# Modifying the Reeb vector field by multplying by a function

Given a contact 3-manifold $(M,\omega)$ and its Reeb vector field $R$ and contact structure $\Delta$, I want to understand in some sense 'how large' is the set of Reeb vector fields supported by $\Delta$ which is the set of Reeb vector fields $R_f$ associated to every $f\omega$ for $f$ positive function.

1- More specifically I want to understand whether if multiplying by a non-zero function $f$ can I obtain a a contact structure whose Reeb vector field $R_f$ lies in a fixed 2 dimensional subbundle of $TM$.

2- A more concrete question related to this would be given a 2 dimensional subbundle of $TM$ can I always find a contact structure whose Reeb vector field is parallel to this 2-dimensional subbundle (in the case of 1 dimensional subbundles the answer is negative as shown in this topic: In a contact manifold, is every tranverse 1-foliation given by some Reeb vector field?)

Any reference where people study such properties of Reeb vector fields would also be welcome.

• How can a 3-manifold be symplectic? – José Figueroa-O'Farrill Feb 12 '15 at 22:51
• @JoséFigueroa-O'Farrill: I imagine that the OP mean 'contact', rather than 'symplectic'. – Robert Bryant Feb 12 '15 at 23:43
• @RobertBryant: I'd agree. It certainly makes the question more sensible! – José Figueroa-O'Farrill Feb 13 '15 at 9:42

## 2 Answers

The answer to your first question depends on the $2$-dimensional subbundle of $TM$. (I don't have an answer to your second question, which is harder.)

Suppose that $\Delta$ is a contact structure on $M^3$ with $\Delta$ defined by a $1$-form $\omega$ such that $\omega\wedge\mathrm{d}\omega\not=0$. Let $X$ be the Reeb vector field for $\omega$, i.e., $\omega(X) = 1$ and $\iota(X)(\mathrm{d}\omega) = 0$. (Here, $\iota(X)$ means the interior product with $X$.)

If $f$ is a nonzero function on $M$, and $\tilde\omega = f\omega$, then the Reeb vector field for $\tilde\omega$ is $$\tilde X = \frac1{f}\ X + F,$$ where $F$ is the unique vector field that satisfies $\omega(F)=0$ and $$\iota(F)(\mathrm{d}\omega) = \frac{\mathrm{d}f - \mathrm{d}f(X)\ \omega}{f^2}.$$

Now, suppose that one has a $2$-plane field on $M$ defined as the kernel of a nowhere vanishing $1$-form $\theta$ on $M$. Then one can write $\theta$ uniquely in the form $\theta = a\,\omega + \iota(B)(\mathrm{d}\omega)$ where $a$ is a smooth function on $M$ and $B$ is a vector field that satisfies $\omega(B) = 0$. To know whether $\tilde X$ lies in the kernel of $\theta$, one just evaluates to get $$0 = \theta(\tilde X) = \frac{a}{f} + \iota(F)\bigl(\iota(B)(\mathrm{d}\omega)\bigr) = \frac{a}{f} - \iota(B)\bigl(\iota(F)(\mathrm{d}\omega)\bigr) = \frac{a}{f} - \frac{\mathrm{d}f(B)}{f^2}$$ Thus, the condition on $f$ is the first order linear PDE $$\mathrm{d}f(B) - af = 0.$$ Since $f$ is nowhere vanishing, we can assume that $f$ is positive and write it in the form $f = e^u$ for some smooth function $u$ on $M$, so that the equation becomes $$\mathrm{d}u(B) = a.$$ Note that one can't have $B=0$ at any point where there is a solution because that would force $a$ to vanish there, which would make $\theta$ vanish there. Thus, $B$ must be nowhere vanishing if there is to be a solution, i.e., $\theta\wedge\omega$ must be nonvanishing. That's one obvious condition.

However, there are more global conditions: For example, if $B$ has a closed integral curve and $a$ is positive along that integral curve, then there cannot be any solution $u$ to the above equation along this curve, because $u$ would have to be strictly increasing along this closed curve. Since it is easy to construct a $B$ with a closed integral curve on such a $3$-manifold and then choose $a$ to be a positive function on $M$, there will always be $2$-plane fields for which there is no solution to your problem.

Another global condition comes directly from the formula for $\tilde X$. Note that, at any critical point of $f$ (i.e., where $\mathrm{d}f$ vanishes), one must have $F$ vanish as well, in which case $\tilde X$ will be a multiple of $X$. Thus, if your $2$-plane field $\theta=0$ is always transverse to the line field spanned by $X$ (i.e., if $a$ is nowhere vanishing), then there can't be a solution in this case either.

• Based on your answer I was able to derive some conditions which are more geometric, but I am still not sure if they admit global solutions. If you can have a look an comment I would be very greatful. Moreover this also means that this is always locally possible right? (contrary to the case where you want the Reeb vector field to be exactly on a specified line bundle which turns out to be a system of PDE). – Avicenna Mar 22 '15 at 10:52
• As long as the $2$-plane field you are trying to get $X$ to lie in is distinct from the contact $2$-plane field (i.e., in the above notation $\theta\wedge\omega$ is nowhere vanishing), the equation $\mathrm{d}u(B)-a = 0$ is always locally solvable for $u$. In fact, the solution is trivial in flow box coordinates for $B$: If $x = (x^i):U\to\mathbb{R}^3$ are $B$-flowbox coordinates on an open set $U\subset M$, i.e., $B = \partial_1$ in $U$, then, in $U$, the solution for $u$ is $$u(x^1,x^2,x^3)=v(x^2,x^3) +\int_0^{x^1}a(t,x^2,x^3)\,\mathrm{d}t$$ for any arbitrary function $v$ of two variables. – Robert Bryant Mar 22 '15 at 16:41
• (cont.)... Thus, in this 'generic' case, the obstructions are all necessarily global. – Robert Bryant Mar 22 '15 at 16:45
• Thanks, any directions or ideas as to how to understand whether if a global solutions is possible? My first attempt was to express the PDE as much as possible using geometric objects. But I have never though about global solutions to PDE before so any keywords or referenes or anything would be most welcome. – Avicenna Mar 22 '15 at 17:51
• @Avicenna: In general, I think that the global theory is not good. You are asking how to tell when a given function $a$ on $M^3$ is globally the Lie derivative of some other function $u$ by a given (nowhere vanishing) vector field $B$. The dynamics of the vector field $B$ can be very complicated, which strongly influences whether or not there are global solutions. I already mentioned obvious things such as when $B$ has closed orbits or when $a$ is everywhere positive, but much more complicated and delicate things can happen in this situation, and general theorems are few and far between. – Robert Bryant Mar 23 '15 at 10:36

Based on the answer of Robert Bryant, I was able to derive some more conditions which look a bit more geometric. As above $\omega$ is the contact form, $\Delta$ the contact structure, $X$ is the reeb vector field and $\theta$ is the other 1-form which defines the plane distribution $P$. I will assume to start with that $P$ and $\Delta$ are distinct everywhere. I will work local coordinates. So they intersect in a one dimensional line bundle spanned by some vector field $Y$ so that $\theta(Y) = \eta(Y)=0$.

I derived two conditions.

First one:

Now I denote by $\eta_f = f\eta$ and its Reeb vector field $R_f$. Now since $d\eta_f(R_f,\cdot)=0$, one has that $\theta(R_f)=0$ if and only if $d\eta_f = \theta \wedge \beta_f$ for some everywhere non-vanishing 1-form $\beta_f$. This means $d\eta_f|_P=0$. Taking $Y$ as above and $X$ any other vectorfield in $P$, expanding $d\eta_f = df \wedge \eta + f d\eta$ and evaluating this 2-form on $(X,Y)$ one gets the equation:

$$\mathcal{L}_Y(ln(f)) = \eta(\mathcal{L}_Y(\frac{X}{\eta(X)}))$$

This is the first equation, I don't know if it admits a global solution.

Second one:

We know that $\theta \wedge d\eta_f=0$ by above. We let $Z$ be a vectorfield inside $\Delta$ distinct from $Y$. Then we evaluate this on $(Z,Y,R)$ to get

$$\mathcal{L}_Y(ln(f)) = \frac{\theta(R)}{\theta(Z)}d\eta(Z,Y)$$

In this case the quantity $d\eta(Z,Y)$ can be related to how the contact structure $\Delta$ twists along the integral curves of $Y$ and I have hopes somehow that right hand side can be written as the differential of a function but I am not sure because here the ugly term is $\theta(R)$ which may vanish at some points.