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Proof of dimension of the tangent space

  1. May 21, 2015 #1
    I am attaching a picture of a proof from the book "general relativity" by wald. This is supposed to show that the tangent space of an n dimensional manifold is also n dimensional. I have two questions.

    In equation 2.2.3 couldn't the function be anything at a since the (x-a) term is 0?

    How is the equality in equation 2.2.5 justified, I'm just not seeing it
     

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  2. jcsd
  3. May 21, 2015 #2

    micromass

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    I'm sorry, I cannot read the picture attached. Can't you just type it?
     
  4. May 21, 2015 #3
    It's hard to write them out on my phone, so I'll try another picture
     

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  5. May 21, 2015 #4
    Is that better?
     
  6. May 21, 2015 #5

    martinbn

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    No, it has to be exacly as written. Otherwise 2.2.2. will not be true.
     
  7. May 21, 2015 #6
    I'm assuming 2.2.2 follows from the gradient theorem, in which case 2.2.3 should really be an integral, otherwise it's not an equality but the linear approximation.
     
  8. May 22, 2015 #7

    Fredrik

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    A proof very similar to Wald's is included in Isham. Start reading on page 79. The proof is on pages 82-84.

    Last time I discussed this proof with someone, I posted my version of it. It's in post #14 here.
     
  9. May 22, 2015 #8

    Fredrik

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    The theorem says that the function is smooth, i.e. differentiable an arbitrary number of times. Differentiable functions are continuous. The value of a continuous function at a specific point in its domain is completely determined by its values in the rest of its domain. For example, if ##f:\mathbb R\to\mathbb R## is continuous then ##f(0)=\lim_{n\to\infty}f\big(\frac 1 n\big)##.

    Assuming that you meant the last equality, this is how:

    First term: f(p) is a constant function (the map that takes an arbitrary q to f(p)), so v(f(p))=0.

    Second term: The sum is of the form ##\sum_\mu (A_\mu-A_\mu)B_\mu## so every term is zero.

    Third term: For all smooth functions f and all constant functions g, we have v(f-g)=v(f)-v(g)=v(f).
     
  10. May 28, 2015 #9

    mathwonk

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    an n manifold is by definition locally isomorphic (i.e. diffeomorphic)) to R^n, hence (by chain rule) each tangent space is isomorphic to the tangent space to R^n at the image point. Do you know that R^n is the tangent space to R^n at each point? That would do it.
     
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