Middle-dimensional surgery kernel (Ex)
(Created page with "<wikitex>; The goal of the following exercise is to prove the following statement which will be indispensible in defining the surgery obstruction. {{beginthm|Proposition}} ...") |
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− | The goal of | + | 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 | + | The statement essentially follows from the technical lemma which you are asked to prove. |
+ | {{beginthm|Lemma|{{citeD|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. | Let $R$ be a ring with involution and $C=C_\ast$ a finite chain complex of finitely generated projective (left) $R$-modules. | ||
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'''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 $H_n(C)$ 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. | ||
+ | {{endthm|Lemma}} | ||
− | The proposition is given as lemma 4.19 in {{citeD|Lück2001}}, however the proof is incomplete. Alternatively a good proof can be found in {{citeD|Wall1999}} and a more detailed one in | + | The proposition is given as lemma 4.19 in {{citeD|Lück2001}}, however the proof is incomplete. Alternatively a good proof can be found in {{citeD|Wall1999}} and a more detailed one in {{citeD|Ranicki2002}}. |
</wikitex> | </wikitex> | ||
[[Category:Exercises]] | [[Category:Exercises]] |
Revision as of 10:40, 19 March 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 be a degree 1 normal map from a -dimensional (resp. -dimensional manifold to a geometric Poincaré complex, inducing the isomorphism . Denote by the homology surgery kernel -module. If is -connected the kernel module 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 be a ring with involution and a finite chain complex of finitely generated projective (left) -modules.
1) If for , for some integer then the -module is finitely generated and
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].