Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 17 (Differential Geometry)

Work with real vector bundles; for complex bundles replace the inner products below by positive-definite Hermitian forms. A smooth rank-$r$ vector bundle is a smooth total space $E$ with a smooth projection $\pi :E\to M$, a real vector space structure on each fiber $E_p$, and an open cover with vector bundle trivializationsEach ${\Phi }_i$ is a diffeomorphism over $U_i$ and is linear on fibers. The overlap maps have the form $(p,v)\mapsto (p,A_{ji}(p)v)$, where $A_{ji}:U_i\cap U_j\to G{L}_r(\mathbb{R})$ is smooth. The rank of a vector bundle may be specified separately on each connected component.

A smooth fiber metric is a positive-definite symmetric bilinear form $h_p$ on each fiber, with smoothly varying coefficients in every local frame. Equivalently it is a smooth section of ${\mathrm{Sym}}^2{E}^{\ast }$ which is positive definite at every point. Each trivialization supplies a local fiber metric $h_i$ by pulling back the Euclidean inner product.

A smooth manifold is Hausdorff and second-countable, hence paracompact. Choose a smooth, locally finite partition of unity $\{{\varphi }_i\}$ subordinate to a trivializing cover, with ${\varphi }_i\ge 0$, $\mathrm{supp}{\varphi }_i\subset U_i$, and ${\sum }_i{\varphi }_i=1$. DefineExtend each weighted term by zero outside its trivializing set; its support lies inside that set, so the extension is smooth. Local finiteness makes the sum smooth. At a point $p$ and for $0\ne v\in E_p$, every nonzero term ${\varphi }_i(p)h_i(v,v)$ is positive and at least one such term occurs. Thus $h_p(v,v)>0$. This proves that every smooth vector bundle admits a smooth fiber metric.

Let $r=\mathrm{rank}F$ and $s=\mathrm{rank}E$. Around any $p$, choose a local frame $f_1,\dots ,f_r$ of the vector subbundle $F$. Choose additional smooth sections $e_{r+1},\dots ,e_s$ of $E$ which complement these vectors at $p$. They can be obtained by constant-coordinate sections in a vector bundle trivialization. The determinant of the resulting frame is nonzero at $p$, hence on a sufficiently small neighborhood. Thusis an adapted local frame of $E$ near $p$.

On the fiberwise quotient, the classes $[e_{r+1}],\dots ,[e_s]$ form a basis. Give the quotient vector bundle its local coordinates byAn overlap between adapted frames has a smooth transition function of a vector bundle represented by an invertible block upper triangular matrix,The zero lower-left block expresses that both sets of first $r$ frame vectors span $F$. Its lower-right block $D$ is smooth and invertible. Passing to quotient coordinates removes the upper block, leaving precisely the transition $D$. These lower-right blocks satisfy the cocycle identity because the original transition matrices do. Consequently they define a compatible smooth bundle atlas of rank $s-r$. This proves $E/F$ is a smooth vector bundle, and its fiberwise quotient map $q:E\to E/F$ is a smooth vector bundle morphism. The construction is independent of the adapted frames: any two choices have the same type of compatible transition. The usual quotient topology agrees with this atlas, since in adapted coordinates $q$ is an open linear projection on each fiber.

Choose the smooth fiber metric supplied by part (a), and let $P_p:E_p\to F_p$ be orthogonal projection. These projections vary smoothly. Indeed, in a local frame of $E$ write its metric as a positive-definite matrix $G(p)$, and let the columns of $A(p)$ be a frame of $F$. ThenThe matrix $A^{t}GA$ is positive definite, so its inverse is smooth. This formula proves smoothness of the intrinsic projections in every chart. It also gives $P^2=P$ and $\mathrm{im}P=F$.

The orthogonal complement $F^{\perp }=\ker P$ is a smooth vector subbundle: applying $I-P$ to a complement frame gives a frame of its fibers, of constant rank $s-r$, after shrinking the chart. Fiberwise,and the restriction $q{|}_{F^{\perp }}:F^{\perp }\to E/F$ is a smooth fiberwise bijective vector bundle morphism. In local frames its matrix is invertible, with smooth inverse, so it is a vector bundle isomorphism.

More explicitly, the desired orthogonal splitting of a vector subbundle isIts inverse is $(f,[v])\mapsto f+(I-P)v$. Replacing $v$ by $v+w$ with $w\in F_p$ leaves $(I-P)v$ unchanged, so the inverse is well-defined. Both maps are smooth and linear on fibers, and their compositions are identities. The splitting exists globally but depends on the chosen fiber metric.

A smooth differential form of degree $p$ is a smooth section of ${\Lambda }^p{T}^{\ast }M$, soIt is equivalently a smooth alternating covariant $p$-tensor field. In a coordinate chart $x^1,\dots ,x^n$ it has a unique expressionwith smooth coefficients. At $p=0$ these are the smooth functions, and for $p>n$ the space is zero.

Define the exterior derivative in a chart byTo check that it is well-defined, let $y^1,\dots ,y^n$ be other smooth coordinates. The expression $d{y}^a={\sum }_i({\partial }_i{y}^a)d{x}^i$ satisfiesbecause the coefficients are symmetric and the wedge products antisymmetric. The coordinate formula also satisfies the graded Leibniz rule, and for a function $b$ the chain rule gives $d_{x}b={\sum }_a(\partial b/\partial y^a)d{y}^a=d_{y}b$. Applying these two facts to $\omega ={\sum }_I{b}_{I}d{y}^I$ yields $d_x\omega ={\sum }_I{d}_y{b}_I\wedge d{y}^I=d_y\omega$. Thus the local operators agree on overlaps and define a global linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$.

A second application in any chart givesagain by symmetry of mixed derivatives and antisymmetry of the wedge product of differential forms. Consequently every exact differential form is closed. Put ${\Omega }^{-1}(M)=0$ and define the de Rham cohomology byThe identity $d^2=0$ is exactly what puts the denominator inside the numerator, making this quotient meaningful. Since ${\Omega }^p(M)=0$ for $p>\dim M$, $H_{\mathrm{dR}}^p(M)=0$ in those degrees.

The two components of a disjoint union are open and closed. Restriction therefore gives an isomorphismIts inverse assigns the first form on $U$ and the second on $V$; it is smooth because there are no overlaps on which compatibility must be imposed. The exterior derivative is local, so this is an isomorphism of cochain complexes.

A form on $M$ is a closed differential form exactly when both restrictions are closed. It is an exact differential form exactly when both restrictions are exact: primitives on $U$ and $V$ join into a global primitive. Taking the quotient therefore yields the de Rham cohomology of a finite disjoint union:The same argument covers degree zero, with the space of degree-minus-one forms taken to be zero.

The direct-sum assertion fails in both degrees when the pieces overlap. Take genuine open covers, so the failure does not depend on any subtle choice of smooth structures on subsets.

For degree zero, let $M=\mathbb{R}$, $U=(-\infty ,1)$ and $V=(-1,\infty )$. By degree-zero de Rham cohomology, $H_{\mathrm{dR}}^0$ consists of locally constant functions. Each of $M,U,V$ is connected and nonempty, soThey are not isomorphic as real vector spaces. The restriction map is $c\mapsto (c,c)$; the overlap forces equality instead of allowing two independent constants.

For degree one, let $M=S^1$, and remove distinct points $a,b$ to form $U=S^1\setminus \{a\}$ and $V=S^1\setminus \{b\}$. Each is diffeomorphic to an open interval. Every one-form $f(t)dt$ on an interval has the smooth primitive ${\int }_{t_0}^{t}f(s)ds$, so both pieces have zero first de Rham cohomology. On the circle, the form $\eta =xdy-ydx$ is closed because all two-forms on a one-dimensional manifold vanish. The parameterization $(x,y)=(\cos t,\sin t)$ givesAn exact form has zero integral around this loop: the integral of $dF$ is the difference of the endpoint values of the periodic function $F$. Thus $[\eta ]\ne 0$, while the proposed right side is zero, disproving the isomorphism.

In fact $H_{\mathrm{dR}}^1(S^1)\cong \mathbb{R}$. For a periodic one-form $f(t)dt$, subtract $cdt$ where $c=(2\pi {)}^{-1}{\int }_0^{2\pi }f(t)dt$. The remainder has the periodic primitive ${\int }_0^t(f(s)-c)ds$, and hence is exact. This also explicitly identifies the nonzero class used above.

A left-invariant vector field $X$ on a Lie group satisfiesor equivalently $(L_g{)}_{\ast }X=X$ for every $g$. For $\xi \in T_{e}G$, defineThis is smooth because multiplication is smooth and its differential depends smoothly on the base points. The identity $L_g\circ L_h=L_{gh}$ gives $(d{L}_g{)}_h{X}_h^{\xi }=X_{gh}^{\xi }$, so $X^{\xi }$ is a left-invariant vector field. Conversely, left invariance forces any such field to have this form with $\xi =X_e$. Evaluation at $e$ and the construction above are inverse linear maps. HenceThe Lie bracket of vector fields is their commutator as derivations on smooth functions: $[X,Y]f=X(Yf)-Y(Xf)$. It is another vector field because the cross terms cancel in the Leibniz rule for its action on a product. The bracket is preserved by a diffeomorphism: conjugating the derivation operators by pullback of a smooth function preserves their commutator. Thereforefor left-invariant $X,Y$. These fields are consequently closed under the bracket.

Transport the bracket to the tangent space byBilinearity and antisymmetry come from the commutator. The Jacobi identity follows by expanding the three nested commutators of derivation operators: their six terms cancel in pairs. Thus $T_{e}G$ is a Lie algebra, and the displayed correspondence is a Lie algebra isomorphism onto the algebra of left-invariant vector fields.

Let $e$ be the identity. For a left-invariant differential form $\omega$ and a left-invariant vector field $X$, one hasThe value is therefore independent of $g$, even if the Lie group is disconnected. The same calculation applies to another left-invariant field $Y$.

The standard exterior derivative of a one-form evaluated on vector fields isBoth differentiated functions are constant in the present situation. ThusThis identity is a direct consequence of left invariance and the stated general formula for the exterior derivative.

The correspondence in part (a) respects Lie brackets. If the Lie algebra is abelian, any two left-invariant vector fields satisfy $[X,Y]=0$ everywhere, not just at the identity. Part (b) therefore gives $d\omega (X,Y)=0$ for every such pair.

At a point $g$, the map $(d{L}_g{)}_e:T_{e}G\to T_{g}G$ is an isomorphism. Thus every tangent vector at $g$ is the value of a left-invariant vector field. The two-form $d\omega$ consequently vanishes on every pair of tangent vectors at every point, provingNo connectedness assumption on the Lie group is needed: the argument uses its left translations and its abelian Lie algebra.

A connection on $TM$ is a metric connection for $g$ if for all smooth vector fields $X,Y,Z$,Equivalently, its induced covariant derivative of the metric satisfies $\nabla g=0$. This is also called a metric-compatible connection.

The torsion tensor of an affine connection isThe connection is symmetric, or a torsion-free connection, exactly whenIn a coordinate frame, write ${\nabla }_{{\partial }_i}{\partial }_j={\sum }_k{\Gamma }_{ij}^k{\partial }_k$. Since coordinate fields commute, the condition is ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$; this is the reason for the word symmetric.

The Levi-Civita connection is the unique affine connection which is both a metric connection and a torsion-free connection. Uniqueness is encoded by the Koszul formulaIt follows by adding the metric-compatibility identities with derivative fields $X,Y$, subtracting the one with derivative field $Z$, and replacing differences of covariant derivatives by Lie brackets. Nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from all its pairings with $Z$. The existence and uniqueness of the Levi-Civita connection supplies existence; no existence proof is needed here.

In Cartesian coordinates on Euclidean space, defineThis is an affine connection by the ordinary product rule. Its connection coefficients in this coordinate frame are all zero. The component formula for the Lie bracket of vector fields gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, so it is torsion-free. Also, since the Euclidean metric has constant coefficients,This is exactly metric compatibility. By uniqueness it is the Euclidean Levi-Civita connection. In particular constant vector fields are parallel, while the derivative of a general field is given by the boxed formula.

In coordinates, write ${\nabla }_{{\partial }_i}{\partial }_{\ell }={\Gamma }_{i\ell }^k{\partial }_k$. The Riemannian curvature two-form is defined by the matriceswith repeated indices summed. It assigns to $X,Y$ the fiberwise endomorphism $R(X,Y)$ whose entries are $R^k{}_{\ell ij}{X}^i{Y}^j$. We can prove both coordinate independence and the required intrinsic identity as follows.

Define an operator on vector fields by $C(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. It is alternating in $X,Y$. For a smooth function $f$, the identities $[fX,Y]=f[X,Y]-Y(f)X$ and ${\nabla }_Y(fZ)=Y(f)Z+f{\nabla }_{Y}Z$ givethe two terms involving $Y(f){\nabla }_{X}Z$ cancel. Alternation gives linearity over smooth functions in $Y$ as well. Expanding in the final argument yieldsThus $C$ is tensorial in all three arguments, defining a global endomorphism-valued two-form.

For coordinate fields ${\partial }_i,{\partial }_j$, their bracket is zero. Expand $[{\nabla }_{{\partial }_i},{\nabla }_{{\partial }_j}]Z$ using $Z=Z^{\ell }{\partial }_{\ell }$. The second derivatives of $Z^{\ell }$ cancel by equality of mixed partials. The terms containing first derivatives of $Z^{\ell }$ cancel in pairs, and the remaining coefficient of $Z^{\ell }{\partial }_k$ is exactly $R^k{}_{\ell ij}$ above. Hence the coordinate definition equals the global tensor $C$, establishing coordinate independence and provingThis convention for the Riemann curvature tensor is valid for any affine connection; in particular it applies to the Levi-Civita connection.

Take $\mathbf{n}$ to be the unit normal, so $P(v)=v-⟨v,\mathbf{n}⟩\mathbf{n}$ is orthogonal projection onto $TM$. With $D$ the Euclidean Levi-Civita connection, the proposed derivative isIt is a smooth tangent field, and independence of the extensions is given. Linearity follows from linearity of $D$ and $P$. Moreover, for $f\in C^{\infty }(M)$, take local extensions of $f,X,Y$ to obtainsince $PY=Y$. Thus it is an affine connection on $M$.

For tangent fields $Y,Z$, the normal component of $DY$ pairs to zero with $Z$. Differentiating the ambient inner product along $X$ givesHence it is a metric connection for the induced Riemannian metric. The ambient torsion vanishes, soHere the restriction of the ambient bracket is the intrinsic bracket and is tangent to $M$: derivations of functions restricted to $M$ give the same bracket, or this follows in submanifold coordinates. Thus the projected derivative is also a torsion-free connection. Uniqueness now provesThis is the projected ambient connection. The projection formula requires a unit normal. If a nonunit normal is used, its normal term must instead be $⟨DY,\mathbf{n}⟩\mathbf{n}/⟨\mathbf{n},\mathbf{n}⟩$. For example, on the unit circle, taking $\mathbf{n}=2p$ and differentiating its unit tangent along itself gives $DY=-p$; the unnormalized printed expression would be $3p$, which is not tangent. Normalizing the normal removes that ambiguity.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhich satisfies $\nabla (fs)=df\otimes s+f\nabla s$ for every smooth function $f$ and section $s$. Here ${\Omega }^1(M;E)=\Gamma (T^{\ast }M\otimes E)$. Equivalently, writing ${\nabla }_{X}s=(\nabla s)(X)$, it is linear over smooth functions in $X$, linear over $\mathbb{R}$ in $s$, and obeysChoose a trivializing open cover and a local frame $e_{i1},\dots ,e_{ir}$ on each $U_i$. Differentiating the coefficient functions gives a local connection on a vector bundle:No compatibility of these local connections is assumed. Choose a smooth locally finite partition of unity $\{{\varphi }_i\}$ subordinate to the cover and defineextending each weighted term by zero outside $U_i$. Its support lies inside $U_i$, so each extension is smooth, and local finiteness makes the whole sum smooth. The operation is linear over smooth functions in $X$. For the Leibniz rule,using ${\sum }_i{\varphi }_i=1$. This proves every smooth vector bundle admits a connection by the construction of a vector bundle connection by a partition of unity.

Fix the coefficient-column convention: write the local frame as a row $e=(e_1,\dots ,e_r)$ and a section as $s=eu$ for a column $u$ of functions. Define the connection matrix $\theta =({\theta }_{ij})$ byEach entry is a smooth one-form. Define the curvature form of a connection byThe same Leibniz cancellations as for tangent-bundle curvature show that this is linear over smooth functions in $X,Y,s$. Thus it is a two-form with values in the endomorphisms of $E$. Its curvature matrix $\Theta =({\Theta }_{ij})$ is specified by $R^E(X,Y)e_j={\sum }_i{e}_i{\Theta }_{ij}(X,Y)$.

The covariant exterior derivative on coefficient forms is $d+\theta \wedge$. Squaring it on $u$ givesTherefore the Cartan curvature matrix equation isMatrix-valued wedge products here combine matrix multiplication with the wedge product of differential forms, in the stated order.

Using $d^2=0$ and the graded Leibniz rule,On the other hand, associativity of the endomorphism-valued exterior product givesThe cubic terms cancel, proving the requested Bianchi identity

Let another local frame be $e^{\prime }=eA$ with $A:U\to G{L}_r(\mathbb{R})$ smooth. Since the coefficient columns satisfy $u=A{u}^{\prime }$, the Leibniz rule gives the change of frame of a vector-bundle connection:The curvature form of a connection is linear over smooth functions in its section argument, so its matrix transforms homogeneously:ConsequentlyThe entries of $A$ are functions, of degree zero, so ordinary cyclic invariance of trace applies without a graded sign. The local forms therefore glue to a globally defined two-form $\alpha =\mathrm{tr}\Theta$, the trace of vector-bundle curvature.

Apply the Bianchi identity proved in part (b):For the first term,The middle interchange has sign $(-1{)}^{2\cdot 1}=1$, and the next equality simply renames $i,j$. ThusOne can also check closedness locally: the off-diagonal terms in $\mathrm{tr}(\theta \wedge \theta )={\sum }_{i,j}{\theta }_{ij}\wedge {\theta }_{ji}$ cancel in pairs, while the diagonal terms vanish. Hence $\alpha =d(\mathrm{tr}\theta )$ in each frame, and $d^2=0$ gives the same conclusion.