Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 15 (Differential Geometry)

The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincare lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.

A Lie group is a finite-dimensional smooth manifold with a group structure for which multiplication and inversion are smooth. The group defined here is the compact symplectic group $Sp(n)=U(2n)\cap Sp(2n,\mathbb{C})$, rather than the full complex symplectic group. Also, the displayed matrix expression requires $B$ to be $2n\times 2n$; the printed $n\times n$ size is incompatible with $J$.

Since $J^2=-I$, we have $J^{-1}=-J$. The matrix exponential commutes with conjugation, as follows term by term from its absolutely convergent power series. ConsequentlyTo construct logarithm charts for the compact symplectic group, consider the real vector spaceDifferentiating the defining identities at $I$ gives precisely these conditions. Conversely if $X\in \mathfrak{k}$, $e^X$ is unitary andso $e^X\in Sp(n)$. These are the infinitesimal conditions of the compact symplectic Lie algebra.

Near $I$, the convergent matrix logarithm series $\log (I+C)={\sum }_{m\ge 1}(-1{)}^{m+1}{C}^m/m$ is smooth and inverse to the matrix exponential near zero. These local inverses respect transpose, conjugate transpose, and conjugation; also $\log (A^{-1})=-\log A$ when both matrices are sufficiently close to $I$. Shrink their neighborhoods accordingly. If $A\in Sp(n)$ there, unitarity givesThe symplectic identity is equivalent to $A^t=J^{-1}{A}^{-1}J$, so, putting $X=\log A$, it gives $X^t=-J^{-1}XJ=JXJ$, equivalently $XJ+J{X}^t=0$. Thus this local logarithm restricts to a bijection between a neighborhood of $I$ in $Sp(n)$ and an open neighborhood of zero in $\mathfrak{k}$. Its inverse is the restricted matrix exponential. These restrictions are manifold charts; left multiplication translates them to every $A_0\in Sp(n)$ via $A\mapsto \log (A_0^{-1}A)$. The chart overlaps are smooth compositions of multiplication, exponential and logarithm. The subspace topology is Hausdorff and second countable, inherited from the finite-dimensional matrix space, so these charts give a smooth manifold.

Closure under products and inverses follows from $AJ{A}^t=J$ and unitarity. Matrix multiplication is polynomial in real and imaginary entries, and inversion on the unitary group is $A\mapsto A^{\ast }$, a real linear operation. Their restrictions are smooth in the charts just constructed. Hence $Sp(n)$ is a Lie group without needing a closed-subgroup theorem.

Write $X$ in $n\times n$ blocks. The two infinitesimal conditions giveThe skew-Hermitian matrix $P$ has $n^2$ real parameters; the complex symmetric matrix $Q$ has $n(n+1)/2$ complex parameters, hence $n(n+1)$ real parameters. ThereforeFor $n=1$, any two-by-two matrix satisfies $AJ{A}^t=(\det A)J$, so the group is $SU(2)$. Explicitly,The rows are orthonormal and the determinant is one; conversely unitarity and determinant one force this form. This is the SU(2) as the three-sphere parametrization. The map and its inverse, extraction of the first row, are smooth. Thus $Sp(1)$ is diffeomorphic to $S^3$.

A fiber metric on a real vector bundle $E\to M$ is a smoothly varying positive-definite inner product on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a locally finite family of subsets of supports, and $\mathrm{supp}{\rho }_i\subset U_i$; the usual Hausdorff second-countable smooth manifold hypotheses ensure this theorem applies. Transfer the Euclidean inner product to each local vector bundle trivialization, obtaining $h_i$, and setEach weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every vector bundle trivialization. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a fiber metric, with no orientability or triviality assumption.

A vector bundle morphism covering the identity is a smooth map $F:E^{\prime }\to E^{\prime \prime }$ that preserves base points and is a linear map on each fiber. Its induced map on the module of smooth sections is $s\mapsto F\circ s$, and is $C^{\infty }(M)$-linear. We prove the converse by constructing bundle morphisms from maps of smooth sections.

First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a smooth bump function $\chi$ supported there with $\chi =1$ near $x$. Then $\chi s=0$, so $\chi \alpha (s)=\alpha (\chi s)=0$, and hence $\alpha (s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.

Choose a local frame $e_1,\dots ,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j{e}_j$ on $V$. A second bump extends each $a_j$ to a global smooth function ${\tilde{a}}_j$ agreeing near $x$. By locality and $C^{\infty }(M)$-linearity,Every fiber vector $v\in E_x^{\prime }$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha (s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are linear maps, and locally their matrix columns are the smooth sections $\alpha (t_j)$ in a frame of $E^{\prime \prime }$. Thus $F$ is smooth, is a vector bundle morphism, and satisfies $\alpha (s)=F\circ s$. Fiberwise evaluation also proves uniqueness.

Finally apply the fiber metric construction to the tangent bundle. A Riemannian metric $g$ defines the musical isomorphismPositive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence $TM$ and $T^{\ast }M$ are isomorphic as real smooth vector bundles on every such manifold. The isomorphism depends on the chosen metric and is not canonical.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

Use the given orientation throughout, and the usual convention that the manifold has no boundary. In an oriented manifold chart the metric volume form isIf $y$ is another oriented chart and $A=\partial x/\partial y$, then $g_y=A^t{g}_{x}A$, so $\sqrt{\det g_y}=(\det A)\sqrt{\det g_x}$ because $\det A>0$. This is exactly the transformation of the top exterior product. The formulas therefore agree on overlaps and define a smooth positive volume form. Equivalently it takes value one on any positively oriented orthonormal tangent frame.

The metric induces an inner product on $p$-forms by making the increasing exterior products of an orthonormal coframe orthonormal. The Hodge star operator is the unique pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ satisfyingFor an increasing multi-index $I$, $\ast e^I$ is the complementary wedge with the sign making $e^I\wedge \ast e^I={\omega }_g$. Swapping the blocks of $p$ and $n-p$ factors introduces $(-1{)}^{p(n-p)}$. ThereforeIn particular $\ast 1={\omega }_g$ and $\ast {\omega }_g=1$.

For compactly supported smooth $f$, the $(n-1)$-form $f\ast \alpha$ has compact support. Stokes theorem and the graded Leibniz rule giveA top form $\tau$ equals $(\ast \tau ){\omega }_g$. Consequently this Hodge integration by parts for one-forms becomes exactlyCompact support of $f$ is enough; $\alpha$ need not itself have compact support.

Define the codifferential on $p$-forms by $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ and the Hodge Laplacian by $\Delta =d\delta +\delta d$. A harmonic differential form is a smooth form in $\ker \Delta$. On functions this is the positive Laplace-Beltrami operator, $\Delta f=\delta df=-{\mathrm{div}}_g{\mathrm{grad}}_{g}f$. This sign convention is required by the product identity; it is the negative of the $\mathrm{div}\mathrm{grad}$ convention also commonly used for the Laplace-Beltrami operator.

For a $p$-form $\beta$, the square of the Hodge star operator and the codifferential formula giveApplying the second formula to $d\beta$ and the first to $\delta \beta$ givesThus Hodge star commutes with the Hodge Laplacian. Since $\ast$ is invertible, $\beta$ is harmonic if and only if $\ast \beta$ is harmonic. This holds without compactness; we have not used the generally false noncompact implication that a harmonic form must be closed and coclosed.

For a smooth function $h$ and one-form $\alpha$, the same graded Leibniz rule gives $\delta (h\alpha )=h\delta \alpha -\ast (dh\wedge \ast \alpha )=h\delta \alpha -⟨dh,\alpha {⟩}_g$. Apply this to $d(f_1{f}_2)=f_{2}d{f}_1+f_{1}d{f}_2$ to obtain the product rule for the positive Laplace-Beltrami operator

The Hodge decomposition theorem states that on a compact oriented boundaryless Riemannian manifold, smooth forms have the $L^2$-orthogonal decompositionand each de Rham cohomology class has a unique harmonic differential form representative. To spell out the last conclusion, a harmonic form is closed and coclosed because $⟨\Delta \eta ,\eta ⟩=\Vert d\eta {\Vert }^2+\Vert \delta \eta {\Vert }^2$. If a closed form decomposes as $h+du+\delta v$, then $d\delta v=0$; integration by parts gives $\Vert \delta v{\Vert }^2=⟨v,d\delta v⟩=0$. Thus it represents $h$. A harmonic exact form has zero norm by adjointness, proving uniqueness. The analytic existence of the decomposition is the stated Hodge decomposition theorem.

Now let $M$ be compact, connected and oriented. A harmonic function satisfies $0=⟨f,\Delta f⟩=\Vert df{\Vert }^2$, so it is constant. The Hodge star operator identifies ${\mathcal{H}}^0(M)$ with ${\mathcal{H}}^n(M)$, hence ${\mathcal{H}}^n(M)=\mathbb{R}{\omega }_g$. A Riemannian metric exists by Question 3 even if none was initially chosen. Using the harmonic representative of each class, we obtain the top de Rham cohomology of a compact connected oriented manifoldThe class is nonzero also directly from Stokes theorem, since ${\int }_M{\omega }_g>0$ while every exact top form has zero integral. Boundarylessness matters: a compact interval has zero first de Rham cohomology.