Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4

Let $E_{rs}$ be the matrix unit with its only nonzero entry in row $r$, column $s$. Forone hasThereforeThe $n^2$ matrix units form an eigenbasis, with $E_{rs}$ having eigenvalue $r-s$. The zero eigenspace consists exactly of the diagonal matrices and has dimension $n$. The rank-nullity theorem now gives

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Fordirect multiplication givesWith respect to the ordered standard basis $(E_{11},E_{12},E_{21},E_{22})$, its matrix is therefore

This nilpotent operator satisfies ${\varphi }_A^3=0$, while ${\varphi }_A^2\ne 0$. Moreover,These ranks determine one nilpotent block of size three and one of size one. Hence its Jordan normal form is

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The quotient topology on $\tilde{X}$ isSince inverse images preserve arbitrary unions and finite intersections,Also ${\pi }^{-1}(\varnothing )=\varnothing$ and ${\pi }^{-1}(\tilde{X})=X$. The displayed collection therefore satisfies all the topology axioms.

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If $f:\tilde{X}\to Y$ is continuous, then $f\circ \pi$ is continuous because the quotient map $\pi$ is continuous by definition.

Conversely, suppose $f\circ \pi$ is continuous. For every open set $V\subseteq Y$,is open in $X$. The definition of the quotient topology then says that $f^{-1}(V)$ is open in $\tilde{X}$. HenceThis is the universal property of the quotient topology.

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No. Take the Hausdorff space $X=\mathbb{R}$ and declare all nonzero points equivalent, leaving $0$ in its own class. The quotient has two points,The nonzero class is open because its inverse image $\mathbb{R}\setminus \{0\}$ is open. However, the only saturated set containing $0$ but not the other class is $\{0\}$, which is not open. Thus every neighbourhood of $[0]$ is the whole quotient, so the two quotient points cannot have disjoint neighbourhoods. Therefore $\tilde{X}$ need not be Hausdorff.

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The hypothesis says that $f$ has a zero of order $k$ at $a$, so writewith $g$ holomorphic near $a$. Differentiating givesChoose $\delta >0$ so small that the closed disc $\overline{D}(a,\delta )$ lies in the domain, $g$ never vanishes there, and the parenthesized factor never vanishes there. Thus $a$ is the only critical point of $f$ in that disc.

On the boundary circle, $f$ is nonzero. SetIf $0<|b|<\epsilon$, then $|b|<|f(z)|$ on the boundary. Rouché's theorem applied to $f$ and $f-b$ says that $f-b$ has the same number of zeros in $D(a,\delta )$ as $f$, namely $k$, counted with multiplicity.

None of these zeros is $a$ because $b\ne 0$, and none is a critical point because $f^{\prime }$ has no other zero in the disc. Every zero of $f-b$ is therefore simple. Hence there are exactly

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The one-dimensional Time-dependent Schrodinger equation for a wavefunction in a real potential isMultiplying this equation by ${\Psi }^{\ast }$, multiplying its complex conjugate by $\Psi$, and subtracting gives the probability continuity equationIf $\Psi$ and its first derivative decay sufficiently rapidly as $x\to \pm \infty$, then the probability current $j$ vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yieldsThis conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put $\beta =\hslash t/m$. Sinceits squared modulus isThe Gaussian integral then giveswhich is independent of time, as required.

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In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

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Use the nodes in the order $0,1,2,3$. Their first divided differences are $4,-2,6$, their second divided differences are $-3,4$, and their third divided difference is $7/3$. The Newton interpolation polynomial is thereforeThis is the unique interpolating polynomial of degree at most three through the four data points.

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Appending the node $-2$ to the existing Newton interpolation polynomial givesAt the new node,The condition $p_4(-2)=10$ therefore givesHence

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For the simple random walk on the integer line, a return to the origin is possible only at an even time. At time $2n$, exactly $n$ of the increments must be $+1$, so the return probability isThe supplied factorial bounds, equivalently the order estimate in the Stirling formula, giveConsequently ${\sum }_n{p}_{2n}$ diverges. By the recurrence criterion by return probabilities, the origin is a recurrent state; translation invariance then makes the whole walk recurrent.

For three independent walks, the probability that all three are at the origin at time $2n$ isThis P-series is convergent, so the first of the Borel-Cantelli lemmas says that simultaneous returns occur only finitely often with probability one. Therefore the requested probability is

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A Hermitian form is a sesquilinear form $H:V\times V\to \mathbb{C}$ such thatUse the convention that $H$ is linear in its first argument and conjugate-linear in its second. Its matrix in the basis $(v_1,\dots ,v_n)$ isIf $v={\sum }_i{x}_i{v}_i$ and $w={\sum }_j{y}_j{v}_j$, then

Since $w_i={\sum }_j{p}_{ij}{v}_j$, the coordinate row of each new basis vector is a row of $P$. Direct substitution givesso the new matrix is

The subspace $N=V^{\perp }$ is the radical of $H$. In coordinates it is the kernel of $A$, and the rank-nullity theorem gives

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The symmetric matrix of the quadratic form in the standard basis isbecause the off-diagonal entries contribute twice to $x^{T}Ax$.

Apply Gram-Schmidt orthogonalization for a symmetric bilinear form toThese vectors are pairwise orthogonal for $B(u,v)=u^{T}Av$, andThus the form is diagonal in the basis $(u_1,u_2,u_3)$, with matrixIt follows that its rank is $3$ and its signature is

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By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.

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For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

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Fix $x\in {\mathbb{R}}^n$ and a coordinate direction $e_i$. The fundamental theorem of calculus givesFor sufficiently small $h$, the points $(t,x+sh{e}_i)$ lie in a fixed compact set. The continuity of $D_{i}g$ makes it uniformly continuous there, so the right-hand side converges to $D_{i}g(t,x)$ uniformly in $t$. We may consequently pass the limit through the finite integral:

To prove continuity, if $x_k\to x$, joint continuity makes $D_{i}g(t,x_k)\to D_{i}g(t,x)$ uniformly for $t\in [0,1]$ once the $x_k$ lie in a compact neighbourhood of $x$. Hence $D_{i}G(x_k)\to D_{i}G(x)$. This proves the stated differentiation under the integral sign result and the continuity of every partial derivative.

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For fixed $(x_1,x_2)$, apply the given one-variable identity to the smooth functionThe chain rule givesThereforewhereEvery derivative of the integrand is continuous. Repeated differentiation under the integral sign therefore shows that $h$ is smooth. This is the Second-order Hadamard lemma.

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An abstract smooth surface is a Hausdorff second-countable topological space with an atlas of charts to open subsets of ${\mathbb{R}}^2$ whose transition maps are smooth. It is orientable when it has such an atlas for which every transition map has positive Jacobian determinant. A map $f:S_1\to S_2$ is a smooth map when every coordinate representationis smooth wherever it is defined.

The map $a:C\to C$ is a smooth involution with no fixed point: the equationwould force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such thatThen the quotient projection restricts to a homeomorphismTransporting a smooth chart ${\varphi }_p:U_p\to {\mathbb{R}}^2$ across this homeomorphism gives the quotient chartOn an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the smooth quotient by a free finite group action, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a local diffeomorphism, in particular smooth.

To test orientability, parametrize the cylinder byThe involution acts in these coordinates aswhose derivative has determinant $-1$. Thus $a$ is an orientation-reversing diffeomorphism of the cylinder.

If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi \circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore

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The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives

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Let $y_{\epsilon }=y+\epsilon \eta$, where the variation $\eta$ satisfies $\eta (a)=\eta (b)=0$. Differentiating the functional at $\epsilon =0$ gives its first variationUsing integration by parts and the fixed endpoint conditions,If this vanishes for every admissible $\eta$, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation

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Forone has $L_y=0$ andA circle of radius $R$ centred at $(0,c)$ on the $y$-axis satisfiesOn any arc that does not cross $y=c$,which is constant. Hence $d{L}_{y^{\prime }}/dx=0=L_y$, so the Euler-Lagrange equation holds and $\delta I=0$. These circles are the coordinate-swapped form of a Geodesic in the Poincare half-plane model.

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Here the integrandhas no explicit $x$-dependence. The Beltrami identity therefore givesfor a positive constant $R$. Thuswhose nonvertical solutions are the semicircles

The endpoint $(a,a)$ and the prescribed value at $x=b$ both lie onbecauseThe required positive arc is consequentlyThe condition $b<2a$ keeps the endpoint above the $x$-axis. The sketch is the descending part of the upper semicircle of radius $a$ centred at $(a,0)$, from $(a,a)$ toThis is precisely a Geodesic in the Poincare half-plane model.

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Fordirect differentiation givesandThus ${\theta }_t=D{\theta }_{xx}$.

Its total mass is found from the Gaussian integral:The unit-mass solution converging to the Dirac delta function as $t\downarrow 0$ is therefore the heat kernel

For the second initial condition, the heat-kernel solution is the convolutionWith $u=(x-y)/(2\sqrt{Dt})$ and the definition of the error function, this becomesThis is the heat equation with interval-indicator initial data for $a=1$.

At $t=0$ the graph is the rectangle of height one on $[-1,1]$. For every $t>0$ it is smooth, positive and even, with its maximum at $x=0$. As time increases the graph broadens and its maximum falls, while its total area remains $2$; pointwise it tends to zero as $t\to \infty$.

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The orbital angular momentum commutation relations giveConsequentlyThe same commutation relations imply $[L^2,L_x]=[L^2,L_y]=0$, and henceThese are the defining identities for the angular momentum ladder operators.

Ifthen, whenever $L_{\pm }\varphi \ne 0$,Thus $L_{\pm }\varphi$ has quantum numbers $(\ell ,m\pm 1)$.