Completely bounded homomorphisms of the Fourier algebra revisited

Lemma 2.1 ([4, Lemma 1.2])

Let α : Y ⊆ H → G be a map. Then

if and only if 𝛼 is piecewise affine.

Let 𝑋 be a set and 𝒮 a collection of subsets of 𝑋. We say that 𝒮 is a relative ring of subsets when 𝒮 is closed under finite unions, intersections, and relative complements, in the sense that if A , B ∈ S , then A ∖ B ∈ S . This means that ∅ ∈ S , but perhaps 𝑋 is not in 𝒮.

Let 𝐺 be a group and 𝛼 a collection of subsets of 𝐺. Let R ⁢ ( α ) be the relative ring of subsets generated by 𝛼 and all left translations of elements of 𝛼. Let R 2 ⁢ ( α ) be the relative ring of subsets generated by 𝛼 and all two-sided translates of 𝛼, that is, sets of the form s ⁢ A ⁢ t , where A ∈ α , s , t ∈ G .

Let 𝐺 be a group, and let Y ∈ Ω ⁢ ( G ) . Then there are subgroups H 1 , H 2 , … , H n in R 2 ⁢ ( { Y } ) such that Y ∈ R ⁢ ( { H 1 , … , H n } ) .

Let 𝑋 be a set, let 𝔄 be a collection of subsets of 𝑋, and let 𝐵 be in the relative ring generated by 𝔄. Then 𝐵 is the finite (disjoint if we wish) union of sets of the form

for some n ≥ 1 , m ≥ 0 , and some A i , B j ∈ A .

Proof

Let 𝔅 be the relative ring generated by 𝔄, so B ∈ B . Let B ′ be the collection of finite unions of sets of the form (3.1) so that A ⊆ B ′ ⊆ B . We shall prove that B ′ is a relative ring of sets so that B ′ = B , as claimed.

By definition, B ′ is closed under unions. If ( P k ) k = 1 t and ( Q l ) l = 1 s are of the form (3.1), set P = ⋃ k P k and Q = ⋃ l Q l ; then

and clearly P k ∩ Q l is of the form (3.1). So B ′ is closed under intersections. Furthermore,

Thus, to show that P ∖ Q ∈ B ′ , it suffices to show that, say P ∩ ( X ∖ Q 1 ) ∈ B ′ . Let Q 1 = ⋂ i = 1 n A i ∩ ⋂ j = 1 m ( X ∖ B j ) so that

As each P ∩ B j ∈ B ′ and B ′ is closed under unions, it remains to show that, say P ∖ A 1 ∈ B ′ , but P ∖ A 1 = ⋃ k = 1 t P k ∖ A 1 , so in fact, it remains to show that, say P 1 ∖ A 1 ∈ B ′ . If P 1 = ⋂ i = 1 n ′ C i ∩ ⋂ j = 1 m ′ ( X ∖ D j ) , then

and so indeed P 1 ∖ A 1 ∈ B ′ .

Having shown that B = B ′ , it is easy that each member of 𝔅 is a disjoint union of sets of the form (1) because, for any sets ( A i ) , we have that

Let 𝐺 be a group, and let H 1 , … , H n be subgroups. There are subgroups K 1 , … , K n ′ , each a subgroup of some H i , and with [ K i : K i ∩ K j ] = 1 or ∞ for all i , j , and with each H i a finite union of cosets of the K j . If the family of subgroups { H i } is closed under taking intersections, then the family { K i } is a subfamily of { H i } .

Proof

If, for some 𝑖, [ H 1 : H 1 ∩ H i ] is not 1 or ∞, then L = H 1 ∩ H i is a subgroup of H 1 of finite index, and so H 1 can be covered by finitely many translates of 𝐿. We can hence replace H 1 by 𝐿; notice that [ L : L ∩ H i ] = 1 . We claim further that, by replacing H 1 by 𝐿, if previously [ H 1 : H 1 ∩ H j ] ∈ { 1 , ∞ } , then we do not change this property.

Indeed, if [ H 1 : H 1 ∩ H j ] = 1 , then H 1 = H 1 ∩ H j , and so also L ⊆ H j , so [ L : L ∩ H j ] = 1 . If [ H 1 : H 1 ∩ H j ] = ∞ , then let K = H 1 ∩ H j so that L ∩ H j = H 1 ∩ H i ∩ H j = L ∩ K , and hence we have L ∩ K ≤ L ≤ H 1 . Towards a contradiction, suppose that [ L : L ∩ H j ] < ∞ , so [ L : L ∩ K ] < ∞ . As 𝐿 is of finite index in H 1 , we can find t i ∈ H 1 with H 1 = ⋃ i = 1 m t i ⁢ L , and we can find r j in 𝐿 with L = ⋃ j = 1 m ′ r j ⁢ ( L ∩ K ) . Thus

and so certainly H 1 = ⋃ i , j t i ⁢ r j ⁢ K , so [ H 1 : H 1 ∩ H j ] ≤ m m ′ < ∞ , a contradiction.

By performing this argument finitely many times, we may suppose that, for each 𝑖, [ H 1 : H 1 ∩ H i ] ∈ { 1 , ∞ } . We now look at H 2 and apply the same argument, and so forth. This process could lead to repeats, and so possibly n ′ ≤ n .

For the final remark, note that, by construction, each K i is an intersection of some of the H j , and thus { K i } is a subfamily of the { H j } . ∎

Let 𝐺 be a group. 𝐺 cannot be written as the finite union of cosets of infinite index.

Proof

Towards a contradiction, suppose that 𝐺 is the finite union of cosets of subgroups H 1 , … , H k , where [ G : H i ] = ∞ for each 𝑖. By Lemma 3.4, there are subgroups of these, say K 1 , … , K k ′ with [ K i : K i ∩ K j ] = 1 or ∞ for all i , j , and still with each K i of infinite index in 𝐺. As each H j is the finite union of cosets of some K i , it follows that 𝐺 can be covered by finitely many cosets of the K i , say C 1 , … , C n .

As K 1 is of infinite index in 𝐺, there is some coset of K 1 which is not a member of our covering, say K = s ⁢ K 1 . Then K ∩ K i is a coset of K 1 ∩ K i or is empty for each 𝑖. However, as 𝐺 is covered by the C i , also 𝐾 is covered by { K ∩ C i } . If C i is a coset of H 1 , then K ∩ C i = ∅ , so we conclude that 𝐾 is covered by finitely many cosets of the subgroups K 2 , K 3 , … , K k ′ . By translating by s - 1 , also K 1 is covered by finitely many cosets of the subgroups K 2 , K 3 , … , K k ′ .

We now complete the argument by using induction on the number of subgroups, k ′ . If we have only one subgroup, that is, k ′ = 1 , the result is trivially true. The previous paragraph then provides the induction step. ∎

Let 𝐺 be a group. Every member of Ω ⁢ ( G ) is either empty or is a finite disjoint union of sets of the form

where each E i is a coset, E i ⊆ E 0 , and E i has infinite index in E 0 for each i > 0 .

Proof

Let Y ∈ Ω ⁢ ( G ) , so in particular, 𝑌 is in the ring generated by some subgroups H 1 , … , H n and their translates. By adding in intersections, we may suppose that the finite family { H i } is closed under intersections. By Lemma 3.4, we may suppose that [ H i : H i ∩ H j ] = 1 or ∞ for each i , j . By using this lemma, the resulting family { H i } may no longer be closed under intersections, but if 𝐻 is an intersection of some of the H i , then 𝐻 is at least a finite union of cosets of some H j .

By Lemma 3.3, 𝑌 is the disjoint union of say 𝑛 sets of the form E 0 ∖ ⋃ i = 1 m E i , where E 0 is a coset of some finite intersection of the H j , and each E i is a coset of some H j . Hence E 0 is a finite union of cosets of the H j , and so (by maybe increasing 𝑛) we may suppose that E 0 actually is a coset of some H j . As before, by replacing E i by E 0 ∩ E i , we may suppose that E i is a subcoset of E 0 . As [ H i : H i ∩ H j ] = ∞ unless H i ⊆ H j , if E i has finite index in E 0 , then E i and E 0 must be cosets of the same subgroup, and as E i ⊆ E 0 , we must have E 0 = E i , a case which may be ignored. ∎

Let 𝐺 be a group, and let H 1 , … , H n be subgroups of infinite index in 𝐺. Let ( K i ) i = 1 N be cosets of the H j , and let C = ⋃ i K i . There is 𝑚 and t 1 , … , t m ∈ G with ⋂ i t i ⁢ C = ∅ .

Proof

Notice that

where the union is over all functions j : { 1 , … , m } → { 1 , … , N } . Thus we need to find t i so that ⋂ i = 1 m t i ⁢ K j ⁢ ( i ) = ∅ for any such function 𝑗. In what follows, suppose that K j is the coset s j ⁢ H k ⁢ ( j ) for each 𝑗.

Set t 1 = e , the identity. We claim that there is t 2 with K j ∩ t 2 ⁢ K j = ∅ for all 𝑗. Indeed, if not, then for each t 2 , there is 𝑗 with s j ⁢ H k ⁢ ( j ) = t 2 ⁢ s j ⁢ H k ⁢ ( j ) (as cosets are either equal or disjoint). Equivalently, for each t 2 , there is 𝑗 with s j - 1 ⁢ t 2 ⁢ s j ∈ H k ⁢ ( j ) , so t 2 ∈ s j ⁢ H k ⁢ ( j ) ⁢ s j - 1 . As each subgroup s j ⁢ H k ⁢ ( j ) ⁢ s j - 1 is of infinite index in 𝐺, this shows that 𝐺 is covered by a finite union of subgroups of infinite index, a contradiction.

Suppose we have chosen t 1 , … , t p with the property that t 1 ⁢ K j ⁢ ( 1 ) ∩ ⋯ ∩ t p ⁢ K j ⁢ ( p ) is only (possibly) non-empty when the j ⁢ ( i ) are all distinct. We have already shown this is true for p = 2 . To show that the claim holds for p + 1 , it suffices to find t p + 1 with t i ⁢ K j ∩ t p + 1 ⁢ K j = ∅ for all i ≤ p and all 𝑗. This is sufficient, for if we have j ⁢ ( 1 ) , … , j ⁢ ( p + 1 ) not all distinct, say j ⁢ ( k ) = j ⁢ ( k ′ ) for k < k ′ , then if k ′ ≤ p , we know already that ⋂ i = 1 p t i ⁢ K j ⁢ ( i ) = ∅ , while if k ′ = p + 1 , then by construction, t k ⁢ K j ⁢ ( k ) ∩ t p + 1 ⁢ K j ⁢ ( k ) = ∅ , so certainly ⋂ i = 1 p + 1 t i ⁢ K j ⁢ ( i ) = ∅ .

To find t p + 1 , we again proceed by contradiction and suppose that, for each t ∈ G , there are some i , j with t i ⁢ K j = t ⁢ K j so that t i ⁢ s j ⁢ H k ⁢ ( j ) = t ⁢ s j ⁢ H k ⁢ ( j ) , so s j - 1 ⁢ t i - 1 ⁢ t ⁢ s j ∈ H k ⁢ ( j ) . That is, t ∈ ⋃ i , j t i ⁢ s j ⁢ H k ⁢ ( j ) ⁢ s j - 1 which is again a finite union of cosets of infinite index, which cannot cover 𝐺.

So, by induction, our claim holds for all 𝑝. Thus ⋂ i = 1 p t i ⁢ K j ⁢ ( i ) can only possibly be non-empty when the j ⁢ ( i ) are all distinct, but there are only 𝑁 many choices, and so if p > N , we have shown our claim. ∎

Let 𝐺 be a group, 𝐻 a subgroup, and H 1 , … , H n subgroups with [ H : H ∩ H i ] = ∞ for each 𝑖. Let 𝐶 be a union of finitely many cosets of the H i . There are 𝑚 and t 1 , … , t m ∈ H with ⋂ i C ⁢ t i = ∅ .

Proof

The proof is similar to the previous one; but note here we need to choose t i in 𝐻 not 𝐺. Let our cosets be C j for j = 1 , … , N . Again, we need to find t i with C j ⁢ ( 1 ) ⁢ t 1 ∩ ⋯ ∩ C j ⁢ ( m ) ⁢ t m = ∅ for any choices j ⁢ ( i ) . Set t 1 = e .

Suppose we have t 1 , … , t p so that if j ⁢ ( 1 ) , … , j ⁢ ( p ) are not distinct, then

Proceeding by induction, t p + 1 ∈ H needs to satisfy that C j ⁢ t i ∩ C j ⁢ t p + 1 = ∅ for all i , j . If no such t p + 1 exists, then for all t ∈ H , there are some i , j with

If C j = s ⁢ H k , say, then s ⁢ H k ⁢ t i ∩ s ⁢ H k ⁢ t ≠ ∅ , so there are a , b ∈ H k with s ⁢ a ⁢ t i = s ⁢ b ⁢ t , so a ⁢ t i = b ⁢ t , so t ⁢ t i - 1 = b - 1 ⁢ a ∈ H k , so t ∈ H k ⁢ t i . Thus H ⊆ ⋃ i , k H k ⁢ t i , but as each t i ∈ H , we have H ∩ H k ⁢ t i = ( H ∩ H k ) ⁢ t i , and so H = ⋃ i , k ( H ∩ H k ) ⁢ t i . As each H ∩ H k is of infinite index in 𝐻, this is a contradiction. ∎

With notation as above, if H i is not contained in any other H j , then there is a subgroup 𝐻 which contains H i as a finite-index subgroup (possibly H = H i ) such that H ∈ R 2 ⁢ ( { Y } ) and such that Y ∈ R ⁢ ( { H } ∪ { H j : j ≠ i } ) .

Proof of Theorem 3.2

We can form a directed acyclic graph (DAG, see [6] for example) with vertices the subgroups H i , and with a directed edge from H i to H j when H i ⊇ H j and there is no other H k with H i ⊇ H k ⊇ H j . We shall say that H i is top level if H i is not contained in any other H j , that is, there is no edge into H i . The depth of a DAG is the length of the longest directed path in the DAG.

For each top level H i , let 𝐻 be given by Proposition 3.9. For any i ≠ j as [ H i : H i ∩ H j ] = 1 or ∞, also [ H : H ∩ H j ] = 1 or ∞; see Lemma 3.15 below. Also, if H i ∩ H j is the finite union of cosets of H k , then so is H ∩ H j ; see Lemma 3.15 below. Hence we may replace H i by 𝐻 and not change any of our assumptions. Do this for all top level H i so that Y ∈ R ⁢ ( { H i } ) and each top-level H i is in R 2 ⁢ ( { Y } ) .

We shall give a proof by induction on the depth of the DAG. Notice that Proposition 3.9, and the previous paragraph, shows that the result is true when the DAG has depth 0 (that is, all subgroups are top level).

Let 𝐻 be some top level H i . Let Y = ⨆ i L i as before (see (3.3)), and reorder these so that E 0 ( i ) is a coset of 𝐻 for i ≤ n 0 , and not for i > n 0 . Define

From the form that each L i is written in, namely that E j ( i ) is a coset of infinite index in E 0 ( i ) , it is clear that E j ( i ) is not a coset of 𝐻 for j ≥ 1 . From this, it follows that Y 0 ∈ R ⁢ ( { H ∩ H i } ∖ { H } ) . For each 𝑖, either H ∩ H i = H which we remove, or H ∩ H i is a union of cosets of some H k , where necessarily H k ⊆ H . Thus actually Y 0 ∈ R ⁢ ( { H k : H k ⊊ H } ) .

Now the family { H k : H k ⊊ H } forms a subgraph of our DAG; in fact, it is DAG “underneath” 𝐻 (all the vertices which have a path leading to them from 𝐻). Thus it is of smaller depth, so by induction, there are subgroups H j ′ in R 2 ⁢ ( { Y 0 } ) with Y 0 ∈ R ⁢ ( { H j ′ } ) . Notice that, as Y , H ∈ R 2 ⁢ ( { Y } ) , also Y 0 ∈ R 2 ⁢ ( { Y } ) , and so also each H j ′ ∈ R 2 ⁢ ( { Y } ) .

Next, reorder so that H i is top level for i ≤ n 0 , and not for i > n 0 , so if i > n 0 , the subgroup H i is contained in some H j with j ≤ n 0 . For i ≤ n 0 , let A i be the union of the sets E 0 ( k ) which are cosets of H i . Set B i = A i ∖ Y , and set C = Y ∖ ⋃ i ≤ n 0 A i . We have shown that, for each 𝑖, there are subgroups H j ′ in R ⁢ ( { Y } ) with B i ∈ R ⁢ ( { H j ′ } ) .

We now consider 𝐶. As each E 0 ( i ) is a coset of some H j , either E 0 ( i ) ⊆ A j for some 𝑗, or otherwise E 0 ( i ) is a coset of some H j which is not top level. Since each E j ( i ) is a subcoset of E 0 ( i ) , we see that 𝐶 is contained in

and so C ∈ R ⁢ ( { H i : H i ⁢ not top level } ) . The DAG given by removing all top level subgroups has smaller depth, and so again by induction, there are subgroups H j ′′ in R 2 ⁢ ( { C } ) with C ∈ R ⁢ ( { H j ′′ } ) . Notice that

and so each H j ′′ ∈ R 2 ⁢ ( { Y } ) .

Let 𝛼 be the collection of all the H j ′ , the H j ′′ , and the top level H i , so each subgroup in 𝛼 is in R 2 ⁢ ( { Y } ) . Then A i , B i , C ∈ R ⁢ ( α ) . As A i ∖ B i = A i ∩ Y , we see that

Thus also Y ∈ R ⁢ ( α ) , which completes the proof. ∎

Given the subgroups ( H i ) , and A ⊆ G some subset, we shall say that a coset s ⁢ H j is big in 𝐴 if A ∩ s ⁢ H j cannot be covered by finitely many cosets of subgroups in { H i : i ≠ j } .

Let ( H i ) be subgroups as above.

1. If A ⊆ B and s ⁢ H j is big in 𝐴, then it is big in 𝐵.

2. With A = ⋃ i = 1 n A i , we have that s ⁢ H j is big in 𝐴 if and only if it is big in some A i .

Proof

1 is clear. For 2, if s ⁢ H j is not big in any A i , so s ⁢ H j ∩ A i can be covered, and hence so can 𝐴 (as 𝐴 is a finite union); the converse follows as A i ⊆ A for each 𝑖. ∎

With Y = ⨆ i L i as before, we have that s ⁢ H 1 is big in 𝑌 if and only if some L i is of the form E 0 ∖ ⋃ j E j with E 0 = s ⁢ H 1 . Furthermore, in this case, the choice of 𝑖 is unique.

Proof

If L i = s ⁢ H 1 ∖ ⋃ E j and yet s ⁢ H 1 is not big in L i , then s ⁢ H 1 ∩ L i = L i can be covered by finitely many cosets of infinite index in H 1 . Union these cosets with the { E j : j > 0 } and we have covered all of s ⁢ H 1 by finitely many cosets of infinite index in H 1 , a contradiction. So s ⁢ H 1 is big in L i and hence big in 𝑌, by Lemma 3.11 1.

If s ⁢ H 1 is big in 𝑌, then by Lemma 3.11 2, we have that s ⁢ H 1 is big in some L i . If L i = E 0 ∖ ⋃ j E j , then towards a contradiction, suppose that E 0 is not s ⁢ H 1 . If E 0 is some other coset of H 1 , then s ⁢ H 1 ∩ E 0 = ∅ , so certainly s ⁢ H 1 is not big in L i . So E 0 is a coset of some other subgroup H j , and so s ⁢ H 1 ∩ H j is either empty (again, not possible) or is a coset of H 1 ∩ H j which has infinite index in H 1 . Then s ⁢ H 1 ∩ L i is contained in a coset of infinite index in s ⁢ H 1 and so is covered, so s ⁢ H 1 is not big in L i , a contradiction.

To show uniqueness, let each L i have the form (3.3). If E 0 ( i ) = E 0 ( j ) = s ⁢ H 1 , then if i ≠ j , then as L i and L j are disjoint, we must have that

is all of s ⁢ H 1 , a contradiction as these are cosets of infinite index. ∎

B = R ⁢ ( { C } ) .

Proof

To get a handle on ℬ, we need some information about R ⁢ ( { Y } ) . By Lemma 3.3, every A ∈ R ⁢ ( { Y } ) is of the form

where n ≥ 1 and s i , t j ∈ G . (This follows as if 𝔄 is left-invariant, so will be R ⁢ ( A ) .) We wish to know when s ⁢ H 1 is big in 𝐴, in terms of the form Y = ⨆ L i as in equation (3.3).

Let us think about these two parts individually. Consider

As argued in the “uniqueness” part of Lemma 3.12, either E 0 ( i ) = t ⁢ E 0 ( j ) = s ⁢ H 1 , or E 0 ( i ) ∩ t ⁢ E 0 ( j ) ∩ s ⁢ H 1 is either empty or is a coset of infinite index in s ⁢ H 1 . It then follows from Lemma 3.12 that s ⁢ H 1 is big in Y ∩ t ⁢ Y if and only if there are (unique) i , j with E 0 ( i ) = t ⁢ E 0 ( j ) = s ⁢ H 1 . A similar argument now shows that s ⁢ H 1 is big in ⋂ s i ⁢ Y if and only if, for each 𝑖, there is a (necessarily unique) 𝑗 with s i ⁢ E 0 ( j ) = s ⁢ H 1 .

By Lemma 3.11 2, we see that s ⁢ H 1 is big in ⋃ t j ⁢ Y if and only if there is some 𝑗 with s ⁢ H 1 big in t j ⁢ Y , if and only if, by Lemma 3.12, there is 𝑗 and a (necessarily unique) 𝑘 with s ⁢ H 1 = t j ⁢ E 0 ( k ) .

Let B 0 = ⋂ s i ⁢ Y and B 1 = ⋃ t j ⁢ Y , so A = B 0 ∖ B 1 .

• If s ⁢ H 1 is not big in B 0 , then it is not big in 𝐴 by Lemma 3.11 1.

• If s ⁢ H 1 is big in B 0 and not big in B 1 , then as B 0 = A ∪ B 1 , also s ⁢ H 1 is big in 𝐴 by Lemma 3.11 2.

• If s ⁢ H 1 is big in both B 0 and B 1 , then from above, we know that s ⁢ H 1 ∩ B 0 is equal to s ⁢ H 1 with a finite union of cosets of infinite index removed, while s ⁢ H 1 ∩ B 1 contains a set of the form s ⁢ H 1 with a finite union of cosets of infinite index removed. Thus s ⁢ H 1 ∩ ( B 0 ∖ B 1 ) is contained in a finite union of cosets of infinite index, so s ⁢ H 1 is not big in A = B 0 ∖ B 1 .