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This is the Title this has been installed into v5 PhL 1.11.15 6.5 The alt, Alt and Sym operators One can regard the sum shown in (6.4.12) as being an operation "alt" performed on the tensor T to produce another tensor A which is totally antisymmetric. If we define [alt(X)]ii...i ≡ ΣP (-1)S(P) XP(i)P(i)...P(i) (6.5.1) then (6.4.12) becomes Aii...i = [alt(T)]ii...i or A = alt(T) . (6.5.2) It is useful to define another version of the alt operator which has a scaling factor 1/k! , [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (6.5.3) so that Alt(X) = (1/k!)alt(X) . (6.5.4) It is shown in (A.1.1) that alt(X) and Alt(X) are totally antisymmetric tensors for any tensor X. If X is already antisymmetric, we find that alt(X) = k! X // since all k! terms in (6.5.1) are the same X = antisym (6.5.5) Alt(X) = X . X = antisym (6.5.6) We showed in (6.4.14) that T = (1/k!) A . (6.5.7) so (6.5.2) can also be written T = Alt(T) . // (1/k!) A = (1/k!) alt(T) (6.5.8) One can also define a symmetrizing operator Sym in this manner, where the (-1)S(P) is omitted, [Sym(X)]ii...i ≡ ΣP XP(i)P(i)...P(i) (6.5.9) If X is already symmetric, we find Sym(X) = X X = symmetric (6.5.10) Fact: The operators Sym and Alt are both projection operators, meaning that Sym2 = Sym and Alt2 = Alt when applied to any tensor X. (6.5.11) Proof: Alt(Alt(X)) = Alt(A) // where A ≡ Alt(X) = antisymmetric = A // by (6.5.6) = Alt(X) // see two lines above Sym(Sym(X)) = Sym(S) // where S ≡ Sym(X) = symmetric = S // by (6.5.10) = Sym(X) // see two lines above Fact: The Sym and Alt projection operators are orthogonal : Sym(Alt(X)) = 0 Alt(Sym(X)) = 0 (6.5.12) These seemingly reasonable claims are proven in Appendix A.4. We can define a third projection operator this way Else() ≡ 1 - Alt() - Sym() // projection operator Else(X) = X - Alt(X) - Sym(X) . // applied to X (6.5.13) One can then decompose an arbitrary tensor X into three pieces X = Alt(X) + Sym(X) + Else(X) = A + S + E (6.5.14) Then Alt(X) = Alt(Alt(X)) + Alt(Sym(X)) + Alt(Else(X)) = Alt(X) + 0 + Alt(Else(X)) Alt(Else(X)) = 0 Alt(E) = 0 . (6.5.15) Sym(X) = Sym(Alt(X)) + Sym(Sym(X)) + Sym(Else(X)) = 0 + Sym(X) + Sym(Else(X)) Sym(Else(X)) = 0 Sym(E) = 0 . (6.5.16) This verifies that the "else" piece E of a tensor has neither an antisymmetric nor a symmetric component. Examples: For a rank-2 tensor Xab we find Aab = (Xab - Xba) Sab = (Xab + Xba) Eab = Tab - Aab - Sab = 0 (6.5.17) so the leftover else piece Eab is null. On the other hand, for a rank-3 tensor Xabc, Aabc = (Xabc - Xacb + Xcab - Xcba + Xbca - Xbac) Sabc = (Xabc + Xacb + Xcab + Xcba + Xbca + Xbac) Eabc = Xabc - Aabc - Sabc = Xabc - (Xabc - Xacb + Xcab - Xcba + Xbca - Xbac) - (Xabc + Xacb + Xcab + Xcba + Xbca + Xbac) = Xabc - (Xabc + Xcab + Xbca ) = Xabc - (Xcab + Xbca ) (6.5.18) so in this case the leftover piece Eabc is not null. Notice that Xcab = Xabc if X is either symmetric or antisymmetric Xbca = Xabc if X is either symmetric or antisymmetric and for this reason Eabc = 0 if X is either symmetric or antisymmetric.