MA455 Manifolds Solutions 1 May 2008

1. (i) Given real numbers a < b, find a diffeomorphism (a, b) → R.Solution: For example first map (a, b) to (0, π/2) and then map (0, π/2) diffeomorphically to R using the functiontan.(ii) Find a diffeomorphism (0,∞) → R.Solution: Use log : (0,∞) → R.(iii) Let f : R → R be a smooth map, and let graph(f) be the set {(x, y) ∈ R× R : y = f(x)}. Show that the mapx 7→ (x, f(x)) is a diffeomorphism from R to graph(f).Solution: The map x 7→ (x, f(x)) is clearly smooth and has smooth (new) inverse obtained as the restriction tograph(f) of the smooth (old) projection (x, y) 7→ x.(iv) Let C0 be the cylinder S1 × (0,∞) ⊂ and let C be the cylinder S1 × R (you can draw both in R

3). Finddiffeomorphisms C0 → R

2 \ {0} and C → R2 \ {0}.

Solution: C0 = {(x, y, z) ∈ R2 : x2 + y2 = 1, 0 < z < ∞}. Define a smooth map φ : C0 → R2 \ {(0, 0)} by

φ(x, y, z) = (zx, zy). I leave you to find its inverse. The diffeomorphism of (ii) enables you to define a diffeo

C0 = (0,∞)× S1 ψ−→ R× S1 = C, and now φ ◦ ψ−1 is a diffeo C → R2 \ {(0, 0)}.

2. (i) Let ∆n be the regular n-simplex

{(t1, . . ., tn+1) ∈ Rn+1 :∑i

ti = 1, ti ≥ 0 for all i}.

Draw ∆2.Solution:

(ii) Let Bn be the set{(x1, . . ., xn)} ∈ Rn : 0 ≤ x1 ≤ · · · ≤ xn ≤ 1}.

Find a diffeomorphism Bn → ∆n. Hint: given (x1, . . ., xn) ∈ Bn, there is a natural choice of an (n + 1)-tuple ofnon-negative numbers with sum equal to 1.Solution: (x1, . . ., xn) 7→ (x1, x2 − x1, . . ., xn − xn−1, 1− xn).

3. Show that if U ⊂ Rn is open, f : U → Rn is smooth, and det(dxf) 6= 0 for all x ∈ U , then f(U) is open in Rn.

Note that in particular this holds if f : U → f(U) is a diffeomorphism.

Solution: Use the inverse function theorem.

4. Let X0 be the subset of R2 consisting of the three lines {x1 = 0}, {x2 = 0} and {x1 = x2}. Let `1, `2 and `3 beany three distinct lines through (0, 0) in R2, and let X = `1 ∪ `2 ∪ `3. Find a diffeomorphism — indeed, a linearisomorphism — taking X0 to X.Solution: Let v1 and v2 be non-zero vectors lying on `1 and `2 respectively. Define a linear isomorphism R

2 → R2

by sending (1, 0) to v2, (0, 1) to v1 and extending linearly. This maps {x1 = 0} to `1, {x2 = 0} to `2 and {x1 = x2}to a third line. A linear isomorphism whose matrix in the basis v1, v2 is diagonal will re-scale the two axes so that

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this line becomes the line `3.

5. Show that if X ⊂ Y ⊂ Rn are manifolds and i : X → Y is inclusion then for each x ∈ X, dxi : TxX → TxY isalso inclusion. It’s obvious if you use the right definition of derivative.Solution: Use the definition of dxi as the map sending γ′(0) to i ◦ γ′(0) (version (i) of definition 1.13).

6. i) Let f : X × Y → X be the projection. Show that d(x,y)f : T(x,y)X × Y → TxX is also projection.ii)Fixing any y ∈ Y gives an injection mapping f : X → X×Y , defined by f(x) = (x, y). Show that dxf(x̂) = (x̂, 0).iii) Let f : X → Y, g : X ′ → Y ′ be smooth maps, and define f × g : X×Y → X ′×Y ′ by f × g(x, y) = (f(x), g(y)).Show that d(x,y)(f × g) = dxf × dyg.

Solution: (i) Suppose that X ⊂ RN and Y ⊂ RP . The projection X × Y → X is the restriction to X × Y of theprojection π : RN × RP → R

N . So its derivative is the restriction to T(x,y)X × Y = TxX × TyY of the derivativeof π. But π is linear, so d(x,y)π = π.(ii) Use Definition 1.13(i).(iii) Check first that this is true for smooth (old) maps. Then if F and G are smooth (old) extensions of f and grespectively, F ×G is a smooth (old) extension of f×. So d(x,y)f ×g is the restriction to T(x,y)X×Y = TxX×TyYof d(x,y)F ×G, which you have checked is equal to dxF × dyG.

7. (i) Prove that Rk and R` are not diffeomorphic if k 6= `.Solution: Suppose g : R` → R

k is a smooth inverse to f — that is, g ◦ f = idRk , f ◦ g = id

R` . Then for each

x ∈ Rk and y ∈ R`, we have df(x)g ◦ dxf = dxidRk = id

Rk and dg(y)f ◦ dyg = dxid

R` = id

R` . Thus both derivatives

are linear isomorphisms, and so k = `.(ii) Prove that smooth manifolds M,N of dimension k, ` cannot be diffeomorphic if k 6= `.Solution: Let x ∈ M and y = f(x) ∈ N , and let φ and ψ be charts around x and y respectively. If f is adiffeomorphism then so is ψ ◦ f ◦ φ−1. Now the argument of (i) applies, and shows k = `.

8. Prove that if Mk ⊂ Rn and N ` ⊂ R

p are smooth manifolds then M × N ⊂ Rn+p is a smooth manifold of

dimension k + `.Solution: If (x, y) ∈ M ×N , let φ : U1 → V1 be a chart on M around x and let φ2 : U2 → V2 be a chart on Naround y. Then U1×U2 is a neighbourhood of (x, y) in M×N and φ1×φ2 : U1×U2 → V1×V2 is a diffeomorphism.As V1 × V2 is open in Rk × R` = R

m+`, M ×N is a manifold of dimension k + `.

9. A Lie group is a manifold G which is also a group, for which the operations of

multiplication:

{p : G×G → G

(g1, g2) 7→ g1g2

and inversion:

{i : G → G

g 7→ g−1

are smooth maps. Show that Gl(n,R) := {invertible n× n real matrices under multiplication} ⊂ Matn(R) = Rn2

is a Lie group.Solution: Gl(n,R) is an open subset of matrix space, which is naturally identified with Rn

2. The group operations

are matrix multiplication and matrix inversion. Each entry in the product of two matrices is a sum of productsof entries of the two, and thus is a polynomial. And polynomials are smooth. The entries in A−1 are quotientsof polynomials in the entries of A, with nowhere-vanishing denominator. Hence these too are smooth functions ofthe entries of A. So both group operations are smooth (old).

10∗. Let V be a k-dimensional vector subspace of Rn. Show that V is a smooth manifold diffeomeorphic to Rk,and that all linear maps on V are smooth. For each v ∈ V , find a “natural” isomorphism V ' TvV .

Solution: Choose a basis v1, . . ., vk for V . The linear map φ sending v ∈ V to its expression as a linearcombination of the vi extends to a linear map from R

n to Rk. All linear maps Rn → Rk are smooth. φ has

inverse (α1, . . ., αk) 7→∑

i αivi. So we can take φ as chart.

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The term “natural” is of course rather vague. In my opinion the following construction deserves the term:given v ∈ V and x ∈ V , let γ(t) = x+ tv. Then v = γ′(0) ∈ TxV. Perhaps the exercise should say “TxV = V ”.

11∗. i) If M ⊆ N are manifolds of the same dimension, show that M is open in N (Hint: use exercise 3 and theinverse function theorem.)ii) Show that S1 × S1 is not diffeomorphic to any manifold M contained in R2. (Hint: S1 × S1 is compact.)

Solution: Let i : M → N be inclusion. Then dxi is also inclusion. As the dimM = dimN , dxi is an isomorphism.The inverse function theorem tells us that there is a neighbourhood U of x in M and a neighbourhood V of i(x)in N such that i : U → V is a diffeomorphism. Of course U must equal V ; because V is open in N, so is U , andthus M is open in N .(ii) If S1×S1 were diffeomorphic to some M ⊂ R2 then by (i), M would be open in R2. It would also be compact,and therefore closed in R2. The only subsets of R2 whoich are both open and closed are ∅ and R2, neither of whichis diffeomorphic to S1 × S1.

12∗. If G is a Lie group, the tangent space TeG is known as the Lie algebra of G. Show that the following areLie groups, and determine their Lie algebras: i) Sl(n,R) = {A ∈ Gl(n,R)|det(A) = 1} ii) 0(k, n − k) = {A ∈Gl(n,R)|AtIk,n−kA = Ik,n−k}, where Ik,n−k is the n × n matrix with k 1’s followed by (n − k) -1’s down thediagonal, and 0’s elsewhere.Solution: (i) By Exercise 11 below, for each invertible matrix A, dA det is surjective. In particular 1 is a regularvalue of det, so Sl(n,R) is at least a manifold. Since the group operations are the restriction of the smooth (old)group operations in those of the ambient manifold Gl(n,R), they are smooth (new) and so Sl(n,R) is a Lie group.Its tangent space at I is equal to the kernel of dI det. The formula given in the solution to Exercise 11 shows thatthis is equal to the set of all matrices with trace 0.(ii) Let f : Matn×n(R) → Symn(R) be the map sending A to AtIk,n−kA. Then O(k, n− k) = f−1(Ik,n−k), and itis necessary to show that Ik,n−k is a regular value. This is easily done by a small modification of the argument in1.28(ii) in the Lecture Notes, as is the calculation of TeO(k, n− k).

13∗. i) Find the derivative of the map det : M(n,R) → R, where M(n,R) is the vector space of n × n matriceswith real entries. Don’t attempt to differentiate a complicated polynomial in the entries; instead, use the fact thatthe determinant is linear in each row of the matrix.ii) Prove that the set of all 2 × 2 matrices of rank 1 is a three-dimensional submanifold of R4 = space of 2 × 2matrices. Hint: prove that the determinant function is a submersion at every point of the manifold of non-zero2× 2 matrices.Solution: (i) Represent a each matrix as an n-tuple of columns: A = (a1, . . ., an), B = (b1, . . ., bn), etc. Then

dA det(B) = limh → 0

det(A+ hB)− detAh

= limh → 0

{det(a1 + hb1, . . ., an + hbn)− det(a1 + hb1, . . ., an−1 + hbn−1, an)h

+det(a1 + hb1, . . ., an−1 + hbn−1, an)− det(a1 + hb1, αn−2 + hbn−2, an−1, an)

h+ · · ·

+det(a1 + hb1, a2, . . ., an)− det(a1, . . ., an)

h

}Each summand is the difference of the determinants of two matrices differing only in one column. By linearity inthe i-th column,

det(a1, . . ., a′i, . . ., an)− det(a1, . . ., a

′′i , . . ., an) = det(a1, . . ., a

′i − a′′i , . . ., an)

and so dA det(B) =

limh → 0

{det(a1 + hb1, . . ., an−1 + hbn−1, hbn)h

+ · · ·+ det(a1 + hb1, . . ., ai−1 + hbi−1, hbi, ai+1, . . ., an)h

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+· · ·+ det(hb1, a2, . . ., an)h

}= lim

h → 0

{det(a1 + hb1, . . ., an−1 + hbn−1, bn) + · · ·+ det(a1 + hb1, . . ., ai−1 + hbi−1, bi, ai+1, . . ., an)

+· · ·+ det(b1, a2, . . ., an)}

= det(a1, . . ., an−1, bn) + · · ·+ det(a1, . . ., ai−1, bi, ai+1, . . ., an) + · · ·+ det(b1, a2, . . ., an)}

= det(a1, . . ., an−1, bn) + · · ·+ det(a1, . . ., ai−1, bi, ai+1, . . ., an) + · · ·+ det(b1, a2, . . ., an).

(ii) If A has rank 1 then A 6= 0. Suppose a1 6= 0. Then there is some λ ∈ R such that a2 = λa1. Using the formulafrom (i), it is now easy to find B such that dA det(B) 6= 0.

14.(i) Let i : Gl(n,R) → Gl(n,R) be the inversion map, i(A) = A−1. Find dAi. Hint: Proposition 1.17 of thelecture notes gives the answer when A is the identity matrix In. For a general A, apply the chain rule to thecommutative diagram

(Gl(n,R), In) i−→ (Gl(n,R), In)`A ↓ ↓ rA−1

(Gl(n,R), A) i−→ (Gl(n,R), A−1)

in which `A is left-multiplication by A, `A(B) = AB, and rA−1 is right-multiplication by A−1, rA−1(C) = CA−1;note that both `A and rA−1 are the restriction to Gl(n,R) of linear maps on the ambient space of all matrices(= R

n2).

Solution: The commutativity of the diagram says:

i ◦ `A = rA−1 ◦ i. (1)

Evaluating both sides on a matrix B, this is simply the familiar fact that

(AB)−1 = B−1A−1.

Applying the chain rule to (1) and taking derivatives at In on both sides we get

dAi ◦ dIn`A = dInrA−1 ◦ dIni. (2)

Now the two maps `A and rA−1 are both the restrictions to Gl(n,R) of maps which are linear on the ambientvector space (of all n × n matrices). Hence the derivative (at any point) of `A is just `A itself, and similarly thederivative of rA−1 is rA−1 itself. So (2) becomes

dAi ◦ `A = rA−1 ◦ dIni = −rA−1

(the last equality because we know from 1.17 (i) in the Lecture Notes that dIni is just multiplication by −1). As`A is a linear isomorphism with inverse `A−1 , we conclude

dAi = −rA−1 ◦ `A−1 ,

or, in other words, for any element B ∈ TAGl(n,R) = Matn×n(R), we have

dAi(B) = −A−1BA−1.

15∗. (Stack of records theorem). Suppose that M and N are smooth manifolds of the same dimension, with Mcompact, f : M → N is a smooth map, and that y is a regular value of f . Show that f−1(y) is finite, and thatthere exists a neighbourhood V of y in Y such that f−1(V ) is a disjoint union of open sets of X, each of which ismapped diffeomorphically to V by f . Does either statement still hold if we drop the requirement that X be compact?

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Solution: By the inverse function theorem, for each x ∈ f−1(y), there exists an open neighbourhood Ux of x onwhich f is a diffeomorphism. The union of the Ux cover the compact set f−1(y), so there is a finite subcover. Aseach Ux only contains one preimage of y, there can be only finitely many of the x’s. Number them x1, . . ., xn, andwrite Ui in place of Uxi . Shrink the Ui until Ui ∩Uj = ∅ if i 6= j. For each i, Vi := f(Ui) is an open neighbourhoodof y. Let V = ∩iVi, and let Wi = f−1(V ) ∩ Ui. Then f−1(V ) = W1 ∪ · · · ∪Wn, the Wi are mutually disjoint, andf : Wi → V is a diffeomorphism for each i.

IfM is not compact, f−1(y) may not be finite - consider, for example, the exponential map f : R → S1 f(x) = e2πix.For each x ∈ f−1(y) there is are neighbourhoods Ux and Vx of x and y such that fx : Ux → Vx is a diffeo, as in theproof above, but now the intersection of all of the Vx will not necessaarily be open. In the case of the exponentialmap f : R → Y , the stack of records theorem still holds, but it’s easy to modify the domain so that it fails. Takey = 1, for example. Then f−1(y) = Z ⊂ R. Now remove from the domain the points n+ 1/n for n ∈ Zr {0}, andwrite M denote R with all these points removed. The map exp : M → S1 is still a submersion, so at each pointn ∈ f−1(1) there is a neighbourhood Un of x and a neighbourhoood Vn of 1 such that f : Un → Vn is a diffeo.Since n+ 1/n /∈ Un, exp(1 + 1/n /∈ Vn. We have exp(1 + 1/n) = exp(1/n), and the sequence exp(1/n) tends to 1,so ∩nVn is not a neighbourhood of 1.

16. With the hypotheses of the previous exercise, suppose also that f is a local diffeomorphism and that Y isconnected. Show that f is onto, and indeed that any 2 points in Y have the same number of preimages in X.

Solution: “f is a local diffeomorphism” means that the derivative of f is everywhere an isomorphism, so theconclusion of the previous exercise holds for all y ∈ Y . It follows that the number of preimages is locally constant:if y has n preimages, so does every point y′ ∈ V (notation as in previous exercise). For each n, the set Vn of pointswith n preimages is therefore open. Suppose Vn 6= ∅. Let V ′n =

⋃m6=n Vm. As a union of open sets, V ′n also is

open, and now if it is not empty then Y = Vn ∪ V ′n is the union of two disjoint non-empty subsets, contradictingits connectedness.

17. Prove that the following are not manifolds: i) The union of the two coordinate axes in R2; ii) the double cone{(x, y, z) ∈ R3|x2 + y2 = z2}. iii) The single cone {(x, y, z) ∈ R3|x2 + y2 = z2, z ≥ 0}

Solution: (i) Many arguments are possible here. For example, removal of the point 0 breaks the set into four con-nected components, something that cannot happen to any manifold. An alternative argument is that if M1 ⊂ R2

is any manifold and x ∈ M then by the inverse function theorem 1.16, in some neighbourhood of x in M , eitherthe projection to the x1 axis, or the projection to the x2 axis is a local diffeomorphism, and in particular is 1-1.Neither is 1-1 in this case.(ii) The double cone is disconnected by removing the vertex; no 2-manifold is disconnected by removing a singlepoint. An argument like the second one in (i) also can be used here.(iii) Many planes through (0, 0, 0) meet the single cone in only one point. By contrast, if M is a smooth surfacein R3 and x ∈ M then any plane through x distinct from the affine plane x + TxM is transverse to M at x, andtherefore meets M in a curve in the neighbourhood of x.

18∗. Prove that if X ⊆ Rn is a (smooth) (n−1)-dimensional manifold and x ∈ X then for any (n−1)-dimensionalvector subspace V of Rn with V 6= TxX, the intersection of the affine subspace x + V with X is smooth and ofdimension n− 2 in some neighbourhood of x.

Solution: If V 6= TxX then V + TxX = Rn, since TxX is a hyperplane and V is not contained in it. Hence

x+ V is transverse to X at x, and, therefore, in some neighbourhood of x, (x+ V ) ∩X is an (n− 2)-dimensionalmanifold.

Section C

19∗. The tangent bundle of a manifold M ⊆ Rn is the set {(x, x̂) ∈M × Rn|x̂ ∈ TxM}.(i) Show that TM is a manifold.

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(ii) Show that TS1 is diffeomorphic to S1 × R.(iii) Show that if V ⊂ Rn is a vector subspace then TV is naturally diffeomorphic to V × V (see exercise 7).

Solution: (iii) For each x ∈ V there is an iso ix : V → TxV . described above in Exercise 10. The map V ×V → TVsending (x, v) to (x, ix(v)) is a diffeo.

20. Prove that the set of m× n matrices of rank r is a manifold (in Matm×n(R) = Rmn) and find its dimension.Hint: Show first that the set of matrices of rank ≥ r is open in Matm×n(R) (recall that rank A ≥ r⇔ A has anr × r submatrix with non-zero determinant). Next, suppose rank(A) ≥ r, and after permuting rows and columns,that A has the form (

B CD E

)(3)

where B is nonsingular of size r × r. Postmultiply by the nonsingular matrix(I −B−1C0 I

)to prove that rank(A) = r if and only if E −DB−1C = 0.Solution: Denote the set of matrices of rank r by

∑r. The map f ,(

B CD E

)7→ E −DB−1C,

is defined on the open set U of matrices with top left r × r corner invertible. It is a submersion. Hence its zerolocus,

∑r ∩U , is a smooth manifold of codimension (m− r)(n− r) (the dimension of the target space of f).

Now every point of∑

r has some r×r minor invertible. On the open set U ′ of matrices for which this particularminor is non-zero, it is possible to define a map to the space of (m − r) × (n − r) matrices, of exactly the samenature as f , whose zero locus is

∑r ∩U ′. Hence

∑r ∩U ′ is also smooth. As every point of

∑r lies in some such

open set, the proof is complete.

21∗. Let f be the embedding S1 → S1×S1 given by eiθ 7−→ (e2iθ, e3iθ), and let g be the diffeomorphism S1×S1 → T 2

described in the lecture. Sketch the image of g ◦ f . What is it?Solution It’s a trefoil knot.

22. Is it possible to find an atlas for T ⊂ R3 consisting of 2 charts?

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Solution: There is an atlas mapping each of two open sets in T to an annulus in R2. The domains of the twocharts are described most easily in S1 × S1:

U1 = {(x1, yy, x2, y2) ∈ S1 × S1 : y1 > −1/2}

U2 = {(x1, yy, x2, y2) ∈ S1 × S1 : y1 < 1/2}.

23. Prove that if U ⊂ Rk is open and f : U → Rp is not smooth then graph(f) ⊂ Rk×Rp is not a (smooth) manifold.

24. RP2 is the identification space obtained from S2 by identifying antipodal points. (Recall that if X is a

topological space and ∼ is an equivalence relation on X, then the identification topology on the quotient spaceX/ ∼ is defined by declaring a set in X/ ∼ to be open if and only if its preimage in X is open.) Show that the

mapping S2 f→ R6 given by f(x, y, z) = (x2, xy, y2, xz, yz, z2) gives rise to a homeomorphism from RP

2 to a smoothmanifold X ⊆ R6.

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