The n-dimensional hyperboloid and hyperbolic geometry
Ken Li, Dennis Merino, and Edgar N. Reyes Southeastern Louisiana University Hammond, LA 70402
Abstract Let H be the hyperboloid model for n-dimensional hyperbolic geometry. In this paper, using relatively elementary methods, we establish the well-known fact that the exponential map of H is surjective. This is proved by describing the geodesics in H.As a corollary, we show that the geodesic are planar curves and when n = 2 the Gaussian curvature of H is −1.
1 Introduction to the Hyperboloid Model
Let n ≥ 2 be an integer, and consider the (n +1)-by-(n +1)matrix 0 . −In . In,1 = . 0 0 ··· 0 1 where In is the identity n-by-n matrix. The matrix In,1 induces a real bilinear form whichwedenoteby∗. That is, for any two real vectors x =(x1,x2, ..., xn,xn+1)and y =(y1,y2, ..., yn,yn+1) we associate a real number according to the rule n T x ∗ y = xIn,1y = − xiyi + xn+1yn+1. (1.1) i=1
The set n+1 H = x ∈ IR : x ∗ x =1,xn+1 > 0 . is the hyperboloid model for n-dimensional hyperbolic geometry. Another model for hyper- bolic geometry is the n-disk
n 2 2 D = {(u1, ..., un) ∈ IR : u1 + ... + un < 1}.
1 Consider the mapping Π from D into H defined by
1 2 Π(u)= 2u1, ..., 2un, |u| +1 (1.2) 1 −|u|2
2 2 where u =(u1, ..., un) ∈ D and |u| = u1 + ... + un. The inverse of Π is given by x1 xn Π−1(x)= , ..., . (1.3) xn+1 +1 xn+1 +1 The mapping Π−1 is the stereographic projection that associates to each point x in H the point u in D on the plane xn+1 = 0 that lies on the line through x and (0, ..., 0, −1). Since Π is a homeomorphism and D is connected, H is a connected subset of IRn+1. n+1 A vector v =(v1, ..., vn+1) ∈ IR is a tangent vector at x in H if there exists a differentiable mapping γ defined on an open interval containing zero into IRn+1 whose image lies in H such that γ(0) = x and γ (0) = v. We denote the set of tangent vectors at x by
TxH.Wecallγ a curve. Similarly, for the n-disk, the set of tangent vectors at u ∈ D is denoted by TuD.
Lemma 1 If x ∈ H,then n+1 TxH = v ∈ IR : v ∗ x =0 (1.4) is an n-dimensional real vector space.
Proof: If γ is a curve in H,then
γ(t) ∗ γ(t)=1.
Differentiating both sides of the previous equation, we obtain
γ (t) ∗ γ(t) = 0 (1.5)
If γ(0) = x and γ (0) = v, then equation (1.5) reduces to v ∗ x =0whent =0. n+1 Let R be the right side of (1.4). Then R is subspace of IR .WejustprovedTxH ⊆R. Since H (realized as an inverse image of a point) is an n-dimensional submanifold of IRn+1, then TxH is an n-dimensional real vector space see [3, Theorem 1.38]. Since x ∗ x =1,we n+1 see that R is a proper subspace of IR . Hence, R = TxH. 2
Lemma 2 The bilinear form on TxH given by
v, wx ≡−(v ∗ w),v,w∈ TxH (1.6) is positive-definite.
2 Proof: The following proof follows closely the one provided in [1, pp. 9-10], which we include for completeness. Let v =(v1, ..., vn+1) =(0 , ..., 0) and x =(x1, ..., xn+1).
Clearly, v, vx > 0ifvn+1 =0.
We assume vn+1 = 0. Using the Cauchy-Schwarz inequality, we obtain
2 2 2 2 2 (x1 + ... + xn)(v1 + ... + vn) ≥ (x1v1 + ... + xnvn)
2 =(xn+1vn+1) Since x ∗ v =0 2 2 2 = vn+1 1+x1 + ... + xn Since x ∗ x =1 2 2 2 2 = vn+1 + vn+1 x1 + ... + xn .
Then 2 2 2 2 2 2 2 2 (x1 + ... + xn)(v1 + ... + vn) − vn+1(x1 + ... + xn) ≥ vn+1 or equivalently 2 2 2 2 2 2 (x1 + ... + xn)(v1 + ... + vn − vn+1) ≥ vn+1.
2 Since vn+1 > 0, we find v, vx > 0. This proves Lemma 2. 2
Let t SO(n, 1) = A ∈ SL(n +1,IR):A In,1A = In,1 where SL(m, IR) is the group of real m-by-m matrices with determinant one.
t Lemma 3 AmatrixA satisfies A In,1A = In,1 if and only if n 2 2 a) − aj,s + an+1,s = −1fors = n +1, j=1 n 2 2 b) − aj,n+1 + an+1,n+1 =1,and j=1 n c) − aj,saj,t + an+1,san+1,t =0ifs = t j=1
t Proof: Simply equate the (s, t)-entries of the two sides in A In,1A = In,1. 2
Consider the special point N =(0, ...0, 1) ∈ H.GivenA ∈ SO(n, 1) and the fact that H is connected, we find NAt ∈ H if and only if xAt ∈ H for all x ∈ H since the image of a connected set under a continuous map is connected. Then G = A ∈ SO(n, 1) : NAt ∈ H (1.7)
3 is a closed subgroup of SO(n, 1). The group G acts on the hyperboloid H by the mapping
(A, x) ∈ G × H → xAt ∈ H.
The next lemma shows that this action is transitive.
Lemma 4 If x ∈ H,thenNAt = x for some A ∈ G.
Proof: Choose any v1 ∈ TxH such that that v1 x =1.Note,x and v1 are linearly n+1 t t independent vectors in IR since x ∗ v1 =0.Ifn = 1, then the 2-by-2 matrix A =[v1,x ] t t t satisfies A I1,1A = I1,1 by Lemma 3. Also, AN = x .Sincedet(A)=±1, we could assume det(A) = 1 for we could multiply the first column of A by −1 if needed. If n ≥ 2, there exists v2 ∈ TxH such that v2 and v1 are linearly independent vectors in n+1 IR . Using the Gram-Schmidt orthogonalization process, let