# I Equivalent paths to the Christoffel symbols

#### snoopies622

Summary
I'm wondering under what circumstances deriving Christoffel symbols in two different ways will produce the same results.
I've noticed that for both the surface of a sphere and a paraboloid, one arrives at the same Christoffel symbols whether using

$$\Gamma^i_{kl} = \frac {1}{2} g^{im} ( \frac {\partial g_{mk} }{\partial x^l} + \frac {\partial g_{ml}}{\partial x^k} - \frac {\partial g_{kl}} {\partial x^m} )$$ which assumes $\nabla g=0$, or by embedding them in (flat) three dimensional space and taking the inner product of the partial derivatives of the basis vectors and the contravariant forms of those basis vectors, as in

$$\Gamma ^k_{ij} = \frac { \partial \bf{e}_i}{ \partial x^j} \cdot { \bf {e} ^k}$$

and I'm wondering why this is the case, and if it is always so. Is it because I'm tacitly assuming $\nabla g=0$ in the three-space as well?

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#### snoopies622

I'm still fascinated with this $\nabla g =0$ matter, when it's true and when it isn't. I know that like anything else in mathematics one can either take it as an axiom or derive it from other axioms, but sometimes in the latter case I don't see where it comes from.

For instance, if I find the Christoffel symbols for the surface of a sphere using the method shown in this video (rather like the second approach I mentioned above)

there is no explicit mention of $\nabla g =0$, but the results are the same as if I had started with that assumption and then calculated them using the first equation above, which is derived directly from it.

It's clear to me that one ordinarily assumes $\nabla g =0$ in a flat 3-space with Cartesian coordinates, but I don't quite see why that by itself would imply that it's also true in any 2-space embedded inside it. Not all things carry over like that. For instance, a zero curvature tensor in the flat 3-space doesn't imply a zero curvature tensor in the embedded sphere.

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