Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 58 (Applications of Differential Geometry to Physics)

Assume the level set is regular and non-null. The latter hypothesis is necessary: a null hypersurface has a degenerate induced metric and hence no ordinary intrinsic Levi-Civita connection determined by that metric. Normalize the normal vector and define the non-null hypersurface projection byHere $h$ denotes the induced metric restricted to tangent vectors, while its ambient extension annihilates $n$. The intrinsic covariant derivative is $D_{X}Y=P{\nabla }_{X}Y$ for tangent vector fields. Projection of the ambient torsion-free connection and metric compatibility shows that $D$ is the Levi-Civita connection of $h$. Define the extrinsic curvature by $K(X,Y)=g({\nabla }_{X}n,Y)$, or $K_{ab}=P^c{}_a{P}^d{}_b{\nabla }_c{n}_d$. It is symmetric: differentiating the defining function twice and projecting removes the derivative of its normalization, leaving a multiple of its symmetric Hessian. This sign convention makes the scalar second fundamental form $g(n,{\nabla }_{X}Y)=-K(X,Y)$.

Let $S$ be the tangent linear map characterized by $h(SX,Y)=K(X,Y)$. Differentiating $g(n,n)=\epsilon$ and $g(n,Y)=0$ gives the Gauss formulaFix the Riemann curvature tensor conventionApplying the Gauss formula twice givesSubtracting the expression with $X,Y$ interchanged, and then subtracting the bracket derivative, provesThus the Gauss–Codazzi equations for a non-null hypersurface give the intrinsic Riemann curvature tensor asAll free indices here are tangent to the hypersurface. The normal component of the same calculation isAfter antisymmetrizing and using $D_{X}Y-D_{Y}X=[X,Y]$, this yields the curved-ambient Codazzi equationThe extrinsic curvature measures how tangent planes turn in the ambient manifold; the Gauss equation says that this turning contributes to intrinsic curvature even when the ambient Riemann curvature tensor is zero.

For the quadric in Minkowski spacetime, write its position vector as $X^a$. On the hypersurface, $g(X,X)=1$ and differentiating this identity in a tangent direction gives $g(X,Y)=0$. Hence $n=X$ is a unit normal and a spacelike vector, $\epsilon =1$, and ${\nabla }_{Y}n=Y$ in flat Cartesian coordinates. Therefore $K=h$. The ambient Riemann curvature tensor vanishes, soThis is unit-radius de Sitter spacetime. Reversing the normal changes $K$ to $-K$ but does not change this curvature. An opposite overall convention for the Riemann curvature tensor would reverse both curvature signs below.

To retain a four-dimensional Lorentzian metric while reversing the curvature, take a flat ambient metric with two negative directions and the quadricNow the position vector is a unit normal with negative squared norm, so $\epsilon =-1$ while $K=h$ remains true. The Gauss equation givesThis is unit-radius Anti-de Sitter spacetime. If a positive-definite induced metric is acceptable, an even simpler alternative is to keep the original one-time ambient Minkowski metric and change the quadric level to $g(X,X)=-1$. Each sheet then gives four-dimensional hyperbolic space with the same negative curvature, but its induced signature differs from that of de Sitter spacetime.

Choose a local orthonormal coframe $e^A$ with $g={\eta }_{AB}{e}^A{e}^B$. The Levi-Civita connection is represented by the connection 1-forms satisfyingThe gamma matrices satisfy the Clifford algebra relations $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}I$. The matrices $\frac{1}{4}[{\gamma }^A,{\gamma }^B]$ represent the infinitesimal orthogonal transformations on spinor fields; lifting the frame connection to this representation defines the spinor covariant derivativeBoth frame indices are summed, so each antisymmetric pair occurs twice. Under a local Lorentz transformation with spin lift $S$, the spin connection transforms as ${\Omega }_{\mu }\mapsto S{\Omega }_{\mu }{S}^{-1}-({\partial }_{\mu }S)S^{-1}$ and $ϵ\mapsto Sϵ$; consequently ${\nabla }_{\mu }ϵ\mapsto S{\nabla }_{\mu }ϵ$. This establishes covariance. Global construction requires a compatible spin structure (and the appropriate orientation and time orientation); the formulas themselves work on a local orthonormal frame patch.

The supplied gamma matrices have squares $-I,I,I$ and anticommute pairwise, so $\eta =\mathrm{diag}(-1,1,1)$. Work on a connected regular patch with $r>0$ and $V\ne 0$. Since the metric depends only on $V^2$, we may use its positive smooth square root $V>0$. TakeTheir exterior derivatives are $d{e}^0=-V^{\prime }{e}^0\wedge e^1$, $d{e}^1=0$, and $d{e}^2=(V/r)e^1\wedge e^2$. Solving Cartan's first structure equation and imposing Lorentzian connection-form antisymmetry givesIn particular ${\omega }_{01}=-V{V}^{\prime }dt$ and ${\omega }_{12}=-Vd\theta$. The parallel spinor equations are thereforeDifferentiate the angular equation with respect to $r$ and use ${\partial }_rϵ=0$. This gives $V^{\prime }{\gamma }^1{\gamma }^2ϵ=0$. Since $({\gamma }^1{\gamma }^2{)}^2=-I$, this matrix is invertible. A parallel spinor that vanishes at any point vanishes everywhere on the connected domain, by uniqueness of parallel transport. Thus a nonzero parallel spinor implies $V^{\prime }=0$ throughout the patch. There are no nonzero parallel spinors unless $V$ is constant. The zero spinor field is always a solution, so “nonzero” is essential in interpreting the printed assertion. A smooth original $V$ of either sign is constant on such a patch exactly when $|V|$ is constant; $V=0$ is excluded because the given metric is not defined there.

Write the constant as $c=|V|>0$, and introduce $T=ct$, $\rho =r/c$, and $\varphi =c\theta$. The metric becomesA circle at small $\rho$ has circumference $cP\rho$ if $\theta$ has period $P$, whereas its proper radius is $\rho$. Smoothness requires circumference divided by radius to tend to $2\pi$, not a multiple of $2\pi$. Conversely, when $cP=2\pi$, the coordinates $X=\rho \cos \varphi$, $Y=\rho \sin \varphi$ exhibit a smooth flat spatial disk. HenceFor a prescribed period $2\pi$, this requires $c=1$; otherwise there is a conical singularity at the apex.

For constant $V=c$, the time and radial equations make the components independent of $t,r$. The angular equation has all solutionswhere ${ϵ}_0$ is an arbitrary constant two-component spinor field and ${\gamma }^1{\gamma }^2=-{\gamma }^0$. There are two independent real solutions for real spinor fields, or two independent complex solutions if complex spinors are used. At the smooth angular period, these polar-frame components change sign. This is exactly the polar-frame parallel spinor transformation under the spin lift of a full rotation: the rotating polar orthonormal frame does not extend through the origin. The spin structure extending over the disk uses this antiperiodic polar trivialization; in a Cartesian orthonormal frame the same parallel spinors have constant components and extend smoothly. Requiring periodic components in the rotating frame would select the other spin structure on the punctured disk, which does not extend over its center.

Use matrix multiplication together with the wedge product of differential forms, suppressing wedge symbols in this calculation. Set $F=dA+A^2$ and $a=\delta A$. The gauge field strength varies as $\delta F=da+Aa+aA$, the covariant exterior derivative of $a$. The matrix trace has graded cyclicity, $\mathrm{Tr}(\alpha \beta )=(-1{)}^{pq}\mathrm{Tr}(\beta \alpha )$ for forms of degrees $p,q$. Thus the Chern-Simons 5-form can be expanded asFor example, the two cross terms in $F^{2}A$ both become $\mathrm{Tr}(A^{3}dA)$; combining the remaining two terms gives the displayed coefficients. This expansion keeps track of the noncommuting products rather than treating $A$ as a scalar one-form.

Put $B=dA$. Vary the first term and use the graded exterior derivative to move derivatives off $a$:Here $dB=0$. For the next term use $d(A^3)=B{A}^2-ABA+A^{2}B$. Graded cyclicity then givesFinally all five contributions to $\delta \mathrm{Tr}(A^5)$ are equal under the matrix trace, giving $5\mathrm{Tr}(a{A}^4)$. Combining these results proves the variation of the Chern-Simons 5-form:Consequently, on an oriented five-manifold $N$,The boundary term vanishes for a closed manifold, for variations supported in its interior, or when the tangential boundary connection is held fixed. A different boundary variational problem would require boundary conditions or a boundary action as well.

Write $A=A^{\alpha }{T}_{\alpha }$ in a basis of the Lie algebra. Independent compactly supported variations of each component yield the bulk Euler-Lagrange equationsEquivalently, with $d_{\alpha \beta \gamma }=\frac{1}{2}\mathrm{Tr}[T_{\alpha }\{T_{\beta },T_{\gamma }\}]$, these are $d_{\alpha \beta \gamma }{F}^{\beta }\wedge F^{\gamma }=0$. The symmetry of the two curvature two-forms accounts for the anticommutator. Writing simply $F\wedge F=0$ as a full matrix equation requires an additional assumption: variations must span a matrix algebra paired nondegenerately by the matrix trace. For a smaller Lie algebra, only its trace-paired projection is constrained. If the trace is literally taken in the adjoint representation of a semisimple Lie algebra, its symmetric cubic invariant vanishes, and these bulk equations are identically zero. Indeed, the adjoint generators are skew with respect to the nondegenerate invariant Killing form; transposition in this pairing reverses the sign of the symmetrized trace of three generators. The usual nontrivial five-dimensional theory therefore needs an invariant cubic trace that does not vanish; saying that the gauge field transforms in the adjoint alone does not specify such a trace.

Under a Yang-Mills gauge transformation, use $A^g=g^{-1}Ag+g^{-1}dg$, so $F^g=g^{-1}Fg$. For any Lie-algebra element $T$,Since $gT{g}^{-1}$ is still in the Lie algebra, the right-hand side vanishes whenever the original field satisfies every equation. Every gauge transform of a bulk solution remains a bulk solution. This conclusion follows from covariance of the field equations; it does not require the action itself to be exactly invariant under all large gauge transformations or under transformations changing prescribed boundary data.

Work on a regular patch with a smooth nonzero conformal factor. Its sign is constant there, so replace it by $|\Omega |$ if necessary and assume $\Omega >0$. Put $u=\log \Omega$, and let all partial derivatives, index raising on $u_i,{\Omega }_i$, and the operators $\Delta ,|\nabla \cdot {|}^2$ in the following formulas use the flat Cartesian metric $\delta$. The orthonormal coframe $E^i=\Omega d{x}^i$ hasThe proposed connection 1-forms reduce toLowering the first frame index with $\delta$ makes these antisymmetric, which establishes metric compatibility. Moreover,Thus Cartan's first structure equation has zero torsion. Uniqueness of the metric-compatible torsion-free connection identifies these forms as the Levi-Civita connection; this also explains why merely guessing a skew connection would not have sufficed.

To calculate the Riemann curvature tensor, use Cartan's second structure equation. Direct differentiation and multiplication giveIntroduce the Schouten tensor in Cartesian coordinate components,The preceding calculation groups into the curvature 2-formsKeep the convention $R_{ijkl}=g({\partial }_i,R({\partial }_k,{\partial }_l){\partial }_j)$, with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$, as in the first solution. Converting frame indices to coordinate indices gives the curvature of a conformally Euclidean metric:These are coordinate components. Components in the dual orthonormal frame are ${\Omega }^{-4}{R}_{ijkl}$, hence the same expression with the overall factor ${\Omega }^{-2}$. This distinction is important when contracting the Riemann curvature tensor.

Contract with $g^{ik}={\Omega }^{-2}{\delta }^{ik}$. In four dimensions the Ricci tensor becomes $R_{jl}=2S_{jl}+{\delta }_{jl}{\sum }_k{S}_{kk}$, and thereforeTaking the remaining metric trace yields the Ricci scalarThe cancellation of the gradient-squared terms in the last expression is special to dimension four.

For the Tolman wormhole, compute on $r>0$:Substitution into the Ricci tensor and Ricci scalar givesFor example, the coordinate radial eigenvalue is $-12\lambda /(r^2+\lambda {)}^2$, while each of the three tangential eigenvalues is $4\lambda /(r^2+\lambda {)}^2$. Their sum vanishes, but none vanishes when $\lambda \ne 0$. The Vacuum Einstein equations with zero cosmological constant require $R_{ij}-\frac{1}{2}R{g}_{ij}=0$ and hence $R_{ij}=0$. The metric is scalar-flat but not a vacuum solution for any nonzero $\lambda$; only $\lambda =0$ gives flat vacuum space. A nonzero cosmological constant does not rescue it: its trace would force $R=4\Lambda$, so $R=0$ would first force $\Lambda =0$.

The smooth wormhole interpretation uses $\lambda >0$; $r=0$ is another asymptotically flat end, not an included origin, as inversion $\rho =\lambda /r$ shows. For $\lambda <0$, the conformal factor vanishes at $r=\sqrt{-\lambda }$, where the metric is degenerate. The local non-vacuum calculation remains valid on each regular region away from that sphere, but such a metric is not a smooth wormhole across it.

Let a dot denote differentiation with respect to an affine geodesic parameter $s$. The five-dimensional geodesic Lagrangian isBecause $z$ is cyclic, its Euler-Lagrange equation gives the conserved quantityThis is the conserved momentum per unit five-dimensional mass associated with the Killing vector field ${\partial }_z$. For the other coordinates, differentiate $\partial L/\partial {\dot{x}}^i={\gamma }_{ij}{\dot{x}}^j+k{A}_i$. Using $\dot{k}=0$ givesThe first two terms combine into ${\gamma }_{ij}({\ddot{x}}^j+{\Gamma }^j{}_{lm}{\dot{x}}^l{\dot{x}}^m)$, with the Christoffel symbols of $\gamma$. Define the four-dimensional electromagnetic field tensor $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i$ and raise its index with $\gamma$. The complete reduced geodesic equations areThis is the Kaluza-Klein geodesic Lorentz force. The Christoffel symbols describe the gravitational motion in general relativity in the four-dimensional metric tensor, while the right-hand side is antisymmetric in its lowered indices and gives the relativistic Lorentz force. In particular,Thus the force preserves the four-dimensional velocity norm, as a Lorentz force must. A particle with $k=0$ is neutral and follows an ordinary four-dimensional geodesic. Nonzero compact momentum gives electric charge, with its sign determined by the direction of motion around the circle.

To identify the actual charge-to-mass ratio, suppose the five-dimensional particle has mass $M>0$ and $s$ is its five-dimensional proper time, so $g_5(\dot{X},\dot{X})=-1$. Then $\gamma (\dot{x},\dot{x})=-(1+k^2)$. Four-dimensional proper time obeys $d\tau /ds=\sqrt{1+k^2}$; since $k$ is constant, $u^i=d{x}^i/d\tau$ satisfiesHere $q$ is in the geometric normalization of the supplied potential $A$. The relation $m^2=M^2+p_z^2$ also follows directly from the five-dimensional mass shell, and continues to identify the effective mass when the higher-dimensional particle is massless. The distinction between the two proper times prevents mistaking $k$ itself for the physical four-dimensional charge-to-mass ratio. A rescaling of $A$ to a canonically normalized electromagnetic potential correspondingly rescales $q$.

For the wave calculation, take a scalar field of five-dimensional mass $M$ in units $\hslash =c=1$, obeying $({\square }_5-M^2)\Psi =0$. In the coordinate order $(x^i,z)$, direct inversion of the metric tensor givesThe determinant follows by the triangular change of coframe from $(d{x}^i,dz)$ to $(d{x}^i,dz+A_{i}d{x}^i)$, whose determinant is one. Write $J=\sqrt{|\det \gamma |}$. Since all coefficients are $z$-independent, the scalar covariant wave operator isExpanding the right-hand side verifies both cross terms $-2A^i{\partial }_i{\partial }_z\Psi$, the term $-J^{-1}{\partial }_i(J{A}^i){\partial }_z\Psi$, and the coefficient $1+A_i{A}^i$ of ${\partial }_z^2\Psi$.

Single-valuedness under the compact coordinate period requires the Fourier seriesFor the $n$th Kaluza-Klein mode, put $q_n=n/R$ and $D_i={\partial }_i-i{q}_n{A}_i$. The five-dimensional Klein-Gordon equation reduces toThis is the charged four-dimensional Klein-Gordon equation: the divergence expression includes both the spacetime connection and the gauge connection. Its interpretation as electric charge can also be checked directly. Under the fibre coordinate change $z^{\prime }=z-\chi (x)$, the metric retains its form with $A^{\prime }=A+d\chi$, and the scalar Fourier coefficient changes by ${\psi }_n^{\prime }=e^{i{q}_n\chi }{\psi }_n$. Hence $D_i^{\prime }{\psi }_n^{\prime }=e^{i{q}_n\chi }{D}_i{\psi }_n$, exactly the transformation law for a field of charge $q_n$.

The Kaluza-Klein charge quantization is thereforePositive and negative integers give opposite charges, and the zero mode is neutral. Restoring $\hslash$ gives compact momenta $p_z=n\hslash /R$, so their fundamental spacing is $\hslash /R$. These are charges in the metric's geometric normalization. If $A_i=\kappa {\mathcal{A}}_i$ for a physically normalized electromagnetic potential ${\mathcal{A}}_i$, the physical charge unit is $\kappa /R$ in natural units, or $\kappa \hslash /R$ with $\hslash$ restored. Determining $\kappa$ requires an electromagnetic action normalization not specified by the metric ansatz alone.