Lie bracket

Definition

Si $X$ e $Y$ son dos campos de vectores en una variedad $M$ definimos un nuevo campo de vectores $[X,Y]$ mediante la expresión

Hay que demostrar que es un campo.

Visualization

Lie bracket of two vector fields, $[u,v]$, measures how is the gap when we try to draw a rectangle along flow lines of $u$ and $v$. Let' see:

Let's go to a local chart $U$ with coordinates $x_i$. If $u=u^i\partial x_i$ and $v=v^i\partial x_i$ we have that

On the other hand, consider a point $P$, in the local chart. Let's call $\phi$ to the flow of $v$ and $\varphi$ to the flow of $u$. If we move a little amount $ϵ$ from $P$ following $v$ we arrive to a point $R$ that can be approximated by:

Let's call $R^{\prime }=P+ϵv_p$. Now we can move along the flow of $u$. If we moved from $R$ we would arrive to, say, $S$ (the perfect final position), but if we begin in $R^{\prime }$ we will arrive to $S^{\prime }$. But since we are approximating, we actually arrive to a certain $S^{″}$ in the following way:

where we take $S^{″}=R^{\prime }+ϵu_{R^{\prime }}$

But observe that

If we begin our "whole approximated journey" from $P$ but beginning with the flow of $u$ instead, we would arrive to a certain $T^{″}$, such that:

So the vector $T^{″}-S^{″}$ (remember we are in a local chart) is

So $[u,v]$ measures the failure to close a parallelogram, in some sense.

Here is a picture from TRTR, by Penrose.

Related: commutativity of flows

Idea

(I think is the same as above, review and delete if needed)

To see the idea of Lie bracket, we are going to consider a local chart around a point $P$, and to use the Taylor expansion theorem in two ways: for a curve in ${\mathbb{R}}^n$ and for a map ${\mathbb{R}}^n\to {\mathbb{R}}^n$:

On the other hand, see TFG Adrián Ruíz for another approach of the Lie bracket as generating a zero velocity curve (lema 4.2. y lema 4.3.)

Useful formula

Also we have the following formula for the Lie bracket (Wikipedia, Lie bracket of vector fields):

where we are denoting the flows of the vector fields by ${\mathrm{\Phi }}_t^X$.

Proof

TFG Adrián Ruíz, teorema 4.3.

Coordinates expression for the Lie bracket

If $u=u^i\partial x_i$ and $v=v^i\partial x_i$ we have that

Other remarks

• It can be shown that is the same as Lie derivative of vector fields.

• It is bilinear, and also satisfies

• Alternativity: $[x,x]=0$

• Jacobi identity: $[x,[y,z]]=[[x,y],z]+[y,[x,z]$

• If we abstract the properties of the Lie bracket we arrive to the notion of Lie algebras.

• See also relation of Lie derivative, covariant derivative and torsion.

• Lie bracket has to do with Clairaut's theorem: $[{\partial }_{x_i},{\partial }_{x_j}]=0$.

• It implies the commutativity of flows.

________________________________________

________________________________________

________________________________________

Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es