# Calculate $r^*\omega$ and show that $\int _{[c,d]} r^*\omega = \int _{[a,b]} \omega$.

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?

EDIT:

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?

• $r^*$ is the pullback, see for instance math.stackexchange.com/questions/1343431/… – mcd Mar 14 '17 at 16:19
• @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. – Barbara Mar 14 '17 at 16:27
• @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? – Barbara Mar 14 '17 at 16:34
• @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! – Barbara Mar 14 '17 at 21:34
• 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 $p$. – mcd Mar 15 '17 at 9:39