Show that the path integral of the pullback of a 1-form is equal to the integral of the 1-form over the transformation image of the curve

Question

This is from Edwards's Advanced Calculus of Several Variables, Exercise V2.15, page 321.

Let $\omega$ be a continuous 1-form and $\gamma:[a,b]\to\mathbb{R}^{2}$ be $\mathscr{C}^{1}.$ Show that

$$ \int_{\gamma}F^{\ast}\omega=\int_{F\circ\gamma}\omega. $$

Based on discussions elsewhere in the text, I assume $F:\mathbb{R}_{uv}^{2}\to\mathbb{R}_{xy}^{2}$ is sufficiently well-behaved. So it is reasonable to write our 1-form as

$$ \omega=\mathcal{P}\mathbf{d}x+\mathcal{Q}\mathbf{d}y, $$

where $\mathcal{P},\mathcal{Q}:\mathbb{R}_{xy}^{2}\to\mathbb{R}.$ The pullback reparameterizes $\mathcal{P},\mathcal{Q}$ such that $F^{\ast}\mathcal{P},F^{\ast}\mathcal{Q}:\mathbb{R}_{uv}^{2}\to\mathbb{R}.$ It also rewrites $\mathbf{d}x,\mathbf{d}y$ so that they take vectors in $\mathbb{R}_{uv}^{2}$ as arguments. That is

$$ F^{\ast}\mathbf{d}x=\left(\frac{\partial\mathit{x}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{x}}{\partial v}\mathbf{d}v\right), $$

$$ F^{\ast}\mathbf{d}y=\left(\frac{\partial\mathit{y}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{y}}{\partial v}\mathbf{d}v\right). $$

So that the pullback of $\omega$ is

$$ F^{\ast}\omega=\mathcal{P}\circ F\left(\frac{\partial\mathit{x}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{x}}{\partial v}\mathbf{d}v\right)+\mathcal{Q}\circ F\left(\frac{\partial\mathit{y}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{y}}{\partial v}\mathbf{d}v\right) $$

From this it is intuitively obvious that the assertion of the exercise holds. That is, at every point along the path $\gamma$ the components of the 1-form $\omega$ are evaluated using the image $F\left(\gamma\left(t\right)\right)$, and $\omega$ operates on the image of the velocity vector, which is the velocity vector of the image curve.

But I'm not sure how to state this in a rigorous way.

The definition Edwards gives for the path integral of a 1-form is

$$ \int_{\gamma}\omega=\int_{a}^{b}\omega_{\gamma\left(t\right)}\left(\gamma^{\prime}\left(t\right)\right)dt. $$

The only thing that comes to mind is appeal to the Riemann sum definition of the integral.

How might the assertion of the exercise be demonstrated rigorously?

Answer 1

I think it is better to do this using the general definitions, of the line integral and pullback. Going to coordinates clutters everything up.

$\int_{\gamma}\omega:=\int_{a}^{b}\omega_{\gamma\left(t\right)}\left(\gamma^{\prime}\left(t\right)\right)dt=\int_{[a,b]}\gamma^*\omega\Rightarrow \int_{F\circ\gamma}\omega=\int_{[a,b]}(F\circ \gamma)^* \omega=\int_{[a,b]}\gamma^*F^* \omega=\int_{\gamma}F^*\omega.$

To do the calculation rigorously from scratch, (with the understanding that I do not know how Edwards introduces these ideas) note that

$1).\ \gamma'(t_0)$ operates on smooth functions $f$ as follows: $\gamma'(t_0)f:=\left(\frac{df\circ \gamma(t)}{dt}\right )\big |_{t=t_0}$. So to distinguish the operator from the ordinary derivative, we write the latter with a dot above the functional symbol. So, $\overset{.}{\gamma}(t)$ means the ordinary derivative of $\gamma$ at $t$.

$2).\ $ The symbol $x$ is the $\textit{function}:(x,y)\mapsto x$. That is, the projection onto the first coordinate. Similarly for $y$.

$3).\ $ We have $\gamma(t)=(\gamma^1(t),\gamma^2(t))=(x,y)$ and the differentials $dx$ and $dy$ operate on $\gamma'(t)$ as follows: $dx(\gamma'(t))=x\circ \gamma(t)=\gamma^1(t)$ and similarly for $dy$.

$4).\ $ the action of $F^*$ on $\omega$ will be clear in what follows.

To start the calculation, we have $\omega_{\gamma(t)}=P(\gamma(t))dx+Q(\gamma(t))dy.$ Now

$F^*\omega_{\gamma(t)}=P(F\circ\gamma(t))d(x\circ F)+Q(F\circ\gamma(t))d(y\circ F)=$

$P(F\circ\gamma(t))dF^1+Q(F\circ\gamma(t))dF^2$ so

$F^*\omega_{\gamma(t)}(\gamma'(t))=P(F\circ\gamma(t))dF^1(\gamma'(t))+Q(F\circ\gamma(t))dF^2(\gamma'(t))=$

$P(F\circ\gamma(t))\overset{.}{(F\circ\gamma})^1(t)+Q(F\circ\gamma(t))\overset{.}{(F\circ\gamma)^2}(t))$ so

$\int_{\gamma}F^*\omega=\int_{a}^{b}F^*\omega_{\gamma\left(t\right)}\left(\gamma^{\prime}\left(t\right)\right)dt=\int_{a}^{b}P(F\circ\gamma(t))\overset{.}{(F\circ\gamma}^1)(t)+Q(F\circ\gamma(t))\overset{.}{(F\circ\gamma)^2}(t))dt$

On the other hand,

$\omega_{F\circ\gamma(t)}(F\circ\gamma)'(t)=P(F\circ\gamma(t))dx(F\circ\gamma)'(t)+Q(F\circ\gamma(t))dy(F\circ\gamma)'(t)=$

$P(F\circ\gamma(t))(F\circ\gamma)'(t)x+Q(F\circ\gamma(t))(F\circ\gamma)'(t))y=$

$P(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^1(t)+Q(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^2(t))$

so

$\int_{F\circ\gamma}\omega=\int_{a}^{b}\omega_{F\circ\gamma\left(t\right)}\left(F\circ\gamma)^{\prime}\left(t\right)\right)dt=\int_{a}^{b}(P(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^1(t)+Q(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^2(t)))dt.$

The items in blockquotes are equal, which finishes the proof.

Answer 2

Here is a short proof of the statement up top:

Theorem: Let $F:M \to N$ be smooth and $\gamma:[a,b] \to M$ a smooth curve segment. If $\omega$ is a smooth cotangent vector field on $N$, prove $$\int_\gamma F^* \omega = \int_{F \circ \gamma} \omega.$$ Proof: Note we have, \begin{align*} \int_{F \circ \gamma} \omega&=\int_{[a,b]} (F \circ \gamma)^* \omega && \text{definition}\\ &=\int_{[a,b]} \gamma^* F^* \omega && \text{property of $*$}\\ &=\int_\gamma F^* \omega && \text{definition} \end{align*} as needed. $\blacksquare$