Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 47 (Solitons and Instantons)

For the usual canonical scalar field theory, subtract the vacuum energy so that the potential $U$ is nonnegative, and write the static energy in $d$ space dimensions asAssume finite energy and sufficient decay to justify the scaling variation, with a fixed vacuum at infinity. Under Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$,A static solution must be stationary under this admissible variation. Thus the Derrick virial identity isFor $d\ge 3$ both terms are nonpositive, and the coefficient of $T$ is strictly negative. Therefore $T=W=0$: the field is a constant vacuum. This proves the Derrick theorem under its hypotheses. Negative potentials, additional derivative terms, gauge fields or time dependence can invalidate this particular obstruction; it is not a theorem about arbitrary energy functionals.

A non-topological soliton can instead be supported by a conserved ordinary Noether charge. For a complex scalar field with global $U(1)$ invariance, take $\varphi =e^{i\omega t}f(x)$ and let $I=\int f^2$. In the normalization with kinetic density $|\partial \varphi {|}^2/2$, its charge is $Q=\omega I$. At fixed charge, eliminating $\omega$ givesThe charge-constrained scalar-field scaling now readsThe positive scaling power of the charge term can balance the others even for $d\ge 3$. Such a Q-ball is time-dependent although its energy density is stationary, and the fixed-charge variational problem is different from the static one.

Suitable potentials permit this mechanism in any spatial dimension. For example, suppose $U(f)>0$ for $f\ne 0$ and the ratio below attains a strictly positive minimum:Such a potential can favor a large region of nearly constant nonzero amplitude. At its preferred amplitude, the bulk terms minimize at $I\approx |Q|/{\omega }_0$, with energy about ${\omega }_0|Q|$. A wall has surface energy of order $|Q{|}^{(d-1)/d}$, lower order than the bulk term for every fixed $d\ge 1$. For large charge the energy can therefore be below the free-particle threshold $m|Q|$. This explains how a localized fixed-charge solution may exist and resist dispersal. Existence and stability still require a suitable potential and charge range; not every complex-field theory or every member of a rotating family is a constrained minimum.

For the particular one-dimensional profile, substituting the rotating field into the equation givesMultiplying by $f^{\prime }$ shows thatBoth $f$ and $f^{\prime }$ vanish at infinity, so the integration constant is zero. HenceHere $U(f)=f^2/2-f^4/4$. This potential is not globally bounded below; the existence of the displayed localized solutions alone does not prove their nonlinear stability.

Use metric signature $(+,-)$ and setThe Lorentz boost of a Q-ball givesThe peak follows $x=x_0+vt$. Because the equation is Lorentz invariant and the field is a scalar, this is again a solution.

The canonical stress-energy tensor gives the energy and signed physical momentum at fixed time:The minus sign agrees with the scalar-field momentum flux in the chosen signature. For the boosted field,andThe first integral gives $U(f)=\frac{1}{2}[(f^{\prime }{)}^2+{\omega }^2{f}^2]$. Changing variable to $\xi$, so $dx=d\xi /\gamma$, yieldsThus the rest-energy normalization and the relativistic energy-momentum relation are

Choose orientation ${ϵ}_{1234}=1$ and the anti-Hermitian basis in the question. The Pauli matrix commutator identity gives $[e_a,e_b]={ϵ}_{abc}{e}_c$. Define the gauge curvature and adjoint covariant derivative byFor a compactly supported variation $a_{\mu }=\delta A_{\mu }$,Antisymmetry and the invariant Lie-algebra inner product therefore giveCovariant integration by parts is valid because
$⟨X,[A,Y]⟩=-⟨[A,X],Y⟩$. Thus the Euler-Lagrange field equations are

The Euclidean Hodge star operator on two-forms isThe Anti-self-dual Yang-Mills equations mean $\boxed{\ast F=-F}$.

The gauge-theory Bianchi identity isTo prove it, act on the defining representation with ${\nabla }_{\mu }={\partial }_{\mu }+A_{\mu }$. Its commutator is multiplication by $F_{\mu \nu }$. The operator Jacobi identity for ${\nabla }_{\mu },{\nabla }_{\nu },{\nabla }_{\rho }$ becomes multiplication by the displayed cyclic sum, which must vanish. Equivalently, substitution of $F=dA+A\wedge A$ into $D_{A}F$ cancels the ordinary derivatives, leaving the Lie-algebra Jacobi identity. Contracting the cyclic identity with the alternating tensor gives $D_{\mu }(\ast F{)}_{\mu \nu }=0$. If $\ast F=-F$, this is exactly $D_{\mu }{F}_{\mu \nu }=0$, proving the first-order-to-second-order implication without assuming the field equations.

LetAll squared component sums below include both orders of each antisymmetric pair, as does the printed energy. Since the Hodge star preserves this norm,Consequently $V\ge |J|/2$. An anti-self-dual field has $J=-2V\le 0$ and saturates the first bound. Among all finite-energy potentials with the same value of $J$, it therefore has the absolute minimum $-J/2$. For positive $J$, the saturating sign is self-duality instead; there cannot be a nonzero anti-self-dual solution in that sector. This is the Yang-Mills instanton Bogomolny bound in the paper's normalization.

For the topological meaning, regard $F=\frac{1}{2}{F}_{\mu \nu }^a{e}_{a}d{x}^{\mu }\wedge d{x}^{\nu }$ as a matrix-valued two-form. Since $\mathrm{tr}(e_a{e}_b)=-{\delta }_{ab}/2$,In the Second Chern number convention used here,With the standard instanton boundary condition, the connection approaches a pure gauge at infinity and extends to a bundle over the compactified four-sphere. This makes $k$ an integer, the instanton number; reversing the orientation or using the opposite charge convention reverses its sign. The square-completion argument itself does not require quantization or a chosen framing.

One can see the boundary interpretation directly. The Chern-Simons 3-form obeysFor an asymptotic potential $A=-dg{g}^{-1}$, the flatness identity $dA=-A\wedge A$ givesThis is the winding number of $g:S_{\infty }^3\to SU(2)\cong S^3$, with the corresponding orientation convention. Smooth variations preserving that boundary class cannot change it. Thus the integral in the question labels the topological sector of the Yang-Mills instanton, and in an anti-self-dual sector the energy is $\boxed{V=32{\pi }^{2}k}$ with $k\ge 0$ in our convention.

Put $h=|\Phi {|}^2$, $D_i={\partial }_i-i{A}_i$, $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$, and $W(h)=h(1-h{)}^2/4$. Because the inverse-density Abelian Higgs model contains $B^2/h$, its finite-energy admissible class matters at zeros of the Higgs field. First derive the formal equations on the open set where $h>0$; on the first-order solutions below the potentially singular quotients extend across the zeros.

For variations $\psi =\delta \Phi$ and $a_i=\delta A_i$,Covariant integration by parts in the scalar term, and ordinary integration by parts in the magnetic term, givewhere ${ϵ}_{12}=1$. HereThus the Euler-Lagrange field equations areEquivalently the gauge equation is ${\partial }_j(F_{ji}/h)+j_i=0$. The current sign follows from the explicit choice $D=\partial -iA$.

For the Bogomolny bound for the inverse-density Abelian Higgs model, computeusing $[D_1,D_2]\Phi =-iB\Phi$. Expanding the first square givessoAlso,The current boundary integral vanishes by the assumed rapid decay. Combining the identities therefore givesFor fixed positive flux the minimum is $\boxed{V_{\min }=\pi N}$, attained when the Bogomolny equations areThe quotient at a zero is understood through the finite-energy limit, not by dividing a literal nonzero numerator by zero.

These equations imply the second-order equations directly. Applying $D_1-i{D}_2$ to the first gives $(D_i{D}_i+B)\Phi =0$. On the second equation,which is exactly the scalar equation. The first equation also implies $j_1=-\frac{1}{2}{\partial }_{2}h$ and $j_2=\frac{1}{2}{\partial }_{1}h$. Since $B/h=(1-h)/2$, the gauge equations follow too. All these coefficients are regular across zeros for smooth first-order fields.

For the scalar reduction, write $\Phi =e^{u/2+i\chi }$ away from its zeros. Separating the real and imaginary parts of the first Bogomolny equation givesThus, with the oriented planar Hodge star $\ast d{x}^1=d{x}^2$, $\ast d{x}^2=-d{x}^1$,away from the zeros. A zero $p_j$ of multiplicity $n_j$ has $u=2n_j\log |x-p_j|+\text{smooth}$ and phase winding $2\pi n_j$. The multiplicities are positive: locally the first-order equation is a covariant Cauchy–Riemann equation, and a nonvanishing integrating factor turns its solutions into holomorphic functions. Distributionally,Consequently the logarithmic equation for inverse-density Abelian Higgs vortices isThe delta terms are essential for multi-vortex reconstruction; omitting them would lose the specified zeros and flux.

Here is an explicit regular reconstruction, which also removes all artificial singularities. Given distinct locations $p_j$ and integer multiplicities, putSince $\Delta u_0=4\pi {\sum }_j{n}_j{\delta }_{p_j}-g$, solving the smooth nonlinear elliptic problem for $v=u-u_0$ amounts toLet $z=x^1+i{x}^2$ and let $z_j$ represent $p_j$. A smooth solution with the topological decay reconstructsThese formulas are smooth at every prescribed zero and agree with the phase formula off the zeros. Taking a curl and using the equation for $v$ verifies $B=e^u(1-e^u)/2$; substitution verifies the other first-order equation. At infinity the phase winds $N$ times, producing flux $2\pi N$. Thus multi-vortex construction reduces to solving one scalar elliptic boundary-value problem for specified positions and multiplicities, followed by these explicit formulas.

For the topological branch, $u\le 0$: a positive interior maximum would have $\Delta u\le 0$, whereas the equation there gives $\Delta u=e^u(e^u-1)>0$. At infinity, the linearized scalar equation is $\Delta u-u=0$. Comparison barriers and elliptic estimates give the decaying tail for regular topological solutions, and hence the required decay of $B,D\Phi$ and $1-h$. The nonlinear elliptic problem determines $v$; the reconstruction formulas do not constitute an arbitrary superposition of single-vortex profiles.

For a gauge-field configuration, spatial symmetry means symmetry up to gauge transformation: the pullback of the fields by a spatial isometry lies in their original gauge orbit. In the anti-Hermitian convention,The potential and charged fields need not be strictly fixed in a particular gauge; their gauge-invariant observables must be fixed.

For example, the Abelian potential $A=Bxdy$ describes a uniform magnetic field. Translating $x$ by $a$ adds $Bady=d(Bay)$, a gauge change, although the potential itself changes. For an axially symmetric vortex $\Phi =f(r)e^{iN\theta }$, $A=a(r)d\theta$, rotation through $\beta$ multiplies $\Phi$ by $e^{iN\beta }$; a constant $U(1)$ transformation compensates. These illustrate why literal invariance of the representative is too strong.

For the 't Hooft-Polyakov monopole, a spatial rotation $R\in SO(3)$ has an $SU(2)$ lift $g_R$ with adjoint action $R$. The two lifts differ by a central sign and act identically on adjoint fields. The hedgehog ansatz for a monopole satisfiesThe second identity follows from invariance of ${ϵ}_{iaj}$ under simultaneous rotations of all three indices. Thus the pulled-back fields are obtained by the constant gauge transformation $g_R$; applying $g_R^{-1}$ restores the representative. In particular $|\Phi {|}^2$ and the energy density are radial even though ${\Phi }^a=f(r)x^a/r$ is not literally invariant under rotating space alone.

For the field equations, use the invariant inner product with $[e_a,e_b]={ϵ}_{abc}{e}_c$. Since $B_i^a{B}_i^a=\frac{1}{2}{F}_{ij}^a{F}_{ij}^a$, a variation $a_j=\delta A_j$, $\psi =\delta \Phi$ givesHere the scalar kinetic contribution to the gauge variation uses
$⟨D_j\Phi ,[a_j,\Phi ]⟩=⟨[\Phi ,D_j\Phi ],a_j⟩$. The Euler-Lagrange field equations for the energy actually printed areThe PDF has $\lambda (1-|\Phi {|}^2)$, without a square. This is not the usual nonnegative monopole potential with vacuum $|\Phi |=1$. If the usual $\lambda (1-|\Phi {|}^2{)}^2$ were intended, only the scalar equation would change toThe final requested proof at $\lambda =0$ is identical for either interpretation; we do not silently alter the printed energy.

To derive the covariant derivative of the radial fields, first differentiate the Higgs profile:The commutator term isAdding the two terms gives exactly

The radial eigenvalue of the supplied magnetic tensor is $2\alpha +r^2{\alpha }^2$, while that of this covariant derivative is $f^{\prime }$. Its transverse coefficients are respectively $r{\alpha }^{\prime }+2\alpha$ and $f/r+r\alpha f$. Consequently $B=-D\Phi$ reduces toEquivalently, the hedgehog monopole equations in radial profile variables with $K=1+r^2\alpha$ areThese are two equations rather than an overdetermined system: radial and transverse projections span the two invariant tensor structures. For a regular monopole in this hedgehog sector, smoothness at the origin gives $f(0)=0$, $K(0)=1$, while the normalized vacuum conditions give $f\to 1$, $K\to 0$ at infinity; solving the ODEs is not needed here.

Finally, prove that the Bogomolny monopole equations imply Yang-Mills-Higgs equations, without restricting to the radial ansatz. The three-dimensional gauge-theory Bianchi identity is $D_i{B}_i=0$. If $B_i=-D_i\Phi$, thenwhich is the scalar equation at $\lambda =0$. For the gauge equation,In the second line antisymmetry eliminates the symmetric derivative part; the curvature commutator is $[D_i,D_k]\Phi =[F_{ik},\Phi ]$. The next equality uses
${ϵ}_{ijk}{ϵ}_{ik\ell }=-2{\delta }_{j\ell }$ and $F_{ik}={ϵ}_{ik\ell }{B}_{\ell }$. These are exactly both field equations at $\lambda =0$, so every smooth Bogomolny solution satisfies them, not merely those of the hedgehog form.

For finite-energy monopoles, the associated energy argument is also transparent:The gauge-theory Bianchi identity converts $⟨B_i,D_i\Phi ⟩$ into this boundary term. At fixed asymptotic magnetic charge, $B=-D\Phi$ saturates the Bogomolny bound. This complements the local differential proof above and explains the energy-minimizing role of the first-order equations.