Riemannian Geometry, Key to Homework #1

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1 Riemannian Geometry Key to Homework # Let σu v sin u cos v sin u sin v cos u < u < π < v < π be a parametrization of the unit sphere S {x y z R 3 x + y + z } Fix an angle < θ < π and consider the parallel in short we just write u θ on the unit sphere αt sin θ cos t sin θ sin t cos θ < t < π i Sketch the curve α ii Calculate the normal curvature of α Solution: ii Method : By the calculation n σ u σ v σ u σ v ±sin u cos v sin u sin v cos u On the other hand α t sin θ sin t sin θ cos t arc-length parameterization is It is not unit-speed its Hence αs sin θ coss/ sin θ sin θ sins/ sin θ α s sin θ coss/ sin θ sins/ sin θ Write αs σus vs we find Hence the restriction of n to αs is us θ vs s sin θ ns ±sin θ coss/ sin θ sin θ sins/ sin θ cos θ By the formula the normal curvature of α is κ n s ns α s ± cos s/ sin θ + sin s/ sin θ ±

2 Method : We can also use Theorem 4 ie 45 to calculate the normal curvature where v α t where α must be a unit-speed curve To do so we need to calculate the second fundmental form e f g By a simple calculation we have e f g sin u On the other hand as we did above after re-parameterized by arc-length parameter we get αs sin θ coss/ sin θ sin θ sins/ sin θ Write αs σus vs we find Thus u s v s / sin θ Hence us θ vs s sin θ κ n s IIα s α s eu s +fu sv s+gv s sin θ sin θ Since the principal normal vector can have two directions we kave κ n s ± Show that the normal curvature of any curve on a sphere of radius r is ±/r Solution: Let M {x y z x + y + z r } Then its unit normal is just its normalized position vector ie n ± x + y + z x y z ± x y z r Let αt be a curve on the sphere M then the restriction of n to αt is nt ± r αt Therefore by the formula the normal curvature of α is κ n nt α t ± r αt α t Since α is on the sphere α α r By differentiate it on both sides we have α α Differentiating it on both sides again yields α α + α α Since α is unit-speed α α Hence α α Therefore κ n ± r αt α t ± r

3 3 Show that if a curve on a surface has zero normal and geodesic curvature everywhere then it is part of a straight line Solution: Since κ κ n + κ g The condition that a curve on a surface has zero normal and geodesic curvature everywhere implies that its curvature is identically zero So by the theorem it is part of a straight line 4 Find the Gauss curvature mean curvature principal curvatures and the corresponding principal directions of the following surfaces a σu v au + v bu v 4uv where a and b are constant b The cylinder: σu v a cos u a sin u v Solution: a σ u a b 4v σ v a b 4u σ uu σ vv σ uv 4 The unit normal is n σ u σ v σ u σ v The first fundamental form is bu + v av u ab 4b u + v + 4a u v + a b / E σ u σ u a +b +6v F σ u σ v a b +6uv G σ v σ v a +b +6u The second fundamental form is e n σ uu f n σ uv The Gauss curvature is the mean curvature is 4ab 4b u + v + 4a u v + a b / g n σ vv EG F 6a u v + 6b u + v + 4a b K eg f EG F 4a b 4b u + v + 4a u v + a b H eg ff + ge EG F aba b + 6uv 4b u + v + 4a u v + a b 3/ To calculate the principal curvatures note that E F F G EG F G F F E 3

4 So the matrix of the shape operator S P with respect to the basis {σ u σ v } is A F I F II EG F 3/ G F F E 8ab 8ab 8ab EG F 3/ F G E F Write µ 8ab then EG F 3/ deta λi µf λ µ EG Setting deta λi ie µf λ µ EG we get µf λ ±µ EG Hence the eigenvaluesprincipal curvatures are κ µf + EG κ µf EG where µ E F G are given as above To get the principal directions for κ µf + EG we solve A κ Iv ie µ EG µg µe µ EG ξ η We get we need normalize it to get a unit-vector!!! ξ η + E G E G G E+G E E+G Similarily ξ Hence the principal directions are where µ η G E+G E E+G G E G E e E + G σ u E + G σ v e E + G σ u + E + G σ v 8ab and E F G are given as above EG F 3/ ii By direct calculation the first fundamental form is E a F G 4

5 and the second fundamental form is e a f g Hence the matrix of the shape operator S P with respect to the basis {σ u σ v } is A F I F II /a Solving the equation deta λi we get the principal curvatures κ κ /a To get the principal directions for κ we solve A κ Iv ie /a ξ We get we need normalize it to get a unit-vector!!! ξ η so e σ v Similarily for κ /a we get ξ η so e σ u η The Gauss curvature is K κ κ and the mean curvature is H κ + κ a 5 Calculate the Christoffel symbols for the surface z fx y Solution: Let σu v u v fu v Then σ u f u σ v f v Hence E + f u F f u f v G + f v E u f u f uu E v f u f uv F u f uu f v +f u f uv F v f uv f v +f u f vv G u f v f uv G v f v f vv Hence Γ f u f uu + fu + fv Γ Γ f u f uv + fu + fv Γ Γ f u f vv + fu + fv f v f uu + fu + fv Γ Γ f v f uv + fu + fv 5

6 Γ f v f vv + fu + fv 6 Assume that the the surface σu v has its first fundamental form as E Prove its Gauss curvature K 4 + u + v F G 4 + u + v Solution: Use the following formula directly: K EG Ev EG v + Gu EG u 6