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Chapter 7: Problem 56

Show that the differential of a differential is equal to zero: \(d d=0\).

Short Answer

Expert verified
Question: Prove that the second derivative of any differential form is equal to zero. Answer: The second derivative of a differential form refers to the result of applying the exterior derivative operation twice, denoted as \(d(d\omega)\). By examining the properties of the exterior derivative and the wedge product, we demonstrated that for a scalar function, the second derivative is equal to zero. This result can be generalized to any differential form by using similar techniques, ultimately showing that \(d^2 \omega = d(d\omega) = 0\).

Step by step solution

01

Introduce the exterior derivative

The exterior derivative is an operation that, when applied to a differential form, produces a new differential form of higher degree. The exterior derivative of some differential form, say \(\omega\), is denoted as \(d\omega\).
02

Establish the properties of the exterior derivative

The exterior derivative has the following properties: 1. Linearity: \(d(\alpha\omega + \beta\eta) = \alpha d\omega + \beta d\eta\) 2. Leibniz Rule: \(d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^k \omega \wedge d\eta\) for \(\omega\)- a k-form and \(\eta\)- an l-form 3. \(d(d\omega) = 0\) (this is the property we want to prove)
03

Consider a 0-form example

Let \(\omega\) be a 0-form (a scalar function \(f\)). Then, the first exterior derivative, \(df\), is a 1-form, and its second exterior derivative, \(d(d f)\), is a 2-form.
04

Compute the second derivative of the example

In this case, we have \(d(d f) = d \left(\frac{\partial f}{\partial x_i} dx^i\right)\). By applying the Leibniz Rule, we have: \(d(d f) = \frac{\partial^2 f}{\partial x_j \partial x_i} dx^j \wedge dx^i\). Since the order of the indices matters in the wedge product (i.e., \(dx^j \wedge dx^i = - dx^i \wedge dx^j\)), we can see that the term with the second derivatives of \(f\) cancel each other out: \(d(d f) = \frac{\partial^2 f}{\partial x_j \partial x_i} dx^j \wedge dx^i - \frac{\partial^2 f}{\partial x_j \partial x_i} dx^i \wedge dx^j = 0\). By showing that the second derivative \(d^2 f\) is equal to 0 for a scalar function, we have demonstrated one case of the property \(d(d\omega) = 0\).
05

Extend the result to general differential forms

We showed that the second derivative of a scalar function is zero. This concept can be generalized for any differential form through a more detailed analysis of the properties of the exterior derivative and the wedge product. Using the techniques used in this example, the result can be extended to show that the second derivative of any differential form is indeed equal to zero: \(d d\omega = 0\).

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Most popular questions from this chapter

Show that the product of monomials is associative: $$ \left(\omega^{k} \times \omega^{r}\right) \wedge \omega^{m}-\omega^{2} \times\left(\omega^{2} \times \omega \omega^{m}\right) $$ and skew-commutative: $$ \omega^{k} \wedge \omega^{l}=(-1)^{k} \omega^{l} \wedge \omega^{k} . $$

Prove the formulas for differentiating a sum and a product: $$ d\left(\omega_{1}+\omega_{2}\right)=d \omega_{1}+d \omega_{2} . $$ and $$ d\left(\omega^{k} \wedge \omega^{\prime}\right)=d \omega^{k} \wedge \omega^{\prime}+(-1)^{k} \omega^{k} \wedge d \omega^{l} . $$

Show that, under the isomorphisms established above, the exterior product of 1 -forms becomes the vector product in \(\mathbb{R}^{3}\), i.e., that $$ \omega_{\mathbf{A}}^{1} \wedge \omega_{\mathrm{B}}^{\mathrm{H}}=\omega_{\left|A_{,} \mathbf{B}\right|}^{2} \text { for any } \mathbf{A}, \mathbf{B} \in \mathbb{R}^{3} \text {. } $$ In this way the exterior product of 1 -forms can be considered as an extension of the vector product in \(\mathbb{R}^{3}\) to higher dimensions. However, in the \(n\)-dimensional case, the product is not a vector in the same space; the space of 2-forms on \(R^{n}\) is isomorphic to \(R^{n}\) only for \(n=3\).

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 \(\omega_{1} \wedge \cdots \wedge \omega_{k}\) is a \(k\)-form.
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