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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 \(\omega_{1} \wedge \omega_{2}\) really is a 2 -form.

Show that, if two of the indices \(i_{1}, \ldots, i_{k}\) are the same, then the form \(x_{i_{1}} \wedge \cdots \wedge x_{i_{k}}\) is zero.

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.
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