From d16551e9067805c08bf21ec6771707ec841b21b9 Mon Sep 17 00:00:00 2001 From: Alba Larraya Sancho Date: Fri, 15 May 2026 13:02:50 +0200 Subject: [PATCH] ejemplo d=2 --- generaldegree.tex | 49 ++++++++++++++++++++++++++++++++++++++++------- 1 file changed, 42 insertions(+), 7 deletions(-) diff --git a/generaldegree.tex b/generaldegree.tex index 2da864e..e612788 100644 --- a/generaldegree.tex +++ b/generaldegree.tex @@ -235,7 +235,7 @@ We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, In order to see this, we can write an explicit expression of $\bar{r}$. For a point $x\in\CC$ the action of $\bar{r}$ is the projection over the real axis, as seen in figure \ref{realprojection}. \begin{figure}[h] \begin{center} -\begin{tikzpicture}[scale=0.7] +\begin{tikzpicture}[scale=0.6] \draw (-5,0) to (5,0); \draw (0,5) to (0,-1); \draw[fill] (3,2) circle (2pt) node[above]{$x_r+ix_i$}; @@ -254,7 +254,7 @@ We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC \begin{figure}[h] \begin{center} -\begin{tikzpicture}[scale=0.7] +\begin{tikzpicture}[scale=0.6] \draw (-5,0) to (5,0); \draw (5,1.5) to (-5,-1.5); \draw (5,3.2) to (-5, -3.2); @@ -542,7 +542,7 @@ We have then defined $\Delta^d$, the simplicial complex with which we can identi \end{array} \end{equation*} -Where the map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x,y,z)\longrightarrow[x:y:z]$. Since we are passing to the projective plane, two points $(x\xi_1, y\xi_2, z\xi_3)$ and $(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)$ will be identified, and therefore, so will $X_{(x\xi_1, y\xi_2, z\xi_3)}$ and $X_{(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)}$. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices. +Where the map $\Delta^d\longrightarrow\Delta^d_\text{proj}$ is given by $(x,y,z)\longrightarrow[x:y:z]$. Since we are passing to the projective plane, two points $(x\xi_1, y\xi_2, z\xi_3)$ and $(x\xi_1\xi, y\xi_2\xi, z\xi_3\xi)$ will be identified, and therefore, so will $X_{(\xi_1, \xi_2, \xi_3)}$ and $X_{(\xi_1\xi, \xi_2\xi, \xi_3\xi)}$. As for the points where one coordinate is $0$, say, for example the first one, since the group action is then by $\mu^3_2/\mu_2$, we write the elements of $\mu^3_2/\mu_2$ as $(1,\xi_1,\xi_2)$, where the $1$ is on the coordinate that is $0$. Then, passing to the projective space, we will say that $L^x_{[1:\xi_1,\xi_2]}$ and $L^x_{[1:\xi'_1,\xi'_2]}$ are the same if $[1:\xi'_1,\xi'_2]=[1:\xi_1\xi,\xi_2\xi]$ for some $\xi\in\mu_2$. The rest of $1-$simplices would be identified similarly. Lastly, for the $0-$simplices, since we have $2$ zero coordinates, we have that all $0-$simplices with the same nonzero coordinate will be identified to each other. As a result, $\Delta^d_\text{proj}$ will only have $3$ $0-$simplices, $3d$ $1-$simplices and $d^2$ $2-$simplices. \begin{equation} \begin{array}{rll} @@ -557,8 +557,34 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d] \ 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 $\phi(\Delta^2)=\Delta$. - 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 $S_1$. - + 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 $X_{(-1,-1,-1)}$. + + The face maps of $X_{(1,1,1)}$ are: + \begin{equation} + \begin{array}{lllll} + \delta^2_0(X_{(1,1,1)}) & = & L^x_{(1,1,1)} & = & \{(0,\sqrt{1-t},\sqrt{t})|t\in[0,1]\}\\ + \delta^2_1(X_{(1,1,1)}) & = & L^y_{(1,1,1)} & = & \{(\sqrt{1-t},0, \sqrt{t})|t\in[0,1]\}\\ + \delta^2_2(X_{(1,1,1)}) & = & L^z_{(1,1,1)} & = & \{(\sqrt{1-t},\sqrt{t},0)|t\in[0,1]\}\\ + \end{array} + \end{equation} + + By applying the group action as shown before we get the faces of the rest of $2-$simplices. + + 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 $L^z_{(-1,-1,1)}$. + +The face maps of $L^z_{(1,1,1)}$, $L^y_{(1,1,1)}$ and $L^x_{(1,1,1)}$ are: +\begin{equation} +\begin{array}{lllll} + \delta^1_0(L^z_{(1,1,1)}) & = & V^y_{(1,1,1)} & = & (1,0,0)\\ + \delta^1_1(L^z_{(1,1,1)}) & = & V^z_{(1,1,1)} & = & (0,1,0)\\ + \delta^1_0(L^y_{(1,1,1)}) & = & V^x_{(1,1,1)} & = & (1,0,0)\\ + \delta^1_1(L^y_{(1,1,1)}) & = & V^z_{(1,1,1)} & = & (0,0,1)\\ + \delta^1_0(L^x_{(1,1,1)}) & = & V^x_{(1,1,1)} & = & (0,1,0)\\ + \delta^1_1(L^x_{(1,1,1)}) & = & V^y_{(1,1,1)} & = & (0,0,1)\\ +\end{array} +\end{equation} + +Lastly, the $0-$simplices will be $V^x_{(1,1,1)}=(1,0,0)$, $V^x_{(-1,1,1)}=(-1,0,0)$, $V^y_{(1,1,1)}=(0,1,0)$, $V^y_{(1,-1,1)}=(0,-1,0)$, $V^z_{(1,1,1)}=(0,0,1)$ and $V^z_{(1,1,-1)}=(0,0,-1)$. In Figure \ref{fig:S2} we can see the $\Delta-$complex that we associate to $S_2$. %The vertices of $\Delta$, $v_0$, $v_1$ and $v_2$, can be identified to $(1,0,0)$, $(0,0,1)$ and $(0,1,0)$ respectively. The preimages of these points are: %\begin{equation} %\begin{array}{rll} @@ -581,7 +607,7 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d] \ %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}[h] + \begin{figure}[!hp] \begin{center} \begin{tikzpicture}[scale=0.7] \draw[gray, thin] (-5,0) to (5,0); @@ -636,7 +662,16 @@ C_0(\Delta^d_\text{proj}) & = & \ZZ[\mu^3_d/\mu_d]/\ZZ[\mu^2_d] \ \end{tikzpicture} \end{center} \caption{Division in simplices of $S_2$} - \label{S2} + \label{fig:S2} \end{figure} + + +Now, we give a description of $\Delta^2_\text{proj}$. First, for the $0-$simplices, we have that $V^x_{[1,1,1]}=V^x_{[-1,1,1]}=[1:0:0]$, $V^y_{[1,1,1]}=V^y_{[1,-1,1]}=[0:1:0]$ and $V^z_{[1,1,1]}=V^z_{[1,1,-1]}=[0:0:1]$. + +We will have $6$ $1-$simplices, obtained from the $12$ of $\Delta^d$ by identifying: $L^x_{[1:1:1]}=L^x{[1:-1:-1]}$, $L^x_{[1:-1:1]}=L^x{[1:1:-1]}$, $L^y_{[1:1:1]}=L^y{[1:-1:-1]}$, $L^y_{[1:-1:1]}=L^y{[1:1:-1]}$, $L^z_{[1:1:1]}=L^z{[1:-1:-1]}$ and $L^z_{[1:-1:1]}=L^z{[1:1:-1]}$. + +Lastly, we will have $4$ $2-$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 $X_{[-1:1:1]}=X_{[1:-1:-1]}$. + +The face maps will be obtained by passing to the projective the face maps of $\Delta^d$. \end{example} \end{document} \ No newline at end of file