Añadir abstract y retoques.

generaldegree.tex

@@ -33,22 +33,27 @@ decorations.markings, calc}
 \maketitle
 
 \begin{abstract}
+We give a explicit construction of a $2$- dimensional $\Delta$-complex that is a strong deformation retract of the complement of a Fermat curve of degree $d$ in the complex projective space.
 
+The retraction is performed in several steps, using the branched cover structure of the Fermat curves to the degree $1$ case.
 \end{abstract}
 \section{Definition of the problem}
-Given $C$ a proyective curve in $\PP^2$ over $\CC$, we know that the complement of the curve has the homotopy type of a CW-complex of real dimension $2$, our problem is how to find an effective model for said CW-complex.
+Given $C$ a proyective curve in $\CC\PP^2$, we know that the complement of the curve has the homotopy type of a CW-complex of real dimension $2$, our problem is how to find an effective model for said CW-complex.
+
+In particular, we will focus on the case of smooth curves.
 \section{Smooth curves}
-Given a smooth curve of degree $d$ in $\PP^2$, we know it is isomorphic to a curve of the form
+Given a smooth curve of degree $d$ in $\PP^2$, we know it can be moved to a curve of the form
 \begin{equation}
-	C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\}
+	C_d=\{[x:y:z]\in\CC\PP^2 | x^d+y^d+z^d=0\}
 \end{equation}
+by an ambient isotopy.
 Therefore, we will work with $C_d$, from now on.
 
 We will denote:
 \begin{eqnarray}
-	M_d&=&\{(x,y,z)\in\CC | x^d+y^d+z^d=1\} \\
+	M_d&=&\{(x,y,z)\in\CC^3 | x^d+y^d+z^d=1\} \\
 	S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
-	K &=&\{[x:y:z]\in\PP^2 | x^d, y^d, z^d\in\RR_{\geq0}\} 
+	K &=&\{[x:y:z]\in\CC\PP^2 \setminus M_2 | x^d, y^d, z^d\in\RR_{\geq0}\} 
 \end{eqnarray}
 
 We then have a commutative diagram:
@@ -59,6 +64,7 @@ We then have a commutative diagram:
 		K & \longhookrightarrow & \PP^2\setminus C_d \\
 	\end{array}
 \end{equation*}
+where the vertical maps consist on taking equivalence classes.
 
 We are going to see that the inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. To do this we first need the following lemma:
 \begin{lemma}
@@ -82,6 +88,7 @@ M_1\times{[0,1]}\rar["r_1"]&M_1\cap\RR^3\\
 (P, t)\rar[mapsto]& t \Re P + (1 - t) \Im P.
 \end{tikzcd}
 \end{equation}
+where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively.
 
 Note that for $(x,y,z)\in M_1$, we already have that
 \begin{equation}
@@ -90,19 +97,19 @@ Note that for $(x,y,z)\in M_1$, we already have that
 	x_i+y_i+z_i & = & 0 \\
 \end{array}
 \end{equation}
-and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x,y,z,1)=(x_r,y_r,z_r)\in M_1\cap\RR^3$ for all $(x,y,z)\in M_1$. $r_1$ is then a strong retraction of $M_1$ to $M_1\cap\RR^3$.
+and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x,y,z,1)=(x_r,y_r,z_r)\in M_1\cap\RR^3$ for all $(x,y,z)\in M_1$. $r_1$ is then a strong retraction from $M_1$ to $M_1\cap\RR^3$.
 \end{proof}
 
 \begin{lemma}
-	$M_1\cap\RR^3$ is a strong deformation retract of $S_1$.
+	$S_1$ is a strong deformation retract of $M_1\cap\RR^3$.
 \end{lemma}
 \begin{proof}
-	This map is a well-defined deformation retract since $M_1$ has real equations. Note that $M_1\cap\RR^3$
+	This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations. Note that $M_1\cap\RR^3$
 $M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is 
 \[
 S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
 \] 
-i.e, $S_1$ is the triangular section defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described in Figure~\ref{fig:triangulo}. We will retract each of them separately:
+i.e, $S_1$ is the triangular section defined by the intersection of this plane with the first octant, delimited by the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described in Figure~\ref{fig:triangulo}. We will retract each of them separately:
 
 
 \begin{figure}[h]
@@ -166,7 +173,7 @@ The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we defin
 		(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
 	\end{array}
 \end{equation}
-Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
+Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ respectively. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
 \begin{equation}
 P_{x,y,z}=\left\{\begin{array}{lr}
 	(x,y,z)&(+,+,+) \\
@@ -212,14 +219,17 @@ Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we w
 	$S_d$ is a strong deformation retract of $M_d$.
 \end{theorem}
 \begin{proof}
-We have the following diagram:
+We want to produce the following diagram:
 \begin{equation*}
 	\begin{array}{ccccc}
-		M_d & \longrightarrow & M_d\cap\{x^d, y^d, z^d\in\RR\}&\longrightarrow& S_d \\
+		M_d & \rightleftharpoons & M_d\cap\{x^d, y^d, z^d\in\RR\}&\rightleftharpoons& S_d \\
 		\downarrow & & \downarrow & & \downarrow\\
-		M_1 & \longrightarrow & M_1\cap\RR^3 & \longrightarrow & S_1 \\
+		M_1 & \rightleftharpoons & M_1\cap\RR^3 & \rightleftharpoons & S_1 \\
 	\end{array}
 \end{equation*}
+where the horizontal arrows represent retractions, and the vertical arrows are the maps $(x,y,z)\mapsto (x^d,y^d,z^d)$. That is, we want to lift the retractions built for $M_1$ and $S_1$ to $M_d$ and $S_d$ via this branched cover.
+
+
 We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, we can write it as $r_1((x,y,z),t)=(\bar{r}(x,t),\bar{r}(y,t),\bar{r}(z,t))$ with $\bar{r}(x,t)+\bar{r}(y,t)+\bar{r}(z,t)=1$ whenever $(x,y,z)\in M_1$. Then, it suffices to find $G(x,t)$ such that $G(x,t)^d=\bar{r}(x^d,t)$. 
 
 
@@ -322,7 +332,7 @@ We can the write the lifting of $r_1$ as $\tilde{r}_1((x,y,z),t)=(G(x,t), G(y,t)
 
 
 Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d, y^d, z^d\in\RR\}$ into $S_d$.
-For a point $(x\xi_1,y\xi_2,z\xi_3)\in M_d\cap\{x^d, y^d, z^d\in\RR\}$, the corresponding point in $M_1\cap\RR^3$ is $(x^ d,y^d,z^d)$ and therefore the goal points of the retraction $r_2$ will be
+For a point $(x\xi_1,y\xi_2,z\xi_3)\in M_d\cap\{x^d, y^d, z^d\in\RR\}$ (where $(x,y,z)$ are real and $\xi_i$ are $d$-th roots of unity), the corresponding point in $M_1\cap\RR^3$ is $(x^ d,y^d,z^d)$ and therefore the goal points of the retraction $r_2$ will be
 \begin{equation}
 P_{x,y,z}=\left\{\begin{array}{lr}
 	(x^d,y^d,z^d)&(+,+,+) \\
@@ -336,7 +346,7 @@ P_{x,y,z}=\left\{\begin{array}{lr}
 \end{equation}
 In the $(+,+,+)$ case, $r_2$ is the constant path, and therefore the lift $\tilde{r}_2$ will also be the constant path.
 
-In the $(-,-,+)$ case, $r_2((x^d,y^d,z^d), t)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x\xi_1,y\xi_2,z\xi_3)$, for $\xi_1$, $\xi_2$ and $\xi_3$ some $d-$th roots of unity. Since $x,y$ are negative real numbers, the retraction will push them to $0$, while we will push $z\xi_3$ to $\xi_3$. We can write this as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3), t)=(\sqrt[d]{(1-t)x^d}\xi_1,\sqrt[d]{(1-t)y^d}\xi_2,\sqrt[d]{(1-t)z^d+t}\xi_3)$. For the cases $(+,-,-)$ and $(-,+,-)$ it is equivalent.
+In the $(-,-,+)$ case, $r_2((x^d,y^d,z^d), t)=(1-t)(x^d,y^d,z^d)+(0,0,t)$. Then, $\tilde{r}_2$ will start at $(x\xi_1,y\xi_2,z\xi_3)$, for $\xi_1$, $\xi_2$ and $\xi_3$ some $d-$th roots of unity. Since $x,y$ are negative real numbers, the retraction will push them to $0$, while we will push $z\xi_3$ to $\xi_3$. We can write this as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3), t)=(\sqrt[d]{(1-t)x^d}\xi_1,\sqrt[d]{(1-t)y^d}\xi_2,\sqrt[d]{(1-t)z^d+t}\xi_3)$, where the $d$'th roots take values in the corresponding sector of each coordinate. For the cases $(+,-,-)$ and $(-,+,-)$ it is equivalent.
 
 Lastly, for the case $(+,+,-)$ and it's permutations, we can write the lifting as $\tilde{r}_2((x\xi_1,y\xi_2,z\xi_3),t)=(\sqrt[d]{(1-t+\frac{t}{1-z^d})x^d}\xi_1,\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}\xi_2,\sqrt[d]{(1-t)z^d}\xi_3)$. We can see by summing the coordinates to the power $d$ that they sum $1$, so the retraction doesn't leave $M_1$.
 
@@ -351,12 +361,12 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
 \end{proof}
 
 \begin{theorem}
- The strong retraction from $M_d$ to $S_d$ induces a strong retraction, and therefore an homotopy equivalence, between $\PP^2\setminus C_d$ and $K$.
+ The strong retraction from $M_d$ to $S_d$ induces a strong retraction, and therefore an homotopy equivalence, between $\CC\PP^2\setminus C_d$ and $K$.
 \end{theorem}
 \begin{proof}
-	The class representatives we choose in $\PP^2\setminus C_d$ are such that the coordinates to the $d$ power sum $1$. Since this choice is not unique, we are now going to see that if we apply the retraction $r=r_2\circ r_1$ to two different representatives, they are sent to the same class by $r$.
+	The path followed by each point in $\CC\PP^2\setminus C_d$ . in the retraction will be given by the path followed by a representative in $M_d$ as it is retracted to $S_d$. The class representatives we choose in $\CC\PP^2\setminus C_d$ will be such that the coordinates to the $d$ power add up to $1$. Since this choice is not unique, we are now going to see that if we apply the retraction $r=r_2\circ r_1$ to two such representatives, they are sent to the same class by $r$.
 	
-	Let $p=[x:y:z]\in\PP^2\setminus C_d$ and $\lambda p=[\lambda x:\lambda y: \lambda z]$ be two representatives of some point $p$, since both representatives must be such that $x^d+y^d+z^d=1$, necessarily $\lambda$ is a $d-$th root of unity.
+	Let $p=[x:y:z]\in\CC\PP^2\setminus C_d$ and $\lambda p=[\lambda x:\lambda y: \lambda z]$ be two representatives of some point $p$, since both representatives must be such that $x^d+y^d+z^d=1$, necessarily $\lambda$ is a $d-$th root of unity.
 	
 	First, we see how $r_1$ acts on these representatives.
 	\begin{equation}
@@ -409,7 +419,7 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
 	Lastly, if $[x^d: y^d: z^d]$ is in $(+,+,+)$, the retraction is the identity, and therefore both representatives will stay the same and therefore go to the same class. 
 \end{proof}
 
-We have now seen that there is a homotopy equivalence between $\PP^2\setminus C_d$ and $K$. Therefore, to get a CW-complex structure on $\PP^2\setminus C_d$, it suffices to get it for $K$.
+We have now seen that there is a homotopy equivalence between $\CC\PP^2\setminus C_d$ and $K$. Therefore, to get a CW-complex structure on $\CC\PP^2\setminus C_d$, it suffices to get it for $K$.
 
 
 We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertices, $2$ edges and $1$ face, then, we can identify $S_1$ with $\Delta$, and we get a diagram:
@@ -439,7 +449,7 @@ First, we give a detailed description of $\Delta$. We will denote:
 	v_2 & = & (0,0,1) & & \\
 	l_0 & = & [v_1,v2] & = & \{(1-t)v_1+tv_2|t\in[0,1]\}\\
 	l_1 & = & [v_0,v2] & = & \{(1-t)v_0+tv_2|t\in[0,1]\}\\
-	l_0 & = & [v_0,v1] & = & \{(1-t)v_0+tv_1|t\in[0,1]\}\\
+	l_2 & = & [v_0,v1] & = & \{(1-t)v_0+tv_1|t\in[0,1]\}\\
 	X & = & [v_0,v_1,v_2]
 \end{array}
 \end{equation}