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[几何] 计算自由度

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hbghlyj posted 2024-3-29 10:45 |Read mode
repositorio.unican.es/xmlui/bitstream/handle/ … 971D6F902?sequence=1
平行四边形ABCD,$A(0,0), B(1,0), C\left(t_1, t_2\right), D\left(d_1, d_2\right)$.
由$D C \px A B$、$A D \px B C$得$$d_1 t_2=d_2\left(t_1-1\right), d_2=t_2$$
自由点$E\left(t_3, t_4\right)$,
取点$F\left(f_1, f_2\right)$使A是EF中点,得$$t_3=-f_1, t_4=-f_2$$
同理取点I使B是FI中点$$f_1+i_1=2,f_2=-i_2$$取点$K\left(k_1, k_2\right)$使C是IK中点,得$$k_1+i_1=2 t_1,k_2+i_2=2 t_2$$
我们得到8个多项式等于0的方程组:
$$d_1 t_2-d_2\left(t_1-1\right), d_2-t_2, t_3+f_1, t_4+f_2, f_1+i_1-2, f_2+i_2, k_1+i_1-2 t_1, k_2+i_2-2 t_2$$
用Maple计算:
  1. HilbertDimension(<(d1)*(t2)-d2*(t1-1),(d2-t2),t3+f1, t4+f2,f1+i1-2, f2+i2, k1+i1-2*t1, k2+i2-2*t2 >,{t1,t2,t3,t4, d1,d2, f1,f2,i1,i2,k1,k2});
Copy the Code
输出4
Screenshot 2024-03-29 021713.png

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original poster hbghlyj posted 2024-3-29 10:48
自由变量数为4($t_1, t_2, t_3, t_4$)
问题:为什么HilbertDimension等于自由变量数呢

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original poster hbghlyj posted 2024-3-29 10:51
Last edited by hbghlyj 2024-12-8 16:10

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original poster hbghlyj posted 2024-3-30 23:29
Maple的HilbertDimension应该就是Dimension of an algebraic variety
维基给出了很多等价定义,其中一个是:

    The degree of the Hilbert polynomial of A.
    The degree of the denominator of the Hilbert series of A.

This allows, through a Gröbner basis computation to compute the dimension of the algebraic set defined by a given system of polynomial equations. Moreover, the dimension is not changed if the polynomials of the Gröbner basis are replaced with their leading monomials, and if these leading monomials are replaced with their radical (monomials obtained by removing exponents).

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original poster hbghlyj posted 2024-3-30 23:33
hbghlyj 发表于 2024-3-30 15:29
the dimension is not changed if the polynomials of the Gröbner basis are replaced with their leading monomials
这个在UTM Ideals Varieties and Algorithms第458页:
Untitled.png
也在40HilbertFunctionIntro.pdf中讲到:
Theorem (MaCaulay) Let $I \subseteq k\left[x_1, \ldots, x_n\right]$ be an ideal and let $>$ be a graded order on $k\left[x_1, \ldots, x_n\right]$. Then the monomial ideal $\langle\operatorname{LT}(I)\rangle$ has the same affine Hilbert function as $I$.


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