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## Main Question or Discussion Point

I'm looking at lecture notes on General Relativity by Sean M. Carroll, and after defining the Riemannanian tensor in the usual theorem, the extent to which the partial derivatives of a vector field fail to commute, it says ' having defined curvature tensor as something which characterizes the connection, in GR we are most concerned with the Christoffel connection. In this case the connection is derived from the metric, and the associated curvature may be thought of as that of the metric itself'.

I'm having trouble understanding this.

Also, I think, 'in GR we are mot concerned with the Christoffel connection' comes from having the fundamental theorem of Riemannian geometry holding , so we get that simple formula relating the connection to the metric. The conditions of the metric in this theorem are that it be 'symmetric, differentiable and non-degenerate'.

So, the unextended Schwarzschild metric has a singulaity at r=2GM, it's determinant is in indeterminate form, and is degenerate. So doesn't this mean that at a curavture singularity, because the conditions for the fundamental theorem of Riemannian geometry to hold are not satisfied, the connection can not be given by the metric, and if I have interpreted the above statement correctly, the curvature computed from the Riemannian tensor will not be that associated with the metric? What does it represent then?

I'm having trouble understanding this.

**What does the curvature describe otherwise? I've only heard of it being the metric itself.**Also, I think, 'in GR we are mot concerned with the Christoffel connection' comes from having the fundamental theorem of Riemannian geometry holding , so we get that simple formula relating the connection to the metric. The conditions of the metric in this theorem are that it be 'symmetric, differentiable and non-degenerate'.

So, the unextended Schwarzschild metric has a singulaity at r=2GM, it's determinant is in indeterminate form, and is degenerate. So doesn't this mean that at a curavture singularity, because the conditions for the fundamental theorem of Riemannian geometry to hold are not satisfied, the connection can not be given by the metric, and if I have interpreted the above statement correctly, the curvature computed from the Riemannian tensor will not be that associated with the metric? What does it represent then?

**Thanks.**