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Calculating tangent spaces via curves.

  1. Oct 4, 2011 #1
    In my experience, whenever we want to calculate the tangent space to a smooth manifold, we usually proceed as follows.

    Let M be a smooth manifold and p in M. Let [itex] \gamma: \mathbb R \to M [/itex] be a smooth curve such that [itex] \gamma(0) = p [/itex] and [itex] \gamma'(0) = X [/itex]. We then use some defining quality of M to get a condition on X.

    For example, for the sphere [itex] S^2 = \{ x \in \mathbb R^3 : x\cdot x - 1 = 0\} [/itex] we can proceed as [itex] \left.\frac{d}{dt}\right|_{t=0} (\gamma(t) \cdot \gamma(t) -1) = 2 \gamma(0)\cdot \gamma'(0)= 2 p \cdot X = 0 [/itex] which implies that [itex] T_pS^2 = \{ X \in \mathbb R^3: p \cdot X = 0 \} [/itex] which is what we classically expect. Similarly, I've seen this technique used with all sorts of Lie groups to find Lie algebras.

    However, I'm not certain exactly why this technique is justified. It seems to make intuitive sense, but I cannot quite convince myself that this is a rigorous method.

    I think it might be related to the fact that if [itex] \gamma [/itex] is a smooth curve, then [itex] \gamma'(t_0) = \gamma_*\left( \left. \frac{d}{dt}\right|_{t=0} \right) [/itex] but I can't quite make the logical jump there. Does anybody have any insight?
     
  2. jcsd
  3. Oct 5, 2011 #2
    The functions that you talk of are often called "germs". You can form an equivalence relation on the set of smooth functions on the manifold by identifying any two which agree on some open set of the point p in question.

    I can't quite remember where you go from here, but there is a method of constructing the tangent space quite directly from these germs.

    [edit: I found this, which seems to explain it quite well:
    http://people.math.gatech.edu/~ghomi/LectureNotes/LectureNotes5G.pdf ]
     
    Last edited: Oct 5, 2011
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