Open mappings of locally compact groups

To set the scene, we recall a few standard examples and counterexamples of such morphisms ϕ : G → H .

1. Let Z p be the compact additive group of 𝑝-adic integers and ℝ the additive group of real numbers. Let 𝐷 be the discrete subgroup { ( m , m ) : m ∈ Z } of G : = Z p × R . Finally, define H : = G / D , the compact 𝑝-adic solenoid T p (see [6, Example 1.28 and Exercise E1.11]), and let ϕ : G → H be the quotient morphism. Then 𝜙 is a surjective open morphism of locally compact groups mapping the identity component G 0 = { 0 } × R ≅ R onto the dense arc component properly contained in the compact connected group H = H 0 (see [6, Exercise E1.11 (iii) and (iv)]).

2. Let 𝐻 be any nondiscrete locally compact group and let 𝐺 be H d , the underlying group of 𝐻 with the discrete topology. Then the identity map ϕ : G → H is a nonopen bijective morphism.

3. Let 𝐺 be any locally compact noncompact abelian group, 𝐻 its Bohr compactification, and ϕ : G → H the natural standard morphism (see [6, Chapter 8, Part 7]). Then 𝜙 is a nonopen injective morphism with dense image properly contained in 𝐻. This is related to the previous example by Fourier duality.

Let ϕ : G → H be a morphism of locally compact groups. If ϕ ⁢ ( G 1 ) is open in 𝐻 for all G 1 in O ⁢ ( G ) , then 𝜙 is an open mapping.

If 𝜙 is open, then trivially ϕ ⁢ ( G 1 ) is open in 𝐻 for all open subgroups G 1 of 𝐺, and so the theorem may also be stated as an equivalence.

Proof

Take a symmetric open subset 𝑈 of 𝐺; then G 1 : = ⟨ U ⟩ is an open subgroup, and hence also closed, and G 1 is 𝜎-compact if 𝑈 is also relatively compact.

To show that ϕ : G → H is open, it is necessary and sufficient to show that the restricted surjective morphism ϕ | G 1 : G 1 → ϕ ⁢ ( G 1 ) is open. Indeed, if 𝜙 is open, then ϕ | G 1 is trivially open; conversely, every open set 𝑈 in 𝐺 is a union ⋃ x ∈ U x ⁢ U x , where U x is an open subset of G 1 ; if ϕ | G 1 is open, then ϕ ⁢ ( x ⁢ U x ) is open, and so ϕ ⁢ ( U ) is open.

The Open Mapping Theorem for locally compact groups [6, Exercise EA1.21] uses Baire category to show that, if ϕ : G → H is a surjective morphism of locally compact groups and 𝐺 is 𝜎-compact, then 𝜙 is open. We apply this result to ϕ | G 1 and the theorem is proved. ∎

Assume that ϕ : G → H is a morphism of locally compact groups and define γ : G / G 0 → H / H 0 as above. Then the following are equivalent:

1. 𝜙 is open;

2. γ : G / G 0 → H / H 0 is open and H 0 = ⋂ G 1 ∈ O ⁢ ( G ) ϕ ⁢ ( G 1 ) .

Proof

Assume that (1) holds. For all G 1 ∈ O ⁢ ( G ) , we have ϕ ⁢ ( G 1 ) ∈ O ⁢ ( H ) and hence H 0 ⩽ ϕ ⁢ ( G 1 ) . Further, if G 0 ⩽ G 1 ⩽ G and G 1 / G 0 ∈ O ⁢ ( G / G 0 ) , then we have G 1 ∈ O ⁢ ( G ) , and so

and γ ⁢ ( G 1 / G 0 ) is open in H / H 0 , that is, 𝛾 is open. Further, ϕ − 1 ⁢ ( H 1 ) ∈ O ⁢ ( G ) for all H 1 ∈ O ⁢ ( H ) , and so

It follows that (2) holds.

Next, assume that Γ is open and H 0 ⩽ ϕ ⁢ ( G 1 ) for all G 1 ∈ O ⁢ ( G ) . If G 1 ∈ O ⁢ ( G ) , then G 1 / G 0 ∈ O ⁢ ( G / G 0 ) , so γ ⁢ ( G 1 / G 0 ) ∈ O ⁢ ( H / H 0 ) . Since H 0 ⩽ ϕ ⁢ ( G 1 ) ,

whence ϕ ⁢ ( G 1 ) ∈ O ⁢ ( G ) . By Theorem 2, 𝜙 is open, as claimed. ∎

Let 𝐺 and 𝐻 be locally compact groups, and let ϕ : G → H be a continuous bijection of 𝐺 onto 𝐻. Assume that ϕ ⁢ ( G 1 ) ∈ O ⁢ ( H ) for all G 1 ∈ O ⁢ ( G ) . Then 𝜙 is an isomorphism of locally compact groups, that is, G ≅ H .

Proof

By Theorem 2, the bijective morphism 𝜙 is open and thus is an isomorphism. ∎

Let ϕ : G → H be a morphism of locally compact groups. If 𝜙 is open, then ϕ ⁢ ( U ) has nonzero measure in 𝐻 for all nonempty open subsets 𝑈 of 𝐺. If 𝜙 is not open, then ϕ ⁢ ( V ) has null measure in 𝐻 for all 𝜎-compact subsets 𝑉 of 𝐺.

Proof

If 𝑈 and 𝑉 are nonempty relatively compact (or 𝜎-compact) open sets in 𝐺, then 𝑉 may be covered by finitely (or countably) many translates of 𝑈 and vice versa, and ϕ ⁢ ( V ) may be covered by finitely (or countably) many translates of ϕ ⁢ ( U ) and vice versa. It follows that ϕ ⁢ ( U ) has null measure if and only if ϕ ⁢ ( V ) has null measure.

In light of this observation and the fact that the Haar measure of a nonempty open set is never 0, it suffices to prove that if 𝑈 is a nonempty relatively compact open set in 𝐺 and ϕ ⁢ ( U ) has positive Haar measure in 𝐻, then ϕ ⁢ ( U ) is open in 𝐻.

Given any z ∈ U , z − 1 ⁢ U is a relatively compact open neighbourhood of the identity in 𝐺, and by [3, p. 18], it is possible to find a relatively compact open neighbourhood 𝑉 of the identity in 𝐺 that is symmetric ( V − 1 = V ) such that V 2 ⊆ z − 1 ⁢ U . It will suffice to show that ϕ ⁢ ( V 2 ) contains an open neighbourhood 𝑊 of the identity in 𝐻, for then ϕ ⁢ ( z ) ⁢ W is an open neighbourhood of ϕ ⁢ ( z ) in 𝐻 that is contained in ϕ ⁢ ( z ⁢ V 2 ) , and hence in ϕ ⁢ ( U ) .

Since 𝑉 is relatively compact and 𝜙 is continuous, ϕ ⁢ ( V ) is relatively compact in 𝐻 and has finite Haar measure, which is not 0 by our initial observation, since the measure of ϕ ⁢ ( U ) is not 0. Then the convolution on 𝐻 of the characteristic function of ϕ ⁢ ( V ) with itself, given by the formula

is a continuous function on 𝐻 (see [3, (20.16)]) that does not vanish at the identity but vanishes outside ϕ ⁢ ( V 2 ) , and { y ∈ H : χ ϕ ⁢ ( V ) ∗ χ ϕ ⁢ ( V ) ⁢ ( y ) ≠ 0 } is the desired set 𝑊. ∎

Let ϕ : G → H be a morphism of locally compact groups. If ϕ ⁢ ( G 1 ) is open in 𝐻 for all compactly generated G 1 in O ⁢ ( G ) , then 𝜙 is an open mapping.

Proof

Assume that 𝜙 is not open, and take a nonempty relatively compact symmetric open subset 𝑈 of 𝐺. Then 𝑈 generates an open subgroup ⟨ U ⟩ of 𝐺, equal to ⋃ n ∈ N U n . Since U n is a nonempty relatively compact open subset of 𝐺 for all positive integers 𝑛 and 𝜙 is not open, each ϕ ⁢ ( U n ) has measure 0 by Lemma 6, whence ϕ ⁢ ( ⟨ U ⟩ ) has measure 0, and so ϕ ⁢ ( ⟨ U ⟩ ) cannot be open. ∎

Let ϕ : G → H be a morphism of locally compact groups, and suppose that there is a compact subset 𝐾 of 𝐺 such that ϕ ⁢ ( K ) has positive Haar measure in 𝐻. Then 𝜙 is open.

For profinite groups G , H , the following conditions are equivalent:

1. 𝜙 is open;

2. S ⁢ ( ϕ ) is open.

Proof

Proposition 5 in [2] shows that (1) implies (2).

For the converse, let 𝑂 be an open subgroup of 𝐺. In view of [1, Proposition 32], S ⁢ ( O ) is open in S ⁢ ( G ) and therefore S ⁢ ( ϕ ) ⁢ ( S ⁢ ( O ) ) is open in S ⁢ ( H ) . As S ⁢ ( ϕ ) ⁢ ( S ⁢ ( O ) ) = S ⁢ ( ϕ ⁢ ( O ) ) , also by [1, Proposition 32], ϕ ⁢ ( O ) is open in 𝐻, and so, by Theorem 2, 𝜙 is open. ∎