By Cvitanovic P.

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**Example text**

0 2 0 λ 2 † 0 0 , λi = λ j . . CM C = .. . 0 . . λ2 λ3 . . 0 0 .. . . 44) i=1 (the characteristic equations will be discussed in sect. 6). In the matrix C(M − λ2 1)C † the eigenvalues corresponding to λ2 are replaced by zeroes: λ1 − λ2 λ1 − λ2 λ1 − λ2 0 . , 0 λ3 − λ2 λ3 − λ2 .. and so on, so the product over all factors (M − λ2 1)(M − λ3 1) . . with exception of the (M − λ1 1) factor has non-zero entries only in the subspace associated with λ1 : 1 0 0 0 1 0 0 0 0 1 0 C (M − λj 1)C † = (λ1 − λj ) .

P −1 p −2 n−p+1 p ... bp−1 . 20) p The number of independent components of fully antisymmetric tensors is given by dA = tr A = = n! (n−p)! 0, ... bp−1 a = = n n−p+1n−p+2 ... p p−1 1 , n≥p . 21) GroupTheory December 10, 2002 53 PERMUTATIONS ... For example, for 2-index antisymmetric tensors the number of independent components is n(n − 1) . 22) dA = 2 Tracing (p − k) pairs of indices yields p ... ... = . 23) ... ... (n − k)! (n − p)! ... 1 1 The antisymmetrization tensor Aa1 a2 ... b2 b1 has non-vanishing components, only if all lower (or upper) indices differ from each other.

0= ... 31) 1 2 3 ... n +1 For example, for − n = 2 : 0= 0 = δad bc − δbd + ac + δcd ab . 32) with cd and using n=2: = ac bc = δab . 33) GroupTheory December 10, 2002 55 PERMUTATIONS ... k! (n − l)! n! 34) ... ... b2 b1 . p! 4 DETERMINANTS = M , ... M= ... a2 a1 = Mab11 Mab22 . . 37) where a b . c b a Maa Mbb . . Mdd = ... . 39) − (p − 1) ... 1 = p ... The subscript p on tr p (. ) distinguishes the traces on [np × np ] matrices Mαβ from the [n × n] matrix trace tr M . 19) to obtain .