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What is meant when one says that one has chosen a basis of fields on the manifold with ``anolonomy"?

I get the feeling that it is a choice of basis with non-trivial structure constants say $C^{k}_{ij}$

Now in some papers people seem to state that once such a basis is chosen one can write the Christoffel symbols in terms of the structure constants by the following equation,

I am writing the equation with all indices lowered (as written in the original paper)

$\Gamma_{ijk} = \frac{1}{2}(C_{ijk} - C_{ikj} - C_{jki})$

I suppose they are assuming that the connection is chosen to be torsion free since the above equation satisfies torsion free-ness condition.

But just being torsion free doesn't seem to be enough to derive the above relationship.

This relationship is seen in the papers in the context of choosing a veirbein on homogeneous spaces equipped with a Riemannian metric.

I would like to know from where and how does the above equation come.

In general it is surely impossible that the structure constants determine the connection!

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  • $\begingroup$ There is an additional assumption allowing you to write down the Christoffel symbols in terms of the $C$s. What you have written is half of the Koszul formula for the Levi-Civita connection. You are missing the terms where the metric is differentiated. Hence it seems to me that you are assuming that the vector fields are such that their scalar product is constant. $\endgroup$ Commented Feb 26, 2010 at 12:29
  • $\begingroup$ Notice, by the way, that the $C$s are most definitely not constants in the general case. $\endgroup$ Commented Feb 26, 2010 at 12:30

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Maybe I should elaborate on my comment. On a riemannian manifold $(M,g)$ there exists a unique metric-compatible torsion-free affine connection. It's the Levi-Civita connection and one can prove its existence and uniqueness constructively by giving a formula, known as the Koszul formula. This formula is given by $$2 g(\nabla_XY,Z) = X g(Y,Z) + Y g(Z,X) - Z g(X,Y) - g(Y,[X,Z]) - g([Y,Z],X) - g(Z,[X,Y]) $$ where $X,Y,Z$ are vector fields on $M$. Now take a local orthonormal frame $(e_i)$ for the tangent bundle. Orthonormality says that $g(e_i,e_j)$ is constant, whereas because they are a frame $[e_i,e_j] = C_{ij}^k e_k$ for some functions $C_{ij}^k$. (They will not be constant in general.) If you now apply the Koszul formula to the elements in this frame you get your expression, where $$C_{ijk} = g([e_i,e_j],e_k).$$

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  • $\begingroup$ Thanks for the answer. If one uses the Koszul's equation then one trivially gets the relation but the confusion started because the paper where I was seeing a use of this didn't ever state that they were on a metric compatible or even a torsion free connection. Hence I was wondering how they got the relation. I suppose it is enough to have the basis to be orthogonal for the relation to work out. right? Or probably one can weaken the requirement to just needing a basis whose inner product is constant? I suppose assuming the basis to be a vierbein is not necessary. $\endgroup$
    – Anirbit
    Commented Feb 27, 2010 at 8:16
  • $\begingroup$ If by vierbein you mean four-dimensional, then indeed it is not necessary. Also the frame need not be orthonormal, it is enough for the inner products of the elements of the frame to be constant. I am using that this is the Levi-Civita connection, though. If the connection is not metric compatible or is not torsion-free there is no longer a unique connection, so one does not expect that there should exist a formula which has no input other than a choice of frame. $\endgroup$ Commented Feb 27, 2010 at 12:57
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You are correct, a nonholonomic (or anholonomic) basis of the tangent bundle is a set of vector fields $X_i$, $i = 1, \ldots, n$, with nonvanishing structure constants: $[X_i, X_j] = C_{ij}^k X_k$. I'm not sure where this terminology comes from, since in classical mechanics nonholonomicity means exactly the opposite (i.e. the Lie bracket is not closed).

I haven't seen the formula you wrote down. The indices seem to be in the wrong places in the second and third term, could you check this?

Not sure if this helps, but on a Lie group at least there is a connection $\nabla$ defined in terms of the Lie bracket of left invariant vector fields as

$\nabla_X Y = \frac{1}{2} [X, Y]$

So the Christoffel symbols for this metric are just the structure constants of the Lie group. This is due to Milnor.

I guess this could be extended to the case of homogeneous manifolds, but at some stage something has got to give since the Lie bracket is not a connection (not tensorial in the first argument). So in that case, maybe your term 2 and 3 are the necessary correction terms.

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  • $\begingroup$ Thanks for your reply. I have corrected the issue with the indices. I am aware of the result of Milnor that you are referring to. That works for any semi-simple lie group (best if it is compact) since in that case the group comes with a natural bi-invariant metric and the Riemann-Christoffel connection for that satisfies the equation you wrote. $\endgroup$
    – Anirbit
    Commented Feb 26, 2010 at 8:59
  • $\begingroup$ The equation which I have asked about seems to be a generalization of this formula of Milnor. But I have no clue how to derive that and when is it valid. Are you sure that the "anholonomy" which you talk of is the same thing as the "anolonomy" I have mentioned? $\endgroup$
    – Anirbit
    Commented Feb 26, 2010 at 9:00
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    $\begingroup$ Is that really due to Milnor? I would have imagined that connection had been known since the beginning of time---which is, more or less, the time of Cartan, Killing and friends. $\endgroup$ Commented Feb 26, 2010 at 13:38
  • $\begingroup$ You're right, this probably goes back to the early days of differential geometry. Milnor "just" fleshed it out considerably. $\endgroup$
    – jvkersch
    Commented Feb 26, 2010 at 16:27

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