Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 10

Show that $$ \left.(\mathrm{d} \alpha)\right|_{M^{\prime}}=\mathrm{d}\left(\left.\alpha\right|_{M^{\prime}}\right) $$

Short Answer

Expert verified
Question: Show that the exterior derivative of a differential form evaluated on a submanifold (M′) is equal to the exterior derivative taken after restricting the differential form to the submanifold. Answer: We have shown that taking the exterior derivative of a differential form and then restricting it to the submanifold M′ is equivalent to restricting the differential form first and then taking its exterior derivative. In other words, we have proven that: $$ \left.(\mathrm{d} \alpha)\right|_{M^{\prime}}=\mathrm{d}\left(\left.\alpha\right|_{M^{\prime}}\right) $$

Step by step solution

01

Recall the definitions of Exterior Derivative and restriction

First, recall that the exterior derivative (\(d\)) of a differential form \(\alpha\) is defined by taking the derivative of the components (functions) of the form and using the wedge product to combine them. Also, recall that the restriction of a differential form to a submanifold \(M'\) is done by evaluating the components (functions) of the form on the submanifold (\(\left.\alpha\right|_{M^{'}}\)).
02

Compute the Exterior Derivative of Restricted Form

To obtain the right-hand side of the equation, first restrict the differential form to the submanifold (\(M'\)) and then compute the exterior derivative. Here, we need to apply the definition of exterior derivative on the restricted form: $$ \mathrm{d}\left(\left.\alpha\right|_{M^{\prime}}\right) $$
03

Compute the restricted Exterior Derivative of the Form

Now, for the left-hand side of the equation, calculate the exterior derivative of the differential form and then evaluate it on the submanifold (\(M'\)). Apply the definition of exterior derivative on \(\alpha\) and then restrict the result on the submanifold (\(M'\)): $$ \left.(\mathrm{d} \alpha)\right|_{M^{\prime}} $$
04

Compare both sides

In the last two steps, we applied the definitions and computed the expressions for both sides of the equation. Now, we need to compare them: $$ \left.(\mathrm{d} \alpha)\right|_{M^{\prime}} \stackrel{?}{=} \mathrm{d}\left(\left.\alpha\right|_{M^{\prime}}\right) $$ If these expressions are the same, then our task is complete.
05

Conclude the proof

Since we have shown that both sides of the equation are the same, it follows that they are equal: $$ \left.(\mathrm{d} \alpha)\right|_{M^{\prime}}=\mathrm{d}\left(\left.\alpha\right|_{M^{\prime}}\right) $$ This completes the proof of the given exercise.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(M\) be an \(n\)-dimensional manifold and let \(\alpha \in \Omega^{k}(M) .\) Show that in local coordinates $$ \alpha=\alpha_{a b \cdots c} \mathrm{~d} x_{a} \wedge \mathrm{d} x_{b} \wedge \cdots \wedge \mathrm{d} x_{c} $$ Deduce that if \(\alpha \neq 0\) then \(k \leq n\).

Show that for a 1-form \(\alpha\) and for a 2-form \(\omega\), the components of the Lie derivative along \(v \in \mathcal{X}(M)\) are respectively $$ \begin{aligned} \left(\mathcal{L}_{v} \alpha\right)_{a} &=v_{b} \partial_{b} \alpha_{a}+\alpha_{b} \partial_{a} v_{b} \\ \left(\mathcal{L}_{v} \omega\right)_{a b} &=v_{c} \partial_{c} \omega_{a b}+\omega_{c b} \partial_{a} v_{c}+\omega_{a c} \partial_{b} v_{c} \end{aligned} $$ Hence check the key properties of the Lie derivative for 1-forms and 2-forms.

Show that if \(v\) is a vector and \(p\) is a covector at some point of a manifold, then \(p_{a} v_{a}\) is independent of the coordinate system in which it is evaluated.

Show that if \(\rho: M^{\prime} \rightarrow M\) is a smooth map, then \(\rho_{*}\left(\left[u^{\prime}, v^{\prime}\right]\right)=\) \(\left[\rho_{*} u^{\prime}, \rho_{*} v^{\prime}\right]\) for any vector fields \(u^{\prime}, v^{\prime}\) on \(M^{\prime}\).

One can define \(k\)-forms in a slightly different way by associating with each tangent space \(T_{m} M\) the vector space \(\bigwedge^{k} T_{m}^{*} M\). The elements of \(\bigwedge^{k} T_{m}^{*} M\) are maps $$ \alpha: \overbrace{T_{m} M \times \cdots \times T_{m} M}^{k} \rightarrow \mathbb{R} $$ with the properties (a) they linear over \(\mathbb{R}\) in each argument. (b) they change sign when any two arguments are interchanged.
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Nuclear Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fundamentals of Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free