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Let \begin{equation} \omega:\mathbb{R}^n\rightarrow(\mathbb{R}^n)^* \end{equation} be a continuous differential form \begin{equation} \omega(\boldsymbol{x})=\sum_{i=1}^n a_i(\boldsymbol{x})dx_i \quad \forall \boldsymbol{x}\in\mathbb{R}^n \end{equation} and:

\begin{equation} \psi:[a,b]\rightarrow\mathbb{R}^n \end{equation} a regular curve.

We can define the line integral of $\omega$ over $\psi$:

\begin{equation} \int_{\psi}\omega=\int_a^b \sum_{i=1}^{n}a_i(\psi(t))\psi'_i(t)dt \end{equation}

Now, let \begin{equation} \boldsymbol{X}=F(\boldsymbol{x}) \end{equation} be an invertible coordinate transformation.

Let $\Gamma=F(\psi)$ be the image curve and consider $\tilde{\omega}(\boldsymbol{X})=\omega(F^{-1}(\boldsymbol{X}))$ (thus $\omega$ as a function of the new coordinates). Is it true that

\begin{equation} \int_{\Gamma} \tilde{\omega}=\int_{\psi}\omega \end{equation}

? If it is true, how do you prove it? Is there a theorem for it?

Mr. Feynman
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    First, you are missing the subscripts on the coefficients $a$. Second, your equation is not correct. You need to pull back $\omega$ by the mapping $F$; this does not just mean composing the coefficient functions with $F$. – Ted Shifrin May 11 '21 at 17:09
  • Added the missing index, sorry. My notation might be wrong or defective. What I meant by $\omega{F^-1(\boldsymbol(X)}$ was not just composing the coefficients but also expressing the elements of the dual basis $dx_i$ with respect to $dX_i$. Is this what you meant? – Mr. Feynman May 11 '21 at 18:23
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    Your notation is very wrong. Learn about the pullback, written $F^*\omega$. So, yes, you have to write $x_i = F_i(X)$, $dx_i = dF_i$. – Ted Shifrin May 11 '21 at 18:49
  • I see .I found what I was looking for here: https://math.stackexchange.com/questions/3429004/show-that-the-path-integral-of-the-pullback-of-a-1-form-is-equal-to-the-integral Please note that $F$ in the link is $F^{-1}$ in my question. – Mr. Feynman May 11 '21 at 19:49

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