The infinite-dimensional torus
1 Locally compact abelian groups
Let denote the positive integers.
If , , are compact abelian groups, we define their direct product to be the cartesian product
with the coarsest topology such that the projection maps are continuous (namely the product topology), with which the direct product is a compact abelian group. We write
We shall be interested especially in the compact abelian group , and we call the infinite-dimensional torus.
If , , are discrete abelian groups, their direct sum, denoted by
consists of those elements of the cartesian product such that the set is finite. Let be the restriction of to . We give the direct sum the finest topology such that the inclusion maps , defined by
are continuous. With this topology, the direct sum is a discrete abelian group. We write
We shall be interested especially in the discrete abelian group , and in the infinite direct sum . (I don’t know how significant an object it is, but I mention that the abelian group is called the Baer-Specker group.)
When speaking about or in a locally compact abelian group, it is unambiguous that this symbol denotes the identity element of the group, because there is only one distinguished element in a locally compact abelian group. Often we denote the identity element of a compact abelian group by and the identity element of a discrete abelian group by .
If are locally compact abelian groups, it is straightforward to check that the cartesian product
with the product topology is a locally compact abelian group. We call this both the direct product and the direct sum and write
2 Dual groups
If is a locally compact abelian group, denote by its dual group, that is, the set of continuous group homomorphisms . For and we write
has the initial topology induced by , with which it is a locally compact abelian group. If is compact then is discrete, and if is discrete then is compact.
Theorem 1.
Suppose that are locally compact abelian groups. Then the dual group of is isomorphic as a topological group to .
We prove in the following theorem that for discrete abelian groups, the dual group of a direct sum is the direct product of the dual groups.11 1 Karl H. Hofmann and Sidney A. Morris, The Structure of Compact Groups, second ed., p. 12, Proposition 1.17. Cf. Walter Rudin, Fourier Analysis on Groups, p. 37, §2.2.3. In particular, this shows that the dual group of is . Then by the Pontryagin duality theorem22 2 Walter Rudin, Fourier Analysis on Groups, p. 28, Theorem 1.7.2. we get that the dual group of is .
Theorem 2.
Suppose that , , are discrete abelian groups and let
Then , defined by
is an isomorphism of topological groups. Here, and are the projection maps.
Proof.
The definition of makes sense because is finite and hence is finite. For and ,
showing that and hence that is a homomorphism. Suppose that . For each and each ,
where is the inclusion map. This is true for all , so is the identity element of . And this is true for all , so is the identity element of . Therefore is one-to-one. Suppose that . Define as follows: for each , take . Then satisfies , hence is onto and is therefore a group isomorphism.
A continuous bijection from a compact topological space to a Hausdorff space is a homeomorphism, so to prove that is a homeomorphism it suffices to prove that is continuous. has the initial topology induced by , which are maps , so by the universal property of the initial topology, to prove that is continuous it suffices to prove that for each ,
is continuous . For , let , which is a finite set. For each , it is straightforward to check that the map is continuous . Hence the map
is continuous , being a product of finitely many continuous functions , and this completes the proof. ∎
Let be a locally compact abelian group. If is a finite subset of and for each , we call the function defined by
a trigonometric polynomial on . Suppose that is a compact abelian group. Its dual group separates points in ; this is not immediate and is proved using the inversion theorem for the Fourier transform.33 3 Walter Rudin, Fourier Analysis on Groups, p. 24, §1.5.2. The set of trigonometric polynomials on is a self-adjoint algebra that contains the constant functions, so the Stone-Weierstrass theorem then tells us that it is dense in the Banach algebra . Because is separable, it follows that if is countable then is separable. In particular, any closed subgroup of is a compact abelian group whose dual group one checks to be countable, so is separable.
A compact Hausdorff space is metrizable if and only if the Banach algebra is separable.44 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 353, Theorem 9.14. We established in the previous paragraph that if is a compact abelian group with countable dual group then the trigonometric polynomials are dense in the Banach algebra . Therefore, every compact abelian group with countable dual group is metrizable. In particular, and all its closed subgroups are metrizable. In fact, it is proved in Rudin that for a compact abelian group, (i) being metrizable, (ii) having a countable dual group, and (iii) being isomorphic as a topological group to a closed subgroup of are equivalent.55 5 Walter Rudin, Fourier Analysis on Groups, p. 38, §2.2.6.
3 and
Let and be the projection maps and let be the inclusion map.
For and ,
where for each , and .
Let be the Haar measure on such that . Because the dual group of is , for any the Fourier transform of is the function defined by
4 Kronecker sets
Suppose that is a locally compact abelian group and that is a subset of , which we give the subspace topology. is called a Kronecker set if for every continuous and every , there is some such that
We first prove the following lemma from Rudin.66 6 Walter Rudin, Fourier Analysis on Groups, p. 104, Lemma 5.2.8.
Lemma 3.
If , then the set of polynomials with integer coefficients and constant term is dense in the real Banach algebra of continuous functions .
Proof.
Let be the closure in of the set of polynomials with integer coefficients and constant term. Because , separates points in and for every there is some such that . It is straightforward to check that is closed under addition and multiplication. If we show that , it will follow that is an algebra over , and then by the Stone-Weierstrass theorem we will get that is dense in , and hence equal to as is closed.
Let , let be prime, and define
Using that is prime, by the binomial theorem it follows that is a polynomial with integer coefficients and constant term. Partitioning into intervals of length , lies in one of these intervals and hence there is some integer such that . For ,
Hence as . is an integer so for each , is a polynomial with integer coefficients and constant term, so this shows that , completing the proof. ∎
An arc in a topological space is a homeomorphic image of a compact subset of of nonzero length. The following theorem shows that there is an arc in that is a Kronecker set.77 7 Walter Rudin, Fourier Analysis on Groups, p. 103, Theorem 5.2.7.
Theorem 4.
contains an arc that is a Kronecker set.
Proof.
Let , define by
and let be the image of under . Assign the subspace topology inherited from , and suppose that is continuous. One proves that there is a continuous function that satisfies
Let , and by Lemma 3, let be a polynomial with integer coefficients such that . Define by for and otherwise. For ,
using the fact that for . Hence, for every there is some such that
showing that is a Kronecker set. ∎
5 Subgroups
Suppose that is a locally compact abelian group. For each , let be defined by , which is a homeomorphism, and let be defined by , which is also a homeomorphism. If is an open set in and is a subset of , then
which is open because is open for each . Furthermore, if and are both compact sets in then is compact in and is the image of under the continuous map hence is compact.
By a neighborhood of a point in a topological space we mean a set such that lies in the interior of the set, in other words, a set that contains an open neighborhood of the point. The collection of all neighborhoods of a point is a filter, and a neighborhood base at is a filter base for the neighborhood filter of . In a locally compact Hausdorff space, every point has a neighborhood base consisting of compact neighborhoods of .
Let be , which is continuous. If is a neighborhood of in , then is a neighborhood of in . A base for the product topology on consists of sets of the form where are open sets in , so there are open sets in such that . Each of and are then open neighborhoods of in , so is also an open neighborhood of in , and then is open in and
Hence , i.e. , and is open because is open. Therefore, for every neighborhood of in a locally compact abelian group, there is some that is an open neigborhood of and that satisfies .
Suppose that is a locally compact abelian group. A subset of is called symmetric if . If is a compact neighborhood of then contains an open neighborhood of . The set is an open neighborhood of and the set is compact (an intersection of compact sets in a Hausdorff space is compact) and contains , hence is a compact symmetric neighborhood of that is contained in . It follows that in a locally compact abelian group, there is a neighborhood base at consisting of compact symmetric neighborhoods of .
Suppose that is an abelian group and that is a subgroup of . We define the quotient group be the collection of cosets of , which is an abelian group where we define
Let be the projection map, which is a homomorphism with .
We are now equipped to define quotient groups in the category of locally compact abelian groups. Suppose that is a locally compact abelian group and that is a closed subgroup of . We assign the final topology induced by the projection map (namely, the quotient topology). For , there is a compact neighborhood of in ; that is, there is a compact set and an open set such that . Because is continuous, is compact, and because is open, is open, so is a compact neighborhood of in . Therefore is locally compact. It remains to prove that is Hausdorff and that addition and negation are continuous to prove that is a locally compact abelian group. Suppose that are distinct elements of , i.e. . The set is closed because is closed, and so is an open neighborhood of , and hence is an open neighborhood of such that is disjoint from . Because is an open neighborhood of there is an open neighborhood of such that . Furthermore, there is a compact symmetric neighborhood of , , contained in . If then there are and such that , and then . But because is symmetric and so , so , and , so , contradicting that and are disjoint. Therefore and are disjoint, and their images under are then disjoint neighborhoods of and in , showing that is Hausdorff. It is straightforward to prove that addition and negation are continuous in , and therefore is a locally compact abelian group.
If is a closed subgroup of a locally compact abelian group , the annihilator of , denoted , is the set of all such that
For each , the map is continuous so the inverse image of under this map is closed. is the intersection of all these inverse images hence is closed, and is a closed subgroup because it is apparent that is a subgroup of . It can be proved that is the dual of the quotient group and that the quotient group is the dual of .88 8 Walter Rudin, Fourier Analysis on Groups, p. 35, Theorem 2.1.2.
The following lemma shows that we can extend continuous characters on a closed subgroup to the entire group.99 9 Walter Rudin, Fourier Analysis on Groups, p. 36, Theorem 2.1.4.
Lemma 5.
Suppose that is a closed subgroup of a locally compact abelian group . If , then there is some whose restriction to is equal to .
Proof.
, so there is some such that for all , . ∎
Suppose that is a locally compact abelian group. It can be proved that if is a compact open set in and , then contains a compact open subgroup of .1010 10 Walter Rudin, Fourier Analysis on Groups, p. 41, Lemma 2.4.3.
We are now equipped to prove the following theorem.1111 11 Walter Rudin, Fourier Analysis on Groups, p. 47, Theorem 2.5.6.
Theorem 6.
Suppose that is a compact group. is connected if and only if having finite order implies that .
Proof.
Assume that is not connected. Then there is a clopen subset that is neither nor . Because is compact, both and are compact and open, and one of them, call it , contains . Because is a compact open set containing , contains a compact open subgroup of , and because . Because is open, the singleton in the quotient group is an open set, and therefore is discrete. But is compact and is the image of under the projection map, so is compact. Hence is finite. The dual of is , which is a subgroup of . Because contains more than one element (as ), contains some , and has finite order because it is contained in the finite subgroup .
Assume that has order finite order and that . Every element of has finite order and , so is not connected. But if were connected then , a continuous image of , would be connected, hence is not connected. ∎
Lemma 7.
Suppose that is a locally compact abelian group. If is an open subgroup of , then is closed.
Proof.
is a subgroup of , which gives us
Because each set is open, this shows that is closed. ∎
6 Measures
Suppose that is a -algebra on a set . If is a complex measure on we denote by its total variation, which is a finite positive measure on .1212 12 Walter Rudin, Real and Complex Analysis, third ed., p. 117, Theorem 6.2 and p. 118, Theorem 6.4. The total variation norm of is .
Suppose that is a Hausdorff space with Borel -algebra and that is a complex Borel measure on . We say that is outer regular if for each ,
inner regular if for each ,
and tight if for each ,
(Because we demand that be Hausdorff, a compact set is closed and hence belongs to the Borel -algebra of ; compact sets need not belong to the Borel -algebra of a topological space that is not Hausdorff.) We remark that the words “inner regular” often means what we call tight. We say that is regular if it is both outer regular and tight, and we also remark that calling a measure regular often means being outer regular and what we call inner regular. What we call a regular complex Borel measure means precisely what Rudin means by these words in Fourier Analysis on Groups, and using Rudin’s notation we define
It is a fact that a complex Borel measure on a metrizable space is outer regular and inner regular,1313 13 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 436, Theorem 12.5. and that a complex Borel measure on a Polish space is regular.1414 14 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 438, Theorem 12.7.
Suppose that and are locally compact Hausdorff spaces and that and . It is a fact that there is a unique element of , denoted , such that for any and ,
We call the product measure of and .
Suppose that is a locally compact abelian group with addition . For , we define the convolution of and to be the pushforward of the product by ,
and it can be proved that , that convolution is commutative and associative, and that .1515 15 Walter Rudin, Fourier Analysis on Groups, p. 13, Theorem 1.3.2; Karl Stromberg, A note on the convolution of regular measures, Math. Scand. 7 (1959), 347–352. Then, with convolution as multiplication and using the total variation norm, is a unital commutative Banach algebra, with unity .
For , the Fourier transform of is the function defined by
One proves that is bounded and uniformly continuous, and we define
7 Idempotent measures
If is a locally compact abelian group and , we say that is idempotent if , and we denote the set of idempotent elements of by . Because the Fourier transform of a convolution is the product of the Fourier transforms, for we have if and only if . But is equivalent to having range contained in , so for , we have that if and only if is the characteristic function of some subset of . For , we write
Suppose that is an open subgroup of . Then is closed, and the fact that is open implies that the singleton containing the identity in is open and hence that is a discrete abelian group. Denoting the annihilator of by , which is a closed subgroup of , the quotient group is the dual group of and hence is compact. Let be the Haar measure on such that . Taking , . If then
If then there is some such that , and then
showing that , and because this implies that . Therefore, .
If , then with
we have and .
8 Sidon sets
Let be a compact abelian group and let . A function is called an -function if implies that . An -polynomial is a trigonometric polynomial on that is an -function.
We call a subset of a Sidon set if there is some such that for every -polynomial on ,
We shall use the following lemma later.1616 16 Walter Rudin, Fourier Analysis on Groups, p. 121, Theorem 5.7.3.
Lemma 8.
Suppose that is a discrete abelian group that is the dual group of a compact abelian group . If is a Sidon set with constant , then every bounded -function on satisfies
9 Dirichlet series
Define by , i.e. the sum of the entries of , which makes sense because any element of has only finitely many nonzero entries.
Let be those such that for all , and let . In other words, the elements of are those one coordinate of which is and all other coordinates of which are . The proof of the following theorem is from Rudin.1717 17 Walter Rudin, Fourier Analysis on Groups, p. 224, Theorem 8.7.9.
Theorem 9.
If and for all , then
Proof.
is a continuous group homomorphism, and is an open subgroup of , because is discrete. Because is a coset of this open subgroup, there is some such that is the characteristic function of , and this satisfies . Define by
whose Fourier transform is . If then or . In the first case and in the second case , and hence implies that , namely, is an -function. Also, it is apparent from the definition of that .
Suppose that is an -polynomial. Hence there is a finite subset of such that implies that , and thus there are , , such that
, so any element of has one entry , say , and all other entries , so
Define by taking for each , and all other entries of to be ; this makes sense because if and then . For this , . But it is apparent that , so
This shows that is a Sidon set with . Therefore by Lemma 8, because is a bounded -function on we get . But is the characteristic function of and , so
proving the claim. ∎
Following Rudin, we use the above theorem to prove a theorem about Dirichlet series due to Bohr.1818 18 Walter Rudin, Fourier Analysis on Groups, pp. 224–225. See also Maxime Bailleul and Pascal Lefèvre, Some Banach spaces of Dirichlet series, arxiv.org/abs/1311.3845
Theorem 10 (Bohr).
If
and for all such that , then
Proof.
For , let such that , where are the primes and where are the projection maps; so far we have denoted these projection maps by , rather than using , but the symbol has such a strong association with the primes that we change notation here. The map is a bijection , and we write . We shall use the fact that the image of the primes under this bijection is .
Let be a complex number in the half-plane of convergence of and write . Then,
Defining by
we have, as ,
One checks that the function defined by satisfies the conditions of Theorem 9, and thus gets
I do not see why . However, granted this, the claim follows. ∎
10 Descriptive set theory
If is a compact metric space, is a Polish space with the uniform metric . We denote by the group of homeomorphisms of , which one proves is a set in . Because is a set in a Polish space, it is a Polish space with the subspace topology. A homeomorphism of is said to be minimal if there is no proper closed subset of that is invariant under , and is called distal if implies that there is some such that for all , . It has been proved (Beleznay-Foreman) that the collection of minimal distal homeomorphisms of is a Borel -complete set in .1919 19 Alexander S. Kechris, Classical Descriptive Set Theory, p 262, Theorem 33.22.
11 Further reading
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