Consider the 2-form $$\sigma=\frac{x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2}{(x_1^2+x_2^2+x_3^2)^{3/2}}.$$ I need to show if it is exact or not. Suppose it is exact, then there exists a 1-form

$$\alpha=f_1dx_1+f_2dx_2+f_3dx_3,$$ such that $$d\alpha=\displaystyle\sum_i df_i\wedge dx_i=\sigma.$$ Now I use $$df_i=\frac{\partial f_i}{\partial x_1}dx_1+\frac{\partial f_i}{\partial x_2}dx_2+\frac{\partial f_i}{\partial x_3}dx_3,$$ then $d\alpha$ becomes $$\left(\frac{\partial f_2}{\partial x_1}-\frac{\partial f_1}{\partial x_2}\right)dx_1\wedge dx_2+\left(\frac{\partial f_3}{\partial x_2}-\frac{\partial f_2}{\partial x_3}\right)dx_2\wedge dx_3+\left(\frac{\partial f_1}{\partial x_3}-\frac{\partial f_3}{\partial x_1}\right)dx_3\wedge dx_1.$$ I do not know how to proceed, is this even the right method?

  • 3
    $\begingroup$ Try to use the fact that $\sigma$ is 'radial'. Do you know differential calculus in spherical coordinates? $\endgroup$
    – Andrea
    Dec 30, 2011 at 22:46

2 Answers 2


Generally it's easier to prove that a form is NOT exact.

For instance, an exact form is necessarily closed. So, if your form was exact it would be closed. And to verify this later property is just a matter of differentiation -which is easier than looking for an antiderivative, as you seem to be trying. Of course, if $d\sigma = 0$, this says nothing about being exact or not.

Another idea: if your form was exact, $\sigma = d\alpha$, for some $\alpha$, then, by Stokes' theorem, its integral over a closed surface $S$ (that is, with no boundary, or empty boundary, $\partial S = \emptyset$) would be zero:

$$ \int_S d\alpha = \int_{\partial S} \alpha = 0 \ . $$

So you could try to find a closed surface $S$ such that $\int_S \sigma \neq 0$.

Where to look for such a surface? Well, there is a "meta criterion" for this: since this is, presumably, an exercise in a beginner's differential geometry book, it can NOT be too far away from your knowledge. :-) So, I would try with the first closed surface that came to my mind without hesitation.

Or also, looking at that denominator, you could think of a surface where it becomes really, really, and I mean really, simple. (For instance, an sphere of radius 1.)


It's not exact. To see this, as suggested by Andrea, we use the spherical coordinates: $$(x_1,x_2,x_3)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),0<\theta<2\pi,0<\phi<\pi.$$ Restricted to $\mathbb{S}^2$, we have $r=1$, i.e. $$(x_1,x_2,x_3)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$ which implies that $$dx_1|_{\mathbb{S}^2}=-\sin\phi\sin\theta\,d\theta+\cos\phi\cos\theta\,d\phi,$$ $$dx_2|_{\mathbb{S}^2}=\sin\phi\cos\theta\,d\theta+\cos\phi\sin\theta\,d\phi,$$ $$dx_3|_{\mathbb{S}^2}=-\sin\phi\,d\phi.$$ This gives $$dx_1\wedge dx_2\big|_{\mathbb{S}^2}=-\sin\phi\cos\phi\,d\theta\wedge d\phi,$$ $$dx_2\wedge dx_3\big|_{\mathbb{S}^2}=-\sin^2\phi\cos\theta\,d\theta\wedge d\phi,$$ $$dx_3\wedge dx_1\big|_{\mathbb{S}^2}=-\sin^2\phi\sin\theta\,d\theta\wedge d\phi.$$

Combining all these, we can express $\sigma$ in terms of spherical coordinates: $$\sigma\big|_{\mathbb{S}^2}=\Big(\frac{x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2}{(x_1^2+x_2^2+x_3^2)^{3/2}}\Big)\Big|_{\mathbb{S}^2}$$ $$=\big(x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2\big)\big|_{\mathbb{S}^2}$$ $$=-\sin^3\phi\cos^2\theta\,d\theta\wedge d\phi-\sin^3\phi\sin^2\theta\,d\theta\wedge d\phi-\sin\phi\cos^2\phi\,d\theta\wedge d\phi$$ $$=-\sin\phi\,d\theta\wedge d\phi.$$

Suppose that $\sigma$ is exact, then $\sigma=df$ for some one-form $f$. By Stoke's theorem, we have $$\int_{\mathbb{S}^2}\sigma=\int_{\mathbb{S}^2}df=\int_{\partial \mathbb{S}^2}f=0$$ where the last equality follows from the fact that $\partial \mathbb{S}^2=\varnothing$, i.e. $\mathbb{S}^2$ is an manifold with no boundary. However, from the above calculation, $$\int_{\mathbb{S}^2}\sigma=-\int_{0}^\pi\int_0^{2\pi}\sin\phi\,d\theta d\phi=-4\pi, $$ which is a contradiction.

  • $\begingroup$ Note: this form is exactly negative of the area form on the unit sphere. $\endgroup$ Aug 18, 2019 at 17:16

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