Past exam of the short options course of the University of Oxford / 2017 (A13752W1, ASO)

Source paper: released examination archive, member 2016/T/a13752w1.pdf. The original cover dates this examination to 2017-06-19.

An intact copy in the Mathematics Part A release, member FOI-20210708-1 - Data/a13752w1 (2).pdf, restores printed pages 2–3. All nine questions, comprising 51 leaf subparts, are answered below. The earlier source provenance above is retained.

For distinct odd primes $p,q$, quadratic reciprocity statesHere the Legendre symbol records whether a nonzero congruence class is a quadratic residue. The first supplementary law for quadratic reciprocity and second supplementary law for quadratic reciprocity giveThus minus one is a square exactly when $p\equiv 1(\mathrm{mod}4)$, and two is a square exactly when $p\equiv 1$ or $7(\mathrm{mod}8)$.

To test the specified congruence class, factor $111=3\cdot 37$. Since $2017\equiv 1(\mathrm{mod}4)$, multiplicativity of the Legendre symbol and quadratic reciprocity yieldThere is no sign change on swapping $19$ and $37$ because $37\equiv 1(\mathrm{mod}4)$. Finally $18=2\cdot 3^2$, so multiplicativity of the Legendre symbol gives $(18/19)=(2/19)=-1$, since $19\equiv 3(\mathrm{mod}8)$. Therefore

For an odd prime $p$, Euler's criterion statesFor $p\nmid a$, the Legendre symbol is $1$ for a quadratic residue and $-1$ for a quadratic nonresidue; if $p\mid a$, both sides are zero.

The product is a single power because the arithmetic progression of exponents has sum $2017\cdot 2018/2$:Since $2017\equiv 1(\mathrm{mod}8)$, the second supplementary law for quadratic reciprocity gives $(2/2017)=1$. Hence Euler's criterion gives $2^{1008}\equiv 1(\mathrm{mod}2017)$. Reduce the exponent using $1009=1008+1$ and $2017=2\cdot 1008+1$:The least nonnegative remainder is therefore $2$.

Euler's totient function counts the integers in a complete complete residue system that are coprime to the modulus:If the prime factorization is $n={\prod }_{j=1}^r{p}_j^{e_j}$ with distinct prime numbers $p_j$, Euler's totient product formula isIn particular, $\varphi (2017)=2016$. Adjoining the prime factor $2$ changes neither the odd factor nor its count of units, since $\varphi (2)=1$. The product formula givesThis is a positive integer distinct from $2017$, as required.

The Chinese remainder theorem says that for pairwise coprime positive moduli $m_1,\dots ,m_r$ and arbitrary integers $a_1,\dots ,a_r$, the simultaneous modular congruenceshave a solution, unique modulo $M={\prod }_i{m}_i$.

Choose $2017$ distinct prime numbers $q_1,\dots ,q_{2017}$. Their prime powers $q_i^{2017}$ are pairwise coprime. Apply the Chinese remainder theorem toFor the resulting integer $n$, each $n+i$ is divisible by $q_i^{2017}$, the required power of an integer greater than one. Adding a sufficiently large multiple of ${\prod }_i{q}_i^{2017}$ makes $n$ positive without changing any modular congruence. Thus such a positive $n$ exists, indeed infinitely many do. This is the general construction of consecutive integers divisible by prescribed prime powers.

Use the fact that $K$ is a normal subgroup to move a factor of $K$ past a factor of $H$. For $h_1,h_2\in H$ and $k_1,k_2\in K$,whileThe identity element belongs to $HK$. Hence $HK$ is a subgroup, by the subgroup test. This proves the product of a subgroup with a normal subgroup construction; $H$ need not be a normal subgroup.

For each $h\in H$, conjugation maps $K$ to itself and respects its multiplication. The inverse of $k\mapsto h^{-1}kh$ is $k\mapsto hk{h}^{-1}$, so both are group automorphisms of $K$.

The printed formula requires a composition convention. For the printed ${\varphi }_h(k)=h^{-1}kh$, ordinary function composition givesIndeed, $(h_1{h}_2{)}^{-1}k(h_1{h}_2)=h_2^{-1}(h_1^{-1}k{h}_1)h_2$. Thus the displayed assignment is a group antihomomorphism with the usual multiplication in the automorphism group. It is a group homomorphism if the automorphisms are written as right operators and multiplied in the corresponding reverse order; this is precisely why right conjugation reverses composition.

With the usual left function composition, the corresponding group homomorphism is insteadDirect expansion gives ${\psi }_{h_1{h}_2}(k)=h_1(h_{2}k{h}_2^{-1})h_1^{-1}=({\psi }_{h_1}\circ {\psi }_{h_2})(k)$. This states the intended conjugation action without silently reversing the order in the printed formula.

Write the group order as $|G|=p^{a}m$, where $p$ is a prime number and $p\nmid m$. A Sylow p-subgroup is a subgroup of group order $p^a$, the largest power of $p$ dividing $|G|$.

The three Sylow theorems assert:

In particular, a unique Sylow p-subgroup is a normal subgroup, since all its conjugates must equal it.

The Sylow theorems give $n_7\mid 10$ and $n_7\equiv 1(\mathrm{mod}7)$. Among the divisors $1,2,5,10$, only $1$ satisfies the modular congruence. HenceUniqueness and the conjugacy assertion of the Sylow theorems make $K$ a normal subgroup.

The Sylow theorems also give $n_5\mid 14$ and $n_5\equiv 1(\mathrm{mod}5)$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a normal subgroup of $G$ as well.

By Lagrange's theorem, $H\cap K=\{1\}$, since its group order divides both $5$ and $7$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $H$ by the normal subgroup property of $H$, and in $K$ by the normal subgroup property of $K$. It is therefore the identity element, so all elements of $H$ commute with all elements of $K$. This is the fact that normal subgroups of coprime order commute.

The subgroup $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have prime orders, they are cyclic groups. Write $H=⟨h⟩$ and $K=⟨k⟩$. The product of commuting elements of coprime order has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a normal subgroup because both its factors are normal subgroups; equivalently, it has subgroup index two. Therefore

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

Two projective triangles are in perspectivity from a point $O$ when their three corresponding-vertex projective lines $P{P}^{\prime }$, $Q{Q}^{\prime }$, and $R{R}^{\prime }$ all pass through $O$.

Desargues's theorem: perspectivity from a point implies perspectivity from a line. More explicitly, if $P{P}^{\prime },Q{Q}^{\prime },R{R}^{\prime }$ are concurrent, then the three corresponding-side intersectionsare collinear. The usual statement assumes nondegenerate projective triangles and well-defined, distinct corresponding-side intersections. The projective line containing $X,Y,Z$ is the axis of perspectivity from a line. All intersections are projective: parallel lines in an affine chart meet on its projective line at infinity.

The dual projective space is $\mathbb{P}(({\mathbb{R}}^3{)}^{\ast })$. A projective line with coefficients $[\alpha :\beta :\gamma ]$ becomes a projective point in this dual plane; a projective point $[a:b:c]$ becomes the dual projective line $a\alpha +b\beta +c\gamma =0$.

The point-line duality reverses incidence:Thus the dual statement is that the dual point $l^{\ast }$ lies on the dual line $P^{\ast }$. A join of projective points becomes an intersection of dual projective lines, and the converse holds too. In particular, collinear projective points correspond to concurrent dual projective lines.

The dual of Desargues's theorem is its converse: if the intersections of the three pairs of corresponding sides are collinear, then the three lines joining corresponding vertices are concurrent. In the same nondegenerate setting, perspectivity from a line therefore implies perspectivity from a point.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.

The corresponding side equations, computed using projective joins and meets by cross products, areSubtracting the two equations in each row and substituting back gives the three projective pointsFor example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfysince $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a projective line. Thus the corresponding-side intersections are collinear, explicitly confirming Desargues's theorem.

The diagram uses the affine chart $x+y+z\ne 0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the projective line at infinity. The apparent parallelism of $QR$ and $Q^{\prime }{R}^{\prime }$ records their common vertical point at infinity, which also belongs to the axis.

The map is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatEquivalently, $f(x+h)=f(x)+Lh+o(\Vert h\Vert )$. This linear map is unique and is the Fréchet derivative $D{f}_x$. Its matrix in the standard bases is the Jacobian matrix, whose $(i,j)$ entry is $\partial f_i/\partial x_j$ at $x$. A function is differentiable on its domain if this condition holds at every point.

A sufficient condition is that the derivative $D{f}_x:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is surjective, equivalently that its Jacobian matrix has rank $m$. Then the implicit function theorem makes the zero set near $x$ a $C^1$ embedded submanifold of dimensionThe rank condition persists nearby because the derivative is continuous. This is the local form of the regular level set theorem; only the rank at the specified point is needed for this local conclusion.

Expand the increment of the quadratic form, using $A^T=A$:The last term obeys $|h^{T}Ah|\le \Vert A\Vert \Vert h{\Vert }^2=o(\Vert h\Vert )$. Thus $q$ is differentiable andThis is the quadratic form gradient. The derivative is a linear functional of the increment $h$, whereas the gradient is its representing vector under the Euclidean inner product.

Since $A$ is invertible, the quadratic form gradient vanishes exactly at the origin. At every nonzero point of $C_{\alpha }$, the scalar derivative is surjective, so the regular level set theorem gives a smooth submanifold of dimension $n-1$.

If $\alpha \ne 0$, the origin is absent. Therefore the largest relatively open smooth subset is all of $C_{\alpha }$.

If $\alpha =0$, the nontriviality assumption forces the form to be an indefinite quadratic form: an invertible definite form would have zero set consisting only of the origin. Every nonzero point is regular. To show that the origin really fails to be a manifold point, use Sylvester's law of inertia to reduce to $|u{|}^2-|w{|}^2=0$. With positive and negative coordinate vectors $e_i,f_j$, the null vectors $e_i+f_j$ and $e_i-f_j$ span ${\mathbb{R}}^n$. Each line in one of these directions lies in the cone.

If a relatively open manifold neighborhood of the origin existed, its tangent space would contain all these line directions, so would have dimension $n$. Such an embedded manifold would be open in ${\mathbb{R}}^n$, impossible for the zero set of a nonzero quadratic form. ConsequentlyThis gives the smooth locus of a nondegenerate real quadratic level set, including maximality at the singular vertex.

Writing the coordinates as $(x,y,z)$, the level equation is $x^2+y^2-z^2=\alpha$. Therefore:

The cone vertex is an instance of the double cone singularity. The tips of the two-sheeted hyperboloid are regular because the gradient $(2x,2y,-2z)$ is nonzero there.

Use the angular-frequency Fourier transform conventionFor $f,f^{\prime }\in L^1(\mathbb{R})$ with the usual local absolute-continuity hypothesis, the Fourier transform of a derivative isFor example, this applies when $f$ is $C^1$ and both displayed functions are integrable. These assumptions make integration by parts valid and the endpoint terms vanish.

Split the integral at the origin and reflect the negative half:Thus the Fourier transform of a two-sided exponential is

With this convention, the Fourier inversion theorem states that if $f,\hat{f}\in L^1(\mathbb{R})$, thenalmost everywhere and at every continuity point of $f$. Apply this to the continuous function in part (ii):The denominator is even. Replacing $x$ by $-t$ and renaming the integration variable givesBoth integrable members of this transform pair satisfy the inversion hypotheses.

Complete the square: $x^2+2x+2=(x+1{)}^2+1$. In the Fourier transform, the substitution $u=x+1$ givesThe phase $e^{is}$ is important. Taking the real part at $s=a$ gives the cosine integral, since changing the sign in the exponential does not change its real part:

For integrable functions $u,v$, the convolution theorem in the chosen convention isLet $F(s,y)$ be the Fourier transform of $f(\cdot ,y)$ in the horizontal coordinate. Transforming Laplace's equation givesFor $s\ne 0$, exclude the growing exponential and obtain $F(s,y)=2e^{-|s|y}/(1+s^2)$. At $s=0$, continuity in frequency gives $F(0,y)=2$; a constant zero-frequency transform is compatible with spatial decay of $f$.

By the transform pair in part (a), the inverse transform of $e^{-|s|y}$ is the Poisson kernel for the upper half-plane, $P_y(x)=y/[\pi (x^2+y^2)]$. ThusThe Poisson extension of exponential boundary data is harmonic, approaches the continuous boundary data and decays at spatial infinity. For uniqueness, apply the maximum principle for harmonic functions to the difference of two solutions on increasingly large half-discs: it is zero on the real boundary and uniformly small on the distant semicircle.

Use the squaring map from a quadrant to the upper half-plane:By conformal invariance of harmonicity, the answer isOn the horizontal axis it gives $f(x^2,0)=e^{-x^2}$, and on the vertical axis it gives $f(-y^2,0)=e^{-y^2}$. Also $|(x+iy{)}^2|=x^2+y^2$, so spatial infinity maps to spatial infinity and the required decay follows. The two boundary values agree at the origin.

Apply a constant Lagrange multiplier to the isoperimetric constraint, using the augmented functional $J-\lambda (I-1)$. Its integrand isThe Euler-Lagrange equation gives $-2qy-2\lambda wy-2(p{y}^{\prime }{)}^{\prime }=0$, henceThe multiplier rule applies because the constraint derivative in direction $y$ is $D{I}_y[y]=2I[y]=2$, so the constraint derivative is not zero. The prescribed endpoint values remove boundary variations.

Multiply the Sturm-Liouville equation by $y$ and integrate. The endpoint values and integration by parts giveSince $I[y]=1$,Thus the multiplier equals the stationary value of $J$ at the extremal, as in stationary values of a quadratic functional with a quadratic constraint. It is also the corresponding Rayleigh quotient.

Multiply the equation for $y_n$ by $y_m$ and that for $y_m$ by $y_n$, then subtract. Integrating the resulting identity yieldsAll endpoint terms vanish because both functions satisfy the zero Dirichlet boundary conditions. Since the multipliers differ,This is the weighted orthogonality of Sturm-Liouville eigenfunctions.

Part (a)(ii) now givesIf $\lambda =0$, continuity makes $y^{\prime }$ identically zero. Thus $y$ is constant, and the boundary condition forces $y=0$, contradicting ${\int }_0^{\pi }{y}^2=1$. Hence

The extremal equation is $y^{\prime \prime }+\lambda y=0$. For $\lambda >0$, its general solution is $A\cos (\sqrt{\lambda }x)+B\sin (\sqrt{\lambda }x)$. The left endpoint forces $A=0$, while the right endpoint and $B\ne 0$ require $\sin (\sqrt{\lambda }\pi )=0$. ThereforeIn particular, each extremal has the requested Fourier sine series form $y={\sum }_{k\ge 1}{a}_k\sin (kx)$. More strongly, for an individual extremal only the coefficient at $k=n$ can be nonzero. An arbitrary combination of distinct modes is an admissible function, but does not solve this one-eigenvalue equation.

The sine functions obey ${\int }_0^{\pi }\sin (nx)\sin (mx)dx=(\pi /2){\delta }_{nm}$. The Parseval identity for the Fourier sine series givesConsequently

For $y=\pm \sqrt{2/\pi }\sin x$, both the constraint and endpoint values hold, the extremal equation has $\lambda =1$, and direct integration gives $J[y]=1$.

For any admissible function, its derivative has Fourier cosine series coefficients $n{a}_n$ and zero constant coefficient. This follows by integration by parts: ${\int }_0^{\pi }{y}^{\prime }\cos (nx)dx=n{\int }_0^{\pi }y\sin (nx)dx$, and ${\int }_0^{\pi }{y}^{\prime }=0$. Thus the Parseval identity givesFor finite sums the same identity follows directly from the cosine product formula in the hint; the $L^2$ identity justifies the general case. Equality holds only if $a_n=0$ for all $n\ge 2$, and normalization then fixes $a_1=\pm \sqrt{2/\pi }$. ThereforeThis is the minimum Dirichlet energy at fixed L2 norm on an interval, equivalently the sharp Wirtinger inequality for this interval.

A tree is a connected graph with no cycles. Equivalently, between any two of its vertices there is exactly one graph path. A single vertex with no edges is also a tree.