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}
@@ -515,30 +558,7 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^z_{[1:1:1]}> \
 	We start with $S_2$. First, we can observe that, from all points in $\CC$, the only ones whose square power is real are either the real numbers of the pure imaginary numbers. And from these, the later have negative square power. Therefore, we can conclude that $S_2=\{(x,y,z)\in\RR^3| x^2+y^2+z^2=1\}$, that is, $S_2$ is the real three dimensional sphere. We now see the simplicial complex $\Delta^2$ that can be associated to it such that $\phi(\Delta^2)=\Delta$. 
 	
 	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 $S_1$.	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
-	
+		
 	%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]\}\\