Vector Fields, Flows, and Lie Brackets

A smooth vector field assigns a tangent vector to every point, and its integral curves organize those infinitesimal directions into local motions. The Lie bracket $[X,Y]$ measures the leading failure of the flows generated by $X$ and $Y$ to commute. It is therefore not merely a component formula: it is the observable displacement left by performing two infinitesimal motions in opposite orders.

Required background. Calculus and Linear Algebra for Relativity — use ordinary differential equations, directional derivatives, the chain rule, and commutators of linear operators.

Vector fields and local flows

A vector field is a smooth section $X:M\to TM$, written in a chart as $X=X^i(x)\partial_i$. An integral curve through $p$ solves

$$\dot\gamma_p(t)=X_{\gamma_p(t)},\qquad \gamma_p(0)=p.$$

Local existence and uniqueness for ordinary differential equations produce a local flow $\Phi_t^X(p)=\gamma_p(t)$, defined for sufficiently small $t$ depending on $p$. Wherever both sides exist,

$$\Phi_0^X=\operatorname{id}, \qquad \Phi_{t+s}^X=\Phi_t^X\circ\Phi_s^X.$$

A field is complete if its flow exists for every real $t$. Completeness is global and must not be inferred from a smooth component formula. On $\mathbb R$, $X=x^2\partial_x$ has solution $x(t)=x_0/(1-tx_0)$, which reaches infinity at finite $t=1/x_0$ for $x_0>0$. The local-flow construction and this distinction are treated in Lee 2013, Chs. 8–9, pp. 174–248.

The Lie bracket as a derivation

Vector fields act on smooth functions as derivations. Their operator commutator is again a derivation:

$$[X,Y]f=X(Yf)-Y(Xf).$$

Consequently $[X,Y]$ is a vector field. In coordinates,

$$[X,Y] = \left(X^j\partial_jY^i-Y^j\partial_jX^i\right)\partial_i.$$

Although partial derivatives appear, their non-tensorial second-derivative terms cancel. The bracket is antisymmetric, bilinear over constants, obeys the Jacobi identity, and satisfies

$$[X,fY]=f[X,Y]+X(f)Y.$$

The final identity also shows that the bracket is not $C^\infty(M)$-linear in either argument, so it is not a tensor. It differentiates the fields whose relative variation it measures. A coordinate-independent development from derivations and flows appears in Wendl 2021–2022, §§6.2–6.4, pp. 50–54, PDF.

Noncommuting flows in the plane

Take $M=\mathbb R^2$ with

$$X=\partial_x, \qquad Y=x\partial_y.$$

Their flows are

$$\Phi_t^X(x,y)=(x+t,y), \qquad \Phi_s^Y(x,y)=(x,y+sx).$$

Performing the $X$ motion and then the $Y$ motion gives

$$\Phi_s^Y\!\circ\Phi_t^X(x,y) =(x+t,y+s(x+t)),$$

whereas reversing the order gives $\Phi_t^X\!\circ\Phi_s^Y(x,y)=(x+t,y+sx)$. Their difference is $st\,\partial_y$ to leading order—in this example, exactly. Direct calculation agrees:

$$[X,Y]=[\partial_x,x\partial_y]=\partial_y.$$

This is the geometric experiment encoded by the bracket. The discrepancy is second order in the two travel times because either motion alone returns no comparison loop. The sign depends on which ordered composition is subtracted, so a commutator-loop formula must always state its ordering.

If $[X,Y]=0$, the local flows commute wherever the compositions are defined. Conversely, commuting local flows have zero bracket. This equivalence is local; it says neither field is complete and does not guarantee a global action of $\mathbb R^2$.

Naturality and limits of the construction

For a diffeomorphism $F:M\to N$, the pushed-forward fields obey

$$F_*[X,Y]=[F_*X,F_*Y].$$

Thus the bracket does not depend on a chart. It also requires no metric or connection. Later, a torsion-free connection satisfies $[X,Y]=\nabla_XY-\nabla_YX$, but that formula derives the bracket from extra structure rather than defining it.

Two cautions are useful. First, $[X,Y]_p$ depends on how the fields vary near $p$, not only on the pair $X_p,Y_p$. Second, a zero bracket means local commutation, not that the vectors are linearly independent or generate coordinates everywhere. Either field may vanish, and orbit dimension can change. These limitations become important when flows are used to describe symmetries and integrable distributions.

A pointwise counterexample

The bracket cannot be reconstructed from two tangent vectors at one point. On the real plane, compare the fields $Y_1=x\partial_y$ and $Y_2=0$ at a point on the vertical axis. They have the same value there. With $X=\partial_x$, however, the bracket with the first field equals $\partial_y$, while the bracket with the second vanishes. The derivative of the field in a neighborhood supplies information absent from its pointwise value.

This example also diagnoses why the bracket is not a torsion tensor in disguise. A connection can combine two non-tensorial covariant derivatives so that their antisymmetric difference differs from the bracket by torsion, but the bracket itself is defined before that choice. In a coordinate basis the basis fields commute; in a general frame they need not. Nonzero frame commutators then record anholonomy rather than spacetime curvature.

For computation, one should check three descriptions against each other when possible: the derivation commutator acting on a test function, the component formula, and the endpoint difference of ordered flows. Agreement among them catches the two common mistakes—dropping derivatives of coefficients and reversing the commutator-loop sign.

The Jacobi identity has a geometric interpretation as consistency among three nested infinitesimal motions. Algebraically, it follows because vector fields act as linear operators on functions and operator commutators satisfy Jacobi. This offers another independent check: a proposed bracket formula that fails Jacobi cannot describe the commutator of flows. The identity later makes vector fields into a Lie algebra and controls the structure coefficients of noncoordinate frames.

Exercises

For $X=\partial_x$ and $Y=x^2\partial_y$ on $\mathbb R^2$, compute $[X,Y]$. Use the exact flows to compare an $X$ step of duration $t$ followed by a $Y$ step of duration $s$ with the reverse ordering. The bracket vanishes on $x=0$; does that make the flows commute on a neighborhood of that line?

Solution

The component formula gives

$$[X,Y]=\partial_x(x^2)\partial_y=2x\partial_y.$$

The exact compositions are

$$\Phi_s^Y\circ\Phi_t^X(x,y)=(x+t,y+s(x+t)^2),$$

and

$$\Phi_t^X\circ\Phi_s^Y(x,y)=(x+t,y+sx^2).$$

The first endpoint exceeds the second by $(2xst+st^2)\partial_y$. Dividing by $st$ and taking $s,t\to0$ gives $2x\partial_y$, as required. At $x=0$, the leading second-order term vanishes but the finite-step discrepancy $st^2\partial_y$ does not. Commuting flows require the bracket to vanish throughout an open region, not merely at the starting point.

References

• Lee, John M. Introduction to Smooth Manifolds. 2nd ed., Graduate Texts in Mathematics 218, Springer, 2013, Chs. 8–9, pp. 174–248.
• Wendl, Chris. Differential Geometry I and II. Humboldt-Universität zu Berlin lecture notes, 2021–2022, §§6.2–6.4, pp. 50–54. Course PDF.