I am having great difficulty calculating what seem to be very simple pullbacks. Right now, I am trying to integrate the 1-form defined by $\omega = dy$ parameterized by $f:[0,1] \rightarrow \mathbb{R}^{2}$ given by $f(t)=(0,t)$. Any help would be appreciated. Here's what I've done so far:
First, we need to pull back the function along $f$, because we can only integrate a k-form in k-dimensional space. The definition of the pullback when $\omega=\Sigma_{1\leq i_{1} \lt …\lt i_{k}\leq n}gdx_{i_{1}}\wedge\ldots\wedge dx_{i_{k}}$ is $f^{*}\omega=\Sigma_{1\leq i_{1} \lt …\lt i_{k}\leq n}g(f)df_{i_{1}}\wedge\ldots\wedge df_{i_{k}}$.
So, for this particular $\omega$, we have $g=1$, but I'm not sure what the value of each $dg_{i}$ should be, and how they should relate to the $dy$ in the $\omega$. I got this far:
$$f^*\omega = g(f)df$$
$$= (0,t)(\frac{\partial f_1}{\partial t},\frac{\partial f_2}{\partial t})^{T}dt $$
$$=\frac{\partial f_2}{\partial t}dt$$
$$=dt$$
Is this correct? Even if I somehow stumbled upon the correct answer, I don't really understand what's going on here. Also, how would this change in the more general case? For example, let's say I'm trying to integrate $\omega_{o}=f(x,y)dy + g(x,y)dx$ along a curve parameterized by $f:[0,1]\rightarrow \mathbb{R}^2$ with $f=(x(t), y(t))$. How would I do that? Also, how does this generalize to higher dimensional manifolds and ambient spaces?
Best Answer
Working with exterior derivatives is VERY simple once you learn the basic rules:
As long as you keep these rules in mind, things become very simple, and it reduces to an essentially robotic calculation.
The "mantra" to keep in mind is that pullback is essentially just substituting things in every single place.
Example 1.
Consider the manifold $M=\Bbb{R}$, $N=\Bbb{R}^2$, where we introduce global coordinates (i.e use the identity chart) $t$ on $M$ and $(x,y)$ on $N$. The meaning of this is that $x:N=\Bbb{R}^2\to\Bbb{R}$ is the function defined as $x(a,b):=a$ and $y:N=\Bbb{R}^2\to\Bbb{R}$ is $y(a,b):=b$. They are the coordinate projections (though it also common practice to use $(x,y)$ to denote a point of $N$). Consider now the differential $1$-form $\omega=dy$ on $\Bbb{R}^2$ and $f:M=\Bbb{R}\to N=\Bbb{R}^2$ defined as $f(p)=(0,p)$. Then, by the rules outlined above, \begin{align} f^*\omega&= f^*(dy)=d(f^*y)=d(y\circ f) \end{align} Now what is $y\circ f$? Well, for any $p\in\Bbb{R}$, we have $(y\circ f)(p)=y(0,p)=p$. THis is the identity function on $\Bbb{R}$. In other words, $y\circ f=t$. Therefore, $f^*\omega=dt$ (so you have to correct answer, but your intermediate steps do not make sense).
Example 2.
Consider the manifold $M=\Bbb{R}$, $N=\Bbb{R}^2$, where we introduce global coordinates $t$ on $M$ and $(x,y)$ on $N$. Let $f,g:N\to\Bbb{R}$ be smooth functions and consider the $1$-form $\omega=f\,dx+g\,dy$, where it is common to write $\omega=f(x,y)\,dx+g(x,y)\,dy$. Now, suppose we have a smooth mapping $\gamma:M=\Bbb{R}\to N=\Bbb{R}^2$. Then, again calculating pullbacks is simple if you use the rules above: \begin{align} \gamma^*(\omega)&=\gamma^*(f\,dx+g\,dy)\\ &=\gamma^*(f\,dx)+ \gamma^*(g\,dy)\\ &=(\gamma^*f)\cdot (\gamma^*dx)+ (\gamma^*g)\cdot (\gamma^*dy)\\ &=(f\circ \gamma)\cdot d(x\circ \gamma)+ (g\circ \gamma) \cdot d(y\circ \gamma)\\ &= (f\circ \gamma)\cdot d(\gamma_1)+ (g\circ \gamma)\cdot d(\gamma_2)\\ &=(f\circ \gamma)\cdot \gamma_1'\,dt+ (g\circ \gamma)\cdot \gamma_2'\,dt\\ &=[(f\circ \gamma)\cdot \gamma_1'+ (g\circ \gamma)\cdot \gamma_2']\,dt \end{align}
In slightly more classical notation, one overuses the symbols $x,y$, where by abuse of notation, we write $x(t)$ rather than the more correct $(x\circ \gamma)(t)=\gamma_1(t)$, and $y(t)$ in place of $(y\circ \gamma)(t)=\gamma_2(t)$. So, the above equality is often written as \begin{align} \gamma^*\omega&=[f(x(t),y(t))\cdot x'(t)+ g(x(t),y(t))\cdot y'(t)]\,dt \end{align}
This is a general calculation for pullbacks of $1$-forms on $\Bbb{R}^2$.
Example 3.
So far we dealt with $1$-forms on $\Bbb{R}^2$. Let us consider slightly more complicated situations. I'll deal with concrete calculations since you already have the general formula for pullbacks in your second paragraph, but you're not sure what that actually means.
So, consider $M=\Bbb{R}^4$ where we use global coordinates $(\xi,\eta,\zeta,\tau)$ and $N=\Bbb{R}^6$ with global coordinates $(x,y,z,u,v,w)$. Now consider the smooth differential $3$-form on $N$ defined by \begin{align} \omega=x\,dx\wedge dy\wedge du+ uv\,dz\wedge dv\wedge dw \end{align} This is just some random thing I pulled out of no where; it's obviously not the most general $3$-form on $N$, but it's just some particular example I came up with.
Now, consider the smooth mapping $f:M\to N$ defined as $f(\xi,\eta,\zeta,\tau)=(\sin \xi, e^{\eta}, \cosh(\zeta\tau), \tau^3, \xi^2,\eta\zeta)$. Now, our goal is to calculate the pullback $f^*\omega$. This will be a $3$-form on $M$. You can now grind through the rules for exterior derivatives, using linearity, and that pullback commutes with exterior derivatives and that pullback of functions is just composition. If you do all this, you'll find \begin{align} f^*\omega&= (x\circ f)\,d(x\circ f)\wedge d(y\circ f)\wedge d(u\circ f)+ (u\circ f)(v\circ f)\,d(z\circ f)\wedge d(v\circ f)\wedge d(w\circ f)\\ &=(\sin \xi)\,d(\sin \xi)\wedge d(e^{\eta})\wedge d(\tau^3)+ (\tau^3)(\xi^2)\,d(\cosh(\zeta\tau))\wedge d(\xi^2)\wedge d(\eta\zeta) \end{align} In other words, to calculate the pullback $f^*\omega$, you look at the formula for $\omega$ and everywhere you see $x,y,z,u,v,w$ you replace them with their compositions by $f$, which are $\sin \xi, e^{\eta}, \cosh(\zeta\tau), \tau^3, \xi^2,\eta\zeta$ respectively, and then you simplify the resulting mess, remembering that $d$ acting on functions works just as you would expect: \begin{align} f^*(\omega)&=(\sin \xi)\,(\cos\xi\,d\xi)\wedge (e^{\eta}\,d\eta)\wedge (3\tau^2\,d\tau)\\ &\,\,+(\tau^3)(\xi^2)\,\left[\tau\sinh(\zeta\tau)\,d\zeta+\zeta\sinh(\zeta\tau)\,d\tau\right]\wedge (2\xi\,d\xi)\wedge ((d\eta)\cdot \zeta+ \eta\cdot d\zeta), \end{align} where for example to calculate $d(\cosh(\zeta\tau))$, we use the fact that it equals $\frac{\partial}{\partial\xi}(\cosh(\zeta\tau))\,d\xi+\frac{\partial}{\partial\eta}(\cosh(\zeta\tau))\,d\eta+ \frac{\partial}{\partial\zeta}(\cosh(\zeta\tau))\,d\zeta+ \frac{\partial}{\partial\tau}(\cosh(\zeta\tau))\,d\tau$. Now in the above expression for $f^*\omega$, you just group all the terms together as much as you can: \begin{align} f^*\omega&=3\tau^2e^{\eta}\sin(\xi)\cos(\xi)\,d\xi\wedge d\eta\wedge d\tau\\ &+2\xi^3\zeta\tau^4\sinh(\zeta\tau)\,d\zeta\wedge d\xi\wedge d\eta\\ &+\text{stuff}\,(d\zeta\wedge d\xi\wedge d\zeta)\\ &+2\xi^3\zeta^2\tau^3\sinh(\zeta\tau)\,d\tau\wedge d\xi\wedge d\eta\\ &+ 2\xi^3\eta\zeta\tau^3\sinh(\zeta\tau)\,d\tau\wedge d\xi\wedge d\zeta \end{align} Note that the term $d\zeta\wedge d\xi\wedge d\zeta=0$ due to the repeated $d\zeta$ term and the alternating nature of wedge products. Hence, the final answer is (assuming I didn't make any mistakes) \begin{align} f^*\omega&=3\tau^2e^{\eta}\sin(\xi)\cos(\xi)\,d\xi\wedge d\eta\wedge d\tau\\ &+2\xi^3\zeta\tau^4\sinh(\zeta\tau)\,d\zeta\wedge d\xi\wedge d\eta\\ &+2\xi^3\zeta^2\tau^3\sinh(\zeta\tau)\,d\tau\wedge d\xi\wedge d\eta\\ &+ 2\xi^3\eta\zeta\tau^3\sinh(\zeta\tau)\,d\tau\wedge d\xi\wedge d\zeta. \end{align} From here it is up to you to "reorder" the wedges as you wish.