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A four-handle is an index-four handle. In a four-dimensional handle decomposition it is a four-ball attached along its whole boundary, a three-sphere. It caps a spherical boundary component left by earlier handle attachments.

Every closed connected orientable three-manifold is obtained by integer Dehn surgery on a finite framed link in the three-sphere. Compare a Heegaard splitting with a standard splitting of the sphere, factor their gluing difference using the Lickorish-Dehn theorem, and realize each twist by unit surgery on a distinct level of a collar of the splitting surface.

A knot is the image $K=f(S^1)$ of a smooth embedding $f:S^1\hookrightarrow {\mathbb{R}}^3$, considered up to ambient isotopy. The circle $S^1$ supplies a closed loop with no free ends; the smooth embedding condition rules out self-intersections and requires a nonzero tangent vector in a regular parametrization. The equivalent convention $S^1\hookrightarrow S^3$ uses the three-sphere: removing a point outside the knot identifies the remaining space with three-dimensional Euclidean space by stereographic projection.

More precisely, knots $K_0,K_1$ are equivalent if there is a smooth family of diffeomorphisms $F_s$ of the ambient space, for $0\le s\le 1$, such that $F_0$ is the identity and $F_1(K_0)=K_1$. Every intermediate $F_s(K_0)$ remains a smooth embedding. This ambient isotopy allows bending, stretching and moving a loop, but neither cutting it nor letting strands cross through each other. We compare the embedded images rather than their parametrizations; an oriented knot additionally remembers a direction of travel.

For a worked example, start with $U(t)=(\cos t,\sin t,0)$, where $t$ is taken modulo $2\pi$, and useBoth scale factors stay positive, so each $F_s$ is a diffeomorphism. At $s=1$ the circle becomes the ellipse $(2\cos t,\frac{1}{2}\sin t,0)$, yet its knot type is still the unknot. Exact lengths and angles are therefore irrelevant to knot type; self-intersection is the obstruction that the deformation must avoid.

Spatial geometry. An embedding of the three-sphere isTaking differentials and collecting terms gives the induced metric tensorPut $\bar{r}=A\sin \chi$. Then $d\bar{r}=A\cos \chi d\chi$, soThe areal radius $\bar{r}$ is not a global radial coordinate on the entire three-sphere: its maximum occurs at the equator and two hemispheres share each smaller value. Use separate charts there; the coordinate singularity at $\bar{r}=A$ is not a physical singularity.

For Euclidean space and hyperbolic space of curvature $-1/A^2$, respectively,For the latter, $\bar{r}=A\sinh \chi$ gives $d{\ell }_{-}^2=A^2[d{\chi }^2+{\sinh }^2\chi d{\Omega }^2]$. In all three cases the induced two-metric at fixed $\bar{r}=\bar{R}$ is ${\bar{R}}^{2}d{\Omega }^2$. Its surface area is thereforeThis equality compares the same areal radius, not the same proper radial distance. For a proper radial distance $\ell$, the corresponding areas are $4\pi A^2{\sin }^2(\ell /A)$, $4\pi {\ell }^2$, and $4\pi A^2{\sinh }^2(\ell /A)$.

Homogeneity, isotropy and redshift. The FRW metric has constant-time spatial slices of constant curvature $k/a^2(t)$, with $k=+1,0,-1$ selecting the three geometries above. Their isometries act transitively on points and isotropically about every point, implementing the cosmological principle. Multiplication by one common time-dependent scale factor preserves these properties. With the source's $k=\pm 1$ normalization, $r$ is dimensionless and $a$ has units of length; using a dimensionless scale factor instead rescales the spatial curvature parameter. The origin of the spherical coordinates is arbitrary; it does not identify a physical centre. Cosmic time is the proper time of the comoving observers, and the absence of time-space cross terms makes their worldlines orthogonal to the homogeneous slices.

For a radial null geodesic between comoving endpoints,The next wave crest travels between the same endpoints. Subtracting the two path integrals, for wave periods short compared with the expansion time, gives $\delta t_0/a_0=\delta t_E/a_E$. Thus the cosmological redshift isBoth observers' clocks are cosmic proper-time clocks. Peculiar velocities and local gravitational-potential shifts have been left out of this homogeneous calculation.

Angular size and brightness. For a galaxy small compared with its distance, the angular diameter distance is $D_A=a_{E}R$ and its angular diameter isHere the emitted proper diameter is measured transverse to the line of sight; radial proper distance is a different integral when $k\ne 0$. Ignoring absorption, lensing caustics and intrinsic evolution, photon-number conservation spreads an isotropic-equivalent luminosity across the sphere $4\pi a_0^2{R}^2$. Each photon loses energy by $1+z$, and its arrival interval is stretched by the same factor. The received bolometric flux isThe disc's proper projected area is ${\mathcal{A}}_d=\pi d^2/4$, and its observed solid angle is ${\Omega }_d\simeq {\mathcal{A}}_d/D_A^2$. Therefore its surface brightness isThis is cosmological surface-brightness dimming. The normalization uses $L$ as isotropic-equivalent luminosity along the viewing direction; another disc emission law changes that normalization but not the redshift power. This is a bolometric result. In a specified observed frequency band the emitted spectrum is sampled at a shifted frequency and requires a spectral correction.

The distinction in telescope aperture and resolved surface brightness matters here. A larger telescope collects more photons and improves angular resolution, but cannot restore the incoming radiance lost by cosmological dimming. For aperture diameter $D$ and a diffraction resolution element of solid angle ${\Omega }_{\mathrm{res}}\sim (\lambda /D{)}^2$, the collected power from uniform resolved brightness scales asindependent of $D$. This is consistent with conservation of etendue. It explains why increasing aperture while keeping every gain in resolution does not increase counts per native resolution element from a smooth extended source. It does not make larger telescopes useless: at a fixed angular bin the counts grow as $D^2$, so a photon-noise-limited or uniform-sky-noise-limited signal-to-noise ratio grows roughly as $D\sqrt{t_{\exp }}$. They also resolve smaller structures and improve point-source sensitivity. Aperture, exposure and angular binning help measure a dim galaxy, without changing its intrinsic $(1+z{)}^{-4}$ brightness penalty.

Use $c=1$ and signature $+---$. The positively curved FLRW metric in the usual radial chart isSet $r=\sin \chi$ and introduce conformal time by $dt=ad\eta$. Then $d{r}^2/(1-r^2)=d{\chi }^2$ in the chart, and the spherical polar coordinate $\chi \in [0,\pi ]$ extends the spatial description to the full three-sphere. ThusThe $r$ coordinate itself is not one-to-one across the equator, which is why the new polar coordinate is useful.

For $\Lambda =0$ and $k=1$, the Friedmann equation is ${\dot{a}}^2=(8\pi G/3)\rho a^2-1$. Since $a^{\prime }=a\dot{a}$, it becomesFor pressureless matter, the background continuity equation gives $\rho a^3={\rho }_0{a}_0^3$. Define the conserved normalization $M=(4\pi /3){\rho }_0{a}_0^3$. It makes the equation $(a^{\prime }{)}^2=2GMa-a^2$. The closed matter-dominated Friedmann solution is the cycloidIndeed $a^{\prime }=GM\sin \eta$ and $2GMa-a^2=(GM{)}^2{\sin }^2\eta$. Integrating $dt/d\eta =a$ gives the second expression, with the Big Bang at $t=\eta =0$. It expands to $a=2GM$ at $\eta =\pi$ and recollapses at $\eta =2\pi$. Here $M$ is the constant appearing in the Friedmann equation; it is not the total mass of the entire closed spatial slice, whose volume is $2{\pi }^2{a}^3$.

Every matrix in $SU(2)$ has the unique formThe four real coordinates of $(\alpha ,\beta )$ give a smooth bijection with smooth inverse onto the unit three-sphere. Matrix multiplication identifies this model with the unit quaternions. Its round metric is invariant under left and right multiplication by unit quaternions. The actionhas kernel $\{(1,1),(-1,-1)\}$: a kernel pair must have $a=b$ from $q=1$ and must commute with all quaternions. Its derivative is injective, and both dimensions are six, givingThe full isometry group is $O(4)$. Quaternion conjugation reverses orientation and supplies the additional component. Thus an arbitrary isometry has either the displayed left-right form or that form composed with conjugation.

For the Anti-de Sitter spacetime model, writeThe hyperboloid with this quadratic form equal to $-1$ is exactly $SL(2,\mathbb{R})$. Its induced Lorentzian metric is preserved by $M\mapsto AM{B}^{-1}$ for $A,B\in SL(2,\mathbb{R})$. At the identity the metric is $g(U,V)=\frac{1}{2}\mathrm{tr}(UV)$ on traceless matrices, so it is the Killing metric divided by eight. This proves the geometric identification, including the metric rather than only the underlying manifold.

A trivial left-right action again forces $A=B$ to be a scalar commuting with all determinant-one matrices, hence $A=B=\pm I$. The differential is injective and both dimensions are six, soThis is the existing left-right double cover of SO0(2,2). The full hyperboloid isometry group is $O(2,2)$. Its other components can be obtained by also using $M\mapsto M^T$ and $M\mapsto CM{C}^{-1}$, $C=\mathrm{diag}(1,-1)$. Transposition reverses one timelike coordinate; conjugation by $C$ reverses one timelike and one spacelike coordinate, so these generate the four component classes. All isometries of these constant-curvature quadrics come from ambient orthogonal transformations: a value and an orthonormal derivative at one point determine an isometry, and the ambient group realizes every such choice.

Here ${\mathrm{AdS}}_3$ means the hyperboloid with periodic time. The physically common causally unwrapped spacetime is its universal cover, modeled by the universal covering Lie group of $SL(2,\mathbb{R})$, and is not itself $SL(2,\mathbb{R})$. Its isometries lift the above local symmetries, with the global covering and central identifications changed accordingly.

Rescale the ambient coordinates by $q_i=a_i{r}_i$. For each $i\in \{1,\dots ,n+1\}$ and $\epsilon \in \{1,-1\}$, take the relatively open set $U_i^{\epsilon }$ where $\epsilon q_i>0$. Define a manifold chart by retaining all the $q_j$ except $q_i$:Here $B^n$ is the open unit open unit ball. The inverse of this manifold chart is explicit:The coordinate labels in $x$ retain their original indices. All the $a_i$ are nonzero, so this inverse exists; the strictly positive radicand makes it smooth on the whole open open unit ball. Projection and this inverse are continuous, so each manifold chart is a homeomorphism onto an open subset of ${\mathbb{R}}^n$.

These $2(n+1)$ manifold charts cover the ellipsoid, because at every point at least one $q_i$ is nonzero. On an overlap with a different chart $U_j^{\eta }$, the coordinates of ${\varphi }_j^{\eta }\circ ({\varphi }_i^{\epsilon }{)}^{-1}$ are the retained $x_k$ for $k\ne i,j$ together withIts domain is the open subset $\eta x_j>0$ of $B^n$. The displayed functions, and the reverse transition functions obtained by exchanging $i,j$, are smooth. Transitions between identical charts are the identity; opposite-sign charts with the same omitted coordinate do not overlap. The ambient subspace topology is Hausdorff and second countable, as it is inherited from Euclidean space. Thus this coordinate-projection atlas of an ellipsoid is a smooth atlas, proving the ellipsoid is a smooth manifold of dimension $n$. The rescaling $r\mapsto q$ also gives a diffeomorphism with the unit sphere $S^n$.

For the three-sphere, use its identification with SU(2) rather than separate local constructions. With the Pauli matrices ${\tau }_i$, put $T_i=-i{\tau }_i$. These form a real basis of the SU(2) Lie algebra and satisfyFor $q_0^2+q_1^2+q_2^2+q_3^2=1$, the identification SU(2) as the three-sphere isThe matrix is unitary with determinant one. Conversely, every SU(2) matrix has this form, and both the correspondence and its inverse are smooth.

The left translations $L_g(h)=gh$ are diffeomorphisms. Define three left-invariant vector fields byMatrix multiplication makes these vector fields smooth globally. Since $(d{L}_g{)}_e$ is an isomorphism from the SU(2) Lie algebra to $T_g\mathrm{SU}(2)$, the three vectors are linearly independent at every point, and each is nowhere zero. This is the parallelization of a Lie group by left translations.

Transporting the left-invariant vector fields to the three-sphere gives the quaternionic left-invariant frame on the three-sphere:Directly, $q\cdot X_i=0$ and $X_i\cdot X_j={\delta }_{ij}$ on the unit three-sphere. Thus the vectors are tangent and form a global orthonormal frame of a vector bundle. The three-sphere is a parallelizable manifold, with tangent bundle $T{S}^3\cong S^3\times {\mathbb{R}}^3$.

The degree of a map between oriented manifolds measures how many times the domain covers the target, with signs recording the local local orientation of a manifold. Let $M,N$ be connected, oriented closed manifolds of the same dimension $n>0$. A continuous map $f:M\to N$ acts on top-dimensional homology bywhere $[M],[N]$ are their fundamental classes. Connectedness and the choices of local orientation of a manifold identify $H_n(N;\mathbb{Z})$ with $\mathbb{Z}$. Reversing the orientation of either manifold changes the sign; reversing both does not.

For a smooth map, Sard theorem supplies a regular value $y$. Its inverse image is discrete and, by compactness, finite. At each $x\in f^{-1}(y)$ the differential is an isomorphism; let its sign be $+1$ or $-1$ according to whether it preserves or reverses the chosen local orientations. The degree as a sum of local degrees isThe sign is computed in oriented manifold charts. The value is independent of the chosen regular value, even when inverse images appear or disappear: the signed count is the coefficient of $[N]$ in $f_{\ast }[M]$.

There is a useful local-density expression for the same topological degree. If $\omega$ is a volume form with ${\int }_N\omega =1$, thenThis degree by integration of a pullback volume form follows first by choosing a smooth top-form supported in a small neighborhood of a regular value, where the inverse branches contribute their orientation signs. Any other normalized top-form differs from it by an exact form: integration identifies $H_{\mathrm{dR}}^n(N)$ with $\mathbb{R}$. The integral of its pullback difference vanishes by Stokes theorem. In particular, for any top-form $\eta$, ${\int }_M{f}^{\ast }\eta =\mathrm{deg}(f){\int }_N\eta$.

A homotopy $H:M\times [0,1]\to N$ preserves this integral, since $d\omega =0$ and Stokes theorem givesThus topological degree is a homotopy invariant. It is multiplicative under composition, because the induced maps on homology compose: $\mathrm{deg}(g\circ f)=\mathrm{deg}(g)\mathrm{deg}(f)$. The identity has degree one, a constant map has degree zero for $n>0$, and an orientation-reversing diffeomorphism has degree minus one. An orientation-preserving finite covering map has degree equal to its number of sheets. Nonzero topological degree forces surjectivity, since an omitted point would be a regular value with an empty inverse image.

For the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, positive or negative. This is its winding number, computable as $(2\pi i{)}^{-1}\int f^{-1}df$. The antipodal map of $S^n$ has degree $(-1{)}^{n+1}$: its extension $-I$ on the ambient $(n+1)$-dimensional vector space has that determinant sign and respects the outward-normal convention. A holomorphic map $z\mapsto z^k$, $k\ge 1$, on the Riemann sphere has degree $k$, whereas its complex conjugate has degree $-k$. These examples show how orientation, rather than simply the number of inverse images, determines the integer.

For maps $S^n\to S^n$, topological degree gives the complete homotopy classification ${\pi }_n(S^n)\cong \mathbb{Z}$. The degree does not classify general manifold maps: the identity of the torus and the map induced by the integer matrix $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but have different induced maps on $H_1(T^2;\mathbb{Z})$ and so are not homotopic. A nonzero-degree map $S^n\to S^n$ cannot extend continuously to $B^{n+1}$, because such an extension would make the boundary map null-homotopic. In the smooth setting, Stokes theorem gives the same obstruction by applying it to the pulled-back normalized volume form.

The hypotheses can be adjusted, but must be stated. For connected oriented noncompact manifolds, a proper map has a degree defined using compactly supported top-forms, and it is invariant under proper homotopies. For manifolds with boundary one uses relative fundamental classes and maps of pairs, or fixes appropriate boundary conditions. Without an integral orientation one can still count inverse images modulo two, obtaining a mod-two degree. The integer integral formula used below assumes the oriented setting.

In classical field theory, these ideas turn continuous fields into quantized topological charges. Suppose a field on ${\mathbb{R}}^d$ approaches one fixed target value at spatial infinity. The one-point compactification makes it a map $\varphi :S^d\to \mathcal{V}$. When the target $\mathcal{V}$ is an oriented closed $d$-manifold, its topological degree labels topological sectors. More generally the sectors are described by homotopy groups; an integer degree is available only when the domain and target have the appropriate dimensions and orientations. Smooth time evolution preserving the boundary condition is a homotopy, so it cannot change the integer. A change requires a singular field, escape from the allowed target, or a change at the boundary.

A normalized closed target $d$-form gives the pullback-volume representation of a topological current. On spacetime, put $\alpha ={\varphi }^{\ast }\omega$. Since $d\alpha ={\varphi }^{\ast }(d\omega )=0$, its dual current is identically conserved, andis independent of time when there is no flux at infinity. This conservation law follows from geometry without using the field equations; it need not arise from a continuous symmetry through Noether theorem.

A concrete example is the O3 nonlinear sigma model in two spatial dimensions. Its unit-vector field $\mathbf{n}$ approaches a constant at infinity, defining $S^2\to S^2$. The normalized area form of the target gives the degree charge of an O3 sigma-model lump:For the energy normalization $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, the identities $\mathbf{n}\cdot {\partial }_i\mathbf{n}=0$ giveThis is the Bogomolny degree bound for the O3 sigma model. Choosing the sign appropriate to $Q$ makes the square nonnegative; vanishing of the square gives first-order Bogomolny equations and a sigma-model lump saturating the bound. With the oriented stereographic projectionthe maps $w=z^k$ have $Q=k$ and $E=4\pi k$. Their conjugates have $Q=-k$ with the same energy. Holomorphic rational maps have positive degree equal to their degree as rational maps; taking a reciprocal does not reverse the orientation. Antiholomorphic dependence reverses it.

The Skyrme model supplies a three-dimensional example. A field $U:{\mathbb{R}}^3\to \mathrm{SU}(2)$ with $U\to I$ at infinity is a map $S^3\to \mathrm{SU}(2)\cong S^3$. Take $T_i=-i{\tau }_i$ and $U^{-1}dU={\theta }^i{T}_i$, with ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$ positive. Since $\mathrm{tr}(T_i{T}_j{T}_k)=-2{ϵ}_{ijk}$, the normalized target volume form isThe integral is one on the unit three-sphere. Consequently the Skyrme baryon number as a mapping degree isThis is the topological baryon number in the Skyrme model; the sign has been fixed by the stated orientation and anti-Hermitian generator convention.

A topological charge alone does not guarantee a stable finite-size solution. The degree and energetic stability of a field configuration concern different properties. For a three-dimensional configuration of size $R$, the two-derivative energy scales as $R$, so it can decrease by shrinking while the topological degree remains fixed for every $R>0$. The limit can be singular. The Skyrme term, with four derivatives, scales as $R^{-1}$ and can balance the shrinking tendency. This is the role of Derrick theorem in distinguishing topological obstruction from energetic stability.

For defects, the relevant boundary map can instead be the sphere surrounding a core. A vacuum manifold equal to $S^1$ gives the integer winding number of a vortex; a vacuum manifold $S^2$ gives the degree of a surrounding $S^2$ for a magnetic monopole. This vacuum-boundary degree as a defect charge obstructs extending the normalized vacuum field through the enclosed ball. A nonzero integer therefore forces the field to leave the vacuum manifold somewhere in the core. This construction does not require the field to take one constant value in every direction at infinity.

Degree also appears in four-dimensional gauge theory through a boundary transition function. For an anti-Hermitian SU(2) gauge connection on ${\mathbb{R}}^4$, write $F=dA+A\wedge A$ and assume finite-action boundary behavior $A\to g^{-1}dg$ on the large bounding three-sphere. In the second-Chern conventionthe identity $d\mathrm{tr}(A\wedge dA+\frac{2}{3}{A}^3)=\mathrm{tr}(F\wedge F)$ and the Maurer-Cartan equation giveThis boundary winding representation of Yang-Mills topological charge relates the Second Chern number to the degree of $g:S^3\to \mathrm{SU}(2)$. The Chern-Simons 3-form turns the bulk integral into the boundary winding integral. Conventions which define the instanton number with the opposite trace sign reverse $k$; the integer quantization is unchanged. A Yang-Mills theta term weights a sector by $e^{i\vartheta k}$, giving periodicity $\vartheta \mapsto \vartheta +2\pi$. Thus the same topological degree that counts oriented inverse images also labels field sectors and expresses their quantized charges as integrals of local densities.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

Use units $c=1$ throughout. Set $\chi =|\mathbf{x}|$, so the spatial Euclidean metric is $d{\mathbf{x}}^2=d{\chi }^2+{\chi }^{2}d{\Omega }^2$. On either hemisphere of the embedded three-sphere, the constraint gives $u^2=R^2-{\chi }^2$ and henceIntroduce the dimensionless comoving coordinate $r=\chi /R$. ThenThis is the positively curved Friedmann-Lemaître-Robertson-Walker metric, with $a(t)$ the physical curvature radius. An equally valid convention retains $r=\chi$, takes $a=b$, and sets $K=R^{-2}$; only the coordinate normalization differs. The chart with $r\le 1$ describes a hemisphere, not the entire closed slice. Writing $r=\sin \psi$ gives $d{\ell }^2=R^2(d{\psi }^2+{\sin }^2\psi d{\Omega }^2)$, with $0\le \psi \le \pi$, displaying the full spherical spatial slices of a closed FLRW universe.

In cosmic time, all three Friedmann-Lemaître-Robertson-Walker metrics can be writtenwhereThe closed spatial slice is a three-sphere of radius $a(t)$, soThe complete flat and open spatial slices are noncompact and have

For a homogeneous fluid in the Spatially flat FLRW metric, the equations in part (a) reduce to the cosmological perfect-fluid continuity equationwhile homogeneity makes the spatial relativistic Euler equation automatic. The Einstein field equations give the flat Friedmann equation and Friedmann acceleration equation,equivalently $\ddot{a}/a=-4\pi G(\rho +3P)/3$.

For this charged scalar field, $\sqrt{-g}=r^3\sin \theta \cos \theta$. On $e^{-i\omega t}$, the gauge-covariant derivative of a charged scalar field supplies the shifted frequencyThe wave equation becomesSeparation of variables with constant $\lambda$ givesandThe angular equation is the axisymmetric spherical harmonic equation on the three-sphere; regular solutions have $\lambda =\ell (\ell +2)$ with $\ell =0,2,4,\dots$, the even values selected by the absence of $\varphi$ and $\psi$ dependence.

A closed simply connected three-manifold is homeomorphic to the three-sphere. Now a theorem, this identifies every homotopy three-sphere with the standard sphere and removes homotopy-sphere ambiguities in three-dimensional connected-sum classification. Smooth structures are also unique in dimension three.

A closed connected orientable smooth manifold of dimension three is prime when a decomposition as a connected sum forces at least one summand to be the three-sphere. Unlike an irreducible three-manifold, it may contain a nonseparating embedded two-sphere.

A simply connected space is path connected and has trivial fundamental group: every closed loop contracts to a point. The three-sphere is simply connected, whereas Real projective space ${\mathbb{RP}}^3$ has fundamental group ${\mathbb{Z}}_2$.

The SO(3) group consists of real three-by-three orthogonal matrices with determinant one. It acts by rotation in three dimensions. Its group manifold is ${\mathbb{RP}}^3$, since the Adjoint double cover from SU(2) to SO(3) identifies antipodal points of the three-sphere.

A round three-sphere of radius $R$ embedded by $|\mathbf{x}{|}^2+u^2=R^2$ has metric $R^2\{d{r}^2/(1-r^2)+r^{2}d{\Omega }^2\}$ for $r=|\mathbf{x}|/R$. Multiplying by $b(t{)}^2$ gives a scale factor $a=Rb$ and normalized positive spatial curvature of an FLRW universe $K=1$. This radial chart covers a hemisphere; $r=\sin \chi$ gives the global polar description.

A smooth, piecewise-linear or locally flat embedded two-sphere in the three-sphere has two complementary components, each with closure homeomorphic to a three-ball. Equivalently a tame embedded two-sphere in ${\mathbb{R}}^3$ bounds a ball on its compact side. Local flatness is essential; arbitrary wild sphere embeddings are excluded.

For nonzero coprime integers $p,q$, the torus knot $T_{p,q}$ lies on a standard unknotted torus in the three-sphere and winds $p$ times meridionally and $q$ times longitudinally. Here a meridian of a solid torus goes around the small cross-section of the doughnut, while the longitudinal direction goes around its central hole. The integers count signed turns, so changing one sign gives the mirror of a link of the torus knot. In this article $p$ is always the meridional count and $q$ the longitudinal count.

An explicit parametric curve in three-dimensional Euclidean space iswhere $R>r>0$. Its cylindrical radius $\rho =\sqrt{x^2+y^2}$ satisfies $(\rho -R{)}^2+z^2=r^2$, the equation of the supporting torus. Its meridional angle is $pt$ and its longitudinal angle is $qt$: this checks the convention directly rather than relying on the picture. Its squared speed isIf two parameter values give the same point, their difference $\delta$ satisfies $p\delta ,q\delta \in 2\pi \mathbb{Z}$. Bézout's identity for the coprime integers $p,q$ then gives $\delta \in 2\pi \mathbb{Z}$. Thus the loop closes and is an injective function on the parameter circle, with nonzero derivative; it is a smooth embedding.

For example, $R=2$, $r=0.7$, $p=2$, $q=3$ produces a trefoil knot. During one circuit the meridional angle changes by $4\pi$ and the longitudinal angle by $6\pi$. The supporting torus is a guide for construction, not a constraint on later ambient isotopies: the knot may move off it. This family makes arithmetic visible as geometry: the coprime integers determine whether a single traversal visits the whole loop without repetition.

The coprime integers hypothesis matters. If the greatest common divisor is $d=\gcd (|p|,|q|)>1$, the same parametric curve retraces a single component $d$ times; it does not automatically draw $d$ different components. The corresponding torus link is instead the full set of points on the torus whose angles satisfy $q\varphi -p\theta \in 2\pi \mathbb{Z}$, with meridional angle $\varphi$ and longitudinal angle $\theta$. It has $d$ disjoint components, each with winding counts $(p/d,q/d)$. For instance, the torus link $T_{2,2}$ consists of two parallel $(1,1)$ loops and is a Hopf link, whereas substituting $(2,2)$ into ${\gamma }_{p,q}$ simply traces one of those loops twice.

The unknot $U$ is the knot type represented by the planar circleEquivalently, an unknot is a knot that bounds a smoothly embedded disk in the three-sphere. The disk must itself be embedded; merely filling a loop with a map that is allowed to intersect itself does not establish that the knot is an unknot. Flattening a neighborhood of an embedded spanning disk explains why this criterion recovers the planar circle.

A complicated drawing can still represent an unknot. Starting with a crossing-free knot diagram of $U$, insert a small kink by a type I Reidemeister move. The resulting knot diagram has one crossing, but removing the kink exhibits the same unknot. Repeating in separate small regions gives knot diagrams of the unknot with arbitrarily many crossings, although the crossing number of a knot is $c(U)=0$. A drawing's apparent complexity is therefore not a certificate of knotting.

The normalized Alexander polynomial of a knot and Jones polynomial of the unknot are both $1$, and its knot group is the infinite cyclic group $\mathbb{Z}$, because the knot exterior is a solid torus. These are useful comparison values. In particular, a nonconstant Alexander polynomial of a knot certifies that a knot is not an unknot, but the converse fails: nontrivial untwisted Whitehead doubles also have Alexander polynomial of a knot $1$.