can someone provide explicit charts for non-standard differentiable structures on, for instance $\mathbb{R}^4$ (or some other manifold)?
explicit "exotic" charts
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0I'm a little confused by the nature of your question. Are you interested in simple examples of smooth manifolds with "non-standard" differentiable structures, or are you interested in clutching constructions with charts? I don't see how the Brieskorn example satisfies your original question -- neither Brieskorn nor Kervaire-Milnor provided explicit chart constructions for such manifolds. What are you looking for, precisely? – 2011-03-29
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1i wanted explicit charts (if any are known). i added the other part to maybe drum up interest in the question or to expand it to include constructions (since no one had written anything). – 2011-03-30
2 Answers
Here is a Kirby Diagram for an exotic $\mathbb R^4,$ taken from Gompf and Stipsicz's book, "4-Manifolds and Kirby Calculus." It's not given in the form of an atlas, but it is a nice explicit description.
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5Great! You've given me motivation to finally go find out what those diagrams actually mean :) – 2011-04-20
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2@Mariano: You're constructing the 4-manifold by taking $\mathbb R^3 \times (-\infty,0]$ then attaching handles on $\mathbb R^3 \times \{0\}$ according to the link diagram. The unknotted dotted circles encode a way of attaching a 1-handle. The possibly knotted circles with number decorations indicate a 2-handle attachment. These conventions are given in detail in Gompf and Stipsicz book, sections 5.1 through 5.4. – 2011-07-24
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0@Ryan Heya, I'm an undergrad so I'm betting the book you mentioned is beyond me at the moment, but I'm _really_ interested in this kind of stuff and I'd really like a better intuition for what this diagram means. For the numbered knots I'm imagining attaching the boundary of a 2-cell to them and then thickening the whole cell. How exactly do you attach a 1-handle to the dotted knots? And how could this possibly be homeomorphic to $\mathbb{R}^4$? Isn't $\mathbb{R}^3 \times (-\infty,0]$ already distinct from $\mathbb{R}^4$?? – 2018-07-15
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1@EbenCowley: Your intuition for the numbered circles is correct. For the dotted circles, you actually push a hemisphere into the $\mathbb R^3\times (-\infty,0]$ and delete a neighborhood of this hemisphere. (The dotted components are always unknots so you can do this.) It is an interesting exercise to show that this has the effect of adding a $1$-handle. – 2018-07-16
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0@CheerfulParsnip I'm not really sure what you mean by "pushing a hemisphere into the manifold," or how the unknots could correspond to a neighborhood of the hemisphere. Sorry if this is terribly elementary, I've never actually taken a topology course. – 2018-07-16
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1@EbenCowley: So the unknot sits inside $\mathbb R^3\times\{0\}$. You can imagine capping it off with a disk where the interior is inside $\mathbb R^3\times(\infty,0)$. If you remove a neighborhood of this disk, it cuts out a neighborhood of the unknot inside the $\mathbb R^3\times\{0\}$ slice. – 2018-07-17
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0Oh I think I understand now. By "neighborhood of the disk" you mean an open set in $\mathbb{R}^3 \times (0, \infty)$ which contains the hemisphere that caps off the unknot, and we're _removing_ this open set, whereas with the numbered circles we're actually adding to the manifold. Is this right? I'm not sure how this could have the effect of adding a 1-handle. – 2018-07-17
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0@EbenCowley: yes exactly. Deleting a neighborhood of the disk is diffeomorphic to adding a $1$-handle. It's analogous to how if you look at $(-\infty,0]\times \mathbb R$ and you remove a neighborhood of an arc that hangs down from $\{0\}\times\mathbb R$, this is diffeomorphic to adding a $0$-handle to your original picture. – 2018-07-17
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0@CheerfulParsnip Ah, alright. I suppose I just can't visualize why it _wouldn't_ be like adding a 0-handle in higher dimensions. I'll have to take up the exercise of showing that the addition and deletion are diffeomorphic at some point in the future. – 2018-07-18
Not really helpful I guess, but I am interested as well in this subject so: A while ago I heard Matthew Baker (from Georgia Tech, on an entirely different subject, namely the Berkovich Projective Line of non-archimedean fields) describing a technique knows as the observers' topology (i.e. take a point in the space, look around and describe what you see). It would be interesting to know if one could see a difference with another differentiable structure on $\mathbb{R}^4$. It doesn't give you an explicit atlas though.
This ties in with Exotic Manifolds from the inside as well. I am afraid that this doesn't give you a straight answer as well, but a hint of where to look further.
By the way, did you know that only a small portion of the exotic $\mathbb{R}^4$'s can be represented by Kirby diagrams directly (by varying things in the diagram, you get a countably many non-diffeomorphic copies I guess [though I haven't seen a proof of this]). At hinsight, Bob Gompf has proved that there are uncountably many of exotic $\mathbb{R}^4$'s, so this doesn't help much.
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0what do you mean when you say only a small portion of exotic $\mathbb R^4$'s can be represented diagrammatically? Do you mean: a) there are some with no Kirby digram, or b) You can't draw the whole diagram because it's infinite? – 2011-07-24
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0There are some with no Kirby diagram (the thing on this page is also infinite). I am at work now with no reference books, but I think there was a line on that in either Gompf/Stipsicz (4-manifolds and Kirby Calculus) or/and Scorpan (The wild would of ...) – 2011-07-25
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0Reading a little futher I am not so sure of myself (or my knowldge about Kirby Diagrams is somewhat outdated). It boils down to the question: Are there exotic $\mathbb{R}^4$'s wihout an (obviously infinite) handlebody structure. – 2011-07-25
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0I'm not so sure what happens for noncompact manifolds. It could be that you're right that there exist smooth manifolds with no diagram. This is something I'd like to understand. – 2011-07-25
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1I am at home and on page 366 of "Gompf and Stpsicz" it says that "The known large exotic $\mathbb{R}^4$'s require infinitely many 3-handles in any handly decompositions, but – 2011-07-25
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1and there is presently no clue as to how one might draw explicit handle diagrams of them". They wrote that in 1999, so there is chance that something has changed. If I have time, I will try and get some answers via this site, by asking this as a question, and not as a comment. – 2011-07-25
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1Thanks. To me that means there are Kirby diagrams for these exotic $\mathbb R^4$s, but nobody knows how to actually construct them. – 2011-07-25