Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 3

Use the Riemann sphere convention for evaluating a Möbius transformation at a pole and at infinity. When the three prescribed points are finite, takeIts zero is $w_2$, its pole is $w_1$, and substitution at $w_3$ gives one. The remaining three cases areAll coefficient determinants are nonzero because the points are distinct. There is only one such Möbius transformation: the composite of any two candidate maps, one inverted, fixes $\infty ,0,1$; fixing infinity makes it affine, and fixing zero and one makes it the identity.

For this question use the cross-ratio normalizationIn the finite case this is $(w_3-w_1)(w_4-w_2)/[(w_3-w_2)(w_4-w_1)]$, with the corresponding limits if a point is infinite. It is finite and distinct from zero and one, since $w_4$ is distinct from the other three points and $f$ is a bijection. Ordering conventions for the cross-ratio differ; the definition here is fixed by the specified images, rather than by importing another ordering formula.

A generalized circle means either a circle or a straight line completed by infinity. Let $K$ be the unique generalized circle through the first three points. Its image under $f$ is the unique generalized circle through $\infty ,0,1$, namely $\mathbb{R}\cup \{\infty \}$. Therefore $w_4\in K$ exactly when $f(w_4)$ is real. Conversely a real $f(w_4)$ lies on that extended real line, whose inverse image is $K$. This proves the real cross-ratio criterion for a generalized circle in both directions.

A subgroup $H$ of a group $G$ is a normal subgroup when $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. The quotient group $G/H$ has the left cosets as elements, with $(gH)(kH)=gkH$, identity $H$ and inverse $(gH{)}^{-1}=g^{-1}H$. Normality makes this multiplication independent of representatives.

The first isomorphism theorem for groups says that for a group homomorphism $\phi :G\to K$, its kernel of a group homomorphism is normal andThe image is taken with the induced group operation; surjectivity onto all of $K$ is not required.

For the given upper triangular matrices, take diagonal entries:The target is a direct product of groups, with multiplication of nonzero real numbers in each coordinate. Indeed the product of two such matrices has diagonal $(a{a}^{\prime },d{d}^{\prime })$, proving the group homomorphism property. Any pair of nonzero diagonal entries is attained by a diagonal matrix, so $\phi$ is surjective. Its kernel of a group homomorphism consists precisely of the matrices with $a=d=1$, namely $H$. Hencewith coordinatewise multiplication, not addition on ${\mathbb{R}}^2$.

The boundary ray has $y/x=3/5$, and its intersection with the hyperbola is $(5/4,3/4)$. Under the given change of variables, $x^2-y^2=r^2$ and $y/x=\tanh \theta$. Thus the region becomesThe hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so the Jacobian determinant isIt is positive in the interior. The coordinate degeneracy at $r=0$ is a boundary set of area zero and does not affect the change of variables formula. ThereforeHere $\cosh (\log 2)=5/4$. The PDF specifies two line segments and one hyperbolic arc; the duplicated $y=0$ line in the TeX is not an extra boundary condition.

Use the Einstein summation convention. Substitute the given basis transformation into the expansion of the vector:By linear independence of the basis vectors, $v_j=R_{ij}{v}_i^{\prime }$. In matrix notation $v=R^T{v}^{\prime }$. A rotation matrix is orthogonal, so multiplication by $R$ givesThe placement of the transpose follows from the specified convention $e_i^{\prime }=R_{ij}{e}_j$. It must not be guessed from a different convention for an active rotation of a vector.

The quotient theorem for Cartesian tensors here says: if contracting $T_{ij}$ with every vector $v_j$ gives the components $w_i=T_{ij}{v}_j$ of a vector in every right-handed orthonormal basis, then $T$ obeys the second-order tensor transformation law. Conversely that transformation law guarantees the contraction is a vector. One may equivalently test that $u_i{T}_{ij}{v}_j$ is a scalar for every pair of vectors $u,v$.

By part (a), $v^{\prime }=Rv$ and $w^{\prime }=Rw$. The contraction property in the new basis saysSince this holds for every test vector, $T^{\prime }R=RT$. Using $R^{-1}=R^T$ givesFor the converse, $T^{\prime }{v}^{\prime }=RT{R}^{T}Rv=RTv=Rw$, exactly the vector law. For the equivalent scalar formulation, invariance of $u^{T}Tv$ for every $u,v$ gives $u^T{R}^T{T}^{\prime }Rv=u^{T}Tv$, forcing the same matrix identity.

The quantifier “every” matters: an arbitrary nonzero matrix can annihilate one fixed vector, so a single successful contraction cannot establish tensoriality. The stated bases test proper rotations; transformation under orthogonal reflections would require an additional hypothesis if that were wanted. “Second-order” counts tensor indices, rather than the matrix rank of a particular component matrix.

Restrict the quotient projection $\pi :G\to G/N$ to $H$. This is a group homomorphism withIts image is a nontrivial subgroup of the prime-order quotient group, because $H$ is not contained in $N$. By Lagrange's theorem for finite groups, the image has order $p$ and is all of $G/N$. The first isomorphism theorem for groups therefore gives

For the second assertion, partition $C$ into conjugacy classes for $N$. Because $N$ is normal, conjugation by $G$ permutes these smaller classes: $g{\mathrm{Cl}}_N(x)g^{-1}={\mathrm{Cl}}_N(gx{g}^{-1})$. This group action is transitive since any two elements of $C$ are conjugate in $G$. Elements of $N$ fix every $N$-class, so the action factors through $G/N$.

If $m$ is the number of smaller classes, the orbit-stabilizer theorem for this transitive action gives $m\mid |G/N|=p$. ThusDistinct conjugacy classes are disjoint because they are orbits of a group action. This proves conjugacy class splitting in a prime-index normal subgroup without assuming all $G$-conjugations are already conjugations by $N$.

Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

For a group action, the orbit of a group action of $x$ is $Gx=\{gx:g\in G\}$, and its stabilizer subgroup is $G_x=\{g:gx=x\}$. The orbit-stabilizer theorem gives $|Gx||G_x|=|G|$: the map $g{G}_x\mapsto gx$ is a well-defined bijection from stabilizer cosets to the orbit.

Within one orbit, the stabilizer subgroups are conjugate, since $G_{hx}=h{G}_x{h}^{-1}$. Their sizes are therefore equal. Summing stabilizer sizes over a single orbit gives $|Gx||G_x|=|G|$, and summing over all $n$ orbits yieldsNow count the pairs $(g,x)$ fixed by the action in two ways. Holding $x$ fixed gives $|G_x|$ choices of $g$, whereas holding $g$ fixed gives $|\mathrm{Fix}(g)|$ choices of $x$. ThusCombining the two identities proves Burnside lemma:The finiteness hypotheses justify these cardinality sums. If $X$ is empty, every sum and the orbit count are zero, so the formula still applies.

Centre the cube at the origin with faces having normals $\pm e_1,\pm e_2,\pm e_3$. Every rotational symmetry fixes the centre and permutes these normals, so its rotation matrix is a signed permutation matrix of determinant one. Conversely any such matrix permutes the cube’s coordinates up to signs and preserves the cube. For each of $3!$ coordinate permutations, four of the eight choices of signs give determinant one. Consequently the rotational symmetry group of a cube hasThis counts the rotations directly, without an unverified stabilizer calculation. The TeX’s $2^k$ is damaged transcription: the PDF prints 24.

Let $X$ be the $3^6$ face colourings. A colouring fixed by a rotation must be constant on each permutation cycle of faces, giving $3^c$ fixed colourings if the rotation has $c$ face cycles. The rotation types are:

There are three face axes, each with two quarter-turns and one half-turn; six edge axes, each with one half-turn; and four vertex axes, each with two nonidentity third-turns. These rotations are distinct and, with the identity, total 24, so the table is exhaustive. Face-axis rotations fix their two axial faces; the remaining four faces cycle or pair. Edge half-turns pair all faces, while vertex third-turns form two triples.

Burnside lemma now givesThe three colours are labelled, orthogonal reflections are excluded, and no condition requires every colour to appear.

With rightmost-first composition, a permutation cycle satisfiesTracing each $a_j$ verifies the equality, and every other letter is fixed. Thus every cycle is a product of transpositions; decomposing a permutation into disjoint permutation cycles proves that every element of $S_n$ is such a product. The identity is the empty product.

To establish that parity is well defined, use the permutation matrix $P_{\sigma }$ defined by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Its determinant is $\pm 1$, and $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. A transposition swaps two columns of the identity matrix and has determinant $-1$. Therefore if $\sigma$ is expressed as a product of $r$ transpositions,The left side depends only on $\sigma$, so any two decompositions have the same parity. DefineMultiplicativity of the determinant makes this the sign homomorphism. Its kernel of a group homomorphism is the alternating groupIt is a normal subgroup consisting of the even permutations. For $n\ge 2$ the sign map is surjective because a transposition has sign $-1$, so $[S_n:A_n]=2$. For $n=1$ both groups are trivial.

Write $c=(12\cdots n)$, with $n\ge 2$. Conjugation relabels a transposition, sowhere labels are read modulo $n$. In particular the subgroup contains $(12),(23),\dots ,(n-1n)$. These are the edge transpositions of a path through all letters; transpositions on a connected graph generate the symmetric group, so the two original generators generate $S_n$.

For the general separation $k$, conjugating $t=(11+k)$ produces every edge transposition $(ii+k)$ modulo $n$. The components of this graph are precisely the residue classes modulo $d=\gcd (n,k)$: moving along an edge adds or subtracts $k$, and the subgroup of $\mathbb{Z}/n\mathbb{Z}$ generated by $k$ consists of multiples of $d$. Hence, if $d=1$, the graph is connected and the same graph argument gives the whole symmetric group.

For completeness, the graph argument can be proved without assuming a generating-set theorem. On a simple path $v_0,\dots ,v_{\ell }$, put $t_j=(v_{j-1}{v}_j)$. ThenThis is a conjugate of the last edge transposition; connectivity therefore supplies every transposition, and part (a) supplies every permutation.

If $d>1$, there are $d$ residue blocks of size $n/d>1$, since $1\le k<n$. The cycle $c$ permutes these blocks, and $t$ swaps two letters in the same block. Both preserve this block system, so their generated subgroup does too. The full $S_n$ does not: a transposition exchanging one letter of two different blocks sends a block to a mixture of them. Thus the subgroup is proper. This provesThus a cycle and a transposition generate the symmetric group exactly at coprime separation. The obstruction is preservation of a partition, not failure of transitivity: the $n$-cycle already acts transitively even when $d>1$.

Assume $B$ is continuously differentiable. Apply the chain rule to each component with $z_j=t{x}_j$:Contracting the first identity with $x_j$ provesHere $\nabla$ differentiates with respect to $x$ at fixed $t$, while the time derivative holds $x$ fixed. The equality follows directly even at $t=0$; no division by $t$ is needed. It holds componentwise for the vector field, so it is not restricted to scalar functions.

For a twice continuously differentiable vector potential $A$, write the curl with the Levi-Civita symbol:The second derivatives are symmetric in $i,j$, whereas ${ϵ}_{ijk}$ is antisymmetric, so the contraction is zero. Equivalently, writing out the three components shows that each mixed derivative cancels its reversed-order partner. ThusThis is divergence of a curl is zero. The regularity hypothesis is what licenses commutation of the mixed partial derivatives; it must not be omitted for a singular potential without specifying a weaker derivative interpretation.

Use the usual setting of a continuously differentiable solenoidal vector field defined on all radial segments from the origin, including the origin. More generally an open star-shaped set containing zero suffices. Differentiation under the integral sign givesApply divergence and curl of a cross product with the second field $x$, using $\nabla \cdot x=3$ and $(D\cdot \nabla )x=D$:The scaling identity in part (a) now converts the last expression to a one-variable derivative:Continuity at the origin ensures the lower endpoint is zero. Hence the radial vector potential of a solenoidal vector field isThe order of the cross product matters: $x\times D$ would produce $-B$. This construction is not valid on every punctured or multiply connected domain. For example $B=x/|x{|}^3$ is solenoidal away from zero, but its radial integral diverges at zero and it has nonzero flux through a surrounding sphere. The stated formula therefore implicitly needs the regular radial-domain hypothesis, rather than just local zero divergence.

Let $a$ be any constant vector and take $u=\varphi a$. The product rule for divergence gives $\nabla \cdot u=a\cdot \nabla \varphi$. The divergence theorem on a bounded region with piecewise smooth boundary, oriented outward, therefore yieldsSince $a$ is arbitrary, equality of all components provesHere $\varphi$ is continuously differentiable on a neighbourhood of the closed region.

For the side of this right circular cone, the parameter tangents are $x_r=(\cos \theta ,\sin \theta ,\sqrt{3})$ and $x_{\theta }=(-r\sin \theta ,r\cos \theta ,0)$. Their cross product in the outward order isThe sign is outward because the solid right circular cone lies at smaller cylindrical radius for fixed height. Reversing the parameter order reverses the oriented surface element.

To check the integral identity, the closed boundary must include the top Euclidean disk $z=1$, radius $1/\sqrt{3}$, as well as the curved side. For $\varphi =z^2$, horizontal components cancel on integrating $\theta$. The side contribution isOn the top Euclidean disk $\varphi =1$ and $d\mathbf{S}=e_{z}dA$, so its contribution is $(\pi /3)e_z$. The total is $(\pi /6)e_z$. Independently, the cross-section of the solid at height $z$ has area $\pi z^2/3$, and $\nabla \varphi =2z{e}_z$, givingThus the two sides agree. The curved side alone is not a closed surface and does not satisfy this volume identity. The apex has zero area; alternatively one can truncate at height $\epsilon$ and let $\epsilon \downarrow 0$, with the extra boundary contribution vanishing.

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.

Let $F=|\mathbf{F}|>0$ and choose an orientation of a regular curve following the vector field. Its unit tangent vector is $t=\sigma \mathbf{F}/F$, where $\sigma =\pm 1$ is constant along a connected regular segment. Arc length differentiation along that curve is $d/ds=(\sigma \mathbf{F}/F)\cdot \nabla$. ThereforeTaking the cross product with $t$ removes the term parallel to $\mathbf{F}$:Where the Frenet frame exists, $t\times t^{\prime }=\kappa (t\times n)=\kappa b$. ThusThe printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent curvature of an integral curve of a vector field, $\kappa =|\mathbf{F}\times (\mathbf{F}\cdot \nabla )\mathbf{F}|/F^3$. This scalar formula also applies at zero curvature, where $b$ itself may be undefined.

The directional derivative along the specified vector field givesUse the magnitude formula for the curvature of an integral curve of a vector field:It vanishes on the plane $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the vector field vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.

As an independent geometric check, an integral curve of a vector field satisfies $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=-z$, giving $(x,y,z)=(a{e}^u,b{e}^u,c{e}^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary curvature of a space curve formula $|r^{\prime }\times r^{\prime \prime }|/|r^{\prime }{|}^3$ gives the same expression.

Work with real-valued smooth functions on the bounded region enclosed by $S$, with the oriented surface element pointing outward. Set $\psi =\varphi -{\varphi }_1$. It is a harmonic function in $V$ and has zero Dirichlet boundary conditions on $S$. Applying Green's first identity to $\psi$ with itself givesBoth terms on the right vanish by the boundary condition and harmonicity. The integrand is nonnegative and continuous, so $\nabla \psi =0$ throughout $V$. Consequently $\psi$ is constant on each connected component; every bounded component meets the prescribed boundary, where that constant is zero. ThereforeThis is the energy proof of Uniqueness of the Dirichlet problem. Connectedness of $V$ is not essential if the boundary values are prescribed on every component. Boundedness, or suitable decay and integrability at infinity, is needed for the boundary/energy argument; the finite enclosed-volume interpretation is used here.

Put $h=u-\varphi -C$. Then $h=0$ on $S$, and the constant does not change a gradient, so $\nabla u=\nabla \varphi +\nabla h$. Expand the unnormalized Dirichlet energy:The cross term is zero by Green's first identity:Thus the useful stronger identity isEquality holds exactly when $h$ has zero gradient. It is then constant on every connected component and zero on that component’s boundary, hence $h=0$. Therefore equality holds precisely when $u=\varphi +C$ throughout $V$. This is the Dirichlet principle; an additive boundary constant does not alter the minimizing energy. The energy here omits the optional factor $1/2$, which does not affect minimizers or the inequality.

Let $E(t)={\int }_V|\nabla w{|}^{2}dV$, using real-valued smooth $w$ and a fixed bounded region. Differentiation under the integral sign and Green's first identity yieldThe boundary term vanishes because $w_t=0$ on $S$, and the heat equation gives $w_t=\Delta w$ in $V$. ConsequentlyThe integrand is continuous and nonnegative. Thus at any fixed time, equality is equivalent to $\Delta w=0$ everywhere in $V$, also giving $w_t=0$ there at that time. This proves Dirichlet energy dissipation for the heat equation.

The boundary condition fixes the boundary temperature in time; it is not a zero normal derivative condition. No sign condition on ${\partial }_{n}w$ is used. The stated smoothness licenses the time derivative, spatial integration by parts and boundary trace; without that regularity an appropriate weak energy argument would be required.