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The special orthogonal group in five dimensions isIts Lie algebra consists of real skew-symmetric matrices. There are $5\cdot 4/2=10$ independent entries above the diagonal, so $\dim SO(5)=10$. Equivalently, the orthogonality equations impose fifteen independent constraints on twenty-five matrix entries; the determinant condition chooses a component without changing the dimension.

Fix the fifth coordinate. The matrices $\mathrm{diag}(S,1)$, $S\in SO(4)$, form an explicit subgroup. Under this subgroup the defining vector space splits as ${\mathbb{R}}^4\oplus \mathbb{R}{\mathbf{e}}_5$, so its branching rule is $5\downarrow SO(4)=4\oplus 1$. An element of the so5 Lie algebra can be written uniquely asConjugation by $\mathrm{diag}(S,1)$ sends $B$ to $SB{S}^{-1}$ and $w$ to $Sw$. The first summand is the six-dimensional Adjoint representation of a Lie algebra of $\mathfrak{so}(4)$, and the second is its four-dimensional vector representation. Hence the SO5 to SO4 branching gives

For the left SU(2) subgroup of SO(4), identify ${\mathbb{R}}^4$ with the quaternions. Left multiplication by a unit quaternion is a real orthogonal transformation and gives an embedded SU(2) group. More generally, $q\mapsto aq{b}^{-1}$ gives the double cover $SU(2{)}_L\times SU(2{)}_R\to SO(4)$ with kernel $\{(1,1),(-1,-1)\}$. After complexifying, the vector representation is $(2,2)$ and the Adjoint representation is $(3,1)\oplus (1,3)$. Restricting to the left factor turns the right factor into a multiplicity space. ThereforeThese are decompositions into complex irreducible representations; the real $4$ is the underlying real representation of a quaternionic doublet. Combining the branching rules gives

Write a weight as $\lambda =(x,y)$. The integrality conditions for the B2 root system give $2x\in \mathbb{Z}$ and $y-x\in \mathbb{Z}$. Hence $x=m/2$, $y=m/2+n$, with $m,n\in \mathbb{Z}$. ThusThis is the B2 weight lattice, with an integer square lattice and a second square lattice shifted by $(1/2,1/2)$. The eight roots of a root system areThe short roots lie on the coordinate axes and the long roots on the diagonals. The positive roots for the given simple-root choice are $\alpha$, $\beta$, $\alpha +\beta$, $2\alpha +\beta$.

The integrality conditions determine the weight lattice of the Lie algebra, equivalently of the simply connected Spin group $\mathrm{Spin}(5)$. For the global special orthogonal group $SO(5)$, a $2\pi$ rotation in either coordinate plane is the identity, so a genuine group representation requires integer $x,y$. Thus the half-integer coset contains spin representations that do not descend to $SO(5)$. Both representations requested here have integer weights, so their diagrams are unaffected by this distinction.

The root system of the displayed $SO(4)$ subgroup is $\{\pm (1,1),\pm (1,-1)\}$. The two orthogonal pairs give its two commuting $\mathfrak{su}(2)$ factors. Choose the left factor to have root $\delta =(1,1)$; exchanging the two diagonal pairs exchanges left and right. Its coroot pairs with a weight asThis demonstrates the diagonal-root SU(2) embedding in SO(5) directly. The short-axis root $\alpha$ would instead give $2x$, and therefore a different subgroup: on the vector representation it would produce a triplet and two singlets rather than two doublets and a singlet.

For the vector representation, simultaneously rotate the first and second coordinate planes. Over $\mathbb{C}$, the two planes give opposite pairs of weights, while the fifth coordinate is fixed. Hence its weight diagram iswith every weight multiplicity equal to one. Evaluating $x+y$ gives $+1$ twice, $-1$ twice and $0$ once, exactly two SU(2) representations of dimension two and one singlet.

For the Adjoint representation, the root-space decomposition has one one-dimensional space for each of the eight roots and a two-dimensional zero-weight Cartan subalgebra. ThusThe coroot values $x+y$ have multiplicities $1,2,4,2,1$ at $-2,-1,0,1,2$. One zero-weight state joins the $\pm 2$ states to make a triplet; the $\pm 1$ states form two doublets, leaving three zero-weight singlets. This verifies the earlier branching rule and accounts for all ten dimensions.

Let $V(\varphi )=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, where ${\varphi }^2={\sum }_i{\varphi }_i^2$. Its stationary-point equation is $(m^2+\lambda {\varphi }^2){\varphi }_i=0$.

For $m^2>0$, the unique minimum is $\varphi =0$. The quadratic mass term is $m^2{\delta }_{ij}$, so there are $n$ real scalar fields, all of mass $\sqrt{m^2}$, and the global special orthogonal group $SO(n)$ remains unbroken.

For $m^2<0$, write $v^2=-m^2/\lambda$. The vacuum manifold is the sphere ${\varphi }^2=v^2$. Choose $⟨\varphi ⟩=v{e}_n$. The transformations preserving this vacuum expectation value rotate the first $n-1$ components, giving $SO(n)\to SO(n-1)$. At this vacuum the potential's Hessian matrix isWriting $\varphi =({\pi }_1,\dots ,{\pi }_{n-1},v+\eta )$, the radial mode $\eta$ therefore hasThese are one massive radial scalar and $n-1$ physical Goldstone bosons. The Goldstone theorem applies to the broken global generators; the flat angular directions of the vacuum manifold explain their zero masses. The spontaneous symmetry breaking is a choice of vacuum, not a change in the invariant Lagrangian. For $n=1$ there are no continuous angular directions: only the radial scalar remains, and the potential's discrete $\varphi \mapsto -\varphi$ symmetry is broken.

For every real $\alpha$, direct substitution gives the axially equivariant monomial rational map identityThe first is a spatial rotation through $\alpha$ about the third axis compensated by a target rotation through $n\alpha$, hence by an isospin rotation of the Skyrmion. The second pairs the spatial half-turn about the first axis with a target half-turn about that axis. For $n>1$, these generate the continuous axial rotation group together with perpendicular half-turns, often written $D_{\infty }$; it is the group of rotations preserving an axis in the three-dimensional special orthogonal group. Under rotations about the axis, the pure spatial stabilizer of the map itself includes the cyclic group $C_n$, whereas the larger symmetry just described is combined spatial-target equivariance. Also $R(\overline{z})=\overline{R(z)}$ gives a paired reflection symmetry of its angular density.

The angular Jacobian of a rational map isIt depends only on latitude. Put $u=\log |z|$; then $J_R=n^2{\cosh }^{2}u/{\cosh }^2(nu)$. Since $\cosh (nu)\ge \cosh u$ for $n\ge 1$, the maximum is $n^2$ at $u=0$, the equator. For $n>1$ it vanishes at the two poles, producing an axially symmetric ring of angular Skyrme baryon density. For $n=1$, $R$ is the identity: every spatial rotation is compensated by the same target rotation, and the rational map approximation for Skyrmions becomes the fully spherically symmetric Skyrmion hedgehog ansatz. The charge and uses areThe degree-one construction captures the symmetry of the unit Skyrmion exactly; its radial profile still has to be solved. The degree-two map supplies the appropriate toroidal two-Skyrmion symmetry and useful initial data. Higher monomials give axial charge-$n$ competitors, but axial symmetry need not minimize the energy: the usual lowest-charge examples already include the tetrahedral three-Skyrmion and the cubic four-Skyrmion. Topological charge determines neither shape nor energy optimality by itself.

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 multi-indices $\alpha ,\beta$, defineThese seminorms define the Fréchet space topology of the Schwartz space: ${\phi }_j\to \phi$ exactly when every $p_{\alpha ,\beta }({\phi }_j-\phi )$ tends to zero. An equivalent increasing family isA tempered distribution is a continuous linear functional on this space. Equivalently, $|⟨T,\phi ⟩|\le C{q}_m(\phi )$ for some $C,m$. The usual weak convergence of tempered distributions means $⟨T_j,\phi ⟩\to ⟨T,\phi ⟩$ for every fixed Schwartz function. The strong dual topology instead requires uniform convergence on every subset of the Schwartz space that is a bounded set in a topological vector space; the Fourier maps below are continuous in both topologies.

Using $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, differentiation under the integral and integration by parts giveThe omitted coefficient has modulus one. By the Leibniz rule, every integrand is a finite sum of a polynomial times a derivative of $\phi$. Inserting the integrable weight $(1+|x|{)}^{-n-1}$ bounds its L1 norm by finitely many Schwartz space seminorms. In particular $q_m(\hat{\phi })\le C_m{q}_{m+n+1}(\phi )$. Thus the Fourier transform maps $\mathcal{S}$ continuously into itself.

For completeness, the Fourier inversion theorem follows here by Gaussian regularization. The inverse transform of $e^{-t|\xi {|}^2}$ is $g_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$. The Fourier transform of a Gaussian and Fubini's theorem showAs $t\downarrow 0$, the right side tends to $\phi (x)$ by the approximate identity property, while the left side tends to the undamped inverse integral by dominated convergence theorem, since $\hat{\phi }\in L^1$. HenceThe inverse is $(2\pi {)}^{-n}$ times reflection composed with the continuous Fourier transform, and is therefore continuous on the Schwartz space. This proves a continuous linear isomorphism with continuous inverse.

Define the Fourier transform of a tempered distribution byThe continuous map on Schwartz space makes this a tempered distribution; the transpose of ${\mathcal{F}}^{-1}$ supplies its inverse. Pointwise convergence of pairings proves weak continuity. For the strong dual topology, the transform of a bounded set of Schwartz functions is bounded, so uniform convergence of pairings on bounded sets proves continuity of both Fourier maps there too.

Now let $R\in SO(n)$ be a rotation matrix in the special orthogonal group and write ${\rho }_{R}T=T\circ R$. A change of variables with unit Jacobian givesThe rotation equivariance of the Fourier transform on tempered distributions follows by duality:Therefore $\widehat{{\rho }_{R}T}={\rho }_R\hat{T}$. Applying this identity and the inverse Fourier transform gives the two directions:This is precisely preservation of radial tempered distributions. For $n>1$, $SO(n)$ acts transitively on spheres, so an invariant smooth function is an ordinary radial function.

Use the energy of a curve normalizationFor a variation $F(s,t)$ put $T={\partial }_{t}F$ and $V={\partial }_{s}F$. The Levi-Civita connection is torsion free, so ${\nabla }_{s}T={\nabla }_{t}V$. Differentiating energy once gives $E^{\prime }=[⟨V,T⟩{]}_a^b-\int ⟨V,{\nabla }_{t}T⟩$. Along a geodesic, ${\nabla }_{t}T=0$. Differentiating again, commuting covariant derivatives, and integrating by parts gives the second variation of geodesic energy:The curvature convention is $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, consistent with the specified positive sectional curvature. Indeed ${\nabla }_s{\nabla }_{t}V={\nabla }_t{\nabla }_{s}V+R(V,T)V$, and $⟨R(V,T)V,T⟩=-⟨R(V,T)T,V⟩$. The endpoint term vanishes for fixed endpoints or for periodic variations of a closed geodesic. The integral is the Riemannian index form $I(V,V)$.

Let $\dim M=2m$. Parallel transport around the closed geodesic preserves the metric and orientation, so lies in the special orthogonal group. It fixes the nonzero tangent $T$, and its restriction to $T^{\perp }$ is an orientation-preserving orthogonal map of odd dimension $2m-1$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues occur in conjugate pairs, while an odd-dimensional real orthogonal map with determinant one must have an eigenvalue $+1$. Choose a nonzero fixed vector $v$ normal to $T$ and parallel-transport it along the curve. It gives a nonzero smooth periodic normal field $V$ with ${\nabla }_{t}V=0$.

For the exponential variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$, the endpoints match periodically. Strictly positive sectional curvature givesThus $E({\gamma }_s)<E(\gamma )$ for small nonzero $s$. By the Cauchy-Schwarz inequality,where the final equality uses the geodesic's constant speed. This proves the instability of a closed geodesic in positive even-dimensional curvature.

For an embedded closed geodesic, sufficiently small variations remain embeddings, hence give a smooth isotopy with strictly shorter curves. For a nonembedded closed geodesic the construction gives a smooth deformation through immersions; an isotopy class of embeddings is not literally defined for such a curve. The stated isotopy conclusion therefore uses the usual embedded-curve interpretation, while the shorter-loop variation holds without it.

The orthogonal group isThe special orthogonal group is its determinant-one subgroup

Every eigenvalue of an orthogonal matrix has modulus one. A real three-by-three matrix has at least one real eigenvalue, and nonreal eigenvalues occur in a complex conjugate pair. If $Q\in SO(3)$ has such a pair $e^{\pm i\theta }$, their product is one, so the remaining eigenvalue is $\det Q=1$. If all three eigenvalues are real, each is $\pm 1$ and their product is one; an odd number of three signs with product one must include $+1$. Hence every element of $SO(3)$ has an eigenvector of eigenvalue one and represents a rotation in three dimensions about its span.

It is false that every element of $O(3)$ is either a rotation or a plane reflection. For example,is an improper orthogonal transformation combining a rotation with a reflection. Its eigenvalues are $e^{\pm i\theta }$ and $-1$, whereas a plane reflection has eigenvalues $1,1,-1$ and a rotation has determinant one.

Let ${\mathrm{Sym}}_n$ be the vector space of real symmetric matrices and considerAt $R\in O(n)$,This derivative is surjective: for $S\in {\mathrm{Sym}}_n$, choosing $H=RS/2$ gives $D_R\Phi (H)=S$. Thus $I$ is a regular value and the orthogonal group $O(n)={\Phi }^{-1}(I)$ is a submanifold of codimensionThe determinant takes only the values $\pm 1$ on $O(n)$, so the special orthogonal group $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore

The tangent space is the kernel of $D_R\Phi$:where the Special orthogonal Lie algebra $\mathfrak{so}(n)$ is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.

Because the bilinear form from part (ii) is invariant, $\rho (SU(2))\subseteq O(3)$. The mapis continuous. Since $SU(2)$ is connected and $\det \rho (I_2)=1$, it is identically one. Hence

To find the kernel, suppose $U$ commutes with every traceless skew-Hermitian matrix. Commutation withforces $U$ to be diagonal. Commutation withthen forces its two diagonal entries to agree. Thus $U=z{I}_2$. Since $U\in SU(2)$,and therefore

It remains to prove surjectivity. The derived representation at the identity isIf this vanishes for every $A$ in the SU(2) Lie algebra, the same commutant calculation makes $X$ scalar; tracelessness then gives $X=0$. Thus $d{\rho }_{I_2}$ is injective. Its domain and codomain are both three-dimensional real Lie algebras, so it is an isomorphismThe inverse function theorem now shows that the image of $\rho$ contains a neighbourhood of the identity in $SO(3)$. Therefore the image is an open subgroup. The special orthogonal group $SO(3)$ is connected, since every rotation can be deformed continuously to the identity by reducing its rotation angle. A connected topological group has no proper open subgroup, soTogether with the kernel calculation, this is the Adjoint double cover from SU(2) to SO(3).

Write $\xi =V_1$, $\eta =\varphi$, and $u_j=d^{j}u/d{x}^j$. On the jet space of a scalar ordinary differential equation, definewhere the total derivative operator isThe order-$N$ prolongation of a vector field is

In the coordinates $(x,u)$, the given action of the special orthogonal group isDifferentiating at $s=0$ gives the infinitesimal generator of a Lie point symmetryso $\xi =u$ and $\eta =-x$. The recursive formula givesandTherefore the second prolongation of the rotation generator for plane graphs is

For ${\Delta }_0=u_{xx}$,which vanishes when ${\Delta }_0=0$. Forwe obtainwhich likewise vanishes on the equation. This is the infinitesimal invariance criterion for both equations.

Geometrically, $u_{xx}=0$ describes straight lines. The signed curvature of a plane graph isso the second equation says $\kappa =1$ and describes consistently oriented arcs of unit circles. Rotation preserves straight lines, circles, and signed curvature, explaining both invariances.

The special orthogonal group in three dimensions isDifferentiating $A(t{)}^{T}A(t)=I$ at the identity shows that its Lie algebra is the space of skew-symmetric matrices. A convenient basis iswith $[J_a,J_b]={ϵ}_{abc}{J}_c$.

The infinitesimal action of $\exp (t{J}_a)$ on a point $x\in {\mathbb{R}}^3$ is $J_{a}x=e_a\times x$. It therefore generates the vector fieldFundamental vector fields for this left action form an antihomomorphism with the stated convention, and direct differentiation givesso their span is closed under the Lie bracket of vector fields.

The brackets $\{x,y\}=z$, $\{y,z\}=x$, and $\{z,x\}=y$ define the rotational Lie-Poisson structure on R3. With the convention that a Hamiltonian vector field acts by $X_H(f)=\{f,H\}$,Hence the required Hamiltonians are simplyThe quadratic functionsatisfies $\{F,x_a\}=0$ for all $a$, so it is a Casimir function of a Poisson manifold. Its nonzero regular level sets $F=R^2$ are spheres. The Poisson tensor has rank two there and is tangent to each level set, so it inverts to a symplectic form; each sphere is a symplectic leaf. Rotations preserve both $F$ and the alternating tensor ${ϵ}_{abc}$, hence preserve the restricted Poisson tensor and its inverse symplectic form. The $SO(3)$ action therefore restricts to a symplectic action on every sphere $S_R^2\subset {\mathbb{R}}^3$.

The special orthogonal group isIfits first column is a unit vector, say $(\cos \theta ,\sin \theta {)}^T$. The second column is the unique unit vector perpendicular to it that gives positive determinant, namely $(-\sin \theta ,\cos \theta {)}^T$. Thusso $A$ is rotation through $\theta$ about the origin.

For $A\in SO(3)$, its real characteristic polynomial of odd degree has a real eigenvalue. Every eigenvalue of an orthogonal matrix has modulus one, so every real eigenvalue is $\pm 1$. Nonreal eigenvalues occur in conjugate pairs whose product is one, while $\det A=1$; if all eigenvalues are real, their product likewise forces at least one to be $+1$. Hence $A$ fixes a nonzero vector $u$. The plane $u^{\perp }$ is $A$-invariant, and the restriction to it is an orientation-preserving orthogonal map. By the $SO(2)$ result it is a planar rotation. Therefore $A$ is a rotation about the axis $\mathbb{R}u$.

In the CP1 spinor representation,where the ${\sigma }_i$ are Pauli matrices. The transformation $z\mapsto e^{i\alpha (x,t)}z$ leaves $\mathbf{n}$ unchanged. For $U\in SU(2)$,with $R$ in the special orthogonal group $SO(3)$, so $z\mapsto Uz$ induces $n_i\mapsto R_{ij}{n}_j$.

A rotationally invariant first-order term is the Ferromagnetic Wess–Zumino termUnder the phase redundancy it changes as ${\mathcal{L}}_{\mathrm{WZ}}\mapsto {\mathcal{L}}_{\mathrm{WZ}}-\kappa \dot{\alpha }$. The change is a total derivative, so the action has the required invariance.

Consider $F(A)=A^{T}A$ with values in the symmetric matrices. At the identity,which is onto the $n(n+1)/2$-dimensional space of symmetric matrices. The same rank holds at every orthogonal matrix. Thus $O(n)=F^{-1}(I)$ is a manifold, and its determinant-one component, the special orthogonal group, has dimensionThe tangent equation at the identity is $X^T+X=0$, so

The special orthogonal group isIts Lie algebra consists of antisymmetric $5\times 5$ matrices, determined by the entries above the diagonal. Therefore

Zero Dirichlet form of the Kac collision operator forces invariance under every pair rotation; strong continuity upgrades almost every angle to every angle. Coordinate-plane rotations generate the special orthogonal group, whose action is transitive on each sphere for $N\ge 2$. Averaging over normalized Haar measure therefore identifies the invariant $L^2$ functions with radial functions. Conversely radial functions are fixed by every pair rotation.

For $n>1$, a tempered distribution $T$ on ${\mathbb{R}}^n$ is radial when $T\circ R=T$ for every $R\in SO(n)$, where $⟨T\circ R,\phi ⟩=⟨T,\phi \circ R^t⟩$. This agrees with the usual radial function definition for regular smooth distributions, because the special orthogonal group acts transitively on spheres.

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.