Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M^m$ be a smooth manifold with boundary. We may form a new topological manifold by "adjoining a handle to $M$", I.e. by choosing a smooth map $h : \partial \mathbb{D}^{\mu} \times \mathbb{D}^{\lambda} \rightarrow \partial M$ ($\mu + \lambda = m$) and taking our new manifold to be $M' := M \cup_h \mathbb{D}^{\mu} \times \mathbb{D}^{\lambda}$.

A priori $M'$ is just a topological manifold with boundary, however there are many well-known ways to "smooth the corners" of $M'$ (I.e. to endow $M'$ with a smooth structure). My question is "what is the right uniqueness statement?" i.e. how can I formally state that all off these different ways to smooth corners result in diffeomorphic manifolds.

To give a sense for the sort of result I'm hoping for I'll say that I have in mind a statement that begins something like "up to diffeomorphism there is a unique smooth atlas for $M'$ such that the inclusion $M \subset M'$ is a smooth imbedding and the inclusion $\mathbb{D}^{\mu} \times \mathbb{D}^{\lambda} \subset M'$ is a smooth imbedding of a manifold with corners." That statement is probably nonsense but I just wanted to give a sense for the sort of thing I'd like to be true. Thanks for considering my question.

share|improve this question
2  
Here are some sketch notes from a course on surgery at IU, in case they might be helpful to you: indiana.edu/~jfdavis/notes/m623/… –  Neal Apr 22 '13 at 23:22
    
Another thread on the same topic: math.stackexchange.com/questions/346333/… –  Ryan Budney Apr 22 '13 at 23:39
1  
@Neal : Thanks. This looks really helpful. –  Tim kinsella Apr 22 '13 at 23:54
    
I think the uniqueness statement you describe is false. For example, adjoin a 7-handle to a 7-sphere (so you just get another 7-sphere). There are now 28 different "smoothings", since there are 28 non-diffeomorphic structures on the 7-sphere. –  MartianInvader Apr 23 '13 at 0:08
    
@MartianInvader: the question is asking for conditions for uniqueness, there is no assertion of a condition for uniqueness in the question. Also, it's not clear what your "example" refers to, as handle attachment does not allow one to attach handles to closed manifolds. Perhaps you mean attach a 7-handle to a 7-dimensional disc? –  Ryan Budney Apr 23 '13 at 0:17

1 Answer 1

up vote 7 down vote accepted

In the literature nowadays there's two approaches to this. One is to avoid even considering your question, by making the handle-attachment construction one that lives purely in the smooth category (smooth manifolds and smooth manifolds with boundary allowed, but never mentioning manifolds with corners). Instead of viewing attaching a $\mu$-handle to an $n$-manifold $M$ as a gluing construction where you attach a $D^\mu \times D^\lambda$ along $S^{\mu -1} \times D^\lambda$ where $\mu + \lambda = n$.

This formalism is laid-out rather nicely in Kosinski's book Differentiable Manifolds. The idea is to take an unknotted $S^{\mu-1}$ in the boundary of a $D^n$. You can use a trivial regular (i.e. half a tubular) neighbourhood of the sphere in $D^n$ to glue the disc to your manifold $M$, provided you have a similar sphere with trivial normal bundle in $\partial M$.

But to literally answer your question about what people mean by uniqueness when you do construct a genuine manifold-with-corners, that is fairly fussy business. People try to avoid these constructions because if you use them, say, in the proof of the h-cobordism theorem, you have to worry about the order you perform multiple smoothings in, and whether or not they give you the same manifold if you do something that requires you to switch the order of smoothings. It becomes a major headache.

The statement is this. $M'$ is a smooth manifold with corners, in particular there are the two smooth inclusions $M \to M'$ and $D^\mu \times D^\lambda \to M'$. $M'$ itself is not a smooth manifold with boundary because of the singular stratum corresponding to $S^{\mu - 1} \times S^{\lambda-1}$.

A smoothing of $M'$ is a smooth manifold with boundary, let's call it $W$. $W$ is equipped with some maps:
$$i_M : M \to W$$ and $$i_D : D^{\mu}\times D^{\lambda} \to M$$ $i_M$ has to be smooth. The key part is that if we restrict $i_D$ to either $$D^\mu \times D^\lambda \setminus S^{\mu-1}\times S^{\lambda-1}$$ or $$D^\mu \times S^{\lambda-1}$$ or $$S^{\mu-1} \times D^\lambda$$ we get smooth maps, all diffeomorphisms onto their images.

We demand that $W$ is the union of the images of both $i_M$ and $i_D$, and that the images intersect along the image of your map $h$. Moreover, we demand the transition map is whichever gluing map you started with.

edit:

This concerns putting together a proof that the above data actually determines the smooth manifold $W$ uniquely up to diffeomorphism.

A standard theorem in manifold theory is that every manifold embeds in a Euclidean space of dimension $2n+1$ provided the manifold is of dimension $n$. A variant of this theorem is that up to smooth isotopy, the embedding is unique provided the Euclidean space has dimension $2n+2$ or larger.

Smooth manifolds with cubical corners are objects which are locally modelled on relatively open subsets of $[0,\infty)^n$. $[0,\infty)^n$ is called the model n-dimensional corner, let's call it $C_n$. Handle attachments are manifolds of this type. The analogue of the above theorem holds, that all n-manifolds with cubical corners embed in $C_n \times \mathbb R^{n+1}$, moreover you can demand that the embedding is "neat" in that all the strata are preserved, much like neat embeddings of manifolds with boundary. So handle attachments embed in $[0,\infty)^2 \times \mathbb R^{2n}$ provided the manifold is $n$-dimensional. The nice thing about this formalism is you have a universal space, so you can simply smooth the corners in this universal space to start, i.e. choose map $[0,\infty)^2 \to \mathbb R \times [0,\infty)$ which is a smooth diffeomorphism away from the origin, and such that the $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist and are non-zero at the origin. Uniqueness follows from showing that this map is essentially unique (provided its orientation-preserving). It's proof by universal example.

share|improve this answer
    
Thanks Ryan. Do you happen to know a reference for the second approach? In particular one containing a proof that these data determine a unique Diffeomorphism class? –  Tim kinsella Apr 22 '13 at 23:25
    
Off the top of my head I don't know where this would be written up in detail. It's one of those things where this is a convenient part of something that is much more general so there are likely proofs out there in much detail but at a level of generality that might not be appetizing. I'll give you a sketch of a proof that works in more generality, in my next edit. –  Ryan Budney Apr 22 '13 at 23:50
    
Thanks. If you care to mention the full generality or where I could read about it that would also be great, even if I'm unlikely to understand it at this point. –  Tim kinsella Apr 22 '13 at 23:55
    
I did a quick search but I'm not finding any references, sorry. I imagine if you Google long enough you'll find somebody's thesis that includes details like this. –  Ryan Budney Apr 23 '13 at 0:25
    
Great answer, Ryan. Thanks a bunch. –  Tim kinsella Apr 23 '13 at 0:27

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.