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Curves and tangent vectors in a manifold setting

  1. Mar 17, 2015 #1

    CAF123

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    Consider the following definition: (##M## denotes a manifold structure, ##U## are subsets of the manifold and ##\phi## the transition functions)

    Def: A smooth curve in ##M## is a map ##\gamma: I \rightarrow M,## where ##I \subset \mathbb{R}## is an open interval, such that for any chart ##(U,\phi)##, the map ##\phi \circ \gamma : I \rightarrow \mathbb{R}^n## is smooth.

    My first question is, why do we define a smooth curve in this way? In particular, why is the map ##\phi \circ \gamma## a good object to consider? The only thing that comes to mind is that now we have a function defined from ##\mathbb{R} \rightarrow \mathbb{R}^n## so differentiation is well defined and thus one may introduce the concept of a tangent vector (as below).

    Now let ##f: M \rightarrow \mathbb{R}## be a smooth function on ##M## and ##\gamma: I \rightarrow M##, smooth curve as before. Then ##f \circ \gamma : I \rightarrow \mathbb{R}## is smooth. Hence we take a derivative to find the rate of change of ##f## along the curve ##\gamma##: $$\frac{d}{dt}f(\gamma(t)) = [(f \circ \phi^{-1}) \circ (\phi \circ \gamma)]'(t) = \sum_{i=1}^n \left(\frac{\partial (f \circ \phi^{-1})}{\partial x^i}\right)_{\phi(\gamma(t))} \frac{d}{dt} x^i(\gamma(t))$$

    My next question is to simply understand how this equation comes about. I can see it is some application of the chain rule but I am struggling with the precise details of the equation, mostly in how the final equality comes about and the subscript on the ##\partial (f \circ \phi^{-1})/\partial x^i## term. Many thanks!
     
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  3. Mar 17, 2015 #2

    Orodruin

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    It allows us to do essentially everything on a smooth curve on a manifold that you can do with a smooth curve in Rn. (You can also check that the definition is coordinate independent.) To quote one of my undergrad teachers (linear algebra, not differential geometry): "It is not silly, it is a definition. It may be a silly definition, but you simply have to live with it."

    ##f\circ \phi^{-1}## is a function from the coordinate chart ##U## to ##\mathbb R## and it therefore has a dependence on the coordinates ##x^i##, which is what you differentiate this function with respect to. The full function you have is a composition of a function from ##\mathbb R## to ##\mathbb R^n## and a function from ##\mathbb R^n## to ##\mathbb R## and you therefore can apply the chain rule (##x^i(\gamma(t))## are simply the coordinate functions evaluated at the curve for some ##t##).
     
  4. Mar 17, 2015 #3

    Fredrik

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    Smoothness of functions between subsets of ##\mathbb R^n## is defined as "differentiable infinitely many times". When we define smoothness of functions between manifolds, we want the new kind of smoothness to be related to the old. Since a manifold is covered by the domains of a bunch of smoothly compatible coordinate systems (such that ##x\circ y^{-1}## is smooth for all coordinate systems x,y), it's very natural to define ##f:M\to N## to be smooth when ##x\circ f\circ y^{-1}## is smooth for all coordinate systems x on N and y on M.

    The chain rule can be written ##(f\circ g)_{,i}(x)=f_{,j}(g(x)) g^j{}_{,i}(x)## when ##f## is real-valued. If the domain of ##g## is a subset of ##\mathbb R##, we can simplify this to ##(f\circ g)'(x)=f_{,j}(g(x))(g^j)'(x)##.
     
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