- #1
CJ2116
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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!
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!
