# What is the Lipschitz constant of the differential of the matrix exponential $\mathfrak{so}(3)\to \mathrm{SO}(3)$

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?

• 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. Commented Oct 4, 2020 at 15:07

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.