Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S$ be a surface and let $f:S\rightarrow S$ be a diffeomorphism. We define the mapping torus $M_f$ of the pair $(S,f)$ to be the quotient

$$(S\times I) /\sim \quad \text{ where } \ (1,x) \sim (0,f(x))$$

i.e., we "glue the cylinder"$S\times I$ along $f$"

I'm given a contact differential form $w$ defined on $S$, and I'm asked to check that the form $w$ descends to a contact form on the mapping torus of $(S,f)$.

Now I know that a necessary condition for any function $f$ to pass to the quotient given a quotient map $q$ is for the function to be constant on quotient classes of $q$, but, do I need some other condition, say, on the pullback of $w$ by the quotient map $q:S\rightarrow M_f$?

share|improve this question
    
Is $w$ really defined in a surface (2-dimensional?). If so, how can it be a contact form? –  40 votes Jul 21 '13 at 6:01

1 Answer 1

up vote 1 down vote accepted

on the pullback of $w $ by the quotient map $q:S→M_f$

First, the quotient map is from $S\times I$ to $M_f$, not from $S$. Second, you can't pull back $w$ by the quotient map because $w$ lives on the domain of the quotient map.

What you want to show is that there is a differential form $\zeta$ on $M_f$ such that the pullback of $\zeta$ by $q$ is $w$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.