Basis one-form and basis vector confusion 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.
 A: It is only a matter of notation. There are a few different ways to look at things, the following is one of those ways.
Take a coordinate chart $\varphi:U\to\mathbb{R}^n$. Then $\varphi$ consists of $n$ real valued functions which we write as $x^1,\ldots,x^n$. The derivative of the $i$th function is $dx^i$, and this is how we get the local frame of the cotangent bundle $dx^1,\ldots,dx^n$.
On the other hand, the chart $\varphi$ also induces a local frame of the tangent bundle, which we write as $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}.$
Now, there are two standard notations for the derivative of $f$ in the direction $X$ - one of them is $X(f)$ and the other is $df(X)$. Using the former, we can write$$\frac{\partial}{\partial x^i}(x^j)=\delta_i^j,$$whereas the latter yields$$dx^j\left(\frac{\partial}{\partial x^i}\right)=\delta_i^j.$$Both equations mean the same.
A: 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 $\delta$.
A: I like Amitai Yuval's answer, but let me try to say something a little more basic.  It's necessarily a bit vague, but maybe it will give you the idea of what's going on.
The coordinates $x^1, \dots, x^n$ define functions on your manifold.  In a physics context, $x^1$, $x^2$ and $x^3$ may represent position with respect to a particular set of axes whilst $x^4$ labels time.  Tangent vectors are little arrows at points in your space, and we have a nice basis at each point given by the arrows pointing along each of the coordinate directions.  In our example these would correspond to arrows in the directions of the $x^1$-, $x^2$- and $x^3$-axes, and a fourth arrow which points forwards in time but has no spatial component.  We write these vectors as $\partial/\partial x^a$ or $\partial_a$, but this is just notation for now.
A one-form is 'a thing you plug vectors into': you feed it a vector and it spits out a number which depends linearly on the input.  For each $a$ we can define a one-form $\mathrm{d}x^a$ to be the one-form which gives $1$ when you input $\partial_a$ and $0$ when you input $\partial_b$ for $b \neq a$.  Again this is just notation.  In other words, the statement $\mathrm{d}x^a(\partial_b)=\delta^a_b$ is basically a tautology.  Back to our example, if you plug a vector $v$ into $\mathrm{d}x^4$ then what you get out is the time component of $v$.
Now, given a vector $v$ at a point $p$ and a function $f$ defined near $p$ you can take the directional derivative of $f$ in the direction $v$, which you could write as $v(f)$ (although personally I don't like this notation).  If you take $v$ to be $\partial_a$ then this directional derivative really just is the partial derivative $\partial f/\partial x^a$.  This is why the notation $\partial_a$ or $\partial/\partial x^a$ is sensible.  So in our example $f$ could be something like density and then $\partial_4(f)$ represents the time-derivative of the density.
Thinking of the vector $v$ in the expression $v(f)$ as variable, we get a thing which takes vectors and spits out numbers (given the vector $v$ it outputs $v(f)$).  In other words it defines a one-form: the differential (or derivative) of $f$, which we write as $\mathrm{d}f$.  Happily, the one-form we defined to be $\mathrm{d}x^a$ purely as a piece of notation really just is the differential of the function $x^a$.
So in terms of directional derivatives it is true that $\partial_a(x^b)=\delta^b_a$.  Whilst in terms of feeding vectors into one-forms it is true that $\mathrm{d}x^a(\partial_b)=\delta^a_b$.  By the above comment about differentials, relating directional derivatives to one-forms, these statements are really saying the same thing.
The expression you write as $\partial_\nu \mathrm{d}x^\mu$ is a bit ambiguous.  It definitely doesn't mean 'differentiate $\mathrm{d}x^\mu$'.  The sensible interpretation (which makes it equal to $\delta^\mu_\nu$) is that you're feeding the one-form into the vector using double-duality.  But this is probably a little too abstract at present.
