Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 3

For a line through the origin making angle $\theta$ with the positive $x$-axis, its reflection matrix isIts determinant is $-1$. An orthogonal matrix has determinant $\pm 1$. If its determinant is $-1$, write its first column as $(\cos \phi ,\sin \phi {)}^T$; orthonormality and the determinant then force the second column to be $(\sin \phi ,-\cos \phi {)}^T$. The matrix is $S_{\phi /2}$, hence is one Euclidean reflection. If its determinant is $+1$, the second column is $(-\sin \phi ,\cos \phi {)}^T$, so it is a rotation matrix. Direct multiplication givesThe identity may be taken as an empty product or as two identical reflections. Thus at most two origin-line reflections suffice in every case.

For a nonzero vector $b=dn$ with $d=|b|$ and unit normal $n$, reflection in the affine line $n\cdot x=c$ is $s_c(x)=x+2(c-n\cdot x)n$. Consequently $s_{d/2}{s}_0(x)=x+dn=x+b$. A translation is therefore a product of two parallel-line reflections. Combining this with the given decomposition of a Euclidean isometry as translation followed by an orthogonal map proves that every plane isometry is a finite product of affine-line reflections. The affine lines need not pass through the origin. In fact three suffice: if an isometry $f$ sends zero to $b\ne 0$, reflection in the perpendicular bisector of $0,b$ takes $b$ to zero. Composing that reflection with $f$ gives an isometry fixing zero, to which the two-reflection linear result applies. This is the planar finite reflection decomposition of a Euclidean isometry.

For an example that genuinely needs three, take the glide reflectionIt has no fixed point, so it is not one reflection. Its orthogonal part has determinant $-1$, whereas the product of two reflections has determinant $+1$, so it is not two reflections; it is not the identity either. Explicitly it is reflection in $y=0$ followed by reflection in $x=0$ and then in $x=1/2$. Thus its minimum reflection count is exactly three.

Lagrange's theorem says that for a subgroup $H$ of a finite group $G$,To prove it, the left cosets $gH$ partition $G$. Indeed every element lies in its own coset; if $gH$ and $kH$ intersect, write $gh=k{h}^{\prime }$ and deduce $g=k{h}^{\prime }{h}^{-1}$, so $gH=kH$. Multiplication by $g$ is a bijection from $H$ to $gH$, and each coset therefore has $|H|$ elements. Summing over the $[G:H]$ distinct cosets proves the formula.

For the failure of the converse to Lagrange's theorem, take the alternating group $A_4$, of order $4!/2=12$, and the divisor $6$. Suppose it had a subgroup $H$ of order six. Its index would be two. An index-two subgroup is normal: for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, while for $g\in H$ they are $H$. Hence $A_4/H$ would be a group of order two.

Every three-cycle in $A_4$ has order three. Its image in the quotient has cube equal to the identity and, since the quotient has order two, must itself be the identity. Thus all three-cycles lie in $H$. There are eight of them: choose the omitted point in four ways and one of two cyclic orders on the other three points. Together with the identity this would give at least nine elements of $H$, contradicting $|H|=6$. Therefore $A_4$ has no subgroup of order six, although six divides its order.

A continuously differentiable vector field is an irrotational vector field when its curl vanishes:For the global-potential construction, assume that the domain is connected and simply connected, or more generally that all closed-curve line integrals of $\mathbf{F}$ vanish. Then define the potential of a conservative vector field byTwo paths with the same endpoints differ by a closed curve. On a simply connected domain that curve can be spanned, or contracted through piecewise smooth spanning surfaces, inside the domain. Stokes theorem gives zero circulation because the curl is zero. The value of $V$ is therefore path-independent and $V({\mathbf{x}}_0)=0$. To compute a derivative, append a short coordinate segment to a path ending at $\mathbf{x}$. Then $V(\mathbf{x}+h{\mathbf{e}}_i)-V(\mathbf{x})=-{\int }_0^h{F}_i(\mathbf{x}+s{\mathbf{e}}_i)ds$, so ${\partial }_{i}V=-F_i$.

The domain qualification is necessary and is implicit in the potential request. Curl-free alone on an arbitrary domain does not suffice: $\mathbf{F}=(-y,x,0)/(x^2+y^2)$ on the complement of the $z$-axis has zero curl, but its circulation around a unit circle is $2\pi$. It has no single-valued global potential. The intended construction is valid on all of ${\mathbb{R}}^3$ for a smooth curl-free field, and on any other domain satisfying the condition above.

For the given spherical-coordinate field, the angular components have $r{F}_{\theta }=\cos \theta \cos \varphi$ and $r{F}_{\varphi }=m\sin \varphi$, independent of $r$, so the $\theta$ and $\varphi$ curl components vanish. The radial component isIt vanishes identically exactly when $m=-1$. In that case integrate $F_{\theta }=-r^{-1}{V}_{\theta }$ to obtain $V=-\sin \theta \cos \varphi +C(\varphi )$. The azimuthal equation $F_{\varphi }=-(r\sin \theta {)}^{-1}{V}_{\varphi }$ forces $C^{\prime }=0$, and $F_r=0$ removes any radial dependence. At the given north-pole point the first term is zero, so normalization fixes the constant to zero:The Cartesian expression shows that this potential and the field $\mathbf{F}=\nabla (x/r)$ are smooth for $r>0$, including the spherical-coordinate poles. The origin is excluded because the specified field is singular there; that exclusion does not obstruct the displayed global potential.

For Cartesian components, ${\partial }_j{x}_i={\delta }_{ij}$, the Kronecker delta. Differentiating $r^2=x_i{x}_i$ givesPut $s=k_{\ell }{x}_{\ell }$. Using the product rule and ${\partial }_j{r}^{-q}=-q{x}_j{r}^{-q-2}$ gives the velocity gradientThis is the rate of strain and vorticity of a Stokeslet, apart from the overall physical normalization. Its symmetric and antisymmetric parts of a matrix areHere $E_{ji}=E_{ij}$ and $W_{ji}=-W_{ij}$, and $d=E+W$. Taking the trace gives the divergenceFor the curl, use the Levi-Civita symbol. Because $d_{kj}={\partial }_j{u}_k$, the symmetric part makes no contribution, andThus $\nabla \times \mathbf{u}=2\mathbf{k}\times \mathbf{x}/r^3$. All derivative identities hold away from the singular origin.

For a group action of a finite group $G$ on a set $X$, let $G_x=\{g:gx=x\}$ be the stabilizer subgroup and $Gx=\{gx:g\in G\}$ the orbit. The orbit-stabilizer theorem isTo prove it, the map from left cosets to the orbit, $g{G}_x\mapsto gx$, is well-defined because multiplying $g$ on the right by an element fixing $x$ does not change $gx$. It is surjective by the definition of the orbit. If $gx=hx$, then $h^{-1}g\in G_x$, so $g{G}_x=h{G}_x$; hence it is injective. The cardinality identity now follows from Lagrange's theorem.

Place the octahedron's vertices at $\pm e_1,\pm e_2,\pm e_3$. A symmetry fixes their centre, so it is an orthogonal map and must take an opposite pair of vertices to another opposite pair. Its matrix is thus a signed permutation matrix. Conversely every signed permutation of these coordinate directions preserves the vertices, edges and faces. Therefore the full octahedral symmetry group hasThese same signed permutation matrices preserve the cube with vertices $(\pm 1,\pm 1,\pm 1)$, so the group is also the symmetry group of a cube. The opposite-vertex lines are the three coordinate axes, so $|D|=3$. The action on them is transitive, giving a line stabilizer of order $48/3=16$.

For the line $\mathbb{R}{e}_3$, the precise axis stabilizer in the full octahedral symmetry group consists of the block matricesThe eight possible $A$ are the symmetries of a square in the perpendicular plane, a dihedral group of order eight; the independent sign $\epsilon$ reverses the selected axis. ThusHere $D_8$ denotes the order-eight square-symmetry group. The selected line is unoriented: its two endpoints may be exchanged. Fixing an individual endpoint instead would give only the order-eight subgroup with $\epsilon =1$.

For a permutation $\sigma$ of a finite set, start with a point $x$. Repeatedly applying $\sigma$ eventually returns to $x$: a repeated point in the finite sequence can be cancelled backwards using injectivity. The distinct successive points before that return form a cycle. If any point has not yet been used, repeat the construction there. The resulting orbits are disjoint and cover the set, so their cycles, including one-cycles for fixed points, give a cycle decomposition of a permutation. Disjoint cycles commute because they act on different points.

Conjugation relabels a cycle:Thus conjugate permutations have the same cycle type. Conversely, if two decompositions have equal numbers of cycles of each length, pair their equally long cycles and define $\gamma$ to take corresponding entries, in order, to one another. Since the cycles partition the set, this defines a bijection, and the displayed identity proves the conjugacy. Hence cycle type characterizes conjugacy in the symmetric group.

For $\tau =(ab)$, use the convention that the rightmost permutation acts first. If $a,b$ are in different cycles, write those cycles asLeft multiplication by $\tau$ changes the arrow into $a$ to an arrow into $b$, and the arrow into $b$ to an arrow into $a$. Following these arrows gives the single cycleNo other cycles change, so $\ell (\tau \sigma )=\ell (\sigma )-1$. If $a,b$ are in the same cycle, write it in that last form; the same two arrow changes split it into the two cycles in the preceding display. Thus $\ell (\tau \sigma )=\ell (\sigma )+1$. Empty lists are allowed and produce one-cycles. This proves transposition changes the cycle count by one in all cases:

Use the existing cross-ratio conventionValues involving infinity are defined by limits. More uniformly, represent each point by a nonzero homogeneous vector $u=(u_0,u_1)$ and put $D(u,v)=u_0{v}_1-u_1{v}_0$. Then the same value is $D(z,a_1)D(a_0,a_2)/[D(z,a_2)D(a_0,a_1)]$, interpreted in the Riemann sphere. Rescaling any representative cancels out. The first three points are distinct, so numerator and denominator cannot simultaneously vanish. The fourth point may coincide with one of them, giving the values $1,0,\infty$ at $a_0,a_1,a_2$, respectively.

With $a_0,a_1,a_2$ fixed, this is a nonconstant Möbius transformation $f$. For finite first three points, write $C=(a_0-a_2)/(a_0-a_1)\ne 0$; thenThe determinant of this fractional-linear map is nonzero because $a_1\ne a_2$. Its inverse is defined at every point of the sphere, including poles and infinity, so each $w$ has exactly one preimage. The same conclusion follows directly from the homogeneous formula when an $a_i$ is infinity.

Möbius transformations preserve generalized circles. To verify the fact used here, affine complex changes preserve ordinary circles and lines, while a generalized circle has an equation $A|z{|}^2+Bz+\overline{B}\overline{z}+D=0$ with $A,D$ real. Substitution $z=1/w$ and multiplication by $|w{|}^2$ gives another equation of the same form. Every non-affine Möbius map is a composition of affine maps and $z\mapsto 1/z$, so the claim follows. Consequentlyis a generalized circle. It contains the three distinct $a_i$, so it is their unique ordinary circle, or their line together with infinity when they are collinear. This is the real cross-ratio criterion for a generalized circle.

Let $c(w)=\overline{w}$, with $c(\infty )=\infty$. The defining equation for $J$ now gives the cross-ratio construction of generalized-circle reflection:This is well-defined by bijectivity of $f$, and $J^2=f^{-1}{c}^{2}f$ is the identity. Moreover $J(z)=z$ precisely for $z\in S$.

If the first three points lie on the extended real line, the normalization map has real coefficients, so $f(\overline{z})=\overline{f(z)}$. Hence $J(z)=\overline{z}$. For a second triple $b_0,b_1,b_2$ on the same generalized circle, write its normalization as $g=Mf$. Each $f(b_i)$ is real or infinity, and the unique Möbius map taking this real triple to $1,0,\infty$ has real coefficients. Thus $M$ commutes with $c$, givingTherefore $J$ depends only on the generalized circle, not on the three selected points. For an ordinary circle with centre $q$ and radius $R$, it is inversion in a circle, $J(z)=q+R^2/(\overline{z}-\overline{q})$; for a straight line it is the corresponding Euclidean reflection, extended to infinity.

A subgroup $K$ is a normal subgroup of $G$ when $gK{g}^{-1}=K$ for every $g\in G$.

For a group homomorphism $\varphi :G\to H$, its kernel contains the identity and, for $x,y$ in the kernel, $\varphi (x{y}^{-1})=\varphi (x)\varphi (y{)}^{-1}=e$. Hence the kernel of a group homomorphism is a subgroup. Also $\varphi (gk{g}^{-1})=\varphi (g)e\varphi (g{)}^{-1}=e$ for $k$ in the kernel; applying this to $g$ and $g^{-1}$ gives equality under conjugation. Thus the kernel is always normal.

The image contains the identity and is closed under products and inverses, so the image of a group homomorphism is always a subgroup of the target. It need not be normal: include $C_2=\{e,(12)\}$ into $S_3$. Conjugation by $(123)$ takes $(12)$ to $(23)$, outside that image. Thus the image is not always normal.

For matrices over either $\mathbb{Z}$ or ${\mathbb{F}}_2={\mathbb{Z}}_2$, multiplication stays in the same coefficient ring, and determinants multiply, so the determinant-one matrices are closed. Associativity is inherited from matrix multiplication, the identity has determinant one, and the inverse isIts entries are still in the same ring and its determinant is one. This proves that both sets are special linear groups under multiplication. Over ${\mathbb{F}}_2$, minus and plus agree, and the same formula applies.

Reduction of integers modulo two preserves addition and multiplication. Therefore reducing entries preserves the determinant condition and satisfies $\varphi (AB)=\varphi (A)\varphi (B)$, proving the stated map is a group homomorphism.

Its target acts on the three nonzero vectors of ${\mathbb{F}}_2^2$. The action is faithful: a matrix fixing all three vectors fixes the two standard basis vectors and so is the identity. Thus its image, and in particular the image of $\varphi$, is isomorphic to a permutation subgroup of $S_3$. In fact SL2 over F2 as a permutation group identifies the entire image. There are $3$ choices for the first nonzero column and $2$ for a second independent column, hence six invertible matrices. Every invertible matrix has determinant one over ${\mathbb{F}}_2$, so the faithful action is onto $S_3$.

To show the integer reduction map reaches all of them, takeOn $\{e_1,e_2,e_1+e_2\}$, the reduction of $S$ exchanges $e_1,e_2$ and fixes their sum, while the reduction of $T$ fixes $e_1$ and exchanges the other two. These two transpositions generate $S_3$. Hence

Put $\delta =w-u$, which vanishes on the boundary. Expanding the Dirichlet energy givesThe cross term vanishes by integration by parts: it is ${\int }_{\partial S}\delta {\partial }_{n}uds-{\int }_S\delta \Delta udA=0$, using both the zero boundary values of $\delta$ and Laplace's equation for $u$. Thus the Dirichlet principle gives the stronger identityWith the usual regular boundary, equality forces $w=u$.

In the unit disc, take boundary data $\cos \theta$. The harmonic function $u=r\cos \theta =x$ realizes these data and has constant gradient of length one, so its energy is $\pi$. For $w=g(r)\cos \theta$, the polar gradient givesThe energy principle therefore yields the sharp radial Dirichlet energy inequality

There is a small regularity point at the centre: finite energy need not make the polar expression classically smooth there. The bound still holds under the stated finite-energy condition. Indeed for $0<s<t$,The tails tend to zero, so $g^2$ has a limit at zero. That limit must be zero, since a positive limit would make $\int g^2/r$ diverge. Completing the square on $[\epsilon ,1]$ and letting $\epsilon \to 0$ givesThis proves the finite-energy extension directly, and also shows that equality holds exactly for $g(r)=r$.

For a twice differentiable vector potential $\mathbf{B}$, commuting partial derivatives in $\mathbf{J}=\nabla \times \mathbf{B}$ gives the necessary conditionThe potential is not unique: for any sufficiently smooth scalar $\psi$, $\mathbf{B}+\nabla \psi$ has the same curl, since the curl of a gradient is zero. This is the usual gradient freedom of a vector potential. A divergence-free condition alone need not suffice globally on every domain; here only its necessity is requested.

Stokes theorem states that for an oriented surface with compatible boundary orientation,For the integrand in this problem, a convenient globally smooth vector potential isThe boundary of the upper ellipsoid is the ellipse $z=0$. The outward normal on the upper surface induces counterclockwise orientation viewed from above, so parameterize it as $\mathbf{r}(\theta )=(a\cos \theta ,b\sin \theta ,0)$, $0\le \theta \le 2\pi$. Along this curve the first component of $\mathbf{B}$ is zero and $dz=0$, leavingSince the last angular integral is $3\pi /4$,The result is independent of the vertical semiaxis $c$. As a sign check, the field is divergence-free, so closing the cap with the downward-oriented base disc makes its flux the positive integral of $x^2$ over that disc, which has the same value.

The integrand is symmetric in $i,j$, so $T$ is a symmetric tensor. The sphere and its area element are invariant under rotations, and the integrand transforms covariantly when $\mathbf{x},\mathbf{y}$ are rotated together. For fixed $\mathbf{y}$, rotations about its direction therefore leave $T$ unchanged. In axes with $\mathbf{y}$ along the third axis, the mixed transverse-axial entries vanish and the two transverse eigenvalues are equal. Thus it is an axisymmetric symmetric rank-two tensor,where $\alpha ,\beta$ depend on $a,n$ but not on the direction of $\mathbf{y}$. The restriction $n>-1$ makes the integral convergent near $\mathbf{x}=\mathbf{y}$: the integrand is of size distance${}^{2n}$ and the local area element has one further radial power.

Take the polar axis along $\mathbf{y}$. Because $|\mathbf{x}|=|\mathbf{y}|=a$,The area element is $a^2\sin \theta d\theta d\varphi$. Hence with $s=1-\cos \theta$,

For the second scalar integral, take the trace. It contracts the two chord-vector factors into their squared length:Put $K=\pi (2a{)}^{2n+2}$. The tensor form gives $3\alpha +\beta a^2=K/(n+1)$, while its axial contraction gives $\alpha +\beta a^2=K/(n+2)$. Solving yields the weighted chord tensor of a sphere coefficientsBecause $a>0$, this is an isotropic tensor exactly when $\beta =0$. Therefore $n=-1/2$ is the unique permitted isotropic exponent.

The divergence theorem states that for a continuously differentiable vector field on a bounded region with outward unit normal $\mathbf{n}$ and suitably regular, possibly piecewise smooth, boundary,For a differentiable homogeneous function, differentiate $f(k\mathbf{x})=k^{n}f(\mathbf{x})$ at $k=1$ to get Euler theorem for homogeneous functions:Only positive $k$ near one is needed for this derivative, so no interpretation of noninteger powers at negative $k$ is required. Applying the product rule gives $\nabla \cdot (f\mathbf{x})=3f+\mathbf{x}\cdot \nabla f=(n+3)f$. The volume integral of a homogeneous function is thereforeThe condition $n\ne -3$ is needed for division; at that degree the divergence theorem instead gives only zero net boundary flux. The field $f\mathbf{x}$ must also be regular on the region, or singularities must be treated by excision and limiting boundary terms. Neither issue arises for the polynomial in the final calculation.

Write $r=\sqrt{x^2+y^2}$. In the cone, $0\le z\le \alpha$ and $0\le r\le z/\alpha$. The polynomial $f=z^4+{\alpha }^4{r}^4$ has degree four. Its direct volume integral isOn the lateral cone the outward normal is proportional to $\alpha {\mathbf{e}}_r-{\mathbf{e}}_z$, so $\mathbf{x}\cdot \mathbf{n}$ is proportional to $\alpha r-z=0$. Thus the lateral contribution vanishes. The cap is $z=\alpha$, $0\le r\le 1$, with outward normal ${\mathbf{e}}_z$, soMultiplying by $1/(n+3)=1/7$ reproduces $4\pi {\alpha }^5/21$, verifying the formula. The cap edge and apex have zero surface measure, and the piecewise smooth version of the divergence theorem applies.