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Chapter 7: Problem 12

Show that the forms $$ x_{i_{1}} \wedge \cdots \wedge x_{i_{k}} \text {, where } 1 \leq i_{1}

Short Answer

Expert verified
Question: Prove that the given \(k\)-forms are linearly independent. Answer: We showed that assuming the basic \(k\)-forms as linearly dependent led to a contradiction when wedging the sum with \(x_n\). Therefore, the given basic \(k\)-forms must be linearly independent.

Step by step solution

01

Familiarize with wedge products and basic \(k\)-forms

Recall that the wedge product (\(\wedge\)) is a binary operation on the space of differential forms. It is anti-symmetric and associative, meaning: - \(x_i \wedge x_j = - x_j \wedge x_i\) for all \(i,j\) - \(x_i \wedge (x_j \wedge x_k) = (x_i \wedge x_j) \wedge x_k\) for all \(i,j,k\) The given exercise deals with basic \(k\)-forms which are of the form $$ x_{i_{1}} \wedge \cdots \wedge x_{i_{k}} \text {, where } 1 \leq i_{1}<i_{2}<\cdots<i_{A} \leq n. $$
02

Assume they are linearly dependent and set up a linear combination

Suppose that these basic \(k\)-forms are linearly dependent. If so, there must exist non-zero coefficients \(c_{i_1,\ldots, i_k}\) such that: $$\sum_{1\leq i_1<\cdots<i_k\leq n} c_{i_1,\ldots, i_k} x_{i_1} \wedge \cdots \wedge x_{i_k} = 0$$
03

Take a set of indices \((i_1^*,\ldots,i_k^*)\) as an example

Without loss of generality, consider the sets of indices with the highest \(i_k^*\). That is, \(i_k^* = n\). Now, let's consider wedging this equation with \(x_n\) to exploit antisymmetry: $$\sum_{1\leq i_1<\cdots<i_k\leq n} c_{i_1,\ldots, i_k} x_{i_1} \wedge \cdots \wedge x_{i_k} \wedge x_n = 0 \wedge x_n = 0$$
04

Consider the term with the selected set of indices and others

In this sum, the terms can be split into two types: one with \(i_k'= n\) (our chosen highest index) and others with \(i_k'< n\). For the term with \(i_k'=n\) and arbitrary \(i_j'\)s, we have: $$c_{i_1^*,\ldots, i_{k-1}^*,n}(x_{i_1^*} \wedge \cdots \wedge x_{i_{k-1}^*} \wedge x_n) \wedge x_n = 0$$ since \(x_n \wedge x_n = 0\) due to the antisymmetric property. For terms with \(i_k'<n\), we have: $$(x_{i_1'} \wedge \cdots \wedge x_{i_k'})\wedge x_n = - (x_{i_1'} \wedge \cdots \wedge x_{i_k'}\wedge x_n \wedge x_n) = x_{i_1'} \wedge \cdots \wedge x_{i_k'} \wedge x_n$$ since again, \(x_n \wedge x_n = 0\).
05

Identify a contradiction

Following from Step 4, the entire equation becomes: $$\sum_{1\leq i_1<\cdots<i_k\leq n} c_{i_1,\ldots, i_k} x_{i_1} \wedge \cdots \wedge x_{i_k} \wedge x_n = c_{i_1^*,\ldots,i_{k-1}^*,n} (x_{i_1^*} \wedge \cdots \wedge x_{i_{k-1}^*} \wedge x_n) \wedge x_n + \sum_{1\leq i_1<\cdots<i_{k-1}<i_k<n} c_{i_1,\ldots, i_k} (x_{i_1} \wedge \cdots \wedge x_{i_k} \wedge x_n) = 0$$ However, the expression on the left is linearly dependent on the highest index \(x_n\), while the expression on the right doesn't have any terms containing \(x_n\). This is a contradiction, as there is no combination of coefficients \(c_{i_1,\ldots, i_k}\) that can make the equation hold.
06

Conclude that basic \(k\)-forms are linearly independent

Since our assumption of linear dependence led to a contradiction, the set of basic \(k\)-forms must be linearly independent.

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Most popular questions from this chapter

Show that the boundary of the boundary of any chain is zero: \(\partial \partial c_{k}=0\).

Let \(D_{1}\) and \(D_{2}\) be two compact, convex polyhedra in the oriented \(k\)-dimensional space \(R^{k}\) and \(f: D_{1} \rightarrow D_{2}\) a differentiable map which is an orientation-preserving diffeomorphism \({ }^{55}\) of the interior of \(D_{1}\) onto the interior of \(D_{2}\). Then, for any differential \(k\)-form \(\omega^{k}\) on \(D_{2}\). $$ \int_{D_{1}} f^{*} \omega^{k}=\int_{D_{2}} \omega^{k} $$

Calculate the value of the forms \(\omega_{1}=d x_{2} \wedge d x_{3}, \omega_{2}=x_{1} d x_{3} \wedge d x_{2}\), and \(\omega_{3}=d x_{3} \wedge d r^{2}\left(r^{2}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)\), on the pair of vectors \(\xi=(1,1,1), \eta=(1,2,3)\) at the point \(\mathbf{x}=(2,0,0)\).

Show that the operation of exterior product of I-forms gives a multi-linear skewsymmetric mapping $$ \left(\omega_{1}, \ldots, \omega_{k}\right) \rightarrow \omega_{1} \wedge \ldots \wedge \omega_{k} $$ In other words, $$ \left(\lambda^{\prime} \omega_{1}^{\prime}+\lambda^{\prime \prime} \omega_{1}^{\prime}\right) \wedge \omega_{2} \wedge \cdots \wedge \omega_{k}=\lambda^{\prime} \omega_{1}^{\prime} \wedge \omega_{2} \wedge \cdots \wedge \omega_{k}+\lambda^{\prime \prime} \omega_{1}^{n} \wedge \omega_{2} \wedge \cdots \wedge \omega_{k} $$ and $$ \omega_{i_{1}} \wedge \cdots \wedge \omega_{l k}=(-1)^{\prime} \omega_{1} \wedge \cdots \wedge \omega_{k,} $$ where $$ v= \begin{cases}0 & \text { if the permutation } i_{1}, \ldots, i_{k} \text { is even, } \\ 1 & \text { if the permutation } i_{1}, \ldots, i_{k} \text { is odd. }\end{cases} $$

. Show that the exterior square of a 1-form, or, in general, of a form of odd order, is equal to zero: \(\omega^{k} \wedge \omega^{k}=0\) if \(k\) is odd.
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