Problems with the interpretation of the Torsion tensor and the Lie Bracket

  • #1
PhysicsObsessed
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Hi,
I've been doing a course on Tensor calculus by Eigenchris and I've come across this problem where depending on the way I compute/expand the Lie bracket the Torsion tensor always goes to zero. If you have any suggestions please reply, I've had this problem for months and I'm desperate to solve it.

I tried computing an actual example on a spherical manifold with some simple vector fields u and v to see if that would help clarify the issue, however it didn't turn out. Though I believe the problem lies with the u(v) being equal to ∇_u(v), which shouldn't be the case.

I screenshotted the problem below, but I'll add some clarification on the notation here:
  • vector / vector field: any letter with a harpoon on top
  • partial derivative operator (w respect to coordinate variables): del
  • covariant derivative: nabla symbol
  • connection coefficients / Christoffel symbols: capital gamma
  • contravariant component: the index will be a superscript (see the vector field components u^i)
  • covariant component: the index will be a subscript (see the basis vectors / partial derivatives del_i)
  • Torsion tensor: T (capital T)
  • Lie bracket: [] (square brackets)
  • Also, I'm using index notation / Einstein notation to represent summations
1676867199560.png
 
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Answers and Replies

  • #3
PhysicsObsessed
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Can you show us your work and use Latex to do it?

There is a reference page on PF with Latex formatting guidelines.

https://www.physicsforums.com/help/latexhelp/
Alright, yeah, I've never used Latex before so it might take a while but I'll give it a go, you mean the example calculation right? Thanks for the link btw.
 
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  • #4
martinbn
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Though I believe the problem lies with the u(v) being equal to ∇_u(v), which shouldn't be the case.
##u(v)## is not ##\nabla_u(v)##. Assuming it implies the torsion is zero. In fact it is stronger than that. Torsion zero is equivalent to ##u(v)-v(u)=\nabla_u(v)-\nabla_v(u)##.

##u(v)## is not a vector field, it is a second order operator.
 
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  • #5
PhysicsObsessed
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##u(v)## is not ##\nabla_u(v)##. Assuming it implies the torsion is zero. In fact it is stronger than that. Torsion zero is equivalent to ##u(v)-v(u)=\nabla_u(v)-\nabla_v(u)##.

##u(v)## is not a vector field, it is a second order operator.
Hi, thanks a lot,
I thought the error might be there, although I'm really new to this area could you explain what you mean by a second order operator? Do you mean something like ##\frac{\partial^2}{\partial x^i\partial x^j}##?
 
  • #6
martinbn
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Do you mean something like ##\frac{\partial^2}{\partial x^i\partial x^j}##?
Yes. And the reason that the Lie bracket is a vector field is because all the mixed second derivatives are equal and will cancel.
 
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  • #7
strangerep
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  • connection coefficients / Christoffel symbols: capital gamma
Christoffel symbols are symmetric in their lower indices, hence their torsion is always zero.
 
  • #8
PhysicsObsessed
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Christoffel symbols are symmetric in their lower indices, hence their torsion is always zero.
Hi, thanks for the help; though I was under the impression that this only applied in some cases, like for the connection coefficients of the Levi-Civita connection, but not in every case.
 
  • #9
martinbn
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Christoffel symbols are symmetric in their lower indices, hence their torsion is always zero.
You reserve the name "Christoffel symbols" only for symmetric connections.
 
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  • #10
strangerep
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[...] I was under the impression that this only applied in some cases, like for the connection coefficients of the Levi-Civita connection, but not in every case.
Yes, but you specified "Christoffel symbols" explicitly.
 
  • #11
martinbn
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Yes, but you specified "Christoffel symbols" explicitly.
Yes, but some use it for any connection.
 
  • #12
AboveSky
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You reserve the name "Christoffel symbols" only for symmetric connections.
I would even reserve this name for the specific connection implied by the metric.
 
  • #13
PhysicsObsessed
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Yes, but you specified "Christoffel symbols" explicitly.
I would even reserve this name for the specific connection implied by the metric.
Oh, I wasn't aware, thanks for the comment, I'll make sure to be more careful in the future. But yes the problem is meant to be for an arbitrary connection, not specifically for the Levi-Civita connection.
 

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