Whitehead torsion IV (Ex)

Proposition 0.1. Let $<wikitex>; The aim of this exercise is to prove the following: {{beginthm|Proposition}} Let $(W; M, M')$ be an [[Wikipedia:h-cobordism|h-coborism]] between closed, connected k$-manifolds with with finite abelian fundamental groups of odd order. If $k \geq 3$ then $M$ and $M'$ are s-cobordant. {{endthm}} You may also wish to possible extensions of this proposition. == Comments == The following results from \cite{Milnor1966} will be helpful. Recall that the canonical involution on the group ring $\Zz[\pi]$ of a finitely generated group $\pi$ induces a conjuation on the [[Wikipedia:Whitehead torsion|Whitehead group]]. {{beginthm|Lemma| \cite{Milnor1966|Lemma 6.7} }} If $\pi$ is finite abelian, then every element of $\text{Wh}(\pi)$ is self-conjugate. {{endthm}} {{beginthm|Theorem| \cite{Milnor1966|Duality Theorem} }} For any orientable h-cobordism $(W; M, M')$ of dimension $n$ we have $$ \tau(W, M') = (-1)^{n-1} \hat \tau(W, M) $$ where $\hat \tau$ denotes the conjugate of $\tau$. {{endthm}} Now if $i \colon M \to W$ and $i' \colon M' \to W$ denote the inclusions, compute the Whitehead torsion of the homotopy equivalence $(i')^{-1} \circ i \colon M \to M'$. Finally, you may use the following theorem of Wall and Bak. {{beginthm| | }} Let $\pi$ be a finite group of odd order, then $L_{1}^s(\Zz[\pi]) = L_3^s(\Zz[\pi]) = 0$. {{endthm}} </wikitex> [[Category:Exercises]] == References == {{#RefList:}}(W; M, M')$ be an h-coborism between closed, connected $2k$-manifolds with with finite abelian fundamental groups of odd order. If $k \geq 3$ then $M$ and $M'$ are s-cobordant.

Lemma 2.1 [Milnor1966, Lemma 6.7] . If $\pi$ is finite abelian, then every element of $\text{Wh}(\pi)$ is self-conjugate.

Theorem 2.2 [Milnor1966, Duality Theorem] . For any orientable h-cobordism $(W; M, M')$ of dimension $n$ we have

$$\displaystyle \tau(W, M') = (-1)^{n-1} \hat \tau(W, M)$$

where $\hat \tau$ denotes the conjugate of $\tau$.

2.3. Let $\pi$ be a finite group of odd order, then $L_{1}^s(\Zz[\pi]) = L_3^s(\Zz[\pi]) = 0$.