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Proposition 7.1.22. A Hausdorff group \( G \) is complete if and only if for every embedding \( j : G \hookrightarrow H \) into a Hausdorff group \( H \) the subgroup \( j\left( G\right) \) of \( H \) is closed. | Proof. Assume that there exists an embedding \( j : G \hookrightarrow H \) into a Hausdorff group \( H \) such that \( j\left( G\right) \) is not a closed subgroup of \( H \) . Then there exists a net \( {\left\{ {y}_{\alpha }\right\} }_{\alpha \in A} \) in \( j\left( G\right) \) converging to some element \( h \in H \... | Yes |
Lemma 7.1.26. Let \( G \) be a Hausdorff group. Every left (respectively, right) Cauchy net of \( G \) with a convergent subnet is convergent. | Proof. Let \( {\left\{ {g}_{\alpha }\right\} }_{\alpha \in A} \) be a left Cauchy net of \( G \) and let \( {\left\{ {g}_{\beta }\right\} }_{\beta \in B} \) be a subnet convergent to \( x \in G \), where \( B \) is a cofinal subset of \( A \) . Let \( U \) be a neighborhood of \( {e}_{G} \) in \( G \) and \( V \) a sym... | Yes |
Lemma 7.2.2. Let \( G \) be a Hausdorff group.\n\n(a) For a filter \( \mathcal{F} \) on \( G \), the following conditions are equivalent:\n\n\( \left( {\mathrm{a}}_{1}\right) \) the filter \( \mathcal{F} \) is Cauchy;\n\n\( \left( {\mathrm{a}}_{2}\right) \) the filter \( {\mathcal{F}}^{-1} \mathrel{\text{:=}} \left\{ {... | Proof. The verification of (a)-(c) is a straightforward application of the definitions.\n\n(d) Pick \( U \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) and \( V \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) with \( {VV} \subseteq U \) . According to \( \left( {\mathrm{a}}_{4}\right) \), there exist \( H \in \mathcal{H} ... | Yes |
Proposition 7.2.6. Let \( G \) be a Hausdorff group and \( \mathcal{F} \) a Cauchy filter on \( G \) . The following conditions are equivalent:\n\n(a) \( \mathcal{F} \) is minimal;\n\n(b) if \( {\mathcal{F}}_{1} \) is a Cauchy filter on \( G \) such that \( {\mathcal{F}}_{1} \subseteq \mathcal{F} \), then \( {\mathcal{... | Proof. (a) \( \Rightarrow \) (b) Let \( F \in \mathcal{F} \) . By the minimality of \( \mathcal{F} \), there exist \( {F}^{\prime } \in \mathcal{F} \) and \( U \in \mathcal{V}\left( {e}_{G}\right) \) such that \( U{F}^{\prime }U \subseteq F \) . Moreover, there exists \( {F}_{1} \in {\mathcal{F}}_{1} \) such that \( {F... | Yes |
Lemma 7.2.7. Let \( G \) be a topological subgroup of a Hausdorff group \( H \) . Let \( \mathcal{F} \) be a Cauchy filter on \( H \) such that the restriction \( \mathcal{F}{ \upharpoonright }_{G} \mathrel{\text{:=}} \{ G \cap U : U \in \mathcal{F}\} \) is a filter base on \( G \) . Then \( \mathcal{F}{ \upharpoonrigh... | Proof. To verify that \( \mathcal{F}{ \upharpoonright }_{G} \) is Cauchy, let \( V \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) ; so \( V = U \cap G \) for some \( U \in {\mathcal{V}}_{H}\left( {e}_{H}\right) \) . Let \( B \in \mathcal{F} \) such that \( B{B}^{-1} \cup {B}^{-1}B \subseteq U \) . Then \( C = B \cap G \... | Yes |
Lemma 7.2.8. Let \( G \) be a topological group. If \( \mathcal{F},\mathcal{H} \) are minimal Cauchy filters on \( G \), then also \( \mathcal{F} \cdot \mathcal{H} \) and \( {\mathcal{F}}^{-1} \) are minimal Cauchy filters on \( G \) . | Proof. We already noticed that \( \mathcal{F} \cdot \mathcal{H} \) and \( {\mathcal{F}}^{-1} \) are Cauchy filters in Lemma 7.2.2. The proof that they are minimal is straightforward. | No |
Lemma 7.2.9. If \( G \) is a topological subgroup of a Hausdorff group \( H \) and \( h \in {\bar{G}}^{H} \), then \( \mathcal{F} = {\mathcal{V}}_{H}\left( h\right) { \upharpoonright }_{G} \) is a minimal Cauchy filter on \( G \) . Consequently, a Hausdorff group \( G \) is complete if and only if every minimal Cauchy ... | Proof. The first assertion follows from Lemma 7.2.7 applied to \( \mathcal{F} = {\mathcal{V}}_{H}\left( h\right) \) . To prove the second assertion, in view of Proposition 7.2.3 we only need to check that if every minimal Cauchy filter on \( G \) converges then \( G \) is complete. To this end, argue by contradiction a... | Yes |
Proposition 7.2.11. Let \( G \) be a Hausdorff linearly topologized group and let \( \left\{ {{N}_{i} : i \in I}\right\} \) be a base of \( {\mathcal{V}}_{G}\left( {e}_{G}\right) \) consisting of open normal subgroups of \( G \) . Then the completion \( \widetilde{G} \) of \( G \) is isomorphic to the inverse limit \( ... | Proof. Since \( \mathop{\bigcap }\limits_{{i \in I}}{N}_{i} \) is trivial, there is a natural embedding of \( G \) in the product \( P = \mathop{\prod }\limits_{{i \in I}}G/{N}_{i} \) of the discrete quotients \( G/{N}_{i} \) . Clearly, \( P \) is complete, by Proposition 7.1.9. Hence, the closure \( {\bar{G}}^{P} \) i... | Yes |
Lemma 7.2.13. Closed subgroups and continuous homomorphic images (provided they are Hausdorff and linearly topologized) of a linearly compact abelian group \( G \) are linearly compact. | Proof. Obviously, closed subgroups of linearly compact groups are linearly compact.\n\nAssume that \( H \) is a Hausdorff linearly topologized abelian group and that \( f : G \rightarrow \) \( H \) is a continuous surjective homomorphism. If \( \mathcal{F} \) is a filter base of closed cosets of subgroups of \( H \), t... | Yes |
Theorem 7.2.14. A linearly compact abelian group \( G \) is complete. | Proof. To see that \( G \) is complete, it suffices to check that every minimal Cauchy filter \( \mathcal{F} \) on \( G \) converges. From the definition of minimal Cauchy filter and the fact that \( G \) is linearly topologized, we deduce that \( \mathcal{F} \) has a base consisting of cosets of open subgroups. Now th... | Yes |
Every power \( {\mathbb{T}}^{I} \) of \( \mathbb{T} \), as well as every closed subgroup of \( {\mathbb{T}}^{I} \), is compact. | It becomes clear in the sequel that this is the most general instance of a compact abelian group: every compact abelian group is topologically isomorphic to a closed subgroup of a power of \( \mathbb{T} \) (see Corollary 11.5.2). | No |
For every abelian group \( G \), the group \( {G}^{ * } = \operatorname{Hom}\left( {G,\mathbb{T}}\right) \) of all characters of \( G \) is closed in the product \( {\mathbb{T}}^{G} \). | In fact, considering the projections \( {\pi }_{x} : {\mathbb{T}}^{G} \rightarrow \mathbb{T} \) for every \( x \in G \) ,\n\n\[ \n{G}^{ * } = \mathop{\bigcap }\limits_{{h, g \in G}}\left\{ {f \in {\mathbb{T}}^{G} : f\left( {h + g}\right) = f\left( h\right) + f\left( g\right) }\right\} \n\]\n\n\[ \n= \mathop{\bigcap }\l... | Yes |
Lemma 8.1.3. Let \( G \) be an abelian group and \( N = {\left\{ {\chi }_{\alpha }\right\} }_{\alpha \in A} \) a net in \( {G}^{ * } \) . Then there exist \( \chi \in {G}^{ * } \) and a subnet \( S = {\left\{ {\chi }_{{\alpha }_{\beta }}\right\} }_{\beta \in B} \) of \( N \) such that \( {\chi }_{{\alpha }_{\beta }}\le... | Proof. By Example 8.1.1, the group \( {\mathbb{T}}^{G} \) endowed with the product topology is compact. Since \( {G}^{ * } \) is a topological subgroup of \( {\mathbb{T}}^{G} \), there exist \( \chi \in {\mathbb{T}}^{G} \) and a subnet \( S = {\left\{ {\chi }_{{\alpha }_{\beta }}\right\} }_{\beta \in B} \) of \( N \) t... | Yes |
For \( n \in {\mathbb{N}}_{ + } \) the set \( U\left( n\right) \) of all \( n \times n \) unitary matrices over \( \mathbb{C} \) (a matrix is unitary if its inverse coincides with its conjugate transposed) is a subgroup of \( {\mathrm{{GL}}}_{n}\left( \mathbb{C}\right) \). | As a subset of \( {\mathbb{C}}^{{n}^{2}}, U\left( n\right) \) is closed and bounded. So, \( U\left( n\right) \) is compact, by Example B.5.6. It is easy to see that \( U\left( 1\right) \cong \mathbb{S} \). | No |
Example 8.1.5. (a) Obviously, every discrete group is locally compact. Vice versa, every countable locally compact group \( G \) is discrete; indeed, \( G \) is of second category by Theorem B.5.20, and since \( G \) is countable, \( G \) is discrete by Corollary 2.1.7. | Vice versa, every countable locally compact group \( G \) is discrete; indeed, \( G \) is of second category by Theorem B.5.20, and since \( G \) is countable, \( G \) is discrete by Corollary 2.1.7. | Yes |
The Hilbert space \( \left( {{\ell }_{2},\parallel - \parallel }\right) \) of square summable real sequences is not locally compact. | Indeed, the closed unit disk is not compact: since \( {\ell }_{2} \) is metrizable, it is enough to observe that the sequence \( {\left\{ {e}_{n}\right\} }_{n \in \mathbb{N}} \) of the vectors of the canonical base has no Cauchy subsequences (so no convergent subsequences), as \( \begin{Vmatrix}{{e}_{n} - {e}_{m}}\end{... | Yes |
Lemma 8.2.1. Let \( G \) be a topological group and \( C, K \) closed sets of \( G \). (a) If \( K \) is compact, then both \( {CK} \) and \( {KC} \) are closed. | Proof. (a) Let \( {\left\{ {x}_{\alpha }\right\} }_{\alpha \in A} \) be a net in \( {CK} \) such that \( {x}_{\alpha } \rightarrow {x}_{0} \in G \) . One has to show that \( {x}_{0} \in {CK} \) . For every \( \alpha \in A \), there exist \( {y}_{\alpha } \in C \) and \( {z}_{\alpha } \in K \) such that \( {x}_{\alpha }... | Yes |
Lemma 8.2.2. Let \( G \) be a topological group and \( K \) a compact normal subgroup of \( G \) . Then the canonical projection \( q : G \rightarrow G/K \) is closed. | Proof. Let \( C \) be a closed set of \( G \) . As \( {q}^{-1}\left( {q\left( C\right) }\right) = {CK} \) is closed by Lemma 8.2.1(a), we may conclude that \( q\left( C\right) \) is closed. | Yes |
Lemma 8.2.3. Let \( H \) be a closed normal subgroup of a topological group \( G \). (a) If \( G \) is compact, then \( G/H \) is compact. | Proof. (a) is obvious, since the canonical projection \( q : G \rightarrow G/H \) is continuous. | No |
Lemma 8.2.5. Let \( G \) be a locally compact group, \( H \) a closed normal subgroup of \( G \), and \( q : G \rightarrow G/H \) the canonical projection. Then:\n\n(a) \( G/H \) is locally compact, too;\n\n(b) if \( C \subseteq G/H \) is compact, there exists \( K \subseteq G \) compact with \( q\left( K\right) = C \)... | Proof. Let \( U \) be an open neighborhood of \( {e}_{G} \) in \( G \) with compact closure.\n\n(a) Consider the open neighborhood \( q\left( U\right) \) of \( {e}_{G/H} \) in \( G/H \) . By the continuity of \( q \) , \( q\left( \overline{U}\right) \subseteq \overline{q\left( U\right) } \) and \( q\left( \overline{U}\... | Yes |
Proposition 8.2.6. A locally compact group \( G \) is Weil complete. | Proof. Let \( U \) be a neighborhood of \( {e}_{G} \) in \( G \) with compact closure and let \( {\left\{ {g}_{\alpha }\right\} }_{\alpha \in A} \) be a left Cauchy net of \( G \) . Then there exists \( {\alpha }_{0} \in A \) such that \( {g}_{\alpha }^{-1}{g}_{\beta } \in U \) for every \( \alpha ,\beta \geq {\alpha }... | Yes |
Proposition 8.2.7. The character and the pseudocharacter of a locally compact group G coincide. | Proof. Clearly, \( \psi \left( G\right) \leq \chi \left( G\right) \) . Let \( U \) be an open neighborhood of \( {e}_{G} \) such that \( \bar{U} \) is compact. To prove that \( \chi \left( G\right) \leq \) \( \psi \left( G\right) \), pick a family \( \mathcal{B} = \left\{ {{V}_{i} : i \in I}\right\} \) of neighborhoods... | Yes |
Any infinite compact group \( G \) satisfies \( \psi \left( G\right) = \chi \left( G\right) = w\left( G\right) \) . | Proof. We are going to prove that \( d\left( G\right) \leq \chi \left( G\right) \) . Let \( \mathcal{B} \) be a local base at \( {e}_{G} \) of cardinality \( \chi \left( G\right) \) consisting of symmetric sets. For every \( U \in \mathcal{B} \), choose a finite subset \( {F}_{U} \) of \( G \) such that \( G = {F}_{U}U... | Yes |
Lemma 8.3.2. If \( G \) is a compactly generated group, then \( G \) is \( \sigma \) -compact. | Proof. There exists a compact subset \( K \) of \( G \) such that \( G = \mathop{\bigcup }\limits_{{n \in {\mathbb{N}}_{ + }}}{\left( K \cup {K}^{-1}\right) }^{n} \) . Since \( K \) is compact, \( {\left( K \cup {K}^{-1}\right) }^{n} \) is compact for every \( n \in {\mathbb{N}}_{ + } \) . | Yes |
Corollary 8.3.3. A locally compact group is a normal space. | Proof. By hypothesis and Lemma 8.3.2, \( G \) contains a \( \sigma \) -compact open subgroup \( N \) . Then \( N \) is Lindelöff by Lemma B.5.18, so a normal space by Theorem B.5.10(b). Now \( G = \mathop{\bigsqcup }\limits_{{g \in G}}{gN} \) is a normal space as well. | Yes |
Lemma 8.3.4. Let \( G \) be a locally compact group.\n\n(a) If \( K \) is a compact subset of \( G \) and \( U \) is an open set of \( G \) such that \( K \subseteq U \), then there \( \\textit{exists an open neighborhood}\\;V\\;{\\text{ofe}}_{G}\\; \) in \( \\;G\\; \) such that \( \\;\\left( {KV}\\right) \\cup \\left(... | Proof. (a) By Lemma 8.2.1(c), there exists an open neighborhood \( V \) of \( {e}_{G} \) in \( G \) such that \( \\left( {KV}\\right) \\cup \\left( {VK}\\right) \\subseteq U \) . Since \( G \) is locally compact, one can choose \( V \) with compact closure. Thus, \( K\\bar{V} \) is compact by Lemma 8.2.1(b). Since \( {... | Yes |
Theorem 8.4.1 (Open mapping theorem). Let \( G, H \) be locally compact groups and \( f : G \rightarrow H \) a continuous homomorphism. If \( G \) is \( \sigma \) -compact and \( f \) is surjective, then \( f \) is open. | Proof. Let \( U \) be a neighborhood of \( {e}_{G} \) in \( G \) . There exists a symmetric open neighborhood \( V \) of \( {e}_{G} \) in \( G \) such that \( \bar{V}\bar{V} \subseteq U \) and \( \bar{V} \) is compact. Since \( G = \mathop{\bigcup }\limits_{{x \in G}}{xV} \) and \( G \) is Lindelöff by Lemma B.5.18, th... | Yes |
The question when an infinite abelian group \( G \) may carry a minimal group topology has been studied thoroughly. | It is known that none of the groups \( {\mathbb{Q}}^{n},\mathbb{Z}\left( {p}^{\infty }\right) ,\mathbb{Z}{\left( {p}^{\infty }\right) }^{n},\mathbb{Z}\left( {p}_{1}^{\infty }\right) \oplus \cdots \oplus \mathbb{Z}\left( {p}_{n}^{\infty }\right) \), where \( n \in {\mathbb{N}}_{ + } \) and \( p,{p}_{1},\ldots ,{p}_{n} \... | Yes |
Theorem 8.5.2. Let \( G \) be a locally compact group. Then:\n\n(a) if \( G \) is hereditarily disconnected, every neighborhood of \( {e}_{G} \) contains a compact open subgroup of \( G \) ; | Proof. (a) By Vedenissov theorem B.6.10, there is a neighborhood base \( \mathcal{O} \) at \( {e}_{G} \) consisting of compact symmetric clopen sets. Let \( U \in \mathcal{O} \) . Then, by Lemma 3.1.1(a),\n\n\[ U = \bar{U} = \bigcap \{ {UV} : V \in \mathcal{O}, V \subseteq U\} \]\n\nwhere every set \( {UV} \) is compac... | Yes |
Let \( p \) be a prime and \( G = {\mathbb{Q}}_{p} \rtimes \mathbb{Z} \) such that \( \mathbb{Z} \cong \left\{ {{p}^{n} : n \in \mathbb{Z}}\right\} \) acts on \( {\mathbb{Q}}_{p} \) by multiplication by \( p \), and the subgroup \( O = {\mathbb{Q}}_{p} \rtimes \{ 1\} \) of \( G \) is taken to be open carrying its natur... | Nevertheless, the only compact open normal subgroups of \( G \) are those containing \( O = {\mathbb{Q}}_{p} \rtimes \{ 1\} \). Indeed, if \( V \) is a compact open subgroup of \( G \), there exists \( n \in \mathbb{N} \) such that \( {U}_{n} \subseteq V \). Since the normal closure of \( {U}_{n} \) (i. e., the smalles... | No |
Corollary 8.5.4. Let \( G \) be a locally compact group. Then \( o\left( G\right) = Q\left( G\right) = c\left( G\right) \) . So, \( G \) is hereditarily disconnected if and only if it is totally disconnected. | Proof. It is always true that \( c\left( G\right) \subseteq Q\left( G\right) \subseteq o\left( G\right) \) . By Theorem 8.5.2(b), \( c\left( G\right) \) is the intersection of open subgroups, so \( c\left( G\right) \supseteq o\left( G\right) \) . | Yes |
Corollary 8.5.9. A quotient of a hereditarily disconnected locally compact group \( G \) is hereditarily disconnected. | Proof. Let \( N \) be a closed normal subgroup of \( G \) . It follows from Theorem 8.5.2(a) that \( G \) has a local base at \( {e}_{G} \) formed by compact open subgroups. This yields that the quotient \( G/N \) has the same property. In particular, \( G/N \) is hereditarily disconnected, too. | Yes |
Corollary 8.5.10. Let \( G, H \) be locally compact groups and \( f : G \rightarrow H \) a continuous surjective homomorphism. If \( G \) is \( \sigma \) -compact, then:\n\n(a) \( f\left( {c\left( G\right) }\right) = c\left( H\right) \), provided \( c\left( G\right) \) is compact;\n\n(b) if \( G \) is hereditary discon... | Proof. (a) Since \( c\left( G\right) \) is a compact normal subgroup of \( G, f\left( {c\left( G\right) }\right) \) is a compact (so, closed) normal subgroup of \( H \), as \( f \) is surjective. The group \( G/c\left( G\right) \) is hereditarily disconnected by Corollary 8.5.9 and \( f \) induces a continuous surjecti... | Yes |
For every topological group \( G \), there exist a compact group \( {bG} \) and \( a \) continuous homomorphism \( {b}_{G} : G \rightarrow {bG} \) with \( \overline{{b}_{G}\left( G\right) } = {bG} \) such that for every continuous homomorphism \( f : G \rightarrow K \), where \( K \) is a compact group, there exists a ... | Proof. Let \( \left\{ {{N}_{j} : j \in J}\right\} \) be the family of all kernels of continuous homomorphisms \( G \rightarrow C \) where \( C \) is a compact group. For every \( j \in J \), let \( {q}_{j} : G \rightarrow G/{N}_{j} \) be the canonical projection and let \( {\mathcal{F}}_{j} = \left\{ {{\tau }_{\left( j... | Yes |
Lemma 9.1.1. Let \( n \in {\mathbb{N}}_{ + }, H \) a topological abelian group, and \( \varepsilon > 0 \) . Then every map \( f : {B}_{\varepsilon }\left( 0\right) \rightarrow H \) such that \( f\left( {x + y}\right) = f\left( x\right) + f\left( y\right) \) whenever \( x, y \in {B}_{\varepsilon /2}\left( 0\right) \) ca... | Proof. Put \( U = {B}_{\varepsilon /2}\left( 0\right) \) . For \( x \in {\mathbb{R}}^{n} \), there exists \( m \in {\mathbb{N}}_{ + } \) such that \( \frac{1}{m}x \in U \), and we put\n\n\[ \n{f}^{\prime }\left( x\right) = {mf}\left( {\frac{1}{m}x}\right) \n\]\n\nTo see that this definition is correct, assume that \( \... | No |
Lemma 9.2.4. If \( H \) is a closed subgroup of \( {\mathbb{R}}^{n}, L \cong \mathbb{R} \) is a one-dimensional subspace of \( {\mathbb{R}}^{n} \), and \( H \cap L \neq \{ 0\} \), then denoting by \( p : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}/L \) the canonical projection, \( p\left( H\right) \) is a closed subg... | Proof. If \( n = 1 \), then \( {\mathbb{R}}^{n}/L \) is trivial, so we are done.\n\nAssume that \( n > 1 \) and consider the nonzero closed subgroup \( {H}_{1} = H \cap L \) of \( L \cong \mathbb{R} \) . If \( {H}_{1} = L \) (i. e., \( L \subseteq H \) ), then the assertion follows from Theorem 3.2.8(b). Now assume tha... | Yes |
Corollary 9.2.6. For every \( n \in {\mathbb{N}}_{ + } \), the only compact subgroup of \( {\mathbb{R}}^{n} \) is \( \{ 0\} \) . | Proof. Let \( K \) be a compact subgroup of \( {\mathbb{R}}^{n} \) . By Theorem 9.2.2, \( K = V \times D \), where, for some \( s, m \in \mathbb{N}, V \cong {\mathbb{R}}^{s} \) is a linear subspace of \( {\mathbb{R}}^{n} \) and \( D \cong {\mathbb{Z}}^{m} \) is a discrete subgroup of \( {\mathbb{R}}^{n} \) . The compac... | Yes |
Lemma 9.3.1. Let \( H \) be a discrete subgroup of \( {\mathbb{R}}^{n} \). If the elements \( {v}_{1},\ldots ,{v}_{m} \) of \( H \) are independent, then they are also \( \mathbb{R} \)-linearly independent. | Proof. Let \( D = \left\langle {{v}_{1},\ldots ,{v}_{m}}\right\rangle \cong {\mathbb{Z}}^{m} \), and let \( V \cong {\mathbb{R}}^{k} \) be the linear subspace of \( {\mathbb{R}}^{n} \) generated by \( H \). We need to prove that \( k \geq m \). We can assume without loss of generality that \( V = {\mathbb{R}}^{n} \) (i... | Yes |
Proposition 9.3.2. For a discrete subgroup \( H \) of \( {\mathbb{R}}^{n}, H \) is free and \( {r}_{0}\left( H\right) \leq n \) . | Proof. Since by Lemma 9.3.1 there are at most \( n \) independent vectors in \( H \), we have \( m : = {r}_{0}\left( H\right) \leq n \) . Then there exist \( m \) independent vectors \( {v}_{1},\ldots ,{v}_{m} \) of \( H \) . By Lemma 9.3.1, the vectors \( {v}_{1},\ldots ,{v}_{m} \) are also \( \mathbb{R} \) -linearly ... | Yes |
Lemma 9.3.3. Let \( H \) be a discrete subgroup of \( {\mathbb{R}}^{n} \) and \( L \cong \mathbb{R} \) a one-dimensional linear subspace of \( {\mathbb{R}}^{n} \) with \( H \cap L \neq \{ 0\} \) . Then, denoting by \( p : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{n}/L \) the canonical projection, \( p\left( H\right) \... | Proof. If \( n = 1 \), then \( L = \mathbb{R} \), so this case is trivial. Assume \( n > 1 \) in the sequel. Since \( \{ 0\} \neq {H}_{1} = H \cap L \) is a discrete subgroup of \( L \cong \mathbb{R} \), we conclude that \( {H}_{1} = \langle a\rangle \) is cyclic, by Proposition 3.1.11. Making use of an appropriate lin... | Yes |
Corollary 9.4.1. A quotient of \( {\mathbb{R}}^{n} \) is isomorphic to \( {\mathbb{R}}^{k} \times {\mathbb{T}}^{m} \), where \( k + m \leq n \). In particular, a compact quotient of \( {\mathbb{R}}^{n} \) is isomorphic to \( {\mathbb{T}}^{m} \) for some \( m \leq n \). | Proof. Let \( H \) be a closed subgroup of \( {\mathbb{R}}^{n} \). By Theorem 9.2.2, \( H = V \times D \), with \( V \cong {\mathbb{R}}^{s} \), \( D \cong {\mathbb{Z}}^{m} \) discrete and \( s + m \leq n \). Let \( {V}_{1} \) be the linear subspace of \( {\mathbb{R}}^{n} \) spanned by \( D \). Pick a complementing line... | Yes |
Every closed subgroup \( C \) of \( {\mathbb{R}}^{n} \times {\mathbb{T}}^{k} \) is elementary locally compact. Consequently, every closed subgroup of \( {\mathbb{T}}^{S} \) is elementary compact. | Let \( q : {\mathbb{R}}^{n + k} \rightarrow {\mathbb{R}}^{n} \times {\mathbb{T}}^{k} \) be the canonical projection. Then \( H = {q}^{-1}\left( C\right) \) is a closed subgroup of \( {\mathbb{R}}^{n + k} \) . By Theorem 9.2.2, \( H \) is a direct product \( H = V \times D \) with \( V \cong {\mathbb{R}}^{s} \) and \( D... | Yes |
Lemma 9.4.4. Let \( H,{H}_{1} \) be subgroups of \( {\mathbb{R}}^{n} \). Then:\n\n(a) \( {H}^{ \dagger } \) is closed subgroup of \( {\mathbb{R}}^{n} \) and the correspondence \( H \mapsto {H}^{ \dagger } \) is monotone decreasing; | Proof. The map \( {\mathbb{R}}^{n} \times {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) defined by \( \left( {x, y}\right) \mapsto \left( {x \mid y}\right) \) is continuous.\n\n(a) Let \( {q}_{0} : \mathbb{R} \rightarrow \mathbb{T} = \mathbb{R}/\mathbb{Z} \) be the canonical projection. For every \( a \in {\mathbb{R}}^{n}... | Yes |
Proposition 9.4.5. For every subgroup \( H \) of \( {\mathbb{R}}^{n},\bar{H} = {\left( {H}^{ \dagger }\right) }^{ \dagger } \) . In particular, \( H \) is dense in \( {\mathbb{R}}^{n} \) if and only if \( {H}^{ \dagger } = \{ 0\} \) . | Proof. By Lemma 9.4.4(b), \( {\left( \bar{H}\right) }^{ \dagger } = {H}^{ \dagger } \), so we can assume without loss of generality that \( H = \bar{H} \) is closed. According to Theorem 9.2.2, there exist a base \( \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) of \( {\mathbb{R}}^{n} \) and \( k \leq n \) such that \( H =... | Yes |
Proposition 9.4.6. Let \( {v}_{1},\ldots ,{v}_{n} \in \mathbb{R} \) . Then, for \( v = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \in {\mathbb{R}}^{n} \), the subgroup \( \langle v\rangle + {\mathbb{Z}}^{n} \) of \( {\mathbb{R}}^{n} \) is dense if and only if \( {v}_{0} = 1,{v}_{1},\ldots ,{v}_{n} \in \mathbb{R} \) are li... | Proof. Assume that \( {v}_{0} = 1,{v}_{1},\ldots ,{v}_{n} \in \mathbb{R} \) are linearly independent and let \( H = \langle v\rangle + {\mathbb{Z}}^{n} \) . Then \( {H}^{ \dagger } \subseteq {\mathbb{Z}}^{n} = {\left( {\mathbb{Z}}^{n}\right) }^{ \dagger } \) . Therefore, some \( z \in {\mathbb{Z}}^{n} \) belongs to \( ... | Yes |
Corollary 9.4.7. Let \( {q}_{0} : \mathbb{R} \rightarrow \mathbb{T} \) be the canonical projection. For \( n \in {\mathbb{N}}_{ + } \) and \( {v}_{1},\ldots ,{v}_{n} \in \) \( \mathbb{R} \) such that \( 1,{v}_{1},\ldots ,{v}_{n} \in \mathbb{R} \) are \( \mathbb{Q} \) -linearly independent in \( \mathbb{R},\left\langle ... | Proof. By Proposition 9.4.6, with \( v = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \in {\mathbb{R}}^{n} \), the subgroup \( H = \langle v\rangle + {\mathbb{Z}}^{n} \) of \( {\mathbb{R}}^{n} \) is dense. Consider the canonical projection \( \pi : {\mathbb{R}}^{n} \rightarrow {\mathbb{T}}^{n} \cong {\mathbb{R}}^{n}/{\mathb... | Yes |
Theorem 9.4.8. The group \( {\mathbb{T}}^{\mathfrak{c}} \) is monothetic. | Proof. Let \( B \) be a Hamel base of \( \mathbb{R} \) on \( \mathbb{Q} \) that contains 1 and let \( {B}_{0} = B \smallsetminus \{ 1\} \) ; in particular, \( \left| {B}_{0}\right| = \left| B\right| = \mathfrak{c} \) . To see that the element \( x = {\left( {x}_{b}\right) }_{b \in {B}_{0}} \in {\mathbb{T}}^{{B}_{0}} \)... | Yes |
Let \( G \) be the free product of a cyclic group \( A = \langle a\rangle \) of order 2 and an infinite cyclic group \( C = \langle b\rangle \) . Every element of \( G \smallsetminus C \) can be uniquely written as a product\n\n\[ w = {b}^{{n}_{0}} \cdot a \cdot {b}^{{n}_{1}} \cdot a \cdot {b}^{{n}_{2}}\cdots a \cdot {... | Set\n\n\[ Y = \left\{ {w \in G:k \in {\mathbb{N}}_{ + }\text{ and }{n}_{k} = 0\text{ in (10.1)}}\right\} \;\text{and}\;X = G \smallsetminus Y.\]\n\nNote that \( Y = {Xa} \), so that \( X = {Ya} \), too. Thus, both \( X \) and \( Y \) are right big, since \( G = \) \( X \cup {Xa} = Y \cup {Ya} \) . Let us see now that n... | Yes |
Lemma 10.1.4. (a) Assume that \( {B}_{j} \) is a left big subset of the group \( {G}_{j} \), for \( j \in \{ 1,\ldots, n\} \) . Then \( {B}_{1} \times \cdots \times {B}_{n} \) is a left big subset of \( {G}_{1} \times \cdots \times {G}_{n} \) . | Proof. (a) and \( \left( {\mathrm{b}}_{2}\right) \) follow directly from the definition. | No |
Proposition 10.1.5. Let \( G \) be a group and \( B \) a left big subset of \( G \) . Then:\n\n(a) for every subgroup \( H \) of \( G,{B}^{-1}B \cap H \) is a big subset of \( H \) ;\n\n(b) for every \( a \in G \), there exists \( n \in {\mathbb{N}}_{ + } \) such that \( {a}^{n} \in {B}^{-1}B \) . | Proof. (a) Let \( F \) be a finite subset of \( G \) such that \( {FB} = G \) . For \( f \in F \), if \( {fB} \cap H \neq \varnothing \), choose \( {a}_{f} \in {fB} \cap H \) and let \( E = \left\{ {{a}_{f} : f \in F,{fB} \cap H \neq \varnothing }\right\} \) . For every \( h \in H \), there exists \( f \in F \) such th... | Yes |
Lemma 10.1.8. Let \( G \) be an infinite group and \( S \) a subset of \( G \). (a) If \( S \) is finite, then \( S \) is left and right small. | Proof. (a) obviously follows from (e). | No |
Lemma 10.2.3. If \( f : G \rightarrow H \) is a continuous surjective homomorphism of topological groups, then \( H \) is totally bounded whenever \( G \) is totally bounded. If \( G \) carries the initial topology off and \( H \) is totally bounded, then also \( G \) is totally bounded. In particular, \( G \) is total... | Proof. To prove the first assertion, it suffices to recall that the homomorphic image of a left big subset under a surjective homomorphism is left big, by Lemma 10.1.4(b,). The second assertion follows from the fact that the open sets of \( G \) are preimages of the open sets of \( H \) . So, Lemma 10.1.4( \( {\mathrm{... | Yes |
Proposition 10.2.4. If \( \left\{ {{G}_{i} : i \in I}\right\} \) is a family of topological groups, then \( G = \mathop{\prod }\limits_{{i \in I}}{G}_{i} \) is totally bounded if and only if each \( {G}_{i} \) is totally bounded. | Proof. If \( G \) is totally bounded, then each \( {G}_{i} \) is totally bounded by Lemma 10.2.3.\n\nAssume that each \( {G}_{i} \) is totally bounded and let \( U \) be a nonempty open set of \( G \) . Then there exist a finite subset \( J \) of \( I \) and a nonempty open set \( V \) of \( {G}_{J} = \mathop{\prod }\l... | Yes |
Proposition 10.2.5. All subgroups of totally bounded groups are totally bounded. In particular, all subgroups of compact groups are precompact. | Proof. Let \( H \) be a subgroup of \( G \) . If \( U \in {\mathcal{V}}_{H}\left( {e}_{G}\right) \), there exists \( W \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) such that \( U = W \cap H \) . Pick \( V \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) such that \( {V}^{-1}V \subseteq W \) . Since \( V \) is left big in... | Yes |
Theorem 10.2.6. (a) A Hausdorff group \( G \) having a dense precompact subgroup \( H \) is necessarily precompact. | Proof. (a) For every \( U \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \), choose an open \( V \in {\mathcal{V}}_{G}\left( {e}_{G}\right) \) with \( {VV} \subseteq U \) . By the precompactness of \( H \), there exists a finite subset \( F \) of \( H \) such that \( H = F\left( {V \cap H}\right) \) . Then \( G = {HV} = F\... | Yes |
Lemma 10.2.7. For a topological group \( G \), the following are equivalent:\n\n(a) \( G \) is not totally bounded;\n\n(b) G has a left small nonempty open set;\n\n(c) G has a right small nonempty open set. | Proof. (b) \( \Leftrightarrow \) (c) since a subset \( S \) of \( G \) is left small if and only if \( {S}^{-1} \) is right small, while \( \left( \mathrm{b}\right) \Rightarrow \left( \mathrm{a}\right) \) is a consequence of Lemma 10.1.8(c).\n\n\( \left( a\right) \Rightarrow \left( b\right) \) If \( U \in {\mathcal{V}}... | No |
Lemma 10.2.8. If \( G \) is a totally bounded group, then for every \( U \in \mathcal{V}\left( {e}_{G}\right) \), there exists \( V \in \mathcal{V}\left( {e}_{G}\right) \) such that \( {g}^{-1}{Vg} \subseteq U \) for all \( g \in G \) . | Proof. Let \( W \in \mathcal{V}\left( {e}_{G}\right) \) be symmetric and such that \( {WWW} \subseteq U \) . By hypothesis, \( G = {FW} \) for some finite subset \( F \) of \( G \) . For every \( a \in F \), pick \( {V}_{a} \in \mathcal{V}\left( {e}_{G}\right) \) such that \( {a}^{-1}{V}_{a}a \subseteq W \) , and let \... | Yes |
Corollary 10.2.11. Let \( G \) be a locally compact group such that \( {M}_{G} \) is dense in \( G \) . Then \( G \) is compact, connected and \( w\left( G\right) \leq \mathfrak{c} \) . | Proof. The compactness of \( G \) and \( w\left( G\right) \leq \mathfrak{c} \) follow, respectively, from Theorem 10.2.9 and Exercise 5.4.4 (the weaker hypothesis \( {M}_{G} \neq \varnothing \) is sufficient). Now assume that \( G \) is not connected. Then \( G \) has a nontrivial proper open subgroup \( U \), by Theor... | Yes |
Proposition 10.2.13. For every topological group \( \left( {G,\tau }\right) \), there exists the finest totally bounded group topology \( {\tau }^{ + } \) on \( G \) coarser than \( \tau \) . | Proof. Let \( \left\{ {{\tau }_{i} : i \in I}\right\} \) be the family of all totally bounded group topologies on \( G \) coarser than \( \tau \), and let \( {\tau }^{ + } = \sup \left\{ {{\tau }_{i} : i \in I}\right\} \) . Then \( \left( {G,{\tau }^{ + }}\right) \) is topologically isomorphic to the diagonal subgroup ... | No |
Proposition 10.2.14. For every topological group \( \left( {G,\tau }\right) \), the quotient group\n\n\[ \n{G}^{ + } \mathrel{\text{:=}} G/{\overline{\left\{ {e}_{G}\right\} }}^{\tau + } = \mathfrak{h}\left( {G,{\tau }^{ + }}\right) \n\]\n\nequipped with the quotient topology of \( {\tau }^{ + } \) is precompact, and e... | Proof. The precompactness of the quotient \( {G}^{ + } \) with the quotient topology of \( {\tau }^{ + } \) follows from Lemma 10.2.3. Let \( {\tau }_{1} \) be the initial topology of \( G \) with respect to \( f : G \rightarrow P \) . According to Proposition 10.2.5, we may assume that \( f \) is surjective. Then \( {... | Yes |
Theorem 10.2.15. For every topological group \( G \), there exists a topological isomorphism \( i : {bG} \rightarrow \widetilde{{G}^{ + }} \) such that \( i \circ {b}_{G} = {j}_{G} \) . | Proof. In view of Theorem 8.6.1, it suffices to prove that \( {j}_{G} : G \rightarrow \widetilde{{G}^{ + }} \) has the universal property of \( {b}_{G} : G \rightarrow {bG} \) . Let \( f : G \rightarrow K \) be a continuous homomorphism, where \( K \) is a compact group. By Proposition 10.2.14, \( f \) factorizes throu... | Yes |
Proposition 10.2.16. If \( G \) is an abelian group, \( \delta > 0 \), and \( {\chi }_{1},\ldots ,{\chi }_{s} \in {G}^{ * } \) with \( s \in {\mathbb{N}}_{ + } \) , then \( U\left( {{\chi }_{1},\ldots ,{\chi }_{s};\delta }\right) \) is a big subset of \( G \) . Moreover, for every \( a \in G \), there exists \( n \in {... | Proof. Let \( h : G \rightarrow {\mathbb{S}}^{s}, x \mapsto \left( {{\chi }_{1}\left( x\right) ,\ldots ,{\chi }_{s}\left( x\right) }\right) \), and let\n\n\[ B = \left\{ {\left( {{z}_{1},\ldots ,{z}_{s}}\right) \in {\mathbb{S}}^{s} : \left| {\operatorname{Arg}\left( {z}_{i}\right) }\right| < \frac{\delta }{2}\text{ for... | Yes |
Theorem 10.3.2 (Gel’fand-Raïkov theorem). For every locally compact group \( G \) and \( a \in G \smallsetminus \left\{ {e}_{G}\right\} \), there exists an irreducible unitary representation \( V : G \rightarrow U\left( \mathcal{H}\right) \) of \( G \) by unitary operators of some Hilbert space \( \mathcal{H} \) such t... | The proof of this theorem can be found in [174, 22.12]. | No |
Theorem 10.3.3 (Peter–Weyl–van Kampen theorem). Let \( G \) be a compact group. For every \( a \in G \smallsetminus \left\{ {e}_{G}\right\} \), there exist \( n \in {\mathbb{N}}_{ + } \) and a continuous homomorphism \( f : G \rightarrow U\left( n\right) \) such that \( f\left( a\right) \neq {e}_{G} \) . | The proof of the Peter–Weyl–van Kampen theorem can be found in [174, 22.13]. | No |
Corollary 10.3.4. If \( G \) is a compact group, then \( G \) is isomorphic to a (closed) subgroup of some power \( {\mathbb{U}}^{I} \) of the group \( \mathbb{U} \) . | Proof. In view of Theorem 10.3.3, there exist a nonempty set \( I \) and a family \( \left\{ {{f}_{i} : G \rightarrow }\right. \) \( \mathbb{U} : i \in I\} \) of continuous homomorphisms that separate the points of \( G \) . Then the diagonal map determined by \( \left\{ {{f}_{i} : G \rightarrow \mathbb{U} : i \in I}\r... | Yes |
Proposition 10.3.5. The Bohr compactification of a topological group \( G \) can be equivalently defined by the property \( \left( \mathrm{{Bc}}\right) \). | Proof. First, we verify that if a continuous homomorphism \( h : G \rightarrow K \), where \( K \) is a compact group, has the property (Bc) (i. e., every continuous homomorphism \( f : G \rightarrow U\left( n\right) \) for some \( n \in \mathbb{N} \) factorizes through \( h \) by means of a unique continuous homomorph... | Yes |
Corollary 10.3.7 ([174, 22.17]). A topological abelian group \( G \) is MAP if and only if the continuous characters \( G \rightarrow \mathbb{T} \) separate the points of \( G \), while \( G \) is minimally almost periodic if and only if \( \widehat{G} \) is trivial. | Using Peter-Weyl-van Kampen theorem 10.3.3, in the abelian case one can prove that every locally compact abelian group is MAP. The proof of this fact (see Theorem 11.6.3) requires several ingredients that we develop in Chapter 11. | No |
Theorem 10.5.1 ([239]). Minimal abelian groups are precompact. | A simplified proof of this theorem can be found in [99, Theorem 2.7.7]. Combining it with Exercise 8.7.6, one can deduce that minimal abelian groups are precisely the dense essential subgroups of compact abelian groups. A significant generalization of Theorem 10.5.1 was recently obtained by Banakh [15]. | No |
Proposition 11.1.1. Let \( G \) be a finite abelian group and \( \varphi ,\chi \in {G}^{ * }, x, y \in G \) . Then:\n\n(a) \( \left( {\varphi \mid \chi }\right) = \left\{ \begin{array}{ll} 1 & \text{ if }\varphi = \chi \\ 0 & \text{ if }\varphi \neq \chi \end{array}\right. \)\n\n(b) \( \frac{1}{\left| {G}^{ * }\right| ... | Proof. (a) If \( \varphi = \chi \), then \( \chi \left( x\right) \overline{\chi \left( x\right) } = \chi \left( x\right) \chi {\left( x\right) }^{-1} = 1 \) . If \( \varphi \neq \chi \), there exists \( z \in G \) such that \( \varphi \left( z\right) \neq \chi \left( z\right) \) . Therefore, the equalities\n\n\[ \matho... | No |
Proposition 11.1.3. Let \( G \) be a finite abelian group and \( f \in {\mathbb{C}}^{G} \) with Fourier coefficients \( \left\{ {{c}_{\chi } : \chi \in {G}^{ * }}\right\} \). Then:\n\n(a) \( {G}^{ * } \) is an orthonormal base of \( {\mathbb{C}}^{G} \);\n\n(b) \( f\left( x\right) = \mathop{\sum }\limits_{{\chi \in {G}^... | Proof. (a) According to Proposition 11.1.1(a), \( {G}^{ * } \) is an orthonormal set of cardinality \( \left| G\right| \) in \( {\mathbb{C}}^{G} \), hence an orthonormal base.\n\n(b) Since \( {G}^{ * } \) is an orthonormal base of \( {\mathbb{C}}^{G} \) by item (a), we obtain for \( f \in {\mathbb{C}}^{G} \) that \( f ... | Yes |
Corollary 11.1.4. Let \( G \) be a finite abelian group, \( f : G \rightarrow {\mathbb{R}}_{ \geq 0} \) a function, and \( E = \{ x \in \) \( G : f\left( x\right) > 0\} \) . Then, for \( g \mathrel{\text{:=}} f * f \) and \( x \in G \) :\n\n(a) \( g\left( x\right) > 0 \) if and only if \( x \in {E}_{\left( 2\right) } \... | Proof. (a) For \( x \in G \), by definition \( g\left( x\right) > 0 \) if and only if there exists \( y \in E \) with \( x + y \in E \) , that is, \( x \in E - E = {E}_{\left( 2\right) } \) .\n\n(b) follows from Proposition 11.1.3(d). | Yes |
Lemma 11.2.3. Let \( A \) be an abelian group and \( {\left\{ {A}_{n}\right\} }_{n \in {\mathbb{N}}_{ + }} \) a sequence of finite subsets of \( A \) such that, for every \( a \in A \) , \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\left| \left( {A}_{n} - a\right) \cap {A}_{n}\right| }{\left| {A}_{n}\right| ... | Proof. Let \( {a}_{1},\ldots ,{a}_{k} \in A \) be such that \( A = \mathop{\bigcup }\limits_{{i = 1}}^{k}\left( {{a}_{i} + V}\right) \) . Let \( \varepsilon > 0 \) . By hypothesis, there exists \( N > 0 \) such that, for every \( n \geq N \) and every \( i \in \{ 1,\ldots, k\} \) , \[ \left| {\left( {{A}_{n} - {a}_{i}}... | Yes |
Corollary 11.2.6. For a subset \( E \) of an infinite abelian group \( A \), the following conditions are equivalent:\n\n(a) \( E \) contains \( {V}_{\left( 8\right) } \) for some big subset \( V \) of \( A \) ;\n\n(b) for every \( n \in {\mathbb{N}}_{ + } \) , \( E \) contains \( {V}_{\left( 2n\right) } \) for some bi... | Proof. (a) \( \Rightarrow \) (c) follows from Følner lemma 11.2.5, while (c) \( \Rightarrow \) (b) follows from Corollary 10.2.17 and Proposition 10.2.16, and (b) \( \Rightarrow \) (a) is obvious. | Yes |
For a subgroup \( H \) of an abelian group \( G \), the Bohr topology \( {\mathfrak{B}}_{G/H} \) of \( G/H \) coincides with the quotient topology of the Bohr topology \( {\mathfrak{B}}_{G} \). | Proof. Let \( q : G \rightarrow G/H \) be the canonical projection. The quotient topology \( {\overline{\mathfrak{B}}}_{G} \) of the Bohr topology \( {\mathfrak{B}}_{G} \) is a precompact group topology on \( G/H \) (as \( H \) is closed in \( {G}^{\# } \) by Proposition 3.1.9). Hence, \( {\overline{\mathfrak{B}}}_{G} ... | Yes |
Lemma 11.3.1 (Prodanov lemma). Let \( G \) be a topological abelian group, \( U \) an open set of \( G, f : U \rightarrow \mathbb{C} \) a continuous function, and \( M \) a convex closed set of \( \mathbb{C} \) . Let \( k \in {\mathbb{N}}_{ + } \) , \( {\chi }_{1},\ldots ,{\chi }_{k} \in {G}^{ * } \), and \( {c}_{1},\l... | Proof. Assume that \( {\chi }_{k} \in {G}^{ * } \) is not continuous; then it is not continuous at 0 . Consequently, there exists a net \( {\left\{ {x}_{\gamma }\right\} }_{\gamma \in A} \) in \( G \) such that \( {x}_{\gamma } \rightarrow 0 \) and \( {\left\{ {\chi }_{k}\left( {x}_{\gamma }\right) \right\} }_{\gamma \... | Yes |
Corollary 11.3.3. For every topological abelian group \( G, C\left( G\right) \cap \mathfrak{A}\left( {G}_{d}\right) = \mathfrak{A}\left( G\right) \) . | In other words, as far as continuous complex-valued functions are concerned, in the definition of \( \mathfrak{A}\left( G\right) \) it is irrelevant whether one approximates via (linear combinations of) continuous or discontinuous characters. | No |
Theorem 11.3.5 (Følner theorem). Let \( G \) be a topological abelian group. If \( k \in {\mathbb{N}}_{ + } \) and \( E \) is a subset of \( G \) such that \( k \) translates of \( E \) cover \( G \), then for every \( U \in {\mathcal{V}}_{G}\left( 0\right) \) there exist \( {\chi }_{1},\ldots ,{\chi }_{m} \in \widehat... | Proof. We can assume, without loss of generality, that \( U \) is open. By Følner lemma 11.2.5, there exist \( {\varphi }_{1},\ldots ,{\varphi }_{m} \in {G}^{ * } \) such that \( {U}_{G}\left( {{\varphi }_{1},\ldots ,{\varphi }_{m};\frac{\pi }{2}}\right) \subseteq {E}_{\left( 8\right) } \) ; our aim is to replace these... | No |
Corollary 11.3.6. Let \( G \) be a topological abelian group, \( g \in {\mathfrak{A}}_{0}\left( G\right) \), and \( M \) a convex closed set of \( \mathbb{C} \) . If \( c \in \mathbb{C} \) is such that \( g\left( x\right) + c \in M \) for every \( x \in G \), then \( c \in M \) . In particular, if \( \varepsilon \geq 0... | Proof. Assume first that \( g \in {\mathfrak{X}}_{0}\left( G\right) \) . Suppose that for every \( x \in G, g\left( x\right) = \mathop{\sum }\limits_{{j = 1}}^{k}{c}_{j}{\chi }_{j}\left( x\right) \) for some \( {c}_{1},\ldots ,{c}_{k} \in \mathbb{C} \) and nonconstant \( {\chi }_{1},\ldots ,{\chi }_{k} \in \widehat{G} ... | Yes |
Corollary 11.3.7. Let \( G \) be an abelian group and \( {\chi }_{0},{\chi }_{1},\ldots ,{\chi }_{k} \in {G}^{ * } \) pairwise distinct characters. Then \( {\chi }_{0},{\chi }_{1},\ldots ,{\chi }_{k} \) are linearly independent. | Proof. Let \( {c}_{0},{c}_{1},\ldots ,{c}_{k} \in \mathbb{C} \) be such that \( \mathop{\sum }\limits_{{i = 0}}^{k}{c}_{i}{\chi }_{i}\left( x\right) = 0 \) for every \( x \in G \) . Then \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\chi }_{i}\left( x\right) {\chi }_{0}{\left( x\right) }^{-1} + {c}_{0} = 0 \) for every... | Yes |
Corollary 11.3.8. Let \( G \) be an abelian group and \( {\chi }_{0},{\chi }_{1},\ldots ,{\chi }_{k} \in {G}^{ * } \) pairwise distinct characters. Then \( \begin{Vmatrix}{\mathop{\sum }\limits_{{j = 1}}^{k}{c}_{j}{\chi }_{j} - {\chi }_{0}}\end{Vmatrix} \geq 1 \) for every \( {c}_{1},\ldots ,{c}_{k} \in \mathbb{C} \) . | Proof. Let \( \varepsilon = \begin{Vmatrix}{\mathop{\sum }\limits_{{j = 1}}^{k}{c}_{j}{\chi }_{j} - {\chi }_{0}}\end{Vmatrix} \) . Then, for every \( x \in G \) ,\n\n\[ \left| {\mathop{\sum }\limits_{{j = 1}}^{k}{c}_{j}{\chi }_{j}\left( x\right) - {\chi }_{0}\left( x\right) }\right| \leq \varepsilon \]\n\n(11.25)\n\nAc... | Yes |
Corollary 11.3.9. Let \( G \) be an abelian group, \( H \) a subgroup of \( {G}^{ * } \), and let \( \chi \in {G}^{ * } \) be such that there exist \( k \in {\mathbb{N}}_{ + },{\chi }_{1},\ldots ,{\chi }_{k} \in H \), and \( {c}_{1},\ldots ,{c}_{k} \in \mathbb{C} \) with\n\n\[ \begin{Vmatrix}{\mathop{\sum }\limits_{{j ... | Proof. We assume without loss of generality that \( {\chi }_{1},\ldots ,{\chi }_{k} \) are pairwise distinct. Assume for a contradiction that \( \chi \neq {\chi }_{j} \) for all \( j \in \{ 1,\ldots, k\} \) . Then Corollary 11.3.8 applied to \( \chi ,{\chi }_{1},\ldots ,{\chi }_{k} \) yields \( \begin{Vmatrix}{\mathop{... | Yes |
Proposition 11.3.10. Let \( G \) be an abelian group. Then \( H = \left( \widetilde{G,{\mathcal{T}}_{H}}\right) \) for every subgroup \( H \) of \( {G}^{ * } \) . | Proof. Obviously, \( H \subseteq \left( \widehat{G,{\mathcal{T}}_{H}}\right) \) . Now let \( \chi \in \left( \widehat{G,{\mathcal{T}}_{H}}\right) \) . For every fixed \( \varepsilon > 0 \), the set \( O = \{ a \in \mathbb{S} : \left| {a - 1}\right| < \varepsilon \} \) is an open neighborhood of 1 in \( \mathbb{S} \) . ... | Yes |
Theorem 11.4.2. Let \( \\left( {G,\\tau }\\right) \) be a topological abelian group. The following conditions are equivalent:\n\n(a) \( \\tau \) is totally bounded;\n\n(b) the neighborhoods of 0 in \( G \) are big subsets;\n\n(c) \( \\tau = {\\mathcal{T}}_{\\widehat{\\left( G,\\tau \\right) }} \) . | Proof. (a) \( \\Rightarrow \) (b) This is the definition of totally bounded topology.\n\n(b) \( \\Rightarrow \) (c) If \( H = \\left( \\widehat{G,\\tau }\\right) \) then \( {\\mathcal{T}}_{H} \\leq \\tau \) . Let \( U, V \) be open neighborhoods of 0 in \( \\left( {G,\\tau }\\right) \) such that \( {V}_{\\left( {10}\\r... | Yes |
Theorem 11.4.4. For an abelian group \( G \), let\n\n\[ \mathcal{D}\left( G\right) = \left\{ {H \leq {G}^{ * } : H}\right. \text{separates the points of}G\} \text{,}\]\n\n\[ \mathcal{P} = \{ \tau : \tau \text{ precompact group topology on }G\} .\n\nThen the following is an order-preserving bijection:\n\n\[ T : \mathcal... | Proof. Corollary 11.4.3 yields that \( {\mathcal{T}}_{H} \in \mathcal{P} \) for every \( H \in \mathcal{D}\left( G\right) \) and that \( T \) is surjective.\n\nBy Proposition 11.3.10, \( {\mathcal{T}}_{{H}_{1}} = {\mathcal{T}}_{{H}_{2}} \) for \( {H}_{1},{H}_{2} \in \mathcal{D}\left( G\right) \) yields \( {H}_{1} = {H}... | Yes |
Corollary 11.4.5. If \( G \) is an infinite abelian group and \( H \) a subgroup of \( {G}^{ * } \) that separates the points of \( G \), then \( w\left( {G,{\mathcal{T}}_{H}}\right) = \chi \left( {G,{\mathcal{T}}_{H}}\right) = \left| H\right| \) . | Proof. Let \( \kappa = \chi \left( {G,{\mathcal{T}}_{H}}\right) \) and note that \( \kappa \) cannot be finite. (Indeed, otherwise \( \kappa = 1 \) and \( {\mathcal{T}}_{H} \) must be discrete, so being also precompact, this would imply that \( G \) is finite, a contradiction.) According to Exercise 5.4.2, \( \kappa \l... | Yes |
Corollary 11.5.2. If \( K \) is a compact abelian group, then \( K \) is topologically isomorphic to \( a \) (closed) subgroup of \( {\mathbb{T}}^{\widehat{K}} \) . | Proof. Since \( \widehat{K} \) separates the points of \( K \) by Corollary 11.5.1, the diagonal map determined by all characters in \( \widehat{K} \) defines a continuous injective homomorphism \( {\Delta }_{\widehat{K}} : K \hookrightarrow \) \( {\mathbb{T}}^{\widehat{K}} \) . By the compactness of \( K \) and the op... | Yes |
Corollary 11.5.3. If \( \left( {K,\tau }\right) \) is a compact abelian group and \( H \) is a subgroup of \( \widehat{K} \) that separates the points of \( K \), then \( H = \widehat{K} \) . | Proof. By Corollary 11.5.1, \( \tau = {\mathcal{T}}_{\bar{K}} \) . Since \( {\mathcal{T}}_{H} \leq {\mathcal{T}}_{\bar{K}} \) by Theorem 11.4.4 and \( {\mathcal{T}}_{H} \) is Hausdorff, \( {\mathcal{T}}_{H} = {\mathcal{T}}_{\widehat{K}} = \tau \), as \( \tau \) is compact and due to the open mapping theorem (Theorem 8.... | Yes |
Proposition 11.5.4. Let \( K \) be a compact abelian group and \( U \) an open neighborhood of 0 in \( K \) . Then there exists a closed subgroup \( C \) of \( K \) such that \( C \subseteq U \) and \( K/C \) is an elementary compact abelian group. In particular, \( K \) is an inverse limit of elementary compact abelia... | Proof. By Corollary 11.5.1, the topology on \( K \) is \( {\mathcal{T}}_{\widehat{K}} \), hence there exists a finite subset \( F \) of \( \widehat{K} \) such that \( C = \mathop{\bigcap }\limits_{{\chi \in F}}\ker \chi \subseteq U \) . Define \( g = \mathop{\prod }\limits_{{\chi \in F}}\chi : K \rightarrow {\mathbb{T}... | Yes |
Corollary 11.5.5. A compact abelian group \( K \) is NSS precisely when \( K \) is an elementary compact abelian group. | Proof. If \( K \) is NSS, then \( K \) must be elementary compact, by Proposition 11.5.4. To prove the inverse implication, note first that \( \mathbb{T} \) is NSS. Moreover, the class of NSS abelian groups is stable under taking finite products and subgroups, by Exercise 11.7.4(a). Thus, all powers \( {\mathbb{T}}^{n}... | No |
Theorem 11.6.3. If \( G \) is a locally compact abelian group, then \( \widehat{G} \) separates the points of \( G \) . | Proof. Let \( V \) be a compact neighborhood of 0 in \( G \) . Take \( x \in G \smallsetminus \{ 0\} \) . Then \( {G}_{1} = \langle V \cup \{ x\} \rangle \) is an open (it has nonempty interior) compactly generated subgroup of \( G \) . In particular, \( {G}_{1} \) is locally compact. By Proposition 11.6.1, there exist... | Yes |
Corollary 11.6.4. If \( G \) is a locally compact abelian group, then every compact subgroup \( K \) of \( G \) is dually embedded. | Proof. Clearly, \( H = \left\{ {\chi \in \widehat{K} : \exists \xi \in \widehat{G},\xi { \upharpoonright }_{K} = \chi }\right\} \) is a subgroup of \( \widehat{K} \) . By Theorem 11.6.3, \( \widehat{G} \) separates the points of \( G \), so \( H \) separates the points of \( K \) . Now apply Corollary 11.5.3 to conclud... | Yes |
Corollary 11.6.5. Let \( G \) be a nontrivial locally compact abelian group.\n\n(a) Then \( G \) is connected if and only if \( \chi \left( G\right) = \mathbb{T} \) for every nontrivial \( \chi \in \widehat{G} \) . | Proof. (a) If \( G \) is connected and \( \chi \in \widehat{G} \) is nontrivial, then \( \chi \left( G\right) \) is a nontrivial connected subgroup of \( \mathbb{T} \), hence \( \chi \left( G\right) = \mathbb{T} \) (see Example 6.1.8(b)). If \( G \) is not connected, then \( G \) has a proper open subgroup \( H \), by ... | Yes |
Corollary 11.6.7. For a compact abelian group \( G \), the following are equivalent:\n\n(a) \( G \) is profinite;\n\n(b) \( G \) is hereditarily disconnected;\n\n(c) \( G \) is topologically torsion (i. e., \( G = G \) !);\n\n(d) \( \widehat{G} \) is torsion;\n\n(e) \( \chi \left( G\right) \neq \mathbb{T} \) for every ... | Proof. (a) \( \Leftrightarrow \) (b) is Corollary 8.5.7,(b) \( \Leftrightarrow \) (e) is Corollary 11.6.5(b),(b) \( \Rightarrow \) (c) was proved in Exercise 8.7.11.\n\n(b) \( \Leftrightarrow \) (d) The image \( \chi \left( G\right) \) under a continuous character \( \chi \) of \( G \) is a compact, hence closed, subgr... | No |
Corollary 11.6.8. For a compact abelian group \( G \) and a prime \( p \), the following conditions are equivalent:\n\n(a) \( G \) is pro-p-finite;\n\n(b) \( G \) is topologically p-torsion (i. e., \( G = {G}_{p} \) );\n\n(c) \( \widehat{G} \) is p-torsion;\n\n(d) \( \chi \left( G\right) \subseteq \mathbb{Z}\left( {p}^... | Proof. (a) \( \Rightarrow \) (b) This was proved in Exercise 8.7.12.\n\n(b) \( \Rightarrow \) (d) Pick any \( \chi \in \widehat{G} \) . Then every \( x \in K \) being topologically \( p \) -torsion implies that \( \chi \left( x\right) \) is a topologically \( p \) -torsion element of \( \mathbb{T} \), so \( \chi \left(... | No |
Corollary 11.6.9. For a locally compact abelian group \( G \), one has \( t{d}_{p}\left( G\right) = {G}_{p} \) for every prime \( p \) . | Proof. Fix a prime \( p \) . In view of Remark 5.3.7, always \( t{d}_{p}\left( G\right) \subseteq {G}_{p} \) . To prove the opposite inclusion, let \( x \in {G}_{p} \) ; so \( {p}^{n}x \rightarrow 0 \) in \( G \) . If \( \langle x\rangle \) is finite, then \( x \in {t}_{p}\left( G\right) \subseteq t{d}_{p}\left( G\righ... | Yes |
Proposition 11.6.10. Let \( G \) be a topological abelian group.\n\n(a) If \( G \) is connected, then \( \widehat{G} \) is torsion-free. | Proof. (a) For every nonzero \( \chi \in \widehat{G} \), the image \( \chi \left( G\right) \) is a nontrivial connected subgroup of \( \mathbb{T} \), so we deduce that \( \chi \left( G\right) = \mathbb{T} \) (see Example 6.1.8(b)). Hence, \( \widehat{G} \) is torsion-free. | Yes |
Theorem 11.6.11. For a locally compact abelian group \( \left( {G,\tau }\right) \), its Bohr modification \( {G}^{ + } = \) \( \left( {G,{\tau }^{ + }}\right) \) has the same compact sets as \( \left( {G,\tau }\right) \) . | A proof of this theorem in the discrete case will be given in Theorem 13.4.9. | No |
Lemma 12.1.2. For a topological group \( G \), a function \( f \in {C}^{ * }\left( G\right) \) is almost periodic if and only if every infinite sequence of translates \( {\left\{ {f}_{{b}_{m}}\right\} }_{m \in \mathbb{N}} \) off admits a Cauchy subsequence. | Proof. If \( f \) is almost periodic, then \( {K}_{f} = \overline{\left\{ {f}_{a} : a \in G\right\} } \) is compact by hypothesis, so every sequence in \( {K}_{f} \) admits a convergent subsequence.\n\nAssume now that every infinite sequence of translates \( {\left\{ {f}_{{b}_{n}}\right\} }_{n \in \mathbb{N}} \) of \( ... | Yes |
Let \( f : \mathbb{R} \rightarrow \mathbb{C} \) be a function. One says that \( a \in \mathbb{R} \) is a period of \( f \) if \( f\left( {x + a}\right) = f\left( x\right) \) for every \( x \in \mathbb{R} \) (i. e., \( {f}_{a} = f \) ). Clearly, if \( a \in \mathbb{R} \) is a period of \( f \), then also \( {ka} \) is a... | It is easy to see that a periodic function \( f : \mathbb{R} \rightarrow \mathbb{C} \) has period \( a \in \mathbb{R} \smallsetminus \{ 0\} \) if and only if \( f \) factorizes through the canonical projection \( q : \mathbb{R} \rightarrow \mathbb{R}/\langle a\rangle \), that is, there exists a function \( {f}^{\prime ... | Yes |
Let \( G \) be a topological abelian group and \( \chi \in \widehat{G} \). For \( \varepsilon > 0 \), one has \( \left| {\chi \left( {x + a}\right) - \chi \left( x\right) }\right| = \left| {\chi \left( a\right) - 1}\right| \) for every \( x \in G \), hence \[ T\left( {\chi ,\varepsilon }\right) = \{ a \in G : \left| {\... | So, there exists \( \delta > 0 \) such that \( {U}_{G}\left( {\chi ;\delta }\right) \subseteq T\left( {\chi ,\varepsilon }\right) \). Since \( {U}_{G}\left( {\chi ;\delta }\right) \) is big by Proposition 10.2.16, also \( T\left( {\chi ,\varepsilon }\right) \) is big, and Proposition 12.1.7 yields that \( \chi \) is al... | No |
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