Differential as generalized directional deriv (Munkres Analysis on Manifolds)

[mathmonkey]

Homework Statement

Let $A$ be open in $\mathbb{R}^n$; let $\omega$ be a k-1 form in $A$. Given $v_1,...,v_k \in \mathbb{R}^n$, define
$h(x) = d\omega(x)((x;v_1),...,(x;v_k)),$
$g_j(x) = \omega (x)((x;v_1),...,\widehat{(x;v_j)},...,(x;v_k)),$
where $\hat{a}$ means that the component $a$ is to be omitted.

Prove that $h(x) = \sum _{j=1}^k (-1)^{j-1} Dg_j (x) \cdot v_j .$

Homework Equations

The problem is broken into 3 parts:
(a) Let $X = \begin{bmatrix} v_1 ... v_k \end{bmatrix}$. For each $j$ let $Y_j = \begin{bmatrix}v_1 ... \hat{v}_j ... v_k \end{bmatrix}$. Given $(i, i_1,...,i_{k-1})$, show that

$detX(i,i_1,...,i_{k-1}) = \sum _{j=1}^k (-1)^{j-1}v_{ij}detY_j(i_1,...,i_{k-1}).$
(b) Verify the theorem in the case $\omega = fdx_I$.
(c) Complete the proof.

The Attempt at a Solution

I'm stuck on part (b), however. By the definition given in the text, if $\omega = fdx_I$ then $d\omega = df \wedge dx_I$. I'm not quite sure how to link the result of part (a) to prove part (b). If anyone can shed any light on this problem I'd be really grateful! Thanks.

 

[Kreizhn]

Sorry, but I am not familiar with your notation. What is $(x;v_i )$? Also, what is $v_{ij}$? The $j^{th}$ component of $v_i$? What is $Dg_j$? The Jacobian? The gradient (which is possible under the identification of $T_p^* \mathbb R^n \cong \mathbb R^n$? What is $Dg_j \cdot v$? Is this the standard Euclidean product? Is $I$ a multi-index or a typo?

I will assume that $Dg_j$ is the gradient so that $Dg_j \cdot v_j$ is the directional derivative.

I'm not sure what you are and are not allowed to use, but if $\omega = f \ dx_I$ then you are correct that $d\omega = df \wedge dx_I$. Thus for two vector fields $v,w$ we have that
$$\begin{align*} d \omega &= df \wedge dx_I(v,w) \\ &= df(v) dx_I(w) - dx_I(v)df(w) \\ &= w_i Df\cdot v - v_i Df\cdot w. \end{align*}$$
With the second equality occurring by definition of the wedge product. Appropriate substitution of your vectors yields the desired equality. Perhaps induction will now work?

Edit: Had to do some craziness with a misplaced tex wrapper.