The $ 1 $-form $ \frac{x\,\mathrm dy - y\,\mathrm dx}{x^2 + y^2} $ in polar coordinates

Question

I have a silly question about the pullback of $ 1 $-forms. Take two smooth manifolds $ M $ and $ N $ and some smooth map $ F\colon M\to N $ between them. Take some $ 1 $-form $ \eta $ on $ N $. Suppose that with respect to some coordinates $ (y^1,\dots,y^m) $ on an open subset $ V $ of $ N $ we can write $$ \eta{\restriction_V} = \sum_{i = 1}^m \eta_i\mathrm dy^i $$ for some smooth functions $ \eta_1,\dots,\eta_m\colon V\to \mathbb R $.

Then $$ F^*\eta{\restriction_{F^{-1}(V)}} = \sum_{i = 1}^m (\eta_i\circ F)\,\mathrm d(y^i\circ F) $$ by the properties of the pullback map $ F^*\colon \Omega_N^1(V)\to \Omega_M^1(F^{-1}(V)) $.

Now, I was playing with the standard example $$ \eta = \frac{x\,\mathrm dy - y\,\mathrm dx}{x^2 + y^2} $$ on the manifold $ X = \mathbb R^2\setminus\{\text{non positive $ x $ semi-axis}\} $. According to Lee's Introduction to Smooth Manifolds Example 11.28, one can find the polar coordinate expression of $ \eta $ simply by computing the pullback $$ \eta = \mathrm{id}_X^*(\eta) $$ under the assumption that "the coordinates on the domain of $ \mathrm{id}_X $ are polar, and the coordinates on the codomain of $ \mathrm{id}_X $ are Cartesian" (semi-quoe from Lee's book). It's not clear to me what it means.

If I compute $ \mathrm{id}_X^*(\eta) $ I obtain (see above) $$ \mathrm{id}_X^*(\eta) = \Bigl(\frac{x}{x^2 + x^2}\circ \mathrm{id}_X\Bigr)\,\mathrm d(y\circ \mathrm{id}_X) - \Bigl(\frac{y}{x^2 + x^2}\circ \mathrm{id}_X\Bigr)\,\mathrm d(x\circ \mathrm{id}_X) $$ which is different from the result $$ \mathrm{id}_X^*(\eta) = \Bigl(\frac{\cos\theta}{r}\Bigr)\,\mathrm d(r\sin\theta) - \Bigl(\frac{\sin\theta}{r}\Bigr)\,\mathrm d(r\cos\theta) = \dots $$ advocated by Lee.

What am I missing?

Answer 1

You have the $1-$form

$$\omega=\frac{x\,dy+ y \, dx}{x^2+y^2}.$$

Use $x=r \cos \theta, y = r \sin \theta, x^2+y^2=r^2$, so you have

\begin{align} \omega&=\frac{x\,dy+ y \, dx}{x^2+y^2}\\ &=\frac{r \cos\theta\, d(r \sin \theta)+r \sin \theta \, d(r \cos \theta)}{r^2}\\ &=\frac{2\cos\theta\sin\theta\, dr+r(\cos^2\theta-\sin^2\theta)\,d\theta}{r} \end{align}

hope this helps!! And of course you can use the trig identity for $2\cos\theta\sin\theta$..and for $d($something$)$ you apply product rule. Lee does an example of swapping to polar I believe in chapter 14 and he does what I did.

Answer 2

OK, after a bit of thought I think I can answer myself. I'll do it in gory detail hoping it can be of some help to other people struggling with these ideas and most importantly with the classical notation.

Our goal is to write the $ 1 $-form $$ \eta = \frac{x\,\mathrm dy - y\,\mathrm dx}{x^2 + y^2} = \frac{x}{x^2 + y^2}\mathrm dy + \frac{-y}{x^2 + y^2}\mathrm dx $$ defined on the space $$ X = \mathbb R^2\setminus{\text{non positive $ x $ semi-axis}} $$ with respect to the local (indeed, global) frame $ \{\mathrm dr,\mathrm d\theta\} $ induced by the coordinates $ \phi = (\phi^1,\phi^2) = (r,\theta) $ on $ X $, where $$ \phi^{-1}\colon \left]0,+\infty\right[\times \left]-\pi,\pi\right[\xrightarrow{{\cong}} X\text{,}\qquad \phi(r,\theta) = \begin{pmatrix}r\cos\theta\\ r\sin\theta\end{pmatrix} $$ is the usual polar-coordinates parametrization of $ X $.

First of all, a little comment on the notation $$ \frac{x}{x^2 + y^2}\text{,}\qquad \frac{-y}{x^2 + y^2} \tag{*}\label{components} $$ used to denote the components $ f_1,f_2\in \mathscr C^\infty(X) $ such that $$ \eta = f_2\mathrm dy + f_1\mathrm dx\text{.} $$ When we work with $ X $ we always have in mind the chart $ (X,\epsilon) $ where $ \epsilon\colon X\to X $ is just the identity. Then we will denote with $ x $ the component $ \epsilon^1\colon X\to \mathbb R $ and with $ y $ the component $ \epsilon^2\colon X\to \mathbb R $ so that $$ x(a,b) = \epsilon^1(a,b) = a\text{,}\qquad y(a,b) = \epsilon^2(a,b) = b $$

for every point $ p = (a,b)\in X $. At this point we have two ways to look at $ \eqref{components} $: they can be thought of as the usual sloppy way "$ f(x) $" to denote the function $ f $ as employed in most elementary treatments; or they can be thought as a quotient in the algebra $ \mathscr C^\infty(X) $ of smooth functions $ X\to \mathbb R $ of the functions $ x $ and $ y $ defined above. This second viewpoint is the one we will employ here.

Now, J. M. Lee in his book suggest that we take the pullback $$ \mathrm{id}_X^*(\eta) = \eta $$ of $ \eta $ by the identity map $ \mathrm{id}_X\colon X\to X $ under the assumption that in the expression $ \mathrm{id}_X\colon X\to X $ the first $ X $ is equipped with the coordinates $ \phi = (\phi^1,\phi^2) = (r,\theta) $, while the second is equipped with the standard coordinates $ (x,y) $.

Consider then the square $$ \require{AMScd} \begin{CD} X @>\mathrm{id}_X>> X\\ @V{\phi = (r,\theta)}VV @VV{\epsilon = (x,y)}V\\ \left]0,+\infty\right[\times \left]-\pi,\pi\right[ @>\widehat{\mathrm{id}_X}>> X \end{CD} $$ where at the bottom there is the coordinate expression of $ \mathrm{id}_X $ with respect to the given choice of coordinates. Then $$ \widehat{\mathrm{id}_X} = \phi^{-1}\qquad\leadsto \qquad \mathrm{id_X} = \phi^{-1}\circ \phi $$ and $$ \mathrm{id}_X^*(\eta) = \Bigl(\frac{x}{x^2 + y^2}\circ \phi^{-1}\circ \phi\Bigr)\,\mathrm d(y\circ \phi^{-1}\circ\phi) + \Bigl(\frac{-y}{x^2 + y^2}\circ \phi^{-1}\circ\phi\Bigr)\,\mathrm d(x\circ \phi^{-1}\circ\phi)\text{.} $$

Let's pause for a moment and examine the expression $$ \frac{x}{x^2 + y^2}\circ \phi^{-1}\circ \phi\text{.} $$ For example let's take some $ p = (a,b)\in X $ and let's evaluate the above function at $ p $. We obtain $$ \frac{x}{x^2 + y^2}\circ \phi^{-1}\circ \phi(a,b) = \frac{\cos\theta(a,b)}{r(a,b)} $$ so that we can write simply $$ \frac{x}{x^2 + y^2}\circ \phi^{-1}\circ \phi = \frac{\cos\theta}{r} $$ (remember the second point of view advocated above; here $ \cos\theta $ is really a composition of functions).

At this point we're ready to conclude that $$ \mathrm{id}_X^*(\eta) = \frac{\cos\theta}{r}\,\mathrm d(r\sin \theta) - \frac{\sin\theta}{r}\,\mathrm d(r\cos \theta) = \mathrm d\theta $$ as it should be.