Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 14 (Differential Geometry)

Let $(x^i)$ and $(y^a)$ be overlapping local coordinates at $p$. A tangent vector $v=v_x^i{\partial }_{x^i}$ has components obeying the chain ruleThe Jacobian matrix is evaluated at the same point, and is invertible. For a smooth curve $a$ with $a(0)=p$, define ${\dot{a}}_x^i(0)=\frac{d}{dt}{x}^i(a(t)){|}_{t=0}$. Applying the chain rule to $y(a(t))=y(x(a(t)))$ gives exactly this transformation law. Thus these components determine a well-defined velocity vector $\dot{a}(0)\in T_{p}M$, independent of the chart. Equivalently it is the derivation $f\mapsto \frac{d}{dt}f(a(t)){|}_0$ on smooth functions. The two curve symbols used in the printed first paragraph refer to this same curve.

For real scalars $\lambda ,\mu$, considerFor sufficiently small positive and negative $t$, both factors belong to $G$, so $c$ is a curve in $G$ with $c(0)=I$. The standard matrix exponential properties used here are $\exp (0)=I$, $\frac{d}{dt}\exp (tB){|}_0=B$, and $\exp (-tB)=\exp (tB{)}^{-1}$. The product rule gives $c^{\prime }(0)=\lambda B_1+\mu B_2$. Every such linear combination is therefore a velocity at the identity.

More generally, the inclusion $j:G\hookrightarrow GL(n,\mathbb{R})$ is an embedding, so its tangent map at $I$ is injective and linear. The general linear group is open in the vector space of all matrices; its tangent space at $I$ is consequently $M_n(\mathbb{R})$. We identifyEvery tangent vector is the velocity of a curve in a manifold chart, so this subspace is precisely the set of the corresponding matrix velocities. The product construction is consistent with its vector-space operations.

Now use the stated logarithm-chart assumption. Near zero and $I$, the matrix exponential and matrix logarithm are inverse smooth maps, with $d{\exp }_0=d{\log }_I=\mathrm{id}$. Hence for any $B\in \mathfrak{g}$, $\exp (tB)$ belongs to $G$ for small $t$: $tB$ lies in the neighborhood which is the image of the logarithm. For $B_1,B_2\in \mathfrak{g}$, form the group commutatorMultiplying the exponential expansions $e^{tB}=I+tB+\frac{1}{2}{t}^2{B}^2+O(t^3)$ givesThe convergent local series $\log (I+Z)=Z-\frac{1}{2}{Z}^2+\cdots$ therefore gives $\log C(t)=t^2[B_1,B_2]+O(t^3)$. For small nonzero $t$, the matrix $t^{-2}\log C(t)$ belongs to the linear subspace $\mathfrak{g}$. A finite-dimensional linear subspace is closed, so taking its limit provesThis is the matrix commutator from a logarithm chart argument. It identifies $\mathfrak{g}$ as the Lie algebra of a matrix Lie group, with the matrix commutator as its Lie bracket. In particular, no potentially nonsmooth square-root change of parameter is needed to turn the second-order commutator into a velocity.

The exterior derivative is the real-linear degree-one operator $d:{\Omega }^k\to {\Omega }^{k+1}$ characterized by its usual derivative on functions, the graded Leibniz ruleand $d^2=0$. In local coordinates,These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with pullback of a differential form.

On the unit ball let $V=x^i{\partial }_i$ be the radial vector field, and let $F_t(x)=tx$. For $k\ge 1$ define the radial homotopy operator byHere the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega ={\int }_0^1{t}^{-1}{F}_t^{\ast }({\iota }_V\omega )dt$, where ${\iota }_V$ is the interior product. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to ${\Omega }^{k-1}$; set $h_0=0$.

For $t>0$, differentiation of the coordinate pullback and Cartan's magic formula giveFor completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $d{x}^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one provesFor positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega =0$ and $k\ge 1$, it follows that $\omega =d(h_k\omega )$. A chart about any point contains such a ball, proving the Poincaré lemma. In degree zero the formula is instead $h_{1}df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This degree-zero correction to radial homotopy specifies the necessary positive-degree convention in the printed request.

To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to ${\mathbb{R}}^2$, where the same star-shaped homotopy applies. A closed differential one-form $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. ThusThis proves directly that closed one-forms on the two-sphere are exact.

For the Hodge decomposition theorem, assume a compact oriented Riemannian manifold without boundary. The Hodge star operator is characterized by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ and obeys $\ast \ast \alpha =(-1{)}^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta =(-1{)}^{n(k+1)+1}\ast d\ast$. Define the Hodge Laplacian $\Delta =d\delta +\delta d$ and the space ${\mathcal{H}}^k=\ker \Delta$ of harmonic differential forms. Integration by parts, with no boundary term, givesHence a form is harmonic exactly when it is both closed and coclosed. The Hodge decomposition theorem states that ${\mathcal{H}}^k$ is finite-dimensional and that there is the orthogonal decompositionIf a closed form has decomposition $\alpha =h+d\beta +\delta \gamma$, then $⟨\alpha ,\delta \gamma ⟩=⟨d\alpha ,\gamma ⟩=0$. Orthogonality consequently forces $\delta \gamma =0$, so its de Rham cohomology class is represented by $h$. An exact harmonic form $h=d\beta$ has $\Vert h{\Vert }^2=⟨\delta h,\beta ⟩=0$, proving uniqueness. Thus

Every top-degree form is $\omega =f{\mathrm{vol}}_g$, and $d\omega =0$ for dimensional reasons. Its Hodge star is $f$, so the formula for $\delta$ shows that $\delta \omega =0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the harmonic top-degree forms on a closed oriented manifold form the spaceAny such smooth manifold admits a Riemannian metric; the assertion is valid for every choice of that metric.

Take a locally finite trivializing cover $(U_{\alpha })$ of the real vector bundle $E$, and a subordinate smooth partition of unity $({\rho }_{\alpha })$. Each trivialization identifies its fibers with ${\mathbb{R}}^m$ and hence supplies a smooth Euclidean inner product $g_{\alpha }$ there. DefineEach weighted term extends by zero outside its chart, since the weight has support within that chart. Local finiteness makes the sum smooth in every bundle chart. At each $b$ at least one weight is positive; for $v\ne 0$ every nonzero contribution to $g_b(v,v)$ is positive. Thus $g$ is a smooth positive-definite fiber metric, which is precisely an inner product varying smoothly with the fibers.

A G-structure on a vector bundle is a reduction of its frame bundle from $GL(m,\mathbb{R})$ to $G$: a principal $G$-subbundle $P\subseteq \mathrm{Fr}(E)$ whose extension of structure group is the full frame bundle. Equivalently, it is a choice of smooth local frames with transition matrices in $G$, and the reduction consists of the right $G$-orbits of these frames. The adapted frames are part of the structure, not merely a statement that an individual fiber has a basis.

Given a fiber metric, the Gram-Schmidt process transforms any smooth local frame into Euclidean orthonormal frames. At each step the squared norm to be divided out is positive, so the divisions and positive square roots depend smoothly on the base point. Two such frames differ by an orthogonal matrix. All the orthonormal frames therefore form a principal $O(m)$ reduction, an orthogonal structure on a real vector bundle.

Conversely, let $P$ be an $O(m)$ reduction. A frame $u:{\mathbb{R}}^m\to E_b$ in $P_b$ definesReplacing $u$ by $uq$ for $q\in O(m)$ leaves this rule unchanged, since $q$ preserves the Euclidean inner product. It is consequently independent of the adapted frame and defines a positive-definite fiber metric. Its coefficients are smooth in local adapted frames. These constructions are mutually inverse. An $O(m)$-structure is equivalent to a smooth fiberwise inner product.

For $SO(m)$, the additional condition is an orientation of a vector bundle. Such a reduction supplies consistently oriented orthonormal frames, since its transition determinants are $+1$. Conversely an orientation together with a fiber metric selects the oriented Euclidean orthonormal frames and gives an $SO(m)$ reduction. ThusIt is not true for every real bundle. The Möbius line bundle is obtained from $[0,2\pi ]\times \mathbb{R}$ by $(2\pi ,v)\sim (0,-v)$. If it had an orientation, hence a nowhere-zero continuous section, its coefficient on the interval would satisfy $f(2\pi )=-f(0)$ while staying nonzero. The intermediate value theorem makes that impossible. Since $SO(1)=\{1\}$, it has no $SO(1)$ reduction. For each $m\ge 1$, its direct sum with a trivial rank-$(m-1)$ bundle has transition matrix $\mathrm{diag}(-1,1,\dots ,1)$ and determinant line equal to the Möbius line. It is also nonorientable and has no $SO(m)$-structure. This obstruction concerns the bundle's orientation, independently of the existence of a metric.

A connection on a vector bundle specifies how to compare nearby fibers without identifying them globally. Several equivalent descriptions expose its algebra, local calculations and geometry.

The first description is a vector bundle covariant derivative, a real-linear mapEquivalently ${\nabla }_{X}s$ is linear over smooth functions in the vector field $X$, and obeys ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$ in the section. The derivative term distinguishes a connection from a bundle homomorphism. The Leibniz rule also makes it local: use a smooth cutoff equal to one near a point to see that its value depends only on the section near that point. Thus a connection restricts consistently to all open subsets.

The second description is a family of connection matrices in local frames. Write a frame as a row $e=(e_1,\dots ,e_m)$, so a section is $s=eu$ with coefficient column $u$. The connection is thenwhere $A$ is a matrix of one-forms determined by $\nabla e_j={\sum }_i{e}_i{A}^i{}_j$. If $e^{\prime }=eg$, the same section has $u=g{u}^{\prime }$. Expanding the Leibniz rule givesHence the change of frame of a vector-bundle connection isConversely, matrices of one-forms satisfying this transformation rule define $\nabla (eu)=e(du+Au)$ in each frame. The displayed calculation proves that these local formulas agree on overlaps; they therefore glue to a connection and obey its Leibniz rule. This proves equivalence of the first two descriptions in both directions.

A third description is parallel transport. Along a smooth curve $\gamma :[a,b]\to B$, a parallel section $s(t)=e(\gamma (t))u(t)$ satisfiesExistence and uniqueness for linear ordinary differential equations supply a unique solution for each initial fiber vector, smoothly dependent on that vector and on smooth families of curves. The frame transformation law makes these local solutions glue. The resulting map $P_{\gamma ,a,b}:E_{\gamma (a)}\to E_{\gamma (b)}$ is linear and invertible; reversing the curve gives its inverse, and concatenating curves composes the maps. It is generally dependent on the path, not just its endpoints.

One can recover the covariant derivative from this transport. For $\gamma (0)=b$ and $\dot{\gamma }(0)=X$, let $P_{0,t}$ be its transport from $0$ to $t$. ThenTo verify the formula rather than assume it, use a local frame and a fundamental solution matrix $P(t)$ with $P(0)=I$ and $\dot{P}=-A(\dot{\gamma })P$. Differentiation gives $(P^{-1}u{)}^{\prime }{|}_0=\dot{u}(0)+A_b(X)u(0)$, exactly the local covariant-derivative formula. Conversely the connection reconstructed from transport defined by this equation has the same differential equation and, by uniqueness, the same transport. Thus this viewpoint is also equivalent for the transport rules arising from a connection.

A fourth description is a linear horizontal distribution on a vector bundle. In a trivialization $E{|}_U=U\times {\mathbb{R}}^m$, putProjection maps this subspace isomorphically to $T_{b}B$, and it complements the vertical tangent space. Under the coordinate change $v=g{v}^{\prime }$, differentiation gives $\dot{v}=(dg)(X)v^{\prime }+g{\dot{v}}^{\prime }$. Thus the horizontal equation transforms into ${\dot{v}}^{\prime }=-A^{\prime }(X)v^{\prime }$, with exactly the same frame-change law. The subspaces consequently glue independently of trivializations. Their fiberwise linear dependence on $v$ is essential: an arbitrary horizontal complement can describe a nonlinear connection on the total space and need not define a vector-bundle connection. Conversely a horizontal complement with this linearity has a unique local expression of the displayed form, recovering the matrices $A$. Its horizontal lifts are precisely the parallel sections just described.

One can also express the geometry on the frame bundle as a principal connection. In local frame-bundle coordinates $(b,g)$, its Lie-algebra-valued form isIt is equivariant under right translation, $R_h^{\ast }\omega ={\mathrm{Ad}}_{h^{-1}}\omega$, and equals $\xi$ on the vertical vector generated by $\xi$. These properties define a principal connection; its kernel is the horizontal distribution on the frame bundle. Pulling back by a local frame section gives $A$, and its frame-change rule is once again the same formula. This identifies the principal viewpoint with the earlier descriptions.

Connections are not a vector space with a distinguished origin. The difference of two connections is linear over smooth functions in both $X$ and $s$, because their Leibniz terms cancel. It is a one-form with values in $\mathrm{End}E$. Conversely adding any such form to a connection gives another connection. Thus the collection is an affine space of vector-bundle connections modeled on ${\Omega }^1(B;\mathrm{End}E)$. The local curvature in the usual vector-bundle convention is $F=dA+A\wedge A$, representing ${\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$. It records the infinitesimal failure of path-independent parallel transport. Question 6 uses the opposite curvature sign, which will be stated explicitly there.

Fix $p$ in a Riemannian manifold $(M,g)$. The preliminary results used are the existence of the Levi-Civita connection, which is metric-compatible and torsion-free; local existence, uniqueness and smooth dependence for the geodesic ordinary differential equation; and the inverse function theorem. The local geodesic equation isso these ordinary differential equation results apply to the initial data $(p,v)$. On an open neighborhood of $0\in T_{p}M$, the solution ${\gamma }_v$ with ${\gamma }_v(0)=p$, ${\dot{\gamma }}_v(0)=v$ is defined up to time one, and the exponential map is ${\exp }_p(v)={\gamma }_v(1)$.

The geodesic equation is invariant under affine rescaling of the parameter. Uniqueness therefore gives ${\gamma }_v(t)={\exp }_p(tv)$ whenever these expressions are defined. In particularThe inverse function theorem now makes ${\exp }_p$ a diffeomorphism from a sufficiently small ball about zero in $T_{p}M$ onto a neighborhood of $p$. Choose a $g_p$-orthonormal basis $(e_i)$ and define coordinates byThese are geodesic coordinates, also called normal coordinates. They exist near every point. Radial coordinate rays $x^i(t)=t{v}^i$ are affinely parametrized geodesics issuing from $p$, and $g_{ij}(p)={\delta }_{ij}$. Substitution of all such rays into the geodesic equation at $t=0$ gives ${\Gamma }^i{}_{jk}(p)v^j{v}^k=0$ for all $v$. Since the connection is torsion-free, these coefficients are symmetric in $j,k$; polarization gives ${\Gamma }^i{}_{jk}(p)=0$. Metric compatibility then gives ${\partial }_k{g}_{ij}(p)=0$. These statements concern the center $p$; they do not make every translated coordinate line a geodesic.

The precise Gauss lemma isfor $v$ in such a sufficiently small neighborhood and every $w\in T_{p}M$. Thus the exponential map preserves the radial pairing; in particular it takes the radial direction orthogonally to the tangent directions of spheres about zero.

To prove it, form the smooth geodesic variationAll the curves with $s$ fixed are geodesics, so $D_{t}T=0$. The torsion-free Levi-Civita connection gives $D_{t}S=D_{s}T$ for this two-parameter map. This equality follows also by writing both derivatives in coordinates: the mixed partial derivatives agree and the Christoffel coefficients are symmetric. Metric compatibility yieldsThe constant-speed property permitted in the question gives $g(T,T)=g_p(v+sw,v+sw)$ for every $t$. Thus at $s=0$ the last expression is $g_p(v,w)$. Since $F(0,s)=p$, we have $S(0,s)=0$, and integrating in $t$ gives $g(T,S)(t,0)=t{g}_p(v,w)$. At $t=1$, the two variation derivatives are $T(1,0)=(d{\exp }_p{)}_{v}v$ and $S(1,0)=(d{\exp }_p{)}_{v}w$, proving the lemma. Its radial-length assertion follows by taking $w=v$, and its orthogonality assertion by taking $w\perp v$. Only constant speed is used here; the PDF correctly states constancy of the length of the velocity vector, whose dot is lost in the TeX aid.

Let $D$ be the Levi-Civita connection of the Riemannian metric. We use the curvature sign specified in the question, the negative of the usual vector-bundle curvature convention in Question 4. Extend $D$ to the covariant exterior derivative on tangent-bundle-valued forms and define the curvature two-form by $R=-D^2$. In a local frame with connection matrix $A$, it is the endomorphism-valued two-form $-(dA+A\wedge A)$. The transformation law for $A$ makes this transform by conjugation, so it is an intrinsic tensor. It measures the infinitesimal failure of parallel derivatives to commute, with this chosen sign.

For a tangent-bundle-valued one-form $\eta$, the covariant exterior derivative isApply this to the one-form $DZ$, for which $(DZ)(Y)=D_{Y}Z$. This calculation proves, rather than just renames, the operator expressionIt is linear over smooth functions in $Z$: the extra first- and second-derivative terms in $R(X,Y)(fZ)$ cancel by $[X,Y]f=X(Yf)-Y(Xf)$. The connection's linearity in its vector-field argument similarly shows linearity over smooth functions in $X$ and $Y$. Thus $R(X,Y)$ is an endomorphism of the tangent bundle, not a differential operator on its sections, and $R$ is a well-defined endomorphism-valued two-form.

The torsion-free identity is $D_{X}Y-D_{Y}X=[X,Y]$. Expand the cyclic curvature sum and use this identity to obtainThe last equality is the Jacobi identity for the Lie bracket of vector fields. This proves the first Bianchi identity

Lower the output index with the metric and define the Riemann curvature tensor byThis is a covariant four-tensor, because the curvature endomorphism is tensorial in all three inputs. Skew-symmetry of the curvature two-form immediately gives $R_{ji,kl}=-R_{ij,kl}$.

Metric compatibility gives the second skew-symmetry. Apply the scalar operator $[X,Y]-XY+YX$, which is zero, to $g(Z,W)$. Expanding each derivative using $Dg=0$ cancels the mixed products of first derivatives and leavesThus $R_{ij,lk}=-R_{ij,kl}$.

For a direct proof of pair interchange, choose the geodesic coordinates of Question 5 at an arbitrary point $p$. There $\Gamma (p)=0$ and $\partial g(p)=0$. Write $D_{{\partial }_i}{\partial }_k={\Gamma }^a{}_{ik}{\partial }_a$. The operator formula gives, at $p$,The Levi-Civita formula is $g_{al}{\Gamma }^a{}_{jk}=\frac{1}{2}({\partial }_j{g}_{kl}+{\partial }_k{g}_{jl}-{\partial }_l{g}_{jk})$. Differentiating at $p$, the products involving first derivatives of the metric and Christoffel coefficients vanish. The terms ${\partial }_i{\partial }_j{g}_{kl}$ cancel, leavingHere a comma denotes ordinary coordinate derivatives at the normal-coordinate center. Swapping $(i,j)$ with $(k,l)$ leaves this expression unchanged, by symmetry of $g$ and commutation of its second partial derivatives. This is the pair symmetry from the normal-coordinate curvature formula. Since $p$ was arbitrary and the equality is tensorial, it holds in every frame. We have proved all the requested symmetries:Changing the overall curvature sign would change the displayed coordinate formula's sign, but would not change the first Bianchi identity or any of these algebraic symmetries.