Vector Fields and Lie Algebra

Vector Fields on Manifolds

Given a differentiable manifold , a vector field on is a section of the tangent bundle. Explicitly, is an assignment that sends each point to a tangent vector . The space of all vector fields is denoted by . or .

e.g. .

Vector Fields are Derivations

One can see from the derivation definition of tangent vectors that a vector field can be identified as a derivation that satisfies the equation:

The space of vector fields has a rich structure:

Proposition

The space of vector fields is a -vector space.

Proof It suffices to check that the scaler multiplication is valid. Indeed, where the dot denotes the scalar multiplication in . It is clear that this satisfies distributivity and the associativity.

Proposition

The space of vector fields is a module over the ring .

Proof We can define the the scaler multiplication of a vector field by a smooth function as follows: where the dot denotes the scalar multiplication in . Clearly this gives a smooth vector field, and thus makes a module over .

Pullback Vector Fields

Pullback Vector Field

Suppose is a local diffeomorphism. The pullback of a vector field by is the pullback section of the tangent bundle. That is

Proof

Vector Fields on Lie Groups

Left-Invariant Vector Field

If is a Lie group and is a vector in , where is the identity element in . Suppose is the left multiplication map that sends to . Then we can use the maps to define a natural vector field on , for each . We set

The resulting vector field is called a left-invariant vector field.

e.g.

• Let Lie group , with the group operation of complex number multiplication. Fix some , the left-invariant vector field is given by

Lie Bracket of Vector Fields

Lie Bracket of Vector Fields

Let . Then the Lie bracket of and is the vector field defined (as a derivation) by
The Lie bracket is thus the commutator of and as operators on smooth functions.

Properties of Lie Bracket

The following hold for the Lie bracket of vector fields:

1. Bilinearity: .
2. Skew-symmetry: .
3. Jacobi Identity: .

That is, forms a real Lie algebra.

Proof Bilinearity and skew-symmetry follow from the definition directly. We now check Jacobi identity. For all , there holds

Corollary

Let be a smooth manifold and let be an immersed submanifold with or without boundary in . If and are smooth vector fields on that are tangent to , then is also tangent to .

Lie Algebra of Lie Groups

Every Lie group has a canonical Lie algebra , which has two natural representations, either as the tangent space at the identity element of , or as the space of left-invariant vector fields on . We will see that the two representations are isomorphic.

Proposition

Let be a Lie group, and suppose and are smooth left-invariant vector fields on . Then is also left-invariant.

Proposition

The left-invariant vector fields on a Lie group form a Lie algebra.

Proof It suffices to check

Theorem

The set of all left-invariant vector fields on a Lie group is isomorphic to the tangent space at the identity element of , via evaluation at . In other words, every tangent vector extends uniquely to a left-invariant vector field.

Proof

Lie Algebra of a Lie Group

The Lie algebra of a Lie group , often denoted , is up to isomorphic, the Lie algebra of left-invariant vector fields on , or the tangent space at the identity element of .

Remark

Elements of the Lie algebra represent “infinitesimal transformations” or “directions of motion” within the group starting from the identity.

Exponential Map

For any Lie group with Lie algebra , the exponential map is defined by where is the unique integral curve of the left-invariant vector field such that , or equivalently, is the one-parameter subgroup generated by .

e.g. The Lie algebra of , denoted , is isomorphic to . The exponential map of satisfies

Proposition

Let be a Lie group with associated Lie algebra . For any , is the one-parameter subgroup of generated by . That is, for all .

Proof

Fundamental Vector Field

Suppose is a Lie group with Lie algebra , acting on a smooth manifold with the action that . The fundamental vector field on generated by the infinitesimal action of , denoted by (or ), is defined at each point as

e.g. Since the -torus , its Lie algebra is isomorphic to . From the previous example of , we can deduce that the exponential map of an -torus takes the form Consider a torus acting on by for nonzero constants . . Then for any and , there holds