Poincaré duality II (Ex)

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Latest revision as of 14:56, 1 April 2012

For a simplicial complex $<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 $$ \tau: C_n(X) \to (C(X)\otimes C(X))_n $$ $$ \sigma \mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q} $$ and the partial evaluation map is defined as $$ E: C^r(X)\otimes C_p(X)\otimes C_q(X) \to C_q(X) $$ $$ a\otimes z \otimes w \mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \ 0, & \mathrm{otherwise} \end{array}\right. ~. $$ 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 $p$ -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

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

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. == References == {{#RefList:}} [[Category:Exercises]]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.

Poincaré duality II (Ex)

Latest revision as of 14:56, 1 April 2012

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@@ Line 3: / Line 3: @@
 The Alexander-Whitney diagonal approximation is given by
-$$
+$$\begin{aligned}
-  \tau: C_n(X) \to (C(X)\otimes C(X))_n $$
+  \tau: C_n(X) &\to (C(X)\otimes C(X))_n \\
-$$ \sigma \mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q}
+ \sigma &\mapsto \sum_{p+q=n}{_p\sigma\otimes\sigma_q}
+\end{aligned}
 $$
 and the partial evaluation map is defined as
 $$
-  E: C^r(X)\otimes C_p(X)\otimes C_q(X) \to C_q(X)
+\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 \\
-  a\otimes z \otimes w \mapsto \left\{ \begin{array}{cc} a(w)\otimes z, & r=q \\
 , & \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))$$ and this descends to a well defined product on (co)homology.
@@ Line 19: / Line 20: @@
 Consider $S^1$ as a simplicial complex with three $0$-simplices and three $1$-simplices. Compute explicitly, using the Alexander-Whitney diagonal approximation, the map $$ -\cap[S^1]: C^{1-*}(S^1) \to C_*(S^1)$$ thus verifying that $S^1$ has Poincaré duality.
 </wikitex>
-== References ==
+<!-- == References ==
-{{#RefList:}}
+{{#RefList:}} -->
 [[Category:Exercises]]
+[[Category:Exercises with solution]]