MA455 Manifolds Solutions 1 May 2008

MA455 Manifolds Solutions 1 May 2008 1. (i) Given real numbers a < b, find a diffeomorpism (a, b) R. Solution: For example first map (a, b) to (0, π/2) and ten map (0, π/2) diffeomorpically to R using te function tan. (ii) Find a diffeomorpism (0, ) R. Solution: Use log : (0, ) R. (iii) Let f : R R be a smoot map, and let grap(f) be te set {(x, y) R R : y = f(x). Sow tat te map x (x, f(x)) is a diffeomorpism from R to grap(f). Solution: Te map x (x, f(x)) is clearly smoot and as smoot (new) inverse obtained as te restriction to grap(f) of te smoot (old) projection (x, y) x. (iv) Let C 0 be te cylinder S 1 (0, ) and let C be te cylinder S 1 R (you can draw bot in R 3 ). Find diffeomorpisms C 0 R 2 \ {0 and C R 2 \ {0. Solution: C 0 = {(x, y, z) R 2 : x 2 + y 2 = 1, 0 < z <. Define a smoot map φ : C 0 R 2 \ {(0, 0) by φ(x, y, z) = (zx, zy). I leave you to find its inverse. Te diffeomorpism of (ii) enables you to define a diffeo C 0 = (0, ) S 1 ψ R S 1 = C, and now φ ψ 1 is a diffeo C R 2 \ {(0, 0). 2. (i) Let n be te regular n-simplex {(t 1,..., t n+1 ) R n+1 : i t i = 1, t i 0 for all i. Draw 2. Solution: (ii) Let B n be te set {(x 1,..., x n ) R n : 0 x 1 x n 1. Find a diffeomorpism B n n. Hint: given (x 1,..., x n ) B n, tere is a natural coice of an (n + 1)-tuple of non-negative numbers wit sum equal to 1. Solution: (x 1,..., x n ) (x 1, x 2 x 1,..., x n x n 1, 1 x n ). 3. Sow tat if U R n is open, f : U R n is smoot, and det(d x f) 0 for all x U, ten f(u) is open in R n. Note tat in particular tis olds if f : U f(u) is a diffeomorpism. Solution: Use te inverse function teorem. 4. Let X 0 be te subset of R 2 consisting of te tree lines {x 1 = 0, {x 2 = 0 and {x 1 = x 2. Let l 1, l 2 and l 3 be any tree distinct lines troug (0, 0) in R 2, and let X = l 1 l 2 l 3. Find a diffeomorpism indeed, a linear isomorpism taking X 0 to X. Solution: Let v 1 and v 2 be non-zero vectors lying on l 1 and l 2 respectively. Define a linear isomorpism R 2 R 2 by sending (1, 0) to v 2, (0, 1) to v 1 and extending linearly. Tis maps {x 1 = 0 to l 1, {x 2 = 0 to l 2 and {x 1 = x 2 to a tird line. A linear isomorpism wose matrix in te basis v 1, v 2 is diagonal will re-scale te two axes so tat

tis line becomes te line l 3. 5. Sow tat if X Y R n are manifolds and i : X Y is inclusion ten for eac x X, d x i : T x X T x Y is also inclusion. It s obvious if you use te rigt definition of derivative. Solution: Use te definition of d x i as te map sending γ (0) to i γ (0) (version (i) of definition 1.13). 6. i) Let f : X Y X be te projection. Sow tat d (x,y) f : T (x,y) X Y T x X is also projection. ii)fixing any y Y gives an injection mapping f : X X Y, defined by f(x) = (x, y). Sow tat d x f(ˆx) = (ˆx, 0). iii) Let f : X Y, g : X Y be smoot maps, and define f g : X Y X Y by f g(x, y) = (f(x), g(y)). Sow tat d (x,y) (f g) = d x f d y g. Solution: (i) Suppose tat X R N and Y R P. Te projection X Y X is te restriction to X Y of te projection π : R N R P R N. So its derivative is te restriction to T (x,y) X Y = T x X T y Y of te derivative of π. But π is linear, so d (x,y) π = π. (ii) Use Definition 1.13(i). (iii) Ceck first tat tis is true for smoot (old) maps. Ten if F and G are smoot (old) extensions of f and g respectively, F G is a smoot (old) extension of f. So d (x,y) f g is te restriction to T (x,y) X Y = T x X T y Y of d (x,y) F G, wic you ave cecked is equal to d x F d y G. 7. (i) Prove tat R k and R l are not diffeomorpic if k l. Solution: Suppose g : R l R k is a smoot inverse to f tat is, g f = id R k, f g = id R l. Ten for eac x R k and y R l, we ave d f(x) g d x f = d x id R k = id R k and d g(y) f d y g = d x id R l = id R l. Tus bot derivatives are linear isomorpisms, and so k = l. (ii) Prove tat smoot manifolds M, N of dimension k, l cannot be diffeomorpic if k l. Solution: Let x M and y = f(x) N, and let φ and ψ be carts around x and y respectively. If f is a diffeomorpism ten so is ψ f φ 1. Now te argument of (i) applies, and sows k = l. 8. Prove tat if M k R n and N l R p are smoot manifolds ten M N R n+p is a smoot manifold of dimension k + l. Solution: If (x, y) M N, let φ : U 1 V 1 be a cart on M around x and let φ 2 : U 2 V 2 be a cart on N around y. Ten U 1 U 2 is a neigbourood of (x, y) in M N and φ 1 φ 2 : U 1 U 2 V 1 V 2 is a diffeomorpism. As V 1 V 2 is open in R k R l = R m+l, M N is a manifold of dimension k + l. 9. A Lie group is a manifold G wic is also a group, for wic te operations of { { p : G G G i : G G multiplication: and inversion: (g 1, g 2 ) g 1 g 2 g g 1 are smoot maps. Sow tat Gl(n, R) := {invertible n n real matrices under multiplication Mat n (R) = R n2 is a Lie group. Solution: Gl(n, R) is an open subset of matrix space, wic is naturally identified wit R n2. Te group operations are matrix multiplication and matrix inversion. Eac entry in te product of two matrices is a sum of products of entries of te two, and tus is a polynomial. And polynomials are smoot. Te entries in A 1 are quotients of polynomials in te entries of A, wit nowere-vanising denominator. Hence tese too are smoot functions of te entries of A. So bot group operations are smoot (old). 10. Let V be a k-dimensional vector subspace of R n. Sow tat V is a smoot manifold diffeomeorpic to R k, and tat all linear maps on V are smoot. For eac v V, find a natural isomorpism V T v V. Solution: Coose a basis v 1,..., v k for V. Te linear map φ sending v V to its expression as a linear combination of te v i extends to a linear map from R n to R k. All linear maps R n R k are smoot. φ as inverse (α 1,..., α k ) i α iv i. So we can take φ as cart.

{ det(a1 + b 1,..., a n 1 + b n 1, b n ) lim + + det(a 1 + b 1,..., a i 1 + b i 1, b i, a i+1,..., a n ) 0 Te term natural is of course rater vague. In my opinion te following construction deserves te term: given v V and x V, let γ(t) = x + tv. Ten v = γ (0) T x V. Peraps te exercise sould say T x V = V. 11. i) If M N are manifolds of te same dimension, sow tat M is open in N (Hint: use exercise 3 and te inverse function teorem.) ii) Sow tat S 1 S 1 is not diffeomorpic to any manifold M contained in R 2. (Hint: S 1 S 1 is compact.) Solution: Let i : M N be inclusion. Ten d x i is also inclusion. As te dim M = dim N, d x i is an isomorpism. Te inverse function teorem tells us tat tere is a neigbourood U of x in M and a neigbourood V of i(x) in N suc tat i : U V is a diffeomorpism. Of course U must equal V ; because V is open in N, so is U, and tus M is open in N. (ii) If S 1 S 1 were diffeomorpic to some M R 2 ten by (i), M would be open in R 2. It would also be compact, and terefore closed in R 2. Te only subsets of R 2 woic are bot open and closed are and R 2, neiter of wic is diffeomorpic to S 1 S 1. 12. If G is a Lie group, te tangent space T e G is known as te Lie algebra of G. Sow tat te following are Lie groups, and determine teir Lie algebras: i) Sl(n, R) = {A Gl(n, R) det(a) = 1 ii) 0(k, n k) = {A Gl(n, R) A t I k,n k A = I k,n k, were I k,n k is te n n matrix wit k 1 s followed by (n k) -1 s down te diagonal, and 0 s elsewere. Solution: (i) By Exercise 11 below, for eac invertible matrix A, d A det is surjective. In particular 1 is a regular value of det, so Sl(n, R) is at least a manifold. Since te group operations are te restriction of te smoot (old) group operations in tose of te ambient manifold Gl(n, R), tey are smoot (new) and so Sl(n, R) is a Lie group. Its tangent space at I is equal to te kernel of d I det. Te formula given in te solution to Exercise 11 sows tat tis is equal to te set of all matrices wit trace 0. (ii) Let f : Mat n n (R) Sym n (R) be te map sending A to A t I k,n k A. Ten O(k, n k) = f 1 (I k,n k ), and it is necessary to sow tat I k,n k is a regular value. Tis is easily done by a small modification of te argument in 1.28(ii) in te Lecture Notes, as is te calculation of T e O(k, n k). 13. i) Find te derivative of te map det : M(n, R) R, were M(n, R) is te vector space of n n matrices wit real entries. Don t attempt to differentiate a complicated polynomial in te entries; instead, use te fact tat te determinant is linear in eac row of te matrix. ii) Prove tat te set of all 2 2 matrices of rank 1 is a tree-dimensional submanifold of R 4 = space of 2 2 matrices. Hint: prove tat te determinant function is a submersion at every point of te manifold of non-zero 2 2 matrices. Solution: (i) Represent a eac matrix as an n-tuple of columns: A = (a 1,..., a n ), B = (b 1,..., b n ), etc. Ten det(a + B) det A d A det(b) = lim 0 { det(a1 + b 1,..., a n + b n ) det(a 1 + b 1,..., a n 1 + b n 1, a n ) = lim 0 + det(a 1 + b 1,..., a n 1 + b n 1, a n ) det(a 1 + b 1, α n 2 + b n 2, a n 1, a n ) + + det(a 1 + b 1, a 2,..., a n ) det(a 1,..., a n ) Eac summand is te difference of te determinants of two matrices differing only in one column. By linearity in te i-t column, and so d A det(b) = det(a 1,..., a i,..., a n ) det(a 1,..., a i,..., a n ) = det(a 1,..., a i a i,..., a n )

+ + det(b 1, a 2,..., a n ) { = lim det(a 1 + b 1,..., a n 1 + b n 1, b n ) + + det(a 1 + b 1,..., a i 1 + b i 1, b i, a i+1,..., a n ) 0 + + det(b 1, a 2,..., a n ) = det(a 1,..., a n 1, b n ) + + det(a 1,..., a i 1, b i, a i+1,..., a n ) + + det(b 1, a 2,..., a n ) = det(a 1,..., a n 1, b n ) + + det(a 1,..., a i 1, b i, a i+1,..., a n ) + + det(b 1, a 2,..., a n ). (ii) If A as rank 1 ten A 0. Suppose a 1 0. Ten tere is some λ R suc tat a 2 = λa 1. Using te formula from (i), it is now easy to find B suc tat d A det(b) 0. 14.(i) Let i : Gl(n, R) Gl(n, R) be te inversion map, i(a) = A 1. Find d A i. Hint: Proposition 1.17 of te lecture notes gives te answer wen A is te identity matrix I n. For a general A, apply te cain rule to te commutative diagram (Gl(n, R), I n ) i (Gl(n, R), I n ) l A r A 1 (Gl(n, R), A) i (Gl(n, R), A 1 ) in wic l A is left-multiplication by A, l A (B) = AB, and r A 1 is rigt-multiplication by A 1, r A 1(C) = CA 1 ; note tat bot l A and r A 1 are te restriction to Gl(n, R) of linear maps on te ambient space of all matrices (= R n2 ). Solution: Te commutativity of te diagram says: Evaluating bot sides on a matrix B, tis is simply te familiar fact tat i l A = r A 1 i. (1) (AB) 1 = B 1 A 1. Applying te cain rule to (1) and taking derivatives at I n on bot sides we get d A i d In l A = d In r A 1 d In i. (2) Now te two maps l A and r A 1 are bot te restrictions to Gl(n, R) of maps wic are linear on te ambient vector space (of all n n matrices). Hence te derivative (at any point) of l A is just l A itself, and similarly te derivative of r A 1 is r A 1 itself. So (2) becomes d A i l A = r A 1 d In i = r A 1 (te last equality because we know from 1.17 (i) in te Lecture Notes tat d In i is just multiplication by 1). As l A is a linear isomorpism wit inverse l A 1, we conclude d A i = r A 1 l A 1, or, in oter words, for any element B T A Gl(n, R) = Mat n n (R), we ave d A i(b) = A 1 BA 1. 15. (Stack of records teorem). Suppose tat M and N are smoot manifolds of te same dimension, wit M compact, f : M N is a smoot map, and tat y is a regular value of f. Sow tat f 1 (y) is finite, and tat tere exists a neigbourood V of y in Y suc tat f 1 (V ) is a disjoint union of open sets of X, eac of wic is mapped diffeomorpically to V by f. Does eiter statement still old if we drop te requirement tat X be compact?

Solution: By te inverse function teorem, for eac x f 1 (y), tere exists an open neigbourood U x of x on wic f is a diffeomorpism. Te union of te U x cover te compact set f 1 (y), so tere is a finite subcover. As eac U x only contains one preimage of y, tere can be only finitely many of te x s. Number tem x 1,..., x n, and write U i in place of U xi. Srink te U i until U i U j = if i j. For eac i, V i := f(u i ) is an open neigbourood of y. Let V = i V i, and let W i = f 1 (V ) U i. Ten f 1 (V ) = W 1 W n, te W i are mutually disjoint, and f : W i V is a diffeomorpism for eac i. If M is not compact, f 1 (y) may not be finite - consider, for example, te exponential map f : R S 1 f(x) = e 2πix. For eac x f 1 (y) tere is are neigbouroods U x and V x of x and y suc tat f x : U x V x is a diffeo, as in te proof above, but now te intersection of all of te V x will not necessaarily be open. In te case of te exponential map f : R Y, te stack of records teorem still olds, but it s easy to modify te domain so tat it fails. Take y = 1, for example. Ten f 1 (y) = Z R. Now remove from te domain te points n + 1/n for n Z {0, and write M denote R wit all tese points removed. Te map exp : M S 1 is still a submersion, so at eac point n f 1 (1) tere is a neigbourood U n of x and a neigbouroood V n of 1 suc tat f : U n V n is a diffeo. Since n + 1/n / U n, exp(1 + 1/n / V n. We ave exp(1 + 1/n) = exp(1/n), and te sequence exp(1/n) tends to 1, so n V n is not a neigbourood of 1. 16. Wit te ypoteses of te previous exercise, suppose also tat f is a local diffeomorpism and tat Y is connected. Sow tat f is onto, and indeed tat any 2 points in Y ave te same number of preimages in X. Solution: f is a local diffeomorpism means tat te derivative of f is everywere an isomorpism, so te conclusion of te previous exercise olds for all y Y. It follows tat te number of preimages is locally constant: if y as n preimages, so does every point y V (notation as in previous exercise). For eac n, te set V n of points wit n preimages is terefore open. Suppose V n. Let V n = m n V m. As a union of open sets, V n also is open, and now if it is not empty ten Y = V n V n is te union of two disjoint non-empty subsets, contradicting its connectedness. 17. Prove tat te following are not manifolds: i) Te union of te two coordinate axes in R 2 ; ii) te double cone {(x, y, z) R 3 x 2 + y 2 = z 2. iii) Te single cone {(x, y, z) R 3 x 2 + y 2 = z 2, z 0 Solution: (i) Many arguments are possible ere. For example, removal of te point 0 breaks te set into four connected components, someting tat cannot appen to any manifold. An alternative argument is tat if M 1 R 2 is any manifold and x M ten by te inverse function teorem 1.16, in some neigbourood of x in M, eiter te projection to te x 1 axis, or te projection to te x 2 axis is a local diffeomorpism, and in particular is 1-1. Neiter is 1-1 in tis case. (ii) Te double cone is disconnected by removing te vertex; no 2-manifold is disconnected by removing a single point. An argument like te second one in (i) also can be used ere. (iii) Many planes troug (0, 0, 0) meet te single cone in only one point. By contrast, if M is a smoot surface in R 3 and x M ten any plane troug x distinct from te affine plane x + T x M is transverse to M at x, and terefore meets M in a curve in te neigbourood of x. 18. Prove tat if X R n is a (smoot) (n 1)-dimensional manifold and x X ten for any (n 1)-dimensional vector subspace V of R n wit V T x X, te intersection of te affine subspace x + V wit X is smoot and of dimension n 2 in some neigbourood of x. Solution: If V T x X ten V + T x X = R n, since T x X is a yperplane and V is not contained in it. Hence x + V is transverse to X at x, and, terefore, in some neigbourood of x, (x + V ) X is an (n 2)-dimensional manifold. Section C 19. Te tangent bundle of a manifold M R n is te set {(x, ˆx) M R n ˆx T x M. (i) Sow tat T M is a manifold.

(ii) Sow tat T S 1 is diffeomorpic to S 1 R. (iii) Sow tat if V R n is a vector subspace ten T V is naturally diffeomorpic to V V (see exercise 7). Solution: (iii) For eac x V tere is an iso i x : V T x V. described above in Exercise 10. Te map V V T V sending (x, v) to (x, i x (v)) is a diffeo. 20. Prove tat te set of m n matrices of rank r is a manifold (in Mat m n (R) = R mn ) and find its dimension. Hint: Sow first tat te set of matrices of rank r is open in Mat m n (R) (recall tat rank A r A as an r r submatrix wit non-zero determinant). Next, suppose rank(a) r, and after permuting rows and columns, tat A as te form ( ) B C (3) D E were B is nonsingular of size r r. Postmultiply by te nonsingular matrix ( I B 1 C ) 0 I to prove tat rank(a) = r if and only if E DB 1 C = 0. Solution: Denote te set of matrices of rank r by r. Te map f, ( ) B C E DB 1 C, D E is defined on te open set U of matrices wit top left r r corner invertible. It is a submersion. Hence its zero locus, r U, is a smoot manifold of codimension (m r)(n r) (te dimension of te target space of f). Now every point of r as some r r minor invertible. On te open set U of matrices for wic tis particular minor is non-zero, it is possible to define a map to te space of (m r) (n r) matrices, of exactly te same nature as f, wose zero locus is r U. Hence r U is also smoot. As every point of r lies in some suc open set, te proof is complete. 21. Let f be te embedding S 1 S 1 S 1 given by e iθ (e 2iθ, e 3iθ ), and let g be te diffeomorpism S 1 S 1 T 2 described in te lecture. Sketc te image of g f. Wat is it? Solution It s a trefoil knot. 22. Is it possible to find an atlas for T R 3 consisting of 2 carts?

Solution: Tere is an atlas mapping eac of two open sets in T to an annulus in R 2. Te domains of te two carts are described most easily in S 1 S 1 : U 1 = {(x 1, y y, x 2, y 2 ) S 1 S 1 : y 1 > 1/2 U 2 = {(x 1, y y, x 2, y 2 ) S 1 S 1 : y 1 < 1/2. 23. Prove tat if U R k is open and f : U R p is not smoot ten grap(f) R k R p is not a (smoot) manifold. 24. RP 2 is te identification space obtained from S 2 by identifying antipodal points. (Recall tat if X is a topological space and is an equivalence relation on X, ten te identification topology on te quotient space X/ is defined by declaring a set in X/ to be open if and only if its preimage in X is open.) Sow tat te mapping S 2 f R 6 given by f(x, y, z) = (x 2, xy, y 2, xz, yz, z 2 ) gives rise to a omeomorpism from RP 2 to a smoot manifold X R 6.