Tube Structures of Co-rank 1 with Forms Defined on Compact Surfaces

Abstract

We study the global solvability of a locally integrable structure of tube type and co-rank 1 by considering a linear partial differential operator ${\mathbb {L}}$ associated to a general complex smooth closed 1-form c defined on a smooth closed n-manifold. The main result characterizes the global solvability of ${\mathbb {L}}$ when $n=2$ in terms of geometric properties of a primitive of a convenient exact pullback of the form $\mathfrak {Im}(c)$ as well as in terms of homological properties of $\mathfrak {Re}(c)$ related to small divisors phenomena. Although the full characterization is restricted to orientable surfaces, some partial results hold true for compact manifolds of any dimension, in particular, the necessity of the conditions, and the equivalence when $\mathfrak {Im}(c)$ is exact. We also obtain informations on the global hypoellipticity of ${\mathbb {L}}$ and the global solvability of ${\mathbb {L}}^{n-1}$—the last non-trivial operator of the complex when M is orientable.

Appendix

1.1 A Proof of Arnold’s Lemma

Let $M^\dag$ be a surface (possibly with a non-empty boundary made up of piecewise smooth curves) on which the closed 1-form $b^\dag$ is defined. Lemma 4 in [1] implies:

Lemma A.1

Let G be a connected component of $M^{\dag }\setminus \Sigma (b^\dag ).$ Then

In order to make the text as self-contained as possible, we offer a more detailed proof of the version for surfaces that we need.

Our strategy is to apply Stokes’ Theorem to a convenient set whose boundary consists of piecewise smooth curves and is sufficiently close to G. As before, we will consider a smooth triangulation h of $M^\dag$ extending a smooth triangulation $h_0$ of its boundary, that is, $h^{-1}\circ h_0$ is a homeomorphism carrying simplices linearly. Given $j\in {\mathbb {Z}}^+$, we will again subdivide $h_n(T_0)$ into “triangles” of side length smaller than $2^{-j}.$

Set $\Sigma _0$ for the subset of $\Sigma (b^\dag )$ of points where $b^\dag$ vanishes to order greater than or equal to 2, that is, writing locally

$$\begin{aligned} b^\dag =\partial _1B^\dag (x,y)\,dx+\partial _2B^\dag (x,y)\,dy, \end{aligned}$$

a point $p=(x_0,y_0)\in \Sigma (b^\dag )$ belongs to $\Sigma _0$ if and only if

$$\begin{aligned} \partial _{11} B^\dag (x_0,y_0)=\partial _{12} B^\dag (x_0,y_0)=\partial _{22} B^\dag (x_0,y_0)=0. \end{aligned}$$

Consider $q\in {\overline{G}}\setminus G$ that is not in $\Sigma _0.$ We can assume that $\partial _{11}B^\dag (q)\ne 0.$ In a sufficiently small triangle $\Delta$ containing q, $\Delta \cap \Sigma (b^\dag )=\Delta \cap (\Sigma (b^\dag )\setminus \Sigma _0)\subset \Delta \setminus \Gamma$, where $\Gamma$ is the graph of a smooth function by the implicit function theorem. Note that $\Delta \setminus \Gamma$ divides $\Delta$ into two components. There are two possibilities: (i) both components are contained in G; (ii) one of them does not intersect G.

Let $\Sigma _{1}$ be the set of points for which (i) holds. Denote by ${\mathcal {T}}^1$ the family of triangles that either contain these points or intersect $\Sigma _0.$

We say that $q\in \Sigma _{2}$ if (ii) holds, and denote by ${\mathcal {T}}^2$ the family of triangles that contain a point q of this type.

Also, denote by ${\mathcal {T}}^0$ the set of triangles intersecting $G\cap \partial M^\dag .$

Now define

$$\begin{aligned} G^\dag \doteq \bigcup _{T\in {\mathcal {T}}^0\cup {\mathcal {T}}^1} T \cup \bigcup _{T\in {\mathcal {T}}^2} \overline{G}\cap T. \end{aligned}$$

Therefore, for j big enough we can distinguish three subsets of $\partial G^\dag .$ The subset $X_0=\partial M^\dag \cap G^\dag$ will be a finite union of closed intervals, which are union of sides of triangles in ${\mathcal {T}}^0.$

The subset $X_1$, in turn, consists of sides of some triangles in ${\mathcal {T}}^1$ intersecting $\Sigma _0.$

Finally, the subset $X_2$ is locally represented by a smooth curve $\Gamma$, containing $\Sigma _{2}$, on which $b^\dag$ vanishes.

We conclude that $G^\dag$ has a piecewise smooth boundary and, by applying Stokes’ Theorem, we obtain

$$\begin{aligned} 0=\int _{G^\dag } db^\dag =\int _{\partial G^\dag } b^\dag =\sum _{k=0}^2\int _{X_k} b^\dag . \end{aligned}$$

Writing $J_k\doteq \int _{X_k}b^\dag$, $k=0,1,2$, and observing that $J_2=0$, we see that in order to prove $(\clubsuit )$ we need to show that

$$\begin{aligned} J_0\rightarrow \int _{\partial M^\dag \cap G} b^\dag \text { and }J_1\rightarrow 0\,\ \text {as } j\rightarrow \infty . \end{aligned}$$

For the first limit, notice that we can estimate the difference $\int _{\partial M^\dag \cap G} b^\dag -J_0$ by considering the integral of $|b^\dag |$ on sides of triangles $T\in {\mathcal {T}}^0$ in $\partial M^\dag$ containing points of $\Sigma (b^\dag ).$ On such a triangle, Taylor’s Theorem asserts that $\Vert b^\dag \Vert _{\infty }\leqslant C_02^{-j}.$ If $\ell$ is any side of T, then

$$\begin{aligned} \int _\ell |b^\dag |\leqslant C_02^{-2j}. \end{aligned}$$

Since the number of such sides is bounded by ${{\text {vol}}(\partial M^\dag )2^j}/{{\tilde{c}}}$, the difference goes to 0.

Now, consider $T\in {\mathcal {T}}^1.$ Such a triangle contains a point on which $b^\dag$ vanishes to order 2, and thus $\Vert b^\dag \Vert _{\infty }\leqslant C_12^{-2j}$ on T. If $\ell$ is any side of T, then

$$\begin{aligned} \left| \int _\ell b^\dag \right| \leqslant C_12^{-3j}. \end{aligned}$$

The number of sides of triangles in ${\mathcal {T}}^1$ is bounded by $C'_12^{2j}$, as in (3.3), and then we have

$$\begin{aligned} \int _{X_1} b^\dag \rightarrow 0, \quad j\rightarrow \infty , \end{aligned}$$

which finishes the proof. $\square$

1.2 On a Technical Property of Semilevel Sets

Let ${\widetilde{M}}$ be a surface and ${\widetilde{B}}$ a real smooth function defined on ${\widetilde{M}}.$ Suppose that

the sublevel sets $\Omega _r=\{x\in {\widetilde{M}}:{\widetilde{B}}(x)<r\}$ are connected for every $r\in {{\mathbb {R}}}.$

If $p,q\in {\widetilde{M}}$ with ${\widetilde{B}}(q)<{\widetilde{B}}(p)\doteq a$ for every $0<\varepsilon <1$, $p,q\in \Omega _{a+\varepsilon }$ and there is a smooth curve $\gamma _\varepsilon$ contained in $\Omega _{a+\varepsilon }$ joining p and q. The following technical question is relevant to this paper:

Question

Is it possible to choose $\gamma =\gamma _\varepsilon$ such that its length satisfies $|\gamma _\varepsilon |\leqslant C\varepsilon ^{-k}$, where $C>0$ and $k\in {\mathbb {Z}}^+$ do not depend on $\varepsilon$?

The answer is known to be true when ${\widetilde{M}}$ is compact. The following example shows that this is not the case in general when ${\widetilde{M}}$ is not compact.

Example A.2

Set ${\widetilde{B}}(x,y)\doteq f(x)g(y)$, $(x,y)\in {{\mathbb {R}}}^2$ , where

$$\begin{aligned} f(x)&=\left( \frac{x^2}{3}-1\right) x+\frac{2}{3},\end{aligned}$$

(7.1)

$$\begin{aligned} \frac{1}{g(y)}&=-\ln \left( \frac{1}{2+y^2}\right) . \end{aligned}$$

(7.2)

The polynomial f has a double root at $x=1$, which is also a local minimum. The other critical point occurs at $x=-\,1$, which is a local maximum ($f(-\,1)=4/3$). In particular, f is strictly increasing on $(-\infty ,-\,1)$ and on $(1,\infty )$, and strictly decreasing on $(-\,1,1).$

The sublevel sets $\omega _r$ of f satisfy the following rules:

• If $r\geqslant f(-\,1)=4/3$, $\omega _r$ is an interval of the type $(-\infty ,s).$

• If $r\leqslant 0$, $\omega _r$ is an interval of the type $(-\infty ,s)$ with $s< -\,2$ (note that $f(-\,2)=0$).

• If $0<r<4/3$, $\omega _r$ is the union of two intervals $I_1=(-\infty ,s)$, $s<-\,1$, $I_2=(\alpha ,\beta )$, $-\,1< \alpha<\beta <\infty .$

Hence, in the latter case, $\omega _r$ will not be connected. Now, we claim that for every $r\in {{\mathbb {R}}}$, $\Omega _r$ is connected. In fact, the critical set of ${\widetilde{B}}$ is $\{(-\,1,0)\}\cup (\{1\}\times {{\mathbb {R}}})$, and $(-\,1,0)$ is a local maximum. Notice that the maximum value of ${\widetilde{B}}$ on the strip $\{-\,2<x<1\}$ is attained at $(-\,1,0)$ (${\widetilde{B}}$ converges uniformly to 0 on the strip when $|y|\rightarrow \infty$, or $x\rightarrow 1$, or $x\rightarrow -\,2$).

The function f is strictly increasing on the half-plane $\{x>1\}$, so it is foliated by the level curves of ${\widetilde{B}}$, which are graphs of $x=f^{-1}(c/g(y))$, $c>0$, converging to the level curve $\{x=1\}$ when $c\rightarrow 0^+.$ A similar analysis holds for the half-plane $\{x<-\,2\}$ and the level curves $x=f^{-1}(c/g(y))$, $c<0.$ The vertical strip $\{-\,2<x<1\}$ is foliated by closed curves around $(-\,1,0)$ and given by the equation $f(x)=c/g(y)$, $0<c<4g(0)/3={\widetilde{B}}(-\,1,0).$ When $c\rightarrow 0^+$, such curves converge to the pair $\{x=-\,2\}\cup \{x=1\}.$

When $r>{\widetilde{B}}(-\,1,0)$, the sets $\Omega _r$ are of the type $x<h(y)$ where $x=h(y)$ is a level curve in the half-plane $\{x>1\}$ (an analogous situation happens when $r\leqslant 0$). For the values of $0<r\leqslant {\widetilde{B}}(-\,1,0)$, the sets $\Omega _r$ are path-connected but not simply connected and to join a pair of points in $\Omega _r$ one has to go around the hole.

We turn now to the above question. Take $p=(1,0)$ and $q=(-\,2,0)$ (${\widetilde{B}}(p)={\widetilde{B}}(q)=0$), and given $\varepsilon >0$, we have that $p,q \in \Omega _\varepsilon .$ Nonetheless, a path $\gamma$ joining $p=\gamma (0)$ and $q=\gamma (1)$ must circumvent the closed curve ${\widetilde{B}}(x,y)=\varepsilon$ in the strip $\{-\,2<x<1\}.$ Therefore, for a parameter $s=s_0$, we have $\gamma (s_0)=(-\,1,y(s_0))$ with ${\widetilde{B}}(-\,1,y(s_0))=(4/3)g(\gamma (s_0))<\varepsilon$, which implies that

$$\begin{aligned} 2+y(s_0)^2\geqslant \exp [4/(3\varepsilon )]. \end{aligned}$$

As the length $|\gamma |$ satisfies $|\gamma |^2>y(s_0)^2+1$, its growth is not tempered when $\varepsilon \rightarrow 0.$

Remark A.3

The arguments in this paper show that the answer to the question is positive in the special case in which ${\widetilde{M}}$ is the minimal covering associated with a real closed form b, and ${\widetilde{B}}$ is a primitive of the pullback of b to ${\widetilde{M}}.$

Remark A.4

Notice that in this example the sublevel sets of${\widetilde{B}}$ are all connected but some of the superlevel sets are not. It is possible, however, to produce a ${\widetilde{B}}$ with connected semilevel sets by defining a more elaborated function g.