Quaternions from complex numbers

In the plane, one imaginary unit is enough for rotation. As shown in the complex-numbers section, multiplication by $e^{i\theta}$ turns a point through $\theta$. A three-dimensional rotation also needs an axis. Quaternions extend complex numbers to represent both the angle and the axis. Robot software often uses them because four numbers store a rotation compactly and make composition inexpensive. This presentation follows appendix B.3 of Lynch and Park (Kevin M. Lynch, 2017) and the discussion of the group $S^3$ by Solà, Deray and Atchuthan (Solà et al., 2018).

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Three imaginary units

A quaternion consists of one real part and three imaginary parts: $$q = w + x\,i + y\,j + z\,k ,$$ Each imaginary unit squares to $-1$: $$i^2 = j^2 = k^2 = ijk = -1 .$$ These rules give the pairwise products $ij = k$, $jk = i$ and $ki = j$, along with $ji = -k$, $kj = -i$ and $ik = -j$. Reversing the order changes the sign. Quaternion multiplication does not commute, just as rotations about different axes do not commute. Complex multiplication cannot represent this behavior, but quaternion multiplication can. Following Lynch and Park, this section places the real part first and writes $q = (w, x, y, z)$. The real part $w$ is the scalar part, and $(x, y, z)$ is the vector part. If $w = 0$, the quaternion is pure. Pure quaternions represent ordinary 3-D vectors.

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A unit axis behaves like i

Let the unit vector $u = (u_x, u_y, u_z)$ specify the rotation axis. Write it as the pure quaternion $$\hat u = u_x\,i + u_y\,j + u_z\,k .$$ When you square it, the mixed terms cancel in pairs because $ij = -ji$. The remaining terms give $$\hat u^2 = -\left(u_x^2 + u_y^2 + u_z^2\right) = -1 .$$ A unit axis squares to $-1$, as $i$ does. You can extend the planar formula $e^{i\theta} = \cos\theta + i\sin\theta$ to any axis. A rotation by $\theta$ about the axis $u$ has the unit quaternion $$q = \cos\frac{\theta}{2} + \hat u\,\sin\frac{\theta}{2} \qquad\text{that is,}\qquad q = \Big(\cos\frac{\theta}{2},\; u\,\sin\frac{\theta}{2}\Big)$$ (Lynch and Park eq. B.9). The sum of the squared components is $\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2} = 1$, hence the term unit quaternion. The rotation formula below explains the half angle.

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Rotating a vector

To rotate the vector $p = (p_x, p_y, p_z)$, first represent it by the pure quaternion $p_x i + p_y j + p_z k$. Then multiply it on both sides: $$p' = q\, p\, q^* , \qquad q^* = w - x\,i - y\,j - z\,k .$$ The conjugate $q^*$ reverses the sign of the vector part, much as complex conjugation reverses the sign of $iy$. For a unit quaternion, the conjugate is also the inverse, $q^* q = 1$. Multiplication by $q$ on the left and its conjugate on the right doubles the angle held in $q$. The quaternion must therefore store $\theta/2$. Consider a quarter turn. Rotate $p = (1, 0, 0)$ by $90^\circ$ about the $z$ axis. In this case, $q = \cos 45^\circ + k \sin 45^\circ = \frac{1}{\sqrt 2}(1 + k)$, and direct multiplication gives $$q\, i\, q^* = \tfrac{1}{2}(1 + k)\, i\, (1 - k) = \tfrac{1}{2}(i + ki)(1 - k) = \tfrac{1}{2}(i + j)(1 - k) = \tfrac{1}{2}(i - ik + j - jk) = \tfrac{1}{2}(i + j + j - i) = j .$$ The calculation maps $(1, 0, 0)$ to $(0, 1, 0)$, a quarter turn about $z$.

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The double cover

Replace $q$ by $-q$. In the product $p' = (-q)\,p\,(-q)^*$, the two minus signs cancel. Thus $q$ and $-q$ produce the same rotation. Every rotation corresponds to these two unit quaternions. Unit quaternions satisfy $w^2 + x^2 + y^2 + z^2 = 1$. These points form the 3-sphere $S^3$, a sphere in four dimensions. They also form a group under quaternion multiplication. Since $q$ and $-q$ represent one rotation, $S^3$ covers SO(3) twice. You obtain SO(3) from $S^3$ by identifying every pair of opposite points. Solà et al. therefore call $S^3$ a double cover of SO(3). Both groups have the same tangent space, so angular velocities, small rotations and uncertainties have the same form in either group. The double cover affects numerical work. When a filter or controller compares two quaternions, it should treat $q$ and $-q$ as equal. Interpolation between orientations should follow the shorter path. This may require changing the sign of one quaternion first.

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Using quaternions

Composition. If you apply $q_1$ first and $q_2$ second, the combined rotation is $q_2 q_1$. For $q = (q_0, q_1, q_2, q_3)$ and $p = (p_0, p_1, p_2, p_3)$, the components of $qp$ are $$qp = \begin{bmatrix} q_0 p_0 - q_1 p_1 - q_2 p_2 - q_3 p_3 \\ q_0 p_1 + p_0 q_1 + q_2 p_3 - q_3 p_2 \\ q_0 p_2 + p_0 q_2 - q_1 p_3 + q_3 p_1 \\ q_0 p_3 + p_0 q_3 + q_1 p_2 - q_2 p_1 \end{bmatrix}$$ (Lynch and Park eq. B.15). This quaternion product corresponds to the matrix product $R_q R_p$. To a rotation matrix. The unit quaternion $(q_0, q_1, q_2, q_3)$ produces the matrix $$R = \begin{bmatrix} q_0^2 + q_1^2 - q_2^2 - q_3^2 & 2(q_1 q_2 - q_0 q_3) & 2(q_0 q_2 + q_1 q_3) \\ 2(q_0 q_3 + q_1 q_2) & q_0^2 - q_1^2 + q_2^2 - q_3^2 & 2(q_2 q_3 - q_0 q_1) \\ 2(q_1 q_3 - q_0 q_2) & 2(q_0 q_1 + q_2 q_3) & q_0^2 - q_1^2 - q_2^2 + q_3^2 \end{bmatrix}$$ (Lynch and Park eq. B.12). Each entry is quadratic in $q$, so replacing $q$ by $-q$ leaves $R$ unchanged. Cost. Composing two rotation matrices requires 27 multiplications and 18 additions. Composing two quaternions requires 16 multiplications and 12 additions. You can correct small numerical errors by rescaling $q$ to length 1. A rotation matrix that has drifted must be re-orthogonalized. Compared with other representations.

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References

• Kevin M. Lynch, F. (2017). Modern Robotics.
• Solà, J., Deray, J., Atchuthan, D. (2018). A micro Lie theory for state estimation in robotics. arXiv [cs.RO].

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