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Models of Hyperbolic Geometry

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Models of Hyperbolic Geometry Models of Hyperbolic Geometry Matthew D. Staley San Francisco State University Abstract: Hyperbolic geometry naturally arises from the context of Spe- cial Relativity. Spacetime is modeled as a 4 dimensional real manifold. Using the Lorenztian inner product on this manifold, we obtain the geometry of a hyperbola. We will investigate the nature of real and complex hyperbolic ge- ometry in a more general setting. We are mostly concerned with the Poincare Disc Model, its tangent space, and whether or not we have a conformal map- ping. 1. Real Hyperbolic Models n+1 We begin our study in the ambient space R , with x = (x1; : : : ; xn; xn+1) 2 n+1 n+1 n+1 2 X 2 R . Denote the Euclidean Norm for x 2 R as jjxjj = xk . k=1 Definition 1. Let x; y 2 Rn+1. The Lorentzian inner product is defined to be < x; y >(n;1) = x1y1 + ··· + xnyn − xn+1yn+1: This is a symmetric bilinear form and is allowed to take on negative values, unlike usual inner-products. In fact, Hyperbolic Space is defined for such constant negative values. The (n; 1) denotes the signature. Definition 2. Real Hyperbolic-n space is the set n n+1 n+1 HR = fx 2 R : < x; x >(n;1)= −1g ⊂ R : 2 MATTHEW D. STALEY .. Figure 1: Minkowski Spacetime 3 Example 1. When n = 3 we have HR as a model of Minkowski Spacetime. The Lorentzian inner product equation becomes 2 2 2 2 x1 + x2 + x3 − x4 = −1: 3 We can think of (x1; x2; x3) 2 R as spacial coordinates and x4 2 R as time. If we fix one of the spacial coordinates at the \origin", say x1 = 0 (such as a 2 2 2 fixed observer), then the resulting equation becomes x2 + x3 − x4 = −1. This is just a hyperboloid in two sheets whose boundary is the \light cone", which is where < x; x >(n;1)= 0. (See Figure 1 above, Source: Wikipedia) n We now wish to model HR for any n. We are only interested in the positive, or upper hyperboloid sheet, of the hyperbolic space. There are several ways to do this, such as the upper half plane model and stereographic projection. The following Klein Disc Model is a very intuitive approach to n modeling HR using projections. MODELS OF HYPERBOLIC GEOMETRY 3 1.1. The Klein Disc Model n 2 n We start with a vector x 2 BR = fjjxjj < 1g. Think of BR embedded n+1 n n+1 into R at xn+1 = 1. We want to map x into H ⊂ R by projecting n n from the origin, through B and into HR. We need to scale the vector ap- n propriately so that x 7! (λx1; : : : ; λxn; λ) 2 HR. To find the λ that works, we simply plug it into the Lorenztian inner product equation and solve for it: 2 2 2 (λx1) + ··· + (λxn) − λ = −1 λ2(jjxjj2 − 1) = −1 −1 λ2 = jjxjj2 − 1 1 λ2 = : 1 − jjxjj2 We want the upper hyperboloid, so we take the positive square root to obtain 1 λ = : p1 − jjxjj2 n n Definition 3. Let Φ: BR ! HR by ! x x x 1 x 7! 1 ; 2 ;:::; n ; : p1 − jjxjj2 p1 − jjxjj2 p1 − jjxjj2 p1 − jjxjj2 See Figure 2 on next page. 1.1.1. The Columns of DΦ Note that Φ(x) is now a vector sitting in Rn+1, and each vector component h i is a function, say f, of n variables. We want to compute DΦ = @fi for @xj i = 1; : : : ; n + 1 and j = 1; : : : ; n. Thus DΦ is an (n + 1) × (n) matrix. 4 MATTHEW D. STALEY Rn+1 n HR n xn+1 = 1 BR 0 .. Figure 2: Klein Model xi Proposition 1. Denote fi(x1; : : : ; xn) = fi = p for i = 1; : : : ; n and 1−||xjj2 1 fn+1 = p . Then we have the following: 1−||xjj2 @fi xixj 1. For i 6= j, = 3=2 : @xj (1 − jjxjj2 ) 2 2 @fi 1 + xi − jjxjj 2. For i = j, = 3=2 : @xi (1 − jjxjj2 ) @fn+1 xj 3. For i = n + 1, and j = 1; : : : ; n, = 3=2 : @xj (1 − jjxjj2 ) Proof. These derivatives easily follow from the usual quotient and chain rule formulas from calculus as follows: MODELS OF HYPERBOLIC GEOMETRY 5 1. For i 6= j, ! @f @ x 0 − x 1 (1 − jjxjj2)−1=2(−2x ) i = i = i 2 j p 2 @xj @xj 1 − jjxjj2 1 − jjxjj −x (−x )(1 − jjxjj2)−1=2 x x = i j = i j : 1 − jjxjj2 (1 − jjxjj2)3=2 2. For i = j, ! p 1 @f @ x 1 − jjxjj2 − x (1 − jjxjj2)−1=2(−2x ) i = i = i 2 j p 2 @xi @xj 1 − jjxjj2 1 − jjxjj 1 1 − jjxjj2 + x2 1 + x2 − jjxjj2 = i = i : p1 − jjxjj2 1 − jjxjj2 (1 − jjxjj2)3=2 3. For i = n + 1, and j = 1; : : : ; n @fn+1 @ 2 −1=2 1 2 −3=2 xj = (1−||xjj ) = − (1−||xjj ) (−2xj) = 2 3=2 : @xj @xj 2 (1 − jjxjj ) n Definition 4. The Tangent Space at point p 2 HR will be the span of the column vectors of DΦ, that is: n Tp HR = spanf f1; : : : ; fn : fi 2 DΦg: 1.1.2. The Klein Disc Model is Not Conformal Proposition 2. Denote the Ith column of DΦ by α and the J th column by β. Then for I 6= J; < α; β >(n;1) 6= 0. 6 MATTHEW D. STALEY Proof. Let us write out what α and β look like as column vectors: 0 1 0 1 x1xI x1xJ B . C B . C B . C B . C B C B C B C B C B xI−1xI C B xI−1xI C B C B C B1 + x2 − jjxjj2C B x x C B I C B I J C B C B C B xI+1xI C B xI+1xJ C B . C B . C B . C B . C 2 3=2 B . C 2 3=2 B . C 1 − jjxjj α = B C ; 1 − jjxjj β = B C : B xJ−1xI C B xJ−1xJ C B C B C B C B 2 2C B xJ xI C B1 + xJ − jjxjj C B C B C B xJ+1xI C B xJ+1xJ C B C B C B . C B . C B . C B . C B C B C B x x C B x x C @ n I A @ n J A xI xJ Then using the Lorentzian inner product on α and β, 2 3 1 − jjxjj < α; β >(n;1) 2 2 2 2 2 = xI xJ (x1 + ··· + xI−1 + 1 + xI − jjxjj + xI+1 + ::: 2 2 2 2 + xJ+1 + 1 + xJ − jjxjj + xJ+1 + ··· + xn − 1) 2 2 2 = xI xJ jjxjj + 1 + 1 − 1 − jjxjj − jjxjj 2 = xI xJ 1 − jjxjj : x x < α; β > = I J ) (n;1) (1 − jjxjj2) Note that this value is dependent on the last coordinate of the column vectors MODELS OF HYPERBOLIC GEOMETRY 7 chosen. These two are only zero whenever x = (0;:::; 0; 0). But this is not in the domain of Φ, and so < α; β >(n;1)6= 0 for any α; β 2 DΦ. Corollary 3. The columns of DΦ are not perpendicular, and so angles are distorted. Thus the Klein Disc model is not conformal. 1.2. Poincare Disc Model n In this model, we want to map a vector x 2 BR, this time centered at n xn+1 = 0, to HR. To do this, we project from the point (0;:::; 0; 1) n n through BR to a point tx 2 HR, where t > 0. Thus we have that n y = (tx1; : : : ; txn; t − 1) 2 HR if and only if 2 2 2 (tx1) + ··· + (txn) − (t − 1) = −1 t2 jjxjj2 − t2 + 2t − 1 = −1 t2 jjxjj2 − 1 = −2t 2 t = : 1 − jjxjj2 Definition 5. Let Ψ: Bn+1 ! Hn by R R 2x 2x 1 + jjxjj2 Ψ(x) = 1 ;:::; n ; : 1 − jjxjj2 1 − jjxjj2 1 − jjxjj2 See Figure 2 on next page. 1.2.1. The columns of DΨ As before, we have a function of n variables for each vector component in h i Ψ(x). Then we can compute DΨ = @fi for i = 1; : : : ; n+1 and j = 1; : : : ; n. @xj So DΨ will be an (n + 1) × (n) matrix. Again, there are three cases for the entries: 8 MATTHEW D. STALEY Rn+1 n HR Bn R 0 (0;:::; 0; 1) .. Figure 3: Poincare Disc Model 2xi Proposition 4. Denote fi(x1; : : : ; xn) = fi = 1−||xjj2 for i = 1; : : : ; n and 2 f = 1+jjxjj . n+1 (1−||xjj2)2 @fi 4xixj 1. For i 6= j, = 2 . @xj (1 − jjxjj2) 2 2 @fi 2 + 4xi − 2jjxjj 2. For i = j, = 2 : @xi (1 − jjxjj2) @fn+1 4xj 3. For i = n + 1 and j = 1; : : : ; n, = 2 : @xj (1 − jjxjj2) Proof. 1. For i 6= j; @fi @ 2xi 0 − (2xi)(−2xj) 4xixj = 2 = 2 2 = 2 : @xj @xj 1 − jjxjj (1 − jjxjj ) 1 − jjxjj MODELS OF HYPERBOLIC GEOMETRY 9 2. For i = j, 2 2 2 @ 2xi (1 − jjxjj )(2) − (2xi)(−2xi) 2 + 4xi − 2jjxjj 2 = 2 2 = 2 2 : @xi 1 − jjxjj (1 − jjxjj ) (1 − jjxjj ) 3. For i = n + 1 and j = 1; : : : ; n, 2 2 2 @fn+1 @ 1 + jjxjj (1 − jjxjj )(2xj) − (1 + jjxjj )(−2xj) = 2 = 2 2 @xj @xj 1 − jjxjj (1 − jjxjj ) 2x − 2jjxjj2 + 2x + jjxjj2 4x = j j = j : (1 − jjxjj2)2 (1 − jjxjj2)2 n Definition 6.
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