Levi-Civita Connection & Riemannian Geometry for GR

[binbagsss]

Conventional GR is based on the Levi-Civita connection.

From the fundamental theorem of Riemann geometry - that the metric tensor is covariantly constant, subject to the metric being symmetric, non-degenerate, and differential, and the connection associated is unique and torsion-free - the connection can be determined by the metric from a relativity simple formula.

BUT , doesn't whether the metric is symmetric or not, depend upon the choice of coordinates when specifying the metric. E.g- Schwarzschild metric given in spherical polar coordinates is diagonal, and so symmetric.
However, the extended Schwarzschild metric in Eddington-Finkelstein coordinates is not diagonal, contains a dudr term, but not a drdu term. Does this mean that for the metric in Eddington-Finkelstein coordinates , the connection can no longer be simply computed from the metric?

Thanks in advance.

 

[Matterwave]

The metric is always symmetric. This is in the definition of a metric.

When people write out the metric in the familiar $ds^2=-dt^2+...$ form, they are actually writing out the contraction $ds^2=g_{ij}dx^i dx^j$. But since $dx^idx^j=dx^jdx^i$ the $g_{ij}$ and $g_{ji}$ terms just get added together so you don't see both terms, just the combination $g_{ij}+g_{ji}=2g_{ij}$.

 

[binbagsss]

The metric is always symmetric. This is in the definition of a metric.

Is this true for any geometry and not just Riemannian?

 

[Nugatory]

Is this true for any geometry and not just Riemannian?

If the metric were not symmetric, then the dot product of two vectors would not be commutative.

 

[lavinia]

Mathematically, inner products are symmetric by definition. In a coordinate system you are merely expressing the inner product ( in fact any tensor) in terms of the coordinates. You are not changing it in any way.

 

[ChrisVer]

If the metric were not symmetric, then the dot product of two vectors would not be commutative.

It would be anti-commutative , like Grassmann variables...
But wouldn't that make a very weird universe?? It blows my mind to think that r \theta = - \theta r

 

[Matterwave]

It would be anti-commutative , like Grassmann variables...
But wouldn't that make a very weird universe?? It blows my mind to think that r \theta = - \theta r

It wouldn't have to be "anti-commutative" unless it were totally anti-symmetric. A general tensor will have both symmetric and anti-symmetric parts. Usually the term "anti-commutative" means that variables anti-commute $\{a,b\}=0$.

Anyways, there is a structure that IS anti-commutative that sometimes we use for manifolds, and that is called a symplectic structure. The basic quantity of interest there; however, is not called a metric, but the symplectic form. We use this for, e.g. geometric Hamiltonian mechanics.