3
$\begingroup$

In the past decade, theory of Kuranishi structures on moduli space of pseudo-holomorphic curves has been in the center of debates between some mathematicians in the field of symplectic geometry.

Kenji Fukaya is teaching a course on this subject at Simons Center and the course is being recorded. Here is the link to the first lecture. http://scgp.stonybrook.edu/archives/10004

In order to clarify some comments and claims, for myself and may be others, I will gradually post some questions on mathoverflow. Hopefully, I will gather the results of these discussions in a Lecture note.

1) At some point (min 33-34), John Morgan comments that for an arbitrary compact subset of Euclidean space, $K\subset \mathbb{R}^m$, and for small enough open set $U\subset \mathbb{R}^m$, $U\cap K$ can be realized as the zero set of some smooth map $f\colon U \to \mathbb{R}^n$. How is the proof? For example if this compact set if the contour set in $\mathbb{R}$.

Comment: zero locus of smooth maps + extra structure, are the bulding blocks of Kuranishi spaces

2) A Delign-Mumford stack is somehow a category itself. Then, is there a category of Deligne-Mumford stacks that includes fiber products, ...?

Comment: A Dream of Kuranishi theory is to build a category out of Kuranihsi spaces.

$\endgroup$
1
  • $\begingroup$ Mohammed: If you drop by sometime, I will be happy to tell you about the 2-category of Deligne-Mumford stacks. Best, Jason $\endgroup$ Commented Feb 3, 2014 at 19:21

2 Answers 2

7
$\begingroup$

#1 is a classical statement: it suffices to have $K$ closed and $U=\mathbb{R}^m$. Here is a proof I can think of right now. For each point $x\in V:=\mathbb{R}^n\setminus K$ there is a smooth function $f_x\colon\mathbb{R}^m\to[0,1]$ such that $f_x(x)=1$ and $f_x|_K=0$. It can be chosen with compact support, so that all derivatives are bounded. The sets $f_x\ne0$ form an open covering of $V$; pick a countable subcovering and consider the series $ \sum a_if_i$ with $a_i\to0$ sufficiently fast. It really is classical, but I don't remember the reference; it is usually given as a homework in topology courses :)

$\endgroup$
6
  • $\begingroup$ en.wikipedia.org/wiki/Urysohn's_lemma IIRC... $\endgroup$ Commented Feb 3, 2014 at 20:26
  • $\begingroup$ @DimaPasechnik: It's close but not quite the same :) Important is the smoothness, which makes no sense in Urysohn's setting. $\endgroup$ Commented Feb 3, 2014 at 20:30
  • 1
    $\begingroup$ OK: math.stackexchange.com/questions/198748/… $\endgroup$ Commented Feb 3, 2014 at 20:34
  • $\begingroup$ @DimaPasechnik: Yep! As I said, I learned it as a homework myself, and now I'm giving this as a homework in my classes :) $\endgroup$ Commented Feb 3, 2014 at 20:41
  • $\begingroup$ @ Dima: I did not go through all the details of question\198748 but is it it clear that the resulting f is not equal to 1 elsewhere. $\endgroup$ Commented Feb 3, 2014 at 21:31
3
$\begingroup$

ad (1) : this is a corollary to the classical Theorem on the existence of a $C^\infty$ partition of unity (subordinate to a given open cover of a smooth manifold) see F.W. Warner, Foundations of Differentiable Manifolds and Lie Groups, Scott, Foresman and Cie 1971,Thm. 1.11, p. 10 and Corollary on p.11. Here $K$ does not need be compact, but only a closed subset of a smooth manifold $M$.

$\endgroup$
1
  • 1
    $\begingroup$ See also the classic text on differential topology by M.W. Hirsch (by the same title) theorem 2.1 on p. 43 and corollary = ercx. 1 on p.55 $\endgroup$ Commented Feb 3, 2014 at 21:47

You must log in to answer this question.

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