[Math] Metric Tensor and Curvature

differential-geometrymanifolds

I have a doubt about the metric tensor and curvature. My point is: last year I was talking to a mathematician that I know and he said to me that the metric tensor is something that we define on a manifold like an inner product is defined on a vector space.

What he said is: given a vector space $V$ we can define any inner product that we want, it just must comply with the axioms. The same happens with the metric tensor: given a manifold $M$ we can define any metric tensor, it just needs to assign to each point a valid inner product on the tangent space.

But hat brought me some doubts: the $2$-sphere is a manifold $S^2$, and it has curvature. This is obvious geometrically: we can see that $S^2$ cannot be considered flat in any way. However, the curvature tensor which gives the curvature, is defined in terms of the metric. If we can define any metric that we want, then how we're going to recover the "true" curvature of the sphere ?

In other words, I think that there wouldn't be a problem to assign to the $S^2$ as a metric tensor the tensor field that at each point assign just the euclidean inner product.

I thought that the metric tensor was derived from the charts, but this mathematician said that it's not, it's just defined at our will, having to satisfy the inner product properties at any point. Can someone point out a good reference for me to read more about how to obtain the metric and so on ? Or even give some example ?

Thanks in advance. And sorry if the question is silly. Also, if it's not well explained, just tell me and I'll rewrite it.

Best Answer

The short answer is: You can't give $\mathbb{S}^2$ a flat metric, because the metric must be assigned to each point in a smooth manner. The different ways in which this can be done are subject to topological constraints. The most famous and elementary example of this is the Gauss-Bonnet theorem, which says that for any closed surface $S$ and any smooth Riemannian metric $g$ on $S$, its sectional curvature $\kappa_g$ must satisfy $$\int_S \kappa_g = 2\pi\chi(S).$$ (Here, $\chi(S)$ denotes the Euler characteristic of $S$.)

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