Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 (Differential Geometry)

A smooth map $i:N\to M$ is a smooth embedding when it is an injective immersion and a homeomorphism onto its image with the subspace topology. The unit sphere is the inverse image of the regular value $1$ under $x\mapsto |x{|}^2$, so the preimage theorem makes it a smooth submanifold of ${\mathbb{R}}^{n+1}$ and its inclusion an immersion. It is injective, and a continuous injection from the compact sphere into the Hausdorff Euclidean space is a homeomorphism onto its image. Thus the inclusion is an embedding.

Products of spheres can be embedded by iterated spinning. Start with a round $S^{n_1}\subset {\mathbb{R}}^{n_1+1}$ translated into the half-space whose last coordinate $r$ is positive. If a compact $m$-manifold $N$ is embedded bythenis an injective immersion; compactness again makes it an embedding. Iterating with $q=n_2,\dots ,n_k$ embeds $S^{n_1}\times \cdots \times S^{n_k}$ in ${\mathbb{R}}^{n+1}$, where $n={\sum }_i{n}_i$.

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The de Rham cohomology isThe Poincare lemma says that every closed positive-degree differential form is locally exact, and is exact on every star-shaped open subset of Euclidean space.

Let $n>1$ and let $\alpha$ be a closed one-form on $M$. The hypothesis gives $\alpha =dg$ on $M\setminus B$. Choose a slightly larger coordinate ball $B^{\prime }$ around $B$. The Poincare lemma gives $\alpha =dh$ on $B^{\prime }$. The annulus $B^{\prime }\setminus B$ is connected when $n>1$, so $d(g-h)=0$ there and $g-h$ is constant. Adjusting $h$ by this constant makes $g$ and $h$ agree on the overlap, and they glue to a global primitive of $\alpha$. Hence $H^1(M)=0$. For $n=1$ the claim fails: remove a closed proper interval from $S^1$. Its complement is an interval and has vanishing first de Rham cohomology, whereas $H^1(S^1)\cong \mathbb{R}$.

Finally choose a nowhere-vanishing $n$-form $\omega$ on $S^n$ and a coordinate $t$ on the finite interval $I$. Every $(n+1)$-form on $S^n\times I$ is $a(x,t)\omega \wedge dt$. Fix $t_0\in I$ and putSince $d_{S^n}A\wedge \omega =0$ for dimensional reasons,Every top-degree form is exact, so $H^{n+1}(S^n\times I)=0$ without using the de Rham theorem.

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The wedge product of differential forms is the alternating tensor productFor nonzero $\eta \in {\Lambda }^{n-1}(({\mathbb{R}}^n{)}^{\ast })$, choose a volume form $\mu$. There is a nonzero vector $v$ with $\eta ={\iota }_v\mu$. Extend $v$ to a basis and use its dual coframe; then $\eta$ is a scalar multiple of $e^2\wedge \cdots \wedge e^n$, hence equals $\xi \wedge {\eta }_0$. The zero form is immediate.

To integrate a top form on a compact oriented $n$-manifold, choose a finite oriented atlas and a subordinate partition of unity; integrate each compactly supported coordinate expression and sum. A smooth map pulls forms back byIf $F^{\ast }\omega =\omega$ for a nowhere-zero top form, $F$ preserves its orientation. The change-of-variables theorem gives

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The metric on covectors is induced by the inverse matrix $g^{-1}$, and on $p$-forms by the determinant pairingThe Riemannian volume form is the unique positive top form taking value one on every oriented orthonormal frame. The Hodge star operator is uniquely determined byNondegeneracy of the wedge pairing proves existence and uniqueness pointwise, and the smooth metric dependence makes $\ast$ a well-defined smooth bundle map.

On compactly supported forms, Stokes theorem and the graded Leibniz rule givewhere $m=\dim M$; this is the formal $L^2$ adjoint of $d$. The Hodge Laplace-Beltrami operator is

For $g_f=e^{2f}g$, the covector metric scales by $e^{-2f}$, the $p$-form metric by $e^{-2pf}$, and the volume form by $e^{mf}$. ThereforeIf $f$ is constant, the two star factors in the codifferential contribute $e^{-2f}$, so ${\delta }_f=e^{-2f}\delta$. Since $d$ is metric-independent,

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If $E$ has local frame transition matrices $g_{ab}$ and the cotangent bundle has transitions $J_{ab}^{-T}$, then $E\otimes {\Lambda }^r{T}^{\ast }B$ has local trivializations with transitionswhich satisfy the cocycle condition. Thus it is a well-defined tensor product of vector bundles.

A connection on a vector bundle is a linear mapsatisfying $\nabla (fs)=df\otimes s+f\nabla s$. Contracting with a vector field gives the covariant derivative ${\nabla }_{X}s$. In a local frame, $\nabla =d+A$ for a matrix-valued one-form $A$; under a frame change $g$ the matrix transforms asIts covariant exterior derivative is defined byand locallyThis formula and the graded Leibniz rule show that definitions in different frames agree.

The curvature form of a connection is $F(A)=d_A^2$. Locally,Its covariant derivative satisfies the Bianchi identityIndeed, substituting $F=dA+A\wedge A$, using $d^2=0$, and applying the graded Leibniz rule leaves equal and opposite terms.

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The induced dual connection is uniquely defined bywhich immediately gives the required pairing identity. Ifthen

A connection on $TB$ is symmetric, or torsion-free, whenequivalently ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in coordinates. With $\mathrm{Alt}B(X,Y)=B(X,Y)-B(Y,X)$,

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A horizontal lift of $\gamma$ from $p$ is a curve $\tilde{\gamma }$ in $E$ projecting to $\gamma$, starting at $p$, and tangent to the horizontal distribution of the connection. In a local frame write $\tilde{\gamma }(t)=(\gamma (t),v(t))$. Horizontality is the linear ordinary differential equationwhose initial-value theorem gives local existence and uniqueness; successive trivializations continue the lift.

A geodesic satisfies ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$, andfor the geodesic with initial velocity $v$. Since $d({\exp }_p{)}_0=1$, the inverse function theorem makes ${\exp }_p$ a diffeomorphism near zero; its inverse gives normal coordinates. A geodesic sphere is ${\exp }_p(\{v:|v|=r\})$ inside such a normal neighborhood.

The Gauss lemma statesFor the variation ${\gamma }_s(t)={\exp }_p(t(v+sw))$, let $J={\partial }_s{\gamma }_s{|}_{s=0}$. The coordinate vector fields commute, so metric compatibility and constant geodesic speed giveSince $J(0)=0$, evaluation at $t=1$ proves the formula. In particular radial and spherical directions are orthogonal.

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Choose a normal ball on which ${\exp }_p$ is a diffeomorphism, and take $\epsilon$ smaller than half its radius. For $|X|<\epsilon$, $t\mapsto {\exp }_p(tX)$ is a geodesic of length $|X|$. If a competing curve remains in the normal ball, write it in polar form. The Gauss lemma makes radial and angular velocities orthogonal, so its speed is at least the absolute radial speed and its length is at least $|X|$. A curve leaving the larger normal ball already accumulates more than $|X|$ in radial variation. Thus the radial geodesic minimizes length.

At $q$, both $X$ and the geodesic sphere $\Sigma$ are hypersurfaces. If their tangent hyperplanes were distinct, they would be transverse. The transverse intersection theorem would then make $X\cap \Sigma$ a submanifold of dimensionFor $m\ge 3$ this has positive dimension near $q$, contradicting that the intersection is the singleton $\{q\}$. Therefore $T_{q}X=T_q\Sigma$.

The conclusion fails in dimension two because a transverse intersection is zero-dimensional and may be isolated. In the Euclidean plane, let $\Sigma$ be the unit circle, $q=(1,0)$, and let $X=\{(x,0):1/2<x<3/2\}$. Then $X\cap \Sigma =\{q\}$, but $T_{q}X$ is horizontal whereas $T_q\Sigma$ is vertical.

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