Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3

Let $G$ act by left multiplication on the set $G/H$ of $n$ left cosets. This gives a group action homomorphismIts kernelis a normal subgroup of $G$, and the first isomorphism theorem gives $G/K\cong \mathrm{im}\phi$. By Lagrange's theorem, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the orbit-stabilizer theorem shows that $n$ divides $|G/K|$; in particular,

Now suppose $G$ is nonabelian and simple. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\phi$ embeds $G$ into $S_n$. Composing with the sign homomorphism givesIts kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so

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For a geodesic triangle with interior angles $\alpha ,\beta ,\gamma$, the local Gauss-Bonnet theorem isbecause its geodesic sides have zero geodesic curvature.

Triangulate a closed oriented surface into $F$ geodesic triangles, with $E$ edges and $V$ vertices. Summing the local formula, the angles around each vertex total $2\pi$, soSince every triangular face has three edges and every edge belongs to two faces, $3F=2E$. Hencewhich is the global Gauss-Bonnet theorem.

For the sphere $S_r$, the unit normal is $N(p)=p/r$. Its shape operator is, up to the conventional sign, $(1/r)I$ on each tangent plane. Both principal curvatures therefore have magnitude $1/r$, and the Gaussian curvature is

An octant has one eighth of the sphere's area:Thus ${\int }_{T}KdA=\pi /2$. Its three great-circle sides meet at three right angles, soThe two sides of the local Gauss-Bonnet formula agree directly.

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Computing the Laplacian givesandThus Laplace's equation holds exactly when

For $z=x+iy$,soOne suitable analytic function is therefore

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Forthe Euler-Lagrange equation isorThe general solution of this inhomogeneous linear differential equation isThe condition $y(0)=-1$ gives $C+D=0$, and the condition at $x=1$ gives $C=1$. Therefore

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The function $f$ has zero mean and its cosine coefficients vanish:Its sine coefficients areso

Away from the corners, $F^{\prime }=f$. Differentiating the Fourier series shows that $n{B}_n=a_n$ and $-n{A}_n=b_n$. Also,Consequentlyand

Evaluating the series for $f$ at $\theta =\pi /2$ gives the Leibniz formula for pi,Evaluating the continuous Fourier series for $F$ at $\theta =0$ givesand hence

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The states ${\chi }_1$ and ${\chi }_2$ are orthonormal energy eigenstates with hydrogenic energiesThe Born rule therefore gives

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At time $t$ the state isDirect evaluation of the radial integral gives the matrix elementsandIt follows thatThus $R(t)$ oscillates sinusoidally about $39a/8$. Its angular frequency is

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Writing $R=x^2+y^2$, direct differentiation givesso $\nabla \cdot \mathbf{u}=0$. Likewise,so the two-dimensional vorticity ${\partial }_{x}v-{\partial }_{y}u$ vanishes. The flow is therefore incompressible and irrotational away from the origin.

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With the conventiona stream function isup to an additive constant. Differentiation recovers both velocity components.

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For a velocity potential $\varphi$, one requires $u={\partial }_x\varphi$ and $v={\partial }_y\varphi$. Integration givesagain up to an additive constant.

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States $i$ and $j$ communicate when each is accessible from the other: there exist $m,n\ge 0$ withThis is an equivalence relation, and its equivalence classes are the communicating classes. A class is closed when no positive-probability transition leaves it.

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Reading the directed edges with positive transition probability, the communicating classes areState $3$ can move to $2$ or $4$, and state $4$ can move to $2$ or $5$, so their singleton classes are not closed. Every transition from $1,2,5,$ or $6$ remains within $\{1,2,5,6\}$, so

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Let $h_i={\mathbb{E}}_i{T}_6$. Then $h_6=0$, and first-step analysis on the closed class givesSolving these linear equations gives

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The rank-nullity theorem says that for a linear map $T:V\to Z$ with finite-dimensional domain,

The direct sum $U\oplus W$ consists of pairs $(u,w)$ with componentwise vector-space operations. The canonical inclusions areFor subspaces $U,W\le V$, defineThis is surjective, andThe First isomorphism theorem for vector spaces therefore gives

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Row reduction of the matrices whose columns are the displayed spanning vectors givesRow-reducing the matrix formed from all six columns givesThe dimension formula for a sum of subspaces now yields

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Expand $\det (A+tB)$ by multilinearity of the determinant in its columns. The coefficient of $t^j$ is a sum of determinants obtained by choosing $j$ columns from $B$ and the remaining columns from $A$. Since the columns of $B$ span a space of dimension $k$, every choice of more than $k$ columns from $B$ is linearly dependent. All coefficients of $t^j$ for $j>k$ therefore vanish, so

If $k=n$, then $B$ is invertible and the coefficient of $t^n$ is $\det B\ne 0$. Hence the degree is exactly $n$.

For a zero-polynomial example with $k=1$, takeThen $\mathrm{rank}B=1$, while $\det (A+tB)=0$ for every $t$.

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Assume that $p$ is a prime element. If $p=ab$, then $p$ divides $ab$, so $p$ divides $a$ or $b$. If $p\mid a$, write $a=pc$; cancellation in the integral domain gives $1=bc$, so $b$ is a unit. The other case is symmetric. Hence every prime element is irreducible.

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Suppose $p$ is irreducible andThen $p=ab$ for some $b\in R$. Irreducibility says that either $a$ is a unit, in which case $(a)=R$, or $b$ is a unit, in which case $(a)=(p)$. Thus no proper ideal lies strictly between $(p)$ and $R$, so

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For any commutative ring with identity, the maximal ideal quotient criterion states that an ideal $I$ is maximal exactly when $R/I$ is a field. Applying it to $I=(p)$ proves

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Every field is an integral domain, so if $R/(p)$ is a field then it is an integral domain.

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By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.

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A topological space is connected when it is not the union of two disjoint nonempty open subsets. It is path-connected when every two points $x,y$ can be joined by a path, meaning a continuous map $\gamma :[0,1]\to X$ with $\gamma (0)=x$ and $\gamma (1)=y$.

To prove that $[0,1]$ is connected, suppose it had a separation $[0,1]=U\cup V$ with $0\in U$. LetWhichever of $U$ or $V$ contains $c$, its relative openness supplies an interval about $c$ that contradicts either the definition of the supremum or the fact that all points immediately below $c$ lie in $U$. Thus no separation exists.

If $X$ is path-connected and $X=U\cup V$ were a separation, choose $u\in U$ and $v\in V$. A path from $u$ to $v$ would make $[0,1]$ the union of the disjoint nonempty open sets ${\gamma }^{-1}(U)$ and ${\gamma }^{-1}(V)$, contradicting the connectedness just proved. Hence

Now let $X\subseteq {\mathbb{R}}^n$ be open. Every point $x\in X$ lies in an open ball contained in $X$, and an open ball is path-connected by straight line segments. It follows that every path component of $X$ is open. If there were more than one path component, one component and the union of all the others would separate $X$. Consequently a connected open subset of Euclidean space is path-connected. The converse follows from the preceding implication.

The same argument answers the locally Euclidean question affirmatively. Every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$, hence a path-connected open neighbourhood after restricting to a sufficiently small ball. Thus $X$ is locally path-connected, its path components are open, and connectedness forces there to be only one.

For the final example, putThe set $Y$ is path-connected: each vertical segment $C_n$ meets the horizontal segment $A$. The segment $B$ is also path-connected and lies in the closure of $Y$, because $(1/n,y)\to (0,y)$ for every $y\in [1/2,1]$. A connected set together with a connected subset of its closure is connected, so

It is not path-connected. The image of the first coordinate of any path in $X$ lies inbut, while the path has positive height, its first coordinate lies in the totally disconnected set $\{0\}\cup \{1/n:n\ge 1\}$. A path beginning on $B$ cannot leave $x=0$ before reaching height zero, and $X$ contains no point $(0,y)$ with $0\le y<1/2$. Hence no path joins $B$ to $Y$, and

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For a smooth curve $\gamma (t)$ on a Riemannian surface, its energy of a curve isChoose a local parameterization $\mathbf{X}(u,v)$ and write $\gamma (t)=\mathbf{X}(u(t),v(t))$. With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by $E,F,G$.

If a straight line segment lies in the surface, parameterize it byThen $\ddot{\gamma }=0$, so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.

For the one-sheeted hyperboloidputTwo distinct ruling lines through $P$ areIndeed, the direction $d_{\pm }$ satisfiesso substitution shows that every point of the line lies in $H$. These are geodesics by the straight-line argument.

A third geodesic is the meridian through $P$. Choose $s_0$ with $\sinh s_0=z_0$; thenIts acceleration is ${\gamma }_m$, which is normal to $H$, so it is a geodesic. If $z_0\ne 0$, these give the required three distinct subsets.

If $z_0=0$, there is also the equatorial circle. Writing $P=(\cos {\theta }_0,\sin {\theta }_0,0)$,Its acceleration is normal to $H$ along $z=0$, so it is a fourth geodesic distinct from the meridian and the two rulings.

Finally, write $\rho =\sqrt{x^2+y^2}=\sqrt{1+z^2}$. Clairaut's relation gives the conserved quantityChoose initial data in $z>0$ withSince $|\sin \psi |\le 1$, every point of the resulting geodesic satisfies $\rho \ge c$, and thereforeContinuity keeps the geodesic in the component $z>0$, and the stated completeness assumption defines it for every real time. The continuum of choices of $c$ supplies infinitely many such geodesics.

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Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

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Let $Q_k=P^{(k)}$. Differentiating Legendre's differential equation $k$ times and using the Leibniz rule givesThe induction step follows by differentiating this equation once: the derivative of the coefficient $-2(k+1)x$ contributes the additional term $-2(k+1)Q_k^{\prime }$. ThusFor fixed $\lambda$, this is negative for every sufficiently large $k$.

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Multiplication by $(1-x^2{)}^k$ puts the equation in self-adjoint form:Multiply by $Q_k$ and integrate over $[-1,1]$. Regularity and the vanishing factor at both endpoints remove the boundary term in integration by parts, leavingIf $Q_k$ is not identically zero, the integral on the right without ${\lambda }_k$ is positive, whereas the left side is nonnegative. Therefore

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Part (a) makes ${\lambda }_k<0$ for all sufficiently large $k$, while part (b) would make ${\lambda }_k\ge 0$ whenever $P^{(k)}$ were nonzero. Hence some derivative of $P$ vanishes identically, so $P$ is a polynomial.

Let its degree be $n$. Then $P^{(n)}$ is a nonzero constant. Substituting $k=n$ into the differentiated equation annihilates both derivative terms and givesThus ${\lambda }_n=0$, andThe nonzero regular solutions are therefore constant multiples of the Legendre polynomial of degree $n$.

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In a region with no current, the Ampère-Maxwell equation and Faraday's law areThereforewhere the last line uses the divergence and curl of a cross product. Thus Poynting's theorem takes the formThe vector $\mathbf{S}$ is the Poynting vector.

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The inverse Lorentz transformation of electromagnetic fields, from the primed rest frame to the unprimed frame, givesandEquivalently, $\mathbf{B}=\gamma \mathbf{v}\times {\mathbf{E}}^{\prime }/c^2$ for $\mathbf{v}=v{\mathbf{e}}_x$.

Using $1/{\mu }_0={ϵ}_0{c}^2$ and ${\gamma }^2=(1-v^2/c^2{)}^{-1}$,

The primed field is static, so $w$ depends on $t,x$ throughand has no explicit $t^{\prime }$ dependence. The chain rule givesConsequentlyThus the field-energy profile is transported rigidly with velocity $v{\mathbf{e}}_x$.

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For the parallel flow $\mathbf{u}=(u(y,t),0)$ with no pressure gradient, the Navier-Stokes equations reduce to the diffusion equationSubstitution ofgivesThe boundary velocities require $f(L_0)=1$ and $f(-L_0)=-1$. The problem is antisymmetric under $y\mapsto -y$, so $f$ is odd. Withthe unique odd solution is