Middle-dimensional surgery kernel (Ex)

From Manifold Atlas
(Difference between revisions)
Jump to: navigation, search
m
m
(2 intermediate revisions by 2 users not shown)
Line 2: Line 2:
The goal of this exercise is to prove the following statement which will be indispensible in defining the surgery obstruction.
The goal of this exercise is to prove the following statement which will be indispensible in defining the surgery obstruction.
{{beginthm|Proposition}} Let $f:M\rightarrow X$ be a degree 1 normal map from a $2k$-dimensional (resp. $(2k+1)$-dimensional manifold to a geometric Poincaré complex, inducing the isomorphism $f_\ast:\pi_1(M)\cong\pi_1(X)=:\pi$. Denote by $K_i(M)=K_i(\widetilde{M})$ the homology surgery kernel $\mathbb{Z}[\pi]$-module. If $f$ is $k$-connected the kernel module $K_k(M)$ is finitely generated and stably free.
+
{{beginthm|Proposition}} Let $f:M\rightarrow X$ be a degree 1 normal map from a $2k$-dimensional (resp. $(2k+1)$-dimensional) manifold to a geometric Poincaré complex, inducing the isomorphism $f_\ast:\pi_1(M)\cong\pi_1(X)=:\pi$. Denote by $K_i(M)=K_i(\widetilde{M})$ the homology surgery kernel $\mathbb{Z}[\pi]$-module. If $f$ is $k$-connected the kernel module $K_k(M)$ is finitely generated and stably free.
{{endthm|Proposition}}
{{endthm|Proposition}}
The statement essentially follows from the technical lemma which you are asked to prove.
The statement essentially follows from the technical lemma which you are asked to prove.
Line 12: Line 12:
'''2)''' If in addition for $j>n$, $H^j(C)=0$ for the same integer $n$ then $H_n(C)$ and $H^n(C)$ are dual, and there are isomorphisms$$H_n(C)\oplus\sum_{i\in\mathbb{Z}}C_{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C_{n+2j}, $$$$H^n(C)\oplus\sum_{i\in\mathbb{Z}}C^{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C^{n+2j},$$where under further assumption that $C$ is a chain complex of free modules, the latter isomorphism implies that $H_n(C)$ and $H^n(C)$ are stably free.
+
'''2)''' If in addition for $j>n$, $H^j(C)=0$ for the same integer $n$ then there are isomorphisms$$H_n(C)\oplus\sum_{i\in\mathbb{Z}}C_{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C_{n+2j}, $$$$H^n(C)\oplus\sum_{i\in\mathbb{Z}}C^{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C^{n+2j},$$where under further assumption that $C$ is a chain complex of free modules, the latter isomorphism implies that $H_n(C)$ and $H^n(C)$ are stably free and hence $H_n(C)$ and $H^n(C)$ are dual.
{{endthm|Lemma}}
{{endthm|Lemma}}
Line 18: Line 18:
</wikitex>
</wikitex>
[[Category:Exercises]]
[[Category:Exercises]]
+
[[Category:Exercises without solution]]

Latest revision as of 14:51, 1 April 2012

The goal of this exercise is to prove the following statement which will be indispensible in defining the surgery obstruction.

Proposition 0.1. Let f:M\rightarrow X be a degree 1 normal map from a 2k-dimensional (resp. (2k+1)-dimensional) manifold to a geometric Poincaré complex, inducing the isomorphism f_\ast:\pi_1(M)\cong\pi_1(X)=:\pi. Denote by K_i(M)=K_i(\widetilde{M}) the homology surgery kernel \mathbb{Z}[\pi]-module. If f is k-connected the kernel module K_k(M) is finitely generated and stably free.

The statement essentially follows from the technical lemma which you are asked to prove.

Lemma 0.2 [Ranicki2002, Lemma 10.26]. Let R be a ring with involution and C=C_\ast a finite chain complex of finitely generated projective (left) R-modules.

1) If for i<n, H_i(C)=0 for some integer n then the R-module H_n(C) is finitely generated and
\displaystyle H^n(C)\rightarrow H_n(C)^\ast, \quad  f\mapsto (x\mapsto f(x)).
is an isomorphism.


2) If in addition for j>n, H^j(C)=0 for the same integer n then there are isomorphisms
\displaystyle H_n(C)\oplus\sum_{i\in\mathbb{Z}}C_{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C_{n+2j},
\displaystyle H^n(C)\oplus\sum_{i\in\mathbb{Z}}C^{n+2i+1}\cong\sum_{j\in\mathbb{Z}}C^{n+2j},
where under further assumption that C is a chain complex of free modules, the latter isomorphism implies that H_n(C) and H^n(C) are stably free and hence H_n(C) and H^n(C) are dual.

The proposition is given as lemma 4.19 in [Lück2001], however the proof is incomplete. Alternatively a good proof can be found in [Wall1999] and a more detailed one in [Ranicki2002].

Personal tools
Namespaces
Variants
Actions
Navigation
Interaction
Toolbox