# Coordinate Basis in Wald's GR

1. Feb 18, 2013

### cpsinkule

In Wald's GR he makes use of a coordinate basis consisting of ∂/∂x$^{n}$ where n runs over the coordinates, and I understand his argument that ∂f/∂x$^{n}$ are tangent vectors, but I can't wrap my head around the operator ∂/x$^{n}$ spanning a tangent space of a manifold. Any clarification on this would be appreciated. He also states that dx$^{m}$∂/∂x$^{n}$=δ$^{m}$$_{n}$, but I don't see how this is true, either. I know fully well that for them to be a basis of the dual and vector space respectively they must satisfy that condition, but I don't see how he arrives at that. Is it just by definition that dx∂/∂x=δ?

2. Feb 19, 2013

### Fredrik

Staff Emeritus
$dx^\mu$ is a 1-form, i.e. a cotangent vector field, so it associates a cotangent vector $(dx^\mu)_p$ which each point p in its domain. This cotangent vector is such that
$$(dx^\mu)_p(v)=v(x^\mu)$$ for all tangent vectors v at p. This is just the definition of d. The right-hand side can be evaluated by expanding v in the basis $\left\{\left.\frac{\partial}{\partial x^\mu} \right|_p\right\}$.

If X is a vector field, $dx^\mu(X)$ is a map from a subset of the manifold into ℝ. For each p in that set, we have
$$dx^\mu(X)(p) =(dx^\mu)_p X_p=X_p(x^\mu).$$ The first equality holds just because $dx^\mu$ is a 1-form, and the second is the definition of d. What do you get when you insert $X=\frac{\partial}{\partial x^\mu}$ into this formula?

I think the best answer to the question of why the vector space spanned by $\left\{\left.\frac{\partial}{\partial x^\mu} \right|_p\right\}$ is called "the tangent space at p" is that this vector space is isomorphic to a vector space whose members can be interpreted as tangent vectors in a natural way. See this post.

A tip for the next time: If you're asking about something that looks like a textbook problem, post in one of the homework forums. If you're asking a general question like this, which doesn't look like a textbook problem, post in the appropriate math forum. In this case, that would be "topology & geometry".

3. Feb 19, 2013

### voko

Which part in theorem 2.2.1 do you not understand?

The other thing you are confused about is indeed by definition. See the remark in brackets at the end of page 21.