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Dual Quaternions

Rigid transformations may be blended efficiently by the use of dual quaternions. This can be used for skinning 3D models on an articulated rigid body1. Before reading these notes you’ll probably want to get familiar with regular quaternions and rigid transformations first.

Dual Numbers

Consider numbers of the form:

\[a = a_0 + \epsilon a_\epsilon\]

for some magic constant \(\epsilon\) satisfying \(\epsilon^2 = 0\), with sum/product distributive as usual. \(\epsilon\) may be seen as spanning infinitesimal numbers, i.e. cancelling anything of order 2 and more (which forms the basis of synthetic differential geometry). For instance, consider the product of dual numbers:

\[\block{a + \epsilon \dd a}\block{b + \epsilon \dd b} = a b + \epsilon \block{a.\dd b + \dd a.b}\]

The above gives exactly the formula for the derivative of a product. In fact, any analytic function \(f\) extends naturally to dual numbers:

\[f\block{a + \epsilon \dd a} = f\block{a} + \epsilon \dd f(a).\dd a\]

In short, dual numbers provide an algebra for tangent vectors.

Dual Inverse

From the usual formula for the derivative of the inverse, we get:

\[\block{x + \epsilon \dd x}^{-1} = \inv{x} - \epsilon \frac{\dd x}{x^2}\]

Dual Quaternions

Again, consider numbers of the form:

\[q + \epsilon \dd q\]

with \(q, \dd q \in \HH\) quaternions and the dual unit as before. Again, functions on quaternions extend naturally on dual quaternions by differentiation.

Dual Quaternion Norm

From the usual norm formula:

\[\norm{q + \epsilon \dd q} = \norm{q} + \epsilon \frac{q^T\dd q}{\norm{q}}\]

Unit Dual Quaternions

If we normalize a dual quaternion \(q + \epsilon \dd q\) by extending the quaternion normalization, we will obtain a real part with unit norm. Therefore, the corresponding dual part will be a tangent vector to the unit quaternion sphere \(S^3\):

\[\begin{align} q + \epsilon \dd q &= q + \epsilon q.\omega^b \\ &= q + \epsilon \omega^s.q\\ \end{align}\]

where the dual part is left-(resp. right)-trivialized using the body velocity \(\omega^b\) (resp. spatial velocity \(\omega^s\)), taken in the Lie algebra \(\mathfrak{s^3}\simeq\RR^3\) of pure imaginary quaternions. (see Quaternions and Lie Groups for details).

In summary: normalizing regular quaternions extends to dual quaternions by differentiation as before, giving so-called unit dual quaternions. The real part is a unit regular quaternion, while the dual part is a tangent vector at the real part. This tangent vector has body/spatial coordinates in \(\RR^3\), which can be used to encode translations.

Connection with Rigid Transformations

The spatial derivative of the unit quaternion product gives (as for any Lie group):

\[\dd^s a.b = \dd^s a + \Ad_a \dd^s b\]

where in this case \(\Ad_a\) is the rotation corresponding to quaternion \(a\). The above formula is exactly the translation part of the rigid composition of \(\block{a, \dd^s a} \simeq \mat{\Ad_a & \dd^s a\\ 0 & 1}\) with \(\block{b, \dd^s b} \simeq \mat{\Ad_b & \dd^s b\\ 0 & 1}\). Therefore, if we encode our rigid transformations as spatial derivatives of unit quaternions (expressed as dual quaternions accordingly), we can obtain the composition of rigid transformations from the product of unit dual quaternions:

\[\block{a + \epsilon \dd^s a. a}\block{b + \epsilon \dd^s b.b} = a.b + \epsilon \block{ \dd^s a + \Ad_a \dd^s b}ab\]

In other words, there is a nice Lie group homomorphism between unit dual quaternions and rigid transformations.

Dual Quaternion Normalization

Now, the whole point of using unit dual quaternions is the cheap projection operator on \(SE(3)\) by means of dual quaternion normalization: from a series of unit dual quaternions \(g_i = q_i + \epsilon t_i q_i\) we may blend them however we like to obtain some (possibly non-unit) dual quaternion \(\tilde{g} = f\block{g_i}\), which we can then normalize to obtain a unit dual quaternion, and a corresponding rigid transformation:

\[g = \frac{\tilde{g}}{\norm{\tilde{g}}}\]

From the usual formula, dual quaternion normalization is:

\[\frac{g}{\norm{g}} = \frac{q}{\norm{q}} + \epsilon \frac{1}{\norm{q}}\block{I - \frac{q q^T}{\norm{q}^2}}\dd q\]

Expanding the dual part gives:

\[\begin{align} \frac{1}{\norm{q}}\block{I - \frac{q q^T}{\norm{q}^2}}\dd q &= \block{\dd q.\inv{q} - \frac{q^T \dd q}{\norm{q}^2}} \frac{q}{\norm{q}}\\ &= \frac{1}{\norm{q}^2}\block{\dd q.\bar{q} - q^T \dd q} \frac{q}{\norm{q}}\\ &= \frac{1}{\norm{q}^2}\mathrm{Im}\block{\dd q.\bar{q}} \frac{q}{\norm{q}}\\ &= \mathrm{Im}\block{\dd q.\inv{q}} \frac{q}{\norm{q}}\\ \end{align}\]

As expected, the dual part is that of a unit dual quaternion i.e. a pure imaginary quaternion representing the translation, times the real part. The translation part of the blended rigid transform is the imaginary part of the blended quaternion spatial velocity, in the sense of the full multiplicative quaternion Lie group \(\HH\) (watch out: the Lie algebra is the whole quaternion space \(\HH\) in this case).

Blending Algorithm

From the above, the general algorithm for dual quaternion blending goes as follows:

  1. encode rigid transformations \((q, t)_i\) as unit dual quaternions \(g_i = q_i + \epsilon \underbrace{t_i q_i}_{\dd q_i}\)

  2. blend real/dual parts2 to obtain a dual quaternion \(g = q + \epsilon \dd q\)

  3. normalize the real part to obtain the blended rotation, and compute the imaginary part of the spatial velocity \(\mathrm{Im}\block{\dd q.\inv{q}}\) to obtain the blended translation.

Jacobian Matrix

For a linear blending \(q = \sum_i \alpha_i q_i\), the dual part is:

\[\dd q = \sum_i \alpha_i \dd q_i = \sum_i \alpha_i t_i q_i\]

Let \(a=\dd q, b=q\), the blended rotation is \(\frac{b}{\norm{b}}\) and the blended translation is \(t=\mathrm{Im}\block{a \inv{b}}\). As before, we have:

\[\begin{align} \dd^s \frac{b}{\norm{b}}.\dd b &= \mathrm{Im}\block{\dd b.\inv{b}}\\ &= \mat{0 && I} R_{\inv{b}} \dd b\\ \end{align}\]

for the rotation part, where \(R_q\) is the matrix form of the right-multiplication by \(q\). Now, the translation part is a little messier:

\[\begin{align} \dd \block{a \inv{b}} &= \dd a.\inv{b} - a.\inv{b}.\dd b.\inv{b}\\ &= R_{\inv{b}}\block{\dd a - L_{a \inv{b}}\dd b} \\ \\ \dd a &= \sum_i \alpha_i \block{\dd t_i q_i + t_i \dd q_i} \\ &= \sum_i \alpha_i \block{\dd t_i q_i + t_i \omega^s_i q_i} \\ &= \sum_i \alpha_i \block{\dd t_i + t_i \omega^s_i}q_i \\ &= \sum_i \alpha_i R_{q_i}\block{\mat{0^T\\I} \dd t_i + L_{t_i}\mat{0^T\\I}\omega^s_i} \\ \\ \dd b &= \sum_i \alpha_i \dd q_i \\ &= \sum_i \alpha_i \omega^s_i q_i \\ &= \sum_i \alpha_i R_{q_i}\mat{0^T\\I}\omega^s_i\\ \end{align}\]

In the end, the \(i\)-th translation Jacobian block is:

\[\begin{align} \ddd{t}{t_i} &= \alpha_i \mat{0 && I}R_{\inv{b}} R_{q_i}\mat{0^T\\ I} \\ &= \alpha_i \mat{0 && I}R_{q_i\inv{b}}\mat{0^T\\ I} \\ \\ \ddd{t}{\omega^s_i} &= \alpha_i \mat{0 && I}R_{\inv{b}}\block{R_{q_i} L_{t_i}\mat{0^T\\ I} - L_{a\inv{b}}R_{q_i}\mat{0^T\\ I}}\\ &= \alpha_i \mat{0 && I}R_{\inv{b}} R_{q_i} \block{L_{t_i} - L_{a\inv{b}}}\mat{0^T\\ I}\\ &= \alpha_i \mat{0 && I}R_{q_i\inv{b}} L_{t_i - a\inv{b}}\mat{0^T\\ I}\\ \end{align}\]

and the \(i\)-th rotation block is:

\[\begin{align} \ddd{^s\frac{b}{\norm{b}}}{\omega^s_i} &= \alpha_i \mat{0 && I} R_{\inv{b}} R_{q_i} \mat{0^T\\I} \\ &= \alpha_i \mat{0 && I} R_{q_i\inv{b}} \mat{0^T\\I} \\ \end{align}\]

Notes & References

  1. Kavan, Ladislav, et al. “Skinning with dual quaternions.” Proceedings of the 2007 symposium on Interactive 3D graphics and games. ACM, 2007. ↩

  2. For the formula to make sense, the dual blending must be the derivative of the real blending. Otherwise, the blended dual part will not be a tangent vector at \(q\). ↩