# On the Homomorphisms of the Lie GroupsSU(2) and S3

Authors: Fatma Özdemir; Hasan Özekes
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

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 to𝕊1.


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## Body

## 1. Introduction

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 [15].In [16–18], it was shown that the Heisenberg group ℍ is nilpotent, and any arbitrary nilpotent subgroup of SU(2) is conjugate to a subgroup of U(1), which is identified with the set of diagonal matrices in SU(2). 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(1)(1x01),x∈ℝ
form a linear group which is isomorphic to ℝ1. The Lie groups ℝ1 and 𝕊1 are related, for the mapping ρ:ℝ1→𝕊1 defined by ρ(x)=e2πix is a continuous homomorphism from ℝ1 onto 𝕊1 [10]. The Lie algebras ℝ1 and 𝕊1 are trivially isomorphic. In this work, we search if there is a similar relationship between the only other sphere Lie group 𝕊3 and the linear group of matrices which is diffeomorphic to ℝ3 [17].The subgroupℍ of 𝔾𝕃(3,ℝ) formed by the matrices of type
(2)(1xz01y001),x,y,z∈ℝ
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, ℍ={(x,y,z):x,y,z∈ℝ}, the multiplicative operation of elements can be expressed as
(3)(x,y,z)·(x′,y′,z′)=(x+x′,y+y′,z+z′+xy′).The identity element of this group is(0,0,0), and the inverse of an element (x,y,x) is (-x,-y,-z).This Lie group is diffeomorphic toℝ3 [17].Moreover, the Lie groupsSU(2) and 𝕊3 are isomorphic, and they are diffeomorphic as manifolds. Since the respective Lie algebras h and su(2) of ℍ and SU(2) 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 h and su(2) as Lie algebras.Consider the following matrices:(4)U1=-12(0ii0),U2=12(0-110),U3=12(-i00i)
which span the Lie algebra su(2) of the Lie group SU(2), and also consider the following matrices:
(5)V1=(010000000),V2=(000001000),V3=(001000000)
which span the Lie algebra h of the Heisenberg group ℍ.Using the exponential map fromh to ℍ we may convert our equations in h to equations in ℍ. Letting V¯i=exp(Vi) for each i, we see that
(6)V¯1=(110010001),V¯2=(100011001),V¯3=(101010001).The standard generating set for the Heisenberg group is{V¯1,V¯2,V¯1-1,V¯2-1}, and the group has the relation
(7)V¯1V¯2V¯1-1V¯2-1=V¯2V¯1-1V¯2-1V¯1=V¯1-1V¯2-1V¯1V¯2=V¯2-1V¯1V¯2V¯1-1=V¯3.In the above, the notationsV¯1-1 and V¯2-1 denote the inverses of the elements V¯1 and V¯2 in the Heisenberg group.From (4) and (5), we find that
(8)[U1,U2]=U3,[U2,U3]=U1,[U3,U1]=U2,[V1,V2]=V3,[V2,V3]=[V3,V1]=0.It was shown in ([13, 16]) that there is no nontrivial Lie algebra homomorphism from su(2) to h. Here, we study Lie algebra homomorphisms from h to su(2). We observe that there exists a nontrivial Lie algebra homomorphism from h to su(2). Moreover, we describe all Lie algebra homomorphisms from h to su(2).


## 2. Homomorphisms fromℍ to SU(2)

Letφ:h→su(2) be a Lie algebra homomorphism. Suppose that
(9)φ(Vi)=∑j=13‍αijUj,i=1,2,3
for some real numbers αij.By using (8) and the following commutators:
(10)φ[V1,V2]=[φV1,φV2],φV3=[φV1,φV2],φ[V1,V3]=[φV1,φV3],φ[V1,V3]=0,φ[V2,V3]=[φV2,φV3],φ[V2,V3]=0,
we obtain the system of equations of coefficients in the real constants αij's
(11)α12α23-α13α22=α31,α13α21-α11α23=α32,α11α22-α12α21=α33,α12α33-α13α32=0,α13α31-α11α33=0,α11α32-α12α31=0,α21α32-α22α31=0,α23α31-α21α33=0,α22α33-α23α32=0.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 αij is nonzero.(1)
α22=α12α21/α11,α23=α13α21/α11,α31=α32=α33=0,  α11≠0, α12, α13, α21 arbitrary.(2)
α23=α13α22/α12,α11=α21=α31=α32=α33=0,  α12≠0,  α13, α22 arbitrary.Also, in the system of (11), if α11≠0 and α12=0, then the first set of solutions is obtained. If α12≠0 and α11=0, then the second set of solutions is obtained. Moreover, if α11 and α12 are both 0, then (11) has trivial solution.In the set of solutions, we considerα22 and α23 in terms of the other arbitrary constants α11, α12, α13, and α21 and obtain that the group homomorphism is spanned by elements {U1,U2,U3} or {U2,U3}. If another configuration, say α11, α13, was chosen, then we see that the homomorphism would be spanned by the same elements.Sinceφ is a Lie algebra homomorphism, we observe from solution sets 1 and 2 that the subalgebra of su(2) is only generated by a1U1+a2U2+a3U3 or aU2+bU3 and not by a1U1+a2U3.As we will use the constantsαij frequently, to simplify the notations we put a1, a2, a3, and b for the coefficients α11, α12, α13, and α21, respectively, and, a, b, and c for α12, α13, and α22.Hence, for the first set of solutions,φ has the following form:
(12)φ(V1)=a1U1+a2U2+a3U3,φ(V2)=ba1φ(V1),a1≠0,φ(V3)=0.We note that the rank ofφ is one, and thus, φ(h) is a one-dimensional Lie subalgebra of su(2) generated by a1U1+a2U2+a3U3.For the second set of the solutions,φ is of the following form:
(13)φ(V1)=aU2+bU3,φ(V2)=caφ(V1),a≠0,φ(V3)=0.Here, we note again that the rank ofφ is one, and thus, φ(h) is a one-dimensional Lie subalgebra of su(2) generated by aU2+bU3.It is known from ([8, 18]) that if Φ:ℍ→SU(2) is a homomorphism, then the following diagram commutes:
(14)
where dΦ is the differential of Φ, 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, φ=dΦ for some Φ, and we will determine Φ.By using (5), for any element of  V=c1V1+c2V2+c3V3∈h we have
(15)V=(0c1c300c2000),V2=(00c1c2000000),V3=(000000000),c1,c2,c3∈ℝ.By using (15), we obtain exp(V) as
(16)exp(V)=1+V+12V2+13!V3+⋯=(1c1c3+c1c2201c2001).To guarantee that(17)exp(V)=(1xz01y001)=(x,y,z)
we must take V=xV1+yV2+(z-xy/2)V3∈h. Here, we put (x,y,z) for the matrix (2) for any real x, y, and z.For the first set of solutions, we obtain(18)φ(V)=φ(xV1+yV2+(z-xy2)V3)=(x+ba1y)(a1U1+a2U2+a3U3).Thus, the kernel ofφ is the plane x+(b/a1)y=0 in ℝ3.ForV∉Kerφ, we obtain
(19)expφ(V)=IcosA(x+ba1y)+1A(x+(b/a1)y)sin(A(x+ba1y))φ(V),
where I is the 2×2 identity matrix, and A=(1/2)a12+a22+a32.In this case,Φ is of the following form:
(20)(x,y,z)⟶IcosA(x+ba1y)+1Asin(A(x+ba1y))(a1U1+a2U2+a3U3).For the second set of solutions, we obtain(21)φ(V)=(x+cay)(aU2+bU3).From (21), the kernel of φ is the plane x+(c/a)y=0 in ℝ3.ForV∉Kerφ, we have
(22)expφ(V)=IcosA(x+cay)+1A(x+(c/a)y)sin(A(x+cay))φ(V),
where I is the 2×2 identity matrix, and A=(1/2)a2+b2.In this case,Φ is of the form
(23)(x,y,z)⟶IcosA(x+cay)+1Asin(A(x+cay))(aU2+bU3).Hence, we can state the following theorem.Theorem 1.
Any nontrivial homomorphism from the Heisenberg group to the 3-sphere is one of (20) and (23).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 consideringΦ as a map from ℝ3 to ℝ4, we can write
(24)Φ:(x,y,z)⟶(u1,u2,u3,u4),where(25)u1=cosAt,u2=-a32AsinAt,u3=-a22AsinAt,u4=-a12AsinAt,t=x+ba1y.Then, we find that the image of the planesPt in ℍ, with the equation x+b/a1=t, is a periodic (closed) curve in 𝕊3.(b)
In the first case, every homomorphismΦ from ℍ to SU(2) depends on four parameters, namely, a1, a2, a3, and b with a1≠0.We now concentrate on the parametersa1 and b, since they are involved in the kernel of Φ. To each kernel x+(b/a1)y=0, we associate a point (b,-a1) in the xy-plane. Furthermore, we normalize the vector (b,-a1) as (b/b2+a12,-a1/b2+a12) which can also be considered as a point of 𝕊1 in the xy-plane.By considering the principal value ofarctan(-b/a1), we define a topology on the set ℍ′ of all homomorphisms from ℍ to SU(2) as follows.For anyΦ1(a1,a2,a3,b)∈ℍ′, Φ2(a1′,a2′,a3′,b′)∈ℍ′ is in the ϵ-neighborhood of Φ1, if and only if |arctan(-b/a1)-arctan(-b′/a1′)|<ϵ, |a2-a2′|<ϵ, and |a3-a3′|<ϵ.Now let us define an equivalence relation onℍ′. For Φ1, Φ2∈ℍ′, Φ1 is equivalent to Φ2 if and only if Φ1 and Φ2 have the same kernel. Denote the set of equivalence classes by ℍ~. We define a multiplication on ℍ~ such that, for any Φ~1 and Φ~2 in ℍ~, Φ~1·Φ~2 denotes the element of ℍ~ whose kernel is the plane obtained by the product of the elements of 𝕊1 corresponding to Φ1 and Φ2. This multiplication makes ℍ~ into a group which is isomorphic to 𝕊1.(c)
The set of the kernels of all the homomorphisms fromℍ to 𝕊3 is a subset of the Grassmann manifold of 2 planes in ℝ3. It is known that the Grassmann manifold of 2 planes in ℝ3 is diffeomorphic to 𝕊1. Any point p of 𝕊1 corresponds to the plane which is orthogonal to the normal of 𝕊1 at p and which contains the z-axis. Hence, there exists a 1-1 correspondence between the set of the kernels of all the homomorphisms from ℍ to 𝕊3 and the equator 𝕊1 of 𝕊2.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 group𝕊1.


## 3. Conclusion

In this paper, our aim is to construct all homomorphisms between the Heisenberg group and the3-sphere which is isomorphic to SU(2). In the literature, it has been shown that no nontrivial Lie algebra homomorphism from su(2) to h exists ([8, 11, 17]). So, a natural question arises: is there a homomorphism from h to su(2)? Here, we answer this question completely.We here observe that there are nontrivial homomorphisms from the Heisenberg group to the3-sphere, and they can be only in the form of (20) or (23).Also, we use these maps to define a topology in order to construct an equivalence relation, and we show that quotient space is isomorphic to𝕊1.



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*Source: 645848-2013-05-08.xml*