ISEGORIA / MATH ENCYCLOPEDIA
Lie groups and Lie algebras: continuous symmetry
Exponentials of rotations, the double cover, and why the bracket measures failure to commute.
Before you begin: Linear algebra and group theory
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The exponential map
A rotation is the exponential of an infinitesimal one. Choose an axis and an angle, then add terms of the power series one at a time. The true rotation moves a point around a circle; each truncated series lands off that circle, and the gap closes as terms are added. Drag in the figure to turn the view.
Worked example. For a unit axis \(\mathbf n\), \(K\mathbf v=\mathbf n\times\mathbf v\) and \(K^3=-K\), so the whole series collapses to Rodrigues’ formula. The trace of \(R\) is \(1+2\cos\theta\).
Watch out. A truncated series is not a rotation: it stretches lengths, so \(R^{\mathsf T}R\neq I\). Only the full sum lies in the group.
Why does a skew-symmetric K give an orthogonal exponential?
Because \((e^{\theta K})^{\mathsf T}=e^{\theta K^{\mathsf T}}=e^{-\theta K}\), which is the inverse of \(e^{\theta K}\).
2. The double cover: 720° to come home
Turn an object about a fixed axis. Its orientation returns after 360°, but the unit quaternion that produced it has only reached −1; it needs a second full turn to return to +1. The ribbon tied to the wall records the same fact. At 720°, drag Untwist: the double twist straightens while the block stays still. At 360° no such move exists.
Worked example. Both \(q\) and \(-q\) give the same rotation, so the map \(SU(2)\to SO(3)\) is two to one. At \(\theta=360^\circ\), \(q=-1\) yet \(R_q=I\).
Watch out. The ribbon picture is a faithful statement about loops in \(SO(3)\), whose fundamental group is \(\mathbb Z/2\); it is not a claim that physical ribbons carry spin.
Which loop in SO(3) can be shrunk to a point?
The 720° loop. The 360° loop cannot, because its lift to \(SU(2)\) ends at \(-1\) instead of closing.
3. The bracket measures failure to commute
Rotate by ε about x, then y, then undo x, then undo y. If rotations commuted you would return to the start. You do not: the loop leaves a gap, and the leftover rotation is a turn of about ε² about the z axis (clockwise seen from above). The log–log plot shows the gap shrinking with slope 2.
Worked example. For the rotation generators of \(\mathfrak{so}(3)\), \([X,Y]=Z\): the Lie algebra of rotations is \(\mathbb R^3\) with the cross product. The product is read right to left, so the first rotation is \(e^{\varepsilon X}\).
Watch out. The \(\varepsilon^2\) law is asymptotic. At large \(\varepsilon\) the third-order terms tilt the leftover axis away from \(z\), which the readout shows.
Why is the gap second order rather than first?
All first-order terms cancel because each step is undone; the first survivor is the commutator term \(\varepsilon^2[Y,X]\).