Lie Bracket and Cross-Product

[nigelscott]

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] = J_YX - J_XY where the J's are the Jacobean matrices.

I computed J_YX - J_XY 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.

 

[fresh_42]

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.

 

[nigelscott]

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?

 

[fresh_42]

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.

 

[George Jones]

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.

 

[nigelscott]

OK. Thanks to you both. I think I understand it now.