Why is the Lie Bracket the same as the Cross Product for a 2 Sphere in R3?

In summary, the Lie bracket of any two vector fields on a manifold is itself a vector field on that manifold. This is demonstrated in the YouTube video and the stackexchange link, although the examples used are different (considering vector fields on ##S^2## and ##\mathbb{R}^3##, respectively). The cross product is only relevant in the context of vector fields on ##\mathbb{R}^3##, not on ##S^2##.
  • #1
nigelscott
135
4
Homework Statement
Prove that for a 2 sphere in R R[SUP]3[/SUP] the Lie bracket is the same as to cross product.
Relevant Equations
Vector: X = (y,-x,0); Y = (0,z-y)

[X,Y] = J[SUB]Y[/SUB]X - J[SUB]X[/SUB]Y where the J's are the Jacobean matrices.
Prove that for a 2 sphere in R3 the Lie bracket is the same as the cross product using the vector: X = (y,-x,0); Y = (0,z-y)

[X,Y] = JYX - JXY where the J's are the Jacobean matrices.

I computed JYX - JXY to get (-z,0,x). I computed (y,-x,0) ^ (0,z,-y) and obtained (xy,y2,yz) = (z,0,x) using Wolfram but I can't figure out why the sign of the x-component is different. Any help would be appreciated.
 
Physics news on Phys.org
  • #2
It is a bit confused what you have written there. Jacobi matrix of what? Which transformations of the 2-sphere do you consider?

One way to solve the exercise is:
  1. Show that ##\{\,(1,0,0),(0,1,0),(0,0,1)\,\}## with the cross product defines a Lie algebra.
  2. There are up to isomorphism only three three dimensional Lie algebras: the Abelian, the Heisenberg algebra and a simple one. As they all have a different product space, the commutator ideal, they can easily be distinguished, and the cross product can only be the simple one.
  3. Now which version aka basis of the simple three dimensional Lie algebra you want to compare the cross product with is a matter of taste. You can choose between ##\mathfrak{sl}(2,\mathbb{R}), \mathfrak{so}(3,\mathbb{R}), \mathfrak{su}_{\mathbb{R} } (2,\mathbb{C})##.
Hence the answer to your question is: It depends on which isomorphism. i.e. basis you choose.
 
Last edited:
  • #3
Ok. Thanks for your response. The example I am using is from this video here starting at 12 mins and continuing here. Here he talks about tangents to the sphere with the Lie bracket being another tangent to the sphere which is at odds with the cross product which would produce a vector normal to the tangent plane. When I go through the same procedure using the example in https://math.stackexchange.com/questions/1326501/question-about-lie-bracket-and-cross-product I get a consistent result. I wonder if I am confusing vectors with vector fields?
 
  • #4
Any normal vector on a sphere is a tangent vector at a different spot. I'm not sure which Lie group is associated with the cross product, I assume ##SO(3)## fits best. But the sphere itself is two dimensional and corresponds to the quotient group ##SO(3)/SO(2)##. Maybe the ##3-##sphere ##SO(4)/SO(3)## works one-to-one.
 
  • #5
nigelscott said:
Ok. Thanks for your response. The example I am using is from this video here starting at 12 mins and continuing here. Here he talks about tangents to the sphere with the Lie bracket being another tangent to the sphere which is at odds with the cross product which would produce a vector normal to the tangent plane. When I go through the same procedure using the example in https://math.stackexchange.com/questions/1326501/question-about-lie-bracket-and-cross-product I get a consistent result. I wonder if I am confusing vectors with vector fields?
The YouTube video and the the stackexchange link consider vector fields on different manifolds.

In order to give an illustrative example of the general property that the Lie bracket of any two vector fields on manifold ##M## is itself a vector field on manifold ##M##, the lecturer in the YouTube video considers two specific vector fields, ##X## and ##Y##, on the specific manifold ##S^2##. The lecturer explicitly calculates ##\left[ X , Y \right]##, and finds that, as expected, ##\left[ X , Y \right]## is a vector field on ##S^2##. As far as I can see, this has nothing to do with the cross product of two vectors.

At the stackexchange link, vector fields on the manifold ##\mathbb{R}^3## are considered. The set of all smooth vector fields forms an infinite-dimensional vector space. The stackexchange link considers a specific 3-dimensional subspace, and relates this 3-dimensional subspace of vector fields on ##\mathbb{R}^3## (not ##S^2##) to cross products.
 
  • Like
Likes WWGD
  • #6
OK. Thanks to you both. I think I understand it now.
 

Related to Why is the Lie Bracket the same as the Cross Product for a 2 Sphere in R3?

What is a Lie Bracket?

A Lie Bracket, also known as a commutator, is a mathematical operation used to determine the degree of non-commutativity between two elements in a mathematical structure, such as a Lie algebra. It is denoted by [x, y] and is defined as the difference between the product of x and y and the product of y and x.

What is the significance of the Lie Bracket in mathematics?

The Lie Bracket is significant in mathematics as it measures the extent to which a mathematical structure is non-commutative. It is also used to define the Lie algebra, a fundamental concept in the field of algebraic structures.

What is a Cross-Product?

A Cross-Product, also known as a vector product, is a mathematical operation that produces a vector that is perpendicular to two given vectors in three-dimensional space. It is denoted by x and is defined as the product of the magnitudes of the two vectors and the sine of the angle between them.

What are some applications of the Lie Bracket and Cross-Product?

The Lie Bracket and Cross-Product have various applications in mathematics, physics, and engineering. They are used in differential geometry, quantum mechanics, and computer graphics, to name a few. In physics, they are used to determine the moment of a force and in engineering, they are used in mechanics and electromagnetism.

What is the difference between the Lie Bracket and Cross-Product?

The Lie Bracket and Cross-Product are two different mathematical operations that serve different purposes. The Lie Bracket measures the non-commutativity of elements in a mathematical structure, while the Cross-Product determines a vector perpendicular to two given vectors. They also have different notations and definitions.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
19
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
696
  • Calculus and Beyond Homework Help
Replies
5
Views
452
  • Linear and Abstract Algebra
Replies
9
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
906
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
916
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top