Analysis and algebra on differentiable manifolds


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Analysis and algebra on differentiable manifolds

  1. 1. Chapter 3Integration on ManifoldsAbstract After giving some definitions and results on orientability of smooth man-ifolds, the problems treated in the present chapter are concerned with orientationof smooth manifolds; especially the orientation of several manifolds introduced inthe previous chapter, such as the cylindrical surface, the Möbius strip, and the realprojective space RP2. Some attention is paid to integration on chains and integrationon oriented manifolds, by applying Stokes’ and Green’s Theorems. Some calcula-tions of de Rham cohomology are proposed, such as the cohomology groups of thecircle and of an annular region in the plane. This cohomology is also used to provethat the torus T 2 and the sphere S2 are not homeomorphic. The chapter ends withan application of Stokes’ Theorem to a certain structure on the complex projectivespace CPn.Dans le domain des paramètres a1,...,ar d’un groupe continu quelconqued’ordre r, il existe en effet un élément de volume qui se conserve par unetransformation quelconque du groupe des paramètres (...) Le premier groupedes paramètres, par example, est formé de l’ensemble des transformationsque laissent invariantes r expressions de Pfaff ω1,...,ωr ; l’élément de vol-ume dτ est ω1ω2 ···ωr . Si l’on désigne par S0 une transformation fixe dugroupe et si l’on pose S0Sa = Sb, à un domaine (a) correspond un domain (b)de même volume. Or il existe des groupes continus dont le domain est ferméet de volume total fini; si alors on parte d’un function quelconque des vari-ables et qu’on fasse l’integral des functiones transformées par les differentstransformations du groupe, on obtient une function invariante par le groupe.11“In the domain of the parameters a1,...,ar of any continuous group of order r, there existsindeed a volume element which is preserved under any transformation of the group of parameters(. . . ) The first group of parameters consists, for instance, of the set of transformations preservingr Pfaff expressions ω1,...,ωr ; the volume element dτ is ω1ω2 ···ωr . Denoting by S0 a fixedtransformation of the group and putting S0Sa = Sb, to a domain (a) corresponds a domain (b) withthe same volume. Now, there are continuous groups with closed domain and finite total volume;hence starting with any function of the variables and integrating the functions obtained by thedifferent transformations of the group, one obtains a function invariant under the group.”P.M. Gadea et al., Analysis and Algebra on Differentiable Manifolds,Problem Books in Mathematics, DOI 10.1007/978-94-007-5952-7_3,© Springer-Verlag London 2013129
  2. 2. 130 3 Integration on ManifoldsÉLIE CARTAN, “Les tenseurs irreductibles et les groupes linéaires simpleset semi-simples,” Boll. Sc. Math. 49 (1925), p. 131. Oeuvres Complètes, vol. I,part I, Gauthier-Villars, avec le concours du C.N.R.S, Paris, 1952, p. 532.(Reproduced with kind permission from Dunod Éditeur, Paris. Not for re-useelsewhere.)The extension to manifolds involves two steps: first, we define integralsover the entire manifold M of suitable exterior n-forms and second, for thoseM which have a predetermined volume element (e.g. Riemannian manifolds),integrals of functions over domains are defined. All the standard properties ofintegrals follow readily from the corresponding facts in the Euclidean space.As an illustration of the use of integration on manifolds an application is madeto compact Lie groups. It is shown that by averaging a left-invariant metric ona compact group one may obtain a bi-invariant Riemannian metric. With thesame techniques—due to Weyl—it is shown that any representation of a com-pact group as a matrix group acting on a vector space leaves invariant someinner product on that vector space, from which it follows that any invariantsubspace has a complementary invariant subspace.WILLIAM M. BOOTHBY, An Introduction to Differentiable Manifolds andRiemannian Geometry, 2nd Revised Ed., Academic Press, 2003, p. 222. (Withkind permission from Elsevier.)3.1 Some Definitions and Theorems on Integration on ManifoldsDefinitions 3.1 Let V be a real vector space of dimension n. An orientation of Vis a choice of component of ΛnV {0}.A connected differentiable manifold M of dimension n is said to be orientable ifit is possible to choose in a consistent way an orientation on T ∗p M for each p ∈ M.More precisely, let O be the “0-section” of the exterior n-bundle ΛnM∗, that is,O =p∈M0 ∈ ΛnT ∗p M .Then since ΛnT ∗p M {0} has exactly two components, it follows that ΛnT ∗M {O}has at most two components. It is said that M is orientable if ΛnT ∗M {O} has twocomponents; and if M is orientable, an orientation is a choice of one of the twocomponents of ΛnT ∗M {O}. It is said that M is non-orientable if ΛnT ∗M {O}is connected.Let M and N be two orientable differentiable n-manifolds, and let Φ : M → Nbe a differentiable map. It is said that Φ preserves orientations or that it isorientation-preserving if Φ∗ : TpM → TΦ(p)N is an isomorphism for every p ∈ M,and the induced map Φ∗ : ΛnT ∗N → ΛnT ∗M maps the component ΛnT ∗M {O}determining the orientation of N into the component ΛnT ∗M {O} determiningthe orientation of M. Equivalently, Φ is orientation-preserving if Φ∗ sends orientedbases of the tangent spaces to M to oriented bases of the tangent spaces to N.
  3. 3. 3.1 Some Definitions and Theorems on Integration on Manifolds 131Proposition 3.2 Let M be a connected differentiable manifold of dimension n. Thenthe following are equivalent:(i) M is orientable;(ii) There is a collection C = {(U,ϕ)} of coordinate systems on M such thatM =(U,ϕ)∈CU and det∂xi∂yj> 0 on U ∩ Vwhenever (U,x1,...,xn) and (V,y1,...,yn) belong to C ;(iii) There is a nowhere-vanishing differential n-form on M.Theorem 3.3 (Stokes’ Theorem I) Let c be an r-chain in M, and let ω be a C∞(r − 1)-form defined on a neighbourhood of the image of c. Then∂cω =cdω.Theorem 3.4 (Green’s Theorem) Let σ(t) = (x(t)),y(t)), t ∈ [a,b], be a sim-ple, closed plane curve. Suppose that σ is positively oriented (that is, σ|(a,b) isorientation-preserving) and let D denote the bounded, closed, connected domainwhose boundary is σ. Let f = f (x,y) and g = g(x,y) be real functions with con-tinuous partial derivatives ∂f/∂x, ∂f/∂y, ∂g/∂x, ∂g/∂y on D. ThenD∂g∂x−∂f∂ydx dy =σfdxdt+ gdydtdt.Definition 3.5 Let M be a differentiable manifold. A subset D ⊆ M is said to bea regular domain if for every p ∈ ∂D there exists a chart (U,ϕ) = (U,x1,...,xn)centred at p such thatϕ(U ∩ D) = x ∈ ϕ(U) : xn0 .Theorem 3.6 (Stokes’ Theorem II) Let D be a regular domain in an oriented n-dimensional manifold M, and let ω be a differential (n − 1)-form on M such thatsupp(ω) ∩ ¯D is compact. ThenDdω =∂Dω.Definitions 3.7 A differential r-form α on M is said to be closed if dα = 0. Itis called exact if there is an (r − 1)-form β such that α = dβ. Since d2 = 0, everyexact form is closed. The quotient space of closed r-forms modulo the space ofexact r-forms is called the rth de Rham cohomology group of M:HrdR(M,R) = {closed r-forms}/{exact r-forms}.
  4. 4. 132 3 Integration on ManifoldsIf Φ : M → N is differentiable, then Φ∗ : Λ∗N → Λ∗M transforms closed(resp., exact) forms into closed (resp., exact) forms. Hence Φ induces a linear mapΦ∗: HrdR(N,R) → HrdR(M,R).3.2 Orientable Manifolds. Orientation-Preserving MapsProblem 3.8 Prove:(i) The product of two orientable manifolds is orientable.(ii) The total space of the tangent bundle over any manifold is an orientable mani-fold.Solution(i) A C∞ manifold M is orientable if and only if (see Proposition 3.2(ii)) there isa collection Φ of coordinate systems on M such thatM =(U,ϕ)∈ΦU and det∂xi∂yj> 0 on U ∩ Vwhenever (U,x1,...,xn) and (V,y1,...,yn) belong to Φ.Suppose M1 and M2 are orientable. Denote by (U1,xi1) and (U2,xj2 ) twosuch coordinate systems on M1 and M2, respectively. With a little abuse ofnotation (that is, dropping the projection maps pr1 and pr2 from M1 × M2 ontothe factors M1 and M2), we can write the corresponding coordinate systems onM1 ×M2 as (U1 ×U2,xi1,xj2 ). As the local coordinates on each factor manifolddo not depend on the local coordinates on the other one, the Jacobian matrix ofthe corresponding change of charts of the product manifold M1 × M2 can beexpressed in block form asJ =J1 00 J2=⎛⎜⎝∂xi1∂yk100∂xj2∂yl2⎞⎟⎠.Since detJ1 and detJ2 are positive, we have detJ > 0.Alternatively, the question can be solved more intrinsically as follows: Giventhe non-vanishing differential forms of maximum degree ω1 and ω2 determin-ing the respective orientations on M1 and M2, it suffices to consider the formω = pr∗1 ω1 ∧ pr∗2 ω2 on M1 × M2.(ii) Let M be a differentiable n-manifold and let π be the projection map of the tan-gent bundle T M. For any coordinates {xi} on an open subset U ⊂ M, denote by{xi,yi} = {xi ◦ π,dxi} the usual coordinates on π−1(U). Let {x i} be anotherset of coordinates defined on a open subset U ⊂ M such that U ∩ U = ∅. The
  5. 5. 3.2 Orientable Manifolds. Orientation-Preserving Maps 133change of coordinates x i = x i(xj ) on U ∩ U induces the change of coordi-nates on π−1(U ∩ V ) given byx i= x ix1,...,xn, y i=nj=1∂x i∂xjyj, i = 1,...,n.The Jacobian matrix of this change of coordinates isJ =⎛⎝∂x i∂xj 0∂2x i∂xk∂xj yk ∂x i∂xj⎞⎠.Since detJ = det(∂x i∂xj )2 > 0, it follows that T M is orientable.Problem 3.9 Prove that if a C∞ manifold M admits an atlas formed by two charts(U,ϕ), (V,ψ), and U ∩ V is connected, then M is orientable. Apply this result tothe sphere Sn, n > 1, with the atlas formed by the stereographic projections fromthe poles (see Problem 1.28).Solution Let ϕ = (x1,...,xn) and ψ = (y1,...,yn) be the coordinate maps. Ifdet(∂xi/∂yj ) = 0 on U ∩ V and U ∩ V is connected, we have either (a) det(∂xi/∂yj ) > 0 for all U ∩ V ; or (b) det(∂xi/∂yj ) < 0 for all U ∩ V . In the case (a), itfollows that M is orientable with the given atlas. In the case (b), we should onlyhave to consider as coordinate maps ϕ = (x1,...,xn) and ψ = (−y1,y2,...,yn).For Sn, n > 1, considering the stereographic projections, we have the coordinatedomainsUN = x1,...,xn+1∈ Sn: xn+1= 1 ,US = x1,...,xn+1∈ Sn: xn+1= −1 .AsUN ∩ US = x1,...,xn+1∈ Sn: xn+1= ±1 = ϕ−1N Rn {0}is connected, we conclude that Sn is orientable.Problem 3.10 Study the orientability of the following C∞ manifolds:(i) A cylindrical surface of R3, with the atlas given in Problem 1.30.(ii) The Möbius strip, with the atlas given in Problem 1.31.(iii) The real projective space RP2, with the atlas given in Problem 1.81.Solution(i) The Jacobian matrix J of the change of the charts given in Problem 1.30 alwayshas positive determinant; in fact, equal to 1. Thus the manifold is orientable.
  6. 6. 134 3 Integration on Manifolds(ii) For the given atlas, the open subset U ∩ V decomposes into two connectedopen subsets W1 and W2, such that on W1 (resp., W2) the Jacobian of thechange of coordinates has positive (resp., negative) determinant. Hence M isnot orientable.(iii) With the notations in (ii) in Problem 1.81, we have in the case of RP2 threecharts (U1,ϕ1), (U2,ϕ2) and (U3,ϕ3), such that, for instance,ϕ1(U1 ∩ U2) = ϕ1 x1,x2,x3: x1= 0,x2= 0 = t1,t2∈ R2: t1= 0= V1 ∪ V2,where V1 = {(t1,t2) ∈ R2 : t1 > 0} and V2 = {(t1,t2) ∈ R2 : t1 < 0} are con-nected. The change of coordinates on ϕ1(U1 ∩ U2) is given byϕ2 ◦ ϕ−11 t1,t2= ϕ2 1,t1,t2=1t1,t2t1,and the determinant of its Jacobian matrix is easily seen to be equal to−1/(t1)3, which is negative on V1 and positive on V2. Hence, RP2 is not ori-entable.Problem 3.11 Consider the mapϕ : R2→ R2, (x,y) → (u,v) = x ey+ y,x ey+ λy , λ ∈ R.(i) Find the values of λ for which ϕ is a diffeomorphism.(ii) Find the values of λ for which the diffeomorphism ϕ is orientation-preserving.Solution(i) Suppose thatx ey+ y = x ey+ y , x ey+ λy = x ey+ λy . ( )Subtracting, we have (1 − λ)y = (1 − λ)y . Hence, for λ = 1, we have y = y .And from any of the two equations ( ), we deduce that x = x .The map ϕ is clearly C∞ and its inverse map, given byy =u − v1 − λ, x =λu − vλ − 1eu−vλ−1 ,is a C∞ map if and only if λ = 1. Thus ϕ is a diffeomorphism if and only ifλ = 1.(ii) Consider the canonical orientation of R2 given by dx ∧ dy, or by du ∧ dv. Wehavedet∂u∂x∂u∂y∂v∂x∂v∂y= detey x ey + 1ey x ey + λ.
  7. 7. 3.3 Integration on Chains. Stokes’ Theorem I 135Therefore,du ∧ dv =∂(u,v)∂(x,y)dx ∧ dy = ey(λ − 1)dx ∧ dy.That is, ϕ is orientation-preserving if λ > 1.3.3 Integration on Chains. Stokes’ Theorem IFor the theory relevant to the next problem and others in this chapter, see, amongothers, Spivak [1] and Warner [2].Problem 3.12 Compute the integral of the differential 1-formα = x2+ 7y dx + −x + y siny2dy ∈ Λ1R2over the 1-cycle given by the oriented segments going from (0,0) to (1,0), thenfrom (1,0) to (0,2), and then from (0,2) to (0,0).Solution Denoting by c the 2-chain (with the usual counterclockwise orientation)whose boundary is the triangle above, by Stokes’ Theorem I (Theorem 3.3), we have∂cα =cdα = −8cdx ∧ dy = −8102(1−x)0dy dx = −8.Problem 3.13 Deduce from Green’s Theorem 3.4:(i) The formula for the area of the interior D of a simple, closed positively orientedplane curve [a,b] → (x(t),y(t)) ∈ R2:A(D) =Ddx dy =12bax(t)dydt− y(t)dxdtdt.(ii) The formula of change of variables for double integrals:DF(x,y)dx dy =ϕ−1DF x(u,v),y(u,v)∂(x,y)∂(u,v)dudv,corresponding to the coordinate transformation ϕ : R2 → R2, x ◦ ϕ = x(u,v),y ◦ ϕ = y(u,v).Solution(i) It follows directly from Green’s Theorem 3.4 by letting g = x, f = −y in theformula mentioned there.
  8. 8. 136 3 Integration on Manifolds(ii) First, we let f = 0, ∂g/∂x = F in Green’s formula. Then, from the formula forchange of variables and again from Green’s Theorem 3.4, we obtainDF(x,y)dx dy =∂Dg dy = ±ϕ−1(∂D)ϕ∗(g dy)= ±ϕ−1(∂D)(g ◦ ϕ)∂y∂uu (t) +∂y∂vv (t) dt= ±ϕ−1(∂D)(g ◦ ϕ)∂y∂ududt+ (g ◦ ϕ)∂y∂vdvdtdt= ±ϕ−1D∂∂u(g ◦ ϕ)∂y∂v−∂∂v(g ◦ ϕ)∂y∂ududv,( )where one takes + if ϕ preserves orientation and − if not. Moreover∂∂u(g ◦ ϕ)∂y∂v=∂g∂x◦ ϕ∂x∂u+∂g∂y◦ ϕ∂y∂u∂y∂v+ (g ◦ ϕ)∂2y∂u∂v,∂∂v(g ◦ ϕ)∂y∂u=∂g∂x◦ ϕ∂x∂v+∂g∂y◦ ϕ∂y∂v∂y∂u+ (g ◦ ϕ)∂2y∂v∂u.Hence,∂∂u(g ◦ ϕ)∂y∂v−∂∂v(g ◦ ϕ)∂y∂u= (F ◦ ϕ)∂x∂u∂y∂v−∂x∂v∂y∂u= (F ◦ ϕ)∂(x,y)∂(u,v).Substituting this equality in ( ), we haveDF(x,y)dx dy = ±ϕ−1D(F ◦ ϕ)∂(x,y)∂(u,v)dudv.Problem 3.14 Let c2 be a 2-chain in R2 and f ∈ C∞R2. Prove that∂c2∂f∂ydx −∂f∂xdy = 0if f satisfies the Laplace equation∂2f∂x2+∂2f∂y2= 0.
  9. 9. 3.4 Integration on Oriented Manifolds. Stokes’ Theorem II 137Solution From Stokes’ Theorem I, we have∂c2∂f∂ydx−∂f∂xdy =c2d∂f∂ydx−∂f∂xdy = −c2∂2f∂x2+∂2f∂y2dx∧dy = 0.Problem 3.15 Consider the 1-chaincr,n : [0,1] → R2 {0}, cr,n(t) = x(t) = r cos2πnt, y(t) = r sin2πnt ,for r ∈ R+, n ∈ Z+.Prove that cr,n is not the boundary of any 2-chain in R2 {0}.Solution Let θ be the angle function on C = cr,n([0,1]). Then, dθ is a globallydefined differential 1-form on C, and we havecr,ndθ =cr,ndarctanyx= 2πn.Suppose cr,n = ∂c2 for a 2-chain c2 ∈ R2 {0}. Then, from Stokes’ Theorem I, itfollows thatcr,ndθ =c2d(dθ) = 0,thus leading us to a contradiction.3.4 Integration on Oriented Manifolds. Stokes’ Theorem IIProblem 3.16 Given on R3 the differential formω = z − x2− xy dx ∧ dy − dy ∧ dz − dz ∧ dx,compute D i∗ω, where i denotes the inclusion map ofD = (x,y,z) ∈ R3: x2+ y21, z = 0in R3.Solution We haveDi∗ω = −Dx2+ xy dx ∧ dy.Taking polar coordinatesx = ρ cosθ, y = ρ sinθ, ρ ∈ (0,1), θ ∈ (0,2π),
  10. 10. 138 3 Integration on Manifoldsone has∂(x,y)∂(ρ,θ)= detcosθ −ρ sinθsinθ ρ cosθ= ρ.Therefore, for D0 = D {[0,1) × {0}}, one hasDi∗ω = −D0x2+ xy dx ∧ dy = −D0ρ2cos2θ + sinθ cosθ ρ dρ ∧ dθ= −2π010ρ3cos2θ + sinθ cosθ dρ dθ−142π01 + cos2θ2+sin2θ2dθ= −π4.Problem 3.17 Let (u,v,w) denote the usual coordinates on R3. Consider theparametrisation (see Remark 1.4)u =12sinθ cosϕ, v =12sinθ sinϕ, w =12cosθ +12, ( )where θ ∈ (0,π), ϕ ∈ (0,2π), of the sphereS2= (u,v,w) ∈ R3: u2+ v2+ w −122=14.Let N = (0,0,1) be its north pole and let π : S2 {N} → R2 be the stereographicprojection onto the plane R2 ≡ w = 0. Let vR2 = dx ∧ dy be the canonical volumeform on R2 and let vS2 = 14 sinθ dθ ∧ dϕ be the volume form on S2 above. Writeπ∗vR2 in terms of vS2 .Remark The 2-formvS2 =14sinθ dθ ∧ dϕis called the canonical volume form on S2 because one has σ(X,Y) = 1, X,Y ∈X(S2), for {X,Y,n} an orthonormal basis of R3, where n denotes the exterior (i.e.pointing outwards) unit normal field on S2.Solution The given stereographic projection is the restriction to S2 {N} of themapπ : R3 {w = 1} → R2, (u,v,w) →u1 − w,v1 − w,
  11. 11. 3.4 Integration on Oriented Manifolds. Stokes’ Theorem II 139whose Jacobian matrix is11−w 0 u(1−w)20 11−wv(1−w)2.Henceπ∗vR2 = π∗(dx ∧ dy)= π∗dx ∧ π∗dy=11 − wdu +u(1 − w)2dw ∧11 − wdv +v(1 − w)2dw=1(1 − w)2du ∧ dv +v(1 − w)3du ∧ dw −u(1 − w)3dv ∧ dw. ( )Thus, substituting ( ) into ( ), we obtain after an easy computationπ∗vR2 = π∗vR2 = −sinθ(1 − cosθ)2dθ ∧ dϕ = −4(1 − cosθ)2vS2 .Problem 3.18 Compute the integral of ω = (x − y3)dx + x3 dy along S1 applyingStokes’ Theorem II.Solution Let D (resp., ¯D) be the open (resp., closed) unit disk of R2, and let D0 =D {[0,1) × {0}}. Applying Stokes’ Theorem II, we haveS1ω =∂ ¯Dω =¯Ddω =D0dω =D03 x2+ y2dx ∧ dy.Taking polar coordinates, we have as in Problem 3.16 thatS1ω =D03ρ3dρ ∧ dθ = 32π010ρ3dρ dθ =3π2.Problem 3.19 Let f be a C∞ function on R2, and D a compact and connectedsubset of R2 with regular boundary ∂D such that f |∂D = 0.(i) Prove the equalityDf∂2f∂x2+∂2f∂y2dx ∧ dy = −D∂f∂x2+∂f∂y2dx ∧ dy.(ii) Deduce from (i) that if ∂2f∂x2 + ∂2f∂y2 = 0 on D, then f |D = 0.
  12. 12. 140 3 Integration on ManifoldsSolution(i) By Stokes’ Theorem II, we haveDf∂2f∂x2+∂2f∂y2+∂f∂x2+∂f∂y2dx ∧ dy =∂Dψ,where ψ is a differential 1-form so that dψ is equal to the 2-form in the left-hand side. One solution is given byψ = −f∂f∂ydx + f∂f∂xdy.Since f |∂D = 0, we have∂Dψ = 0,from which the wanted equality follows.(ii) If f = −∂2f∂x2 − ∂2f∂y2 = 0, by the equality we have just proved, one hasD∂f∂x2+∂f∂y2dx ∧ dy = 0,that is, |df | being the modulus of df , we have D |df |2 dx ∧ dy = 0; thus f isconstant on D, but since f |∂D = 0 we have f |D = 0.Problem 3.20 Let α = 12πx dy−y dxx2+y2 ∈ Λ1(R2 {0}).(i) Prove that α is closed.(ii) Compute the integral of α on the unit circle S1.(iii) How does this result show that α is not exact?(iv) Let j : S1 → R2 be the canonical embedding. How can we deduce from (iii)that j∗α is not exact?Solution(i) Immediate.(ii) Parametrise S1 (see Remark 1.4) as x = cosθ, y = sinθ, θ ∈ (0,2π). ThenS1α =12π2π0cos2θ + sin2θ dθ = 1.(iii) If it were α = df for a given function f , applying Stokes’ Theorem II, it wouldbeS1α =S1df =∂S1f = 0,contradicting the result in (ii).
  13. 13. 3.5 De Rham Cohomology 141(iv) Let us suppose that j∗α is exact, i.e. j∗α = df . Then we would haveS1j∗α =j(S1)α =S1α = 1.On the other hand, as j∗ d = dj∗, and denoting by ∅ the empty set, we wouldhaveS1j∗α =S1j∗df =S1dj∗f =∂S1j∗f =∅j∗f = 0,but this contradicts the previous calculation.Problem 3.21 Considerα =x dy ∧ dz − y dx ∧ dz + zdx ∧ dy(x2 + y2 + z2)3/2∈ Λ2R3 {0} .(i) Prove that α is closed.(ii) Compute S2 α.(iii) How does this prove that α is not exact?Solution(i) Immediate.(ii) Consider the parametrisation of S2 given (see Remark 1.4) byx = cosθ cosϕ, y = cosθ sinϕ, z = sinθ,θ ∈ (−π/2,π/2), ϕ ∈ (0,2π), which covers the surface up to a set of measurezero. We have α|S2 = −cosθ dθ ∧ dϕ andS2α =2π0π2− π2−cosθ dθ dϕ = −4π.(iii) If α = dβ, by Stokes’ Theorem II, it would beS2α =S2dβ =∂S2β = 0,which contradicts the result in (ii).3.5 De Rham CohomologyProblem 3.22 Prove that the de Rham cohomology groups of the circle areHidR S1,R =R, i = 0,1,0, i > 1.The relevant theory is developed, for instance, in Warner [2].
  14. 14. 142 3 Integration on ManifoldsSolution One has H0dR(S1,R) = R because S1 is connected. Since dimS1 = 1, onehas HidR(S1,R) = 0 if i > 1.As for H1dR(S1,R) = R, every 1-form on S1 is closed. Now, let ω0 be the re-striction to S1 of the differential form (−y dx + x dy)/(x2 + y2) on R2 {(0,0)}.We locally have ω0 = dθ, θ being the angle function. Hence dθ is non-zero at everypoint of S1. (In spite of the notation, dθ is not exact, cf. Problem 3.20.) Hence, ifω is any 1-form on S1, then we have ω = f (θ)dθ, where f is differentiable andperiodic with period 2π. To prove this, we only have to see that there is a constantc and a differentiable and periodic function g(θ) such thatf (θ)dθ = c dθ + dg(θ).In fact, if this is so, integrating we havec =12π2π0f (θ)dθ.We then defineg(θ) =θ0f (t) − c dt,where c is the constant determined by the previous equality. One clearly has that gis differentiable. Finally, we must see that it is periodic. Indeed,2g(θ + 2π) = g(θ) +θ+2πθf (t) − c dt= g(θ) +θ+2πθf (t)dt −2π0f (t)dt (f is periodic)= g(θ).Problem 3.23 Compute the de Rham cohomology groups of the annular regionM = (x,y) ∈ R2: 1 < x2 + y2 < 2 .Hint Apply the following general result: if two maps f,g : M → N between twoC∞ manifolds are C∞ homotopic, that is, if there exists a C∞ map F : M ×[0,1] → N such that F(p,0) = f (p), F(p,1) = g(p) for every p ∈ M, then themapsf ∗: HkdR(N,R) → HkdR(M,R), g∗: HkdR(N,R) → HkdR(M,R),are equal for every k = 0,1,....The relevant theory is developed, for instance, in Warner [2].
  15. 15. 3.5 De Rham Cohomology 143Solution Let N = S1(3/2) be the circle with centre at the origin and radius3/2 in R2. Let j : N → M be the inclusion map and let r be the retractionr : M → N, p → 32 (p/|p|). Then, r ◦ j : N → N is the identity on S1(3/2). Themap j ◦ r : M → M, p → 32 (p/|p|), although not the identity of M, is homotopicto the identity. In fact, we can define the homotopy byH : M × [0,1] → M, (p,t) → tp + (1 − t)32p|p|.Thus, for k = 0,1,2, we havej∗: HkdR(M,R) → HkdR S1(3/2),R , r∗: HkdR S1(3/2),R → HkdR(M,R),so, applying the general result quoted in the hint, we haver∗◦ j∗= (j ◦ r)∗= identity on HkdR(M,R),j∗◦ r∗= (r ◦ j)∗= identity on HkdR S1(3/2),R .Hence, j∗ and r∗ are mutually inverse and it follows thatHkdR(M,R) ∼= HkdR S1(3/2),R . ( )Consequently, H0dR(M,R) = R (as one can also deduce directly since M is con-nected). In fact, there are no exact 0-forms, and the closed 0-forms (that is, thedifferentiable functions f such that df = 0) are the constant functions, since M isconnected.As dimS1(3/2) = 1, from the isomorphism ( ) we obtain HkdR(M,R) = 0,k 2.Finally, H1dR(M,R) ∼= H1dR(S1(3/2),R) = R, henceHkdR(M,R) =R, k = 0,1,0, k > 1.Problem 3.24(i) Prove that every closed differential 1-form on the sphere S2 is exact.(ii) Using de Rham cohomology, conclude that the torus T 2 and the sphere are nothomeomorphic.Hint Consider the parametrisation (see Remark 1.4)x = (R + r cosθ)cosϕ, y = (R + r cosθ)sinϕ, z = r sinϕ,R > r, θ,ϕ ∈ (0,2π),of the torus T 2, and take the restriction to T 2 of the differential form ω = x dy−y dxx2+y2on R3 {z-axis}.The relevant theory is developed, for instance, in Warner [2].
  16. 16. 144 3 Integration on ManifoldsSolution Let ω be a closed 1-form on the sphere. We shall prove that it is exact. LetU1 and U2 be the open subsets of S2 obtained by removing two antipodal points,respectively. Then, writing ωi = ω|Ui , since Ui is homeomorphic to R2, there existfunctions fi : Ui → R, such that ωi = dfi. As U1 ∩ U2 is connected, one has f1 =f2 + λ on U1 ∩ U2, for λ ∈ R. The function f : S2 → R defined by f |U1 = f1,f |U2 = f2 + λ is differentiable and df = ω.To prove that T 2 and S2 are not homeomorphic, we only have to find a closed1-form on the torus which is not exact. Let j : T 2 → R3 {0} be the canonicalinjection map. The formω =x dy − y dxx2 + y2on R3 {0} is closed. Since d ◦ j∗ = j∗ ◦ d, the form j∗ω on T 2 is also closed. Tosee that ω is not exact, by Stokes’ Theorem, we only have to see that there existsa closed curve γ on the torus such that γ j∗ω = 0. In fact, let γ be the parallelobtained taking θ = 0 in the parametric equations above of the torus. Hence,γj∗ω =2π0dϕ = 2π = 0.Problem 3.25 Let z0,...,zn be homogeneous coordinates on the complex projec-tive space CPn, and let Uα be the open subset defined by zα = 0, α = 0,...,n. Letus fix two indices 0 α < β n. Set uj = zj /zα on Uα, vj = zj /zβ on Uβ.We define two differential 2-forms ωα on Uα and ωβ on Uβ, by settingωα =1ij duj ∧ d¯ujϕ−j,k uj ¯uk duk ∧ d¯ujϕ2,ωβ =1ij dvj ∧ d¯vjψ−j,k vj ¯vk dvk ∧ d¯vjψ2,where ϕ = nj=0 uj ¯uj , ψ = nj=0 vj ¯vj . Prove:(i) ωα|Uα∩Uβ = ωβ|Uα∩Uβ .(ii) There exists a unique differential 2-form ω on CPn such that ω|Uα = ωα, forall α = 0,...,n.(iii) dω = 0.(iv) ω ∧(n)··· ∧ ω is a volume form.(v) ω is not exact.Remark Let a = [ω] be the (real) cohomology class of ω. It can be proved thata generates the real cohomology ring of CPn; specifically, that H∗dR(CPn,R) ∼=R[a]/(an+1).
  17. 17. 3.5 De Rham Cohomology 145Solution(i) On Uα ∩ Uβ one has vj = uj /uβ, and hence ϕ = ψ uβ ¯uβ. We havedvk∧ d¯vj=uβ duk − uk duβ(uβ)2∧¯uβ d¯uj − ¯uj d¯uβ(¯uβ)2=1(uβ)2(¯uβ)2uβ¯uβduk∧ d¯uj− uβ¯ujduk∧ d¯uβ− uk¯uβduβ∧ d¯uj+ uk¯ujduβ∧ d¯uβ,and substituting into the expression of ωβ, on Uα ∩ Uβ we obtainiωβ =uβ ¯uβϕj1(uβ)2(¯uβ)2uβ¯uβduj∧ d¯uj− uβ¯ujduj∧ d¯uβ− uj¯uβduβ∧ d¯uj+ uj¯ujduβ∧ d¯uβ−(uβ)2(¯uβ)2ϕ2j,kuj ¯ukuβ ¯uβ1(uβ)2(¯uβ)2uβ¯uβduk∧ d¯uj− uβ¯ujduk∧ d¯uβ− uk¯uβduβ∧ d¯uj+ uk¯ujduβ∧ d¯uβ.Since the sum of the first and fifth summands above is iωα, and moreover, thefourth and eighth summands are easily seen to cancel, we haveiωβ = iωα−1ϕuβ ¯uβjuβ¯ujduj∧ d¯uβ+ uj¯uβduβ∧ d¯uj+1ϕ2uβ ¯uβj,kuj¯ukuβ¯ujduk∧ d¯uβ+ uk¯uβduβ∧ d¯uj. ( )Consider the last summand. Interchanging the indices j and k, since j uj ¯uj= k uk ¯uk = ϕ, this summand can be written as1ϕ2uβ ¯uβϕj¯ujuβduj∧ d¯uβ+kuk¯uβduβ∧ d¯uk,which is the opposite to the second summand in ( ). Hence ωβ = ωα on Uα ∩Uβ.(ii) Because of (i), we only need to prove that ωα takes real values. In fact,¯ωα = −1ij d¯uj ∧ dujϕ−j,k ¯uj uk d¯uk ∧ dujϕ2,and permuting the indices j and k in the second summand, we obtain ¯ωα = ωα.
  18. 18. 146 3 Integration on Manifolds(iii) On Uα we easily getidω = idωα= −j,k1ϕ2ukduj∧ d¯uj∧ d¯uk+ ¯ukduj∧ d¯uj∧ duk−j,k1ϕ2¯ukduj∧ duk∧ d¯uj+ ujd¯uk∧ duk∧ d¯uj+2ϕ3dϕ ∧j,kuj¯ukduk∧ d¯uj. ( )The first two summands at the right-hand side of ( ) cancel. The third sum-mand vanishes, asdϕ =huhd¯uh+ ¯uhduhyieldsdϕ ∧j,kuj¯ukduk∧ d¯uj= −dϕ ∧j,kujd¯uj∧ ¯ukduk= −dϕ ∧jujd¯uj∧ dϕ −jujd¯uj.(iv) As in (iii), we havej,kuj¯ukduk∧ d¯uj=k¯ukduk∧jujd¯uj.Setν =1ϕj¯ujduj, μ =1ϕjduj∧ d¯uj.Then iω = μ − ν ∧ ¯ν. Thusinωn= μn−n1μn−1∧ ν ∧ ¯ν.Now,μn=n!ϕndu1∧ d¯u1∧ ··· ∧ dun∧ d¯un
  19. 19. References 147(we suppose that α = 0, so that only the coordinates u1, ¯u1,...,un, ¯un areeffective), andμn−1=(n − 1)!ϕn−1nk=1du1∧ d¯u1∧ ··· ∧ duk ∧ d¯uk ∧ ··· ∧ dun∧ d¯un.Hencenμn−1∧ ν ∧ ¯ν =n!ϕn+1nk=1du1∧ d¯u1∧ ···∧ duk ∧ d¯uk ∧ ··· ∧ dun∧ d¯un∧ ¯ukduk∧ ukd¯uk=n!ϕn+1nk=1uk¯ukdu1∧ d¯u1∧ ··· ∧ dun∧ d¯un=n!(ϕ − 1)ϕn+1du1∧ d¯u1∧ ··· ∧ dun∧ d¯un,fornk=0uk¯uk= u0¯u0+nk=1uk¯uk= 1 +nk=1uk¯uk= ϕ.Thusinωn=n!ϕn+1du1∧ d¯u1∧ ··· ∧ dun∧ d¯un,which does not vanish on U0, hence on CPn, as the same argument holds forany α = 0,...,n.(v) Immediate from (iv) and Stokes’ Theorem.References1. Spivak, M.: Calculus on Manifolds. Benjamin, New York (1965)2. Warner, F.W.: Foundations of Differentiable Manifolds and Lie Groups. Graduate Texts inMathematics. Springer, Berlin (2010)Further Reading3. Boothby, W.M.: An Introduction to Differentiable Manifolds and Riemannian Geometry, 2ndrevised edn. Academic Press, New York (2002)4. Godbillon, C.: Éléments de Topologie Algébrique. Hermann, Paris (1971)5. Hicks, N.J.: Notes on Differential Geometry. Van Nostrand Reinhold, London (1965)
  20. 20. 148 3 Integration on Manifolds6. Lee, J.M.: Manifolds and Differential Geometry. Graduate Studies in Mathematics. Am. Math.Soc., Providence (2009)7. Lee, J.M.: Introduction to Smooth Manifolds. Graduate Texts in Mathematics, vol. 218.Springer, New York (2012)8. Petersen, P.: Riemannian Geometry. Springer, New York (2010)9. Spivak, M.: Differential Geometry, vols. 1–5, 3rd edn. Publish or Perish, Wilmington (1999)10. Sternberg, S.: Lectures on Differential Geometry, 2nd edn. AMS Chelsea Publishing, Provi-dence (1999)11. Tu, L.W.: An Introduction to Manifolds. Universitext. Springer, Berlin (2008)