Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 19 (Differential Geometry)

A smooth vector field on a smooth manifold $M$ is a smooth section $X:M\to TM$ of its tangent bundle. It acts on smooth functions by $Xh(p)=d{h}_p(X_p)$. In a manifold chart, $X={\sum }_i{X}^i{\partial }_i$. The Lie bracket of vector fields is the commutator of their actions on functions:Expanding in coordinates makes all second derivatives cancel and givesThus the commutator is again a smooth vector field, not a second-order operator. The coordinate expression represents the intrinsic derivation on functions, so agrees on overlaps.

For a diffeomorphism $f:M\to N$, its pushforward of a vector field isFor every smooth function $h$ on $N$, the chain rule gives $((f_{\ast }X)h)\circ f=X(h\circ f)$. Applying this twice yieldsSmooth functions determine a tangent vector, and $f$ is surjective, so

On a Lie group, a left-invariant vector field satisfies $(L_g{)}_{\ast }X=X$ for every left translation $L_g(h)=gh$. Given $\xi \in T_{e}G$, setThis field is smooth by smoothness of multiplication. The equality $L_a{L}_g=L_{ag}$ proves left invariance. Conversely, left invariance forces $X(g)=(d{L}_g{)}_{e}X(e)$, so $X_{\xi }$ is the unique left-invariant field with value $\xi$ at $e$. This is a linear isomorphism between $T_{e}G$ and the space of left-invariant vector fields.

The pushforward identity applied to $L_g$ shows that $[X_{\xi },X_{\eta }]$ is again left invariant. Consequentlydefines a Lie bracket on $T_{e}G$. Bilinearity and antisymmetry follow from the operator commutator. Its Jacobi identity follows by expanding the three nested commutators: the six triple compositions cancel in pairs. These properties transfer through the linear isomorphism, making $T_{e}G$ the Lie algebra of $G$.

For the general linear group $G=\mathrm{GL}(n,\mathbb{R})$, use matrix-entry coordinates $x_{pq}$. It is an open subset of $M_n(\mathbb{R})$, so every tangent space is naturally $M_n(\mathbb{R})$. Since $L_g(h)=gh$, its differential sends $A$ to $gA$, and henceFor $A,B\in M_n(\mathbb{R})$,and reversing $A,B$ gives $(gBA{)}_{pq}$. Thus $[X_A,X_B]=X_{AB-BA}$; evaluating at the identity gives the matrix commutator

If a vector bundle $E\to M$ has rank $k$, form the exterior power of a vector bundle fibrewise: its fibre at $p$ is ${\Lambda }^r{E}_p$. A vector bundle trivialization over $U$ induces one with fibre ${\Lambda }^r{\mathbb{R}}^k$. Transition matrices $A_{\alpha \beta }(p)$ induce ${\Lambda }^r{A}_{\alpha \beta }(p)$, which are smooth and obey the cocycle law because exterior powers preserve composition. They glue to a smooth bundle of rank $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$; for $r>k$ it is the zero bundle. At $r=k$ it is the determinant line bundle.

A real line bundle is trivial precisely when it has a nowhere-zero smooth section: a section $s$ gives the bundle isomorphism $(p,t)\mapsto ts(p)$ from $M\times \mathbb{R}$. Apply this to ${\Lambda }^n{T}^{\ast }M$. Suppose first that an oriented atlas is given. Its local coordinate formsare positive multiples of one another on every overlap. Choose a subordinate smooth partition of unity ${\rho }_{\alpha }\ge 0$. Extend ${\rho }_{\alpha }{\nu }_{\alpha }$ by zero outside its chart and set $\nu ={\sum }_{\alpha }{\rho }_{\alpha }{\nu }_{\alpha }$. Local finiteness makes this smooth. At any point at least one weight is positive, and all summands are nonnegative multiples of the same positive top form. Therefore $\nu$ never vanishes, proving triviality of the line bundle.

Conversely let $\nu$ be a nowhere-zero section of ${\Lambda }^n{T}^{\ast }M$. In each sufficiently small connected chart write $\nu =a_{\alpha }d{x}_{\alpha }^1\wedge \cdots \wedge d{x}_{\alpha }^n$. Its coefficient has constant nonzero sign. If necessary reverse the first coordinate, making $a_{\alpha }>0$. On overlaps the equality of the two expressions givesso each transition determinant is positive. These charts cover $M$, provingFor dimension zero both assertions hold with the empty exterior product equal to one.

For the orientability of a vector bundle total space, write $\pi :E\to M$, with rank $r$ and orientable base. The vertical tangent vectors at $v\in E_p$ identify with $E_p$ by differentiating translation in that fibre. The differential of $\pi$ therefore gives the exact sequence of vector bundles on $E$Taking determinant line bundles givesTo see the determinant identification directly, wedge a vertical frame with any lifts of a base frame; altering a lift by a vertical vector does not change that top wedge. Since $M$ is orientable, $\det TM$ is trivial. Consequently $\det TE\cong {\pi }^{\ast }\det E$. If $\det E$ is trivial, so is its pullback, making the total space orientable. Conversely, if $E$ is orientable as a manifold, its tangent determinant is trivial; restricting this isomorphism to the zero section gives triviality of $\det E$. The tangent and cotangent determinant lines are dual, so either tests orientability. We obtain

For a cotangent bundle the base need not be orientable. If base coordinates change from $x$ to $y=y(x)$ with Jacobian $J$, fibre coordinates of the same covector change from $\xi$ to $\eta =J^{-T}\xi$. The full coordinate change on $T^{\ast }M$ has block JacobianThus every induced cotangent coordinate change preserves orientation, even when the base change reverses it. These charts form an oriented atlas on the total space, proving every cotangent bundle is orientable. Equivalently the canonical symplectic form of the cotangent bundle has a nowhere-zero top exterior power, as in cotangent bundle orientation.

In coordinates, write a differential form as $\eta ={\sum }_I{\eta }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Its exterior derivative isFor $r=0$ this is the ordinary differential $df$. The chain rule, together with cancellation of symmetric second derivatives against the exterior product, proves that these formulas agree under coordinate changes. They give an $\mathbb{R}$-linear map $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ with $d^2=0$ and $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^r\alpha \wedge d\beta$ when $\alpha$ has degree $r$.

For a differential one-form $\omega ={\sum }_j{\omega }_{j}d{x}^j$, evaluation givesExpanding $X(\omega (Y))-Y(\omega (X))$ gives these coefficient derivatives plus ${\sum }_j{\omega }_j{\sum }_i(X^i{\partial }_i{Y}^j-Y^i{\partial }_i{X}^j)$. The latter is $\omega ([X,Y])$ by the Lie bracket of vector fields formula. Hence

A connection on a vector bundle is an $\mathbb{R}$-linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $D(f\sigma )=df\otimes \sigma +fD\sigma$. Evaluation on a vector field gives $D_X\sigma$, which is $C^{\infty }$-linear in $X$ and satisfies $D_X(f\sigma )=X(f)\sigma +f{D}_X\sigma$. The covariant exterior derivative extends this to bundle-valued forms byThis respects $(f\alpha )\otimes \sigma =\alpha \otimes (f\sigma )$: applying the rule on either side gives the same extra term $df\wedge \alpha \otimes \sigma$. It therefore defines an operator on bundle-valued forms independently of their local tensor representations. In a local frame with connection matrix $A$, its coefficient-column formula is $d^E=d+A\wedge$.

Define the curvature form of a connection by $R\sigma =(d^E{)}^2\sigma$. The extra terms in $(d^E{)}^2(f\sigma )$ cancel, so this equals $f(d^E{)}^2\sigma$. Equivalently the local formula iswhich acts fibrewise on the section coefficients. Thus curvature is a smooth endomorphism-valued two-form, a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(E)$.

For an $E$-valued one-form $a$, applying the ordinary one-form formula in a frame givesTaking $a=D\sigma$ proves the required curvature identityThe bracket correction is what makes the right side tensorial in the vector fields and the section.

Now define the endomorphism bundle connection by $({\tilde{D}}_X\alpha )(s)=D_X(\alpha s)-\alpha (D_{X}s)$. This expression is $C^{\infty }$-linear in $s$: the two terms involving $X(f)\alpha (s)$ cancel when $s$ is replaced by $fs$. It is therefore a smooth bundle endomorphism. It is $C^{\infty }$-linear in $X$, andThese identities prove it is a connection on a vector bundle. In a local frame it reads ${\tilde{D}}_X\alpha =X(\alpha )+[A(X),\alpha ]$.

For the full commutator identity for an endomorphism connection, expand once more:Subtracting the expression with $X,Y$ interchanged cancels both middle terms. ThusHere $D_X,D_Y$ are differential operators on sections, and the square brackets are their operator commutators. In general they are not fibrewise bundle endomorphisms; the curvature operator becomes fibrewise linear after subtracting $D_{[X,Y]}$.

Subtracting ${\tilde{D}}_{[X,Y]}\alpha$ from this identity gives the curvature of an endomorphism bundle connection:Consequently $\tilde{R}=0$ precisely when every $R_p(X,Y)$ commutes with every endomorphism of $E_p$. Arbitrary such endomorphisms can be extended locally in a frame, so this is a fibrewise assertion. A matrix commuting with all diagonal matrix units is diagonal; commuting also with all off-diagonal matrix units forces all its diagonal entries to coincide. The centre of the full endomorphism algebra therefore consists of scalar matrices. For rank $k>0$, define the smooth two-form $\omega =k^{-1}\mathrm{tr}R$. We obtain the scalar-curvature criterion for a flat endomorphism connectionThe converse follows immediately because scalar endomorphisms commute with all endomorphisms. For the zero bundle both curvatures vanish and one can take $\omega =0$. Scalar curvature here means the scalar endomorphism-valued curvature of this bundle connection, not the scalar trace curvature of a Riemannian manifold.

A vector field along a curve $\gamma :[a,b]\to M$ is a smooth section of the pullback tangent bundle ${\gamma }^{\ast }TM$: it assigns $V(t)\in T_{\gamma (t)}M$ smoothly. It need not be the restriction of one ambient vector field, since its values at different times with the same image point may differ.

Given a Koszul connection, use a local frame $e_j$ of $TM$ and write $V(t)={\sum }_j{v}^j(t)e_j(\gamma (t))$. Its covariant derivative along a curve isThe first term differentiates the time-dependent coefficients, and the second uses the connection on the local frame. Under a change of frame, differentiating the change matrix supplies exactly the extra term in the connection transformation, so the expression is independent of the frame. This definition also works when $\dot{\gamma }(t)=0$. In coordinate components it isA field is parallel when $D_{t}V=0$.

Thus a parallel field solves the linear system $\dot{v}=-A(t)v$, with smooth coefficient matrix $A(t)$. Given its value at one time, existence and uniqueness for a linear ordinary differential equation determine it on each coordinate segment. To obtain it on all of $[a,b]$, cover the compact parameter interval by finitely many frame neighborhoods and subdivide so that each closed subinterval lies in one such neighborhood. The coefficients are bounded on each subinterval, and the standard estimate $\Vert v(t)\Vert \le \Vert v(t_0)\Vert \exp (\int \Vert A(s)\Vert ds)$ prevents finite-time blowup. The solutions match uniquely at the subdivision points. This proves existence and uniqueness of the requested global parallel field.

Define parallel translation by ${\tau }_t(v)=V_v(t)$, where $V_v(a)=v$ and $D_t{V}_v=0$. Linearity of the equation and uniqueness make ${\tau }_t$ linear. Solving the same equation backwards gives its inverse; equivalently, a solution zero at time $t$ must be identically zero. Therefore

A metric connection for a Riemannian metric satisfies $\nabla g=0$, orPulling this identity along the curve givesFor parallel fields the right side vanishes, provingFor a piecewise smooth path compose the parallel transport maps on its smooth segments. On a closed path their domain and codomain are the same tangent space, so the resulting map is orthogonal. Its determinant need not be positive on a nonorientable surface.

An explicit reflection example is the reflection holonomy along a closed projective geodesic. Give ${\mathbb{RP}}^2$ the round metric induced from the sphere by the antipodal quotient. The Levi-Civita connection descends because the antipodal map is an isometry, so the quotient map is a local isometry carrying geodesics and parallel fields to their counterparts. The great-circle arcprojects to a smooth closed geodesic: $c(t+\pi )=-c(t)$, including all derivatives under the quotient identification. Along it the fieldsare parallel. The first is the velocity of a great-circle geodesic. The second is tangent, unit length, and orthogonal to the first. Metric compatibility makes its covariant derivative orthogonal to both frame vectors: differentiate its constant length and its zero inner product with the parallel first vector. Its derivative therefore vanishes in the two-dimensional tangent plane. At $t=\pi$, $E_1(\pi )=-E_1(0)$ and $E_2(\pi )=E_2(0)$. Identifying the endpoint with the start through the derivative of the antipodal map, namely $-I$, fixes the returned $E_1$ and negates the returned $E_2$. HenceThis is a reflection fixing the geodesic direction.

Finally orient the unit sphere outward and traverse a spherical triangle with its interior on the left. Let its angles be $\alpha ,\beta ,\gamma$. Its parallel transport preserves both the metric and orientation, so is a rotation by some angle $\delta$ in the initial tangent plane. On each geodesic side the unit tangent is parallel. At a vertex with angle $\alpha$, the actual forward tangent turns left by $\pi -\alpha$ relative to the transported incoming tangent. Tracking the tangent around all three sides and making those three corner turns returns it to its initial direction.

Rotations of oriented two-dimensional tangent planes commute with orientation-preserving parallel transport. Consequently the total angle in this tracking procedure is the transport angle plus the three corner turns:This determines parallel transport around a spherical triangle asBy the spherical excess formula, this is rotation by the enclosed area on the unit sphere. Reversing traversal negates the angle. For an octant triangle all three angles are $\pi /2$, giving a positive $\pi /2$ rotation.

Use the Riemannian curvature two-form conventionand define its four-covariant Riemann curvature tensor by $R(X,Y,Z,W)=g(R(X,Y)W,Z)$. This makes $R(X,Y,X,Y)$ positive on an orthonormal two-plane of the unit sphere. The full algebraic symmetries areTogether with multilinearity, these are antisymmetry in each pair, interchange of pairs, and the first Bianchi identity.

Define the Ricci tensor by the basis-independent contractionfor any orthonormal basis $(e_a)$. The trace is invariant under change of basis, and contraction of the smooth curvature tensor gives a smooth covariant two-tensor. Pair interchange gives $R(e_a,Y,e_a,W)=R(e_a,W,e_a,Y)$, so $\mathrm{Ric}(Y,W)=\mathrm{Ric}(W,Y)$. Thus the Ricci tensor is a symmetric covariant 2-tensor.

The sectional curvature of the two-plane $\sigma =\mathrm{span}(u,v)$ isBoth numerator and denominator multiply by the square of the determinant under a change of basis of $\sigma$, so this is a plane invariant. The directional Ricci curvature of $u\ne 0$ is $\mathrm{Ric}(u,u)/g(u,u)$. For an orthonormal basis beginning with $u/|u|$, it equals the sum of the sectional curvatures of the planes containing that basis vector.

In dimension three, suppose this directional value is $c$ at the point under consideration. For any orthonormal triple $(e_1,e_2,e_3)$, put $K_{ij}=K(\mathrm{span}(e_i,e_j))$. Thenso all three are $c/2$. Every two-plane can be the span of the first two members of such a triple. ConsequentlyThis is a pointwise conclusion, as required; it uses sectional curvatures from Ricci curvature in dimension three and asserts no spatial constancy.

For the product, the canonical identification $T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N$ defines the product Riemannian metricIt is smooth in product charts and positive definite, since a nonzero vector has a nonzero component in at least one factor. Those components are orthogonal. The coordinate connection computation in the subparts below givesTaking the trace in orthonormal bases of the two factors therefore gives ${\mathrm{Ric}}_{M\times N}={\mathrm{Ric}}_M\oplus {\mathrm{Ric}}_N$.

For the final example choose the two round unit spheres. Each has sectional curvature one and Ricci tensor equal to its metric. Hence the product has $\mathrm{Ric}=g$, so every directional Ricci curvature is one. A two-plane tangent to one factor has sectional curvature one, whereas a mixed two-plane has sectional curvature zero by (iii). This gives constant Ricci curvature without constant sectional curvature:Both kinds of plane occur at every point, so the sectional curvatures fail to be constant even at a single point.

Use product coordinates $x^i$ on $M$ and $y^{\alpha }$ on $N$. The lifted vector fields have the form $X={\sum }_i{X}^i(x){\partial }_{x^i}$ and $Y={\sum }_{\alpha }{Y}^{\alpha }(y){\partial }_{y^{\alpha }}$. The coefficients of each depend only on its own coordinate set, and the coordinate partial derivatives commute. The Lie bracket of vector fields formula therefore givesThus $\boxed{[X,Y]=0}$ for fields lifted separately from the two factors. The lift hypothesis matters: arbitrary product-manifold fields can have coefficients depending on both factors.

The product Riemannian metric has block matrixIn the given Christoffel symbols formula, every mixed coefficient vanishes. Off-diagonal metric entries are zero, ${\partial }_{y^{\alpha }}{g}_{ij}=0$, and ${\partial }_{x^i}{g}_{\alpha \beta }=0$. These facts give ${\Gamma }_{i\alpha }^A=0$ for every output index $A$, as well as ${\Gamma }_{ij}^{\alpha }=0$ and ${\Gamma }_{\alpha \beta }^i=0$. The remaining pure blocks are precisely the Christoffel symbols of the respective factor metrics.

For the lifted fields of (i), ${\partial }_{x^i}{Y}^{\alpha }=0$. ConsequentlyThus $\boxed{{\nabla }_{X}Y=0}$. The same calculation gives ${\nabla }_{Y}X=0$; differentiating two fields from one factor gives the lifted factor covariant derivative. This identifies the product Levi-Civita connection with the product affine connection and proves the curvature splitting used in the root solution.

Since $Y$ is lifted from $N$, its covariant derivative ${\nabla }_{Y}Y$ is also lifted from $N$. Part (ii) therefore gives ${\nabla }_X({\nabla }_{Y}Y)=0$. It also gives ${\nabla }_{X}Y=0$, and part (i) gives $[X,Y]=0$. Substituting into the Riemannian curvature two-form definition yieldsWith the four-slot convention fixed above, this provesFor nonzero lifted vectors from different factors they are orthogonal, so the sectional-curvature denominator is positive. Hence the corresponding mixed plane has sectional curvature zero, providing the mixed-plane value in the $S^2\times S^2$ example.