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generaldegreeHypersurfaces.tex

@@ -169,7 +169,7 @@ Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
 \begin{equation*}
 \begin{array}{lllll}
 S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
-&=&\{x_0e_0+\dots+x_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
+&=&\{x_0e_0+\dots+x_ne_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
 \end{array}
 \qedhere
 \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{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}
@@ -380,8 +380,8 @@ 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 $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
 
 
-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}
@@ -848,7 +848,7 @@ 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()