Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 (Solitons, Instantons and Geometry)

A classical field-theory soliton is a smooth, spatially localized, finite-energy solution which retains its identity under time evolution and is stable against small perturbations, commonly because of a topological charge or a balance between energy terms with different scaling behavior.

For a static field writewhereUnder the Derrick scaling ${\varphi }_{\lambda }(\mathbf{x})=\varphi (\lambda \mathbf{x})$, a change of variables givesA static solution must be stationary under this variation, so the Derrick virial identity isAll three energies are nonnegative. For $D\ge 4$, every coefficient is nonpositive and the coefficient of the strictly positive $E_2$ of any nonconstant field is negative. The identity is impossible. Thus, when the quartic-gradient term is available,For $D=2$ or $3$, the $E_4$ term has the opposite sign to at least one other term and the Derrick theorem does not rule out a soliton. If $\kappa =0$ from the outset, the identity reduces to $(2-D)E_2-D{E}_0=0$, recovering the stronger standard obstruction for $D\ge 2$.

The potential factorizes asIt is even and nonnegative, with three degenerate vacua at $\varphi =-2,0,2$. Its two intervening maxima occur at $\varphi =\pm 2/\sqrt{3}$ and have height $256/27$.

A finite-energy static solution in one dimension satisfiesMultiplication by ${\varphi }^{\prime }$ and use of the vacuum boundary conditions gives the first integralAny path from a negative vacuum to a positive vacuum must pass through the intermediate vacuum $\varphi =0$. There both $U$ and ${\varphi }^{\prime }$ vanish. The Picard-Lindelof theorem then forces a solution reaching $(\varphi ,{\varphi }^{\prime })=(0,0)$ at finite $x$ to remain there. Equivalently, the first-order orbit approaches $\varphi =0$ only as $|x|\to \infty$. This intermediate-vacuum obstruction to a kink means that a single kink cannot connect $-2$ to $2$; it splits into two elementary kinks at infinite separation.

For the kink in a phi-six model, choose the sector from $\varphi =0$ to $\varphi =2$ and the increasing sign of the Bogomolny equation:For $y={\varphi }^2$ this becomes the logistic differential equationAfter translating the centre to $x_0$, its solution isIt tends to $0$ as $x\to -\infty$ and to $2$ as $x\to +\infty$. Spatial reflection and $\varphi \mapsto -\varphi$ generate the other kink and antikink sectors.

Because the first-order equation saturates the Bogomolny bound, its mass is

Away from a zero of the Higgs field, writeSeparating real and imaginary parts of $(D_x+i{D}_y)\varphi =0$ givesThus $B=-\Delta u/2$ away from zeros. Combining this with $B=(1-e^u)/2$ givesIf $\varphi$ has a zero of multiplicity $N$ at the origin, its vortex number is the winding number $N$ and $u=2N\log r+O(1)$. Since $\Delta \log r=2\pi {\delta }^{(2)}(\mathbf{x})$ as a distributional identity, the complete Taubes equation isThe boundary conditions for a single $N$-vortex areThe latter is the finite-energy condition $|\varphi |\to 1$.

Suppose $u$ were positive somewhere. Since it tends to zero at infinity and to $-\infty$ at its vortex zero, it would attain a positive interior maximum away from the origin. At such a maximum the second-derivative test gives $\Delta u\le 0$, whereas the smooth Taubes equation givesThis contradiction with the maximum principle for subharmonic functions provesConsequently the Higgs magnitude of a vortex satisfies $|\varphi |=e^{u/2}\le 1$ everywhere.

For a rotationally symmetric vortex, the equation away from the origin isInsertThe prescribed zero fixes $\alpha =2N$. Since $N\ge 1$, $e^u=e^{\beta }{r}^{2N}[1+o(1)]$, whileMatching the singular and constant terms with $e^u-1=-1+o(1)$ givesThe undetermined $\beta$ is fixed by matching this local expansion to $u\to 0$ at infinity.

For a conformal rescaling of a Riemannian metric $h=\Omega (d{x}^2+d{y}^2)$, the Taubes equation in flat coordinates isThe first solution $u$ has $\Omega =1$ and vortex number $N$. For $\tilde{g}=e^{u}g$, the second solution therefore obeyswhere we used ${\Delta }_{\tilde{g}}=e^{-u}{\Delta }_g$ and converted the covariant delta distribution to the flat coordinate measure. Adding the equation for $u$ givesThus $u+\tilde{u}$ is again a flat-metric Taubes equation solution. Its vortex divisor is the union of the two divisors, with multiplicities added at coincident zeros, and its total vortex number isThis is vortex composition by conformal rescaling.

On an oriented Euclidean four-space with its metric volume form, the Hodge star operator is defined byfor forms $\alpha$ and $\beta$ of the same degree. In $n$ Euclidean dimensions it satisfiesHence on two-forms in four dimensions,The normalization of $\mathrm{vol}$ as the metric volume form is essential; reversing its orientation reverses a single Hodge star but leaves its square unchanged.

Because ${}^{\star }{}^{\star }=1$, the exterior algebra of two-forms has the orthogonal decompositionwhere ${}^{\star }H=H$ for a self-dual differential form and ${}^{\star }G=-G$ for an anti-self-dual differential form. The Hodge star is self-adjoint, soIt follows that

For an $SU(n)$ connection, use the positive normand define the Second Chern number byOrthogonality givesTherefore the Euclidean Yang-Mills actionobeys the Yang-Mills instanton Bogomolny boundEquality holds precisely when $F_{-}=0$ or $F_{+}=0$, according to the sign of $k$; these are the self-dual and anti-self-dual Yang-Mills instantons. Other trace and orientation conventions may reverse $k$ but leave the absolute-value bound unchanged.

Let $D_i={\partial }_i+A_i$ and $D_{\tau }={\partial }_{\tau }+A_{\tau }$. In temporal gauge, $A_{\tau }=0$, soChoose the orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d\tau$ and the anti-self-duality convention matching the question. The three independent components of $F=-{}^{\star }F$ areConsequently the Anti-self-dual Yang-Mills equations in temporal gauge areThey identify anti-self-dual Yang-Mills fields with a first-order flow of three-dimensional gauge connections.

Let $\zeta \in {\mathbb{CP}}^1$ be the spectral parameter. Define the two covariant differential operatorsThe auxiliary systemis compatible exactly when $[L(\zeta ),M(\zeta )]=0$ for every $\zeta$. The constant and quadratic coefficients giveand their equivalent conjugate equations, while the linear coefficient givesSince $F_{i\tau }=-F_{\tau i}$, these are preciselyThusis a Lax pair for the anti-self-dual Yang-Mills equations. In temporal gauge one simply sets $D_{\tau }={\partial }_{\tau }$.