I am interested in the following situation: Given any $n>1$ suppose I have a codimension-1 foliation of $R^n\_{++}$ (i.e. the subset of strictly positive $n$-vectors) arising from an $(n-1)$-dimensional, involutive, plane field $F$. Further, suppose that the normal to the subspace spanned by the vectors of each $F\_p$ is either strictly positive for all $p \in R^n\_{++}$ or strictly negative for all $p \in R^n\_{++}$ (so I take it to be the former).


My understanding is that this situation can arise from considering the inverse images of $f(p),p \in R$, where f is a smooth real-valued function on $R^n\_{++}$ (with surjective differential everywhere). My question is: Does such a foliation *have* to arise in this manner? Does there always exist such an $f$? The reason I ask is because I would like to be able to construct a complete ordering of the leaves of this type of foliation; if my question has an answer in the affirmative then I can do it by means of $f$. Otherwise, I would have to find a different way (say by considering "positive" line segments going from one leaf to another). I've looked at a paper by Novikov dealing with partial orderings of codimension-1 foliations but the foliation itself is too general for my needs.


Any references to material dealing with such a question would be welcome (as would any corrections to my understanding of the subject matter).



Codimension-1 foliations of Euclidean space with strictly positive normal bundle
I believe the answer is yes. It's enough to assume that the last coordinate of each normal is positive. Then each leaf is a graph $x\_n=x\_n(x\_1,\ldots, x\_{n-1})$ on some domain (depending on the leaf) in $\mathbb R^{n-1}$. This function must go to infinity at the boundary of the domain (if the boundary is nonempty) which implies that all leaves are proper. If they are proper then each leaf separates $\mathbb R^n$. This seems clear but there is also a reference (see section 2 in [this paper](https://link.springer.com/article/10.1007%2FBF02621903)). This allows you to get a linear ordering on the space of leaves given your condition on the normals.


Edit: sorry my example claiming that the leaf space might be non Hausdorff was wrong so I removed it. You always get that in this situation the leaf space is Hausdorff. Given any two leaves $F\_1<F\_2$ the space of leaves between them $\{F\_1<F<F\_2\}$ is open and such sets can separate any two leaves. Then the space of leaves in $\mathbb R$ and the projection from $\mathbb R^n$ to that $\mathbb R$ is the desired function $f$. I am not sure what can be said about the regularity of this function however.


Edit: The question was changed to a foliation on the first quadrant $\mathbb R^n\_{++}$. This case easily reduces to the case of a foliation on $\mathbb R^n$ considered above. Take a increasing diffeomorphism $\phi: (0,\infty)\to \mathbb R$. Then use $\Phi:\mathbb R^n\_{++}\to\mathbb R^n$ given by $\Phi(x\_1,\ldots, x\_n)=(\phi(x\_1),\ldots, \phi(x\_n))$. Push forward the hyperplane field to $\mathbb R^n$ using this map. The assumption on the normals is preserved.


