Covariant derivative of Lie-Bracket in normal orthonormal frame

[holy_toaster]

Hi there,

I was doing some calculations with tensors and ran into a result which seems a bit odd to me. I hope someone can validate this or tell me where my mistake is.

So I have a normal orthonormal frame field $\{E_i\}$ in the neighbourhood of a point $p$ in a Riemannian manifold $(M,g)$, i.e. Riemannian normal coordinates about $p$ such that $\nabla_{E_i}E_k=0$ and subsequently $[E_i,E_k]=0$ at the point $p$ for all $i,k$. As $[E_i,E_k]=\sum_j c^j_{ik}E_j$ the structure functions $c^j_{ik}$ also vanish at $p$.

Now I compute $$\nabla_{E_k}[E_i,E_k]=[E_k,[E_i,E_k]]+\nabla_{[E_i,E_k]}E_k=\sum_{jl}c^l_{ik}c^j_{lk}E_lE_j+\nabla_{[E_i,E_k]}E_k=0$$ at $p$, where the first summand vanishes because of $c^j_{ik}=0$ and the second summand vanishes because the covariant derivative $\nabla_XY$ is tensorial in $X$ so it vanishes if $X=0$.

In total that means that in an normal orthonormal frame about a point $p$ not only all the covariant derivatives and the Lie-Bracket of the basis vectors vanish at $p$, but also the covariant derivative of the Lie-Bracket. Does that make sense? Or am I mistaken here?

 

[Sina]

Wouldn't it be simpler to write $[E_i,E_k]=\sum_j c^j_{ik}E_j$ in the covariant derivative so that

$\nabla_{E_k}[E_i,E_k] = \nabla_{E_k} \sum_j c^j_{ik}E_j = \sum_j (E_k(c^j_{ik})E_j + c^j_{ik} \nabla_{E_k}(E_j) )$

Now in the nbd of p you assumed that the functions c are all zero and constant so their derivatives should vanish too

 

[holy_toaster]

Aha, yes there appear derivatives of the structure functions $c^i_{jk}$. But the Riemannian normal coordinates about $p$ can always be chosen such that these derivatives vanish at $p$, right?

 

[Sina]

Well on Riemann manifolds, you can find a covering such that on each cover the frame is orthonormal (however the covers do not have to transform well so they might not form an atlas). Now if the structure functions of the frame are identically zero on a chosen nbd of p, then shouldn't the derivatives also vanish through out the whole nbd?

As for the point case, if you have a riemannian metric, around any point p you can find coordinates (this time indeed coordinate not just a chart) such that the curvature is identically zero at the point p and roughly proportional to distance to p on other points around p.

 

[holy_toaster]

As for the point case, if you have a riemannian metric, around any point p you can find coordinates (this time indeed coordinate not just a chart) such that the curvature is identically zero at the point p and roughly proportional to distance to p on other points around p.

Now that seems rather strange to me. As I always understood that curvature is exactly the quantity, which we can NOT gauge to zero at some point p by choosing normal coordinates about p. We get coordinate lines which are geodesics and have orthonormal tangent vectors at p, as well as vanishing Christoffel symbols at p which lead to vanishing covariant derivatives of the tangent vectors at the coordinate lines at p. Furthermore, the Lie brackets of the tangent vectors to the coordinate lines vanish in the neighbourhood of p. But as curvature contains derivatives of the Christoffel symbols it is generically not zero at p, in no coordinate system, except for the case when the manifold is flat at p.

 

[Sina]

Note that after the transformation mentioned, the curvature vanishes only at a point not on a nbd around it. See section 7.7 here

http://people.sissa.it/~percacci/lectures/genrel/07-riemannian.ps

 

[holy_toaster]

Sorry, I can't find anything in there, which supports this claim of vanishing curvature at p. Note that curvature is a tensor, whereas Christoffel symbols are not tensors. If the components of a tensor vanish at a point in one coordinate system, they vanish in all coordinate systems, hence the tensor would be zero. But Christoffel symbols are components of a connection, which obey a different law of transformation between coordinates, so they may vanish at point in some coordinates but not in others.

 

[Sina]

Wait you are right sorry curvature does not vanish the christofel symbols do. I have mistaken it with the concept of locally inertial frames.

 

[joebohr]

Aha, yes there appear derivatives of the structure functions $c^i_{jk}$. But the Riemannian normal coordinates about $p$ can always be chosen such that these derivatives vanish at $p$, right?

What do you mean by the structure functions of a tensor?

 

[Sina]

not of the tensor of the vector fields. it is a generalization of the concept of structure constants of a lie algebra-which are really genuine constants due to property of invariant vector fields. These are functions however instead of being constants.

since the lie bracket is an operation from the space of vector fields to space of vector fields then lie bracket of any two basis elements (element orthonormal frames) can be given as linear combinations of other basis elements. the coefficients of linear combination are structure functions (or constants for lie algebras).