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Tangent vector to curve - notational confusion.

  1. May 5, 2014 #1
    Given a curve ##\gamma: I \to M## where ##I\subset \mathbb{R}## and ##M## is a manifold, the tangent vector to the curve at ##\gamma(0) = p \in M## is defined in some modern differential geomtery texts to be the differential operator
    $$V_{\gamma(0)}= \gamma_* \left(\frac{d}{dt}\right)_{t=0}.$$
    However, quite often when I read differential geometry texts I encounter expressions like
    $$V_{\gamma(0)}= \left(\frac{d}{dt}\right)_{t=0} \gamma(t).$$

    What is the meaning of the latter expression, and what is the relation between the two?

    The only relation I can think of is that in a coordinate chart ##(U,\phi)## on ##M## and for ##f \in C(M)## we have
    $$V_{\gamma(0)}f= \gamma_* \left(\frac{d}{dt}\right)_{t=0}f = \left(\frac{d}{dt}\right)_{t=0} f\circ \gamma(t) = \left(\frac{d}{dt}\right)_{t=0} f\circ \phi^{-1} \circ (\phi \circ \gamma(t)) = \frac{dx^\mu}{dt}(\gamma(0)) \left(\frac{\partial}{\partial x^\mu}\right)_{x^\mu(\gamma(0))} f\circ \phi^{-1}$$
    where I have defined ##\phi\circ \gamma(t) = x^\mu(\gamma(t))##. Is the latter expression only meant to be a shorthand for
    $$V_{\gamma(0)} = \frac{dx^\mu}{dt}(\gamma(0)) \left(\phi^{-1}_*\frac{\partial}{\partial x^\mu}\right)_{\gamma(0)}?$$
  2. jcsd
  3. May 6, 2014 #2


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    I think the books are just making different identifications with the tangent vector. Any 1-1 isomorphism between a linear vector space will do. The first expression is when the books identify a tangent vector as a direction derivative operator on smooth functions on the manifold. One can easily prove that the directional derivative operators form a linear vector space over the real (complex) numbers.

    The second one is identifying the tangent vector with a literal tangent to a curve. This is basically identifying the vector with the actual n-tuple of numbers this derivative will produce once you have an arbitrary chart defined around ##\gamma(0)##. Since the numbers themselves depend on the chart, but the actual tangency conditions do not, then one identifies a tangent vector with an equivalence class of these such tangents (if the derivatives of two different curves are equal at point ##\gamma(0)## with respect to one chart, they will be equal with respect to all well defined charts there).

    One can also identify the tangent vector as an equivalence class of curves even. As long as one can make a 1-1 isomorphism, and prove that the underlying class of objects behaves as linear vectors over the real (complex) numbers, one is free to make any identification one wishes.
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