9
$\begingroup$

I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates: $$ R(u) := \exp(u_\times) $$ with $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ is the cross-product matrix of vector $u$.

The directional derivative of $R(u)$ in the direction $Y$ is: $$ [D_u R]Y = [T(u)Y]_\times R(u) $$ for any vector $Y\in \mathbb{R}^3$, where $$ T(u) := \int_0^1R(su)ds $$

Both $R$ and $T$ are Lipschitz continuous with constants $1$ and $\tfrac{1}{2}$ respectively: $$ \|R(u)-R(v)\| \le|u-v| \\ \|T(u)-T(v)\| \le \tfrac{1}{2}|u-v| $$ for any $u$ and $v$, where I use the operator norm (subordinate norm) of the Euclidean norm.

To find the convergence bounds of Newton's iterations for the numerical method I'm using (conditions of Kantorovich) I need to estimate a bound on the second derivative (which is really hard to compute explicitly as far as I know), or the Lipschitz constant of the differential. Experimentally (using a program) I found that it is $1$. How to prove this?

$\endgroup$
1
  • 1
    $\begingroup$ It is probably easier to work with the double cover $SU(2)$, i.e. the unit quaternions, since the exponential is then the usual one applied to imaginary quaternions. Then use the fact that $SU(2)\to SO(3)$ is a local isometry. $\endgroup$
    – Ben McKay
    Commented Oct 4, 2020 at 15:07

1 Answer 1

2
$\begingroup$

Found the proof! It's done using the integral definition of $T$: $$ T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n R\left(\tfrac{i}{n}v\right) $$ So for any vectors $X$ and $Y$: \begin{align*} &\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\ &\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \end{align*} and we only need to prove that each summand is less than $|u-v| |X| |Y|$: \begin{align*} & \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\ &= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\ & \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\ &+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\ &\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\ &= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr| \end{align*} where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question). Since $X$ and $Y$ are arbitrary, $$ \left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v| $$ in the subordinate norm.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.