TITLE: Basis one-form and basis vector confusion
QUESTION [6 upvotes]: Still trying to teach myself some basic differential geometry in relation to general relativity. I've read that, in relation to basis vectors $e_{\nu}=\partial_{\nu}$ and basis one-forms $\omega^{\mu}=dx^{\mu}$,
  then $$e_{\nu}\omega^{\mu}=\partial_{\nu}dx^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\nu}}=\delta_{\nu}^{\mu}.$$
 Fair enough, but why does the partial derivative operator $\partial_{\nu}$
  (which I thought meant $\frac{\partial}{\partial x^{\nu}}$
 , where the slot is for an arbitrary function) when applied to $dx^{\mu}$
  give $\frac{\partial x^{\mu}}{\partial x^{\nu}}$
  and not $\frac{\partial\left(dx^{\mu}\right)}{\partial x^{\nu}}$? Or does $\frac{\partial x^{\mu}}{\partial x^{\nu}}=\frac{\partial\left(dx^{\mu}\right)}{\partial x^{\nu}}$? But if that's the case what does $\partial_{\nu}x^{\mu}$ give? It can't also be correct that
$$\partial_{\nu}x^{\mu}=\frac{\partial x^{\mu}}{\partial x^{\nu}}=\delta_{\nu}^{\mu},$$
 but I can't see why not. I'm confused. Thanks.

REPLY [4 votes]: I found this question because I was having the same problem. I had to figure it out and I believe i did. Way too late for the original questioner, but I respond now for people with the same question. I think this is the solution:
If you were reading somebody's book who uses the same notation as Carroll's book on GR, you probably saw three different d's. Carroll uses $d$ for differentials, $\partial{}$ for partial derivatives and the letter d for a one-form. Note the italics $d$ versus regular d. Schutz uses the ~ with the d and writes $\tilde{\text{d}}$ for a one form, and I like that better.
Now, what I think you saw was  d$x_i (\frac{\partial{}}{dx^i}) =\delta^j_i\;$ and what it meant was $\;\tilde{\text{d}}x_i (\frac{\partial{}}{dx^i}) =\delta^j_i$ and not $\;dx_i (\frac{\partial{}}{dx^i}) =\delta^j_i$. That is, what was being said is that a one-form basis takes the vector basis and produces $\delta$.