Try
The quaternion q=(1+i+j+k)/2 will rotate the three coordinate axes cyclically. Its negative will give exactly the same spatial rotation. Predict whether this means q and −q are the same quaternion, or instead means the map from quaternions to rotations loses some information.
We will construct that map and determine precisely what it loses. A spatial vector (x,y,z) will be represented by the imaginary quaternion xi+yj+zk. We use right-handed coordinates, so i×j=k in the corresponding vector notation. Multiplication order matters: ij=k but ji=−k. The order-eight construction already used eight special quaternion matrices; here we construct all real quaternions and their unit group.
Predict
Write a quaternion as a+bi+cj+dk with real coefficients, and assign it the complex matrix
M(q) = [[z,w],[−conjugate(w),conjugate(z)]], z=a+bi, w=c+di.
In the entries on the right, i is the usual complex square root of −1. This convention fixes the signs of the quaternion units. Compute M(i)M(j) and M(j)M(i). Then compare M(q)⁎ with M(a−bi−cj−dk), where ⁎ means conjugate transpose.
Predict why an injective matrix realization can establish associativity without expanding all three-factor quaternion products. Also distinguish a matrix’s determinant from its entries: the quaternion q and the complex matrix M(q) are two descriptions of the same four real coordinates, not arbitrary complex 2 by 2 matrices.
Experiment
For q=(1+i+j+k)/2, compute q i q̄, q j q̄ and q k q̄, where q̄ reverses the signs of the imaginary coefficients. Use the unit relations i²=j²=k²=−1 and ij=k, jk=i, ki=j, reversing the sign when the order reverses. Check your results against the table.
| Spatial input | q v q̄ | (−q) v conjugate(−q) | Matrix column |
|---|
| i, or (1,0,0) | j | j | (0,1,0) |
| j, or (0,1,0) | k | k | (0,0,1) |
| k, or (0,0,1) | i | i | (1,0,0) |
Thus the rotation matrix has rows (0,0,1), (1,0,0), (0,1,0). It fixes the axis through (1,1,1), and the right-handed angle about its positive direction is 2π/3. The quaternion has scalar part 1/2=cos(π/3) and imaginary part (i+j+k)/2=sin(π/3)(i+j+k)/√3: its angle parameter is half the spatial angle.
Both q and −q have four nonzero coordinates and are different elements. The cancellation of the two minus signs in conjugation explains why their rotations agree. We will prove that no other ambiguity occurs.
Understand
Construct an associative algebra. Matrices of the displayed form are closed under real linear combinations and multiplication. Indeed, multiplying the pairs (z,w) and (z′,w′) gives top entries zz′−w conjugate(w′) and zw′+w conjugate(z′); the lower entries again have the required conjugate form. The assignment of four real coefficients is injective. Transfer matrix multiplication to these coefficients. Associativity, distributivity and the identity now follow from matrix algebra. Direct multiplication of the three unit matrices gives exactly the unit relations used in Experiment.
Conjugation q̄=a−bi−cj−dk corresponds to M(q)⁎. Hence conjugation reverses products: conjugate(pq)=q̄p̄. Multiplying q by q̄ gives the scalar a²+b²+c²+d², in either order. Define |q| as the nonnegative square root of this real number. Also det M(q)=|q|², so determinants give |pq|²=|p|²|q|² and therefore |pq|=|p||q|. Every nonzero q has inverse q̄/|q|². In particular the unit quaternions, those with |q|=1, form a group.
Identify SU(2). Each unit quaternion matrix satisfies M(q)⁎M(q)=I and det M(q)=1, so it lies in SU(2), the group of complex unitary 2 by 2 matrices of determinant 1. Conversely, let the first column of such a matrix be (z,u). Its unit orthogonal second column must be λ(−conjugate(u),conjugate(z)) with |λ|=1: this vector spans the one-dimensional orthogonal complement of the first column. Its determinant is λ(|z|²+|u|²)=λ. Determinant 1 forces λ=1. Set w=−conjugate(u); the resulting matrix is M(q) for exactly one unit quaternion. Thus M is a group isomorphism from unit quaternions onto SU(2).
Derive the spatial action. For imaginary vectors u,v, the unit rules give uv=−u·v+u×v, with dot and cross products in ℝ³. By real bilinearity,
(a+u)(b+v) = (ab−u·v) + (av+bu+u×v).
For unit q=a+u define R_q(v)=qv q̄ on imaginary v. Its conjugate is −qv q̄, so it is imaginary, and it is real-linear in v. Norm multiplicativity gives |R_q(v)|=|v|. Expanding the product, using u×(u×v)=u(u·v)−|u|²v, yields
R_q(v) = (a²−|u|²)v + 2(u·v)u + 2a(u×v).
The vector triple-product identity follows by expanding its three coordinates; no quaternion commutativity was assumed. If u≠0, put n=u/|u|. The formula fixes n. On its perpendicular plane it is (a²−|u|²)I+2a|u|J, where J(v)=n×v and J²=−I on that plane. Its plane determinant is (a²−|u|²)²+4a²|u|²=(a²+|u|²)²=1. Together with the fixed axis this proves det R_q=+1, not just orthogonality. If u=0, q=±1 and R_q is the identity. This direct argument needs no connectedness theorem.
Writing q=cos(θ/2)+n sin(θ/2), the same expression becomes the Rodrigues formula
R_q(v) = cos θ v + (1−cos θ)(n·v)n + sin θ(n×v).
This also proves the half-angle convention and the right-handed sign. Every unit quaternion admits such an expression, with the purely real cases handled separately.
Prove the quotient statement. Conjugation gives R_pq=R_p R_q, with the rightmost rotation applied first. If R_q is identity, q commutes with every imaginary vector. Comparing qv and vq in the product formula gives 2u×v=0 for every v, forcing u=0. The unit condition then gives q=±1. Conversely both scalars act trivially, so the kernel is exactly ❴1,−1❵.
The preceding spatial rotation theorem states and proves that every member of SO(3) is an axis rotation, including the identity. Choose its axis n and angle θ; Rodrigues’ formula supplies a preimage cos(θ/2)+n sin(θ/2). Thus the map is onto. The group isomorphism theorem yields SU(2)/❴±I❵≅SO(3). Every rotation has exactly two quaternion preimages: equal images imply p⁻¹q is in the kernel. This is an algebraic quotient and a two-element fiber calculation. A topological covering assertion would need additional local-topological proof, which is not supplied here.
Apply
Change the unit quaternion to r=(1+2i+2j)/3. Verify its norm, find its axis and the cosine of its spatial angle, and construct its rotation matrix. Separately classify all lifts of the identity and of the half-turn about the positive k-axis.
Check the changed matrix and the angle boundaries
The squared norm is(1+4+4)/9=1. The axis is (1,1,0)/√2. The spatial cosine is a²−|u|²=−7/9 and the sine is 2a|u|=4√2/9. Substitution into the formula gives rows (1/9,8/9,4/9), (8/9,1/9,−4/9), (−4/9,4/9,−7/9). Its columns are orthonormal and its determinant is 1; multiplying it by (1,1,0) fixes that vector. The quaternion −r gives the same matrix.
The identity has lifts ±1. The k-axis half-turn has matrix diag(−1,−1,1) and lifts ±k. A half-turn does not have infinitely many lifts: choosing the opposite axis changes k to −k, the same two possibilities already supplied by the kernel. Its lift has order four because k²=−1, even though the spatial rotation has order two.
Continue
Compare the perpendicular-plane operator J(v)=n×v with real skew-adjoint rotation blocks. As optional algebraic depth, the product formula gives the quaternion commutator uv−vu=2u×v for imaginary vectors. The factor two is consistent with the half-angle: differentiating conjugation by cos t+n sin t at t=0 gives 2n×v, whereas conjugation by cos(t/2)+n sin(t/2) has derivative n×v. These derivative statements follow directly from the displayed formulas; they do not assert a general Lie-group theorem. For a rotation certificate, record the convention, a unit quaternion, its computed action, and the unavoidable sign ambiguity.
