# $d\phi = \sum \frac{\partial \phi}{\partial x_i}dx_i.$

Just a work out for a very tautological question that I am very uncertain about.

If $\phi: X \to \mathbb{R}$ is a smooth function, $d\phi_x: T_x(X) \to \mathbb{R}$ is a linear map at each point $x$. Thus $x \to d\phi_x$ defines a $1$-form $d\phi$ on $X$, called the differential of $\phi$.

The coordinate functions $x_1, \dots, x_k$ on $\mathbb{R}^k$ yield $1$-form $dx_1, \dots, dx_k$ on $\mathbb{R}^k$. Check $dx_1, \dots, dx_k$ have the specific action $1$-forms of coordinate function $dx_i(z)(a_1, \dots, a_k) = a_i$. Show that if $\phi$ is a smooth function on $\mathbb{R}^k$, then $$d\phi = \sum \frac{\partial \phi}{\partial x_i}dx_i.$$

So I am asked to show two function are the same. So the general method would be show they have the same domain, and for each element in the domain, they are sent to the same place.

$d \phi$ is defined by $$d\phi_x: T_x(X) \to \mathbb{R}.$$

And according to $1$-forms of coordinate function $dx_i(z)(a_1, \dots, a_k) = a_i$. $dx_i$s are coordinate functions on the tangent space $T_x{X}$. Hence $\forall v \in T_x(X)$, $$\sum \frac{\partial \phi}{\partial x_i}dx_i(v) =\Big( \frac{\partial \phi}{\partial x_1}dx_1 + \cdots + \frac{\partial \phi}{\partial x_k}dx_k\Big)(v).$$

Coordinate functions are certainly linear, so $$\sum \frac{\partial \phi}{\partial x_i}dx_i(v) = \frac{\partial \phi}{\partial x_1}dx_1 (v)+ \cdots + \frac{\partial \phi}{\partial x_k}dx_k(v) = \frac{\partial \phi}{\partial x_1} v_1+ \cdots + \frac{\partial \phi}{\partial x_k}v_k.$$

And that's it - granting this is exactly what $d\phi_x$ is.

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There must be a coordinate-free definition of the map $d\phi_x \colon T_x(X) \to \mathbb{R}$. Otherwise, it would be a pointless exercise. – Daniel Fischer Jul 29 '13 at 20:58
There's just one key ingredient you're missing here - you have not yet defined $d\phi_x$. You've written $d\phi_x: T_x(X) \rightarrow \mathbb{R}$, but of course that doesn't actually tell you what the values of the function are. – Christopher A. Wong Jul 29 '13 at 21:02
Thanks @ChristopherA.Wong. The problem is, I didn't see $d\phi_x$ defined on Guillemin and Pollack's Differential Topology. I cited exactly what is given. Perhaps the fact that $d\phi_x: T_x(X) \to \mathbb{R}$ is a linear map at each point $x$ gives me $d\phi_x = \frac{\partial \phi}{\partial x_1} v_1+ \cdots + \frac{\partial \phi}{\partial x_k}v_k$? – WishingFish Jul 29 '13 at 21:11
Hi @DanielFischer, I still couldn't get the point... But I updated some related definitions... – WishingFish Jul 29 '13 at 21:13
$d\phi_x(v)$ is the derivative of $\phi$ (along the manifold) in the direction of $v$, at the point $x$. – Christopher A. Wong Jul 29 '13 at 21:30

You are almost there. All you have to do, as commenters have pointed out, is use the coordinate-free definition of $d\phi$.
Since $d\phi_p(v) = v_p(\phi)$, and in coordinates the vector field $v_p = \sum_i v_i(p)\partial_i$, we have that \begin{align*}d\phi_p(v) &= v_p(\phi) \\ &= \sum_i v_i(p)\partial_i\phi \\ &= \sum_i v_i(p) \frac{\partial \phi}{\partial x^i} \\ &= \sum_i \frac{\partial \phi}{\partial x^i} dx_i(v)(p) \end{align*} where in that last equality we have used that the coordinate 1-form $dx_i$ returns the $i^{th}$ coordinate of a vector field $v$.
Thanks Neal, I guess this is what I really suppose to know, but could you explain why $d\phi_p(v) = v_p(\phi)$...? – WishingFish Jul 29 '13 at 22:28
@WishingFish That's the (a) coordinate-free definition of $d\phi$. – Neal Jul 29 '13 at 23:06