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Use the orientation selected by the volume form, and let ${\mathrm{vol}}_g$ be its normalized Riemannian volume form. The Euclidean metric induces an inner product on the exterior algebra of covectors. The Hodge star operator is the unique map from $p$-forms to $(4-p)$-forms satisfyingFor an oriented orthonormal coframe $e^1,\dots ,e^4$, it sends a basis wedge to the complementary wedge with the sign of the permutation that restores $e^1\wedge \cdots \wedge e^4$. Applying the Hodge star operator twice exchanges blocks of $p$ and $4-p$ covectors, soHere the supplied volume is interpreted in the usual metric-normalized sense. If instead one defines the operator with an arbitrary unnormalized $\mathrm{vol}=f{\mathrm{vol}}_g$, that operator is $f{\ast }_g$ and its square on two-forms is $f^2$. The usual self-duality statements use the metric-normalized Hodge star operator, with the supplied volume specifying orientation.

The projections $P_{\pm }=(1\pm \ast )/2$ give the Hodge splitting of Euclidean two-forms:Both spaces have dimension three. With $e^4=dt$, bases areThe Hodge star operator is an orthogonal involution on two-forms and therefore is self-adjoint. For a self-dual two-form $H$ and an anti-self-dual two-form $G$,ConsequentlyThis is the wedge orthogonality of opposite-duality two-forms.

For the Yang-Mills action, take an anti-Hermitian special unitary group connection and the fundamental matrix trace, so the positive invariant pairing on its Lie algebra is $⟨X,Y⟩=-\mathrm{tr}(XY)$. Write its gauge curvature as $F=F_{+}+F_{-}$ using the Hodge splitting of Euclidean two-forms, and defineCross terms vanish by the wedge orthogonality of opposite-duality two-forms. ThusWith this anti-Hermitian convention the Second Chern number isFor example, this normalization follows by expanding $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. Interpreting the integral as an integer Second Chern number on ${\mathbb{R}}^4$ assumes the usual decay and gauge behavior that allow extension over the point at infinity. The following norm inequality itself does not require integrality:The Yang-Mills instanton Bogomolny bound is saturated precisely when one component vanishes: $F=\ast F$ or $F=-\ast F$. In the stated convention the self-dual case has $k\le 0$ and the anti-self-dual case has $k\ge 0$. Reversing orientation, or defining topological charge with the opposite sign, reverses this assignment while leaving the absolute-value bound unchanged.

For the self-dual Yang-Mills equations in temporal gauge, choose ${\mathrm{vol}}_g=d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge dt$ and ${ϵ}_{123}=1$. The relevant Hodge star operator identities areWriting $F=F_{ti}dt\wedge d{x}^i+\frac{1}{2}{F}_{jk}d{x}^j\wedge d{x}^k$, the self-dual Yang-Mills equations $F=\ast F$ giveIn temporal gauge, $A_t=0$, so the gauge curvature component is $F_{ti}={\partial }_t{A}_i-{\partial }_i{A}_t+[A_t,A_i]={\partial }_t{A}_i$. HenceThe minus sign follows from placing $dt$ last in the orientation and first in the mixed curvature component.

Write $w$ for the complex coordinate. The inverse stereographic projection iswith $\infty$ mapped to the north pole. A rotation $R_z(\theta )$ about the third axis sends $w$ to $e^{i\theta }w$, represented byFor the rotation $R_y(\theta ):(x,y,z)\mapsto (x\cos \theta +z\sin \theta ,y,-x\sin \theta +z\cos \theta )$, substitution in the stereographic projection givesBoth displayed matrices are unitary with determinant one, hence belong to the special unitary group.

Use the rotation-generation result that every $R\in SO(3)$ can be written $R_z(\alpha )R_y(\beta )R_z(\gamma )$, with angles chosen suitably. This is the Euler-angle decomposition, which is allowed here. Composition of Möbius transformations corresponds to multiplication of their matrix representatives, soThe formulas extend at poles in the Riemann sphere. This realizes sphere rotations as special-unitary Möbius transformations. The two representatives $U$ and $-U$ define the same Möbius transformation, as expected for the double covering of the special orthogonal group.

For a complex finite-dimensional simple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal commuting subalgebra of elements whose adjoint maps are semisimple. Equivalently in this setting it is a nilpotent self-normalizing subalgebra. Its dimension is the rank of a semisimple Lie algebra. Simultaneous diagonalization of its Adjoint representation gives the root-space decompositionA root of a root system is a nonzero linear functional $\alpha$ for which this root space is nonzero. For a complex semisimple algebra each root space is one-dimensional. A Cartan-Weyl basis consists of a basis $H_i$ of $\mathfrak{h}$ and one nonzero root vector $E_{\alpha }$ for every root.

The general Lie brackets have the formFor the opposite-root bracket, use the Killing form to define $h_{\alpha }$ by $\kappa (h_{\alpha },H)=\alpha (H)$. Its invariant bilinear form on a Lie algebra property givesOne may normalize the root vectors so that the pairing is one. If instead one uses a coroot as the opposite-root bracket, the root-vector normalization changes accordingly. In particular, root evaluation coordinates cannot simply be used as coefficients in a nonorthonormal Cartan basis.

For the matrix calculation take $N\ge 2$. The complexification of a Lie algebra of the special unitary group Lie algebra is the special linear Lie algebra ${\mathfrak{sl}}_N(\mathbb{C})$: traceless complex matrices. Its Cartan subalgebra consists of traceless diagonal matrices. Write $E_{jk}={\mathcal{T}}^{(j,k)}$ for the matrix units. The given Cartan basis is $H_i=E_{ii}-E_{i+1,i+1}$, $1\le i<N$, and the other basis elements are $E_{jk}$ with $j\ne k$.

The matrix-unit identity $E_{jk}{E}_{lm}={\delta }_{kl}{E}_{jm}$ givesThus all the roots, expressed as evaluation vectors in this precise Cartan basis, areThey are the functionals $e_j-e_k$ on traceless diagonal matrices; there are $N(N-1)$ of them. The corresponding root vector is $E_{jk}$. Together with $N-1$ Cartan generators, they give $N^2-1$ basis elements. The simple roots can be chosen as ${\alpha }_{i,i+1}$, whose evaluation vectors are the rows of the type-$A_{N-1}$ Cartan matrix, with $2$ on the diagonal and $-1$ on adjacent entries. These vectors are evaluations on $H_i$, not coordinates in an orthonormal realization of the root system.

To express every bracket strictly in the chosen basis, introduce the abbreviationThen all pairs are covered byHere both input root vectors have distinct row and column indices. The first case is the only one producing diagonal matrix units, and the displayed sum of $H_p$ resolves them completely into the chosen Cartan basis. Reversing the order gives the negative bracket. This also shows explicitly that the two nonzero non-Cartan cases have structure constants $+1$ and $-1$, and verifies the required root-addition rule.

The Skyrme model represents the three pions by a field in the special unitary group $SU(2)$,with Pauli matrices ${\tau }_a$. Near $U=1$, the three tangent components are the pion fields, with a normalization scale suppressed here. Define $L_{\mu }=U^{†}{\partial }_{\mu }U$. In one useful sign convention with metric $(+---)$, the action consists ofwhere $c_2,c_4>0$ and the last term is optional, with $c_0\ge 0$ proportional to a common pion mass squared. The first term is a nonlinear sigma model kinetic term, and the second is the four-derivative term of the Skyrme model. For a static configuration of size $s$, the two derivative contributions scale as $E_2\propto s$ and $E_4\propto 1/s$, whereas a mass term scales as $s^3$. This Derrick scaling explains why the four-derivative term can stabilize a finite size instead of allowing collapse.

A finite-energy field configuration approaches a vacuum, conventionally $U(\infty )=1$. Compactifying space makes $U$ a map $S^3\to SU(2)\simeq S^3$. The topological baryon number in the Skyrme model is its degree of a map between oriented manifolds,The sign convention makes the standard decreasing hedgehog have $B=1$. Smooth evolution with the vacuum boundary condition preserves this integer. A Skyrmion is a localized soliton in such a sector; the unit soliton, after quantization, models a nucleon, and higher positive charges model multi-baryon systems. The topological conservation law is distinct from an ordinary Noether charge of isospin. Classical pion fields are bosonic, so obtaining fermionic nucleons also needs the quantum-statistics choice discussed below.

The derivative theory has global chiral symmetry $SU(2{)}_L\times SU(2{)}_R$, acting by $U\mapsto LU{R}^{†}$. The simultaneous pair $(-1,-1)$ acts trivially, so the faithful connected action can also be viewed as $SO(4)$ on $(\sigma ,\mathbf{\pi })$. In the massless theory the choice of vacuum breaks this to the vector subgroup. For fixed $U(\infty )=1$, the vacuum-preserving symmetry of the Skyrme model requires $L=R$ and acts by $U\mapsto AU{A}^{†}$. This is isospin, effectively $SO(3)$ because $A$ and $-A$ act the same way. A usual common pion-mass term explicitly preserves only this vector subgroup; full chiral symmetry is then an approximate massless-limit symmetry, not an exact symmetry of that term. Independent axial rotations change the vacuum and are not extra localized rigid-rotor coordinates in a sector with fixed boundary vacuum.

The space-time symmetry is the Poincaré group, comprising translations and Lorentz transformations, including spatial rotations. Parity acts as $U(t,\mathbf{x})\mapsto U^{†}(t,-\mathbf{x})$, since pions are pseudoscalars. For a static finite-energy solution, translations change its position; spatial rotations and isorotations change its orientation. These transformations generate collective coordinates, but a particular field can be unchanged by certain combined transformations.

For the usual low-charge minimum branches of the standard model, the relevant shapes and density symmetries are the following. These are not claims about every field of a given degree, all excited solutions or arbitrary modified pion potentials.

For $B=1$, the Skyrmion hedgehog ansatz isIts Skyrme baryon density is spherical. Indeed $B=-(2/\pi ){\int }_0^{\infty }{f}^{\prime }{\sin }^{2}fdr=1$. A spatial rotation rotates the pion direction, so the field itself is invariant under a compensating isorotation, not under every spatial rotation alone. Its proper combined stabilizer is a diagonal $SO(3)$.

For $B=2$, the toroidal two-Skyrmion has a ring-shaped density with an axial hole and full density symmetry $D_{\infty h}$. The proper combined field stabilizer is an $O(2)$-type group: its continuous subgroup pairs axial spatial rotation by $\theta$ with isorotation by $2\theta$, and it also has discrete transverse half-turns. The angular approximation $R(z)=z^2$ makes the axial factor two explicit. A static two-baryon minimum is consequently not simply two separate round unit lumps.

For $B=3$, the tetrahedral three-Skyrmion has a tetrahedral shell with four face-hole directions and full density group $T_d$. Its proper rotational group is $T\cong A_4$, the tetrahedral symmetry group of order 12. The actual field symmetries again pair these rotations with isorotations.

For $B=4$, the cubic four-Skyrmion has a cubic shell with six face-hole directions. Its full density group is $O_h$, the symmetry group of a cube, and its proper group is $O\cong S_4$, the rotational symmetry group of a cube of order 24. Cubic and octahedral symmetry name the same point group; this particular density shape is cubic. Reflections in these full density groups should not be confused with the proper rotation-isorotation group used for angular-momentum quantization.

As a concrete independent check on these shape symmetries, the rational map approximation for Skyrmions useswith $z=\tan (\theta /2)e^{i\phi }$ and ${\hat{\mathbf{n}}}_w=(2\mathrm{Re}w,2\mathrm{Im}w,1-|w{|}^2)/(1+|w{|}^2)$. Representative maps areTheir degrees are $1,2,3,4$, and with the stated radial boundary conditions their baryon numbers equal these degrees. The angular Jacobian of a rational map is $J_R=[(1+|z{|}^2)|R^{\prime }|/(1+|R{|}^2){]}^2$. It vanishes at branch directions. $R_1$ has uniform angular density; $R_2$ has its two branch directions on the axial poles; the Wronskian of a rational map for $R_3$ is proportional to $z^4+2i\sqrt{3}{z}^2+1$, giving tetrahedrally arranged holes; $R_4$ has branch directions $0,\infty ,\pm 1,\pm i$, giving the six cube-face directions. Also $R_4(iz)=1/R_4(z)$ pairs a spatial quarter-turn with an isospin half-turn. These give illustrative approximate fields, not exact analytic solutions or a proof of global energy minimality.

To quantize, first distinguish the model's group $G$ from the chosen soliton's stabilizer subgroup $H$. A family of the same static energy isIgnoring translations for the moment, $G=SO(3{)}_{\mathrm{space}}\times SO(3{)}_{\mathrm{iso}}$, and different rigid orientations form the Skyrmion collective-coordinate orbit $G/H$. The connected stabilizer dimensions are $3,1,0,0$ for the four shapes, so their orientation-orbit dimensions are $3,5,6,6$; adding translations gives $6,8,9,9$. For the unit hedgehog, treating rotation and isorotation as six independent modes would double-count its locked orientation. For higher charges, arbitrary separation of constituent lumps is not an exact flat moduli space in this non-Bogomolny theory.

Let the orbit coordinates depend slowly on time. Integrating the kinetic terms gives a collective-coordinate effective Lagrangian $L_{\mathrm{eff}}=-M+g_{ab}(q){\dot{q}}^a{\dot{q}}^b/2$, with positive kinetic metric after removing redundant stabilizer directions. Its angular blocks are the spatial, isospin and mixed inertia tensors. Collective-coordinate quantization produces a rotor Hamiltonian on this orbit, with translations giving center-of-mass momentum. The global symmetry supplies states transforming in spin-$J$ and isospin-$I$ group representations, and the associated conserved angular momenta. Inertia tensors determine energy splittings; group symmetry alone does not specify those tensors or their numerical energies.

The topology and the static stabilizer supply further collective-rotation constraints for a Skyrmion. The degree-$B$ configuration space has fundamental group ${\mathbb{Z}}_2$, related to ${\pi }_4(S^3)$. Choosing the nontrivial Finkelstein-Rubinstein constraints character gives fermionic unit baryons. Wavefunctions live on the appropriate cover, and a lifted stabilizer operation obeysThe sign depends on whether the actual field-configuration loop is contractible, not just on whether the density looks symmetric. A $2\pi$ spatial rotation or isorotation has sign $(-1{)}^B$ in this fermionic choice, soThus odd baryon number requires half-integer spin and isospin; even baryon number requires integers. Discrete or continuous combined stabilizer symmetries impose additional restrictions on the allowed pairs and body-fixed rotor states. They must not simply be omitted, or replaced by trivial invariance under every density symmetry. The trivial character is a possible bosonic quantization but would not produce a fermionic nucleon. Improper field symmetries, implemented using the model parity transformation, can additionally constrain parity labels; scalar-density reflection symmetry alone is insufficient to infer those labels.

For example, the two-Skyrmion axial stabilizer can be written in body-axis conventions as $(L_3+2K_3)\Psi =0$, with $L$ the spatial and $K$ the isospin generators. A transverse spatial half-turn accompanied by an isospin half-turn has the nontrivial sign in the Finkelstein-Rubinstein constraints. The integer-spin scalar state $J=I=0$ is therefore excluded despite even $B$. A spin-one, isospin-zero state with zero axial body projection has the required minus sign under that half-turn; the spin-zero, isospin-one channel can also satisfy the constraints. Their relative energies require the inertia tensors.

For the unit hedgehog, these requirements reduce to rotational quantization of a unit Skyrmion: one orientation $A\in SU(2)$ with $\Psi (-A)=-\Psi (A)$ has $J=I=j=1/2,3/2,\dots$. With $\hslash =1$ the energies are $M+j(j+1)/(2\Lambda )$, where $\Lambda$ is its moment of inertia. Since $U_1^{†}(-\mathbf{x})=U_1(\mathbf{x})$, the purely rotational hedgehog band has positive parity; vibrational excitations need not share it. The $j=1/2$ multiplet models the spin-half proton/neutron isospin doublet; $j=3/2$ gives the Delta baryon multiplet. The familiar lowest rotor assignments for $B=2,3,4$ are respectively $(J,I)=(1,0),(1/2,1/2),(0,0)$. These assignments incorporate the corresponding field-stabilizer constraints and standard inertia ordering, not a prediction from density shape alone. Deformations, vibrational modes, radiation and binding dynamics lie beyond the rigid approximation. The two symmetry roles are therefore

The internal direct product theorem states that if $H,K⊴G$, $H\cap K=\{1\}$, and $HK=G$, then $G\cong H\times K$. Indeed, for $h\in H,k\in K$, the commutator $[h,k]$ lies in both $H$ and $K$, so it is trivial. The map $(h,k)\mapsto hk$ is consequently a homomorphism; the intersection condition makes it injective and $HK=G$ makes it surjective. Conversely, the two factors in a direct product satisfy these conditions.

In odd dimension, $-I$ is central and has determinant $-1$. Every $Q\in O(3)$ has a unique expression $Q=(-I{)}^{\epsilon }R$ with $R\in SO(3)$, soBy contrast, $SO(2)\times C_2$ is abelian, while reflections in $O(2)$ conjugate a rotation to its inverse. Thus $O(2)\ncong SO(2)\times C_2$.

The center of the unitary group consists exactly of scalar unitary matrices:The determinant gives a surjective homomorphismwhose kernel is the special unitary group $SU(2)$.

Nevertheless $U(2)$ is not isomorphic to $SU(2)\times \mathbb{T}$. The center of $U(2)$ has exactly one nonidentity element of order two, namely $-I$. The center of the proposed product is $\{\pm I\}\times \mathbb{T}$, which has three nonidentity elements of order two. Since an isomorphism preserves the center and element orders,

The special unitary group in two dimensions isLet $V_1={\mathbb{C}}^2$ be its defining group representation, and for $n\ge 0$ letthe homogeneous polynomial representation of SU2 on homogeneous polynomials of degree $n$ in two variables. It has degree of a representation $n+1$. The representations $V_0,V_1,V_2,\dots$ are pairwise nonisomorphic, continuous, finite-dimensional irreducible representations, and the classification of finite-dimensional representations of SU2 says that this is the complete list.

Every $A\in SU(2)$ is conjugate to $\mathrm{diag}(z,z^{-1})$ with $|z|=1$. The monomial basis of $V_n$ gives the character of the homogeneous polynomial representation of SU2Writing $z=e^{i\theta }$, this is $\sin ((n+1)\theta )/\sin \theta$, interpreted by continuity when $\sin \theta =0$. The Clebsch-Gordan decomposition for SU2 is

A compact topological group is a group $G$ equipped with a compact Hausdorff topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous.

The special unitary group isIt is closed in the finite-dimensional real vector space $M_n(\mathbb{C})$ because the equations $U^{\ast }U=I$ and $\det U=1$ are closed conditions. It is bounded because every column of a unitary matrix has norm one. The Heine-Borel theorem therefore makes it compact. Matrix multiplication is continuous, and inversion is continuous because $U^{-1}=U^{\ast }$. Hence $SU(n)$ is a compact topological group.

Let $N=(0,\dots ,0,1)$ and $S=(0,\dots ,0,-1)$ be the poles of $S^n$. The open sets $U_N=S^n\setminus \{N\}$ and $U_S=S^n\setminus \{S\}$ cover the sphere. Stereographic projection gives coordinate chartsfrom these sets to ${\mathbb{R}}^n$. Their inverses areOn the overlap, the transition map iswhich is a smooth diffeomorphism of ${\mathbb{R}}^n\setminus \{0\}$. These two compatible charts make $S^n$ a smooth manifold of dimension $n$.

Now let $G$ be an $m$-dimensional Lie group and choose a basis $e_1,\dots ,e_m$ of its tangent space $T_{e}G$ at the identity. Definewhere $L_g$ is Left translation on a Lie group. Smoothness of multiplication makes each $E_i$ a smooth left-invariant vector field, and invertibility of $(d{L}_g{)}_e$ makes $E_1(g),\dots ,E_m(g)$ a basis of $T_{g}G$ at every point. Thus the $E_i$ form a global frame and every Lie group is a parallelizable manifold.

The columns of a matrix in the special unitary group $SU(2)$ are orthonormal and its determinant is one. Consequently every element has the unique formThe pair $(z_1,z_2)\in {\mathbb{C}}^2\cong {\mathbb{R}}^4$ therefore identifies $SU(2)$ diffeomorphically with $S^3$. The Lie-group construction then proves that $S^3$ is parallelizable; this is the SU(2) as the three-sphere identification.

Another example is $S^1$, which is the Lie group $U(1)$. Explicitly, at $(x,y)\in S^1$ the vectoris smooth, tangent, and nowhere zero, so it is a global one-vector frame. Thus $S^1$ is another parallelizable sphere.

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.

The special unitary group is $SU(2)=\{U\in M_2(\mathbb{C}):U^{†}U=I,\det U=1\}$. Differentiating these equations at the identity shows that its Lie algebra isFor $T_i=-i{\sigma }_i/2$, the Pauli-matrix identity $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$ givesso the structure constants in this basis are $c^k{}_{ij}={ϵ}_{ijk}$.

The real Lie algebra of the special unitary group consists of traceless skew-Hermitian matrices. Differentiating unitarity and determinant one gives the two constraints; exponentiating any such matrix proves their sufficiency. The commutator is again skew-Hermitian and traceless. Its real dimension is $n^2-1$, and its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$.

Under the stereographic projection $w=(x+iy)/(1-z)$, rotations about the third axis have representatives $\mathrm{diag}(e^{i\theta /2},e^{-i\theta /2})$, and rotations about the second axis have representativesThese belong to the special unitary group $SU(2)$. Euler-angle generation of the special orthogonal group $SO(3)$ shows that every sphere rotation becomes a Möbius transformation represented in $SU(2)$. Representatives $U$ and $-U$ induce the same transformation.

The four-by-four unitary matrices of determinant one form this special unitary group. Its two-index antisymmetric representation has complex dimension six and the real structure of the SU4 exterior square realizes it on a six-dimensional real space.

For $n\ge 2$, the special unitary group acts transitively on the complex unit sphere. Extend any unit vector to an orthonormal basis, then adjust the phase of a remaining column to give determinant one. The stabilizer subgroup of the first vector consists of special unitary transformations of its orthogonal complement. Thus the sphere is a homogeneous quotient, not just a set with a norm-preserving action.