EDIT: changes to the question are in bold.

Suppose I have a smooth ($C^1$) codimension one foliation $\mathscr{F} \subset P$, the open subset of $R^n$ consisting of points with all positive components. The leaves of $\mathscr{F}$ are connected and proper and each is the graph of a function $f$ defined on some domain $D$ contained in $Q$, the open subset of $R^{n-1}$ consisting of points with all positive components ($D$ may depend on the leaf). Further, at every point $p \in P$, the normal to the tangent plane of the leaf $L$ passing through $p$, at that point, is non-negative with last coordinate strictly positive. If I now define the diagonal $\Delta \equiv R^n \bigcap \{ x: x_1 = x_2 = \cdots = x_n \} \subset P$ then I know that through every point $t$ of $\Delta$ passes a leaf $L$ (integral surface) of the foliation and, furthermore, $\Delta$ is transversal to $L$ (because of the normal condition).

Can I then conclude that $\Delta$ passes through every leaf of $\mathscr{F}$? If not, does there exist some transversal in the positive direction passing through every leaf? I have researched this question; i.e. the existence of a global transversal for a foliation of this type, every way I know. The paper "Foliations and global inversion" by E. Cabral Bereira seems to suggest (Lemma 4.3) that we can always find a such a curve by "standard means", but I may not be applying that result correctly. But Danny Calegari (Section 1.2 of Chapter 4 of notes on his book) states that there only exists a finite collection of transversals which together pass through every leaf of a foliation, which of course is not what I need.

I would appreciate any ideas you might have for further research.

Edit: After discussion with Sam Nead and some more thought it is also apparent that if we denote by $\delta$ the (orthogonal) projection of $\Delta$ into $Q$ then $\Delta$ does not intersect a leaf $L$ only if the domain of $L$, say $D_L$, does not intersect $\delta$. However, more is necessary as the closure of $D_L$ cannot be contained entirely in $Q \setminus \delta$ as $L$ must separate $P$ into two connected components.

  • $\begingroup$ You are using $D$ both for the "domain of definition of $f$" and for the diagonal. You might want to fix this. Also, with the new version of the question, isn't the answer trivially "no"? This is because there may be leaves whose domain of definition is far from the diagonal. $\endgroup$
    – Sam Nead
    Jul 19, 2022 at 10:24
  • $\begingroup$ @Sam Nead Thanks for pointing this out, edits made. Yes, I guess really the question is is it possible for D to not intersect the diagonal given the properties of the leaves. In particular, it is known that each leaf L separates P into two connected components. That plus the fact we KNOW through each point of the diagonal passes a leaf and all leaves have positive normals might be enough to get the result. But I can't seem to prove it, which is unfortunate, since I need a way to index the leaves so I can obtain a "monotonic" projection onto the leaf space. $\endgroup$
    – user167131
    Jul 19, 2022 at 11:29
  • $\begingroup$ The only "propert[y] of the leaves" that can save us is "properness". What do you mean by "properness" in this context? $\endgroup$
    – Sam Nead
    Jul 19, 2022 at 18:35
  • $\begingroup$ @Sam Nead Each leaf is the image of a proper map. In this case, I believe that amounts to f being a proper map from D to R, which I take to mean the inverse image of every compact subset K in R is compact in D. Equivalently, for every sequence of points in D that approach the boundary the sequence of values of f must "escape to infinity". There is a result that says in this context a leaf is proper if and only if it is closed. $\endgroup$
    – user167131
    Jul 19, 2022 at 18:57
  • 1
    $\begingroup$ @Sam Nead: I appreciate the hint, and I WANT to do this as much as possible by myself, but I can't for the life of me see why the assumption that the leaf (graph) is proper would imply that d + (t,t,....,t) is in D whenever d is. Properness is satisfied (IIRC) if f goes to infinity at a point on the boundary of D (that is not in the boundary of P). Further, isn't d + (t,t,...,t) going to be parallel to the diagonal in Q, so will never meet it? I must be missing something. Also, your comments/answers have been extremely useful and I appreciate your willingness to help. $\endgroup$
    – user167131
    Jul 20, 2022 at 22:12

1 Answer 1


Trying to answer the new version of the question.

We claim that, yes, the diagonal $\Delta$ meets all leaves.

Let $R$ be the plane spanned by $(0, 0, 0, \ldots, 0, 1)$ and $(1, 1, 1, \ldots, 1, 1)$. Note that $R$ contains the ray $\Delta$. Let $\Gamma = R \cap Q$ - that is, $\Gamma$ is the orthogonal projection of $\Delta$ "down" to $Q$.

Suppose that $L$ is a leaf of the given foliation $\mathcal{F} \subset P$. Let $D$ be the image of the orthogonal projection of $L$ into the hyperplane $Q$. We are told that the "inverse" of this projection is a continuous map $f \colon D \to L$. Let $x$ be any point of $D$. Define $R_x$ to be the two-dimensional plane which (1) contains $x$ and (2) is parallel to $R$. Define $D_x = D \cap R_x$ and $L_x = L \cap R_x$. Finally define $f_x = f|D_x \colon D_x \to L_x$. It is an exercise to show that $f_x$ is non-strictly monotonically decreasing. So we have shown the following.

Lemma: If $D$ meets $\Gamma$, then $L$ meets $\Delta$.

[Now things become a bit vague.]

If $D$ has no boundary, then $D = Q$ and we are done. Suppose that the boundary $\partial D$ is non-empty and "nice" (a smooth codimension-one submanifold of $Q$, properly embedded in $Q$). Let $y$ be a point of $\partial D$ and let $n_y$ be the unit vector at $y$ which (1) is normal to $\partial D$, (2) is tangent to $Q$, and (3) points into $D$. Note that the last coordinate of $n_y$ is zero.

Note that $f$ either blows up, or vanishes, at $y$. I claim that if $f$ blows up at $y$ then all coordinates of $n_y$ (except the last) are positive. On the other hand, if $f$ vanishes at $y$ then all coordinates of $n_y$ (except the last) are negative. Thus $\partial D$ has a "convexity property", and so meets $\Gamma$, and we win.

Below we answer a previous version of the question - see the edit history.

Yes, $\Delta$ meets all leaves. Let $R$ be the plane spanned by $(0, 0, 0, \ldots, 0, 1)$ and $(1, 1, 1, \ldots, 1, 1)$. Note that $R$ contains $\Delta$. Every leaf meets $R$ and its intersection with $R$ is a graph over $Q \cap R$. By your hypotheses these graphs are (non-strictly) monotonically decreasing. So by the intermediate value theorem they meet $\Delta$.

  • $\begingroup$ I see exactly what you're saying, thank you, but I left out a detail that may affect your answer. Sorry, I do this all the time. In fact each leaf is a graph of a function defined on some DOMAIN D in Q, not necessarily on all of Q. I think that will affect things as then we can't be sure that each leaf meets R. I'll edit the original question. $\endgroup$
    – user167131
    Jul 18, 2022 at 21:18
  • $\begingroup$ Thanks for all your work, I know I needed to add complicating detail(s) to the original question, making things harder. I follow your latest response up to (2) in the next to last paragraph. I thought you were working in (n-1) dimensions, with Q the containing manifold. If so then in what sense is n_y tangent to Q? I also don't follow your last paragraph, but I'm sure this is my fault. I will work at it and see if I can figure out exactly what's going on. $\endgroup$
    – user167131
    Jul 21, 2022 at 19:57
  • $\begingroup$ I am always thinking of $Q$ as embedded in $P$. So "tangent to $Q$" means that the final coordinate (the n^th coordinate) vanishes. $\endgroup$
    – Sam Nead
    Jul 21, 2022 at 20:17
  • $\begingroup$ I see about the normals but cannot figure out exactly what you mean by "convexity property" or why that would imply that D intersects Gamma. Could you offer an additional hint? (BTW, I have modified my original question to reflect my latest thoughts). Again, appreciate your help and could understand if you are getting bored. :) $\endgroup$
    – user167131
    Jul 23, 2022 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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