# Definition of one-forms and their action

1. Mar 18, 2014

### maverick280857

Hi,

While reading Sean Carroll's book, I came across the following statement:

Okay so this has me confused. Perhaps I am nitpicking, but isn't $f$ a scalar function, i.e. a 0-form? So shouldn't he really be saying "why shouldn't $df$ be considered the one-form..."?

If f is a scalar, then df (as he defines it) is

$$df = \frac{\partial f}{\partial x^\mu} dx^\mu$$

the 1-form (one way of thinking of d is the exterior derivative of a 0 form, which produces a 1 form). Now, the components of df in the coordinate basis are indeed dependent on the $\epsilon$ neighborhood of any point where you take the derivative. But in the limit of such an $\epsilon$ going to 0, the derivative (as a 1-form) does become local.

However what is the problem if the 1 form has a limiting expression that does depend on the neighborhood?

Am I missing something?

2. Mar 19, 2014

### pervect

Staff Emeritus
I'm not following what Sean means here either, I'm afraid. My impression was that you had a vector space and it generated naturally a dual space, but which one you called the vector space and which one you called the dual space was somewhat a matter of convention.

However, it's definitely conventional to call the tangent space the vector space.

Once you have defined the tangent space as the vector space, the dual of the vector space is a map from the vector space to a scalar. It turns out that the dual vector is also a one-form, as an n-form is an anti-symmetric map from n vectors to a scalar.

3. Mar 19, 2014

### maverick280857

Thanks for the reply pervect. I am not sure I understand the distinction he's trying to put forth though about the alleged locality of the one-form as opposed to that of a derivative. If I consider the exterior derivative of a 0 form, I generate a 1 form, which is (very loosely speaking) similar to a 'differential'. So why does a 1 form have to be dependent only on the value at that point?