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- Thread starter nigelscott
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My understanding is that if I pick a point on a manifold that in the limit can be considered as being flat, I can use cartesian coordinates. If I uses cartesion coordinates the derivative of the metric tensor is 0 and covariant derivative become the ordinary derivative. So doesn't this imply that the Riemann tensor is not defined at that point? What am I not understanding?

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To talk about curvature, you need more than a point, you need a neighborhood around the point. This is the same as in freshman calculus. If I tell you that the function f passes through the origin, that doesn't help you to calculate its curvature (second derivative) at the origin.My understanding is that if I pick a point on a manifold that in the limit can be considered as being flat, I can use cartesian coordinates.

The connection between the tangent plane and the Riemann tensor would be that the Riemann tensor is an operator that works with vectors taken from the tangent plane. If you take a vector from the tangent plane and transport it around a little parallelogram, it's changed by the time it comes back to the start. There are four vectors from the tangent plane involved here: (1) the original vector, (2) the change in the original vector, (3) and (4) two vectors that describe the parallelogram. This is why the Riemann tensor has four indices.

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