Lie dragging preserves commutation proof.

In summary: Lie derivative is zero, then it must also be zero. Therefore, ##[V_{(a)},V_{(b)}]=0## on M, as desired. In summary, to prove that ##[V_{(a)},V_{(b)}]=0## on M, we can use the bilinearity of the Lie bracket and the fact that the Lie derivative of each vector field is zero. This allows us to conclude that the Lie derivative of the bracket is also zero, and therefore the bracket itself must be zero on M.
  • #1
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Hi! A proof of Frobenius' theorem (in Schutz Geometrical Methods) uses the fact that if a set of vector fields ##{V_{(a)}}## commute on some submanifold S of an ambient manifold M, and one use an additional vector field ##Z## to Lie transport/drag the set ##{V_{(a)}}## around the manifold M, then the set ##{V_{(a)}}## will also commute away from S.
I want to prove this. This is where I stand: Since ##\mathcal{L}_Z V_{(a)} = 0## per definition of lie transportation we have

$$\mathcal{L}_Z [V_{(a)},V_{(b)}] = [ \mathcal{L}_Z V_{(a)},V_{(b)}] + [ V_{(a)}, \mathcal{L}_Z V_{(b)}] =0$$

But does ##\mathcal{L}_Z [V_{(a)},V_{(b)}]= 0## on M somehow imply that ##[V_{(a)},V_{(b)}]=0## on M? I'm not sure how to conclude this proof.
 
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Hello! That's a great question. To prove that ##[V_{(a)},V_{(b)}]=0## on M, we can use the fact that the Lie bracket is a bilinear operation. This means that it satisfies the following properties:

1) ##[aV_{(1)} + bV_{(2)},V_{(3)}] = a[V_{(1)},V_{(3)}] + b[V_{(2)},V_{(3)}]## for any real numbers a and b.

2) ##[V_{(1)},aV_{(2)} + bV_{(3)}] = a[V_{(1)},V_{(2)}] + b[V_{(1)},V_{(3)}]## for any real numbers a and b.

Using these properties, we can rewrite the equation as:

$$\mathcal{L}_Z [V_{(a)},V_{(b)}] = [ \mathcal{L}_Z V_{(a)},V_{(b)}] + [ V_{(a)}, \mathcal{L}_Z V_{(b)}] = a[\mathcal{L}_Z V_{(1)},V_{(b)}] + b[\mathcal{L}_Z V_{(2)},V_{(b)}] + a[V_{(1)},\mathcal{L}_Z V_{(b)}] + b[V_{(2)},\mathcal{L}_Z V_{(b)}]$$

Since we know that ##\mathcal{L}_Z V_{(a)} = 0## and ##\mathcal{L}_Z V_{(b)} = 0##, we can substitute these values into the equation and get:

$$\mathcal{L}_Z [V_{(a)},V_{(b)}] = a[0,V_{(b)}] + b[0,V_{(b)}] + a[V_{(1)},0] + b[V_{(2)},0] = 0$$

This means that ##\mathcal{L}_Z [V_{(a)},V_{(b)}] = 0## for any real numbers a and b. But we also know that ##[V_{(a)},V_{(b)}]## is a vector field, so if its
 

1. What is "lie dragging" in relation to preserving commutation?

"Lie dragging" refers to the physical phenomenon in which the rotation of a massive object causes the spacetime around it to be dragged along in the same direction. In the context of preserving commutation, it means that the rotation of an object does not affect the order in which two operations are performed.

2. How does lie dragging preserve commutation?

Lie dragging preserves commutation by ensuring that the order of operations remains the same regardless of the rotation of an object. This is because the rotation causes the spacetime around it to be dragged along, so the relative positions of objects and the order of operations remain unchanged.

3. Can you provide an example of lie dragging preserving commutation?

A classic example of lie dragging preserving commutation is the Foucault pendulum experiment. The rotation of the Earth causes the pendulum to appear to rotate, but the order in which the pendulum swings back and forth remains unchanged.

4. What is the significance of lie dragging preserving commutation?

Lie dragging preserving commutation is significant because it is a fundamental aspect of general relativity. It helps explain how the rotation of massive objects affects the spacetime around them and how this can have an impact on the order of operations.

5. Are there any practical applications of lie dragging preserving commutation?

While there are no direct practical applications of lie dragging preserving commutation, it has important implications for our understanding of the universe and can be used in calculations and predictions related to general relativity.

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