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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 645848, 5 pages
On the Homomorphisms of the Lie Groups and
1Department of Mathematics, Faculty of Science and Letters, Istanbul Technical University, Maslak, 34469 Istanbul, Turkey
2Department of Mathematics, Faculty of Science and Letters, Okan University, 34959 Istanbul, Turkey
Received 14 February 2013; Revised 1 April 2013; Accepted 7 April 2013
Academic Editor: Nail Migranov
Copyright © 2013 Fatma Özdemir and Hasan Özekes. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We first construct all the homomorphisms from the Heisenberg group to the 3-sphere. Also, defining a topology on these homomorphisms, we regard the set of these homomorphisms as a topological space. Next, using the kernels of homomorphisms, we define an equivalence relation on this topological space. We finally show that the quotient space is a topological group which is isomorphic to .
Discrete and continuous forms of the Heisenberg group have been studied in mathematics and physics such as analysis [1–3], geometry [4–6], topology [3, 7], and quantum physics [8–14]. An introductory review can be also found in .
In [16–18], it was shown that the Heisenberg group is nilpotent, and any arbitrary nilpotent subgroup of is conjugate to a subgroup of , which is identified with the set of diagonal matrices in . Since groups can be considered as metric spaces, it leads us to examine if there exists any geometry in these groups.
It is known that the matrices form a linear group which is isomorphic to . The Lie groups and are related, for the mapping defined by is a continuous homomorphism from onto . The Lie algebras and are trivially isomorphic. In this work, we search if there is a similar relationship between the only other sphere Lie group and the linear group of matrices which is diffeomorphic to .
The subgroup of formed by the matrices of type is called three-dimensional Heisenberg group. It is convenient to denote the elements of this group by three-tuples of numbers. Using this convention, that is, , the multiplicative operation of elements can be expressed as
The identity element of this group is , and the inverse of an element is .
This Lie group is diffeomorphic to .
Moreover, the Lie groups and are isomorphic, and they are diffeomorphic as manifolds. Since the respective Lie algebras and of and are three-dimensional real vector spaces, they are isomorphic as real vector spaces. But they are not isomorphic as Lie algebras, because there is no Lie algebra isomorphism between a compact and a noncompact Lie algebra. However, there may be a Lie algebra homomorphism between them. We try to find the relationship between and as Lie algebras.
Consider the following matrices: which span the Lie algebra of the Lie group , and also consider the following matrices: which span the Lie algebra of the Heisenberg group .
Using the exponential map from to we may convert our equations in to equations in . Letting for each , we see that
The standard generating set for the Heisenberg group is , and the group has the relation
In the above, the notations and denote the inverses of the elements and in the Heisenberg group.
It was shown in ([13, 16]) that there is no nontrivial Lie algebra homomorphism from to . Here, we study Lie algebra homomorphisms from to . We observe that there exists a nontrivial Lie algebra homomorphism from to . Moreover, we describe all Lie algebra homomorphisms from to .
2. Homomorphisms from to
Let be a Lie algebra homomorphism. Suppose that for some real numbers .
By using (8) and the following commutators: we obtain the system of equations of coefficients in the real constants 's
By using the Mathematica program, we may solve the system (11). We write below only two of the non-trival solutions, because a group homomorphism induced by any of the other solutions will be equal to a group homomorphism induced by one of the solutions given below. Here, a nontrivial solution means a solution in which at least one of the constants is nonzero.(1), , , , arbitrary.(2), , , arbitrary.
In the set of solutions, we consider and in terms of the other arbitrary constants , , , and and obtain that the group homomorphism is spanned by elements or . If another configuration, say , , was chosen, then we see that the homomorphism would be spanned by the same elements.
Since is a Lie algebra homomorphism, we observe from solution sets 1 and 2 that the subalgebra of is only generated by or and not by .
As we will use the constants frequently, to simplify the notations we put , , , and for the coefficients , , , and , respectively, and, , , and for , , and .
Hence, for the first set of solutions, has the following form:
We note that the rank of is one, and thus, is a one-dimensional Lie subalgebra of generated by .
For the second set of the solutions, is of the following form:
Here, we note again that the rank of is one, and thus, is a one-dimensional Lie subalgebra of generated by .
It is known from ([8, 18]) that if is a homomorphism, then the following diagram commutes: (14) where is the differential of , and it is a Lie algebra homomorphism. It is also known that for the matrix groups the exponential map is given by the exponentiation of matrices. In our notation, for some , and we will determine .
By using (5), for any element of we have
By using (15), we obtain as
To guarantee that we must take . Here, we put for the matrix (2) for any real , , and .
For the first set of solutions, we obtain
Thus, the kernel of is the plane in .
For , we obtain where is the identity matrix, and .
In this case, is of the following form:
For the second set of solutions, we obtain
From (21), the kernel of is the plane in .
For , we have where is the identity matrix, and .
In this case, is of the form
Hence, we can state the following theorem.
We now observe the following properties for the first type of homomorphisms. It can be shown that same observations are valid for the second set of solutions.(a) By considering as a map from to , we can write where Then, we find that the image of the planes in , with the equation , is a periodic (closed) curve in .(b) In the first case, every homomorphism from to depends on four parameters, namely, , , , and with . We now concentrate on the parameters and , since they are involved in the kernel of . To each kernel , we associate a point in the -plane. Furthermore, we normalize the vector as which can also be considered as a point of in the -plane. By considering the principal value of , we define a topology on the set of all homomorphisms from to as follows. For any , is in the -neighborhood of , if and only if , , and . Now let us define an equivalence relation on . For , , is equivalent to if and only if and have the same kernel. Denote the set of equivalence classes by . We define a multiplication on such that, for any and in , denotes the element of whose kernel is the plane obtained by the product of the elements of corresponding to and . This multiplication makes into a group which is isomorphic to .(c) The set of the kernels of all the homomorphisms from to is a subset of the Grassmann manifold of 2 planes in . It is known that the Grassmann manifold of 2 planes in is diffeomorphic to . Any point of corresponds to the plane which is orthogonal to the normal of at and which contains the -axis. Hence, there exists a 1-1 correspondence between the set of the kernels of all the homomorphisms from to and the equator of .
Thus, we state the following theorem.
Theorem 2. The set of all homomorphisms from the Heisenberg group to the 3-sphere is isomorphic (up to a certain equivalence relation concerning kernels) with the topological group .
In this paper, our aim is to construct all homomorphisms between the Heisenberg group and the -sphere which is isomorphic to . In the literature, it has been shown that no nontrivial Lie algebra homomorphism from to exists ([8, 11, 17]). So, a natural question arises: is there a homomorphism from to ? Here, we answer this question completely.
Also, we use these maps to define a topology in order to construct an equivalence relation, and we show that quotient space is isomorphic to .
The authors would like to thank the referee for his/her helpful comments and suggestions.
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