1 Differentiable manifolds and smooth maps. (Solutions)

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1 1 Differentiable manifolds and smooth maps Solutions Last updated: February Problem 1 a The projection maps a point P x y S 1 to the point P u 0 R 2 the intersection of the line NP with the x-axis Parametric equation of the line: P t tn +1 tp 1 tx t+1 ty 1 tx t1 y+y and the intersection with the x-axis is given by t y 1 y or 1 t 1 Hence 1 y u x 1 y valid for y 0 ie for all points of S 1 except for the north pole N 0 1 To find the inverse map ie for a given u R find x y S 1 it is possible to express x via u and y from the above formulas together with the equation of the circle x 2 + y 2 1 We obtain x 1 yu and plugging that into the equation of the circle we arrive at 1 y 2 u 2 + y 2 1 or 1 y 2 u 2 1 y 2 We can divide by 1 y because y 1 for P N so obtain 1 yu 2 1+y Finally we get 1 y 2 and y u2 1 This gives the desired answer: u 2 u 2 +1 b We have x 2u u y u2 1 u u cos θ 1 sin θ because x y cos θ sin θ on S 1 The function u uθ is smooth because sin θ 1 on S 1 \ {N} When 1 0 or 1 0 is chosen as a center of the projection instead of N further simplification is possible by the introduction of the half-angle Problem 2 a The coordinates are defined for all points P of S 2 except for N because the straight line NP is undefined if P N b The projection maps a point P x y z on S 2 R 3 to the point P u v 0 the intersection of the line NP with the plane z 0 Parametric equation of the line: P t tn + 1 tp 1 tx 1 ty t + 1 tz 1 tx 1 ty t1 z + z and the intersection with the plane z 0 is for t z or 1 t 1 Hence we have the formulas 1 z 1 z u x 1 z v y 1 z valid for z 0 ie for all points of S 2 except for the north pole itself as it should be by the geometrical meaning It is clear geometrically that 1

2 the intersection point P can be an arbitrary point of the plane ie the coordinates u v can be arbitrary and there is a one-to-one correspondence between S 2 \ {N} and R 2 To find the inverse map ie for given u v find x y z S 2 it is possible to express x and y via u v and z from the above formulas together with the equation of the sphere x 2 +y 2 +z 2 1 We obtain x 1 zu y 1 zv and plugging that into the equation of the sphere we arrive at 1 z 2 u 2 + v 2 + z 2 1 or 1 z 2 u 2 + v 2 1 z 2 We can divide by 1 z because z 1 for P N so obtain 1 zu 2 +v 2 1+z 2 Finally we get 1 z and z u2 +v 2 +1 This gives the desired u 2 +v 2 +1 u 2 +v 2 +1 answer: x 2u u 2 + v y 2v u 2 + v z u2 + v 2 1 u 2 + v As expected these formulas make sense for arbitrary u v R 2 Problem 3 For the south pole taken as the center of the projection everything is similar to the results obtained above see the solution to Problem 2 Since the north pole N and the south pole S exchange places under the transformation z z all the formulas can be obtained from the corresponding formulas for the stereographic projection from the north pole by a change of signs Answers: a The coordinates u S v S are defined for all points of S 2 except for the point S which is the center of projection b We have u S x 1 + z v S y 1 + z and conversely x 2u S u S 2 + v S y 2v S u S 2 + v S z 1 u S 2 v S 2 u S 2 + v S c Both coordinate systems u N v N and u S v S are defined on S 2 \{N S} To find the change of coordinates u N u N u S v S v N v N u S v S we substitute x y z as functions of u S v S to the expressions for u N and v N For example for u N we obtain u N x 2u S u S 2 +v S z u S 2 +v S u S 2 +v S 2 u S u S 2 +v S because 1 z 1 1 u S 2 v S 2 2u S 2 +v S 2 2 u S 2 +v S 2 +1 u S 2 +v S And similarly 2 +1 for v N So the change of coordinates u N u N u S v S v N v N u S v S has the form u N u S u S 2 + v S 2 v N v S u S 2 + v S 2 These formulas are defined for all u S v S except for u S v S 0 0 which are the coordinates of the north pole N in the u S v S -system Conversely the change of coordinates u S u S u N v N v S v S u N v N is given by u S u N u N 2 + v N 2 v S 2 v N u N 2 + v N 2

3 These formulas are defined for u v 0 0 which are the coordinates of the south pole S in the u v coordinate system Problem 4 equation Consider the unit sphere S n R n+1 It is specified by the x x n 2 + x n Let x x 1 x n R n and denote x n+1 z We write P x z for P x 1 x n+1 R n+1 We take the point N 0 1 ie as the center of the projection There are two maps: F : S n \ {N} R n and F 1 : R n S n \ {N} If NP is the line through N and P x z S n \ {N} and P u 0 is the intersection of NP with the z 0 plane then F : P u and F 1 : u P Here u u 1 u n is a point in R n To find the explicit formulas we write a point P t on the line NP as P t tn + 1 tp 1 tx t + 1 tz The last coordinate vanishes when t z hence 1 t 1 We obtain for F : x z u 1 z 1 z u x 1 z For the inverse map we solve for x and z having in mind the relation x 2 + z 2 1 We have x u1 z hence we arrive at the quadratic equation denoting 1 z w: u 2 w w 2 1 or 1 + x 2 w 2 2w 0 from which we obtain the unique solution w 2/ u since w 1 z 0 and therefore x 2u u z u 2 1 u We have arrived at two mutually inverse maps establishing a bijection between S n \ {N} and R n The map F 1 : R n S n \ {N} can be viewed as a chart for S n We can change notation and denote it ϕ N : F 1 So we have ϕ N : R n N Sn \ {N} ϕ N : u N ϕ 1 N : x z u N 2uN u N u N 2 1 u N x 1 z We have introduced the subscript N as a label of the chart By considering the south pole S 0 1 as the center of the projection instead of N we can construct the chart ϕ S : R n S Sn \ {N} in the same way The formulas differ by the sign of the last coordinate only since N and 3

4 S are swapped under z z: 2uS ϕ S : u S u S u S u S ϕ 1 S : x z u S x 1 + z These two charts make an atlas because S n \ {N} S n \ {S} S n On the intersection S n \ {N} S n \ {S} S n \ {N S} we can calculate the changes of coordinates ie the expression of u N via u S and vice versa Note that S n \{N S} corresponds to R n \{0} at each chart One can find directly: u N u S u S 2 check! This is a smooth map even analytic therefore we have a structure of a smooth manifold for S n Problem 5 The state of the planar pendulum is entirely defined by the position of its moving end in the plane R 2 Since the length is fixed the end is always at a constant distance from the origin O ie is on the circle of center O and certain radius R It can be identified with the standard unit circle by a change of scale All positions on the circle are possible Therefore the configuration space of the planar pendulum is the manifold S 1 Problem 6 Solution for arbitrary n We have RP n {x 1 : : x n+1 } where x 1 : : x n+1 stands for the equivalence class of a non-zero vector x 1 x n+1 R n+1 Two non-zero vectors are considered equivalent if they are proportional Define subsets U k : {x 1 : : x n+1 x k 0} For convenience introduce copies of R n denoting them R n k as subsets of R n+1 where the kth coordinate equals 1: R n k : {u 1 k 1 u n+1 k } here 1 stands on the kth place We used subscript for denoting the number of chart and put it in brackets to avoid confusion Define maps ϕ k : R n k U k ϕ k u 1 k 1 u n+1 k : u1 k : : 1 : : u n+1 k The inverse maps are given by the formulas x ϕ 1 k x1 : : x n+1 1 xn+1 1 xk x k 4

5 The overlap of two charts U k U l {x 1 : : x n+1 x k 0 and x l 0} corresponds to the open subsets {u l k 0} Rn k and {uk l 0} Rn l To find the transition functions changes of coordinates ϕ kl ϕ 1 k ϕ l : {u l R n l u k l 0} {u k R n k u l k 0} we write down the equation ϕ k u k ϕ l u l ie u 1 k : : 1 : : u n+1 k u1 l : : 1 : : u n+1 l where 1 appears at the kth place in the LHS and at the lth place in the RHS from where we obtain where ϕ kl : u 1 l 1 u n+1 l u 1 k 1 u n+1 k u i k ui l u k l for all i Two special cases of this formula are i k where we get u k k 1 as it should be and i l where u l k 1 since u l u k l 1 l In small dimensions we have two charts for RP 1 and three charts for RP 2 For RP 1 the charts are ϕ 1 : R U 1 where U 1 {x 1 : x 2 x 1 0} ϕ 1 : x 1 1 : x 1 and ϕ 2 : R U 1 where U 2 {x 1 : x 2 x 2 0} ϕ 2 : x 2 x 2 : 1 and the changes of coordinate are x 2 1 x 1 x 1 1 x 2 For RP 2 the charts are ϕ 1 : R 2 U 1 where U 1 {x 1 : x 2 : x 3 x 1 0} ϕ 1 : x 1 y 1 1 : x 1 : y 1 ϕ 2 : R 2 U 2 where U 2 {x 1 : x 2 : x 3 x 2 0} ϕ 2 : x 2 y 2 x 2 : 1 : y 2 and ϕ 3 : R 2 U 3 where U 3 {x 1 : x 2 : x 3 x 3 0} ϕ 3 : x 3 y 3 x 3 : y 3 : 1 so that the changes of coordinates are 1 x 1 y 1 y 2 x 2 x 2 1 x 2 y 2 y 1 x 1 x 1 1 x 3 y 3 x 1 y 1 y 1 y3 x 3 x3 y 3 x2 y 2 1 x 3 1 y 3 1 y 2 5

6 For local coordinates on RP 1 and RP 2 we used a traditional notation such as x or x y with a superscript indicating the chart number instead of the notation u i k used for general RP n Problem 7 For CP n everything is literally the same as for RP n with C n instead of R n in the coordinate charts Note that C n may be regarded as R 2n by replacing each complex coordinate by its real and imaginary parts The resulting changes of coordinates are complex-analytic Problem 8 The general reason why the identifications of RP 1 with S 1 and CP 1 with S 2 arise is that in suitable atlases the descriptions of the corresponding manifolds in terms of local coordinates and their transformations are exactly the same a For RP 1 consider the standard atlas described above There are two charts; denote the single coordinate in either of them by u 1 and u 2 Both variables take all real values and the transformation of coordinate is defined for non-zero values: u 1 1 u 2 For S 1 consider the stereographic atlas Again there are two 1-dimensional charts with the coordinates u N and u S resp that take all real values; and the change of coordinate has exactly the same form as for RP 1 : u N 1 u S Therefore we may identify RP 1 and S 1 by the identification of coordinates: u 1 u N and u 2 u S which is consistent with the changes of coordinates More formally we may view these formulas as defining a diffeomorphism RP 1 S 1 An alternative and equally good choice is to set u 2 u N hence u 1 u S Such two identifications differ by the antipodal symmetry of S 1 which swaps the two poles b For CP 1 consider the standard atlas A point w 1 : w 2 CP 1 has the complex coordinate w 1 w2 in the chart where w 1 0 or the w 1 complex coordinate w 2 w1 in the chart where w 2 0 On the intersection w 2 w 2 1 w 1 For S 2 consider the stereographic atlas see Problems 2 and 3 We may assemble the coordinates u N v N into a complex number w u N + iv N and u S v S into a complex number w u S + iv S Then the change of coordinates is w w w 1/ w Up to the complex conjugation w 2 w w it is the same as the change of coordinates on CP 1 Therefore we may introduce an identification of CP 1 with S 2 for example by the identification of coordinates w w 2 w1 hence w w w w There are three other equally good choices corresponding to swapping the poles of S 2 and applying the complex conjugation to CP 1 c In terms of homogeneous coordinates on RP 1 and CP 1 which are not local coordinates! and the standard coordinates x y and x y z on the ambient Euclidean spaces R 2 S 1 and R 3 S 2 resp the formulas are as follows: S 1 RP 1 x y x : 1 y RP 1 S 2 CP 1 x y z x + iy : 1 z CP 1 6

7 and RP 1 S 1 x 1 : x 2 2x 1 x 2 x y x x 2 x1 2 x x x 2 2 CP 1 S 2 w 1 : w 2 2w1 w 2 x + iy z w w 2 w1 2 w w w 2 2 To get the first set one writes x 1 : x 2 x 1 : 1 u : 1 x : 1 so x 2 1 y finally x 1 : x 2 x 1 y and similarly for the complex case To get the second set one has to plug u x1 or u+iv w1 into the inverse formulas for x 2 w 2 the stereographic projection and get rid of the denominators In particular for cos θ : sin θ RP 1 we get as the image x y sin 2θ cos 2θ S 1 It is instructive to express the diffeomorphisms RP 1 S 1 and CP 1 S 2 in this way which is independent of local coordinates! Problem 9 a Consider the special linear group { } a b SL2 g det g 1 c d For 2 by 2 matrices the equation det g 1 can be explicitly resolved wrt each of the matrix entries Consider the subsets U a {g SL2 a 0} and U b {g SL2 b 0} Define open sets V a : {a b c R 3 a 0} and V b : {a b d R 3 b 0} There are maps ϕ a : V a U a ϕ b : V b U b a b a b c c d a b a b d c d where d 1 + bc a where c ad 1 b They are invertible the inverses are just the projections g a b c and g a b d and hence can be taken as charts for SL2 Note that U a U b SL2 because a and b cannot simultaneously vanish on SL2 On the intersection U a U b both a and b are non-zero and the change of coordinates is given by the smooth formulas ϕ 1 a ϕ b : a b d a b c a b ad 1 b defined on an open subset of V b Hence SL2 is a manifold of dimension 3 b For SLn n > 2 one can guess that it is a manifold of dimension n 2 1 To prove that it is possible to follow closely the above proof for n 2 by considering the first row expansion of det g recall that the coefficients of the entries g 1k are ±M k where M k is the minor obtained by deleting the first row and the kth column of g One can introduce analogs of the charts above 7

8 and show that they make a smooth atlas for SLn There is however an alternative proof which we shall consider later Problem 10 A k-dimensional linear subspace L in R n can be specified practically in two ways: either as the span of k linearly independent vectors from R n which make a basis in L or as the solution space of a system of n k independent linear equations ie of rank n k Sometimes one way is more convenient and sometimes there is no preference We use the notation G k R n for the set of all k-dimensional linear subspaces in R n a A 2-plane through the origin in R 3 can be conveniently specified by an equation ax + by + cz 0 where at least one coefficient is non-vanishing Since multiplying the equation by a non-zero factor does not change the plane we may identify the 2- planes with the equivalence classes a : b : c ie with the points of the projective space RP 2 In particular G 2 R 3 which we identified with RP 2 is a manifold of dimension 2 b To study 2-planes in R 4 it is more convenient to describe them using bases Suppose linearly independent vectors a 1 and a 2 span a 2-plane in R 4 We may write them as a matrix 2 by 4 a11 a A 12 a 13 a 14 a 21 a 22 a 23 a 24 of rank 2 That means that there is a non-vanishing 2 by 2 minor Two pairs of independent vectors a 1 a 2 and b 1 b 2 span the same subspace L if and only if b 1 b 2 is obtained from a 1 a 2 by an invertible linear transformation: b1 g11 a 1 + g 12 a 2 g11 g 12 a1 g 21 a 1 + g 22 a 2 g 21 g 22 a 2 b 2 In the matrix form we may express that as A ga where g GL2 and A and A specify the same subspace L if and only if A ga Therefore the set of 2-planes linear subspaces in R 4 can be identified with the set of equivalence classes of 2 4 matrices A of rank 2 as above wrt the equivalence relation A ga for all g GL2 The analogy with the projective space is clear: one takes matrices with linearlyindependent rows instead of non-zero vectors and multiplies/ divides them by invertible square matrices instead of non-zero numbers In detail charts for G 2 R 4 can be introduced as follows Define U 34 G 2 R 4 as the set of planes specified by matrices A for which the square matrix made of the third 8

9 and fourth columns of A is invertible this does not change under the left multiplication by g GL2 There is a chart [ ] ϕ: Mat2 2 U 34 G 2 R 4 b11 b B 12 b11 b [A] b 21 b 22 b 21 b the square brackets denote the equivalence class for which 1 ϕ 1 b11 b : U 34 Mat2 2 [A] B 12 a13 a 14 a11 a 12 b 21 b 22 a 23 a 24 a 21 a 22 is the inverse map We may put extra labels on ϕ and B denoting them as ϕ 34 and B 34 and introduce similar charts ϕ 12 ϕ 13 ϕ 23 and ϕ 24 corresponding to other choices of columns of A giving invertible 2 2 submatrices It is clear that the union of all images of these charts is the whole G 2 R 4 because of the rank condition satisfied by the matrices A So it is an atlas As for the changes of coordinates they are given by algebraic operations with matrix entries hence are smooth We conclude that G 2 R 4 is a smooth manifold of dimension c In general the Grassmann manifold G k R n is indeed a manifold of dimension k n k The equivalence classes of matrices A Matk n of rank k can be considered as homogeneous coordinates A A A ga for g GLk A coordinate chart can be obtained by dividing A from the left by an invertible k k submatrix with fixed column numbers j 1 < < j k so that the genuine inhomogeneous coordinates in each such chart are given by the entries of a k n k matrix There are n k such charts corresponding to the n k choices of k columns in A The alternative description of k-dimensional subspaces using n k independent equations gives rise to the identification G k R n G n k R n The complex case is similar 9

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