# Every differential form $\omega$ of degree 1 in the sphere $S^m\subset\mathbb{R}^{m+1}$ that is closed is also exact.

I am in a series of self-studies on differential forms on $$m$$-dimensional surfaces in Euclidean space. I'm two days thinking about the exercise below. The book that proposes this exercise gives no hint of how to solve it. And it seems to me that there is nothing in the text of the book that can help.

This exercise is in a chapter preceding the chapter on the Stokes' theorem ($$\int_{\partial M} \omega = \int_{M} d\omega$$). So I am trying a solution that does not use Stokes' theorem.

Exercise. For every differential form of class $$C^k$$ in sphere $$S^m$$ there exists a differential $$r$$-form $$\Omega$$ of class $$C^k$$ in $$\mathbb{R}^{m + 1}-\{0\}$$ such that $$\Omega|_{S^m} = \omega$$. If $$m>1$$, conclude that every closed differential form of degree $$1$$ in the sphere $$S^m$$ is the differential of a function $$f:S^m \to \mathbb{R}$$. In particular, any closed form of degree $$1$$ in $$S^m$$ ($$m>1$$) must be nullified by at least two points.

My job. Let $$\omega$$ be a differential form of class $$C^k$$ and degree $$r$$ in sphere $$S^m = \{(x_1,\ldots, x_{m + 1}) \in \mathbb{R}^{m + 1}: x_1^2 +\ldots + x_{m + 1}^2 = 1\}.$$ Let us obtain a differential form $$\Omega$$ of class $$C^k$$ and degree $$r$$ in $$\mathbb{R}^{m + 1}$$ such that $$\Omega|_{S^m}=\omega$$. Let $$x \in \mathbb{R}^{m+1}-\{0\}$$ be fixed. Given any other vector $$v \in \mathbb{R}^m$$ there exists a unique vector $$P_x(v)$$ perpendicular to $$x$$ such that $$v=c_v\cdot x + P_x(v)$$ where $$c_v$$ is a constant that depends on $$v$$. Note that $$P_x(v)\in T_xS^m$$. Since the projection $$P_x$$ is a linear application we can define $$\Omega(x)(v_1,\ldots,v_r)=\omega({x}/{|x|})(P_x(v_1),\ldots,P_x(v_r)).$$

I believe that this first part of the exercise I did correctly.Now let's go to the second part of the exercise which is proof that $$\omega$$ is exact.

For every point $$x$$ of the sphere $$S^m$$ there exists a parametrization $$\varphi: \overline{U} \to \tilde{U}$$ with $$x = \varphi (u)$$ from which we can obtain the base $$\left\{\frac{\partial\varphi(u)}{\partial u_1},\ldots,\frac{\partial\varphi(u)}{\partial u_m} \right\}$$ of $$T_xS ^ m$$ and its dual base $$\{du_1,\ldots, du_m\}$$ of $$(T_xS^m)^\ast$$. In addition, there are coordinate functions $$a_j^{\varphi}:\overline{U}\to \mathbb{R}$$ such that $$\omega(x)(v) = a_1^{\varphi}(u) du_1\cdot v+\ldots+ a_m^{\varphi} du_m\cdot v$$ for all $$x=\varphi(u)\in \tilde{U}$$ and all $$v\in T_xS^m$$. On the assumption that $$\omega$$ is closed we have $$d\omega(x)(v_1,v_2) =\sum_{i for all $$x=\varphi(u)\in \tilde{U}$$ and all $$v_1,v_2\in T_xS^m$$. This means that for every point $$x\in S^m$$ there is a parametrization $$\varphi$$, as $$\varphi(u)=x$$ for some $$u \in \overline{U}$$, which provides coordinate functions $$a_j^{\varphi}:\overline{U}\to\mathbb{R}$$ of $$\omega$$ that satisfies for all $$i in $$\{1,\ldots,m\}$$ $$\frac{\partial a_j^{\varphi}}{\partial u_i}(u)-\frac{\partial a_i^{\varphi}}{\partial u_j}(u)=0$$ for all $$u\in \overline{U}$$.

I think of using the coordinate functions $$a_i^{\varphi}$$ to define a vector field in $$\overline{U}$$ or $$\tilde{U}$$ and achieve some geometric result with line integrals. But I have no idea how to define this field of vectors. Maybe I have not noticed any geometrical property of the sphere that can be used.

• Have you learned anything about deRham cohomology and how to compute it? – Ted Shifrin Jan 8 at 20:19
• @TedShifrin No, the book I'm studying is of introductory character. – MathOverview Jan 8 at 22:22
• Thanks for putting your work in your question and where you got stuck: +1 from me – Neal Jan 8 at 22:56

## 1 Answer

Now that I've read your post more carefully, I see that the intent is to prove bare-hands that a closed $$1$$-form $$\Omega$$ on $$\Bbb R^{m+1}-\{0\}$$ (assuming $$m\ge 2$$) is exact. Even though the region is not star-shaped, you can make a standard calculus argument by defining a potential function $$f(x) = \int_{x_0}^x \Omega$$ and checking that the integral is path-independent by virtue of the fact that $$\Omega$$ is closed. (This would be easiest with Stokes's Theorem, but can be reduced to direct checking with rectangles in planes and checking that Green's Theorem holds there.) You can restrict yourself to polygonal paths with edges parallel to the coordinate axes, and then it's easy to check as well that $$df = \Omega$$.

Although the problem didn't tell you that you could assume that $$\Omega$$ is closed, it's easy enough to arrange this by taking $$\Omega = r^*\omega$$, with $$r(x)=x/\|x\|$$ (and this is smooth on $$\Bbb R^{m+1}-\{0\}$$) and using the fact that $$d\Omega = dr^*\omega = r^*(d\omega) = r^*0 = 0$$.