Definition 1.1 (Curve). A continuous function α : [a, b] R 3 α(t) = (x(t), y(t), z(t)), and each component is continuous.

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1 1 Space Curves Definition 1.1 (Curve). A continuous function α : [a, b] R 3 α(t) = (x(t), y(t), z(t)), and each component is continuous. is a curve Definition 1. (Differentiable Curve). α is a differentiable curve if x, y, z are differentiable. Definition 1.3 (Plane Curve). Curves α : [a, b] R are called plane curves. Definition 1.4 (Regular Curve). A curve is regular if the vector α (t) 0, that is, α (t) 0 Theorem 1.1. The arclength along a curve is s = b a α (t) Proof. α(t i ) α(t i 1 ) is a first approximation. By the mean value theorem, this is equal to α (t c ) t i t i 1. If we take the limit as (t i t i 1 ) 0, then we get b a α (t) Definition 1.5 (Unit Tangent Vector). Let α(t) be a regular curve. Then the unit tangent vector is T (t) = α (t) α (t) Definition 1.6 (Curvature). Let T (s) be the unit tangent vector parametrized by arclength. Then dt = k(s) n(s) with k(s) > 0 k(s) = dt is called the curvature, with n(s) as the unit normal vector. Definition 1.7 (Dot Product). Let u, v R 3. u v = x 1 x + y 1 y + z 1 z Lemma 1.. d ( u v) = u v + u v Corollary 1.3. Given a unit vector, ie u = 1 for t I, then d u to u(t) is orthogonal Theorem 1.4. Let α(s) be parametrized by arclength. Then α (s) = 1. Corollary 1.5. α and α are orthogonal. Definition 1.8 (Singular Points of the First Kind). Places where k(s) = 0 are singular points of the first kind. Definition 1.9 (Bi-normal Vector). The bi-normal vector b(s) is defined as b(s) = T (s) n(s) where signifies the cross product. By this definition, the binormal vector is orthogonal to both the unit tangent vector and the unit normal vector, and is itself a unit vector. Definition 1.10 (Frenet Trihedron). The Frenet Trihedron is the set of orthonormal vectors b(s), T (s), n(s). Lemma 1.6. b (s) = τ(s)n(s) Proof. b (s) = d T (s) n(s) = T (s) n(s) + T (s) n (s) T (s) = k(s)n(s), so b (s) = T (s) n (s), and so b (s) is perpendicular to both T (s) and b(s) (because b(s) is a unit vector) so b (s) = τ(s)n(s). 1

2 Definition 1.11 (Torsion). τ(s) is called the torsion of the curve. Lemma 1.7. If τ(s) 0, then α lies in a plane. Proof. First, fix the plane Π. Pick a point α(s 0 ) on the curve and let that be the origin and let b 0 be the binormal vector, which is constant because τ 0. Our curve lies in the plane Π. Let f(s) = α(s) b 0 f(0) = 0, so if the function has a zero derivative, then we are done. f (s) = α (s) b 0 + α(s) 0 = α (s) b 0, but, as α (s) = T (s), f (s) = 0. Theorem 1.8 (Frenet Relations). The Frenet Relations are 1. dt = k(s)n(s). db = τ(s)n(s) 3. dn = k(s)t (s) τ(s)b(s) Proof. The first two Frenet Relations are either previously defined or proved. As dn dn is perpendicular to n(s), it is = a 1(s)T (s) + a (s)b(s). n T = α 1 (T n) T n = a 1 T n = a 1 kn n = a 1 a 1 (s) = k(s) n b = α (b n) b n = a b n = a τn n = a a (s) = τ(s) Theorem 1.9 (Fenchel-Milner Theorem). Take a simple, closed, space curve c. c k(s) π and if k(s) = π then the curve is a circle. Also, if c is c knotted, then k(s) 4π c Theorem 1.10 (Fundamental Theorem for Curves). Given k(s) > 0 and τ(s), then!α(s) such that it has curvature k(s) and torsion τ(s), up to a rigid motion, that is, Ax + b where A is a matrix such that A t A = AA t = I, b is a vector, and x is an arbitrary point in space. Proof. We are given k(s) > 0 and τ(s), some function which is continuous. We will construct a curve α(s) such that α(s) passes through p and whose unit tangent vector points in the direction v 0. We have the Frenet Relations, and if we let W (s) = T (s) n(s), and with b(s) this we can rewrite the Frenet Relations, with an initial condition, as: dw = 0 k(s) 0 k(s) 0 τ(s) 0 τ(s) 0 W W (0) = v 0 v 0 v 0 v 0 This is a system of linear ordinary differential equations, and so a solution exists, and α(s) = s 0 T (u)du + p Now, we attempt to prove that this solution is unique.

3 We will translate the starting points of two such curves so that they coincide, rotate them so that their initial Frenet Trihedrons coincide, and use the fact that under rigid motions, k(s), τ(s) and arclength are invariant. Let M be a rigid motion such that Mx = M x 1 x x = A x 1 x x + Then, consider Mα(s) = β(s). Then, α, β have the same k(s), τ(s). β (s) = Aα (s), so β β = β = Aα Aα = A t Aα α = Iα α = α Consider (T, n, b) for α 1 and (T, n, b) for α, with T (0) = T (0), n(0) = n(0), b(0) = b(0) We claim that having the same T, n, b at s = 0 implies that they are the same for all s. Let f(s) = T T + n n + b b Note, f(0) = 0, so we must show that f(s) 0. So, if f (s) ( = 0, then f(s) = c, but as f(0) = 0, we must have c = 0. f (s) = T T, T T + n n, n n + b b, b b ) And, by the Frenet Relations, this is 0. Thus, α 1(s) = T (s) = T (s) = α (s), so, by integration, α 1 = α + c, but α 1 (0) = α (0), so α 1 = α So, now we have a global result. Theorem 1.11 (Isoperimetric Inequality). Let Γ be a simple, closed, plane curve. Let Γ enclose a region D with finite area A. Let length of Γ be denoted Γ = L. Then, 4πA L, and equality hol if and only if Γ is a circle. Lemma 1.1 (Geometric-Arithmetic Mean Inequality). ab a+b Lemma 1.13 (Cauchy-Schwartz Inequality). u, v u v Theorem 1.14 (Green s Theorem in the Plane). Given P (x, y) and Q(x, y) and a simple closed curve Γ enclosing a region D, then ( Q P dx + Qdy = x P ) dxdy y Γ Now, we are ready to prove the Isoperimetric Inequality. D Proof. Step 1 is to slide two parallel lines towar D until the fixed point of contact. Step is to set up coordinates and a circle of radius r. Step 3 is to Parametrize Γ by arclength s so Γ = (x(s), y(s)) = α(s) so α (s) = 1 The circle C is also parametrized as (x(s), y(s) Step 4 is to apply Green s Theorem A = Γ x dy = Γ x and πr = C y dx So, A + πr = C 0 ( x dy y dx ) L 0 x dy c 1 c c y dx 3

4 By the Cauchy-Schwartz Inequality, this is equal to L 0 (x + y ) 1/ ( dx + dy )1/ L = r = rl 0 By the Geometric-Arithmetic Mean Inequality,we have that Aπr A + πr rl (1), and so Aπr r L 4 4Aπ L. Assume that 4Aπ = L. Then, we need that Γ is a circle. So, we show that if 4πA = L, then r = f(l, A). From (1), we have that if L = 4Aπ, then Aπr = rl, then r = L± L 4πA And so, A + πr = L 0 π (xy yx ) = = L π L 0 ( x + y ) 1/ ( x + y ) 1/ Thus, (xy yx ) = (x + y )(x + y ) (xx + yy ) = 0 So, xx + yy = 0, then we divide by x y to get x y = y x = a. So, y = ax and x = ay. This means that y = a x and x = a y. So if we add the two equations together, we get that x + y = a (x + y ) = a Thus, x + y = ±a, meaning that a = ±r, so x y = y x = ±r So, x = ±ry, and my changing the directions of the parallel lines, we can just switch x and y, so y y 0 = ±rx. Therefore, (y y 0 ) + x = r Regular Surfaces We want to be able to do multivariable calculus on a curved surface. But first we need to discuss defining coordinates on them. Because there are no natural coordinates, we need to find the interesting geometric quantities that are coordinate invariant. One such property is Gauss Curvature. Definition.1 (Regular Surfaces). A set S in R 3 such that for every point in S we can find a ball, V, in R 3 and a map Φ from an open set U in R to V S such that 1. Φ : U V S is in C 1. Φ 1 : V S U is in C 0 3. dφ is one to one, that is, it has full rank (in the case of surfaces, rank dφ = ) It is clear that dφ = x u y u z u x v y v z v Regular Surfaces generalize to n Manifol, and are the special case where n = 4

5 Theorem.1. A surface is regular if (x u, y u, z u ) (x v, y v, z v ) 0, that is, Φ u Φ v 0 Theorem.. A graph is a regular surface, that is, z = f(x, y) y = g(x, z) and x = h(y, z) Proof. Assume the graph is of the form z = f(x, y). f C 1, so f f x and y exist and are continuous. U is the domain of f. So, let Φ = (u, v, f(u, v)). Then x(u, v) = u, y(u, v) = v and z(u, v) = f(u, v). As these are all in C 1, their partials exist, so Φ u and Φ v exist and are continuous. By simple calculation, we have Φ u Φ v 0, so z is regular. Corollary.3. The Sphere, S, is a regular surface. Consider F : W R 3 R and t = F (x, y, z) for (x, y, z) W. Definition. (Critical Point). Any point (x 0, y 0, z 0 ) with F (x 0, y 0, z 0 ) = 0 is called a critical point. Definition.3 (Critical Value). The value at a critical point is a critical value. Definition.4 (Regular Point). A noncritical point is a regular point. Definition.5 (Regular Value). The value at a regular point is a regular value. Theorem.4. Consider the level at S a = {(x, y, z) : F (x, y, z) = a}. If a is a regular value, then S a is a regular surface. Corollary.5. T, ie the two dimensional genus one torus, is a regular surface. Theorem.6 (Inverse ( Function ) Theorem). Let F : U R R : (x, y) fx f (f(x, y), g(x, y)). If y is invertible, then F has a local inverse. g x g y Theorem.7. Let S be a regular surface. Let Φ be a parametrization for p S. Φ : U V S, p V S, and with the first and third conditions of Φ verified, then if Φ 1 exists, it is continuous. Theorem.8. Given a regular surface S and any point p S, there is a neighborhood V, p V S such that the piece of the surface in V may be viewed as a graph of the type z = f(x, y) or y = g(x, z) or x = h(y, z) Proof. Our surface is regular, so we can assign local coordinates (u, v) U. Φ(u, v) = (x(u, v), y(u, v), z(u, v)) We know Φ u Φ v 0, so let us assume it is the z component that is nonzero. x x Then, u y u v y v 0 Consider Ψ : (u, v) (x(u, v), y(y, v)) with U R By the inverse function theorem, Ψ 1 exists. Definition.6 (Tangent Plane). T p S =tangent plane= {all tangent vectors of S at p} 5

6 Theorem.9. Let T be a tangent vector to S. Then, there exists an a in R such that [ x u (u x 0, v 0 ) v (u ] 0, v 0 ) y u (u y 0, v 0 ) v (u a = T 0, v 0 ) Proof. Consider (u 0, v 0 ) + t v = α(t), α (0) = v Consider Φ(α(t)) = β(t) Φ(α(0)) = Φ(u 0, v 0 ) = p β (0) = dφ(α (0)) = dφ( v) So, the set of tangent vectors T p S {dφ(p) ( ) a } b Let w T p S. w = β (0), where β(t) is a C 1 curve on S with β(0) = 0 Consider α(t) = Φ 1 (β(t)) d Φ(α(t)) = d β(t) ( a So, dφ = dφ(α b) (0)) = β (0) = w Remark - T p S is a dimensional vector space. Definition.7 (Coordinate Curves). Let p S and (u 0, v 0 ) with Φ(u 0, v 0 ) = p. Call this the center point. Consider α(t) = (u 0, v 0 + t) and β(t) = (u 0 + t, v 0 ) Then, Φ(α(t)), Φ(β(t)) are the coordinate curves through p. Remark.1. So, Φ(α(t)) = (x(u 0, v 0 + y), y(u 0, v 0 + y), z(u 0, v 0 + y)), then d Φ(α(t)) t=0 = (x v (u 0, v 0 ), y v (u 0, v 0 ), z v (u 0, v 0 )) = Φ v (u 0, v 0 ) And similarly with β and u. So, a basis for T p S is Φ v, Φ u at (u 0, v 0 ). Definition.8 (Diffeomorphism). Given regular surfaces S 1, S, assume there is a map F : S 1 S. If F is a bijection and, given p S, let u, v be local coordinates etcetera, and Ψ 1 F Φ(u, v) is differentiable and the differential is invertible, then F is a diffeomorphism. ( G1 G ) df = u u G 1 v G v So, we can say that if F is a diffeomorphism, we have df : T p (S 1 ) T F (p) (S ) Definition.9 (First Fundamental Form). Let S be a regular surface. Let E = Φ u Φu, F = Φ u Φ v, G = Φ v Φ v. Then, = E du + F du dv + G dv, and so we define = Edu + F dudv + Gdv as the first fundamental form. Definition.10 (Orthogonal Parametrization). A parametrization is orthogonal if Φ u, Φ v = 0 Definition.11 (Isothermal Coordinates). Coordinates are called isothermal if = E(u, v) ( du + dv ) Theorem.10. On a regular surface, isothermal coordinates always exist. Theorem.11. da = EG F dudv and does not depend on the coordinates. 6

7 Remark.. For a graph z f(x, y), f C 1, = (1 + f x)dx + f x f y dxdy + (1 + f y )dy =.1 Rhomb Lines, or, Loxodromes [ 1 + f ] 1/ dxdy (Theory of discontinuous groups, loxodromic/automorphic forms.) Definition.1 (Hyperbolic Metric). The metric on the Poincare half plane is = dx +dy y, and is called the hyperbolic metric. Theorem.1 (Hilbert, 1960s Efimov). If K δ < 0 an the -Manifold is complete, then one cannot embed it in R 3 Definition.13 (Rhomb Line/Loxodrome). A curve in S is called a rhomb line, or loxodrome, if and only if it intersects every meridian at a constant angle β. 3 The Gauss Map Definition 3.1 (Unit Normal of a surface). Let S be a regular surface. We define the unit normal as N(u, v) = Φu Φv Φ u Φ v Definition 3. (Gauss Map). The Gauss Map is N : S S : p N(p) and dn : T p S T N(p) S T p S Definition 3.3 (Endomorphism). A homomorphism from a group into itself is an endomorphism. Theorem 3.1. The differential of the Gauss Map is an endomorphism. Definition 3.4 (En). On the punctured sphere, the punctures are called en. Theorem 3.. The map dn : T p S T p S is a self-adjoint linear map. Proof. It is clearly linear, so all we need is dn(v), w) = v, dn(w) We have the equations and dn(φ u ), Φ u = Φ u, dn(φ u ) dn(φ v ), Φ v = Φ v, dn(φ v ) Thus, we only need to show that dn(φ u ), Φ v = Φ u, dn(φ v ), that is N u, Φ v = Φ u, N v We not that f(u, v) = N, Φ u = 0, so N v, Φ u + N, Φ uv = 0 and N u, Φ v + N, Φ uv = 0. Subtracting one from the other gives the result. [ ] a11 a Definition 3.5 (Quadratic Form). If A, is a matrix 1, then we say a 1 a Av, v = Q(v, v) is a quadratic form. Definition 3.6 (The Second Fundamental Form). II p (w) = dn(w), w is called the second fundamental form. If w = xφ u + yφ v then II p (w) = ex + fxy + gy where e = N u, Φ u, f = N u, Φ v, g = N v, Φ v. 7

8 Definition 3.7 (Normal Curvature). If w = 1, p S and w T p S, then we call II p (w) the normal curvature. So, II p (w) = k cos θ where k is the curvature of a curve on the surface and θ hte angle between N(u, v) and the normal to the curve at p. Theorem 3.3 (Meusnier). Curves passing through p and having the same tangent vector have the same normal curvature. Proof. Let α(s) be a curve α : I S in arclength parametrization, with α(0) = p, α (0) = w. Note, that N(s), α (s) = 0. Taking the derivative, we get dn, α (s) + N(s), α (s) = 0, so k cos θ = dn(s), α (s) = dn(w), w) = II p (w) [ ] [ ] cos θ x Lemma 3.4. Let w = =. Consider Q(cos θ, sin θ). sin θ y Then f(θ) = a cos θ + b cos θ sin θ + c sin θ, if max x [0,π] f(θ) = f(0) then b = 0. Proof. f(θ) is periodic, so we look at it on [ π, π]. Then f (0) = 0, so f (0) = b cos θ θ=0 = b = 0 If this is the case, then Q(x, y) = ax + cy, so f(θ) = a cos θ + c sin θ and f(0) = a. Note f(π/) = c a. Notice now that f(θ) = a cos θ + c sin θ c cos θ + c sin θ = c, so c is the minimum value, and is at a right angle to a. Application Consider w T p S. w = cos θφ u + sin θφ v. dn(w), w = e cos θ + f cos θ sin θ + g sin θ = f(θ), where e = dn(φ u ), Φ u = N u, Φ, f = N u u, Φ and g = N v v, Φ v. Assume ( that max θ [0,π] ) ( f(θ) ) = f(θ ( 0 ). Rotate ) the axes so that θ 0 = 0. cos θ0 sin θ So 0 u u = sin θ 0 cos θ 0 v v, so w = cos θ 0 Φ u +sin θ 0 Φ v. That is, w = Φ u if θ 0 = 0. dn(w), w = e, so we let w be orthogonal to w in T p S. Now we claim that w and w are principal directions. That is, dn(w) = k 1 w and dn( w) = k w. k 1, k are called the principal curvatures. Proof. Note that {w, w} is a basis for T p S. So we consider dn(w), w. Recall that w = Φ u and w = Φ v. We claim that dn(w), w = 0. b = 0 = f so f = dn(φ u ), Φ v = 0. dn(φ u ) = c 1 Φ u + c Φ v, so 0 = dn(φ u ), Φ v = c 1 Φ u, Φ v + c Φ v, Φ v. We will now check that w, w are the principal directions: Step 1: Let us call Φ u = e 1, Φ v = e. Choose the direction in the tangent plane such that the quadratic form f(θ) = e cos θ + f cos θ sin θ + g sin θ has a maximum in the direction θ = 0. Then w(θ) = e 1 cos θ + e sin θ gives dn(w), w = e cos θ + f sin θ cos θ + g sin θ. Step : At p consider w, the orthogonal direction. What is f for the pair {w, w}? f = 0 dn(φ u ), Φ v = dn(e 1 ), e = dn(w), w = 0. Step 3: Note that dn(w) = c 1 w + c w, thus 0 = dn(w), w = c 1 w, w + c w, w = c. 8

9 Definition 3.8 (Line of Curvature). A line of curvature C is a curve on S such that for all points on C, the tangent direction is principal. Theorem 3.5 (O. Rodriguez). A curve α(t) is a line of curvature iff dn = k(t)α (t), where N(t) = N(α(t)). Definition 3.9 (Gauss Curvature). The Guass curvature at p S is K = k 1 k. Definition 3.10 (Mean Curvature). The mean curvature if H = k1+k. Definition 3.11 (Classification of Points). A point p S is said to be elliptic if K(p) > 0, hyperbolic if K(p) < 0, parabolic if K(p) = 0 but either k 1 or k is not 0, and planar if k 1 = k = Euler Formula Let e 1, e be principal directions, and pick v T p S such that v = 1. Then v = cos θe 1 + sin θe. That is, {e 1, e } is an orthonormal basis of T p S. Recall: dn(v), v) =normal curvature k n = K cos φ. So we take dn(e 1 cos θ + e sin θ), e 1 cos θ + e sin θ = dn(e 1 ) cos θ + dn(e ) sin θ, e 1 cos θ + e sin θ =. k 1 cos θe 1 + k sin θe, e 1 cos θ + e sin θ = k 1 cos θ + k sin θ = k n Definition 3.1 (Umbilic Points). A point p S is said to be umbilic iff k 1 (p) = k (p) At an umbilic point, every direction is principal. Theorem 3.6. Let S be a connected surface with every point umbilic. Then S is either a piece of a sphere or a flat plane. Proof. N u = dn(φ u) = k(u, v)φ u, N v = dn(φ v) = k(u, v)φ v. Now we look at dn(φu) v dn(φv) u. So 0 = k v Φ u k u Φ v, so k v Φ u = k u Φ v. Thus, k u = k v = 0, so k = c. By connectedness, we have k = c everywhere. Now, if k = 0, then N v = N If k = c 0, then f(u, v) = Φ(u, v) + 1 c u = 0, so N(u, v) = N 0, so S is flat. f N(u, v). Thus u = Φ u + 1 N c u = 0 = 0. So f(u, v) = v 0, adn v 0 = Φ(u, v) + 1 c N v 0 and f v = Φ v + 1 N c v Φ(u, v) = 1 c N, so v 0 Φ(u, v) = 1 c, so S is a part of a sphere. Theorem 3.7 (Liebman). If S is a surface for which K(p) = c > 0, then S is a part of a sphere. Definition 3.13 (Asymptotic Direction). A direction v T p S is an Asymptotic Direction iff dn(v), v = 0. Definition 3.14 (Asymptotic Curve). A curve γ(t) S is called an asymptotic curve iff γ (t) is an asymptotic direction for all t I. Remark: If a point is elliptic, then there are no asymptotic directions. 9

10 Definition 3.15 (Minimal Surface). A Minimal Surface is a surface satisfying k 1+k = 0 for all p S. Theorem 3.8 (Osserman). If S is minimal and complete, and S \ N(S) is open, then S is the plane. Definition 3.16 (Conjugate directions). Let w 1, w T p S. We say w 1, w are conjugate directions iff dn(w 1 ), w = dn(w ), w 1 = 0. Theorem 3.9. If w is an asymptotic direction, then w is conjugate to itself. Theorem If e 1, e are principal, then they are conjugate. 3. Dupin Indicatrix Given p S, D = {w T p S : II p (w) = ±1}. Theorem D is the union of conics. Proof. Let e 1, e be principal directions, then w = r cos θe 1 + r sin θe. II p (w) = dn(w), w = r cos θk 1 e + r sin θk e, r cos θe 1 + r sin θe = r cos θk 1 + r sin θk = ±1 Case 1: k 1 > k > 0. Then r cos θk 1 + r sin θk = 1. Let ξ = r cos θ and η = r sin θ. Then ξ k 1 + η k = 1, which is the equation of an ellipse. Case : k 1 > 0 > k. Then k 1 ξ k 1 η = ±1, so we get k 1 ξ k 1 η = 1 and k 1 ξ k η = 1, a pair of hyperbolas. 3.3 Gauss Map in Local Coordinates Note: dn(φ u ) = N u = a 11Φ u +a 1 Φ v T p S, dn(φ v ) = N v = a 1Φ u +a Φ v T p S. ( ) a11 a This gives a matrix 1 with eigenvalues k a 1 a 1, k and eigenvectors the principal directsions. Note det A = k 1 k the Gauss curvature, and tr A/ = k1+k, the Mean Curvature. N u, Φ u = e = a 11 E + a 1 F N u, Φ v = f = a 11 F + a 1 G N v, Φ u = f = a 1 E + a F N v, Φ v = g = a 1 F + a G ) ( e f The above equations tell us that f g so we get the following theorem: ( a11 a = 1 a 1 a F G ) ( E F ), Theorem 3.1. ( e f f g ) ( E F F G ) 1 = ( ) a11 a 1 a 1 a 10

11 ( ) a11 a The Gauss Curvature is then K(p) = det 1. Thus K(p) = a 1 a eg f EG F. Now, recall that dn(v) = k 1 v = k 1 Iv, and dn(w) ( = k w. ) So (dn + a11 a 1, dn+ki = a 1 a ki)v = 0, and so, in the basis {Φ u, Φ v }, we have dn = ( a11 + k a 1 ), and so det(dn + ki) = (a 11 + k)(a + k) a 1 a 1 = a 1 a + k k kh + K where H is the mean curvature and K is the Gaussian curvature. Thus, k = H ± H K, and so, k 1 k k1 + k. 3.4 Equations of Weingarten ( ) 1 ( ) E F 1 G F = F G EG F F E And so, we have a 11 = a 1 = a 1 = a = ff eg EG F ef fe EG F gf fg EG F ff ge EG F Thus, H = eg ff +ge (EG F ). Theorem If (u, v) is a parametrization, f = F = 0 and p is not an umbilic point, then the coordinate curves are lines of curvature. Proof. Recall the above matrix equation. Φ(u(t), v(t)) = γ(t), γ = u Φ u + v Φ v. curvature. dn(γ ) = λ(t)γ along a line of dn(γ ) = dn(u Φ u + v Φ v ) = u dn(φ u ) + v dn(φ v ) = λu Φ u + λv Φ v = u (a 11 Φ u + a 1 Φ v ) + v (a 1 Φ u + a Φ v ), so we must have a 11 u + a 1 v = λu and a 1 u + a v = λv. ff eg That is, EG F u gf fg + EG F v = λ(t)u ef fe and EG F u ff ge + EG F v = λ(t)v are the equations of lines of curvature in general. Assume that f = F = 0, v(t) = c, then e E = λ(t). Now we assume that the coordinate curves are lines of curvature. As we are at a nonumbilic point, we have Φ u, Φ v = F = 0, and so the lines of curvature are e/e = λ 1 (t) and f/g = 0. so f = 0. Theorem A necessary and sufficient condition for coordinate curves to be asymptotic curves in a neighborhood of a hyperbolic point if e = g = 0. 11

12 Proof. II p (w) = dn(γ ), γ, γ (t) = w = u Φ u + v Φ v, so II p (w) = e(u ) + fu v + g(v ), and we need this to be 0. Assume u(t) = t and v(t) = c. Then e = g = 0. Assume e = g = 0, then we have fu v = 0. Since eg f < 0, we have f 0, so u v = 0, giving us the conclusion. Theorem 3.15 (S. Bernstein 1916). If S is a minimal surface such that it is a graph and is defined x, y R, then S is a plane. This theorem was extended in 1966 by F. Almgren and E. DiGiorgi to three dimensions, J. Simon in 1969 to n 7 and E. Bombieri, E. DiGiorgi and Ginsti proved, in 1970, that it is false for n 8. Theorem If K > 0 at p S, then the surface sits to one side of the tangent plane at p. If K < 0 at p S, then the surface sits on both sides. Proof. Φ(0, 0) = p. (Φ u (u, v) Φ(0, 0)) N at (0, 0) is ( uφu + vφ v + u Φ uu + uvφ uv + v Φ vv +... ) N = u Φ uu, N + uv Φ uv, N + v Φ vv, N = eu + fuv + gv +error It is positive at p, so all is on one side. The other part goes similarly. Theorem Let M be a compact surface, M =. Then at least one elliptic point. Remark: If M is compact and not homeomorphic to S, then M will in fact have both elliptic and hyperbolic points. Proof. Pick x 0 / M and consider a sphere S R (x 0 ). Claim: the first points touched as the sphere is contracted are elliptic. Let f(q) = q x 0 for q M. f(q) achieves maxima at x 1,.... Let α(s) M be any curve such that α(0) = x 1. Then consider g(s) = α(s) x 0 = α(s) x 0, α(s) x 0. Note g (0) = 0 and g (0) 0. g (0) = α (0), α(0) x 0 = α (0), x 1 x 0 = 0, so x 1 x 0 is normal to M at x 1. g (s) = α (s), α(s) x 0, g (s) = α (s), α(s) x 0 + α (s), α (s). g (0) = α (0), x 1 x 0 + 0, so α (0), N(0) 1. We recall that II p (α (s)) = N s, α (s), where N s = s (N(α(s))). But N s, α = N, α, so II p (x 1 ) 1, so the principal curvatures are less than zero, so elliptic. 4 Intrinsic Geometry of Surfaces Let S 1, S be surfaces, ϕ : S 1 S a diffeomorphism. Take p S 1 and a coordinate neighborhood of p. Look agt Φ 1 ϕ Φ 1 : V R W R : ( (u, v) (f(u, ) v), g(u, v)) fu f where f u, f v, g u, g v exists and are continuous, with v invertible for g u g v all (u, v) V. 1

13 Then ϕ is a diffeomorphism and if f, g C then it is a smooth diffeomorphism. Definition 4.1 (Isometry). An isometry is a map ϕ : S 1 S such that ϕ is a diffeomorphism which preserves the inner product, that is, v, w TpS 1 = dϕ(v), dϕ(w) Tϕ(p) S. Remark: if dϕ(w), dϕ(w) = w w T p S, then dϕ(w 1 ), dϕ(w ) = w 1, w. This is becuase dϕ(w) = w gives dϕ(w 1 + w ), dϕ(w 1 + w ) = w 1 + w, w 1 + w. Thus, dϕ(w 1 ), dϕ(w 1 ) + dϕ(w ), dϕ(w ) + dϕ(w 1 ), dϕ(w ) = w 1 + w + w 1, w. Definition 4. (Local Isometry). Given two regular ( surfaces ) and a map ϕ : fu f S 1 S such that ϕ is a smooth map, Jac(ϕ) = v is invertible for g u g v all (u, v) V, and p S 1, V 1 S a neighborhood of p and a corresponding neighborhood V of ϕ(p) in S such that ϕ : V 1 V is an isometry. Theorem 4.1. Let Φ : U S 1, Φ : U S. Then E = Ẽ, F = F, and G = G iff Φ 1 Φ : U U is a local isometry. Proof. Let v T p0 S 1 and w = aφ u + bφ v. Let α(t) S 1, with d t=0φ(α(t)) = w. d t=0 Φ(α(t)) = a Φ u + b Φ v = dψ(w) dψ(w) = dψ(w), dψ(w) = a Φ u + b Φ v = a ( Φ u Φu ) + ab( Φ u Φ v ) + b ( Φ v Φ v ) = a Ẽ + ab F + b G. As Ẽ = E, F = F and G = G, we have a E + abf + b G = w. The converse is trivial. Definition 4.3 (Conformal Mapping). Let ϕ : S 1 S be a diffeomorphism. If dϕ(w 1 ), dϕ(w ) Tϕ(p) S = λ (p) w 1, w TpS 1, then ϕ is conformal. Theorem 4.. Two regular surfaces are locally conformal. Theorem 4.3 (Chow). Assume H CP n a complex analytic manifold. Then H is an algebraic variety. 4.1 Christoffel Symbols Consider a regular surface S, Φ a local chart and Φ uu = aφ u + bφ v + cn = (Φ uu Φ uu, N N) + Φ uu, N N. So c = e from the second fundamental form, and a and b are called Christoffel Symbols. Definition 4.4 (Christoffel Symbols). We write Φ uu = Γ 1 11Φ u + Γ 11Φ v + en, Φ uv = Γ 1 1Φ u + Γ 1Φ v + fn, and, Φ vv = Γ 1 Φ u + Γ Φ v + gn and we call Γ k ij the Christoffel symbols. As Φ uv = Φ vu, we have Γ k ij = Γk ji, and so there are six Christoffel symbols. 13

14 Lemma 4.4. There are formulas for the Christoffel symbols in terms of E, F and G and their derivatives. Proof. By taking inner products, we get equations of the form Φ uu, Φ u = 1 E u = Γ 1 11E + Γ 11F, etcetera. By Cramer s Rule, we can solve for the Γ k ij. Now, Gauss s equation implies the Theorema Egregium, and we will also demonstrate Codazzi s Equation. Theorem 4.5 (Gauss s Equation). Assume that E 0. Then we have EK = (Γ 11) u (Γ 11) v + Γ 1 1Γ 11 + (Γ 1) Γ 11Γ Γ 11Γ 1 Proof. Observe that (Φ uu ) v = (Φ uv ) u, (Φ vv ) u = (Φ uv ) v and N uv = N vu. So we have (Φ uu ) v (Φ uv ) u = 0, we can write this as A 1 Φ u +B 1 Φ v +C 1 N = 0, and since Φ u, Φ v, N are linearly independent, A 1 = B 1 = C 1 = 0. So we start with Φ uu = Γ 1 11Φ u + Γ 11Φ v + en and Φ uv = Γ 1 1Φ u + Γ 1Φ v + fn, and so (Φ uu ) v = (Γ 1 11) u Φ u + (Γ 11) v Φ v + e v N + Γ 1 11Φ uv + Γ 11Φ vv + en v, and (Φ uv ) u = (Γ 1 1) u Φ u + (Γ 1) v Φ v + f u N + Γ 1 1Φ uu + Γ 1Φ uv + fn v. We focus on B 1, the coefficient of Φ v, and it is ( (Γ 11 ) v (Γ 1) u + Γ 1 11Γ 1 + Γ 11Γ Γ 1 1Γ 11 Γ 1Γ ) 1 Φv = 0 Corollary 4.6 (Theorema Egregium). The Gauss Curvature of a surface is invariant under a local isometry. Proposition 4.7 (Codazzi-Mainardi Equations). e v f u = eγ f(γ 1 Γ 1 11) gγ 11 and f v g u = eγ 1 + f(γ Γ 1 1) gγ 1 Theorem 4.8 (Bonnet s Theorem). Given E, F, G, e, f, g and EG F > 0 which satisfy the Gauss and Codazzi-Mainardi Equations, then there exists a map Φ : (u, v) U R R 3 such that Φ defines a regular surface with first fundamental form Edu + F dudv + Gdv and second fundamental form edu + fdudv + gdv. Additionally, this surface is unique up to rigid motion. Definition 4.5 (Orthogonal Parametrization). A parametrization is orthogonal iff F = Parallel Transport Given a tangent vector field on U S for each point, p U, there exists w(p) T p (S). Let α : [ ɛ, ɛ] U be a regular curve such that α(0) = p and α (0) = v T p S. Consider w(α(t)), and also dw/. The covariant derivative Dw (or v w, or D v w) is the orthogonal projection of dw back to the tangent plane. Note that this defines a map D v : T p S T p S Let Φ(u(t), v(t)) = α(t). Then w(u, v) = a(u, v)φ u + b(u, v)φ v, and dw = (a u u + a v v )Φ u + a(φ uu u + Φ uv v ) + (b u u + b v v )Φ v + b(φ uv u + Φ vv v ). This, after some manipulation, is equal to (a u u + a v v + aγ 1 11u + aγ 1 1v + bu Γ bv Γ 1 )Φ u + 14

15 (b u u + b v v + aγ 11u + aγ 1v + bu Γ 1 + bv Γ )Φ v + cn) The Levi-Civita Connection is our covariant derivative, and we set D Y X D X Y = [X, Y ]. So D v w doesn t depend on u or v, just the tangent directions. From this formula, all that matters is the value at zero of u, v, u, v. The curve α(t) is irrelevant, and things only depend on the first fundamental form D x i x j = Γ k ij Φ u k. D u Φ u+v Φ v w = D u Φ u w + D v Φ v w = u D Φu w + v D Φv w Definition 4.6 (Parallel). the tangent vector field w(t) is said to be parallel along α(t) iff D α (t)w = 0 Theorem 4.9. Let X(t), Y (t) be two parallel vector fiel along α(t). X(t), Y (t), X(t), Y (t) are all constants. Then d dx dy dx Proof. X(t), Y (t) =, Y + X,. As Y is tangent, we have, Y = D α X, Y and so D α X, Y + X, D α Y = d X, Y. By hypothesis, this is zero, and so X, Y is constant. Definition 4.7 (Geodesic). A curve α(t) in S is a geodesic iff D α (t)α (t) = 0. Remark 4.1. As D α (t)α (t) = 0, we have α (t) = c, and so t/c = s is the arclength, so α(t/c) = β(t) is the arc length parameterization. β (t) = α (t)/c β = 1 c α = c/c = 1 If X(0) = w and D α X = 0, then this differential equation is solved by the solution to a + au Γ 11 + av Γ bu Γ bv Γ 1 = 0 b + au Γ 11 + av Γ 1 + bu Γ 1 + bv Γ = 0 Set a = u, b = v. Then X = a(t)φ u + b(t)φ v, and we have a system of equations called the geodesic equations: u + (u ) Γ u v Γ (v ) Γ 1 = 0 v + (u ) Γ 11 + u v Γ 1 + (v ) Γ = 0 For a surface of revolution, this simplifies to u + u v f /f = 0 and v ff (u ) = 0. Theorem If Ψ is a lcoal isometry between S 1 and S, then geodesics are mapped to geodesics. Proof. Consider the geodesic equations for (u(t), v(t)). A local isometry will preserve the first fundamental form. Thus, a local isometry will preserve the geodesic equations. That is, the curve has the same image in R and the Christoffel symbols are the same. 15

16 Theorem 4.11 (Picard). Given a point p 0 S and v T p0 S, there exists a unique geodesic α(t), α(0) = p 0 and α (0) = v for t ( ɛ, ɛ). Proof. This follows from the existence and uniqueness theorem for nd order nonlinear ODEs. Theorem 4.1. If S 1, S are tangent along Γ, then covariant derivative along Γ is the same for S 1 and S. Theorem Let F : S S and assume that F is an isometry. Assume that F ix(f ) = {p F (p) = p} is a regular curve. Then F ix(f ) is a geodesic. Proof. Let p 0 F ix(f ) and γ (p 0 ) = v. By previous theorem there is a geodesic α(s) such that α(0) = p 0, α (0) = v. σ = F α is also a geodesic as F is an isometry. Moreover, σ(0) = p 0 and σ (0) = df (α (0)). So F (γ) = γ and so df (α (0)) = α (0) = v. Thus, α = σ, and so α F ix(f ), so α = γ. We are working towar the local Gauss-Bonnet Theorem, and Clairant s Theorem/Relation. 4.3 Algebraic Value of the Covariant Derivative Assume that we fix an orientation for U S, we have determined N. Let w(t) T S a tangent vector field with w = 1. We define [ ] Dw = [Dv w] = λ(t), where Dw = λ(t)n w. Note that N w is always a unit tangent vector. Definition 4.8 (Geodesic Curvature). Let [ α(s) ] be a regular curve parameterized by arclength. Geodesic curvature is k g =. Remark 4.. k g = 0 iff α(s) is a geodesic. [ Dw So ] given [ a regular curve α(t), and v(t) = w(t) = 1, v, w T S then Dv ] = dφ with φ the angle between v and w. k = kn + kg. Lemma Let α : I S be a regular curve. Let v(t), w(t) be a fmaily of unit tangent vectors along α. Then [ ] [ Dw Dv ] = dφ where φ(t) is one determination of the angle from v to w. Proof. Construct v = N(t) v(t). Now w(t) = cos φ(t)v(t) + sin φ(t) v(t). Also w (t) = sin φφ v + cos φv + cos φφ v + sin φ v. Note that w, w = 0 as w = 1. Form w = N w and [ Dw So λ(t) = dw cos φ v sin φv. So we have, N w = dw Dα ] = λ(t) where Dw = λ(t)n w., w and N w = w = cos φn v +sin φn v = sin φφ v + cos φv + cos φφ v + sin φ v, cos φ v sin φv = cos φ v, v + cos φφ + sin φφ sin φ v, v = v, V + φ And so [ ] Dw = v, v + φ = [ ] Dv + φ. 16

17 Lemma Assume (u, v) are orthogonal parameters for S. Let w(t) be a unit tangent vector field along Φ(u(t), v(t)) = α(t). Then [ ] [ ] Dw 1 = dv G u EG E du v + dφ where φ(t) is the angle between Φ u and w(t). Proof. Choose v(t) = Φ u / E. We compute dv dv, then find, N v = [ ] Dv. dv dv, N v = Φ uu du + Φ uv dv d 1 + Φ u E E = 1 E Φ uuu + Φ uv v, N Φ u = 1 E Φ uu, N Φ u u + 1 E Φ uv, N Φ v v = = G E Γ 11 Φv G, N Φ u E u + G E [ Γ 11 u + Γ 1v ] G Φv E Γ 1, N Φ u v G E And so the result hol. Definition 4.9 (Triangle). A triangle T S is a region such that 1. T is homeomorphic to a disc. T must be a piecewise regular simple closed curve having three vertices Definition 4.10 (Triangulable). A surface is said to be triangulable iff there exists a family F of triangeles {T i } such that 1. S = T i. If T i T j =, then they only share a common vertex or edge. Theorem 4.16 (Radó). A regular surface is traingulable. One can define F E + V for a triangulation to be the Euler characteristic. This is a topological invariant. Fact: In the traingulation, you can always arrange it so that each triangle is in a local chart. We define the genus g to be the number of holes or handles attached to a sphere to get the surface, and χ(s) = g. Theorem For a surface of revolution 1. Meridians are geodesics. Parallels of lattitude with f (u ) = 0 are geodesics. 3. Other geodesics intersect parallels with angel θ, and c = f(u )θ is constant. 17

18 Proof. The geodesic equations are u 1 + f f u 1u = 0 and u ff (u 1) = On a meridian, u 1 = a is constant, and u = t. So then we have a + f f a t = 0, and 0 ff 0 = 0, which are both zero, and so the meridians are geodesics.. On a parallel, u 1 = t, u = a, and so we need to have ff = 0, and f 0, so f = Let (u(t), v(t)) be a geodesic with arclength parameterization. That is, α(t) = Φ(u(t), v(t)) and α (t) = 1 = Φ u u +Φ v v. On a circle of latitude, β(t) = (f(a) cos t, f(a) sin t, g(a)) and β (t) = ( f(a) sin t, f(a) cos t, 0). So β (a) = f(a), and so cos θ = α β (t) (t) β (t). Thus, cos θ = (Φ uu + Φ v v ) (sin t, cos t, 0). So cos θ = (Φ u1 u 1 + Φ u u ) Φu 1 f(a) = Φu 1 Φu 1 f(a) u 1 = fu 1, and so cos θ = fu 1, which menas we have f cos θ = f u 1. By the geodesic equations, we have that u 1 + f (u ) f(u ) u 1u = 0, so f u 1 + f fu 1u = 0 = (f u 1). Lemma If a geodesic has been parameterized by arclength, then du 1 = cdu f(u )(f (u ) c ) 1/ Definition 4.11 (Simple Region). A simple region R S is a region R which is homeomorphic to a disc in R and R is piecewise regular and simple. Theorem 4.19 (Local Gauss-Bonnet). Consider a simple region R S. Assume a positive orientation for R = Γ 1 + Γ Γ n. Then n k g (s) + Γ i i=1 R Kdσ + n θ i = π i=1 where θ k is the exterior angle between Γ k and Γ k+1. Theorem 4.0 (Global Gauss-Bonnet). Given any region R with boundary R (which may be empty), then n k g (s) + Γ i i=1 R Kdσ + n θ i = πχ(r) i=1 Corollary 4.1. If S is a compact surface, then Kdσ = πχ(s) S We will prove the local result first, and we will need the following: Theorem 4. (Turning Tangents). (φ(ti+1 ) φ(t i )) + θ i = π Now we prove Local G-B 18

19 Proof. First we recall that we can choose [( an ) orthogonal ( ) coordinate ] system = Edu + Gdv G. So now K = EG u E + EG v 1 EG Now we assume that R is in our orthogonal coordinate chart. So then Kdσ = K(u, v) EGdudv = 1 ( ( ) Gu + ( )) Ev dudv R R R u EG v EG ( ) Then, by Green s Theorem, we have that 1 EG G u dv R + Ev du EG Recall that [ ] ( Dw = 1 dv Gu EG + E ) v du + dφ where φ is the angle from Φ u to w(s) along α(s). Apply this to w = α, then Kdσ = 1 R u ( Gu EG dv v G v du EG ) R = 1 k g (s) dφ Γi Γ i n φ n i=1 Γ i s = kdσ + k g (s) R i=1 Γ i And so φ(t i+1 ) φ(t i ) = R Kdσ + n i=1 Γ i k g (s), and so φ(t i+1 ) φ(t i ) + θ i = π = R Kσ + n i=1 k g(s) + θ i. Theorem 4.3. Let R S, and let R be bounded by C 1,..., C n piecewise regular curves. Let θ i be the external angles at the vertices v i on C i. Then n k g (s) + Kdσ + θ i = πχ(r) C i R i=1 where χ(r) = F E + V for any triangulation and C i has positive orientation. Proof. We traingulate R into T i with T i = R. For each T i we have 3 3 k g (s) + Rdσ + θ ik = π (1) k=1 C ik T i i k =1 If we sum (1) over all the triangles, we get n i=1 C i k g (s) + R Kdσ + F 3 i=1 k=1 θ ik = πf. We now must analyze F 3 i=1 k=1 θ ik. We rewrite as interior angles, θ = π φ ik with φ ik the interior angle, and so F 3 i=1 k=1 (π φ ik) = F i=1 (π 3 k=1 φ ik). This is equal to 3πF F 3 i=1 k=1 φ ik. Call E e the number of exterior edges and E I the number of interior edges, and define V e and V I similarly. Fact: 3F = E I + E e. And so 3πF = πe I + πe e, which means we have πe I + πe e F 3 i=1 k=1 φ ik = πe πe e φ ik, and as E e = V e, we have πe πv e φ ik = πe πv e πv ext φ ik. The exterior triangulation splits into two pieces. So we note that ext φ ik = vert of T i + vert of C i = πv et + π θ i = πv et +πv ec θ i = πv e θ i, and so the formula gives πe πv + θ i, and so we establish the theorem. 19

20 Corollary 4.4. If S is compact and S =, then Kdσ = πχ(s) S Corollary 4.5. A compact surface with positive curvature is homeomorphic to S. Proof. Kdσ > 0 χ(s) = S Corollary 4.6. Let K S 0, S compact and S =, then S Kdσ. 0