15
$\begingroup$

Morse functions on a manifold $M$ are defined as smooth maps $f:M \rightarrow \mathbb{R}$, such that at the critical points we can find local coordinates so that $$f(x_1,\dots,x_n)=-x_1^2-x_2^2-\dots-x^2_{i}+x^2_{i+1}+\dots+x_n^2.$$ There is an interpretation of Morse functions as "generic" functions. More precisely, one gives a stratification of $J^2(M,\mathbb{R})$---the second jet space of maps from $M$ to $\mathbb{R}$, and Morse functions are those functions whose second jet is transverse to the strata. By the Thom transversality theorem, the resulting space of functions is dense and open in $C^\infty(X,\mathbb{R})$.

The space of Morse functions is not connected. To make it connected one can introduce generalized Morse functions, where the critical points have a local description as above or as follows $$f(x_1,\dots,x_n)=-x_1^2-x_2^2-\dots-x^2_{i}+x^2_{i+1}+\dots+x_{n-1}^2+x_n^3.$$ We call the latter a birth-death singularity. The result in this case is that a "generic" smooth homotopy $M\times I\rightarrow \mathbb{R}$ is a Morse function at the endpoints and it is a Morse function at all values of the parameter, $t\in I$, except for finitely many values, where the function is a generalized Morse function.

What I am interested in is a generalization of Cerf's result, where the parameterizing space $I$ is replaced by an n-simplex. Here is my question.

Is there a "nice" description of a generic family of smooth functions of the form $M\times \Delta^n\rightarrow \mathbb{R}$?

By "nice" I mean having finitely many critical points and the critical points have local descriptions similar to the ones given above.

The goal roughly is to construct a simplicial set, where $n$-simplices are the generic smooth maps $M\times\Delta^n\rightarrow \mathbb{R}$. The hope is that given the transversality results one should be able to show that the simplicial set is contractible, since it would amount to extending a generic function $M\times\partial\Delta^n\rightarrow \mathbb{R}$ to a generic functions $M\times\Delta^n\rightarrow \mathbb{R}$, which we would do by first extending by arbitrary smooth function (the space of all smooth functions is contractible) and then deform it something generic relative to a neighborhood of the boundary.

$\endgroup$
2
  • 4
    $\begingroup$ There's quite a variety of work on topics like this. Have you looked at Cerf's work? There's also recent work of people like Igusa and Eliashberg arxiv.org/abs/1108.1000 . Does that suffice for your purposes? $\endgroup$ Oct 3, 2014 at 3:58
  • $\begingroup$ The simplicial set construction analogous to the one mentioned above is possible if one uses the framed Morse functions. The work of Eliashberg and Mishachev is directly applicable in this situation. I am trying to see whether there are other options of constructing a contractible space of "nice" functions into $\mathbb{R}$. My hope is that one of these options would generalize to $\mathbb{R}^n$. $\endgroup$ Oct 3, 2014 at 4:32

1 Answer 1

5
$\begingroup$

This is what catastrophe theory does, at least for small $n$, $n\leq 10$. Volume 1 of the book by Arnold, Gussein-Zade and Varchenko on singularities has a nice description of this theory; see especially Part 2 of that book.

$\endgroup$
4
  • $\begingroup$ Based on your response, my impression is that there is no general theory that works for arbitrary $n$. Is that right? $\endgroup$ Oct 3, 2014 at 14:34
  • $\begingroup$ That is correct. $C^\infty(M)$ comes with a Whitney like stratification and essentially the question boils down to understanding the codimension $n$-strata and the lower codimension strata containing in their closure a given codimension $n$-stratum. This turns out to be very difficult. $\endgroup$ Oct 3, 2014 at 14:38
  • $\begingroup$ Could one say something to the effect, that for a function in the codimension n-stratum $f:M\rightarrow\mathbb{R}$, the critical points are isolated or that the preimages have measure 0? $\endgroup$ Oct 3, 2014 at 15:02
  • $\begingroup$ I don't know. You have to check that book. I don't have it with me. $\endgroup$ Oct 4, 2014 at 12:36

Your Answer

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