Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 131 (Riemannian Geometry)

For a connected Riemannian manifold $(M,g)$, its Riemannian distance iswhere the infimum is over the piecewise smooth curves from $p$ to $q$. Connectedness of a smooth manifold implies path connectedness, so this set of curves is nonempty.

The Gauss lemma says that the differential of ${\exp }_p$ preserves the radial inner product: for $v,w\in T_{p}M$,Consequently radial geodesics from $p$ are orthogonal to the images of tangent vectors to spheres centred at the origin in $T_{p}M$. In a sufficiently small normal neighbourhood of $p$, this impliesevery competing curve has length at least the total variation of its radial coordinate, and the radial geodesic has that length.

The axioms $d_g(p,q)\ge 0$, symmetry, and the triangle inequality follow directly from length and concatenation. Certainly $d_g(p,p)=0$. If $q\ne p$, choose a normal ball $B_g(p,r)$ that does not contain $q$. Every curve from $p$ to $q$ first meets its boundary, and its initial part has length at least $r$ by the Gauss lemma. Hence $d_g(p,q)\ge r>0$. Thus $d_g(p,q)=0$ if and only if $p=q$.

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Choosewhere $r_p$ is a normal radius at $p$. Take piecewise smooth curves $c_j$ from $p$ to $q$ such that $L_g(c_j)\to d_g(p,q)$. Each $c_j$ first leaves the normal ball $B_g(p,\delta )$ at a point $p_j$. The Gauss lemma gives $d_g(p,p_j)=\delta$. The geodesic sphereis compact, because the tangent-space sphere is compact and ${\exp }_p$ is defined on it. After taking a convergent subsequence, let $p_j\to p_0$.

The part of $c_j$ after $p_j$ has length at least $d_g(p_j,q)$, so continuity of the Riemannian distance givesThe triangle inequality gives the reverse inequality. Therefore

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The metric $g$ is geodesically complete when every maximal affinely parametrized geodesic is defined on all of $\mathbb{R}$; equivalently, ${\exp }_p$ is defined on every $T_{p}M$ for every $p\in M$.

The Hopf-Rinow theorem says that for a connected Riemannian manifold, the following are equivalent: geodesic completeness; completeness of the Riemannian distance $d_g$; compactness of every closed bounded subset; and the existence, between every two points, of a length-minimizing geodesic. It is enough in the exponential-map formulation that ${\exp }_p$ be defined on all of $T_{p}M$ for one point $p$.

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The pointwise inequality $\hat{g}(X,X)\ge g(X,X)$ impliesEvery $d_{\hat{g}}$-Cauchy sequence $(x_j)$ is therefore $d_g$-Cauchy. Since $g$ is geodesically complete, the Hopf-Rinow theorem makes $(M,d_g)$ a complete metric space, so $x_j\to x$ in $d_g$ for some $x\in M$.

On a coordinate neighbourhood with compact closure around $x$, smooth positive-definite Riemannian metrics are uniformly equivalent. Thus there is $C>0$ such thatthere. For all sufficiently large $j$, a short $g$-geodesic from $x$ to $x_j$ stays in this neighbourhood, and henceThus $(M,d_{\hat{g}})$ is complete. Another application of Hopf-Rinow shows that $\hat{g}$ is geodesically complete.

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Let $\gamma :[0,\ell ]\to M$ be a unit-speed geodesic, let $F(s,t)$ be a smooth variation with fixed endpoints, and letbe its variation vector field. Then $V(0)=V(\ell )=0$. If $V^{\perp }=V-⟨V,T⟩T$ is its component normal to $\gamma$, the second variation of Riemannian arc length isHere $D_t={\nabla }_T$ is the covariant derivative along $\gamma$, $R$ is the Riemann curvature tensor, and $I$ is the Riemannian index form. Fixed endpoints remove the boundary term. The normal projection removes a tangential change of parametrization, which does not change length to second order.

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The Bonnet-Myers theorem states that if a complete connected $n$-dimensional Riemannian manifold satisfiesfor some $k>0$, thenIn particular, $M$ is compact and has finite fundamental group.

By the Hopf-Rinow theorem, points $p,q\in M$ are joined by a unit-speed length-minimizing geodesic $\gamma :[0,\ell ]\to M$. Choose a parallel orthonormal frame $E_1,\dots ,E_{n-1}$ normal to $T=\dot{\gamma }$ and setThe endpoint-vanishing fields $V_i$ arise from fixed-endpoint variations. Since $\gamma$ minimizes length, its Riemannian index form is nonnegative on each $V_i$. Summing the second variation of Riemannian arc length givesUsing the Ricci curvature bound and integrating ${\sin }^2$ and ${\cos }^2$ yieldsso $\ell \le \pi /\sqrt{k}$. Taking the supremum over $p,q$ proves the diameter bound. Hopf-Rinow now makes the closed bounded space $M$ compact. Finally, the same bound applies to the complete universal cover; a compact universal cover has finite fibres over $M$, so ${\pi }_1(M)$ is finite.

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Give $S^2\times \mathbb{R}$ the Riemannian product of the unit round metric and the Euclidean metric. It is complete and has infinite diameter. The round sphere has scalar curvature $2$, while the line has scalar curvature $0$; scalar curvature is additive under Riemannian products, soThus this manifold has a strictly positive uniform lower bound on scalar curvature but violates the conclusion of the Bonnet-Myers theorem. Its Ricci curvature vanishes in the $\mathbb{R}$ direction, showing precisely why a scalar-curvature bound is insufficient.

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An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

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Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.

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On $p$-forms in dimension $n$, the defining identity for the Hodge star operator givesFor $n=4$ and $p=2$, therefore, ${\ast }^2=1$. For every $\alpha \in {\Omega }^2(N)$ defineThen $\ast {\alpha }_{+}={\alpha }_{+}$, $\ast {\alpha }_{-}=-{\alpha }_{-}$, and $\alpha ={\alpha }_{+}+{\alpha }_{-}$. The two eigenspaces of the involution $\ast$ have zero intersection, which proves uniqueness. They are respectively the spaces of self-dual and anti-self-dual two-forms.

Now suppose $N$ is compact and let $\beta$ be an exact three-form, say $\beta =d\theta$. Apply the Hodge decomposition theorem to the two-form $\theta$:Set $a=\delta \psi$. Then $da=d\theta =\beta$ and $\delta a={\delta }^2\psi =0$. For a two-form in dimension four, $\delta =-\ast d\ast$, so $d\ast a=0$. The self-dual formsatisfiesThus every exact three-form is the exterior derivative of a self-dual two-form.

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A line in a Riemannian manifold is a unit-speed geodesic $\gamma :\mathbb{R}\to M$ that minimizes globally:for all $s,t\in \mathbb{R}$. A connected noncompact manifold is disconnected at infinity if some compact set $K$ has a complement with at least two unbounded connected components.

Choose points $p_j$ and $q_j$ in two such components withThe Hopf-Rinow theorem supplies a length-minimizing geodesic ${\gamma }_j$ from $p_j$ to $q_j$. Its image must meet $K$, since otherwise it would connect the two different components of $M\setminus K$. Reparametrize so that ${\gamma }_j(0)=x_j\in K$. After taking a subsequence, compactness gives $x_j\to x\in K$ and the unit tangent vectors ${\dot{\gamma }}_j(0)$ converge to some unit $v\in T_{x}M$.

Both endpoint parameters tend to infinity because their distances from $K$ do. Smooth dependence of geodesics on initial data therefore makes ${\gamma }_j$ converge on every compact parameter interval to the complete geodesicEvery finite segment of every ${\gamma }_j$ minimizes length. Passing to the limit gives $d_g(\gamma (s),\gamma (t))=|s-t|$, so $\gamma$ is a line.

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The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product

The Hadamard-Cartan theorem states that if a complete simply connected Riemannian manifold has nonpositive sectional curvature, then for every point $p$ its exponential mapis a diffeomorphism. In particular, the manifold is diffeomorphic to Euclidean space and is contractible.

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Suppose for a contradiction that $Y\times \mathbb{R}$ carries a complete Ricci-flat metric. Since $Y$ is closed, the two subsets $Y\times (A,\infty )$ and $Y\times (-\infty ,-A)$ are different unbounded components outside the compact set $Y\times [-A,A]$. Thus $Y\times \mathbb{R}$ is disconnected at infinity and, by part (a), contains a line in a Riemannian manifold.

Its Ricci curvature is zero, so the Cheeger-Gromoll splitting theorem gives an isometryThe product Ricci tensor shows that $N$ is a complete three-dimensional Ricci-flat manifold. By the allowed fact, $N$ is flat, and hence so is $Y\times \mathbb{R}$.

The universal cover of a complete flat manifold is complete, simply connected, and has zero sectional curvature. The Hadamard-Cartan theorem therefore identifies it diffeomorphically with ${\mathbb{R}}^4$, so it is contractible. On the other hand, the universal cover of the product iswhich deformation retracts onto $\tilde{Y}$. It is contractible only if $\tilde{Y}$ is contractible, contrary to the hypothesis. Hence no such complete Ricci-flat metric exists.

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