= Paper 3 {scope} https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2024/paperia_3_2024.pdf = 1D {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=1D} A is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equation $$ \alpha|z|^2+\beta z+\overline\beta\,\overline z+\gamma=0, $$ where $\alpha,\gamma\in\mathbb R$; the case $\alpha=0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w|^2$ to obtain $$ \gamma|w|^2+\overline\beta w+\beta\overline w+\alpha=0, $$ which is again such an equation. The generators, and hence every Möbius transformation, therefore preserve . Now let $$ f(z)=\frac{az+b}{cz+d}. $$ If $c\ne0$, 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 in question is the , consisting of $z\mapsto az+b$ with $a\ne0$. It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne0$, then $T_b$ is affine but $$ IT_bI(z)=\frac{z}{1+bz} $$ has a finite pole and is not affine. Hence conjugation by $I$ does not preserve the . Solved by gpt-5.6-sol high. = 2D {parent=Paper 3} {scope} {title2=Groups} = i {parent=2d} {scope} = Solution {parent=i} The is $$ SO(n)=\{A\in M_n(\mathbb R):A^TA=I,\ \det A=1\}. $$ If $$ A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SO(2), $$ its first column is a unit , say $(\cos\theta,\sin\theta)^T$. The second column is the unique unit perpendicular to it that gives positive , namely $(-\sin\theta,\cos\theta)^T$. Thus $$ A=\begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}, $$ so $A$ is rotation through $\theta$ about the origin. For $A\in SO(3)$, its real characteristic of odd degree has a real . Every of an orthogonal has one, so every real is $\pm1$. Nonreal occur in conjugate pairs whose product is one, while $\det A=1$; if all are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero $u$. 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 $\mathbb Ru$. Solved by gpt-5.6-sol high. = ii {parent=2d} {scope} = Solution {parent=ii} Choose an angle $2\pi\alpha$ with $\alpha\notin\mathbb Q$. Rotation through that angle has infinite order, so the it generates is isomorphic to $\mathbb Z$. The 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 $\alpha=\sqrt2$ and $\beta=\sqrt3$. Define $$ \phi:\mathbb Z^2\longrightarrow SO(2), \qquad \phi(m,n)=R_{2\pi(m\alpha+n\beta)}. $$ This is a homomorphism. If $\phi(m,n)=I$, then $m\alpha+n\beta\in\mathbb Z$, and the stated rational independence forces $m=n=0$. Thus $\phi$ is injective and its image is isomorphic to $\mathbb Z^2$. Solved by gpt-5.6-sol high. = 3B {parent=Paper 3} {scope} {title2=Vector Calculus} = i {parent=3b} {scope} = Solution {parent=i} In the Cartesian , writing $F=(F_1,F_2,F_3)$, $$ \boxed{\nabla\cdot F +=\partial_iF_i +=\frac{\partial F_1}{\partial x} ++\frac{\partial F_2}{\partial y} ++\frac{\partial F_3}{\partial z}} $$ defines the , while the is $$ \boxed{(\nabla\times F)_i +=\epsilon_{ijk}\partial_jF_k}. $$ Equivalently, $$ \nabla\times F= +(\partial_yF_3-\partial_zF_2)e_x ++(\partial_zF_1-\partial_xF_3)e_y ++(\partial_xF_2-\partial_yF_1)e_z. $$ Solved by gpt-5.6-sol high. = ii {parent=3b} {scope} = Solution {parent=ii} Use the and the product rule. First, $$ \begin{aligned} \nabla\cdot(F\times G) +&=\partial_i(\epsilon_{ijk}F_jG_k)\\ +&=\epsilon_{ijk}(\partial_iF_j)G_k ++\epsilon_{ijk}F_j\partial_iG_k\\ +&=G\cdot(\nabla\times F)-F\cdot(\nabla\times G). \end{aligned} $$ For the second identity, the contraction $$ \epsilon_{ijk}\epsilon_{klm} +=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl} $$ gives $$ \begin{aligned} +[\nabla\times(F\times G)]_i +&=\epsilon_{ijk}\partial_j(\epsilon_{klm}F_lG_m)\\ +&=\partial_j(F_iG_j-F_jG_i)\\ +&=F_i\partial_jG_j-G_i\partial_jF_j ++G_j\partial_jF_i-F_j\partial_jG_i. \end{aligned} $$ In notation this is $$ \boxed{ +\nabla\times(F\times G) +=F(\nabla\cdot G)-G(\nabla\cdot F) ++(G\cdot\nabla)F-(F\cdot\nabla)G}. $$ These are the identities. Solved by gpt-5.6-sol high. = iii {parent=3b} {scope} = Solution {parent=iii} The first identity in part (ii), with $F$ and $G$ interchanged, says $$ \nabla\cdot(G\times F) +=F\cdot(\nabla\times G)-G\cdot(\nabla\times F), $$ so the proposed identity is missing its final term. For a concrete counterexample, take $$ G=e_x, \qquad F=ye_z. $$ Then $\nabla\times G=0$, so the proposed left side is zero, whereas $$ G\times F=-ye_y, \qquad \nabla\cdot(G\times F)=-1. $$ Thus the identity is false. Solved by gpt-5.6-sol high. = iv {parent=3b} {scope} = Solution {parent=iv} For $F=(x,y,z)$, $$ \boxed{\nabla\cdot F=1+1+1=3} $$ and all cross-partial terms in the curl vanish, so $$ \boxed{\nabla\times F=0}. $$ Solved by gpt-5.6-sol high. = 4B {parent=Paper 3} {scope} {title2=Vector Calculus} = Solution {parent=4B} For a regular curve, the definition gives $$ \kappa=\left|\frac{dT}{ds}\right| +=\frac{|\gamma'\times\gamma''|}{|\gamma'|^3}. $$ It measures the rate at which the unit tangent turns per unit arclength. Here $$ \gamma'(t)=(-2\sin2t,\sqrt5,2\cos2t), \qquad \gamma''(t)=(-4\cos2t,0,-4\sin2t). $$ Thus $|\gamma'|=3$, $|\gamma''|=4$, and $\gamma'\cdot\gamma''=0$. Hence $$ |\gamma'\times\gamma''|=|\gamma'||\gamma''|=12, \qquad \boxed{\kappa=\frac{12}{27}=\frac49}. $$ The from $t=0$ is $s=3t$, so the curvature remains the constant $\kappa(s)=4/9$ for $0\leq s\leq3\pi$. Therefore the is $$ \boxed{\int_\gamma\kappa\,ds +=\frac49(3\pi)=\frac{4\pi}{3}}. $$ The curve $\widetilde\gamma$ is a unit circle traversed once. Its curvature is one and its length is $2\pi$, so without further calculation $$ \boxed{\int_{\widetilde\gamma}\kappa\,ds=2\pi}. $$ The 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. Solved by gpt-5.6-sol high. = 5D {parent=Paper 3} {scope} {title2=Groups} = i {parent=5d} {scope} = Solution {parent=i} For $\sigma\in S_n$, let $P_\sigma$ be its permutation and define the by $$ \operatorname{sgn}(\sigma)=\det P_\sigma\in\{\pm1\}. $$ Since $P_{\sigma\tau}=P_\sigma P_\tau$, $$ \operatorname{sgn}(\sigma\tau) =\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau), $$ so this is a homomorphism. A transposition interchanges two columns of the identity and therefore has $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\operatorname{sgn}(\sigma)=(-1)^r$; because the depends only on $\sigma$, the parity of $r$ is independent of the chosen expression. Solved by gpt-5.6-sol high. = ii {parent=5d} {scope} = Solution {parent=ii} 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 are $$ 1,\quad15,\quad20,\quad12,\quad12. $$ A normal 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 number $$ 1+\text{a nonempty proper subset sum} $$ divides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the . Solved by gpt-5.6-sol high. = iii {parent=5d} {scope} = Solution {parent=iii} If a homomorphism $A_5\to\{\pm1\}$ were surjective, its kernel would be a normal of index two and hence order $30$. This contradicts the simplicity of $A_5$. Therefore no such surjective homomorphism exists. Solved by gpt-5.6-sol high. = iv {parent=5d} {scope} = Solution {parent=iv} Let $\phi:A_5\times A_5\to\{\pm1\}$ be a homomorphism. Its restrictions to $A_5\times\{1\}$ and $\{1\}\times A_5$ cannot be nontrivial, because any nontrivial of $\{\pm1\}$ is the whole and part (iii) rules out such a surjection. Both restrictions are therefore trivial. Since $$ (g,h)=(g,1)(1,h), $$ $\phi$ itself is trivial and in particular is not surjective. Solved by gpt-5.6-sol high. = 6D {parent=Paper 3} {scope} {title2=Groups} = i {parent=6d} {scope} = Solution {parent=i} From $bab^{-1}=a^{-1}$ we obtain $ba=a^{-1}b$. Moving every occurrence of $b$ to the right and replacing $b^2$ by $a^n$ shows that every element has the form $$ a^k\quad\hbox{or}\quad a^kb, \qquad 0\leq k<2n. $$ The element $b$ cannot lie in $\langle a\rangle$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $\langle a\rangle$ and $\langle a\rangle b$ are disjoint and each has $2n$ elements. Hence every $n$-dicyclic has order $4n$. Existence is explicit. Put $\zeta=e^{\pi i/n}$ and take $$ a=\begin{pmatrix}\zeta&0\\0&\zeta^{-1}\end{pmatrix}, \qquad b=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. $$ Then $a$ has order $2n$, $$ b^2=-I=a^n, \qquad bab^{-1}=a^{-1}. $$ The $2n$ diagonal $a^k$ and the $2n$ off-diagonal $a^kb$ are distinct, so they form a of order $4n$. Solved by gpt-5.6-sol high. = ii {parent=6d} {scope} = Solution {parent=ii} Five examples are $$ C_{12},\qquad C_6\times C_2, \qquad A_4,\qquad D_{12},\qquad \operatorname{Dic}_3, $$ where $D_{12}$ denotes the symmetry of a hexagon and $\operatorname{Dic}_3$ is the $3$-dicyclic . The first two are abelian, while the last three are not. Of the first two, only $C_{12}$ is cyclic. Among the nonabelian , $A_4$ has no element of order $6$; both of the others do. The dihedral $D_{12}$ has seven involutions, whereas $\operatorname{Dic}_3$ has the unique involution $a^3$ and also has elements outside $\langle a\rangle$ of order four. These invariants prove that all five are pairwise non-isomorphic. Solved by gpt-5.6-sol high. = 7D {parent=Paper 3} {scope} {title2=Groups} = i {parent=7d} {scope} = Solution {parent=i} states that every $G$ is isomorphic to a of the symmetric on its underlying set. For $g\in G$, define $$ L_g:G\to G, \qquad L_g(x)=gx. $$ Each $L_g$ is a permutation, and $L_gL_h=L_{gh}$, 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 $S_n$. Solved by gpt-5.6-sol high. = ii {parent=7d} {scope} = Solution {parent=ii} The says that if a prime $p$ divides $|G|$, then $G$ contains an element of order $p$. Consider $$ X=\{(x_1,\ldots,x_p)\in G^p:x_1x_2\cdots x_p=1\}. $$ The 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, then $$ x_2\cdots x_px_1=x_1^{-1}(x_1\cdots x_p)x_1=1. $$ Every orbit has size one or $p$. The fixed points are exactly the constant tuples $(x,\ldots,x)$ satisfying $x^p=1$. Since $|X|\equiv0\pmod p$, 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 $p$. Solved by gpt-5.6-sol high. = iii {parent=7d} {scope} = Solution {parent=iii} Let $n=p^k$ and take $G=C_{p^k}$. If it embedded in $S_{n-1}$, that symmetric 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 . Solved by gpt-5.6-sol high. = iv {parent=7d} {scope} = Solution {parent=iv} Because $n$ is not a prime power, distinct primes $p $H,K$ of orders $p,q$. Combine the two s to obtain $$ G\longrightarrow S_{n/p}\times S_{n/q} \longrightarrow S_{n/p+n/q}. $$ The 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 $p\geq2$, $q\geq3$, and $n\geq6$, $$ \frac np+\frac nq \leq\frac{5n}{6} \leq n-1. $$ Adding fixed points embeds this symmetric into $S_{n-1}$. Thus every of non-prime-power order $n$ is a of $S_{n-1}$, as summarized by . Solved by gpt-5.6-sol high. = 8D {parent=Paper 3} {scope} {title2=Groups} = a {parent=8d} {scope} = Solution {parent=a} The states that for a finite $G$ acting on a set $X$ and $x\in X$, $$ \boxed{|G|=|Gx|\,|G_x|}. $$ Define $G/G_x\to Gx$ by $gG_x\mapsto gx$. This is well-defined because $gG_x=hG_x$ 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. Solved by gpt-5.6-sol high. = b {parent=8d} {scope} = i {parent=b} {scope} = Solution {parent=i} The 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 gives $$ \boxed{|G|=8\cdot6=48}. $$ This begins the three stabilizer descriptions in the . Solved by gpt-5.6-sol high. = ii {parent=b} {scope} = Solution {parent=ii} There are twelve edges and the action on them is transitive, so an edge stabilizer has order $$ \frac{48}{12}=4. $$ For 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 $$ \boxed{G_e\cong C_2\times C_2}. $$ Solved by gpt-5.6-sol high. = iii {parent=b} {scope} = Solution {parent=iii} There are four main diagonals and the action is transitive, so the stabilizer of one has order $$ \frac{48}{4}=12. $$ For the diagonal spanned by $(1,1,1)$, the stabilizing signed permutation are $P$ and $-P$ with $P$ a permutation . Hence $$ \boxed{G_d\cong C_2\times S_3}. $$ The action is not faithful: both $I$ and the central inversion $-I$ fix every main diagonal setwise. In fact its kernel is exactly $\{I,-I\}$. Solved by gpt-5.6-sol high. = iv {parent=b} {scope} = Solution {parent=iv} The action on the four diagonals identifies $$ G\cong C_2\times S_4, $$ where 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, $$ \boxed{\text{all elements of order three in }G\text{ are conjugate}.} $$ 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 $$ \boxed{\text{elements of order two in }G\text{ need not be conjugate}.} $$ Solved by gpt-5.6-sol high. = 9B {parent=Paper 3} {scope} {title2=Vector Calculus} = a {parent=9b} {scope} = Solution {parent=a} Suppose $\phi_1,\phi_2$ solve the Dirichlet problem and put $u=\phi_1-\phi_2$. Then $\nabla^2u=0$ in $V$ and $u=0$ on $S$. The applied to $u\nabla u$ gives Green's identity $$ \int_V|\nabla u|^2\,dV =\int_Su\,\partial_nu\,dS- \int_Vu\nabla^2u\,dV=0. $$ Thus $\nabla u=0$, 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 does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard distinction. Solved by gpt-5.6-sol high. = b {parent=9b} {scope} = Solution {parent=b} For any rotation $R$ about the centre, $\phi_R(x)=\phi(Rx)$ satisfies the same equation and the same Dirichlet data because $V,f,g$ are rotationally invariant. Uniqueness from part (a) gives $\phi_R=\phi$ for every $R$. Thus $\phi$ is constant on spheres and is a of $r$ alone. Solved by gpt-5.6-sol high. = c {parent=9b} {scope} = i {parent=c} {scope} = Solution {parent=i} For a radial in three dimensions, the gives $$ \nabla^2\phi=\frac1{r^2}(r^2\phi')'. $$ Here regularity at zero selects $$ r^2\phi'(r)=\int_0^rs^2(s-1)e^s\,ds =e^r(r^3-4r^2+8r-8)+8. $$ A further integration gives $$ \phi(r)=\frac{e^r(r^2-5r+8)-8}{r}+C, $$ whose apparent singularity is removable, with $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the $\phi(1)=3$ gives $C=11-4e$. Therefore $$ \boxed{\phi(r)= \frac{e^r(r^2-5r+8)-8}{r}+11-4e}. $$ Solved by gpt-5.6-sol high. = ii {parent=c} {scope} = Solution {parent=ii} The radial equation is $$ (r^2\phi')'=-\frac1r. $$ Its solutions satisfying decay at infinity have the form $$ \phi(r)=\frac{\log r+1+C}{r}. $$ The boundary value $\phi(1)=1$ gives $C=0$, so $$ \boxed{\phi(r)=\frac{1+\log r}{r}}. $$ It tends to zero as $r\to\infty$ and directly satisfies $\nabla^2\phi=-r^{-3}$. Solved by gpt-5.6-sol high. = iii {parent=c} {scope} = Solution {parent=iii} Regularity at zero gives $$ r^2\phi'(r)=\int_0^r\frac{s^2}{1+s^2},ds =r-\arctan r. $$ Integrating once more, $$ \phi(r)=\frac12\log(1+r^2)+\frac{\arctan r}{r}+C, $$ where $\arctan r/r$ has removable one at zero. At $r=1$, the nonconstant terms sum to $\tfrac12\log2+\tfrac\pi4$. Hence $$ \boxed{ \phi(r)=\frac12\log(1+r^2)+\frac{\arctan r}{r} -1-\frac12\log2-\frac\pi4}. $$ Solved by gpt-5.6-sol high. = 10B {parent=Paper 3} {scope} {title2=Vector Calculus} = i {parent=10b} {scope} = Solution {parent=i} Parametrize the upward-oriented paraboloid by $$ X(r,\theta)=(r\cos\theta,r\sin\theta,1-r^2), \quad0\leq r\leq1,\quad0\leq\theta\leq2\pi. $$ Then $$ X_r\times X_\theta=(2r^2\cos\theta,2r^2\sin\theta,r). $$ On the surface, $$ F=(r^2\cos^2\theta,-r^2\sin^2\theta,(1-r^2)^4). $$ The cubic trigonometric terms integrate to zero over $0\leq\theta\leq2\pi$, so the is $$ \begin{aligned} \int_SF\cdot dS &=2\pi\int_0^1r(1-r^2)^4\,dr\\ &=\boxed{\frac\pi5}. \end{aligned} $$ Solved by gpt-5.6-sol high. = ii {parent=10b} {scope} = Solution {parent=ii} 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. Also $$ \nabla\cdot F=2x-2y+4z^3. $$ The $x$ and $y$ terms integrate to zero by symmetry. The therefore gives $$ \begin{aligned} \int_SF\cdot dS &=\int_0^{2\pi}\int_0^1\int_0^{1-r^2} 4z^3r\,dz\,dr\,d\theta\\ &=2\pi\int_0^1r(1-r^2)^4\,dr =\boxed{\frac\pi5}. \end{aligned} $$ Solved by gpt-5.6-sol high. = 11B {parent=Paper 3} {scope} {title2=Vector Calculus} = i {parent=11b} {scope} = Solution {parent=i} In components, $$ [\nabla\times\nabla\phi]_i =\epsilon_{ijk}\partial_j\partial_k\phi=0, $$ because the is antisymmetric in $j,k$ while equality of mixed makes $\partial_j\partial_k\phi$ symmetric. Hence $$ \boxed{\nabla\times\nabla\phi=0}. $$ Solved by gpt-5.6-sol high. = ii {parent=11b} {scope} = Solution {parent=ii} Similarly, $$ \nabla\cdot(\nabla\times A) =\partial_i(\epsilon_{ijk}\partial_jA_k) =\epsilon_{ijk}\partial_i\partial_jA_k=0, $$ since the last are symmetric in $i,j$. Thus $$ \boxed{\nabla\cdot(\nabla\times A)=0}. $$ Solved by gpt-5.6-sol high. = iii {parent=11b} {scope} = Solution {parent=iii} The states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$, $$ \int_S(\nabla\times F)\cdot n\,dS =\oint_{\partial S}F\cdot dr. $$ A field is if its line is path-independent, equivalently if every closed line vanishes. If $F=\nabla\phi$, the gives $$ \int_\gamma F\cdot dr=\phi(\gamma(1))-\phi(\gamma(0)). $$ This is zero for every closed curve, so every globally defined field is conservative. Solved by gpt-5.6-sol high. = iv {parent=11b} {scope} = Solution {parent=iv} The first field is $$ B_1=\nabla\bigl(e^{x^2}\cos(yz)\bigr), $$ so its around $C$ is zero and it is conservative with potential $e^{x^2}\cos(yz)$. Along $C$, $dr=(-\sin s,\cos s,0)ds$ and $$ B_2=(-\sin s,\cos s,0), $$ so $$ \oint_CB_2\cdot dr=\int_0^{2\pi}1\,ds=\boxed{2\pi}. $$ Thus $B_2$ is not conservative and cannot be a on the stated punctured region. Finally, $$ B_3=\nabla\log(x^2+y^2+z^2). $$ Its around $C$ is zero, and it is conservative with the displayed potential. Therefore the three are respectively $$ \boxed{0,\quad2\pi,\quad0}, $$ and exactly $B_1,B_3$ are conservative fields. Solved by gpt-5.6-sol high. = 12B {parent=Paper 3} {scope} {title2=Vector Calculus} = i {parent=12b} {scope} = Solution {parent=i} A of rank $n$ is a multilinear map of $n$ 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. Solved by gpt-5.6-sol high. = ii {parent=12b} {scope} = Solution {parent=ii} A tensor is if interchanging any two indices changes its sign. Consequently every component with repeated indices vanishes. In $\mathbb R^3$: * for $n=2$, the general tensor is $A_{ij}=a_k\epsilon_{kij}$, with three free components; * for $n=3$, it is $A_{ijk}=a\epsilon_{ijk}$; * for $n>3$, every component repeats an index, so the tensor is zero. These are all possibilities. Solved by gpt-5.6-sol high. = iii {parent=12b} {scope} = Solution {parent=iii} A tensor is if its components are unchanged under every proper orthogonal change of . The general isotropic rank-four tensor is $$ \boxed{T_{ijkl} =a\delta_{ij}\delta_{kl} +b\delta_{ik}\delta_{jl} +c\delta_{il}\delta_{jk}}, $$ where $a,b,c$ are . Solved by gpt-5.6-sol high. = iv {parent=12b} {scope} = Solution {parent=iv} 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 of the corresponding permutation is exactly the product of the two permutation signs. Hence $$ \boxed{ \epsilon_{ijk}\epsilon_{lmn} =\begin{vmatrix} \delta_{il}&\delta_{im}&\delta_{in}\\ \delta_{jl}&\delta_{jm}&\delta_{jn}\\ \delta_{kl}&\delta_{km}&\delta_{kn} \end{vmatrix}}. $$ There are $3!=6$ nonzero components of $\epsilon_{ijk}$, each equal to $\pm1$, so $$ \boxed{\epsilon_{ijk}\epsilon_{ijk}=6}. $$ Solved by gpt-5.6-sol high. = v {parent=12b} {scope} = Solution {parent=v} Contracting two antisymmetric tensors gives the nonzero isotropic rank-four tensor $$ \boxed{ \epsilon_{ijm}\epsilon_{klm} =\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}}. $$ A nonzero isotropic rank-five tensor is, for example, $$ \boxed{T_{ijklm}=\epsilon_{ijk}\delta_{lm}}. $$ 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 are $$ \binom52=10 $$ such placements: $$ \epsilon_{ijk}\delta_{lm}, \epsilon_{ijl}\delta_{km}, \epsilon_{ijm}\delta_{kl}, \epsilon_{ikl}\delta_{jm}, \epsilon_{ikm}\delta_{jl}, \epsilon_{ilm}\delta_{jk}, $$ $$ \epsilon_{jkl}\delta_{im}, \epsilon_{jkm}\delta_{il}, \epsilon_{jlm}\delta_{ik}, \epsilon_{klm}\delta_{ij}. $$ Thus the unrestricted isotropic rank-five ansatz requires ten component coefficients, one for each delta-epsilon placement. Solved by gpt-5.6-sol high.