The Lie Algebra, Twists, and the Adjoint

A rigid body's configuration lives on the group $SE(3)$, a curved manifold. Its velocity does not: at any pose the set of admissible velocities is a flat vector space, the tangent space. The tangent space at the identity is the Lie algebra $\mathfrak{se}(3)$, and its elements are exactly the twists from the last article. The adjoint is the linear map that carries a twist from one frame's version of that flat space to another's.

The Lie algebra is the tangent space at the identity

Take a curve $T(t) \in SE(3)$ through the identity, $T(0) = I$. Its derivative $\dot T(0)$ is a tangent vector at $I$, and it always has the form

$$\dot T(0) = \hat{\mathcal{V}} = \begin{bmatrix} \hat\omega & v \\ 0 & 0 \end{bmatrix}, \qquad \hat\omega = \begin{bmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{bmatrix}.$$

The hat map $\wedge$ sends a twist $\mathcal{V} = (v, \omega)$ to this $4\times4$ matrix. The set of all such matrices is $\mathfrak{se}(3)$, a $6$-dimensional vector space, and the matrix exponential $T = \exp(\hat{\mathcal{V}})$ integrates a constant twist back into a screw motion on the group. So a twist is not just a stacked pair of $3$-vectors: it is a bona fide tangent vector, an element of the algebra whose bracket encodes how infinitesimal motions fail to commute.¹

The adjoint moves a twist between frames

The same physical motion has different coordinates in different frames. Let $T_{AB} = (R, p) \in SE(3)$ be the pose of frame $B$ measured in frame $A$. A twist written in $B$ transports to $A$ by the adjoint $\mathrm{Ad}_{T_{AB}}$, a linear map on twists whose action on $\mathcal{V} = (v, \omega)$ is

$$\mathcal{V}_A = \mathrm{Ad}_{T_{AB}}\, \mathcal{V}_B, \qquad \mathrm{Ad}_{T}\begin{bmatrix} v \\ \omega \end{bmatrix} = \begin{bmatrix} R\,v + \omega \times p \\ R\,\omega \end{bmatrix}.$$

As a matrix this is the $6\times6$ Plucker transform. Geometrically $\mathrm{Ad}_T$ does two things: it rotates each part into the new axes through $R$, and it adds the $\omega \times p$ coupling that appears because the two frame origins are offset by $p$, exactly the rigid-body relation $v_P = v_O + \omega \times r$ now promoted to act on the whole $6$D twist at once. (Sign and stacking conventions for the adjoint vary between references; we fix the ordering $(v, \omega)$ and the coupling $+\,\omega \times p$ throughout, and specialize to the plane below.)

The dual map $\mathrm{Ad}_T^{\top}$ transports wrenches the other way, so motions and forces stay a matched pair. This single change-of-frame is the engine of Featherstone's articulated-body algorithm: velocities are pushed outward from base to tips and forces gathered back inward, each hop a Plucker transform ${}^{B}X_A$ that is precisely $\mathrm{Ad}$ between adjacent link frames.² The $6\times6$ transform introduced as spatial algebra is the adjoint.

A worked transport in SE(2)

Drop to the plane, where $\omega$ is a scalar and $J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ is the $90^\circ$ rotation that realizes the cross product, $\omega \times p = \omega\, J p$. With the twist ordered $(v_x, v_y, \omega)$ the adjoint collapses to

$$\mathrm{Ad}_{T} = \begin{bmatrix} R & J p \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos\phi & -\sin\phi & -p_y \\ \sin\phi & \cos\phi & p_x \\ 0 & 0 & 1 \end{bmatrix}, \qquad \mathrm{Ad}_T \mathcal{V} = \begin{bmatrix} R v + \omega\, J p \\ \omega \end{bmatrix}.$$

Take a body frame rotated by $\phi = 90^\circ$ and offset to $p = (1, 0)$, carrying the body twist $\mathcal{V}_B = (v, \omega)$ with $v = (1, 0)$ and $\omega = 1$. The rotation sends $R(90^\circ) = J$, so the pieces are

$$R v = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \qquad \omega\, J p = 1 \cdot \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},$$

and the full matrix product reads

$$\mathcal{V}_A = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \left( (0, 2),\ 1 \right).$$

The linear part is not $R v = (0,1)$: the spin about an origin one unit away adds the coupling $\omega \times p = (0,1)$, pushing $v_A$ to $(0,2)$. Slide $p$ back to the origin and that second term vanishes; the adjoint becomes a pure rotation of the parts.

Transporting a twist with the SE(2) adjointinteractive

φ = 90°ω = 1.00pₓ = 1.00p_y = 0.00v_Bₓ = 1.00v_B_y = 0.00

V_B linear

(1.00, 0.00)

R v_B

(0.00, 1.00)

ω × p

(0.00, 1.00)

V_A linear

(0.00, 2.00)

ω (both)

1.00

Because $\mathrm{Ad}_T$ is linear and invertible, transporting a twist there and back is exact: $\mathrm{Ad}_{T}^{-1} = \mathrm{Ad}_{T^{-1}}$. That is what lets a recursive solver shuttle spatial velocities and forces across dozens of joints without ever leaving the flat, well-behaved world of the Lie algebra.

Footnotes

1. Sola, Deray, Atchuthan, A micro Lie theory for state estimation in robotics (2018). ↩

2. Featherstone, Rigid Body Dynamics Algorithms (2008). ↩