On the Heisenberg Group

HEISENBERG GROUP OVERVIEW

ARMIN SCHIKORRA

The Heisenberg group is R3 with an interesting metric structure. Usually one denotes points in the Heisenberg-Group by (x, y, t) ∈ R3.

1. Horizontal space

The horizontal space at a point (x, y, t) tells us the permissible directions. Usually it is described as a kernel of a one-form α = dt + 2 x1dx2 − x2dx1 .

At a point (x, y, t) the horizontal vectors v = (v1, v2, v3) are exactly those which satisfy  v1   +2y   2    3 2 1 0 =  v  ·  −2x  = v + 2(xv − yv ) v3 1

The horizontal vectors form a two-dimensional vector space in R3 with basis (X,Y ) where  1   0      X :=  0  ,Y :=  1  2y −2x and v = (v1, v2, v3) is horizontal if and only if v = v1X + v2Y q 2 2 The lenght of a (horizontal!) vector v is deﬁned as |v1| + |v2|

2. Curves

A horizontal curve γ :[a, b] → R3 with derivativeγ ˙ (t) satisﬁes (2.1) 0 =γ ˙ 3(t) + 2(γ1(t)γ ˙ 2(t) − γ2(t)γ ˙ 1(t)) The same equation can also be written as γ1(t)γ ˙ 1(t) ! (2.2) 0 =γ ˙ 3(t) + 2 det γ2(t)γ ˙ 2(t) 1 HEISENBERG GROUP OVERVIEW 2

The lenght of a (horizontal!) curve is deﬁned as Z b Z b q 1 2 2 2 L(γ) := |γ˙ (t)|H dt = |γ˙ (t)| + |γ˙ (t)| dt a a It is often useful to consider projections into the (x, y)-coordinate, which we will denote by π : R3 → R2 with π(x, y, t) := (x, y). Observe that by the fundamental theorem of calculus, whenever we have a curveγ ˜ = (˜γ1, γ˜2):[a, b] → R2 in the two-dimensional plane then we can set for any ta ∈ R Z t 3 1 2 2 1 γ (t) := ta + 2(˜γ (s) γ˜˙ (t) − γ˜ (s) γ˜˙ (s))ds a and ﬁnd that γ := (˜γ1, γ˜2, γ3):[a, b] → R3 is a horizontal curve starting at γ(a) = 1 2 (˜γ (a), γ˜ (a), ta) – this procedure is called Lifting. We can use this Lifting for example in the following way:

Letγ ˜ = (˜γ1, γ˜2) : [0, 1] → R2 be a curve in the two-dimensional plane, withγ ˜(0) = (0, 0). We can lift this to a horizontal curve γ := (˜γ1, γ˜2, γ3) : [0, 1] → R3 which satisﬁes γ(0) = (0, 0, 0), recall that γ3(t) is then given by Z t γ3(t) := 2(˜γ1(s) γ˜˙ 2(t) − γ˜2(s) γ˜˙ 1(s))ds 0 In particular, Z 1 γ3(1) = 2(˜γ1(s) γ˜˙ 2(t) − γ˜2(s) γ˜˙ 1(s))ds 0 Now ifγ ˜ is a closed curve without self-intersection (in particular γ(0) = γ(1)) then Z 1 1 1 2 2 1 (˜γ (s) γ˜˙ (t) − γ˜ (s) γ˜˙ (s))ds = Area(A) 2 0 where A is the area enclosed by γ. This is the Sector formula of Leibniz, (I found only a German link)

3. Scaling and rotation

3 2 Let r > 0 then for a point p = (x, y, t) ∈ R we set δr(p) := (rx, ry, r t). Observe that if 3 3 γ :[a, b] → R is horizontal, then (δr ◦ γ):[a, b] → R is still horizontal. Moreover, 0 0 |(δrγ) (t)|H = r|γ (t)|H , and thus

L(δrγ) = rL(γ). Let P be a rotation matrix of R2, i.e. for some α ∈ R let sin(α) − cos(α) ! P = . cos(α) sin(α) HEISENBERG GROUP OVERVIEW 3

3 We then deﬁne the rotation of a vector v = (v1, v2, v3) ∈ R , rot(v; P ) as ! ! v1 1 2 1 2 rot(v; P ) := P , v3 ≡ sin(α)v − cos(α)v , cos(α)v + sin(α)v , v3 . v2

One can check (the determinant formula as in (2.2) is here easier) that if v = (v1, v2, v3) is a horizontal vector at the point p = (p1, p2, p3), then for any rotation P the vector rot(v; P ) is horizontal at the point rot(p; P ). This implies if we have a horizontal curve γ :[a, b] → R3 and set for a ﬁxed rotation P , η(t) := rot (γ(t),P ) then η is a horizontal curve from rot (γ(a),P ) to rot (γ(b),P ), with the same lenght, L(η) = L(γ).

4. Group structure

Let (x, y, t), (x0, y0, t0) ∈ R3 then we deﬁne the group multiplication (x, y, t) ∗ (x0, y0, t0) = (x + x0, y + y0, t + t0 + 2(x0y − xy0)) Observe

• (x, y, t)∗(−x, −y, −t) = (0, 0, 0), we thus say for p = (x, y, t) that p−1 = (−x, −y, −t) is the inverse. • this group does not commute: in general p ∗ q 6= q ∗ p (careful!) • Let γ :[a, b] → R3 be a horizontal curve and p ∈ R3, then 3 p ∗ γ :[a, b] → R is a horizontal curve: Indeed, set η(t) := p ∗ γ(t), then

1 2 3 1 2 η(t) = (p1 + γ (t), p2 + γ (t), p3 + γ (t) + 2(γ (t)p2 − p1γ (t)) so 1 2 3 1 2 η˙(t) = (γ ˙ (t), γ˙ (t), γ˙ (t) + 2(γ ˙ (t)p2 − p1γ˙ (t)) so by (2.1)

1 2 1 2 1 2 η˙(t) = (γ ˙ (t), γ˙ (t), 2(γ ˙ (t)(p2 + γ (t)) − (p1 + γ (t))γ ˙ (t)) Thus, 0 =η ˙3(t) + 2(η1(t)η ˙2(t) − η˙1(t)η2) which means that η satisﬁes (2.1), i.e. η is horizontal. • If γ :[a, b] → R3 is horizontal and for p ∈ R3 we set η(t) := p ∗ γ(t), then

|γ˙ (t)|H = |η˙(t)|H . In particular, γ and η have the same length! HEISENBERG GROUP OVERVIEW 4

5. Geodesics (shortest curves) • It suﬃces to compute shortest curves (“geodesics”) between the origin (0, 0, 0) and a point (x, y, z). If we can do this, then to compute the shortest geodesics between p and q we compute the geodesic γ between (0, 0, 0) and p−1 ∗ q, then η := p ∗ γ is the shortest geodesic between p and q. • The shortest geodesic between (0, 0, 0) and (x, y, 0) is γ(t) := (tx, ty, 0) for t ∈ [0, 1]. • The shortest geodesic between (0, 0, 0) and (0, 0,T ) is non-unique: One example is the spiral

q q  |T | |T | T η(t) :=  √ sin(t), √ (cos(t) − 1), (t − sin(t)) t ∈ [0, 2π] 4π 4 π 2π

(This weird formula comes from the fact that the projection onto the xy plane must be a circle, then the t-entry can be computed from (2.1)) For α ∈ [0, 2π] also any R2 × {0}-rotation is a geodesic between (0, 0, 0) and (0, 0,T ):

1 2 1 2 3 ηα(t) := sin(α)η (t) − cos(α)η (t), cos(α)η (t) + sin(α)η (t), η (t)

• The shortest geodesic between (0, 0, 0) and (x, y, T ) (with (x, y) 6= (0, 0) and T 6= 0) lies on one of these spirals geodesics (to some other point on the t-axis). Since (x, y) 6= (0, 0) there is a unique rotation P such that √ x ! x2 + y2 ! = P y 0

√ Set R := x2 + y2. t If R 6= 0 and we have a geodesic γ(t) from (0, 0, 0) to (1, 0, R2 ) then

η(t) := rot(δR(γ(t)),P )

is a geodesic from (0, 0, 0) to (x, y, t). So we need to compute a formula for geodesics from (0, 0, 0) to (1, 0,T ). The idea is that the geodesic from (0, 0, 0) to (1, 0,T ) is part of the (unique) geodesic between (0, 0, 0) and (0, 0, τ) for τ > T (assume that T > 0) passing through (1, 0,T ). After projecting this situation into x, y-axis R2, we are left with the task of computing τ as the area of the circle that passes through (0, 0) and 2 (0, 1) with area of the circle intersected with R− being t. See (2.16) and (2.17) in [Haj lasz and Zimmerman, 2015], HEISENBERG GROUP OVERVIEW 5

According to wikipedia, wikipedia we can compute the (red) area A of the circula 1 segment (which will need to equal to 4 T ) as follows: R2 A = (θ − sin(θ)). 2 We will parametrize it in h ∈ R, then we have 1 R2 = + h2 4 1 1 tan(θ/2) = ⇔ θ = 2 arctan 2h 2h Consequently, for h ∈ R the corresponding A˜(h) is given by 1 + h2 1 1 A˜(h) = 4 2 arctan − sin(2 arctan ) 2 2h 2h This is at least consistent with the notion that π A(0+) = 1/2Area( ball of radius 1/2) = 8 This formula works only for h > 0, so we need to set  1 2 4 +h 1 1  2 2 arctan 2h − sin(2 arctan 2h ) h > 0 A(h) := 1 +h2  2 4 1 1 π(1/4 + h ) − 2 2 arctan −2h − sin(2 arctan −2h ) h < 0 According to Wolfram alpha A˜ : [0, ∞) → [0, π/8] is one-to-one; so A : R → [0, +∞) with limh→∞ A(h) = 0 and limh→−∞ A(h) = +∞. Observe that according to Wolfram alpha, A˜0(h) = 2h cot−1(2h − 1) In particular for any T > 0 there exists h ∈ R such that 4A(h) = T , h = A−1(T/4). So our geodesic is the lifting of the curve along the sphere centered at q the point (1/2, h) of radius 1/4 + h2. The corresponding height of the lifted curve HEISENBERG GROUP OVERVIEW 6

is τ = 4π(1/4 + h2). The curve in the projected version is given byγ ˜ : [0, 2π] → R2 with R sin(t) ! γ˜(t) = P R(1 − cos(t)) Here P is the rotation matrix that takes (0,R) into (1/2, h).

6. references

References [Hajlasz and Zimmerman, 2015] Hajl asz, P. and Zimmerman, S. (2015). Geodesics in the Heisenberg group. Anal. Geom. Metr. Spaces, 3(1):325–337.

Email address: [email protected]