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I don't understand the principal idea of Semi-Riemannian Manifold. Why is that if I have a metric tensor $g$ on a smooth manifold $M$ that is a symmetric nondegenerate $(0, 2)$-tensor field on $M$ of constant index (metric tensor), I can define a scalar product on every tangent space?

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This is going to be a generalized scalar product $g(X,Y)$ that may take negative values. This kind of thing is useful in relativity theory, for example. But it is not a familiar Euclidean scalar product!

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  • $\begingroup$ Agreed. From engineering point of view, sometimes, it is impossible to define a bi-invariant Riemannian metric on some Lie groups of our interest, then one could go for a bi-invariant semi or pseudo Riemannian metric. A lot of symmetric spaces I deal with are unfortunately of this type. $\endgroup$ – Troy Woo Aug 20 '14 at 14:34

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