Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
Mohammed: If you drop by sometime, I will be happy to tell you about the 2-category of Deligne-Mumford stacks. Best, Jason –  Jason Starr Feb 3 at 19:21
    
Thanks Jason, I will. –  Mohammad F. Tehrani Feb 3 at 19:35

2 Answers 2

#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 :)

share|improve this answer
    
    
@DimaPasechnik: It's close but not quite the same :) Important is the smoothness, which makes no sense in Urysohn's setting. –  Alex Degtyarev Feb 3 at 20:30
1  
    
@DimaPasechnik: Yep! As I said, I learned it as a homework myself, and now I'm giving this as a homework in my classes :) –  Alex Degtyarev Feb 3 at 20:41
    
@ 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. –  Mohammad F. Tehrani Feb 3 at 21:31

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 $M$.

share|improve this answer
1  
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 –  Raphael Albrecht Feb 3 at 21:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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