1.5. Exponential Coordinates

Exponential Coordinates

Exponential coordinates provide the inverse of Rodrigues’ Formula: given a rotation matrix $R\in SO(3)$, extract the axis $\omega$ and angle $\theta$ such that $R=e^{\hat{\omega }\theta }$. This is the logarithmic map from the Lie group $SO(3)$ back to its Lie algebra $so(3)$.

The forward and inverse maps form a pair:

$\text{exp}:so(3)\to SO(3),\hat{\omega }\theta \mapsto e^{\hat{\omega }\theta }$
$\text{log}:SO(3)\to so(3),R\mapsto \hat{\omega }\theta$

Definition: The vector $\omega \theta \in {\mathbb{R}}^3$, where $\Vert \omega \Vert =1$ and $e^{\hat{\omega }\theta }=R$, is called the exponential coordinate of $R$.

The existence of this decomposition for every $R$ is guaranteed by Euler’s Rotation Theorem: every rotation in 3D can be represented as a rotation by some angle $\theta \in [0,\pi ]$ about a fixed unit axis $\omega \in {\mathbb{R}}^3$.

The key relationship linking $R$ to $\theta$ is:

$\text{tr}(R)=1+2\cos \theta$

Combined with the antisymmetric part $R-R^T=2\sin \theta \hat{\omega }$, this gives a constructive algorithm for recovering $(\omega ,\theta )$ in all cases.

Intuition

The exponential map connects two spaces that both have 3 degrees of freedom but very different geometry:

$so(3)$ — the Lie algebra	$SO(3)$ — the Lie group
$3\times 3$ skew-symmetric matrices	$3\times 3$ orthogonal matrices with $\det =1$
A flat vector space ($\cong {\mathbb{R}}^3$)	A curved 3D manifold
Additive (you can add skew matrices)	Multiplicative (you compose rotations)
Tangent space at the identity of $SO(3)$	The rotation group itself

The exponential map “wraps” the flat tangent space around the curved group, just like $e^{i\theta }$ wraps the real line around the unit circle. The log map “unwraps” it back.

$SO(3)$ can be visualized as a solid ball of radius $\pi$: each point at distance $\theta$ from the center in direction $\omega$ represents the rotation $e^{\hat{\omega }\theta }$. The center is the identity, and antipodal points on the surface ($\theta =\pi$) are identified because $(\omega ,\pi )\sim (-\omega ,\pi )$.

Properties

(Surjectivity / Euler’s Theorem) The exponential map is surjective: every $R\in SO(3)$ has at least one preimage. The proof is constructive — the three cases below give an explicit algorithm.

(Non-injectivity) The map is not injective globally: $(\omega ,\theta )$ and $(\omega ,\theta +2\pi )$ give the same rotation. Even restricting to $\theta \in [0,\pi ]$ and $\Vert \omega \Vert =1$, the map is only nearly bijective — the remaining ambiguity is at $\theta =\pi$, where $(\omega ,\pi )$ and $(-\omega ,\pi )$ coincide.

The Three Cases of the Log Map

Since $\text{tr}(R)=1+2\cos \theta$, the trace determines $\theta$, and there are three regimes:

(Case 1: $\text{tr}(R)=3$) $\theta =0$, $R=I$. No rotation occurred. $\omega$ is undefined (any axis suffices for a zero-angle rotation).

(Case 2: $\text{tr}(R)\in (-1,3)$) $\theta \in (0,\pi )$ — the generic case. Solve for both $\theta$ and $\omega$:

$\theta ={\cos }^{-1}\left(\frac{\text{tr}(R)-1}{2}\right)$
${\omega }_1=\frac{r_{32}-r_{23}}{2\sin \theta },{\omega }_2=\frac{r_{13}-r_{31}}{2\sin \theta },{\omega }_3=\frac{r_{21}-r_{12}}{2\sin \theta }$
This works because the antisymmetric part of Rodrigues’ formula isolates $\hat{\omega }$: the matrix $R-R^T=2\sin \theta \cdot \hat{\omega }$.

(Case 3: $\text{tr}(R)=-1$) $\theta =\pi$. Since $\sin \theta =0$, the Case 2 formulas divide by zero. Instead, Rodrigues’ formula simplifies to $R=I+2{\hat{\omega }}^2$, so $R+I=2\omega {\omega }^T$. This is rank-1: every nonzero column points in the $\pm \omega$ direction — normalize any one to find $\omega$ (up to sign).

(Summary)

$\text{tr}(R)$	$\theta$	$\omega$	Notes
$3$	$0$	undefined	$R=I$
$(-1,3)$	${\cos }^{-1}\left(\frac{\text{tr}(R)-1}{2}\right)$	$\frac{1}{2\sin \theta }(R-R^T{)}^{\vee }$	Generic, unique
$-1$	$\pi$	normalize nonzero col of $R+I$	Sign ambiguity

(Trace Derivation) From the explicit Rodrigues expansion with $\omega =[{\omega }_1,{\omega }_2,{\omega }_3{]}^T$, the diagonal entries of $R$ are ${\omega }_i^2(1-\cos \theta )+\cos \theta$. Summing and using ${\omega }_1^2+{\omega }_2^2+{\omega }_3^2=1$:

$\text{tr}(R)=(1-\cos \theta )+3\cos \theta =1+2\cos \theta$

Examples

Identity Matrix

$R=I$, $\text{tr}(R)=3$. Case 1: $\theta =0$, $\omega$ undefined. Confirmed by Rodrigues’: $e^{\hat{\omega }\cdot 0}=I$ for any $\omega$.

90° about $z$

Solution $R=\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right],\text{tr}(R)=0+0+1=1$

Case 2: $\theta ={\cos }^{-1}\left(\frac{1-1}{2}\right)={\cos }^{-1}(0)=\frac{\pi }{2}$.

${\omega }_1=\frac{0-0}{2}=0,{\omega }_2=\frac{0-0}{2}=0,{\omega }_3=\frac{1-(-1)}{2}=1$

So $\omega =[0,0,1{]}^T$ and $\theta =\pi /2$. Correct: this is a 90° rotation about $z$.

180° about $x$

Solution $R=\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right],\text{tr}(R)=1-1-1=-1$

Case 3: $\theta =\pi$. Compute $R+I=\left[\begin{array}{ccc}2& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$.

First column is $[2,0,0{]}^T$. Normalize: $\omega =[1,0,0{]}^T$. This is a 180° rotation about $x$.

Implementation

import numpy as np
 
def log_map(R):
   """R in SO(3) -> (omega, theta) exponential coordinates"""
   tr = np.trace(R)
   if np.isclose(tr, 3.0):                          # Case 1
       return np.zeros(3), 0.0
   elif np.isclose(tr, -1.0):                        # Case 3
       theta = np.pi
       RpI = R + np.eye(3)
       col = RpI[:, np.argmax(np.linalg.norm(RpI, axis=0))]
       omega = col / np.linalg.norm(col)
       return omega, theta
   else:                                              # Case 2
       theta = np.arccos((tr - 1) / 2)
       omega_hat = (R - R.T) / (2 * np.sin(theta))
       omega = np.array([omega_hat[2,1], omega_hat[0,2], omega_hat[1,0]])
       return omega, theta