Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

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Fix $x\in X$. By assumption there is a neighbourhood $U$ of $x$ on which $f_n\to f$ with uniform convergence. Each restriction $f_n{|}_U$ is continuous, and a uniform limit of continuous functions is continuous. Thus $f{|}_U$ is continuous, in particular at $x$. Since $x$ was arbitrary, $f$ is continuous on $X$.

Now let $K\subseteq X$ be compact and let $\epsilon >0$. For each $x\in K$, choose a neighbourhood $U_x$ on which convergence is uniform. The sets $U_x$ cover $K$, so compactness gives a finite subcoverFor each $j$, choose $N_j$ such thatTaking $N={\max }_j{N}_j$, every $y\in K$ belongs to one of these finitely many neighbourhoods, and thereforeThus $f_n\to f$ uniformly on every compact subset. This proves the local uniform convergence on compact subsets principle.

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The domains are not conformally equivalent. Suppose that $F:\mathbb{C}\setminus \{0\}\to \{w:0<|w|<1\}$ were a conformal equivalence. Since $F$ is bounded near zero, the Riemann removable singularity theorem extends it holomorphically across zero. The extension is a bounded entire function, so the Liouville theorem makes it constant, contradicting bijectivity. This is the punctured plane is not conformally equivalent to the punctured unit disc obstruction.

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The domains are not conformally equivalent. If $F:\mathbb{C}\to \mathbb{H}$ were a conformal equivalence, composing it with a Cayley transform from the upper half-plane $\mathbb{H}$ to the unit disc would produce a bounded nonconstant entire function. This contradicts the Liouville theorem, proving that the complex plane is not conformally equivalent to the upper half-plane.

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The domains are conformally equivalent. The Möbius transformationmaps the upper half of the unit disc onto the first quadrant: its diameter $(-1,1)$ maps to the positive real axis and its upper semicircle maps to the positive imaginary axis. Squaring maps the first quadrant bijectively and conformally onto the upper half-plane. Thereforeis the required equivalence, as recorded by the conformal equivalence between the upper half-disc and upper half-plane.

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Write the expectation value asDifferentiating and retaining the possible explicit time dependence of the observable givesThe Schrodinger equation and its adjoint arewhere the Hamiltonian operator is Hermitian. Substitution yieldsThis is the proof of Ehrenfest theorem from the Schrodinger equation.

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For a particle in the scalar potential energy $U(x)$,The canonical commutation relation $[x,p]=i\hslash$ givesThe Ehrenfest theorem with $O=x$ therefore givesSince $[p^2,p]=0$ and direct action on a wavefunction gives $[U(x),p]=i\hslash U^{\prime }(x)$,Finally $H$ has no explicit time dependence and $[H,H]=0$, so

The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that $⟨U^{\prime }(x)⟩\simeq U^{\prime }(⟨x⟩)$. These conclusions are summarized by the classical equations from Ehrenfest theorem.

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Let $\theta$ be the angle between $r$ and the positive $z$-axis. By superposition of the electric potential of point charges,For $r>d$, the Legendre polynomial expansion givesThereforeThe three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is $Q(N-M-1)$ and the electric dipole moment is $Qd(M-1)\hat{z}$. This is the multipole expansion of three collinear charges.

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The monopole vanishes exactly whenand the dipole vanishes exactly whenSince $N$ and $M$ are positive integers, both vanish precisely for

Assume now that the monopole has been cancelled. If $Qd$ is held fixed while $d\to 0$, the quadrupole coefficient $Q{d}^2=(Qd)d$ tends to zero, whereas the dipole coefficient $(M-1)Qd$ has a finite limit. For $M\ne 1$ this is a point-dipole limit; for $M=1$ the limiting external potential vanishes.

If instead $Q{d}^2$ is held fixed, the dipole coefficient isIt diverges unless $M=1$. When $M=1$ and $N=2$, both lower multipoles vanish and the quadrupole coefficient has a finite nonzero limit. These are the dipole and quadrupole scaling limits of three collinear charges.

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Here $N=2$ and $M=1$. At $(x,0,0)$, the $z$-components of the fields of the two outer charges cancel by reflection symmetry. By Coulomb's law, the remaining field isMultiplying by the test charge $-Q$ givesThe quantity in parentheses is positive. Thus for $x>0$ the force points in the negative $x$ direction, while for $x<0$ it points in the positive $x$ direction. It is always attractive toward the origin, as in the transverse force from a collinear electric quadrupole.

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The nonzero orthogonal polynomials $Q_0,\dots ,Q_n$ have distinct degrees and form an orthogonal basis of ${\mathcal{P}}_n$. DefineFor each $j\le n$,Thus the residual $f-p_n^{\ast }$ is orthogonal to all of ${\mathcal{P}}_n$.

For any $p\in {\mathcal{P}}_n$, writeThe two terms are orthogonal, so the Pythagorean theorem in an inner-product space givesEquality holds only for $p=p_n^{\ast }$. This proves the formula and uniqueness of the least-squares polynomial in an orthogonal-polynomial basis.

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Part (a) shows that $f-p_n^{\ast }$ is orthogonal to $p_n^{\ast }\in {\mathcal{P}}_n$. Sincethe Pythagorean theorem in an inner-product space immediately givesThis is the norm decomposition associated with orthogonal projection onto a finite-dimensional subspace.

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Let $g(0)=0$. First-step analysis givesand, because state $3$ moves to $2$ or remains at $3$ with equal probabilities,The last equation gives $g(3)=2+g(2)$. Substitution into the second gives $g(2)=4+g(1)$, and the first then gives $g(1)=3+g(1)/2$. HenceThese are the expected hitting times of the absorbing state.

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Let $h(i)$ be the probability of reaching state $3$ before state $0$. The boundary values are $h(0)=0$ and $h(3)=1$. First-step analysis at states $1$ and $2$ givesTherefore $h(1)=h(1)/4+1/4$, so the requested hitting probability is

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Let $u(i)$ be the expected number of visits to state $3$ before absorption, counting the present state when $i=3$. Its reward equations areThe last equation gives $u(3)=2+u(2)$, so the second nontrivial equation gives $u(2)=2+u(1)$. Hence $u(1)=1+u(1)/2$, andThus the expected occupation count is two; together with the previous parts, this is the four-state reflecting absorbing random walk calculation.

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For ordered bases $\mathcal{B}=(v_1,\dots ,v_n)$ of $V$ and $\mathcal{C}=(w_1,\dots ,w_m)$ of $W$, the matrix representation of a linear map $\gamma :V\to W$ is the matrix whose $j$th column consists of the $\mathcal{C}$-coordinates of $\gamma (v_j)$. Thus

The operators $\alpha ,\beta :V\to V$ are conjugate linear operators when there is an isomorphism $s:V\to V$ such thatIn one basis their matrices therefore satisfy $[\beta ]=[s{]}^{-1}[\alpha ][s]$, so they are similar matrices.

For invertible $\beta$, the mapis linear, and its inverse is $A\mapsto \beta A{\beta }^{-1}$; hence it is a linear isomorphism of $\mathrm{End}(V)$. If ${\beta }^{\prime }=s^{-1}\beta s$, define ${\Phi }_s(A)=s^{-1}As$. A direct substitution givesso ${\varphi }_{{\beta }^{\prime }}$ and ${\varphi }_{\beta }$ are conjugate. This is the conjugation operator on an endomorphism space.

It remains to compute the Jordan normal form over $\mathbb{C}$. By the preceding conjugacy, we may put $\beta$ in Jordan form.

Ifthen the matrix units $E_{ij}$ are eigenvectors becauseThusThis includes the scalar case $\lambda =\mu$, when ${\varphi }_{\beta }$ is the identity.

Otherwise $\beta$ has one size-two Jordan block. Multiplying $\beta$ by a nonzero scalar does not change ${\varphi }_{\beta }$, and conjugating within its Jordan class allows us to use $J=I+N$ with $N=E_{12}$. Put $D={\varphi }_J-I$. Direct multiplication givesHence $D^3=0$, $\mathrm{rank}D=2$, and $\mathrm{rank}{D}^2=1$. The nilpotent Jordan blocks of $D$ therefore have sizes three and one. Adding the identity givesThese two cases are the Jordan normal form of conjugation on two-by-two matrices.

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Letbe primitive. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.

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A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

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Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

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DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.

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Letbe reflection in the unit circle, and represent a Möbius transformation $g$ byA direct calculation shows that $JgJ$ is represented byThus $gJ=Jg$ exactly when $M$ and $\tau (M)$ differ by a nonzero scalar. Applying the conjugate-linear involution $\tau$ twice shows that this scalar has modulus one. Rescaling $M$ by a suitable complex scalar then makes $M=\tau (M)$. Consequently all commuting maps, and only those maps, have the formThis is the Möbius maps commuting with reflection in the unit circle classification.

For such a map,It follows that $|g(z)|<1$ whenever $|z|<1$ precisely whenThe inverse has the same property, so these and only these maps preserve the unit disc.

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The Poincare disc model isIts geodesics through the origin are the Euclidean diameters. To see directly that the radial segment from $0$ to $r{e}^{i{\theta }_0}$ minimizes length, write an arbitrary joining curve as $z(t)=r(t)e^{i\theta (t)}$. Its hyperbolic length satisfiesThe radial segment has constant $\theta$ and monotone $r$, so equality holds. Henceand every radial diameter is length minimizing. Rotational symmetry and uniqueness for the geodesic equation show that these are all geodesics through $0$.

Given any geodesic and a point $p$ on it, a disc-preserving maptakes $p$ to the origin. By part (a), $T$ is a hyperbolic isometry and commutes with $J(z)=1/\overline{z}$. The transformed geodesic is a diameter, hence a generalized circle invariant under $J$. Its inverse image is therefore also a Generalized circle under a Möbius transformation invariant under $J$. Thus every hyperbolic geodesic is the part in $\mathbb{D}$ of a Euclidean line or circle preserved by reflection in the unit circle; equivalently, it is a diameter or a circle orthogonal to the unit circle. This is the Geodesics of the Poincare disc description.

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Rotate the disc so that $P=r\in (0,1)$ lies on the positive real axis. From part (b), the perpendicular hyperbolic line $\ell$ is a Euclidean circle orthogonal to the unit circle, with centre $c>1$ on the real axis and radius $R=c-r$. Orthogonality of the two circles givesCombining the two equations yields

The radial distance formula from part (b) saysThe hyperbolic double-angle identities now giveThereforeThis is the Euclidean circle representing a perpendicular hyperbolic line.

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Use the hyperboloid model with Minkowski inner productPut the vertex with angle $\theta$ at $v=(1,0,0)$, and place the endpoints of its adjacent sides of lengths $a$ and $b$ atThe geodesic through $P$ perpendicular to $vP$ has spacelike unit normalwhile the geodesic through $Q$ perpendicular to $vQ$ has spacelike unit normal

These two geodesics form the remaining two sides of the quadrilateral. Their angle equals the angle between their normals in the tangent plane at their intersection. The third right angle therefore givesDividing by $\cosh a\cosh b$ proves the Lambert quadrilateral identity

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Because the matrix-valued Laplace transform is taken componentwise, for every $i,j$ we haveThe sum is finite and $A$ is constant, so it may be moved outside the integral. Therefore

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Taking the Laplace transform of the differential equation and usinggivesHenceWhenever $s$ is not an eigenvalue of $A$, the matrix $sI-A$ is invertible, and thusThis is the Laplace-transform solution of a constant-coefficient vector ODE.

Set $g=0$. The given homogeneous solution is $y(t)=e^{tA}{y}_0$, so the preceding formula saysfor every initial vector $y_0$. Equality on every vector gives the matrix identityon the common domain of convergence and in particular away from the eigenvalues of $A$. This is the Laplace transform of a matrix exponential.

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The coefficient matrix has eigenvector $u_{+}=(1,1{)}^T$ with eigenvalue $3$ and eigenvector $u_{-}=(1,-1{)}^T$ with eigenvalue $-1$. WriteSinceand $y(0)=(1,-2{)}^T=-\frac{1}{2}{u}_{+}+\frac{3}{2}{u}_{-}$, the system diagonalises toUsing an integrating factor givesThe unique fastest term is thereforeConsequently $e^{-nt}y(t)$ diverges for $n<3$, tends to zero for $n>3$, and for $n=3$ tends to the finite nonzero vectorThus the only requested integer isThis is an instance of dominant eigenmode in a forced linear system.