Inmechanics andgeometry, the3D rotation group, often denotedSO(3), is thegroup of allrotations about theorigin ofthree-dimensionalEuclidean space${\displaystyle {\mathbb{R}}^3}$ under the operation ofcomposition.^[1]

By definition, a rotation about the origin is a transformation that preserves the origin,Euclidean distance (so it is anisometry), andorientation (i.e.,handedness of space). Composing two rotations results in another rotation, every rotation has a uniqueinverse rotation, and theidentity map satisfies the definition of a rotation. Owing to the above properties (along composite rotations'associative property), the set of all rotations is agroup under composition.

Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation. Rotations are not commutative (for example, rotatingR 90° in the x-y plane followed byS 90° in the y-z plane is not the same asS followed byR), making the 3D rotation group anonabelian group. Moreover, the rotation group has a natural structure as amanifold for which the group operations aresmoothly differentiable, so it is in fact aLie group. It iscompact and has dimension 3.

Rotations arelinear transformations of${\displaystyle {\mathbb{R}}^3}$ and can therefore be represented bymatrices once abasis of${\displaystyle {\mathbb{R}}^3}$ has been chosen. Specifically, if we choose anorthonormal basis of${\displaystyle {\mathbb{R}}^3}$, every rotation is described by anorthogonal 3 × 3 matrix (i.e., a 3 × 3 matrix with real entries which, when multiplied by itstranspose, results in theidentity matrix) withdeterminant 1. The group SO(3) can therefore be identified with the group of these matrices undermatrix multiplication. These matrices are known as "special orthogonal matrices", explaining the notation SO(3).

The group SO(3) is used to describe the possible rotational symmetries of an object, as well as the possible orientations of an object in space. Itsrepresentations are important in physics, where they give rise to theelementary particles of integerspin.

Besides just preserving length, rotations also preserve theangles between vectors. This follows from the fact that the standarddot product between two vectorsu andv can be written purely in terms of length (see thelaw of cosines):$${\displaystyle \mathbf{u}\cdot \mathbf{v}=\frac{1}{2}\left(\Vert \mathbf{u}+\mathbf{v}{\Vert }^2-\Vert \mathbf{u}{\Vert }^2-\Vert \mathbf{v}{\Vert }^2\right).}$$ 

It follows that every length-preserving linear transformation in${\displaystyle {\mathbb{R}}^3}$  preserves the dot product, and thus the angle between vectors. Rotations are often defined as linear transformations that preserve the inner product on${\displaystyle {\mathbb{R}}^3}$ , which is equivalent to requiring them to preserve length. Seeclassical group for a treatment of this more general approach, whereSO(3) appears as a special case.

Every rotation maps anorthonormal basis of${\displaystyle {\mathbb{R}}^3}$  to another orthonormal basis. Like any linear transformation offinite-dimensional vector spaces, a rotation can always be represented by amatrix. LetR be a given rotation. With respect to thestandard basise₁,e₂,e₃ of${\displaystyle {\mathbb{R}}^3}$  the columns ofR are given by(Re₁,Re₂,Re₃). Since the standard basis is orthonormal, and sinceR preserves angles and length, the columns ofR form another orthonormal basis. This orthonormality condition can be expressed in the form

${\displaystyle R^{\mathsf{T}}R=R{R}^{\mathsf{T}}=I,}$ 

whereR^T denotes thetranspose ofR andI is the3 × 3identity matrix. Matrices for which this property holds are calledorthogonal matrices. The group of all3 × 3 orthogonal matrices is denotedO(3), and consists of all proper and improper rotations.

In addition to preserving length, proper rotations must also preserve orientation. A matrix will preserve or reverse orientation according to whether thedeterminant of the matrix is positive or negative. For an orthogonal matrixR, note thatdetR^T = detR implies(detR)² = 1, so thatdetR = ±1. Thesubgroup of orthogonal matrices with determinant+1 is called thespecialorthogonal group, denotedSO(3).

Thus every rotation can be represented uniquely by an orthogonal matrix with unit determinant. Moreover, since composition of rotations corresponds tomatrix multiplication, the rotation group isisomorphic to the special orthogonal groupSO(3).

Improper rotations correspond to orthogonal matrices with determinant−1, and they do not form a group because the product of two improper rotations is a proper rotation.

The rotation group is agroup underfunction composition (or equivalently theproduct of linear transformations). It is asubgroup of thegeneral linear group consisting of allinvertible linear transformations of thereal 3-space${\displaystyle {\mathbb{R}}^3}$ .^[2]

Furthermore, the rotation group isnonabelian. That is, the order in which rotations are composed makes a difference. For example, a quarter turn around the positivex-axis followed by a quarter turn around the positivey-axis is a different rotation than the one obtained by first rotating aroundy and thenx.

The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of theCartan–Dieudonné theorem.

The finite subgroups of${\displaystyle \mathrm{S}\mathrm{O}(3)}$  are completelyclassified.^[3]

Every finite subgroup is isomorphic to either an element of one of twocountably infinite families of planar isometries: thecyclic groups${\displaystyle C_n}$  or thedihedral groups${\displaystyle D_{2n}}$ , or to one of three other groups: thetetrahedral group${\displaystyle \cong A_4}$ , theoctahedral group${\displaystyle \cong S_4}$ , or theicosahedral group${\displaystyle \cong A_5}$ .

Every nontrivial proper rotation in 3 dimensions fixes a unique 1-dimensionallinear subspace of${\displaystyle {\mathbb{R}}^3}$  which is called theaxis of rotation (this isEuler's rotation theorem). Each such rotation acts as an ordinary 2-dimensional rotation in the planeorthogonal to this axis. Since every 2-dimensional rotation can be represented by an angleφ, an arbitrary 3-dimensional rotation can be specified by an axis of rotation together with anangle of rotation about this axis. (Technically, one needs to specify an orientation for the axis and whether the rotation is taken to beclockwise orcounterclockwise with respect to this orientation).

For example, counterclockwise rotation about the positivez-axis by angleφ is given by

 

Given aunit vectorn in${\displaystyle {\mathbb{R}}^3}$  and an angleφ, letR(φ, n) represent a counterclockwise rotation about the axis throughn (with orientation determined byn). Then

• R(0,n) is the identity transformation for anyn
• R(φ,n) =R(−φ, −n)
• R(π + φ,n) =R(π − φ, −n).

Using these properties one can show that any rotation can be represented by a unique angleφ in the range 0 ≤ φ ≤π and a unit vectorn such that

• n is arbitrary ifφ = 0
• n is unique if 0 <φ <π
• n is unique up to asign ifφ =π (that is, the rotationsR(π, ±n) are identical).

In the next section, this representation of rotations is used to identify SO(3) topologically with three-dimensional real projective space.

The Lie group SO(3) isdiffeomorphic to thereal projective space${\displaystyle {\mathbb{P}}^3(\mathbb{R}).}$ ^[4]

Consider the solid ball in${\displaystyle {\mathbb{R}}^3}$  of radiusπ (that is, all points of${\displaystyle {\mathbb{R}}^3}$  of distanceπ or less from the origin). Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball. Rotations through an angle 𝜃 between 0 andπ (not including either) are on the same axis at the same distance. Rotation through angles between 0 and −π correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations throughπ and through −π are the same. So weidentify (or "glue together")antipodal points on the surface of the ball. After this identification, we arrive at atopological spacehomeomorphic to the rotation group.

Indeed, the ball with antipodal surface points identified is asmooth manifold, and this manifold isdiffeomorphic to the rotation group. It is also diffeomorphic to thereal 3-dimensional projective space${\displaystyle {\mathbb{P}}^3(\mathbb{R}),}$  so the latter can also serve as a topological model for the rotation group.

These identifications illustrate that SO(3) isconnected but notsimply connected. As to the latter, in the ball with antipodal surface points identified, consider the path running from the "north pole" straight through the interior down to the south pole. This is a closed loop, since the north pole and the south pole are identified. This loop cannot be shrunk to a point, since no matter how it is deformed, the start and end point have to remain antipodal, or else the loop will "break open". In terms of rotations, this loop represents a continuous sequence of rotations about thez-axis starting (by example) at the identity (center of the ball), through the south pole, jumping to the north pole and ending again at the identity rotation (i.e., a series of rotation through an angleφ whereφ runs from 0 to2π).

Surprisingly, running through the path twice, i.e., running from the north pole down to the south pole, jumping back to the north pole (using the fact that north and south poles are identified), and then again running from the north pole down to the south pole, so thatφ runs from 0 to 4π, gives a closed loop whichcan be shrunk to a single point: first move the paths continuously to the ball's surface, still connecting north pole to south pole twice. The second path can then be mirrored over to the antipodal side without changing the path at all. Now we have an ordinary closed loop on the surface of the ball, connecting the north pole to itself along a great circle. This circle can be shrunk to the north pole without problems. Theplate trick and similar tricks demonstrate this practically.

The same argument can be performed in general, and it shows that thefundamental group of SO(3) is thecyclic group of order 2 (a fundamental group with two elements). Inphysics applications, the non-triviality (more than one element) of the fundamental group allows for the existence of objects known asspinors, and is an important tool in the development of thespin–statistics theorem.

Theuniversal cover of SO(3) is aLie group calledSpin(3). The group Spin(3) is isomorphic to thespecial unitary group SU(2); it is also diffeomorphic to the unit3-sphereS³ and can be understood as the group ofversors (quaternions withabsolute value 1). The connection between quaternions and rotations, commonly exploited incomputer graphics, is explained inquaternions and spatial rotations. The map fromS³ onto SO(3) that identifies antipodal points ofS³ is asurjectivehomomorphism of Lie groups, withkernel {±1}. Topologically, this map is a two-to-onecovering map. (See theplate trick.)

In this section, we give two different constructions of a two-to-one andsurjectivehomomorphism of SU(2) onto SO(3).

The groupSU(2) isisomorphic to thequaternions of unit norm via a map given by^[5] restricted to$a^2+b^2+c^2+d^2=|\alpha {|}^2+|\beta {|}^2=1$  where$q\in \mathbb{H}$ ,$a,b,c,d\in \mathbb{R}$ ,$U\in \mathrm{SU}(2)$ , and${\displaystyle \alpha =a+bi\in \mathbb{C}}$ ,${\displaystyle \beta =c+di\in \mathbb{C}}$ .

Let us now identify${\displaystyle {\mathbb{R}}^3}$  with the span of${\displaystyle \mathbf{i},\mathbf{j},\mathbf{k}}$ . One can then verify that if${\displaystyle v}$  is in${\displaystyle {\mathbb{R}}^3}$  and${\displaystyle q}$  is a unit quaternion, then$${\displaystyle qv{q}^{-1}\in {\mathbb{R}}^3.}$$ 

Furthermore, the map${\displaystyle v\mapsto qv{q}^{-1}}$  is a rotation of${\displaystyle {\mathbb{R}}^3.}$  Moreover,${\displaystyle (-q)v(-q{)}^{-1}}$  is the same as${\displaystyle qv{q}^{-1}}$ . This means that there is a2:1 homomorphism from quaternions of unit norm to the 3D rotation groupSO(3).

One can work this homomorphism out explicitly: the unit quaternion,q, with is mapped to the rotation matrix 

This is a rotation around the vector(x,y,z) by an angle2θ, wherecosθ =w and|sinθ| = ‖(x,y,z)‖. The proper sign forsinθ is implied, once the signs of the axis components are fixed. The2:1-nature is apparent since bothq and−q map to the sameQ.

The general reference for this section isGelfand, Minlos & Shapiro (1963). The pointsP on the sphere

${\displaystyle \mathbf{S}=\left\{(x,y,z)\in {\mathbb{R}}^3:x^2+y^2+z^2=\frac{1}{4}\right\}}$ 

can, barring the north poleN, be put into one-to-one bijection with pointsS(P) =P' on the planeM defined byz = −⁠1/2⁠, see figure. The mapS is calledstereographic projection.

Let the coordinates onM be(ξ,η). The lineL passing throughN andP can be parametrized as

${\displaystyle L(t)=N+t(N-P)=\left(0,0,\frac{1}{2}\right)+t\left(\left(0,0,\frac{1}{2}\right)-(x,y,z)\right),t\in \mathbb{R}.}$ 

Demanding that thez-coordinate of${\displaystyle L(t_0)}$  equals−⁠1/2⁠, one finds

${\displaystyle t_0=\frac{1}{z-\frac{1}{2}}.}$ 

We have${\displaystyle L(t_0)=(\xi ,\eta ,-1/2).}$  Hence the map

${\displaystyle \{\begin{array}{l}S:\mathbf{S}\to M\\ P=(x,y,z)⟼P^{\prime }=(\xi ,\eta )=\left(\frac{x}{\frac{1}{2}-z},\frac{y}{\frac{1}{2}-z}\right)\equiv \zeta =\xi +i\eta \end{array}}$ 

where, for later convenience, the planeM is identified with the complex plane${\displaystyle \mathbb{C}.}$ 

For the inverse, writeL as

${\displaystyle L=N+s(P^{\prime }-N)=\left(0,0,\frac{1}{2}\right)+s\left(\left(\xi ,\eta ,-\frac{1}{2}\right)-\left(0,0,\frac{1}{2}\right)\right),}$ 

and demandx² +y² +z² =⁠1/4⁠ to finds =⁠1/1 +ξ² +η²⁠ and thus

${\displaystyle \{\begin{array}{l}{S}^{-1}:M\to \mathbf{S}\\ P^{\prime }=(\xi ,\eta )⟼P=(x,y,z)=\left(\frac{\xi }{1+{\xi }^2+{\eta }^2},\frac{\eta }{1+{\xi }^2+{\eta }^2},\frac{-1+{\xi }^2+{\eta }^2}{2+2{\xi }^2+2{\eta }^2}\right)\end{array}}$ 

Ifg ∈ SO(3) is a rotation, then it will take points onS to points onS by its standard actionΠ_s(g) on the embedding space${\displaystyle {\mathbb{R}}^3.}$  By composing this action withS one obtains a transformationS ∘ Π_s(g) ∘S⁻¹ ofM,

${\displaystyle \zeta =P^{\prime }⟼P⟼{\mathrm{\Pi }}_s(g)P=gP⟼S(gP)\equiv {\mathrm{\Pi }}_u(g)\zeta ={\zeta }^{\prime }.}$ 

ThusΠ_u(g) is a transformation of${\displaystyle \mathbb{C}}$  associated to the transformationΠ_s(g) of${\displaystyle {\mathbb{R}}^3}$ .

It turns out thatg ∈ SO(3) represented in this way byΠ_u(g) can be expressed as a matrixΠ_u(g) ∈ SU(2) (where the notation is recycled to use the same name for the matrix as for the transformation of${\displaystyle \mathbb{C}}$  it represents). To identify this matrix, consider first a rotationg_φ about thez-axis through an angleφ,

 

Hence

${\displaystyle {\zeta }^{\prime }=\frac{x^{\prime }+i{y}^{\prime }}{\frac{1}{2}-z^{\prime }}=\frac{e^{i\varphi }(x+iy)}{\frac{1}{2}-z}=e^{i\varphi }\zeta =\frac{e^{\frac{i\varphi }{2}}\zeta +0}{0\zeta +e^{-\frac{i\varphi }{2}}},}$ 

which, unsurprisingly, is a rotation in the complex plane. In an analogous way, ifg_θ is a rotation about thex-axis through an angleθ, then

${\displaystyle w^{\prime }=e^{i\theta }w,w=\frac{y+iz}{\frac{1}{2}-x},}$ 

which, after a little algebra, becomes

${\displaystyle {\zeta }^{\prime }=\frac{\cos \frac{\theta }{2}\zeta +i\sin \frac{\theta }{2}}{i\sin \frac{\theta }{2}\zeta +\cos \frac{\theta }{2}}.}$ 

These two rotations,${\displaystyle g_{\varphi },g_{\theta },}$  thus correspond tobilinear transforms ofR² ≃C ≃M, namely, they are examples ofMöbius transformations.

A general Möbius transformation is given by

${\displaystyle {\zeta }^{\prime }=\frac{\alpha \zeta +\beta }{\gamma \zeta +\delta },\alpha \delta -\beta \gamma \ne 0.}$ 

The rotations,${\displaystyle g_{\varphi },g_{\theta }}$  generate all ofSO(3) and the composition rules of the Möbius transformations show that any composition of${\displaystyle g_{\varphi },g_{\theta }}$  translates to the corresponding composition of Möbius transformations. The Möbius transformations can be represented by matrices

 

since a common factor ofα,β,γ,δ cancels.

For the same reason, the matrix isnot uniquely defined since multiplication by−I has no effect on either the determinant or the Möbius transformation. The composition law of Möbius transformations follow that of the corresponding matrices. The conclusion is that each Möbius transformation corresponds to two matricesg, −g ∈ SL(2,C).

Using this correspondence one may write

 

These matrices are unitary and thusΠ_u(SO(3)) ⊂ SU(2) ⊂ SL(2,C). In terms ofEuler angles^{[nb 1]} one finds for a general rotation

one has^[6]

For the converse, consider a general matrix

 

Make the substitutions

 

With the substitutions,Π(g_α,β) assumes the form of the right hand side (RHS) of (2), which corresponds underΠ_u to a matrix on the form of the RHS of (1) with the sameφ,θ,ψ. In terms of the complex parametersα,β,

 

To verify this, substitute forα.β the elements of the matrix on the RHS of (2). After some manipulation, the matrix assumes the form of the RHS of (1).

It is clear from the explicit form in terms of Euler angles that the map

${\displaystyle \{\begin{array}{l}p:\mathrm{SU}(2)\to \mathrm{SO}(3)\\ \pm {\mathrm{\Pi }}_u(g_{\alpha \beta })\mapsto g_{\alpha \beta }\end{array}}$ 

just described is a smooth,2:1 and surjectivegroup homomorphism. It is hence an explicit description of theuniversal covering space ofSO(3) from theuniversal covering groupSU(2).

Associated with everyLie group is itsLie algebra, a linear space of the same dimension as the Lie group, closed under a bilinear alternating product called theLie bracket. The Lie algebra ofSO(3) is denoted by${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  and consists of allskew-symmetric3 × 3 matrices.^[7] This may be seen by differentiating theorthogonality condition,A^TA =I,A ∈ SO(3).^{[nb 2]} The Lie bracket of two elements of${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  is, as for the Lie algebra of every matrix group, given by the matrixcommutator,[A₁,A₂] =A₁A₂ −A₂A₁, which is again a skew-symmetric matrix. The Lie algebra bracket captures the essence of the Lie group product in a sense made precise by theBaker–Campbell–Hausdorff formula.

The elements of${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  are the "infinitesimal generators" of rotations, i.e., they are the elements of thetangent space of the manifold SO(3) at the identity element. If${\displaystyle R(\varphi ,\mathit{n})}$  denotes a counterclockwise rotation with angle φ about the axis specified by the unit vector${\displaystyle \mathit{n},}$  then

${\displaystyle \mathrm{\forall }\mathit{u}\in {\mathbb{R}}^3:{\frac{\mathrm{d}}{\mathrm{d}\varphi }|}_{\varphi =0}R(\varphi ,\mathit{n})\mathit{u}=\mathit{n}\times \mathit{u}.}$ 

This can be used to show that the Lie algebra${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  (with commutator) is isomorphic to the Lie algebra${\displaystyle {\mathbb{R}}^3}$  (withcross product). Under this isomorphism, anEuler vector${\displaystyle \mathit{\omega }\in {\mathbb{R}}^3}$  corresponds to the linear map${\displaystyle \tilde{\mathit{\omega }}}$  defined by${\displaystyle \tilde{\mathit{\omega }}(\mathit{u})=\mathit{\omega }\times \mathit{u}.}$ 

In more detail, most often a suitable basis for${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  as a3-dimensional vector space is

 

Thecommutation relations of these basis elements are,

${\displaystyle [{\mathit{L}}_x,{\mathit{L}}_y]={\mathit{L}}_z,[{\mathit{L}}_z,{\mathit{L}}_x]={\mathit{L}}_y,[{\mathit{L}}_y,{\mathit{L}}_z]={\mathit{L}}_x}$ 

which agree with the relations of the threestandard unit vectors of${\displaystyle {\mathbb{R}}^3}$  under the cross product.

As announced above, one can identify any matrix in this Lie algebra with an Euler vector${\displaystyle \mathit{\omega }=(x,y,z)\in {\mathbb{R}}^3,}$ ^[8]

 

This identification is sometimes called thehat-map.^[9] Under this identification, the${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  bracket corresponds in${\displaystyle {\mathbb{R}}^3}$  to thecross product,

${\displaystyle \left[\hat{\mathit{u}},\hat{\mathit{v}}\right]=\widehat{\mathit{u}\times \mathit{v}}.}$ 

The matrix identified with a vector${\displaystyle \mathit{u}}$  has the property that

${\displaystyle \hat{\mathit{u}}\mathit{v}=\mathit{u}\times \mathit{v},}$ 

where the left-hand side we have ordinary matrix multiplication. This implies${\displaystyle \mathit{u}}$  is in thenull space of the skew-symmetric matrix with which it is identified, because${\displaystyle \mathit{u}\times \mathit{u}=\mathbf{0}.}$ 

InLie algebra representations, the group SO(3) is compact and simple of rank 1, and so it has a single independentCasimir element, a quadratic invariant function of the three generators which commutes with all of them. The Killing form for the rotation group is just theKronecker delta, and so this Casimir invariant is simply the sum of the squares of the generators,${\displaystyle {\mathit{J}}_x,{\mathit{J}}_y,{\mathit{J}}_z,}$  of the algebra

${\displaystyle [{\mathit{J}}_x,{\mathit{J}}_y]={\mathit{J}}_z,[{\mathit{J}}_z,{\mathit{J}}_x]={\mathit{J}}_y,[{\mathit{J}}_y,{\mathit{J}}_z]={\mathit{J}}_x.}$ 

That is, the Casimir invariant is given by

${\displaystyle {\mathit{J}}^2\equiv \mathit{J}\cdot \mathit{J}={\mathit{J}}_x^2+{\mathit{J}}_y^2+{\mathit{J}}_z^2\propto \mathit{I}.}$ 

For unitary irreduciblerepresentationsD^j, the eigenvalues of this invariant are real and discrete, and characterize each representation, which is finite dimensional, of dimensionality${\displaystyle 2j+1}$ . That is, the eigenvalues of this Casimir operator are

${\displaystyle {\mathit{J}}^2=-j(j+1){\mathit{I}}_{2j+1},}$ 

wherej is integer or half-integer, and referred to as thespin orangular momentum.

So, the 3 × 3 generatorsL displayed above act on the triplet (spin 1) representation, while the 2 × 2 generators below,t, act on thedoublet (spin-1/2) representation. By takingKronecker products ofD^1/2 with itself repeatedly, one may construct all higher irreducible representationsD^j. That is, the resulting generators for higher spin systems in three spatial dimensions, for arbitrarily largej, can be calculated using thesespin operators andladder operators.

For every unitary irreducible representationsD^j there is an equivalent one,D^−j−1. All infinite-dimensional irreducible representations must be non-unitary, since the group is compact.

Inquantum mechanics, the Casimir invariant is the "angular-momentum-squared" operator; integer values of spinj characterizebosonic representations, while half-integer valuesfermionic representations. Theantihermitian matrices used above are utilized asspin operators, after they are multiplied byi, so they are nowhermitian (like the Pauli matrices). Thus, in this language,

${\displaystyle [{\mathit{J}}_x,{\mathit{J}}_y]=i{\mathit{J}}_z,[{\mathit{J}}_z,{\mathit{J}}_x]=i{\mathit{J}}_y,[{\mathit{J}}_y,{\mathit{J}}_z]=i{\mathit{J}}_x.}$ 

and hence

${\displaystyle {\mathit{J}}^2=j(j+1){\mathit{I}}_{2j+1}.}$ 

Explicit expressions for theseD^j are,

 

wherej is arbitrary and${\displaystyle 1\le a,b\le 2j+1}$ .

For example, the resulting spin matrices for spin 1 (${\displaystyle j=1}$ ) are

 

Note, however, how these are in an equivalent, but different basis, thespherical basis, than the aboveiL in the Cartesian basis.^{[nb 3]}

For higher spins, such as spin⁠3/2⁠ (${\displaystyle j=\frac{3}{2}}$ ):

 

For spin⁠5/2⁠ (${\displaystyle j=\frac{5}{2}}$ ),

 

The Lie algebras${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  and${\displaystyle \mathfrak{s}\mathfrak{u}(2)}$  are isomorphic. One basis for${\displaystyle \mathfrak{s}\mathfrak{u}(2)}$  is given by^[10]

 

These are related to thePauli matrices by

${\displaystyle {\mathit{t}}_i⟷\frac{1}{2i}{\sigma }_i.}$ 

The Pauli matrices abide by the physicists' convention for Lie algebras. In that convention, Lie algebra elements are multiplied byi, the exponential map (below) is defined with an extra factor ofi in the exponent and thestructure constants remain the same, but thedefinition of them acquires a factor ofi. Likewise, commutation relations acquire a factor ofi. The commutation relations for the${\displaystyle {\mathit{t}}_i}$  are

${\displaystyle [{\mathit{t}}_i,{\mathit{t}}_j]={\epsilon }_{ijk}{\mathit{t}}_k,}$ 

whereε_ijk is the totally anti-symmetric symbol withε₁₂₃ = 1. The isomorphism between${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  and${\displaystyle \mathfrak{s}\mathfrak{u}(2)}$  can be set up in several ways. For later convenience,${\displaystyle \mathfrak{s}\mathfrak{o}(3)}$  and${\displaystyle \mathfrak{s}\mathfrak{u}(2)}$  are identified by mapping

${\displaystyle {\mathit{L}}_x⟷{\mathit{t}}_1,{\mathit{L}}_y⟷{\mathit{t}}_2,{\mathit{L}}_z⟷{\mathit{t}}_3,}$ 

and extending by linearity.

The exponential map forSO(3), is, sinceSO(3) is a matrix Lie group, defined using the standardmatrix exponential series,

${\displaystyle \{\begin{array}{l}\exp :\mathfrak{s}\mathfrak{o}(3)\to \mathrm{SO}(3)\\ A\mapsto e^A=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}{A}^k=I+A+\frac{1}{2}{A}^2+\cdots .\end{array}}$ 

For anyskew-symmetric matrixA ∈ 𝖘𝖔(3),e^A is always inSO(3). The proof uses the elementary properties of the matrix exponential

${\displaystyle {\left(e^A\right)}^{\mathsf{\text{T}}}{e}^A=e^{A^{\mathsf{\text{T}}}}{e}^A=e^{A^{\mathsf{\text{T}}}+A}=e^{-A+A}=e^{A-A}=e^A{\left(e^A\right)}^{\mathsf{\text{T}}}=e^0=I.}$ 

since the matricesA andA^T commute, this can be easily proven with the skew-symmetric matrix condition. This is not enough to show that𝖘𝖔(3) is the corresponding Lie algebra forSO(3), and shall be proven separately.

The level of difficulty of proof depends on how a matrix group Lie algebra is defined.Hall (2003) defines the Lie algebra as the set of matrices

${\displaystyle \left\{A\in \mathrm{M}(n,\mathbb{R})|e^{tA}\in \mathrm{SO}(3)\mathrm{\forall }t\right\},}$ 

in which case it is trivial.Rossmann (2002) uses for a definition derivatives of smooth curve segments inSO(3) through the identity taken at the identity, in which case it is harder.^[11]

For a fixedA ≠ 0,e^tA, −∞ <t < ∞ is aone-parameter subgroup along ageodesic inSO(3). That this gives a one-parameter subgroup follows directly from properties of the exponential map.^[12]

The exponential map provides adiffeomorphism between a neighborhood of the origin in the𝖘𝖔(3) and a neighborhood of the identity in theSO(3).^[13] For a proof, seeClosed subgroup theorem.

The exponential map issurjective. This follows from the fact that everyR ∈ SO(3), since every rotation leaves an axis fixed (Euler's rotation theorem), and is conjugate to ablock diagonal matrix of the form

 

such thatA =BDB⁻¹, and that

${\displaystyle B{e}^{\theta L_z}{B}^{-1}=e^{B\theta L_z{B}^{-1}},}$ 

together with the fact that𝖘𝖔(3) is closed under theadjoint action ofSO(3), meaning thatBθL_zB⁻¹ ∈ 𝖘𝖔(3).

Thus, e.g., it is easy to check the popular identity

${\displaystyle e^{-\pi L_x/2}{e}^{\theta L_z}{e}^{\pi L_x/2}=e^{\theta L_y}.}$ 

As shown above, every elementA ∈ 𝖘𝖔(3) is associated with a vectorω =θu, whereu = (x,y,z) is a unit magnitude vector. Sinceu is in the null space ofA, if one now rotates to a new basis, through some other orthogonal matrixO, withu as thez axis, the final column and row of the rotation matrix in the new basis will be zero.

Thus, we know in advance from the formula for the exponential thatexp(OAO^T) must leaveu fixed. It is mathematically impossible to supply a straightforward formula for such a basis as a function ofu, because its existence would violate thehairy ball theorem; but direct exponentiation is possible, andyields

 

where$c=\cos \frac{\theta }{2}$  and$s=\sin \frac{\theta }{2}$ . This is recognized as a matrix for a rotation around axisu by the angleθ: cf.Rodrigues' rotation formula.

GivenR ∈ SO(3), let${\displaystyle A=\frac{1}{2}\left(R-R^{\mathrm{T}}\right)}$  denote the antisymmetric part and let$\Vert A\Vert =\sqrt{-\frac{1}{2}\mathrm{Tr}\left(A^2\right)}.$  Then, the logarithm ofR is given by^[9]

${\displaystyle \log R=\frac{{\sin }^{-1}\Vert A\Vert }{\Vert A\Vert }A.}$ 

This is manifest by inspection of the mixed symmetry form of Rodrigues' formula,

${\displaystyle e^X=I+\frac{\sin \theta }{\theta }X+2\frac{{\sin }^2\frac{\theta }{2}}{{\theta }^2}{X}^2,\theta =\Vert X\Vert ,}$ 

where the first and last term on the right-hand side are symmetric.

${\displaystyle SO(3)}$  is doubly covered by the group of unit quaternions, which is isomorphic to the 3-sphere. Since theHaar measure on the unit quaternions is just the 3-area measure in 4 dimensions, the Haar measure on${\displaystyle SO(3)}$  is just the pushforward of the 3-area measure.

Consequently, generating a uniformly random rotation in${\displaystyle {\mathbb{R}}^3}$  is equivalent to generating a uniformly random point on the 3-sphere. This can be accomplished by the following$${\displaystyle (\sqrt{1-u_1}\sin (2\pi u_2),\sqrt{1-u_1}\cos (2\pi u_2),\sqrt{u_1}\sin (2\pi u_3),\sqrt{u_1}\cos (2\pi u_3))}$$ 

where${\displaystyle u_1,u_2,u_3}$  are uniformly random samples of${\displaystyle [0,1]}$ .^[14]

SupposeX andY in the Lie algebra are given. Their exponentials,exp(X) andexp(Y), are rotation matrices, which can be multiplied. Since the exponential map is a surjection, for someZ in the Lie algebra,exp(Z) = exp(X) exp(Y), and one may tentatively write

${\displaystyle Z=C(X,Y),}$ 

forC some expression inX andY. Whenexp(X) andexp(Y) commute, thenZ =X +Y, mimicking the behavior of complex exponentiation.

The general case is given by the more elaborateBCH formula, a series expansion of nested Lie brackets.^[15] For matrices, the Lie bracket is the same operation as thecommutator, which monitors lack of commutativity in multiplication. This general expansion unfolds as follows,^{[nb 4]}

${\displaystyle Z=C(X,Y)=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+\cdots .}$ 

The infinite expansion in the BCH formula forSO(3) reduces to a compact form,

${\displaystyle Z=\alpha X+\beta Y+\gamma [X,Y],}$ 

for suitable trigonometric function coefficients(α,β,γ).

It is worthwhile to write this composite rotation generator as

${\displaystyle \alpha X+\beta Y+\gamma [X,Y]\underset{\mathfrak{s}\mathfrak{o}(3)}{=}X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+\cdots ,}$ 

to emphasize that this is aLie algebra identity.

The above identity holds for allfaithful representations of𝖘𝖔(3). Thekernel of a Lie algebra homomorphism is anideal, but𝖘𝖔(3), beingsimple, has no nontrivial ideals and all nontrivial representations are hence faithful. It holds in particular in the doublet or spinor representation. The same explicit formula thus follows in a simpler way through Pauli matrices, cf. the2×2 derivation for SU(2).