Talk:Whitehead torsion IV (Ex)
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For the Whitehead torsion of $(i')^{-1}\circ i:M\rightarrow M'$ we have, using the quoted results of Milnor, | For the Whitehead torsion of $(i')^{-1}\circ i:M\rightarrow M'$ we have, using the quoted results of Milnor, | ||
$$\tau\left((i')^{-1}\circ i\right)=-\tau(W,M')+\tau(W,M)=(-1)^{2k+1}\hat\tau(W,M)+\tau(W,M)=-\tau(W,M)+\tau(W,M)=0$$ | $$\tau\left((i')^{-1}\circ i\right)=-\tau(W,M')+\tau(W,M)=(-1)^{2k+1}\hat\tau(W,M)+\tau(W,M)=-\tau(W,M)+\tau(W,M)=0$$ | ||
− | So on the boundary $M$ of $W$ we have for $f|_M=(i')^{-1}\circ i: | + | So on the boundary $M$ of $W$ we have for $f|_M=(i')^{-1}\circ i:M\rightarrow M'$ that $\tau(f|_{M})=0$. |
Since the odd-dimensional L-groups $L_{2k+1}^s(\Zz\pi)$ vanish for finite groups $\pi$ of odd order there is no obstruction to do surgeries in the interior of $W$ to turn it into an s-cobordism without changing the boundary. Thus $M$ and $M'$ are s-cobordant. | Since the odd-dimensional L-groups $L_{2k+1}^s(\Zz\pi)$ vanish for finite groups $\pi$ of odd order there is no obstruction to do surgeries in the interior of $W$ to turn it into an s-cobordism without changing the boundary. Thus $M$ and $M'$ are s-cobordant. | ||
</wikitex> | </wikitex> |
Revision as of 13:36, 29 March 2012
Let a Morse function with . We define by .
For the Whitehead torsion of we have, using the quoted results of Milnor,
So on the boundary of we have for that . Since the odd-dimensional L-groups vanish for finite groups of odd order there is no obstruction to do surgeries in the interior of to turn it into an s-cobordism without changing the boundary. Thus and are s-cobordant.