# Confused about representation of an inertial frame by a one-form

I'm self-studying general relativity and I'm using Schutz's text. A lot is starting to click but some things are pretty confusing, especially this part which seems pretty important.

There's a Section 3.3 talking about notation for derivatives in which is given: $x^\alpha_{,\beta}\equiv \delta^\beta_\alpha$. Fine so far. Then by comparison with an earlier equation given we have this implying: $$\tilde{\mathrm{d}}x^\alpha := \tilde{\omega}^\alpha.\tag{3.20}$$

My understanding is that $\tilde{\omega}$ represents the one-form basis. So, is $\tilde{\mathrm{d}}x^\alpha$ just notation defining now the $\alpha$-th basis of the one-form? Why the new notation (if I'm even right about that).

Then he says we can use this to write, for any function $f$: $\tilde{\mathrm{d}}f = {\partial f \over\partial x^\alpha}\tilde{\mathrm{d}}x^\alpha.$

warning not to confuse $\tilde{\mathrm{d}}f$ with a differential, as it's a tensor. Unfortunately, I'm very confused by this. Is this because $f$ is a function (one-form) which acts on $\tilde{\mathrm{d}}x^\alpha$? What really is the meaning of $\tilde{\mathrm{d}}f$?

This comes up again in the next chapter when introducing 4-vectors with special fluids. It's shown that an inertial frame can be represented by a one-form, e.g. $E = \langle\tilde{\mathrm{d}}t, \vec{p}\rangle = p^0$ This is the inner product of $\tilde{\mathrm{d}}t$ and $\vec{p}$ correct? Again though, what is the meaning of $\tilde{\mathrm{d}}t$?

First of all, the reader is probably thinking: What's with all the tildes? Well, Schutz in Section 3.3 adopts the notation that $\raise.8ex\tilde{}$ above a symbol denotes a one-form, just as $\vec{}$ above a symbol denotes a vector.
What really is the meaning of $\tilde{\mathrm{d}}f$?
The one-form/co-vector $\tilde{\mathrm{d}}f$ acts on vectors $\vec{A}~=~A^{\mu}e_{\mu}$, which are identified with first order differential operators $\vec{A}~=~A^{\mu}\partial_{\mu}$, which in turn act on functions $f$, so $$\langle \tilde{\mathrm{d}}f, \vec{A}\rangle~:=~\vec{A}[f]~=~A^{\mu}\partial_{\mu}f.$$