A compact topological 4-manifold is a CW complex if and only if it is triangulable.

Key Explanation:

• Triangulability vs. CW Complexes: A triangulable manifold can be homeomorphically identified with a simplicial complex, which is a type of CW complex. Conversely, for topological manifolds, the existence of a CW structure (i.e., being homeomorphic to a CW complex) implies triangulabilit...

Short answer Exactly when it is smoothable (equivalently, triangulable/PL). In particular, a compact topological 4–manifold is homeomorphic to a CW complex if and only if its Kirby–Siebenmann class κ(M)∈H^4(M;Z/2) vanishes. Thus Freedman’s E8–manifold (with κ≠0) is not a CW complex.

Why

• If a topological n–manifold M is a CW complex, then the manifold condition forces the CW structure to be “reg...

When is a Compact Topological 4-Manifold a CW Complex?

This is a fascinating question in 4-dimensional topology, where many standard results from other dimensions break down.

Background Analysis

The question already provides two key pieces of information:

1. Freedman's -manifold is non-triangulable (as proven in the Akbulut-McCarthy text)
2. Kirby showed that a compact 4-manifold has...

Let M be a compact topological 4–manifold (with or without boundary).
The following four statements are equivalent.

(1) M is homeomorphic to a CW–complex.
(2) M is triangulable (i.e. it is homeomorphic to the underlying space of a simplicial complex).
(3) M admits a PL–structure.
(4) The Kirby–Siebenmann obstruction
ks(M) ∈ H⁴(M; ℤ₂)
vanishes.

In particular, a compact 4–ma...

For dimension 4 the question has not been settled – no‐one presently knows a necessary and sufficient condition that tells exactly when a compact topological 4–manifold is (homeomorphic to) a CW–complex. What is known can be summarized as follows.

1. The “easy” direction (already mentioned in the question).
   • If a 4–manifold is smoothable, Kirby showed that it possesses a finite handle de...

This is an excellent question that gets to the heart of the subtle differences between the categories of topological, piecewise-linear (PL), and smooth manifolds, especially in dimension 4.

The short and direct answer is:

Every compact topological 4-manifold is homeomorphic to a finite CW complex.

In fact, a more general statement is true: every topological manifold (even non-compact ones) ...

Below is the rough picture. One shows that in dimension four the only thing that can go wrong with a topological 4–manifold is exactly the same “Kirby–Siebenmann obstruction’’ which in higher dimensions is known to obstruct PL‐ or smooth‐structures. In fact one has two equivalent characterizations:

1. (Handle/CW ⇔ smooth/PL.) By work of Kirby and Siebenmann (and in four‐space further elabor...

<think> Okay, the user is asking about when a compact topological 4-manifold can be given a CW complex structure. This is a deep question in geometric topology, specifically concerning 4-manifolds.

First, I recall that the question references Freedman's E8 manifold, which is a key example of a topological 4-manifold that isn't triangulable. That's important because triangulability relates to bei...

This is an excellent question that touches on some deep results in 4-manifold topology. Let me provide a comprehensive answer.

The Answer

Every compact topological 4-manifold admits a CW complex structure.

This follows from a general theorem that applies to all topological manifolds:

Theorem (Kirby-Siebenmann): Every topological manifold of any dimension admits a CW complex structur...