Question

On the manifold \(\mathbb{R}^{n}\) compute the exteror derivative \(\mathrm{d}\) of the differential form $$ \alpha=\sum_{t=1}^{n}(-1)^{i-1} x^{i} \mathrm{~d} x^{\prime} \wedge \cdots \wedge d x^{l-1} \wedge d x^{\prime+1} \wedge \cdots \wedge d x^{n}. $$ Do the same for \(\beta=r^{-n} \alpha\) where \(r^{2}=\left(x^{1}\right)^{2}+\cdots+\left(x^{n}\right)^{2}\).

Short answer

The exterior derivative of \(\alpha\) is zero, \(d\alpha=0\), and the exterior derivative of \(\beta\) is \(-n r^{-n-1} dr \wedge \alpha\), where \(dr = \sum_{i=1}^{n} x^i dx^i\).

Step-by-step solution

Compute the Exterior Derivative of \(\alpha\)

The exterior derivative of a differential form \(\alpha = A dx \) is defined as \(d\alpha = dA \wedge dx\). Applying this formula to \(\alpha\), we get\[d\alpha = \sum_{i=1}^{n}(-1)^{i-1} dx^i \wedge dx^{\prime} \wedge \cdots \wedge d^2x^{l-1} \wedge d^2x^{\prime+1} \wedge \cdots \wedge d^2x^{n}\]However, taking the exterior derivative twice gives zero, \(d^2=0\), so\[d\alpha = 0\]

Calculate \(r^{-n}\) and its Differential

We are given \(r^2 = (x^1)^2 + \cdots + (x^n)^2\), hence \(r = \sqrt{(x^1)^2 + \cdots + (x^n)^2}\). Therefore, \(r^{-n} = \frac{1}{(x^1)^n + \cdots + (x^n)^n}. The differential of \(r^{-n}\) is given by\[d(r^{-n}) = -n r^{-n-1} dr\]where \(dr = \sum_{i=1}^{n} x^i dx^i\).

Compute the Exterior Derivative of \(\beta\)

\(\beta = r^{-n}\alpha \) is the product of a function and a differential form, so the rule for the exterior derivative gives\[d\beta = d(r^{-n}) \wedge \alpha + r^{-n} d\alpha\]but we've calculated in Step 1 that \(d\alpha = 0\), hence\[d\beta = d(r^{-n}) \wedge \alpha\]Substituting for \(d(r^{-n})\) and forgetting the zero part, we get\[d\beta = -n r^{-n-1} dr \wedge \alpha\]where, as noted above, \(dr = \sum_{i=1}^{n} x^i dx^i\).

Key concepts

Exterior Derivative

The exterior derivative is a core concept in differential geometry, used to generalize the notion of derivatives from calculus to intricate mathematical objects known as differential forms. It is an operator denoted by \(d\), which acts on differential forms and increases their degree by one.

• When applying the exterior derivative on a 0-form, or a scalar function \(f\), the result is its differential, \(df\), resembling the gradient in vector calculus.
• For a differential form \(\alpha\) of degree \(k\), \(d\alpha\) becomes a differential form of degree \(k+1\).
• The important rule \(d^2 = 0\) signifies that the exterior derivative applied twice yields zero, capturing the core principle behind the notion of "closed forms", as seen in de Rham cohomology.

Understanding the exterior derivative is crucial in assessing how forms change across a manifold, yet maintaining the property that doubling the operation nullifies it as seen in our step-by-step solutions.

Manifold

A manifold is a fundamental concept essential to modern geometry and physics. Manifolds are spaces that, at a small enough scale, resemble Euclidean space. They offer a general framework to analyze complex geometric shapes and structures.

• Euclidean space \(\mathbb{R}^{n}\) is a simple example of a manifold, where every point has a neighborhood that looks like \(\mathbb{R}^{n}\) itself. However, manifolds can have far more complex topology.
• Manifolds are the natural setting where differential forms and exterior calculus are applied, permitting the extension of concepts like integration and differentiation into higher dimensions.
• In the exercise, \(\mathbb{R}^{n}\) is a manifold where the exterior derivative is evaluated, allowing the exploration of forms within this multi-dimensional framework.

Learning how to operate within a manifold equips one with the ability to handle various advanced mathematical and physical concepts, as manifolds underlie a large majority of problems in modern theoretical science.

Wedge Product

The wedge product is a critical operation in the realm of differential forms, denoted by \(\wedge\). It combines differential forms to produce new ones, offering an analog to the cross and dot products but applicable in higher dimensions and spaces.

• For two differential forms \(\alpha\) of degree \(p\) and \(\beta\) of degree \(q\), the wedge product \(\alpha \wedge \beta\) produces a new form of degree \(p+q\).
• The wedge product is antisymmetric, implying \(\alpha \wedge \beta = -\beta \wedge \alpha\), a characteristic particularly relevant in multidimensional integration and determinants.
• In the exercise, the wedge product assembles components of \(\alpha\) and \(\beta\), ensuring that the computation of the exterior derivatives maintains the antisymmetric condition required for these forms.

Mastering the wedge product enables one to construct and deconstruct differential forms across manifolds, a pivotal skill in navigating the complexities of exterior calculus and geometry.