About the definition of topological manifold using closed sets

[cianfa72]

TL;DR

How to define the notion of topological vs smooth manifold using closed sets.

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in $\mathbb R^n$.

It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on $\mathbb R^n$ ? I'm positive for this.

Perhaps the definition of smooth manifold would be problematic, though.

 

[fresh_42]

You can always replace an open set by its closed complement. Of course, this is not really a difference.

All terms in this context are about local phenomena, continuity, and differentiability. These are inevitably connected to open neighborhoods. If you make those neighborhoods closed, then you immediately get into trouble with singletons.

 

[wrobel]

How to define the notion of topological vs smooth manifold using closed sets.

for what purpose, if I may ask?

 

[cianfa72]

for what purpose, if I may ask?

Nothing. I asked it just to better understand how topological vs differential elements enter the definition of manifold (topological vs differentiable).

 

[cianfa72]

All terms in this context are about local phenomena, continuity, and differentiability. These are inevitably connected to open neighborhoods. If you make those neighborhoods closed, then you immediately get into trouble with singletons.

Sorry, can you better explain what is the trouble with singletons set ?

 

[martinbn]

It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on $\mathbb R^n$ ? I'm positive for this.

How?

 

[fresh_42]

Sorry, can you better explain what is the trouble with singletons set ?

A singleton is a one-point set $\{p\}.$ This is a closed set (in the usual topologies that are required to perform calculus). However, there is no way to define continuity or differentiability at $p$ without having a neighborhood of it.

The most basic concept of a manifold uses continuity. But continuity means a function is a morphism in TOP. It says that you cannot gain an open set by a morphism that wasn't already in the topology (of open sets) beforehand. That's why you need open sets. You can rephrase these by closed sets, but that is a void improvement, as it only means to switch to the complements.

You will also get problems with bounded manifolds. The boundary is closed. Again, a problem for continuity unless you restrict the entire setup to the boundary itself. Same with the boundaries of possibly closed neighborhoods. You cannot define continuity there, only one-sided. But how would you patch them if only one-sided continuity is defined?

Felix Hausdorff [12] introduced an axiomatic concept of neighborhoods in 1914 [13].

Source: https://www.physicsforums.com/insights/the-many-faces-of-topology/
where you can find the references [12] and [13].

 

[cianfa72]

How?

Define the topologies involved by using axioms for closed sets (i.e. arbitrary intersections and finite unions instead of vice versa). Define continuous map by using the preimage of closed sets and then homeomorphism, chart accordingly. From a topological manifold perspective I believe it makes perfect sense (basically we're just recasting the definitions using closed sets).

Let's move on to the definition of differentiable/smooth manifold. Consider the representative of a (continuous) function in a chart (that's a closed set in $\mathbb R^n$). Would it make sense to talk about differentiability for it ?

 

[martinbn]

Define the topologies involved by using axioms for closed sets (i.e. arbitrary intersections and finite unions instead of vice versa). Define continuous map by using the preimage of closed sets and then homeomorphism, chart accordingly. From a topological manifold perspective I believe it makes perfect sense (basically we're just recasting the definitions using closed sets).

There many problems. Give the definition that you think is appropriate.

 

[cianfa72]

There many problems. Give the definition that you think is appropriate.

Use this -- Definition using closed sets. The definition of chart $(U, \phi)$ is based on closed sets using homemorphism on the image. The transition maps $\varphi \circ \phi^{-1}: \phi (U \cap V) \to \varphi (V)$ are well defined, i.e. they are homemorphisms (intersection of closed sets is closed and we can use subspace topology on them).

 

[martinbn]

Use this -- Definition using closed sets. The definition of chart $(U, \phi)$ is based on closed sets using homemorphism on the image. The transition maps $\varphi \circ \phi^{-1}: \phi (U \cap V) \to \varphi (V)$ are well defined, i.e. they are homemorphisms (intersection of closed sets is closed and we can use subspace topology on them).

Let's take our topological space to be the plane $M=\mathbb R^2$. How do you make it a topological manifold using closed sets? Every point in $M$ is contained in closed line segment, which is homeomorphic to a line segment in $\mathbb R^1$. Does it mean that our space $M$ is a one dimensional topological manifold? The line segments are also homeomorphic to line segments in $\mathbb R^n$ for any $n$. Does it mean that $M$ is a $n$ dimensional topological manifold?

 

[cianfa72]

Let's take our topological space to be the plane $M=\mathbb R^2$. How do you make it a topological manifold using closed sets? Every point in $M$ is contained in closed line segment, which is homeomorphic to a line segment in $\mathbb R^1$. Does it mean that our space $M$ is a one dimensional topological manifold? The line segments are also homeomorphic to line segments in $\mathbb R^n$ for any $n$. Does it mean that $M$ is a $n$ dimensional topological manifold?

Ah yes, well spotted. The same issue doesn't occur when using open sets to define neighborhoods. So, by using closed sets, we no longer get a well defined notion of dimension for the manifold.

As far as I can tell, the key point here is that an open set of $\mathbb R^n$ can't be homeomorphic to a subset of euclidean space with dimension $m \neq n$.

 

[fresh_42]

As far as I can tell, the key point here is that an open set of $\mathbb R^n$ can't be homeomorphic to a subset of euclidean space with dimension $m \neq n$.

The key point is that you can choose whatever small neighborhood around a point and still have the same behavior. A closed set might immediately put you outside such a neighborhood. An open set has this problem only if we enlarge the neighborhood. However, we are not interested in global properties. We need local properties, i.e., getting smaller neighborhoods, not larger ones. Try to define differentiability in $\mathbb{R}^2$ if we only consider the closed sets $\coprod_{c\in \mathbb{R}}\{(x,y)\,|\,y=x+c\}.$ Good luck!

 

[lavinia]

A way to construct topological manifolds from closed subsets of Euclidean space is through triangulations. For example, one can imagine triangulating a surface then cutting out each triangular region to obtain a set of solid triangles. Each solid triangle is homeomorphic to a closed disk in the plane. By remembering how the edges of these triangles are pasted to each other, one can reassemble the surface. In this way, one imagines constructing surfaces by reshaping closed disks into solid triangles then gluing them together.

This idea extends to triangulable 3- manifolds constructed from closed solid pyramids and in general n-manifolds constructed from the n-dimenionsal analogues of solid triangles.

Most manifolds that one can think of are triangulable. All surfaces can be made from solid triangles. All smooth manifolds are triangulable. But there are exceptions in dimensions 4 and above. So this method of construction isn't completely general.

 

[lavinia]

Perhaps the definition of smooth manifold would be problematic, though.

One intuitive idea of defining a differentiable structure on a triangulated manifold is to embed the manifold in Euclidean space and then smooth out the vertices and edges to eliminate all of the creases and sharp points. For instance a tetrahedron can be made into a smoothly embedded sphere by rounding off the four vertices and bending the four triangles along the edges so that their tangent planes line up.

The rounded off manifold is differentiable if it has a well defined tangent plane at every point and if these planes vary continuously from point to point.

While I don't know much about smoothing theory, here is an idea of how the idea of rounding of triangulable manifolds might lead to the idea of a differentiable structure.

In a triangulated n-manifold that can be rounded out, each set of n-dimensional solid triangles that share a common vertex form a triangulated closed n-dimensional ball. After the rounding, these triangulated closed n-balls become closed smooth n- balls and their interiors ,if they form an open cover of the manifold, will form a smooth atlas for the manifold. So a smoothable triangulated manifold already contains a choice of an atlas and that atlas becomes a differential structure when edges and vertices are rounded out.

In this way of defining a differentiable structure, a differentiable atlas of coordinates charts would not be a starting point but rather an after effect of smoothing the edges and vertices of a manifold that is made up of generalized solid triangles, (closed disks). So one can think of a smooth manifold as a rounded triangulable manifold.

A couple of points: First of all not all triangulated manifolds can be smoothed. So however such a manifold is embedded in Euclidean space, there is not way to round it out. Second, a topological manifold that has a smoothable triangulation may have other triangulations that can not be smoothed. Third, the same triangulated manifold may be smoothable in different ways to produce non-diffeomorphic smooth manifolds. The case of the 7 sphere is interesting because if one embeds the 7 dimensional tetrahedron in Euclidean 8 space there is not enough room to smooth it into an exotic 7 sphere. One can only get the standard 7 sphere. However in 9 dimensions some fo the 28 exotic spheres can be made by smoothing out the tetrahedron but I think not all of the, I think to get all of them you need at least 11 dimensions.

Comment: Here is a proof of Whitelhead's Theorem, one of the foundational theorems connecting triangulations to smooth structures. It shows that any two smooth triangulations of a smooth manifold are combinatorially equivalent. This shows that differential structures on a smooth manifold are in some sense characterized by their
smooth triangulations.

Note that for the case of the 7 sphere all smooth triangulations are combinatorally equivalent.

Ihttps://www.math.ias.edu/~lurie/937notes/937Lecture5.pdf