SO(2): the whole story in one dimension

Angle, unit complex number, rotation matrix

A planar rotation by $\theta$ can be stored three ways, and each has its own composition rule:

They agree because multiplying complex numbers multiplies their lengths and adds their angles. The matrix $R(\theta)$ is just "multiply by $z$" written as a linear map of $(x, y)$. The angle representation is the only one with a catch: the "mod $2\pi$" is a seam you have to remember to handle. The complex and matrix forms have no seam, because they are the circle, not a line wrapped onto it. That is a pattern the rest of the series keeps: parameters have seams, the group itself does not.

Differentiate a path through I

Take a curve of rotations starting at the identity, $R(t) = R(\omega t)$, turning at rate $\omega$. Its velocity at $t=0$ is

Every velocity through $I$ is a multiple of the one matrix $E$, the generator. The set of all of them, $\{\omega E : \omega\in\mathbb{R}\}$, is the Lie algebra $\mathfrak{so}(2)$: a 1-dimensional vector space of skew-symmetric matrices ($E^\top = -E$). Two small operators move between the number $\omega$ and the matrix $\omega E$. They look pedantic here and are indispensable in 3D:

The demo approximates $\dot R(0)$ by a finite difference $\bigl(R(\omega h) - I\bigr)/h$. As $h$ shrinks, the matrix converges to $\omega E$: the diagonal (the $\cos$ part) fades, and the skew off-diagonal (the $\sin$ part) survives. The error shrinks in proportion to $h$.

A power series that becomes cos and sin

Given a velocity in the algebra, where does the rotation end up after unit time at that constant rate? The answer is the exponential map. For matrices it is defined by the same power series as for numbers:

The key fact is $E^2 = -I$: the generator behaves exactly like $i$. The even powers are $\pm I$ and the odd powers are $\pm E$, so the series splits into the two familiar ones:

We write $\operatorname{Exp}(\theta) = \exp(\theta^\wedge)$ (capital E) for the version that takes the plain number. Each term of the series is a small arrow turned $90^\circ$ from the last, so the partial sums trace a spiral that closes in on the point at angle $\theta$. Push $\theta$ past $\pi$ and it still converges, only more slowly. Exp is defined on the whole line, which is why it can wrap the line around the circle again and again.

Many preimages, one chosen answer

Exp wraps the whole real line around the circle infinitely many times: $\operatorname{Exp}(\theta) = \operatorname{Exp}(\theta + 2\pi k)$ for every integer $k$. So its inverse, the logarithm, has to choose. The standard choice is the principal branch: the unique preimage in $(-\pi, \pi]$.

Near the identity Log is perfectly smooth. At the point exactly opposite, $\theta = \pm\pi$, it jumps by $2\pi$: two rotations a hair apart on either side of $180^\circ$ get logs of about $+\pi$ and $-\pi$. That is the cut locus. It cannot be avoided, only kept away from. It will reappear in 3D as the sphere of radius $\pi$ (Part 5). The practical rule it gives is to always take Log of something close to the identity, which is exactly what $Y\ominus X = \operatorname{Log}(X^{-1}Y)$ does when $X$ and $Y$ are close.

The two operators every algorithm uses

Exp and Log act at the identity. To act at any point $X$, compose with it:

In $SO(2)$ this is angle arithmetic with wrap-around: $\oplus$ adds and wraps, and $\ominus$ subtracts and wraps into $(-\pi, \pi]$. Because $SO(2)$ is commutative, it makes no difference whether you compose on the right, $X\operatorname{Exp}(\tau)$, or on the left, $\operatorname{Exp}(\tau)X$. In 3D it will, and Part 8 is devoted to that difference. The demo interpolates from $X$ to $Y$ by $X \oplus t\,(Y\ominus X)$, which always takes the short way round, and compares it with naive interpolation of the angle numbers, which sometimes takes the long way.

Gauss–Newton on the circle

A robot takes $N$ noisy compass readings $z_i$ of a true heading. The least-squares estimate minimizes the sum of squared geodesic residuals:

Linearize by perturbing $X$ in its tangent space, $X \oplus \delta$. Then $r_i(X\oplus\delta) \approx r_i(X) - \delta$, so the Jacobian is $J_i = -1$. The Gauss–Newton step solves the normal equations $(J^\top J)\,\delta = -J^\top r$, which here gives $\delta = \tfrac1N\sum_i r_i$, and the update is $X \leftarrow X\oplus\delta$. Put the truth near north so the readings straddle the seam. The Lie-group solver converges in one or two steps. The naive solver, which does least squares on the raw angle numbers, lands on the wrong side of the circle.

Every object, and what it becomes in 3D

Everything above has a counterpart for 3D rotations and poses. In $SO(2)$ two of the objects, the adjoint and the Jacobians, are trivially $1$, because the group is commutative. They exist to correct for non-commutativity, so they only come alive in $SO(3)$ and $SE(3)$. That is the precise sense in which the circle "has nothing that can go wrong".