1. Introduction and Preliminaries
In 1936, Birkhoff and Kakutani independently proved a significant theorem:
A Hausdorff group is homeomorphic with a metric space, if and only if satisfies the first countability axiom. They also showed that this group has a right invariant metric. The theorem then became known as Birkhoff–Kakutani’s metrization theorem for groups [
1,
2]. A metric
d on a semigroup
is called left-invariant if
and right-invariant if
whenever
. The metric
d is said to be invariant if it is both right and left-invariant. In 1950, V. L. Klee studied invariant metrics on groups to solve a problem of Banach [
3]. In this article, we focus on normed groups closely related to the groups with invariant metrics and which recently played a role in the theory of topological groups [
4,
5,
6,
7]. Some may use the term “length function” instead of “norm” for groups [
8,
9]. Recently, new approaches have been presented to the theory of normed groups, such as probabilistic normed groups [
10,
11]; see [
4] for a broader discussion about the history of normed groups. We start with some preliminaries as required in the paper.
Definition 1 ([
4])
. Let be a group with identity element e. A function is called a group-norm if the following holds for all :- 1.
;
- 2.
;
- 3.
.
Then, equipped with a group-norm is said to be a normed group.
Definition 2 ([
4])
. Let be a normed group and and be a sequence in . Then,- 1.
The sequence converges to s if for every , , there exists a positive integer depending on ϵ such that for every . We denote this by .
- 2.
The sequence is called a Cauchy sequence if for every , there exists a positive integer depending on ϵ such that for every .
- 3.
The normed group is called complete if any Cauchy sequence in converges to an element of ; i.e., it has a limit in group .
- 4.
A Banach group is a normed group that is complete with respect to the metric
Definition 3 ([
12])
. Let be a group with identity element e. The order of an element is the smallest such that . If no such n exists, v is said to have infinite order. An abelian group is said to be torsion-free if no element other than the identity e is of finite order. Let
be a normed group. For
, the
s-conjugate norm is defined by
Note that the group-norm is abelian iff the norm is preserved under conjugacy [
13]. It is obvious that each norm on an abelian group is an abelian norm. The following example shows how non-trivial cases can be considered.
Example 1. Let be the nonabelian dihedral group . A matrix representation of this group is given by The relationsholds for integers i and j such that and both and are computed modulo 3. Note that is the identity element, and , for each . Now let be defined bywhere denotes the trace of matrix V. Note that is abelian because This shows that there is an abelian norm on a non-abelian group.
Definition 4 ([
14])
. Let be a group. An element is said to be divisible by if has a solution x in . A group is called infinitely divisible if each element in is divisible by every positive integer.A group norm is -homogeneous if for each Remark 1. For a divisible element v in a -homogeneous normed group , let ; then, and as , we have .
Definition 5 ([
4])
. Let and be normed groups. A map is called continuous if for every there exists such a that implies 2. Inner Product Groups
In this section, we introduce the notion of the inner product on abelian groups. Here, we derive and investigate in detail the induced norm properties such as Cauchy–Schwartz inequality for groups and the Parallelogram law for inner product groups. Besides, we show that every inner product group is a normed torsion-free group. Note that in this paper, the identity element of groups is denoted by e.
Definition 6. A semi-inner product on a group with identity element e is a function that associates a real number with each pair of elements v and s in in such a way that the following axioms are satisfied for all elements v, s and z in :
- 1.
( Symmetry);
- 2.
(Distributivity);
- 3.
( Positivity ).
A group with a semi-inner product is called a semi-inner product group.
Note that combining (1) and (2) gives the equation
Example 2. - 1.
Let denote the discrete Heisenberg group and let be defined bywhere and . Then is a semi-inner product group. - 2.
Let . Define the group as matrices by Note that the group operation is given by the matrix multiplication. Then, bywhere and are elements in , this group is a semi-inner product group. - 3.
Let be a finite abelian group with for all , where denotes the order of an element . For elements v and s in , with orders m and n, respectively, if m and n are co-prime, then has order [15]. Then, for all ,is a semi-inner product on . It is obvious that and . The principle of distributivity remains to be proven.
Proposition 1. Let be a semi-inner product group. Here are some elementary properties of semi-inner product:
- 1.
- 2.
- 3.
- 4.
The definition of inner products usually comes immediately after defining semi-inner products; i.e., if after three axioms of Definition 6, implies , we expect that is an inner product for group . However, there is an additional condition for groups with an inner product. In other words, we show that the definition of our inner product imposes an abelian structure on groups.
Definition 7 ([
12])
. Let be a group. The subgroup of generated by the set is called a commutator subgroup of and denoted by . The elements are called commutators. Theorem 1 ([
12])
. Group is abelian if and only if . In fact,
provides an indicator for measuring differences between group
and an abelian group. Now, let
be an semi-inner product group and suppose that
implies
. We show that
. Since for
Then, . This shows that all commutators of are equal to e. So, we define inner products on abelian groups as follows:
Definition 8. Let be a semi-inner product abelian group, in which for , implies . Then is called an abelian inner product group.
In this paper, it is supposed that every inner product group is abelian. Therefore, when we talk of an inner product group or a Hilbert group, we mean that there is an inner product on an abelian group.
Example 3. - 1.
Let denote the group of positive real numbers with multiplication as the group operation. By will be an inner product group. - 2.
Let , then is an abelian group by real multiplication. By setting is an inner product group. - 3.
Let . Then, with a binary operationand by setting an inner product is an inner product group. - 4.
Let . Then, with a binary operationand by setting an inner product is an inner product group.
Proposition 2. Let be an inner product group. For such that for all , then .
Proof. For , let for all . Therefore, for all , and we have . So, , which means . □
The proofs of the next two lemmas are straightforward, and thus we omit them.
Lemma 1. Let be an abelian group and be an inner product group. If is a group monomorphism, thenis also an inner product on . Lemma 2. Let and be inner product groups.
Then, the group withis also an inner product group by Definition 9. Let be an inner product group. We define the map for all by Theorem 2. Let ) be an inner product group. Then, Proof. Let
v and
s be arbitrary in
. If
, then the inequality is true. For
and
, the positivity of the inner product shows that
Let
,
and
. Then, the equation becomes
. This is a quadratic equation for
with real coefficients. Since this polynomial takes only non-negative values, its discriminate
must be non-positive
□
Inequality (
1) is called Cauchy–Schwartz inequality for groups.
Proposition 3. Let be an inner product group. Then, Proof. Recall that every inner product group is abelian; then,
□
Theorem 3. Let be an inner product group. Then, it is a normed group with norm for all .
Proof. The positivity principle of the norm is clear. From Proposition 1, we have
whenever
. Then,
Now, we show that
satisfies the triangle inequality. For all
, we have
So, the Cauchy–Schwartz inequality for groups implies that
□
Example 4. Let be the abelian group of matrices of the form , where , and The group operation is given by the multiplication of matrices. It follows that . Now, we define Therefore, it is obvious that is a normed group with
Corollary 1. Let be an inner product group. Then, the group norm induced by the inner product is -homogeneous; i.e., .
Proof. It follows immediately from the fact that
, since
□
Theorem 4. Let be an inner product group. Then, is a torsion-free abelian group.
Proof. Let
such that
and
. Then, the
-homogeneous property of the induced group norm implies that
Since , then and . Hence, is a torsion-free abelian group. □
Lemma 3 (Parallelogram Law)
. Let be an inner product group. Then, Theorem 5. Let be an abelian normed group. The norm is induced by an inner product iff the parallelogram law holds in .
Proof. Let
be an abelian normed group whose norm satisfies the parallelogram law. Put
We wish to show that is an inner product on . We begin by observing that for all , so is non-negative.
Since
the function
is also a symmetric function. The final step in this proof is showing that
satisfies the distributivity property. We have
Exchanging
v and
s in the last equation gives
Replacing
z by
, we have
Since
for all
, we get
This shows that the defined satisfies all three criteria of an inner product. Therefore, is an inner product group. □
Lemma 4. Let be an inner product group with induced norm . Then, is continuous.
Proof. Let
and for all
put
and
. Then, we have
From which it follows that is continuous. □
Definition 10. Let and be normed groups. An isomorphism (bijective group homomorphism) is said to be isometric if for all .
Proposition 4. Let and are inner product groups. Then, an isomorphism is isometric if and only if Proof. Suppose that the given identity holds. Then,
For the converse, suppose that
is isometric, and
, then
□
3. Hilbert Groups and a Riesz Representation Theorem
In this section, we intend to prove a Riesz Representation Theorem for groups. However, we first need to define the notion of orthogonality and midconvexity and prove related theorems and results on Hilbert groups.
Definition 11. Let and be normed groups. Suppose that is an arbitrary function. Define Then, α is a possibly infinite number that is called bounded if is finite.
Remark 2. Suppose that and are normed groups and is a bounded homomorphism. Then, α is continuous.
We denote the set of all bounded homomorphisms from into the by . Clearly, , under the pointwise multiplication , is also a normed group.
Lemma 5. Let and be normed groups. If is Banach group, then so is .
Proof. Suppose that
is a Cauchy sequence in the group
. Let
. Since
it follows that
is also a Cauchy sequence in the group
. Now, define
by
To see that is bounded, note that is bounded. Hence, there exists an such that for each .
Moreover, since
then,
.
Now, let and choose an so that if , then .
Let
with
. Since
for each
, we have
In particular,
in
. □
Let be a topological abelian group. We denote the set of all continuous homomorphisms by . Note that in this content, we consider as an additive group.
Lemma 6. Let be an inner product group. Fix and define by . Then, and .
Proof. , so
and
. Since
then
So . □
Definition 12. Let be an inner product group. Then, is called a Hilbert group if is complete with respect to the norm induced by the inner product.
Example 5. - 1.
Let . Then with the inner product is a Hilbert group.
- 2.
Let . Then with the inner product is a Hilbert group.
Moreover, the connected component of identity in an abelian Lie group is an infinitely divisible group [
16]. So, considering Example 5, let
be the connected component of identity in
with the same inner product. Then,
is an infinitely divisible Hilbert group.
Definition 13. Two elements v and s of an inner product group are said to be orthogonal, written , if . We say that subsets A and B are orthogonal if for each and . The orthogonal complement of a subset A is the set of elements such that u is orthogonal to all elements in A.
Lemma 7. Let A be a subset of Hilbert group . Then, the orthogonal complement of A is a closed subgroup of and .
Proof. Let
be a Hilbert group and
. If
, then for arbitrary
, we have
Thus, is a subgroup of .
To show that is closed, let be a sequence in such that converges to b. We show .
Hence, .
To prove , let . Then, and . □
Definition 14 ([
4])
. Let be a group. A subset C of is called -convex (or midconvex), if for every there exists an element , denoted by , such that . Lemma 8. Let be a Hilbert group and A be a non-empty, closed and -convex subset of . Then, A contains a unique element of the smallest norm; i.e., Proof. There exists a sequence
in
A such that
By applying the Parallelogram law to
and
, we obtain
Since
, given
if
N is large enough, then for
By
-convexity of
A, we have
, so
Combining these estimates gives
Since
is complete and
A is closed,
. Moreover,
Thus,
v exists, and if
v and
are two elements in
A with
, then
Therefore, . □
Lemma 9. Let A be a closed and -convex subgroup of Hilbert group . Then, . This mens that any has a unique decomposition where and .
Proof. Let
A be a closed and
-convex subgroup of
. If
, then
and there is nothing to show. Consider
and set
where
B is non-empty, because
. So, for
and
in
B, we have
In addition, every sequence in B is of the form , where is a sequence in A. Thus, converges in B if and only if converges in A. It follows that B is a closed subset of .
Now, we define to be the element of the smallest norm in ; this exists as a result of Lemma 8. Put ; it is clear that . Then, .
To prove the uniqueness of this, let for some elements and . Then, . Since and , we have . □
Definition 15. Let be a group and be a sequence of rational numbers that converges to . For each and , define Remark 3. Let be an inner product group. We know that for each and . Then for each and , Since for each Thus, . We wish to show that for each and .
Because of the density of in , for each , there is a sequence of that converges to r. Let and ; then, for each , Thus, for each , we have Since , then Theorem 6. Let be an infinitely divisible Hilbert group and be a continuous epimorphism. Then, there is a unique element in such that for every .
Proof. Let . If then , and thus assume that is a non-zero epimorphism. Because is continuous, then M is a closed -convex subgroup of . As we suppose that , so . Therefore, there is v in such that .
Now, if
and
, then
Therefore,
and we have
Thus, if , then for all .
If
such that
for all
, then
In particular, and so . □