Equivalent paths to the Christoffel symbols

[snoopies622]

TL;DR

I'm wondering under what circumstances deriving Christoffel symbols in two different ways will produce the same results.

I've noticed that for both the surface of a sphere and a paraboloid, one arrives at the same Christoffel symbols whether using

\Gamma^i_{kl} = \frac {1}{2} g^{im} ( \frac {\partial g_{mk} }{\partial x^l} + \frac {\partial g_{ml}}{\partial x^k} - \frac {\partial g_{kl}} {\partial x^m} )<br /> which assumes <br /> <br /> \nabla g=0, or by embedding them in (flat) three dimensional space and taking the inner product of the partial derivatives of the basis vectors and the contravariant forms of those basis vectors, as in

\Gamma ^k_{ij} = \frac { \partial \bf{e}_i}{ \partial x^j} \cdot { \bf {e} ^k}

and I'm wondering why this is the case, and if it is always so. Is it because I'm tacitly assuming <br /> <br /> \nabla g=0 in the three-space as well?

 

[snoopies622]

I'm still fascinated with this \nabla g =0 matter, when it's true and when it isn't. I know that like anything else in mathematics one can either take it as an axiom or derive it from other axioms, but sometimes in the latter case I don't see where it comes from.

For instance, if I find the Christoffel symbols for the surface of a sphere using the method shown in this video (rather like the second approach I mentioned above)



there is no explicit mention of \nabla g =0, but the results are the same as if I had started with that assumption and then calculated them using the first equation above, which is derived directly from it.

It's clear to me that one ordinarily assumes \nabla g =0 in a flat 3-space with Cartesian coordinates, but I don't quite see why that by itself would imply that it's also true in any 2-space embedded inside it. Not all things carry over like that. For instance, a zero curvature tensor in the flat 3-space doesn't imply a zero curvature tensor in the embedded sphere.

 

[romsofia]

I'm not sure what your question is, you can use one forms/exterior derivatives to arrive at curvature.
That is... the connection one form $w_{ij} = \hat{e_j} \cdot d \hat{e_i}$ where $d \hat{e_i}$ tells us how our basis vectors move, this is independent of the coordinate system we are in.

You can relate this to your christoffel symbols by the equation $w^i_j = \Gamma^i_{jk} \sigma^k$

For more details, reference: http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/connections
http://physics.oregonstate.edu/coursewikis/GDF/book/gdf/components
Not sure if these is what you're asking, but both those sections have a lot of good material in them.

 

[snoopies622]

Thanks for the links, romsofia. I briefly read about exterior derivatives years ago but I don't remember if I understood them even then. Will give this matter some more thought.

 

[lavinia]

I'm still fascinated with this ∇g=0∇g=0 matter, when it's true and when it isn't. I know that like anything else in mathematics one can either take it as an axiom or derive it from other axioms, but sometimes in the latter case I don't see where it comes from.

∇g=0 seems to be a statement of fact which is either true or not. How is it an axiom? For instance it is true for the dot product in Euclidean space with respect to the directional derivative as the affine connection. One can prove it by direct calculation. What are the axioms in that case? The same is true for Minkowski geometry. It is a theorem that the Minkowski metric is covariantly constant with respect to the directional derivative. It is not an axiom.

It's clear to me that one ordinarily assumes ∇g=0∇g=0 in a flat 3-space with Cartesian coordinates, but I don't quite see why that by itself would imply that it's also true in any 2-space embedded inside it. Not all things carry over like that. For instance, a zero curvature tensor in the flat 3-space doesn't imply a zero curvature tensor in the embedded sphere.

Again this is a theorem. In this case, it is for the induced affine connection on the 2 dimensional submanifold of 3 space and the dot product restricted to its tangent planes. The calculation is straight forward.

 

[snoopies622]

The calculation is straight forward.

Thanks lavinia, that's good news for me!

 

[lavinia]

Thanks lavinia, that's good news for me!

BTW; Do you know what the induced affine connection is for a surface in 3 space?

 

[snoopies622]

No.

 

[lavinia]

No.

Given a vector field tangent to the surface and a tangent vector at a point, orthogonally project the ordinary directional derivative of the vector field onto the tangent plane to the surface at that point.

A theorem called the Nash Embedding Theorem says that every Levi-Civita connection on a Riemannian manifold - Riemannian meaning that the metric is positive definite - arises in this way although the ambient Euclidean space may be high dimensional. For instance, for the flat torus one must embed it in four dimensional Euclidean space to retrieve its connection.

This theorem gives an extrinsic way to think of the Levi-Civita connection. It is the projection of the ordinary directional derivative onto the tangent plane of the manifold. One can think of the manifold as a constraint that restricts curves to lie upon it. A geodesic then is a curve whose acceleration is perpendicular to the surface. A particle moving along a geodesic will not feel acceleration tangent to the manifold. One thinks of the geodesic as a curve whose only constraint is the force that holds it onto the manifold. I think - not sure - that Gauss called these curves of minimal constraint.

For instance the acceleration vector of a particle moving at constant speed along circle on a sphere points to the center of the circle. This vector will have a zero tangential component only if the circle is a great circle.

I do not know if there is a similar theorem for non-definite metrics for instance the metric of Space-Time.

 

[snoopies622]

Thank you, this exactly answered my question!