Dual vector is the covariant derivative of a scalar?

[NSH]

Homework Statement

In Wald's text on General Relativity he makes an assertion that I'm not sure why it is allowed mathematically. Here's the basic setup:

Let $\omega_{b}$ be a dual vector, $\nabla_{b}$ and $\tilde{\nabla}_{b}$ be two covariant derivatives and $f\in\mathscr{F}$. Then we may let $\omega_{b}=\nabla_{b}f=\tilde{\nabla}_{b}f$

This is in chapter 3 on curvature between equations 3.1.7 and 3.1.8...

Homework Equations

If it is relevant he is using this assertion to show:

$\nabla_{a}\omega_{b}=\tilde{\nabla}_{a}\omega_{b}-C^{c}_{ab}\nabla_{c}f$

implies
$\nabla_{a}\nabla_{b}f=\tilde{\nabla}_{a}\tilde{ \nabla}_{b}f-C^{c}_{ab}\nabla_{c}f$

The Attempt at a Solution

I know how to plug in his assertion I just don't get why the heck it is allowed? I've tried reading up on dual vector spaces but I haven't found what I'm looking for... any help would be appreciated.

 

[mighty2000]

Thank you for bringing this up. I understand your confusion about Wald's assertion and I will try my best to explain it to you.

Firstly, let's define what a dual vector is. A dual vector is a linear functional that maps a vector to a scalar. In other words, it takes in a vector and gives out a number. In contrast, a vector takes in a scalar and gives out a vector. In the context of General Relativity, dual vectors are used to describe the covariant derivative of a vector field.

Now, let's look at the assertion in question. Wald is saying that the dual vector ωb can be equated to the covariant derivative of the function f with respect to two different covariant derivatives, namely, \nabla_{b} and \tilde{\nabla}_{b}. This may seem strange at first, but it is allowed because both \nabla_{b} and \tilde{\nabla}_{b} are acting on the same function f. In other words, the dual vector ωb is defined by the covariant derivative of the function f, and since both \nabla_{b} and \tilde{\nabla}_{b} are acting on the same function, they will give the same result.

To understand this better, let's look at an example. Consider the function f(x) = x^2. The covariant derivative of this function with respect to the coordinate x is simply 2x. Now, let's say we use a different coordinate y, then the covariant derivative of f with respect to y is also 2x. This is because both coordinates are acting on the same function f(x) = x^2. In this case, we can say that the dual vector ωb is equal to both \nabla_{b}f and \tilde{\nabla}_{b}f.

I hope this explanation helps to clarify why Wald's assertion is allowed mathematically. If you have any further questions, please don't hesitate to ask. Keep up the good work with your studies!