Problem 23
Question
Show that if a group \(G\) acts freely and properly discontinuously on a Hausdorff space \(X\), then the action is a covering space action. (Here 'properly discontinuously' means that each \(x \in X\) has a neighborhood \(U\) such that \(\\{g \in G | U \cap g(U) \neq \varnothing\\}\) is finite.) In particular, a free action of a finite group on a Hausdorff space is a covering space action.
Step-by-Step Solution
Verified Answer
A free and properly discontinuous group action on a Hausdorff space is a covering space action because it provides disjoint, homeomorphic neighborhoods across the space.
1Step 1: Understanding Group Action
To solve this problem, we must first understand what it means for a group \( G \) to act on a space \( X \). A group action of \( G \) on \( X \) can be described by a function \( G \times X \to X \) which satisfies two conditions: for all \( x \in X \), \( e \cdot x = x \) where \( e \) is the identity element, and for all \( g, h \in G \), \( g \cdot (h \cdot x) = (gh) \cdot x \). In our scenario, the action is free, meaning if \( g \cdot x = x \) for any \( x \in X \) and \( g \in G \), then \( g \) must be the identity e.
2Step 2: Properly Discontinuous Action
A properly discontinuous action means each point \( x \in X \) has a neighborhood \( U \) where only finitely many elements \( g \in G \) act to take points of \( U \) into \( U \), i.e., the set \( \{ g \in G \mid U \cap g(U) eq \varnothing \} \) is finite. This finiteness condition ensures that different 'translates' of \( U \) under the action of \( G \) don't overlap too much, which will be crucial for establishing a covering space property.
3Step 3: Hausdorff Property
The Hausdorff property is critical in topological spaces, stating that any two distinct points have disjoint neighborhoods. This property will be useful in ensuring that open sets remain 'nice' when transitioned under the group action. It helps in establishing local homeomorphisms in covering space actions.
4Step 4: Definition of Covering Space Action
To demonstrate that the action is a covering space action, we must show that there exists a space \( Y \) such that each point has an open neighborhood evenly covered by \( X \), where each point in \( Y \) has disjoint neighborhoods mapping homeomorphically into \( X \). This translates to \( G \) acting such that it evenly covers \( X \) with sets that map homeomorphically to the quotient space \( X/G \).
5Step 5: Construction of Evenly Covered Neighborhoods
For each \( x \in X \), choose a neighborhood \( U \) such that \( \{ g \in G \mid U \cap g(U) eq \varnothing \} \) is finite, as per properly discontinuous action. Since \( G \) acts freely, the intersection \( U \cap g(U) = \emptyset \) for all but possibly one \( g \), the identity. Thus, \( \{ gU \}_{g \in G} \) are all disjoint.
6Step 6: Homeomorphic Mapping
Map each subset \( gU \) homeomorphically back to \( U \), establishing that locally each \( U \) looks 'like' all its translates \( gU \), confirming the local homeomorphic condition necessary for a covering space.
7Step 7: Conclusion
Since \( G \) acts freely and properly discontinuously, we can cover \( X \) with disjoint translates that map homeomorphically onto neighborhoods in \( X/G \), proving that the action is indeed a covering space action.
Key Concepts
Group ActionHausdorff SpaceProperly Discontinuous Action
Group Action
A group action provides a way for groups to interact with other mathematical objects, like spaces. Imagine a group, denoted as \( G \), acting on a set \( X \). Here, the action is defined through a function: \( G \times X o X \). This function has some important properties that ensure it behaves well:
- For every element \( x \) in \( X \), if you "act" on it with the identity element of \( G \), denoted as \( e \), it remains unchanged. Mathematically, this is stated as \( e \cdot x = x \).
- If you have two elements, \( g \) and \( h \), in \( G \), and you "act" first by \( h \) and then by \( g \), it should be the same as acting by their product, \( gh \). Formally, \( g \cdot (h \cdot x) = (gh) \cdot x \).
Hausdorff Space
In topology, a Hausdorff space is one of the most fundamental concepts. The main feature of a Hausdorff space is its 'separate-ness'—a unique characteristic crucial for understanding complex topological properties. The key idea is that any two different points in a Hausdorff space can be entirely separated by neighborhoods. This means:
- For any pairs of distinct points \( x \) and \( y \) in the space, you can find open neighborhoods \( U \) of \( x \) and \( V \) of \( y \) such that \( U \cap V = \emptyset \). They don’t 'touch' each other.
Properly Discontinuous Action
When considering group actions on spaces, the idea of "properly discontinuous" actions ensures that the group behavior doesn't create chaotic overlaps. With a properly discontinuous action, each point in the space \( X \) has a neighborhood that doesn't overlap too much with its 'translates' under the group's actions. Here's what this means:
- For a point \( x \) in \( X \), you can find an open neighborhood \( U \) such that when you consider different group elements \( g \) affecting \( U \), the set \( \{ g \in G \mid U \cap g(U) eq \emptyset \} \) is finite.
- This means there are only finitely many group elements where \( U \) still shares points with its "move" under these group elements.
Other exercises in this chapter
Problem 21
Let \(X\) be the space obtained from a torus \(S^{1} \times S^{1}\) by attaching a Möbius band via a homeomorphism from the boundary circle of the Möbius band t
View solution Problem 22
Given covering space actions of groups \(G_{1}\) on \(X_{1}\) and \(G_{2}\) on \(X_{2},\) show that the action of \(G_{1} \times G_{2}\) on \(X_{1} \times X_{2}
View solution Problem 25
Let \(\varphi: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation \(\varphi(x, y)=(2 x, y / 2) .\) This generates an action of \(Z\) on \(
View solution Problem 27
For a universal cover \(p: \tilde{X} \rightarrow X\) we have two actions of \(\pi_{1}\left(X, x_{0}\right)\) on the fiber \(p^{-1}\left(x_{0}\right),\) namely t
View solution