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?

  • 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


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.


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