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[函数] 轮换对称 分解为 完全对称+完全反对称

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hbghlyj posted 2025-1-30 19:38 |Read mode
假设函数 $T$ 满足 $T(a,b,c)=T(c,a,b)$ 对所有 $a,b,c$ 成立。

令 $S(a,b,c)=\frac{T(a,b,c)+T(b,a,c)}2,A(a,b,c)=\frac{T(a,b,c)-T(b,a,c)}2$,则 $T(a,b,c)=S(a,b,c)+A(a,b,c)$,求证:
  • $S(a,b,c)$ 是完全对称的:$S(a,b,c)=S(a,c,b)=S(b,a,c)=S(b,c,a)=S(c,a,b)=S(c,b,a)$
  • $A(a,b,c)$ 是完全反对称的:$A(a,b,c)=-A(a,c,b)=-A(b,a,c)=A(b,c,a)=A(c,a,b)=-A(c,b,a)$

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original poster hbghlyj posted 2025-1-30 20:22
如何推广到n元

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