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The simply connected Bertotti-Robinson geometry is the product of the universal cover of two-dimensional anti-de Sitter spacetime and a two-sphere with equal radius $m>0$. In metric signature $(-,+,+,+)$,Its radial Penrose diagram is an infinite conformal strip of two-dimensional anti-de Sitter spacetime. The metric tensor $-{\lambda }^{2}d{t}^2+m^2{\lambda }^{-2}d{\lambda }^2+m^{2}d{\Omega }_2^2$ is a patch of it, with $\lambda =m(\sin T-\sin X)/\cos X$. Its Killing vector field ${\partial }_t$ extends across alternating degenerate Killing horizons $\lambda =0$. Both factors are geodesically complete, giving a maximal analytic extension of this patch without periodic time identification.

For $\mathbf{d}\ne 0$, put $\hat{\mathbf{d}}=\mathbf{d}/|\mathbf{d}|$. The lower-band orthogonal projector is $P_{-}=(I-\hat{\mathbf{d}}\cdot \mathbf{\sigma })/2$. Applying the Pauli matrix multiplication law to $i\mathrm{Tr}(P_{-}[{\partial }_{k_x}{P}_{-},{\partial }_{k_y}{P}_{-}])$ givesThus the Chern number is the signed number of times the Brillouin zone covers a two-sphere under $\hat{\mathbf{d}}$. The lower energy eigenstate's actual Bloch vector is $-\hat{\mathbf{d}}$. The Berry curvature has the displayed positive sign for this lower band and the stated Berry connection; switching to the upper band changes its sign.

A useful local calculation exposes what each valley contributes. Write $p=\hslash v|\mathbf{q}|$. The linear Dirac Hamiltonian givesEach unbounded continuum cone gives a half-integer in this calculation. It is not a separately realizable half-integer Chern number of an isolated lattice energy band. The full honeycomb lattice supplies both valleys and the large-momentum completion. Starting with a trivial large-$|M|$ state, each single-valley mass crossing changes the lattice integer by the corresponding unit. There are no other band touchings. This establishes the exact lattice resultfor $t_2>0$, excluding the boundaries. This sign belongs to our explicit bond vectors, orientation and occupied-band convention.

For an original worked example, choose $t_1=1$, $t_2=0.08$, $\varphi =\pi /2$ and $M=0$. Then $\mu \simeq 0.415692$, $(m_{+},m_{-})=(-\mu ,+\mu )$, both valley gaps are $2\mu \simeq 0.831384$, and $C=-1$. Choosing instead $t_2=0$, $M=\mu$ gives exactly the same valley-gap magnitudes but $C=0$. The first state's two valley contributions reinforce; the second state's cancel.

An independent eigenvector-overlap calculation on periodic $31\times 31$, $61\times 61$ and $121\times 121$ momentum grids gives $C=-1$ for the first example and $C=0$ for the second. Reversing $\varphi$ in the first example gives $C=+1$. Direct integration of the analytic lattice Berry curvature agrees. For the overlap calculation one can remove the embedding phase from $f$, replacing it by $1+e^{i\mathbf{k}\cdot {\mathbf{a}}_1}+e^{i\mathbf{k}\cdot {\mathbf{a}}_2}$; this makes the cell-basis Hamiltonian periodic and prevents a false boundary-seam contribution. The local Berry curvature can depend on this basis convention; the integrated Chern number agrees. All mass-zero parameters are excluded before evaluating any integer.

There are exactly four types: the two-sphere, real projective plane, torus, and Klein bottle. Four sides admit three pairings and four direction choices for each pairing. Up to cyclic relabelling, labels are adjacent, $a{a}^{s}b{b}^t$, or alternating, $ab{a}^s{b}^t$. For adjacent labels, the vertex counts for $(s,t)=(-1,-1)$, a mixed-sign pair, and $(1,1)$ are $3,2,1$. For alternating labels they are $1,1,2$. There are two edge classes and one face, so Euler characteristic is $V-1$. Orientability occurs exactly when $s=t=-1$. The classification theorem for surfaces identifies every case, leaving no additional closed surfaces from a square. Examples are $a{a}^{-1}b{b}^{-1}$ for the sphere, $aab{b}^{-1}$ for the projective plane, $ab{a}^{-1}{b}^{-1}$ for the torus and $ab{a}^{-1}b$ for the Klein bottle.

The CP1 spinor representation writes a unit vector as $n_i=z^{†}{\sigma }_{i}z$ with $z^{†}z=1$. The phase redundancy $z\mapsto e^{i\alpha }z$ identifies normalized spinors that represent the same point of the two-sphere.

For a finite CW complex, $b_0$ counts connected components: its connected components are its path components, as recorded by the zeroth homology group. For a finite graph regarded as a one-dimensional CW complex, no higher homology groups occur, so $b_1=E-V+b_0$ counts independent graph cycles, and $\chi =b_0-b_1$. For a connected planar surface obtained by removing $h$ disjoint open disks from the interior of a closed disk, $(b_0,b_1,b_2)=(1,h,0)$ and $\chi =1-h$; the annulus is $h=1$.

The word "hole" needs its dimension. A two-sphere encloses a cavity but has $b_1=0$ and $b_2=1$. A closed orientable surface of genus $g$ instead has $(b_0,b_1,b_2)=(1,2g,1)$ and $\chi =2-2g$: increasing the genus by one adds two independent one-dimensional homology classes. In higher dimensions, further homology groups enter with alternating signs; there is no universal "components minus handles" rule.

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.

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 closed differential form of degree one on the two-sphere has primitives on its north-punctured and south-punctured charts by the Poincaré lemma. The overlap is connected, so their difference is constant. Adjust one constant and glue the primitives. Thus every closed one-form is an exact differential form, proving the first de Rham cohomology vanishes.

The orbifold fundamental group is the group of deck transformations of the universal orbifold cover. It retains the local finite group data that the ordinary fundamental group of the underlying topological space forgets.^[ref]

For an oriented closed topological surface of genus $g$ with cone points of orders $m_1,\dots ,m_r$, the group presentation isEach local cone relation says $c_i^{m_i}=1$; it does not guarantee that $c_i$ has order exactly $m_i$ in the resulting global group. For the teardrop orbifold, a two-sphere with one cone point of order $m>1$, the remaining relation is $c_1=1$, so its orbifold fundamental group is trivial despite nontrivial local isotropy.^[ref]

For a complete hyperbolic surface orbifold, the universal cover is the hyperbolic plane. A torsion-free finite-index subgroup then acts freely and gives a smooth surface cover. The hyperbolic covering model is what identifies local cone stabilizers with actual finite subgroups in this case.

For a closed connected orientable surface, its orientable genus is the number of torus summands in its connected sum classification, with the two-sphere assigned genus zero. The Euler characteristic is $2-2g$, so it determines this number within the orientable family.

The metric tensor in a basis ${\mathbf{e}}_a$ is $g_{ab}={\mathbf{e}}_a\cdot {\mathbf{e}}_b$. Differentiating this inner product with the product rule and using the assumed basis-derivative relation yieldsThus the coefficients satisfy metric compatibility. For a coordinate basis with zero torsion tensor, symmetrizing this relation in $a,c$, adding the corresponding two equations and subtracting the third givesThey are therefore the Christoffel symbols of the Levi-Civita connection. Metric compatibility by itself does not establish zero torsion tensor; that also uses the coordinate basis and its commuting coordinate derivatives.

For an embedded curved surface, the stated ambient basis-derivative relation needs a qualification: the Gauss formula is ${\partial }_c{\mathbf{e}}_a={\Gamma }_{ac}^b{\mathbf{e}}_b+h_{ac}\mathbf{n}$, where $\mathbf{n}$ is a unit normal and $h_{ac}$ is the second fundamental form. The normal term has zero inner product with each tangent basis vector, so the displayed identity for ${\partial }_c{g}_{ab}$ still holds. For a concrete counterexample to dropping that term, on a two-sphere one has ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-R\mathbf{n}$, whereas every ${\Gamma }_{\theta \theta }^b$ is zero. Thus the intrinsic connection describes the tangential projection of the ambient derivative.

In standard spherical coordinates, parametrize the two-sphere byThe coordinate basis transforms from the fixed Cartesian basis according to ${\mathbf{e}}_a=(\partial x^i/\partial q^a){\mathbf{e}}_i$. ThereforeTheir inner products are ${\mathbf{e}}_{\theta }^2=R^2$, ${\mathbf{e}}_{\theta }\cdot {\mathbf{e}}_{\varphi }=0$ and ${\mathbf{e}}_{\varphi }^2=R^2{\sin }^2\theta$. Thus the induced metric isThe source's intermediate display places an extra $R^2$ outside entries already labelled $g_{\theta \theta },\dots$. Interpreting those entries as the actual metric tensor components would double-count the factor: for example, $R=2$ gives $g_{\theta \theta }=4$ directly from ${\mathbf{e}}_{\theta }^2$, not $16$. Its final matrix is the correct induced metric; the entries of its intermediate matrix must instead denote dimensionless coefficients.

A nondegenerate Lorentzian metric requires $M\ne 0$; take $M>0$ as the radius convention. The original coordinates describe $\lambda >0$, with $\lambda =0$ a candidate boundary rather than an ordinary point of that chart. The change of coordinate givesSubstituting and cancelling the two terms proportional to $d{\lambda }^2/{\lambda }^2$ producesThe $(v,\lambda )$ block of this metric tensor has determinant $-M^2$ and inverse metricAll coefficients are smooth through $\lambda =0$, and the block has one negative and one positive eigenvalue. Together with the positive round two-sphere factor this is a regular Lorentzian metric. It extends to a neighborhood with either sign of $\lambda$. The usual polar-coordinate problem at the poles of the two-sphere is removed by separate angular charts. Therefore $\lambda =0$ is only a coordinate singularity of the original chart. This is the degenerate-horizon patch of Bertotti-Robinson spacetime.

Put $d{\Omega }_2^2=d{\theta }^2+{\sin }^2\theta d{\phi }^2$, the standard metric tensor on the unit sphere. From $t=v+m/\lambda$ one has $dt=dv-m{\lambda }^{-2}d\lambda$. The apparently divergent terms cancel:The $(v,\lambda )$ metric tensor block has determinant $-m^2\ne 0$, independently of $\lambda$. Its coefficients are real analytic, so this formula defines a regular Lorentzian metric onThe polar-coordinate degeneracies at the poles are handled by ordinary two-sphere charts. Thus $\lambda =0$ is a coordinate singularity of the original chart and is crossed by this spacetime extension.

First use the original null coordinates $u=t+m/\lambda$, $v=t-m/\lambda$. They obey $u-v=2m/\lambda$, and the radial metric tensor becomesApply the given substitutions $u=\tan (U/2)$ and $v=-\cot (V/2)$. The identitiesgive the analytic extension of a spacetimeThe poles of the individual tangent and cotangent substitutions are now ordinary coordinate lines wherever $\cos ((U-V)/2)\ne 0$. Set $T=(U+V)/2$ and $X=(U-V)/2$. Allowing these coordinate functions to continue across every such line givesThis is the universal cover of two-dimensional anti-de Sitter spacetime of radius $m$, times a two-sphere of radius $m$: the simply connected Bertotti-Robinson spacetime. In defining this maximal analytic extension, $T$ ranges over the whole real line without periodic identification. A periodic identification would instead create closed timelike curves and would be an additional global choice.

For clarity, the original patch is embedded by the branches $-\pi <U<\pi$ and $0<V<2\pi$. The inequality $u>v$ is equivalent on these branches to $|U-V|<\pi$, so the original patch occupiesThe first spacetime extension in part (a) occupies the larger band $0<T-X<2\pi$, with $|X|<\pi /2$. Its finite-$v$ Killing horizon is the interior line $T+X=\pi$. The edges $T-X=0,2\pi$ of that band are further regular Poincaré horizons; continuing through them produces the full conformal strip of two-dimensional anti-de Sitter spacetime.

The original coordinate $\lambda$ has a smooth continuation throughout the strip:Consequently the complete set of zero-level Killing horizons isThese are two alternating families of parallel null lines; different families meet only at the excluded conformal boundaries, never inside the spacetime. The continuation of the original Killing vector field isThe field never vanishes in the interior, including on the Killing horizons.

To obtain the Penrose diagram, suppress the two-sphere and multiply by the conformal factor ${\cos }^{2}X/m^2$. The resulting radial metric tensor is $-d{T}^2+d{X}^2$, so the diagram is an infinitely tall strip of width $\pi$, with null lines at 45 degrees and timelike edges $X=\pm \pi /2$. Generically,At the special edge points where the two families of Killing horizons end, the displayed formula is a $0/0$ limit and $\lambda$ has no unique limiting value. Thus the infinity labels on the diagram refer to the open edge segments, not those vertices. Each interior point of the diagram represents a two-sphere of fixed radius $m$; neither edge is a centre where that two-sphere shrinks.

Finally, these timelike edges are conformal boundaries, not omitted regular points of a further spacetime extension. One way to establish maximality is geodesic completeness. The two-dimensional anti-de Sitter spacetime factor embeds locally as $Z\cdot Z=-m^2$ in a flat space with metric signature $(-,-,+)$. Along an affinely parametrized geodesic with constant $ϵ=\dot{Z}\cdot \dot{Z}$, the ambient equation isIts oscillatory, linear or hyperbolic solutions for negative, zero or positive $ϵ$ exist for all real affine parameters and lift to the universal cover. The two-sphere is also geodesically complete, and geodesics of the product split into geodesics of its factors. Hence $\tilde{g}$ is geodesically complete. A proper regular spacetime extension would furnish a geodesic reaching its boundary in finite affine parameter, contradicting that completeness. This proves that the displayed simply connected analytic spacetime is maximal.

The two-dimensional Schoenflies theorem says that a simple closed piecewise linear curve in $S^2$ separates it into two components whose closures are piecewise linear balls of dimension two. In particular, a homeomorphism of the curve to the equator extends to a piecewise linear homeomorphism of the whole sphere.

A piecewise linear ball is a space piecewise linearly homeomorphic to a standard simplex. Write $D=B_1\cap B_2$. The complementary disk to $D$ in each $\partial B_j$ is a disk by the two-dimensional Schoenflies theorem. Thus each pair $(\partial B_j,D)$ is piecewise linearly homeomorphic to a sphere with a chosen hemisphere. Choose these boundary maps to agree on $D$. Extend them to the interiors by coning from an interior point, after compatible subdivisions. The resulting maps identify $B_1$ and $B_2$ with the two halves of a standard ball, glued along its equatorial disk. ConsequentlyThe same boundary parametrization proves a useful companion fact: if a ball is indented by removing a smaller ball attached along a boundary disk, the closure of what remains is again a ball. In the standard boundary-disk model, move the indented disk through a collar neighbourhood to a flat disk.

A handle decomposition is a filtration obtained by attaching handles $D^r\times D^{n-r}$ along $\partial D^r\times D^{n-r}$ in the boundary of the preceding stage. The number $r$ is the handle index; index-zero handles are disjoint balls, and index-$n$ handles close the remaining spherical boundary components. Starting with a triangulation of $M$, thicken its vertices, then its edges, and successively its higher-dimensional simplexes. In a barycentric subdivision these thickenings have precisely the indicated handle and attaching-region product structures. Adding collars between successive attachments gives a collared handle decomposition.

For a critical level handle embedding, assign successive distinct heights $t_j$ to the handles. At height $t_j$ the horizontal slice contains the $j$th handle and its attaching region. Between handle heights, the embedded boundary of the previously attached handles travels along its product collar; there is no topological change there. The lower boundary collar meets the attaching region of the next handle, and the upper collar leaves its new boundary. The stated ambient isotopy therefore separates all topology changes into horizontal handles and all intervening motion into collars. For an embedded surface, the local changes are the appearance of a disk, attachment of a band, and disappearance of a disk.

Here is the three-dimensional conclusion, with the distinction between the bounded and unbounded regions made explicit: a piecewise linear embedded $S^2$ in ${\mathbb{R}}^3$ bounds a piecewise linear three-ball; in the one-point compactification $S^3$, both complementary closures are three-balls. The unbounded closure in ${\mathbb{R}}^3$ itself is not a compact ball.

Put the embedded sphere in critical level handle embedding position. At a regular horizontal level its intersection with the plane is a finite disjoint union of simple closed curves. An innermost such curve bounds a planar disk whose interior misses the sphere. Cut the sphere along this curve and insert two slightly displaced copies of the disk. This replaces one sphere by two spheres. Repeat at regular levels between the critical heights, using innermost curves first. The number of curves at these chosen levels decreases at each operation, so the process stops. Each resulting sphere then lies in a slab containing at most one horizontal handle.

These terminal spheres bound balls directly. With no handle the surface is a disk transported through a collar and capped at its ends, the boundary of $D^2\times I$. A minimum or maximum merely supplies one of those caps. At a band level there are two planar boundary circles on one side and one on the other. Cap the two circles, straighten each planar disk using the two-dimensional Schoenflies theorem, and straighten the joining band in its planar collar. If the circles are not nested, this gives two balls joined along a boundary disk; if they are nested, it gives a ball with one boundary-attached ball removed. The gluing three-balls along a boundary disk argument and its indentation version show in both cases that the terminal sphere bounds a ball. This also explains why the planar position of the band matters: no assertion about an arbitrary knotted band is being made.

Reverse the disk surgeries. Suppose the two resulting spheres bound balls $A$ and $B$. Their disjoint boundaries imply that the balls are disjoint or one contains the other. Shrink the small displacement of the two surgery caps back to the original planar disk. In the disjoint case, the pre-surgery sphere bounds the union of two balls along that disk. In the nested case, it bounds the larger ball with the smaller boundary-attached ball removed. Both are balls by the preceding elementary facts. Induction through the finitely many surgeries proves the Three-dimensional Schoenflies theorem. To treat the other region in $S^3$, remove a point inside the ball just constructed and apply the same argument in the resulting copy of ${\mathbb{R}}^3$.

A 3-manifold is an irreducible three-manifold when every piecewise linear embedded two-sphere bounds a ball. Suppose $p:{\mathbb{R}}^3\to M$ is a universal cover, and lift an embedded sphere $S$ to $\tilde{S}$. Since $S$ is simply connected, its lift maps homeomorphically to $S$. By the Three-dimensional Schoenflies theorem, $\tilde{S}$ bounds a compact ball. The inverse image of $S$ is locally finite, so only finitely many other lifted spheres lie inside this ball. Choose an innermost one, bounding a ball $B$ whose interior meets no lift of $S$.

The interior of $B$ maps to one component $U$ of $M\setminus S$. This restriction is a covering map onto $U$: a path in $U$ starting in the image lifts into $\mathrm{int}B$, since it cannot cross a lifted boundary sphere. The covering is proper, because a compact subset of $U$ has preimage closed in the compact ball and cannot meet $\partial B$. Near $S$, the boundary map $\partial B\to S$ is one-to-one, so this connected covering has one sheet. The map on $B$ is therefore a homeomorphism onto $U\cup S$. Thus every embedded sphere in $M$ bounds a ball, and $M$ is an irreducible three-manifold.

We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

There is an orientation convention in this statement: the two volume forms must be positive for one fixed orientation, as is usual in Moser theorem for volume forms. Without that convention, the stated necessity is false. For example, on an oriented two-sphere, take a positive volume form ${\beta }_0$ and an orientation-reversing diffeomorphism $r$, and set ${\beta }_1=r^{\ast }{\beta }_0$. Then $(r^{-1}{)}^{\ast }{\beta }_1={\beta }_0$ but $\int {\beta }_1=-\int {\beta }_0$. We prove the intended result with a common orientation.

If ${\varphi }^{\ast }{\beta }_1={\beta }_0$, positivity forces $\varphi$ to preserve the orientation, and integration of the pullback of a differential form gives equality of the integrals.

For sufficiency, top-degree de Rham cohomology gives a differential form $\alpha$ of degree $n-1$ with ${\beta }_1-{\beta }_0=d\alpha$. Put ${\beta }_t=(1-t){\beta }_0+t{\beta }_1$. Positivity makes every ${\beta }_t$ a volume form. The interior product of a differential form gives an linear isomorphismThus there is a unique smooth time-dependent vector field $X_t$ satisfying ${\iota }_{X_t}{\beta }_t=-\alpha$. Since $M$ is compact, its local flow exists throughout $0\le t\le 1$ and defines a smooth isotopy ${\varphi }_t$ starting at the identity. By Cartan's magic formula, and since a top-degree differential form has zero exterior derivative,HenceThe zero-dimensional case, a single point, is immediate from equality of the integrals.

A ferromagnetic vacuum breaks the special unitary group $SU(2)$ to the subgroup of rotations about its magnetization, so its vacuum manifold is the two-sphere $S^2$. The massless field is therefore a unit vector $\mathbf{n}(x,t)$ with ${\mathbf{n}}^2=1$.

The only rotational scalar linear in $\dot{\mathbf{n}}$ that can be formed directly from $\mathbf{n}$ is $\mathbf{n}\cdot \dot{\mathbf{n}}=\frac{1}{2}{\partial }_t({\mathbf{n}}^2)=0$. A local one-form $\mathbf{A}(\mathbf{n})\cdot \dot{\mathbf{n}}$ can describe the required Berry phase, but no choice of $\mathbf{A}$ is globally nonsingular and strictly rotationally invariant on $S^2$. One therefore needs coordinate patches, an extension, or a redundant spinor parametrization.

There are exactly four homeomorphism types: the two-sphere, real projective plane, torus, and Klein bottle.

For completeness, enumerate all pairings. Four sides have three pairings; each pair can be matched in either direction, giving twelve signed pairings before symmetries. Up to cyclic relabelling, the repeated labels are either adjacent, of form $a{a}^{s}b{b}^t$, or alternating, of form $ab{a}^s{b}^t$, where $s,t\in \{1,-1\}$. The second adjacent pattern $abba$ rotates to the first. Endpoint identification gives the following table:Here the quotient always has two edge classes and one face, so its Euler characteristic is $V-2+1=V-1$. To reproduce the vertex count, label consecutive square vertices $0,1,2,3$. For paired sides $i,j$, a same-direction pairing identifies $i$ with $j$ and $i+1$ with $j+1$; a reversed-direction pairing identifies $i$ with $j+1$ and $i+1$ with $j$, with indices modulo four. Taking the generated equivalence classes gives exactly the displayed values of $V$.

The quotient is orientable precisely when each label is traversed once positively and once negatively around the boundary: then the face orientation extends across every paired edge. A same-direction pair obstructs such an orientation. Thus the two $\chi =0$ possibilities are distinguished as the orientable torus and nonorientable Klein bottle; the other rows are fixed by Euler characteristic and the classification theorem for surfaces.

Explicit square representatives areAll edge interior points have disk neighbourhoods by joining two half-disks, and each identified vertex has a cyclic sequence of corner sectors, again a disk neighbourhood. There are no leftover boundary edges. The enumeration therefore describes closed surfaces, and the twelve signed pairings leave no additional cases.