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generaldegreeHypersurfaces.tex

@@ -81,7 +81,7 @@ The retraction is performed in several steps, using the branched cover structure
 \section{Definition of the problem}
 Given $C$ an affine hypersurface in $\CC^2$ (i.e. a curve), Libgober gave in \cite{libgober1986} a construction of a $2$-dimensional CW-complex that is a deformation retract of its complement. However, it is not clear how to extend that construction to the projective case (that is a curve in $\CC\PP^2$).
 
-In this work, we give a different construction to produce a specific $\Delta$-complex that is a deformation retract of the complement of a Fermat hypersurface in $\CC\PP^n$. Since any smooth hypersurface is isotopic to a Fermat hypersurface, this particular case solves the problem for a general smooth hypersurface.
+In this work, we give a different construction to produce a specific $\Delta$-complex that is a deformation retract of the complement of a Fermat hypersurface in $\CC\PP^n$. Since any smooth hypersurface is isotopic to a Fermat hypersurface, this particular case solves the problemfor a general smooth hypersurface.
 
 
 \section{Smooth hypersurfaces}
@@ -145,11 +145,21 @@ and the restriction of $r_{1t}$ to $M_1\cap\RR^{n+1}$ is the inclusion.
 %where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively.
 
 Note that for $(x_0,\dots,x_n)\in M_1$, we already have that
+<<<<<<< HEAD
+\begin{equation*}
+\begin{array}{lll}
+	\sum x^r_j & = & 1 \\
+	\sum x^i_j & = & 0 \\
+\end{array}
+\end{equation*}
+and therefore $r_1(x_0,\dots,x_n,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x_0,\dots,x_n,1)=(x^r_0,\dots,x^r_n)\in M_1\cap\RR^{n+1}$ for all $(x_0,\dots,x_n)\in M_1$. 
+=======
 \begin{align*}
 	\sum_{j=0}^n x^r_j & =  1 \\
 	\sum_{j=0}^n  x^i_j & =  0 \\
 \end{align*}
 and therefore $r_1(x_0,\dots,x_n,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x_0,\dots,x_n,1)=(x^r_0,\dots,x^r_n)\in M_1\cap\RR^{n+1}$ for all $(x_0,\dots,x_n)\in M_1$.
+>>>>>>> 2a87fee69dfb1e6fdd6e608f9b4ee957cc3f5a74
 Let us summarize it.
 
 
@@ -171,17 +181,18 @@ Note that $M_1\cap\RR^{n+1}$
 is a hypersurface in $\RR^{n+1}$, in Figure~\ref{fig:triangulo} we can see the example $n=2$. 
 
 \begin{lemma}
-The space $S_1$ is a $n$-simplex.
+	The space $S_1$ is a $n$-simplex.
 \end{lemma}
 \begin{proof}
-Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
-\begin{align*}
-S_1&=\left\{(x_0,\dots,x_n)\in\RR^{n+1}\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\} \\
-&=\left\{x_0e_0+\dots+x_ne_n\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\}=\Delta_n
-\qedhere
-\end{align*}
+	Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
+	\begin{align*}
+	S_1&=\left\{(x_0,\dots,x_n)\in\RR^{n+1}\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\} \\
+	&=\left\{x_0e_0+\dots+x_ne_n\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\}=\Delta_n
+	\qedhere
+	\end{align*}
 \end{proof}
 
+
 \begin{figure}[ht]
 \begin{center}
 \begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
@@ -234,13 +245,13 @@ S_1&=\left\{(x_0,\dots,x_n)\in\RR^{n+1}\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\
 \end{center}
 \end{figure}
 
-As we can see in Figure \ref{fig:triangulo}, the hyperplanes defined by $\{x_i=0\}$ divide the space $S_1$ in $2^{n+1}-1$ regions. We denote each of these regions as $(\pm,\dots,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ would be $(+,\dots,+)$.
-Note that there is no region $(-,\dots,-)$.
+As we can see in Figure \ref{fig:triangulo}, the hypersurfaces defined by $\{x_i=0\}$ divide the space in $2^{n+1}-1$ sections. We denote each of these sections as $(\pm,\dots,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ would be $(+,\dots,+)$.
+Note that there is no section $(-,\dots,-)$.
 
 The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the rest of the parts, we define a retraction:
 \begin{equation*}
 \begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
-(M_1\cap\RR^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\\
+(M_1\cap\RR^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\times [0,1]\\
 (Q, t)\rar[mapsto]& t P_Q + (1 - t) Q,
 \end{tikzcd}
 \end{equation*}
@@ -320,31 +331,33 @@ m(x,t)=+\sqrt{x_r^2+(1-t)^2x_i^2} \text{ and } \theta(x,t)=\arctan\frac{(1-t)x_i
 \]
 We can divide $\CC$ in $2d$ sectors of angle $\frac{\pi}{d}$ as in Figure~\ref{divisionC}, where each of the sections will be homeomorphic to either the upper half complex plane or to the lower half one.
 
+
 \begin{figure}[h]
-\begin{center}
-%\begin{tikzpicture}[scale=0.6]
-%	\draw (-5,0) to (5,0);
-%	\draw (5,1.5) to (-5,-1.5);
-%	\draw (5,3.2) to (-5, -3.2);
-%	\draw (4.7,5) to (-4.7,-5);
-%	\draw (2.8,5) to (-2.8,-5);
-%	\draw (0.9,5) to (-0.9,-5);
-%	\draw (-0.9,5) to (0.9,-5);
-%	\draw (-2.8,5) to (2.8,-5);
-%	\draw (-4.7,5) to (4.7,-5);
-%	\draw (-5,3.2) to (5, -3.2);
-%	\draw (-5,1.5) to (5,-1.5);
-%\end{tikzpicture}
-%
-\begin{tikzpicture}
-\foreach \x in {0, ..., 16}{
-	\draw (0,0) to (360/17*\x:3);}
-\end{tikzpicture}
+	\begin{center}
+		%\begin{tikzpicture}[scale=0.6]
+		%	\draw (-5,0) to (5,0);
+		%	\draw (5,1.5) to (-5,-1.5);
+		%	\draw (5,3.2) to (-5, -3.2);
+		%	\draw (4.7,5) to (-4.7,-5);
+		%	\draw (2.8,5) to (-2.8,-5);
+		%	\draw (0.9,5) to (-0.9,-5);
+		%	\draw (-0.9,5) to (0.9,-5);
+		%	\draw (-2.8,5) to (2.8,-5);
+		%	\draw (-4.7,5) to (4.7,-5);
+		%	\draw (-5,3.2) to (5, -3.2);
+		%	\draw (-5,1.5) to (5,-1.5);
+		%\end{tikzpicture}
+		%
+		\begin{tikzpicture}
+		\foreach \x in {0, ..., 5}{
+			\draw (0,0) to (360/6*\x:3);}
+		\draw[->] (1.5,0) to[out=90, in=350] (0.6,1.1) node[ xshift=1.0cm, yshift=-0.2cm]{$\pi/d$};
+		\end{tikzpicture}
 		
 		
-\end{center}
-\caption{Division of the complex plane}
-\label{divisionC}
+	\end{center}
+	\caption{Division of the complex plane}
+	\label{divisionC}
 \end{figure}
 
 Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$, which we will denote $f_k$. The lifting of $r_1$ will then be given by the $G$ previously described. This $G$ will act by taking the point $(x_0,\dots,x_n)\in M_d$ and mapping it to the upper or lower complex plane via the inverse homeomorphisms $f^{-1}_k$, then taking $r_1$ to take it to the real line, and finally bringing it back to its original sector with the homeomorphism~$f_k$.
@@ -389,8 +402,8 @@ G(z,t)^d\dar[equal, equal]&=&m(z,t)^d\exp{\left(i\arctan((1-t)\tan d\theta)\righ
 We can then write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
 
 
-Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d_0, \ldots , x^d_n\in\RR\}$ into $S_d$.
-For a point $P=(x_0\xi_0,\dots,x_n\xi_n)\in M_d\cap\{x^d_0, \ldots , x^d_n\in\RR\}$ (where $(x_0,\dots,x_n)$ are real and $\xi_i$ are $d$-th roots of unity), the corresponding point in $M_1\cap\RR^{n+1}$ is $(x^d_0,\dots,x^d_n)$ and therefore the goal point of the retraction $r_2$ will be
+Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d_0, y^d, x^d_n\in\RR\}$ into $S_d$.
+For a point $P=(x_0\xi_0,\dots,x_n\xi_n)\in M_d\cap\{x^d_0, y^d, x^d_n\in\RR\}$ (where $(x_0,\dots,x_n)$ are real and $\xi_i$ are $d$-th roots of unity), the corresponding point in $M_1\cap\RR^{n+1}$ is $(x^d_0,\dots,x^d_n)$ and therefore the goal point of the retraction $r_2$ will be
 \begin{equation}
 P^d_Q=\left(\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}, \dots,\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)} \right)
 \end{equation}
@@ -434,6 +447,23 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
 	If $[x^d_0: \dots: x^d_n]$ is in $(+,\dots,+)$, the retraction is the identity, and therefore both representatives will stay the same and therefore go to the same class.
 	
 	For the rest of the  regions, we have
+<<<<<<< HEAD
+\begin{equation}
+\begin{array}{lll}
+\tilde{r}_2([x_0\xi_0:\dots:x_n\xi_n], t)& =& \left[\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right]
+\end{array}
+\end{equation}
+
+
+Then, for the point $\lambda p$ we get 
+\begin{equation}
+\begin{array}{lll}
+\tilde{r}_2([\lambda x_0\xi_0:\dots:\lambda x_n\xi_n], t)&=&\left[\sqrt[d]{(1-t)\lambda^d x^d_0+t\frac{\lambda^d x^d_0\chi_0}{1-\sum \lambda^d x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)\lambda^d x^d_n+t\frac{\lambda^d x^d_n\chi_n}{1-\sum \lambda^d x^d_j(1-\chi_j)}}\xi_n\right]\\
+&=&\left[\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right]\\
+\end{array}
+\end{equation}
+
+=======
 \begin{equation*}
 \tilde{r}_2([x_0\xi_0:\dots:x_n\xi_n], t) =\left[H(x_0,\chi_0)\xi_0:\dots:H(x_n,\chi_n) \xi_n\right]
 \end{equation*}
@@ -446,6 +476,7 @@ Then, for the point $\lambda p$ we get
 \tilde{r}_2([\lambda x_0\xi_0:\dots:\lambda x_n\xi_n], t)&=\left[H(\lambda x_0, \chi_0)\xi_0:\dots:H(\lambda x_n, \chi_n)\xi_n\right]\\ 
 &=\left[H( x_0, \chi_0)\xi_0:\dots:H(x_n, \chi_n)\xi_n\right].
 \end{align*}
+>>>>>>> 2a87fee69dfb1e6fdd6e608f9b4ee957cc3f5a74
 So we see that $\tilde{r}_2$ also takes both representatives will end to the same point.
 \end{proof}
 
@@ -483,10 +514,17 @@ First, we give a detailed description of $\Delta$. We will denote the $0$-simpli
 	X & =  [v_0,\dots,v_n]
 \end{align*}
 The face maps of $X$ and $X^{(j_0,\dots,j_i)}$ are:
+<<<<<<< HEAD
+\begin{equation}
+	\delta^i_k(X^{(j_0,\dots,j_i)}) =[v_{j_0},\dots,\hat{v},\dots,v_{j_i}] 
+\end{equation}
+where $\hat{v}$ indicates that we take out the vertix in position $k$, for $k\leq i$.
+=======
 \begin{equation*}
 	\delta^i_k(X^{(j_0,\dots,j_i)}) =[v_{j_0},\dots,\hat{v},\dots,v_{j_i}]
 \end{equation*}
 where $\hat{v}$ indicates that we take out the vertex in position $k$, for $k\leq i$.
+>>>>>>> 2a87fee69dfb1e6fdd6e608f9b4ee957cc3f5a74
 an the boundary maps are given by:
 \begin{equation*}
 \partial^i =\sum (-1)^{k}\delta^i_k
@@ -515,10 +553,20 @@ X_{(1,\dots,\xi_i, 1, 1)}\cap\{x_j=1,\,\forall j\neq i\}.
 \]
 We can remark that $X_{(\xi_0, \dots, \xi_n)}=(\xi_0, \dots, \xi_n)\cdot X_{(1,\dots,1)}$, and therefore that we have a group action of $\mu^{n+1}_d$ over $S_d$.
 Let
+<<<<<<< HEAD
+\begin{equation}
+\begin{array}{rll}
+	X^{(j_0,\dots,j_i)}_{(1,\dots,1)} & = & X_{(1, \dots, 1)}\cap\{x_{j_k}\neq0 \forall k=0,\dots i\text{ and }x_{i}=0\text{ for the rest}\}\\
+	X^j_{(1,\dots,1)} & = & X_{(1, \dots, 1)}\cap\{x_i=0\forall i\neq j\}\\
+	
+\end{array}
+\end{equation}
+=======
 \begin{align*}
 	X^{(j_0,\dots,j_i)}_{(1,\dots,1)} & =  X_{(1, \dots, 1)}\cap\{x_{j_k}\neq0,\, \forall k=0,\dots i\text{ and }x_{i}=0\text{ for the rest}\}\\
 	X^j_{(1,\dots,1)} & =  X_{(1, \dots, 1)}\cap\{x_i=0,\,\forall i\neq j\}\\
 \end{align*}
+>>>>>>> 2a87fee69dfb1e6fdd6e608f9b4ee957cc3f5a74
 We can describe the chain complexes. They are abelian groups but they are also $R_d$-complexes
 where $R_d$ is the group algebra $\ZZ[\mu_d^{n+1}]$. For the $n$-chains we have
 \[
@@ -544,6 +592,7 @@ a free abelian group of rank~$(n+1)d$.
 %\end{array}
 %\end{equation}
 
+
 The face maps are $R_d$-morphisms defined as 
 \begin{equation*}
 \begin{aligned}
@@ -843,7 +892,7 @@ Delta_Complex = DeltaComplex_hypersurface(4,6)
 Once we have gotten the $\Delta$-complex, we can compute the fundamental group or the cohomology of the space, for example.
 \begin{python}
 Dc = SimplicialSet(Delta_Complex)
-dc_punto = Dc.set_base_point(Dc.all_n_simplices(0)[0])
+dc_punto = it commit Dc.set_base_point(Dc.all_n_simplices(0)[0])
 dc_punto.fundamental_group()
  -> Finitely presented group < e0 | e0^6 >
 dc_punto.cohomology()