- #1

maverick280857

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

Carroll said:The canonical example of a one-form is the gradient of a function [itex]f[/itex], denoted [itex]df[/itex] as in (1.52). Its action on a vector [itex]d/d\lambda[/itex] is exactly the directional derivative of the function:

[tex]df\left(\frac{d}{d\lambda}\right) = \frac{df}{d\lambda}[/tex]

It's tempting to ask, "why shouldn't the function [itex]f[/itex] itself be considered the one-form, and [itex]df/d\lambda[/itex] its action?". The point is that a one-form, like a vector, exists only at the point it is defined, and does not depend on information at other points on M. If you know a function in some neighborhood of a point, you can take its derivative, but not just from knowing its value at the point; the gradient, on the other hand, encodes precisely the information necessary to take the directional derivative along any curve through [itex]p[/itex], fulfilling its role as a dual vector.

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

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

[tex]df = \frac{\partial f}{\partial x^\mu} dx^\mu[/tex]

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 [itex]\epsilon[/itex] neighborhood of any point where you take the derivative. But in the limit of such an [itex]\epsilon[/itex] 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?