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I have a frame that varies along a curve $\gamma$ : the frame consists in the tangent vector of the curve plus some constant non orthogonal vectors.

I need to compute $\nabla_{\dot\gamma}{\dot\gamma}$. Apparently, the Christoffel symbols in that case are computed with :

$$\omega^i{}_{k\ell}=\frac{1}{2}g^{im} \left( g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m} + c_{mk\ell}+c_{m\ell k} - c_{k\ell m} \right)\,$$

where $c_{k\ell m}=g_{mp} {c_{k\ell}}^p\ $ are the commutation coefficients of the basis; that is, $[\mathbf{u}_k,\mathbf{u}_\ell] = c_{k\ell}{}^m \mathbf{u}_m\,\ $


I have several questions regarding the commutation coefficients:

  • I read that the commutator is defined by $[X,Y]=\nabla_X Y - \nabla_Y X$ : isn't it a catch 22? how is it possible to compute that commutator, knowing that the connection $\nabla$ is obtained by the Christoffel symbols that uses the commutator that uses the connection that uses [...] ?
  • For one of my applications, my tangent space is made of symmetric matrices (and the manifold has a fancy metric $g$) : in that case, is the commutator only defined by $[X,Y]=X*Y-Y*X$ with $*$ the standard matrix multiplication ?
  • For another application, my tangent space are vectors in $R^N$ and my space is Euclidean. Of course, in that case there are probably easier ways to handle that, but I'd still like to understand what would the commutator be in that case.
  • In the definition $[\mathbf{u}_k,\mathbf{u}_\ell] = c_{k\ell}{}^m \mathbf{u}_m\,\ $ , is-it necessary to solve a linear system to get the coefficients $c_{k\ell}{}^m$ or can I just project $[\mathbf{u}_k,\mathbf{u}_\ell]$ on $\mathbf{u}_m$ using the metric $g$ (my basis is not orthogonal, so I guess the linear system is needed?).


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up vote 2 down vote accepted

This "commutator" $[X,Y]$ is in fact the Lie bracket of two vector fields $X$ and $Y$ seen as derivations on the algebra $C^\infty(M)$ of smooth functions on the manifold $M$.

This Lie bracket is appropriately defined as an operator on $C^\infty(M)$ given by $$ [X,Y]f := X(Y\,f) - Y(X\,f) \tag{1} $$ where $f \in C^\infty(M)$.

It turns out to be a derivation again, and so represents a vector field $[X,Y]$.

Now the trick is that one can show that $$ [X,Y] = \nabla_X Y - \nabla_Y X \tag{2} $$ for any torsion-free connection $\nabla$, in particular, for the Levi-Civita connection that you are using in your calculations.

This is an example of a Lie algebra with the bracket defined by (1). Another example that you have mentioned is the algebra of square matrices or order $n$ with the bracket given by the commutator of the matrices. It is not necessarily related to the first example, of course.

It must be emphasized that in definition (1) we do not use any metric or connection at all, only a smooth structure on the manifold, that is the choice of charts that we may have chosen to identify $M$ locally with pieces of $\mathbb{R}^n$.

Simply put, just compute the directional derivatives!

Remark. The formula that you give in the quotation is the expression for the Christoffel symbols of the Levi-Civita connection. You can prove it as an easy consequence of the so-called Koszul formula.

(All references are in the appropriate articles in Wikipedia)

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thanks! It's still not entierely crystal clear for me : why can't I use the matrix multiplication version instead of the directional derivative ? does it gives something different ? Also, if I use directional derivative, in my case, I know how to compute $XY$ if $X$ is the tangent of my curve and $Y$ a constant vector (in that case, it is just $0$ as the vector is constant), but how can I compute $YX$ ? I need to differentiate the tangent vector of my curve in a direction that is not along the curve... should I assign an arbitrary value ? Also, what about my last item ? Thanks!! – WhitAngl Dec 19 '12 at 7:35
note that as I understood, I need to extend continuously my curve to produce a continuous vector field in its neighborhood. Now, I need to program that, and dealing with "arbitrary extensions" is not something really concrete.... Should I just set $YX$ to a random value (let's say $0$) if $Y$ is an arbitrary vector and $X = \dot\gamma$ ? Thanks! – WhitAngl Dec 19 '12 at 22:16
if you don't mind talking at some point, I created a chatroom (the previous one has been frozen) :… – WhitAngl Dec 20 '12 at 0:45
@WhitAngl Sure I will join your chat. Sorry, being busy a bit. – Yuri Vyatkin Dec 20 '12 at 2:15

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