Problem 14
Question
Prove that cvery closed 1-form on \(S^{2}\) is exact. Show that this statement does not, extend to 2 -forms by showing that the 2 -form $$ \alpha=r^{-3 / 2}(x \mathrm{~d} y \wedge \mathrm{d} z+y \mathrm{~d} z \wedge \mathrm{d} x+z \mathrm{~d} x \wedge d y) $$ is closed, but has non-vanishing integral on \(S^{2} .\)
Step-by-Step Solution
Verified Answer
We have proven that every closed 1-form on \(S^{2}\) is exact by contradiction. In addition, we showed that this statement does not extend to 2-forms by proving the specific 2-form, \(\alpha\), is closed but has a non-vanishing integral over \(S^{2}\), thus is not exact.
1Step 1: Understanding the definitions
First, we need to understand what the terms 'closed' and 'exact' mean in the context of differential forms. A differential form is closed if its exterior derivative is equal to zero. A form is exact if there exists a form of one degree less such that it's exterior derivative is our original form. Our first task is to prove that every closed 1-form on \(S^2\) is exact. Considering the Poincare lemma which states that every closed form is exact in contractible domains, and since \(S^2\) is not contractible, its application is not straightforward. So we need a specific treatment for \(S^2\).
2Step 2: Prove Closed 1-forms are exact on \(S^2\)
We can prove that all closed 1-forms on \(S^2\) are exact by contradiction. Assume there is a closed but not exact 1-form \(\omega\) on \(S^2\), this would mean that the integral of \(\omega\) over some non-contractible cycle (like the equator of \(S^2\)) is non-zero. We can cut \(S^2\) along this equator and get a topological disk, which is contractible, and over this region, being closed, \(\omega\) must be exact by the Poincare lemma. So its integral over the boundary (our equator) must be zero, which contradicts our initial assumption.
3Step 3: Prove the 2-form not extending the property
For the 2-form, \(\alpha\), first we need to prove it is closed by showing that its exterior derivative \(d\alpha\) is zero. Then to show it is not exact, we need to recognize that \(\alpha\) can be viewed as the volume form of the standard metric on \(S^{2}\). It is known that its integral over the whole \(S^{2}\) is not zero, specifically, it equals \(4\pi\). So \(\alpha\) is not exact as its integral over \(S^{2}\) is not zero unlike exact forms.
Key Concepts
Closed FormExact FormPoincare LemmaIntegral on Manifolds
Closed Form
In differential geometry, a *closed form* is an important concept related to differential forms. To call a form closed means that its exterior derivative is zero. In simple terms, if you take the exterior derivative of a closed form, you end up with zero. This is akin to having a function with a zero derivative—it's in a state of equilibrium. Closed forms often represent conserved quantities in physics, like in the context of vector fields where they correspond to irrotational fields.
- For a form \( \omega \) to be closed: \( d\omega = 0 \)
Exact Form
An *exact form* is a step further than a closed form. If a differential form is exact, it means that there exists another form of one degree less such that applying the exterior derivative to this form results in the original form. Essentially, all exact forms are closed, but not all closed forms are exact.In practical terms:
- If \( \omega \) is exact, \( \omega = d\eta \) for some form \( \eta \).
Poincare Lemma
The *Poincare Lemma* is a significant theoretical tool in differential geometry. It states that on a contractible domain—a space that can be continuously "shrunk" to a point—every closed form is also exact.However, this becomes intriguing when dealing with spaces that are not contractible, like the 2-sphere \( S^2 \). The lemma itself cannot be directly applied here, so it requires alternate reasoning. For a closed form on such non-contractible spaces, proving exactness can involve clever topological arguments.
- Poincare Lemma: "Every closed form is exact on a contractible domain."
Integral on Manifolds
In the world of differential geometry, integrating a form on a manifold explores quite literally "summing" the form's effects over that space. When discussing whether a form is exact or not, examining the integral of the form over a manifold can be revealing.For the 2-form \( \alpha \) as discussed on the sphere \( S^2 \), it was shown to be closed, but not exact because its integral over the sphere yielded a non-zero value. This breaks the condition set by the definition of exact forms, as exact forms would integrate to zero over closed manifolds.
- If \( \alpha \) is exact, the integral \( \int_{S^2} \alpha = 0 \)
Other exercises in this chapter
Problem 10
For any pair of subspaces \(H\) and \(K\) of the extenor algebra \(\Lambda^{*}(M)\), set \(H \wedge K\) to be the vector subspace spanned by all \(\alpha \wedge
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Show that for any set of real numbers \(a_{1} \ldots, a_{k}\) there exists a closed \(r\)-form \(\alpha\) whose periods \(\int_{C} \alpha=a_{r}\),
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Show that the Maxwell 2-form satisfies the identities $$ \begin{gathered} \varphi \wedge * \varphi=* \varphi \wedge \varphi=4\left(B^{2}-\mathbf{E}^{2}\right) \
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Show that in \(n\) dimensions, if \(V\) is a regular \(n\)-domain with boundary \(S=\partial V\), and we set \(\alpha\) to be an \((n-1)\)-form with components
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