- #1
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Hi,
I'm confused about what differentiation on smooth manifolds means. I know that a vector field [tex] v [/tex] on a manifold [tex] M[/tex] is a function from [tex] C^{\infty}(M) [/tex] to [tex] C^{\infty}(M) [/tex] which is linear over [tex] R [/tex] and satisfies the Leibniz law. This should be thought of, I'm told, as a 'derivation' on smooth functions on the manifold, i.e. differentiation. Intuitively, a vector field is a field of little arrows on the manifold, and its action on a function is just the directional derivative of the function in the direction of the tangent vector at that point.
What I don't quite understand is what it means to differentiate on a space without any notion of distance. To take a simpler example, consider a vector space [tex] R^2 [/tex] (not the euclidean vector space with metric, [tex] E^2 [/tex]). What is a smooth function on a vector space? What is differentiation of smooth functions on a vector space without any other structure?
I'm confused about what differentiation on smooth manifolds means. I know that a vector field [tex] v [/tex] on a manifold [tex] M[/tex] is a function from [tex] C^{\infty}(M) [/tex] to [tex] C^{\infty}(M) [/tex] which is linear over [tex] R [/tex] and satisfies the Leibniz law. This should be thought of, I'm told, as a 'derivation' on smooth functions on the manifold, i.e. differentiation. Intuitively, a vector field is a field of little arrows on the manifold, and its action on a function is just the directional derivative of the function in the direction of the tangent vector at that point.
What I don't quite understand is what it means to differentiate on a space without any notion of distance. To take a simpler example, consider a vector space [tex] R^2 [/tex] (not the euclidean vector space with metric, [tex] E^2 [/tex]). What is a smooth function on a vector space? What is differentiation of smooth functions on a vector space without any other structure?