Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 55 (Applications of Differential Geometry to Physics)

Treat the orthonormal coframe in spacetime and the Lorentz connection 1-form as independent variables, with nondegenerate coframe and ${\omega }_{ab}=-{\omega }_{ba}$. Fix the curvature conventionThe second expression is Cartan's first structure equation; the first is Cartan's second structure equation. Use compactly supported variations, or boundary data that make the variation's surface term vanish.

For a connection variation, $\delta R^{ab}=D\delta {\omega }^{ab}$, where $D$ is the covariant exterior derivative. Since $D{ϵ}_{abc}=0$, its graded product rule givesFor a coframe variation, relabeling the dummy indices makes the three variations of the cubic term identical. HenceThe independent variations therefore give ${ϵ}_{abc}{T}^c=0$ and ${ϵ}_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions the alternating symbol identifies an antisymmetric index pair with a single index, so these equations are equivalent toThese are the equations of first-order three-dimensional gravity. The first selects the Levi-Civita connection; the second fixes constant sectional curvature $K=-3\lambda$. The factor of three follows from varying the coefficient exactly as printed.

On a static coordinate patch where $f>0$, choose the Lorentz frame metric ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and coframePut $A=f^{\prime }/(2\sqrt{f})$ and $B=\sqrt{f}/r$. Exterior differentiation gives $d{E}^0=-A{E}^0\wedge E^1$, $d{E}^1=0$, and $d{E}^2=B{E}^1\wedge E^2$. Solving the equation of vanishing torsion forms givesHere antisymmetry is in the two raised or two lowered frame indices; mixed time-space components have equal, rather than opposite, signs. This is the Lorentzian connection-form antisymmetry needed to retain the curvature signs.

Using Cartan's second structure equation, the nonzero curvature 2-forms areFor example, $d(A{E}^0)=-(\sqrt{f}{A}^{\prime }+A^2)E^0\wedge E^1=-f^{\prime \prime }{E}^0\wedge E^1/2$, while the $02$ form is the product ${\omega }^0{}_1\wedge {\omega }^{12}=-AB{E}^0\wedge E^2$. Thus the field equations require $f^{\prime \prime }/2=3\lambda$ and $f^{\prime }/(2r)=3\lambda$. Integrating the latter gives $f=C+3\lambda r^2$; the former is then satisfied automatically. The origin condition fixes $C=1$, yieldingAs a sign and normalization check, the Ricci tensor is $R_{\mu \nu }=-6\lambda g_{\mu \nu }$ and the scalar curvature is $R=-18\lambda$. The coframe calculation applies initially in the region with $f>0$; if $\lambda <0$, $f$ vanishes at $r=1/\sqrt{-3\lambda }$, delimiting that static chart without changing the derived function.

Absorb the gauge coupling into a matrix-valued connection 1-form, and usefor an adjoint-valued $p$-form $\alpha$. The matrix trace supplies the usual invariant nondegenerate pairing of the Lie algebra, and the spacetime metric defining the Hodge star is fixed during the gauge-field variation. Since $\delta F=D\delta A$ and the two-form Hodge pairing is symmetric,For variations fixed at the boundary, the Yang-Mills equations are thereforeThey accompany, rather than replace, the gauge-theory Bianchi identity $DF=0$, which follows directly by substituting $F=dA+A\wedge A$ and using $d^2=0$.

Fix the Yang-Mills gauge transformation conventionTaking $g=1+\eta +O({\eta }^2)$ givesThe curvature variation also follows from $D(D\eta )=[F,\eta ]$. If the group parameter is defined with the opposite sign, both displayed infinitesimal variations change sign together.

The Hodge star acts only on spacetime indices, so $\delta (\ast F)=[\ast F,\eta ]$. ConsequentlyThis proves exact off-shell gauge invariance, with no surviving boundary term, even for a gauge parameter nonzero at the boundary. It does not require the Yang-Mills equations.

One can check the surface terms explicitly in the integrated variation. The Lie-algebra-valued four-form $[F,\ast F]=F\wedge \ast F-\ast F\wedge F$ vanishes: writing $F=F^i{T}_i$, the coefficients $F^i\wedge \ast F^j$ are symmetric in $i,j$, and multiply the antisymmetric commutator $[T_i,T_j]$. Thus $D^2\ast F=[F,\ast F]=0$, and the gauge substitution in the variation givesThe last equality is Stokes theorem on the closed boundary; equivalently the cancellation is already pointwise in the preceding derivation. Keeping only the first surface term from an arbitrary variation would incorrectly produce a gauge boundary term.

For the additional Yang-Mills theta term, denote its coefficient-free integral by $P={\int }_M\mathrm{tr}(F\wedge F)$. Its gauge variation is again pointwise zero:Under an arbitrary connection variation, its boundary variation of the Yang-Mills theta term isThe bulk term vanishes by the gauge-theory Bianchi identity. Therefore adding a constant multiple of $P$ does not change the bulk Yang-Mills equations. With fixed boundary potential, it also leaves the usual variational problem unchanged. If boundary variations are allowed, it changes the boundary term: $I+\kappa P$ has surface variation $2{\int }_{\partial M}\mathrm{tr}(\delta A\wedge (\ast F+\kappa F))$, so compatible boundary conditions or a boundary action must account for that change.

Locally $\mathrm{tr}(F\wedge F)=d\mathrm{tr}(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$, the unnormalized Chern-Simons 3-form identity. That three-form need not be globally defined on a nontrivial bundle, so absence of a bulk variation does not force $P$ itself to vanish. Its exact gauge invariance is the curvature-conjugation calculation above, independently of the choice of local gauge potential.

Interpret $L=-\square$ as the minimally coupled scalar positive Laplace-Beltrami operator on the smooth closed manifold. Writing $dV$ for its metric volume measure (the Riemannian volume form on an oriented chart), integration by parts gives $⟨u,Lu⟩={\int }_M|\nabla u{|}^{2}dV\ge 0$. The compact elliptic spectral theorem supplies an orthonormal basis $u_j$ with $L{u}_j={\lambda }_j{u}_j$, real nonnegative eigenvalues tending to infinity and finite multiplicities. Solve this eigenvalue problem to obtain the spectral zeta functionThe zero eigenvalues are excluded, not assigned inverse powers. Their multiplicity $n_0$ is the number of connected components: $Lu=0$ implies $\int |\nabla u{|}^{2}dV=0$, so $u$ is constant on each component.

An equivalent calculation uses the spectral expansion of the Riemannian heat kernel,Since ${\int }_0^{\infty }{t}^{s-1}{e}^{-t\lambda }dt=\Gamma (s){\lambda }^{-s}$ for $\lambda >0$, the Mellin transform givesThe reduced trace decays exponentially at large $t$. At small $t$, subtract a finite number of terms of the heat kernel expansion, integrate those terms explicitly, and then integrate the remainder. Increasing the number of subtractions extends the formula successively to further left half-planes, producing the meromorphic continuation of ${\zeta }_L$ rather than substituting $s=0$ in its divergent defining series.

For completeness, the needed curvature coefficient comes from the heat-kernel transport equations. In normal coordinates $\xi$ centered at $y$, the Riemannian volume form has density $J(\xi )=1-\frac{1}{6}{R}_{ab}(y){\xi }^a{\xi }^b+O(|\xi {|}^3)$. The leading Gaussian amplitude is thereforeSubstitution of the Gaussian amplitude series into the heat equation gives the first diagonal transport relation $u_1(y,y)={\square }_x{u}_0(x,y){|}_{x=y}=R(y)/6$. The curvature coefficient of the scalar heat kernel is consequentlyIn dimension two this impliesSet $A_0=\mathrm{Area}(M)/(4\pi )$ and $A_1={\int }_{M}RdV/(24\pi )$. Splitting the Mellin integral at $t=1$ and subtracting these two heat terms gives, near $s=0$,where $H$ is holomorphic for $\mathrm{Re}s>-1$. As $1/\Gamma (s)=s+O(s^2)$, only the second term survives at zero. Thus the zero-mode correction to the spectral zeta value isThe Gauss-Bonnet theorem, with $R$ twice the Gaussian curvature, makes this $\boxed{{\zeta }_L(0)=\chi (M)/6-n_0}$. For a connected surface $n_0=1$, so the answer is $\chi (M)/6-1$; a sphere gives $-2/3$, and a flat torus gives $-1$.

The purely local heat coefficient is $A_1=\int RdV/(24\pi )$. Sometimes a formal formula for $\zeta (0)$ quotes this coefficient with zero modes left implicit. For the actual positive-spectrum spectral zeta function defined above, the compact scalar problem necessarily has the additional subtraction $-n_0$. A strictly positive shifted operator is a different problem and has its own heat coefficients.

Use coordinate indices $\mu ,\nu$ and Lorentz-frame indices $A,B$, and take ${\eta }_{AB}=\mathrm{diag}(-1,1,1,1)$. Locally choose four independent one-forms $e^A=e^A{}_{\mu }d{x}^{\mu }$ satisfyingOne construction begins with a smooth timelike vector field in a sufficiently small neighborhood, normalizes it, and applies the Gram-Schmidt process to its spacelike orthogonal complement. The dual orthonormal coframe in spacetime gives the vierbein; equivalently its matrix factorizes the metric as $g=e^T\eta e$. The inverse matrix gives the frame $e_A=e_A{}^{\mu }{\partial }_{\mu }$ with $e_A{}^{\mu }{e}^B{}_{\mu }={\delta }_A^B$. This is a local construction: a global frame is not guaranteed by a metric, and a globally defined spinor field requires a compatible spin structure.

The metric does not determine a unique vierbein. A local Lorentz transformation changes the coframe components byThe coordinate index is unchanged in this frame transformation. Restrict to the proper orthochronous Lorentz group for its connected spin lift; orientation-reversing transformations require further structure and are not part of this local spin-group discussion.

Let constant gamma matrices obey the Clifford algebra $\{{\gamma }^A,{\gamma }^B\}=2{\eta }^{AB}$, and write ${\gamma }^{AB}=\frac{1}{2}[{\gamma }^A,{\gamma }^B]$. A spinor field transforms by the spin lift of the Lorentz transformation, not by its vector matrix:For $\Lambda =1+\alpha +O({\alpha }^2)$, with ${\alpha }_{AB}=-{\alpha }_{BA}$, the Clifford algebra identityshows explicitly thatIndeed the commutator of ${\gamma }^C$ with this infinitesimal spin matrix is ${\alpha }^C{}_B{\gamma }^B$. A finite transformation connected to the identity is obtained by exponentiating these generators along a path. Its lift has two possible signs, so spinors are not ordinary Lorentz vectors: a $2\pi$ spatial rotation acts by $-1$, and $4\pi$ by $+1$. Consistent transition lifts define the spin structure; the metric alone does not select it.

For the metric connection, take the Levi-Civita connection and determine the spin connection from the tetrad postulateIt is equivalently determined by $d{e}^A+{\omega }^A{}_B\wedge e^B=0$ and ${\omega }_{\mu AB}=-{\omega }_{\mu BA}$. Explicitly, in terms of the inverse frame,Thus both the coordinate connection and the Lorentz connection 1-form are determined by the metric and chosen frame; no independent torsion has been supplied.

An ordinary derivative fails covariance because ${\partial }_{\mu }(S\psi )=S{\partial }_{\mu }\psi +({\partial }_{\mu }S)\psi$. Seek ${\nabla }_{\mu }\psi =({\partial }_{\mu }+{\Omega }_{\mu })\psi$. Covariance requiresThe coframe's connection obeys ${\omega }^{\prime }=\Lambda \omega {\Lambda }^{-1}-d\Lambda {\Lambda }^{-1}$. Applying the spin representation of the same Lorentz algebra, whose infinitesimal generator was just determined, suppliesThis proves the transformation law rather than choosing the coefficient arbitrarily. It is the spinor covariant derivative induced by the spin connection. With curved matrices ${\gamma }^{\nu }(x)=e_A{}^{\nu }{\gamma }^A$, the same coefficient and tetrad postulate giveso differentiation is compatible with Clifford multiplication and $\{{\gamma }^{\mu }(x),{\gamma }^{\nu }(x)\}=2g^{\mu \nu }(x)$.

Finally, expand the commutator of these derivatives:The spin representation preserves the Lorentz algebra bracket. Consequently $d\Omega +\Omega \wedge \Omega =\frac{1}{4}{R}_{AB}{\gamma }^{AB}$, where $R^A{}_B=d{\omega }^A{}_B+{\omega }^A{}_C\wedge {\omega }^C{}_B$. In components the spinor curvature identity isFor the full second derivative, ${\nabla }_{\mu }({\nabla }_{\nu }\psi )$ also includes $-{\Gamma }_{\mu \nu }^{\rho }{\nabla }_{\rho }\psi$, because its first derivative has a covector index. This term cancels on antisymmetrization for the torsion-free connection. With unit-weight brackets, $X_{[\mu \nu ]}=(X_{\mu \nu }-X_{\nu \mu })/2$, the requested expression is thereforeThis convention distinguishes the antisymmetrized second derivative from the full commutator by its factor of two. In a noncoordinate orthonormal frame, an operator commutator must additionally subtract the derivative along the frame-vector commutator; using the full tensorial second derivative incorporates that correction.