generaldegreeHypersurfaces.tex

@@ -145,21 +145,11 @@ 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.
 
 
@@ -192,7 +182,6 @@ is a hypersurface in $\RR^{n+1}$, in Figure~\ref{fig:triangulo} we can see the e
 \end{align*}
 \end{proof}
 
-
 \begin{figure}[ht]
 \begin{center}
 \begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
@@ -245,13 +234,13 @@ is a hypersurface in $\RR^{n+1}$, in Figure~\ref{fig:triangulo} we can see the e
 \end{center}
 \end{figure}
 
-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 $(-,\dots,-)$.
+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 $(-,\dots,-)$.
 
 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}\times [0,1]\\
+(M_1\cap\RR^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\\
 (Q, t)\rar[mapsto]& t P_Q + (1 - t) Q,
 \end{tikzcd}
 \end{equation*}
@@ -331,7 +320,6 @@ 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]
@@ -349,9 +337,8 @@ We can divide $\CC$ in $2d$ sectors of angle $\frac{\pi}{d}$ as in Figure~\ref{d
 %\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]{$\pi/d$};
+\foreach \x in {0, ..., 16}{
+	\draw (0,0) to (360/17*\x:3);}
 \end{tikzpicture}
 
 
@@ -402,8 +389,8 @@ 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 $M_d\cap\{x^d_0, y^d, x^d_n\in\RR\}$ 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
+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
 \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}
@@ -447,23 +434,6 @@ 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*}
@@ -476,7 +446,6 @@ 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}
 
@@ -514,17 +483,10 @@ 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
@@ -553,20 +515,10 @@ 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
 \[
@@ -592,7 +544,6 @@ 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}
@@ -892,7 +843,7 @@ 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 = it commit Dc.set_base_point(Dc.all_n_simplices(0)[0])
+dc_punto = Dc.set_base_point(Dc.all_n_simplices(0)[0])
 dc_punto.fundamental_group()
  -> Finitely presented group < e0 | e0^6 >
 dc_punto.cohomology()