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So I know that any n-dimensional manifold can be embedded in the Euclidean space of dimension 2n.
Does this mean that there is some canonical embedding that allows us to define the derivative on the manifold in an unambiguous way?

pmal
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  • This is definitely a math question, but it's also not clear to new what you're asking, so I hesitate to migrate. – DanielSank Oct 15 '16 at 02:07
  • I posted it here because it came to me while studying Yang's mill theory. What is it that is unclear? – pmal Oct 15 '16 at 02:08
  • Well, there are several kinds of derivative so I'm just not sure what you're asking about when you ask for an unambiguously defined derivative. Could you maybe specify what you mean by derivative in a simple case (like on $\mathbb{R}^n$) and perhaps with an example so we know what you're getting at? – DanielSank Oct 15 '16 at 02:17
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    There is NO canonical embedding. However one may still define derrivatives which are independant of the choice of coordinates. First one can define the Lie-derivative. One can further define a covariant derivative (a choice of which can be used to distinguish some embeddings over others). You should be able to find a lot of material on both topics. In any case it is modern to think of manifolds as not being embedded but just described as an abstract topological space endowed with charts. – Adomas Baliuka Oct 15 '16 at 02:18
  • For any neighborhood in the manifold there's a chart where we can define a derivative to be the $R^n$ one. Since there are infinitely many functions that can be used as charts, the coordinates that define this derivative are not unique... but this doesn't mean its definition is ambiguous. – QuantumBrick Oct 15 '16 at 04:34

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