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Question about theorem 2.2.1 in Wald's General Relativity

  1. Sep 9, 2012 #1
    Hi everyone, first of all I have been a lurker here for years and have benefited greatly from many of the discussions in the math and physics sections. Thanks, I have received a lot of helpful information from these forums!

    I have been working through Wald's General Relativity book and I am having trouble following the reasoning behind one part of a theorem. From page 15, the theorem and part of the proof is (For those who don't have the book):

    Let M be an n-dimensional manifold. Let [itex]p \in M[/itex] and let [itex]V_p[/itex] denote the tangent space at p. Then dim [itex]V_p=n[/itex]

    Proof We shall show that dim [itex]V_p=n[/itex] by constructing a basis of [itex]V_p[/itex], i.e. by finding n linearly independent tangent vectors that span [itex]V_p[/itex]. Let [itex]\psi : O \rightarrow U\subset R^n[/itex] be a chart with [itex]p\in O[/itex]. If [itex]f\in \mathfrak{F}[/itex], then by definition [itex]f\circ \psi^{-1}:U\rightarrow R[/itex] is [itex]C^{\infty}[/itex]. For [itex]\mu=1,...,n[/itex] define [itex]X_{\mu}:\mathfrak{F}\rightarrow R[/itex] by
    $$X_{\mu}(f)=\frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1})\bigg|_{\psi (p)}$$
    $$\vdots$$

    I can't seem to figure out how the term [itex]\frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1})\bigg|_{\psi (p)}[/itex] is a mapping from [itex]\mathfrak{F}\rightarrow R[/itex]. [itex]f\circ \psi^{-1}[/itex] was defined to be a mapping from [itex]U\rightarrow R[/itex]. In other words, I don't see why these last two terms should be equal. I think I am missing something obvious here. Is there maybe some sort of chain rule argument?

    Thanks, any pointer in the right direction would be greatly appreciated!
     
  2. jcsd
  3. Sep 10, 2012 #2

    haushofer

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    You have [itex]f o \ \psi^{-1}(\psi(p)) = f(p) [/itex].
     
    Last edited: Sep 10, 2012
  4. Sep 10, 2012 #3
    Wow, that was more embarrassingly obvious than I thought!:blushing:

    Thanks for the reply!
     
  5. Sep 10, 2012 #4

    haushofer

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    Don't worry. It is easy to drown in all those formalities ;)
     
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