# Equivalence of area forms on the sphere

This is Problem 23.5 from Tu's An Introduction to Manifolds.

Prove that the area form $$\omega$$ on $$S^2$$ in Example 23.11 is equal to the orientation form $$x dy \wedge dz - y dx \wedge dz + z dx \wedge dy$$ of $$S^2$$.

The hint says to take the exterior derivative of $$x^2 + y^2 + z^2 = 1$$ to obtain a relation among the $$1$$-forms $$dx, dy, dz$$ on $$S^2$$. Then show for example that for $$x \neq 0$$, one has $$dx \wedge dy = (z/x)dy \wedge dz$$. However, I do not see how to solve this problem using this hint. I would appreciate any help.

Following the path suggested, you get $$x\,dx+y\,dy+z\,dz = 0$$ on $$S^2$$. So, for example, when $$z\ne 0$$, we have $$dz = -\dfrac xz\,dx - \dfrac yz\,dy$$. Thus, \begin{align*} x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy &= (x\,dy-y\,dx)\wedge dz + z\,dx\wedge dy \\ &= (x\,dy-y\,dx)\wedge \big(-\frac xz\,dx+\frac yz\,dy\big) + z\,dx\wedge dy \\ &= \big(\frac{x^2}z +\frac{y^2}z\big)dx\wedge dy + z\,dx\wedge dy \\ &= \frac{x^2+y^2+z^2}z dx\wedge dy = \frac{dx\wedge dy}z, \end{align*} as required.