how to get A from T in Ch 10 REVIEWED

New Theorem for Appendix A Well, it turns out there is no new theorem, but what there is is a clearer understanding of something I got wrong! The "solution" is shown at the very bottom and is now entered into wedge Section 10.1. The issue is this: given A in an ordered sum for a k-form, how would you produce a viable T for the symmetric version of the k-form. My answer is a simple Boolean deal, earlier answers were just plain wrong. Existing Theorem near 8.4.7: 1. If T^ = ΣITI λ^I where you are given some TI coefficients, then you can also write T^ = Σ'I [k!AltI(TI)] λ^I where you have converted from a symmetric sum to an ordered sum. If you then define AI ≡ k!AltI(TI) Then you can write the second sum as T^ = Σ'I AI λ^I . In this scenario, by definition the tensor AI is totally antisymmetric and TI has no symmetry. [ ok ] 2. I don't seem to deal with the reverse scenario which is this. Suppose you are given T^ = Σ'I fI λ^I (1) where fI is perhaps not totally antisymmetric. How can you write this as T^ = ΣI TI λ^I ? (2) In other words, how do you express TI in terms of fI ?? We do know this fact is true T^ = Σ'I AI λ^I AI ≡ k!AltI(TI) (3) [ incorrect conjecture] But we cannot compare (1) with (3) and conclude that fI = AI because AI is known TA whereas fI may not be TA. So this is a cleanly stated problem, what is the solution? Remember that there is nothing wrong with fI here having arbitrary symmetry!!! This always arises for example when you do the external derivative. I would like to show that this is valid T^ = Σ'I fI λ^I T^ = Σ'I [k!AltI(fI)] λ^I (4) [ wrong ] Then I could say AI ≡ k!AltI(fI) Now I would have a TA coefficient AI . I would then conjecture that TI = (1/k!) AI was a viable solution for TI for this reverse problem, because then k!AltI(TI) = k!(1/k!)AltI(AI) = AI which agrees with the relationship going the other direction. [ all wrong! ] 3. Can I prove conjecture (4)?? Let's first try a simple case and see if the conjecture is even true! T^ = Σi<i fii λi ^ λi T^ = Σi<i[2!AltI(fI)]λi ^ λi = Σi<i2 [fii - fii]λi ^ λi Can I show these two expressions are the same? Work on the second line: Σi<i [fii - fii]λi ^ λi = Σi<i fiiλi ^ λi - Σi<i fiiλi ^ λi = Σi<i fiiλi ^ λi + Σi<i fiiλi ^ λi = Σi<i fiiλi ^ λi + Σi<i fiiλi ^ λi = Σii fiiλi ^ λi = ΣI fI λ^I But then my conjecture implies that T^ = Σi<i fii λi ^ λi = Σii fiiλi ^ λi which certainly does not seem right. Suppose n = 2, then the left side is LHS = f12 λ1 ^ λ2 RHS = f12 λ1 ^ λ2 + f21 λ2 ^ λ1 = [f12- f21] λ1 ^ λ2 and these are NOT the same, so my conjecture is wrong. Start Over Statement of the Problem: You are given T^ = Σ'I fI λ^I where fI is NOT TA Find TI such that T^ = ΣI TI λ^I Comment: In Σ'I fI λ^I the coefficient fI is only "sensed" for order I. We should be able to replace fI by any gI as long as fI = gI for ordered I. For example, fij is only sensed for i<j. Suppose I try gij = fij + 27 fji Then g12 = f12 + 27 f21 and this does not work unless we declare that fij vanishes for non-ordered indices. Let's try that approach; We are given T^ = Σ'I fI λ^I and we know that all fI appearing in this equation have I = ordered. Suppose we extend fI in this way FI = Then we have T^ = Σ'I fI λ^I = ΣI FI λ^I = ΣI TI λ^I [ this is the correct solution ] so here then is a viable solution to my problem Solution #1 Problem: Given Σ'I fI λ^I for some arbitrary ordered fI, how can you write this as ΣI TI λ^I ? Solution: Select TI = It is a sort of Boolean θ function type solution. It might not be the only solution. In some other solution, you would have to arrange for the non-ordered summation components to vanish. I have now entered this solution into Section 10.1 of wedge doc.