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

I've been asked for a reference for the following construction and since I didn't know one, I thought I'd ask here if anyone did.

Consider two smooth manifolds with boundary of the same dimension, $M$ and $N$. Suppose that we have a submanifold $S$ of the boundaries $\partial M$ and $\partial N$ of codimension $0$ ($S$ may have boundary as well). Then glue $M$ and $N$ together along $S$, smoothing out the corners (corresponding to the boundary of $S$) if necessary. To do this properly, one would have to add a collar to each of $M$ and $N$ which is "broken" at the boundary of $S$ (thus making manifolds-with-corners) and glue those.

Is there a standard reference for this that works through the details?

share|improve this question
1  
mathoverflow.net/questions/67809/… asks the same thing, I think. It's a good question- Kosinski does it to some extent, and the Jim Davis reference which Igor Rivin gave also gives some details, but I don't know somewhere that all the details, including things like the isotopy extension theorem, properly appear in a unified framework. –  Daniel Moskovich Nov 30 '11 at 14:08
    
Daniel, I hadn't seen that one (I did check the suggestions, but that didn't come up). I'll need to take a look in Kosinski's book to see if that covers this case because as written, the questions are not duplicates: this situation has the added wrinkle of not gluing along the whole boundary (or a component of the boundary). (That said, if Kosinski does have it then I have no issue with close-as-duplicate.) –  Andrew Stacey Nov 30 '11 at 14:30
    
My goto book for such things is, as Daniel mentioned, Kosinski. Look at chapter 6 section 5 of Kosinski. –  Kelly Davis Nov 30 '11 at 16:40
1  
Although this isn't specifically in Kosinski, it is a small variant of the ideas there. Given a non-closed co-dimension 0 submanifold of the boundary, there are special collars so that VI.5 still works exactly as stated. –  Ryan Budney Nov 30 '11 at 17:48
    
Have you looked at Douady's articles in Seminaire Cartan vol 14 . –  Mohan Ramachandran Dec 4 '11 at 18:55
show 2 more comments

3 Answers

This was a bit too long for a comment, so I am posting it as an answer. You are sort of asking two things:

  1. How to turn your manifolds M and S into an appropriate manifold with corners together with an appropriate notion of collar?

  2. How to then glue these to obtain a new manifold?

To do (1) you'll need some assumptions on S, probably including compactness. In many cases though it might be clear that you can choose such collars. In that case you might be interested in Theorem 3.5 from my 2009 dissertation (arXiv:1112.1000, page 140). There I show that even if the collars are not specified, the glued manifold is still unique up to (non-canonical) diffeomorphism fixing S and restricting to the identity outside a neighborhood of S. In fact the construction shows that there is a canonical contractible family of these diffeomorphisms (and so there is a canoncial isotopy class of diffeomorphisms).

I used this to build one version of the 2-category of cobordisms, where you need to glue along parts of the boundary in the manner you describe, but where you also don't want to mod out by diffeomorphisms too early.

When S is a component of the boundary, you can find this result here:

James R. Munkres, Elementary differential topology, Lectures given at Massachusetts Institute of Technology, Fall, vol. 1961, Princeton University Press, Princeton, N.J., 1966.

I basically adapted this proof to cover the case of gluing manifolds along a portion of the boundary.

share|improve this answer
    
That looks very useful, I'll pass those along. –  Andrew Stacey Dec 8 '11 at 20:19
add comment

This seems to be discussed here. (particularly page 6)

share|improve this answer
    
That seems to be the desired result, but it also seems to be lacking a few details! Such as proof or reference to proof. Nonetheless, the statement is more detailed than I have so far so it's definitely progress. –  Andrew Stacey Nov 30 '11 at 14:33
3  
Well, you COULD just write to Jim Davis, and ask for a reference (and then share it with us...) –  Igor Rivin Nov 30 '11 at 14:37
    
Igor: I did, and have. –  Andrew Stacey Dec 8 '11 at 13:14
1  
You are a wise man! –  Igor Rivin Dec 8 '11 at 14:35
add comment

I did email Jim Davis and I have permission to post his reply (I'll summarise it). He taught a course which needed this result (which the notes that Igor Rivin links to are from). Being unable to find the precise statement (or proof) in the literature, he proved himself. He has ambitions to flesh out those notes to something fuller.

Since Kosinski's book was mentioned in the comments, it is perhaps worth pointing out that it contains the statement (p14):

Complications arise when more than one handle has been attached. When this happens some proofs have to rely strongly on the technique known as vigorous hand waving.

so for a precise statement/proof, then it would appear that Kosinski's book has to be discounted. (I should be honest and say that I haven't checked Kosinski's book myself; the original request was from a colleague and he has checked the book and is not happy with what is in there. The above quote was highlighted by Jim Davis.)

share|improve this answer
    
Thanks! This quote is precisely addressed here: mathoverflow.net/questions/70248/… It relates to the proof of the Morse Lemmas. Palais's proof was further simplified by Fukui. BTW, Kosinski does the "one handle attachment case" extremely nicely, and must be the best reference out there for this. But I don't think that this relates to your question, which is much more basic. I think it's a separate issue. –  Daniel Moskovich Dec 8 '11 at 14:30
2  
@Andrew: I think that quote is in reference to the "smoothing the corners" approach to viewing handle attachments as attaching $D^n \timd D^m$ along $(\partial D^n)\times D^m$. But the point of the quote is that if you think of it in a slightly different way, as Kosinski does, you don't have to deal with smoothing the corners to begin with. –  Ryan Budney Dec 8 '11 at 14:38
    
(Could the anonymous down-voter please explain their reasoning.) –  Andrew Stacey Dec 9 '11 at 8:03
    
I am the masked down-voter. My reasons are as follows: First, your statement is hearsay, which would not be so bad if it were correct. Second, in the first edition of Kosinski, there is no such statement on page 14. Third, in the second edition of Kosinski, there is also no such statement on page 14. Fourth, the first and second editions of Kosinski do have such a statement on page 142. However, this statement is related to how "other books" treat attaching handles. It is not a statement on how Kosinski's book treats attaching handles. (See Ryan's remark above.) –  Kelly Davis Dec 9 '11 at 11:55
    
Kelly, thank you for that comment. That is so much more useful to me than a drive-by down-voting. Your first reason is somewhat dubious, given the comments on Igor's answer and the setting (that I'm asking for a reference for someone else who doesn't use MO so almost everything to do with this question is "hearsay"). Your second reason, however, is completely valid and is something that everyone reading what I wrote should also read - which is why you should have commented in the first place! –  Andrew Stacey Dec 9 '11 at 13:05
show 1 more comment

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.