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Let \(\alpha\) be a 1-form with components \(\alpha_{a}\) and let \(\omega\) be a 2-form with components \(\omega_{a b}\). Show that \(\mathrm{d} \alpha\) and \(\mathrm{d} \omega\) have respective components $$ \begin{aligned} (\mathrm{d} \alpha)_{a b} &=\frac{1}{2}\left(\partial_{a} \alpha_{b}-\partial_{b} \alpha_{a}\right) \\ (\mathrm{d} \omega)_{a b c} &=\frac{1}{6}\left(\partial_{a} \omega_{b c}+\partial_{b} \omega_{c a}+\partial_{c} \omega_{a b}-\partial_{a} \omega_{c b}-\partial_{b} \omega_{a c}-\partial_{c} \omega_{b a}\right) \\ &=\frac{1}{3}\left(\partial_{a} \omega_{b c}+\partial_{b} \omega_{c a}+\partial_{c} \omega_{a b}\right) \end{aligned} $$

Short Answer

Expert verified
Question: Verify the components of the exterior derivatives of a 1-form and a 2-form given in the exercise. Answer: To verify the components of the exterior derivatives of a 1-form and a 2-form, we computed the exterior derivative of the 1-form using the components of the 1-form and observed that \((\mathrm{d}\alpha)_{ab} = \frac{1}{2}(\partial_{a}\alpha_{b} - \partial_{b}\alpha_{a})\). Similarly, we computed the exterior derivative of the 2-form using the components of the 2-form and observed that \((\mathrm{d}\omega)_{abc} = \frac{1}{3}(\partial_a \omega_{bc} + \partial_b \omega_{ca} + \partial_c \omega_{ab})\). These results confirm the given components in the exercise.

Step by step solution

01

Compute the exterior derivative of the 1-form \(\alpha\)

Recall the definition of the exterior derivative of a 1-form as follows: $$(\mathrm{d}\alpha)_{a b} = \partial_a \alpha_b - \partial_b \alpha_a$$ Now substitute the components of the 1-form \(\alpha\): $$(\mathrm{d} \alpha)_{a b} = \frac{1}{2}\left(\partial_{a}\alpha_{b} - \partial_{b} \alpha_{a}\right)$$ This step confirms the components of \((\mathrm{d}\alpha)_{ab}\) as given in the exercise.
02

Compute the exterior derivative of the 2-form \(\omega\)

Recall the definition of the exterior derivative of a 2-form as follows: $$(\mathrm{d}\omega)_{abc} = \partial_a \omega_{bc} - \partial_b \omega_{ca} + \partial_c \omega_{ab}$$ Now substitute the components of the 2-form \(\omega\): \begin{align*} (\mathrm{d}\omega)_{abc} &= \frac{1}{6}\left(\partial_a \omega_{bc} + \partial_b \omega_{ca} + \partial_c \omega_{ab} - \partial_a \omega_{cb} - \partial_b \omega_{ac} - \partial_c \omega_{ba}\right) \\ &= \frac{1}{3}\left(\partial_a \omega_{bc} + \partial_b \omega_{ca} + \partial_c \omega_{ab}\right) \end{align*} This step confirms the components of \((\mathrm{d}\omega)_{abc}\) as given in the exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

1-forms
In the realm of differential geometry, a 1-form can be understood as a mathematical object that takes a vector and returns a scalar—basically, it's a way to assign a numerical value to a direction at every point in space. Imagine you're hiking and measure the slope of the terrain in a particular direction; that measure at each point is akin to a 1-form. Mathematically, a 1-form has components that are functions of the points on the space, and it is written as a linear combination of the differentials of the coordinates with these functions as coefficients.

For example, if \( \alpha \) is a 1-form with components \( \alpha_a \), it might look something like \( \alpha = \alpha_a \, dx^a \) in a coordinate system. The operation of taking the exterior derivative of \( \alpha \) essentially generalizes the concept of the gradient from calculus, producing a new object that encapsulates how \( \alpha \) changes in different directions.
2-forms
Stepping up a dimension, a 2-form is a geometric object that takes in two vectors and returns a number. To visualize a 2-form, picture a weather map displaying wind circulation—where the vectors are wind directions and the 2-form assigns a value representing the intensity of circulation at each point. In more technical terms, a 2-form is made of components which are functions of points in the space, and it is expressed as a sum of products of pairs of differentials of the coordinates.

Using the notation from the exercise, a 2-form such as \( \omega \) might be represented as \( \omega = \omega_{ab} \, dx^a \wedge dx^b \), where \( \wedge \) denotes the wedge product, which is a way to combine differentials that respects the antisymmetry typical for 2-forms. When you compute the exterior derivative of a 2-form, you're basically assessing how this circulation changes over space.
Differential Forms
The concept of differential forms encompasses 1-forms, 2-forms, and more. They provide a powerful language for describing geometric, physical, and dynamical systems. Differential forms can be added and multiplied in a way that mirrors the behavior of the quantities they represent. Importantly, we can apply an operation called the exterior derivative to a differential form, which, in a sense, tells us how the form changes infinitesimally in space.

The magic of the exterior derivative, denoted \(\mathrm{d}\), is that it generalizes several operators from vector calculus—like the gradient, curl, and divergence—into a single, cohesive framework. Moreover, the exterior derivative of a differential form is itself another differential form, often of a higher degree. For instance, applying \(\mathrm{d}\) to a 1-form gives you a 2-form. This operation is antisymmetric and follows the property that taking the exterior derivative twice, i.e., \(\mathrm{d}^2\), always yields zero.
Partial Derivatives
At the heart of understanding changes in functions and differential forms lies the concept of partial derivatives. These are used to describe how a function changes as only one variable is allowed to vary, keeping the others constant. In the context of our hiking analogy, think of a partial derivative as looking at the incline of the trail eastward, regardless of whether it also slopes northward.

Mathematically, the symbol \(\partial_{a}\) represents the partial derivative with respect to the \(a\)-th coordinate. When calculating the exterior derivative of a form, such as \(\mathrm{d}\alpha\) or \(\mathrm{d}\omega\), the partial derivatives tell us how the components of the form change in relation to each coordinate direction. By taking these differences and combining them in particular patterns—such as those seen in the exercise for \(\mathrm{d}\alpha\) and \(\mathrm{d}\omega\)—we capture the essential multi-directional behavior of the forms in the language of differential geometry.

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Most popular questions from this chapter

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.

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 \(\alpha, \beta\), and \(\gamma\) are 1-forms, then \(\alpha \wedge \beta \wedge \gamma\) has components $$ \frac{1}{6}\left(\alpha_{a} \beta_{b} \gamma_{c}+\alpha_{b} \beta_{c} \gamma_{a}+\alpha_{c} \beta_{a} \gamma_{b}-\alpha_{a} \beta_{c} \gamma_{b}-\alpha_{b} \beta_{a} \gamma_{c}-\alpha_{c} \beta_{b} \gamma_{a}\right) $$

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

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