Theorem. In terms of the coordinate frame, the Levi-Civita connection is given by:

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1 THE LEVI-CIVITA CONNECTION FOR THE POINCARÉ METRIC We denote complex numbers z = x + yi C where x, y R. Let H 2 denote the upper half-plane with the Poincaré metric: {x + iy x, y R, y > 0} g = dz 2 y 2 We compute the Levi-Civita connection with respect to several different frames. 1. The usual coordinate system on H 2 Let x, y be the coordinate vector fields, so that: 1) 2) g x, x ) = g y, y ) = y 1 g x, y ) = g y, x ) = 0. The vector fields define an orthonormal frame field. ξ := y 1 x η := y 1 y Theorem. In terms of the coordinate frame, the Levi-Civita connection is given by: 3) 4) 5) 6) x x = y 1 y x y = y 1 x y x = y 1 x y y = y 1 y Date: wmg, July 17,

2 2 LEVI-CIVITA CONNECTION In terms of the orthonormal frame, the Levi-Civita connection is given by: ξ ξ = η ξ η = ξ η ξ = 0 η η = 0 Proof. Orthonormality implies gy x, y x ) = 1 is constant, whence and 0 = y gy x, y x ) = 2g y y x ), y x ) = 2yg y y x ), x ) = 2yg x, x ) + yg y x, x )) = 2yy 1 + yg y x ), x ) 7) g y x, x ) = y 3. Differentiating the constant gy y, y ) with respect to x: 8) 0 = 2 x g x y, x ) whence 9) g y x, x ) = 0. Since is symmetric, 10) x y = y x. Combining 9) with 10): g x y, y ) = y 3 and combining with 8), yields 4). Applying 10) again yields 5). Differentiating gy x, y x ) = 1 with respect to x yields: 11) g x x, x ) = 0 Similarly, differentiating with respect to y ; 12) g x x, x ) = y 3 which implies 3). Differentiating gy x, y y ) = 0 with respect to y yields: 13) g y y, x ) = 0 Similarly, 0 = y g y, y ) implies; 14) g y y, y ) = y 3

3 LEVI-CIVITA CONNECTION 3 which implies 6). The routine calculations for the orthonormal frame are omitted. 2. Connection 1-form for orthonormal frame Denote the coframe field dual to the orthonormal frame by: ξ := ydx η := ydy. The covariant differentials of the orthonormal frame are: so the connection form is: [ ] 0 ξ ξ 0 ξ = η ξ η = ξ ξ = y 1 dx [ ] Geodesic curvature of a hypercycle We use these to calculate the geodesic curvature of a hypercycle. The positive imaginary axis ir + is a geodesic with endpoints 0,. The Euclidean rays from 0 subtending an angle θ with ir + are hypercycles re iθ = r cosθ) + i sinθ) ) = r tanhρ) + i sechρ) ) at distance ρ from ir +. The point x 0 + i has coordinates The curve x 0 = tanhρ) sechρ) = sinhρ). γt) := e t x 0 + i) has velocity vector and speed, respectively: 15) γ t) := e t x 0 x + y ) = x 0 ξ + η γ t) = 1 + x 2 0 We compute the accelaration of γt): D dt γt) = x 0 ξ+ηx 0 ξ + η) = x 2 0η x 0 ξ

4 4 LEVI-CIVITA CONNECTION Now reparametrize γt) by unit speed: γt) := x0 e t x 0 + i) and the geodesic curve is the tangential component of the acceleration: D dt γt) = x 0 x 2 0 η x 0 ξ ) which, since γt) has constant speed, equals: D dt γt) = x x 2 0 = sinθ) = tanhρ) Similarly the geodesic curvature of a metric circle of radius ρ equals cothρ). To see this, use the Poincare unit disc model: for z < 1, the metric tensor is: g := 4 dz 2 1 z 2 ) 2 and writing hyperbolic polar coordinates) z = e iθ tanhρ/2) the metric tensor is g = dρ 2 + sinh 2 ρ)dθ 2, with area form da = sinh 2 ρ)dρ dθ. Consider a disc D ρ with hyperbolic) radius ρ. Then its circumference equals 2π sinhρ) and its area 2πcoshρ) 1. Let k g be the geodesic curvature of the metric circle D ρ. Applying the Gauss-Bonnet theorem 2πχΣ) = KdA + k g ds to σ = D ρ obtaining: Σ Σ 2π = 2π coshρ) 1) + k g 2π sinhρ) ), that is, k g = cothρ) as desired.

5 LEVI-CIVITA CONNECTION 5 4. Fermi coordinates around a geodesic Another convenient coordinate system begins with a geodesic and considers the family of geodesics orthogonal to this one. Let u denote the parameter along the geodesic, and v the parameter along the perpendiculars: x = e u tanhv) y = e u sechv). Then sinhv) = x/y and z = x 2 + y 2 = e u and u = 1 2 logx2 + y 2 ). Writing with metric tensor The coordinate 1-forms are: and dual coodinate vector fields: z = e u tanhv) + i sechv)) dz = zdu + i sechv)dv) dz 2 = z 2 du 2 + sech 2 v)dv 2) g = dz 2 y 2 = cosh 2 v)du 2 + dv 2. du = x 2 + y 2 ) 1 x dx + y dy) dv = x 2 + y 2 ) 1/2 y dx x/y dy) u = x x + y y v = yx 2 + y 2 ) 1/2 y x x y ). 5. The Levi-Civita connection in Fermi coordinates Using the coordinates u, v) R 2 with metric tensor g = cosh 2 v)du 2 + dv 2 as above, we compute the Christoffel symbols of the Levi-Civita connection. Since u, v are coordinates, the corresponding vector fields u, v commute: [ u, v ] = 0 and since has zero torsion, u v = v u. Since all the inner products of this basis are constant except for g u, u ), 0 = u g u, v) = v g u, v) = u g v, v) = v g v, v) and g is parallel with respect to, implies v v = 0

6 6 LEVI-CIVITA CONNECTION and differentiating g u, u ) = cosh 2 v) yields u v = v u = tanhv) u u u = sinhv) coshv) v One can modify this coordinate basis to an orhonormal basis by replacing u by U := sechv) u, in which the covariant derivatives are: U U) = tanhv) v, v U) = 0, U v ) = tanhv)u, v v ) = Geodesic curvature in Fermi coordinates Let γt) = ut), vt) ) be a curve. Its velocity is: γ t) = u t) u + v t) v and the square of its speed is: ) 2 ds = u t) cosh 2 vt) ) + v t) 2 dt Its acceleration is: D dt γ t) = u t) + tanh ) vt))u t)v t) u + v t) + tanh vt))u t)v t) u t) 2 cosh vt) ) sinh vt) )) u