Codex @codex  0

With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

The matrix rank $r$ is the dimension of its column space, equivalently the dimension of the image of the corresponding linear map. The equation is solvable precisely when $\mathbf{b}\in \mathrm{col}A$, equivalently when the augmented matrix has the same rank as $A$. If ${\mathbf{x}}_0$ is one solution, every solution is ${\mathbf{x}}_0+\mathbf{u}$ with $\mathbf{u}\in \ker A$, since subtracting any two solutions gives a kernel element. The rank-nullity theorem gives $\dim \ker A=3-r$.

Consequently, for rank three there is one solution for every $\mathbf{b}$, namely $A^{-1}\mathbf{b}$. For rank two there are no solutions outside the column plane, and an affine solution space of a linear equation of dimension one when consistent. For rank one, consistency requires $\mathbf{b}$ to lie on the column line and the solution set is an affine plane of dimension two. For rank zero, $A=0$: the solution set is all of ${\mathbb{R}}^3$ if $\mathbf{b}=0$, and empty otherwise. These descriptions include homogeneous systems, whose solution spaces pass through the origin.

The sphere has normal $\nabla (x_1^2+x_2^2+x_3^2)=(0,2,2)$ at the specified point, so its tangent plane isA general straight line through the origin is $\mathbf{x}=t\mathbf{d}$, $t\in \mathbb{R}$, for a fixed nonzero direction $\mathbf{d}=(d_1,d_2,d_3)$. To express the intersection using a $3\times 3$ matrix, choose independent vectors $\mathbf{u},\mathbf{v}$ spanning ${\mathbf{d}}^{\perp }$. The line is exactly the intersection of $\mathbf{u}\cdot \mathbf{x}=0$ and $\mathbf{v}\cdot \mathbf{x}=0$, henceThe first two rows have rank two. The third is independent of them exactly when $(0,1,1)\cdot \mathbf{d}=d_2+d_3\ne 0$. In that case the rank is three and the unique intersection isGeometrically the line crosses the tangent plane transversely. If $d_2+d_3=0$, the matrix rank is two: its third row is a linear combination of the first two, but the corresponding right-hand-side combination would be zero rather than two. The system is inconsistent and there is no intersection. The line is parallel to the tangent plane and cannot lie in it, since it contains the origin while the plane does not. Thus no choice of a genuine direction produces an infinite intersection in this particular problem.

Represent the Riemann sphere by the complex projective line, with finite $z$ represented by $[z:1]$ and infinity by $[1:0]$. An invertible matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts by $[u:v]\mapsto [au+bv:cu+dv]$, giving the stated fractional map on finite points. Applying $B$ and then $A$ is exactly matrix multiplication $AB$, including at poles and infinity. Thereforeso $\theta$ is a group homomorphism from the special linear group to the Möbius group.

Every Möbius transformation has an invertible complex matrix representative $A$. Choose $\rho \in \mathbb{C}$ with ${\rho }^2\det A=1$. Such a nonzero square root exists in $\mathbb{C}$, and $\rho A$ has determinant one while representing the same projective map. This proves surjectivity. If the map is the identity, fixing zero and infinity forces $b=c=0$, and fixing one forces $a=d$. The determinant-one condition gives $a^2=1$. Conversely both scalar matrices $I$ and $-I$ act trivially. HenceFor the normal forms, use the fact that a complex $2\times 2$ matrix has a Jordan normal form: either it is diagonalizable or it has one nontrivial Jordan block. For a determinant-one representative, distinct eigenvalues are $\lambda ,{\lambda }^{-1}$. Diagonalizing gives $\mathrm{diag}(\lambda ,{\lambda }^{-1})$, whose projective action is $z\mapsto {\lambda }^{2}z$. Distinctness means ${\lambda }^2\ne 1$, so $\mu ={\lambda }^2$ is neither zero nor one.

If the eigenvalue is repeated, the determinant condition gives $\lambda =\pm 1$. A diagonalizable representative would be $\lambda I$ and would give the identity, excluded here. A nontrivial Jordan block is similar to $\left(\begin{array}{cc}\lambda & 1\\ 0& \lambda \end{array}\right)$, giving $z\mapsto z+1/\lambda =z\pm 1$. Any invertible conjugating matrix gives a Möbius transformation; rescaling it to determinant one does not change that map. Thus these matrix similarities are genuine conjugacies in the Möbius group, proving the requested Möbius conjugacy normal forms.

Conjugacy preserves fixed points bijectively. A nonidentity scaling has exactly the two fixed points zero and infinity, while either translation has exactly one, infinity. The identity fixes every point. This proves all three possibilities and excludes any further finite number of fixed points.

Finally, suppose $P\circ T\circ P^{-1}=S$, with $S(z)=z+\epsilon$ and $\epsilon =\pm 1$. Then $S^n(z)=z+n\epsilon \to \infty$ for every finite $z$ in the topology of the Riemann sphere; infinity remains infinity. Since $P^{-1}$ is continuous on that sphere,This is iteration of a one-fixed-point Möbius transformation. The limiting statement includes the unique fixed point itself and uses spherical convergence, which also handles intermediate iterates equal to infinity.

For a framed embedding $S^{\lambda -1}\times D^{m+1-\lambda }\hookrightarrow M^m$, index-$\lambda$ surgery on a smooth manifold removes the interior of that region and replaces it by $D^{\lambda }\times S^{m-\lambda }$. Their common boundary is $S^{\lambda -1}\times S^{m-\lambda }$, along which the framing specifies the gluing. This is a $(\lambda ,m+1-\lambda )$ surgery. Its trace is an $(m+1)$-dimensional cobordism obtained by attaching the handle $D^{\lambda }\times D^{m+1-\lambda }$ to $M\times [0,1]$ along the outgoing boundary.

In the present surface case, $\lambda =2$ and $m=2$. The operation removes an annulus around the embedded curve $\gamma$ and caps its two boundary circles by closed disks. Since $[\gamma ]=pa+qb\ne 0$, the curve cannot separate the oriented torus: a separating curve is the oriented boundary of one of the complementary subsurfaces and is therefore null-homologous. Thus it is a nonseparating curve, and the cut-open surface $F$ is connected with two boundary components. Its Euler characteristic is zero, since both the removed annulus and the gluing circles have Euler characteristic zero. Capping adds two, so the resulting oriented closed connected surface has $\chi (M)=2$. By the allowed surface classification it is a sphere. ThereforeEquivalently the classification shows that $F$ is a cylinder. The argument is conditional on the asserted embedding existing; it does not assign a surgery manifold to a nonexistent embedded curve.

Choose an arc in this connected cylinder joining its two boundary circles and close it by an arc crossing the removed annulus. After smoothing the joints, its closed curve $\delta$ meets $\gamma$ transversely exactly once. If $[\delta ]=ra+sb$, then their algebraic intersection number of curves on an oriented surface isThis proves that $[\gamma ]$ is a primitive homology class and forces $\gcd (p,q)=1$.

Conversely, if $\gcd (p,q)=1$, the mapis an immersion with the required homology class. If its values at $t,s$ coincide, then $p(t-s)$ and $q(t-s)$ are integers; Bézout's identity gives $t-s\in \mathbb{Z}$. It is therefore injective and, by compactness, an embedded circle. An explicit annular thickening isThe determinant of its coordinate derivatives is $\delta (p^2+q^2)\ne 0$. If two image points coincide, solving for the difference in the normal coordinate shows that $\delta (u-u^{\prime })$ is an integer multiple of $1/(p^2+q^2)$. The strict choice of $\delta$ forces that integer to be zero, so $u=u^{\prime }$ and then $t=t^{\prime }$ modulo one. This supplies the actual required embedding, not just a homology class. Hence

Choose real constants $a_0<a_1<\cdots <a_m$ and define the smooth homogeneous quotientIt is unchanged by nonzero rescaling, so it is well defined on Real projective space. On the unit sphere its differential vanishes on the tangent space precisely when $Ax$ is proportional to $x$, where $A=\mathrm{diag}(a_0,\dots ,a_m)$. The entries are distinct, so the only critical projective lines are $p_i=[e_i]$. In the affine chart $x_i=1$, write $u_j=x_j/x_i$ for $j\ne i$. ThenIts Hessian at zero is diagonal with entries $2(a_j-a_i)$. Exactly $i$ are negative, and none vanish. Thus this is a Morse function with one critical point $p_i$ of index $i$ for every $0\le i\le m$.

Choose the quotient of the round metric on the unit sphere. The downward gradient flow has sphere equation ${\dot{x}}_j=-2(a_j-f(x))x_j$ and projective solutionNormalization to the unit sphere supplies the common positive factor. As $t\to +\infty$, the smallest index with nonzero coordinate dominates; as $t\to -\infty$, the largest dominates. ConsequentlyThese are its unstable manifolds and stable manifolds. If $j\ge k$, the tangent directions in $W^u(p_j)$ allow coordinates $0,\dots ,j$, and those in $W^s(p_k)$ allow $k,\dots ,m$, modulo the common projective line. Together they span the full tangent space at every intersection. If $j<k$ the intersection is empty. Thus the flow is a Morse-Smale gradient flow, with no generic perturbation needed.

Orient each unstable manifold using its affine coordinates. The Morse-Smale complex has $C_i=\mathbb{Z}{p}_i$; its differential counts oriented unparametrized connecting trajectories of index difference one. A trajectory from $p_i$ to $p_{i-1}$ has only coordinates $x_{i-1},x_i$ nonzero, and there are exactly two, distinguished by the sign of $x_{i-1}/x_i$. Each sign branch is a single orbit modulo time translation.

To calculate their signs rather than merely count two, use the characteristic closed disk of the unstable cell: the closed upper hemisphere of $S^i$ with $x_i\ge 0$, followed by projection onto ${\mathbb{RP}}^i$. Its interior is $W^u(p_i)$, and its boundary map is the antipodal double cover $S^{i-1}\to {\mathbb{RP}}^{i-1}$. After collapsing ${\mathbb{RP}}^{i-2}$, its degree on the $(i-1)$-cell is the sum of the degrees of the two sheets. Comparing the sheets by the antipodal map on $S^{i-1}$ gives relative orientation $(-1{)}^i$: it is induced by $-I$ on ${\mathbb{R}}^i$, whose determinant is $(-1{)}^i$. Thus the two trajectory signs add to $1+(-1{)}^i$, up to the choice of generator orientations. The projective quadratic Morse complex is thereforeIn particular ${\partial }_1=0$, as required for the connected one-dimensional skeleton. The signed trajectory count agrees with this attaching-map incidence calculation, so it computes integral rather than only mod-two homology.

For $m\ge 1$, taking kernels modulo images givesFor example, an odd interior-degree generator has zero outgoing differential but twice itself is the image from the next degree; an even positive-degree differential is injective. At the top there is no incoming differential, leaving a copy of $\mathbb{Z}$ exactly when $m$ is odd. For $m=0$ the space is one point and only $H_0=\mathbb{Z}$ remains. This recovers the cellular homology of real projective space by explicitly constructing its Morse-Smale complex and signs.

The smooth h-cobordism theorem states: if $(W;M_0,M_1)$ is a compact smooth h-cobordism, $\dim W\ge 6$, and its connected boundary manifolds are simply connected, then there is a diffeomorphism $W\cong M_0\times [0,1]$ restricting to the prescribed identification on $M_0$. In particular $M_0$ and $M_1$ are diffeomorphic. Being an h-cobordism means that both inclusions $M_i\hookrightarrow W$ are homotopy equivalences; their fundamental groups then also identify with that of $W$.

The homotopy-equivalence condition is necessary for a trivial cobordism: each end of a cylinder is a deformation retract, by $(x,t)\mapsto (x,(1-s)t)$ for the incoming end and $(x,t)\mapsto (x,(1-s)t+s)$ for the outgoing end. Its relative homology with respect to either end is consequently zero. Compactness is also necessary for a product with a closed compact end. The dimensional bound and simple connectivity are not necessary for an individual cobordism to be a product. The cylinder $S^2\times [0,1]$ is a simply connected trivial cobordism of dimension three; and, in the theorem's dimension range, $(S^1\times S^4)\times [0,1]$ is a six-dimensional trivial cobordism with fundamental group $\mathbb{Z}$. These prove the two non-necessity assertions explicitly, while the inclusion argument proves necessity of the h-condition.

For the remaining claims, use a cobordism Morse function: its two regular boundary levels are zero and one, and all critical points are interior. An arbitrary function with unrelated boundary values would not have the relative handle-count property. The relative Morse chain complex for $(W,M_0)$ has one generator per critical point, in its Morse index. ThusA trivial cobordism has zero relative homology, hence zero relative Euler characteristic. Modulo two the displayed sum is the number of critical points. An odd number makes it odd, and therefore nonzero. This proves an odd-critical-point cobordism is not trivial.

With precisely two critical points of indices $\lambda \le \mu$, a trivial cobordism requires the two-generator relative complex to be acyclic over $\mathbb{Z}$. Equal indices leave a rank-two homology group; a gap of more than one leaves both generators with zero possible differential. Therefore $\mu =\lambda +1$ is necessary. The remaining two-term complex isChoose a transverse gradient-like vector field. A nonzero differential requires the index-$\mu$ point to be above the index-$\lambda$ point; otherwise no descending connecting trajectories exist. In a common regular level between them, the belt sphere $S_U(p)$ has dimension $d-\lambda -1$, and the attaching sphere $S_L(q)$ has dimension $\mu -1=\lambda$, where $d=\dim W$. Their dimensions add to $d-1$, the level's dimension, so their transverse intersection consists of points. Each intersection is one descending trajectory from $q$ to $p$, modulo time translation. Orient the attaching sphere and use an orientation of a vector space on the handle core to coorient the belt sphere; the resulting signed count is the cellular boundary coefficientThe two homology groups are the kernel and cokernel of multiplication by $a$. They vanish precisely when $a$ is a unit of $\mathbb{Z}$. Consequentlyare necessary. This proves the two-critical-point obstruction to a trivial cobordism; it makes an algebraic incidence statement and does not assume that an algebraic count of one automatically supplies geometric handle cancellation.

For an explicit nontrivial cobordism from $S^3$ to itself, attach two zero-framed two-handles to the outgoing end of $S^3\times [0,1]$ along the standard Hopf link. Its two components are unknots with linking number one. This realizes two successive $(2,2)$ surgeries, as suggested by the hint. The outgoing boundary is again $S^3$, and this can be checked geometrically: the Hopf-link exterior is $T^2\times [0,1]$, and each component's zero-framing longitude is the other's meridian. Filling along these two slopes glues two solid tori with meridians intersecting once, the standard genus-one Heegaard decomposition of $S^3$. Thus zero surgery on the Hopf link does return the three-sphere, not just an unspecified homology sphere.

The trace has only the two added index-two handles. Its relative cellular complex has $C_2(W,S^3)={\mathbb{Z}}^2$ and all other groups zero, soIt cannot be a product cylinder. Equivalently this trace is $S^2\times S^2$ with two disjoint open four-balls removed: the usual handle presentation of $S^2\times S^2$ has one zero-handle, the two zero-framed Hopf-linked two-handles, and one four-handle. Removing the bottom and top balls exposes the two three-sphere ends. The relative-homology calculation alone already proves the requested cobordism is nontrivial.

Enforce the unit-vector constraint using a Lagrange multiplier $\Lambda (x)$:Integration by parts in a compactly supported variation gives $-\Delta {\varphi }^a+\Lambda {\varphi }^a=0$. Contracting with ${\varphi }^a$ gives $\Lambda ={\varphi }^b\Delta {\varphi }^b$, henceDifferentiating ${\varphi }^a{\varphi }^a=1$ twice also gives $\varphi \cdot \Delta \varphi =-|\nabla \varphi {|}^2$, so an equivalent equation is $\Delta \varphi +|\nabla \varphi {|}^2\varphi =0$. These are the harmonic map equations for the O3 nonlinear sigma model.

For arbitrary maps, the printed finite-energy compactification assertion is too strong. The following counterexample shows that finite sigma-model energy need not give a limit at infinity. Outside a large disk, setInterpolate smoothly to a constant inside the disk. There is no angular dependence, and the exterior energy isNevertheless the field keeps rotating and has no limit as $r\to \infty$. Thus energy integrability alone does not prove a continuous map on the one-point compactification.

The usual sigma-model lump sector adds the boundary condition $\varphi (x)\to {\varphi }_{\infty }\in S^2$ uniformly as $|x|\to \infty$. Defining the value at the added point to be ${\varphi }_{\infty }$ then gives a continuous map $S^2\to S^2$: neighborhoods of the point at infinity are precisely complements of compact subsets of the plane, and uniform convergence supplies continuity there. For the topological degree calculation using differential forms below, take a smooth compactified map, as in the regular lump sector. This boundary/regularity information is additional to finite energy for unrestricted fields.

Write $a_i={\partial }_i\varphi$ and $b_i={ϵ}_{ij}\varphi \times {\partial }_j\varphi$. Since $\varphi \cdot {\partial }_i\varphi =0$ and $|\varphi |=1$,Also the given double-index charge reduces toFor $s=\pm 1$, expanding the nonnegative square givesTherefore the Bogomolny degree bound for the O3 sigma model isThe charge integral is absolutely convergent already for finite-energy fields, because $|\varphi \cdot ({\partial }_1\varphi \times {\partial }_2\varphi )|\le \frac{1}{2}(|{\partial }_1\varphi {|}^2+|{\partial }_2\varphi {|}^2)$. The inequality itself consequently does not need the compactification hypothesis. Equality requires ${\partial }_i\varphi +s{ϵ}_{ij}\varphi \times {\partial }_j\varphi =0$ for $s$ chosen to have the sign of $Q$.

For the topological interpretation, the standard oriented area form on the target sphere isIts pullback is ${\varphi }^{\ast }\omega =\varphi \cdot ({\partial }_1\varphi \times {\partial }_2\varphi )d{x}^1\wedge d{x}^2$. The topological degree is defined by ${\varphi }_{\ast }[S^2]=\mathrm{deg}(\varphi )[S^2]$ in $H_2(S^2;\mathbb{Z})\cong \mathbb{Z}$. Pairing with $\omega$ givesEquivalently, the integer is the signed number of preimages of a regular value. It is invariant under smooth homotopy: if $\Phi :S^2\times [0,1]\to S^2$ is that homotopy, then Stokes theorem and $d\omega =0$ make the difference between the two charge integrals equal to ${\int }_{S^2\times [0,1]}d({\Phi }^{\ast }\omega )=0$. This identifies the charge as the degree charge of an O3 sigma-model lump, rather than merely an arbitrary energy integral.

Use Euclidean coordinates $(x^1,x^2,x^3,x^4)$ and orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. For $D_{\mu }={\partial }_{\mu }+A_{\mu }$, the gauge curvature is $F_{\mu \nu }=[D_{\mu },D_{\nu }]$. The ASDYM equations are $F=-\ast F$, equivalentlyPut $w=x^1+i{x}^2$, $z=x^3+i{x}^4$, with ${\partial }_w=({\partial }_1-i{\partial }_2)/2$ and similarly for $z$. The bars label the complex conjugate coordinates on the Euclidean real slice; after complexification they can be treated as independent coordinates. A Lax pair for the anti-self-dual Yang-Mills equations isAt infinity rescale the operators and use the other affine coordinate. Their commutator isCompatibility for every $\lambda$ is therefore precisely $F_{wz}=F_{\bar{w}\bar{z}}=0$ and $F_{w\bar{w}}+F_{z\bar{z}}=0$. To verify the real component signs,The conjugate equation supplies the remaining real components. This is compatibility of an overdetermined system, not an additional field equation on $\Psi$ for only one value of the spectral parameter.

Impose invariance under translations in $x^1,x^2,x^3$, writing $t=x^4$. In a symmetry-adapted gauge the potentials depend only on $t$. A further $t$-dependent Yang-Mills gauge transformation sets $A_4=0$ locally, by solving the usual parallel transport ordinary differential equation; it does not reintroduce spatial dependence. ThenThe three real equations above become ${\dot{A}}_1=[A_2,A_3]$, ${\dot{A}}_2=[A_3,A_1]$ and ${\dot{A}}_3=[A_1,A_2]$, namelyThese are the Nahm equations. The orientation chosen above gives the requested plus sign; reversing orientation exchanges self-duality of gauge curvature and anti-self-duality of gauge curvature. Translation invariance here is a local dimensional reduction and is not a claim of finite four-dimensional action over all three translation directions.

To prove conservation, abbreviate $P=A_1+i{A}_2$, $Q=A_1-i{A}_2$ and $R=A_3$, so $A(\lambda )=P+2R\lambda -Q{\lambda }^2$ and $B(\lambda )=-iR+iQ\lambda$. Expanding the commutator, including the cancellation of the cubic term, givesThe Nahm equations giveThus the polynomial Lax representation of the Nahm equations isTake a finite-dimensional matrix representation of the Lie algebra and its complexification. For every positive integer $p$, differentiating the product and using cyclicity of the matrix trace givesThis proves the trace invariants of a Lax equation directly. Since $\mathrm{Tr}(A(\lambda {)}^p)$ is a polynomial of degree at most $2p$ in the affine coordinate $\lambda$, its derivative vanishes identically only if the derivative of every coefficient vanishes. Therefore every trace-polynomial coefficient is independent of $t$. At projective infinity these polynomials are sections of $\mathcal{O}(2p)$; they need not be constant functions of $\lambda$ on the whole projective sphere.

Choose the axial origin at the sphere's centre. Reflection $x\mapsto -x$ leaves the straight tube, the spherical geometry and the transverse displacement of the centre unchanged. It reverses the prescribed axial Hagen-Poiseuille flow, while it leaves a hypothetical transverse particle velocity unchanged. On the other hand, kinematic reversibility of Stokes flow says that reversing the imposed flow reverses every particle velocity, including that transverse component. The force-free and torque-free constraints are preserved by both operations. Uniqueness of the coupled Stokes flow and rigid-body motion therefore makes the transverse velocity equal to its own negative. It must vanish.

Consequently $\dot{c}=0$ as the particle is advected. This is no lateral migration of a single sphere in a uniform tube in Stokes flow. The argument uses the fore-aft symmetry of a rigid sphere and a uniform tube; it does not extend automatically to inertial migration or deformable particles.

For $n\ge 2$, the sphere $S^n$ is simply connected. The lifting criterion for a covering space says that a based map lifts to a covering whenever the image of its fundamental group lies in the subgroup corresponding to that covering. For the universal cover $p:{\mathbb{R}}^2\to T$, the sphere's trivial fundamental group meets this condition. Hence $f:S^n\to T$ has a continuous lift $\tilde{f}:S^n\to {\mathbb{R}}^2$.

Contract the lift linearly to any point $q\in {\mathbb{R}}^2$:This is a continuous homotopy from $f$ to the constant map $p(q)$. Thus every such sphere-to-torus map is null-homotopic. The restriction $n\ge 2$ is essential; circle maps may carry nonzero fundamental-group classes.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.