Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 50 (Applications of Differential Geometry to Physics)

There is a small domain issue in the printed notation. If ${\mathbb{R}}^3/{\mathbb{R}}^2$ means a quotient vector space by the horizontal plane, the displayed formula does not descend to that quotient vector space: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is stereographic projection, restricted to the unit sphere with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The holomorphic stereographic atlas of the sphere is obtained as follows.

Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus \{N\}$ use $\zeta =(x+iy)/(1-z)$. Its inverse, with $\zeta =u+iv$, isThese formulas give a smooth manifold chart from $U_N$ onto $\mathbb{C}$. On $U_S=S^2\setminus \{S\}$ choose the second manifold chartIts inverse is $x=2\mathrm{Re}\eta /(1+|\eta {|}^2)$, $y=-2\mathrm{Im}\eta /(1+|\eta {|}^2)$ and $z=(1-|\eta {|}^2)/(1+|\eta {|}^2)$, so this is also a smooth manifold chart onto $\mathbb{C}$. The conjugation in this second stereographic projection is essential. On the overlap, $x^2+y^2=1-z^2$, soBoth directions of this transition are holomorphic maps on ${\mathbb{C}}^{\times }$, with nonzero derivative. The two manifold charts cover the sphere, hence define a holomorphic atlas, giving precisely the Riemann sphere. If one instead used $x+iy$ in both manifold charts, the transition would be $1/\bar{\zeta }$ and would not be holomorphic.

Orient the sphere by this holomorphic atlas. The given volume form is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its exterior product gives the same expressionMoreover, $id\zeta \wedge d\bar{\zeta }=2du\wedge dv$, so the normalization of this volume form isThus the printed volume form has half the area of the standard round unit sphere; replacing its integral by $4\pi$ would introduce an erroneous factor of two.

For $k\ge 1$, the holomorphic map $f(\zeta )={\zeta }^k$ extends over infinity, since the target reciprocal coordinate is ${\eta }^k$ when the source reciprocal coordinate is $\eta$. Its pullback of a differential form isUsing $s=r^{2k}$ givesThe degree of a map between oriented manifolds is thereforeFor $k=0$, the formula on the finite manifold chart is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression ${\infty }^0$. Its pullback of a differential form is zero and its degree of a map between oriented manifolds is zero.

For the preimage calculation when $k\ge 1$, choose a regular value $w\in {\mathbb{C}}^{\times }$. There are exactly $k$ distinct roots of ${\zeta }^k=w$. At each root the real Jacobian determinant is $|k{\zeta }^{k-1}{|}^2>0$, so every local contribution to the degree of a map between oriented manifolds is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne 1$ is a regular value with no preimages, giving the same answer zero. This establishes the degree of a power map of the Riemann sphere for every allowed $k$.

The geometric object underlying an $SU(2)$ gauge field is a principal connection on a principal bundle $P\to M$ with structure group SU(2). The base $M$ is spacetime, or an oriented Riemannian manifold in the Euclidean formulation. A choice of local section identifies each fibre with SU(2) and produces a local gauge potential. Such a choice is a choice of internal frame; changing it produces a gauge equivalence of principal connections. The bundle specifies the global gluing of these frames, while the principal connection specifies how frames at neighbouring points are compared.

The SU(2) group consists of $2\times 2$ unitary matrices of determinant one. Its Lie algebra consists of traceless Skew-Hermitian matrices. For definiteness usewhere the ${\sigma }_a$ are Pauli matrices. The commutator fixes all signs below. The positive invariant inner product on this Lie algebra is $-\mathrm{tr}(XY)$. Absorb the coupling into the gauge potential, so the gauge covariant derivative in the defining two-dimensional representation is $D=d+A$ with $A=A^a{t}_a$. A convention using Hermitian Pauli matrices instead moves factors of $i$ and the coupling into $D$; the underlying principal connection is the same geometric data.

A principal connection can be given by a Lie algebra-valued differential one-form $\mathcal{A}$ on $P$ satisfyingHere ${\xi }_P$ is the vertical vector generated by the right group action. The first identity recovers vertical motion and the second makes the construction independent of a fibre frame. Its kernel is the horizontal distribution of a principal connection, complementary to the vertical tangent spaces. A curve is horizontal when $\mathcal{A}$ annihilates its velocity, so a principal connection defines parallel transport along curves in $M$.

For a local section $s$, the local principal connection form is $A=s^{\ast }\mathcal{A}$. Change the section to $s^{\prime }=sh$, with $h:U\to SU(2)$. Differentiating $sh$ has a translated horizontal part and a vertical part from $dh$. The two defining identities for the principal connection therefore giveOn overlapping trivializations this is exactly the compatibility law for the local principal connection forms. In particular, the gauge potentials need not glue as ordinary globally defined differential one-forms on $M$. On a nontrivial principal bundle there need not be any global section with respect to which one could write a single matrix-valued $A$.

An active gauge equivalence of principal connections is generated by a bundle automorphism covering the identity on $M$ and commuting with the right SU(2) action. Its local description has the same transformation formula, but its local group-valued functions must satisfy transition compatibility on overlaps. This distinguishes the structure group SU(2) from the full gauge group of a fixed bundle. Two descriptions related by a passive change of section encode the same principal connection; two connections related by such an automorphism lie in the same physical gauge orbit. The space of principal connections on a fixed bundle is affine: the difference of two connection forms is a differential one-form with values in the adjoint bundle, since their inhomogeneous transformation terms cancel.

The curvature of a principal connection isIt is horizontal and equivariant, hence descends to a differential two-form on $M$ valued in the adjoint bundle. In a local section the gauge curvature becomesThe product $A\wedge A$ combines the exterior product with matrix multiplication. The commutator term is the specifically non-Abelian interaction. Substitution of the transformation of $A$, using $d(h^{-1})=-h^{-1}(dh)h^{-1}$, cancels all derivatives of $h$ and givesThus the gauge curvature transforms tensorially even though the gauge potential does not. Geometrically, the curvature of a principal connection measures the failure of the horizontal distribution of a principal connection to be integrable; physically it is the non-Abelian gauge field strength.

Matter fields are sections of associated vector bundles. A field in the defining representation of SU(2) is a section of $E=P{\times }_{SU(2)}{\mathbb{C}}^2$. Its two local components satisfy ${\psi }^{\prime }=h^{-1}\psi$ under the section convention above. The induced gauge covariant derivative obeysThis covariance is why replacing ordinary derivatives by gauge covariant derivatives makes the kinetic terms compatible with changes of internal frame. Applying the gauge covariant derivative twice yields $D^2\psi =F\psi$, so the gauge curvature also measures the noncommutativity of covariant differentiation. Other group representations give other associated vector bundles and replace $A$ by its image under the differential of the representation.

For an adjoint-valued differential form $\Phi$ of degree $r$, setExpansion of $F=dA+A\wedge A$, using $d^2=0$, proves the gauge-theory Bianchi identityThis is an identity for every principal connection, independent of any field equation. It should be distinguished from the dynamical Yang-Mills equations.

To obtain dynamics, equip $M$ with a Riemannian metric and an orientation, and let $\ast$ be its Hodge star operator. With the Skew-Hermitian normalization above, the positive Euclidean Yang-Mills action isThe matrix trace and the Hodge star operator make this independent of the choice of local section. The minus sign compensates for the negative trace form on Skew-Hermitian matrices. Varying the principal connection gives $\delta F=D_A\delta A$, and thereforeFor compactly supported variations, or on a closed base, integration by parts gives the Yang-Mills equationsIn components these are $D_{\mu }{F}^{\mu \nu }=0$. The gauge-theory Bianchi identity provides the complementary geometric identity $D_{A}F=0$. Coupling matter adds the associated gauge current to the dynamical equation; the transformation law of the gauge covariant derivative ensures covariance of that current as well.

The principal connection also supplies nonlocal observables. Along a path $\gamma$, parallel transport is determined byThe ordering is necessary because the Lie algebra matrices at different points need not commute. The resulting Wilson line transforms as $U_{\gamma }^{\prime }=h(\gamma (1){)}^{-1}{U}_{\gamma }h(\gamma (0))$. For a closed path, its matrix trace is a Wilson loop, independent of the frame at the basepoint. These observables record the holonomy of a connection. A flat principal connection has trivial holonomy of a connection around sufficiently small contractible loops but may have nontrivial holonomy of a connection around noncontractible loops. Thus vanishing local gauge curvature need not eliminate global gauge information.

The global topology becomes particularly visible in four dimensions. For a closed oriented four-dimensional base, Chern-Weil theory gives the Second Chern number of the defining associated vector bundle:The plus sign here uses Skew-Hermitian curvature and the mathematical second-Chern convention: expanding $\det (I+iF/(2\pi ))$ with $\mathrm{tr}F=0$ gives precisely the displayed Second Chern form. A convention defining a physics topological charge with a leading minus sign instead calls that charge $-k$. Neither choice changes the absolute-value action bound.

The Second Chern number is independent of the principal connection on a fixed bundle. Indeed, the gauge-theory Bianchi identity implieswhose integral vanishes on a closed base by Stokes theorem. A nonzero Second Chern number obstructs a global trivialization: for a globally defined $A$, $\mathrm{tr}(F\wedge F)=d\mathrm{tr}(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$, so its integral on a closed base would be zero. On $S^4$, gluing two trivial bundles over four-balls uses a transition map $S^3\to SU(2)$. Since SU(2) as the three-sphere identifies the target with $S^3$, such gluing maps have integer degree of a map between oriented manifolds, illustrating how distinct topological sectors arise.

In four Euclidean dimensions, the Hodge star operator squares to one on differential two-forms. Decompose $F=F_{+}+F_{-}$ with $\ast F_{\pm }=\pm F_{\pm }$. Define the nonnegative squared norms $a_{\pm }=-\int \mathrm{tr}(F_{\pm }\wedge \ast F_{\pm })$. Orthogonality givesEquality means that one component vanishes, giving the self-dual Yang-Mills equations or the Anti-self-dual Yang-Mills equations. Such finite-action solutions are Yang-Mills instantons. They solve the full Yang-Mills equations because $D_A\ast F=\pm D_{A}F=0$. With the second-Chern convention just specified, anti-self-dual curvature has $k\ge 0$ and self-dual curvature has $k\le 0$. On noncompact spacetime one must impose suitable decay or boundary conditions before treating the topological integral as an integer and using the same bound without boundary terms.

Local gauge potentials, matter gauge covariant derivatives, gauge field strength, parallel transport and Second Chern number are different manifestations of one principal connection. The principal bundle preserves the global information that a single local gauge potential cannot capture; the Yang-Mills action then selects its physical dynamics.

Acting on the column $(x,1{)}^T$ gives the faithful Matrix Lie group representationMultiplication and inversion areThus $G$ is the real affine group. Differentiating the matrices at the identity gives its Lie algebraWiththe matrix commutator is $[D,T]=T$, and $[D,D]=[T,T]=0$.

The left Maurer-Cartan form and its right analogue areFor a constant $h$, left translation leaves ${\theta }_L$ unchanged, and right translation leaves ${\theta }_R$ unchanged. Taking their dual vector fields givesThe first pair consists of left-invariant vector fields; the second consists of right-invariant vector fields. These are the invariant frames of the real affine group. Computing their Lie brackets of vector fields yieldsAll brackets of a field with itself vanish. The sign difference is necessary: evaluation at the identity identifies left-invariant vector fields with the matrix Lie algebra as a homomorphism, whereas right-invariant vector fields realize the opposite Lie algebra. Equivalently, $\xi \mapsto -{\xi }_R$ is a homomorphism for the same matrix commutator. There is no convention in which these particular coordinate fields both have the positive structure constant while retaining the usual Lie bracket of vector fields.

The Maurer-Cartan equation gives the same check. If ${\theta }_L=D{\ell }^D+T{\ell }^T$, thenFor ${\theta }_R=D{r}^D+T{r}^T$, insteadThus the left equation is $d{\theta }_L+{\theta }_L\wedge {\theta }_L=0$ and the right equation is $d{\theta }_R-{\theta }_R\wedge {\theta }_R=0$, explaining the opposite bracket signs directly.

Finally, the two given matrix curves are $\exp (ϵT)$ and $\exp (ϵD)$. Their left action on a general group element isDifferentiating at $ϵ=0$ gives $T_R$ and $D_R$ respectively. Hence Left translations on a Lie group are generated by right-invariant vector fields. More generally, the velocity of $\exp (t\xi )g$ is $(d{R}_g{)}_e\xi$, which is right-invariant because left and right multiplication commute. In the other direction, the flows $g\exp (t\xi )$ generate left-invariant vector fields and are right translations on a Lie group.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.