I was working through an exercise in Tu's book on differential geometry, (ex. 19.5) and I'm trying to get the most out of the exercise, and test my understanding, given that I've already computed the answer. I've starred $(*)$ where I'm stuck and collected them at the end for convenience.
It asks to compute the pullback of the form on $\Bbb R^2-\{0\}$ $$\omega=\frac{-y\,\mathrm{d}x+x\,\mathrm{d}y}{x^2+y^2}$$
by defining $c:\Bbb R\to\Bbb R^2-\{0\}$, as $c(t)=(\cos(t),\sin(t))$, and computing $c^*\omega$.
Doing the computation gives $c^*\omega=\mathrm{d}t$ which is a globally defined $1$-form on $\Bbb R$.
I'm trying to understand the outcome of mapping $\Bbb R$ into $\Bbb R^2$ as a circle of radius $1$ and it's relation to forms on the circle.
There's the obvious combination of $x^2+y^2$ which of course simplifies when calculating the pullback onto $\Bbb R$, so it makes a circle a clear choice, but suppose instead we take the same form $\omega$ and thought of $S^1\subset \Bbb R^2$ as a sub manifold and restricted the form, should one expect a similar (same) nice answer? $(*)$
I've seen two ways of doing this restriction:
$1)$ Here pg.84 it defines the restriction as the pullback of the inclusion of a sub manifold, so define $\omega\big|_{S^1}$ as $\iota^*\omega$ where $\iota:S^1\hookrightarrow\Bbb R^2-\{0\}$ as the unit circle say.
$2)$ In Tu's book however it says rather to consider $T_p S^1\subset T_p(\Bbb R^2-\{0\})$ and thus to consider for $X_1\in T_p S^1$ that $\left(\omega\big|_{S^1}\right)_p(X_1)=\omega_p(X_1)$.
So for $1)$ I could think of $S^1$ as being parametrised by an angle $t$ and take the inclusion as more or less the same as $c(t)$, which is why I would expect a similar answer, barring a possible point of discontinuity?? $(*)$
For $2)$ I can't see it working out as nicely because I've to try write a basis for the tangent space at $p\in S^1$ using the tangent space at $p\in \Bbb R^2$, so I could pick four open sets in $S^1$, like Tu does, $$U_x^+=\{(x,y)\in S^1\mid x>0\}\\ U_x^-=\{(x,y)\in S^1\mid x<0\}\\ U_y^+=\{(x,y)\in S^1\mid y>0\}\\ U_y^-=\{(x,y)\in S^1\mid y<0\}$$
A basis for the tangent space $T_p S^1$ for $p\in U_x^{\pm}$ is $\{\frac{\partial}{\partial y}\big|_p\}$, and for $T_p S^1$ with $p\in U_y^{\pm}$, it's $\{\frac{\partial}{\partial x}\big|_p\}$ so $$\omega\big|_{S^1} =\begin{cases}x\,\mathrm{d}y\quad \text{ on } U_x \\ -y\,\mathrm{d}x \quad \text{ on } U_y \end{cases}$$
where $U_x=U_x^+\cup U_x^-$, and $U_y=U_y^+\cup U_y^-$, and setting $x^2+y^2=1$, since I should also restriction the function coefficients of the $1$-form to $S^1$ as well?? $(*)$
Or maybe more correctly $$\omega\big|_{S^1} =\begin{cases}\sqrt{1-y^2}\,\mathrm{d}y\quad \text{ on } U_x \\ -\sqrt{1-x^2}\,\mathrm{d}x \quad \text{ on } U_y \end{cases}$$
Just to collect the $(*)$'s into a question:
- When pulling back $\omega$ to $\Bbb R$ by mapping to $S^1$ we got a globally defined $1$-form, is the restriction of $\omega$ to $S^1$ by a pullback $\iota^*$ globally defined, call it $\mathrm{d}t$, or will there be a point of discontinuity, say if $S^1$ is included by the same map $c(t)$ with $t\in[0,2\pi]$, or for $t\in(0,2\pi)$, can it extend to the missing point?
- Is the method of $2)$ correct? It seems to give a rather different $1$-form since it's now point dependent, and actually maybe it's not well defined on the intersection and that's why?