Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 15 (Differential Geometry)

The exterior derivative is an $\mathbb{R}$-linear operator $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ which agrees with the differential of a smooth function on degree zero, satisfies $d^2=0$, and obeys the graded Leibniz rule

For the uniqueness of the exterior derivative from its axioms, first note that these axioms force locality. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ equal to one near $p$ and supported where $\alpha =0$. Then $0=d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $d\alpha (p)=0$. Thus $d\alpha (p)$ depends only on the form near $p$.

In a coordinate chart write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Since $d(d{x}^i)=d^2{x}^i=0$, the axioms forceThis also proves existence. Define $d$ by that formula in each chart. The product rule for partial derivatives gives the graded Leibniz rule, and symmetry of mixed partial derivatives against antisymmetry of the wedge product of differential forms gives $d^2=0$. On overlaps it still agrees with the differential of every function, by the chain rule, so the preceding uniqueness argument in the second coordinate system makes the formulas agree. They therefore glue to a global exterior derivative.

For a differential one-form $\omega ={\sum }_i{a}_{i}d{x}^i$, the exterior derivative of a one-form evaluated on vector fields follows directly:In $X(\omega (Y))-Y(\omega (X))$, the additional terms are ${\sum }_i{a}_i[X(Y^i)-Y(X^i)]=\omega ([X,Y])$, where the bracket is the Lie bracket of vector fields. Hence

The de Rham cohomology iswith zero denominator in degree zero. The Poincaré lemma states that a closed differential form of positive degree on a star-shaped open subset of Euclidean space is an exact differential form; consequently every closed positive-degree form is locally exact on a smooth manifold.

For the suspension isomorphism for top de Rham cohomology of spheres, put $U=S^n\setminus \{\text{south pole}\}$ and $V=S^n\setminus \{\text{north pole}\}$. Both are diffeomorphic to ${\mathbb{R}}^n$, while $W=U\cap V$ deformation retracts onto the equatorial $S^{n-1}$. The Mayer--Vietoris sequence for de Rham cohomology comes from the short exact sequence of cochain complexesThe final map is surjective: a partition of unity ${\chi }_U+{\chi }_V=1$ lets a form $\eta$ on $W$ be lifted by extending ${\chi }_V\eta$ to $U$ and $-{\chi }_U\eta$ to $V$ by zero. The Poincaré lemma makes $H^{n-1}(U),H^{n-1}(V),H^n(U),H^n(V)$ vanish when $n>1$. Exactness givesIf $j:S^{n-1}\hookrightarrow W$ is the equatorial inclusion, homotopy invariance of de Rham cohomology makes $j^{\ast }$ an isomorphism. Thereforeis injective, indeed an isomorphism. Concretely, choose local primitives $d{\alpha }_U=\omega {|}_U$, $d{\alpha }_V=\omega {|}_V$, and send $[\omega ]$ to $[j^{\ast }({\alpha }_U-{\alpha }_V)]$. Their difference is closed; changing either primitive changes it by an exact form, so this agrees with the cohomological construction.

A $d$-dimensional embedded submanifold $F\subseteq M^m$ has the subspace topology and, near each of its points, a manifold chart identifying it with a coordinate slice ${\mathbb{R}}^d\times \{0\}$. These slice charts for an embedded submanifold give its smooth structure, and its inclusion is a smooth embedding.

A regular value $q\in N^n$ of a smooth map $f:M^m\to N^n$ is one for which $d{f}_p:T_{p}M\to T_{q}N$ is surjective at every $p\in f^{-1}(q)$. If the preimage is nonempty this forces $m\ge n$.

The inverse function theorem used here says that if $G:U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^m$ is smooth and $D{G}_p$ is an invertible matrix, then its restriction to a suitable neighbourhood of $p$ is a diffeomorphism onto an open neighbourhood of $G(p)$, with smooth inverse.

To prove the regular level set theorem, fix $p\in f^{-1}(q)$ and choose coordinates centred at $p,q$. The coordinate representative $\tilde{f}$ has a rank-$n$ Jacobian matrix. After permuting domain coordinates, its first $n$ columns form an invertible matrix. DefineIts Jacobian matrix is block triangular with invertible diagonal blocks, so the inverse function theorem makes $G$ a local coordinate change. In these coordinates, $f^{-1}(q)$ is precisely the slice with first $n$ coordinates zero. Thus it is an embedded submanifold of dimension $m-n$, withAn empty preimage is harmless; the local assertion is then vacuous.

Three equivalent formulations of orientability of a smooth manifold are: an oriented atlas whose transition Jacobian determinants are positive; a smoothly varying orientation of a vector space on each tangent space; and a nowhere-vanishing smooth top-degree differential form. An oriented atlas gives compatible oriented coordinate frames. A partition of unity glues their positive local top forms, and a nonzero top form declares a frame positive precisely when it evaluates positively on that frame. These constructions prove the equivalence.

For the orientability of a regular fibre, write $F=f^{-1}(q)$. The derivative gives a short exact sequence of vector bundlesThis is the reason the normal bundle of a regular fibre is trivial. Choose any fixed positive basis $b_1,\dots ,b_n$ of $T_{q}N$ and an ambient orientation on $M$. Declare a frame $v_1,\dots ,v_{m-n}$ of $T_{p}F$ positive whenis positive in $T_{p}M$, where $d{f}_p({\tilde{b}}_i)=b_i$. Changing a lift adds a tangent vector and leaves the determinant sign unchanged. Local smooth lifts exist because $df$ is surjective, so this rule varies smoothly and agrees on overlaps. It gives a global orientation on $F$. Only an orientation of the single vector space $T_{q}N$ was chosen; $N$ itself need not be orientable.

An orthogonal structure on a real vector bundle $E$ of rank $r$ is a choice of local frames whose transition matrices lie in the orthogonal group $O(r)$. In such frames declareOrthogonal transitions preserve this expression, so it defines a smooth positive-definite fiber metric. Conversely, a fiber metric turns any local frame into a smooth orthonormal frame by the Gram-Schmidt process; the resulting transition matrices are orthogonal. The two constructions are inverse, giving the required equivalence.

An orthogonal local trivialization $E{|}_U\cong U\times {\mathbb{R}}^r$ identifies each fiber isometrically with standard Euclidean space. Equivalently, its coordinate frame satisfies $h(e_i,e_j)={\delta }_{ij}$. This is local and does not assert the existence of a global frame.

Every real vector bundle over a smooth manifold admits a fiber metric. Choose a trivializing open cover and local Euclidean metrics $h_i$. The partition of unity theorem for a Hausdorff second-countable smooth manifold supplies nonnegative smooth functions ${\rho }_i$, with locally finite supports contained in the respective cover members, such that ${\sum }_i{\rho }_i=1$. Setextending each weighted metric by zero off its chart. Local finiteness makes this smooth. For every nonzero fiber vector $v$, at least one positive weight gives $h(v,v)={\sum }_i{\rho }_i{h}_i(v,v)>0$. Thus $h$ is a fiber metric and provides an orthogonal structure.

One definition of an orthogonal connection is that its covariant derivative preserves the fiber metric:for all vector fields $X$ and smooth sections of a vector bundle $s,t$. Another is that its parallel transport along every smooth curve is an isometry between the endpoint fibers.

For the first implication, take parallel sections $s,t$ along a curve. Their covariant derivatives vanish, so the displayed identity makes $h(s,t)$ constant: parallel transport preserves a fibre metric. Conversely, if parallel transport preserves the inner product, choose parallel extensions of arbitrary initial fiber vectors along a curve with prescribed initial tangent $X$. Differentiating their constant inner product gives $({\nabla }_{X}h)(s,t)=0$ at the initial point. Since $X,s,t$ were arbitrary, the connection is metric-compatible.

The equivalent local description is that the connection matrix in an orthonormal frame is skew-symmetric. Writing $\nabla e_j={\sum }_i{A}^i{}_j{e}_i$ in an orthogonal trivialization, differentiate $h(e_i,e_j)={\delta }_{ij}$ to obtainConversely this matrix identity gives the metric-preservation formula by expanding arbitrary sections. It is the orthogonal group connection description of the same condition.

Let $D$ be the ambient Levi-Civita connection and $\nabla$ the one for the induced metric on $M$. The Gauss formula iswhere the second fundamental form $II(X,Y)=(D_{X}Y{)}^{\perp }$ is normal. The tangential part of $D$ is metric-compatible and torsion-free, so the existence and uniqueness of the Levi-Civita connection identifies it with $\nabla$.

For a normal field $\xi$, the Weingarten formula defines the shape operator in a normal direction $A_{\xi }$ and the normal connection byDifferentiate $⟨\xi ,Y⟩=0$ using metric compatibility to getThe symmetry of the second fundamental form follows from the ambient torsion-free condition:since the Lie bracket of vector fields tangent to $M$ is tangent. Thereforeso every normal-direction shape operator is self-adjoint.

The paper uses $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, so use the Gauss equation with reversed curvature convention. Applying the Gauss formula and Weingarten formula twice givesSubtracting in the stated curvature order yieldsPair with a tangent field $W$ and use the defining relation for the shape operator in a normal direction:This is the Gauss equation in the supplied sign convention. If one defines curvature by $[D_X,D_Y]-D_{[X,Y]}$ instead, the two quadratic terms have the opposite signs.

For an orthonormal frame $X,Y$ on a surface, positive Gaussian curvature with the convention here is $K=⟨R^M(X,Y)X,Y⟩$. If the ambient space is Euclidean and $\xi$ is a unit normal, the equation becomesThe Koszul formula determines $\nabla$ from the induced metric, and hence determines its Riemann curvature tensor and $K$. Therefore the Gaussian curvature depends only on that metric, even though its expression as a product of principal curvatures appears extrinsic. This is Theorema Egregium.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.