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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

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

A group homomorphism is a multiplication-preserving map $\varphi :G_1\to G_2$, so $\varphi (gh)=\varphi (g)\varphi (h)$. A group isomorphism is a bijective homomorphism; its inverse is automatically multiplication-preserving as well.

For the real general linear group, the identity matrix has nonzero determinant. If $A,B$ have nonzero determinants, multiplicativity of the determinant gives $\det (AB)=\det A\det B\ne 0$. Associativity is inherited from matrix multiplication, and every such matrix has an inverse, with $\det A^{-1}=1/\det A\ne 0$. Hence the given set $\mathrm{GL}(n,\mathbb{R})$ is a group.

For the special orthogonal group, the identity is included andwhen both factors are special orthogonal. Moreover $A^{-1}=A^T$ is again orthogonal and has determinant one. Thus $\mathrm{SO}(n)$ is a subgroup. The orientation-preserving general linear group is also a subgroup: the identity has positive determinant, products preserve positivity, and inversion replaces the determinant by its positive reciprocal.

Now take an element $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ of the special orthogonal group in two dimensions. Orthogonality and determinant one implyHence $d=a$, $b=-c$, and the unit length of the first column gives $a^2+c^2=1$. There is an angle $\theta$, unique modulo $2\pi$, such that $a=\cos \theta$, $c=\sin \theta$. ThereforeDefine the map from the circle group byIf $z=a+ic$ and $w=p+iq$, matrix multiplication gives $\varphi (z)\varphi (w)=\left(\begin{array}{cc}ap-cq& -(aq+cp)\\ aq+cp& ap-cq\end{array}\right)=\varphi (zw)$. Thus it is a homomorphism. The entries determine $a,c$ uniquely, proving injectivity, and the parametrization proves surjectivity. This establishesThe use of the unit complex number, rather than an unrestricted real angle, removes the $2\pi$ ambiguity: angles add modulo $2\pi$, exactly as unit complex numbers multiply.

The supplied matrix obeysSeparate the even and odd powers in the matrix exponential to obtainIt fixes the $z$ component and is a counterclockwise active rotation viewed from $+z$. This sign differs from the clockwise convention in b, because the source fixes the sign of both $J_3$ and the exponential. Replacing $\theta$ by $-\theta$ yields the clockwise rotation.

With $(J_k{)}_{ij}=-i{ϵ}_{kij}$, the generators of rotations obeyThe commutator measures the leading difference between performing two infinitesimal rotations in opposite orders. For example, with $X=-i\alpha J_1$ and $Y=-i\beta J_2$,The failure to commute is itself an infinitesimal rotation about the perpendicular axis; the Lie algebra closes on the same three generators of rotations.

To pass from this local statement to the full group property, put $A_k=-i{J}_k$. These are real antisymmetric matrices, with $[A_i,A_j]={ϵ}_{ijk}{A}_k$. The Baker--Campbell--Hausdorff formula shows locally that the logarithm of a product of their exponentials remains a linear combination of the $A_k$, because all nested commutators do. Globally, each exponential has $R^{T}R=I$ and $\det R=e^{\mathrm{tr}A}=1$. If $R_1,R_2$ have these properties, thenEvery real three-dimensional orthogonal matrix of determinant $+1$ has an eigenvector with eigenvalue $+1$: its nonreal eigenvalues come in conjugate pairs, and the product of all three is $+1$. On the perpendicular plane it acts as a two-dimensional rotation. It therefore has axis-angle form and can be written $e^{-i\theta {\hat{n}}^k{J}_k}$. Thus the product is again a rotation. The identity is $\theta =0$, the inverse reverses $\theta$, and associativity comes from matrix multiplication. This proves that all these rotations form the special orthogonal group $SO(3)$. The commutator supplies local closure, while the orthogonality and axis-angle argument supplies the necessary global step.

A matrix $R$ belongs to the special orthogonal group in four dimensions precisely whenDifferentiating the orthogonality relation along a path $R(ϵ)=I+ϵX+O({ϵ}^2)$ gives $X^{\mathsf{T}}+X=0$. Thus its Lie algebra consists of real antisymmetric four-by-four matrices. Such a matrix has one free entry for each unordered coordinate pair, giving dimension $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$.

Each displayed generator has a single two-coordinate antisymmetric block. The $K$ matrices cover planes $(1,2),(1,3),(1,4)$; the $L$ matrices cover $(3,4),(2,4),(2,3)$, with their printed orientations. These are all six distinct planes. Therefore the generators are linearly independent and span the full antisymmetric space: they form a basis of $\mathfrak{so}(4)$.

Conversely, exponentiating any real antisymmetric $X$ gives $(e^X{)}^{\mathsf{T}}{e}^X=I$ and $\det (e^X)=e^{\mathrm{tr}X}=1$. This verifies that the infinitesimal matrices generate proper rotations rather than reflections.

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.

For the block-rotation maximal torus in the special orthogonal group $SO(2m+1)$, the Weyl group acts by permutations and independent inversions of $z_i$. Its invariant Laurent polynomials are the symmetric polynomials in $x_i=z_i+z_i^{-1}$, hence the polynomial ring in their elementary symmetric polynomials $e_1,\dots ,e_m$. Restriction of the group representation ring of a compact group is injective because every group element is conjugate into the torus and characters distinguish virtual representations. For the defining complexified representation $V$, the exterior-power characters satisfy ${\sum }_k{\chi }_{{\Lambda }^{k}V}{t}^k=(1+t){\prod }_i(1+x_{i}t+t^2)$. For $k\le m$, its coefficient is $e_k$ plus an integral linear combination of $e_j$ for $j<k$. Induction puts every $e_k$ in the restriction image, proving equality with the invariant ring and giving polynomial generators $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$.

In Euclidean geometry, a rotation about the origin is an orientation-preserving orthogonal linear map, represented by an element of the special orthogonal group. About another center $a$, the map has form $x\mapsto a+R(x-a)$. In three dimensions a nonidentity rotation fixes an axis.

The group of Euclidean isometries preserving orientation of ${\mathbb{R}}^n$. Its elements have the form $x\mapsto Ax+b$, with $A$ in the special orthogonal group and $b\in {\mathbb{R}}^n$. The function composition law is $(A,b)(B,c)=(AB,Ac+b)$, the identity element is $(I,0)$, and the inverse element is $(A^{-1},-A^{-1}b)$. These formulas prove closure and inverses; function composition supplies associativity. For $n=2$, $A$ is a planar rotation.

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.

A real symmetric tensor of the special orthogonal group transforms as $Q\mapsto OQ{O}^T$. If its vacuum expectation value has distinct eigenvalues with multiplicities $m_1,\dots ,m_p$, its stabilizer subgroup is $S(O(m_1)\times \cdots \times O(m_p))$: each transformation preserves each eigenspace and the product of the block determinants is one. Its identity component is ${\prod }_{i}SO(m_i)$. Thus a scalar multiple of the identity preserves the whole group, whereas a tensor with simple spectrum preserves only diagonal signs with total sign one. A tracelessness condition constrains the eigenvalues but does not change this stabilizer formula.