GEOMETRY MID-TERM CLAY SHONKWILER

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1 GEOMETRY MID-TERM CLAY SHONKWILER. Polar Coordinates R 2 can be considered as a regular surface S by setting S = {(, y, z) R 3 : z = 0}. Polar coordinates on S are given as follows. Let U = {(ρ, θ) : ρ > 0, 0 < θ < 2π} let φ : U V S be given by φ(ρ, θ) = (ρ cos θ, ρ sin θ, 0), where V = S\{(, 0, 0) R 3 : 0}. (a) Compute the first fundamental form of S in polar coordinates. Answer: First, we need to compute Φ = φ ρ Φ 2 = φ θ : Φ = (cos θ, sin θ, 0) Φ 2 = ( ρ sin θ, ρ cos θ, 0). Now, the coefficients of the first fundamental form are as follows: E = Φ, Φ = cos 2 θ + sin 2 θ = F = Φ, Φ 2 = ρ sin θ cos θ + ρ sin θ cos θ = 0 G = Φ 2, Φ 2 = ρ 2 sin 2 θ + ρ 2 cos 2 θ = ρ 2. (b) Compute the Christoffel symbols of S in polar coordinates. How does this compare with the computation in the usual coordinate system ψ(, y) = (, y, 0)? Answer: In order to compute the Christoffel symbols, first we need to compute the partials of E, F G: E ρ = 0 E θ = 0 F ρ = 0 F θ = 0 G ρ = 2ρ G θ = 0 Now, plugging these values into the system of equations that lets us determine the Christoffel symbols, we see 0 = 2 E ρ = Φ, Φ = Γ E + Γ2 F = Γ 0 = F ρ 2 E θ = Φ, Φ 2 = Γ F + Γ2 G = Γ2 ρ2 0 = 2 E θ = Φ 2, Φ = Γ 2 E + Γ2 2 F = Γ 2 ρ = 2 G ρ = Φ 2, Φ 2 = Γ 2 F + Γ2 2 G = Γ2 2 ρ2 ρ = F θ 2 G ρ = Φ 22, Φ = Γ 22 E + Γ2 22 F = Γ 22 0 = G θ = Φ 22, Φ 2 = Γ 22 F + Γ2 22 G = Γ2 22 ρ2. From this, then, we conclude that Γ = Γ 2 = Γ 2 = Γ 2 22 = 0

2 2 CLAY SHONKWILER Γ 2 2 = ρ, Γ 22 = ρ. These differ somewhat from the Christoffel symbols in the usual coordinate system, which are all zero, due to the fact that the partials of E, F G in the usual coordinate system are all zero. 2. Maimal Integral Curves On a previous HW we showed that on a compact manifold the maimal integral curves of a smooth vector field are defined for all time t. Show that on the open disk D = {(, y) R 2 : 2 + y 2 = } there eists a smooth vector field X for which the maimal integral curves are not defined for all time t. Hence, the compactness condition on our HW problem is necessary. Eample: Let X q = q for all q D. Specifically, the vector in T p D associated with the point p = (, y) is simply (, y). Now, let p = ( 0, y 0 ) D let α : I D be an integral curve through p of X. Then α(t) = ((t), y(t)) for t I, α(0) = ( 0, y 0 ) ( (t), y (t)) = α (t) = X α(t) = ((t), y(t)) = α(t). Hence, we see that α(t) = ( 0 e t, y 0 e t ). Since the integral curves are unique, we see that curves of this form are the only integral curves on D; specifically, any maimal integral curve must be of this form. Now, if p = 0, a maimal integral curve can be defined for all time t. However, this curve is constant, forever remaining fied at the origin. On the other h, if p 0, then we see that, for all t > t 0 for some t 0 R, ( 0 e t, y o e t ) >, meaning α is not defined for these values of t. Since, as we argued above, all integral curves, including any maimal integral curves, must be of this form, we see that the maimal integral curves of X on D are not defined for all time t. 3. Covering Maps Let S S be two regular surfaces. A map F : S S is said to be a smooth covering map if (a) F is smooth. (b) For each p S there eists a neighborhood O of p such that F (O) = α J V α is the disjoint union of open sets V α such that for each α J we have F : V α O is a diffeomorphism. Such a neighborhood O is said to be evenly covered by F.

3 GEOMETRY MID-TERM 3 Now suppose that F : S S F 2 : S 2 S are smooth covering maps g : S S 2 is a homeomorphism such that F 2 g = F F g = F 2. Show that g : S S 2 is a diffeomorphism. Proof. Let p S let q = F (p) = (F 2 g)(p). Let O be a neighborhood of q that is evenly covered by F let O 2 be a neighborhood of q that is evenly covered by F 2. Then F (O ) = α J V α F2 (O 2 ) = α J V 2α where F : V α O F 2 : V 2α O 2 are diffeomorphisms. Let V β be the V α containing p let V 2β be the V 2α containing g(p). Let O = O O 2, V = V β F (O), V 2 = V 2β F 2 (O). Then F : V O F 2 : V 2 O are diffeomorphisms. Let φ : U R 2 W V φ 2 : U 2 R 2 W 2 V 2 φ : U R 2 W O be coordinate charts on p, g(p) q, respectively, such that F (φ (U )) φ(u) F 2 (φ 2 (U 2 )) φ(u). Then are differentiable. Furthermore, if H := φ F φ : U U H 2 := φ F 2 φ 2 : U 2 U Ũ := H (U ) H 2 (U 2 ), Ũ := H (Ũ) Ũ 2 := H2 (Ũ) then H : Ũ Ũ H 2 : Ũ2 Ũ are diffeomorphisms. Hence, H H 2 is differentiable, meaning that, when the domains of the F s φ s are restricted appropriately, H H 2 = (φ F φ ) (φ F 2 φ 2 ) = φ F φ φ F 2 φ 2 = φ F F 2 φ 2 = φ F F g φ 2 = φ g φ 2 is differentiable. This says precisely that g is differentiable on the appropriate neighborhood of g(p).

4 4 CLAY SHONKWILER A similar argument shows that g is differentiable on the corresponding neighborhood of p. Since our choice of p was arbitary, we see that g is a differentiable homeomorphism with differentiable inverse at all points in S : that is to say, g is a diffeomorphism. 4. Non-Orientable Surfaces In do Carmo ( in class) Gaussian Curvature was developed only for orientable surfaces. Is this necessary? Please discuss. Answer: No, it is not necessary for a surface to be orientable in order for us to determine the Gaussian Curvature at a point p on the surface. To see why this is the case, we first note that every surface covered by a single coordinate system is trivially orientable. Hence, any surface, even a non-orientable surface like the Mobius strip is locally orientable. Therefore, if we take our point p consider some coordinate chart φ : U R 2 V on a neighborhood of p, then we see that we can assign an orientation to this neighborhood of V. Given this orientation N, then, we can certainly define the Gauss map N : V R 3 on this neighborhood V of p. In fact, since V is itself a regular surface, we can define the Gaussian curvature K, which is just the determinant of dn p. Now, this determinant is not dependent on the choice of orientation for V (remember, V is an orientable surface), so this notion of Gaussian curvature at p is well-defined. 5. Regular n-manifolds Let M 2 (R) be the set of all 2 2 matrices SL(2, R) = {A M 2 (R) : det(a) = } the set of all real 2 2 matrices with determinant [ ]. a b (a) M 2 (R) can be identified with R 4 via the map (a, b, c, d). c d Use this identification to show that SL(2, R) is a regular 3-surface. [ ] a b Proof. Let SL(2, R), which is identified with (a, b, c, d) R c d 4. Since the zero matri is not in SL(2, R) we may assume, up to a change of aes, that a 0. Then let V = {(, y, z, w) R 4 : 0} let U = {(, y, z) R 3 : 0. Then V is certainly an open set of R 4 U is an open set of R 3. Now, let φ : U V be given by ( (, y, z), y, z, + yz ). We see that [ y det z +yz ] = + yz yz = + yz yz =,

5 GEOMETRY MID-TERM 5 so the map φ is well-defined. Since 0, we see that φ is smooth since each coordinate function is. Also, 0 0 dφ = (+yz) 2 The top minor of dφ has determinant for all points in U, so dφ is injective. Hence, it only remains to show that φ is a homeomorphism. Certainly, since φ is smooth, φ is continuous, so we need only show that φ eists is continuous. If A = (a, b, c, d) SL ( 2, R) (again, remember we may assume a 0), then = det A = ad bc d = + bc a so φ (a, b, c, d) = (a, b, c). Certainly this is a smooth function, therefore continuous, so we see that φ is, indeed, a coordinate chart on V SL(2, R). Since our choice of (a, b, c, d) was arbitrary (up to change of aes), we conclude that SL(2, R) is indeed a regular 3-surface. [ ] 0 (b) Show that the tangent space of SL(2, R) at I = is given by 0 [ a b T I (SL(2, R)) = {(I; c d z y ] ) : a + d = 0}. That is, T I (SL(2, R)) can be identified with the set of traceless 2 2 matrices over R. This tangent space is an eample of a Lie algebra. Proof. By the theorem given in the statement of the problem, we know that T I (SL(2, R)) = dφ I (R 3 ). In part (a) above we gave the general form for dφ, we need only plug in the coordinates of φ (, 0, 0, ) = (, 0, 0) to that formula. Specifically, we see that, if (, 2, 3 ) R 3, dφ I = = 3 (+0) Now, we identify this vector with the matri ( 2 3 )

6 6 CLAY SHONKWILER We note that + ( ) = 0; when we vary (, 2, 3 ) over all of R 3, we see that, indeed, [ ] T I (SL(2, R)) = dφ I (R 3 a b ) = {(I; ) : a + d = 0}. c d 6. Vector Fields as Differential Operators Let X p T p S be a tangent vector f : U S R be a smooth function defined on an open neighborhood of p. Then X p f d dt (f α) t=0 where α : ( ɛ, ɛ) S is a smooth curve such that α(0) = p α (0) = X p. (a) Let X be a vector field on an open set U of S. Show that X is a smooth vector field if only if for any function f : U R the function q X q f is smooth. Proof. Suppose X is a smooth vector field. This means that, if φ : W R 2 V subsetequ is a coordinate chart on some point p U, X q = a(q)φ q + b(q)φ 2q where a, b : V R are smooth maps. Let alpha : ( ɛ, ɛ) S be a smooth curve such that α(0) = q α (0) = X q. If f : U R is any smooth map, then, by the chain rule, X q f = d dt (f α) t=0 = df α(0) α (0) = df α(0) (a(q)φ q + b(q)φ 2q ), which is certainly a smooth function, since df, a, b, Φ Φ 2 are all smooth. On the other h, suppose X is a vector field q X q f is smooth. Then X q = a(q)φ q + b(q)φ 2q where φ : W R 2 V U is a coordinate chart on a neighborhood of p U. We want to show that a, b : V R are smooth. Let α : ( ɛ, ɛ) U be a smooth curve such that α(0) = q α (0) = X q. Then, X q f = d dt (f α) t=0 = df α(0) α (0) = df α(0) (a(q)φ q +b(q)φ 2q = df q a(q)φ q +df q b(q)φ 2q since df is a linear map. Since q X q f is smooth, it follows that a b are smooth, meaning that X is smooth. (b) Let λ R f, g : U R be smooth functions. Show that for any p S we have X p (λf + µg) = λx p f + µx p g X p (fg) = X p (f)g(p) + f(p)x p (g).

7 GEOMETRY MID-TERM 7 Proof. Let α : ( ɛ, ɛ) S be a smooth curve such that α(0) = p α (0) = X p. Since λf + µg is smooth, we can use the chain rule to see that X p (λf + µg) = d dt ((λf + µg) α) t=0 = d(λf + µg) α(0) α (0) = (d(λf) α(0) + d(µg) α(0) )α (0) = (λdf α(0) + µdg α(0) )α (0) = λdf α(0) α (0) + µdg α(0) α (0) = λ d dt (f α) t=0 + µ d dt (g α) t=0 = λx p f + µx p g. On the other h, using the product rule for differentials the chain rule, X p (fg) = d dt (fg α) t=0 = d(fg) α(0) α (0) = (df α(0) g(α(0)) + f(α(0))dg α(0) )α (0) = df α(0) g(p)α (0) + f(p)dg α(0) α (0) = df α(0) α (0)g(p) + f(p)dg α(0) α (0) = d dt (f α) t=0g(p) + f(p) d dt (g α) t=0 = X p (f)g(p) + f(p)x p (g). (c) Show that for any smooth function f : W S R on a neighborhood of p S any tangent vector X p T p S we have df p (X p ) = X p f. Proof. Let α be a smooth curve such that α(0) = p α (0) = X p. Now, since f is smooth we may use the chain to see that X p f = d dt (f α) t=0 = df α(0) α (0) = df p X p. Hence, we conclude that df p (X p ) = X p f. (d) Let X, Y, Z be smooth vector fields on S. Then the function q Y q, Z q is a smooth function on S. Show that for any p S we have X p Y, Z = X Y (p), Z p + Y p, X Z(p). Proof. Again, let α be a smooth curve such that α(0) = p α (0) = X p. Using the result proved in class that, if X, X 2 are vector fields along α, then d dt X (t), X 2 (t) = DX 2, X 2 + X, DX 2, dt dt

8 8 CLAY SHONKWILER we see that X p Y, Z = d dt ( Y, Z α) t=0 = d dt ( Y (t), Z(t) ) t=0 = DY dt, Z p + Y p, DZ dt = X Y (p), Z p + Y p, X Z(p). 7. Vector Fields & Coordinate Charts Let X be a smooth vector field on S suppose p S is such that X p 0. Show that there eists a coordinate chart φ : U V containing p such that Φ q = X q for all q V. Proof. We show this in two parts. First we prove the result on an open set contained in R 2, then etend that result to any regular surface. Lemma 7.. Let X be a vector field on an open set W R 2 let p U such that X p 0. Then there eists a coordinate chart φ : U V W on a neighborhood of p such that Φ q = X q for all q V. Proof. Choose a coordinate system on R 2 such that p = (0, 0) X p is in the direction of the ais. Let α : Ṽ I W be the local flow at p, Ṽ W, t I let φ be the restriction of α to the rectangle (Ṽ I) {(, y, t) R3 ; = 0}. We can identify {(, y, t) R 3 : = 0} with R 2 in such a way that the t direction is identified with the u direction the y direction is identified with the u 2 direction. The above rectangle, then, is a subset of the plane we see that φ is a map from a subset of R 2 into W. By the definition of local flow, d φ p maps the unit vector in the u direction into X maps the unit vector of the u 2 ais into itself. Therefore, d φ p is nonsingular. By the inverse function theorem, then, it follows that there eists a neighborhood V W of p where φ is defined differentiable. Now, let U = φ (V ) let φ be the restriction of φ to U. Since φ is a differentiable function with differentiable inverse, bijective with nonsingular differential, we see that φ : U V is a coordinate chart on a neighborhood of p. It only remains to show that Φ q = X q for all q V. However, this follows immediately from how we have defined φ. Since the u direction in U is identified with the t direction in the original rectangle lying in the yt-plane, φ (u, u 2 ) = α u t (q, t) = X α(q,t). Hence, if q V (u 0, v 0 ) U such that φ(u 0, v 0 ) = q (that is to say, u 0 v 0 are identified with t 0 I q 0 is the unique point on the y-ais

9 GEOMETRY MID-TERM 9 such that the trajectory through q 0 passes through q), then φ q = φ u (u 0, v 0 ) = α t (q 0, t 0 ) = X α(q0,t 0 ) = X q. Now, we turn to the general case. Let S, X p be as stated in the problem let ψ : U V S be a coordinate chart on a neighborhood of p. Since X is smooth, we know that in the coordinate system, X q = a(q)ψ q + b(q)ψ 2q where a, b : V R are smooth. Consider the vector field X = ãe + be 2 where ã = a ψ, b = b ψ : U R. Then, as in the proof of the local first integral lemma for surfaces that we did in class, X = dψ( X). Now, by the above lemma, there eists a coordinate chart φ : U Ṽ U such that Φ v = X v for all v Ṽ. Define φ := ψ φ : U V = ψ(ṽ ). Then, by the chain rule, dφ = dψ d φ. Specifically, Φ q = dψ v ( Φ v ) = dψ v ( X v ) = X q where v = ψ (q). DRL 3E3A, University of Pennsylvania address: shonkwil@math.upenn.edu