The Heisenberg group in mathematics and physics Vladimir V. Kisil School of Mathematics University of Leeds (England) email: [email protected] Web: http://www.maths.leeds.ac.uk/~kisilv Varna, 2016 Origins of the Heisenberg({Weyl) group Roger Howe said: An investigator might be able to get what he wanted out of a situation while overlooking the extra structure imposed by the Heisenberg group, structure which might enable him to get much more. d The basic operators of differentiation dx and multiplication by x in satisfy to the same Heisenberg commutation relations [Q, P] = I as observables of momentum and coordinate in quantum mechanics. We shall start from the general properties the Heisenberg group and its representations. Many important applications will follow. The Symplectic Form Let n 1 be an integer. For complex vectors z, w Cn, we define: > 2 zw¯ = z1w¯ 1 + z2w¯ 2 + + z w¯ , (1) ··· n n where z = (z1, z2,..., zn), w = (w1, w2,..., wn). The following notion is central for Hamiltonian mechanics. Definition 2 The symplectic form ! on R n is a function of two vectors such that: 2n !(x, y; x0, y0) = xy0 - x0y, where (x, y), (x0, y0) R , (2) 2 Check the following properties: 1 ! is anti-symmetric !(x, y; x0, y0) = -!(x0, y0; x, y). 2 ! is bilinear: !(x, y; αx0, αy0) = α!(x, y; x0, y0), !(x, y; x0 + x00, y0 + y00) = !(x, y; x0, y0) + !(x, y; x00, y00). The Symplectic Form Further properties Exercise 1 Let z = x + iy and w = x0 + iy0 then ! can be expressed through the complex inner product (1) as !(x, y; x0, y0) = -=(zw¯ ). Ç å 2 xx 0 2 The symplectic form on R is equal to det . Consequently it yy 0 vanishes if and only if (x, y) and (x0, y0) are collinear. 3 Let A SL2(R) be a real2 2 matrix with the unit determinant. 2 × Define: Çx~å Çxå Çx~ å Çx å = A and 0 = A 0 . (3) y~ y y~ 0 y0 Then, !(x~, y~; x~0, y~ 0) = !(x, y; x0, y0). Moreover, the symplectic 2 group Sp(2)|the set of all linear transformations of R preserving !|coincides withSL 2(R). The Heisenberg group Definition Now we define the main object of our consideration. Definition An element of the n-dimensional Heisenberg group Hn is 2 +1 (s, x, y) R n , where s R and x, y Rn. The group law on Hn is 2 2 2 given as follows: (s, x, y) (s , x , y ) = (s + s + 1 !(x, y; x , y ), x + x , y + y ), (4) · 0 0 0 0 2 0 0 0 0 where ! the symplectic form. For the Heisenberg group Hn, check that: - 1 The unit is (0, 0, 0) and the inverse (s, x, y) 1 = (-s, -x, -y). 2 It is a non-commutative, the centre of Hn is: n Z = f(s, 0, 0) H , s Rg. (5) 2 2 3 The group law is continuous, so we have a Lie group. Alternative group lawsI for the Heisenberg group 1 1 Introduce complexified coordinates (s, z) on H with z = x + iy. Then the group law can be written as: (s, z) (s , z ) = (s + s + 1 =(z z¯), z + z ). · 0 0 0 2 0 0 3 2 Show that the set R with the group law (s, x, y) (s0, x0, y0) = (s + s0 + xy0, x + x0, y + y0) (6) ∗ 1 is isomorphic to the Heisenberg group H . It is called the polarised Heisenberg group. Hint: Use the explicit form of the homomorphism (s, x, y) (s + 1 xy, x, y). 7! 2 Alternative group lawsII for the Heisenberg group 1 3 Define the map φ : H M3(R) by ! Ö 1 è 1 xs + 2 xy φ(s, x, y) = 01 y . (7) 001 1 This is a group homomorphism from H to the group of3 3 × matrices with the unit determinant and the matrix multiplication as the group operation. Write also a group homomorphism from the polarised Heisenberg group to M3(R). 4 Expand the above items from this Exercise to Hn. The Weyl algebra The key idea of analysis is a linearization of complicated object in small neighbourhoods. Applied to Lie groups it leads to the Lie algebras. The Lie algebra of the Heisenberg group h1 is also called Weyl algebra. From the general theory we know, that h1 is a three-dimensional real 1 3 vector space, thus, it can be identified as a set with the group H ∼ R itself. There are several standard possibilities to realise h1: 1 Generators X of one-parameter subgroups: x(t) = exp(Xt), t R. 2 2 Tangent vectors to the group at the group unit. 3 Invariant first-order vector fields (differential operators) on the group. There is the important exponential map between a Lie algebra and respective Lie group. The exponent function can be defined in any topological algebra as the sum of the following series: X (tX)n exp(tX) = . n1=0 n! Generators of subgroups and the exponential map 1 Matrices from (7) are created by the following exponential map: Ö è Ö 1 è 0 xs 1 xs + 2 xy exp 00 y = 01 y . (8) 000 001 Thus h1 isomorphic to the vector space of matrices in the left-hand 1 side. We can define the explicit identification exp : h1 H by (8), ! 1 which is also known as the exponential coordinates on H . 2 Define the basis of h1: Ö è Ö è Ö è 001 010 000 S = 000 , X = 000 , Y = 001 . (9) 000 000 000 1 Write the one-parameter subgroups of H generated by S, X and Y. Invariant vector fields A (continuous) one-parameter subgroup is a continuous group homomorphism F from (R, +) to Hn: F(t + t0) = F(t) F(t0). · We can calculate the left and right derived action at any point g Hn: 2 d F t g d g F t ( ( ) ) ( ( )) · and · . (10) dt t=0 dt t=0 1 1 Check that the following vector fields on H are left (right) invariant: Sl(r) = @ , Xl(r) = @ - 1 y@ , Yl(r) = @ + 1 x@ . (11) ± s ± x 2 s ± y 2 s Show, that they are linearly independent and, thus, are bases of the Lie algebra h1 (in two different realizations). 1 2 Find one-parameter groups of right (left) shifts on H generated by these vector fields. CommutatorI The principal operation on a Lie algebra, besides the linear structure, is the Lie bracket|a bi-linear, anti-symmetric form with the Jacoby identity: [[A, B], C] + [[B, C], A] + [[C, A], B] = 0. In the above exercises, as in any algebra, we can define the Lie bracket as the commutator [A, B] = AB - BA: for matrices and vector fields through the corresponding algebraic operations in these algebras. 1 Check that bases from (9) and (11) satisfy the Heisenberg commutator relation [Xl(r), Yl(r)] = Sl(r) (12) and all other commutators vanishing. More generally: ( ) ( ) ( ) ( ) [A, A0] = !(x, y; x0, y0)S, where A 0 = s 0 S + x 0 X + y 0 Y, (13) and ! is the symplectic form. CommutatorII 2 Show that any second (and, thus, any higher) commutator [[A, B], C] on h1 vanishes. This property can be stated as \the Heisenberg group is a step2 nilpotent Lie group". 3 Check the formula 1 exp(A) exp(B) = exp(A + B + [A, B]), where A, B h1. (14) 2 2 The formula is also true for any step2 nilpotent Lie group and is a particular case of the Baker{Campbell–Hausdorff formula. Hint: In the case of H1 you can use the explicit form of the exponential map (8). 4 Define the vector space decomposition h1 = V0 V1, such that V0 = [V1, V1]. (15) ⊕ Automorphisms of the Heisenberg group Erlangen programme suggest investigate invariants under group action. This recipe can be applied recursively to groups themselves. Transformations of a group which preserve its structure are called group 1 automorphisms. The following are automorphisms of H : 1 1 Inner automorphisms or conjugation with (s, x, y) H : 2 -1 (s0, x0, y0) (s, x, y) (s0, x0, y0) (s, x, y) (16) 7! · · = (s0 + !(x, y; x0, y0), x0, y0) = (s0 + xy0 - x0y, x0, y0). Çx~å Çxå 2 Symplectic maps (s, x, y) (s, x~, y~), where = A with A 7! y~ y from the symplectic group Sp(2) ∼ SL2(R), see Exercise3. 3 Dilations: (s, x, y) (r2s, rx, ry) for a positive real r. 7! 4 Inversion: (s, x, y) (-s, y, x). 7! The last three types of transformations are outer automorphisms. The group of all automorphisms of the Heisenberg group Show that 1 3 1 Automorphism groups of H and h1 coincide as groups of maps of R onto itself. Hint: Use the exponent map and the relation (14). The crucial step is a demonstration that any automorphism of H1 is a linear map of R3. 2 All transforms from the previous slide, viewed as automorphisms of h1, preserve the decomposition (15). 1 3 Every automorphism of H can be written uniquely as composition of a symplectic map, an inner automorphism, a dilation and a power ( mod 2) of the inversion from the previous slide. Hint: Any automorphism is a linear map (by the previous item) of R3 which maps the centre Z to itself. Thus it shall have the form (s, x, y) (cs + ax + by, T(x, y)), where a, b and c are real and T is a 7! linear map of R2.