On the order of indices of the Christoffel symbol of the 1st kind.

[ric peregrino]

Homework Statement

The order of indices of the Christoffel symbol of the 1st kind seems to vary from source to source. Is there a preference, and if so why?

Relevant Equations

Christoffel symbol of the 1st kind.

The 1st definition of the Christoffel symbol of the 1st kind I came across was from Einstein's "The Meaning of Relativity", 1953, pg. 71, eq. 69:

$$[ij,k]=\frac{1}{2}\left(g_{ik,j}+g_{jk,i}-g_{ij,k}\right)$$

This is the same as I found in Barry Spain's "Tensor Calculus", 1953, but differs from other sources, for example wiki:

$$[ij,k]=\frac{1}{2}\left(g_{ki,j}+g_{kj,i}-g_{ij,k}\right)$$

I've taken the liberty to use i, j, and k to make the difference here obvious. I realize that with a symmetric metric, that perhaps the order of these 3 indices may not matter, but I wonder if the order may matter for an asymmetric metric? I've tried getting some answers on my own, and had limited success, lastly on stackexchange, where it seems the gatekeepers there are a tough crowd, and they closed my question on this with out any answer there, or any good answers to others' questions there concerning an asymmetric metric, mostly just responses that amount to "the metric HAS to be symmetric".

I had pointed out that Einstein and Kaufman published a paper in 1953 ( https://www.jstor.org/stable/1969690?origin=crossref) that included an asymmetric metric. If I read that correctly I may have learned a few things:

1. in 4d, there can be 2 unique components for the anti-symmetric part of such an asymmetric metric.
2. if the anti-symmetric part of the metric is non-singular at one point in space time, then it will be so for all points.
3. I didn't see a simple relation between the resulting metric compatible connection, and such a metric.
4. An asymmetric metric doesn't seem to immediately solve any issues. Nor does an asymmetric field, aka torsion.

I think another issue with my question at stackexchange was that I went big up front, and then when I tried to narrow it down to just this question here, I had proposed a preferred order that differs from the original, and from the wiki definitions, and made unsubstantiated claims about this proposed order. Any way, I've gone on long enough, and hope I got the root question across about the differing published orders of indices of the Christoffel symbol of the 1st kind, considering an asymmetric metric, is there a preference and if so why?

Cheers,
Ric

 

[anuttarasammyak]

The both are all right by symmetry of metric tensor, i.e.
$$g_{ij}=g_{ji}$$

 

[Orodruin]

The metric tensor is symmetric by definition. Einstein and others considered a general tensor formalism with the hope that it would lead to unification since the electromagnetic field tensor is rank 2 and antisymmetric. This however turned out to be fruitless. It is not something that is really considered much today as far as I am aware.

The both are all right by symmetry of metric tensor, i.e.
$$g_{ij}=g_{ji}$$

OP knows this, they have stated as much.

 

[ric peregrino]

My proposed order is: $$[ij,k]=\frac{1}{2}\left(g_{ik,j}+g_{kj,i}-g_{ij,k}\right)$$

The both are all right by symmetry of metric tensor, i.e.
$$g_{ij}=g_{ji}$$

Indeed, with a symmetric metric it doesn't matter. What I'm asking is does it matter with an asymmetric metric.

 

[anuttarasammyak]

As far as
$$ds^2=g_{ij}dx^i dx^j$$
holds, and
$$dx^j dx^i=dx^i dx^j$$
I have no idea how we can introduce non symmetric metric tensors.

 

[ric peregrino]

To be sure, I'm asking about asymmetric metrics, which are neither symmetric nor antisymmetric, but have both symmetric and antisymmetric components, and though the antsymmetric part does not contribute to ds^2 as anuttarasammyak points out, or affect geodesics, it does contribute to torsion.

 

[anuttarasammyak]

$$g_{ij}=s_{ij}+a_{ij}$$
$$ds^2=s_{ij}dx^idx^j+a_{ij}dx^idx^j$$
changing dummy indeces i and j
$$ds^2=s_{ji}dx^jdx^i+a_{ji}dx^jdx^i$$
Adding these two
$$2ds^2=(s_{ij}+s_{ji})dx^idx^j+(a_{ij}+a_{ji})dx^idx^j=2s_{ij}dx^idx^j$$
Antisymmetric components disappear.

 

[Orodruin]

$$g_{ij}=s_{ij}+a_{ij}$$
$$ds^2=s_{ij}dx^idx^j+a_{ij}dx^idx^j$$
changing dummy indeces i and j
$$ds^2=s_{ji}dx^jdx^i+a{ji}dx^jdx^i$$
Adding these two
$$2ds^2={s_{ij}+s_{ji})dx^idx^j+(a_{ij}+a_{ji})dx^idx^j=2s_{ji}dxjidxij$$
Antisymmetric components disappear.

Again, this is not what the OP is asking about. They have said so already in the OP and clarified again.

There are connections with non-zero torsion that are metric compatible and formalisms where the connection is a priori separate from the metric. It then turns out to be Levi-Civita from the equations of motion in the basic scenario.

How this connection would be affected by the introduction of an antisymmetric part to the metric I will leave unsaid as I have not studied this in detail.

I believe it is however fair to assume that Christoffel symbols of the first kind assume a symmetric metric, ie, the metric that goes into the definitions of the line element etc. To those, the antisymmetric part would not contribute, but they would generally not be equal to the connection coefficients.