[组合] 一些恒等式求极大线性无关组

hbghlyj

四元的时候是$(g_1-g_2)(g_3-g_4),(g_1-g_3)(g_4-g_2),(g_1-g_4)(g_2-g_3)$
它们的和为0,所以极大线性无关组的阶数为2。
六元的时候：
求$M=\{(g_a-g_b)(g_c-g_d)(g_e-g_f)\mid a,b,c,d,e,f是1,2,3,4,5,6的排列,且a<b,c<d,e<f\}$的极大线性无关组
一个因子出现三次的话就去掉(因为剩下的4个元出现三次必然线性相关)
得到$(g_1-g_2)(g_3-g_4)(g_5-g_6),(g_1-g_3)(g_2-g_4)(g_5-g_6),(g_1-g_3)(g_2-g_5)(g_6-g_4),(g_1-g_4)(g_2-g_5)(g_6-g_3),(g_1-g_5)(g_2-g_3)(g_4-g_6),(g_1-g_5)(g_2-g_4)(g_6-g_3),(g_1-g_6)(g_2-g_3)(g_4-g_5)$
因为$(g_1-g_2)(g_3-g_4)(g_5-g_6)+(g_1-g_3)(g_2-g_4)(g_5-g_6)+2(g_1-g_3)(g_2-g_5)(g_6-g_4)-2(g_1-g_4)(g_2-g_5)(g_6-g_3)+(g_1-g_5)(g_2-g_3)(g_4-g_6)+2(g_1-g_5)(g_2-g_4)(g_6-g_3)+(g_1-g_6)(g_2-g_3)(g_4-g_5)=0$
而且\[(g_1-g_2)(g_3-g_4)(g_5-g_6)+(g_1-g_3)(g_2-g_5)(g_6-g_4)-(g_1-g_4)(g_2-g_5)(g_6-g_3)+(g_1-g_5)(g_2-g_3)(g_4-g_6)+(g_1-g_5)(g_2-g_4)(g_6-g_3)=0\]
所以最终是5阶。一个极大线性无关组是：
$(g_1-g_2)(g_3-g_4)(g_5-g_6),(g_1-g_3)(g_2-g_4)(g_5-g_6),(g_1-g_3)(g_2-g_5)(g_6-g_4),(g_1-g_4)(g_2-g_5)(g_6-g_3),(g_1-g_5)(g_2-g_3)(g_4-g_6)$
一般情况,共$(n-1)!!=3·5·\ldots·(n-1)$个,极大线性无关组的阶数是多少呢？

hbghlyj

是否和共形不变量有关？
四元的时候,交比是共形不变量.
六元的时候,算两个交比乘积，消去一个因式,就得到六元的那个比值了.
一般情形？？？

hbghlyj

1. f[a_, b_, c_, d_] = {(a - b) (c - d), (a - c) (b - d), (b - c) (a - d)}
   
2. t = {u, v, w, x, y, z}
   
3. f = Function[Evaluate[t],Evaluate[Flatten[Table[(t[[1]] - t[[i]]) f @@ Table[t[[j]], {j, Complement[Range[2, 6], {i}]}], {i, 2, 6}]]]]
   
4. s = Table[Subscript[a, i], {i, 1, 15}]
   
5. MatrixRank[CoefficientArrays[Complement[Flatten[CoefficientList[Total[Table[s[[i]] f[g1, g2, g3, g4, g5, g6][[i]], {i, 15}]],{g1, g2, g3,g4, g5, g6}]], {0}], Evaluate[s]][[2]]]

复制代码

效果图：

2020-8-16 17:15

hbghlyj

2020-8-16 17:21

八元的时候是14
十元的时候是42
猜测：Catalan数列