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Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

For independent edge vectors $u=B-A$, $v=C-A$, $w=D-A$, the displacement $x=O-A$ solves $x\cdot u=|u{|}^2/2$, $x\cdot v=|v{|}^2/2$, $x\cdot w=|w{|}^2/2$. The reciprocal cross-product basis gives the displayed formula. Its squared distances to all four vertices are equal. It need not lie inside the tetrahedron.

Suppose a nonconstant holomorphic map $f:X\to {\mathbb{P}}^1$ from a connected compact Riemann surface has degree of a holomorphic map two and exactly four distinct branch values of a holomorphic map. The valency theorem forces each of their fibres to contain one ramification point with $e_p=2$, a simple ramification point of a holomorphic map. HenceThis determines the genus without recognizing a torus in a drawing.

A concrete realization is the compact Riemann surface obtained by adding the two points at infinity to the even-degree hyperelliptic model given by the algebraic curveProjection to $z$ has degree of a holomorphic map two. At a simple zero $a$ of $P$, $w$ is a local coordinate and $z-a=w^2/P^{\prime }(a)+O(w^4)$, so $e_p=2$. At infinity set $u=1/z$ and $s=w/z^2$: then $s^2=(1-4u^2)(1-u^2)$. There are two points with $u=0$, $s=\pm 1$, each unramified since $u$ is a local coordinate. The four finite roots are therefore all the branch values of a holomorphic map.^[ref]

Choose a triangulation of a surface on the Riemann sphere with these four branch values of a holomorphic map as vertices, combinatorially the boundary of a tetrahedron. Its counts are $V=4,E=6,F=4$. Upstairs, each vertex has one preimage and each open edge and face has two lifts: $V_X=4,E_X=12,F_X=8$, giving Euler characteristic $4-12+8=0$. This is a cell schematic, not a geometric immersion of $X$. The Triangulation proof of the Riemann-Hurwitz formula applies the same counting idea after refining a general triangulation of a surface around the branch values of a holomorphic map.^[ref]

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

A finite element mesh partitions a domain into simple regions called elements, such as intervals, triangles or tetrahedra. In a conforming partition, distinct element interiors are disjoint, their closures cover the domain closure, and any nonempty intersection is a common face, edge or vertex of both elements. Chen's author notes give this compatibility condition for simplicial meshes.^[ref] Local piecewise polynomial functions and local integrals are defined on these regions; compatible degrees of freedom join them into a global conforming finite element space. The assembled stiffness matrix and mass matrix then inherit the local overlap pattern.

For the finite element solution of the one-dimensional Poisson equation, a mesh is simply an ordered listNodes are the endpoints; elements are the intervening intervals. Four equal elements have five nodes but only three unknown nodal values after imposing the zero Dirichlet boundary conditions. A nonuniform partition such as $0,1/8,1/4,1/2,1$ has element lengths $1/8,1/8,1/4,1/2$ and $h=1/2$, so counting elements alone does not measure the largest unresolved scale.

Refining a finite element mesh replaces some or all elements by smaller ones. For continuous piecewise linear functions in one dimension, retaining all old nodes gives nested conforming finite element spaces $V_h\subset V_{h/2}$ when every interval is bisected. The old approximation remains available, and the new interior nodes allow additional changes of slope. With a symmetric coercive bilinear form and exact integration, best approximation from Galerkin orthogonality means that the energy norm error cannot increase under such nested refinement.

For the finite element solution of the one-dimensional Poisson equation, uniform refinement from four to eight elements increases the unknown count from three to seven. The exact formulas give maximum errors $0.0078125$ and $0.001953125$, and energy norm errors approximately $0.0721688$ and $0.0360844$. Refinement near a region of strong curvature can spend unknowns where the linear interpolation error matters most; convergence in a general problem still needs approximation and regularity assumptions such as those in the finite element interpolation estimate.

A fundamental normal surface has a nonzero admissible coordinate vector which is not the sum of two nonzero admissible vectors. Fixing the allowed quadrilateral type in every tetrahedron reduces the problem to the Hilbert basis of a rational cone. There are finitely many choices of quadrilateral types and finitely many indecomposable vectors for each choice.

The center of the sphere tangent to all four faces is the positive face-area-weighted combination of the opposite vertices. If $V$ is the volume and $S$ the total face area, its distance from each face is $3V/S$. Indeed its weight for vertex $A$ is $S_A/S$ and the corresponding altitude is $3V/S_A$, so their product is the common distance. Positive weights place it inside the tetrahedron.

Relative to a triangulation, a normal surface meets every tetrahedron in disjoint normal disks: four types of triangles cutting off vertices and three types of quadrilaterals separating opposite pairs of edges. A normal disk meets any edge at most once; its face arcs join distinct edges. Two different quadrilateral types cannot coexist in a tetrahedron.

For $t$ tetrahedra, count the seven disk types in each tetrahedron. Across each glued face the counts of each of its three normal arc types must agree. These are homogeneous linear equations with integer coefficients. An admissible solution also has at most one nonzero quadrilateral type in each tetrahedron. Matching the arcs in their boundary order reconstructs the normal surface up to an isotopy preserving the tetrahedra.

Place one vertex of a tetrahedron at the origin and denote the other vertex vectors by $a,b,c$. Orthogonality of two pairs of opposite edges of a tetrahedron forces $a\cdot b=a\cdot c=b\cdot c$. It follows that the third pair is orthogonal too. The three sums of squared lengths of opposite edges all equal $|a{|}^2+|b{|}^2+|c{|}^2-2a\cdot b$. This gives a dot product test for the special geometry without solving for the face planes.

CO and CO${}_2$ are discrete molecules, whereas the other oxides gain stability from extended bonding or ionic lattices. carbon's compact 2p orbitals overlap effectively with oxygen 2p orbitals, allowing strong multiple bonds that satisfy the valence requirements within small molecules. Intermolecular forces between CO or CO${}_2$ molecules are much weaker than those internal bonds, so low condensation temperatures and gaseous states at room conditions are unsurprising. CO${}_2$ is nonpolar overall but still has London dispersion forces and interactions between electric quadrupoles; “nonpolar” does not mean no attraction.

For larger Group 14 atoms, overlap with oxygen 2p orbitals is less effective for strong localized multiple bonding. SiO${}_2$ instead forms an extended network of SiO${}_4$ tetrahedra with each oxygen bridging two silicon atoms. Breaking or melting the network involves much more than separating intact SiO${}_2$ molecules. The Si–O electronegativity difference is $3.61-1.92=1.69$, so these are strongly polar covalent bonds rather than evidence for a molecular solid.

For Pb–O the difference is $3.61-1.85=1.76$, and ionic Pb${}^{2+}$/O${}^{2-}$ or Pb${}^{4+}$/O${}^{2-}$ descriptions provide useful starting points, with appreciable covalent contributions. PbO has layered structures with asymmetric oxygen coordination consistent with a stereochemically active 6s${}^2$ pair. A common PbO${}_2$ structure is rutile-like, with PbO${}_6$ octahedra linked through oxygen. Extended lattice interactions account for their solid states.^[ref] The similar Si–O and Pb–O electronegativity differences show why a single difference threshold cannot determine the entire bonding type, coordination number or structure.

A suitable representative structure is a three-dimensional network of corner-sharing CO${}_4$ tetrahedra, with C–O single bonds. Each carbon is four-coordinate; every bridging oxygen is bonded to two carbons. Sharing four oxygens between two tetrahedra gives $4/2=2$ oxygens per carbon, so the network retains empirical formula CO${}_2$ without containing separate O=C=O molecules.

High pressure favours a compact, highly coordinated network through its smaller molar volume: at fixed temperature, increasing pressure changes relative Gibbs energies according to $\mathrm{d}(\Delta G)=\Delta V\mathrm{d}p$. Heating helps overcome the large reconstructive barrier to breaking and redistributing molecular pi bonding and establishing an extended single-bond network. It should not be argued that high temperature by itself favours the lower-entropy network. Nor is merely freezing molecular CO${}_2$ sufficient: dry ice remains a molecular solid.

The sketch shows connectivity and local tetrahedral coordination, not a unique predicted crystal structure. Real amorphous carbonia need not have perfect periodic order or exclusively one coordination environment. The original carbonia experiments combined high-pressure preparation with infrared, Raman and diffraction evidence for loss of molecular character and formation of an amorphous network.^[ref]

Neutral counting assigns five electrons to cyclopentadienyl ligand and two to CO. Thus each $\mathrm{FeCpCO}$ fragment has $8+5+2=15$ valence electrons and contributes three skeletal electrons; each $\mathrm{CoCpCO}$ fragment has sixteen and contributes four:For A the six-pair four-vertex count is the nido count derived from a five-vertex trigonal bipyramid. Removing an axial vertex leaves a tetrahedron: the three equatorial vertices plus the remaining axial vertex. For B the corresponding arachno fragment is obtained by removing both axial vertices from that same parent, leaving a three-metal triangle. Small clusters require specifying the parent and removed vertices; mechanically calling every nido fragment a square pyramid is wrong.

The Fe tetramer is indeed known to have a tetrahedral $\mathrm{F}{\mathrm{e}}_4$ core, with four face-capping CO groups. The carbon atoms of those carbonyls and the iron atoms form interpenetrating tetrahedra, giving a cubane-like Fe–C connectivity. Bridging rather than terminal CO changes the distribution of local donation, not the total electron count above.^[ref]

A normal surface meets each tetrahedron in disjoint disks of the four triangular and three quadrilateral types. Its arcs in a face join distinct edges, each disk meets an edge at most once, and at most one quadrilateral type occurs in a tetrahedron. These last requirements exclude, for example, a disk with an octagonal boundary.

We first establish the surgery operation needed for the sphere system. Suppose a disk $D$ disjoint from the other spheres compresses one sphere $S$. It produces spheres $S^{\prime }$ and $S^{\prime \prime }$. Small parallel copies of the three spheres bound a three-times punctured three-sphere $P$. Denote the old piece on the uncompressed side by $A$, and the two pieces on the compressed side by $B^{\prime }$ and $B^{\prime \prime }$. If both $B^{\prime }$ and $B^{\prime \prime }$ were punctured three-spheres, then $P$ glued to them would also be a punctured three-sphere, contrary to the hypothesis on the original cut manifold. Choose, say, $B^{\prime }$ which is not of this type, and retain $S^{\prime }$ alone in place of $S$.

The other new piece is $A\cup P\cup B^{\prime \prime }$. It too cannot be a punctured three-sphere. If it were, cap its external boundary spheres to make $S^3$. All the spheres separating $A$ from the attached pieces then bound three-balls by the Three-dimensional Schoenflies theorem; the component containing $A$ would be a punctured three-sphere, contradicting the hypothesis on $A$. Thus the sphere surgery preserving nontrivial complementary pieces keeps exactly $k$ spheres and keeps every complementary piece nontrivial. The assumption that all spheres separate ensures that the old and new pieces used here really are distinct sides.

Apply general position to the whole sphere system, avoiding the vertices and making it transverse to edges and faces. We describe a terminating normalization, allowing the replacement operation just proved as well as isotopy. If some component of its intersection with a tetrahedron is not a disk, choose a boundary circle innermost among circles not bounding such disk components. A disk just inside the tetrahedral boundary gives a compression; if intervening disk components occur, first push them into collars of their boundary disks. Perform the compression and retain one new sphere as above. This replaces a non-disk tetrahedral piece by disk pieces without increasing edge intersections. Repetition is finite: every compression divides the original sphere along a separating circle, and the sum of $1-\chi$ over the non-disk planar tetrahedral pieces decreases; disk components can then be put in boundary collars. A closed sphere component wholly in a tetrahedron is impossible, because it would cut off a three-ball, contrary to the complementary-piece condition.

Now each tetrahedral piece is a disk. A face circle innermost in a face allows its incident disk to be moved through that face disk, removing the circle. This reduces the number of face intersections without increasing the edge weight. A returning face arc similarly gives an edge bigon across which the surface can be pushed, reducing the number of edge intersections by two. Restore tetrahedral disk pieces by the preceding operations whenever necessary.

It remains to exclude multiple intersections of a disk with an edge, which the absence of returning face arcs alone does not exclude. Suppose a tetrahedral disk meets an edge twice. Among such disks choose a pair of intersections consecutive on the edge and with no other surface intersection between them. Such a pair exists: an intervening disk crosses the edge interval an even number of times, so passing to an innermost intervening pair eventually gives consecutive points. Join them by an arc in the disk. The disk pieces divide the tetrahedron into three-balls, and this arc together with the edge interval lies in the boundary of one of these three-balls. It therefore bounds a bigon with interior disjoint from the surface. Push across the bigon to reduce edge weight by two. At fixed weight the previous moves reduce face and non-disk complexity; the edge weight itself is a nonnegative integer. Hence the whole process terminates. A final tetrahedral disk meets each edge at most once, and its boundary separates either one vertex from three or two vertices from two. It is respectively a normal triangle or quadrilateral. Thus a normal system of exactly $k$ spheres with the required complementary-piece property exists.

For the finite decomposition, we need a bound on such normal systems, not merely an assertion that they have been normalized. Let $f_2$ be the number of faces of the fixed triangulation. In a triangular face cut by normal arcs, all but at most four complementary regions are rectangles between adjacent parallel arcs. Thus at most $4f_2$ complementary three-dimensional components contain a nonrectangular face region. Every remaining component is assembled from product regions between parallel normal disks, so is an interval bundle whose boundary components are spheres. Its connected base is either $S^2$ or ${\mathbb{RP}}^2$. The product $S^2\times I$ is a punctured three-sphere, and is excluded. The other possibility is the twisted interval bundle over ${\mathbb{RP}}^2$, namely ${\mathbb{RP}}^3$ with a three-ball removed. Each such piece contributes a distinct ${\mathbb{F}}_2$ summand to $H_1(M;{\mathbb{F}}_2)$: gluing along spheres does not identify first-homology classes, by the Mayer–Vietoris sequence. Since $k$ separating spheres give $k+1$ components, we obtain the explicit finite bound

For an arbitrary closed connected $X$, first cut out nonseparating spheres. A neighbourhood of one such sphere together with an arc joining its two sides is a punctured $S^2$-bundle over $S^1$; the bundle is the product or the twisted bundle according to the orientation of the attaching identification. Its complement is connected, so this extracts a connected sum factor. The mod-two first Betti number of the remaining capped manifold decreases by one. Hence only finitely many such extractions are possible. The remaining manifolds have only separating spheres.

Whenever a remaining factor is not a prime three-manifold, split it along a sphere into two factors neither homeomorphic to $S^3$. Viewed in the original remaining manifold, all these successive splitting spheres form a disjoint system whose pieces are not punctured three-spheres. The normalization and bound just proved prohibit infinitely many such splittings. This proves existence of a finite prime decomposition, including for $X$ without assuming orientability.

For completeness, both sphere bundles extracted above are prime three-manifolds. Their fundamental group is $\mathbb{Z}$, so a separating sphere and the van Kampen theorem express $\mathbb{Z}$ as a free product in which one capped factor has trivial fundamental group. Its punctured piece lifts homeomorphically to the universal cover $S^2\times \mathbb{R}$, identified with ${\mathbb{R}}^3\setminus \{0\}$. That compact lifted piece has one spherical boundary and must be the bounded side of this sphere in ${\mathbb{R}}^3$. It misses the origin, so the Three-dimensional Schoenflies theorem makes it a three-ball. The corresponding capped factor is $S^3$, as required.

Suppose now that $X$ is orientable. A prime factor which is not an irreducible three-manifold contains a nonseparating sphere: an essential separating sphere would already split it nontrivially. Extracting the corresponding sphere bundle shows that the factor must be $S^2\times S^1$. Thus the other nontrivial primes are irreducible.

Consider two prime decompositions. Represent each by a sphere system, including one nonseparating sphere in each $S^2\times S^1$ factor. The capped pieces are now the irreducible factors and copies of $S^3$. Make the two sphere systems transverse. To remove their intersections, choose a circle innermost on a sphere of the second system and surger the first system along its disk. Retain both surgery spheres temporarily. The old sphere and the new pair bound a three-times punctured sphere; adding the pair and then deleting the old sphere only inserts or removes spherical punctures.

This operation cannot alter an irreducible capped piece: every new sphere in that piece bounds a three-ball, so it can only split off a punctured $S^3$ piece. The other pieces were already punctured $S^3$, where the same conclusion follows from the Three-dimensional Schoenflies theorem. After small displacements the surgery removes the chosen intersection circle and introduces none. Repeat until the sphere systems are disjoint. Their union is a common refinement, and its non-$S^3$ capped irreducible pieces are precisely those of each original decomposition. Hence these factors agree, up to order and orientation-preserving homeomorphism.

Finally, the number $r$ of $S^2\times S^1$ factors is determined bywhere the sum runs over the common irreducible factors. This uses additivity of first homology under connected sum, not a claim that the fundamental group determines a manifold. Therefore the orientable prime decomposition is unique up to order and insertion or deletion of $S^3$ factors.

For $t$ tetrahedra, a normal surface has a vector $x\in {\mathbb{Z}}_{\ge 0}^{7t}$ counting four triangle types and three quadrilateral types in each tetrahedron. On every glued face, the numbers of arcs of each of its three types must agree on the two sides. These give homogeneous integer linear equationsFor closed surfaces the coordinates of disks meeting an unglued boundary face must also vanish. The nonnegativity and matching equations are not the whole admissibility condition: in each tetrahedron at most one of its three quadrilateral coordinates can be nonzero. Conversely, for an admissible vector, arrange the prescribed parallel disks and glue their face arcs in order. This reconstructs an embedded normal surface, unique up to normal isotopy.

A fundamental normal surface corresponds to a nonzero admissible vector which cannot be written as $y+z$ with both $y$ and $z$ nonzero admissible vectors. These indecomposable vectors form a fundamental set: every admissible vector is their finite sum. To find them, fix which quadrilateral type is allowed in each tetrahedron. This leaves the integer points of a rational polyhedral cone $Ax=0,x\ge 0$, with some coordinates set to zero. Compute its Hilbert basis of a rational cone, for example by subdividing into rational simplicial cones and enumerating the lattice points in their fundamental parallelepipeds along with their integral ray generators, then removing decomposable generators. Repeat the finite list of quadrilateral choices, remove duplicates, and remove vectors decomposable in their compatible sector. Finiteness follows because each of these pointed rational cones has a finite Hilbert basis.

Let $E$ be the link exterior. By Alexander duality, for a two-component linkAn embedded connected closed surface separates $S^3$; its mod-two class is nonzero exactly when the two link components lie on different sides. The connectedness qualification matters: two parallel splitting spheres form a disconnected surface with zero total class. We use the criterion only for connected surfaces.

Normalize a splitting sphere by isotopy, disk compressions and removal of components lying inside tetrahedra. A compression replaces a sphere by two spheres whose mod-two classes add to the old class; at least one is still a splitting sphere. The discarded tetrahedral spheres have zero class. Thus there is a normal splitting sphere. Choose one, $S$, of least normal surface weight.

Suppose its vector is decomposable. Realize $S=A+B$ as a Haken sum, choosing a decomposition with the least possible number of intersection curves. Two elementary exchange observations are useful. First, $A$ and $B$ can be taken connected. If their total union has more than two components, perform regular exchanges connecting components until exactly two embedded groups remain, resolving any intersections within each group. Since the final sum is connected, this can be done before the last joining exchange; it uses up at least one of the old intersection curves. The resulting two normal surfaces still have total vector $x(S)$, contradicting minimality. Second, no intersection curve separates both $A$ and $B$. If it did, resolve that curve first: the two separated sides give two groups. Resolve any remaining self-intersections within each group and continue exchanges between groups only while the result remains disconnected. Just before the exchange making the final connected surface, there are again exactly two embedded normal groups with fewer intersection curves. Both assertions follow from the local four-bank resolution at an intersection circle; internal crossings are resolved before regarding a group as one embedded summand.

Every closed embedded surface in $S^3$ is orientable. Additivity of Euler characteristic in the Haken sum givesFor connected orientable closed surfaces the positive Euler characteristic is $2$, and the only zero Euler characteristic is that of a torus. Thus one summand, say $A$, is a sphere and the other $B$ is a torus. Both have smaller weight than $S$. The sphere $A$ cannot split the link by minimality, soEach intersection circle separates the sphere $A$, so the second exchange observation makes it nonseparating on $B$. In particular, these circles are essential parallel curves on the torus.

Cut $A$ along the intersection circles. There are at least two innermost disk patches. Choose one of smallest weight, $D$; then $2w(D)\le w(A)$. Its interior misses $B$, and its boundary is essential on $B$. Compress the torus along two slightly separated copies of $D$. The result is a sphere $C$, and the compression takes place in the exterior, soThus $C$ is a splitting sphere.

This alone is not quite a contradiction if the last weight inequality is equality: $C$ need not yet be normal. At the compression seam it uses the alternative, rather than regular, exchange between the disk patch and the torus. An intersection circle of normal disk sheets must cross a tetrahedral face; it cannot be a closed curve entirely in two disk interiors in one tetrahedron, which can be put in standard position with only intersection arcs. In a face, parallel arcs of the same type do not cross. At a crossing of different normal arc types, the alternative exchange joins two endpoints on the same edge, producing a returning arc. Pushing across its edge bigon strictly reduces weight. Normalize the resulting surface, retaining a nonzero-class sphere at each compression as before. We obtain a normal splitting sphere of weight strictly less than $w(S)$, a contradiction. Therefore

The algorithm is now finite. Triangulate the exterior from a link diagram, enumerate its fundamental admissible vectors with the closed-surface boundary condition, reconstruct their surfaces, and inspect the connected components and Euler characteristics. A fundamental vector yields a connected surface, since otherwise its component vectors decompose it. A connected closed surface in $S^3$ with Euler characteristic two is a sphere. Compute its mod-two class in the finite simplicial chain complex, or equivalently check which link boundary tori lie on either side. Answer yes exactly when such a sphere has nonzero class. The argument proves that every split link is detected, and every detected sphere witnesses splitting. For more than two components, $H_2(E;{\mathbb{F}}_2)\cong {\mathbb{F}}_2^{\ell -1}$ and a nonzero sphere class separates a nonempty proper subset of the components; the same least-weight argument and finite enumeration apply.

Translate one vertex of the tetrahedron to the origin and denote the other vertex vectors by $a,b,c$. Translation preserves edge vectors and their dot products. We may label the two known perpendicular pairs of opposite edges of a tetrahedron as the pairs represented by $a,c-b$ and $b,a-c$. Their orthogonality givesHence all three pairwise dot products have the same value $s$. The third pair satisfiesso the third pair of opposite edges is also perpendicular.

Expanding squared lengths using the dot product givesThus the sums of squared opposite-edge lengths are equal. These two conclusions are the basic identities for orthogonal opposite edges of a tetrahedron.

Take the half-turn about the $z$-axis, represented by the rotation matrixIt is an orthogonal matrix with determinant $1$, and it fixes the $z$-axis while reversing both perpendicular coordinates. Direct substitution gives $Ma=b$, $Mb=a$, $Mc=d$ and $Md=c$. A linear map that permutes the four vertices also preserves their convex hull, which is the solid tetrahedron. Hence this rotation matrix leaves $T$ invariant. Its axis joins the midpoints of the two opposite edges $ab$ and $cd$.

Move the origin to the centroid and write the four vertex vectors as $r_1,r_2,r_3,r_4$. Their sum is zero. The midpoints of the first opposite-edge pair have vectors $(r_1+r_2)/2$ and $(r_3+r_4)/2=-(r_1+r_2)/2$, so their distances from the centroid are equal. The same argument applies to the other two opposite-edge pairs.

Choose the three midpoint distances as $u=|r_1+r_2|/2$, $v=|r_1+r_3|/2$, $w=|r_1+r_4|/2$. Expanding the dot products,Identifying these four norms with the vertex distances proves the tetrahedron centroid-to-midpoint sum of squares:The proof uses no orthogonality or regularity assumption on the tetrahedron.

Use vectors in three-dimensional Euclidean space, with the usual dot product and cross product. The following explicit assignments compute the circumcenter of a tetrahedron:General position means $d\ne 0$. The reciprocal cross-product identities give $x\cdot u=|u{|}^2/2$, $x\cdot v=|v{|}^2/2$ and $x\cdot w=|w{|}^2/2$. Consequentlyand likewise for $C,D$. Thus $O$ is the unique point equidistant from the four vertices. It can lie outside the tetrahedron; no inside constraint should be imposed.

For a point $\mathbf{x}$ on the plane, the Cauchy-Schwarz inequality gives $|d|=|\mathbf{x}\cdot \mathbf{n}|\le |\mathbf{x}|$, because $\mathbf{n}$ is a unit vector. Equality is achieved at $\mathbf{x}=d\mathbf{n}$. Thus the distance from the origin is $|d|$.

Let $\delta =\mathbf{p}\cdot \mathbf{n}-d$ and let $\mathbf{q}=\mathbf{p}-\delta \mathbf{n}$ be the perpendicular projection of the sphere's center onto the plane. Every point of the plane has form $\mathbf{x}=\mathbf{q}+\mathbf{u}$ with $\mathbf{u}\cdot \mathbf{n}=0$. The Pythagorean theorem givesIf $|\delta |=r$, the sphere equation therefore forces $\mathbf{u}=0$. There is exactly one contact point, $\mathbf{q}$; it lies on both surfaces.

Write $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The nonzero scalar triple product proves that $\mathbf{a},\mathbf{b},\mathbf{c}$ are linearly independent, so they form a positively oriented basis of ${\mathbb{R}}^3$ and define a nondegenerate tetrahedron.

For the three faces through the origin, choose the inward unit vectorsTheir signs are correct because each has positive inner product with the opposite vertex vector. Write the center as $\mathbf{p}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. It lies on the interior side of every face. Tangency of the interior sphere therefore means signed distance $r$, not $-r$, from each of those three planes. Taking inner products givesSolving proves the center formulaThis is the sphere center from three tetrahedron face distances; the interior hypothesis selects these signs rather than an exterior center.

Finally set $\mathbf{N}=\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}$. The scalar triple product identities give $\mathbf{N}\cdot \mathbf{a}=\mathbf{N}\cdot \mathbf{b}=\mathbf{N}\cdot \mathbf{c}=\tau$. In particular $\mathbf{N}\ne 0$, and the plane through the three vertices isThe second formula follows by normalizing $\mathbf{N}$ to a unit vector and applying the first distance result.