Talk:S-duality I (Ex)

Exercise 0.1 (first part). Any finite CW complex $X$$<wikitex>; {{beginthm|Exercise|(first part)}} Any finite CW complex $X$ has an $S^N$-dual, for $N$ sufficiently large. {{endthm}} {{beginproof}} (tentative: see remarks below) Let $X$ be a finite pointed CW complex with basepoint $x_0$. Embed $X\hookrightarrow S^N$ with regular neighbourhood $M$, so that $$ X \hookrightarrow M \hookrightarrow S^N $$ with $X\hookrightarrow M$ a homotopy-equivalence and $M$ an $N$-dimensional manifold-with-boundary embedded in $S^N$. Now let $$ c\colon S^N \longrightarrow S^N / (S^N \setminus \mathring{M}) = M / \partial M $$ be the collapse map, $$ \Delta\colon M / \partial M \longrightarrow (M\times M) / ((\partial M \times M)\cup (M\times \lbrace x_0 \rbrace)) \;\cong\; (M / \partial M) \wedge M $$ be the map induced by $$ x\mapsto (x,x)\colon (M,\partial M) \longrightarrow (M\times M, (\partial M \times M)\cup (M\times \lbrace x_0 \rbrace)), $$ and let $\alpha$ be the composite $$ S^N \xrightarrow{c} M / \partial M \xrightarrow{\Delta} (M / \partial M) \wedge M \xrightarrow{1\wedge r} (M / \partial M) \wedge X , $$ where $r\colon M\to X$ is a chosen homotopy-inverse for $X\hookrightarrow M$. The map $c$ is degree-1 by construction, so $c_*([S^N]) = [M] \in \widetilde{H}_N(M / \partial M) = H_N(M,\partial M)$. For any $x\in \widetilde{H}^i(M)$, we have the identity $$ x\setminus \Delta_*([M]) \;=\; x\cap [M] \;\in\; \widetilde{H}_{N-i}(M / \partial M) = H_{N-i}(M,\partial M), $$ so by Poincare-Lefschetz duality, $$ -\setminus (\Delta\circ c)_*([S^N]) \;\colon\; \widetilde{H}^i(M) \longrightarrow \widetilde{H}_{N-i}(M / \partial M) $$ is an isomorphism for all $i$. In general for any $X\wedge Y \xrightarrow{1\wedge f} X\wedge Z$ and $a\in \widetilde{H}_N(X\wedge Y)$, $$ f^*(-)\setminus a \;=\; -\setminus (1\wedge f)_*(a) \;\colon\; \widetilde{H}^i(Z) \longrightarrow \widetilde{H}_{N-i}(X). $$ So in our case we have $$ -\setminus \alpha_*([S^N]) \;=\; r^*(-)\setminus (\Delta\circ c)_*([S^N]) \;\colon\; \widetilde{H}^i(X) \longrightarrow \widetilde{H}_{N-i}(M / \partial M) $$ which is an isomorphism by above and the fact that $r$ is a homotopy-equivalence. This witnesses that $M / \partial M$ is an $S^N$-dual of $X$. {{endproof}} {{beginrem|Remark}} This uses reduced homology instead of unreduced homology in the definition of $S$-duality, which is correct (?) as we are using smash products rather than direct products. {{endrem}} {{beginrem|Remark}} It doesn't seem to be immediately clear why the map $\Delta$ defined above is the correct geometric `diagonal' map to use so that the claimed identity $x\setminus \Delta_*(z) = x\cap z$ holds. It seems more natural to use the map $$ M / \partial M \longrightarrow M\times M / (\partial M \times M), $$ but this would witness that $M / \partial M$ is an $S^N$-dual of $X_+$, rather than $X\ldots$ {{endrem}} </wikitex>X$ has an $S^N$$S^N$-dual, for $N$$N$ sufficiently large.

Proof. (tentative: see remarks below) Let $X$$X$ be a finite pointed CW complex with basepoint $x_0$$x_0$. Embed $X\hookrightarrow S^N$$X\hookrightarrow S^N$ with regular neighbourhood $M$$M$, so that

$\displaystyle X \hookrightarrow M \hookrightarrow S^N$ $$\displaystyle X \hookrightarrow M \hookrightarrow S^N$$

with $X\hookrightarrow M$$X\hookrightarrow M$ a homotopy-equivalence and $M$$M$ an $N$$N$-dimensional manifold-with-boundary embedded in $S^N$$S^N$.

Now let

$\displaystyle c\colon S^N \longrightarrow S^N / (S^N \setminus \mathring{M}) = M / \partial M$ $$\displaystyle c\colon S^N \longrightarrow S^N / (S^N \setminus \mathring{M}) = M / \partial M$$

be the collapse map,

$\displaystyle \Delta\colon M / \partial M \longrightarrow (M\times M) / ((\partial M \times M)\cup (M\times \lbrace x_0 \rbrace)) \;\cong\; (M / \partial M) \wedge M$ $$\displaystyle \Delta\colon M / \partial M \longrightarrow (M\times M) / ((\partial M \times M)\cup (M\times \lbrace x_0 \rbrace)) \;\cong\; (M / \partial M) \wedge M$$

be the map induced by

$\displaystyle x\mapsto (x,x)\colon (M,\partial M) \longrightarrow (M\times M, (\partial M \times M)\cup (M\times \lbrace x_0 \rbrace)),$ $$\displaystyle x\mapsto (x,x)\colon (M,\partial M) \longrightarrow (M\times M, (\partial M \times M)\cup (M\times \lbrace x_0 \rbrace)),$$

and let $\alpha$$\alpha$ be the composite

$\displaystyle S^N \xrightarrow{c} M / \partial M \xrightarrow{\Delta} (M / \partial M) \wedge M \xrightarrow{1\wedge r} (M / \partial M) \wedge X ,$ $$\displaystyle S^N \xrightarrow{c} M / \partial M \xrightarrow{\Delta} (M / \partial M) \wedge M \xrightarrow{1\wedge r} (M / \partial M) \wedge X ,$$

where $r\colon M\to X$$r\colon M\to X$ is a chosen homotopy-inverse for $X\hookrightarrow M$$X\hookrightarrow M$.

The map $c$$c$ is degree-1 by construction, so $c_*([S^N]) = [M] \in \widetilde{H}_N(M / \partial M) = H_N(M,\partial M)$$c_*([S^N]) = [M] \in \widetilde{H}_N(M / \partial M) = H_N(M,\partial M)$.

For any $x\in \widetilde{H}^i(M)$$x\in \widetilde{H}^i(M)$, we have the identity

$\displaystyle x\setminus \Delta_*([M]) \;=\; x\cap [M] \;\in\; \widetilde{H}_{N-i}(M / \partial M) = H_{N-i}(M,\partial M),$ $$\displaystyle x\setminus \Delta_*([M]) \;=\; x\cap [M] \;\in\; \widetilde{H}_{N-i}(M / \partial M) = H_{N-i}(M,\partial M),$$

so by Poincare-Lefschetz duality,

$\displaystyle -\setminus (\Delta\circ c)_*([S^N]) \;\colon\; \widetilde{H}^i(M) \longrightarrow \widetilde{H}_{N-i}(M / \partial M)$ $$\displaystyle -\setminus (\Delta\circ c)_*([S^N]) \;\colon\; \widetilde{H}^i(M) \longrightarrow \widetilde{H}_{N-i}(M / \partial M)$$

is an isomorphism for all $i$$i$.

In general for any $X\wedge Y \xrightarrow{1\wedge f} X\wedge Z$$X\wedge Y \xrightarrow{1\wedge f} X\wedge Z$ and $a\in \widetilde{H}_N(X\wedge Y)$$a\in \widetilde{H}_N(X\wedge Y)$,

$\displaystyle f^*(-)\setminus a \;=\; -\setminus (1\wedge f)_*(a) \;\colon\; \widetilde{H}^i(Z) \longrightarrow \widetilde{H}_{N-i}(X).$ $$\displaystyle f^*(-)\setminus a \;=\; -\setminus (1\wedge f)_*(a) \;\colon\; \widetilde{H}^i(Z) \longrightarrow \widetilde{H}_{N-i}(X).$$

So in our case we have

$\displaystyle -\setminus \alpha_*([S^N]) \;=\; r^*(-)\setminus (\Delta\circ c)_*([S^N]) \;\colon\; \widetilde{H}^i(X) \longrightarrow \widetilde{H}_{N-i}(M / \partial M)$ $$\displaystyle -\setminus \alpha_*([S^N]) \;=\; r^*(-)\setminus (\Delta\circ c)_*([S^N]) \;\colon\; \widetilde{H}^i(X) \longrightarrow \widetilde{H}_{N-i}(M / \partial M)$$

which is an isomorphism by above and the fact that $r$$r$ is a homotopy-equivalence. This witnesses that $M / \partial M$$M / \partial M$ is an $S^N$$S^N$-dual of $X$$X$.

$\square$$\square$

Remark 0.2. This uses reduced homology instead of unreduced homology in the definition of $S$$S$-duality, which is correct (?) as we are using smash products rather than direct products.

Remark 0.3. It doesn't seem to be immediately clear why the map $\Delta$$\Delta$ defined above is the correct geometric `diagonal' map to use so that the claimed identity $x\setminus \Delta_*(z) = x\cap z$$x\setminus \Delta_*(z) = x\cap z$ holds. It seems more natural to use the map

$\displaystyle M / \partial M \longrightarrow M\times M / (\partial M \times M),$ $$\displaystyle M / \partial M \longrightarrow M\times M / (\partial M \times M),$$

but this would witness that $M / \partial M$$M / \partial M$ is an $S^N$$S^N$-dual of $X_+$$X_+$, rather than $X\ldots$$X\ldots$