# Questions about surgery theory

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The natural first port of call for quick answers is Mathoverflow.

Below is a list of questions, possibly with answers.

The Atlas also has a chapter Questions for questions which attract longer answers.

## 1 Questions

### 1.1 How can you tell if a space is homotopy equivalent to a manifold?

This is in fact a Mathoverflow question.

### 1.2 Simply connected surgery obstruction groups

How does one prove that $L_{4j}(e)=\Zz$$This page organizes questions and answers about surgery theory. The natural first port of call for quick answers is [http://www.mathoverflow.net/search?q=surgery Mathoverflow]. Below is a list of questions, possibly with answers. The Atlas also has a chapter [[:Category:Questions|Questions]] for questions which attract longer answers. == Questions == === How can you tell if a space is homotopy equivalent to a manifold? === This is in fact a [http://mathoverflow.net/questions/129/how-can-you-tell-if-a-space-is-homotopy-equivalent-to-a-manifold Mathoverflow question]. === Simply connected surgery obstruction groups === ; How does one prove that L_{4j}(e)=\Zz, L_{4j+2}(e)=\Zz_2 and L_{2k+1}(e) = 0 ? Read {{cite|Kervaire&Milnor1963}} and/or {{cite|Browder1972}} and/or {{cite|Ranicki2002|Chapter 12}}. === Handlebody and CW structures on topological 4-manifolds === ; Do topological 4-manifolds have a handlebody structrure? a CW structure? Topological manifolds of dimension \leq 3 have a piecewise linear (in fact a differentiable) structure ({{cite|Moise1952}}), and a fortiori are triangulable. Topological manifolds of dimension \geq 5 have a handlebody structure ({{cite|Kirby&Siebenmann1977}}), and hence a CW structure. All topological manifolds in dimensions \geq 5 can be triangulated if and only if the Kervaire-Milnor-Rohlin surjection \alpha:\theta^H_3\to {\mathbb Z}_2 splits ({{cite|Galewski&Stern1980}}). There do exist non-triangulable topological 4-manifolds, e.g. the Freedman E_8-manifold ({{cite|Akbulut&McCarthy1990}}). Rob Kirby's answer to the handlebody question (28.8.2010): If a TOP handle decomposition, then the first 1-handle attached to a (smooth) 0-handle can be isotoped to have a smooth attaching map, so the result is smooth. Just keep going like this, using the fact that a top embedding of the attaching sphere can be smoothed in dim 3, and the result is a smooth 4-mfd. Some aren't!!! So a nonsmoothable 4-manifold, e.g. the Freedman E_8-manifold, cannot have a handlebody structure. According Allen Hatcher's Algebraic Topology (starting from Corollary A.12), A compact manifold is homotopy equivalent to a CW complex. Restricting to closed manifolds: "for manifolds of dimensions less than 4, simplicial complex structures always exist. In dimension 4 there are closed manifolds that do not have simplicial complex structures, while the existence of CW structures is an open question. In dimensions greater than 4, CW structures always exist, but whether simplicial structures always exist is unknown, though it is known that there are n-manifolds not having simplicial structures locally isomorphic to any linear simplicial subdivision of R^n, for all n ≥ 4. For more on these questions, see {{cite|Kirby&Siebenmann1977}} and {{cite|Freedman&Quinn1990}}." == References == {{#RefList:}} [[Category:Surgery]] [[Category:Questions]]L_{4j}(e)=\Zz$, $L_{4j+2}(e)=\Zz_2$$L_{4j+2}(e)=\Zz_2$ and $L_{2k+1}(e) = 0$$L_{2k+1}(e) = 0$ ?

Read [Kervaire&Milnor1963] and/or [Browder1972] and/or [Ranicki2002, Chapter 12].

### 1.3 Handlebody and CW structures on topological 4-manifolds

Do topological 4-manifolds have a handlebody structrure? a $CW$$CW$ structure?

Topological manifolds of dimension $\leq 3$$\leq 3$ have a piecewise linear (in fact a differentiable) structure ([Moise1952]), and a fortiori are triangulable. Topological manifolds of dimension $\geq 5$$\geq 5$ have a handlebody structure ([Kirby&Siebenmann1977]), and hence a $CW$$CW$ structure. All topological manifolds in dimensions $\geq 5$$\geq 5$ can be triangulated if and only if the Kervaire-Milnor-Rohlin surjection $\alpha:\theta^H_3\to {\mathbb Z}_2$$\alpha:\theta^H_3\to {\mathbb Z}_2$ splits ([Galewski&Stern1980]). There do exist non-triangulable topological 4-manifolds, e.g. the Freedman $E_8$$E_8$-manifold ([Akbulut&McCarthy1990]).

Rob Kirby's answer to the handlebody question (28.8.2010): If a TOP handle decomposition, then the first 1-handle attached to a (smooth) 0-handle can be isotoped to have a smooth attaching map, so the result is smooth. Just keep going like this, using the fact that a top embedding of the attaching sphere can be smoothed in dim 3, and the result is a smooth 4-mfd. Some aren't!!! So a nonsmoothable 4-manifold, e.g. the Freedman $E_8$$E_8$-manifold, cannot have a handlebody structure.

According Allen Hatcher's Algebraic Topology (starting from Corollary A.12), A compact manifold is homotopy equivalent to a CW complex. Restricting to closed manifolds: "for manifolds of dimensions less than 4, simplicial complex structures always exist. In dimension 4 there are closed manifolds that do not have simplicial complex structures, while the existence of CW structures is an open question. In dimensions greater than 4, CW structures always exist, but whether simplicial structures always exist is unknown, though it is known that there are n-manifolds not having simplicial structures locally isomorphic to any linear simplicial subdivision of R^n, for all n ≥ 4. For more on these questions, see [Kirby&Siebenmann1977] and [Freedman&Quinn1990]."