The rotation group SO(2) as a circle

2 \times 2

matrix has four entries, so you can treat it as a point in

\mathbb{R}^4

. Most of these points are not rotations. A rotation matrix has columns of unit length (two constraints) that are perpendicular (one more constraint). Four entries minus three constraints leaves one free number, the angle

\theta

R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

\theta

changes,

R(\theta)

moves along a curve in

\mathbb{R}^4

. That curve is

SO(2)

. This section draws it, shows that it is a circle, and finds its tangent at the identity, which is the Lie algebra

\mathfrak{so}(2)

. These two objects are the first two rows of the Lie cheatsheet.

Sampling the curve

Each angle gives one rotation matrix. Write its entries in a row,

(r_{11}, r_{12}, r_{21}, r_{22})

, and you get a point in

\mathbb{R}^4

. The code also checks that every sample is a rotation.

import matplotlib.pyplot as plt
import numpy as np


def rot(theta):
    c, s = np.cos(theta), np.sin(theta)
    return np.array([[c, -s], [s, c]])


thetas = np.linspace(0, 2 * np.pi, 400)
Rs = np.stack([rot(t) for t in thetas])  # shape (400, 2, 2)
points = Rs.reshape(len(thetas), 4)  # (r11, r12, r21, r22) per row

# Every sample satisfies the rotation constraints
orth_err = np.abs(np.einsum("nji,njk->nik", Rs, Rs) - np.eye(2)).max()
det_err = np.abs(np.linalg.det(Rs) - 1).max()
print(f"max |R^T R - I| = {orth_err:.1e}")
print(f"max |det R - 1| = {det_err:.1e}")

max |R^T R - I| = 2.2e-16
max |det R - 1| = 2.2e-16

The shape of the curve

The point for angle

\theta

(\cos\theta, -\sin\theta, \sin\theta, \cos\theta)

. Two entries repeat the others:

r_{22} = r_{11}

and

r_{12} = -r_{21}

. These two equations keep the curve in a plane. Inside that plane,

r_{11}^2 + r_{21}^2 = 1

, which is a circle. Every point is

\sqrt{2}

from the origin, and the circle closes because

\theta

and

\theta + 2\pi

give the same matrix.

r11, r12, r21, r22 = points.T

print(f"max |r22 - r11|        = {np.abs(r22 - r11).max():.1e}")
print(f"max |r12 + r21|        = {np.abs(r12 + r21).max():.1e}")
print(f"distance from origin   = {np.linalg.norm(points, axis=1).min():.4f} to {np.linalg.norm(points, axis=1).max():.4f}")
print(f"sqrt(2)                = {np.sqrt(2):.4f}")
print(f"closes on itself       = {np.allclose(rot(0), rot(2 * np.pi))}")

max |r22 - r11|        = 0.0e+00
max |r12 + r21|        = 0.0e+00
distance from origin   = 1.4142 to 1.4142
sqrt(2)                = 1.4142
closes on itself       = True

Drawing the curve

You cannot draw

\mathbb{R}^4

. But

r_{22}

always equals

r_{11}

, so you can drop it. The plot uses

(r_{11}, r_{21}, r_{12})

and colors each point by

\theta

. The result is a tilted circle around the origin.

Rotations and reflections

Perpendicular unit columns are not enough to make a rotation. The matrices

F(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix}

have them too, but their determinant is

-1

. Each one is a reflection across a line through the origin. The reflections form a second circle in

\mathbb{R}^4

. Together the two circles make up

O(2)

, the group of all orthogonal

2 \times 2

matrices. Along any path of orthogonal matrices the determinant would have to jump from

+1

-1

to reach a reflection, and it cannot jump. So

SO(2)

is the circle that contains the identity.

One column is enough

The first column of

R(\theta)

(\cos\theta, \sin\theta)

, a point on the unit circle in the plane. The second column is the first one turned by

90°

counterclockwise, so it adds nothing new. Each rotation matches one point on the unit circle, and each point matches one rotation. This is another way to see that

SO(2)

is a circle. The top panel follows the first column as

\theta

grows. The row below shows the whole rotated frame

\{R\mathbf{e}_1, R\mathbf{e}_2\}

at the same angles.

The tangent at the identity

Seen up close, a curve looks like a straight line. At the identity,

\theta = 0

, that line points along the derivative of

R(\theta)

\left.\frac{dR}{d\theta}\right|_{\theta=0} = \begin{bmatrix} -\sin 0 & -\cos 0 \\ \cos 0 & -\sin 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = G

The matrix

G

is skew-symmetric. Every velocity at the identity is

\omega G

for some angular rate

\omega

\mathfrak{so}(2) = \left\{ \omega G : \omega \in \mathbb{R} \right\} = \operatorname{span}\{G\}

\mathfrak{so}(2)

is one-dimensional, like

SO(2)

, and

G

is its single generator. The matrix exponential leads back to the curve:

e^{\theta G} = R(\theta)

. The code below checks the derivative, the skew symmetry and the exponential. Lie I covers tangent spaces and the exponential map in general. Appendix D of Corke (Corke, 2023) repeats this construction for

SO(3)

, which has three generators.

from scipy.linalg import expm

h = 1e-6
dR = (rot(h) - rot(-h)) / (2 * h)  # central difference at theta = 0
print("dR/dtheta at identity:")
print(np.round(dR, 6))
print("skew-symmetric:", np.allclose(dR, -dR.T))

theta = 1.2
print(f"\nexp({theta} * dR) equals R({theta}):", np.allclose(expm(theta * dR), rot(theta)))

dR/dtheta at identity:
[[ 0. -1.]
 [ 1.  0.]]
skew-symmetric: True

exp(1.2 * dR) equals R(1.2): True

References

Corke, P. (2023). Robotics, vision and control: Fundamental algorithms in python.

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Robotics Systems

Motion Algebras

Kinematics

Dynamics

Systems Integration

The rotation group SO(2) as a circle

Sampling the curve

The shape of the curve

Drawing the curve

Rotations and reflections

One column is enough

The tangent at the identity

References

​Sampling the curve

​The shape of the curve

​Drawing the curve

​Rotations and reflections

​One column is enough

​The tangent at the identity

​References

Sampling the curve

The shape of the curve

Drawing the curve

Rotations and reflections

One column is enough

The tangent at the identity

References