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@ -33,14 +33,15 @@ decorations.markings, calc}
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\maketitle
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\begin{abstract}
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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.
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Due to a result by Androtti and Frankel \cite{andreotti19}, we can see that the complement o a complex projective curve has the homotopy type of a $2$-dimensional CW complex. However, no general method has been given to compute explicitly this complex. Here 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.
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The retraction is performed in several steps, using the branched cover structure of the Fermat curves to the degree $1$ case.
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\end{abstract}
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\section{Definition of the problem}
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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.
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Given $C$ a proyective curve in $\CC^2$, Libgober gave in \cite{libgober1986} a construction of a $2-$ dimensional CW-complex that is a deformation retract of its complement. However, it is not clear how to extend that construction to the projective case (that is a curve in $\CC\PP^2$).
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In particular, we will focus on the case of smooth curves.
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In this work, we give a different construction to produce a specific $\Delta$-complex that is a deformation retract of the complement of a Fermat curve (which is equivalent to the case of general smooth curves).
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\section{Smooth curves}
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Given a smooth curve of degree $d$ in $\PP^2$, we know it can be moved to a curve of the form
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\begin{equation}
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@ -53,7 +54,7 @@ We will denote:
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\begin{eqnarray}
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M_d&=&\{(x,y,z)\in\CC^3 | x^d+y^d+z^d=1\} \\
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S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR_{\geq0}\} \\
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K &=&\{[x:y:z]\in\CC\PP^2 \setminus M_2 | x^d, y^d, z^d\in\RR_{\geq0}\}
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K &=&\{[x:y:z]\in\CC\PP^2 \setminus M_2 | x^d, y^d, z^d\in\RR_{\geq0}\}
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\end{eqnarray}
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We then have a commutative diagram:
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@ -80,11 +81,11 @@ $M_1\cap\RR^3$ is a strong deformation retract of $M_1$.
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% \end{array}
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%\end{equation*}
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We are going to define a strong retraction $r_1:M_1\times I \longrightarrow M_1$ such that $r_1(M_1,0)=M_1$ and $r_1(M_1,1)=M_1\cap\RR^3$
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% and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
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% and $r_2:M_1\cap\RR^3\longrightarrow S_1$.
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In particular, we can take $r_1$ that takes the imaginary part to $0$:
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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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M_1\times{[0,1]}\rar["r_1"]&M_1\cap\RR^3\\
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M_1\times{[0,1]}\rar["r_1"]&M_1\cap\RR^3\\
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(P, t)\rar[mapsto]& t \Re P + (1 - t) \Im P.
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\end{tikzcd}
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\end{equation}
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@ -105,10 +106,10 @@ and therefore $r_1(x,y,z,t)\in M_1$ for all $t\in[0,1]$, and in particular, we g
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\end{lemma}
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\begin{proof}
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This map is a well-defined deformation retract since $M_1\cap\RR^3$ has real equations. Note that $M_1\cap\RR^3$
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$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is
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$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is
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\[
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S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
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\]
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\]
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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:
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@ -230,7 +231,7 @@ We want to produce the following diagram:
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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.
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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 $G(x,t)^d=\bar{r}(x^d,t)$.
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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 $G(x,t)^d=\bar{r}(x^d,t)$.
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@ -283,7 +284,7 @@ We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC
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\label{divisionC}
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\end{figure}
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Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$, which we will denote $f_k$. The lifting of $r_1$ will then be given by the $G$ previously described. This $G$ will act by taking the point $(x,y,z)\in M_d$ and mapping it to the upper or lower complex plane via the inverse homeomorphisms $f^{-1}_k$, then taking $r_1$ to take it to the real line, and finally bringing it back to its original sector with the homeomorphism $f_k$.
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Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$, which we will denote $f_k$. The lifting of $r_1$ will then be given by the $G$ previously described. This $G$ will act by taking the point $(x,y,z)\in M_d$ and mapping it to the upper or lower complex plane via the inverse homeomorphisms $f^{-1}_k$, then taking $r_1$ to take it to the real line, and finally bringing it back to its original sector with the homeomorphism $f_k$.
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@ -353,11 +354,11 @@ Lastly, for the case $(+,+,-)$ and it's permutations, we can write the lifting a
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All that is left to see is that the formulas for each case coincide on the intersections of the regions. For the intersection of $(+,+,-)$ and $(-,+,-)$, we have that:
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\begin{equation*}
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\sqrt[d]{(1-t+\frac{t}{1-z^d})y^d}=\sqrt[d]{(1-t+\frac{t}{y^d})y^d}=\sqrt[d]{y^d-ty^d+t}=\sqrt[d]{y^d(1-t)+t}
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\end{equation*}
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\end{equation*}
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The first root corresponds to $\tilde{r}_2$ in $(+,+,-)$ and the last one corresponds to the lifting in $(-,+,-)$, so we see that it coincides. It would be similar in the rest of intersections.
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We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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\end{proof}
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\begin{theorem}
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@ -365,9 +366,9 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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\end{theorem}
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\begin{proof}
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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 $r$.
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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.
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First, we see how $r_1$ acts on these representatives.
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\begin{equation}
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\begin{array}{lll}
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@ -375,7 +376,7 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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\tilde{r}_1(\lambda p,t) & = & [G(\lambda x,t): G(\lambda y,t): G(\lambda z,t)] \\
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\end{array}
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\end{equation}
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If we define $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$ for $x\in\CC$, we get that
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If we define $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$ for $x\in\CC$, we get that
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\begin{equation}
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\theta(\lambda x,t)=\arctan\frac{(1-t)\lambda x_i}{\lambda x_r}=\arctan\frac{(1-t)x_i}{x_r}=\theta(x,t)
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\end{equation}
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@ -386,10 +387,10 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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G(\lambda x,t) & = & +\sqrt[2d]{(\lambda^d x^d)_r^2+(1-t)^2(\lambda^d x^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))} \\
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\end{array}
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\end{equation}
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Since $\lambda$ is a $d-$th root of unity, we get that $G(x,t)=G(\lambda x,t)$. Therefore, we get that $\tilde{r}_1(p,t)=\tilde{r}_1(\lambda p,t)$. It's just left to check that $r_2$ also sends them to the same class.
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For a point $(x,y,z)\in M_d\cap\RR^3$, $\tilde{r}_2$ is defined by the region of $M_1\cap\RR^3$ where $(x^d, y^d, z^d)$ is, and therefore, when passing to the projective, it will depend on where the point $p^d=[x^d: y^d: z^d]$ lies. Since we have that $\lambda$ must be a $d-$th root of unity, we get that both points $p^d$ and $(\lambda p)^d=p$ will lie in the same region.
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Since $\lambda$ is a $d-$th root of unity, we get that $G(x,t)=G(\lambda x,t)$. Therefore, we get that $\tilde{r}_1(p,t)=\tilde{r}_1(\lambda p,t)$. It's just left to check that $r_2$ also sends them to the same class.
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For a point $(x,y,z)\in M_d\cap\RR^3$, $\tilde{r}_2$ is defined by the region of $M_1\cap\RR^3$ where $(x^d, y^d, z^d)$ is, and therefore, when passing to the projective, it will depend on where the point $p^d=[x^d: y^d: z^d]$ lies. Since we have that $\lambda$ must be a $d-$th root of unity, we get that both points $p^d$ and $(\lambda p)^d=p$ will lie in the same region.
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In the case that they lie in the regions $(-,-,+)$, $(-,+,-)$ or $(+,-,-)$, the retraction $\tilde{r}_2$ is, in the first case:
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\begin{equation}
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\tilde{r}_2([x\xi_1:y\xi_2:z\xi_3],t)=\left[\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\right]
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@ -398,9 +399,9 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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\begin{equation}
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\tilde{r}_2([\lambda x\xi_1:\lambda y\xi_2:\lambda z\xi_3],t)=\left[\sqrt[d]{(1-t)(\lambda x)^d}\xi_1:\sqrt[d]{(1-t)(\lambda y)^d}\lambda y\xi_2:\sqrt[d]{(1-t)(\lambda z)^d+t}\xi_3\right]
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\end{equation}
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The rest of these cases would be equivalent.
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Next, for the cases $(+,+,-)$, $(+,-,+)$ or $(-,+,+)$, the retraction has the form, in the first case:
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\begin{equation}
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\tilde{r}_2([x\xi_1,y\xi_2,z\xi_3],t)=\left[\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\right]
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@ -413,10 +414,10 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
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& = &\left[\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\right]
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\end{array}
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\end{equation}
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We can then see that both representatives will go to the same point by the retraction.
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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.
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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.
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\end{proof}
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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 $K$.
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@ -485,7 +486,7 @@ Let $\mu_d=\{\xi| \xi^d=1\}$. We can express $S_d$ as an union
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\begin{equation}
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\begin{array}{rll}
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S_d & = & \{(x,y,z)| x^d+y^d+z^d=1 \text{ and } x^d, y^d, z^d\in\RR_{\geq0}\} \\
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& = & \bigcup_{(\xi_1, \xi_2, \xi_3)\in\mu^3_d} X_{(\xi_1, \xi_2, \xi_3)} \\
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& = & \bigcup_{(\xi_1, \xi_2, \xi_3)\in\mu^3_d} X_{(\xi_1, \xi_2, \xi_3)} \\
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\end{array}
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\end{equation}
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@ -494,7 +495,7 @@ where $X_{(\xi_1, \xi_2, \xi_3)}=\{(x,y,z)| x^d+y^d+z^d=1 \text{ and } \xi_1x, \
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Then, we can see that the faces of the simplex are given by $\mathring{X}_{(\xi_1, \xi_2, \xi_3)}$, while the edges are given by $X_{(\xi_1, \xi_2, 1)}\cap\{z=0\}$, $X_{(\xi_1, 1, \xi_2)}\cap\{y=0\}$, $X_{(1, \xi_1, \xi_2)}\cap\{x=0\}$ and the vertices by $X_{(\xi, 1, 1)}\cap\{y=0,z=0\}$, $X_{(1, 1, \xi)}\cap\{x=0,y=0\}$, $X_{(1, \xi, 1)}\cap\{x=0, z=0\}$.
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We can remark that $X_{(\xi_1, \xi_2, \xi_3)}=(\xi_1, \xi_2, \xi_3)X_{(1,1,1)}$, and therefore that we have a group action of $\mu^3_d$ over $S_d$.
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Let
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Let
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\begin{equation}
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\begin{array}{rll}
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L^z_{(1,1,1)} & = & X_{(1, 1, 1)}\cap\{z=0\}\\
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@ -535,7 +536,7 @@ For the rest of the simplices, due to the group action, the face maps are given
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\begin{array}{lll}
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\delta^2_i(X_{(\xi_1, \xi_2, \xi_3)}) & = & (\xi_1, \xi_2, \xi_3)\delta^2_i(X_{(1,1,1)})\\
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\delta^1_i(L^\bullet_{(\xi_1, \xi_2, \xi_3)}) & = & (\xi_1, \xi_2, \xi_3)\delta^1_i(L^\bullet_{(1,1,1)})\\
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\end{array}
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\end{equation}
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@ -566,10 +567,10 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^z_{[1:1:1]}> \
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\begin{example}{}
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We are going to compute now the simplicial complex for $S_d$ and $K$ when $d=2$, this is, our curve will be $C_D=\{[x:y:z]\in\PP^2|x^2+y^2+z^2=0\}$, $S_2=\{(x,y,z)\in M_2| x^2, y^2, z^2\in\RR_{\geq 0}\}$ and $K=\{[x:y:z]\in\PP^2|x^2,y^2,z^2\in\RR_{\geq0}\}$.
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We start with $S_2$. First, we can observe that, from all points in $\CC$, the only ones whose square power is real are either the real numbers of the pure imaginary numbers. And from these, the later have negative square power. Therefore, we can conclude that $S_2=\{(x,y,z)\in\RR^3| x^2+y^2+z^2=1\}$, that is, $S_2$ is the real three dimensional sphere. We now see the simplicial complex $\Delta^2$ that can be associated to it such that $\phi(\Delta^2)=\Delta$.
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We start with $S_2$. First, we can observe that, from all points in $\CC$, the only ones whose square power is real are either the real numbers of the pure imaginary numbers. And from these, the later have negative square power. Therefore, we can conclude that $S_2=\{(x,y,z)\in\RR^3| x^2+y^2+z^2=1\}$, that is, $S_2$ is the real three dimensional sphere. We now see the simplicial complex $\Delta^2$ that can be associated to it such that $\phi(\Delta^2)=\Delta$.
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In this case, we have that $\mu_2=\{\xi|\xi^2=1\}=\{1,-1\}$. Then, we have that $X_{(1,1,1)}=\{(x,y,z)\in\RR^3_{\geq0} |x^2+y^2+z^2=1\}$, this is, $X_{(1,1,1)}$ is the set of points given by the positive square roots of the points in $S_1$. And the rest of $2-$simplices will be given by the action of $\mu^3_2$ on $X_{(1,1,1)}$, which will give us the $2-$simplices $X_{(1,1,-1)}$, $X_{(1,-1,1)}$, $X_{(-1,1,1)}$, $X_{(1,-1,-1)}$, $X_{(-1,1,-1)}$, $X_{(-1,-1,1)}$ and $X_{(-1,-1,-1)}$.
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The face maps of $X_{(1,1,1)}$ are:
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\begin{equation}
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\begin{array}{lllll}
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@ -577,12 +578,12 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d]<V^z_{[1:1:1]}> \
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\delta^2_1(X_{(1,1,1)}) & = & L^y_{(1,1,1)} & = & \{(\sqrt{1-t},0, \sqrt{t})|t\in[0,1]\}\\
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\delta^2_2(X_{(1,1,1)}) & = & L^z_{(1,1,1)} & = & \{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
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\end{array}
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\end{equation}
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\end{equation}
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By applying the group action as shown before we get the faces of the rest of $2-$simplices.
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For the $1-$simplices, the group acting would be $\mu^3_2/\mu_2$, since one of the coordinates is $0$. Therefore, the $1-$simplices will be $L^x_{(1,1,1)}$, $L^x_{(1,1,-1)}$, $L^x_{(1,-1,1)}$, $L^x_{(1,-1,-1)}$, $L^y_{(1,1,1)}$, $L^y_{(1,1,-1)}$, $L^y_{(-1,1,1)}$, $L^y_{(-1,1,-1)}$, $L^z_{(1,1,1)}$, $L^z_{(1,-1,1)}$, $L^z_{(-1,1,1)}$ and $L^z_{(-1,-1,1)}$.
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The face maps of $L^z_{(1,1,1)}$, $L^y_{(1,1,1)}$ and $L^x_{(1,1,1)}$ are:
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\begin{equation}
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\begin{array}{lllll}
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@ -604,7 +605,7 @@ Lastly, the $0-$simplices will be $V^x_{(1,1,1)}=(1,0,0)$, $V^x_{(-1,1,1)}=(-1,0
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%\phi^{-1}((0,0,1))&=&\{(0,0,1),(0,0,-1)\}
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%\end{array}
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%\end{equation}
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%The edges of the simplex, $[v_0v_1]$, $[v_0v_2]$ and $[v_1v_2]$, can be identified to the points of $S_1$ given respectively by $\{(1-t, 0, t)|t\in[0,1]\}$, $\{(1-t,t,0)|t\in[0,1]\}$ and $\{(0,t,1-t)|t\in[0,1]\}$. Therefore, each edge will have four preimages. In the first case, these preimages are:
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%\begin{equation}
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%\phi^{-1}((1-t,t,0))=\left\{\begin{array}{l}
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@ -613,17 +614,17 @@ Lastly, the $0-$simplices will be $V^x_{(1,1,1)}=(1,0,0)$, $V^x_{(-1,1,1)}=(-1,0
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% \{(-\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\
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% \{(-\sqrt{1-t},-\sqrt{t},0)|t\in[0,1]\}
|
||||
%\end{array}\right.
|
||||
%\end{equation}
|
||||
%For the rest of the edges the preimages by $\phi$ would be analougous. We will denote the edges as a triple of $\{+,-,0\}$ indicating in each coordinate the sign of the root taken.
|
||||
|
||||
%\end{equation}
|
||||
%For the rest of the edges the preimages by $\phi$ would be analougous. We will denote the edges as a triple of $\{+,-,0\}$ indicating in each coordinate the sign of the root taken.
|
||||
|
||||
%Finally, we will have $8$ $2$-faces. For three edges to form a face on the simplex, we need the corresponding vertices of the edges to coincide, that is, the edges with non-zero first coordinate must both take the same sign on said coordinate, and equivalently for the rest of coordinates. Then, the faces are defined by taking the square root of the coordinates of the points in $S_1$ with sign equal to those in the edges.
|
||||
|
||||
|
||||
\begin{figure}[!hp]
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.7]
|
||||
\draw[gray, thin] (-5,0) to (5,0);
|
||||
\draw[gray, thin] (0,-5) to (0,5);
|
||||
|
||||
|
||||
\draw[gray, thin] (-4,-4) to (4,4);
|
||||
\draw[fill, red] (4,0) circle (2pt);
|
||||
\draw[fill, red] (-4,0) circle (2pt);
|
||||
@ -631,42 +632,42 @@ Lastly, the $0-$simplices will be $V^x_{(1,1,1)}=(1,0,0)$, $V^x_{(-1,1,1)}=(-1,0
|
||||
\draw[fill, green] (0,-4) circle (2pt);
|
||||
\draw[fill, blue] (1.41,1.41) circle (2pt);
|
||||
\draw[fill, blue] (-1.41,-1.41) circle (2pt);
|
||||
|
||||
|
||||
\draw (0,0) circle (4cm);
|
||||
|
||||
|
||||
\draw (-4,0) arc (180:360:4 and 1.5);
|
||||
\draw[dashed] (4,0) arc (0:180:4 and 1.5);
|
||||
\draw[dashed] (0,4) arc (90:270:-1.5 and 4);
|
||||
\draw (0,-4) arc (270:90:1.5 and 4);
|
||||
|
||||
|
||||
\draw[pink, thick](-4,0.02) arc (180:250:4 and 1.5);
|
||||
\draw[pink, thick](-4.02,0.02) arc (180:90:4);
|
||||
\draw[pink, thick](0,4) arc (270:90:1.5 and -4);
|
||||
|
||||
|
||||
\draw[purple, thick](-4,-0.02) arc (180:250:4 and 1.5);
|
||||
\draw[purple, thick](-4,-0.02) arc (180: 270:4);
|
||||
\draw[purple, thick] (-1.43,-1.41) arc (200:270:1.5 and 3.9);
|
||||
|
||||
|
||||
\draw[yellow, thick, dashed](-3.98,0.02) arc (180:70:4 and 1.5);
|
||||
\draw[yellow, thick, dashed](-3.98,0.02) arc (180: 90:4);
|
||||
\draw[yellow, thick, dashed] (1.39,1.41) arc (200:270:-1.5 and -3.9);
|
||||
|
||||
|
||||
\draw[orange, thick, dashed](-3.98,-0.02) arc (180:70:4 and 1.5);
|
||||
\draw[orange, thick, dashed](-3.98,-0.02) arc (180: 270:4);
|
||||
\draw[orange, thick, dashed] (1.39,1.41) arc (160:270:-1.5 and 4);
|
||||
|
||||
|
||||
\draw[teal, thick, dashed](4.02,0.02) arc (180:110:-4 and 1.5);
|
||||
\draw[teal, thick, dashed](4.02,0.02) arc (180: 270:-4);
|
||||
\draw[teal, thick, dashed] (1.42,1.41) arc (200:270:-1.5 and -3.9);
|
||||
|
||||
|
||||
\draw[violet, thick](4.02,-0.02) arc (0:110:4 and -1.5);
|
||||
\draw[violet, thick](4.02,-0.02) arc (180: 90:-4);
|
||||
\draw[violet, thick] (-1.39,-1.42) arc (200:270:1.5 and 3.9);
|
||||
|
||||
|
||||
\draw[magenta, thick, dashed](3.98,0.02) arc (180:110:-4 and 1.5);
|
||||
\draw[magenta, thick, dashed](3.98,-0.02) arc (180: 90:-4);
|
||||
\draw[magenta, thick, dashed] (1.43,1.41) arc (160:270:-1.5 and 4);
|
||||
|
||||
|
||||
\draw[lime, thick, dashed](4.02,0.02) arc (0:110:4 and -1.5);
|
||||
\draw[lime, thick, dashed](3.98,0.02) arc (180: 270:-4);
|
||||
\draw[lime, thick, dashed] (0.02,3.98) arc (270:160:1.5 and -4);
|
||||
@ -685,4 +686,26 @@ Lastly, we will have $4$ $2-$simplices, $X_{[1:1:1]}=X_{[-1:-1:-1]}$, $X_{[1:1:-
|
||||
|
||||
The face maps will be obtained by passing to the projective the face maps of $\Delta^d$.
|
||||
\end{example}
|
||||
|
||||
\begin{thebibliography}{99}
|
||||
|
||||
\bibitem{andreotti1959}
|
||||
A. Andreotti and T. Frankel,
|
||||
\textit{The Lefschetz theorem on hyperplane sections},
|
||||
Annals of Mathematics,
|
||||
Vol. 69, No. 3,
|
||||
pp. 713--717,
|
||||
1959.
|
||||
|
||||
\bibitem{libgober1986}
|
||||
A. Libgober,
|
||||
\textit{On the homotopy type of the complement to plane algebraic curves},
|
||||
J. Reine Angew. Math.,
|
||||
Vol. 367,
|
||||
pp. 103--114,
|
||||
1986.
|
||||
DOI: 10.1515/crll.1986.367.103
|
||||
|
||||
\end{thebibliography}
|
||||
|
||||
\end{document}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user