face maps delta complex

generaldegree.tex

@@ -59,7 +59,7 @@ We then have a commutative diagram:
 	\end{array}
 \end{equation*}
 
-The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma:
+We are going to see that the inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. To do this we first need the following lemma:
 \begin{lemma}
 $M_1\cap\RR^3$ is a strong deformation retract of $M_1$.
 \end{lemma}
@@ -411,7 +411,7 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
 We have now seen that there is a homotopy equivalence between $\PP^2\setminus C_d$ and $K$. Therefore, to get a CW-complex structure on $\PP^2\setminus C_d$, it suffices to get it for $K$.
 
 
-We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertices, $2$ eedges and $1$ face, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
+We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertices, $2$ edges and $1$ face, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
 
 \begin{equation*}
 	\begin{array}{ccccc}
@@ -430,6 +430,43 @@ where $\Delta^d$ is a $\Delta-$complex that we are trying to find and identify w
 %The edges of $\Delta$ can be identified with the lines of $S_1$ parametrized as $(1-t,t,0)$, $(1-t,0,t)$ and $(0,t,1-t)$ with $t\in[0,1]$. Then, the preimages of these lines by $\phi$ are of the form $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$ in the first case and analogous in the rest, where we denote $\sqrt[s]{x}$ the positive real $d$-root for $x$ real, and where $\xi_i$ and $\xi_j$ are two any $d$-$th$ roots of unity. For each of these preimages we can define an edge between the corresponding vertices in $\Delta^d$.
 
 %The $2$-faces of the simplicial complex $\Delta^d$ will then be identified to the areas between the lines $(\sqrt[d]{1-t}\xi_i, \sqrt[d]{t}\xi_j, 0)$, $(\sqrt[d]{1-t}\xi_i,0, \sqrt[d]{t}\xi_k)$ and $(0,\sqrt[d]{t}\xi_j, \sqrt[d]{1-t}\xi_k)$.
+First, we give a detailed description of $\Delta$. We will denote:
+\begin{equation}
+\begin{array}{lllll}
+	v_0 & = & (1,0,0) & & \\
+	v_1 & = & (0,1,0) & & \\
+	v_2 & = & (0,0,1) & & \\
+	l_0 & = & [v_1,v2] & = & \{(1-t)v_1+tv_2|t\in[0,1]\}\\
+	l_1 & = & [v_0,v2] & = & \{(1-t)v_0+tv_2|t\in[0,1]\}\\
+	l_0 & = & [v_0,v1] & = & \{(1-t)v_0+tv_1|t\in[0,1]\}\\
+	X & = & [v_0,v_1,v_2]
+\end{array}
+\end{equation}
+
+The face maps of $X$, $l_0$, $l_1$ and $l_2$ are:
+\begin{equation}
+\begin{array}{lllll}
+	\delta^2_0(X) & = & [v_1,v2] & = & l_0\\
+	\delta^2_1(X) & = & [v_0,v2] & = & l_1\\
+	\delta^2_2(X) & = & [v_0,v1] & = & l_2\\
+	\delta^1_0(l_0) & = & v_1 \\
+	\delta^1_1(l_0) & = & v_2\\
+	\delta^1_0(l_1) & = & v_0\\
+	\delta^1_1(l_1) & = & v_2\\
+	\delta^1_0(l_2) & = & v_0\\
+	\delta^1_1(l_2) & = & v_1\\
+\end{array}
+\end{equation}
+Lastly, the border maps are given by:
+\begin{equation}
+\begin{array}{lll}
+\partial^2 & = & \delta^2_2-\delta^2_1+\delta^2_0 \\
+\partial^1 & = & \delta^2_1-\delta^2_0 \\
+\end{array}
+\label{bordermaps}
+\end{equation}
+
+
 $\Delta^d$ will have $3d$ $0-$simplices, $3d^2$ $1-$simplices and $d^3$ $2-$simplices.
 
 
@@ -467,26 +504,32 @@ C_0(\Delta^d) & = & \ZZ[\mu^3_d]/\ZZ[\mu^2_d]<V^z_{(1,1,1)}> \bigoplus \ZZ[\mu^3
 \end{array}
 \end{equation}
 
-The border maps are given by:
+The face maps of $X_{(1,1,1)}$, $L^z_{(1,1,1)}$, $L^y_{(1,1,1)}$ and $L^x_{(1,1,1)}$ are:
 \begin{equation}
 \begin{array}{lll}
-	\partial X_{(1,1,1)} & = & L^z_{(1,1,1)}-L^y_{(1,1,1)}+L^x_{(1,1,1)}\\
-	\partial L^z_{(1,1,1)} & = & V^y_{(1,1,1)}-V^x_{(1,1,1)}\\
-	\partial L^y_{(1,1,1)} & = & V^z_{(1,1,1)}-V^x_{(1,1,1)}\\
-	\partial L^x_{(1,1,1)} & = & V^z_{(1,1,1)}-V^y_{(1,1,1)}\\
+	\delta^2_0(X_{(1,1,1)}) & = & L^x_{(1,1,1)}\\
+	\delta^2_1(X_{(1,1,1)}) & = & L^y_{(1,1,1)}\\
+	\delta^2_2(X_{(1,1,1)}) & = & L^z_{(1,1,1)}\\
+	\delta^1_0(L^z_{(1,1,1)}) & = & V^y_{(1,1,1)} \\
+	\delta^1_1(L^z_{(1,1,1)}) & = & V^z_{(1,1,1)}\\
+	\delta^1_0(L^y_{(1,1,1)}) & = & V^x_{(1,1,1)}\\
+	\delta^1_1(L^y_{(1,1,1)}) & = & V^z_{(1,1,1)}\\
+	\delta^1_0(L^x_{(1,1,1)}) & = & V^x_{(1,1,1)}\\
+	\delta^1_1(L^x_{(1,1,1)}) & = & V^y_{(1,1,1)}\\
 \end{array}
 \end{equation}
 
-The border maps for the rest of simplices are defined by the group action as:
+For the rest of the simplices, due to the group action, the face maps are given by:
 \begin{equation}
-\begin{array}{lllll}
-	\partial X_{(\xi_1, \xi_2, \xi_3)} & =& (\xi_1, \xi_2, \xi_3)\partial X_{(1,1,1)}  & = & (\xi_1, \xi_2, 1)L^z_{(1,1,1)}-(\xi_1, 1, \xi_3)L^y_{(1,1,1)}+(1, \xi_2, \xi_3)L^x_{(1,1,1)}\\
-	\partial L^z_{(\xi_1, \xi_2, 1)} & = & (\xi_1, \xi_2, 1)\partial L^z_{(1,1,1)} & = & (1, \xi_2, 1)V^y_{(1,1,1)}-(\xi_1, 1, 1)V^x_{(1,1,1)}\\
-	\partial L^y_{(\xi_1, 1, \xi_2)} & = & (\xi_1, 1, \xi_2)\partial L^y_{(1,1,1)} & = & (1, 1, \xi_2)V^z_{(1,1,1)}-(\xi_1, 1, 1)V^x_{(1,1,1)}\\
-	\partial L^x_{(1, \xi_1, \xi_2)} & = & (1, \xi_1, \xi_2)\partial L^x_{(1,1,1)} & = & (1, \xi_1, 1)V^z_{(1,1,1)}-(1, 1, \xi_2)V^y_{(1,1,1)}\\
+\begin{array}{lll}
+	\delta^2_i(X_{(\xi_1, \xi_2, \xi_3)}) & = & (\xi_1, \xi_2, \xi_3)\delta^2_i(X_{(1,1,1)})\\
+	\delta^1_i(L^\bullet_{(\xi_1, \xi_2, \xi_3)}) & = & (\xi_1, \xi_2, \xi_3)\delta^1_i(L^\bullet_{(1,1,1)})\\
+	
 \end{array}
 \end{equation}
 
+Then, the border maps are given by (\ref{bordermaps}).
+
 We have then defined $\Delta^d$, the simplicial complex with which we can identify $S_d$. Our last step now is to complete the diagram with a simplicial complex associated to the projective space $K$, which we will call $\Delta^d_\text{proj}$.
 
 \begin{equation*}
@@ -499,7 +542,7 @@ We have then defined $\Delta^d$, the simplicial complex with which we can identi
 	\end{array}
 \end{equation*}
 
-Where the map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x,y,z)\longrightarrow[x:y:z]$. Since we are passing to the projective plane,two points $(x\xi_1, y\xi_2, z\xi_3)$ and $(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)$ will be identified, and therefore, so will $X_{(x\xi_1, y\xi_2, z\xi_3)}$ and $X_{(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)}$. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices.
+Where the map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x,y,z)\longrightarrow[x:y:z]$. Since we are passing to the projective plane, two points $(x\xi_1, y\xi_2, z\xi_3)$ and $(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)$ will be identified, and therefore, so will $X_{(x\xi_1, y\xi_2, z\xi_3)}$ and $X_{(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)}$. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices.
 
 \begin{equation}
 \begin{array}{rll}
@@ -516,29 +559,6 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^z_{[1:1:1]}> \
 	
 	In this case, we have that $\mu_2=\{\xi|\xi^2=1\}=\{1,-1\}$. Then, we have that $X_{(1,1,1)}=\{(x,y,z)\in\RR^3_{\geq0} |x^2+y^2+z^2=1\}$, this is, $X_{(1,1,1)}$ is the set of points given by the positive square roots of the points in $S_1$.	
 		
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 	%The vertices of $\Delta$, $v_0$, $v_1$ and $v_2$, can be identified to $(1,0,0)$, $(0,0,1)$ and $(0,1,0)$ respectively. The preimages of these points are:
 	%\begin{equation}
 		%\begin{array}{rll}
@@ -548,7 +568,7 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^z_{[1:1:1]}> \
 		%\end{array}
 	%\end{equation}
 	
-	T%he edges of the simplex, $[v_0v_1]$, $[v_0v_2]$ and $[v_1v_2]$, can be identified to the points of $S_1$ given respectively by $\{(1-t, 0, t)|t\in[0,1]\}$, $\{(1-t,t,0)|t\in[0,1]\}$ and $\{(0,t,1-t)|t\in[0,1]\}$. Therefore, each edge will have four preimages. In the first case, these preimages are:
+	%The edges of the simplex, $[v_0v_1]$, $[v_0v_2]$ and $[v_1v_2]$, can be identified to the points of $S_1$ given respectively by $\{(1-t, 0, t)|t\in[0,1]\}$, $\{(1-t,t,0)|t\in[0,1]\}$ and $\{(0,t,1-t)|t\in[0,1]\}$. Therefore, each edge will have four preimages. In the first case, these preimages are:
 	%\begin{equation}
 		%\phi^{-1}((1-t,t,0))=\left\{\begin{array}{l}
 		%	\{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\