Question about theorem 2.2.1 in Wald's General Relativity

[CJ2116]

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!

 

[haushofer]

You have [itex]f o \ \psi^{-1}(\psi(p)) = f(p) [/itex].

 

[CJ2116]

Wow, that was more embarrassingly obvious than I thought!

Thanks for the reply!

 

[haushofer]

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

 

[sardar]

Dear fellow scientist,

First of all, I'm glad to hear that you have found helpful information from this forum. We are always happy to assist and share knowledge with others.

Regarding your question about theorem 2.2.1 in Wald's General Relativity, I believe the confusion lies in the notation used. Let's break down the notation and see if that helps clarify things.

First, in the statement of the theorem, we are dealing with a manifold M and a point p \in M. The tangent space at p, denoted V_p, is the set of all tangent vectors at p. So, V_p is a vector space.

Next, we have the function \psi : O \rightarrow U \subset R^n, which is a chart that maps an open set O in the manifold M to an open set U in the Euclidean space R^n. This means that \psi is a coordinate system on M, and \psi(p) is the coordinate of the point p in this system.

Now, we come to the definition of X_{\mu}:\mathfrak{F}\rightarrow R. This notation means that X_{\mu} is a linear map from the space of smooth functions on the manifold, denoted by \mathfrak{F}, to the real numbers, denoted by R. In other words, X_{\mu} takes a smooth function f and maps it to a real number.

To understand the expression \frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1})\bigg|_{\psi (p)}, we need to look at the chain rule. The function f\circ \psi^{-1} is a composition of two functions - f and \psi^{-1}. The chain rule tells us that the derivative of this composition is given by the product of the derivatives of the individual functions. So, in this case, we have:

\frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1}) = \frac{\partial f}{\partial x^{\mu}}\circ \psi^{-1} \cdot \frac{\partial \psi^{-1}}{\partial x^{\mu}}

Now, evaluating this expression at the point \psi(p) gives us:

\frac{\partial}{\partial x^{\mu}}(f\circ \psi^{-1})\bigg|_{\psi (p)}