1
$\begingroup$

I am (numerically) computing the second fundamental form for a curve $\gamma(t)$ embedded in a Riemannian manifold $(M, g)$. I would like to double check if what I am doing is correct.

  • First, define a basis where the first element of the basis is $\dot\gamma(t)$ (which I numerically evaluate) and the other elements are arbitrary (not necessarily orthogonal) fixed vectors in $TM$ (we'll suppose all the vectors span $TM$ and are thus independant). My basis vectors are called $\partial_i$.

  • Compute the Levi-Civita connection $\nabla_{\dot\gamma}\dot\gamma$ using the Christoffel symbols $\Gamma^i_{j,k}$. In particular, I use the standard formula (that uses $\frac{\partial g_{i,j}}{\partial_k}$ etc.) with no particular change due to the non orthogonality of the basis. My final value is $\nabla_{\dot\gamma}\dot\gamma = \sum_i \Gamma^i_{1,1}\partial_i$

  • I project $\nabla_{\dot\gamma}\dot\gamma$ on the normal space $NM$ by computing $\nabla_{\dot\gamma}\dot\gamma - \frac{g(\nabla_{\dot\gamma}\dot\gamma,\dot\gamma)}{g(\dot\gamma,\dot\gamma)}\dot\gamma$

In particular, my second step does not seem to lead to $\nabla_{\dot\gamma}\dot\gamma=0$ for geodesics, which could be either due to my reasoning or other factors (bugs, numerical approximations etc.). Is-there a problem with the method above ? Am-I allowed to use a non orthogonal basis ? My second doubt is about $\nabla_{\dot\gamma}\dot\gamma = \sum_i \Gamma^i_{1,1}\partial_i$ : is it correct ?

Thanks!

$\endgroup$
12
  • $\begingroup$ Vector $\nabla_{\dot\gamma}\dot\gamma - \frac{g(\nabla_{\dot\gamma}\dot\gamma,\dot\gamma)}{g(\dot\gamma,\dot\gamma)}\dot\gamma$ lives not in $NM$ (which is not defined here!) but in the normal bundle along $\gamma$ (which is spanned by the remaining vectors from your frame: $N\gamma = \operatorname{span}(\partial_2,\dots,\partial_n)$) $\endgroup$ Dec 16, 2012 at 22:20
  • $\begingroup$ yep, that's what I meant by $NM$ ($N$ for normal in "normal bundle" like $T$ in $TM$ for "tangent bundle of M") - isn't it standard notation ? thanks! $\endgroup$
    – WhitAngl
    Dec 17, 2012 at 7:27
  • $\begingroup$ Usually $NM$ denotes the normal bundle of an immersed submanifold $M$, so in your case something like $N \gamma$ looks more appropriate, because your $\gamma$ is a submanifold in $M$ in a certain sense (if it is regular and injective). $\endgroup$ Dec 17, 2012 at 9:15
  • $\begingroup$ oh yes, indeed, that makes sense - thanks $\endgroup$
    – WhitAngl
    Dec 17, 2012 at 20:45
  • 1
    $\begingroup$ Oops, I just noticed that you said "I use the standard formula ... with no particular change due to the non orthogonality of the basis". Please refer to section "Connection coefficients in a non holonomic basis" here. This however requires somewhat cleaner treatment that I don't have ready. May be someone can answer? $\endgroup$ Dec 18, 2012 at 9:25

1 Answer 1

2
$\begingroup$

As the discussion in the comments revealed, the major problem that may affect the calculations is that in a frame is not arising from a coordinate system, so called non-holonomic frame, the Christoffel symbols of the Levi-Civita connection of the metric $g$ are given by a formula that is slightly different. See in Wikipedia here.

This formula may be obtained by a generalization of the standard calculation of the Christoffel symbols of the Levi-Civita connection that you can find here, taking into account that the Lie brackets of the elements of the non-holonomic frame need not commute: $$ [\vec{e}_i,\vec{e}_j] := \nabla_{\vec{e}_i}{\vec{e}_j} - \nabla_{\vec{e}_j}{\vec{e}_i} = c_{ij}{}^k \vec{e}_k $$

Remark. The elements of such a frame should not be denoted as $\partial_i$ but rahter as $\vec{e}_i$, $\mathbf{e}_i$ or $\mathbf{u}_i$ etc. Using $\partial_i$ is conventionally reserved for vectors form a coordinate frame.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .