Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 (Solitons, Instantons and Geometry)

The Euler-Lagrange equation isTime-translation invariance gives the conserved energyIndeed, after integrating the spatial term by parts and imposing finite-energy boundary conditions,

The potential is even, so field reflection has $\alpha =-1$; spatial reflection has $\beta =-1$. If $K(x)={\varphi }^{(0,1)}(x)$, the four oriented interpolating solutions, allowing a common translation, areThe first and third are kinks under this orientation convention, and spatial reflection gives their antikinks.

On the sector $0\le \varphi \le 1$, choose the superpotentialThe static energy has the Bogomolny bound completionEquality holds for ${\varphi }^{\prime }=W^{\prime }$, and therefore

The first-order Bogomolny equation isPartial fractions and one integration giveThus

As $x\to -\infty$, the term $-1/\varphi$ dominates the implicit equation, soAs $x\to +\infty$, put $\delta =1-\varphi$. Thenand henceThe profile rises monotonically from $0$ to $1$, with an algebraic left tail and an exponential right tail.

The Hodge star operator is defined byOn two-forms in oriented Euclidean four-space, ${\star }^2=1$ and the wedge product is symmetric. ThereforeFor an anti-self-dual curvature $F=-\star F$, the Yang-Mills instanton action becomes purely topological. Withthe standard positive-action convention gives

Let $D_{\mu }={\partial }_{\mu }+A_{\mu }$. A Lax pair with spectral parameter $\lambda$ isThe coefficients of $[L,M]=0$ at orders $1,\lambda ,{\lambda }^2$ are respectivelywhich are precisely the anti-self-dual Yang-Mills equations.

The equation $F_{wz}=0$ says that the partial connection $D_w,D_z$ is flat. Its compatibility condition therefore guarantees a local $G$-valued solution $g$ ofAfter the associated gauge transformation, both transformed components vanish:

In this gauge, $F_{\bar{w}\bar{z}}=0$ says that the remaining partial connection is flat. Hence locallyorThe remaining curvature equation then becomes

For Maxwell theory the group is Abelian, so write $J=e^{\chi }$. The reduced equation loses its commutators and becomesSince the Euclidean Laplacian isthe anti-self-dual Maxwell equations in this gauge are equivalent to

For a smooth map $\varphi :M_1\to M_2$ between connected oriented closed manifolds of equal dimension and a volume form $\omega$ on $M_2$, the topological degree is defined byThe standard area form on the unit sphere isSince ${\int }_{S^2}\omega =\mathrm{vol}(S^2)$, this gives

For a generic target value $y\in {\mathbb{CP}}^1$, its finite preimages solveWhen $m\ne n$ this has $\max (m,n)$ roots after the missing roots or poles at infinity are counted; when $m=n$, a generic $y$ again gives degree $m$. Holomorphic maps preserve orientation at regular preimages, so every local sign is positive. Hence

For $f(z)=z^m$,The pullback area is thereforeThe same form integrates to $2\pi$ on the target sphere, so

Using $z=x+iy$, the energy isBecause $|\varphi |=1$, both derivatives are tangent to $S^2$ and $|\varphi \times {\partial }_y\varphi |=|{\partial }_y\varphi |$. Completing the square givesThe final integral is $4\pi |\mathrm{deg}\varphi |$ by the topological degree formula. ThusEquality holds exactly when the appropriate first-order Bogomolny equation is satisfied:with the sign chosen to match the degree.