I Induced metric in general case

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gerald V
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TL;DR
Does the familiar Gauss formula for the embedded metric hold if the embedding space itself is intrinsically curved?
The Gauss formula for the induced metric reads ##g_{\mu \nu} = G_{ab}\frac{\partial X^a \partial X^b}{\partial x^\mu \partial x^\nu}##, where ##G## is the metric of the embedding space, the capital ##X## are positions in the embedding space, the little ##x## are positions inside the embedded manifold and the ##\partial## denote partial differentiation (no notion of covariance present).


I have read some literature on embedding, pullbacks and so on, but they are all quite abstract. My only question is whether the above formula equally holds if the embedding space is not equipped with a Minkowski-like metric (in some number of dimensions), in particular if the embedding space itself is intrinsically curved.

From what I know the answer is yes but I would be grateful for confirmation or correction.


Thank you very much in advance.
 
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Here is a sketch of deduction of the Lagrange equations by means of the covariance argument. I believe that it is a suitable substitute for the archaic terminology that is employed in most textbooks. Assume we have ##\nu## particles with masses ##m_1,\ldots,m_\nu## and with position vectors $$\boldsymbol r_i=(x^{3(i-1)+1},x^{3(i-1)+2},x^{3(i-1)+3})\in\mathbb{R}^3,\quad i=1,\ldots,\nu.$$ Thus the position of the system is characterized by a vector...

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