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| -rw-r--r-- | controle-20230412.tex | 62 |
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diff --git a/controle-20230412.tex b/controle-20230412.tex index c7c55e6..65a37c0 100644 --- a/controle-20230412.tex +++ b/controle-20230412.tex @@ -101,7 +101,7 @@ Use of electronic devices of any kind is prohibited. Duration: 2 hours \ifcorrige -This answer key has 6 pages (cover page included). +This answer key has 7 pages (cover page included). \else This exam has 4 pages (cover page included). \fi @@ -591,6 +591,66 @@ Plücker coordinates $w$.) The important part of this question is: how can we compute equations for $Y$ given the equation $h=0$ of $X$? +\begin{answer} +We have seen in question (6) that, so long as $w_{0,3} \neq 0$, the +line $L_w$ defined by $(w_{0,1}{:}w_{0,2}{:}w_{0,3} {:} \penalty0 +w_{1,2}{:}w_{1,3}{:}w_{2,3})$ in $Q$ is the line through +$(w_{0,3}{:}w_{1,3}{:}w_{2,3}{:}0)$ and +$(0{:}w_{0,1}{:}w_{0,2}{:}w_{0,3})$. In other words, $\lambda,\mu$ it +is the line consisting of points $(\lambda w_{0,3} : \lambda w_{1,3} + +\mu w_{0,1} : \lambda w_{2,3} + \mu w_{0,2} : \mu w_{0,3})$. This +line is included in $X$ iff $h(\lambda w_{0,3}, \lambda w_{1,3} + \mu +w_{0,1}, \lambda w_{2,3} + \mu w_{0,2}, \mu w_{0,3}) = 0$ for all +$\lambda,\mu$ in $k^{\alg}$, which just means that, seen as a +polynomial in $\lambda,\mu$ now seen as two \emph{indeterminates}, +this is identically zero. But this is a homogeneous polynomial in +$\lambda,\mu$ whose coefficients are homogeneous polynomials in the +$w_{i,j}$, so equating all these coefficients to $0$ gives homogeneous +equations in the $w_{i,j}$ (coordinates in $\mathbb{P}^5$) for $L_w$ +to be included in $X$. This only holds so long as $w_{0,3} \neq 0$ +(when it is zero, some of the resulting equation will be trivial); +however, at least one $w_{i,j}$ must be zero in any case, so writing +the corresponding equations for all permutations of coordinates, +together with the equation ($\dagger$) of the Plücker quadric itself +(to ensure that the $w_{i,j}$ do correspond to a line in +$\mathbb{P}^3$) gives us the equations of the desired $Y$. + +To illustrate that this is actually algorithmic, the following Sage +code computes the equations for the set $Y$ of lines inside the +“diagonal cubic surface” $X := \{x_0^3 + x_1^3 + x_2^3 + x_3^3 = 0\}$: +{\fontsize{8}{10}\relax +\begin{verbatim} +sage: R.<x0,x1,x2,x3,w01,w02,w03,w12,w13,w23,u,v> = PolynomialRing(QQ,12) +sage: xvars = [x0,x1,x2,x3] +sage: wvars = [[0,w01,w02,w03],[-w01,0,w12,w13],[-w02,-w12,0,w23],[-w03,-w13,-w23,0]] +sage: # Plücker equation: +sage: plucker = w01*w23 - w02*w13 + w03*w12 +sage: # Equation of the surface X: +sage: h = x0^3 + x1^3 + x2^3 + x3^3 +sage: deg = h.degree() +sage: # All possible permutations of (0,1,2,3): +sage: perm4 = [(j0,j1,j2,j3) for j0 in range(4) for j1 in range(4) for j2 in range(4) +....: for j3 in range(4) if len(set([j0,j1,j2,j3]))==4] +sage: # Generate the ideal I of the set Y of lines in X, as above: +sage: I = R.ideal([plucker] + [h.subs(dict([(xvars[j0],wvars[j0][j3]*u), (xvars[j1],w +....: vars[j1][j3]*u+wvars[j0][j1]*v), (xvars[j2],wvars[j2][j3]*u+wvars[j0][j2]*v), ( +....: xvars[j3],wvars[j0][j3]*v)])).coefficient({u:deg-k,v:k}) for k in range(deg) fo +....: r (j0,j1,j2,j3) in perm4]) +sage: # Compute its radical: +sage: I0 = I.radical() +sage: # This really means Y is 0-dimensional in projective space: +sage: I0.dimension() +7 +sage: # This computes its number of geometric points (i.e., geometric lines on X): +sage: hp = I0.hilbert_polynomial() ; hp.leading_coefficient()*factorial(hp.degree()) +27 +\end{verbatim} +\par}\noindent (notation is as above except that $\lambda,\mu$ have +been called \texttt{u},\texttt{v}); the above code proves that thare +are $27$ geometric lines in the surface $\{x_0^3 + x_1^3 + x_2^3 + +x_3^3 = 0\} \subseteq \mathbb{P}^3$ (over $\mathbb{Q}$). +\end{answer} + |