Differential as generalized directional deriv (Munkres Analysis on Manifolds)

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

I will assume that [itex]Dg_j[/itex] is the gradient so that [itex]Dg_j \cdot v_j[/itex] is the directional derivative.

I'm not sure what you are and are not allowed to use, but if [itex]\omega = f \ dx_I[/itex] then you are correct that [itex]d\omega = df \wedge dx_I[/itex]. Thus for two vector fields [itex]v,w[/itex] we have that
[tex] \begin{align*}<br /> d \omega &= df \wedge dx_I(v,w) \\<br /> &= df(v) dx_I(w) - dx_I(v)df(w) \\<br /> &= w_i Df\cdot v - v_i Df\cdot w.<br /> \end{align*}[/tex]
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.
 
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