cambio en un dibujo

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Alba Larraya Sancho 2026-09-24 15:42:04 +02:00
parent 4de6ddea39
commit e0c4a4604e

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@ -169,7 +169,7 @@ Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
\begin{equation*} \begin{equation*}
\begin{array}{lllll} \begin{array}{lllll}
S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\ S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
&=&\{x_0e_0+\dots+x_ne_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\ &=&\{x_0e_0+\dots+x_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
\end{array} \end{array}
\qedhere \qedhere
\end{equation*} \end{equation*}
@ -227,13 +227,13 @@ S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
\end{center} \end{center}
\end{figure} \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 $(+,\dots,+)$. 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 region $(-,\dots,-)$. 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: 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{equation}
\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}] \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, (Q, t)\rar[mapsto]& t P_Q + (1 - t) Q,
\end{tikzcd} \end{tikzcd}
\end{equation} \end{equation}
@ -315,20 +315,15 @@ We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC
\begin{figure}[h] \begin{figure}[h]
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.6] \begin{tikzpicture}[scale=0.6]
\draw (-5,0) to (5,0); \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 (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 (-4.7,5) to (4.7,-5);
\draw (-5,3.2) to (5, -3.2); \draw[->] (3,0) to[out=90, in=350] (1,2) node[ xshift=1.0cm, yshift=-0.2cm]{$\pi/d$};
\draw (-5,1.5) to (5,-1.5);
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
\caption{Division of the complex plane} \caption{Division of the complex plane when $d=3$}
\label{divisionC} \label{divisionC}
\end{figure} \end{figure}
@ -380,8 +375,8 @@ Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since:
We can the write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$. We can the write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
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 $S_d$. 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 $P=(x_0\xi_0,\dots,x_n\xi_n)\in M_d\cap\{x^d_0, \ldots , x^d_n\in\RR\}$ (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 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} \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) 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} \end{equation}
@ -848,7 +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. Once we have gotten the $\Delta$-complex, we can compute the fundamental group or the cohomology of the space, for example.
\begin{python} \begin{python}
Dc = SimplicialSet(Delta_Complex) 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() dc_punto.fundamental_group()
-> Finitely presented group < e0 | e0^6 > -> Finitely presented group < e0 | e0^6 >
dc_punto.cohomology() dc_punto.cohomology()