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With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

Use the unit sphere $X^2+Y^2+Z^2=1$, its north pole $(0,0,1)$, and the equatorial complex plane $Z=0$. The line from the north pole to a sphere point meets that plane atThe north pole maps to $\infty$. The inverse stereographic projection isThe antipodal point has coordinate $w=-(X+iY)/(1+Z)$. Since $X^2+Y^2=(1-Z)(1+Z)$, this is $-1/\bar{z}$ when both sides are finite and nonzero. Zero and infinity correspond to the two poles and are exchanged by the same extended convention. The inverse formula also proves the converse, giving the antipodal stereographic coordinate relation

The requested rotation result is that sphere rotations correspond exactly to Möbius transformationsThe matrices $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ form $SU(2)$, and opposite matrices induce the same transformation. Thus the rotation group $SO(3)$ is identified with $SU(2)/\{\pm I\}$. This is the sphere rotations as special-unitary Möbius transformations result, which need not be proved here.

For a planar circle $|z-c|=r$, $c=a+ib$ and $r>0$, substitute $|z{|}^2=(1+Z)/(1-Z)$ and $\Re (\bar{c}z)=(aX+bY)/(1-Z)$. Multiplication by $1-Z$ gives the circle-plane relation under stereographic projection:The lift lies in this plane, and conversely its intersection with the sphere projects onto the circle. It is a nondegenerate circle on the sphere. The north pole is not on the displayed plane, since substitution there gives two rather than zero.

A great circle is the intersection with a plane through the sphere's centre. Here this holds exactly whenThe unit circle lifts to the equator. If a different circle lifts to a great circle, its plane and the equatorial plane are distinct planes through the origin, so their intersection cuts the sphere at two antipodal equatorial points. Their planar coordinates are $z$ and $-z$ on $|z|=1$.

Conversely, if the planar circle meets the unit circle in $z$ and $-z$, their lifts are antipodal equatorial points. The plane of the lifted circle contains their midpoint, the origin, so that circle is great. Equivalently, adding $|z-c{|}^2=r^2$ and $|-z-c{|}^2=r^2$ gives $r^2=1+|c{|}^2$. The exclusion of the unit circle ensures the two intersection points are a genuine pair rather than coincident circles. This proves the antipodal equatorial intersection criterion for a great circle.

Use the coefficients in the authoritative PDF, including their factor $i$. Write $a=2\sqrt{3}i$ andThe numerator and denominator are coprime: a common zero would, by subtraction, require $z=0$, where both equal one. Therefore the degree of a rational map of the Riemann sphere is four. Direct algebra givesThe domain transformations $z\mapsto iz$ and $z\mapsto (iz+1)/(1-iz)$ are sphere rotations as special-unitary Möbius transformations: respectively a quarter-turn about the third axis and a one-third turn cycling the three coordinate axes. In particular the latter sends $0\mapsto 1\mapsto i\mapsto 0$, the north, first-axis and second-axis directions. They generate the order-24 rotational symmetry group of a cube, isomorphic to the symmetric group $S_4$. Their target transformations are rotations: $w\mapsto 1/w$ is a half-turn about the first target axis and $w\mapsto \omega w$ is a one-third turn about the third target axis. This proves whole-map combined equivariance, not just symmetry of selected roots.

The target rotation image is the order-six dihedral group $D_3$. The spatial half-turns about all three coordinate axes act trivially on the map: $R(-z)=R(z)$ and $R(1/z)=R(z)$ generate a Klein four-group kernel. Thus the full Skyrmion symmetry combines the spatial cubic rotations with compensating isospin rotations, while its angular energy and Skyrme baryon density have pure spatial cubic symmetry.

The critical directions make the geometry explicit. Its Wronskian of a rational map isThe five finite ramification points are $0,\pm 1,\pm i$; infinity supplies the sixth, since in the local coordinate $w=1/z$ the map starts with $R=1+2a{w}^2+O(w^4)$. These are the six coordinate-axis directions, or face centres of a cube. The angular Jacobian of a rational map vanishes there, consistent with a cube-shaped shell whose density is concentrated away from its face centres. Any further rotational equivariance would have to preserve this set, so the spatial proper rotation group is exactly the cubic group already generated above.

There is also a reflection relation $R(\overline{z})=1/\overline{R(z)}$. Together with the proper rotations, it makes the angular density invariant under the full order-48 symmetry group of a cube, usually denoted $O_h$. The target operation in this reflection relation is orientation reversing; it should not be mistaken for a proper isospin rotation. In the full Skyrme model, reflections are expressed using the field parity operation $U(\mathbf{x})\mapsto U^{†}(-\mathbf{x})$ together with a compensating isospin rotation.

This is the cubic charge-four rational-map ansatz:The cubic rational-map ansatz for four Skyrmions is a useful approximation and starting point for the cubic four-Skyrmion, whose lowest spin-zero, isospin-zero quantized state models an alpha particle. A radial minimization and, for precision, unrestricted field relaxation are still required. The TeX's missing $i$ changes this map; the six critical directions alone would not detect the error, because the same Wronskian zero set persists when $a$ is real. The actual rotational equivariance identities are the stronger check.

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.