually taking subnets, let us assume that {eα } is indexed
by an ultrafilter, this is to ensure that the bounded approximate identity converges
σ(A00 , A0 ).
The first part of (1) is easy. Obviously every derivation restricts to give a derivation. Also we may define ψ(i) = limα D(i)(eα ), then
ψ(ij) = lim D(ij)(eα )
α
= lim D(i)(jeα ) + D(j)(eα i)
α
= D(i)(j) + D(j)(i),
as required.
Conversely, given a derivation D : I → I 0 , we extend it to a derivation D̃ : A →
0
A in 2 steps.
Step 1: We extend to a derivation D̄ : I → A0 .
This extension is made be defining
D̄(i)(a) = lim D(i)(a − χ(a)eα ) + χ(a)ψ(i).
α
Note: the limit exists because the derivation is into a ball in a dual module and we
have an ultrafilter. In special cases the limit may exist for any bounded approximate
identity, for example if I 2 is dense in I.
12
B. E. JOHNSON AND M. C. WHITE
We now need to verify that D̄ is a derivation from I to A0
D̄(ij)(a) = lim D(ij)((a − χ(a)eα ) + χ(a)ψ(ij)
α
= lim D(i)(j(a − χ(a)eα )) + lim D(j)((a − χ(a)eα )i) + χ(a)ψ(ij)
α
α
= D(i)(ja − jχ(a)) + D(j)(ai − iχ(a)) + +χ(ij)
= lim D(i)(ja − χ(ja)eα ) + lim D(j)(ai − χ(ai)eα )
α
α
+ χ(a) (ψ(ij) − D(i)(j) − D(j)(i))
= D̄(i)(ja) + D̄(j)(ai)
Step 2: We extend to a derivation D̃ : A → A0 .
This extension is made be defining
D̃(a)(b) = lim D̄(a − χ(a)eα )(b).
α
Note: the limit exists because the derivation is into a ball in a dual module and
we have an ultrafilter. However, for any bounded approximate identity of A and
derivation into the dual of an essential A module, X 0 , this limit exists; consider
D(eα − eβ )(bx) = D((eα − eβ )b)(x) + D(b)(x(eα − eβ )).
This shows that the limit in the definition of D̃ above is a Cauchy net in this case,
without the ultrafilter assumption.
We now need to verify that D̃ is a derivation from A to A0 . We begin by assuming
that i ∈ I and b, c ∈ A.
D̃(ib)(c) = lim D̄(ib − χ(ib)eα )(c)
α
= lim D̄(i(b − χ(b)eα )) + χ(b)D̄(i)(c)
α
= lim D̄(i)((b − χ(b)eα )c) + lim D̄(b − χ(b)eα )(ci) + χ(b)D̄(i)(c)
α
α
= D̄(i)(bc − χ(b)c) + D̃(b)(ci) + χ(b)D̄(i)(c)
= D̄(i)(bc) + D̃(b)(ci)
= lim D̄(i − χ(i)eα )(bc) + D̃(b)(ci)
α
= D̃(i)(bc) + D̃(b)(ci)
Note that for k ∈ I, we have D̄(k)(c) = D̃(k)(c). We will use this simple observation
rather subtly below with k = (a − χ(a)eα )b. Now we verify the derivation law in
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
13
the general case.
D̃(ab)(c) = lim D̄(ab − χ(ab)eα )(c)
α
= lim D̄((a − χ(a)eα )b + χ(a)(eα b − b) + χ(a)(b − χ(b)eα ))(c)
α
= lim D̄((a − χ(a)eα )b)(c) + χ(a)D̃(b)(c)
α
= lim D̃((a − χ(a)eα )b)(c) + χ(a)D̃(b)(c)
α
= lim D̃(a − χ(a)eα )(bc) + D̃(b)(c(a − χ(a)eα )) + χ(a)D̃(b)(c)
α
= lim D̄(a − χ(a)eα )(bc) + D̃(b)(ca − χ(a)c)) + χ(a)D̃(b)(c)
α
= D̃(a)(bc) + D̃(b)(ca)
Thus we have that D̃ : A → A0 is indeed a derivation, which extends D.
It is an obvious question to wonder whether the extension given in the previous
Lemma is unique. It cannot always be unique because the restriction to I may
be trivial and yet the derivation is not trivial. It is clear that this situation arises
when A has point derivations at the character determined by the maximal ideal.
The obvious condition to remove point derivations is to require (I 2 ) = I.
Lemma 5.4. Let A be a Banach algebra with a bounded approximate identity {eα }
and χ a non-trivial character on A which has kernel I and let D be a derivation
D : I → I 0 . Assume further that (I 2 ) = I, then there is at most one extension of
D to a derivation D : A → A0 .
Proof. We show that the derivation D ≡ 0 has only the trivial extension. Let D̃ be
any extension.
D̃(ij)(a) = D̃(i)(ja) + D̃(j)(ai)
= D(i)(ja) + D(j)(ai) = 0
As (I 2 ) = I we have that D̃(i)(a) = 0 for all (i ∈ I, a ∈ A). Next we have
D̃(eα )(ab) = D̃(beα )(a) − D̃(b)(eα a).
But clearly the right hand side tends to zero. Hence limα D̃(eα )(ba) = 0. But as A
has a bounded approximate identity every element of A can be factorized, by the
the Cohen Factorisation Theorem. Hence
D̃(a)(b) = lim D̃((a − χ(a)eα ))(b) + D̃(χ(a)eα )(b)
α
= lim D̃((a − χ(a)eα ))(b)
α
=0
as a − χ(a)eα is in I.
It remains only to note that the difference of any 2 extensions of a derivation is
a derivation extending the zero derivation. Hence any extension is unique.
14
B. E. JOHNSON AND M. C. WHITE
The reader who is fond of long exact sequences will have noticed that Lemma 5.3
and Lemma 5.4 suggest the exactness of the following portion of a long exact
sequence:
→ H1 (I, C0 ) → H1 (I, A0 ) → H1 (I, I 0 ) → H2 (A, C0 ) →
or at least in the special case when H1 (A, C0 ) = 0 and when H1 (I, A0 ) = H1 (A, A0 ).
Now, as we are interested in derivations on the augmentation ideal which is
known to satisfy (I 2 ) = I, we are just left with the question of whether H1 (I, A0 ) =
H1 (A, A0 ) in this case.
Corollary 5.5. Let A be a Banach algebra with a bounded approximate identity
{eα } and χ a non-trivial character on A, with kernel I, such that (I 2 ) = I. Then
A is weakly amenable if and only if I # is.
Proof. Let A be weakly amenable then H1 (A, A0 ) = 0 = H1 (I, A0 ) hence by exactness of the long exact sequence the map H1 (I, I 0 ) → H2 (I, C0 ) is injective. Now
consider the Banach algebra I # we have the short exact sequence
0 → I → I # → C → 0.
There is an induced long exact sequence of cohomology
→ H1 (I, C0 ) → H1 (I, (I # )0 ) → H1 (I, I 0 ) → H2 (I, C0 ) →
as the map H1 (I, I 0 ) → H2 (I, C0 ) is injective, we have that H1 (I, (I # )0 ) = 0,
but we know by Theorem 5.1 (or in fact by the well-known special case) that
H1 (I # , (I # )0 ) = H1 (I, (I # )0 ) = 0, which is just to say that I # is weakly amenable.
The reader may object that the C is not the same in both long exact sequences, in
one it is A/I, in the other it is I # /I. This is true, but we are concerned only with I
modules, and they are isomorphic as I modules. The reader may also wonder about
the induced map in each case, but in each case it is the connecting homomorphism
and can be computed to be just the map of Lemma 1, namely D 7→ φ, where
φ(i, j) = D(i)(j) + D(j)(i).
The converse has essentially that same proof.
Now we give the application which we have been building up to.
Corollary 5.6. Let I be the augmentation ideal of L1 (SL(2, R)), then I is not
weakly amenable, but its unitization I # is weakly amenable.
Proof. It is well-known that the augmentation ideal is a codimension one bi-ideal
in the group algebra L1 (SL(2, R)), which has a bounded approximate identity. It
was shown by GW, in his thesis, that (I 2 ) = I. It was proved by Johnson, [8]
that L1 (G) is weakly amenable for all G. It is shown in Theorem 5.2 that the
augmentation ideal of SL(2, R) is not weakly amenable. Thus by Corollary 5.6 I #
is also weakly amenable.
The reader who is unhappy with the abstract nonsense and prefers the explicit,
but rather opaque calculations of the lemmas, can follow through the proof using
Lemma 5.3 and Lemma 5.4. One takes a derivation D : I # → (I # )0 , this restricts
to a derivation D : I → I 0 for which there exists a function ψ such that ψ(ij) =
D(i)(j) + D(j)(i). Consequently this derivation extends to a derivation D̃ : A →
A0 . We know that A is weakly amenable and hence there is F ∈ A0 such that
D̃(a)(b) = F (ab − ba). We can restrict this function F to an element of I 0 and
extend it to an element of (I # )0 by setting F̃ (1) = 0. Now it is clear that D − dF
A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
15
restrict to zero as a map I → I 0 . Hence we may then assume that D restricts to be
zero on I → I 0 . In this case D(ij)(1) = D(i)(j) + D(j)(i) = 0 and as (I 2 ) = I we
have D(i)(1) = 0. All derivation satisfy D(1)(· ) = 0. Thus D vanishes as a map
I # → (I # )0 and I # is weakly amenable.
6. Multipliers on I0 (G)
Let G be a locally compact group. Then I0 (G) is a closed ideal in L1 (G) of
codimension 1. The ideal I0 (G) has a bounded approximate identity if and only
if G is amenable. In this section we show that some results which hold for L1 (G)
because it has a bounded approximate identity can be extended to I0 (G) even for
non-amenable groups.
Recall below the result of Willis [10].
Proposition 6.1. Every element of I0 (G) can be approximated in norm by linear
combinations of products of elements of elements in I0 (G)
The reader should note that Willis [11] has also proved that this result holds for
any ideal of finite codimension in L1 (G). However the proof is rather deeper.
Proposition 6.2. For µ ∈ M (G) define Tµ : I0 (G) → I0 (G) by Tµ (j) = jµ.
Then Tµ ∈ ∆r (I0 (G)) and the map µ 7→ Tµ is a homomorphism of M (G) onto
∆r (I0 ). If G is not compact then kTµ k = kµk (µ ∈ M (G)). If G is compact then
p(µ) ≤ kTµ k ≥ 21 p(µ), where p is the quotient norm in M (G)/C1.
Proof. It is straightforward to see that µ 7→ Tµ is a norm decreasing homomorphism into ∆(I0 (G)). To prove that it is surjective, let µ ∈ ∆r (I0 (G)). Then
for j1 , j2 ∈ I0 (G), µ ∈ M (G) we have T (µj1 j2 ) = µj1 T (j2 ) = µT (j1 j2 ) so by
Proposition 6.1, T (µj) = µT (j) (µ ∈ M (G), j ∈ I0 (G)). It follows that for
F ∈ I0 )(G)∗ = L∞ (G)/C1 we have T ∗ (F µ) = T ∗ (F )µ. Thus T ∗ maps the essential
right L1 (G) submodule of I0 (G)∗ into itself. The essential right L1 (G)-submodule of
L∞ (G) consists of left uniformly continuous
functions, Clu (G), so T ∗ maps Clu /C1
R
into itself. If µ ∈ J0 then (F, µ) = G F dµ is defined for F ∈ Cb (G)/C1. Let Q now
denote the quotient map from Cb (G) → Cb (G)/C1. For h ∈ G define µk ∈ M (G)
by
(F, muk ) = (T ∗ (qF ), δk − δe ) (F ∈ C0 (G))
Then for h, k ∈ G,
(F, µhk ) = (T ∗ (qF ), δk − δe + δhk − δk )
= (F, µh ) + (T ∗ (qF )δh , µk )
= (F, µh ) + (F h, µk )
= (F, µh + hµk )
Thus h 7→ µh is a bounded crossed homomorphism from G into the left G module
M (G). For ξ ∈ L1 (G), pξ (µ) = kµ ∗ ξk is a uniformly convex seminorm on M (G).
Using the set of these and [7, Proposition 3.7] we see that there is σ ∈ M (G) with
µg = gσ − σ (g ∈ G). Put S = T − Tσ . Then for F ∈ C0 (G), (Tσ∗ (qF ), δk − δe ) =
(σF, δk − δe ) = (F, hσ − σ) = (F, µh ) = (T ∗ (qF ), δk − δe ) Thus S ∗ (qF ) = 0. As S ∗
is w∗ -continuous and C0 (G) is w∗ -dense in L∞ (G) we see that S ∗ = 0 so T = Tσ
and T 7→ Tµ is surjective.
16
B. E. JOHNSON AND M. C. WHITE
Suppose now that G is non-compact. Let µ ∈ M (G) and > 0. There is
a ∈ L1 (G) with kak ≤ 1 and kaµk ≥ kµk −. There is a compact set K with
|aµ|(K) > |µ| −. Take g ∈ G \ KK −1 . Put j = ga − e. Then j ∈ I0 (G) and
kjk ≤ 2. We have |gaµ|(gK) > kµk −2. As K ∩ gK = ∅ and kgaµk ≤ kµk,
|gaµ|(K) < 2. A similar argument shows that |aµ|(gK) < 2. Thus kjµk ≥
|gaµ|(gK) − |aµ|(gK) + |aµ|(K) − |gaµ|(K) > 2 kµk −8 so kTµ k ≥ kµk −4.
Finally suppose that G is compact. We have kTµ k = kTµ+cλ k ≤ kµ + cλk for all
c ∈ C so kTµ k ≤ p(µ). The algebra I0 (G) has an approximate identity {eα } with
keα k ≤ 2. We shall show that for µ ∈ I0 (G), lim inf kTµ eα k ≥ kµk. For a ∈ I0 (G)
aTµ eα = aeα µ → aµ in norm. This holds for all a in L1 (G). Thus for a ∈ L1 (G)
with kak ≤ 1, lim inf kTµ eα k ≥ kaµk and taking the supremum over such elements
a we get lim inf kTµ eα k ≥ µ and so 2 kTµ k ≥ µ for µ ∈ I0 (G). For ν ∈ M (G) put
µ = ν − ν(G)λ ∈ J0 (G) so that 2 kTν k = 2 kTµ k ≥ kµk ≥ p(ν).
Theorem 6.3. Let D : I0 (G) → I0 (G) be a derivation. Then D has a unique
extension to a derivation from L1 (G) to L1 (G).
Proof. Note that L1 (G) and I0 (G) are both semisimple Banach algebras so the
derivations we are considering are automatically continuous. We begin by noting
that all derivations from L1 (G) to itself map into I0 (G). For composing with the
algebra homomorphism λ gives a point derivation on L1 (G). But as I02 equals I0
there are no non-zero point derivations at the character λ.
We consider first of all the case in which G is not compact. Let µ ∈ M (G) and
consider the operator ∆µ on I0 (G) defined by ∆µ j = D(jµ) − D(j)µ (j ∈ I0 (G)).
It is straightforward to check that ∆r is a right multiplier and so there is ∆(µ) ∈
M (G) such that ∆µ (j) = j∆(µ). Again it is straightforward to check that for
j ∈ I0 (G), µ, ν ∈ M (G) we have j∆(µν) = jµ∆(ν) + j∆(µ)ν and so, since the map
σ 7→ Tσ in Proposition 6.2 is injective, ∆(µν) = µ∆(ν) + ∆(µ)ν. For j 0 ∈ I0 (G),
∆j 0 (j) = D(jj 0 ) − D(j)j 0 = jD(j 0 ), so ∆(j 0 ) = D(j 0 ) and ∆ is an extension of D.
For a, b ∈ L1 (G), ∆(ab) = a∆(b) + ∆(a)b ∈ L1 (G) and since every element of
1
L (G) is a product we see that ∆ maps L1 (G) into L1 (G) and so gives the required
extension.
If G is compact we can define ∆µ in the same way. Restricting to J0 (G), there
is a unique element ∆(µ) of J0 (G) with ∆µ (j) = j∆(µ) (j ∈ I0 (G) and the same
argument shows that ∆ is an extension of D to a derivation J0 (G) → J0 (G). Defining ∆(δe ) = 0 extends ∆ to a derivation from M (G) to J0 (G) and the remainder
of the proof is the same as in the non-compact case.
To show uniqueness, suppose that D is a derivation from L1 (G) into L1 (G) with
D = 0 on I0 (G). Then jD(a) = D(ja) − D(j)a = 0 (a ∈ L1 (G), j ∈ I0 (G)). As
D(a) ∈ I0 (G) this implies that D(a) = 0 whether or not G is compact.
References
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A NON-WEAKLY AMENABLE AUGMENTATION IDEAL
17
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London Math. Soc. (2) 26 (1982) 143–154.
Department of Mathematics, University of Newcastle, Newcastle upon Tyne, NE1
7RU, England
E-mail address: [email protected]