Tangent Lie algebras and left-invariant fields
Identify tangent vectors with invariant vector fields and fix the bracket sign through concrete matrix calculations.
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The question for today
What you’ll understand
- Transport an identity tangent vector by left translation.
- Recover the matrix commutator from left-invariant vector fields.
- Distinguish left- and right-invariant bracket conventions.
A Lie group has a tangent space at every point, but translation makes the tangent space at the identity especially informative. A vector there can be transported consistently to every other point, producing an invariant vector field. The commutator of these fields supplies a bilinear operation that records the first noncommutative part of group multiplication. It is not ordinary matrix multiplication, and its sign depends on whether left or right invariance is used. We fix left invariance throughout and check the convention in an explicit affine group.
Use translation to identify tangent spaces
For g in G, left translation sends h to gh and is a diffeomorphism. Its derivative transports a vector X at the identity to a vector at g. Defining the field this way makes it left invariant, and every left-invariant field arises from exactly one identity tangent vector. Thus the finite-dimensional vector space g, the tangent space at the identity, encodes a distinguished family of fields on the entire group rather than a single tangent approximation at one isolated point.
For a matrix group the transported field is gX. Differentiating defining equations identifies which matrices X are tangent. Orthogonality gives Xᵀ+X=0, so the orthogonal Lie algebra consists of skew-symmetric matrices. Determinant one gives trace X=0. These are linear conditions, even though the group constraints are nonlinear. They describe the identity component's infinitesimal directions and do not record how many other components the group has.
The bracket is the commutator of derivations
Vector fields act on smooth functions as derivations. Their bracket is the difference of the two successive derivative operations, and it is again a vector field. The bracket of left-invariant fields remains left invariant, so evaluating it at the identity defines a bracket on g. In a matrix group, differentiating the fields gX and gY gives XY−YX. This operation is antisymmetric and obeys the Jacobi identity, making the tangent space a Lie algebra.
Right translation instead gives fields Xg. With the same derivation convention, the bracket of these right-invariant fields corresponds to minus the matrix commutator. Both descriptions are valid, but mixing them silently changes signs in adjoint formulas and evolution equations. The distinction is also visible in coordinates. It is useful to derive one nonabelian example by acting on coordinate functions rather than trust an abstract name to specify the convention.
Small noncommutative motions are still local data
If the bracket vanishes, two identity generators commute infinitesimally, and their one-parameter subgroups commute. If every bracket vanishes, the connected Lie group is abelian, but a disconnected group can have additional component behavior not seen by that conclusion. The bracket therefore controls a strong local structure while requiring connectedness for some global deductions. A one-dimensional Lie algebra is necessarily abelian because antisymmetry leaves no independent pair with a nonzero bracket.
A Lie algebra basis should be accompanied by its bracket table and the convention used to identify fields. Those constants let one reconstruct brackets of arbitrary linear combinations and check Jacobi numerically or symbolically. A picture of rotating arrows alone does not establish the bracket table. Conversely, a few verified matrix commutators can make the abstract definition tangible and reveal a sign error before it propagates into exponentials, adjoint actions, or symmetry reductions.
Work it through
Compute the affine bracket in two descriptions
For matrices g(a,b)=[[a,b],[0,1]] with a>0, use H=[[1,0],[0,0]] and E=[[0,1],[0,0]]. Calculate their bracket and their left-invariant coordinate fields.
Matrix multiplication gives HE=E and EH=0, hence [H,E]=E. The tangent algebra consists of matrices [[s,t],[0,0]], so H and E form a basis.
Multiplying gH gives [[a,0],[0,0]], corresponding to the coordinate field a∂ₐ. Multiplying gE gives [[0,a],[0,0]], corresponding to a∂ᵦ. Their vector-field bracket is a∂ᵦ because a∂ₐ differentiates the coefficient a in the second field.
For comparison the right fields are Hᴿ=a∂ₐ+b∂ᵦ and Eᴿ=∂ᵦ. Their bracket is −∂ᵦ, confirming the opposite sign for the same matrix generators under right invariance. The coordinate and matrix calculations therefore agree with the chosen convention.
The affine group is already nonabelian in two dimensions: its scale generator changes the translation generator, giving [H,E]=E.
The living notebook
Explore the model
Interactive lab · fields
Recognize a rotation generator as a field
Pause, Reset, and set source strength q=0 and rotation Ω=1. Predict counterclockwise tangential arrows for the field (−y,x). Repeat from Reset with both q=0 and Ω=−1 to reverse them. This plane field illustrates the action of one rotation generator; its schematic display is not a computation of a Lie bracket or a proof about a general group.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. 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 forA generator can act as a spatial vector field, while its bracket with another generator still requires a separate derivation.
Pause & reason
Check your understanding
Your turn
Try it yourself
In sl₂ take H=diag(1,−1), E=E₁₂, and F=E₂₁. Compute [H,E], [H,F], and [E,F].
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Multiplying a diagonal matrix on the left scales rows; multiplying it on the right scales columns.
Compare with the solution
One obtains HE=E and EH=−E, so [H,E]=2E. Similarly HF=−F and FH=F, giving [H,F]=−2F. Finally EF=diag(1,0) and FE=diag(0,1), so [E,F]=H. All three results remain traceless, as required for closure of the Lie algebra. These are brackets of the corresponding left-invariant fields under the fixed convention.
Keep the idea
Take it with you
- Left translation identifies identity tangent vectors with invariant fields.
- The matrix Lie bracket is XY−YX rather than matrix multiplication itself.
- Changing left to right invariance changes the bracket sign under the same identification.
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