Let $ [a,b]\in \mathbb{R} $ be a closed interval. Let $ \omega\in \Omega^1([a,b]) $ be a smooth 1-form (by this we mean that $ \omega $ is the restriction to $ [a,b] $ of a smooth 1-form defined on some open interval containing $ [a,b] $.): this can be written as $$ \omega(t)=g(t)d $$ for some smooth function $ g $ on $ [a,b] $. Define $$ \int _{[a,b]} \omega =\int_{a}^{b} g(t)dt $$ Let $ r:[c,d]\to [a,b] $ be a smooth map such that $ r(c)=a $ and $ r(d)=b $. I want to calculate $ r^*\omega $ and show that $$ \int _{[c,d]} r^*\omega = \int _{[a,b]} \omega. $$ The problem is that I don't understand what $ r^* $ is, and how it acts on $ \omega $. I think it's domain is $ \Omega^1([a,b]) $, but I don't know what it does with these 1-forms. Can someone help me out?


I think I got it now. Since $ \omega(t)=g(t)dt $, the pull-back $ r^* $ on $ \omega $ is $$r^*\omega=r^*\omega(t)=r^*g(t)dt=g(r(t))dt$$ Hence, \begin{align*} \int_{[c,d]}r^*\omega&=\int_{c}^{d}g(r(t))dt\\ &=\int_{a}^b g(t)dt\\ &=\int_{[a,b]} \omega \end{align*} where I have used that $r^*\omega=g(r(t))dt$, $ r(c)=a $ and $ r(d)=b $.

Is it correct?

  • $\begingroup$ $r^*$ is the pullback, see for instance math.stackexchange.com/questions/1343431/… $\endgroup$ – mcd Mar 14 '17 at 16:19
  • $\begingroup$ @mcd. So $r^*\omega$ is just $\omega(r)$? Can I state that without feeding it something? In my book, the pull-back for 1-forms is defined as $$f^*(\omega)(p)(v)=\omega(f(p))(f_*(v)),$$ but it is not clear to me what $p$ and $v$ is in my case. $\endgroup$ – Barbara Mar 14 '17 at 16:27
  • $\begingroup$ @mcd. If $ r^*\omega=\omega(r) $, then I get something like this: $$ \int_{[c,d]}r^*\omega=\int_{[c,d]}\omega(r)=\int_{c}^{d}g(r(t))dt=\int_{a}^b g(t)dt=\int_{[a,b]} \omega $$ but step 2 and 3 is kind of just because I see it works, I don't know if it is correct. Can I just put $ r $ between $ g $ and $ t $ like that, and if yes, why? $\endgroup$ – Barbara Mar 14 '17 at 16:34
  • $\begingroup$ @mcd. I think I figured it out now. Can you look at the edit above and let me know what you think? Thanks a lot! $\endgroup$ – Barbara Mar 14 '17 at 21:34
  • $\begingroup$ No, if $r : M \to N$ then the pullback maps $\Omega^p(N) \to \Omega^p(M)$. In your case $p$ is a point in the interval and $\nu$ is a cotangent vector at $p$. $\endgroup$ – mcd Mar 15 '17 at 9:39

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