Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3

A Möbius transformation is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto 1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equationwhere $\alpha ,\gamma \in \mathbb{R}$; the case $\alpha =0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w{|}^2$ to obtainwhich is again such an equation. The generators, and hence every Möbius transformation, therefore preserve Generalized circle under a Möbius transformation.

Now letIf $c\ne 0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the subgroup in question is the Affine subgroup of the Möbius group, consisting of $z\mapsto az+b$ with $a\ne 0$.

It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne 0$, then $T_b$ is affine buthas a finite pole and is not affine. Hence conjugation by $I$ does not preserve the subgroup.

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The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

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Choose an angle $2\pi \alpha$ with $\alpha \notin \mathbb{Q}$. Rotation through that angle has infinite order, so the subgroup it generates is isomorphic to $\mathbb{Z}$.

The group also contains a copy of ${\mathbb{Z}}^2$. Choose real numbers $\alpha ,\beta$ such that $1,\alpha ,\beta$ are linearly independent over $\mathbb{Q}$, for example $\alpha =\sqrt{2}$ and $\beta =\sqrt{3}$. DefineThis is a homomorphism. If $\varphi (m,n)=I$, then $m\alpha +n\beta \in \mathbb{Z}$, and the stated rational independence forces $m=n=0$. Thus $\varphi$ is injective and its image is isomorphic to ${\mathbb{Z}}^2$.

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In the Cartesian basis, writing $F=(F_1,F_2,F_3)$,defines the divergence, while the curl isEquivalently,

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Use the Levi-Civita symbol and the product rule. First,For the second identity, the contractiongivesIn vector notation this isThese are the divergence and curl of a cross product identities.

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The first identity in part (ii), with $F$ and $G$ interchanged, saysso the proposed identity is missing its final term. For a concrete counterexample, takeThen $\nabla \times G=0$, so the proposed left side is zero, whereasThus the identity is false.

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For $F=(x,y,z)$,and all cross-partial terms in the curl vanish, so

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For a regular curve, the curvature and torsion definition givesIt measures the rate at which the unit tangent turns per unit arclength.

HereThus $|{\gamma }^{\prime }|=3$, $|{\gamma }^{\prime \prime }|=4$, and ${\gamma }^{\prime }\cdot {\gamma }^{\prime \prime }=0$. HenceThe arc-length parametrization from $t=0$ is $s=3t$, so the curvature remains the constant $\kappa (s)=4/9$ for $0\le s\le 3\pi$. Therefore the total curvature is

The curve $\tilde{\gamma }$ is a unit circle traversed once. Its curvature is one and its length is $2\pi$, so without further calculationThe helical curve $\gamma$ devotes part of its unit tangent to the constant vertical direction. Its tangent therefore turns more slowly on the unit sphere than the tangent to the planar circle, which explains the smaller total curvature.

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For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and define the sign of a permutation bySince $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$,so this is a homomorphism. A transposition interchanges two columns of the identity matrix and therefore has determinant $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\mathrm{sgn}(\sigma )=(-1{)}^r$; because the determinant depends only on $\sigma$, the parity of $r$ is independent of the chosen expression.

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The conjugacy classes in $A_5$ consist of the identity, the $15$ double transpositions, the $20$ three-cycles, and two classes of $12$ five-cycles. Thus their sizes areA normal subgroup is a union of conjugacy classes containing the identity, and its order must divide $60$ by Lagrange's theorem. Checking sums of a subset of $15,20,12,12$ shows that no numberdivides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the Simplicity of the alternating group A5.

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If a homomorphism $A_5\to \{\pm 1\}$ were surjective, its kernel would be a normal subgroup of index two and hence order $30$. This contradicts the simplicity of $A_5$. Therefore no such surjective homomorphism exists.

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Let $\varphi :A_5\times A_5\to \{\pm 1\}$ be a homomorphism. Its restrictions to $A_5\times \{1\}$ and $\{1\}\times A_5$ cannot be nontrivial, because any nontrivial subgroup of $\{\pm 1\}$ is the whole group and part (iii) rules out such a surjection. Both restrictions are therefore trivial. Since$\varphi$ itself is trivial and in particular is not surjective.

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From $ba{b}^{-1}=a^{-1}$ we obtain $ba=a^{-1}b$. Moving every occurrence of $b$ to the right and replacing $b^2$ by $a^n$ shows that every group element has the formThe element $b$ cannot lie in $⟨a⟩$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $⟨a⟩$ and $⟨a⟩b$ are disjoint and each has $2n$ elements. Hence every $n$-dicyclic group has order $4n$.

Existence is explicit. Put $\zeta =e^{\pi i/n}$ and takeThen $a$ has order $2n$,The $2n$ diagonal matrices $a^k$ and the $2n$ off-diagonal matrices $a^{k}b$ are distinct, so they form a dicyclic group of order $4n$.

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Five examples arewhere $D_{12}$ denotes the symmetry group of a hexagon and ${\mathrm{Dic}}_3$ is the $3$-dicyclic group.

The first two are abelian, while the last three are not. Of the first two, only $C_{12}$ is cyclic. Among the nonabelian groups, $A_4$ has no element of order $6$; both of the others do. The dihedral group $D_{12}$ has seven involutions, whereas ${\mathrm{Dic}}_3$ has the unique involution $a^3$ and also has elements outside $⟨a⟩$ of order four. These invariants prove that all five groups are pairwise non-isomorphic.

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Cayley theorem states that every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. For $g\in G$, defineEach $L_g$ is a permutation, and $L_g{L}_h=L_{gh}$, so $g\mapsto L_g$ is a homomorphism. If $L_g$ is the identity permutation, then $g=L_g(1)=1$; hence the homomorphism is injective. For finite $G$ of order $n$, this embeds $G$ in $S_n$.

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The Cauchy theorem for groups says that if a prime $p$ divides $|G|$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. Cyclic rotation acts on $X$: if the product is one, thenEvery orbit has size one or $p$. The fixed points are exactly the constant tuples $(x,\dots ,x)$ satisfying $x^p=1$. Since $|X|\equiv 0(\mathrm{mod}p)$, the number of fixed points is divisible by $p$. The identity supplies one, so there is a nonidentity $x$ with $x^p=1$. Its order divides the prime $p$ and is not one, hence is $p$.

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Let $n=p^k$ and take $G=C_{p^k}$. If it embedded in $S_{n-1}$, that symmetric group would contain a permutation of order $p^k$. The order of a permutation is the least common multiple of its cycle lengths. For this least common multiple to be divisible by $p^k$, one cycle length must itself be divisible by $p^k$, requiring at least $p^k=n$ points. This is impossible in $S_{n-1}$. Thus $C_n$ is the required group.

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Because $n$ is not a prime power, distinct primes $p<q$ divide $n$. By Cauchy's theorem, $G$ has subgroups $H,K$ of orders $p,q$. Combine the two coset actions to obtainThe kernel of the first action is contained in $H$, and that of the second is contained in $K$. Their intersection is trivial because $H\cap K=\{1\}$, so the combined action is faithful.

Since $p\ge 2$, $q\ge 3$, and $n\ge 6$,Adding fixed points embeds this symmetric group into $S_{n-1}$. Thus every group of non-prime-power order $n$ is a subgroup of $S_{n-1}$, as summarized by symmetric-group embedding at one less than the group order.

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The orbit-stabilizer theorem states that for a finite group $G$ acting on a set $X$ and $x\in X$,Define $G/G_x\to Gx$ by $g{G}_x\mapsto gx$. This is well-defined because $g{G}_x=h{G}_x$ exactly when $h^{-1}g\in G_x$, equivalently $gx=hx$. The same equivalence proves injectivity, and the definition of the orbit proves surjectivity. Thus $|Gx|=[G:G_x]$, giving the formula.

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The group acts transitively on the eight vertices. At a chosen vertex, an isometry fixing it may permute its three incident edges arbitrarily, and each of the six permutations is realized by an isometry. Thus the vertex stabilizer is isomorphic to $S_3$ and has order six. Orbit-stabilizer givesThis begins the three stabilizer descriptions in the symmetry group of a cube.

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There are twelve edges and the action on them is transitive, so an edge stabilizer has orderFor an edge parallel to the $z$-axis with midpoint $(1,1,0)$, its stabilizer is generated by interchanging $x$ and $y$ and by replacing $z$ by $-z$. These are commuting involutions, so

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There are four main diagonals and the action is transitive, so the stabilizer of one has orderFor the diagonal spanned by $(1,1,1)$, the stabilizing signed permutation matrices are $P$ and $-P$ with $P$ a permutation matrix. HenceThe action is not faithful: both $I$ and the central inversion $-I$ fix every main diagonal setwise. In fact its kernel is exactly $\{I,-I\}$.

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The action on the four diagonals identifieswhere the $C_2$ factor is generated by central inversion. An element of order three must have trivial $C_2$ component and a three-cycle in $S_4$. Since all three-cycles in $S_4$ are conjugate,Elements of order two are not all conjugate. For example, $-I$ is central and therefore has a singleton conjugacy class, while a reflection is a different element of order two. Thus

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Suppose ${\varphi }_1,{\varphi }_2$ solve the Dirichlet problem and put $u={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $S$. The divergence theorem applied to $u\nabla u$ gives Green's identityThus $\nabla u=0$, so $u$ is constant; its zero boundary value makes it zero. The Dirichlet solution is therefore unique.

For homogeneous Neumann data the same calculation again shows that $u$ is constant, but the boundary condition does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard uniqueness of Poisson equation distinction.

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For any rotation $R$ about the centre, ${\varphi }_R(x)=\varphi (Rx)$ satisfies the same equation and the same Dirichlet data because $V,f,g$ are rotationally invariant. Uniqueness from part (a) gives ${\varphi }_R=\varphi$ for every $R$. Thus $\varphi$ is constant on spheres and is a function of $r$ alone.

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For a radial function in three dimensions, the radial Laplacian givesHere regularity at zero selectsA further integration giveswhose apparent singularity is removable, with limit $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the boundary condition $\varphi (1)=3$ gives $C=11-4e$. Therefore

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The radial equation isIts solutions satisfying decay at infinity have the formThe boundary value $\varphi (1)=1$ gives $C=0$, soIt tends to zero as $r\to \infty$ and directly satisfies ${\nabla }^2\varphi =-r^{-3}$.

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Regularity at zero givesIntegrating once more,where $\arctan r/r$ has removable limit one at zero. At $r=1$, the nonconstant terms sum to $\frac{1}{2}\log 2+\frac{\pi }{4}$. Hence

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Parametrize the upward-oriented paraboloid byThenOn the surface,The cubic trigonometric terms integrate to zero over $0\le \theta \le 2\pi$, so the surface integral is

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Close the surface with the unit disk $D$ in the plane $z=0$. Its outward normal is $-e_z$, but $F\cdot (-e_z)=-z^4=0$ there. AlsoThe $x$ and $y$ terms integrate to zero by symmetry. The divergence theorem therefore gives

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In components,because the Levi-Civita symbol is antisymmetric in $j,k$ while equality of mixed partial derivatives makes ${\partial }_j{\partial }_k\varphi$ symmetric. Hence

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Similarly,since the last derivatives are symmetric in $i,j$. Thus

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The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

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The first field isso its integral around $C$ is zero and it is conservative with potential $e^{x^2}\cos (yz)$.

Along $C$, $dr=(-\sin s,\cos s,0)ds$ andsoThus $B_2$ is not conservative and cannot be a gradient on the stated punctured region.

Finally,Its integral around $C$ is zero, and it is conservative with the displayed potential. Therefore the three integrals are respectivelyand exactly $B_1,B_3$ are conservative gradient fields.

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A tensor of rank $n$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. In components it obeys one change-of-basis factor for each of its $n$ indices.

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A tensor is totally antisymmetric tensor if interchanging any two indices changes its sign. Consequently every component with repeated indices vanishes. In ${\mathbb{R}}^3$:

These are all possibilities.

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A tensor is isotropic tensor if its components are unchanged under every proper orthogonal change of basis. The general isotropic rank-four tensor iswhere $a,b,c$ are scalars.

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Both sides are alternating in $i,j,k$ and in $l,m,n$. If either triple repeats an index, both sides vanish. Otherwise each triple is a permutation of $(1,2,3)$; the determinant of the corresponding permutation matrix is exactly the product of the two permutation signs. HenceThere are $3!=6$ nonzero components of ${ϵ}_{ijk}$, each equal to $\pm 1$, so

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Contracting two antisymmetric tensors gives the nonzero isotropic rank-four tensorA nonzero isotropic rank-five tensor is, for example,

More generally, one obtains the rank-five isotropic building blocks by choosing which two of the five indices belong to the Kronecker delta and putting the remaining three into the Levi-Civita symbol. There aresuch placements:Thus the unrestricted isotropic rank-five ansatz requires ten scalar component coefficients, one for each delta-epsilon placement.

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