Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 313 (Applications of Differential Geometry to Physics)

Let $N=(0,\dots ,0,1)$ and $S=(0,\dots ,0,-1)$ be the poles of $S^n$. The open sets $U_N=S^n\setminus \{N\}$ and $U_S=S^n\setminus \{S\}$ cover the sphere. Stereographic projection gives coordinate chartsfrom these sets to ${\mathbb{R}}^n$. Their inverses areOn the overlap, the transition map iswhich is a smooth diffeomorphism of ${\mathbb{R}}^n\setminus \{0\}$. These two compatible charts make $S^n$ a smooth manifold of dimension $n$.

Now let $G$ be an $m$-dimensional Lie group and choose a basis $e_1,\dots ,e_m$ of its tangent space $T_{e}G$ at the identity. Definewhere $L_g$ is left translation on a Lie group. Smoothness of multiplication makes each $E_i$ a smooth left-invariant vector field, and invertibility of $(d{L}_g{)}_e$ makes $E_1(g),\dots ,E_m(g)$ a basis of $T_{g}G$ at every point. Thus the $E_i$ form a global frame and every Lie group is a parallelizable manifold.

The columns of a matrix in the special unitary group $SU(2)$ are orthonormal and its determinant is one. Consequently every element has the unique formThe pair $(z_1,z_2)\in {\mathbb{C}}^2\cong {\mathbb{R}}^4$ therefore identifies $SU(2)$ diffeomorphically with $S^3$. The Lie-group construction then proves that $S^3$ is parallelizable; this is the SU(2) as the three-sphere identification.

Another example is $S^1$, which is the Lie group $U(1)$. Explicitly, at $(x,y)\in S^1$ the vectoris smooth, tangent, and nowhere zero, so it is a global one-vector frame. Thus $S^1$ is another parallelizable sphere.

The special orthogonal group in three dimensions isDifferentiating $A(t{)}^{T}A(t)=I$ at the identity shows that its Lie algebra is the space of skew-symmetric matrices. A convenient basis iswith $[J_a,J_b]={ϵ}_{abc}{J}_c$.

The infinitesimal action of $\exp (t{J}_a)$ on a point $x\in {\mathbb{R}}^3$ is $J_{a}x=e_a\times x$. It therefore generates the vector fieldFundamental vector fields for this left action form an antihomomorphism with the stated convention, and direct differentiation givesso their span is closed under the Lie bracket of vector fields.

The brackets $\{x,y\}=z$, $\{y,z\}=x$, and $\{z,x\}=y$ define the rotational Lie-Poisson structure on R3. With the convention that a Hamiltonian vector field acts by $X_H(f)=\{f,H\}$,Hence the required Hamiltonians are simplyThe quadratic functionsatisfies $\{F,x_a\}=0$ for all $a$, so it is a Casimir function of a Poisson manifold. Its nonzero regular level sets $F=R^2$ are spheres. The Poisson tensor has rank two there and is tangent to each level set, so it inverts to a symplectic form; each sphere is a symplectic leaf. Rotations preserve both $F$ and the alternating tensor ${ϵ}_{abc}$, hence preserve the restricted Poisson tensor and its inverse symplectic form. The $SO(3)$ action therefore restricts to a symplectic action on every sphere $S_R^2\subset {\mathbb{R}}^3$.

For the curvature two-formuse the normalizationfor the Second Chern form. Graded cyclicity of the trace gives $\mathrm{Tr}(A^{\wedge 4})=0$ andOn the other hand,andThus the coefficient in the Chern-Simons 3-form must beand

For the gauge transformation $A^{\prime }=gA{g}^{-1}-dg{g}^{-1}$, use $d(g^{-1})=-g^{-1}(dg)g^{-1}$ and the Maurer-Cartan equation. Expanding $d{A}^{\prime }+A^{\prime }\wedge A^{\prime }$ makes the terms linear and quadratic in $dg{g}^{-1}$ cancel, leavingInvariance of the matrix trace under conjugation then gives

A Yang-Mills instanton on ${\mathbb{R}}^4$ has finite Euclidean action, smooth curvature in the interior, and $F\to 0$ sufficiently rapidly at infinity. Its connection therefore approaches a pure gauge on the asymptotic three-sphere,up to a decaying correction, for a map $g:S_{\infty }^3\to SU(2)$. By Stokes theorem,Writing $\theta =dg{g}^{-1}$ gives $d\theta =\theta \wedge \theta$ and, for $A=-\theta$,Hence the instanton number as a winding number at infinity isUnder $SU(2)\cong S^3$, this integer is the degree of a map between oriented manifolds $g:S^3\to S^3$. Reversing the trace or orientation convention reverses the displayed sign.