Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 15 (Differential Geometry)

Regard $T_{p}M$ as the real vector space of derivations at a point $p$: a vector $v$ acts on germs of smooth functions by a linear map satisfying $v(fh)=f(p)v(h)+h(p)v(f)$. In a coordinate chart $x=(x^1,\dots ,x^n)$, the coordinate derivations $\partial /\partial x^i{|}_p$ form a basis, so $v={\sum }_i{v}^i\partial /\partial x^i{|}_p$, with $v^i=v(x^i)$. To see that these exhaust all derivations, the first-order Taylor theorem locally writes $f-f(p)={\sum }_i(x^i-x^i(p))h_i$ with $h_i(p)={\partial }_{i}f(p)$. The product rule gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$; applying this to the coordinate functions also proves independence. Equivalently one can use velocities of smooth curves through $p$.

Set $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (p,v)=p$. Over a coordinate chart $x:U\to x(U)$ defineGive the disjoint union the topology generated by the inverse images of open sets under these bijections. For overlapping charts $y=f\circ x$, the chain rule gives the tangent coordinate transitionIt is a smooth bijection with smooth inverse $(w,u)\mapsto (f^{-1}(w),D{f}^{-1}(w)u)$. Thus the induced charts form a smooth atlas of dimension $2n$. The chain rule gives the cocycle on triple overlaps, and any additional smooth coordinate chart gives compatible induced charts, so the structure is independent of the starting atlas.

The total space is Hausdorff: different base points can be separated downstairs, and two distinct vectors over one point can be separated in a common product chart. A countable coordinate cover, together with countable bases for its products, gives second countability. Hence this atlas defines a smooth manifold. In every local trivialization, $\pi$ is projection onto the first factor and the fibre change $v\mapsto Df(z)v$ is an invertible linear map depending smoothly on $z$. Fibrewise addition and scalar multiplication are smooth in these product coordinates. Consequently $TM\to M$ is naturally a smooth real vector bundle of rank $n$.

Define the cotangent space $T_p^{\ast }M={\mathrm{Hom}}_{\mathbb{R}}(T_{p}M,\mathbb{R})$ and the cotangent bundle $T^{\ast }M={\coprod }_p{T}_p^{\ast }M$, with projection $(p,\alpha )\mapsto p$. A coordinate chart $x$ gives the dual basis $d{x}^1{|}_p,\dots ,d{x}^n{|}_p$. Write $\alpha ={\sum }_i{\xi }_{i}d{x}^i{|}_p$ and define the induced local trivialization byIf $y=f(x)$ and $\alpha ={\sum }_j{\eta }_{j}d{y}^j$, the chain rule gives $\xi =Df(x{)}^T\eta$. Therefore the cotangent coordinate transition isThe inverse Jacobian matrix is smooth wherever its determinant is nonzero, so this transition and its inverse are smooth. The cocycle follows either from the chain rule or by dualizing the tangent transitions.

Use these product charts to define the topology and smooth atlas. The Hausdorff and second-countability arguments are the same as for the tangent bundle: separate distinct fibres using the base, separate distinct covectors in one product chart, and use a countable coordinate cover. The fibre transitions are smooth linear isomorphisms, so the projection is a smooth vector bundle of rank $n$. Using every compatible base coordinate chart shows that this structure is natural and independent of chart choices.

Yes, the two real smooth vector bundles are isomorphic, although the choice is not canonical. A standard smooth manifold is Hausdorff and second countable, hence admits a smooth locally finite partition of unity. Choose Euclidean Riemannian metrics on coordinate neighborhoods and take their partition-of-unity sum. At each point at least one positive coefficient occurs, so the resulting smooth symmetric tensor $g$ is positive definite.

The resulting musical isomorphism isIt covers the identity of $M$ and is linear on every fibre. In coordinates it is $(x,v)\mapsto (x,G(x)v)$, where $G=(g_{ij})$ is a smooth positive-definite matrix. Its inverse uses the smooth inverse matrix $G^{-1}$, giving the map ${♯}_g$. Thus it is a smooth vector bundle isomorphism.

The Riemannian metric supplies the extra choice: no positive-dimensional real vector space has a linear identification with its dual invariant under all changes of basis. Indeed a scalar change $tI$ acts by $t$ on vectors and $t^{-1}$ on covectors, so an equivariant map $L$ would satisfy $tL(v)=t^{-1}L(v)$ for every $t>0$, forcing $L=0$. An isomorphism therefore exists without being a metric-free canonical one.

For fixed $p$, the covariant derivative is linear in its tangent-vector input, so $L_p:T_{p}M\to T_{p}M$, $v\mapsto ({\nabla }_{v}X)(p)$, is a well-defined linear map. Its trace defines the divergence of a Riemannian vector field. In the given Riemannian orthonormal frame, write $X={\sum }_j{f}_j{E}_j$. The Levi-Civita connection product rule givesAt $p$, the second sum vanishes by the normal orthonormal frame hypothesis. Thus the $i$th diagonal coefficient of $L_p$ is $E_i(f_i)(p)$, andEquivalently one can take ${\sum }_{i}g({\nabla }_{E_i}X,E_i)$; the trace does not depend on the orthonormal basis chosen.

Since the Riemannian orthonormal frame is positively oriented, its dual coframe satisfies ${\omega }_g={\omega }_1\wedge \cdots \wedge {\omega }_n$. The form $\nu$ is the interior product ${\iota }_X{\omega }_g$. The contraction rule for an exterior product givesOne can also obtain the sign by moving $E_i$ from the first argument into slot $i$ of the alternating form, using $i-1$ transpositions. Linearity in $X={\sum }_i{f}_i{E}_i$ proves the interior contraction of the Riemannian volume form formula:The exponents $i-1$ and $i+1$ have the same parity.

Fix any point $p$ and use a normal orthonormal frame there. Such frames exist: in geodesic normal coordinates, $g_{ij}(p)={\delta }_{ij}$, ${\partial }_k{g}_{ij}(p)=0$ and the Christoffel symbols vanish at $p$. Applying the Gram-Schmidt process smoothly to the coordinate fields gives identity coefficients at $p$ and zero first derivatives there, so it produces the required frame. Reverse one frame vector if necessary to match the orientation.

Because the Levi-Civita connection is torsion-free, $[E_j,E_k](p)=({\nabla }_{E_j}{E}_k-{\nabla }_{E_k}{E}_j)(p)=0$. The supplied formula for the exterior derivative of a one-form givesThus $d{\omega }_i(p)=0$, and the graded Leibniz rule implies $d{\theta }_i(p)=0$. Differentiate the formula in part (b):Since $d{f}_i={\sum }_j{E}_j(f_i){\omega }_j$, every summand with $j\ne i$ vanishes because ${\omega }_j$ already occurs in ${\theta }_i$. The remaining term has ${\omega }_i\wedge {\theta }_i=(-1{)}^{i-1}{\omega }_g$, so both signs cancel. HenceThe point was arbitrary, proving the global exterior derivative of contracted volume form identity

Use the standard convention that these compact smooth manifolds have no boundary. The interior product identity for a one-form gives $df\wedge {\iota }_X{\omega }_g=df(X){\omega }_g=X(f){\omega }_g$. For example, in the coframe from part (b), only the matching index survives in this wedge product, with the same pair of cancelling signs as in part (c). Thus the graded Leibniz rule and part (c) giveIntegrate and apply the Generalized Stokes theorem. Since $\partial M=\varnothing$, the integral of $d(f\nu )$ is zero. Rearranging proves integration by parts for Riemannian divergence:If a boundary were allowed, the exact general formula would instead have the additional term ${\int }_{\partial M}f\nu$ on the right; compactness alone does not remove that boundary contribution.

A connection on a vector bundle is a linear operator $\nabla :\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$. Choose coordinates $x^i$ and a local frame $e_1,\dots ,e_m$, and define the connection matrix by $\nabla e_b={\sum }_a{A}^a{}_b\otimes e_a$, with $A^a{}_b={\sum }_i{A}^a{}_{bi}d{x}^i$. For a section $s={\sum }_a{s}^a{e}_a$, the coefficient column satisfiesThe symbol $d$ on the right means the ordinary exterior derivative of the coefficient functions.

Extend uniquely to $E$-valued differential forms by the graded Leibniz ruleIn the same local trivialization, an $E$-valued form is a column $\sigma =({\sigma }^a)$ and the covariant exterior derivative isMoving the one-form $A$ past the scalar $r$-form produces exactly the $(-1{)}^r$ in the graded rule. This explains the sign rather than replacing the rule by an ungraded product formula.

The endomorphism bundle has the induced endomorphism bundle connection characterized on sections by ${\nabla }_X^{\mathrm{End}}(T)s={\nabla }_X(Ts)-T{\nabla }_{X}s$. Its coefficient formula is $dT+AT-TA$. On an endomorphism-valued $r$-form $T$, extend as a graded derivation of the endomorphism-valued exterior product, givingExplicitly its $(a,b)$ entry is $d{T}^a{}_b+{\sum }_c{A}^a{}_c\wedge T^c{}_b-(-1{)}^r{\sum }_c{T}^a{}_c\wedge A^c{}_b$. This version is characterized byThe endomorphism-valued version also satisfies $d_A(T\wedge S)=d_{A}T\wedge S+(-1{)}^{r}T\wedge d_{A}S$. Products here combine composition of endomorphisms with wedge products of coefficient forms; matrix factors must retain their order.

These local formulas are independent of the local frame. For $e^{\prime }=eg$, the change of frame of a vector-bundle connection gives $A^{\prime }=g^{-1}Ag+g^{-1}dg$, ${\sigma }^{\prime }=g^{-1}\sigma$ and $T^{\prime }=g^{-1}Tg$. Using $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $d_{A^{\prime }}{\sigma }^{\prime }=g^{-1}{d}_A\sigma$ and $d_{A^{\prime }}{T}^{\prime }=g^{-1}(d_{A}T)g$. Thus both extended operators are globally defined.

Define the curvature form of a connection by the operatorIt is alternating in $X,Y$. It is also tensorial. For instance, replacing $s$ by $fs$ makes the coefficient of $s$ from derivatives of $f$ equal to $X(Yf)-Y(Xf)-[X,Y]f=0$; the remaining mixed terms $(Xf){\nabla }_{Y}s$ and $(Yf){\nabla }_{X}s$ cancel between the two second derivatives. Thus $F(X,Y)(fs)=fF(X,Y)s$. Replacing $X$ by $fX$, the unwanted term $-(Yf){\nabla }_{X}s$ cancels the opposite term from $[fX,Y]=f[X,Y]-(Yf)X$. Hence $F(fX,Y)=fF(X,Y)$, and the other input follows by alternation.

Consequently $F_p$ depends only on the tangent vectors $X_p,Y_p$ and on $s_p$, giving a smooth global element of ${\Omega }^2(M;\mathrm{End}E)$. In the coefficient-column convention of part (a), applying $d_A$ twice to an arbitrary $E$-valued $r$-form givesFor sections this agrees with the operator definition, so the local Cartan curvature matrix equation isHere $F=\frac{1}{2}{\sum }_{i,j}{F}_{ij}d{x}^i\wedge d{x}^j$, with $F_{ij}=F({\partial }_i,{\partial }_j)$. The same calculation for all degrees provesUnder $e^{\prime }=eg$, applying the covariance of $d_A$ twice gives $F^{\prime }=g^{-1}Fg$. This is exactly the transition rule of the endomorphism bundle, independently confirming that $F$ is a well-defined bundle-valued two-form.

Use the covariant exterior derivative on the endomorphism bundle and remember that $F$ has degree two. The Bianchi identity is $d_{A}F=dF+A\wedge F-F\wedge A=0$. The local formula from part (b) givesThe two lines cancel, establishingBecause this is an endomorphism-valued equation and the operators transform covariantly under a change of local frame, this local calculation proves the global identity.

For a line bundle over $k=\mathbb{R}$ or $\mathbb{C}$, every fibre endomorphism is multiplication by a scalar. Thus $\mathrm{End}E$ is canonically the trivial scalar bundle, even when $E$ itself is nontrivial. In a local frame $A$ is a scalar one-form, so $A\wedge A=0$ and $F=dA$ locally. Also $A\wedge F=F\wedge A$, since their degrees are one and two. The Bianchi identity therefore reduces to the ordinary closedness conditionThis does not assert that the local scalar one-forms $A$ combine into a global one-form.

For two connections on a vector bundle $\nabla ,{\nabla }^{\prime }$, the difference is $C^{\infty }(M)$-linear in the section: the two $df\otimes s$ terms cancel. It is consequently a global endomorphism-valued one-form $a$. In rank one, $a$ is a global ordinary scalar one-form. Locally $A^{\prime }=A+a$, and the scalar cross terms and $a\wedge a$ vanish, givingThis is a global equality, so the two closed differential forms differ by an exact differential form. It proves the connection-independent curvature class of a line bundle:

For every oriented open set $U$ and nowhere-vanishing orientation form $\omega$, let ${\hat{U}}_{\omega }$ contain the points $(p,o_p)$ for which $o_p$ is the orientation determined by ${\omega }_p$. These sets form a basis of a topology for the desired topology on the orientation double cover.

They cover the set of all pairs: every point has an oriented coordinate chart, and changing the sign of its orientation form gives the other orientation. To check the basis intersection axiom, consider ${\hat{U}}_{\omega }$ and ${\hat{V}}_{\eta }$. On $U\cap V$ there is a unique smooth nonzero function $h$ with $\eta =h\omega$. Their intersection is the sheet overSince $W$ is open, this intersection is ${\hat{W}}_{\omega {|}_W}$, another basis set. Thus intersections are open without requiring the two local orientations to agree on every component of the overlap.

The restriction $\pi :{\hat{U}}_{\omega }\to U$ is bijective. Its inverse sends a relatively open set $V\subset U$ to the basis set ${\hat{V}}_{\omega {|}_V}$, so the inverse is continuous. Conversely the projection of the intersection of ${\hat{U}}_{\omega }$ with any other basis set is an open agreement set like $W$, so the projection is continuous as well. Hence each sheet projects homeomorphically onto its oriented open set. In particular,with both sheets open. This establishes the two-sheeted covering space structure.

For a coordinate chart $x:U\to x(U)$ and either orientation sheet over $U$, take the chart $x\circ \pi$ on that sheet. On a nonempty overlap, its transition to $y\circ \pi$ is exactly $y\circ x^{-1}$, restricted to the open agreement set from part (a). It is smooth with smooth inverse. These charts therefore define a smooth structure for which $\pi$ is locally a diffeomorphism.

The topology is Hausdorff: points with distinct base points are separated by inverse images of disjoint base neighborhoods; two points over the same base point are separated by the two disjoint sheets over one coordinate neighborhood. A countable coordinate cover of $M$ gives a countable collection of two-sheeted charts, whose Euclidean chart bases show second countability. Thus the cover is a genuine smooth manifold, not just a collection of compatible charts. The construction using all base charts makes the smooth structure independent of the chosen atlas. It also makes $\pi :{\hat{U}}_{\omega }\to U$ a diffeomorphism for every oriented open $U$, since it is a bijective local diffeomorphism.

At $\hat{p}=(p,o_p)$, the derivative $d{\pi }_{\hat{p}}:T_{\hat{p}}\hat{M}\to T_{p}M$ is a linear isomorphism. Define the orientation of $T_{\hat{p}}\hat{M}$ to be the one mapped to $o_p$. On ${\hat{U}}_{\omega }$ it is represented by ${\pi }^{\ast }\omega$. Where two sheets overlap, their base orientation forms differ by a positive function, so their pullbacks determine the same orientation. This proves the canonical orientation of the cover:The deck involution $(p,o_p)\mapsto (p,-o_p)$ reverses this canonical orientation of the orientation double cover. No global orientation of the base was used.

A connected smooth manifold is path connected because it is locally path connected. Each connected component $C$ of the orientation double cover is itself path connected. Choose a point of $C$ over $p$. For any $q\in M$, lift a path from $p$ to $q$ starting at that chosen point, using the path lifting theorem. The lifted path lies in $C$ and ends above $q$. Therefore every component maps onto $M$.

Fix one fibre, containing exactly two points. Every component meets that fibre, and distinct components meet disjoint subsets of it, so there are at most two connected components.

If $M$ is orientable, a global continuous choice of tangent-space orientation $o$ gives two disjoint sheetsThey are open: on a sufficiently small oriented coordinate neighborhood the global choice agrees consistently with one of the two local sheets. Each projects homeomorphically to the connected base, so these are precisely two connected components.

Conversely suppose the cover is disconnected. It has exactly two components $C_{+},C_{-}$. Both project onto the base and each fibre contains only two points, so each component meets every fibre exactly once. The restriction $\pi :C_{+}\to M$ is consequently a bijective local diffeomorphism, hence a global diffeomorphism. Transporting the canonical orientation of $C_{+}$ through this map or, equivalently, using its smooth inverse section, gives a global orientation of $M$. Thus the connectedness criterion for the orientation double cover is

Let $M$ be a compact Riemannian manifold without boundary. Let $\delta =d^{\ast }$ be the formal $L^2$ adjoint operator of the exterior derivative, and let ${\Delta }_H=d\delta +\delta d$ be the Hodge Laplacian. Write ${\mathcal{H}}^r(M)=\ker {\Delta }_H\cap {\Omega }^r(M)$ for the smooth harmonic differential forms. The Hodge decomposition theorem states that this space is finite-dimensional andan $L^2$-orthogonal direct sum. Thus every smooth $r$-form is uniquely the sum of a harmonic, an exact and a coexact form; its exact and coexact components are unique, though the primitives used to write those components need not be unique. AlsoThe isomorphism sends a harmonic form to its de Rham cohomology class. For example, a closed coexact component $\delta \gamma$ must vanish because $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$; an exact harmonic form also vanishes by orthogonality. These facts explain uniqueness of the harmonic representative of a cohomology class.

On an oriented $n$-manifold the codifferential is $\delta =(-1{)}^{n(r+1)+1}\ast d\ast$ on $r$-forms. The general formulation above also holds on nonorientable Riemannian manifolds: use the globally defined Riemannian volume density for the $L^2$ inner product and its formal adjoint, rather than require a global ordinary volume form. The closed-manifold hypothesis is essential unless suitable boundary conditions are added.

Use the sign convention in the supplied scalar Laplace-Beltrami operator identity, and let $X={\omega }^{♯}$ be the metric-dual vector field. Integrate that Bochner formula for one-forms over the closed Riemannian manifold, with its volume density $d{V}_g$. The integral of a scalar Laplacian is zero: on an oriented manifold this is the Riemannian divergence theorem applied to its gradient, and the same density formula is valid without orientability. HenceBoth terms are nonnegative, because $\mathrm{Ric}\ge 0$. The smooth nonnegative integrand must therefore vanish everywhere: a strictly positive value would remain positive on an open neighborhood of positive volume and make its integral positive. In particular $|\nabla \omega {|}^2=0$ at every point, provingThus harmonic one-forms are parallel under nonnegative Ricci curvature. The same argument also gives $\mathrm{Ric}(X,X)=0$ pointwise. No orientation or connectedness assumption is needed for this step.

By the Hodge decomposition theorem, the first Betti number is $b_1(M)=\dim {\mathcal{H}}^1(M)$. Part (b) makes every element of this space a parallel differential form. For any fixed $p\in M$, consider the linear evaluation mapIt is injective. A parallel form with ${\omega }_p=0$ remains zero under parallel transport along any path from $p$; connectedness of the smooth manifold makes every point reachable by such a path. Equivalently, the derivative of the squared norm of a parallel form is zero in every direction, so its norm is constant on the connected manifold. Either argument gives $\omega =0$ globally. Since $\dim T_p^{\ast }M=n$, the dimension bound for harmonic one-forms under nonnegative Ricci curvature gives

For an explicit obstruction in dimension three, take $N={\Sigma }_2\times S^1$, where ${\Sigma }_2$ is a closed orientable surface of genus two. It is a compact connected smooth three-manifold. A cell decomposition of ${\Sigma }_2$ has one vertex, four one-cells and one two-cell attached by a product of commutators. Its cellular boundary into the one-cells is zero, because each generator has total exponent zero in that attaching word; the boundary of every one-cell is also zero. Therefore $H^1({\Sigma }_2;\mathbb{R})={\mathbb{R}}^4$. The Künneth theorem in degree one and $H^1(S^1;\mathbb{R})=\mathbb{R}$ giveConsequentlyThis is the Betti-number obstruction to nonnegative Ricci curvature. It rules out every Riemannian metric on that underlying manifold, not merely a particular product metric.