aa r X i v : . [ m a t h . M G ] O c t GEODESICS IN THE HEISENBERG GROUP
PIOTR HAJ LASZ AND SCOTT ZIMMERMAN
Abstract.
We provide a new and elementary proof for the structure of geodesics in theHeisenberg group H n . The proof is based on a new isoperimetric inequality for closedcurves in R n . We also prove that the Carnot-Carath´eodory metric is real analytic awayfrom the center of the group. Introduction
The aim of this paper is to provide a detailed and a self-contained presentation of thestructure of geodesics in the Heisenberg groups. While the paper is mostly of expositorycharacter, some results are new. The new results are: the isoperimetric inequality forclosed curves in R n (Theorem 2.4), a new proof of the structure of the geodesics in theHeisenberg group (Theorem 2.1), and the real analyticity of the Carnot-Carath´eodorymetric away from the center of the group (Theorem 3.1). The proof of the isoperimetricinequality (Theorem 2.4) is based on an elementary adaptation of the classical proof of theisoperimetric inequality due to Hurwitz. This new isoperimetric inequality is then used toestablish the structure of geodesics in H n . The proof of the real analyticity of the Carnot-Carath´eodory metric follows from the explicit formulas for the geodesics in H n establishedin Theorem 2.1. We believe that, because of the increasing popularity of the geometrictheory of the Heisenberg groups among young researchers and graduate students, such asurvey paper will be useful.The Heisenberg group H n is C n × R = R n +1 given the structure of a Lie group withmultiplication( z, t ) ∗ ( z ′ , t ′ ) = (cid:16) z + z ′ , t + t ′ + 2Im n X j =1 z j z ′ j (cid:17) = (cid:16) x + x ′ , . . . , x n + x ′ n , y + y ′ , . . . , y n + y ′ n , t + t ′ + 2 n X j =1 ( x ′ j y j − x j y ′ j ) (cid:17) with Lie algebra g whose basis of left invariant vector fields at any p =( x , . . . , x n , y , . . . , y n , t ) ∈ H n is X j ( p ) = ∂∂x j + 2 y j ∂∂t , Y j ( p ) = ∂∂y j − x j ∂∂t , T = ∂∂t , j = 1 , , . . . , n. Mathematics Subject Classification.
Primary 53C17; Secondary 42A05, 53C22.
Key words and phrases.
Heisenberg group, geodesics, Fourier series, isoperimetric inequality.P.H. was supported by NSF grant DMS-1161425.
We call H H n = span { X , . . . , X n , Y , . . . , Y n } the horizontal distribution on H n , and denoteby H p H n the horizontal space at p . An absolutely continuous curve Γ : [0 , S ] → R n +1 issaid to be horizontal if ˙Γ( s ) ∈ H Γ( s ) H n for almost every s ∈ [0 , S ]. It is easy to see thatthe horizontal distribution is the kernel of the standard contact form α = dt + 2 n X j =1 ( x j dy j − y j dx j ) . That is, H p H n = ker α ( p ). Hence it follows that an absolutely continuous curve Γ =( x , . . . , x n , y , . . . , y n , t ) is horizontal if and only if ˙ t ( s ) = 2 P nj =1 ( ˙ x j ( s ) y j ( s ) − x j ( s ) ˙ y j ( s ))for almost every s ∈ [0 , S ]. This means that(1.1) t ( s ) − t (0) = 2 n X j =1 Z s ( ˙ x j ( τ ) y j ( τ ) − x j ( τ ) ˙ y j ( τ )) dτ for every s ∈ [0 , S ]. Suppose γ = ( x , . . . , x n , y , . . . , y n ) : [0 , S ] → R n is absolutelycontinuous. If a value for t (0) is fixed, then (1.1) gives a unique horizontal curve Γ =( γ ( s ) , t ( s )) whose projection onto the first 2 n coordinates equals γ . We call this curve Γ a horizontal lift of γ .If the curve γ : [0 , S ] → R n is closed and Γ = ( γ, t ) is a horizontal lift of γ , then itfollows from Green’s Theorem that the total change in height t ( S ) − t (0) equals − γ j of the curve to each x j y j -plane.Since the curves in the x j y j -planes may have self-intersections, the signed area has to takemultiplicity of the components of the complement of γ j into account. This multiplicity canbe defined in terms of the winding number. We will not provide more details here as thisinterpretation of the change of the height will not play any role in our argument, and it ismentioned here only to give an additional vantage point to the geometry of the problem.We equip the horizontal distribution H H n with the left invariant Riemannian metric sothat the vector fields X j and Y j are orthonormal at every point p ∈ H n . Note that thismetric is only defined on H H n and not on T H n . If v = n X j =1 (cid:16) a j ∂∂x j (cid:12)(cid:12)(cid:12) p + b j ∂∂y j (cid:12)(cid:12)(cid:12) p (cid:17) + c ∂∂t (cid:12)(cid:12)(cid:12) p ∈ H p H n , then it is easy to see that v = n X j =1 a j X j ( p ) + b j Y j ( p )and hence c = 2 P nj =1 ( a j y j ( p ) − b j x j ( p )). Clearly | v | H ≤ | v | E , where | v | H = vuut n X j =1 a j + b j and | v | E = vuut(cid:16) n X j =1 a j + b j (cid:17) + c are the lengths of v with respect to the metric in H p H n and the Euclidean metric in R n +1 respectively. On compact sets in R n +1 the coefficient c is bounded by C | v | H , so for anycompact set K ⊂ R n +1 there is a constant C ( K ) ≥ | v | H ≤ | v | E ≤ C ( K ) | v | H for all p ∈ K and v ∈ H p H n . EODESICS IN THE HEISENBERG GROUP 3
A horizontal curve Γ : [ a, b ] → H n satisfies˙Γ( s ) = n X j =1 ˙ x j ( s ) X j (Γ( s )) + ˙ y j ( s ) Y j (Γ( s )) . Its length with respect to the metric in H H n equals(1.3) ℓ H (Γ) := Z ba | ˙Γ( s ) | H ds = Z ba vuut n X j =1 ( ˙ x j ( s ) + ˙ y j ( s ) ) ds. Notice that the length ℓ H (Γ) of a horizontal curve Γ = ( γ, t ) given by (1.3) equals theusual Euclidean length ℓ E ( γ ) in R n of the projection γ .The Carnot-Carath´eodory metric d cc in H n is defined as the infimum of lengths ℓ H (Γ) ofhorizontal curves connecting two given points. It is well known that any two points can beconnected by a horizontal curve (we will actually prove it), and hence d cc is a true metric.It follows from (1.2) that for any compact set K ⊂ R n +1 there is a constant C = C ( K )such that(1.4) | p − q | ≤ Cd cc ( p, q ) for all p, q ∈ K .If Γ : [ a, b ] → X is a (continuous) curve in any metric space ( X, d ), then the length of Γ isdefined by ℓ d (Γ) = sup n − X i =0 d (Γ( s i ) , Γ( s i +1 )) , where the supremum is taken over all n ∈ N and all partitions a = s ≤ s ≤ . . . ≤ s n = b .In particular, if Γ : [ a, b ] → H n is a curve in the Heisenberg group, then its length withrespect to the Carnot-Carath´eodory metric is(1.5) ℓ cc (Γ) = sup n − X i =0 d cc (Γ( s i ) , Γ( s i +1 )) . It follows immediately from the definition that if Γ is a horizontal curve, then ℓ cc (Γ) ≤ ℓ H (Γ); hence every horizontal curve is rectifiable (i.e. of finite length), but it is not obvi-ous, whether ℓ cc (Γ) = ℓ H (Γ). It is also not clear whether every rectifiable curve can bereparametrized as a horizontal curve. Actually all of this is true and we have Proposition 1.1. In H n we have (1) Any horizontal curve Γ is rectifiable and ℓ cc (Γ) = ℓ H (Γ) . (2) Lipschitz curves are horizontal. (3)
Every rectifiable curve admits a -Lipschitz parametrization (and hence it is hori-zontal with respect to this parametrization). Thus, up to a reparametrization, the class of rectifiable curves coincides with the classof horizontal ones. Proposition 1.1 is well known, but it is difficult to find a good referencethat would provide a straightforward proof. For the sake of completeness, we decided toprovide a proof of this result in the Appendix. For a proof of (1) in more generality, seeTheorem 1.3.5 in [12].
PIOTR HAJ LASZ AND SCOTT ZIMMERMAN A geodesic from p to q is a curve of shortest length between the two points. It is alreadywell known that any two points in the Heisenberg group can be connected by a geodesic.Also, the structure of every geodesic in a form of an explicit parameterization is known.The proofs in the case of H can be found in [2, 5, 7, 11], and the general case of H n istreated in [1, 3, 13].If n = 1, the structure of geodesics can be obtained via the two dimensional isoperimetricinequality (see [2, 5, 11]). Consider a horizontal curve Γ = ( γ, t ) in H connecting the originto some point q = (0 , , T ) with T = 0. The length of Γ equals the length of its projection γ to R (which is a closed curve). Also, by (1.1), the change T in the height of Γ mustequal − γ . Thus the projection of any horizontal curveconnecting 0 to q must enclose the same area | T | /
4, and finding a geodesic which connects0 to q reduces to a problem of finding a shortest closed curve γ enclosing a fixed area.Thus the classical isoperimetric inequality implies that Γ will have smallest length when γ is a circle. Then the t component of Γ is determined by (1.1) and one obtains an explicitparametrization of the geodesics in H connecting the origin to a point on the t -axis. Suchgeodesics pass through all points ( x , y , t ), t = 0 in H . If q = ( x , y , q connecting the origin to q is a geodesic. This describesall geodesics connecting the origin to any other point in H . Due to the left-invariance ofthe vector fields X and Y , parameterizations for geodesics between arbitrary points in H may be found by left multiplication of the geodesics discussed above.This elegant argument, however, does not apply to H n when n > H n are based on the Pontryagin maximum principle [1, 3, 13].In this paper we will provide a straightforward and elementary argument leading to anexplicit parameterization of geodesics in H n (Theorem 2.1). Our argument is based onHurwitz’s proof [9], of the isoperimetric inequality in R involving Fourier series. TheHurwitz argument is used to prove a version of the isoperimetric inequality for closed curvesin R n (Theorem 2.4). This isoperimetric inequality allows us to extend the isoperimetricproof of the structure of geodesics in H to the higher dimensional case H n as seen in theproof of Theorem 2.1. For a related, but different isoperimetric inequality in R n , see [14].As an application of our method we also prove that the Carnot-Carath´eodory metric isreal analytic away from the center of the group (Theorem 3.1). This improves a result ofMonti [12, 13] who proved that this distance is C ∞ smooth away from the center. We alsofind a formula for the Carnot-Carath´eodory distance (Corollary 3.2) that, we hope, willfind application in the study of geometric properties of the Heisenberg groups.The paper is organized as follows. In Section 2 we will state and prove the isoperimetricinequality and the result about the structure of the geodesics in H n , and in Section 3 we usethis structure to show that the distance function in H n is analytic away from the center.In Section 4 we address the problem of comparing different geodesics (there are infinitelymany of them) connecting the origin to a point (0 , , T ) ∈ H n on the t -axis. Finally, in theAppendix we prove Proposition 1.1. Acknowledgements.
The authors would like to thank the referee for valuable commentsthat led to an improvement of the paper.
EODESICS IN THE HEISENBERG GROUP 5 The isoperimetric inequality and the structure of geodesics
Any horizontal curve Γ is rectifiable, and we may use the arc-length parametrization toassume that the speed | ˙Γ | H of Γ : [0 , ℓ H (Γ)] → H n equals 1. Then, we can reparametrize itas a curve of constant speed defined on [0 , , → H n satisfies(2.1) n X j =1 ˙ x j ( s ) + ˙ y j ( s ) = ℓ H (Γ) = ℓ cc (Γ) for almost all s ∈ [0 , H n can be reparametrized as a horizontal curvevia the arc length parameterization (Proposition 1.1), and thus, when looking for lengthminimizing curves (geodesics), it suffices to restrict our attention to horizontal curvesΓ : [0 , → H n satisfying (2.1).Since the left translation in H n is an isometry, it suffices to investigate geodesics con-necting the origin 0 ∈ H n to another point in H n . Indeed, if Γ is a geodesic connecting 0to p − ∗ q , then p ∗ Γ is a geodesic connecting p to q .If q belongs to the subspace R n × { } ⊂ R n +1 = H n , then it is easy to check thatthe straight line Γ( s ) = sq , s ∈ [0 ,
1] is a unique geodesic (up to a reparametrization)connecting 0 to q . Indeed, it is easy to check that Γ is horizontal, and its length ℓ cc (Γ) = ℓ H (Γ) equals the Euclidean length | q | of the segment 0 q because Γ is equal to its projection γ . For any other horizontal curve ˜Γ = (˜ γ, ˜ t ) connecting 0 to q , the projection ˜ γ on R n would not be a segment (since horizontal lifts of curves are unique up to vertical shifts),and hence we would have ℓ cc (˜Γ) = ℓ H (˜Γ) = ℓ E (˜ γ ) > | q | = ℓ cc (Γ) which proves that ˜Γcannot be a geodesic.In Theorem 2.1, we will describe the structure of geodesics in H n connecting the origin toa point (0 , , T ) ∈ R n × R = H n , T = 0, lying on the t -axis. Later we will see (Corollary 2.7)that these curves describe all geodesics in H n connecting 0 to q R n × { } . The geodesicsconnecting 0 to q ∈ R n × { } have been described above. Theorem 2.1.
A horizontal curve Γ( s ) = ( x ( s ) , y ( s ) , t ( s )) = ( x ( s ) , . . . , x n ( s ) , y ( s ) , . . . , y n ( s ) , t ( s )) : [0 , → H n of constant speed, connecting the origin Γ(0) = (0 , , ∈ R n × R = H n to a point Γ(1) = (0 , , ± T ) , T > , on the t -axis is a geodesic if and only if (2.2) x j ( s ) = A j (1 − cos(2 πs )) ∓ B j sin(2 πs ) y j ( s ) = B j (1 − cos(2 πs )) ± A j sin(2 πs ) for j = 1 , , . . . , n and t ( s ) = ± T (cid:18) s − sin(2 πs )2 π (cid:19) where A , . . . , A n , B , . . . , B n are any real numbers such that π P nj =1 ( A j + B j ) = T . Remark 2.2.
Observe that if Γ(1) = (0 , , + T ), the equations (2.2) give a constant-speed parametrizations of negatively oriented circles in each of the x j y j -planes, centeredat ( A j , B j ), and of radius q A j + B j . Each circle passes through the origin at s = 0. The PIOTR HAJ LASZ AND SCOTT ZIMMERMAN signed area of such a circle equals − π ( A j + B j ). Thus the change in height t (1) − t (0) whichis − x j y j -planesequals ( − n X j =1 (cid:0) − π ( A j + B j ) (cid:1) = T. Clearly this must be the case, because Γ connects the origin to (0 , , T ). Any collectionof circles in the x j y j -planes passing through the origin and having radii r j ≥ , , T ) where T = 4 π P nj =1 r j . Inparticular we can find a geodesic for which only one projection is a nontrivial circle (allother radii are zero) and another geodesic for which all projections are non-trivial circles.That suggests that the geodesics connecting (0 , ,
0) to (0 , , T ) may have many differentshapes. This is, however, an incorrect intuition. As we will see in Section 4, all suchgeodesics are obtained from one through a rotation of R n +1 about the t -axis. This rota-tion is also an isometric mapping of H n . The above reasoning applies also to the case whenΓ(1) = (0 , , − T ) with the only difference being that the circles are positively oriented. Remark 2.3.
The parametric equations for the geodesics can be nicely expressed withthe help of complex numbers, see (2.15).The proof is based on the following version of the isoperimetric inequality which is ofindependent interest.In the theorem below we use identification of R n with C n given by R n ∋ ( x, y ) = ( x , . . . , x n , y , . . . , y n ) ↔ ( x + iy , . . . , x n + iy n ) = x + iy ∈ C n . Every rectifiable curve γ admits the arc-length parametrization. By rescaling it, we mayassume that γ is a constant speed curve defined on [0 , Theorem 2.4. If γ = ( x , . . . , x n , y , . . . , y n ) : [0 , → R n is a closed rectifiable curveprametrized by constant speed, then (2.3) L ≥ π | D | , where L is the length of γ and D = D + . . . + D n is the sum of signed areas enclosed bythe curves γ j = ( x j , y j ) : [0 , → R , i.e. D j = 12 Z ( ˙ y j ( s ) x j ( s ) − ˙ x j ( s ) y j ( s )) ds. Moreover, equality in (2.3) holds if and only if there are points
A, B, C, D ∈ R n such that γ has the form (2.4) γ ( s ) = ( C + iD ) + (1 − e +2 πis )( A + iB ) , when L = 4 π D and (2.5) γ ( s ) = ( C + iD ) + (1 − e − πis )( A + iB ) when L = − π D . EODESICS IN THE HEISENBERG GROUP 7
Remark 2.5.
Let A j , B j , C j and D j , j = 1 , , . . . , n be the components of the points A, B, C and D respectively. In terms of real components of γ , (2.4) can be written as(2.6) x j ( s ) = C j + A j (1 − cos(2 πs )) + B j sin(2 πs ) y j ( s ) = D j + B j (1 − cos(2 πs )) − A j sin(2 πs )and (2.5) as(2.7) x j ( s ) = C j + A j (1 − cos(2 πs )) − B j sin(2 πs ) y j ( s ) = D j + B j (1 − cos(2 πs )) + A j sin(2 πs ) . That is, the curves γ j = ( x j , y j ) are circles of radius q A j + B j passing through ( C j , D j )at s = 0. In the case of (2.4) they are all positively oriented and in the case of (2.5) theyare all negatively oriented. In either case, they are parametrized with constant angularspeed. Remark 2.6.
If we have two different circles of the form (2.4) having the same radius,then one can be mapped onto the other one by a composition of translations and a unitarymap of C n . See the proof of Proposition 4.1. The same comment applies to circles of theform (2.5). Proof.
Let γ = ( x , . . . , x n , y , . . . , y n ) : [0 , → R n be a closed rectifiable curve. Bytranslating the curve, we may assume without loss of generality that γ (0) = 0. It sufficesto prove (2.3) along with equations (2.6) and (2.7) (with C = D = 0) which are, as waspointed out in Remark 2.5, equivalent to (2.4) and (2.5). Since the curve has constantspeed, its speed equals the length of the curve, so n X j =1 ˙ x j ( s ) + ˙ y j ( s ) = L . In particular the functions x j and y j are L -Lipschitz continuous and x j (0) = y j (0) = x j (1) = y j (1) = 0. Hence the functions x j , y j extend to 1-periodic Lipschitz functions on R , and so we can use Fourier series to investigate them. We will follow notation used in[6]. For a 1-periodic function f letˆ f ( k ) = Z f ( x ) e − πikx dx, k ∈ Z be its k th Fourier coefficient. By Parseval’s identity, L = n X j =1 Z | ˙ x j ( s ) | + | ˙ y j ( s ) | ds = n X j =1 X k ∈ Z | ˆ˙ x j ( k ) | + | ˆ˙ y j ( k ) | = n X j =1 X k ∈ Z π k (cid:0) | ˆ x j ( k ) | + | ˆ y j ( k ) | (cid:1) (2.8)Note that D = D + . . . + D n = 12 n X j =1 Z ( ˙ y j ( s ) x j ( s ) − ˙ x j ( s ) y j ( s )) ds. PIOTR HAJ LASZ AND SCOTT ZIMMERMAN
Since ˙ x j and ˙ y j are real valued, we have ˙ x j ( s ) = ˙ x j ( s ) and ˙ y j ( s ) = ˙ y j ( s ). Thus we mayapply Parseval’s theorem to this pair of inner products to find D = 12 n X j =1 X k ∈ Z ˆ˙ y j ( k )ˆ x j ( k ) − X k ∈ Z ˆ˙ x j ( k )ˆ y j ( k ) ! = 12 n X j =1 X k ∈ Z πki (cid:16) ˆ x j ( k )ˆ y j ( k ) − ˆ y j ( k )ˆ x j ( k ) (cid:17) = π n X j =1 X k ∈ Z k · (cid:16) ˆ y j ( k )ˆ x j ( k ) (cid:17) , (2.9)since i (¯ z − z ) = 2 Im z . Subtracting (2.9) from (2.8) gives L π − D π = n X j =1 h X k ∈ Z k (cid:16) | ˆ x j ( k ) | + | ˆ y j ( k ) | (cid:17) − k · (cid:16) ˆ y j ( k )ˆ x j ( k ) (cid:17)i = n X j =1 h X k ∈ Z ( k − | k | ) (cid:16) | ˆ x j ( k ) | + | ˆ y j ( k ) | (cid:17) + | k | (cid:16) | ˆ y j ( k ) | − k )Im (cid:16) ˆ y j ( k )ˆ x j ( k ) (cid:17) + | ˆ x j ( k ) | (cid:17) i = n X j =1 h X k ∈ Z ( k − | k | ) (cid:16) | ˆ x j ( k ) | + | ˆ y j ( k ) | (cid:17) + X k ∈ Z | k | (cid:12)(cid:12) ˆ y j ( k ) + i sgn( k )ˆ x j ( k ) (cid:12)(cid:12) i . (2.10)The last equality follows from the identity | a + ib | = | a | − ab ) + | b | which holds forall a, b ∈ C . Since every term in this last sum is non-negative, it follows that L π − D π ≥ L ≥ π D . Reversing the orientation of the curve, i.e. applying the aboveargument to ˜ γ ( t ) = γ (1 − t ) gives L ≥ − π D , so (2.3) follows.Equality in (2.3) holds if and only if either L = 4 π D or L = − π D . We will firstconsider the case L = 4 π D . This equality will occur if and only if each of the two sumscontained inside the brackets in (2.10) equals zero. Since k −| k | > | k | ≥
2, the first ofthe two sums vanishes if and only if ˆ x j ( k ) = ˆ y j ( k ) = 0 for every | k | ≥ j = 1 , , . . . , n .Hence nontrivial terms in the second sum correspond to k = ±
1, and thus this sum vanishesif and only if ˆ y j ( ±
1) = − i sgn( ± x j ( ± j = 1 , . . . , n ,(2.11) ˆ y j (1) = − i ˆ x j (1) and ˆ y j ( −
1) = i ˆ x j ( − . Now since each x j and y j is Lipschitz, their Fourier series converge uniformly on [0 , | k | ≤
1. Thus L = 4 π D ifand only if (2.11) is satisfied and for every s ∈ [0 ,
1] and j = 1 , . . . , n (2.12) x j ( s ) = ˆ x j ( − e − πis + ˆ x j (0) + ˆ x j (1) e πis y j ( s ) = ˆ y j ( − e − πis + ˆ y j (0) + ˆ y j (1) e πis . EODESICS IN THE HEISENBERG GROUP 9
In particular, 0 = x j (0) = ˆ x j ( −
1) + ˆ x j (0) + ˆ x j (1) and hence ˆ x j (0) = − ˆ x j ( − − ˆ x j (1) foreach j = 1 , . . . , n . This together with Euler’s formula gives x j ( s ) = ˆ x j ( − (cid:0) e − πis − (cid:1) + ˆ x j (1) (cid:0) e πis − (cid:1) = − (ˆ x j ( −
1) + ˆ x j (1))(1 − cos(2 πs )) + ( − i ˆ x j ( −
1) + i ˆ x j (1)) sin(2 πs )= − (ˆ x j ( −
1) + ˆ x j (1))(1 − cos(2 πs )) − (ˆ y j ( −
1) + ˆ y j (1)) sin(2 πs ) . The last equality follows from (2.11). Similarly, we have y j ( s ) = − (ˆ y j ( −
1) + ˆ y j (1))(1 − cos(2 πs )) + (ˆ x j ( −
1) + ˆ x j (1)) sin(2 πs ) . If we write A j = − (ˆ x j ( −
1) + ˆ x j (1)) and B j = − (ˆ y j ( −
1) + ˆ y j (1)), then we have(2.13) x j ( s ) = A j (1 − cos(2 πs )) + B j sin(2 πs ) y j ( s ) = B j (1 − cos(2 πs )) − A j sin(2 πs ) . Note that it follows directly from the definition of Fourier coefficients that the numbers A j , B j are real.The case L = − π D is reduced to the above case by reversing the orientation of γ aspreviously described. In that case the curves γ j are given by(2.14) x j ( s ) = A j (1 − cos(2 πs )) − B j sin(2 πs ) y j ( s ) = B j (1 − cos(2 πs )) + A j sin(2 πs ) . We proved that if L = 4 π D , then γ is of the form (2.6) and if L = − π D , then it is ofthe form (2.7). In the other direction, a straightforward calculation shows that any curveof the form (2.6) satisfies L = 4 π D and any curve of the form (2.7) satisfies L = − π D .This completes the proof. (cid:3) Proof of Theorem 2.1.
Suppose first that Γ = ( γ, t ) : [0 , → H n is any horizontal curveof constant speed connecting the origin to the point (0 , , + T ), T >
0. Recall from (2.1)that n X j =1 ˙ x j ( s ) + ˙ y j ( s ) = ℓ cc (Γ) =: L . Thus γ : [0 , → R n is a closed curve of length L parametrized by arc-length. Moreover γ (0) = 0.If D is defined as in Theorem 2.4, it follows from (1.1) that T = 2 n X j =1 Z ( ˙ x j ( s ) y j ( s ) − ˙ y j ( s ) x j ( s )) ds = − D so D < L ≥ πT by Theorem 2.4. Now Γ is a geodesic if and only if L = πT = − π D which is the case of the equality in the isoperimetric inequality (2.3). We proved abovethat this is equivalent to the components of γ satisfying (2.14), and this is the (0 , , + T )case of (2.2). One may also easily check that 4 π P nj =1 ( A j + B j ) = − D = T .Suppose now that Γ : [0 , → H n is any horizontal curve of constant speed connectingthe origin to the point (0 , , − T ), T >
0. Then Γ = ( x ( s ) , y ( s ) , t ( s )) is a geodesic if and only if ˜Γ( s ) = (˜ x ( s ) , ˜ y ( s ) , ˜ t ( s )) = ( x (1 − s ) , y (1 − s ) , t (1 − s ) + T )is a geodesic connecting ˜Γ(0) = (0 , ,
0) and ˜Γ(1) = (0 , , T ) since reversing a curve’sparametrization does not change its length and since the mapping ( x, y, t ) ( x, y, t + T )(the vertical lift by T ) is an isometry on H n . Therefore ˜Γ = (˜ x, ˜ y, ˜ t ) must have the form(2.14). Hence the (0 , , − T ) case of (2.2) follows from (2.14) by replacing s with 1 − s .The formula for the t component of Γ follows from (1.1); the integral is easy to computedue to numerous cancellations. (cid:3) Using the complex notation as in Theorem 2.4, the geodesics from Theorem 2.1 con-necting the origin to (0 , , ± T ), T > s ) = (cid:16)(cid:0) − e ∓ πis (cid:1) ( A + iB ) , t ( s ) (cid:17) where A = ( A , . . . , A n ), B = ( B , . . . , B n ) are such that 4 π | A + iB | = T and t ( s ) = ± T (cid:18) s − sin(2 πs )2 π (cid:19) . Theorem 2.1 and a discussion preceding it describes geodesics connecting the origin topoints either on the t -axis (0 , , ± T ), T > R n × { } . The question now is howto describe geodesics connecting the origin to a point q which is neither on the t -axis norin R n × { } . It turns out that geodesics described in Theorem 2.1 cover the entire space H n \ ( R n × { } ) and we have Corollary 2.7.
For any q ∈ H n which is neither in the t -axis nor in the subspace R n ×{ } there is a unique geodesic connecting the origin to q . This geodesic is a part of a geodesicconnecting the origin to a point on the t -axis.Proof. Let q = ( c , . . . , c n , d , . . . , d n , h ) be such that h = 0 and c j , d j are not all zero. Wecan write q = ( c + id, h ) ∈ C n × R . First we will construct a geodesic Γ q given by (2.15) sothat Γ q ( s ) = q for some s ∈ (0 , q (cid:12)(cid:12) [0 ,s ] will be part of a geodesicconnecting the origin to a point on the t -axis. Then we will prove that this curve is aunique geodesic (up to a reparametrization) connecting the origin to q . Assume that h > h < q that connects (0 , , , , T ), for some T >
0. (If h < , ,
0) to (0 , , − T ).) Itsuffices to show that there is a point A + iB ∈ C n such that the system of equations(2.16) (cid:0) − e − πis (cid:1) ( A + iB ) = c + id, π | A + iB | (cid:18) s − sin(2 πs )2 π (cid:19) = h has a solution s ∈ (0 , A + iB = ( c + id ) / (1 − e − πis ) and hence(2.17) 2 π | c + id | − cos(2 πs ) (cid:18) s − sin(2 πs )2 π (cid:19) = h. This equation has a unique solution s ∈ (0 ,
1) because the function on the left hand sideis an increasing diffeomorphism of (0 ,
1) onto (0 , ∞ ). We proved that, among geodesicsconnecting (0 , ,
0) to points (0 , , T ), T >
0, there is a unique geodesic Γ q passing through q . Suppose now that ˜Γ is any geodesic connecting (0 , ,
0) to q . Gluing ˜Γ with Γ q (cid:12)(cid:12) [ s , we EODESICS IN THE HEISENBERG GROUP 11 obtain a geodesic connecting (0 , ,
0) to (0 , , T ) and hence (perhaps after a reparametriza-tion) it must coincide with Γ q . This proves uniqueness of the geodesic Γ q (cid:12)(cid:12) [0 ,s ] . (cid:3) We will now use the proof of Corollary 2.7 to find a formula for the Carnot-Carath´eodorydistance between 0 and q = ( z, h ), z = 0, h >
0. We will need this formula in the nextsection. Let(2.18) H ( s ) = 2 π − cos(2 πs ) (cid:18) s − sin(2 πs )2 π (cid:19) : (0 , → (0 , ∞ )be the diffeomorphism of (0 ,
1) onto (0 , ∞ ) described in (2.17). LetΓ( s ) = (cid:0)(cid:0) − e − πis (cid:1) ( A + iB ) , t ( s ) (cid:1) be the geodesic from the proof of Corollary 2.7 that passes through q at s ∈ (0 , s is a solution to (2.17) and hence s is a function of q given by s ( q ) = H − ( h | z | − ) . Note that A + iB = z/ (1 − e − πis ), so | A + iB | = | z | p − cos(2 πs )) . Hence vuut n X j =1 ˙ x j ( s ) + ˙ y j ( s ) = L = √ πT = 2 π | A + iB | = 2 π | z | p − cos(2 πs ))where L is the length of Γ and Γ(1) = (0 , , T ). Therefore(2.19) d cc (0 , q ) = Z s vuut n X j =1 ˙ x j ( s ) + ˙ y j ( s ) ds = 2 πs | z | p − cos(2 πs )) . Analyticity of the Carnot-Carath´eodory metric
The center of the Heisenberg group H n is Z = { ( z, h ) ∈ H n | z = 0 } . It is well knownthat the distance function in H n is C ∞ smooth away from the center [12, 13], but throughthe use of (2.2), we will now see that this distance function is actually real analytic. Theorem 3.1.
The Carnot-Carath´eodory distance d cc : R n +1 × R n +1 → R is real analyticon the set (cid:8) ( p, q ) ∈ H n × H n = R n +1 × R n +1 : q − ∗ p Z (cid:9) . Proof.
In the proof we will make a frequent use of a well known fact that a compositionof real analytic functions is analytic, [10, Proposition 2.2.8]. It suffices to prove that thefunction d ( p ) = d cc (0 , p ) is real analytic on H n \ Z . Indeed, w ( p, q ) = q − ∗ p is real analyticas it is a polynomial. Also, d cc ( p, q ) = ( d ◦ w )( p, q ), so real analyticity of d on H n \ Z will imply that d cc is real analytic on w − ( H n \ Z ) = { ( p, q ) ∈ H n × H n | q − ∗ p / ∈ Z } . Define H : ( − , → R as(3.1) H ( s ) = 2 π − cos(2 πs ) (cid:18) s − sin(2 πs )2 π (cid:19) = πs − (2 πs ) + (2 πs ) − . . . − (2 πs ) + (2 πs ) − . . . . Here, we divided by a common factor of (2 πs ) in the two power series on the right handside. That is, the denominator equals (1 − cos(2 πs ))(2 πs ) − which does not vanish on( − , H is real analytic on ( − , s as a complexvariable, we see that H ( s ) is holomorphic (and hence analytic) in an open set containing( − ,
1) as a ratio of two holomorphic functions with non-vanishing denominator.As we pointed out in (2.18), the function H is an increasing diffeomorphism of (0 ,
1) onto(0 , ∞ ). Since it is odd and H ′ (0) = 2 π/ = 0, H is a real analytic diffeomorphism of ( − , R . Again, using a holomorphic function argument we see that H − : R → ( − ,
1) isa real analytic.The function z
7→ | z | − is analytic on R n \ { } (as a composition of a polynomial z
7→ | z | and an analytic function 1 /x ), so the function ( z, h ) h | z | − is analytic in H n \ Z . Hence also s ( q ) = H − ( h | z | − ) is analytic on H n \ Z .Fix q = ( z, h ) ∈ H n \ Z with h >
0. Then by (2.19)(3.2) d ( q ) = 2 πs | z | p − cos(2 πs )) . Since H ( s ) = h | z | − , formula (3.1) yields2 πs = (1 − cos(2 πs )) h | z | − + sin(2 πs ) . Substituting 2 πs in the numerator of the right hand side of (3.2) gives d ( q ) = h p − cos(2 πs ) √ | z | + | z | sin(2 πs ) √ p − cos(2 πs ) = h sin( πs ) | z | − + | z | cos( πs )where we used the trigonometric identities p − cos(2 πs ) = √ | sin( πs ) | = √ πs ) and sin(2 πs )sin( πs ) = 2 cos( πs ) . To treat the case h ≤ s ( q ) = H − ( h | z | − ) for any q = ( z, h ), z = 0.Previously we defined s ( q ) only when h >
0. It is easy to check that the mapping q = ( x, y, t ) = ( z, t ) ¯ q = (¯ z, − t ) = ( x, − y, − t )is an isometry of the Heisenberg group, so d ( q ) = d (¯ q ).If h < q = (¯ z, − h ), then s ( q ) = H − ( h | z | − ) = − H − ( − h | ¯ z | − ) = − s (¯ q )and hence d ( q ) = d (¯ q ) = − h sin( πs (¯ q )) | ¯ z | − + | ¯ z | cos( πs (¯ q )) = h sin( πs ( q )) + | z | cos( πs ( q )) . In the case h = 0, Γ is a straight line in R n from the origin to q , and so d ( q ) = | z | . EODESICS IN THE HEISENBERG GROUP 13
Therefore d ( q ) = h sin( πs ( q )) | z | − + | z | cos( πs ( q )) = h sin( πH − ( h | z | − )) | z | − + | z | cos( πH − ( h | z | − ))for every q = ( z, h ) ∈ H n \ Z , and so d is analytic on H n \ Z . (cid:3) We also proved
Corollary 3.2.
For z = 0 , the Carnot-Carath´eodory distance between the origin (0 , and ( z, h ) , z = 0 equals d cc ((0 , , ( z, h )) = h sin( πH − ( h | z | − )) | z | − + | z | cos( πH − ( h | z | − )) . Classification of non-unique geodesics
Any point (0 , , ± T ), T > t axis can be connected to the origin by infinitelymany geodesics. The purpose of this section is to show that all such geodesics are actuallyobtained from one geodesic by a linear mapping which fixes the t -axis. This map is anisometry of H n and also an isometry of R n +1 . Proposition 4.1. If Γ : [0 , → H n and Γ : [0 , → H n are constant-speed geodesicswith Γ (0) = Γ (0) = (0 , , and Γ (1) = Γ (1) = (0 , , ± T ) with T > , then we canwrite Γ = V ◦ Γ where V is a isometry in H n which fixes the t -coordinate. The map V is also an isometry of R n +1 , specifically a rotation about the t -axis.Proof. Consider geodesics Γ = ( γ , t ) and Γ = ( γ , t ) defined in the statement of theproposition. As in the discussion before Corollary 2.7, we consider γ and γ as functionsinto C n rather than into R n and write γ ( s ) = (cid:0) − e ∓ πis (cid:1) ( A + iB ) , γ ( s ) = (cid:0) − e ∓ πis (cid:1) ( C + iD )where 4 π | A + iB | = 4 π | C + iD | = T . We claim that there is a unitary matrix U ∈ U( n, C )such that U ( A + iB ) = C + iD . Indeed, for any 0 = z ∈ C n , use the Gram-Schmidtprocess to extend { z/ | z |} to an orthonormal basis of C n and define W z to be the matrixwhose columns are these basis vectors. Here, we consider orthogonality with respect tothe standard Hermitian inner product h u, v i C = P nj =1 u j v j . Then W z ∈ U( n, C ) and W z e = z/ | z | where { e , . . . , e n } is the standard basis of C n . Thus the desired operator is U = W C + iD ◦ W − A + iB .Define the linear map V : C n × R → C n × R by V ( z, t ) = ( U z, t ). Since U (cid:0)(cid:0) − e ∓ πis (cid:1) ( A + iB ) (cid:1) = (cid:0) − e ∓ πis (cid:1) ( C + iD ) , for every s ∈ [0 ,
1] and since V fixes the t -component of C n × R , we have V ◦ Γ = Γ .We now prove that V is an isometry on H n . Indeed, suppose p, q ∈ H n and Γ = ( γ, t ) :[0 , → H n is a geodesic connecting them. Then V ◦ Γ = ( U ◦ γ, t ). Since Γ is horizontal, itis easy to check that ˙ t ( s ) = 2 Im h γ ( s ) , ˙ γ ( s ) i C for almost every s ∈ [0 , preserve the standard inner product on C n , and so˙ t ( s ) = 2 Im h γ ( s ) , ˙ γ ( s ) i C = 2 Im h ( U ◦ γ )( s ) , ( U ◦ ˙ γ )( s ) i C = 2 Im (cid:28) ( U ◦ γ )( s ) , dds ( U ◦ γ )( s ) (cid:29) C for almost every s ∈ [0 , V ◦ Γ is horizontal. Also, ℓ H (Γ) = Z p h ˙ γ ( s ) , ˙ γ ( s ) i C ds = Z p h ( U ◦ ˙ γ )( s ) , ( U ◦ ˙ γ )( s ) i C ds = ℓ H ( V ◦ Γ) . Thus d cc ( V p, V q ) ≤ ℓ H (Γ) = d cc ( p, q ). Since U is invertible and U − ∈ U( n, C ), we mayargue similarly to show that d cc ( p, q ) = d cc ( V − V p, V − V q ) ≤ d cc ( V p, V q ), and so V isan isometry on H n . Clearly unitary transformations of C n are also orientation preservingisometries of R n and hence V is a rotation of R n +1 about the t -axis. (cid:3) Appendix
Proof of Proposition 1.1. (1) The components of any horizontal curve Γ are absolutelycontinuous, and so their derivatives are integrable. Thus the inequality ℓ cc (Γ) ≤ ℓ H (Γ) < ∞ yields rectifiability of horizontal curves. It remains to prove that ℓ cc (Γ) ≥ ℓ H (Γ). Extendthe Riemannian tensor defined on the horizontal distribution H H n to a Riemannian tensor g in R n +1 . For example we may require that the vector fields X j , Y j , T are orthonormal atevery point of R n +1 . The Riemannain tensor g defines a metric d g in R n +1 in a standardway as the infimum of lengths of curves connecting two given points, where the length ofan absolutely continuous curve α : [ a, b ] → R n +1 is defined as the integral ℓ g ( α ) = Z ba p g ( ˙ α ( s ) , ˙ α ( s )) ds. This is the same approach that was used to define the Carnot-Carath´eodory metric. Forhorizontal curves Γ, we have ℓ g (Γ) = ℓ H (Γ), and so it is obvious that d g ( p, q ) ≤ d cc ( p, q )since we now take an infimum over a larger class of curves. It is a well known fact inRiemannian geometry that for an absolutely continuous curve α : [ a, b ] → R n +1 ℓ g ( α ) = sup n − X i =0 d g ( α ( s i ) , α ( s i +1 )) , where the supremum is taken over all n ∈ N and all partitions a = s ≤ s ≤ . . . ≤ s n = b as before. Hence if Γ : [ a, b ] → R n +1 is horizontal we have ℓ H (Γ) = ℓ g (Γ) = sup n − X i =0 d g (Γ( s i ) , Γ( s i +1 )) ≤ sup n − X i =0 d cc (Γ( s i ) , Γ( s i +1 )) = ℓ cc (Γ) . (2) If Γ : [ a, b ] → H n is Lipschitz, then by (1.4) it is also Lipschitz with respect to theEuclidean metric in R n +1 . Thus it is absolutely continuous and differentiable a.e. Itremains to show that ˙Γ( s ) ∈ H Γ( s ) H n a.e. We will actually show that this is true wheneverΓ is differentiable at s .Suppose Γ is differentiable at a point s ∈ ( a, b ) and let i ∈ N be such that s + 2 − i ≤ b .Then d cc (Γ( s + 2 − i ) , Γ( s )) ≤ L − i for some L > i ≥ i . By the definition of the EODESICS IN THE HEISENBERG GROUP 15
Carnot-Carath´eodory metric on H n , there is some horizontal curve η i : [0 , − i ] → H n whichconnects Γ( s ) to Γ( s +2 − i ) whose length approximates the distance between them. That is,we can choose a horizontal curve η i = ( x i , y i , t i ) so that η i (0) = Γ( s ), η i (2 − i ) = Γ( s + 2 − i )and ℓ H ( η i ) < L − i . After a reparameterization, we may assume that η i has constantspeed | ˙ η i | H < L on [0 , − i ]. Since η i is horizontal, we can write˙ η i ( τ ) = n X j =1 ˙ x ij ( τ ) X j ( η i ( τ )) + ˙ y ij ( τ ) Y j ( η i ( τ ))for almost every τ ∈ [0 , − i ]. Now Γ( s + 2 − i ) − Γ( s ) = η i (2 − i ) − η i (0) = R − i ˙ η i ( τ ) dτ , andso Γ( s + 2 − i ) − Γ( s )2 − i = 12 − i Z − i n X j =1 (cid:0) ˙ x ij ( τ ) X j ( η i ( τ )) + ˙ y ij ( τ ) Y j ( η i ( τ )) (cid:1) dτ = 12 − i Z − i n X j =1 ˙ x ij ( τ ) (cid:0) X j ( η i ( τ )) − X j (Γ( s )) (cid:1) dτ (5.1) + 12 − i Z − i n X j =1 ˙ y ij ( τ ) (cid:0) Y j ( η i ( τ )) − Y j (Γ( s ))) (cid:1) dτ (5.2) + 12 − i Z − i n X j =1 (cid:0) ˙ x ij ( τ ) X j (Γ( s )) + ˙ y ij ( τ ) Y j (Γ( s )) (cid:1) dτ The images of all the curves η i are contained in a compact subset of R n +1 . Hence (1.2)yields sup τ ∈ [0 , − i ] | η i ( τ ) − Γ( s ) | ≤ ℓ E ( η i ) ≤ C ( K ) ℓ H ( η i ) ≤ C ( K ) L − i +1 → i → ∞ .Here ℓ E ( η i ) stands for the Euclidean length of η i . Observe also that the functions ˙ x ij and˙ y ij are bounded almost everywhere by the speed | ˙ η i | H which is less than 2 L . This andthe continuity of the vector fields X j , Y j immediately imply that the sums (5.1) and (5.2)converge to zero as i → ∞ . Also, by again using the uniform boundedness of the functions˙ x ij and ˙ y ij , we conclude that for some subsequence and all j = 1 , , . . . , n the averages2 i R − i ˙ x ij and 2 i R − i ˙ y ij converge to some constants a j and b j respectively. Denote such asubsequence by i k . Therefore, since ˙Γ( s ) exists we have˙Γ( s ) = lim k →∞ Γ( s + 2 − i k ) − Γ( s )2 − i k = n X j =1 ( a j X j (Γ( s )) + b j Y j (Γ( s ))) ∈ H Γ( s ) H n . This proves (2). Finally (3) follows from a general fact that a rectifiable curve in any metricspace admits an arc-length parametrization with respect to which the curve is 1-Lipschitz,see e.g. [4, Proposition 2.5.9], [8, Theorem 3.2]. (cid:3)
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P. Haj lasz: Department of Mathematics, University of Pittsburgh, 301 ThackerayHall, Pittsburgh, PA 15260, USA, [email protected]