HOMEWORK 4 - GEOMETRY OF CURVES ND SURFCES NDRÉ NEVES DISCLIMER: This homework was very tough and it involves being comfortable with facts you might have seen in other classes but are not so fresh. The good news are: I will have that into consideration when I grade; The exam will not be like this in any way. I just did this to make sure I can distinguish the students. (1) Let S be a compact surface in R 3 with no boundary. (a) Show there is a point p S where the Gaussian curvature is positive. Hint: Consider p 0 R 3 but not is S. Because S is compact then S lies in the interior of B r (p 0 ) (the ball of radius r centered at p 0 ) for all r very large and S does not lie in the interior of B r (p 0 ) for all r very small. Take R = inf{r S B r (p 0 )}. You should argue that B R (p 0 ) S contains a point q; T q S = { x x.(q p 0 ) = 0} The Gaussian curvature at of S at p is greater or equal than R 2. We will follow the hint. Claim 1: B R (p 0 ) S contains a point q. First S B R (p 0 ) because for all x S we have x B R+1/n (p 0 ) for all n N (by definition of R) and so x n N B R+1/n (p 0 ) = B R (p 0 ). Second, if B R (p 0 ) S = then, because S and B R (p 0 ) are closed subsets, there would exist ε so that B R ε (p 0 ) S = which means that S B R ε (p 0 ). This contradicts the definition of R. Claim 2: T q S = { x x.(q p 0 ) = 0}. 1
2 NDRÉ NEVES The function f(x) = x p 0 2 defined on R 3 has the following properties f(x) R 2 for all x S (because S B R (p 0 )); f(q) = R 2 (because q B R (p 0 ) S). It suffices to see that T q S { x x.(q p 0 ) = 0}. If X T q S, then X = (0) for some path in S with (0) = q. But the function f(t) = f (t) has a critical point at the origin because of the two properties I just mentioned. From the chain rule we know f (0) = f(q). (0) and thus f (0) = 0 = f(q). (0) = 0 = (q p 0 ). (0) = 0 = X.(q p 0 ) = 0. This is exactly what we wanted. Claim 3: The Gaussian curvature at of S at p is greater or equal than R 2. Before we proceed lets make sure the idea is clear. If we do a picture of S then S is tangential to B R (p 0 ) at q and lies inside of B R (p 0 ). Thus S has to me more curved than B R (p 0 ) at q. The Gaussian curvature of the latter is R 2 and so S should have Gaussian curvature bigger than R 2. From the technical point of view, the key thing to use is that if u, v are two functions defined on the plane with u(0, 0) = v(0, 0) and u v, then the eigenvalues of Hess (u v) should be nonnegative which should mean that graph(u) is less curved than graph(v). Without loss of generality lets assume that q is the origin and T q S is the xy-plane (we can always achieve that by choosing an appropriate coordinate system). In sum, we have a function u defined on a neighborhood of the origin containing the xy-plane so that a) φ(x, y) = (x, y, u(x, y)) is a chart of S near the origin; b) u(0, 0) = 0, (0, 0) = (0, 0) (this is just saying that q is the origin and T q S the xy-plane); In view of Claim 2 we have that p 0 lies in the z-axis and, again without loss of generality, we can assume that p 0 = (0, 0, R) (the only other possibility is if p 0 = (0, 0, R)). Moreover, in Claim 2 we defined a function f and, if we abuse notation and set f to be f φ, we have c) f(x, y) = x 2 + y 2 + (u(x, y) R) 2 ; d) f(x, y) f(0, 0) = R 2 (this follows from the two conditions for the function f stated in Claim 2). From the lectures we know that the principal eigenvalues at the origin are just the eigenvalues of the Hessian of u at the origin. Finally, and this will be the last simplification, we can assume that the basis chosen for the xy-plane diagonalizes Hess u at the origin,
i.e., HOMEWORK 4 - GEOMETRY OF CURVES ND SURFCES 3 u xx (0, 0) = λ 1, u xy (0, 0) = u yx (0, 0) = 0 u yy (0, 0) = λ 2. Thus from a) and c) we get f xx (0, 0) = 2 2u xx (0, 0)R = 2 2Rλ 1, f xy (0, 0) = 2u xy (0, 0) = 0, f yy (0, 0) = 2 2u yy (0, 0)R = 2 2Rλ 2. From d) we also have that f has a local maximum at the origin and thus Hess f(0, 0) must be nonpositive definite by the Hessian test. But from the computations I just did this is the same as saying that f xx (0, 0) 0 = λ 1 R 1 and f yy (0, 0) 0 = λ 2 R 1. Therefore K(q) = λ 1 λ 2 R 2. (a) Show that if S is not diffeomorphic to the unit sphere then S has points where the Gaussian curvature is positive, zero, and negative. From Gauss-Bonnet we have that Kd = 2πχ(S) = 2π(2 2g). S Now if g > 0, then S Kd 0. Because S has for sure one point with positive Gaussian curvature, then there must be a point where the Gaussian curvature is negative. But K is a continuous function and thus if it has points where is positive and points where is negative, it must have points where is zero (becasue S is connected). (2) Let S be a compact surface in R 3 with no boundary which has positive Gaussian curvature and denote the Gauss map by N : S {x 2 + y 2 + z 2 = 1}. Show that if is a geodesic in S which divides S into a set and another set B (i.e., S = B and = B = ), then area(n()) = area(n(b)). Hint: If you have a diffeomorphism F : S {x 2 + y 2 + z 2 = 1} R 3, what is the formula for area(f ()) in terms of DF and the set? Because K > 0, we have from Gauss-Bonnet that S is diffeomorphic to a sphere. Thus andb are both diffeomorphic to a disc and
4 NDRÉ NEVES = B = is a geodesic. Hence, if ν denotes the interior unit normal to we have Kd + k.νdσ = 2π = Kd = 2π. Likewise B Kd = 2π. I will show now that area(n()) = Kd = 2π and this completes the exercise. The first remark is that the Gaussian map N must be a diffeomorphism. Why? Well, because K = det = detdn, we have from the inverse function theorem that N is locally a diffeomorphism. nd it happens that locally diffeomorphisms from a sphere into a sphere must be global diffeomorphism. The second remark is to show the hint, namely that area(f ()) = detdf d. Suppose that φ : D R 2 is a chart for. Then ψ = F φ is a chart for N = {x 2 + y 2 + z 2 = 1} because F is a diffeomorphism. From the lectures we know that area(ψ(d)) = x y dxdy. D Now let {e 1, e 2 } be a basis for T φ(p) which has df p (e 1 ) = λ 1 e 1 and df p (e 2 ) = λ 2 e 2, i.e., {e 1, e 2 } is an eigenbasis for df. Then and thus x (p) = a 1e 1 + b 1 e 2 y (p) = a 2e 1 + b 2 e 2 x y = a 1b 2 a 2 b 1 e 1 e 2. Furthermore ( ) x (p) = df φ(p) x (p) = λ 1 a 1 e 1 + λ 2 b 1 e 2 and ( ) y (p) = df φ(p) y (p) = λ 1 a 2 e 1 + λ 2 b 2 e 2. Thus x y = λ 1λ 2 a 1 b 2 a 2 b 1 e 1 e 2 = detdf φ(p) x y.
HOMEWORK 4 - GEOMETRY OF CURVES ND SURFCES 5 (This formula was long to derive but it should be highly expected). Therefore we have area(f (φ(d))) = area(ψ(d)) = D x y dxdy. = detdf φ(p) x y dxdy = detdf d. D φ(d) Covering with a finite number of charts we arrive at area(f ()) = detdf d. Using this formula with F = N the Gauss map we obtain area(n()) = detdn d = K d = Kd = 2π. (3) Let S be the cylinder {x 2 + y 2 = 1} in R 3, with the chart φ : [0, 2π] R S, φ(θ, z) = (cos θ, sin θ, z). and let be a simple closed curve in S. (a) Let be a simple closed curve in S for which there is a path α : [0, 1] [0, 2π] R so that α(0) = (0, z 0 ), α(1) = (2π, z 0 ), and = φ α. In other words, loops once around the z-axis. If ν denotes a unit normal vector to show that k. νdσ = 0. If the path α(t) = (θ(t), z(t)), choose z 1 so that z(t) > z 1 for all t [0, 1] and consider the path β(t) = (2πt, z 1 ). Note that φ β is a geodesic and there is a region R in S so that S = φ β. To recognize this last fact just let R = φ(), where is the region in [0, 2π] R below α and above β. Note that R is diffemorphic to a cylinder and thus has zero Euler characteristic. Using Gauss-Bonnet we obtain that Kd + k. νdσ + k. νdσ = 0, R where ν is the interior unit normal. Because K = 0 and φ β is a geodesic we obtain that k. νdσ, φ β
6 NDRÉ NEVES which is what we wanted to show. (a) Let be a simple closed curve in S for which there is a path α : [0, 1] [0, 2π] R so that α(0) = (θ 0, z 0 ) = α(1). In other words, is the boundary of a disc D in S. Show that is not a geodesic. The Euler characteristic of D is one and so by the Gauss-Bonnet Theorem we have Kd + k. νdσ = 2π = k. νdσ = 2π. D Thus, cannot be a geodesic.