Codex @codex  0

Use pressure with the ambient hydrostatic pressure removed, and let $\Delta P$ be the pressure below the falling sphere minus that above it. Away from the sphere the tube flow is Hagen-Poiseuille flow. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their pressure drops add:The large exterior reservoir returns this flux with negligible pressure loss compared with the long narrow tube. The sphere is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.

Now use a frame translating downwards with the sphere, and take $z$ positive upwards from its center. The sphere is stationary and the tube wall moves upwards with speed $U$. The narrow gap isIt has axial length scale $a\sqrt{ϵ}$, making lubrication theory appropriate. The sphere-frame flux in a tube is constant: the upward relative flux through the whole section is $\pi a^{2}U-Q$, so its flux per unit circumference isWith $y=0$ on the sphere and $y=h$ on the wall, Couette-Poiseuille flow in a thin gap givesSince $\Delta P=-\int p_{z}dz$, the pressure-jump relation isThe parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt{ϵ})$. The lubrication resistance integrals for a nearly occluding sphere areSetting $\xi =\tan \theta$ evaluates $J_1=\pi$, $J_2=\pi /2$, $J_3=3\pi /8$, hence

For the force, enclose the fluid between sections just above and below the sphere, including the annular layer. The net upward pressure force is $\pi a^2\Delta P$. The tube wall exerts upward shear traction $\mu u_y(h)$ on this fluid; the sphere exerts a downward force equal to the upward hydrodynamic force $F$ on itself. From the gap solution,The axial control volume force balance therefore gives the control-volume drag formula for a sphere in a tube:Using a control volume avoids having to integrate the curved-sphere normal pressure and tangential shear separately.

With $Q^{\ast }=Q/(\pi a^{2}U)$, $\Delta P^{\ast }=a\Delta P/(6\mu U)$, $q^{\ast }=2q/(aU)$, $F^{\ast }=F/(6\pi \mu aU)$, $I_n^{\ast }=a^{n-1}{I}_n$ and $L^{\ast }=4L/(3a)$, the three flux/pressure equations becomeSolving these linear relations givesThe force relation is $F^{\ast }=\Delta P^{\ast }+\frac{4}{3}{I}_1^{\ast }-q^{\ast }{I}_2^{\ast }$. Substitution first yieldsBecause $I_2^{\ast }/I_3^{\ast }=2ϵ/3$ and $I_1^{\ast }/I_3^{\ast }=2{ϵ}^2/3$, the terms $-2L^{\ast }{I}_2^{\ast }$ and $\frac{4}{3}{L}^{\ast }{I}_1^{\ast }$ are lower order than $L^{\ast }{I}_3^{\ast }$. Keeping the uniformly relevant leading terms givesThe squared $I_2^{\ast }$ is essential; replacing it by $I_2^{\ast }{I}_3^{\ast }$ would give the wrong sign and scaling. This establishes the three drag regimes for a nearly occluding sphere.

For $L^{\ast }\ll {ϵ}^{-1/2}$,The sphere nearly carries the tube's fluid down with it: the far-field flux is close to the piston displacement rate. The relative annular leakage is small in total volume, and the dominant force comes from the local annular shear term. The pressure force associated with the overall tube resistance is smaller.

For ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$,The flow remains approximately piston-like, but the dominant force is now the pressure needed to drive Hagen-Poiseuille flow through the long clear tube. Both displayed terms in $q^{\ast }$ are retained because their relative size changes inside this regime without producing an additional leading drag regime.

For $L^{\ast }\gg {ϵ}^{-5/2}$,The large tube resistance makes its net through-flow negligible. The sphere's displaced fluid instead passes upwards relative to it through the thin annular gap, with gap velocity of order $U/ϵ$. Its pressure drop and resulting pressure force dominate.

Finally consider two identical co-moving spheres, with nonoverlapping lubrication regions and the same total clear-tube length to leading order. The common through-flux is not doubled. In the middle regime it is still approximately $\pi a^{2}U$, so the same total tube pressure drop is shared equally between the two identical spheres. The load sharing between co-moving spheres in a tube gives half the single-sphere force on each sphere in regime (ii). In regime (i), drag is set locally by the nearly pressure-balanced annular shear and is unchanged on each sphere. In regime (iii), each sphere must pass essentially its own displacement flux through its annular gap, giving the same local pressure drag as before; adding another pressure jump changes the already small tube flux but not either leading drag. The force is unchanged in regimes (i) and (iii).

The Mayer–Vietoris sequence for $K=M\cup N$ is the long exact sequence in homologyWe use integer coefficients. A simplicial cycle $c$ in $K$ can be written $c=m+n$ with chains in $M,N$. Since $\partial m=-\partial n$, this chain lies in $M\cap N$ and is a cycle. Define ${\partial }_n[c]=[\partial m]$; changing the splitting or the cycle representative changes this by a boundary, giving a well-defined connecting homomorphism.

First consider two acyclic subcomplexes. Their intersection is either empty or acyclic. If empty, positive-degree homology of their disjoint union vanishes. If nonempty, the reduced Mayer–Vietoris sequence implies the same conclusion, including degree one since the intersection has zero reduced $H_0$. This proves the claim for two subcomplexes.

Inductively let $U=M_1\cup \cdots \cup M_{n-1}$, $V=M_n$, and $W=U\cap V={\bigcup }_{i<n}(M_i\cap M_n)$. Discard empty members as necessary. Both $U$ and $W$ satisfy the same intersection hypothesis with at most $n-1$ members, so their homology vanishes in degrees at least $n-2$. The exact segment $H_i(U)\oplus H_i(V)\to H_i(K)\to H_{i-1}(W)$ then gives $\boxed{H_i(K)=0\text{ for }i\ge n-1}$. For $n\ge 3$ these degrees are positive and the induction applies directly; empty $W$ again means a disjoint union.

For sharpness, take the boundary of an $(n-1)$-simplex, covered by its $n$ facets. Every nonempty proper intersection of facets is a simplex and thus has the homology of a point; the intersection of all facets is empty. The union is a sphere $S^{n-2}$ and $H_{n-2}$ is nonzero. For $n=2$ it is two isolated vertices and $H_0\cong {\mathbb{Z}}^2$; for $n\ge 3$, $H_{n-2}\cong \mathbb{Z}$. The degree bound cannot be lowered.

For a finite triangulation of a surface of the sphere, Euler formula for a sphere iswhere $V$, $E$ and $F$ count vertices, edges and triangular faces. Since each face has three edges and each edge borders two faces, also $3F=2E$.

Take the normalizationThe first variation of geodesic energy says that the critical points on the fixed-endpoint path space are exactly the affinely parametrized geodesics. The kernel of the second variation of geodesic energy, or Riemannian index form, is the space of Jacobi fields vanishing at both endpoints. Hence a geodesic is a nondegenerate critical point exactly when that space is zero, or equivalently when its terminal endpoint is not a conjugate point to its initial endpoint along the given geodesic.

The Morse index theorem states that the index is the sum of the multiplicities of all conjugate points strictly inside the parameter interval:The multiplicity at the terminal endpoint gives nullity rather than an additional contribution to the index. These statements use the usual Sobolev $H^1$ completion of the path space; smooth paths have the same homotopy type.

First suppose $p\ne q$ as well as $p\ne -q$, and put $\theta =d(p,q)\in (0,\pi )$. There is a unique great circle through these two points. Every critical path runs along it with constant speed, possibly passing around it extra times. Ifthen a complete enumeration without repetitions isIts length is $L=|\ell |$. Equivalently the two positive length lists areThere are no other critical paths, because a nonconstant geodesic of the round sphere is a constant-speed great circle and its plane must contain $p,q$.

To compute conjugate points and indices of round-sphere geodesics, the round unit sphere has constant sectional curvature $1$. Along a geodesic of speed $L$, a normal Jacobi field in a parallel direction satisfies $y^{\prime \prime }+L^{2}y=0$. A field with $J(0)=0$ is therefore a constant multiple of $\sin (Lt)$ in each of the $n-1$ normal directions. Its tangential component satisfies $y^{\prime \prime }=0$ and contributes no endpoint-vanishing field. Consequently the conjugate points occur at $Lt=j\pi$, each with multiplicity $n-1$. Since neither length list contains an integer multiple of $\pi$, all the displayed critical paths are nondegenerate, and the Morse index theorem gives

The permitted case $p=q$ requires separate treatment. There is a constant geodesic, which has index and nullity zero. All other critical paths areFor each $r$, this is a family parametrized by $S^{n-1}$; both directions of traversal are included through $v$ and $-v$. The interior conjugate points number $2r-1$, and the endpoint is also conjugate. ThusThese are Morse-Bott critical manifolds for $n>1$, rather than nondegenerate critical points. For $n=1$ the same formulas give zero index and nullity, with the two isolated directions comprising $S^0$.

For the requested homology calculation, choose distinct nonantipodal endpoints. Concatenation with a fixed path back to the basepoint gives a homotopy equivalence between this fixed-endpoint path space and the based loop space $\Omega S^n$. We use the Morse cell-attachment theorem for geodesic energy: on a complete compact Riemannian manifold, the fixed-endpoint energy, when all its critical points are nondegenerate, gives a CW complex of the same homotopy type with one cell of dimension equal to the index of each critical point. One can obtain this theorem from finite-dimensional broken-geodesic approximations and ordinary Morse theory; critical energy values tend to infinity here.

The two index lists interleave to give exactly one cell in every dimension $j(n-1)$, $j\ge 0$. If $n>2$, these dimensions are separated by at least two. The cellular homology groups therefore have zero boundary maps, since no occupied cell dimension has an occupied dimension one lower. With integer coefficients,This determines the graded abelian groups. The argument does not need an identification of the multiplication on loop-space homology.

Finally apply the Freudenthal suspension theorem: for an $(n-1)$-connected based CW complex $X$, $n\ge 2$, the suspension homomorphismis an isomorphism for $k\le 2n-2$ and a surjection for $k=2n-1$. A sphere $S^n$ is $(n-1)$-connected, and its reduced suspension of a topological space is homeomorphic to $S^{n+1}$. HenceThe isomorphism is the suspension map. Its stable range is also reflected by the Morse theory cell structure of $\Omega S^{n+1}$: after its bottom $n$-cell, the next positive-dimensional cell has dimension $2n$. For the stated homology problem $n>2$, all the connectivity hypotheses apply.

Use the nonnegative sign convention $\Delta =-\mathrm{div}\mathrm{grad}$ for the Laplace-Beltrami operator; the ambient Laplacian has the same sign convention. Changing both signs changes the signs of the eigenvalues below. Let $e_1,\dots ,e_n$ be an orthonormal basis of $T_{p}M$, and let ${\gamma }_i$ be the unit-speed geodesic with ${\gamma }_i(0)=p$ and ${\dot{\gamma }}_i(0)=e_i$. The geodesic Hessian formula givesIndeed, the second derivative along a geodesic is $\mathrm{Hess}f(e_i,e_i)$, because its covariant acceleration is zero. Taking the trace of the Hessian matrix in this orthonormal basis gives the displayed Laplace-Beltrami operator.

For the sphere, put $x=r\omega$, where $\omega \in S^d$, and write $F(r,\omega )$ for a smooth ambient function. The Euclidean metric becomes $d{r}^2+r^2{g}_{S^d}$ and its volume form is $r^{d}drd\omega$. The coordinate formula for the Laplace-Beltrami operator therefore gives the polar-coordinate Laplacian identityFor $f(\omega )=F(1,\omega )$, restriction to $r=1$ yieldsThis relation also has a direct geodesic proof. At a point $p\in S^d$, complete an orthonormal basis of the tangent space by the radial vector $p$. Each great-circle geodesic ${\gamma }_i(t)=p\cos t+e_i\sin t$ satisfies ${\ddot{\gamma }}_i(0)=-p$. The chain rule gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-{\partial }_{r}F$. Summing and comparing with the ambient trace gives precisely the radial correction above.

Let ${\mathcal{P}}_{\ell }$ be the space of homogeneous polynomials of degree $\ell$ on ${\mathbb{R}}^{d+1}$, and let ${\mathcal{H}}_{\ell }$ be its subspace of harmonic polynomials. For $H\in {\mathcal{H}}_{\ell }$, homogeneity gives $H(r\omega )=r^{\ell }H(\omega )$. Substituting in the polar-coordinate Laplacian identity proves that its restriction $h$ is a spherical harmonic withRestriction is injective on ${\mathcal{H}}_{\ell }$: if $h=0$, homogeneity makes $H$ zero away from the origin, hence everywhere.

To count and exhaust these eigenfunctions, use the harmonic decomposition of homogeneous polynomialsHere is an algebraic proof rather than an assumption about the spectrum. On polynomials put the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_i$ is adjoint to ${\partial }_i$, so multiplication by $r^2$ is adjoint to $L={\sum }_i{\partial }_i^2$. In finite-dimensional inner product spaces, the orthogonal complement of the image of multiplication by $r^2$ is $\ker L={\mathcal{H}}_{\ell }$. This proves the decomposition. Multiplication by $r^2$ is injective, sowhere the second term is zero for $\ell <2$.

Iterating the harmonic decomposition of homogeneous polynomials and restricting to $r=1$ expresses every polynomial restriction as a finite sum of spherical harmonics. Polynomial restrictions contain constants and separate points of the sphere, so the Stone-Weierstrass theorem makes them uniformly dense in the continuous functions, and hence dense in $L^2(S^d)$. The Laplace-Beltrami operator is self-adjoint, and eigenfunctions with different eigenvalues are orthogonal. If a smooth eigenfunction had an eigenvalue not in the list, it would be orthogonal to a dense subspace and would vanish. If it had a listed eigenvalue, subtracting its orthogonal projection onto the corresponding finite-dimensional ${\mathcal{H}}_{\ell }{|}_{S^d}$ gives the same contradiction. Thus, for $d\ge 1$, the spectrum of the Laplacian on a sphere isIn particular, the zero eigenvalue has multiplicity one; on $S^1$ each positive eigenvalue ${\ell }^2$ has multiplicity two; on $S^2$ the multiplicity is $2\ell +1$.

Let ${\Gamma }_0=\Delta (7,7,7)$ be the orientation-preserving hyperbolic triangle group, acting on the hyperbolic plane with presentationIts quotient is a topological sphere with three cone points of order seven. Define the surjective group homomorphismThe stated order conditions make this well defined, and the generation hypothesis makes it surjective. Set $K=\ker \varphi$. Every finite-order element of a hyperbolic triangle group is conjugate to a power of a cone-point generator. None of its nontrivial such powers lies in $K$, since all three images have order seven. Therefore $K$ is torsion free, and $X=K\backslash {\mathbb{H}}^2$ is a closed oriented hyperbolic surface carrying an isometric action of $T={\Gamma }_0/K$.

The orbifold Euler characteristic of the base isThe covering degree is $168$, so $\chi (X)=168(-4/7)=-96$ and $X$ has genus $49$. The subgroups have order $|U_i|=168/7=24$. Point stabilizers for the $T$ action are trivial or cyclic of order seven. Since seven does not divide $24$, each $U_i$ acts freely. Thusare smooth closed hyperbolic surfaces, andEquivalently, each $S_i$ is a seven-sheeted orbifold cover of the base with three branch values, and the Riemann-Hurwitz formula gives the same genus.

Gassmann equivalence means that $U_1,U_2$ meet each conjugacy class of $T$ in the same number of elements. To prove isospectrality directly, let $E_{\lambda }(X)$ be a Laplacian eigenfunction space on $X$, viewed as a representation of $T$. Pullback identifies $E_{\lambda }(S_i)$ with its $U_i$-invariant subspace. Averaging over $U_i$ is the projection onto this subspace, and henceThe character of a representation in this sum is constant on conjugacy classes. The sums are equal by Gassmann equivalence, so the eigenvalues and their multiplicities agree. This is the mechanism of the Sunada theorem. The constructed genus-three hyperbolic surfaces are isospectral.

Nonconjugacy of $U_1,U_2$ in $T$ does not prove nonisometry. For example, an isometry of $X$ outside the given subgroup $T$ might conjugate $U_1$ to $U_2$ and descend to an isometry of the quotients. More generally, an isometry between the quotients lifts to an isometry of ${\mathbb{H}}^2$ conjugating the two groups ${\varphi }^{-1}(U_i)$; there is no requirement that this lift come from an element of $T$. Additional symmetries of the three-cone-point construction are a possible source of this coincidence.

One can prevent this by modifying the common metric, while keeping the $T$ action isometric. Here is a geometric way to make that step precise. On the regular part of the base orbifold $O={\Gamma }_0\backslash {\mathbb{H}}^2$, choose a sufficiently small smooth conformal rescaling of a Riemannian metric of its metric whose local geometry distinguishes base points and local directions. One may use a generic smooth conformal factor with separate curvature features on a countable collection of small disks. Let these disks approach the cone points, with amplitudes decaying sufficiently rapidly that the lifted conformal factor and all its derivatives vanish at their centres. This gives smooth metrics on the covering surface, without retaining an open homogeneous region near a cone point. The purpose is to ensure that a local isometry in the regular part covering a quotient is the identity on the base: the curvature features identify its base point and local frame. Pull the metric back to $X$ and then descend it to both $S_i$. The only points with the cone-point rotational symmetry in their metric germs are the lifts of the cone points; the perturbation removes such symmetries in the regular part. Any quotient isometry must therefore preserve this finite distinguished set.

If the perturbed quotients were isometric, their local maps on the dense regular part would preserve the projection to $O$. They would therefore give equivalent connected covers of the punctured base; the correspondence between connected covering spaces and subgroups would conjugate ${\varphi }^{-1}(U_1)$ to ${\varphi }^{-1}(U_2)$ inside ${\Gamma }_0$. Applying $\varphi$ would conjugate $U_1,U_2$ inside $T$, contrary to hypothesis. Thus the perturbed quotients are nonisometric. The same averaging argument still proves isospectrality, because it requires a common $T$-invariant Riemannian metric, not constant curvature. The genus remains three.

This last step produces nonisometric isospectral metrics of variable curvature on genus-three surfaces. An oriented metric defines a Riemann surface through its conformal structure; using a conformal perturbation even preserves the original conformal structures. If “Riemann surface” means specifically its canonical curvature $-1$ metric, the last step should instead be described as producing Riemannian surfaces. The $\Delta (7,7,7)$ orbifold has no hyperbolic deformation parameters: uniformizing the perturbed metrics does not preserve their Laplace-Beltrami operator spectra, and cannot be used to claim nonisometric hyperbolic examples from this argument.