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Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

Eight nearest neighbours at cube corners around a central site; fluorite structure cations have this arrangement.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

The boundary of a cube has eight vertices, twelve edges and six square faces. Treating each open face as a two-dimensional topological cell givesThis agrees with its being a two-sphere: stretching the hollow polyhedral surface changes its geometry but not its topology. The interior is excluded from this space.

Now include the solid interior. The cube has the same lower-dimensional topological cells, together with one open three-dimensional topological cell. The cell-count formula for Euler characteristic becomesThe filled cube is contractible, so its homology agrees with that of a point. The term $-1$ is essential: $V-E+F$ by itself is the full count only for a two-dimensional CW complex.

Adding a diagonal within one square face replaces that face by two triangular faces and introduces one edge. Thus the Euler characteristic changes by $-1+1=0$, as in subdivision invariance of Euler characteristic. Including an interior is a different operation: it adds points to the space, rather than giving the same space a finer description. This distinction matters whenever a geometric model could mean either a surface or the solid it encloses.

A freely floating body at rest has no supporting contact or applied holding force. Neglecting surface tension and the buoyancy of air, its upward buoyancy balances its downward force from gravity:For a body of uniform mass density ${\rho }_{\mathrm{body}}$ and total volume $V_{\mathrm{body}}$, its mass is $m={\rho }_{\mathrm{body}}{V}_{\mathrm{body}}$. Substitution into Archimedes' principle givesPartial immersion is possible when ${\rho }_{\mathrm{body}}<{\rho }_f$. A nonuniform body uses its mean mass density $m/V_{\mathrm{body}}$ instead; for a hollow body this volume must count the enclosed space that excludes water. This explains how the amount of displaced water can support a vessel whose walls alone are denser than water. ^[ref]

For the $600\mathrm{kg}{\mathrm{m}}^{-3}$ cube in water of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$, the displaced fraction is $600/1000=0.60$: sixty percent of its volume lies below the surface. With a horizontal top and bottom, the immersed depth $d$ satisfies $V_{\mathrm{displaced}}=Ad$, soIts remaining height is $0.040\mathrm{m}$, and the buoyancy is $1000\times 9.81\times 0.000600=5.886\mathrm{N}$, exactly balancing its downward force from gravity.

Consider the vector field $\mathbf{F}=(p(x),q(x),r(z))$ in the open unit cube, where $p$ and $r$ extend to continuously differentiable functions on $[0,1]$, and $q$ is continuous on $(0,1]$ with ${\int }_0^1|q(x)|dx<\infty$. Even if $q$ is unbounded as $x\downarrow 0$, the normal component on an $x$-face has the finite trace $\pm p(x)$. The two $y$-faces have finite improper integral fluxes $\pm {\int }_0^{1}q(x)dx$ and therefore cancel. The remaining face contributions giveTo justify this use of the divergence theorem, first apply it to $[\epsilon ,1]\times [0,1{]}^2$, on which the field is regular. Its divergence is $p^{\prime }(x)+r^{\prime }(z)$, independent of $q$. Its outward flux is $p(1)-p(\epsilon )+(1-\epsilon )[r(1)-r(0)]$. Taking $\epsilon \downarrow 0$ yields the displayed answer, and absolute convergence makes the two $y$-face limits finite separately. For example, $q(x)=1/[x(1+(\ln x{)}^2)]$ diverges at zero butThis argument defines the normal traces and the face integrals explicitly; it does not assert that the whole vector field has a finite value on the singular face.

Hydrostatic pressure is the pressure distribution in a resting fluid, balancing its body force from gravity. Take $z$ upward and a uniform downward gravitational field $\mathbf{g}=-g{\mathbf{e}}_z$, where $g>0$. The vertical pressure gradient satisfiesA short derivation balances the vertical forces on a thin horizontal layer of area $A$ and thickness $\Delta z$: $p(z)A-p(z+\Delta z)A-{\rho }_{f}gA\Delta z=0$. Dividing by $A\Delta z$ and taking the derivative gives the equation above. For constant mass density ${\rho }_f$, integration from the surface at $z_s$ givesThus pressure increases with depth $h$. For the stipulated water with mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ and gravitational acceleration $9.81\mathrm{m}{\mathrm{s}}^{-2}$, depths $0.20\mathrm{m}$ and $0.30\mathrm{m}$ give gauge pressures $1962\mathrm{Pa}$ and $2943\mathrm{Pa}$. Their difference produces the buoyancy of the example cube. ^[ref]

For a uniform cube of side $s$ and mass $M$, reflection symmetry makes central off-diagonal inertia components zero. Equal coordinate second moments are $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=s^2/12$, so each diagonal component is $M{s}^2/6$. Thus every central axis is a principal axis. A nonspherical shape can have an isotropic central inertia tensor: that tensor measures only quadratic moments of the mass distribution, not the whole shape.

For $0<\epsilon <1/2$, these inequalities define a deformed cube with $2^n$ vertices of a polytope. Choose either the lower or upper bound recursively at each coordinate to obtain a vertex of a polytope; changing one choice gives an adjacent vertex of a polytope. When maximizing $x_n$, smallest-index improving-facet pivoting visits all $2^n$ vertices of a polytope: first traverse the lower last-coordinate face, then move to the upper face and traverse the preceding path in reverse. Thus this geometric version of the Bland pivoting rule can require $2^n-1$ pivots. Rescaling the coordinate $x_i$ to $y_i={\epsilon }^{2(n-i)}{x}_i$ makes the positive rate per unit freed coordinate equal to ${\epsilon }^{-(n-i)}$; Dantzig pivot rule then selects the same exponential path. Euclidean edge-length pricing is a different rule.

For every positive integer $n$, the divisor sum of the Möbius function isThe sum includes all positive integer divisors, but only squarefree divisors contribute. Equivalently, $\mu \ast 1=\epsilon$ under Dirichlet convolution, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. ^[ref]

Each squarefree divisor chooses a subset of the distinct prime factors of $n$, contributing $(-1{)}^r$ when that subset has $r$ elements. Expanding a product therefore givesFor $n>1$ at least one prime factor occurs, so the product is zero. At $n=1$ it is an empty product, equal to one. This is the same alternating cancellation that drives inclusion-exclusion.

For $30=2\cdot 3\cdot 5$, all eight positive integer divisors are squarefree integers, givingPair each subset not containing $2$ with the subset obtained by inserting $2$: $(1,2),(3,6),(5,10),(15,30)$. Each pair cancels. The same nonzero terms occur for $60$, because its squarefree divisors are still the integer divisors of $30$; its other integer divisors have Möbius function value zero.

The Möbius divisor-sum identity is the cancellation behind Möbius inversion and reciprocal Dirichlet series: aggregation over integer divisors has an exact inverse.

Neutral buoyancy means a fully immersed body's upward buoyancy equals its downward force from gravity. For a body of fixed volume in a constant-mass density fluid under a uniform gravitational field, this occurs when its mean mass density equals the surrounding mass density. A $0.10\mathrm{m}$ cube of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ in the stipulated water has mass $1.000\mathrm{kg}$, so both forces are $9.81\mathrm{N}$.

Ideally it can remain at rest at any depth for which it stays fully immersed and clear of boundaries. Raising or lowering it changes the absolute pressure, but neither its displaced volume nor the difference between the vertical pressure forces. There is therefore no vertical restoring force toward a preferred depth in this model. Neutral buoyancy is a vertical force balance, not a claim about resistance to tilting or about a compressible body whose volume changes with pressure.

First count the organic groups and then the unsatisfied oxygen valences. A is the fully condensed T8 silsesquioxane cage, $\mathrm{C}{\mathrm{y}}_8\mathrm{S}{\mathrm{i}}_8{\mathrm{O}}_{12}$, with silicon at the eight vertices of a cube and bridging oxygen on its twelve edges. Each silicon has one cyclohexyl group and three siloxane bridges. Its eight cyclohexyl groups supply exactly ${\mathrm{C}}_{48}{\mathrm{H}}_{88}$.

B is a fully condensed T10 silsesquioxane cage, $\mathrm{C}{\mathrm{y}}_{10}\mathrm{S}{\mathrm{i}}_{10}{\mathrm{O}}_{15}$, for which a pentagonal-prismatic silicon skeleton with an oxygen on every edge is a standard compatible structure. Its ten groups give ${\mathrm{C}}_{60}{\mathrm{H}}_{110}$. The formula establishes full condensation but does not, without more structural data, rule out every other possible ten-vertex cage connectivity.

C is a corner-deficient T7 trisilanol cage, $\mathrm{C}{\mathrm{y}}_7\mathrm{S}{\mathrm{i}}_7{\mathrm{O}}_9(\mathrm{OH}{)}_3$. Seven cyclohexyl groups account for ${\mathrm{C}}_{42}{\mathrm{H}}_{77}$, leaving three hydrogen atoms as Si–OH groups. Remove one silicon corner of the T8 cube and replace the three severed connections by terminal hydroxyl groups. Nine oxygen atoms remain as bridges and three become terminal silanol oxygen atoms, giving precisely twelve oxygen atoms. There are three silicon sites bearing OH, three further equivalent fully condensed sites, and one opposite fully condensed site, consistent with three silicon environments in a 3:3:1 site-count ratio. The less negative resonance is naturally associated with the three silanol-bearing sites; the remaining two resonances belong to the two fully condensed classes. Peak heights alone are not quantitative integrations.

The source figure appears to label the spectrum ${}^{27}\mathrm{Si}$. The chemically appropriate routine experiment here is silicon-29 NMR, using the spin-one-half isotope ${}^{29}\mathrm{Si}$; the figure label is treated as an apparent isotope-label typographical error, not as evidence for another cage.

Each calcium cation has eight fluoride neighbours at the vertices of a cube: cubic coordination. Each fluoride anion has four calcium neighbours in tetrahedral coordination. Both unlike-neighbour Euclidean distances are $\sqrt{3}a/4$, with $a$ the conventional cubic lattice parameter.

Apply the Goldschmidt packing principle at its limiting eight-coordinate geometry: the eight surrounding anions just touch one another and also touch the central cation. The anions form a cube of edge $a/2$, so $2r_a=a/2$ and $r_a+r_c=\sqrt{3}a/4$. Eliminating $a$ givesEquivalently, $r_c/r_a=0.732$. This is a limiting radius ratio rule, rather than a measured ionic-radius ratio. It does not assume that the calcium sublattice consists of mutually touching spheres: the calcium positions have face-centred cubic symmetry, but that symmetry alone does not specify which ions touch. A different construction with touching calcium spheres and fluoride just filling their tetrahedral interstitial sites would instead give $r_a/r_c=\sqrt{3/2}-1=0.225$; that is the opposite contact limit, not the eight-coordinate anion-contact ideal used here.

The nickel-rich $\gamma$ matrix is a disordered substitutional solid solution on a face-centered cubic lattice. The ordered ${\gamma }^{\prime }$ phase has the L12 structure, approximately $\mathrm{N}{\mathrm{i}}_3\mathrm{Al}$: aluminium occupies cube corners and nickel occupies face centres. Its occupied-site geometry is often called ordered FCC, but its chemically labelled Bravais lattice is primitive cubic, with a four-atom basis. A face-centre translation exchanges chemical sublattices and is not a symmetry of the ordered crystal.

Each Al atom has twelve nearest-neighbour Ni atoms at Euclidean distance $a/\sqrt{2}$. Each Ni atom has twelve nearest neighbours too, but these are four Al and eight Ni. The chemical distinction matters for slip. In disordered $\gamma$, the shortest perfect Burgers vector isFor the ordered primitive cubic translation lattice, the shortest order-preserving perfect vector isThis does not imply that $⟨100⟩$ slip must dominate. A vector must lie in a favourable glide plane and have an accessible core structure and activation stress. Matrix dislocations commonly glide on $\{111\}⟨110⟩$ systems. A single $a⟨110⟩/2$ passage through ${\gamma }^{\prime }$ leaves an antiphase boundary; a second matching passage restores order, giving total vector $a⟨110⟩$. Such a superdislocation has a longer total vector than $a⟨100⟩$, yet its constituent dislocations can move on the favourable close-packed planes and interact with the matrix. Paired motion balances line and antiphase-boundary energies, so the shortest abstract lattice translation alone does not select the active slip system.^[ref]

We derive the required Calderón–Zygmund decomposition and apply the preceding two parts. Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, and put $A=\Vert \hat{K}{\Vert }_{\infty }$. The Plancherel theorem givesFor completeness this starts with smooth rapidly decreasing inputs, where $\widehat{K\ast f}=\hat{K}\hat{f}$, and extends by $L^2$ density. It agrees with the pointwise convolution in the question: if $f_j\to f$ in $L^2$, then $\Vert K\ast (f_j-f){\Vert }_{\infty }\le \Vert K{\Vert }_2\Vert f_j-f{\Vert }_2$. Thus there is no need to assume that $K$ itself is integrable.

Fix $f\in L^1({\mathbb{R}}^d)$, let $F=\Vert f{\Vert }_1$, and choose $\lambda >0$. Select the maximal dyadic cubes $Q$ for which $|Q{|}^{-1}{\int }_Q|f|>\lambda$. They exist above every qualifying cube because averages over increasingly large ancestors tend to zero. They are disjoint, their total volume is at most $F/\lambda$, and maximality givesOutside their union, dyadic differentiation gives $|f|\le \lambda$ almost everywhere. One can obtain this differentiation fact from the allowed martingale convergence theorem: on each fixed dyadic cube the averages are conditional expectations of the integrable restriction, their almost-everywhere limit exists, and approximation by dyadic step functions shows in $L^1$ that this limit is $f$.

Let $f_Q=|Q{|}^{-1}{\int }_{Q}f$, define $g=f$ outside the selected cubes and $g=f_Q$ on $Q$, and set $b_Q=(f-f_Q)1_Q$. This constructswith the estimatesThese follow by integrating separately on each selected cube; $|f_Q||Q|\le {\int }_Q|f|$ gives both the good-part norm bound and the last cancellation-error bound.

Let $\Omega ={\bigcup }_Q\hat{Q}$, with the dilation $L=4\sqrt{d}$ from part (ii). Then $|\Omega |\le L^{d}F/\lambda$. On its complement, part (i), followed by part (ii), givesFinally, $|Tf|>\lambda$ implies $|Tg|>\lambda /2$ or $|Tb|>\lambda /2$. The first alternative is controlled by the $L^2$ operator bound, and the second by the last integral outside $\Omega$. HenceThus $T$ has weak type $(1,1)$, with a constant depending only on the dimension, the bounded Fourier transform, and the Hörmander integral kernel condition.

Let $c$ be the center and $\ell$ the side length of $Q$. The zero integral of $f$ lets us subtract a constant kernel value:Choose $L=4\sqrt{d}$. For $x$ outside the concentric cube $\hat{Q}$ of side length $L\ell$, and for $y\in Q$,Thus the domain of integration in $x$ is contained in $\{|x-c|>2|y-c|\}$. After the translations $X=x-c$ and $Y=y-c$, the Hörmander integral kernel condition givesIf $Y=0$ the difference is zero and the same inequality holds. Apply the triangle inequality and Tonelli theorem to obtain the cancellation estimate outside a dilated cube:In particular the constant on the right is the same $C$ as in the hypothesis; only the dilation $L$ depends on the dimension.

Work on the unit sphere with geodesic edges. A spherical lune of angle $\alpha$ has area ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi =2\alpha$. The three great circles through the sides of a spherical triangle partition the sphere into four antipodal pairs of triangles. If the original triangle has area $A$ and the other pair representatives have areas $B,C,D$, then $A+B+C+D=2\pi$. The three lunes at its vertices have areas $A+B,A+C,A+D$, soThis proves the spherical excess formula $\boxed{A=\alpha +\beta +\gamma -\pi }$. Triangulate a convex spherical $k$-gon from one vertex into $k-2$ triangles; summing areas and vertex angles givesFor a decomposition of the whole sphere, angles around every vertex sum to $2\pi$ and each edge belongs to two faces. Summing the polygon area formula over all faces therefore givesNow $F=m+n+p$, $2E=4m+5n+6p$, and $3V=2E$, since three edges meet at each vertex. Substituting into Euler's formula yields $2m+n=12$. Nonnegative integer solutions therefore give at most seven pairs:At least three distinct pairs are realized: a dodecahedron gives $(0,12)$, a pentagonal prism gives $(5,2)$, and a cube gives $(6,0)$. Center each convex polyhedron around an interior origin and radially project its faces onto the sphere; edges become great-circle arcs and the resulting spherical faces are convex, preserving the incidence counts. This supplies the required actual decompositions.

Encode a vertex of a polytope by three bits: bit $j$ records whether the lower or upper bound of coordinate $j$ is tight. Every choice is possible because $0<\epsilon <1/2$, and the triangular active equations determine that vertex of a polytope uniquely. Its three adjacent vertices of a polytope are obtained by changing one bit. The geometric Bland pivoting rule specified here chooses the smallest-index active facet that can be left while increasing $x_3$. It produces the following path; the final column records the facet left on arrival at the next vertex of a polytope:For example, at the origin all three outgoing edges improve the objective, but leaving facet 1 has priority. At bit string $110$, switching bit 2 decreases $x_3$, so the rule switches bit 1, leaving facet 2. At $010$, neither of the earlier-coordinate switches improves $x_3$, so the rule finally leaves facet 5. The upper face reverses the direction in which changes of $x_2$ improve $x_3$, and the remaining three horizontal pivots follow. The objective values in the table are strictly increasing for the stated range of $\epsilon$.

Thus the simplex method takes seven pivots although the optimum is directly adjacent to the starting point. The same Klee-Minty cube in $n$ dimensions has $2n$ inequalities and $2^n$ vertices of a polytope. On its lower last-coordinate face, increasing $x_n$ means increasing $x_{n-1}$, and the first $n-1$ coordinates follow the corresponding lower-dimensional path before the last-coordinate facet can be left. On the upper face, increasing $x_n$ means decreasing $x_{n-1}$, and the smallest-index rule traverses that path in reverse. The reverse-path assertion follows by the same induction, interchanging maximization and minimization on each face. Hence the pivot count satisfies $T_n=2T_{n-1}+1$, $T_1=1$, andThe usual Bland pivoting rule prevents cycling, including degenerate pivots, but this exponential example shows that it does not give a polynomial worst-case pivot bound. For rational fixed $\epsilon$, the constraints have polynomial encoding length. Arithmetic cost per pivot does not remove the exponential obstruction.

For a lower bound that applies to every pivot choice, use $0\le x_i\le 1$ and maximize ${\sum }_{i=1}^n{x}_i$, starting at the origin. Each edge of this cube changes only one coordinate, whereas the unique optimal vertex of a polytope is $(1,\dots ,1)$. Every coordinate must therefore be changed along at least one edge. Every route of adjacent pivots takes at least $n$ pivots, and every strictly improving route takes exactly $n$.

For Dantzig pivot rule, the rate is the objective increase per unit entering nonbasic variable, as in standard reduced cost pricing. Rescale the three coordinates byThe transformed linear program still maximizes $y_3$, now with boundsWhen the active facet for coordinate $i$ is freed, the other active equations give $|d{x}_3/d{x}_i|={\epsilon }^{3-i}$. The positive rates per unit $y_i$ are thereforefor $i=1,2,3$, respectively. Equivalently, use the nonnegative lower- and upper-bound slack variables of these displayed, normalized inequalities: freeing coordinate $i$ changes its entering slack by $|d{y}_i|$. Thus Dantzig pivot rule chooses the lowest coordinate index among improving edges and follows the same seven-pivot path. In $n$ dimensions set $y_i={\epsilon }^{2(n-i)}{x}_i$; the rates ${\epsilon }^{-(n-i)}$ give the same $2^n-1$ pivots. These statements concern coordinate/slack pricing, not objective increase per unit Euclidean edge length, which defines a different pivot rule.

Place the cube at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a signed permutation matrix: there are $2^{3}3!=48$ choices. Its group action on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this centraliser:For the remaining isomorphism, let $R$ be the rotational symmetry group of a cube. Half the signed permutation matrices have determinant $1$, so $|R|=24$. Central inversion has determinant $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap \{I,-I\}=\{I\}$, multiplication gives a direct product of groups:The group action of $R$ on the four body diagonals is faithful. To prove this, choose direction vectors $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span ${\mathbb{R}}^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed vector to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.