Talk:S-duality II (Ex)

The map we use is closely related to the previous exercise so we will copy and paste some terminology used there - note that we will use $X_+$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}X_+$ where $X$$X$ is used there, and have changed the diagonal map appropriately:

Let $X$$X$ be a finite CW complex. Embed $X\hookrightarrow S^N$$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,

$\displaystyle \Delta\colon M / \partial M \longrightarrow (M\times M) / (\partial M \times M) \;\cong\; (M / \partial M) \wedge M_+$ $$\displaystyle \Delta\colon M / \partial M \longrightarrow (M\times M) / (\partial M \times M) \;\cong\; (M / \partial M) \wedge M_+$$

be the map induced by

$\displaystyle x\mapsto (x,x)\colon (M,\partial M) \longrightarrow (M\times M, \partial M \times M),$ $$\displaystyle x\mapsto (x,x)\colon (M,\partial M) \longrightarrow (M\times M, \partial M \times M),$$

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$.

Solution 0.1.

First, take $N=n+k$$N=n+k$ where $n$$n$ is the dimension of $X$$X$. Fix a Thom class $U\in C^k(Th(\nu_X))=C^k(M/\partial M)$$U\in C^k(Th(\nu_X))=C^k(M/\partial M)$ and a class $[X]\in C_n(X)$$[X]\in C_n(X)$ determining the Poincaré duality for $X$$X$.

Take reduced chain complexes (denoted $\tilde{C}$$\tilde{C}$) for the spaces in the composite $\alpha$$\alpha$. We then make a choice of Eilenberg-Zilber map (unique up to chain homotopy) so that

$\displaystyle \tilde{C}((M/\partial M)\wedge M_+)\simeq \tilde{C}(M/\partial M)\otimes \tilde{C}(M_+)\simeq \tilde{C}(M/\partial M)\otimes C(M).$ $$\displaystyle \tilde{C}((M/\partial M)\wedge M_+)\simeq \tilde{C}(M/\partial M)\otimes \tilde{C}(M_+)\simeq \tilde{C}(M/\partial M)\otimes C(M).$$

The chain-level slant map now involves a choice - in general if we have two chains $C$$C$ and $D$$D$ then the slant involves a choice of ordering to form $C\otimes D\simeq Hom(C^{-*},D)$$C\otimes D\simeq Hom(C^{-*},D)$ or $\simeq Hom(D^{-*},C)$$\simeq Hom(D^{-*},C)$. When considering $S$$S$-duality, the choice represents the symmetry that $X=Y^*$$X=Y^*$ iff $Y=X^*$$Y=X^*$. We will need to choose the slant map so that

$\displaystyle \tilde{C}(M/\partial M)\otimes C(M)\simeq Hom(\tilde{C}^{N-*}(M/\partial M),C(M)),$ $$\displaystyle \tilde{C}(M/\partial M)\otimes C(M)\simeq Hom(\tilde{C}^{N-*}(M/\partial M),C(M)),$$

this is followed with the map induced by $1\wedge r$$1\wedge r$ so that the overall slant map isomorphism induced is $-\backslash \alpha_*([S^N]):\tilde{H}^{N-*}(M/\partial M)\to H_*(X)$$-\backslash \alpha_*([S^N]):\tilde{H}^{N-*}(M/\partial M)\to H_*(X)$ (see the previous exercise for a justification of this, albeit for the opposite choice of slant).

Considering the composite $\alpha$$\alpha$ again, note that before we use the smash map $1\wedge r$$1\wedge r$, it is still valid to use the identity for slant product that for any $x\in \tilde{H}^{N-*}(M/\partial M)$$x\in \tilde{H}^{N-*}(M/\partial M)$, we have

$\displaystyle x\setminus \Delta_*(c_*([S^N])) \;=\; x\cap c_*([S^N]) \;\in\; H_{*}(M),$ $$\displaystyle x\setminus \Delta_*(c_*([S^N])) \;=\; x\cap c_*([S^N]) \;\in\; H_{*}(M),$$

as this is still a (relative) definition of cap product on the chain level.

We now want to know how the slant map interacts with the smash map $1\wedge r$$1\wedge r$. In general for any $X\wedge Y \xrightarrow{g\wedge f} W\wedge Z$$X\wedge Y \xrightarrow{g\wedge f} W\wedge Z$ and $a\in \tilde{H}_N(X\wedge Y)$$a\in \tilde{H}_N(X\wedge Y)$, we have a commutative diagram:

$\displaystyle \xymatrix{ C(X\wedge Y)\ar[d]_{(g\wedge f)_*}\ar[r]^-{\simeq}&Hom(C(X)^{-*},C(Y))\ar[d]\\ C(W\wedge Z)\ar[r]^-{\simeq}&Hom(C(W)^{-*},C(Z))}$

and so it is clear that $-\backslash(g\wedge f)_*(a)=f_*(g^*(-)\backslash a)$$-\backslash(g\wedge f)_*(a)=f_*(g^*(-)\backslash a)$.

Hence

$\displaystyle r_*(U\backslash \Delta(c_*([S^N])))=r_*(U\cap c_*([S^N])).$ $$\displaystyle r_*(U\backslash \Delta(c_*([S^N])))=r_*(U\cap c_*([S^N])).$$

But this is $\pm[X]$$\pm[X]$ by the definition of the Spivak normal structure.

Solution 0.2.

This now follows easily from breaking the diagram down in a careful fashion to

$\displaystyle \xymatrix{ C^{n-*}(X)\ar[ddr]^{-\cap [X]}\ar[r]^{r^*}& C^{n-*}(M)\ar[r]^-{-\cup U} &C^{n+k-*}(M,\partial M)\ar[ddl]^{-\backslash \alpha_*([S^N])}\\ &&\\ &C(X)&}$

and noting that following the top side of the diagram round to $C(X)$$C(X)$ gives

$\displaystyle r_*((r^*(-)\cup U)\cap c_*([S^N]))=-\cap r_*(U\cap c_*([S^N]))$ $$\displaystyle r_*((r^*(-)\cup U)\cap c_*([S^N]))=-\cap r_*(U\cap c_*([S^N]))$$

so that the diagram commutes by the first part of the question.