Flows and Lie Derivatives — Mathematics Knowledge System

Overview

Every smooth vector field X on a manifold M generates a one-parameter group of diffeomorphisms — its flow φ_t — by solving the ODE φ̇(t) = X(φ(t)). The Lie derivative L_X measures the rate of change of any tensor field T along this flow: L_X T = d/dt|_{t=0} (φ_t)* T. For functions, L_X f = Xf. For vector fields, L_X Y = [X,Y] (the Lie bracket). This gives a coordinate-free notion of differentiation not requiring any connection.

Intuition

Think of X as a 'wind' that pushes every point of the manifold. After time t, each point has been moved to φ_t(p). The Lie derivative of a tensor T along X asks: how does T change when you drag it along with the wind? Unlike the covariant derivative, the Lie derivative uses only the smooth structure (no metric, no connection needed), making it the most intrinsic notion of differentiation on a manifold.

Formal Definition

Definition

Let X be a smooth vector field on M with flow φ: ℝ × M → M (or a local flow). For a tensor field T of type (r,s), the Lie derivative is:

\frac{d}{dt}\phi_t(p)\Big|_{t=0} = X(p), \quad \phi_0 = \mathrm{id}_M

Flow equation for X

\mathcal{L}_X T = \lim_{t\to 0} \frac{(\phi_{-t})_* T - T}{t} = \frac{d}{dt}\Big|_{t=0} (\phi_t)^* T

Lie derivative of a tensor field T

\mathcal{L}_X f = X(f) = df(X)

Lie derivative of a function

\mathcal{L}_X Y = [X, Y]

Lie derivative of a vector field Y equals the Lie bracket

\mathcal{L}_X \omega = i_X d\omega + d(i_X \omega) = (di_X + i_X d)\omega

Cartan's magic formula (for differential forms)

Theorems

Theorem 1: Lie Derivative as Lie Bracket

\text{For vector fields } X, Y \text{ on } M: \quad \mathcal{L}_X Y = [X,Y] = XY - YX \text{ (as derivations on } C^\infty(M)\text{).}

Theorem 2: Cartan's Formula

\text{For a differential } k\text{-form } \omega: \quad \mathcal{L}_X \omega = i_X d\omega + d(i_X\omega),

Theorem 3: Flow Commutativity

\text{Vector fields } X \text{ and } Y \text{ have commuting flows } (\phi^X_s \circ \phi^Y_t = \phi^Y_t \circ \phi^X_s) \text{ if and only if } [X,Y] = 0.

Worked Examples

1
The flow of X = ∂_x satisfies ∂φ/∂t = X(φ) = (1, 0). With initial condition φ₀(x,y) = (x,y), we get φ_t(x,y) = (x+t, y).
$\phi_t(x,y) = (x+t,\, y)$
2
Compute the Lie bracket [X,Y] = [∂_x, x∂_y]. Apply to a function f: [∂_x, x∂_y]f = ∂_x(x ∂_y f) − x ∂_y(∂_x f) = ∂_y f + x ∂_x∂_y f − x ∂_y ∂_x f = ∂_y f.
$[\partial_x, x\partial_y] = \partial_y f + x\,\partial_{xy}f - x\,\partial_{yx}f = \partial_y f$
3
So L_X Y = [X,Y] = ∂/∂y.
$\mathcal{L}_X Y = \frac{\partial}{\partial y}$
4
Verify via pullback: (φ_t)* Y = (φ_{-t})_* Y. Y at the point φ_t(p) = (x+t, y) is (x+t)∂_y. Pulling back: (φ_{-t})_* [(x+t)∂_y] = (x+t)∂_y. So d/dt|_{t=0} (φ_t)*Y = d/dt|_{t=0} (x+t)∂_y = ∂_y. Consistent.
$\frac{d}{dt}\Big|_{t=0}(x+t)\partial_y = \partial_y$

✓ Answer

The flow of ∂_x is translation: φ_t(x,y) = (x+t, y). The Lie derivative L_{∂_x}(x∂_y) = [∂_x, x∂_y] = ∂_y.

Practice Problems

Mediumfree response

Find the flow of the vector field X = x ∂/∂x + y ∂/∂y on ℝ² and identify what transformation it generates.

Hardproof writing

Prove that L_X[Y,Z] = [L_X Y, Z] + [Y, L_X Z] (the Lie derivative is a derivation for the Lie bracket).

Quiz

Summary

The flow φ_t of a vector field X is the one-parameter family of diffeomorphisms satisfying d/dt φ_t(p) = X(φ_t(p)), φ_0 = id.
The Lie derivative L_X T = d/dt|_{t=0} (φ_t)* T measures how T changes along the flow of X.
For functions: L_X f = Xf. For vector fields: L_X Y = [X,Y]. For k-forms: L_X ω = i_X dω + d(i_X ω) (Cartan's formula).
The Lie bracket [X,Y] = 0 iff the flows of X and Y commute.
The Lie derivative requires no metric or connection — it is a purely smooth-manifold notion.