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@ -169,7 +169,7 @@ Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
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\begin{equation*}
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\begin{equation*}
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\begin{array}{lllll}
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\begin{array}{lllll}
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S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
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S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
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&=&\{x_0e_0+\dots+x_ne_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
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&=&\{x_0e_0+\dots+x_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
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\end{array}
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\end{array}
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\qedhere
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\qedhere
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\end{equation*}
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\end{equation*}
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@ -227,13 +227,13 @@ S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
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\end{center}
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\end{center}
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\end{figure}
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\end{figure}
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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 $(+,\dots,+)$.
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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 $(+,\dots,+)$.
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Note that there is no region $(-,\dots,-)$.
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Note that there is no section $(-,\dots,-)$.
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The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the rest of the parts, we define a retraction:
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The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the rest of the parts, we define a retraction:
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\begin{equation}
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\begin{equation}
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\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
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\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
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(M_1\cap\RR^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\\
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(M_1\cap\RR^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\times [0,1]\\
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(Q, t)\rar[mapsto]& t P_Q + (1 - t) Q,
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(Q, t)\rar[mapsto]& t P_Q + (1 - t) Q,
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\end{tikzcd}
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\end{tikzcd}
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\end{equation}
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\end{equation}
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@ -315,20 +315,15 @@ We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC
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\begin{figure}[h]
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\begin{figure}[h]
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\begin{center}
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\begin{center}
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\begin{tikzpicture}[scale=0.6]
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\begin{tikzpicture}[scale=0.6]
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\draw (-5,0) to (5,0);
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\draw (-5,0) to (5,0);
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\draw (5,1.5) to (-5,-1.5);
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\draw (5,3.2) to (-5, -3.2);
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\draw (4.7,5) to (-4.7,-5);
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\draw (2.8,5) to (-2.8,-5);
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\draw (2.8,5) to (-2.8,-5);
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\draw (0.9,5) to (-0.9,-5);
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\draw (-0.9,5) to (0.9,-5);
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\draw (-2.8,5) to (2.8,-5);
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\draw (-4.7,5) to (4.7,-5);
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\draw (-4.7,5) to (4.7,-5);
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\draw (-5,3.2) to (5, -3.2);
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\draw[->] (3,0) to[out=90, in=350] (1,2) node[ xshift=1.0cm, yshift=-0.2cm]{$\pi/d$};
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\draw (-5,1.5) to (5,-1.5);
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\end{tikzpicture}
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\end{tikzpicture}
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\end{center}
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\end{center}
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\caption{Division of the complex plane}
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\caption{Division of the complex plane when $d=3$}
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\label{divisionC}
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\label{divisionC}
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\end{figure}
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\end{figure}
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@ -380,8 +375,8 @@ Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since:
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We can the write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
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We can the write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
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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 $S_d$.
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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 $S_d$.
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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
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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
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\begin{equation}
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\begin{equation}
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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)
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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)
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\end{equation}
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\end{equation}
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@ -848,7 +843,7 @@ Delta_Complex = DeltaComplex_hypersurface(4,6)
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Once we have gotten the $\Delta$-complex, we can compute the fundamental group or the cohomology of the space, for example.
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Once we have gotten the $\Delta$-complex, we can compute the fundamental group or the cohomology of the space, for example.
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\begin{python}
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\begin{python}
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Dc = SimplicialSet(Delta_Complex)
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Dc = SimplicialSet(Delta_Complex)
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dc_punto = Dc.set_base_point(Dc.all_n_simplices(0)[0])
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dc_punto = it commit Dc.set_base_point(Dc.all_n_simplices(0)[0])
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dc_punto.fundamental_group()
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dc_punto.fundamental_group()
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-> Finitely presented group < e0 | e0^6 >
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-> Finitely presented group < e0 | e0^6 >
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dc_punto.cohomology()
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dc_punto.cohomology()
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