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When simulating rigid bodies, one generally ends up with non-symmetric Hessian matrices due to the curvature of \(SO(3)\) (where Schwartz’ theorem about the symmetry of Hessian does not hold), which do not play well with numerical solvers down the road. More generally, second derivatives are a pain to work with since there are several definitiions to choose from (e.g. Lie derivative, Levi-Civita connection) which do not always agree. Even though there is sometimes a natural choice for some spaces (\(SO(3)\) which is compact and as such enjoys a bi-invariant Riemannian metric), for others there is none (\(SE(3)\) which does not have a bi-invariant Riemannian metric) and it is not always clear what to do in such situations.
Putting theoretical considerations aside, one can simply remark that in practical applications like physics simulation, one generally uses local charts for these spaces anyways (for instance, using body-fixed velocity exponential for rotation integration), therefore an obvious solution is to work inside these charts, which side-steps the problem entirely. This means one must keep track of the local chart all along, together with their derivatives. These notes describe the procedure for exponential/logarithmic charts.
Let \(f: SO(3) \to E\), with \(E\) some finite dimensional vector space. Let \(R\) be the current input, we parametrize \(SO(3)\) locally using the exponential:
\[\begin{align} \hat{f}^s: \quad \RR^3 &\to E \\ \omega &\mapsto f(\exp(\omega)R) = \hat{f}^s(\omega)\\ \end{align}\]We’ll call this \(\hat{f}^s\) the spatial parametrization of \(f\) at \(R\). Similarly, the body-fixed parametrization at \(R\) is:
\[\begin{align} \hat{f}^b: \quad \RR^3 &\to E \\ \omega &\mapsto f(R\exp(\omega)) = \hat{f}^b(\omega)\\ \end{align}\]which at \(\omega=0\) gives:
\[\dd \hat{f}^s(0).\dd \omega = \dd f(R).\hat{\dd \omega} R\]therefore \(\dd \omega\) plays the role of a spatial velocity. So we get:
\[\dd \hat{f}^s(0).\dd \omega = \dd^s f(R).\dd \omega\]Likewise:
\[\dd \hat{f}^b(0).\dd \omega = \dd f(R).R \hat{\dd \omega}\]and this time \(\dd \omega\) plays the role of a body-fixed velocity:
\[\dd \hat{f}^s(0).\dd \omega = \dd^s f(R).\dd \omega\]which at \(\omega=0\) gives:
\[\begin{align} \dd^2 \hat{f}^s(0).\dd \omega_2.\dd \omega_1 &= \dd^2 f(R).\hat{\dd \omega_2} R.\hat{\dd \omega_1} R + \dd f(R).\frac{\hat{\dd \omega_1}\hat{\dd \omega_2} + \hat{\dd \omega_2} \hat{\dd \omega_1}}{2}R \\ &= \dd^2 f(R).\hat{\dd \omega_2} R.\hat{\dd \omega_1} R + \dd f(R).\block{\frac{\dd \omega_1 \dd \omega_2^T + \dd \omega_2 \dd \omega_1^T}{2} - \dd \omega_2^T \dd \omega_1 I}R \end{align}\]Let us now introduce a generalized end-force \(\lambda \in E\) and consider the following pairing:
\[\trace{\lambda^T\dd^2 \hat{f}^s(0).\dd \omega_2.\dd \omega_1}\]\(f(R) = Rx\) for some fixed \(x \in \RR^3\). \(f\) is linear therefore:
\[\dd f(R).\dd R = \dd R.x = f(\dd R)\] \[\dd^2 f = 0\]Given an end-force \(\lambda \in \RR^3\), the geometric stiffness is:
\[\begin{align} \trace{\lambda^T\dd^2 \hat{f}^s(0).\dd \omega_2.\dd \omega_1} &= \trace{\lambda^T.\block{\frac{\dd \omega_1 \dd \omega_2^T + \dd \omega_2 \dd \omega_1^T}{2} - \dd \omega_2^T \dd \omega_1 I}Rx} \\ &= \trace{Rx\lambda^T.\block{\frac{\dd \omega_1 \dd \omega_2^T + \dd \omega_2 \dd \omega_1^T}{2} - \dd \omega_2^T \dd \omega_1 I}} \\ &= \dd \omega_2^T\frac{Rx \lambda^T + \lambda (Rx)^T}{2}\dd \omega_1 - \trace{\lambda^TRx} \dd \omega_2^T \dd \omega_1 \\ &= \dd \omega_2^T\frac{\hat{Rx}\hat{\lambda} + \hat{\lambda}\hat{Rx}}{2}\dd \omega_1 \\ \end{align}\]Let \(R = f\block{\bar{x}}\) for some input \(\bar{x}\), we parametrize the codomain around \(R\) using the logarithm:
\[\begin{align} \hat{f}^s: \quad E &\to \RR^3 \\ x &\mapsto \log\block{f(x)\inv{R}} \\ \end{align}\]Similarly:
\[\begin{align} \hat{f}^s: \quad E &\to \RR^3 \\ x &\mapsto \log\block{\inv{R}f(x)} \\ \end{align}\]which at \(\bar{x}\) gives:
\[\begin{align} \dd \hat{f}^s\block{\bar{x}}.\dd x &= \dd f(x).\dd x.\inv{R} \\ &= \dd^s f(x).\dd x\\ \end{align}\]this corresponds to the spatial velocity.
which at \(x = \bar{x}\) reduces to:
\[\begin{align} \dd^2 \hat{f}^s\block{\bar{x}}.\dd x_2.\dd x_1 &= \dd^2 \log(I).\dd f(x).\dd x_2.\inv{R}. \dd f(x).\dd x_1.\inv{R}\\ &+ \dd^2 f(x).\dd x_2.\dd x_1.\inv{R} \end{align}\](using spatial parametrizations at \(R_p, R_c\))
\[f\block{\omega_p, \omega_c} = \log\block{R_p^T\exp\block{-\omega_p}\exp\block{\omega_c}R_c}\]Let \(g\block{\omega_p, \omega_c} = \exp\block{-\omega_p}\exp\block{\omega_c}\), we obtain the jacobian:
\[\dd f\block{\omega_p, \omega_c}.\dd \omega_p.\dd \omega_c = \dd \log\block{R_p^T g\block{\omega_p, \omega_c}R_c}.R_p^T\block{\dd g\block{\omega_p, \omega_c}.\dd \omega_p.\dd \omega_c}R_c\]where
\[\dd g\block{\omega_p, \omega_c}.\dd \omega_p.\dd \omega_c = -\dd \exp\block{-\omega_p}.\dd \omega_p.\exp\block{\omega_c} + \exp\block{-\omega_p}\dd \exp\block{\omega_c}.\dd \omega_c\] e-mail comments/suggestions