Question

Let \(\omega\) be the two-form in \(R^{3}-\\{0\\}\) given by $$ \omega=\frac{x d y d z+y d z d x+z d x d y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ Show that \(d \omega=0\) but that there is no one-form \(\eta\) such that \(d \eta=\omega\). (Hint: If there were such a one-form, then by Stokes's theorem, with \(T\) being the unit sphere, we would have \(\int_{T} \omega=\int_{T} d \eta=\int_{S} \eta=0\), because the boundary of \(T\) is empty. Then calculate \(\int_{T} \omega\) directly.)

Short answer

Question: Show that for the two-form \(\omega = \frac{x}{(x^2+y^2+z^2)^{3/2}} dy \wedge dz + \frac{y}{(x^2+y^2+z^2)^{3/2}} dz \wedge dx + \frac{z}{(x^2+y^2+z^2)^{3/2}} dx \wedge dy\), we have \(d\omega = 0\). Also, show that there is no one-form \(\eta\) such that \(d\eta = \omega\).

Step-by-step solution

Compute the exterior derivative of the given two-form.#

To show that \(d\omega=0\), we need to compute the exterior derivative of the given two-form. Recall that if a two-form \(\omega=P dy \wedge dz+Q dz \wedge dx+R dx \wedge dy\), then $$ d\omega= \left(\frac{\partial R}{\partial y}- \frac{\partial Q}{\partial z}\right)dx\wedge dy\wedge dz + \left(\frac{\partial P}{\partial z}- \frac{\partial R}{\partial x}\right)dy\wedge dz\wedge dx + \left(\frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}\right)dz\wedge dx\wedge dy. $$ Using this formula with $$ P = \frac{x}{(x^2+y^2+z^2)^{3/2}}, \quad Q = \frac{y}{(x^2+y^2+z^2)^{3/2}}, \quad R = \frac{z}{(x^2+y^2+z^2)^{3/2}}, $$ we can calculate the partial derivatives to compute \(d\omega\).

Perform the partial derivative calculations.#

We need to perform the following calculations: - \(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\) - \(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\) - \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\) Compute each partial derivative and simplify the results. Note that when computing the partials of \(P,Q,\) and \(R\), always use the quotient rule on the fractions, and the chain rule to differentiate the denominator.

Combine results to obtain \(d\omega\).#

Once the calculations from Step 2 are completed, plug the results into the general formula for the exterior derivative of a two-form. Show that the coefficients of each term are equal to 0, thus showing that \(d\omega = 0\).

Calculate the integral \(\int_T\omega\).#

Now we aim to calculate the integral \(\int_{T}\omega\) using the spherical coordinate system. The volume form in spherical coordinates is given by \(\rho^2\sin\phi d\rho d\theta d\phi\). Since we are considering the unit sphere, take \(\rho=1\). Determine the integral \(\int_T\omega\) directly using the spherical coordinate system.

Prove that there is no one-form \(\eta\) such that \(d\eta=\omega\).#

Use the hint given in the problem. By Stokes' Theorem, if there were such a one-form, then we would have \(\int_T\omega=\int_T d\eta=\int_{\partial T}\eta=0\), because the boundary of \(T\) is empty. However, if in Step 4, we find that \(\int_T\omega\neq 0\), then we can conclude that there is no such one-form \(\eta\).

Key concepts

Differential Forms

Differential forms are a way to generalize the concept of functions and vectors in a space. They allow for the description of geometric and physical quantities in a more general, coordinate-independent manner. A key advantage of using differential forms is their ability to integrate over arbitrary manifolds, yielding important results in various fields like physics and geometry.

The given problem involves a two-form \( \omega \), specifically in three-dimensional space minus the origin. Two-forms can be visualized as objects that measure how much a surface gets 'twisted' or 'warped' through a region of space. The expression for the two-form represents a 'field' that assigns an oriented plane and magnitude to each point in space, except at the origin where it is undefined due to a singularity.

The computation of the exterior derivative \( d\omega \) is an operation that takes our given two-form and produces a new three-form. This process is akin to finding the curl of a vector field in classical vector calculus, which measures the infinitesimal rotation at each point of a vector field. A key property of the exterior derivative is that it squares to zero, meaning that \( d(d\omega) = 0 \) for any differential form \( \omega \). This property is connected to the fact that the 'boundary of a boundary' is always zero in topology.

Stokes' Theorem

Stokes' theorem is a powerful tool in calculus that provides a crucial link between integrals over different dimensions. It generalizes several theorems from vector calculus, including the Fundamental Theorem of Calculus, Green's Theorem, and the Divergence Theorem. In its most general form, Stokes' theorem relates the integral of a differential form over the boundary of a manifold to the integral of its exterior derivative over the manifold itself.

The theorem can be stated as \( \int_{\partial M} \omega = \int_{M} d\omega \), where \( \partial M \) is the boundary of the manifold \( M \) and \( \omega \) is a differential form.

In the context of our exercise, Stokes' theorem implies that if there existed a one-form \( \eta \) such that \( d\eta = \omega \) for the given two-form, the integral of \( \omega \) over the unit sphere (\( T \) being our manifold without boundary) should be zero. This is because the boundary of \( T \) is empty, and thus \( \int_{\partial T} \eta = 0 \). However, if the integral of \( \omega \) over \( T \) is not zero, it contradicts the existence of such \( \eta \)—a critical realization that can be used to solve the given problem.

Spherical Coordinates

Spherical coordinates offer a three-dimensional coordinate system which can be extremely helpful in problems involving symmetry around a point, such as spheres. They are an alternative to the more familiar Cartesian (x, y, z) coordinates, better suited to certain shapes and configurations.

Spherical coordinates are defined by three values: the radius \( \rho \) from the origin to a point, the polar angle \( \phi \) (often measured from the positive z-axis), and the azimuthal angle \( \theta \) (usually measured in the x-y plane from the positive x-axis). The volume element in spherical coordinates is given by \( \rho^2 \sin(\phi) d\rho d\theta d\phi \), which becomes especially simplified for integrals over a sphere of radius one.

In the step-by-step solution provided, spherical coordinates are applied to directly calculate the integral of \( \omega \) over the unit sphere \( T \). This method exploits the spherical symmetry of \( \omega \) to simplify the integral, rather than attempting a more complex calculation in Cartesian coordinates. The ability to translate the original form into spherical coordinates is particularly advantageous when dealing with the curvature and geometry of spheres, making this coordinate system a highly valuable tool in many areas of mathematics and physics.