# Proof of dimension of the tangent space

1. May 21, 2015

### hideelo

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. May 21, 2015

### micromass

Staff Emeritus
I'm sorry, I cannot read the picture attached. Can't you just type it?

3. May 21, 2015

### hideelo

It's hard to write them out on my phone, so I'll try another picture

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4. May 21, 2015

### hideelo

Is that better?

5. May 21, 2015

### martinbn

No, it has to be exacly as written. Otherwise 2.2.2. will not be true.

6. May 21, 2015

### hideelo

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.

7. May 22, 2015

### Fredrik

Staff Emeritus
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.

8. May 22, 2015

### Fredrik

Staff Emeritus
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).

9. May 28, 2015

### mathwonk

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.