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A rigid Skyrmion orientation is quantized on a cover of its Skyrmion collective-coordinate orbit. A lifted combined rotation in the static field's stabilizer subgroup identifies the same classical field and imposes ${\hat{D}}^J(R){\hat{D}}^I(A)\Psi ={\chi }_{\mathrm{FR}}\Psi$, where the sign is the chosen Finkelstein-Rubinstein constraints character of that configuration-space loop. Choosing the nontrivial character models unit baryons as fermions. A $2\pi$ spatial rotation or isorotation of charge $B$ then has sign $(-1{)}^B$: deforming the field to separated unit lumps adds the unit-lump loop classes in the fundamental group ${\mathbb{Z}}_2$, while the labelled orbital paths can be contracted in three dimensions. The resulting Finkelstein-Rubinstein constraints sign is the product of the unit-lump signs. Hence spin and isospin are half-integer for odd $B$ and integer for even $B$. Further stabilizer constraints restrict which pairs $(J,I)$ and which body-fixed states occur. The energy operator comes from the inertia tensors on the orbit; symmetry selects allowed states but does not by itself determine their energies. A bosonic choice of the trivial topological character is mathematically possible but does not model a fermionic nucleon.

The mass second-moment tensor about a centre of mass is $I_{jk}=\int y_j{y}_{k}dm$. It measures the shape of a mass distribution and appears in tidal torque theory. It differs from the mechanical inertia tensor, which is ${\mathcal{I}}_{jk}=\mathrm{tr}(I){\delta }_{jk}-I_{jk}$. For a uniform ellipsoid of mass $M$ and semiaxes $a_1,a_2,a_3$, its principal values are $M{a}_j^2/5$.

In a frame centered at the fixed point or centre of mass, a rotating material point has velocity $\dot{r}=\omega \times r$. The identity $|\omega \times r{|}^2=|\omega {|}^2|r{|}^2-(\omega \cdot r{)}^2$ converts the rotational kinetic energy into $T=\frac{1}{2}{\omega }_i{I}_{ij}{\omega }_j$, where the inertia tensor isLikewise integrating $r\times (\omega \times r)$ gives the angular momentum $\boxed{L=I\omega }$. These expressions refer to rotational motion about the chosen origin.

Place the origin at the centre and the axes along the edges. Symmetry makes the off-diagonal entries of the inertia tensor zero. Uniform density gives $⟨x^2⟩=a^2/12$, $⟨y^2⟩=b^2/12$, $⟨z^2⟩=c^2/12$. Consequently the principal moments of inertia areFor example $I_1=\rho \int (y^2+z^2)dxdydz$ gives the first expression directly.

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 Poincare 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

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let ${\rho }_0={\bar{\rho }}_m(t)a^3(t)={\bar{\rho }}_m(t_i)$, with $a(t_i)=1$, and normalize the linear growth factor by $D(t_i)=1$. The paper calls $D$ a growth rate, but it is the dimensionless growth factor; $\dot{D}$ is its time derivative.

Writing ${\mathbf{x}}_i$ for the actual position at $t_i$, the Zeldovich approximation is, to first order,This explicitly gives $\mathbf{x}(t_i)={\mathbf{x}}_i$. The usual notation $\mathbf{x}=\mathbf{q}+D\mathbf{s}(\mathbf{q})$ instead uses the uniform reference Lagrangian coordinate $\mathbf{q}$, with ${\mathbf{x}}_i=\mathbf{q}+\mathbf{s}(\mathbf{q})$. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine $\mathbf{s}$, mass conservation between the initial and current positions givesDemanding $\delta (t)=D(t){\delta }_i$ gives $\nabla \cdot \mathbf{s}=-{\delta }_i$. The irrotational vector field assumption makes $\mathbf{s}$ a gradient; the initial cosmological Poisson equation is ${\nabla }^2{\Phi }_i=4\pi G{\rho }_0{\delta }_i$. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,The evolution of $D$ follows from the linearized peculiar-motion equation $\ddot{\mathbf{x}}+2H\dot{\mathbf{x}}=-\nabla \Phi /a^2$. Since cosmological Poisson equation gives $\Phi (t)=(D/a){\Phi }_i$ to this order, substitution yields the linear growth equationThus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark-matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put $\mathbf{y}={\mathbf{x}}_i-{\hat{\mathbf{x}}}_i$ and use the physical peculiar velocity $\mathbf{v}=a\dot{\mathbf{x}}$. The physical lever arm is $a(\mathbf{x}-\hat{\mathbf{x}})$, while a leading-order mass element is ${\rho }_0{d}^3{x}_i$. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass givesTerms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because ${\int }_{V_L}\mathbf{y}{d}^3{x}_i=0$ at this order.

Apply the divergence theorem componentwise:The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element $d\mathbf{S}$,The original PDF prints a different coefficient, $-\dot{D}/(4\pi G{\bar{\rho }}_m{a}^2)$. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical $V_L$ centered on its barycentre, the outward normal is parallel to $\mathbf{y}$, so $\mathbf{y}\times d\mathbf{S}=0$ pointwise. HenceFor an equipotential surface, ${\Phi }_i={\Phi }_{\ast }$ on ${\Sigma }_L$, andThereforeThe second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:The constant has no force; $\mathbf{g}$ has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensorwe obtainThe second equality swaps the indices $k,l$ and uses the symmetry of both tensors. Define the initial acceleration tidal tensor byThen the formula in the question has exactly the stated sign:If one defines $T$ as the positive Hessian matrix of ${\Phi }_i$ instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's $I$ is a mass second-moment tensor; the mechanical inertia tensor is $\mathrm{tr}(I)1-I$.

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where $I=\mathrm{diag}(I_1,I_2,I_3)$, for example,Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe, $a\propto t^{2/3}$, $D\propto a$, and ${\bar{\rho }}_m\propto a^{-3}$. The initial $I$ and $T$ are time independent, soThis is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

Write the position of a mass element relative to the center of mass as $\mathbf{r}$, so its inertial-frame velocity is $\dot{\mathbf{X}}+\mathbf{\omega }\times \mathbf{r}$. The center-of-mass identity $\int \mathbf{r}dm=0$ removes all cross terms. Define the inertia tensor about the center of mass byThe total angular momentum about the origin and kinetic energy are then

Let $\rho (r)=C{r}^{-1}\sin (\pi r/R)$. Spherical symmetry makes $I=I_{0}1$. The mass normalization isso $C=M/(4R^2)$. A moment about any diameter isTherefore

Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integralvanish because its integrand is odd in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its inertia tensor at the centre has the isotropic form

Forpart (a) and $\lambda =M{\ell }^2/6$ giveThe displayed symmetric matrix has orthogonal eigenvectorswith corresponding eigenvaluesFor a unit angular velocity, $|\mathbf{L}|=|I\mathbf{\omega }|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is

Scale the unit ball by $(u,v,w)\mapsto (au,bv,cw)$. Symmetry makes all products of inertia vanish, while the unit-ball averages satisfy
$⟨u^2⟩=⟨v^2⟩=⟨w^2⟩=1/5$. The inertia tensor about the centre is therefore

A principal axis is an eigenvector direction of the inertia tensor. Choosing three orthogonal principal axes diagonalizes the inertia tensor.