# Classification of open subset of $\mathbb{R}^{3}$ [closed]

There is a theorem which gives a classification of connected open sets of $\mathbb{R}^{2}$. Unfortunately, I don't remember the correct statement, but it looks like this

Theorem ? Let $U\subset\mathbb{R}^{2}=\mathbb{C}$ be a connected open subset, then $U$ is conformally equivalent to $\mathbb{R}^{2}-\cup_{n\in \mathbb{N}}I_{n}$, where $I_{n}$ is either empty or an interval of the form $[z,z+x_{n}]$ where $z$ is a complex number and $x_{n}$ is a positive real number.

conformally equivalent means that there exists a $C^{1}$-diffeomorphism $f:U\rightarrow \mathbb{R}^{2}-\cup_{n\in \mathbb{N}}I_{n}$ such that $df_{u}$ preserves angles for any $u\in U$.

1. I would like to know if the statement is correct ?
2. Does it admit a generalization in dimension 3. I mean a (EDIT Not necessarily conformal ?) classification of open connected subsets of $\mathbb{R}^{3}$
• As far as I know there is no known generalization to $\mathbb R^3$. Such a theorem would be a significant jump in complexity from the 2-dimensional case so I wouldn't hold my breath. Nov 27, 2015 at 21:17
• Even if we restrict to contractible open subsets of $\mathbb R^3$ there is no no diffeomorphism classification in sight. It is unclear what a classification might be couple be. Certainly, there are uncountable many such manifolds, see e.g. Theorem 2 of McMillan's jstor.org/stable/1993684 SOME CONTRACTIBLE OPEN 3-MANIFOLDS. Many end space structures can be realized. As stated the question is meaningless. Nov 27, 2015 at 21:17

The theorem in $C$ is stated incorrectly. The exact result is the following.

Let $D$ be an open connected set in $C$. Then $D$ is conformally equivalent to $\overline{C}\backslash E$, where $E$ is a closed set whose connected components are horizontal segments of the form $[z_j,z_j+x_j],\; x_j\geq 0$. In particular, these components can be points. The sets of components can be uncountable.

For a proof, see Jenkins, Univalent functions and conformal mapping. Springer 1958. The case of finite connectivity is due to Grotsch, and in this case there is uniqueness.

As stated, in general the region with horizontal slits may not be unique. To make it unique, one has to impose an additional condition on the set $E$ stated in Jenkins' book.

In $R^3$ this makes no sense because there are too few conformal mappings. So your question about "classification of open sets in $R^3$ with not necessarily conformal maps" is meaningless. To ask about a classification you should define which regions are considered equivalent.

• Thanks, I did not pretend that the stated theorem is correct! that was an item of my question. For 2) I think it is my fault in some sense! I was just hoping for a suggestion of a good formulation of the question in dimension 3! Igor Belegradek has made a clarification... Nov 27, 2015 at 21:24
• Until the equivalence relation is defined, "classification" makes no sense. Belegradek writes correctly that question 2) is meaningless. Nov 27, 2015 at 21:35
• I'm learning! :) Nov 27, 2015 at 21:45
• "To ask about a classification you should define which regions are considered equivalent": One could interpret the OP's question to be about a classification of open subsets of $\mathbb R^3$ up to diffeomorphism. I don't think that there will be a good answer, but like that the question is not meaningless. Nov 28, 2015 at 0:16
• @MalikYounsi: yes the theorem is due to Grotsch, I wanted to give an English reference. The exact statement of uniqueness part (which I omitted) is different in Jenkins and Grotsch. Nov 28, 2015 at 14:42

The result in two dimensions the OP is thinking of is Grotsch's uniformization by slit domains (I think Grotsch only did finitely many boundary components). Koebe has conjectured that any planar domain is conformally equivalent to a circle domain (the complement of a set of closed disks and points). As far as I know, this is still open in full generality, but He and Schramm showed that this is true when the number of boundary components is at most countable:

Fixed points, Koebe uniformization and circle packings
Zheng-Xu He, Oded Schramm*
Annals of Math, 1993.


As Adam Goucher states, it would be too much to expect a direct analogue in one dimension higher.

• do you have an example of a connected open of $R^{2}$ with non countable number of boundary components ? Nov 27, 2015 at 19:15
• BTW for $R^{3}$ I'm looking for a classification not necessarily conformal Nov 27, 2015 at 19:21
• @google complement of a Cantor set? Nov 27, 2015 at 19:30
• The result on uniformization by slit domains is indeed due to Grotzsch, Uber das Parallelschlitztheorem der konformen Abbildung schlichter Bereiche, Berichte Leipzig (84) (1932) 15-36. The theorem says that any domain in the plane is conformally equivalent to a domain whose complementary components are all points or compact horizontal segments. The number of complementary components might be uncountable. Nov 27, 2015 at 20:27
• I think you can find a proof of this result in Goluzin's book "Geometric theory of functions of a complex variable". I don't have it with me though so I cannot say for sure. Nov 27, 2015 at 20:28
1. You're thinking of the Riemann mapping theorem, which is similar to what you said but with the additional constraint that the domain must be simply-connected.

2. There isn't an analogue in three dimensions, because conformal transformations are too constrained:

https://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)

In particular, the only conformal transformations are the obvious ones (MÃ¶bius maps).

• 1) I am looking for a theorem for ANY connected open subspace of $R^2$. 2) I'm looking for a classification of open connected subspace of $R^3$ not necessarily conformal classification (that is why i put a question mark for conformal in the second question) Nov 27, 2015 at 18:58