Question

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

Short answer

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

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 (\(\left.\alpha\right|_{M^{'}}\)).

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

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 (\(M'\)): $$ \left.(\mathrm{d} \alpha)\right|_{M^{\prime}} $$

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.

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.

Key concepts

Exterior Derivative

The exterior derivative is a fundamental operator in differential geometry. It extends the concept of differentiation to differential forms, allowing us to analyze them in a wider context. For a differential form \(\alpha\) of any degree \(k\), the exterior derivative \(d\) produces a new \(k+1\) form \(d\alpha\). This operation is linear and follows two main rules:

• An exterior derivative applied to a function \(f\) yields the differential form \(df\), analogous to a gradient in vector calculus.
• The Leibniz rule applies, where the derivative of a wedge product of forms \(\alpha \wedge \beta\) is \(d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\text{deg} \alpha}\alpha \wedge d\beta\).

Furthermore, applying the exterior derivative twice results in zero, \(d(d\alpha) = 0\), reflecting its similarity to the curl of a gradient being zero in vector calculus.

Restriction of Differential Forms

When dealing with differential forms on manifolds, one often encounters the need to restrict these forms to submanifolds. Consider a manifold \(M\) and a submanifold \(M'\) within it. The restriction of a differential form \(\alpha\) to \(M'\) is denoted as \(\left.\alpha\right|_{M^{'}}\). This process involves evaluating \(\alpha\) only on the points within \(M'\), effectively reducing its domain.
This is important for many applications in geometry and physics where you need to focus on a specific region or subspace of a greater manifold. When restricting a form, one ensures that the components of the form that do not "fit" the confines of the submanifold are discarded, thus simplifying calculations while retaining the necessary information for that subset.

Submanifolds

Submanifolds play an integral role in differential geometry as they allow us to study properties and behaviors in constrained environments. Imagine a manifold \(M\) as a higher-dimensional space, and see submanifolds \(M'\) as "slices" within that space. If \(M\) is a two-dimensional plane, \(M'\) could be a curve on that plane.
Submanifolds are lower-dimensional manifolds embedded in a higher-dimensional one, and their properties inherit characteristics from both their parent manifold and their reduced dimensionality. In the context of differential forms, any form defined on \(M\) can be restricted to \(M'\), allowing us to analyze phenomena specifically within the submanifold space. This is crucial for understanding localized behaviors within the parent manifold.

Wedge Product

The wedge product is a fundamental operation in the algebra of differential forms. It is akin to the cross product in vector calculations but applies to differential forms of any degree. Given two differential forms, \(\alpha\) and \(\beta\), their wedge product \(\alpha \wedge \beta\) creates a new form that combines their degrees. It is anti-commutative, meaning \(\alpha \wedge \beta = -\beta \wedge \alpha\).
This property suggests that the order of forms in the product matters significantly. Moreover, the wedge product follows the associative law, \((\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma)\), which allows for flexible grouping in calculations.
These properties make the wedge product key in defining complex differential forms and calculations, especially when integrating these forms over submanifolds or other parts of a manifold.