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A mathematician associated with Rodrigues' formula and the Rodrigues rotation formula.

Use the standard ordered basis $({\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3)$ for both maps. The reflection in a hyperplane interchanges the second and third coordinates, so its matrix isOrient the rotation axis along $\mathbf{n}=(1,1,1)/\sqrt{3}$ and use the right-hand convention. The rotation matrix for angle $2\pi /3$ sends ${\mathbf{e}}_1$ to ${\mathbf{e}}_2$, ${\mathbf{e}}_2$ to ${\mathbf{e}}_3$, and ${\mathbf{e}}_3$ to ${\mathbf{e}}_1$. For example, the Rodrigues rotation formula givesMultiplying the rotation matrix by the dilation factor givesDirect multiplication gives $A^2=I$, $B^2=4\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)$ and $B^3=8I$. Hence $B^3=8A^2$. If the opposite axis orientation is chosen, $B=2R^T$ also has cube $8I$. Independently of coordinates, two applications of the reflection give the identity, while three applications of the rotation-dilation give a full turn and dilation by $2^3$. Every basis represents these resulting maps by $I$ and $8I$. Thus the equality even holds when the two matrices are expressed in different bases.

Identify ${\mathbb{R}}^3$ with the traceless Hermitian matrices $X=x\cdot \sigma$, where ${\sigma }_i$ are the Pauli matrices. Then $\det X=-|x{|}^2$. For $U\in SU(2)$ the action $X\mapsto UX{U}^{†}$ preserves trace, Hermiticity and determinant, so gives an orthogonal transformation of ${\mathbb{R}}^3$. Since $SU(2)$ is connected, the determinant of this transformation is $+1$.

An element in the kernel commutes with every ${\sigma }_i$, hence is scalar. Unitarity and determinant one restrict it to $\pm I$. Every three-dimensional rotation is a rotation through some angle $\theta$ around a unit axis $n$. It is obtained byIndeed multiplication with ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ gives the Rodrigues rotation formula for $UX{U}^{†}$. This proves surjectivity, not just equality of Lie-algebra dimensions. ThereforeA full $2\pi$ rotation lifts from $I$ to $-I$, explaining the double cover and the Spin group interpretation.

A positively oriented planar rotation matrix acts byFor a unit vector $n$ in three dimensions, the parallel and perpendicular orthogonal projections areOn the plane perpendicular to $n$, the map $v\mapsto n\times v$ preserves lengths and turns vectors through a positive right angle about $n$. It squares to minus the identity there. Thus $\cos \theta x_{\perp }+\sin \theta (n\times x_{\perp })$ rotates that component, while the parallel component remains fixed. This proves the Rodrigues rotation formulaIts component form uses the Levi-Civita symbol:Since $n_i{n}_i=1$ and ${ϵ}_{iik}=0$, its matrix trace is $3\cos \theta +1-\cos \theta =1+2\cos \theta$.

For the given matrix, the matrix trace is $1+2r$ with $r=1/\sqrt{2}$, so $\cos \theta =1/\sqrt{2}$. In the stated angular range, $\theta =\pm \pi /4$. The antisymmetric entries giveTherefore the two axis-angle pairs areReversing the axis and the angle leaves the Rodrigues rotation formula unchanged, so these describe the same rotation.

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

The two Lie groups have the same local infinitesimal structure, but different global topology. This difference determines which Lie algebra representations integrate to representations of each group.

An element of the SU(2) group is a unitary matrix of determinant one. Orthogonality of its columns and its determinant give the unique formThus its group manifold is the unit three-sphere in ${\mathbb{C}}^2\simeq {\mathbb{R}}^4$: this is SU(2) as the three-sphere. In particular it is compact, connected and simply connected. In terms of the Pauli matrices one may also write $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real coefficients satisfying $a_0^2+{\mathbf{a}}^2=1$.

Differentiate $U(t{)}^{†}U(t)=I$ and $\det U(t)=1$ at $U(0)=I$. The tangent space consists of traceless skew-Hermitian matrices:The Lie bracket of a Matrix Lie group is the matrix commutator. With $T_a=-i{\sigma }_a/2$, the Pauli matrix multiplication law givesThis derives the SU(2) Lie algebra as a three-dimensional real Lie algebra. The Hermitian physics generators $J_a={\sigma }_a/2$ instead obey $[J_a,J_b]=i{ϵ}_{abc}{J}_c$; they are $i$ times the skew-Hermitian tangent generators, so these are consistent conventions.

The SO(3) group consists of real orthogonal matrices with determinant one. Differentiating $R(t{)}^{T}R(t)=I$ at the identity givesThe determinant condition gives no additional infinitesimal constraint because a skew-symmetric matrix already has trace zero. Define $L_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The cross product identity impliesTherefore the SO(3) Lie algebra has the same structure constants and $T_a\mapsto L_a$ is a Lie algebra isomorphism.

The global relation is the Adjoint double cover from SU(2) to SO(3). For $V=\mathbf{v}\cdot \mathbf{\sigma }$, define $R(U)$ byConjugation preserves the real space of traceless Hermitian matrices and its inner product $\frac{1}{2}\mathrm{tr}(VW)=\mathbf{v}\cdot \mathbf{w}$. Hence $R(U)$ is orthogonal. Continuity and connectedness, together with $R(I)=I$, put it in $SO(3)$. Composition of conjugations makes $R$ a group homomorphism. If $R(U)=I$, then $U$ commutes with every Pauli matrix, hence is scalar; unitarity and determinant one leave precisely $U=\pm I$. The differential sends $T_a$ to $L_a$, so it is an isomorphism. More concretely,induces rotation through angle $\theta$ about $\mathbf{n}$, by the Rodrigues rotation formula. Every three-dimensional rotation has such an axis and angle, proving surjectivity. ConsequentlyThe matrices $U$ and $-U$ are antipodal points on the three-sphere, so the SO(3) group manifold is Real projective space ${\mathbb{RP}}^3$. Equivalently, the closed axis-angle ball $\Vert \theta \mathbf{n}\Vert \le \pi$ has opposite boundary points identified. The fundamental group is ${\pi }_1(SO(3))\simeq {\mathbb{Z}}_2$, whereas ${\pi }_1(SU(2))=0$. A $2\pi$ rotation lifts from $I$ to $-I$; a $4\pi$ rotation returns to $I$. Thus the covering is the universal cover and the groups are not globally isomorphic.

For representation theory, specify finite-dimensional complex continuous representations. Compactness permits an invariant Hermitian inner product, obtained by averaging against Haar measure, and therefore complete reducibility. Complexifying either real Lie algebra gives the sl2 Lie algebra. Its finite-dimensional irreducibles are indexed by $n\in {\mathbb{Z}}_{\ge 0}$, have highest weight $n$, and have dimension $n+1$. By integration of a Lie-algebra representation, since $SU(2)$ is simply connected, every such Lie algebra representation integrates uniquely. The resulting homogeneous polynomial representation of SU2 isIn the spin angular momentum notation $j=n/2$, its Hermitian $J_3$ eigenvalues are $j,j-1,\dots ,-j$. The central matrix $-I$ acts on the symmetric power by $(-1{)}^n$, so descent of an SU(2) representation to SO(3) occurs exactly when $n$ is even. HenceFor a reducible representation, every summand must satisfy the descent condition. The spin-one-half doublet is a genuine representation of the covering group but does not define a single-valued representation of $SO(3)$; the spin-one triplet does and is its vector representation. The distinction is topological, rather than a difference in their isomorphic Lie algebras.

Decompose $\mathbf{v}$ by orthogonal projection onto the straight line spanned by $\mathbf{n}$ and its orthogonal complement:The first term is fixed by the rotation. If $\mathbf{w}\ne 0$, the vectors $\mathbf{w}/|\mathbf{w}|$ and $(\mathbf{n}\times \mathbf{w})/|\mathbf{w}|$ form an orthonormal basis of the perpendicular plane, oriented in the positive sense about $\mathbf{n}$. A planar rotation through $\theta$ therefore sendsHere $\mathbf{n}\times \mathbf{w}=\mathbf{n}\times \mathbf{v}$. Adding the fixed parallel component proves the Rodrigues rotation formula in the required form:When $\mathbf{w}=0$, both perpendicular terms vanish, so the same formula fixes $\mathbf{v}$ as required. Positive angles use the right-hand convention about $\mathbf{n}$.

The displayed power series is the matrix exponential $G=e^H$. From $H^3=-{\alpha }^{2}H$, its odd powers and nonconstant even powers sum separately:The scalar power series are $\sin \alpha /\alpha$ and $(1-\cos \alpha )/{\alpha }^2$, with continuous values $1$ and $1/2$ at zero. Substituting $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$ and the preceding formula for $H^2\mathbf{x}$ yieldsThis is the Rodrigues rotation formula: the component parallel to $\mathbf{n}$ is fixed, while the perpendicular component undergoes a rotation through $\alpha$. For $\alpha =0$ the linear operator is the identity operator.