Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 51 (Solitons and Instantons)

The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.

Use the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$. On the Adjoint representation, the adjoint covariant derivative is $D_{\mu }X={\partial }_{\mu }X+[A_{\mu },X]$. Set $a_{\mu }=\delta A_{\mu }$ and $\chi =\delta \Phi$. ThenThe trace obeys $\mathrm{Tr}(X[Y,Z])=\mathrm{Tr}([Z,X]Y)$, and covariant integration by parts has the ordinary boundary term because the trace of a commutator is zero. Thus compactly supported variations giveThe trace pairing on $\mathfrak{su}(2)$ is nondegenerate, so the Euler-Lagrange field equations areIn particular the sign of the scalar mass term follows from the plus sign in the supplied trace Lagrangian density; it should not be guessed from a different metric convention. This is the adjoint Yang-Mills-Higgs trace variation.

For the static Bogomolny-Prasad-Sommerfield monopole, take the usual purely magnetic extension ${\partial }_0{A}_i={\partial }_0\Phi =0$ and temporal gauge $A_0=0$. At $m=0$, the equations reduce towith the $\nu =0$ equation identically zero. We now verify both spatial equations from $B_i=D_i\Phi$ without assuming them. The gauge-theory Bianchi identity gives $D_i{B}_i=0$, hence immediately $D_i{D}_i\Phi =0$. Also $F_{ji}={\epsilon }_{jik}{B}_k$, soHere ${\epsilon }_{jik}{\epsilon }_{jk\ell }=-2{\delta }_{i\ell }$. This proves that the Bogomolny equations imply all static zero-mass Yang-Mills equations and the scalar equation. The specification of $A_0$ matters: the spatial Bogomolny equations by themselves do not constrain an arbitrary extra electric potential, so the implication is for their standard purely magnetic static extension.

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, the Euclidean Yang-Mills equations. This proves the implication locally; finite action would additionally be needed to call the solution an instanton.

Use the complex orientation for $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. Accordingly the three gauge curvature equations areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions ${\partial }_w-\lambda {\partial }_{\bar{z}}$ and ${\partial }_z+\lambda {\partial }_{\bar{w}}$ are mutually orthogonal null vectors for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental matrix $\Psi$ forces the commutator to vanish; a single zero solution would not establish it. This is the Euclidean complex-coordinate anti-self-dual Lax pair.

For an Abelian U(1) connection, all commutators of the connection components vanish, and the gauge curvature equations therefore readLocally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar{\partial }$-closed. The Dolbeault-Poincaré lemma, and its complex conjugate, give scalar functions $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis is the scalar Laplace equation, since for the stated metric $\Delta =2({\partial }_z{\partial }_{\bar{z}}+{\partial }_w{\partial }_{\bar{w}})$. The complex gauge transformation need not remain in $U(1)$.

A reduction using only a unitary gauge transformation is also available. Since $A$ is anti-Hermitian, $A_{\bar{w}}=-\overline{A_w}$ and $A_{\bar{z}}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar{u}$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts gauge curvature is $\partial \bar{\partial }f$. Thus the Abelian anti-self-dual connections from harmonic scalar potentials reduction is local and does not require silently enlarging the real gauge group.

Interpret triviality as holomorphic triviality. Choose a global holomorphic bundle frame $s_1,\dots ,s_k$ for $E$. Let $H_{\alpha }$ have as its columns the coordinates of these sections in the prescribed local frame over $U_{\alpha }$, with the convention that a fiber coordinate vector transforms as $s_{\beta }=F_{\alpha \beta }{s}_{\alpha }$. ThenEach $H_{\alpha }$ is holomorphic and invertible because its columns form a basis in every fiber. This proves the requested holomorphic splitting of transition functions on a trivial bundle. Smooth or topological triviality alone would not justify holomorphic splitting.

For the line bundle on projective twistor space, pull its patching function back to the twistor line $L_x$ using the incidence convention of this problem, ${\omega }^A=x^{A{A}^{\prime }}{\pi }_{A^{\prime }}$, without an extra factor of $i$. Triviality on each line gives nonvanishing linewise holomorphic functions $H_0,H_1$ with $F_{01}=H_1{H}_0^{-1}$. The factors can be chosen as holomorphic functions of $x$ locally: a degree-zero scalar transition function has zero winding on the overlap annulus, hence admits a logarithm there; splitting its positive and negative Laurent series parts gives factors whose coefficients are Cauchy integrals that are holomorphic functions of $x$. The explicit splitting is worked out below. Their only common linewise ambiguity is multiplication by a nonzero function $g(x)$: the ratio is a global holomorphic function on the compact complex projective line, hence constant on that line.

Take ${ϵ}_{0^{\prime }{1}^{\prime }}=1$, ${ϵ}^{0^{\prime }{1}^{\prime }}=-1$ and raise a two-component spinor by left multiplication with ${ϵ}^{A^{\prime }{B}^{\prime }}$. The derivative along an alpha-plane is ${\ell }_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$. It annihilates the unsplit twistor function, becauseThusis holomorphic on the whole line and has projective weight one. A section of $\mathcal{O}(1)$ is linear in the homogeneous coordinates: in an affine coordinate, holomorphic regularity at infinity bounds its growth to at most first order, so the Cauchy integral formula excludes all higher powers. ThereforeThe inverse splitting functions obey $({\ell }_A+G_A)H_{\alpha }^{-1}=0$. Commuting these two equations provesThe spinor decomposition of gauge curvature iswhere both $\phi$ two-component spinors are symmetric. Vanishing of the contraction for every $\pi$ forces ${\phi }_{A^{\prime }{B}^{\prime }}=0$. This is exactly the anti-self-dual Maxwell equations. Since the Abelian gauge curvature is $dA$, it is closed; its duality condition also makes it co-closed, giving the source-free Maxwell equations. Changing both splitting factors by $g(x)$ changes $A$ by $d\log g$, the expected Abelian gauge transformation. This derives the rank-one Penrose-Ward correspondence in the needed local form.

For the exponential patching function, choose a representative $f$ of the Čech cohomology class, of homogeneous degree zero. On the overlap of the two line patches write $f_x(\zeta )=f(x\pi ,\pi )={\sum }_{n\in \mathbb{Z}}{f}_n(x){\zeta }^n$. To compute the gauge potential explicitly, use a constant two-component spinor basis in which ${\iota }_{A^{\prime }}=(1,0)$ and the finite patch ${\pi }_{0^{\prime }}=\zeta ,{\pi }_{1^{\prime }}=1$. Then ${\pi }^{0^{\prime }}=-1$, ${\pi }^{1^{\prime }}=\zeta$ andFor a positively oriented contour in the overlap annulus, split its Laurent series asThe first is holomorphic inside the contour, the second outside including infinity; $H_{\alpha }=e^{h_{\alpha }}$ split $e^f$. Since $(-{\partial }_{B{0}^{\prime }}+\zeta {\partial }_{B{1}^{\prime }})f_x=0$, coefficient comparison gives ${\partial }_{B{0}^{\prime }}{f}_n={\partial }_{B{1}^{\prime }}{f}_{n-1}$. Applying that operator separately to $h_0,h_1$ cancels all nonconstant terms and leavesThus $-A_{B{0}^{\prime }}+\zeta A_{B{1}^{\prime }}=G_B$ yields $A_{B{1}^{\prime }}=0$ and $A_{B{0}^{\prime }}=-{\partial }_{B{1}^{\prime }}{f}_{-1}$. MoreoverSubstitution gives $A_{B{0}^{\prime }}=-(2\pi i{)}^{-1}\oint f_{,B}d\zeta$, which in homogeneous form is preciselyThis is the Abelian Ward splitting by Laurent series. The factors have weights $-1,-1,2$, respectively, so the integrand is invariant under a change of homogeneous representative. The contour avoids ${\iota }_{C^{\prime }}{\pi }^{C^{\prime }}=0$; the basis chosen above places that point at infinity. One can keep the contour fixed for $x$ in a sufficiently small domain, allowing differentiation under the integral.

Finally differentiate the contour potential for an anti-self-dual Maxwell field. By the twistor incidence relation, ${\partial }_{A{A}^{\prime }}{f}_{,B}={\pi }_{A^{\prime }}{f}_{,AB}$. ConsequentlyThe second equality uses the elementary two-spinor identity ${\pi }_{A^{\prime }}{\iota }_{B^{\prime }}-{\pi }_{B^{\prime }}{\iota }_{A^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}({\iota }_{C^{\prime }}{\pi }^{C^{\prime }})$ in the declared convention. The remaining two-component spinor is symmetric in $A,B$, and no primed symmetric gauge curvature part survives. This directly verifies $dA$ is anti-self-dual. In complexified Lorentzian signature duality has Hodge star eigenvalues $\pm i$; here the vanishing primed two-component spinor fixes the stated ASD convention rather than incorrectly imposing the Euclidean equation $\ast F=-F$ on that signature.