S-duality I (Ex)

Latest revision as of 21:01, 28 May 2012

Let $<wikitex>; Let $X$ and $Y$ be a finite pointed CW-complexes. A map $$ \alpha \colon S^N \to X \wedge Y $$ is called an S-duality if the slant product induced by $\alpha$ $$ - \backslash \alpha \colon H^i(X; \Zz) \to H_{N-i}(Y;\Zz) $$ is an isomorphism for all $i$. In this case $X$ and $Y$ are called an $S$-duals of each other. {{beginthm|Exercise}} \label{exrcs:S-duality-property} Show that $S$-duality satisfies the following" *For every finite CW-complex $X$ there exists an $N$-dimensional S-dual, which we denote $X^\ast$, for some large $N \geq 1$. * If $X^\ast$ is an $N$-dimensional $S$-dual of $X$ then $\Sigma X^\ast$ is an $(N+1)$-dimensional $S$-dual of $X$. * For any space $Z$ we have isomorphisms $$ S \co [X,Z] \cong [S^N,Z \wedge Y] \quad \gamma \mapsto S(\gamma) = (\gamma \wedge \id_Y) \circ \alpha,$$ $$ S \co [Y,Z] \cong [S^N,X \wedge Z] \quad \gamma \mapsto S(\gamma) = (\id_X \wedge \gamma) \circ \alpha.$$ * A map $f \co X \rightarrow Y$ induces a map $f^\ast \co Y^\ast \rightarrow X^\ast$ for $N$ large enough via the isomorphism $$ [X,Y] \cong [S^N,Y \wedge X^\ast] \cong [Y^\ast,X^\ast]. $$ * If $X \rightarrow Y \rightarrow Z$ is a cofibration sequence then $Z^\ast \rightarrow Y^\ast \rightarrow X^\ast$ is a cofibration sequence for $N$ large enough. {{endthm}} </wikitex> == References == {{#RefList:}} [[Category:Exercises]] [[Category:Exercises without solution]]X$ and $Y$ be a finite pointed CW-complexes. A map

$$\displaystyle \alpha \colon S^N \to X \wedge Y$$

is called an S-duality if the slant product induced by $\alpha$

$$\displaystyle - \backslash \alpha \colon H^i(X; \Zz) \to H_{N-i}(Y;\Zz)$$

is an isomorphism for all $i$. In this case $X$ and $Y$ are called an $S$-duals of each other.

Exercise 0.1. Show that $S$-duality satisfies the following"

• For every finite CW-complex $X$ there exists an $N$-dimensional S-dual, which we denote $X^\ast$, for some large $N \geq 1$.
• If $X^\ast$ is an $N$-dimensional $S$-dual of $X$ then $\Sigma X^\ast$ is an $(N+1)$-dimensional $S$-dual of $X$.
• For any space $Z$ we have isomorphisms

$$\displaystyle S \co [X,Z] \cong [S^N,Z \wedge Y] \quad \gamma \mapsto S(\gamma) = (\gamma \wedge \id_Y) \circ \alpha,$$

$$\displaystyle S \co [Y,Z] \cong [S^N,X \wedge Z] \quad \gamma \mapsto S(\gamma) = (\id_X \wedge \gamma) \circ \alpha.$$

• A map $f \co X \rightarrow Y$ induces a map $f^\ast \co Y^\ast \rightarrow X^\ast$ for $N$ large enough via the isomorphism

$$\displaystyle [X,Y] \cong [S^N,Y \wedge X^\ast] \cong [Y^\ast,X^\ast].$$

• If $X \rightarrow Y \rightarrow Z$ is a cofibration sequence then $Z^\ast \rightarrow Y^\ast \rightarrow X^\ast$ is a cofibration sequence for $N$ large enough.

[edit] References