DIFFERENTIAL GEOMETRY 1 PROBLEM SET 2 SOLUTIONS

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1 DIFFERENTIAL GEOMETRY 1 PROBLEM SET 2 SOLUTIONS Lee: 2-1c, 2-6, 2-8, 2-10 CFB: 1.4.2:1-5 Remark 0.1. Note that problem was corrected. The errata for the text Cartan For Beginners can be found at CFB V a vector space with nondegenerate inner product,. Let v t be a curve in V such that F t : v t, v t constant. Show that v t v t for all t. Show that the converse is true. First note that F t v t, v t + v t, v t 2 v t, v t. Then F is constant if and only if 0 F t for all t. Thus F is constant if and only if 0 2 v t, v t for all t. Assuming the field doesn t have characteristic 2 we have that 0 2 v t, v t if and only if 0 v t, v t. But this inner product vanishes for all t if an only if v t v t for all t. Therefore F t is constant if and only if v t v t for all t. Date: February 7,

2 2 CFB Suppose c is regular. Let s t t 0 c τ dτ and consider c parameterized by s instead of t. Since s gives the length of the image c : [0, s] E 2, s is called an arclength parameter. Show that in this preferred parameterization, κ s e 2, de 1. ds Recall that c is regular if and only if c t 0. Thus v t c t 0 and hence we may divide by v t without issue. With this in mind we calculate However, de 1 de 1 de 1 ds ds de 1 ds Thus de 1 ds, e 2 κ t. d t 0 de 1 ds c t by the chain rule c t de 1 ds v t λ t e 2. So we have by the fundamental theorem of calculus. de 1 ds λ t v t e 2 κ t e 2 CFB Show that κ t is constant if and only if the curve is an open subset of a line if κ 0 or a circle of radius 1 κ. We treat the line and the circle cases separately. First the line. Claim 0.2. c is an open subset of a line if and only if κ 0.. If c is an open subset of a line then we may parameterize c by c t a 1 t + b 1, a 2 t + b 2. Differentiating we find c t a 1, a 2. But v t c t a a 2 2 and hence e 1 ct a 1, a 2. 1 vt a 2 1 +a2 2 We immediately see that e 1 is constant and hence de 1 0. Since e 2 0 and de 1 λ t e 2 we can conclude e 2 0. Moreover, c is an open subset of a line and hence a 1, a 2 cannot both vanish and so v t 0. Therefore κ t λt 0. vt Suppose κ t 0.

3 Given the definition of κ t we immediately conclude that λ t 0 and hence de 1 0. Applying the definition of e 1 we can conclude that c t is constant and thus is expressible as a 1, a 2 R 2. Integrating gives c t a 1 t + b 2, a 2 t + b 2. 3 Claim 0.3. c is an open subset of a circle of radius 1 κ nonzero. if and only if κ is constant and Supposing c is an open subset of a circle of radius 1. Then since we re after computing κ which is a differential invariant we may translate and rotate the circle without κ affecting κ. Thus c may be parameterized as c t 1 cos at, sin at. Of course κ κ is parameterization invariant and so we may assume, without loss of generality, that a 1. Differentiating we find c t 1 sin t, cos t and v t κ c t 1. κ Thus e 1 t c t sin t, cos t from which it follows that e c t 2 cos t, sin t. Differentiating e 1 we find de 1 cos t, sin t e 2 t Therefore λ t 1 and so κ t λt vt 1 1/κ κ. Suppose that κ is a nonzero constant. Then taking a unit speed parameterization we have de 1 κe 2. Differentiating again we find d2 e 1 κ de 2 2 since κ is constant. However, 0 d 1 d e 2, e 2 2 de 2, e 2 and 0 d 0 d e 1, e 2 de 1, e 2 + de 2, e 1 κ + de 2, e 1. Therefore de 2 κe 1. This result should be familiar to those that have experience with Frenet frames. Therefore, d2 e 1 κ 2 e 2 1. This is a familiar equation and integrating it reveals e 1 Ae iκt+ϕ0. Since e 1 has magnitude 1 we may take A 1. Integrating again to find c t gives c t 1 κ eiκt+ϕ 0.

4 4 CFB Let c t x t, y t be given in coordinates. Calculate κ t in terms of x t, y t, and their derivatives. Note: Below the t dependence will be suppressed to avoid notational clutter. First we compute the speed of this curve and find v c x 2 + y 2. This immediately gives e 1 E 2 y we may take e 2, x v v Next we compute v and find Applying this to compute de 1 de 1 x. x, y v v gives. Since e 1 and e 2 constitute an orthonormal frame in v x x + y y v v x v v, y 2 v y v v 2 x y 2 x y y x y y x Thus λ x y y x v 2 and so κ λ v x y y x x 2 +y 2 3/2. v 2, y x 2 y x x v 3 v 3 e 2 CFB Calculate the function κ t for an ellipse. Characterize the points on the ellipse where the maximum and minimum values of κ t occur. Since κ is a differential invariant we may assume, without loss of generality, that c t may be parameterized as c t x, y c 1 cos t, c 2 sin t with c 1 c 2 > 0. Now if c 1 c 2 then c parameterizes a circle and so by CFB we know that the curvature is constant. Thus we will henceforth take c 1 > c 2. To make use of the results of CFB we compute the following derivatives x c 1 sin t x c 1 cos t y c 2 cos t y c 2 sin t

5 5 Applying the previous problem gives κ t Differentiating κ to find the extrema gives where v x 2 + y 2. Examination of v reveals that 3c 1c 2 c 1 c 2 c 2 1 cos 2 t + c 2 2 sin 2 t 3/2 0 κ 3c 1c 2 2v 7/2 x x + y y 2v 7/2 curvatures occur only for x x y y. Applying our above derivatives this condition reads is defined everywhere and is nonzero. Thus extremal 0 c 2 1 c 2 2 sin t cos t Since c 1 > c 2 we must have sin t 0 or cos t 0 and so t { 0, π, π, } 3π 2 2. An application of the second derivative test reveals that 0 and π are maxima and π and 2 3π are minima. This makes intutive sense, since at t 0 and t π we are along the 2 semi-major axis and the curve is bending the most while at t π and t 3π we are along 2 2 the semi-minor axis where the ellipse is flattest. Lee 2-1c Compute the coordinate representation of F : S 3 S 2 given by F z, w zw + wz, iwz izw, zz ww in stereographic coordinates. Use this result to prove that F is smooth. Note that here we think of S 3 as the subset { w, z C 2 z 2 + w 2 1 } C 2. Identifying C 2 with R 4 via x 1 + ix 2, x 3 + ix 4 x 1, x 2, x 3, x 4 we have f x 1, x 2, x 3, x 4 F x 1 + ix 2, x 3 + ix 4 2x 1 x 3 + 2x 2 x 4, 2x 2 x 3 2x 1 x 4, x x 2 2 x 3 2 x 4 2 Since R 4 and C 2 are diffeomorphic it suffices to show that f is smooth. Next recall that stereographic projection was given by σ 3 x 1,..., x 4 1 x 1,..., x 3 defined on S 3 \ {0, 0, 0, 1} 1 x 4 σ3 1 x 1, x 2, x 3 1 2x 1 x x x 3 2, 2x 2, 2x 3, x x x and σ 3 : S 3 \ 0, 0, 0, 1 was defined by σ 3 x σ 3 x and σ 1 3 σ 1 3 x. We have similar functions on S 2 which will be denoted by σ 2 and σ 2.

6 6 With these functions direct calculation reveals σ 2 f σ3 1 x 1, x 2, x 3 2x 1 x x x 3 2, x2 σ 2 f σ 3 1 x 1, x 2, x 3 2x 1 x x x 3 2, x 2 x x x 3 2 σ 2 f σ 3 1 x 1, x 2, x 3 2x 1 x x x 3 2, x2 σ 2 f σ3 1 x 1, x 2, x 3 2x 1 x x x 3 2, x 2 x x x 3 2 These are all rational functions and hence are smooth on their domains of definition. Therefore, f is smooth. But f is merely F under the C 2 R 4 diffeomorphism and hence F is smooth. Lee 2-6 For any topological space M, let C M denote the algebra of continuous functions f : M R. If F : M N is a continuous map, define F : C N C M by F f f F. a Show that F is a linear map. b If M and N are smooth manifolds, show that F is smooth if and only if F C N C M. c If F : M N is a homeomorphism between smooth manifolds show that it is a diffeomorphism if and only if F restricts to an isomorphism from C N to C M. Claim 0.4. F is a linear map. Let f, g C N and a, b R then F af + bg af + bg F af F + bg F a f F + b g F af f + bf g Since f, g, a, b were arbitrary we can conclude that F is a linear map. Claim 0.5. If M and N are smooth manifolds, show that F is smooth if and only if F C N C M.

7 Suppose F is smooth and f C N. Then F f f F : M R is a composition of smooth maps and hence is smooth by Lee s Lemma 2.4. Thus F f C N for any f C N. In other words F C N C M.. Suppose that F C N C M. Fixing m M and f C N. Since F C N C M we know that g : F f is smooth at m. Furthermore, f is smooth at F m. Then by the inverse function theorem we know that there is a chart U, ϕ on N containing F m on which f locally has a smooth inverse. So restricting to U we know that g F f f F and so F f 1 g. However, on U, f 1 g is smooth and so F is smooth on U and in particular at m. Since this holds for any m M we can conclude that F is smooth. 7 Claim 0.6. If F : M N is a homeomorphism between smooth manifolds show that it is a diffeomorphism if and only if F restricts to an isomorphism from C N to C M. Since F is a diffeomorphism we know that F 1 : N M exists and is smooth. So by part b we can conclude that F 1 C M C N. Furthermore, by a we know that F and F 1 are linear. So it suffices to show that F 1 F 1 and that F rep. F 1 respects the multiplicative structure on C N resp.c M. These results are achieved by direct calculation, let f, g C N then F 1 F f f F F 1 f Id N f F f g f g F f F g F F f F g Thus F 1 F Id C M and F respects the product point-wise multiplication in C N. The calculations for F 1 are completely analogous. Thus F 1 F 1 and F are algebra homomorphisms and hence C N C M.. Suppose that C N C M and that F is a homeomorphism. Then F 1 : N M exists and is continuous. Furthermore, examining the proof we see that F 1 F 1 did not rely on the smoothness of F or F 1 and so we can still conclude that F 1 F 1. But F 1 F 1 : C M C N and F : C N C M are isomorphisms by hypothesis and so F 1 C M C N and F : C N C M. So we can apply b to conclude that F and F 1 are smooth whence F is a diffeomorphism.

8 8 Lee 2-8 Show that ɛ n : R n T n defined in Example 2.8d is a smooth covering map. We first need to show that ɛ n is smooth. If the book is to be believed then ɛ n is a Lie algebra homomorphism and hence is smooth. However, the book asserts this without proof and so we will show smoothness for completeness. To do this note that ɛ n t 1,..., t n ɛ 1 t 1,..., ɛ 1 t n and so it suffices to show that ɛ 1 : R S 1 is smooth. To do this we recall that stereographic projection is given by σ : S 1 \ {0, 1} R and σ : S 1 \ {0, 1} R where Then x1 σ x 1, x 2 1 x 2 2x σ 1 x x 2 + 1, x2 1 x σ x 1, x 2 σ x 1, x 2 σ 1 x σ 1 x σ ɛ 1 t σ ɛ 1 t cos 2πt 1 sin 2πt cos 2πt 1 + sin 2πt which are smooth. Thus ɛ 1 is smooth from which we can conclude that ɛ n is smooth. So it remains to show that ɛ n is a covering map. To do this it again suffices to show that ɛ 1 is a covering map since coordinate-wise ɛ n is given by the ɛ 1. So consider e 2πit S 1 and set δ 1/4. Then ɛ 1 { e 2πiτ t δ < τ < t + δ } n Z t δ + n, t + δ + n Restricting ɛ 1 to one component of this disjoint union t δ + n, t + δ + n we see that ɛ maps it onto {e 2πiτ t δ < τ < t + δ} since e 2πit e 2πit+2nπi. Furthermore, ker ɛ 1 Z and so the restriction of ɛ 1 to t δ + n, t + δ + n is injective. Thus ɛ 1 : R S 1 is a smooth covering map and hence ɛ n : R n T n is a smooth covering map.

9 Lee 2-10Let CP n denote n-dimensional complex projective space, as defined in Problem 1-7. a Show that the quotient map π : C n+1 \ 0 CP n is smooth. b Show that CP 1 is diffeomorphic to S 2. 9 Claim 0.7. π : C n+1 \ 0 CP n is smooth Let f : R 2n+2 C n+1 be the diffeomorphism defined by x 1, y 1,..., x n+1, y n+1 x 1 + iy 1,..., x n+1 + iy n+1. Furthermore, let U i, ϕ i be the charts for CP n defined in problem 1-7. Then we have ϕ i π f x 1, y 1,..., x n+1, y n+1 ϕ i π z 1,..., z n where z j x j + iy j [ ϕ i z 1,..., z n] z 1 z,, zi 1, zi+1 i z i z,..., zn 1 z i But each component of this result is a rational function and hence is smooth on its domain of definition. Since this holds for all i we can conclude that π is smooth. Claim 0.8. CP 1 is diffeomorphic to S 2 Let U 1, ϕ 1 and U 2, ϕ 2 be the charts on CP 1 given by U 1 CP 1 \ {[0 : 1]} and U 2 CP 1 \ {[1 : 0]} with ϕ 1 [z 1 : z 2 ] z 2 z 1 and ϕ 2 [z 1 : z 2 ] z 1 z 2. These are of course the charts from problem 1-7 and so we know that they constitute an atlas. Next let V 1 S 2 \0, 0, 1 and V 2 S 2 \0, 0, 1. Then V 1, ψ 1 and V 2, ψ 2 constitute an atlas were ψ 1 a, b, c a 1 c + i b 1 c ψ 2 a, b, c a 1 + c i b 1 + c This is sterographic projection in complex notation and so there is no need to check that it is a smooth atlas this was done in last weeks homework. We now define f : CP 1 S 2 by f ϕ 1 i z ψ 1 i z. To see that this function is well defined we note that ψ1 1 z 1 ψ2 1 z 2 if and only if z 2 ψ 2 ψ1 1 z 1 z1 1 ϕ 2 ϕ 1 1 z 1 if and only if ϕ 1 z 1 ϕ 1 z 2. While

10 10 ψ 1 i z 1 ψ 1 i z 2 if and only if z 1 z 2 if and only if ϕ 1 i z 1 ϕ 1 i z 2 for all i. Since f is well defined on all of CP 1 we need to show that f is smooth and has a smooth inverse. To do this we note ψ i f ϕ 1 i z z ψ j f ϕi 1 z 1 z Furthermore, f 1 : S 2 CP 1 is clearly given by f 1 ψ 1 i for i j and z 0. z ϕ 1 i z. We know that f 1 is well defined by arguments completely symmetric to those given for f. Moreover, f f 1 Id and f 1 f Id by construction. Finally, we see that f 1 is smooth since ϕ i f 1 ψ 1 i z z ϕ i f 1 ψ 1 j z 1 for i j and z 0 z Thus f : CP 1 S 2 is smooth with smooth inverse and hence these two manifolds are diffeomorphic.