advanced34 min studyLesson 496 in Mathematics

Adjoint and coadjoint actions

Differentiate conjugation and its dual action with a fixed inverse and sign convention.

Before you begin

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The question for today

What you’ll understand

Compute Ad by differentiating conjugation and ad by differentiating Ad.
Use the inverse in the coadjoint definition to preserve the natural pairing.
Verify an affine-group coadjoint transformation and its infinitesimal signs.

Changing the group element used to conjugate a motion changes its generator. The adjoint representation describes that change on the Lie algebra. Its dual counterpart, the coadjoint action, transports linear functionals such as generalized momenta. The inverse in that dual action is essential: ordinary transpose alone would give the wrong composition order and fail the natural pairing identity. We will define both actions explicitly, derive their infinitesimal versions with the left-invariant bracket convention, and verify every sign in a two-dimensional nonabelian example.

Conjugation acts on identity generators

Conjugation by g sends h to ghg inverse, fixes the identity, and is a smooth group automorphism. Its identity derivative is the invertible linear map Ad-g on the Lie algebra. For a matrix group this is simply X↦gXg inverse. Conjugating a one-parameter subgroup gives the exponential of the transformed generator, so this action connects finite symmetry changes with the infinitesimal description without approximating g by a small matrix.

Differentiate Ad-exp-tX at zero to obtain ad-X. Under the fixed left-invariant convention, ad-X sends Y to [X,Y]. Thus the Lie bracket is the infinitesimal adjoint action, and Jacobi expresses that these infinitesimal maps act by derivations. The identity Ad-exp-X equals exp-ad-X is an operator identity on the Lie algebra, distinct from the exponential of a matrix acting on the original representation space. The dimensions of those two spaces need not be the same.

\operatorname{Ad}_gY=gYg^{-1},\qquad \operatorname{ad}_X Y=[X,Y],\qquad \operatorname{Ad}_{\exp X}=e^{\operatorname{ad}_X}

The conjugation formula is the matrix realization. The remaining identities hold for finite-dimensional Lie groups with the stated bracket convention, acting on the Lie algebra.↔ Scroll wide equations horizontally.

The dual action uses the inverse

Let xi be a linear functional on the Lie algebra. Define the coadjoint action of g by evaluating xi on Ad-g-inverse of the test vector. In coordinates this is the inverse transpose of the Ad matrix. The inverse makes it a left action, rather than reversing multiplication order through a bare dual map. It also ensures that transforming a generator and its covector together leaves their pairing unchanged. This natural pairing does not require choosing an inner product.

Differentiating the inverse introduces a minus sign: the infinitesimal coadjoint action of X on xi evaluates on Y as minus xi of [X,Y]. Some texts arrange the notation for dual maps differently, so a symbol ad-star is insufficient by itself to identify the sign. Writing the evaluation formula removes the ambiguity. Adjoint and coadjoint orbits can then be studied as genuine group orbits, with stabilizers and tangent directions derived from these definitions.

\langle\operatorname{Ad}_g^*\xi,Y\rangle=\langle\xi,\operatorname{Ad}_{g^{-1}}Y\rangle,\qquad \langle\operatorname{ad}_X^*\xi,Y\rangle=-\langle\xi,[X,Y]\rangle

Here Ad* denotes the coadjoint left action, explicitly the inverse dual. Consequently ⟨Ad*g ξ,Adg Y⟩=⟨ξ,Y⟩. The infinitesimal star is minus the transpose of ad in dual bases.↔ Scroll wide equations horizontally.

Coordinates are a useful sign audit

The positive affine group has generators H and E satisfying [H,E]=E. Its adjoint action preserves the translation line but can shear the scale generator into that line. Computing the two transformed basis vectors gives a small matrix whose inverse transpose can be checked independently. This example is useful because an incorrect sign or a missing inverse immediately changes a simple numeric pairing, rather than remaining hidden inside abstract notation.

An inner product can identify the algebra with its dual, but that identification need not intertwine Ad and the coadjoint action. For compact groups an Ad-invariant positive inner product can be built by averaging, and then the identification behaves especially well. For the noncompact affine group no such positive invariant form should be assumed. Keep a vector's coordinates and a covector's dual coordinates distinct, even when both are written as columns for matrix multiplication.

Work it through

Check a finite coadjoint map by pairing

For g(a,b)=[[a,b],[0,1]], a>0, use the affine basis H,E and a covector ξ with ξ(H)=p, ξ(E)=q. Find the coadjoint action and evaluate a=2,b=3,p=1,q=4.

Direct conjugation gives Ad-g(H)=H−bE and Ad-g(E)=aE. In the ordered basis H,E, its matrix is $M=\begin{pmatrix}1&0\\-b&a\end{pmatrix}$ . The inverse transpose is $M^{-\mathsf T}=\begin{pmatrix}1&b/a\\0&1/a\end{pmatrix}$ .
Thus the transformed dual coordinates are $(p+bq/a,q/a)$ . At the stated values they become (7,2). The transformation is not multiplication by M transpose; the inverse is responsible for both the scaling and the sign.
Take Y=2H−E. Its transformed coordinates are (2,−8). The new pairing is 7·2+2·(−8)=−2, while the original pairing is 1·2+4·(−1)=−2. Equality verifies the finite coadjoint convention directly.

The result

The inverse-transpose formula preserves the natural pairing and fixes the coadjoint action's signs without relying on a chosen metric.

The living notebook

Explore the model

Interactive lab

Interactive lab · matrix

Inspect an invertible map before forming its dual inverse

Pause, Reset, then set a=2 and b=1 for T=[[2,1],[0,1]]. Predict determinant two. Repeat from Reset with both a=0.5 and b=−0.5 for T⁻¹. The instrument displays these maps on ordinary vectors. A covector instead transforms by the inverse transpose, which must be calculated separately; neither selected triangular map is being claimed to be the affine adjoint matrix above.

Interpolated geometryOriginal unit squareSelected final transform

Area scale · det A: 1.4
Singular values · σmax, σmin: 1.597, 0.876
Condition number · κ₂(A): 1.823

Diagonal scale a1.4Shear b0.6

t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forInversion and dualization are separate operations; a geometric vector display cannot silently substitute for a coadjoint calculation.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Differentiating Ad-exp(−tX) in the dual evaluation introduces the minus sign. That sign is tied to the inverse in the defined coadjoint left action.

Your turn

Try it yourself

Using the affine formula, transform ξ=(2,6) by g(3,−3). Also compute ad*ₕξ and ad*ₑξ in the dual H,E coordinates at the original ξ.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

For the finite action use (p+bq/a,q/a). For the infinitesimal actions evaluate the brackets with H and E separately.

Compare with the solution

The finite result is (2−6,2)=(−4,2). Since [H,H]=0 and [H,E]=E, the infinitesimal coadjoint H-action is (0,−q)=(0,−6). Since [E,H]=−E and [E,E]=0, the E-action is (q,0)=(6,0). These signs agree with differentiating the finite formula along $\exp(tH)=(e^t,0)$ and $\exp(tE)=(1,t)$ .

Keep the idea

Take it with you

Ad differentiates conjugation, and ad differentiates Ad into the Lie bracket.
The coadjoint left action is the inverse dual, preserving simultaneous pairings.
Writing evaluation formulas prevents ambiguity in starred notation and infinitesimal signs.

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