a new paradox

A New Paradox PhL 11.3.15 Review 1.15.16. It was here that I discovered a serious problem in my tensor expansions in the wedge world. I was writing T = Σij Tij ei ^ ej, but by taking the rs element of both sides I was finding that the seemingly arbitrary Tij coefficient object had to be antisymmetric. This has now been globally remedied by the new form shown in (4.3.5) which is T = Σij Tij ei ^ ej where Tij is arbitrary but Tij is antisymmetric. I implemented this change globally, so this paradox is fully resolved. I start off a new Section 7 on dual wedge k, and start with a comparison between the dual and wedge worlds. I did this for a while, then found a contradiction in (4.3.30) which says [T(e)]rs = (Trs - Tsr)/2 which certainly seems wrong! I will pursue this now at 8:30AM. Here is the problem: T = Σij Tij ei ^ ej . (4.3.5) Trs = Σij Tij (ei ^ ej)rs = Σij Tij [(ei)r(ej)s - (ej)r(ei)s]/2 (4.3.29) = Σij Tij [δirδjs - δjrδis]/2 = (Trs - Tsr)/2 The implication here is that Trs = (Trs - Tsr)/2 2Trs = Trs - Tsr Trs = - Tsr I guess I was unaware that Trs was antisymmetric! Simple Case Here is a simple example of my paradox: T = Σij Tij ei^ ej // my assumed symmetric expansion form Look at this special case T = T12 e1^ e2 + T21 e2^ e1 = (T12 - T21) e1^ e2 At this point, T12 and T21 could be arbitrary! I am certainly allowed to write T = Σij Tij ei^ ej even knowing it is redundant. But now I try to take the 12 element of both sides of the equation: [T]12 = (T12 - T21) [e1^ e2]12 = (T12 - T21) [e11e22 - e12e21]/2 = (T12 - T21)(1/2) This then says 2T12 = T12 - T21 so that T12 = - T21 . Suddenly T12 and T21 are related to each other, but at the start they were arbitrary. What happened here?? Try doing this without using the wedge symbol. Then T = Σij Tij [ eiej - ejei]/2 Now I can see that only the antisymmetric part of Tij can contribute to this sum! Write T = Ts + Ta Then by the usual argument I can say, T = Σij Tij [ eiej - ejei]/2 = Σij (Ts+Ta)ij [ eiej - ejei]/2 = ΣijTaij [ eiej - ejei]/2 Thus, T on the left depends only on Ta, cannot depend on Ts. So the symmetric expansion cannot "represent" a tensor T whose components are not antisymmetric. The correct equation is then T = Σij Tij ei^ ej where T is antisymmetric. I could write this and note that only the AS part of T can contribute, so it can only generate an AS T. So how does this affect things? I have to start pondering changes at (4.3.5). That equation is just plain wrong I was unhappy about this anyway. What happens in the higher cases, an area I was unsettled about. T = Σijk Tijk ei^ ej^ ek Suppose I write, T = Ta + Ts + Tx T = Σijk [ Taijk + Tsijk + Txijk ] aijk Write this as Q = Qa + Qs + Qx = Σijk [ Taijk + Tsijk + Txijk ] aijk First, look at the Qa contribution: Qa = Σijk [ Taijk]aijk = "is what it is" and does not vanish Next, look at the Qs contribution Qs = Σijk Tsijk aijk = Σijk Tsjik aijk T is symmetric = Σijk Tsijk ajik // do j↔i = Σijk Tsijk (-aijk) // a is TA = - Σijk Tsijk aijk = - Qs Therefore Qs = 0, no contribution. Finally, consider the X contribution Qx = Σijk Txijk aijk = ???? I don't know what to do next! I fudged this in my writeup. I don't see why this should vanish! How about using this idea: aijk = a εijk // since only one TAS tensor. Then we have Qx = a Σijk Txijk εijk So here is a more basic question. Consider Q = εijkTijk = T123 - T132 + T312 - T321 + T231 - T213 Now suppose we define, 6Aabc ≡ Tabc - Tacb + Tcab - Tcba + Tbca - Tbac Then we know that Q = 6A123 This seems to say that in T = Σijk Tijk ei^ ej^ ek , only Aijk is represented. Well how about this T = Σijk=1n Tijk ei^ ej^ ek = Σijk εijk Tijk e1^ e2^ e3 WRONG! But this fails because εijk then has no meaning since indices take values 1 to n. Back up again to this: T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) 6.4.3 Comparison vj ^ vj ^ .... ^ vj = εjj....j ( v1^ v2^ .....^ vk ) . (6.2.10) (ei ^ ei ^ .... ^ ei) = εii....i (e1 ^ e2 ^ .... ^ ek) ???? In the first line, the jr indices only take values 1 to k, so yes, you can use the ε. In the second line, the ir indices can take values from 1 to n, so NO, you cannot use ε. There are k indices ik , but each index can take a value 1 to n, so you cannot say the above! Well, I do state this in (6.2.10) for a more general case. OK very good. Then I have from above T = [ εijk Tijk] e1^ e2^ e3 Now it seems clear that εijk Tijk = T123 - T132 + T312 - T321 + T231 - T213 = 6Ta123 Pause. This says there is only one basis vector in L3 and it is e1^ e2^ e3 . I am very confused about n and k I think. My first wedge comments are in (4.3), and my first expansion is T = Σij Tij ei ^ ej . I am talking the space L2 but I have been vague there about n and k. Go back to (4.1.7). There I write T ≡ Σij Tij eie'j = element of V2 *************** Start over. T = Σijk=1n Tijk (ei^ ej^ ek) The object (ei^ ej^ ek) is totally antisymmetric in its labels. Object is not a normal rank-3 tensor. Each of the three labels i,j,k can take values from 1 to n. What can be said about the Tijk in terms of symmetry? Well try this Tabc = Σijk=1n Tijk (ei^ ej^ ek)abc = ?? I could write this all out I suppose. But how about, ei^ ej = (1/2)( eiej - ejei) = (1/2) Σ ei^ ej^ ek = (1/3!) ΣP (-1)S(P) eP(i ej ek Need better notation. So try this T = Σiii Tiii (ei^ei^ei) Then write (ei^ei^ei) = (1/3!) ΣP (-1)S(P) eP(i) eP(i) eP(i) Then (ei^ei^ei)abc = (1/3!) ΣP (-1)S(P) [ eP(i) eP(i) eP(i)]abc = (1/3!) ΣP (-1)S(P) [ eP(i)a eP(i)b eP(i)c] = (1/3!) ΣP (-1)S(P) [ δP(i)a δP(i)b δP(i)c] = (1/3!) [ δia δib δic + signed permutations ] Then Tabc = Σiii Tiii (ei^ei^ei)abc = Σiii Tiii (1/3!) ΣP (-1)S(P) [ δP(i)a δP(i)b δP(i)c] = Σiii Tiii (1/3!) [ δia δib δic + signed permutations ] = (1/3!)[ Tabc + signed permutations ] = (1/3!)[ Tabc - Tacb + Tcab - Tcba + Tbca - Tbac] = (1/3!)ΣP (-1)S(P)[ TP(abc)] Maybe rewrite as Tjjj = Σiii Tiii (ei^ei^ei)jjj = Σiii Tiii (1/3!) ΣP (-1)S(P) [ δP(i)j δP(i)j δP(i)j] = Σiii Tiii (1/3!) [ δij δij δij + signed permutations of the ir ] = (1/3!)[ Tjjj + signed permutations ] = (1/3!)[ Tjjj - etc ] = (1/3!)ΣP (-1)S(P)[ TP(j)P(j)P(j)] = [Alt(T)]jjj How about δP(i)j = δiQ(j) Q = P-1 P(i1) = j1 i1 = Q(j1) ? That looks good. So then Tjjj = Σiii Tiii (1/3!) ΣP (-1)S(P) [ δP(i)j δP(i)j δP(i)j] = Σiii Tiii (1/3!) ΣP (-1)S(P) [ δiQ(j) δiQ(j) δiQ(j)] = (1/3!) ΣP (-1)S(P) Σiii Tiii [ δiQ(j) δiQ(j) δiQ(j)] = (1/3!) ΣP (-1)S(P) TQ(j)Q(j)Q(j) Now consider ΣP (-1)S(P) f(Q) where Q = P-1. Certainly (-1)S(P) = (-1)S(Q) and ΣP = ΣQ So then, ΣP (-1)S(P) f(P-1x) = ΣQ (-1)S(Q) f(Qx) = ΣP (-1)S(P) f(Px) the theorem is then that ΣP (-1)S(P) f(Qx) = ΣP (-1)S(P) f(Px) where Q = P-1 Then we end up with Tjjj = (1/3!) ΣP (-1)S(P) TP(j)P(j)P(j) = Alt(T) New Lemma: (ΣP [ΣP(i),P(i),...P(i)]) fii...i = Σi,i,....i [ΣP fP(i)P(i)...P(i)] . Try k = 2: (ΣP [ΣP(i),P(i)]) fii = Σi,iΣP fP(i)P(i) ? LHS = (Σii + Σii) fii = 2 Σiifii = 4 terms RHS = Σi,i [ fii + fii] = Σi,i[ fii + fii] = Σi,ifii + Σi,i fii = same Now in the general Lemma it seems that ΣP(i),P(i),...P(i) = Σi,i,...i since ordering does not matter. Then the theorem states k! Σi,i,...i fii...i = Σi,i,....i [ fii...i + all permutations ] = same So I guess this really is a theorem, but a trivial one. Try this lemma idea here Tabc = Σiii Tiii (ei^ei^ei)abc = Σiii Tiii (1/3!) ΣP (-1)S(P) [ δP(i)a δP(i)b δP(i)c] = (1/3!) ΣP (-1)S(P) Σiii Tiii [ δP(i)a δP(i)b δP(i)c] Question: is this valid δP(i)a = δiP(a) ? P(i1) = a a = no go Try again, Tabc = Σiii Tiii (ei^ei^ei)abc = Σiii Tiii (1/3!) ΣP (-1)S(P) [eP(i) eP(i) eP(i)]abc ************************************************** here I am editing the wedge doc text: ************************************************** So italicizing the coefficient in the symmetric expansion gives me an escape hatch from my problem!! Where would my next repairative edit be?? OK, I am now going to jump up to v5 and start putting in this fix.