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Find \(d \widehat{\omega}\) if $$ \widehat{\boldsymbol{\omega}}=\left(\begin{array}{cc} 0 & -\cot \theta \boldsymbol{\epsilon}^{\varphi} \\ \cot \theta \boldsymbol{\epsilon}^{\varphi} & 0 \end{array}\right), $$ where \((\theta, \varphi)\) are coordinates on the unit sphere \(S^{2}\).

Short Answer

Expert verified
The exterior derivative of the 1-form \(\omega = -\cot(\theta) d\varphi\) is given by \(d\omega = csc^{2}(\theta) d\theta \wedge d\varphi\).

Step by step solution

01

Break down the matrix

We can write our 1-form \(\widehat{\boldsymbol{\omega}}\) in a more familiar form as \(\omega = -\cot(\theta) d\varphi\). This form represents the sum of the components of the original matrix, expressing them in terms of the differentials \(d\theta\) and \(d\varphi\).
02

Use the formula for the exterior derivative to calculate \(d\omega\)

The exterior derivative of a 1-form \(\omega = f(\theta, \varphi) d\varphi\) in spherical coordinates is calculated using the formula \(d\omega = \frac{\partial f(\theta, \varphi)}{\partial \theta} d\theta \wedge d\varphi + \frac{\partial f(\theta, \varphi)}{\partial \varphi}d\varphi \wedge d\varphi\). Applying this formula to \(\omega\), we get \(d\omega = \frac{\partial (-\cot(\theta))}{\partial \theta} d\theta \wedge d\varphi + 0\), because the derivative of \(-\cot(\theta)\) with respect to \(\varphi\) is 0.
03

Simplify the result

After performing the partial derivative calculation and simplifying the result, we get \(d\omega = csc^{2}(\theta) d\theta \wedge d\varphi\). This is a 2-form on the sphere \(S^{2}\) which can be interpreted as giving the differential area element of a region on the sphere.

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Key Concepts

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

Exterior Derivative
The exterior derivative is a powerful tool in differential geometry and calculus on manifolds. It allows us to extend the concept of differentiation to forms, which are geometric objects that generalize functions and vector fields. Given a differential form, the exterior derivative helps to compute its differential, which is essential in understanding the behavior of the form under transformations.

For a 1-form, such as \( \omega = f(\theta, \varphi) d\varphi \),we can compute its exterior derivative using:
  • \( d\omega = \frac{\partial f}{\partial \theta} d\theta \wedge d\varphi + \frac{\partial f}{\partial \varphi} d\varphi \wedge d\varphi \)
The wedge product \(\wedge\) in this formula makes this operator anti-commutative, crucial for encoding the geometry of the space. The exterior derivative yields a form that is one degree higher than the original, transforming a 1-form into a 2-form. This transformation provides insights into the underlying geometry, such as calculating areas, volumes, and more.
1-Form
A 1-form is a type of differential form that corresponds to a linear functional on the tangent space of a manifold, like a sphere. In simple terms, it can be viewed as a generalization of the concept of a vector field; however, instead of associating vectors with each point in a space, it assigns linear functionals.

When expressed in local coordinates, a 1-form is typically written as:
  • \( \omega = f_1(\theta, \varphi) d\theta + f_2(\theta, \varphi) d\varphi \)
This denotes a weighted sum of basic differential forms \(d\theta\) and \(d\varphi\). In the context of our exercise, the 1-form is expressed as \(-\cot(\theta) d\varphi\). 1-forms are essential for performing integration over curves, and the exterior derivative allows us to "lift" this integration into higher dimensions.
Spherical Coordinates
Spherical coordinates are a mathematical system for representing points in three dimensions using three values: radius, polar angle (\(\theta\)), and azimuthal angle (\(\varphi\)). These coordinates are particularly helpful when working with objects that are naturally spherical, such as planets or atoms.

For a unit sphere, the radius is always 1. Thus, the location of any point on the sphere can be uniquely determined just by the angles \(\theta\) and \(\varphi\).
Where:
  • \(\theta\) typically represents the angle from the positive z-axis (polar angle).
  • \(\varphi\) represents the angle from the positive x-axis in the x-y plane (azimuthal angle).
These angles range continuously from \(0\) to \(\pi\) for \(\theta\) and \(0\) to \(2\pi\) for \(\varphi\). When dealing with calculations on the sphere, spherical coordinates simplify the mathematics by aligning more naturally with the sphere's symmetry.
Unit Sphere
The unit sphere, often denoted as \(S^2\), is a set of points equidistant from a central point in three-dimensional space, where the radius is unitary, or 1. This sphere serves as a critical model as it simplifies many mathematical problems due to its symmetry and homogeneity.

It is defined mathematically by the equation:\[ x^2 + y^2 + z^2 = 1 \] where \((x, y, z)\) are the coordinates of any point on the sphere. Using spherical coordinates, any point on this unit sphere is defined with two parameters, \(\theta\) and \(\varphi\), making calculations and analysis more straightforward as explained above.

In differential geometry, the unit sphere is often used to study surfaces, curvature, and angle relationships, providing a foundational understanding of more complex geometrical and topological concepts.

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Most popular questions from this chapter

Let \(E_{i j}\) be the most general second-rank symmetric tensor constructed from the metric and Riemann curvature tensors that is linear in the curvature tensor. (a) Show that $$ E_{i j}=a R_{i j}+b g_{i j} R+\Lambda g_{i j} $$ where \(a, b\), and \(\Lambda\) are constants. (b) Show that \(E_{i j}\) has a vanishing divergence if and only if \(b=-\frac{1}{2} a\). (c) Show that in addition, \(E_{i j}\) vanishes in flat space-time if and only if \(\Lambda=0\)

Find the curvature of the two-dimensional space whose arc length is given by \(d s^{2}=d x^{2}+x^{2} d y^{2}\).

Show that in an \(n\) -dimensional manifold without metric the number of independent components of the Riemann curvature tensor is $$ \frac{n^{3}(n-1)}{2}-\frac{n^{2}(n-1)(n-2)}{6}=\frac{n^{2}\left(n^{2}-1\right)}{3} . $$ If the manifold has a metric, the number of components reduces to $$ \left[\frac{n(n-1)}{2}\right]^{2}-\frac{n^{2}(n-1)(n-2)}{6}=\frac{n^{2}\left(n^{2}-1\right)}{12} $$

Find the geodesics of a manifold whose arc element is \(d s^{2}=d x^{2}+\) \(d y^{2}+d z^{2}\).

The Riemann normal coordinates are given by \(x^{i}=a^{i} t .\) For each set of \(a^{i}\), one obtains a different set of geodesics. Thus, we can think of \(a^{i}\) as the parameters that distinguish among the geodesics. (a) By keeping all \(a^{i}\) (and \(t\) ) fixed except the \(j\) th one and using the definition of tangent to a curve, show that \(\mathbf{n}_{j}=t \partial_{j}\), where \(\mathbf{n}_{j}\) is (one of) the \(\mathbf{n}(\) 's) appearing in the equation of geodesic deviation. (b) Substitute (a) plus \(u^{i}=\dot{x}^{i}=a^{i}\) in Eq. (37.29) to show that $$ R_{i j k}^{m}+R_{j i k}^{m}=3\left(\Gamma_{i k, j}^{m}+\Gamma_{j k, i}^{m}\right) $$ Substitute for one of the \(\Gamma\) 's on the RHS using Eq. (36.45). (c) Now use the cyclic property of the lower indices of the curvature tensor to show that $$ \Gamma_{i j, k}^{m}=-\frac{1}{3}\left(R_{i j k}^{m}+R_{j i k}^{m}\right) $$.

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