DIFFERENTIAL GEOMETRY HW 5

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1 DIFFERENTIAL GEOMETRY HW 5 CLAY SHONKWILER 3. Le M be a complee Riemannian manifold wih non-posiive secional curvaure. Prove ha d exp p v w w, for all p M, all v T p M and all w T v T p M. Proof. Le γ be a geodesic on M wih γ = p, γ = v. Define he Jacobi field J = d exp p v w. Then J = and J = D d d exp p v w d = = d exp p v w+ d d exp p v w = w. = Hence, J, γ = w, v. Now, le M = T p M, which is linearly isomeric o R n wih he usual inner produc, so he secional curvaures are all, which is greaer han or equal o all secional curvaures of M. Le γ = v; hen γ is a geodesic in M and, if we define J = w, hen J is a Jacobi field. Moreover, J, γ = w, v. Therefore, by he Rauch Comparison Theorem, J J for all. In paricular, for =, d exp p v w = J J = w. 4. a: Le C R 2 be a regular curve. Show ha he focal se F C R 2 of C is obained by aking, on he posiive normal n a p C a lengh equal o /k, where k is he curvaure of C a p. Proof. Le xs paramerize C. Recall ha exp, s = x+ns. If he angen T and he normal n form a basis for R 2, hen x d exp = s + n s, x s n, x x s x s + n s, n = s + n s, x s n, n x s + n s, n. n, n Since n, n, d exp is singular and C has a focal poin if and only if x s + n s, x =. s

2 2 CLAY SHONKWILER By he Frene equaions, n s = κst s, so x s + n = T s κst s = κst s s which is zero precisely when κs = : i.e., when = κs, exacly as prediced. Noe ha x s = precisely when κs =, so we see ha he above enirely characerizes he focal poins of C. b: Show ha he focal se of he ellipse x2 + y2 = is given by a 2 b 2 {x, y R 2 ; ax 2/3 + by 2/3 = a 2 b 2 2/3} Proof. Paramerize he ellipse by αs = a cos s, b sin s noe ha, in spie of he noaion, his curve is no paramerized by arc lengh. Then α s = a sin s, b cos s. If we le ns = b cos s, a sin s, hen n is normal o he ellipse. Now, n s = b sin s, a cos s, so d exp = a 2 sin 2 s + b 2 cos 2 s ab b 2 a 2 sin s cos s b 2 cos 2 s + a 2 sin 2 s which has a criical poin when = a2 sin 2 s+b 2 cos 2 s ab. Therefore, he focal locus of he ellipse is given by a cos s a2 sin 2 s + b 2 cos 2 s b cos u, ab αs + a2 sin 2 s + b 2 cos 2 s ns = ab = =, b sin s a2 sin 2 s + b 2 cos 2 s a sin s ab a 2 a 2 sin 2 s b 2 cos 2 s cos s, b2 a 2 sin 2 s b 2 cos 2 s a b b 2 a 2 a cos 3 s, a2 b 2 b sin 3 s. Since b 2 a 2 cos 3 s 2/3 + a 2 b 2 sin 3 s 2/3 = a 2 b 2 2/3, we see ha he focal locus of he ellipse is given by { x, y R 2 ax 2/3 + by 2/3 = a 2 b 2 2/3}. c: Show ha he focal se of he curve cos + sin, sin + cos is he circle cos, sin. sin s

3 DIFFERENTIAL GEOMETRY HW 5 3 Proof. Le αs = cos s + s sin s, sin s + s cos s again, despie noaion, no necessarily paramerized by arc lengh. Then α s = s cos s, s sin s. If we le ns = sin s, cos s, hen n is normal o he spiral. Now, so n s = cos s, sin s, d exp = s 2 + s which has criical poins when = s. Therefore, he focal locus of he spiral is given by αs sns = cos s+s sin s s sin s, sin s+s cos s s cos s = cos s, sin s, as expeced. 6. Wha follows is a sligh generalizaion of Surm s Comparison Theorem. We presen he heorem in geomeric form. Le M 2 be a complee Riemannian manifold of dimension 2, and le γ : [, M 2 be a geodesic. Le J be a Jacobi field along γ wih J = J =,,, and J,,. Then J is a field normal o γ and can be wrien J = fe 2, where e 2 is he parallel ranspor of a uni vecor e 2 T γ M wih e 2 γ. Because J is a Jacobi field, f + Kf =,, where K is he Gaussian curvaure of M 2. Assume ha K L, where L is a difffereniable funcion on [,. Prove ha any soluion of he equaion f + L f = has a zero on [, ], ha is, here exiss [, ] wih f =. Proof. Suppose f is a soluion o he given equaion and ha f for all [, ]. Since f + Kf = and f + L f =, [ = ff + Kf f f f] + L d = ff f f d + = [ ff f f ] = f f ff + K Lf fd f f f f d + K Lf fd. K Lf fd

4 4 CLAY SHONKWILER Now, since f for all [, ], hen eiher f > on his inerval or f < on his inerval. Similarly, since is he firs conjugae poin of γ along γ, f > on, or f < on,. If f > and f > on,, hen, since f = = f, f > and f <. However, since K L, K Lf fd + f f ff f f ff <, conradicing above. The oher cases f > and f <, ec. follow similarly. 7. Le M 2 be a complee Riemannian manifold of dimension wo and le γ : [, M 2 be a geodesic wih γ = p. Le Ks be he Gaussian curvaure of M 2 along γ. Assume ha: 2 Ksds, for all, 4 + in he sense ha he inegral converges and has he bound indicaed. a: Define ω = Ksds + 4 +, and show ha ω + ω 2 K. Proof. Firs, noe ha ω = Ksds Ksds + so, by he Fundamenal Theorem of Calculus, ω = K , Now, 2 ω 2 = Ksds + Ksds so ω + ω 2 2 = K Ksds = K Ksds K = K , Ksds + Ksds 6 + 2

5 DIFFERENTIAL GEOMETRY HW 5 5 b: For, pu ω + ω 2 = L hence L K and define f = exp ωsds,. Show ha Proof. By definiion, f = d ωsds exp d so Therefore, f + L f =, f =. ωsds = ω exp ωsds = ω f, f = ω f + ω 2 f. f + L f = ω f + ω 2 f + ω + ω 2 f =. Also, f = exp =, as desired. c: Observe ha f > and use he oscillaion heorem of Surm o show ha here does no exis a Jacobi field Js on γs wih J = and Js =, for some s,. Therefore, he condiion 2 implies ha here do no exis conjugae poins o p along γ. Proof. Since he exponenial is always posiive, f >. Now, suppose here exiss a Jacobi field Js on γs such ha J = and Js = for some s,. Then, by problem 6 above, since f is a soluion of he equaion f s+ls fs = and Ks Ls, f mus have a zero on [, s ]. However, we jus said ha f is sricly posiive, so his is impossible. Therefore, we conclude ha here mus no exis such a Jacobi field, meaning ha here are no conjugae poins o p along γ. B Le S n denoe he n-sphere of radius in R n+. Consider he produc meric on S 3 S. a: Wha can one say abou he curvaure of his meric? Answer: Le M = S 3 S and, for any p M and V T p M, le us denoe by V 3 he componen of V parallel o S 3 and by V he componen of V parallel o S. Now, for orhonormal X, Y T p M, KX, Y = RX, Y X, Y = RX + X 3, Y + Y 3 X + X 3, Y + Y 3.

6 6 CLAY SHONKWILER Since everyhing in his expression is linear, we can compleely spli up all he erms. Moreover, since S is only one-dimensional, X and Y mus be parallel, so RX, Y. Also, if we hink of M as a bundle over S 3, X is verical and Y 3 is horizonal, so RX, Y 3 and similarly for RY, X 3, RX, X 3, RY, Y 3, ec. Therefore, by rearranging according o he usual formulas, all of he erms in he above expression vanish excep for RX 3, Y 3 X 3, Y 3. Hence, KX, Y = RX 3, Y 3 X 3, Y 3 = X 3 Y 3 2, which varies from o. In essence, we re aking he uni square defined by X and Y, projecing i ono S 3 and whaever he area of he projecion is he secional curvaure of he plane spanned by X and Y. As for scalar curvaure, le z, z 2, z 3, z 4 be an orhonormal basis for T p M, where z, z 2, z 3 T p S 3 and z 4 T p S. Then Kp = Rz i, z j z i, z j ; 2 ij he summands are zero whenever z 4 is involved and oherwise. There are 6 erms involving z 4, so Kp = 6 2 = 2. For Ricci curvaure, he Ricci curvaure in an arbirary direcion is in he S componen and more in he S 3 direcions. Wihou loss of generaliy, assume v T p M is given by v = d 2,,, d where he firs hree erms are he S 3 direcions and he fourh is he S direcion. Complee his o an orhonormal basis: w =,,,, w 2 =,,,, w 3 = d,,, d 2. Then Ric p v = Rv, w v, w + Rv, w 2 v, w 2 + Rv, w 3 v, w 3 3 = w 3 3, w3 v, v v, w3 w 3, v + w3, 2 w3 v, 2 v v, w3 w 2 3, 2 v + w3, 3 w3 v, 3 v v, w3 w 3 3, 3 v = d 2 + d 2 + d 2 d 2 d 2 d 2 3 = 2 3 d2 since only he pars in he S 3 direcion are relevan and S 3 has consan secional curvaure. b: Wha are he geodesics in his meric? Answer: As in any produc manifold, geodesics in M are of he form γ, γ 3, where γ is a geodesic in S and γ 3 is a geodesic in S 3.

7 DIFFERENTIAL GEOMETRY HW 5 7 c: Wha is he firs conjugae locus and cu locus of any poin? Answer: In he following schemaic, he boundary of he picure forms he cu locus of he cener poin p, so C m p = S 3 S. To deermine he conjugae locus, we know ha he conjugae locus of S 3 S is he image of he conjugae locus of he universal cover, S 3 R. As we see in he following schemaic picure, his means he conjugae locus of p is S : C Visualize RP 2 inside CP 2. Answer: Thinking of CP 2 as complex lines in C 3 R 6, we know each complex line mees he uni sphere S 5 in a grea circle, which gives an induced Hopf map h : S 5 CP 2. Now, hink of z i = x i + iy i ; hen x, x 2, x 3, y, y 2, y 3 serve as coordinaes of R 6 ; if we consider he real 3-plane deermined by x, x 2, x 3, hen his 3-plane inersecs he uni 5-sphere in

8 8 CLAY SHONKWILER a grea 2-sphere S 2. Under he map h, anipodal poins on his S 2 are idenified, so hs 2 = RP 2 CP 2. The isomeries of CP 2 cerainly conain U3/e iθ = SU3, since U3 akes complex lines o complex lines and he circle acion akes each complex line o iself. Moreover, complex conjugaion, ha is, he map z, z 2, z 3 z, z 2, z 3 akes complex lines o complex lines, so we guess ha he group of isomeries of CP 2 is SU3 complex conjugaion. Noe ha his group of isomeries is ransiive and isoropic. By he ransiiviy of he isomeries of CP 2, we can look a jus a single poin on RP 2 o invesigae furher; le p = : :. Recall from las week ha KX, Y = + 3 cos 2 φ where cos φ = X, iȳ, where X and Ȳ are horizonal lifs o T S 5 of X, Y T CP 2. Specifying o he case of RP 2, if q h p i.e. q =,, or,,, hen we can jus hink of T S 2 q S 2 as he 2-plane perpendicular o q hough of as a vecor in he 3-plane described above. Leing q =,,, hen T q S 2 is spanned by,, and,,. Hence, cos φ =,,, i,, =,,,,, i =, so he secional curvaure a p is + 3 cos 2 φ =. Now, suppose p CP 2 and X, Y T p CP 2 be orhonormal such ha KX, Y =. Then, since he isomeries of CP 2 are ransiive, here exiss an isomery f of CP 2 such ha f : : = p. Since he isomeries are isoropic, we may as well assume f h,, = X. Thinking of he horizonal par of T q S 5 as he 4-plane perpendicular o he complex line associaed wih hq, hen, since X, iȳ =, f Ȳ = ±,, or ±,, i. In he firs case, he 2-plane spanned by X and Y is angen o frp 2 ; in he second case, i s no. On he oher hand, in he second case, he 2-plane spanned by X and Y is angen o f RP 2 where RP 2 is he RP 2 given as above by inersecing he 3-plane spanned by x, x 2, y 3 wih S 5, so i is angen o an RP 2. Exra Problem A poin in CP 2 is chosen a random, and hen a angen 2-plane a he poin is chosen a random. We know he secional curvaure can be any number beween and 4. Suppose we are old ha i is eiher or 4. Which of hese wo possibiliies is more likely, and why? Answer: Le p CP 2. Then p corresponds o a complex line in C 3. By a suiable roaion of C 3, we may as well assume p corresponds o he complex line given by he firs facor in C 3 = C C C. This plane corresponds o : : CP 2 in homogeneous coordinaes, so we may as well assume p = : :. Now, I like o visualize T p CP 2 as he se of all vecors in C 3 perpendicular o he complex line corresponding o he poin p. Hence, if u, v T p CP 2, hen we can hink of u, v as vecors in C C C 2.

9 DIFFERENTIAL GEOMETRY HW 5 9 Now, we know from problem 2 on HW #4 ha Ku, v = + cos 2 φ where cos φ = ū, i v and ū and v are he horizonal lifs of an orhonormal pair u and v, respecively, o T S 5. For q S 5, we can hink of T q S 5 as he 5-plane perpendicular o q hough of as a vecor in R 6 C 3. Now, since p = : :, we know ha q =,, lies in he fiber over p and, in he above inerpreaion, T q S 5 = {iy, x 2 + iy 2, x 3 + iy 3 C 3 }. The verical par of T q S 5 is jus {iy,, } and he horizonal par is {, x 2 +iy 2, x 3 +iy 3 }. Hence, for u, v T p CP 2 hough of as u, u 2, v, v 2 C 2, ū =, u, u 2 and v =, v, v 2. Now, suppose u, v T p CP 2 are an orhonormal pair such ha Ku, v = or 4, so eiher ū, i v = cos φ = or ū, i v = cos φ =. In oher words, eiher u is parallel o iv in he above visualizaion, or u is perpendicular o iv. Since he space of vecors parallel o iv is -dimensional while he space of vecors perpendicular o iv is 3-dimensional, we expec ha he laer should be much more likely and so Ku, v = should be more likely han Ku, v = 4. To see his rigorously, suppose u = a + ib, a 2 + ib 2. If v = ±iu, hen cos φ = ū, i v = ū, ū =. On he oher hand, supposing b 2 if i is, perform he below calculaion wih v = ±, i, hen if we le v = ±, b b2, we see ha cos φ = ū, i v = + b2 b 2 2 a, b, a 2, b 2, ± = ± + b2 b 2 2 b b =. + a2 a b2 b 2 2,,, b b 2 Also, supposing a 2 if i is, perform he below calculaion wih v = ±,, hen if we le v = ± i, i a a 2, we see ha cos φ = ū, i w 2 = a, b, a 2, b 2, ± + a2 a 2 2 = ± a + a + a2 a 2 2 =.,, a, a 2

10 CLAY SHONKWILER The above hree possibiliies for v are pairwise orhonormal, so any vecor orhonormal o u in T p CP 2 is a linear combinaion of hem. Hence, we see ha he only v T p CP 2 for which Ku, v = 4 are v = ±iu, whereas Ku, v = for any norm linear combinaion of + b2 b 2 2, b b2 and + a2 a 2 2 i, i a a 2. Thus, we conclude ha, if we choose a angen 2-plane o p, hen i is much more likely ha he secional curvaure a ha 2-plane is han 4. DRL 3E3A, Universiy of Pennsylvania address: shonkwil@mah.upenn.edu