20
$\begingroup$

I don't know if this is the right place for this question, if not then please let me know and I will delete it.

In 1974, Steve Smale published an article in the first issue of Journal of Mathematical Economics with the short title "Global Analysis and Economics" which bears on my current research interests. In it, he applies variants of the inverse function theorem/implicit function theorem to a set that is not a manifold and explains that these results "make sense" even in that context. Now please understand, I know who Steve Smale is so I am not saying he's wrong, but this is unlike anything I have seen in my research so I thought someone could help with some thoughts.

In brief, he defines the set $\mathscr{S} \equiv (\bar{P})^m \times S_+$, where $\bar{P}$ is the non-negative orthant of $R^l$ and $S_+$ is the non-negative part of the $l-1$-dimensional sphere $S$. Though his use of the inverse function theorem occurs in a number of places, I will quote one part of the article which should highlight my difficulty:

"We recall that $\psi_u : \mathscr{S} \rightarrow (S)^{m+1}$ is said to be transversal to $\Delta \subset (S)^{m+1}$ provided that for each $s$ with $\psi_u(s) = y \in \Delta$, the image of $D\psi_u(s)$ and $T_y(\Delta)$ (as subspaces of $T_y((S)^{m+1})$) span $T_y((S)^{m+1})$…. Note that although $\mathscr{S}$ is not exactly [italics mine] a manifold, the notion of transversality still makes sense in our context".

"It follows from the implicit function theorem that if $\psi_u$ is transversal to $\Delta$, then $\psi^{-1}_u(\Delta)$ is a submanifold of $\mathscr{S}$ with $\operatorname{codimension} \psi_u^{-1}(\Delta) = \operatorname{codimension} \Delta$. We note also that the notion 'submanifold' of $\mathscr{S}$ is a slight extension of the usual notion. A precise definition would be as follows: Take an open set $\mathscr{O}$ of $\mathscr{S}$ in $(R^l)^m \times S$. Then a 'submanifold' of $\mathscr{S}$ is a subset of $\mathscr{S}$ of the form $V \cap \mathscr{S}$ where $V$ is a submanifold of $\mathscr{O}$." (again, italics mine)."

I hope I haven't violated any copyright, but this article is hard to get hold of online. I have it only because I have hard copies of the first several volumes of JME in my office.

Again, I am posting because I would like to be able to use the IFT et al. on sets like $(\bar{P})^m$, but cannot for the life of me figure out why this works (according to Smale, at least). I realize I haven't yet posed a question, so let me do so now:

Question: In the above passage, the hypothesis is that the function $\psi_u$ is transverse to $\Delta$ only on the "non-manifold" $\mathscr{S}$, yet Smale is saying the corresponding theorem can be applied as though $\mathscr{S}$ were a manifold. Why? At first I thought Smale was making an implicit assumption that $\mathscr{S}$ was contained in some open set $\mathscr{O}$ that would be a manifold and on which $\psi_u$ would be transverse to $\Delta$. Then you could apply the IFT theorems to $\mathscr{O}$ and simply take the intersection of the resulting submanifold with $\mathscr{S}$. But it seems that Smale doesn't require that $\psi_u$ satisfy the transversality condition anywhere on $\mathscr{O}$, just on $\mathscr{S}$.

I understand I may have misunderstood something simple, also that I may have left out vital information (just ask), but I can't for the life of me figure out what is going on.

$\endgroup$
15
  • 3
    $\begingroup$ I think the basic idea comes from the transversality extension theorem. You can make a map transverse with respect to a manifold with boundary by first making it transverse to the boundary, and then (without affecting the intersection with the boundary) making it transverse to the interior. This idea extends to manifolds with cubical corners, which I think is the context of Smale's paper. $\endgroup$ Commented Nov 29 at 23:38
  • 2
    $\begingroup$ The textbook of Guillemin and Pollack does a really nice job of this. Also Hirsch's. Both provide clear-enough proofs so that you can see how to extrapolate to manifolds with cubical corners. $\endgroup$ Commented Nov 30 at 0:16
  • 2
    $\begingroup$ Manifolds with cubical corners are not "standard" smooth manifolds. Yes, they are topological manifolds. But you can talk about smooth structures on manifolds with cubical corners, and it's far more natural than considering them to be topological manifolds. If you read Guillemin and Pollack's discussion of smoothness you see that there is a notion of "smooth structure" for any subset of a smooth manifold. They define smoothness of functions in terms of local smooth extendability. That's how you talk about manifolds with cubical corners being smooth. $\endgroup$ Commented Nov 30 at 2:51
  • 2
    $\begingroup$ I don't know Smale personally but my presumption (from reading many of his papers) is that while this wouldn't be considered elementary, it's the kind of thing people steeped in geometric-topology understand. So if he was ever called on his sloppy writing (often he is not) he knew he had a solid foundation to retreat to. $\endgroup$ Commented Nov 30 at 2:52
  • 6
    $\begingroup$ I worked out the details of transversality for manifolds with corners in my lecture notes on differential geometry. drive.google.com/file/d/1_WxFupPeAeHyLAp6bzaJur2ReHuNZemo/… $\endgroup$
    – Ben McKay
    Commented Nov 30 at 10:19

1 Answer 1

23
$\begingroup$

Having reviewed the things you wrote the right frame for the question should be manifolds with corners (sets whose differentiable structure is locally modelled on open subsets of the orthants in euclidean space).

Hence your needs should be covered by

J. Margalef-Roig, E. Outerelo Dominguez: Differential Topology 1992, 1

I don't think that there is a public version of the book. It develops first the theory of manifolds with corners (and differentiable mappings between them) in the setting of possibly infinite dimensional Banach manifolds. The implicit function theorem which should cover your needs is then

Theorem 2.3.11 in 1.

Let $X,Y,Z$ be differentiable manifolds of class $p \in \mathbb{N}_0 \cup \{\infty\}$ $f\colon X\times Y \rightarrow Z$ a map of class $p$ and $(a,b)\in X\times Y$ a point. Suppose that $T^2_{(a,b)} f \colon T_b Y \rightarrow T_{f(a,b)}Z$ is a linear homeomorphism ($T^2$ is their notation for the derivative with respect to the second component, not the iterated tangent map!) and suppose that there are open neighborhoods $V^a$ of $a$ and $V^b$ of $b$ such that $f(V^a\times (V^b\cap \partial Y))\subseteq \partial Z$.

Then there are open neighborhoods $W^a$ of $a$ , and $W^b$ of $b$ and a unique map $q \colon W^a \rightarrow W^b$ such that $f(x,q(x))=f(a,b)$ for $x\in W^a$. Furthermore, $q(a)=b$; $q$ is of class $p$ on $W^a$; for every $x\in W^a$, $T^2_{(x,q(x))}f$ is a linear homeomorphism and $T_xq = - (T^2_{(x,q(x)}f)^{-1}\circ T^1_{(x,q(x))}f$

There is actually a bit more information on the tangent spaces here which I did not repeat as it references earlier numbers in the book. The proof is what you would expect taking into account the boundary conditions (note in particular the added requirements of mapping the boundary into the boundary beyond the usual condition of the tangent maps).

$\endgroup$
3
  • 3
    $\begingroup$ I feel like this should go on the nLab, but I'm not sure where. Not saying you should do it, either! $\endgroup$
    – David Roberts
    Commented Nov 30 at 11:13
  • 2
    $\begingroup$ It seems more like one should create a reference for the various generalisations of manifolds with boundary. Helge has done lately some cool things with his notion of locally polyhedral manifolds. Michor has his concept of Whitney germs and then there is the very general notion of manifolds with rough boundary. I have a master student looking at some of these things... need to stop now my family starts staring angrily at me for "working on the weekend ". $\endgroup$ Commented Nov 30 at 11:52
  • 2
    $\begingroup$ Appreciate the effort you put in to write this up, very generous of you. I believe it's as close to a complete answer as I will ever get. $\endgroup$
    – user167131
    Commented Nov 30 at 13:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .