Contraction of a Lie Algebra with respect to one of its Subalgebras and its application to Group Representations

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1 Contraction of a Lie Algebra with respect to one of its Subalgebras and its application to Group Representations Maury LeBlanc Department of Mathematics The University of Georgia 1 December 2011

2 Motivation Classical Mechanics is the limiting case of relativistic mechanics. Hence the group of the former, the Galilei group, must be in some sense a limiting case of the relativistic mechanics group, the representations of the former must be limiting cases of the latter s representations. -Inonu and Wigner

3 -1 / 61 Motivation Oftentimes while studying physics, we are forced by our lack of understanding to master first a simple model in hopes of generalizing the result. In the study of curved space-times, it is often convenient first to consider flat Minkowski space-time and then extend to the curved case. In this talk I will investigate one of the tools physicists use to conduct such studies: the concept of contraction. Beginning with a family of manifolds, we can contract this family to a simpler case where we already have a firm understanding of the manifold structure. We will further simplify this process by utilizing Lie algebras.

4 Outline Background Main definitions Examples Main Object of Study: The ax + b Group Group Structure Lie Algebra Structure Contraction of a Lie Algebra with Respect to a Subalgebra R 2 sl2 (C) Heisenberg Algebra ax + b Algebra Constructing a Group Contraction in Unitary Representations Representation of the family Gρ for ρ 0 Contraction in terms of these Representations.

5 Background

6 2 / 61 What is a Lie algebra? Definition A Lie algebra is a vector space, L, over a field with an operation [, ] : L L L called the bracket which satisfies: Bi-linearity [xx] = 0 for all x L [x[yz]] + [y[zx]] + [z[xy]] = 0 for all x, y, z L Remarks: This last condition is the Jacobi Identity and is usually the only condition whose verification requires work. Notice that the first two conditions give anti-commutativity: [yx] = [xy]

7 2 / 61 What is a Lie algebra? Definition A Lie algebra is a vector space, L, over a field with an operation [, ] : L L L called the bracket which satisfies: Bi-linearity [xx] = 0 for all x L [x[yz]] + [y[zx]] + [z[xy]] = 0 for all x, y, z L Remarks: This last condition is the Jacobi Identity and is usually the only condition whose verification requires work. Notice that the first two conditions give anti-commutativity: [yx] = [xy]

8 3 / 61 What is a subalgebra? Definition A subspace s of g is a subalgebra if [xy] s whenever x, y s. A subalgebra is a Lie algebra in its own right relative to the inherited operations.

9 4 / 61 What is a Lie group? Definition A Lie group is a group which is also a differentiable manifold. Therefore in a Lie group, the following maps must be differentiable: the multiplication map (g, h) gh the identity map {e} id (Note: this map is trivially differentiable) the inversion map g g 1

10 5 / 61 How are they connected? Let G be a Lie group. Since the Lie group G has a manifold structure, we can consider the tangent space to G at the id, T e (G). We can define the vector space g = T e (G). For x, y g, we can form left-invariant vector fields X and Y on G with X e = x and Y e = y. The commutation bracket of these two vector fields is another left-invariant vector field: [XY ] = Z Define [xy] := Z e

11 5 / 61 How are they connected? Let G be a Lie group. Since the Lie group G has a manifold structure, we can consider the tangent space to G at the id, T e (G). We can define the vector space g = T e (G). For x, y g, we can form left-invariant vector fields X and Y on G with X e = x and Y e = y. The commutation bracket of these two vector fields is another left-invariant vector field: [XY ] = Z Define [xy] := Z e

12 6 / 61 Example 1: R 2 Example R 2 is an abelian group under addition. It is also a manifold. The multiplication map ((x, y), (x, y )) (x + x, y + y ) and the inversion map (x, y) ( x, y) are differentiable. Hence R 2 is a Lie group. Since T e (R 2 ) = R 2, its Lie algebra is also abelian - i.e. the Lie algebra is spanned by {X, Y } satisfying [XY ] = 0. Notice however that these R 2 s are different objects. The Lie group is equipped with neither a bracket nor a vector space structure.

13 7 / 61 Example 2: SL 2 (C) Example SL 2 (C) is the collection of 2 2 matrices with determinant one. It forms a Lie group under matrix multiplication. Claim The set of all 2 2 complex matrices with trace zero is the Lie algebra of SL 2 (C)

14 7 / 61 Example 2: SL 2 (C) Example SL 2 (C) is the collection of 2 2 matrices with determinant one. It forms a Lie group under matrix multiplication. Claim The set of all 2 2 complex matrices with trace zero is the Lie algebra of SL 2 (C)

15 8 / 61 sl 2 (C) = {x M 2 2 (C) tr(x) = 0} Proof. SL 2 (C) is the set of 2 2 matrices with determinant one. Using the natural correspondence of M 2 2 (C) and C 4, SL 2 (C) can be viewed as the sub-manifold M of ordered quadruples in C 4 for which f (a, b, c, d) = ad bc 1 equals zero. The gradient of f is given by ( d, c, b, a ). The gradient of f at the identity element of SL 2 (C) (which corresponds to (1, 0, 0, 1) in C 4 ) is (1, 0, 0, 1). Since the gradient is orthogonal to the level sets, T (1,0,0,1) (M ) = {(a, b, c, d) C 4 (a, b, c, d) (1, 0, 0, 1) = 0} = {(a, b, c, d) C 4 a + d = 0}. We conclude that sl 2 (C) := T e (SL 2 (C)) = {x M 2 2 (C) trace(x) = 0}.

16 8 / 61 sl 2 (C) = {x M 2 2 (C) tr(x) = 0} Proof. SL 2 (C) is the set of 2 2 matrices with determinant one. Using the natural correspondence of M 2 2 (C) and C 4, SL 2 (C) can be viewed as the sub-manifold M of ordered quadruples in C 4 for which f (a, b, c, d) = ad bc 1 equals zero. The gradient of f is given by ( d, c, b, a ). The gradient of f at the identity element of SL 2 (C) (which corresponds to (1, 0, 0, 1) in C 4 ) is (1, 0, 0, 1). Since the gradient is orthogonal to the level sets, T (1,0,0,1) (M ) = {(a, b, c, d) C 4 (a, b, c, d) (1, 0, 0, 1) = 0} = {(a, b, c, d) C 4 a + d = 0}. We conclude that sl 2 (C) := T e (SL 2 (C)) = {x M 2 2 (C) trace(x) = 0}.

17 8 / 61 sl 2 (C) = {x M 2 2 (C) tr(x) = 0} Proof. SL 2 (C) is the set of 2 2 matrices with determinant one. Using the natural correspondence of M 2 2 (C) and C 4, SL 2 (C) can be viewed as the sub-manifold M of ordered quadruples in C 4 for which f (a, b, c, d) = ad bc 1 equals zero. The gradient of f is given by ( d, c, b, a ). The gradient of f at the identity element of SL 2 (C) (which corresponds to (1, 0, 0, 1) in C 4 ) is (1, 0, 0, 1). Since the gradient is orthogonal to the level sets, T (1,0,0,1) (M ) = {(a, b, c, d) C 4 (a, b, c, d) (1, 0, 0, 1) = 0} = {(a, b, c, d) C 4 a + d = 0}. We conclude that sl 2 (C) := T e (SL 2 (C)) = {x M 2 2 (C) trace(x) = 0}.

18 8 / 61 sl 2 (C) = {x M 2 2 (C) tr(x) = 0} Proof. SL 2 (C) is the set of 2 2 matrices with determinant one. Using the natural correspondence of M 2 2 (C) and C 4, SL 2 (C) can be viewed as the sub-manifold M of ordered quadruples in C 4 for which f (a, b, c, d) = ad bc 1 equals zero. The gradient of f is given by ( d, c, b, a ). The gradient of f at the identity element of SL 2 (C) (which corresponds to (1, 0, 0, 1) in C 4 ) is (1, 0, 0, 1). Since the gradient is orthogonal to the level sets, T (1,0,0,1) (M ) = {(a, b, c, d) C 4 (a, b, c, d) (1, 0, 0, 1) = 0} = {(a, b, c, d) C 4 a + d = 0}. We conclude that sl 2 (C) := T e (SL 2 (C)) = {x M 2 2 (C) trace(x) = 0}.

19 Main Object of Study

20 10 / 61 The ax + b group Example The ax + b group is the collection of affine linear transformations on R, denoted here by G. φ : G Aut(R) is given φ (b,a) (x) = ax + b Natural coordinates are (b, a) with b R and a R 0 I prefer to use coordinates (b, α) with b R and α R by setting e α = a. It is a Lie group under the group law: (b, α) (b, α ) = (b + e α b, α + α ).

21 10 / 61 The ax + b group Example The ax + b group is the collection of affine linear transformations on R, denoted here by G. φ : G Aut(R) is given φ (b,a) (x) = ax + b Natural coordinates are (b, a) with b R and a R 0 I prefer to use coordinates (b, α) with b R and α R by setting e α = a. It is a Lie group under the group law: (b, α) (b, α ) = (b + e α b, α + α ).

22 10 / 61 The ax + b group Example The ax + b group is the collection of affine linear transformations on R, denoted here by G. φ : G Aut(R) is given φ (b,a) (x) = ax + b Natural coordinates are (b, a) with b R and a R 0 I prefer to use coordinates (b, α) with b R and α R by setting e α = a. It is a Lie group under the group law: (b, α) (b, α ) = (b + e α b, α + α ).

23 11 / 61 Structure of the ax + b group Example In order to discover some of the structure of the group, we consider the conjugation action: (b, α) (B, A) (b, α) 1 = (b, α) (B, A) ( be α, α) = (b(1 e A ) + e α B, A) It is clear that the subgroup B = {(b, 0) : b R} is a normal subgroup of G. It follows that G is a solvable group since the subnormal series 1 B G has abelian factor groups. There is a map ψ : R Aut(R) given by ψ α (b) = e α b. G is a non-abelian, simply connected, two-dimensional Lie group which is isomorphic to R R.

24 11 / 61 Structure of the ax + b group Example In order to discover some of the structure of the group, we consider the conjugation action: (b, α) (B, A) (b, α) 1 = (b, α) (B, A) ( be α, α) = (b(1 e A ) + e α B, A) It is clear that the subgroup B = {(b, 0) : b R} is a normal subgroup of G. It follows that G is a solvable group since the subnormal series 1 B G has abelian factor groups. There is a map ψ : R Aut(R) given by ψ α (b) = e α b. G is a non-abelian, simply connected, two-dimensional Lie group which is isomorphic to R R.

25 11 / 61 Structure of the ax + b group Example In order to discover some of the structure of the group, we consider the conjugation action: (b, α) (B, A) (b, α) 1 = (b, α) (B, A) ( be α, α) = (b(1 e A ) + e α B, A) It is clear that the subgroup B = {(b, 0) : b R} is a normal subgroup of G. It follows that G is a solvable group since the subnormal series 1 B G has abelian factor groups. There is a map ψ : R Aut(R) given by ψ α (b) = e α b. G is a non-abelian, simply connected, two-dimensional Lie group which is isomorphic to R R.

26 11 / 61 Structure of the ax + b group Example In order to discover some of the structure of the group, we consider the conjugation action: (b, α) (B, A) (b, α) 1 = (b, α) (B, A) ( be α, α) = (b(1 e A ) + e α B, A) It is clear that the subgroup B = {(b, 0) : b R} is a normal subgroup of G. It follows that G is a solvable group since the subnormal series 1 B G has abelian factor groups. There is a map ψ : R Aut(R) given by ψ α (b) = e α b. G is a non-abelian, simply connected, two-dimensional Lie group which is isomorphic to R R.

27 11 / 61 Structure of the ax + b group Example In order to discover some of the structure of the group, we consider the conjugation action: (b, α) (B, A) (b, α) 1 = (b, α) (B, A) ( be α, α) = (b(1 e A ) + e α B, A) It is clear that the subgroup B = {(b, 0) : b R} is a normal subgroup of G. It follows that G is a solvable group since the subnormal series 1 B G has abelian factor groups. There is a map ψ : R Aut(R) given by ψ α (b) = e α b. G is a non-abelian, simply connected, two-dimensional Lie group which is isomorphic to R R.

28 12 / 61 The ax + b group as a matrix group The ax + b group can be realized as a 2 2 real matrix group. ( e γ : G GL 2 (R) given by (b, α) α ) b 0 1 ( ) 1 0 γ (0, 0) = 0 1 ( γ (b, α) 1 e = γ ( be α, α) = α be α ) 0 1 γ ((b, α) (b, α )) ( = γ (b + e α b, α + α ) e α+α b + e = α b ) 0 1 ( e = α ) ( b e α b ) = γ (b, α) γ (b, α )

29 12 / 61 The ax + b group as a matrix group The ax + b group can be realized as a 2 2 real matrix group. ( e γ : G GL 2 (R) given by (b, α) α ) b 0 1 ( ) 1 0 γ (0, 0) = 0 1 ( γ (b, α) 1 e = γ ( be α, α) = α be α ) 0 1 γ ((b, α) (b, α )) ( = γ (b + e α b, α + α ) e α+α b + e = α b ) 0 1 ( e = α ) ( b e α b ) = γ (b, α) γ (b, α )

30 12 / 61 The ax + b group as a matrix group The ax + b group can be realized as a 2 2 real matrix group. ( e γ : G GL 2 (R) given by (b, α) α ) b 0 1 ( ) 1 0 γ (0, 0) = 0 1 ( γ (b, α) 1 e = γ ( be α, α) = α be α ) 0 1 γ ((b, α) (b, α )) ( = γ (b + e α b, α + α ) e α+α b + e = α b ) 0 1 ( e = α ) ( b e α b ) = γ (b, α) γ (b, α )

31 12 / 61 The ax + b group as a matrix group The ax + b group can be realized as a 2 2 real matrix group. ( e γ : G GL 2 (R) given by (b, α) α ) b 0 1 ( ) 1 0 γ (0, 0) = 0 1 ( γ (b, α) 1 e = γ ( be α, α) = α be α ) 0 1 γ ((b, α) (b, α )) ( = γ (b + e α b, α + α ) e α+α b + e = α b ) 0 1 ( e = α ) ( b e α b ) = γ (b, α) γ (b, α )

32 13 / 61 The Lie algebra of the ax + b group Example In order to discover the Lie algebra of G, it is insightful to use its matrix representation. B := dγ db (0, 0) = ( ) A := dγ dα (0, 0) = ( e α ) (0,0) = (0,0) = ( ) ( ) Hence {B, A} serves as a basis for the tangent space to G at the identity and the Lie algebra g has the relation: ( ) ( ) ( ) ( ) [AB] = AB BA = = B

33 13 / 61 The Lie algebra of the ax + b group Example In order to discover the Lie algebra of G, it is insightful to use its matrix representation. B := dγ db (0, 0) = ( ) A := dγ dα (0, 0) = ( e α ) (0,0) = (0,0) = ( ) ( ) Hence {B, A} serves as a basis for the tangent space to G at the identity and the Lie algebra g has the relation: ( ) ( ) ( ) ( ) [AB] = AB BA = = B

34 13 / 61 The Lie algebra of the ax + b group Example In order to discover the Lie algebra of G, it is insightful to use its matrix representation. B := dγ db (0, 0) = ( ) A := dγ dα (0, 0) = ( e α ) (0,0) = (0,0) = ( ) ( ) Hence {B, A} serves as a basis for the tangent space to G at the identity and the Lie algebra g has the relation: ( ) ( ) ( ) ( ) [AB] = AB BA = = B

35 13 / 61 The Lie algebra of the ax + b group Example In order to discover the Lie algebra of G, it is insightful to use its matrix representation. B := dγ db (0, 0) = ( ) A := dγ dα (0, 0) = ( e α ) (0,0) = (0,0) = ( ) ( ) Hence {B, A} serves as a basis for the tangent space to G at the identity and the Lie algebra g has the relation: ( ) ( ) ( ) ( ) [AB] = AB BA = = B

36 14 / 61 Properties of the Lie algebra g The derived series for g is: g (0) = g g (1) = [g, g] = B g (2) = [g (1), g (1) ] = [BB] = 0. Hence g is a solvable Lie algebra. This is unsurprising as the ax + b group is solvable. The derived series of Lie algebras is the Lie algebra of the derived series of (closed) subgroups. However g is not nilpotent since its descending central series is: g 0 = g g 1 = [g, g] = B g 2 = [g, g 1 ] = [g, B] = B.

37 14 / 61 Properties of the Lie algebra g The derived series for g is: g (0) = g g (1) = [g, g] = B g (2) = [g (1), g (1) ] = [BB] = 0. Hence g is a solvable Lie algebra. This is unsurprising as the ax + b group is solvable. The derived series of Lie algebras is the Lie algebra of the derived series of (closed) subgroups. However g is not nilpotent since its descending central series is: g 0 = g g 1 = [g, g] = B g 2 = [g, g 1 ] = [g, B] = B.

38 Contraction of a Lie Algebra with Respect to a Subalgebra

39 16 / 61 What is a contraction? In this context, a contraction is a deformation of an algebra (or group) into another algebra using a singular transformation. This transformation can be used to model the limiting behavior of one algebra (or group) by effecting the structure of the algebra. A contraction is carried out with respect to a subalgebra (or subgroup); this subalgebra remains unchanged by the contraction.

40 17 / 61 Structure Constants of a Lie algebra A Lie algebra can be described by giving a basis and a set of relations. Definition Given a basis {x i } n i=1 for a Lie algebra g, the relations are given by [x i, x j ] = n k=1 C k ij x k These C k ij are known as the structure constants of g.

41 18 / 61 Contraction with respect to a subalgebra We may contract a Lie algebra with respect to any of its subalgebras by altering the structure constants. For a fixed subalgebra, s, construct a basis {x k } for s. Now extend this basis to a basis for g, {x k } {x µ }. Take c to be the span of {x µ }. We can now write g as a vector space direct sum: g = s c. Notice that c is not usually a subalgebra of g.

42 19 / 61 Contraction with respect to a subalgebra Following the contraction process laid out by Wigner & Inonu, let roman indices refer to basis elements which span s and greek indices refer to basis elements which span c. The new contracted structure constants, c k ij are given by: c k ij = C k ij c µ ij = C µ ij = 0 c k iµ = 0 c ν iµ = C ν iµ c k µν = c ϑ µν = 0 The structure constants of the subalgebra are not affected. Since s is a subalgebra, these structure constants were zero in the uncontracted algebra. This makes c an ideal of the new Lie algebra. This has the effect of abelianizing c.

43 20 / 61 Alternative description of a contraction Alternatively, we can demonstrate how the bracket between two arbitrary elements of g is affected by contraction. Let [, ] o denote the contracted Lie algebra relations and the unadorned bracket denote the original Lie algebra relations. Fix arbitrary s, t s and y, z c where c is the vector space complement of s in g. [st] o = [st] [sy] o = proj c [sy] [yz] o = 0 the projection of the bracket onto its component in c

44 21 / 61 The object described is in fact a Lie algebra Claim The contraction of a Lie algebra produces another Lie algebra. Proof. Bi-linearity and Alternating are immediately inherited from the original Lie algebra. The Jacobi Identity [x[yz] o ] o + [y[zx] o ] o + [z[xy] o ] o = 0 must be verified.

45 22 / 61 The object described is in fact a Lie algebra Proof. We can separate each bracket into its s and c components & write: [x[yz] s ] s + [x[yz] c ] s + [x[yz] s ] c + [x[yz] c ] c + [y[zx] s ] s + [y[zx] c ] s + [y[zx] s ] c + [y[zx] c ] c + [z[xy] s ] s + [z[xy] c ] s + [z[xy] s ] c + [z[xy] c ] c = 0 Thus it must be shown that [x[yz] o s ] o s + [x[yz] o c ] o s + [y[zx] o s ] o s + [y[zx] o c ] o s + [z[xy] o s ] o s + [z[xy] o c ] o s = 0 and that [x[yz] o s ] o c + [x[yz] o c ] o c + [y[zx] o s ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c + [z[xy] o c ] o c = 0

46 22 / 61 The object described is in fact a Lie algebra Proof. We can separate each bracket into its s and c components & write: [x[yz] s ] s + [x[yz] c ] s + [x[yz] s ] c + [x[yz] c ] c + [y[zx] s ] s + [y[zx] c ] s + [y[zx] s ] c + [y[zx] c ] c + [z[xy] s ] s + [z[xy] c ] s + [z[xy] s ] c + [z[xy] c ] c = 0 Thus it must be shown that [x[yz] o s ] o s + [x[yz] o c ] o s + [y[zx] o s ] o s + [y[zx] o c ] o s + [z[xy] o s ] o s + [z[xy] o c ] o s = 0 and that [x[yz] o s ] o c + [x[yz] o c ] o c + [y[zx] o s ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c + [z[xy] o c ] o c = 0

47 23 / 61 The object described is in fact a Lie algebra Proof. If x s and y, z c, then [yz] o = 0, [xy] o s = 0, [zx] o s = 0 Hence the first expression simplifies to: [x[yz] o s ] o s + [x[yz] o c ] o s + [y[zx] o s ] o s + [y[zx] o c ] o s + [z[xy] o s ] o s + [z[xy] o c ] o s = 0 and the second simplifies to: [x[yz] o s ] o c + [x[yz] o c ] o c + [y[zx] o s ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c + [z[xy] o c ] o c = 0 Each of the remaining brackets is the bracket of elements in c and thus equals zero. Hence the Jacobi Identity is satisfied.

48 24 / 61 The object described is in fact a Lie algebra Proof. If x, y s and z c, then [xy] o c = 0, [zx] o s = 0, [yz] o s = 0 Hence the first expression simplifies to: [x[yz] o s ] o s + [x[yz] o c ] o s + [y[zx] o s ] o s + [y[zx] o c ] o s + [z[xy] o s ] o s + [z[xy] o c ] o s = 0 and the second simplifies to: [x[yz] o s ] o c + [x[yz] o c ] o c + [y[zx] o s ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c + [z[xy] o c ] o c = 0 Since [yz] o c and [zx] o c c, [x[yz] o c ] o s = 0 and [y[zx] o c ] o s = 0. [z[xy] o s ] c implies that [z[xy] o s ] o s = 0 Hence the first expression equals zero.

49 24 / 61 The object described is in fact a Lie algebra Proof. If x, y s and z c, then [xy] o c = 0, [zx] o s = 0, [yz] o s = 0 Hence the first expression simplifies to: [x[yz] o s ] o s + [x[yz] o c ] o s + [y[zx] o s ] o s + [y[zx] o c ] o s + [z[xy] o s ] o s + [z[xy] o c ] o s = 0 and the second simplifies to: [x[yz] o s ] o c + [x[yz] o c ] o c + [y[zx] o s ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c + [z[xy] o c ] o c = 0 Since [yz] o c and [zx] o c c, [x[yz] o c ] o s = 0 and [y[zx] o c ] o s = 0. [z[xy] o s ] c implies that [z[xy] o s ] o s = 0 Hence the first expression equals zero.

50 25 / 61 The object described is in fact a Lie algebra Proof. In order to dispense with the second expression, [x[yz] o c ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c we use the fact that the Jacobi Identity was satisfied in the original Lie algebra. Separating the expression as before... [x[yz] s ] c +[x[yz] c ] c +[y[zx] s ] c +[y[zx] c ] c +[z[xy] s ] c +[z[xy] c ] c = 0 Since x, y, [yz] s, [zx] s s, the first, third, and last term are zero. Thus [x[yz] c ] c + [y[zx] c ] c + [z[xy] s ] c = 0 and the proof is complete.

51 25 / 61 The object described is in fact a Lie algebra Proof. In order to dispense with the second expression, [x[yz] o c ] o c + [y[zx] o c ] o c + [z[xy] o s ] o c we use the fact that the Jacobi Identity was satisfied in the original Lie algebra. Separating the expression as before... [x[yz] s ] c +[x[yz] c ] c +[y[zx] s ] c +[y[zx] c ] c +[z[xy] s ] c +[z[xy] c ] c = 0 Since x, y, [yz] s, [zx] s s, the first, third, and last term are zero. Thus [x[yz] c ] c + [y[zx] c ] c + [z[xy] s ] c = 0 and the proof is complete.

52 26 / 61 A more abstract approach Claim Let g be a Lie algebra with subalgebra s. The Lie algebra produced by contracting g with respect to s can be realized as a semi-direct product of Lie algebras. Proof. Consider the vector space direct sum decomposition: g = s c where c is defined as it was previously. Define the map φ : s End vs (g) where φ s (x) = [sx] for s s and x g.

53 27 / 61 A more abstract approach Proof. This map gives an action of s on the vector space g which maps the subalgebra s to itself. Hence there is a vector space action of s on the quotient vector space a := g / s We can consider this vector space a as an abelian Lie algebra. Notice that c = a as vector spaces.

54 28 / 61 A more abstract approach Proof. We now have a Lie algebra a and a Lie algebra action of s on this Lie algebra a. Since a vector space endomorphism on an abelian Lie algebra is automatically a Lie algebra endomorphism, we may form the Lie algebra semi-direct product g o := s a. This new object is isomorphic to g as a vector space and has the desired structure constant changes. Remark: The need to demonstrate that contractions produce Lie algebras was first brought to my attention by Dr. Rothstein. Fortunately he and Dr. Usher then assisted me in these proofs.

55 29 / 61 Contractions - Example 1: R 2 Example R 2 is an abelian Lie algebra with basis {X, Y } and relation [XY ] = 0 Clearly this Lie algebra is already abelian and therefore a contraction can have no effect upon the structure constants. Hence any contraction of the Lie algebra R 2 is trivially R 2.

56 30 / 61 Contractions - Example 2: sl 2 (C) Example sl 2 (C) is the simplest non-abelian Lie algebra so it is our first non-trivial example. sl 2 (C) is a three-dimensional Lie algebra with basis {x, h, y} and relations [xy] = h, [hx] = 2x, and [hy] = 2y

57 31 / 61 Contractions - Example 2: sl 2 (C) Example Let us contract sl 2 (C) with respect to the subalgebra spanned by {x}. [xy] = h Since h a, this bracket relation is unaffected. [hx] = 2x After contracting the bracket of an element of s with an element of a must be in a. We therefore annihilate this bracket. [hx] o = 0 [hy] = 2y After contraction the subalgebra a is abelian; hence this bracket must also be set to zero. [hy] o = 0

58 31 / 61 Contractions - Example 2: sl 2 (C) Example Let us contract sl 2 (C) with respect to the subalgebra spanned by {x}. [xy] = h Since h a, this bracket relation is unaffected. [hx] = 2x After contracting the bracket of an element of s with an element of a must be in a. We therefore annihilate this bracket. [hx] o = 0 [hy] = 2y After contraction the subalgebra a is abelian; hence this bracket must also be set to zero. [hy] o = 0

59 32 / 61 Contractions - Example 2: sl 2 (C) Example Hence after contraction, we produce a new Lie algebra g o with basis {x, h, y} and relations [xy] o = h while [hx] o = [hy] o = 0. This Lie algebra is known as the Heisenberg Algebra which will be one of my later examples. The contraction with respect to the subalgebra spanned by {y} produces the same algebra.

60 32 / 61 Contractions - Example 2: sl 2 (C) Example Hence after contraction, we produce a new Lie algebra g o with basis {x, h, y} and relations [xy] o = h while [hx] o = [hy] o = 0. This Lie algebra is known as the Heisenberg Algebra which will be one of my later examples. The contraction with respect to the subalgebra spanned by {y} produces the same algebra.

61 33 / 61 Contractions - Example 2: sl 2 (C) Example Consider the contraction of sl 2 (C) with respect to the subalgebra spanned by {h}. [xy] o = 0 since a is abelian. [hx] o = [hx] = 2x and [hy] o = [hy] = 2y since these relations conform with a being an abelian ideal.

62 33 / 61 Contractions - Example 2: sl 2 (C) Example Consider the contraction of sl 2 (C) with respect to the subalgebra spanned by {h}. [xy] o = 0 since a is abelian. [hx] o = [hx] = 2x and [hy] o = [hy] = 2y since these relations conform with a being an abelian ideal.

63 34 / 61 Contractions - Example 2: sl 2 (C) Example We are unable to contract with respect to {x, y} as this set does not form a subalgebra. Now we consider the contraction with respect to the subalgebra spanned by {x, h} [hx] o = [hx] = 2x The subalgebra with respect to which the contraction is performed is unaffected. [xy] o = 0 This follows from the requirement that a be an ideal. [hy] o = 2y is compatible with a being an ideal and is unaffected. Again the contraction with respect to the subalgebra spanned by {h, y} has the same effect on the structure constants.

64 34 / 61 Contractions - Example 2: sl 2 (C) Example We are unable to contract with respect to {x, y} as this set does not form a subalgebra. Now we consider the contraction with respect to the subalgebra spanned by {x, h} [hx] o = [hx] = 2x The subalgebra with respect to which the contraction is performed is unaffected. [xy] o = 0 This follows from the requirement that a be an ideal. [hy] o = 2y is compatible with a being an ideal and is unaffected. Again the contraction with respect to the subalgebra spanned by {h, y} has the same effect on the structure constants.

65 34 / 61 Contractions - Example 2: sl 2 (C) Example We are unable to contract with respect to {x, y} as this set does not form a subalgebra. Now we consider the contraction with respect to the subalgebra spanned by {x, h} [hx] o = [hx] = 2x The subalgebra with respect to which the contraction is performed is unaffected. [xy] o = 0 This follows from the requirement that a be an ideal. [hy] o = 2y is compatible with a being an ideal and is unaffected. Again the contraction with respect to the subalgebra spanned by {h, y} has the same effect on the structure constants.

66 35 / 61 Contractions - Example 3: The Heisenberg Algebra Example As was shown above, the Heisenberg Algebra is the contraction of sl 2 (C) with respect to to subalgebra spanned by {x}. It has basis {x, h, y} and non-zero relation [xy] = h Contraction with respect to subalgebra spanned by {x} [xy] o = [xy] = h This is a simple example of the fact that contracting a Lie algebra with respect to the same subalgebra twice has no further effect.

67 35 / 61 Contractions - Example 3: The Heisenberg Algebra Example As was shown above, the Heisenberg Algebra is the contraction of sl 2 (C) with respect to to subalgebra spanned by {x}. It has basis {x, h, y} and non-zero relation [xy] = h Contraction with respect to subalgebra spanned by {x} [xy] o = [xy] = h This is a simple example of the fact that contracting a Lie algebra with respect to the same subalgebra twice has no further effect.

68 36 / 61 Contractions - Example 3: The Heisenberg Algebra Example The Heisenberg Algebra has basis {x, h, y} and relation [xy] = h Contraction with respect to the subalgebra spanned by {h} produces the abelian Lie algebra C 3 since the ideal spanned by {x, y} is abelian after undertaking the contraction.

69 37 / 61 Contractions - Example 4: the ax + b algebra Example The Lie algebra for the ax + b group is two-dimensional with basis {A, B} and relation [AB] = B The contraction with respect to the subalgebra spanned by {A} has no effect upon the Lie algebra structure since [AB] = B which lies in the abelian ideal a as required. However the contraction with respect to the subalgebra spanned by {B} abelianizes the algebra since [AB] o must lie in the abelian ideal spanned by {A}. Thus after contraction with respect to the subalgebra spanned by {B}, the Lie algebra for the ax + b group becomes abelian.

70 37 / 61 Contractions - Example 4: the ax + b algebra Example The Lie algebra for the ax + b group is two-dimensional with basis {A, B} and relation [AB] = B The contraction with respect to the subalgebra spanned by {A} has no effect upon the Lie algebra structure since [AB] = B which lies in the abelian ideal a as required. However the contraction with respect to the subalgebra spanned by {B} abelianizes the algebra since [AB] o must lie in the abelian ideal spanned by {A}. Thus after contraction with respect to the subalgebra spanned by {B}, the Lie algebra for the ax + b group becomes abelian.

71 Constructing a Group Contraction in Unitary Representations

72 39 / 61 A family of Lie groups: G ρ Definition The ax + b group can be naturally generalized by considering a family of Lie groups {G ρ : ρ 0} Using the global coordinates (b, α) ρ, the group law can be written as (b, α) ρ (b, α ) ρ = (b + e ρα b, α + α ) ρ For ρ = 0 the group law deteriorates to (b, α) 0 (b, α ) 0 = (b + b, α + α ) 0. Hence G 0 = R 2 with its usual abelian structure. The ax + b group is G 1

73 40 / 61 A family of Lie groups: G ρ Claim For ρ > 0, G ρ = G1 as Lie groups Proof. Consider the map π : G ρ G 1 given by π((b, α) ρ ) = (b, ρ α) In order to verify this is a homomorphism of groups, it must be shown that π ((b, α) ρ (b, α ) ρ ) = π((b, α) ρ ) π((b, α ) ρ ) π ((b, α) ρ (b, α ) ρ ) = π (b + e ρ α b, α + α ) ρ ) = (b + e ρ α b, ρ (α + α )) = (b, ρ α) (b, ρ α) = π((b, α) ρ ) π((b, α ) ρ )

74 40 / 61 A family of Lie groups: G ρ Claim For ρ > 0, G ρ = G1 as Lie groups Proof. Consider the map π : G ρ G 1 given by π((b, α) ρ ) = (b, ρ α) In order to verify this is a homomorphism of groups, it must be shown that π ((b, α) ρ (b, α ) ρ ) = π((b, α) ρ ) π((b, α ) ρ ) π ((b, α) ρ (b, α ) ρ ) = π (b + e ρ α b, α + α ) ρ ) = (b + e ρ α b, ρ (α + α )) = (b, ρ α) (b, ρ α) = π((b, α) ρ ) π((b, α ) ρ )

75 40 / 61 A family of Lie groups: G ρ Claim For ρ > 0, G ρ = G1 as Lie groups Proof. Consider the map π : G ρ G 1 given by π((b, α) ρ ) = (b, ρ α) In order to verify this is a homomorphism of groups, it must be shown that π ((b, α) ρ (b, α ) ρ ) = π((b, α) ρ ) π((b, α ) ρ ) π ((b, α) ρ (b, α ) ρ ) = π (b + e ρ α b, α + α ) ρ ) = (b + e ρ α b, ρ (α + α )) = (b, ρ α) (b, ρ α) = π((b, α) ρ ) π((b, α ) ρ )

76 41 / 61 What is a group representation? Definition A group representation of a group G on a vector space V is a group homomorphism from G into GL(V ). If Φ : G GL(V ) is a representation, then Φ(g h) = Φ(g) Φ(h) for all g, h G. A representation is called (algebraically) irreducible if the only subspaces of V which are invariant under the action of G are the zero-dimensional subspace and V itself. A representation is unitary if it preserves a norm on V i.e. Φ g (v) = v for every g G and v V. If a representation is injective, then it is called faithful.

77 41 / 61 What is a group representation? Definition A group representation of a group G on a vector space V is a group homomorphism from G into GL(V ). If Φ : G GL(V ) is a representation, then Φ(g h) = Φ(g) Φ(h) for all g, h G. A representation is called (algebraically) irreducible if the only subspaces of V which are invariant under the action of G are the zero-dimensional subspace and V itself. A representation is unitary if it preserves a norm on V i.e. Φ g (v) = v for every g G and v V. If a representation is injective, then it is called faithful.

78 41 / 61 What is a group representation? Definition A group representation of a group G on a vector space V is a group homomorphism from G into GL(V ). If Φ : G GL(V ) is a representation, then Φ(g h) = Φ(g) Φ(h) for all g, h G. A representation is called (algebraically) irreducible if the only subspaces of V which are invariant under the action of G are the zero-dimensional subspace and V itself. A representation is unitary if it preserves a norm on V i.e. Φ g (v) = v for every g G and v V. If a representation is injective, then it is called faithful.

79 41 / 61 What is a group representation? Definition A group representation of a group G on a vector space V is a group homomorphism from G into GL(V ). If Φ : G GL(V ) is a representation, then Φ(g h) = Φ(g) Φ(h) for all g, h G. A representation is called (algebraically) irreducible if the only subspaces of V which are invariant under the action of G are the zero-dimensional subspace and V itself. A representation is unitary if it preserves a norm on V i.e. Φ g (v) = v for every g G and v V. If a representation is injective, then it is called faithful.

80 41 / 61 What is a group representation? Definition A group representation of a group G on a vector space V is a group homomorphism from G into GL(V ). If Φ : G GL(V ) is a representation, then Φ(g h) = Φ(g) Φ(h) for all g, h G. A representation is called (algebraically) irreducible if the only subspaces of V which are invariant under the action of G are the zero-dimensional subspace and V itself. A representation is unitary if it preserves a norm on V i.e. Φ g (v) = v for every g G and v V. If a representation is injective, then it is called faithful.

81 42 / 61 Equivalent representations Definition Two representations of a group G, β 1 and β 2, are equivalent if there exists a vector space isomorphism γ such that β 2 (g) = γ 1 β 1 γ(g) for all g G.

82 43 / 61 Representation of the ax + b group on L 2 (R) Let G 1 be the ax + b group Consider the group homomorphism Φ : G 1 GL(L 2 (R)) where Φ (b,α) ψ(x) = e α 2 +ibx ψ(e α x) for (b, α) G 1 and ψ(x) L 2 (R)

83 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

84 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

85 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

86 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

87 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

88 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

89 44 / 61 Representation of the ax + b group on L 2 (R) We must verify that this is indeed a group homomorphism. ( Φ(b,α) Φ (b,α )ψ ) (x) = e α 2 +ibx ( Φ (b,α )ψ ) (e α x) = e α 2 +ibx e α 2 +ib (e α x) ψ(e α e α x) = e α+α 2 +ibx+ib e αx ψ(e α+α x) = e α+α 2 +i(b+b e α )x ψ(e α+α x) = Φ (b+e α b,α+α )ψ(x) = Φ (b,α)(b,α )ψ(x) Hence Φ is a group homomorphism from G 1 to GL(V ).

90 45 / 61 Φ (b,α) is a unitary operator on L 2 (R) Let Φ (b,α) ψ(x) = e α 2 +ibx ψ(e α x) Claim Φ is a unitary representation of G 1 on L 2 (R) Proof. Φ (b,α) ψ(x) 2 2 = e α 2 +ibx ψ(e α x) 2 e α 2 = 2 +ibx ψ(e α x) 2 dx = e α ψ(e α x) 2 dx = ψ(y) 2 dy = ψ(y) 2 2

91 45 / 61 Φ (b,α) is a unitary operator on L 2 (R) Let Φ (b,α) ψ(x) = e α 2 +ibx ψ(e α x) Claim Φ is a unitary representation of G 1 on L 2 (R) Proof. Φ (b,α) ψ(x) 2 2 = e α 2 +ibx ψ(e α x) 2 e α 2 = 2 +ibx ψ(e α x) 2 dx = e α ψ(e α x) 2 dx = ψ(y) 2 dy = ψ(y) 2 2

92 46 / 61 The only game in town The representation Φ (b,α) on L 2 (R) is practically the only such irreducible infinite-dimensional representation of the ax + b group. There is another (non-equivalent) representation that can be produced by mapping b b. For each s R, there is also a unitary finite-dimensional representation of the ax + b group (G 1 ) on C given by: where Υ s : G 1 GL 1 (C) Υ s (b,α) (z) = eisα z for (b, α) G 1 and z C.

93 46 / 61 The only game in town The representation Φ (b,α) on L 2 (R) is practically the only such irreducible infinite-dimensional representation of the ax + b group. There is another (non-equivalent) representation that can be produced by mapping b b. For each s R, there is also a unitary finite-dimensional representation of the ax + b group (G 1 ) on C given by: where Υ s : G 1 GL 1 (C) Υ s (b,α) (z) = eisα z for (b, α) G 1 and z C.

94 47 / 61 Representation of the G ρ on L 2 (R) for ρ > 0 It is natural to extend Φ to a representation Φ ρ of G ρ on L 2 (R) by using the isomorphism π : G ρ G 1 which maps (b, α) ρ to (b, ρα) Thus there is a representation Φ ρ,(b,α)ρ = Φ (b,ρα) of G ρ on L 2 (R) ( Φρ,(b,α)ρ ψ ) (x) = ( Φ (b,ρα) ψ ) (x) = e ρα 2 +ibx ψ(e ρα x)

95 47 / 61 Representation of the G ρ on L 2 (R) for ρ > 0 It is natural to extend Φ to a representation Φ ρ of G ρ on L 2 (R) by using the isomorphism π : G ρ G 1 which maps (b, α) ρ to (b, ρα) Thus there is a representation Φ ρ,(b,α)ρ = Φ (b,ρα) of G ρ on L 2 (R) ( Φρ,(b,α)ρ ψ ) (x) = ( Φ (b,ρα) ψ ) (x) = e ρα 2 +ibx ψ(e ρα x)

96 48 / 61 An equivalent representation of G ρ Example In order to see the effect on the representation of the contraction ρ 0, it is easier to consider the equivalent representation Π ρ,(b,α)ρ := M 1 if ρ if (x) where M if is multiplication by e ρ ρ function f (x) and ρ > 0. M if ρ Φ ρ,(b,α)ρ M if ρ for a real-valued, differentiable is an operator which maps L 2 (R) to itself isomorphically. 1 Thus Π ρ,(b,α)ρ is equivalent to Φ ρ,(b,α)ρ 1 See slides 63 & 64

97 49 / 61 An equivalent representation of G ρ Example Let ψ L 2 (R) ( Πρ,(b,α)ρ ψ ) (x) = if (x) ( = e ρ ( ) M 1 if Φ ρ,(b,α)ρ M if ψ (x) ρ ρ ) Φ ρ,(b,α)ρ M if ψ (x) ρ ) (e ρα x) ( if (x) = e ρ e ρα 2 +ibx M if ψ ρ if (x) = e ρ e ρα 2 +ibx e if (eρα x) ρ ψ(e ρα x) = e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x)

98 49 / 61 An equivalent representation of G ρ Example Let ψ L 2 (R) ( Πρ,(b,α)ρ ψ ) (x) = if (x) ( = e ρ ( ) M 1 if Φ ρ,(b,α)ρ M if ψ (x) ρ ρ ) Φ ρ,(b,α)ρ M if ψ (x) ρ ) (e ρα x) ( if (x) = e ρ e ρα 2 +ibx M if ψ ρ if (x) = e ρ e ρα 2 +ibx e if (eρα x) ρ ψ(e ρα x) = e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x)

99 49 / 61 An equivalent representation of G ρ Example Let ψ L 2 (R) ( Πρ,(b,α)ρ ψ ) (x) = if (x) ( = e ρ ( ) M 1 if Φ ρ,(b,α)ρ M if ψ (x) ρ ρ ) Φ ρ,(b,α)ρ M if ψ (x) ρ ) (e ρα x) ( if (x) = e ρ e ρα 2 +ibx M if ψ ρ if (x) = e ρ e ρα 2 +ibx e if (eρα x) ρ ψ(e ρα x) = e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x)

100 49 / 61 An equivalent representation of G ρ Example Let ψ L 2 (R) ( Πρ,(b,α)ρ ψ ) (x) = if (x) ( = e ρ ( ) M 1 if Φ ρ,(b,α)ρ M if ψ (x) ρ ρ ) Φ ρ,(b,α)ρ M if ψ (x) ρ ) (e ρα x) ( if (x) = e ρ e ρα 2 +ibx M if ψ ρ if (x) = e ρ e ρα 2 +ibx e if (eρα x) ρ ψ(e ρα x) = e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x)

101 49 / 61 An equivalent representation of G ρ Example Let ψ L 2 (R) ( Πρ,(b,α)ρ ψ ) (x) = if (x) ( = e ρ ( ) M 1 if Φ ρ,(b,α)ρ M if ψ (x) ρ ρ ) Φ ρ,(b,α)ρ M if ψ (x) ρ ) (e ρα x) ( if (x) = e ρ e ρα 2 +ibx M if ψ ρ if (x) = e ρ e ρα 2 +ibx e if (eρα x) ρ ψ(e ρα x) = e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x)

102 50 / 61 lim ρ 0 Π ρ,(b,α) ρ By sending ρ 0, the ax + b group (aka G 1 ) can be contracted to R 2 with its usual abelian structure. ( lim Πρ,(b,α)ρ ψ ) (x) ρ 0 = lim e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) ρ 0 = lim e if (e ρα x)e ρααx e ρα 2 e ibx ψ(e ρα x) ρ 0 = e ixf (x)α e ibx ψ(x) We can therefore define ( Π 0,(b,α)0 ψ ) (x) := e ixf (x)α e ibx ψ(x)

103 50 / 61 lim ρ 0 Π ρ,(b,α) ρ By sending ρ 0, the ax + b group (aka G 1 ) can be contracted to R 2 with its usual abelian structure. ( lim Πρ,(b,α)ρ ψ ) (x) ρ 0 = lim e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) ρ 0 = lim e if (e ρα x)e ρααx e ρα 2 e ibx ψ(e ρα x) ρ 0 = e ixf (x)α e ibx ψ(x) We can therefore define ( Π 0,(b,α)0 ψ ) (x) := e ixf (x)α e ibx ψ(x)

104 50 / 61 lim ρ 0 Π ρ,(b,α) ρ By sending ρ 0, the ax + b group (aka G 1 ) can be contracted to R 2 with its usual abelian structure. ( lim Πρ,(b,α)ρ ψ ) (x) ρ 0 = lim e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) ρ 0 = lim e if (e ρα x)e ρααx e ρα 2 e ibx ψ(e ρα x) ρ 0 = e ixf (x)α e ibx ψ(x) We can therefore define ( Π 0,(b,α)0 ψ ) (x) := e ixf (x)α e ibx ψ(x)

105 50 / 61 lim ρ 0 Π ρ,(b,α) ρ By sending ρ 0, the ax + b group (aka G 1 ) can be contracted to R 2 with its usual abelian structure. ( lim Πρ,(b,α)ρ ψ ) (x) ρ 0 = lim e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) ρ 0 = lim e if (e ρα x)e ρααx e ρα 2 e ibx ψ(e ρα x) ρ 0 = e ixf (x)α e ibx ψ(x) We can therefore define ( Π 0,(b,α)0 ψ ) (x) := e ixf (x)α e ibx ψ(x)

106 50 / 61 lim ρ 0 Π ρ,(b,α) ρ By sending ρ 0, the ax + b group (aka G 1 ) can be contracted to R 2 with its usual abelian structure. ( lim Πρ,(b,α)ρ ψ ) (x) ρ 0 = lim e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) ρ 0 = lim e if (e ρα x)e ρααx e ρα 2 e ibx ψ(e ρα x) ρ 0 = e ixf (x)α e ibx ψ(x) We can therefore define ( Π 0,(b,α)0 ψ ) (x) := e ixf (x)α e ibx ψ(x)

107 51 / 61 Contraction of a Lie group With the variable parameter ρ in hand, we have a simple, intuitive model for the concept of a group contraction. By allowing ρ 0, the ax + b group can be contracted to R 2. Since it has been shown that G ρ = G1 for all ρ > 0, we expect something dramatic to happen at ρ = 0. However Lie groups are often difficult to work with, and a similar but simpler problem can be considered by working with the Lie algebra of G ρ, which will be notated as g ρ.

108 52 / 61 Contraction of a Lie group The Lie algebra is an indispensable tool in studying the matrix Lie groups. On the one hand, Lie algebras are simpler that the matrix Lie groups, because the Lie algebra is a linear space. Thus, we can understand much about Lie algebras just by doing linear algebra. On the other hand, the Lie algebra of a matrix Lie group contains much information about that group. Thus many questions about matrix Lie groups can be answered by considering a similar but easier problem for the Lie algebra. -Brian Hall

109 53 / 61 The representation of g ρ on L 2 (R) for ρ > 0 In order to discover the representation of the Lie algebra on L 2 (R), we differentiate the representation of the Lie group. π b ρ:= d db (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d db (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = ixψ(x) Notice that this operator is independent of ρ

110 53 / 61 The representation of g ρ on L 2 (R) for ρ > 0 In order to discover the representation of the Lie algebra on L 2 (R), we differentiate the representation of the Lie group. π b ρ:= d db (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d db (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = ixψ(x) Notice that this operator is independent of ρ

111 53 / 61 The representation of g ρ on L 2 (R) for ρ > 0 In order to discover the representation of the Lie algebra on L 2 (R), we differentiate the representation of the Lie group. π b ρ:= d db (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d db (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = ixψ(x) Notice that this operator is independent of ρ

112 53 / 61 The representation of g ρ on L 2 (R) for ρ > 0 In order to discover the representation of the Lie algebra on L 2 (R), we differentiate the representation of the Lie group. π b ρ:= d db (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d db (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = ixψ(x) Notice that this operator is independent of ρ

113 53 / 61 The representation of g ρ on L 2 (R) for ρ > 0 In order to discover the representation of the Lie algebra on L 2 (R), we differentiate the representation of the Lie group. π b ρ:= d db (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d db (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = ixψ(x) Notice that this operator is independent of ρ

114 54 / 61 The representation of g ρ on L 2 (R) for ρ > 0 π α ρ := d dα (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d dα (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = [e i f (e ρα x) f (x) i ρ ρ f (e ρα x)e ρα xρ e ρα 2 e ibx ψ(e ρα x) + e i f (e + e i f (e ρα x) f (x) ρ e ρα 2 ρ 2 eibx ψ(e ρα x) ρα x) f (x) ρ e ρα 2 e ibx ψ (e ρα x)e ρα xρ] (0,0) = i ρ f (x)xρψ(x) + ρ 2 ψ(x) + xρψ (x) = ( ixf (x) + ρ( x d dx )) ψ(x)

115 54 / 61 The representation of g ρ on L 2 (R) for ρ > 0 π α ρ := d dα (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d dα (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = [e i f (e ρα x) f (x) i ρ ρ f (e ρα x)e ρα xρ e ρα 2 e ibx ψ(e ρα x) + e i f (e + e i f (e ρα x) f (x) ρ e ρα 2 ρ 2 eibx ψ(e ρα x) ρα x) f (x) ρ e ρα 2 e ibx ψ (e ρα x)e ρα xρ] (0,0) = i ρ f (x)xρψ(x) + ρ 2 ψ(x) + xρψ (x) = ( ixf (x) + ρ( x d dx )) ψ(x)

116 54 / 61 The representation of g ρ on L 2 (R) for ρ > 0 π α ρ := d dα (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d dα (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = [e i f (e ρα x) f (x) i ρ ρ f (e ρα x)e ρα xρ e ρα 2 e ibx ψ(e ρα x) + e i f (e + e i f (e ρα x) f (x) ρ e ρα 2 ρ 2 eibx ψ(e ρα x) ρα x) f (x) ρ e ρα 2 e ibx ψ (e ρα x)e ρα xρ] (0,0) = i ρ f (x)xρψ(x) + ρ 2 ψ(x) + xρψ (x) = ( ixf (x) + ρ( x d dx )) ψ(x)

117 54 / 61 The representation of g ρ on L 2 (R) for ρ > 0 π α ρ := d dα (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d dα (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = [e i f (e ρα x) f (x) i ρ ρ f (e ρα x)e ρα xρ e ρα 2 e ibx ψ(e ρα x) + e i f (e + e i f (e ρα x) f (x) ρ e ρα 2 ρ 2 eibx ψ(e ρα x) ρα x) f (x) ρ e ρα 2 e ibx ψ (e ρα x)e ρα xρ] (0,0) = i ρ f (x)xρψ(x) + ρ 2 ψ(x) + xρψ (x) = ( ixf (x) + ρ( x d dx )) ψ(x)

118 54 / 61 The representation of g ρ on L 2 (R) for ρ > 0 π α ρ := d dα (0,0) ( Πρ,(b,α)ρ ψ ) (x) = d dα (0,0)e i f (e ρα x) f (x) ρ e ρα 2 e ibx ψ(e ρα x) = [e i f (e ρα x) f (x) i ρ ρ f (e ρα x)e ρα xρ e ρα 2 e ibx ψ(e ρα x) + e i f (e + e i f (e ρα x) f (x) ρ e ρα 2 ρ 2 eibx ψ(e ρα x) ρα x) f (x) ρ e ρα 2 e ibx ψ (e ρα x)e ρα xρ] (0,0) = i ρ f (x)xρψ(x) + ρ 2 ψ(x) + xρψ (x) = ( ixf (x) + ρ( x d dx )) ψ(x)

119 55 / 61 The representation of the Lie algebra of G ρ on L 2 (R) The basis for the operators in the Lie algebra representation of g ρ in L 2 (R) for ρ > 0 is: π b ρ = ix π α ρ = ( ixf (x) + ρ( x d dx )) Notice that for ρ > 0, the operators π b 0 and πα 0 do not commute. Note also that these operators are not bounded on L 2 (R)

120 55 / 61 The representation of the Lie algebra of G ρ on L 2 (R) The basis for the operators in the Lie algebra representation of g ρ in L 2 (R) for ρ > 0 is: π b ρ = ix π α ρ = ( ixf (x) + ρ( x d dx )) Notice that for ρ > 0, the operators π b 0 and πα 0 do not commute. Note also that these operators are not bounded on L 2 (R)

121 56 / 61 The representation of the Lie algebra of G ρ on L 2 (R) In order to discover the action after contraction, send ρ 0 π b 0 = ix π α 0 = ixf (x) Notice that after contracting, the operators π b 0 and πα 0 now commute, but remain unbounded on L 2 (R).

122 56 / 61 The representation of the Lie algebra of G ρ on L 2 (R) In order to discover the action after contraction, send ρ 0 π b 0 = ix π α 0 = ixf (x) Notice that after contracting, the operators π b 0 and πα 0 now commute, but remain unbounded on L 2 (R).

123 57 / 61 The representation of the Lie algebra of G ρ on L 2 (R) This corresponds to contracting the Lie group with respect to the B = {(b, 0) : b R} subgroup as we have considered the representation associated to Π ρ,(b,α)ρ := M 1 if ρ = M 1 if ρ Φ ρ,(b,α)ρ M if ρ Φ (b,ρα) M if ρ The contraction of the Lie group with respect to the subgroup A = {(0, α) : α R} has no effect on the representation of the group and subsequently no effect on the representation of the algebra. 2 2 see slide 65

124 57 / 61 The representation of the Lie algebra of G ρ on L 2 (R) This corresponds to contracting the Lie group with respect to the B = {(b, 0) : b R} subgroup as we have considered the representation associated to Π ρ,(b,α)ρ := M 1 if ρ = M 1 if ρ Φ ρ,(b,α)ρ M if ρ Φ (b,ρα) M if ρ The contraction of the Lie group with respect to the subgroup A = {(0, α) : α R} has no effect on the representation of the group and subsequently no effect on the representation of the algebra. 2 2 see slide 65

125 58 / 61 Applications and Future Research I hope to use this concept of contractions to investigate the geometry of curved spacetimes and their associated Hilbert spaces. As flat spacetimes are well-understood, the contraction of a curved spacetime to a flat one may give insight into the physical realm. The techniques described here can help to carry out a study of homogenous spacetimes by their flat spacetime approximations.

126 59 / 61 Thank you Thank you for listening.