Poincaré duality II (Ex)

For a simplicial complex $X$$<wikitex>; For a simplicial complex $X$, define the front $p$-face of an $n$-simplex $\sigma = [v_0\ldots v_n]$ as $_p\sigma:=[v_0\ldots v_p]$ and the back $q$-face as $\sigma_q := [v_{n-q}\ldots v_n].$ The Alexander-Whitney diagonal approximation is given by $$\begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned} $$ and the partial evaluation map is defined as $$ \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned} $$ Recall, we define the cap product on the chain level by $$a\cap z := E(a\otimes \tau(z))$$ and this descends to a well defined product on (co)homology. Consider $S^1$ as a simplicial complex with three <script type="math/tex">X$, define the front $p$$p$-face of an $n$$n$-simplex $\sigma = [v_0\ldots v_n]$$\sigma = [v_0\ldots v_n]$ as $_p\sigma:=[v_0\ldots v_p]$$_p\sigma:=[v_0\ldots v_p]$ and the back $q$$q$-face as $\sigma_q := [v_{n-q}\ldots v_n].$$\sigma_q := [v_{n-q}\ldots v_n].$

The Alexander-Whitney diagonal approximation is given by

$\displaystyle \begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned}$ $$\displaystyle \begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned}$$

and the partial evaluation map is defined as

$\displaystyle \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned}$ $$\displaystyle \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned}$$

Recall, we define the cap product on the chain level by

$\displaystyle a\cap z := E(a\otimes \tau(z))$ $$\displaystyle a\cap z := E(a\otimes \tau(z))$$

and this descends to a well defined product on (co)homology. Consider $S^1$$S^1$ as a simplicial complex with three $0$$0$-simplices and three $1$$1$-simplices. Compute explicitly, using the Alexander-Whitney diagonal approximation, the map

$\displaystyle -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$ $$\displaystyle -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$$

thus verifying that $S^1$$S^1$ has Poincaré duality.

$-simplices and three X, define the front $p$$p$-face of an $n$$n$-simplex $\sigma = [v_0\ldots v_n]$$\sigma = [v_0\ldots v_n]$ as $_p\sigma:=[v_0\ldots v_p]$$_p\sigma:=[v_0\ldots v_p]$ and the back $q$$q$-face as $\sigma_q := [v_{n-q}\ldots v_n].$$\sigma_q := [v_{n-q}\ldots v_n].$

The Alexander-Whitney diagonal approximation is given by

$\displaystyle \begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned}$ $$\displaystyle \begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned}$$

and the partial evaluation map is defined as

$\displaystyle \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned}$ $$\displaystyle \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned}$$

Recall, we define the cap product on the chain level by

$\displaystyle a\cap z := E(a\otimes \tau(z))$ $$\displaystyle a\cap z := E(a\otimes \tau(z))$$

and this descends to a well defined product on (co)homology. Consider $S^1$$S^1$ as a simplicial complex with three $0$$0$-simplices and three $1$$1$-simplices. Compute explicitly, using the Alexander-Whitney diagonal approximation, the map

$\displaystyle -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$ $$\displaystyle -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$$

thus verifying that $S^1$$S^1$ has Poincaré duality.

$-simplices. Compute explicitly, using the Alexander-Whitney diagonal approximation, the map $$ -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$$ thus verifying that $S^1$ has Poincaré duality. [[Category:Exercises]] [[Category:Exercises with solution]]X, define the front $p$$p$-face of an $n$$n$-simplex $\sigma = [v_0\ldots v_n]$$\sigma = [v_0\ldots v_n]$ as $_p\sigma:=[v_0\ldots v_p]$$_p\sigma:=[v_0\ldots v_p]$ and the back $q$$q$-face as $\sigma_q := [v_{n-q}\ldots v_n].$$\sigma_q := [v_{n-q}\ldots v_n].$

The Alexander-Whitney diagonal approximation is given by

$\displaystyle \begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned}$ $$\displaystyle \begin{aligned} \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} \end{aligned}$$

and the partial evaluation map is defined as

$\displaystyle \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned}$ $$\displaystyle \begin{aligned} E: C^r(X)\otimes C_p(X)\otimes C_q(X) &\to C_q(X)\\ a\otimes z \otimes w &\mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\ 0, & \mathrm{otherwise} \end{array}\right. ~. \end{aligned}$$

Recall, we define the cap product on the chain level by

$\displaystyle a\cap z := E(a\otimes \tau(z))$ $$\displaystyle a\cap z := E(a\otimes \tau(z))$$

and this descends to a well defined product on (co)homology. Consider $S^1$$S^1$ as a simplicial complex with three $0$$0$-simplices and three $1$$1$-simplices. Compute explicitly, using the Alexander-Whitney diagonal approximation, the map

$\displaystyle -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$ $$\displaystyle -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$$

thus verifying that $S^1$$S^1$ has Poincaré duality.