Problem 33
Question
Show that if \(M\) is a compact contractible \(n\) -manifold then \(\partial M\) is a homology \((n-1)-\) sphere, that is, \(H_{i}(\partial M ; \mathbb{Z}) \approx H_{i}\left(S^{n-1} ; \mathbb{Z}\right)\) for all \(i.\)
Step-by-Step Solution
Verified Answer
\(\partial M\) has the homology of an \(n-1\)-sphere.
1Step 1: Understand the Problem Statement
We need to show that the boundary of a compact contractible manifold has the same homology as a sphere of dimension \(n-1\). This involves comparing the homology groups of the boundary and the sphere and demonstrating their equivalence.
2Step 2: Define Important Concepts
A compact contractible \(n\)-manifold \(M\) is a space that can be continuously shrunk to a point within itself. The boundary of \(M\), denoted \(\partial M\), is the set of points not lying strictly inside \(M\). We aim to show that the homology groups \(H_i(\partial M ; \mathbb{Z})\) and \(H_i(S^{n-1} ; \mathbb{Z})\) are isomorphic for all \(i\).
3Step 3: Apply the Lefschetz Duality Theorem
According to Lefschetz duality for a compact manifold \(M\) with boundary, we have \(H^i(M, \partial M; \mathbb{Z}) \approx H_{n-i}(M; \mathbb{Z})\). For contractible \(M\), the homology of \(M\) is trivial except when \(i = 0\), where \(H_0(M; \mathbb{Z}) \approx \mathbb{Z}\).
4Step 4: Analyze Homology of \(M\)
Because \(M\) is contractible, \(H_i(M; \mathbb{Z}) = 0\) for all \(i \geq 1\). This affects the dual groups, giving us nontrivial homology for intersecting dimensions. Specifically, \(H_i(M, \partial M; \mathbb{Z}) = 0\) for \(i eq 0\).
5Step 5: Relate to Homology of the Sphere
A sphere \(S^{n-1}\) has homology groups \(H_i(S^{n-1}; \mathbb{Z}) = \mathbb{Z}\) for \(i = 0, n-1\) and 0 otherwise. We need to match this structure with that of \(\partial M\).
6Step 6: Conclude Homology Equivalence
The homology of \(\partial M\) matches that of the \((n-1)\)-sphere because the manifold \(M\) satisfies \(H_{n-i}(M; \mathbb{Z}) = 0\) for \(i eq 0\) due to contractibility. Hence, \(H_i(\partial M; \mathbb{Z}) \approx H_i(S^{n-1}; \mathbb{Z})\) for all \(i\).
Key Concepts
Compact ManifoldHomology SphereLefschetz Duality TheoremContractible Space
Compact Manifold
A compact manifold is a type of topological space with no boundary and finite size. Imagine a surface that is complete without any endpoints or open boundaries. For example:
- A circle is compact because it can be traced without lifting your pen.
- In higher dimensions, a sphere is compact; it’s closed and bounded.
Homology Sphere
A homology sphere shares some properties with a standard sphere, particularly in terms of its homological features. While a traditional sphere, like a circle or a 3D sphere, can be visualized easily, a homology sphere is more abstract:
- Its homology groups match those of a standard sphere.
- Despite this, it might not be topologically equivalent to a sphere.
Lefschetz Duality Theorem
The Lefschetz Duality Theorem provides a powerful principle in algebraic topology, particularly concerning manifolds with boundaries. It connects the homology of a manifold with the homology of its boundary:
- It asserts a form of equivalence between certain homology and cohomology groups.
- For a compact oriented manifold with a boundary, it relates the manifold and its boundary via a homological shortcut.
Contractible Space
A contractible space is one that can be 'shrunk' to a point, essentially making its topology simple and manageable. Conceptually, think about how you could crush a rubber ball to a point without tearing it:
- Contractibility means there is a homotopy—continuous transformation—that takes the space into a single point.
- In terms of algebraic topology, this implies all higher homotopy groups are trivial.
Other exercises in this chapter
Problem 31
Show that if \(M\) is a compact \(R\) -orientable \(n\) -manifold, then the boundary map \(H_{n}(M, \partial M ; R) \rightarrow H_{n-1}(\partial M ; R)\) sends
View solution Problem 32
Show that a compact manifold does not retract onto its boundary.
View solution Problem 28
28\. Show that a nonsingular symmetric or skew-symmetric bilinear pairing over a field \(F,\) of the form \(F^{n} \times F^{n} \rightarrow F,\) cannot be identi
View solution