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The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.

First solve the hinted fixed-gap problem. Let $y$ measure distance across the gap and $H(x)=h+x^2/(2a)$ be its leading local height. The lubrication approximation gives $\mu u_{yy}=p_x$, with $u=0$ at $y=0,H$, soThe constant flux per unit transverse length is $q$. Integrating its pressure gradient gives the pressure-driven flux through a parabolic lubrication gap:With $x=\sqrt{2ah}t$, the integral is $\sqrt{2a}{h}^{-5/2}{\int }_{-\infty }^{\infty }(1+t^2{)}^{-3}dt$. The substitution $t=\tan \chi$ gives ${\int }_{-\pi /2}^{\pi /2}{\cos }^4\chi d\chi =3\pi /8$, and therefore

Now take the sphere speed to be $U>0$ downwards and assume zero net liquid through-flow, so the displaced liquid must pass through the annular gap. Near the equator, the sphere's surface radius is $R-z^2/(2R)+\cdots$, giving $H(z)=d+z^2/(2R)$. In the sphere frame the relative volume flux is $U\pi (R+d{)}^2$; dividing by circumference $2\pi R$ gives $q\simeq UR/2$. The Couette contribution from the moving tube wall is only order $Ud$, smaller by $d/R$, so the fixed-gap pressure result givesThe pressure changes principally within axial distance $\sqrt{Rd}$ around the equator. It is nearly constant on each hemisphere, so its leading force is the pressure difference times projected area, $F_p\simeq \pi R^2\Delta p$. Direct viscous shear contributes a relative order $d/R$ and is smaller. Balance with the buoyant weight:Thus the settling speed of a nearly occluding sphere without through-flow isThe printed speed lacks the factor $1/\pi$. The parabolic-gap integral above fixes the normalization: its $3\pi /8$ cannot be dropped. With standard dynamic viscosity and Stokes drag law conventions, the requested coefficient must be corrected by dividing the printed expression by $\pi$. Also, the bypass result assumes no net through-flow; allowing a piston-like displacement of the entire fluid column would require specifying tube end conditions and a different resistance balance.

A sphere of radius $d$ in an unbounded fluid has sedimentation speed $U_d=2\Delta \rho g{d}^2/(9\mu )$. ThereforeThe large confined sphere falls much more slowly even than this much smaller unconfined sphere.

For radial offset $e<d$, let $ϵ=e/d$. To leading order the narrowest gap varies as $d(\varphi )=d(1+ϵ\cos \varphi )$. All circumferential strips share the same pressure drop and act as parallel channels. Inverting the gap-resistance formula shows that each carries flux proportional to $d(\varphi {)}^{5/2}\Delta p$. At fixed buoyant load, the leading pressure drop is fixed, so the speed ratio isThe eccentricity increases narrow-gap bypass mobility result holds because the gain through the wider side outweighs the loss through the narrower side. Strict convexity of $x^{5/2}$ and the zero mean of cosine prove the inequality. Pairing the positive- and negative-cosine contributions in the derivative proves monotonicity with offset. The broader hint's assertion for every exponent $\alpha >0$ is not correct for $0<\alpha <1$, when concavity reverses the inequality; the required exponent $5/2$ is in the valid convex range.

Reflection of the sphere in the plane $Y=0$ sends $(X,Y,Z)$ to $(X,-Y,Z)$. It preserves dot products, hence the spherical distance between every pair of points. Its stereographic expression isThis is an isometry of the sphere but is not a Möbius transformation: at any finite point, its complex difference quotient is one along real increments and minus one along imaginary increments, so it is not holomorphic. A Möbius transformation is holomorphic away from its pole. The geometric distinction is that this reflection reverses orientation, whereas a Möbius transformation preserves it.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

For the unknot the three-manifolds are $S^3$, $U_0=S^1\times S^2$, and $U_n=L_n$, where $L_n$ denotes positive $n$-surgery with its surgery orientation. This fixes the orientation independently of competing conventions for the name $L(n,1)$.

Use the grading convention $\mathrm{gr}{\Theta }_{S^3}^{+}=0$ and $\mathrm{gr}{\Theta }_{S^3}^{-}=-2$. Let $R=\mathbb{Z}[U]$ and $\hat{R}=\mathbb{Z}[[U]]$. The ordinary minus groups arewhere $0\le k<n$ andThe lens-space absolute grading can be calculated from the standard torus surgery triple. For $0\le k<n$, the reverse trace structure with evaluation $2k-n$ has a small embedded holomorphic triangle with basepoint multiplicity zero and signed count $\pm 1$, taking $y_k$ to ${\Theta }^{-}$. Its degree shift is ${\Delta }_k=(1-(2k-n{)}^2/n)/4$. Thus $(d_k-2)+{\Delta }_k=-2$, which gives the boxed formula for $d_k$. Additional triangles wind around the surgery region and give the higher powers of $U$ calculated below.

Their infinity groups are obtained by inverting $U$: one Laurent tower for $S^3$ and each lens-space structure, two for the torsion structure on $S^1\times S^2$, and $\mathbb{Z}[U,U^{-1}]/(1-U^{|i|})$ in its non-torsion structures. Their plus groups are respectively ${\mathcal{T}}_0^{+}$, ${\mathcal{T}}_{1/2}^{+}\oplus {\mathcal{T}}_{-1/2}^{+}$, and ${\mathcal{T}}_{d_k}^{+}$. The hat groups are $\mathbb{Z}$ in degree zero, $\mathbb{Z}$ in each degree $\pm 1/2$, and $\mathbb{Z}$ in degree $d_k$. These follow directly by straightening the standard genus-one Heegaard diagrams: the lens space has one generator in each structure and no chain differential, while the two bigons in the torsion $S^1\times S^2$ diagram cancel with coherent signs. The sphere adjunction condition eliminates its non-torsion plus and hat groups, but not its ordinary minus group. The Non-torsion Heegaard Floer homology of the sphere-circle product calculation comes from a two-generator chain complex with chain differential $1-U^{|i|}$; its cokernel is the displayed quotient. For the triangle, complete the free minus towers by replacing $R$ with $\hat{R}$. The non-torsion summands disappear after completion, because $1-U^{|i|}$ has an inverse $1+U^{|i|}+U^{2|i|}+\cdots$ in $\hat{R}$.

For $k=0$, choose signs so that the exact triangle becomesThe zero-surgery trace has $\chi =1$, $\sigma =0$, and $c_1^2=0$ in the torsion structure; hence its absolute degree shift is $-1/2$, taking ${\Theta }^{-}$ to $x_{-}$. The auxiliary-generator map takes $x_{+}$ to $y_0$, so its effective grading shift isFor the underlying cobordism $Z_n$ in its torsion structure, $c_1^2=0$, $\chi =1$, $\sigma =0$, and its actual cobordism shift is $-1/2$. Tensoring with the canonical lens generator adds $d_0=(n-1)/4$ under the connected-sum grading convention, explaining the effective shift. For $n=1$ the auxiliary lens space is $S^3$, and this is simply the other degree-$-1/2$ surgery trace map.

The reverse trace components are explicit in every $k$. For $j=k+n\ell$, put $m_j=2j-n$. Its intersection form is $(-n)$, soIndeed $c_1^2=-m_j^2/n$, $\chi =1$, $\sigma =-1$, and matching source and target degrees gives precisely the displayed exponent. The exponent is nonnegative, and the coefficient is a unit: the infinity trace map is an isomorphism, and restriction to the free minus towers leaves the unique grading-compatible monomial.

With a compatible choice of Coherent orientations in Heegaard Floer homology, the sum $C_k$ is multiplication byFor $k=0$, pair $\ell$ with $1-\ell$: their exponents agree and their signs are opposite, so $P_0=0$, in agreement with the first triangle. For $0<k<n$ the term $\ell =0$ is the unique constant term, and every other exponent is positive. Thus $P_k=1+U(\cdots )$ is a unit in $\hat{R}$. The zero-surgery summand for such a $k$ vanishes, and the exact triangle isIt is isomorphic, as an ungraded completed-module triangle, to one with the final map the identity. The sum generally has no single absolute Heegaard Floer grading shift; each individual trace component has shift ${\Delta }_j$. This distinction and the power series explain why replacing the completed triangle by a polynomial-module triangle is not innocuous.

The geometric argument uses the Mixed Heegaard Floer invariant defined in Question 5(iii). First suppose $g(\Sigma )>0$; the genus-zero qualification is discussed below. Write $r=\Sigma \cdot \Sigma \ge 0$. Blow up $r$ distinct points of the surface. Its proper transform $\tilde{\Sigma }$ in $\tilde{M}=M\#r{\overline{\mathbb{CP}}}^2$ has classChoose the extension of the Spin-c structure withSince $E_j^2=-1$, its exceptional evaluations are $-1$, so the Blowup formula for the mixed Heegaard Floer invariant has no extra power of $U$. Thus ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}\ne 0$, andThis calculation is why the self-intersection appears with a plus sign.

It remains to bound the Chern pairing for a square-zero surface $S$. Its oriented normal bundle has Euler number zero and is trivial, so $\partial \nu S=S\times S^1$. Choose an embedded positive-square surface $T$ disjoint from $S$, with positive direction left in its complement, and cut along $N=\partial \nu T$. To see why the choice is possible, impose orthogonality to $[S]$ in the intersection form. A nonzero isotropic class uses at most one positive direction; $b_2^{+}\ge 3$ therefore leaves at least two. Choose an integral positive-square class in that complement, represent it by a surface, and remove algebraically cancelling intersections with $S$ by tubing along $S$. This alters genus but not the class. The neighborhood of $T$ supplies one positive direction and the complement supplies another. Since its normal Euler number is nonzero, $H^1(\nu T)\to H^1(N)$ is onto, proving that the cut is admissible.

Put the outgoing puncture inside $\nu S$, so that the plus Heegaard Floer cobordism map on the side containing $S$ passes through the Floer group of $\partial \nu S$. On the chain level this means splitting the cobordism along that boundary and composing its handle maps; with the fixed determinant class, every intermediate generator has restriction $\mathfrak{t}=\tilde{\mathfrak{s}}{|}_{S\times S^1}$. If the Chern pairing exceeded $2g(S)-2$, the given completed minus group would be zero. As in Question 3, this is the completed convention: for a non-torsion structure, ordinary infinity may retain $(1-U^N)$-torsion, whereas $1-U^N$ is invertible after completion. Such a pairing is positive for $g\ge 1$, so $\mathfrak{t}$ is non-torsion and the completed infinity group ${\mathbf{HF}}^{\infty }(S\times S^1,\mathfrak{t})$ is zero as well. The completed fundamental long exact sequence, whose plus group is the usual plus theory, then makes that plus group zero. The Heegaard Floer cobordism map, and consequently the Mixed Heegaard Floer invariant, would vanish, contradicting nonvanishing. ThereforeSubstituting the preceding blowup calculation provesReversing the orientation of the surface gives the usual absolute-value form of the Adjunction inequality for the mixed Heegaard Floer invariant.

The printed assumptions omit the positive-genus qualification. With the actual Floer groups, the asserted three-dimensional vanishing is false for $S=S^2$: the torsion structure on $S^2\times S^1$ has two nonzero minus towers, although its pairing is $0>-2$. Correspondingly, a small sphere embedded in a ball of a K3 surface has $[S]=0$, $S^2=0$ and Chern pairing zero, while the canonical Mixed Heegaard Floer invariant is nonzero; the desired inequality would read $0\le -2$. Thus the displayed proof establishes the intended positive-genus statement. If the supplied vanishing is instead taken literally as a formal additional axiom for every genus, the same factorization would force the inequality even for a sphere, but that axiom contradicts the actual $S^2\times S^1$ computation.

The map is a homogeneous Laurent polynomial ring homomorphism between the two rank-one infinity towers. Choose homogeneous generators $e_i$ with absolute degrees $q_i$. Its degree isConsequently the general graded-module description isIf the grading does not allow an integer exponent, the map is zero. Negative exponents are allowed in the infinity theory. The notation $F$ here denotes the map called ${\Phi }^{\infty }$ in the question.

The standard topological statements for Heegaard Floer cobordism maps, which require no proof in this part, areIn the second case $a=\pm 1$ and the map is therefore $\pm U^r$ with the exponent determined by the displayed degree formula. Since both boundary components are rational homology spheres, the rational intersection form is nondegenerate, so $b_2^{+}=0$ means that it is negative definite, allowing also the rank-zero form.

The printed hypothesis $b_1(Y_i)=0$ does not imply $b_1(W)=0$. Thus one must not replace the second implication by an unconditional isomorphism assertion for every negative-definite cobordism. A concrete check is $W=(S^1\times S^3)\setminus (B^4\bigsqcup B^4):S^3\to S^3$. It has $b_2^{+}=0$ and $b_1(W)=1$. A one-handle followed by a three-handle gives $e\mapsto e\otimes {\theta }^{+}$ and then kills the ${\theta }^{+}$ summand; hence its infinity map is zero. Its grading shift is one, also incompatible with a nonzero map between the even-graded $S^3$ towers. With only the stated boundary assumption, the general homogeneous Laurent-module description and the two topological implications above are the valid conclusions; determining any remaining coefficient requires the actual cobordism, not just its boundary Betti numbers.

Let $p=|a|$, $q=|b|$ and $r=|ab|$ denote element orders. Remove three points from an oriented sphere. Its fundamental group has peripheral generators ${\gamma }_a,{\gamma }_b,{\gamma }_c$ with ${\gamma }_a{\gamma }_b{\gamma }_c=1$. Send them to $a,b,(ab{)}^{-1}$. Since $a,b$ generate $T$, the kernel defines a connected regular covering of the punctured sphere with degree $|T|$ and deck transformation group $T$.

Complete the lifted punctures by disks. Above the three removed points the monodromy cycles have lengths $p,q,r$, respectively, and the disk projection has the local form $z\mapsto z^p$, $z^q$ or $z^r$. This gives a closed connected oriented topological surface $M$ and an orientation-preserving $T$-action. At the three kinds of filled-in points the stabiliser subgroups are conjugates of $⟨a⟩$, $⟨b⟩$ and $⟨ab⟩$, respectively. Every other point has trivial stabiliser subgroup. Orders equal to one simply correspond to unbranched filled points.

The punctured sphere has Euler characteristic $-1$, so its cover has Euler characteristic $-|T|$. Filling the ends adds $|T|/p+|T|/q+|T|/r$ disks. This proves the formula for the branched surface action from two generators:

For a subgroup $U$, its quotient map is always regular as a branched quotient. It is an ordinary unbranched normal covering map exactly when $U$ acts freely, that is,Necessity follows from the listed stabiliser subgroups. Conversely, these conditions eliminate every possible nontrivial point stabiliser. A free finite action gives a covering space, and its deck transformations are $U$, acting transitively on each fibre; hence the cover is normal. In this caseThe condition $U⊴T$ concerns regularity of the different map $U\backslash M\to T\backslash M$; it is not needed for $M\to U\backslash M$ when the action of $U$ is free.

For the specified order-$96$ group, $p=q=3$. Set $c=ab$ and $z=c^3$. Since $z$ has order two, $c^6=1$ and $c^3\ne 1$, so $|c|$ is either two or six. The order-two alternative would make the displayed formula give $\chi (M)=96(1/3+1/3+1/2-1)=16$. That is impossible for a connected closed oriented topological surface, whose Euler characteristic is $2-2g\le 2$. Thus $|c|=6$, and

Each $U_i$ has order eight. By Lagrange's theorem, it contains no element of order three or six, so it meets all conjugates of $⟨a⟩$ and $⟨b⟩$ trivially. In $⟨c⟩$, the only nonidentity element whose order divides eight is $z=c^3$. Since $z$ is central, it is also the unique such element in every conjugate of $⟨c⟩$, and it is excluded from both $U_i$ by hypothesis. Both subgroup actions are therefore free. The quotients are closed oriented surfaces with

The branched-cover construction supplies a smooth structure, with the stabilisers acting by rotations in the disk charts. Average a smooth Riemannian metric over $T$ to make the whole action isometric. The Gassmann equivalence of $U_1,U_2$ then gives isospectral manifolds: for every upstairs eigenspace with character ${\chi }_{\lambda }$, its downstairs multiplicities are $|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\lambda }(u)$, and these are equal because the two subgroups have equal intersection sizes with every conjugacy class. This proves the spectral assertion for any such invariant metric.

To guarantee nonisometry, choose the invariant metric with curvature markers distinguishing finite-cover quotients. Here are the geometric details, since nonconjugacy alone would not guarantee nonisometry for an arbitrary metric. Choose a small disk in the free locus of the $T$-action, with all its translates disjoint. In that disk choose three noncollinear points in a geodesically convex neighbourhood. Supported conformal smooth bump functions can give them three distinct high values as strict Gaussian curvature maxima, occurring nowhere else in the unmodified metric. Arrange these three maxima as one tight cluster, much closer to each other than to any translated cluster, and copy the same modification to every translate. Curvature and its first few derivatives can be adjusted by the second and higher derivatives of the conformal factor; narrow bumps give high distinguished curvature with small change in metric distances. Further small supported perturbations remove accidental maxima of these exact heights. Thus the resulting metric remains smooth and $T$-invariant, and its marked clusters are intrinsic and uniquely labelled by their three curvature values.

Any proposed quotient isometry $F:U_1\backslash M\to U_2\backslash M$ must carry a whole labelled cluster to another. Lift its restriction near that cluster to the corresponding two marked disks in $M$, and identify these disks using the appropriate group element $t\in T$. The resulting local isometry fixes all three marked points. Its differential at the first point fixes the two independent directions of the unique geodesics to the other two; hence it is the identity, and the local isometry itself is the identity near that point. If $p_i:M\to U_i\backslash M$ are the quotient maps, we therefore haveon an open set. Both sides are local isometries on the connected complete Riemannian surface $M$. A local isometry is determined along each geodesic by its value and differential at the initial point, so the equality propagates to all of $M$.

For $u\in U_1$, this equality implies $p_2\circ tu=p_2\circ t$. Evaluating at a free point of the $T$-action shows $tu{t}^{-1}\in U_2$; equivalently these maps differ by a deck transformation of $p_2$. Thus $t{U}_1{t}^{-1}\subseteq U_2$, and equality of subgroup orders gives equality. This contradicts their assumed nonconjugacy. For this marked invariant metric the two quotients are nonisometric, isospectral Riemannian surfaces of genus two. No constant-curvature requirement is imposed or used.