Extending a differential form from a subset of $S^2$ to $S^2$ and then integrating it

Question

Let $X = \{(x,y,z) \in S^2 \ \mid \ z \neq 0 \} \subset S^2$ and let $$\omega = \frac{1}{z} dx \wedge dy $$ be a differential $2$-form on $X$.

$(i)$ How can we extend $\omega$ from $X$ to a differential $2-$form $\theta$ on $S^2$ such that $\theta |_{X} = \omega$?

$(ii)$ What is then $i_V(\theta)$ (the interior product of $\theta$ into $V$), with $$V = y \cdot \frac{\partial}{ \partial x} - x \cdot \frac{\partial}{\partial y}.$$

$(iii)$ How can we compute $\displaystyle \int_{S^2} (x+y)\cdot \theta$?

I am not really sure how to solve the above questions. This post - Differential Forms on submanifolds - has an answer on how to extend $\omega$ to $\theta$, but it uses a decomposition of and tangent bundle. Is there a way to do $(i)$ using partitions of unity? The problem with partitions of unity, however, is that we have no control over them. Wouldn't using partitions of unity make $(ii)$ and $(iii)$ very difficult to solve (assuming that we can write $\theta = \displaystyle \sum_{i=1}^n (f_i \cdot \omega)$, where $\{f_i \}_{i=1}^n$ is the smooth parittion of unity)?

A nicer way would be to multiply $\omega$ with a bump function that is $1$ on $X$ and $0$ on $S^2 \setminus X$ (although such function does not exist, but maybe we can use some variation of this).

Answer 1

The question is looking for a concrete (and straightforward) solution, not an abstract one. The partition of unity argument is probably better suited to a form defined on a closed submanifold, rather than an open one.

Note that on the sphere we have $x\,dx+y\,dy+z\,dz = 0$ and the well-known area $2$-form $\theta=x\,dy\wedge dz +y\,dz\wedge dx+ z\,dx\wedge dy$. (You should be able to check both of these, if you don't already know them.) Now, eliminating $dz$ from $\theta$ (i.e., working on the open set $z\ne 0$), show that $$x\,dy\wedge dz +y\,dz\wedge dx+ z\,dx\wedge dy = \dfrac{dx\wedge dy}z.$$

(As a side comment, I'll add that many students will remember $\omega$ from their multivariable calculus course (without the wedge, most likely) as the creature one integrates to get surface area on the upper hemisphere.)

EDIT: Since what I suggested in the comments is probably not well-known, I'm going to add it to the record. Using the relation $x\,dx+y\,dy+z\,dz=0$, we check that when $x,y,z\ne 0$ we have $$\underbrace{\frac{dx\wedge dy}z}_{\omega_1} = \underbrace{\frac{dz\wedge dx}y}_{\omega_2} = \underbrace{\frac{dy\wedge dz}x}_{\omega_3}.$$ Since $x^2+y^2+z^2=1$, we have $$\omega=\omega_1=z^2\omega_1+y^2\omega_2+x^2\omega_3 = z\,dx\wedge dy+y\,dz\wedge dx + z\,dx\wedge dy = \theta$$ when $x,y,z\ne 0$. But this form is smooth on the closure, namely on all of $S^2$.

Answer 2

Let me show you a less "magical" way to find the extension to all of $S^2$. To do so, we just have to extend $\omega$ to coordinate charts covering $S^2\setminus X$. We can cover $S^2$ by six coordinate charts, namely the six charts where one of the three coordinates has fixed (nonzero) sign, and we use the other two coordinates for our chart (so for instance, one of these charts is $\{(x,y,z)\in S^2:x>0\}$, and we use $y$ and $z$ as our coordinates on that chart, with $x$ being uniquely determined as $\sqrt{1-y^2-z^2}$). The given definition already gives us a well-defined $2$-form on the charts where $z$ is positive or negative, so we just have to worry about the charts where $x$ or $y$ has fixed sign.

So, for instance, let's look at the chart where $x>0$. In that chart we use $y$ and $z$ as our coordinates, so we want to rewrite the formula for $\omega$ using just $y$ and $z$ and not $x$ and hope that the $z$ in the denominator goes away (since it's the only obstruction to having a well-defined $2$-form). To that end, we use the relation $x\,dx+y\,dy+z\,dz=0$ on $S^2$ (which comes from differentiating $x^2+y^2+z^2=1$) to say $dx=-\frac{y\,dy+z\,dz}{x}$ on our coordinate chart (since $x$ is always nonzero on it). Substituting this into our formula for $\omega$ we get $$\omega=\frac{1}{z}\cdot(-\frac{z}{x} dz\wedge dx)=-\frac{1}{x}dz\wedge dx.$$ Now we see that this obviously extends to a well-defined $2$-form (given by the same formula) on the entire chart where $x>0$, even when $z=0$. The same calculation also works on the chart where $x<0$, and a similar calculation works on the charts where $y>0$ or $y<0$. So we see that $\omega$ extends to a $2$-form on all the charts, and these extensions are obviously compatible on overlaps (because they come from the same formula, or alternatively because $X$ is dense in $S^2$ and so there is at most one continuous extension to any point).