Unitary representations of SL(2, R)

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Milena Bielecka
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1 Unitary representations of SL(, R) Katarzyna Budzik 8 czerwca 018 1/6

2 Plan 1 Schroedinger operators with inverse square potential Universal cover of SL(, R) x + (m 1 4) 1 x 3 Integrating sl(, R) representations to SL(, R) representations 4 Lowest and highest weight representations of SL(, R) on L [0, [ 5 Principal and complementary representations of SL(, R) on L [0, [ L [0, [. /6

3 Schroedinger operator with 1/x potential Consider a formal differential expression: L m = (m x + 1 ) 1 4 x. When does it define a self-adjoint operator? a closed operator? We have corresponding minimal and maximal operators on domains: Dom(L max m ) = { f L [0, [ : L m f L [0, [ } Properties Dom(L min m ) = (C c ]0, [) cl. homogeneous of degree - L min m L max m ( L min m ) = L max m = L min m is Hermitian for m R. 3/6

4 Schroedinger operator with 1/x potential Notice that formally L m x 1 +m = 0 Suppose: then Re m > 1: x 1 +m x 1 +m 1+ Re m = x ξ - smooth cutoff function with compact support x 1 +m ξ Dom(L max m ). Definition For Re m > 1, define operator H m as a restriction of L max m to domain: sp H m = [0, [ Dom(L min m ) Cx 1 +m ξ. Hm = H m = H m is self-adjoint for m R m H m is a 1-parameter holomorphic family of closed operators. 4/6

5 Schroedinger operator with 1/x potential 1 1 < Re m: 1 < Re m < 1: H m = L min m L min m = Lmax m Hm Lmaxm and the codimension of the domain is 1. 5/6

6 Schroedinger operator with 1/x potential H m generates a holomorphic 1-parameter semigroup: ( e t xy Hm (x, y) = )e πt Im x +y t, Re t > 0 t ( e ±i t Hm (x, y) = e ±i π (m+1) xy )e πt Im i x +y t, ± Im t 0 t Kernel of the resolvent: 1 (H m + k ) (x, y) = 1 k { Im(kx)K m(ky) 0 < x < y I m(ky)k m(kx) 0 < y < x, Re k > 0 6/6

7 Universal cover of SL(, R) 7/6 SL(, R) has subgroups: { [ ] } cos φ sin φ SO() = u φ := : φ ] π, π] sin φ cos φ { [ ] } a 0 AN(, R) = t := : n R, a > 0 n KAN decomposition: Lets define a group: and maps: 1 a SL(, R) = SO() AN(, R). SL(, R) := R AN(, R) τ : SL(, R) (φ, t) uφ t SL(, R) φ : SL(, R) (φ, t) φ R SL(, R) is the universal cover of SL(, R): τ : SL(, R) SL(, R) is a surjective homomorphism SL(, R) is a simply connected Lie group.

8 Universal cover of SL(, R) Center of SL(, R): 1 n cover 1 and 1: 1 n := (πn, ( 1) n 1) R AN(, R) SL(, R) 1 n τ ( 1) n 1 SL(, R). 8/6

9 Integrating representations of Lie algebra G - Lie group with Lie algebra g G - universal cover of G Exponential maps: exp : g X e X G ẽxp : g X ẽxpx G Theorem Let π : g gl(v ) be a representation of the Lie algebra g. If G is compact, there exists exactly one representation of G π : G GL(V ) such that π(ẽxpx ) = e π(x ). 9/6

10 Representations of SL(, R) We will define a representation of sl(, R): sl(, R) X π π(x ) unbounded operator on a Hilbert space V which integrates to a representation of SL(, R): i.e. π : SL(, R) h π( h) B(V ) exp sl(, R) X ẽxpx SL(, R) π π(x ) e π(x ) = π ( ẽxpx ) exp π Problems: π(x ) are unbounded operators SL(, R) is not compact = π doesn t have to integrate to a representations of SL(, R). 10/6

11 sl(, R) standard triplet Matrices are called a standard triplet if A +, A, N sl(, R) [A +, A ] = N, [N, A ±] = ±A ±. Introduce standard triplet in sl(, R): [ ] [ ] [ ] 0 A + =, A 0 0 =, N = Then, e s+a+ = [ ] 1 s+, e s A = 0 1 [ ] [ 1 0, e tn e t/ = s 1 ] 0 0 e t/. 11/6 Any h SL(, R) can be written using e s+a+, e s A, e tn, 1 n. Therefore, representation π( h) for h SL(, R) can be written using e s+π(a+), e s π(a ), e tπ(n), π(1 n).

12 SL(, R) representation on L [0, [ 1 Define operators on L [0, [: A := i (x x + xx) dilation operator K := x H m := x + ( m 1 4) 1 x Schroedinger operator Define representation π of sl(, R) on L [0, [ π i A + Hm A i K N i A formally satisfy the commutation relations of sl(, R) the Casimir operator C = 1 (A+A + A A+) + N π(c) = 1 4 m 1 4 π(g) are antihermitian = e π(g) are unitary and bounded. 1/6

13 SL(, R) representation on L [0, [ Kernels of the three operators: 1 π(a +) = i Hm e απ(a+) (x, y) = e i π(m+1) π α Im ( xy α π(a ) = i K 3 π(n) = i A ) e i βk f (x) = e i βx f (x) e βπ(a ) (x, y) = e i βx δ(x y) e ita f (x) = e t/ f (e t x) e γπ(n) (x, y) = e γ 4 δ ( e γ x y ). e i x +y α, Im α 0 13/6

14 SL(, R) representation on L [0, [ SL(, R) = SO() AN(, R) =] π, π] AN(, R) We have a map φ : SL(, R) ] π, π]. Using φ we can partition SL(, R) into sectors: Y := {h : φ(h) ] π, 0[ } = {h : h 1 < 0} Z 0 := {h : φ(h) = 0} = {h : h 1 = 0, h 11 > 0} = AN(, R) Y + := {h : φ(h) ]0, π[ } = {h : h 1 > 0} = Y Z 1 := {h : φ(h) = π} = {h : h 1 = 0, h 11 < 0} = AN(, R) SL(, R) is a disjoint sum of sectors: SL(, R) = Y Z 0 Y + Z 1. 14/6

15 SL(, R) representation on L [0, [ SL(, R) = R AN(, R) We have a map φ : SL(, R) R. Using φ we can partition SL(, R) into sectors: } Ỹ n+ 1 := { h SL(, R) : φ( h) ]nπ, (n + 1)π[ } Z n := { h SL(, R) : φ( h) = nπ 1 n Z n SL(, R) is a disjoint sum of sectors: SL(, R) = we can identify n Z Ỹ n+ 1 Z n. SL(, R) Y Z 0 Y + = Ỹ 1 Z 0 Ỹ 1 SL(, R). 15/6

16 SL(, R) representation on L [0, [ For h Z 0 = AN(, R) so the representation is For h Y + Y (h 1 0) So the representation is given by [ ] a 0 h = = e n a A e log(a)n n 1 a π(h) = e n a ( i K) e log(a)( i A). h = e h 1 A h 1 e h 1A + e h 11 1 A h 1. π(h) = e h 1 h ( i K) 1 e h 1( i Hm) e h 11 1 h ( i K) 1. Kernel of this operator equals ( ) π(h)(x, y) = e i π (m+1)sgn(h 1) xy π h 1 Im h 1 e h i (h 11 x +h y ) 1. 16/6

17 SL(, R) representation on L [0, [ 1 SL(, R) [ ] 1 ±ɛ 1 = lim ɛ = π(1) = s lim e ± i ɛhm ɛ 0 Therefore, kernels of these operators ( ) e ±i π (m+1) xy πɛ Im e ɛ i (x +y ) ɛ 0 δ(x y) = π(1)(x, y) ɛ 17/6 1 ±1 SL(, R) [ ] 1 ɛ 1 ±1 = lim ɛ These matrices correspond to operators with kernels: ( ) e i π (m+1) xy πɛ Im e ɛ i (x +y ). ɛ So the operators converge to π(1 ±1) = e ±iπ(m+1) 1.

18 SL(, R) representation on L [0, [ 1 n SL(, R) 1 n = (πn, ( 1) n 1) R AN(, R), n Z Center of SL(, R): Z( SL(, R)) = {1n : n Z}. Then, from Schur s lemma: π(1 n) = e iπ n (m+1) 1. In particular: m = 1, 3,...: π is also a representation of PSL(, R) and SL(, R) π(1 n) = 1 m = 0,,...: π is also a representation of SL(, R) π(1 n) = ( 1) n 1. We calculated π( h) for all h SL(, R). 18/6

19 sl(, C) representation Lets take a standard triplet in sl(, C): A +, A, N sl(, C) = Csl(, R). Consider space of monomials: W m := { w k : k Z + m }, m C. Define representation X l,m of sl(, C) on W m : A + A l + := w w + lw A A l := w + lw 1 N N l := w w satisfy the commutation relations of sl(, C) Spectrum of N l : N l w k = kw k = sp(n l ) Z + m. 19/6

20 sl(, C) representation Theorem Suppose π is an irreducible representation of sl(, C) π(n ) has an eigenvector with eigenvalue m C. Then, there exists l C s.t. the representation π is equivalent to X l,m (or one of its subrepresentations). Example of a standard triplet in sl(, C): [ ] [ ] [ ] A + = 1 1 i, A i 1 = 1 1 i, N = 1 0 i. i 1 i 0 Notice: (in ) = 1. Connection between this sl(, C) standard triplet and previous sl(, R) standard triplet: N = i (A A+). 0/6

21 Representation of 1 n Fact 1 From Schur s lemma: π(1 n) = e iπ n (m+1) 1. Fact For X sl(, R) such that (X ) = 1: 1 n = ẽxp(πnx ) The representation π is an exponent of representation π: π(1 n) = e πn π(x ). From 1 and X has to satisfy: sp(π(x )) i(m + Z), (X ) = 1 Therefore X = in, N = 1 [ ] 0 i. i 0 1/6

22 Lowest weight representation of SL(, R) We can write N from sl(, C) standard triplet using A +, A from sl(, R) standard triplet: N = i (A A+) Then, the operator corresponding to N : π(n ) = 1 4 (K + Hm) = 1 4 ( x + ( m 1 ) ) 1 4 x + x, which is a Hamiltonian of a harmonic oscillator + potential 1/x. = π(n ) is an operator with spectrum bounded from below = π is equivalent to X l,lw. /6

23 Lowest weight representation of SL(, R) on L [0, [ Theorem The representation (m,lw) of SL(, R) can be represented as bounded operators on L [0, [ with kernels: h Ỹn+ 1 : h Z n : Its generators are h m,lw (x, y) = e iπ(m+1)(n+ 1 ) π h 1 Im ( xy h 1 ) e h i (h 11 x +h y ) 1 h m,lw (x, y) = e iπn(m+1) log( h 11 )/4 e i h 1 h 11 x δ ( h 11 1 x y ). π(a +) = i Hm = i π(a ) = i K = i x ( ( x + m 1 ) ) 1 4 x π(n) = i A = 1 (x x + xx). 4 3/6

24 Lowest and highest weight representations of SL(, R) 4/6 Lowest weight representation generators: A m,lw + = i Hm = i ( x + A m,lw = i K = i x N m,lw = i A N m,lw = i ( A m,lw Am,lw + Highest weight representation generators: N m,hw = i A m,hw + = i Hm A m,hw = i K N m,hw = i A. ( A m,hw A m,hw + = 1 (x x + xx). 4 ( m 1 ) ) 1 4 x ) ( ( 1 = x + m 1 ) ) x + x ) ( ( 1 = x + m 1 ) ) x + x.

25 SL(, R) representation on L [0, [ L [0, [ Define a representation of sl(, R) on L [0, [ L [0, [): A + i Hm ( Hm) A i K ( K) N 1 A A formally satisfy the commutation relations of sl(, R) 5/6

26 Operator H m ( H m ) For 1 < Re m < 1 we have operators (f +, f ) Dom ( L max m L min m ( Lmin m ) ( L max m ) ) : Lmax m ( Lmax m ). where α + +, α +, α +, α C. f + (x) α + +x 1 +m + α + x 1 m f (x) α + x 1 +m + α x 1 m, Definition Define H ω+,ω m condition: as a restriction of L max m ( L max m ) to functions satisfying α + +ω + = α + α + ω = α, where ω +, ω C { }. 6/6 H ω+,ω m, A A, K ( K) generate principal and complementary series.

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