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MA455 Manifolds Exercises III Solutions May 2008 1. Let V and W be oriented vector spaces of the same dimension, and let E and F be positive bases for V and W respectively. Show that an isomorphism f : V → W preserves orientation if E det[f ]E F > 0 and reverses orientation if det[f ]F < 0. Solution: f preserves orientation iff f (E) is a positive basis of W . The change of basis matrix f (E) [I]F is equal to [f ]E F. 2. (i) Example 3.10(i) claims to orient the preimage M := f −1 (t) of a regular value t of a smooth function f : Rn+1 → R, by stipulating that a basis v1 , . . . , vn for Tx M is positive if the basis ∇f (x).v1 , . . . , vn of Rn+1 is positive. Prove that this definition satisfies the compatibility condition for an orientation, 3.7(iii). Solution: If v1 , . . . , vn is a tangent frame field for M on U , then ∇f, v1 , . . . , vn is a tangent frame field for Rn+1 . Hence det(∇f (x), v1 (x), . . . , vn (x)) 6= 0 for all x ∈ U . So by continuity of the determinant (and the connectivity of U ), det(∇f (x), v1 (x), . . . , vn (x)) > 0 for all x ∈ U or det(∇f (x), v1 , . . . , vn ) < 0 for all x ∈ U . In the first case, v1 (x), . . . , vn (x) is a positive basis for Tx M for all x ∈ U , and In the second case, v1 (x), . . . , vn (x) is a negative basis for Tx M for all x ∈ U . (ii) Ditto for the orientation defined in 3.10(ii). Solution: Essentially the same as (i). 3.Prove that the antipodal map S 1 → S 1 preserves orientation but that the antipodal map S 2 → S 2 reverses orientation. Solution: We orient both spheres by using the outward pointing unit normal. Then v is positive basis for Tx S 1 if x, v is positive basis for R2 . As the antipodal map A is the restriction of the linear map multiplication by −1 on the ambient space, its derivative is also multiplication by −1. Multiplying both columns of a 2 × 2 matrix by −1 leaves its determinant unchanged, so =⇒ v + ive basis for Tx S 1 −x, −v + ive basis for R2 =⇒ =⇒ x, v + ive basis for R2 . −v + ive basis for T−x S 1 So A : S 1 → S 1 preserves orientation. On the other hand multiplying all three columns of a 3 × 3 determinant by −1 has the effect of multipplying its determinant by −1, so =⇒ v1 , v2 + basis forTx S 2 −x, −v1 , −v2 − basis forR3 =⇒ =⇒ x, v1 , v2 + basis forR3 −v1 , −v2 − basis forT−x S 2 so A : S 2 → S 2 reverses orientation. 5. Let M m and N n be oriented manifolds, and give M × N and N × M the product orientation. Prove that the map i : M × N → N × M, i(x, y) = (y, x) preserves or reverses orientation according to whether mn is even or odd. Solution: Let v1 , . . . , vm be a positive basis for Tx M and w1 , . . . , wm be a positive basis for Ty N . Then in the product orientation on M × N , (v1 , 0), . . . , (vm , 0), (0, w1 ), . . . , (0, wn ) is a positive basis for T(x,y) M × N . Applying the derivative of the map M × N → N × N , we get the basis (0, v1 ), . . . , (0, vm ), (w1 , 0), . . . , (wn , 0). To transform this to the positive basis (w1 , 0), . . . , (wn , 0), (0, v1 ), . . . , (0, vm ) requires mn transpositions (try it!). Hence the change of basis matrix has determinant equal to (−1)mn . 6. Suppose that M is an orientable manifold, and suppose given two, possibly distinct, orientations of M . (i) Let U be a connected subset of the domain of a chart. Show that either the two orientations agree everywhere on U , or they disagree everywhere on U . Hint: find a frame field on U . (ii) Show that if the two orientations agree at x ∈ M , then they agree on a neighbourhood of x, and if they disagree at x, then they disagree on a neighbourhood of x. (iii) Prove that if M is connected then the two orientations either agree everywhere or disagree everywhere. Hint: by (ii), the set of points where they agree is open, and so is the set of points where they disagree. (iv) Conclude that on a connected orientable manifold there are just two possible orientations. Solution: Suppose given two orientations. Each point x in M has a connected neighbourhood V on which a frame field F is defined. F is either everywhere positive or everywhere negative for each of the two orientations. So either they agree everywhere on V , or they disagree everywhere on V . It follows that the set of points in U where they agree is open, and so is the set of points where they disagree. As U is connected, one or other is all of U . If they differ, each orientation is obtained by reversing the other. So there are exactly two. 7. Do exercise 3.7: (i) Show that if M ⊂ Rn is a smooth manifold then for each x0 ∈ M there is neighbourhood U of x0 such that for all x ∈ U , the orthogonal projection π : RN → Tx0 M restricts to give an isomorphism πx : T x M → T x 0 M . (ii) Show that the local compatibility condition 3.5(iii) is equivalent to the condition that every x0 ∈ M has a neighbourhood U such that for all x ∈ U , πx : Tx M → Tx0 M not only is an isomorphism but also preserves orientation. Solution:(i) Let F = v1 , . . . , vm be a frame field on a neighbourhood of x. Let E = e1 , . . . , em be a basis for Tx0 M . The matrix A(x) = π(v1 (x)· · ·· · ·π(vm (x)) E whose columns are the expressions of the projected vectors π(vi (x)) with respect to the basis E, varies smoothly with x. This is most obvious if E is orthonormal, for then the entries in A(x) are just the dot products vi (x) · ej . When x = x0 , A(x) is invertible. Hence there is a neighbourhood U of x in M on which A continues to be invertible. For x ∈ U , π : Tx M → Tx0 M is an isomorphism. F (x) (ii) A(x) = [dx π]E . 8. (i) Suppose that f : M → N is a map of oriented manifolds of the same dimension, with M connected. Suppose that dx f is an isomorphism for all x ∈ M . Show that either for all x ∈ M dx f preserves orientation, or for all x ∈ M dx f reverses orientation. (ii) Let N be the image of the map f : S 2 → R6 of Example 1.34. Show that N is not orientable by considering the diagram S2 A A AA AA A f AA N / S2 } } }} }} f } }~ where A is the antipodal map. 9. Find an explicit diffeomorphism from some neighbourhood U in B n+1 of the point N = (0, . . . , 0, 1), to an open set in the half-space H n+1 . Solution: Let U = B n+1 ∩ {xn+1 > 0} be the upper half-ball. Define φ : U → H n+1 by s X x2j . φ(x1 , . . . , xn , xn+1 ) = x1 , . . . , xn , xn+1 − 1 − j=1,... ,n 10. Show that if M m is a smooth manifold, and f : M → R has t0 as a regular value, and f −1 (t0 ) 6= ∅, then f −1 ((−∞, t0 ]) and f −1 ([t0 , ∞)) are m-dimensional manifolds with boundary, and each has boundary equal to f −1 (t0 ). Solution: Consider M+ := f −1 ([0, ∞)). The set of points x where f (x) > t0 is already a manifold: since it’s open in M , you can use the existing charts on M , restricting their domains suitably. These will give diffeos of open sets in M+ to open sets in Rm ; composed with a suitable embedding of Rm in H m , such as (x1 , . . . , xm ) 7→ (x1 , . . . , xm−1 , exp(xm )), they become diffeos of open sets in M+ to open sets in H m . In the neighbourhood of a point where f (x) = t0 , use the local normal form of a submersion: there is a chart φ on M around x such that f ◦ φ−1 (x1 , . . . , xm ) = xm . The restriction of φ to M+ gives a diffeomorphism of a neighbourhood of x in M+ to an open set in H m , and sends M+ ∩ f −1 (t0 ) to ∂H m . 11. Show that if M m is a manifold and N n is a manifold with boundary then M × N is an m + ndimensional manifold with boundary, and ∂(M × N ) = M × ∂N . What if both M and N have non-empty boundary? 12. Prove Lemma 3.20. 13. Prove that if f : M m → N n and Z n−m ⊂ N with all manifolds oriented, M compact and Z closed in N , and if g : M 0 → M is a diffeomorphism, then if g preserves orientation everywhere, (f ◦ g · Z)N = (f · Z)N and if g reverses orientation everywhere, (f ◦ g · Z) = −(f · Z). (This is used in the proof of 3.23). Solution: Clearly g determines a bijection (f ◦ g)−1 (Z) → f −1 (Z). Let x0 ∈ (f ◦ g)−1 (Z), let x = g(x0 ) and y = f (x). Let E 0 be a positive base for Tx0 M 0 , write E = dx0 g(E 0 ), and F a positive base for Ty Z. If g preserves orientation, then E is a positive base for Tx M , and sign(dx f (E), F ) = (f · Z)x . As dx f (E) = dx0 (f ◦ g)(E 0 ), this shows that (f ◦ g · Z)x0 = (f · Z)x . If g reverses orientation, then E is a negative base for Tx M , and sign(dx f (E), F ) = −(f · Z)x . This shows that (f ◦ g · Z)x0 = −(f · Z)x . 14. Prove Lemma 3.31: Suppose F : W p → N n is a smooth map, and Z is a submanifold of N of codimension k. If F tZ and ∂F tZ, then F −1 (Z) is a p − k-dimensional manifold with boundary, and ∂(F −1 (Z)) = (∂F )−1 (Z). Hint: after composing F with the inverse of a chart on W , you can replace W by an open set in H p . Use the local normal formm of a submersion. Solution: As in Exercise 10, for points in F −1 (Z) away from ∂W , there is not much to say - just use Corollary 1.29. However in the neighbourhood of a point x0 ∈ (∂F )− −1(Z) = F −1 (Z) ∩ ∂W , we need a new argument. Let φ be a chart on W around x0 and write F̃ = F ◦ φ−1 , x̃0 = φ(x0 ). F̃ , defined on an open set in H m , has a local smooth extension G to a neighbouhhood U of x0 in Rm . Provided U is chosen small enough, GtZ, and so G−1 (Z) is a smooth manifold. Then F̃ −1 (Z) is the intersection of G−1 (Z) with H m . Let f : G−1 (Z) → R be the restriction of the m’th coordinate and is thus f −1 ([0, ∞)). Then F̃ −1 (Z) = f −1 ([0, ∞)). To show that F̃ −1 (Z) is a smooth manifold with boundary equal to ∂ F̃ −1 (Z), all we need do is apply the result of Exercise 7 - after checking that the hypothesis applies, of course. That is, we must show that f is a submersion at x0 . The kernel of dx0 f is the “horizontal” part of Tx̃0 G−1 (Z) - its intersection with {xm = 0}. But this intersection is equal to (dx̃0 ∂ F̃ )−1 (Ty Z), and has dimension one less than the dimension of G−1 (Z), by transversality. This shows that f is a submersion at x0 . 15. (i) Check that the sequence j i 0 → R2 −→ R5 −→ R3 → 0 is exact, when i and j are the linear maps with matrices 1 3 2 −1 0 0 1 0 0 0 0 3 2 0 −7 0 and 1 1 0 1 0 −2 3 0 1 (1) . (ii) Let a1 = (1, 0), a2 = (0, 1) be the standard basis of R2 , and give R2 the standard orientation (so that this basis is positive). Give R4 the standard orientation also. Find three vectors b1 , b2 , b3 ∈ R5 such that i(a1 ), i(a2 ), b1 , b2 , b3 is a positive basis for R5 . Solution: (i) Exactness can be checked by first checking that j ◦ i = 0, and then that dim ker j = dim im i. (ii) The two columns of the matrix of i are i(a1), i(a2 ). The three additional vectors (0, 0, 1, 0, 0), (0, 0, 0, 1, 0) and (0, 0, 0, 0, 1), together with i(a1 ) and i(a2 ), make up a basis for R5 . Its sign is that of the determinant 1 1 0 0 0 2 −1 0 0 0 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 which is negative. So we get a positive basis by interchanging,say, the last two. Thus, we take b1 = (0, 0, 1, 0, 0), b2 = (0, 0, 0, 0, 1), b3 = (0, 0, 0, 1, 0). (iii) Let R3 also have the standard orientation. What is the sign of the short exact sequence (1)? 4 Solution: It is the sign of the basis j(b1 ), j(b2 ), j(b3 ) of R3 . That is, the sign of the determinant of the last three columns of the matrix of j, with the last two columns interchanged. Since 1 0 0 0 0 −7 > 0 0 3 −2 the s.e.s. is positive. (iv) Now orient R2 by taking the standard basis to be positive, and orient R3 by taking as positive the basis c1 = (1, −1, 2), c2 = (−1, −1, 3), c3 = (0, 2, 5). Orient R5 (by exhibiting a positive basis) so that the sequence is positive. Solution: The new orientation of R3 is opposite to the old, since 1 −1 0 −1 −1 2 < 0, 2 3 5 so since the s.e.s. was positive with respect to the old orientation of R3 , to restore positivity now we reverse the orientation of R5 . For example we could take as positive basis i(a1 ), i(a2 ), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0 Another correct procedure would be to find b̃1 , b̃2 , b̃3 in R5 such that j(b̃i ) = ci and take as positive basis for R5 the vectors i(a1 ), i(a2 ), b̃1 , b̃2 , b̃3 . (v) (v) Now orient R5 and R3 by taking as positive the bases b1 = (0, 1, 0, 0, 0), b2 = (0, 0, 1, 0, 0), b3 = (0, 0, 0, −1, 0), b4 = (1, 1, 1, 1, 0), b5 = (1, 2, 3, 4, 5) and c1 = (1, −2, 2), c2 = (2, 0, 0), c3 = (0, 1, 2) respectively. As explained in the lecture notes, there is a unique orientation on R2 making the sequence (1) exact. Is it the standard orientation or the opposite orientation? 17. Suppose that A, B and C are oriented vector spaces, and that 0 → A → B → C → 0 is a positive short exact sequence. Let f : A → A0 , g : B → B 0 and h : C → C 0 be isomorphisms, and suppose A0 , B 0 and C 0 are all oriented (though we do not suppose that f , g and h necessarily preserve orientation). Then the procedure of the previous exercise gives short exact sequences 0 → A0 → B → C → 0, 0 → A → B0 → C → 0 and 0 → A → B → C 0 → 0. Give conditions on f, g and h for these to be positive. When is the exact sequence 0 → A0 → B 0 → C 0 → 0, constructed by using the same procedure three times, positive? Solution: The four sequences can be seen in the diagrams 0 → & A l A0 0 → A → % → B B → → & C → 0 0 → A → & B l B0 → % C → 0 C l C0 → % 0 0 → & A l A0 → B l B0 → C l C0 → % 0 5 → → The sign of 0 → A0 → B → C → 0 is the same as the sign of 0 → A → B → C → 0 if A → A0 preserves orientation, and opposite otherwise. Similarly for the next two cases. The last is the product of three replacements, and has the same sign as 0 → A → B → C → 0 iff an even number of the isomorphisms A → A0 , B → B 0 and C → C 0 preserve orientation. 18. Suppose f : M → N , Z is a submanifold of N , and f tZ, with f −1 (Z) 6= 0. Explain how to construct the short exact sequence 0 → Tx f −1 (Z) → Tx M → Ty N/Ty Z → 0 (used in the definition of the preimage orientation 3.26(iii)). Solution: Begin with dx f Tx M −→ Tf (x) N. The map Tx M → Tf (x) N/Tf (x) Z in the s.e.s is the composite Tx M dx f −→ Tf (x) N ↓ Tf (x) N/Tf (x) Z where the vertical map is projection to the quotient. The formula in 1.44 in the Lecture Notes, defining transversality, shows that this composite is surjective. Moreover, 1.45 shows that Tx f −1 (Z) is its kernel. This completes the proof of exactness. 19. Suppose f : M m → N n is smooth and Z n−m ⊂ N , with M compact, Z closed in N , and all manifolds oriented. Let f˜ : M → M × N be the map x 7→ (x, f (x)), whose image, of course, is the graph of f . (i) Find a natural bijection f −1 (Z) → f˜−1 (M × Z), and show that f t Z if and only if f˜ t M × Z. (ii) Let v1 , . . . , vm be a positive basis for Tx M . Write down the image of this basis under dx f˜. (iii) Show that if we give M × Z and M × N the product orientations, then (f˜ · M × Z)M ×N if m is even (f · Z)N = −(f˜ · M × Z)M ×N if m is odd Solution:(i) The bijection f −1 (Z) → f˜−1 (M × Z) is given by x ↔ (x, f (x)). (ii) The image of v1 , . . . , vm is v1 , dx f (v1 ) , . . . , vm , dx f (vm ) . (iii) Let z1 , . . . , zn−m be a positive basis for Tf (x) Z. Then assuming f (x) ∈ Z, (v1 , 0), . . . , (vm , 0), (0, z1 ), . . . , (0, zn−m is a positive basis for T(x,f (x) M × Z. We have (f˜ · M × Z)x = sign of v1 , dx f (v1 ) , . . . , vm , dx f (vm ) , (v1 , 0), . . . , (vm , 0), (0, z1 ), . . . , (0, zn−m ) as basis of T(x,f (x) M × N . This basis is equivalent to 0, dx f (v1 ) , . . . , 0, dx f (vm ) , (v1 , 0), . . . , (vm , 0), (0, z1 ), . . . , (0, zn−m ) 2 and (−1)m transpositions convert it to (v1 , 0), . . . , (vm , 0), 0, dx f (v1 ) , . . . , 0, dx f (vm ) , (0, z1 ), . . . , (0, zn−m ). 6 By definition of the product orientation on M × N , this basis is positive iff the basis dx f (v1 ), . . . , dx f (vm ), z1 , . . . , zn−m of Tx N is positive, i.e. iff f, Z x = +1. Hence 2 (f · Z)x = (−1)m f˜ · M × Z x 20. Suppose that f : M m → N m is a smooth map, with M compact, and M and N oriented. For a fixed y ∈ N , let iy : M → M × N be the map iy (x) = (x, y). Show that y is a regular value of f if and only if iy t graph(f ). What is the precise relation between deg(f ) and (iy ·graph(f ))M ×N ? Solution: This can be solved by a method very like the last exercise — good practice if you haven’t already done it. There is a slick way, viewing it as a special case of the last exercise: 2 2 deg(f ) = f · {y} N = (−1)m f˜ · (M × {y}) M ×N = (−1)m graph(f ) · (M × {y}) M ×N 2 2 = (−1)mn (−1)m (M × {y}) · graph(f ) M ×N = (−1)mn (−1)m (iy · graph(f ))M ×N . 7