# Riemannian Geometry , Levi-Civita connection, metric form q

## Main Question or Discussion Point

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?

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Matterwave
Gold Member
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 $g_{ij}+g_{ji}=2g_{ij}$.

The metric is always symmetric. This is in the definition of a metric.
Is this true for any geometry and not just Riemannian?

Nugatory
Mentor
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
Gold Member
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
Gold Member
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
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 $\{a,b\}=0$.