Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 131 (Riemannian Geometry)

Use the sign conventionFor the Levi-Civita connection this is the Riemannian curvature two-form, an element of ${\Omega }^2(M;\mathrm{End}(TM))$: $R(X,Y)$ is an endomorphism of the tangent bundle, alternating in $X,Y$, and the expression has tensoriality in all arguments. It is the curvature form of a connection for the tangent-bundle connection.

For the first Bianchi identity, take the cyclic sum in $X,Y,Z$. Torsion-freeness says ${\nabla }_{Y}Z-{\nabla }_{Z}Y=[Y,Z]$, so the double-derivative terms combine to ${\sum }_{\mathrm{cyc}}{\nabla }_X[Y,Z]$. The remaining terms may be cyclically relabelled as $-{\sum }_{\mathrm{cyc}}{\nabla }_{[Y,Z]}X$. Using torsion-freeness once more, followed by the Jacobi identity for vector fields, givesThe Ricci curvature is the tracefor any orthonormal basis. The sectional curvature of the two-plane spanned by independent $u,v$ isThese conventions give positive curvature on a round sphere. In dimension three, put $K_{ij}=K(e_i,e_j)$ and $r_i=\mathrm{Ric}(e_i,e_i)$. Curvature symmetries giveSolving this linear system yields the sectional curvatures from Ricci curvature in dimension three formulaHere $\mathrm{Scal}=r_1+r_2+r_3$ is the scalar curvature. The chosen orthonormal basis need not diagonalize Ricci; its diagonal evaluations already determine these three sectional curvatures.

A local isometry here is a local diffeomorphism whose pullback metric is the source metric. Use the usual convention that a covering map is surjective. Assume the target is connected and the source is nonempty; this is the connected-manifold setting of the intended assertion. For a disconnected target the precise result is a covering of the union of target components met by the map. For example, inclusion $\mathbb{R}\hookrightarrow \mathbb{R}\bigsqcup \mathbb{R}$ is a local isometry from a complete manifold but is not a surjective covering of the whole target.

The correct Hopf-Rinow theorem says that on a connected finite-dimensional Riemannian manifold, completeness for the induced distance is equivalent to geodesic completeness, and to compactness of every closed bounded subset. These conditions imply that every two points are joined by a minimizing geodesic. The last property by itself does not imply completeness: an open Euclidean ball has minimizing segments between every pair but is incomplete.

A local isometry preserves the Levi-Civita connection, by uniqueness of the torsion-free metric connection on each local isometry chart. Hence it preserves affinely parametrized geodesics, andwhenever the source geodesic is defined. This is geodesic preservation by a local isometry.

Suppose $M$ is complete. Given $y\in N$, join $f(x)$ to $y$ by a minimizing target geodesic using target completeness. The source geodesic whose initial velocity is its inverse image under $d{f}_x$ exists for its full parameter interval by source completeness; its image is the target geodesic by uniqueness. Thus $f$ is onto.

Choose a normal ball $U={\exp }_y(B_r(0))$ with ${\exp }_y$ a diffeomorphism on $B_r(0)\subset T_{y}N$. For every $x\in f^{-1}(y)$ put $U_x={\exp }_x(B_r(0))$. Source completeness defines these exponential maps throughout the tangent balls. The displayed exponential identity makes $f:U_x\to U$ a diffeomorphism: injectivity follows from injectivity of the target exponential map, and differentiating the identity shows the source exponential has invertible derivative on the ball, so its image is open.

These sheets are disjoint. If $z$ lay in two, lift the reversed radial target geodesic from $f(z)$ to $y$, starting at $z$. Its initial velocity is uniquely fixed by $d{f}_z$; uniqueness of the source geodesic forces the two sheet centres to be equal. They exhaust $f^{-1}(U)$: for any point over $U$, lift that reversed radial geodesic using source completeness, obtaining a centre over $y$, and then reverse it. Thus $U$ is evenly covered. This proves the complete local isometry is a covering assertion.

Conversely suppose $f$ is a covering. Given any initial tangent vector in $M$, the corresponding target geodesic extends over all $\mathbb{R}$ by target completeness. The path lifting theorem for coverings lifts it over every compact time interval, with uniqueness gluing the lifts to a curve on $\mathbb{R}$. In local covering charts the lift is a geodesic, because the charts are local isometries. It extends the source initial-value geodesic. Therefore $M$ is geodesically complete, and Hopf-Rinow yields metric completeness on each source component. In the stated connected-target setting,

For $v$ in the domain of the exponential map at $p$, let ${\gamma }_v$ be the unique geodesic with initial data $(p,v)$ and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists of initial velocities whose geodesics exist on $[0,1]$; it is open, contains zero, and need not be the whole tangent space. The ordinary differential equation theorem used here says that a smooth vector field has unique maximal integral curves, with an open flow domain and smooth dependence on initial point and time. Applied to the geodesic equation on $TM$, it gives smoothness of the exponential map. Rescaling the parameter gives ${\exp }_p(tv)={\gamma }_v(t)$ and hence $(d{\exp }_p{)}_0(v)=v$. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighbourhood of zero to one of $p$. Coordinates in an orthonormal tangent basis transported by this map are geodesic normal coordinates.

The geodesic sphere of radius $r$ is $S_r(p)=\{x:d(p,x)=r\}$. For small $r>0$, it is ${\exp }_p(\{v:\Vert v\Vert =r\})$, a smooth compact hypersurface; larger distance spheres need not be smooth. The Gauss lemma statesIn particular radial and angular directions are orthogonal and the radial coordinate measures arc length.

For a piecewise smooth path $c$, put $L(c)=\int \Vert \dot{c}(t)\Vert dt$. The Riemannian distance is $d(x,y)=\inf L(c)$ over such paths from $x$ to $y$. Connectedness makes it finite. Reversal and concatenation give symmetry and the triangle inequality; positivity and compatibility with the manifold topology follow locally from geodesic normal coordinates and the Gauss lemma.

Choose $r_0>0$ so that ${\exp }_p$ is a diffeomorphism on a neighbourhood of the closed tangent ball of radius $2r_0$. For $r<r_0$, radial geodesics realize the distance from $p$ on that ball. Indeed, Gauss lemma bounds any path remaining in the normal ball below by its radial change, while a path leaving the radius-$2r_0$ ball already has length at least $2r_0$. Thus the radius-$r$ sphere is the compact exponential image described above.

Given $q\ne p$, take $0<\delta <\min \{r_0,d(p,q)\}$. The continuous function $z\mapsto d(z,q)$ attains its minimum at some $p_0\in S_{\delta }(p)$. Every path from $p$ to $q$ crosses that sphere, since the distance from $p$ is continuous. Splitting at a crossing gives length at least $\delta +{\min }_{S_{\delta }(p)}d(\cdot ,q)$. Taking the infimum over paths, and using the triangle inequality in the other direction, proves the distance splitting through a small geodesic sphere equalityThis compact local argument does not assume completeness or the existence of a globally minimizing geodesic to $q$.

For the Lie-group clause, a bi-invariant Riemannian metric has adjoint-invariant identity inner product. Differentiating this invariance gives $⟨[X,Y],Z⟩+⟨Y,[X,Z]⟩=0$. The Koszul formula for left-invariant vector fields has no metric-derivative terms, and this skew-adjointness reduces it toThis is the Levi-Civita connection of a bi-invariant metric.

Let $\sigma$ be the integral curve through the identity of the left-invariant field with identity value $v$. Uniqueness and left translation show $\sigma (s+t)=\sigma (s)\sigma (t)$ wherever the local curves are defined. The field is complete: a fixed local existence interval about the identity translates to an equally long interval at every point of the group. Restarting the curve before any proposed finite endpoint extends it beyond that endpoint. Thus $\sigma$ exists on $\mathbb{R}$, and uniqueness now gives the group law for every $s,t$.

Since ${\nabla }_{X}X=0$, this integral curve is a geodesic. Conversely any geodesic starting at the identity has the same initial data as one of these curves and agrees with it by uniqueness. Therefore the geodesics of a bi-invariant metric are one-parameter subgroups conclusion isThe zero velocity gives the constant, trivial subgroup. Here ${\exp }_G$ is the Exponential map of a Lie group; for this metric it agrees at the identity with the Riemannian exponential.

Write $T=\dot{\gamma }$ and $D_t={\nabla }_T$. A Jacobi field is a vector field along the geodesic satisfyingThe curvature convention is that used in Question 1. A geodesic variation is a smooth map $F:(-ϵ,ϵ)\times [0,1]\to M$ with $F(0,t)=\gamma (t)$ whose $t$-curves are affinely parametrized geodesics. Torsion-freeness gives $D_s{\partial }_{t}F=D_t{\partial }_{s}F$. Differentiating $D_t{\partial }_{t}F=0$ and commuting covariant derivatives gives the Jacobi equation for ${\partial }_{s}F{|}_{s=0}$.

For the converse, let $a=J(0)$ and $b=D_{t}J(0)$. Choose a curve $p(s)$ with $p(0)=\gamma (0)$ and $p^{\prime }(0)=a$. parallel transport along it identifies its tangent spaces with $T_{\gamma (0)}M$. Set $v(s)=P_s(T(0)+sb)$, so $v(0)=T(0)$ and $D_{s}v(0)=b$. The geodesics with initial data $(p(s),v(s))$ give a variation $F(s,t)$. Smooth dependence on initial conditions and compactness of the original time interval ensure that this family is defined for all $t\in [0,1]$ after shrinking $ϵ$, even if $M$ is incomplete. Its variation field has the initial values $a,b$, so uniqueness for the linear Jacobi equation identifies it with $J$. This proves the realization of Jacobi fields by geodesic variations.

The endpoint-vanishing pointwise normal fields form a vector space, since their conditions and equation are linear. If $\gamma$ is nonconstant, the map $J\mapsto D_{t}J(0)$ is injective because $J(0)=0$ and zero initial derivative force the zero solution. Differentiating $⟨J,T⟩=0$ gives $D_{t}J(0)\perp T(0)$. Hence the endpoint-vanishing normal Jacobi fields have dimension at most n-1 bound isFor a constant geodesic, $D_t^{2}J=0$, and both zero endpoint values force $J=0$, so the bound still holds.

On the unit round $S^n$, let $\gamma (t)=\cos (\pi t)e_0+\sin (\pi t)e_1$. Its speed is $\pi$. The constant ambient vectors orthogonal to $e_0,e_1$ give $n-1$ independent parallel normal fields $E_j(t)$. Since $R(J,T)T={\pi }^{2}J$ on normal fields, the fields $J_j(t)=\sin (\pi t)E_j(t)$ satisfy the Jacobi equation and vanish at both antipodal endpoints. They attain dimension $n-1$.

For the last clause, interpret a closed geodesic as a nonconstant smoothly periodic geodesic. A constant loop has length zero and cannot be shortened. Parametrize ${\gamma }_0$ on $[0,1]$ at constant nonzero speed. parallel transport once around it fixes $T(0)$. Because the manifold is orientable, this transport preserves orientation; on the normal space of dimension $n-1$, which is odd, it lies in $SO(n-1)$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues pair with their conjugates, real eigenvalues are $\pm 1$, and determinant one in odd dimension forces an eigenvalue $+1$.

Transport such a nonzero normal fixed vector around the loop. It gives a periodic parallel normal field $V$, with $D_{t}V=0$ and $V(1)=V(0)$. Its jets also agree at the seam, so it is smooth as a field on the parametrizing circle. For small $s$, $F(s,t)={\exp }_{{\gamma }_0(t)}(sV(t))$ is a smooth variation through closed curves, providing their homotopy to ${\gamma }_0$.

For energy $E(s)=\frac{1}{2}{\int }_0^1|{\partial }_{t}F{|}^{2}dt$, the first variation vanishes at the closed geodesic. The permitted second variation of geodesic energy has no endpoint term for this periodic variation, soThus $E(s)<E(0)$ for small nonzero $s$. The Cauchy-Schwarz inequality gives $L(F_s{)}^2\le 2E(s)$, while constant speed gives $L({\gamma }_0{)}^2=2E(0)$. Consequently the instability of a closed geodesic in positive even-dimensional curvature yieldsThe deformation need not remain a geodesic or an embedded curve.

At each point, use the inverse metric on the cotangent space. It induces an inner product on $p$-forms byEquivalently, wedges of distinct members of an orthonormal coframe form an orthonormal basis. Choose an orientation to define the ordinary global Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩d{\mathrm{vol}}_g$. It maps $p$-forms to $(n-p)$-forms and obeys ${\ast }^2=(-1{)}^{p(n-p)}$. We use real forms; complex forms are obtained by complex-linear extension. Without an orientation an ordinary global star is not available without twisting the target, so orientation is implicit in this first part.

With the codifferential ${\delta }_p=(-1{)}^{n(p+1)+1}\ast d\ast$ for $p>0$ and ${\delta }_0=0$, define the Hodge Laplacian $\Delta =d\delta +\delta d$. On functions it is the nonnegative-sign Laplace-Beltrami operator, $\Delta f=-\mathrm{div}\mathrm{grad}f$. A harmonic differential form satisfies $\Delta \alpha =0$.

For $\alpha$ of degree $p$, the displayed sign formula and star square giveApplying the identities again, at degrees $p-1$ and $p+1$, proves the Hodge star commutes with the Hodge Laplacian relationSince star is invertible, $\alpha$ is harmonic if and only if $\ast \alpha$ is harmonic. This part does not need compactness; zero-degree and top-degree terms are interpreted as zero when their degrees are outside the range.

For the remaining cohomological statements assume a compact oriented manifold without boundary. The Hodge decomposition theorem gives the $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^p=\ker \Delta$ is finite-dimensional and consists of smooth forms. Formal adjointness gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic differential forms are closed differential forms and coclosed differential forms.

If $\Delta f\ge 0$, formal adjointness and $\Delta 1=0$ give ${\int }_M\Delta f=0$. A continuous nonnegative function with zero integral vanishes, so $\Delta f=0$. Then $\Vert df{\Vert }_{L^2}^2=⟨\Delta f,f⟩=0$. Thus the a function with nonnegative Laplacian on a closed manifold is locally constant assertion isIt is globally constant when $M$ is connected. The PDF does not explicitly include connectedness here; on two disjoint circles one may take different constants, so literal global constancy needs that qualification. Boundaryless is the standard manifold convention in this argument; otherwise boundary conditions are necessary.

For a closed differential form $\alpha$, write its Hodge decomposition as $\alpha =h+d\beta +\delta \gamma$. Since $d\alpha =dh=0$, we have $d\delta \gamma =0$. Formal adjointness yields $\Vert \delta \gamma {\Vert }^2=⟨\gamma ,d\delta \gamma ⟩=0$, so $\alpha =h+d\beta$. Thus $h$ represents its de Rham cohomology class. If two harmonic representatives differ by $d\beta$, their difference $h$ has $\Vert h{\Vert }^2=⟨d\beta ,h⟩=⟨\beta ,\delta h⟩=0$. Therefore

Finally let $h\in G$. A connected Lie group admits a smooth path $h_t$ from the identity to $h$. The maps $R_{h_t}$ give a smooth homotopy between the identity and $R_h$. The homotopy invariance of de Rham cohomology says their pullbacks agree on cohomology. The maps $R_{h_t}$ preserve orientation, since their Jacobian signs cannot change from the identity along the path. As an orientation-preserving isometry, $R_h$ commutes on forms with star, exterior differentiation, codifferentiation and consequently the Hodge Laplacian. Hence $R_h^{\ast }\alpha$ is harmonic and represents the same cohomology class as a harmonic $\alpha$. Uniqueness now proves the harmonic forms are fixed by a connected isometric group action conclusionThe argument also applies componentwise when $M$ is disconnected.

Use the nonnegative Hodge-Laplacian convention from Question 4. The Bochner-Weitzenbock formula for one-forms isHere $\nabla$ is the Levi-Civita connection induced on the cotangent bundle, and the rough Laplacian in a local orthonormal frame isIt is the composition of covariant derivative with its formal adjoint. The Ricci endomorphism is defined by $g({\mathrm{Ric}}^{♯}X,Y)=\mathrm{Ric}(X,Y)$ and acts on one-forms by $(\mathrm{Ric}\cdot \alpha )(X)=\alpha ({\mathrm{Ric}}^{♯}X)$. The musical isomorphism defines ${\alpha }^{♯}$ by $g({\alpha }^{♯},X)=\alpha (X)$. The associated scalar formula isThese signs make the integrated rough-Laplacian term $\Vert \nabla \alpha {\Vert }^2$ on a closed manifold.

For a connected manifold, the full Riemannian holonomy group at $p$ is the subgroup of $O(T_{p}M,g_p)$ consisting of parallel transports around all piecewise smooth loops based at $p$. Its natural action on $T_{p}M$ is the holonomy representation; it induces actions on cotangent spaces and all tensor spaces. The holonomy representation is an irreducible representation when it has no nonzero proper invariant subspace.

The fundamental principle of Riemannian holonomy identifies parallel tensor fields with tensors at $p$ fixed by full holonomy. A parallel field returns to its value under every loop. Conversely, transport a fixed tensor along a path from $p$ to each point. Any two paths differ by a loop, so the result is independent of the path; local smooth parallel transport yields a smooth parallel field. Evaluation and construction are inverse. Full holonomy, not merely the contractible-loop subgroup, is required for this global correspondence.

On a compact manifold without boundary and with nonnegative Ricci curvature, a harmonic one-form satisfies the integrated Bochner identityBoth terms are nonnegative; therefore $\nabla \alpha =0$. This proves that harmonic one-forms are parallel under nonnegative Ricci curvature. Integration uses the Riemannian volume density and does not require an orientation.

If $n=\dim M\ge 2$, a nonzero such form would give a nonzero fixed tangent vector via metric duality and hence a proper invariant line, contradicting irreducibility. This is precisely the irreducible holonomy in dimension at least two has no parallel one-form criterion. Since the question permits harmonic representatives of all classes, it gives $b_1(M)=0$, where $b_1$ is the first Betti number.

The covering $\pi :X\times T^k\to M$ is finite: its fibre over a point is a closed discrete subset of the compact total space and hence finite. Use the explicitly permitted equality of the Betti numbers of $M$ and its finite cover, in particular $b_1(X\times T^k)=b_1(M)=0$. This equality is a permission of this question, not a general theorem about finite covers.

The $k$ angular one-forms on $T^k$, pulled back to $X\times T^k$, represent linearly independent de Rham classes: their periods on the $k$ coordinate circles are the standard basis vectors. exact differential forms have zero periods. Thus $b_1(X\times T^k)\ge k$, forcing $k=0$. Now $\pi :X\to M$ is a finite universal covering map because $X$ is simply connected. The fundamental group acts freely and transitively on a fibre, or equivalently loop-lifting identifies its elements with the finite set of possible endpoints upstairs. Consequently the intended conclusion isThe printed statement needs the dimension qualification under the usual definition of irreducibility. Take the standard circle $M=S^1$, with $X$ a point, $k=1$, and the identity covering $X\times T^1\to S^1$. Its Ricci curvature is zero. parallel transport fixes its global unit tangent, so its holonomy representation is the trivial representation on a one-dimensional real vector space, which is irreducible. The allowed Betti-number equality holds for this identity cover, yetThus there is no proof of the unqualified literal assertion in dimension one. If “irreducible holonomy” is instead intended to exclude the one-dimensional trivial representation, the preceding intended proof applies. A connected zero-dimensional manifold is a point and has trivial fundamental group.