"대칭군의 지표(character)에 대한 프로베니우스 공식"의 두 판 사이의 차이

2012년 8월 25일 (토) 06:55 판

\(C_\mathbf{i}}\) : conjugacy class in \(S_{m}\) where \(i_1+2i_2+\cdots mi_m=m\)
\(\left(\sum_{l=1}^{m} x_l\right)^{i_1}\left(\sum_{l=1}^{m} x_l^2\right)^{i_2}\cdots \left(\sum_{l=1}^{m} x_l^m\right)^{i_m}=\sum_{\lambda}\chi_{\lambda}(C_{\mathbf{i}})S_{\lambda}\)
여기서 \(S_{\lambda}\) 는 슈르 다항식(Schur polynomial)

\(S_{(3)}+2S_{(2,1)}+S_{(1,1,1)}=\left(x_1+x_2+x_3\right){}^3\)

\(S_{(3)}+0\cdot S_{(2,1)}-S_{(1,1,1)}=\left(x_1+x_2+x_3\right) \left(x_1^2+x_2^2+x_3^2\right)\)

\(S_{(3)}-1S_{(2,1)}+S_{(1,1,1)}=x_1^3+x_2^3+x_3^3\)

\(\prod_{j}P_{j}(x)^{i_j}=\sum_{\lambda}\chi_{\lambda}(C_{\mathbf{i}})S_{\lambda}\)

@@ 7번째 줄: / 7번째 줄: @@
 <h5>개요</h5>
-* <math>C_\mathbf{i}}</math> : conjugacy class in <math>S_{m}</math> where <math>\mathbf{i}}=(i_1,i_2,\cdots,i_m)</math> and <math>i_1+2i_2+\cdots mi_m=m</math>
+* <math>C_\mathbf{i}}</math> : conjugacy class in <math>S_{m}</math> where  <math>i_1+2i_2+\cdots mi_m=m</math>
 * <math>\left(\sum_{l=1}^{m} x_l\right)^{i_1}\left(\sum_{l=1}^{m} x_l^2\right)^{i_2}\cdots \left(\sum_{l=1}^{m} x_l^m\right)^{i_m}=\sum_{\lambda}\chi_{\lambda}(C_{\mathbf{i}})S_{\lambda}</math><br> 여기서 <math>S_{\lambda}</math> 는 [[슈르 다항식(Schur polynomial)]]<br>
@@ 43번째 줄: / 43번째 줄: @@
 <math>S_{(3)}+2S_{(2,1)}+S_{(1,1,1)}=\left(x_1+x_2+x_3\right){}^3</math>
-<math>S_{(3)}+2S_{(2,1)}+S_{(1,1,1)}=\left(x_1+x_2+x_3\right){}^3</math>
+<math>S_{(3)}+0\cdot S_{(2,1)}-S_{(1,1,1)}=\left(x_1+x_2+x_3\right) \left(x_1^2+x_2^2+x_3^2\right)</math>
-<math>S_{(3)}+2S_{(2,1)}+S_{(1,1,1)}=\left(x_1+x_2+x_3\right){}^3</math>
+<math>S_{(3)}-1S_{(2,1)}+S_{(1,1,1)}=x_1^3+x_2^3+x_3^3</math>