Talk:Structure set (Ex)

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(Added solution for h-structure set)
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'''Solution''':
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'''Solution'''(simple structure set):
We begin with the map $\mathcal{S}^s(M) \to \mathcal{M}^s(M)$, from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things:
We begin with the map $\mathcal{S}^s(M) \to \mathcal{M}^s(M)$, from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things:
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'''Solution''' (homotopy structure set):
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Basically, the proof follows the same line. However in this case equality $[(N_0,f_0)]=[(N_1,f_1)]$ in $\mathcal{S}(M)$ (by definition) is existence of an $h$-cobordism $$(H,\partial_0H,\partial_0H)\colon (W,\partial_0W,\partial1W)\to (M\times I,M\times\{0\},M\times\{1\})$$ satisfying $f_i=\partial_iH\circ g_i\colon N_i\to \partial_iH\to M\times \{i\}$ for some orientation preserving diffeomorphisms $g_i$, $i=0,1$. But the h-cobordism gives us a homotopy equivalence $d\colon N_0\to N_1$ and the same formula for $h\colon M\to M$ applies.
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Analogously, equality $h\cdot [(N,f)] = \cdot [(N',f')]$ in $\mathcal{S}^h(M)$ implies existence of an $h$-cobordism between $N$ and $N'$ extending $h\circ f$ and $f'$.
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Latest revision as of 14:14, 29 August 2013

Solution(simple structure set):

We begin with the map \mathcal{S}^s(M) \to \mathcal{M}^s(M), from the simple structure set of manifolds simply homotopy equivalent to the orbit space and then we show two things:

  • If manifolds N and N' simply homotopy equivalent to M/var/www/vhost/map.mpim-bonn.mpg.de/tmp/AppWikiTex/tex_7IHgNZ are diffeomorphic then their images by the map belong to the same orbit of \mathcal{E}^s(M)-action on \mathcal{S}^s(M).
  • If two elements of \mathcal{S}^s(M) belong to the same orbit of \mathcal{E}^s(M)-action, then they are diffeomorphic.

Let N be a smooth manifold and f\colon N\to M a simple homotopy equivalence. Consider a map which takes N to [(N,f)]\in \mathcal{S}^s(M). Suppose now that N' is a manifold diffeomorphic to N, and f'\colon N'\to M a simple homotopy equivalence (possibly N'=N and f\nsim f'). Then there exists h\colon M\to M such that the following diagram commutes.

\displaystyle  \xymatrix{ N \ar[r]^{f} \ar[d]^d_{\cong} & M \ar@{.>}[d]^{h}\\ N'\ar[r]^{f'}& M }

Map h is given by composition f'\circ d\circ f^{-1} (the homotopy inverse) and hence is a simple homotopy equivalence. The commutativity of the diagram tells us that in \mathcal{S}^s(M) we have the following equalities.

\displaystyle h\cdot [(N,f)] = [(N,h\circ f)]=[(N,f'\circ d\circ f^{-1}\circ f\simeq f'\circ d)]=[(N',f')],

where \cdot denotes the \mathcal{E}^s(M)-action. Therefore [(N,f)] and [(N',f')] belong to the same orbit.

Suppose now, that [(N,f)],[(N',f')]\in \mathcal{S}^s(M) belong to the same orbit of \mathcal{E}^s(M)-action. It means, that there exist a simple homotopy equivalence h\colon M\to M such that
\displaystyle h\cdot[(N,f)]=[(N,h\circ f)]=[(N',f')].
But equality in the simple structure set \mathcal{S}^s(M) is just the existence of a diffeomorphism d\colon N\to N' making the following diagram commute.
\displaystyle  \xymatrix{ N \ar[r]^{f} \ar@{.>}[d]^d & M \ar[d]^{h}\\ N'\ar[r]^{f'}& M }  .

Solution (homotopy structure set):

Basically, the proof follows the same line. However in this case equality [(N_0,f_0)]=[(N_1,f_1)] in \mathcal{S}(M) (by definition) is existence of an h-cobordism
\displaystyle (H,\partial_0H,\partial_0H)\colon (W,\partial_0W,\partial1W)\to (M\times I,M\times\{0\},M\times\{1\})
satisfying f_i=\partial_iH\circ g_i\colon N_i\to \partial_iH\to M\times \{i\} for some orientation preserving diffeomorphisms g_i, i=0,1. But the h-cobordism gives us a homotopy equivalence d\colon N_0\to N_1 and the same formula for h\colon M\to M applies.

Analogously, equality h\cdot [(N,f)] = \cdot [(N',f')] in \mathcal{S}^h(M) implies existence of an h-cobordism between N and N' extending h\circ f and f'.


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