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Chapter 7: Problem 45

Show that the integral depends linearly on the form: $$ \int_{\sigma} \lambda_{1} \omega_{1}+\lambda_{2} \omega_{2}=\lambda_{1} \int_{\sigma} \omega_{1}+\lambda_{2} \int_{\theta} \omega_{2} . $$

Short Answer

Expert verified
Question: Show that the integral of a linear combination of two differential forms \(\omega_{1}\) and \(\omega_{2}\) with scalar constants \(\lambda_{1}\) and \(\lambda_{2}\) depends linearly on the form. Answer: To show this, we found that the integral of the linear combination of the forms, \(\int_{\sigma} \lambda_{1} \omega_{1}+\lambda_{2} \omega_{2}\), is equal to the sum of the scaled integrals of each form, \(\lambda_{1} \int_{\sigma} \omega_{1} + \lambda_{2} \int_{\sigma} \omega_{2}\), confirming the linearity of the integral for these forms.

Step by step solution

01

Integrate the linear combination

Begin by finding the integral of the linear combination of the forms: $$ \int_{\sigma} \lambda_{1} \omega_{1}+\lambda_{2} \omega_{2} $$
02

Use the linearity property of integration

We know that integration is a linear operation. Therefore, we can separate the integrals of the linear combination: $$ \int_{\sigma} (\lambda_{1} \omega_{1}+\lambda_{2} \omega_{2}) = \int_{\sigma} \lambda_{1} \omega_{1} + \int_{\sigma} \lambda_{2} \omega_{2} $$
03

Factor out the scalar constants

Since \(\lambda_{1}\) and \(\lambda_{2}\) are constants, we can factor them out of the integrals: $$ \int_{\sigma} \lambda_{1} \omega_{1} + \int_{\sigma} \lambda_{2} \omega_{2} = \lambda_{1} \int_{\sigma} \omega_{1} + \lambda_{2} \int_{\sigma} \omega_{2} $$
04

Conclusion

We have now shown that the integral of the linear combination of the forms is equal to the sum of the scaled integrals of each form: $$ \int_{\sigma} \lambda_{1} \omega_{1}+\lambda_{2} \omega_{2} = \lambda_{1} \int_{\sigma} \omega_{1} + \lambda_{2} \int_{\sigma} \omega_{2} $$ This result confirms the linearity of the integral for these forms.

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Most popular questions from this chapter

Show that the mapping $$ \left(w_{1}, \omega_{2}\right) \rightarrow w_{1} \wedge w_{2} $$ is bilinear and skew symmetric: $$ \begin{gathered} \omega_{1} \wedge \omega_{2}=-\omega_{2} \wedge \omega_{1} \\ \left(\lambda^{\prime}\left(\omega_{1}^{\prime}+\lambda^{\prime \prime} \omega_{1}^{\prime \prime}\right) \wedge \omega_{2}=\lambda^{\prime} \omega_{1}^{\prime} \wedge \omega_{2}+\lambda^{N} \omega_{1}^{\prime \prime} \wedge \omega_{2}\right. \end{gathered} $$

Find the first Betti number of the torus \(T^{2}=S^{1} \times S^{1}\).

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

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 \(\mathbf{x}=(2,0,0)\).
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