volume element REVIEWED

The Volume Element Issue PhL 5.3.16 Reviewing in red 5/7/16. All items in this doc have been dealt with, no need ever to reread. I did learn various things here for the first time, all written up now in Appendix F. 5.16.16: Includes the sum of squared minors theorem and implications, all in Appendix F. My first shot at this is in the doc "Measure in Sja" where I look at various possibilities but reach no conclusions. One conjecture presented there is this: The volume element of a dV of the tangent space TxM is given by dt1dt2....dtn [ this is now proven in (F.5.2) ] So I have written up a 2-form type integration and I do get this result for the "volume element" dA = K(t) dt1dt2 where K(t)2 = det2 + det2 + det2 Is it possible that this expression can be written K(t)2 = det(RTR) ? Let's first just try this in Maple to see if by any chance there is a match. The R matrix in this case n = 2 and m = 3 is a 3 x 2 matrix. Here are the Maple results: So the answer is YES: det(RTR) IS another way to write the K2(t) factor! So now I have to learn why this works! And show that it works for any tall m x n R matrix! Note that RTR is an nxn matrix. So consider (T is the matrix transpose, NOT the covariant transpose), K(t)2 = det([RTR]**) = ΣP (-1)S(P) [RTR]1P(1) [RTR]2P(2) ... [RTR]nP(n) = ΣP (-1)S(P) [RTR]1P(1) [RTR]2P(2) ... [RTR]nP(n) = ΣP (-1)S(P)Σi1,i2...in=1m (RT)1i1Ri1P(1)(RT)2i2Ri2P(2) ... (RT)nikRikP(n) = ΣP (-1)S(P)Σi1,i2...in=1m Ri11Ri1P(1)Ri22Ri2P(2) ...RinnRikP(n) Now approach from the other direction in order to find the right threading, and I will stick with this simple example at first, K(t)2 = det2 + det2 + det2 = [ ΣP (-1)S(P) R1P(1)R2P(2)]2 + [ ΣP (-1)S(P) R1P(1)R3P(2)]2 + [ ΣP (-1)S(P) R2P(1)R3P(2)]2 There must be some easier way. Here is what I want to prove: Theorem: Given a tall m x n R matrix, then sum of squares of all minors of width n = det(RTR) // proven now in F.6 Proof: I need some way to write out the LHS. I need notation to indicate a minor. How about this: minorI = minor(ri1, ri2, ....rin) where I is an ordered multiindex! This would enumerate all the minors! Then the claim of the theorem is this ; Σ'I [minorI]2 = det(RTR) In the above, I can think of the rs as n rows selected from the R matrix. I can treat those rows as columns of a different matrix which is just the transpose of the matrix of rows, and the notation above that minorI = det(ri1, ri2, ....rin) is saying the minor is the determinant of these n column vectors. Conjectured Lemma: Σ'I [minorI]2 = (1/n!) ΣI [minorI]2 // part of F.6 Proof: ΣI [minorI]2 = Σi1,i2...in=1m [minorI]2 = Σi1≠i2≠...≠in [minorI]2 = (Σi1<i2<..<in + n!-1 other orderings) [minorI]2 = (ΣP [ΣP(i)<P(i)<...<P(i)]) [minorI]2 = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . by (A.9.1) with fii...in = [minorI]2 But fii...in = [minorI]2 is a totally symmetric function of the indices since row swaps don't affect the squared determinant. This = Σi<i<...<i [ΣP [minorI]2] = Σi<i<...<i (n!) [minorI]2 = n! ΣI [minorI]2 QED Meanwhile, I know that minorI = ΣP(-1)S(P) Ri1P(1) Ri2P(2) ....RinP(n) = det(RIZ) [minorI]2 = ΣP(-1)S(P) Ri1P(1) Ri2P(2) ....RinP(n) ΣP'(-1)S(P') Ri1P'(1) Ri2P'(2) ....RinP'(n) Then Σ'I [minorI]2 = Σ1≤1i<i2..<in<m ΣP(-1)S(P) Ri1P(1) Ri2P(2) ....RinP(n) ΣP'(-1)S(P') Ri1P'(1) Ri2P'(2) ....RinP'(n) Then at least I have "an expression" for this object. Each term has 2n powers of R! Meanwhile, det(RTR) = ΣP(-1)S(P) (RTR)1P(1) (RTR)2P(2) ....(RTR)nP(n) and this also has 2n powers of R, so they stand a chance of being equal. Let's now write this thing out in gory detail, where I assume the "matrix transpose" and not the covariant transpose. det(RTR) = ΣP(-1)S(P) Σi1,i2...in=1m Ri11Ri1P(1)Ri22Ri2P(2) ...RinnRikP(n) How might I state the desired theorem in multiindex notation? Σ'I [minorI]2 =Σ'IΣP(-1)S(P) Ri1P(1) Ri2P(2) ....RinP(n) ΣP'(-1)S(P') Ri1P'(1) Ri2P'(2) ....RinP'(n) =Σ'IΣP(-1)S(P) RIP(Z) ΣP'(-1)S(P') RIP'(Z) = ΣP(-1)S(P)ΣP'(-1)S(P') Σ'I RIP(Z) RIP'(Z) det(RTR) = ΣP(-1)S(P) Σi1,i2...in=1m Ri11Ri1P(1)Ri22Ri2P(2) ...RinnRikP(n) = ΣP(-1)S(P) ΣI Ri11Ri1P(1)Ri22Ri2P(2) ...RinnRikP(n) = ΣP(-1)S(P) ΣI (Ri11Ri22...Rinn) (Ri1P(1)Ri2P(2) ...RikP(n)) = ΣP(-1)S(P) ΣI RIZ RIP(Z) So here then is what I want to show, expressed in multiindex notation, ΣP(-1)S(P)ΣP'(-1)S(P') Σ'I RIP'(Z) RIP(Z) = ΣP(-1)S(P) ΣI RIZ RIP(Z) Plan A. I think the rearrangement theorem will be called into action here. [ correct ] Consider the left side, LHS = Σ'I ΣP(-1)S(P)RIP(Z) ΣP'(-1)S(P')RIP'(Z) I can replace P' by P'Q and make no difference, so this says LHS = Σ'I ΣP(-1)S(P)RIP(Z) [ΣP'(-1)S(P'Q)RIP'Q(Z)] Now since P is a fixed quantity in the the last bracket, why not take Q = ?? I don't see a way to do anything here. And if I could do something, I would end up with the freed-up ΣP creating n! which I don't want. So maybe this is not the path. Plan B: Maybe this theorem will prove helpful, [ no, was not helpful ] Fact : ΣI TIJ x^I = Σ'I det(TIJ) x^I if TIJ has factored form (A.8.36) ΣI TJI x^I = Σ'I det(TJI) x^I if TJI has factored form Let's back up and try this new form: Σ'I [minorI]2 = Σ'I[ΣP(-1)S(P) Ri1P(1) Ri2P(2) ....RinP(n)[ minorI] ] = Σ'I[ΣP(-1)S(P) Ri1P(1) Ri2P(2) ....RinP(n) det(RIZ) ] = Σ'I ΣP(-1)S(P) RIP(Z) det(RIZ) Then I need to show that ΣP(-1)S(P) Σ'I RIP(Z) det(RIZ) = ΣP(-1)S(P) ΣI RIP(Z) RIZ Plan C. Go back to minorI = minor(ri1, ri2, ....rin) I = ordered multiindex This was just one way to write the minor. Another way might be this Σ'I [minorI]2 = (1/n!) ΣI [minorI]2 which I have now proven above. Then the theorem I need to prove is this: (1/n!) ΣP(-1)S(P)ΣP'(-1)S(P') ΣI RIP'(Z) RIP(Z) = ΣP(-1)S(P) ΣI RIZ RIP(Z) ? and now I have (1/n!) ready to cancel any n! I can develop by playing with perm sums! How about rewriting the above as (1/n!) ΣP(-1)S(P)ΣP'(-1)S(P') ΣI (RT)P'(Z)I RIP(Z) = ΣP(-1)S(P) ΣI (RT)ZI RIP(Z) ? Now do the ΣI normal matrix multiplications to get (1/n!) ΣP(-1)S(P)ΣP'(-1)S(P') (RTR)P'(Z)P(Z) = ΣP(-1)S(P) (RTR)ZP(Z) ? I may be close here! Let RTR = A just to simplify (1/n!) ΣP(-1)S(P)ΣP'(-1)S(P')AP'(Z)P(Z) = ΣP(-1)S(P) AZP(Z) ? Now in the P' sum on the left replace P' by QP' to get (1/n!) ΣP(-1)S(P)ΣP'(-1)S(QP')AQP'(Z)P(Z) = ΣP(-1)S(P) AZP(Z) ? Now since we are inside the ΣP sum, select Q = P to get (1/n!) ΣP(-1)S(P)ΣP'(-1)S(PP')APP'(Z)P(Z) = ΣP(-1)S(P) AZP(Z) ? Now consider that AIJ = Ai1j1Ai2j2..... so it does have "factored form". Then I can say APP'(Z)P(Z) = AP'(Z)Z using (A.8.31) which leaves me with (1/n!) ΣP(-1)S(P)ΣP'(-1)S(PP')AP'(Z)Z = ΣP(-1)S(P) AZP(Z) ? Now swap dummy names P↔P' on the left (1/n!) ΣP'(-1)S(P')ΣP(-1)S(P'P)AP(Z)Z = ΣP(-1)S(P) AZP(Z) ? (1/n!) ΣP' ΣP(-1)S(P)AP(Z)Z = ΣP(-1)S(P) AZP(Z) ? ΣP(-1)S(P)AP(Z)Z = ΣP(-1)S(P) AZP(Z) ? But in this last statement, both sides are representations of det(A)!!! I think I have done it! So let's now pretend I have a solid proof of this theorem: Σ'I [minorI]2 = det(RTR) this plan is mostly what I did in F.6 I have seen now this works out properly in the case of m = 3 and r = 2 which is a special tall R matrix. How did the left side arise in my 2-form case of Section 10.13? It seems a bit specialized there as I look at it The general case for a k-form says ∫φ αx = ∫φ Σ'I fI(x) dxi ^ dxi ^ ... ^ dxi // αx = Σ'I fI(x) dx^I ≡ ∫[0,1]k Σ'J gJ(t) dtj ^ dtj ^ ... ^ dtj // first definition (pull back) ≡ (!Syntax Error, I!Syntax Error, I ... !Syntax Error, I) Σ'J gJ(t) dtjdtj ... dtj // second definition where gJ(t) = Σ'I fI(φ(t)) det(RIJ) and x = φ(t) , R = (Dφ) . (10.12.1) Write all together as ∫φ αx = (!Syntax Error, I!Syntax Error, I ... !Syntax Error, I)Σ'J Σ'I fI(φ(t)) det(RIJ)dtjdtj ... dtj Now det(RIJ) is in fact a minor of R and RIJ is a particular k x k submatrix of R. The measure here seems to be measure = Σ'J det(RIJ)dtjdtj ... dtj In making a connection to theorem Σ'I [minorI]2 = det(RTR), I think k = n is required, because only then are you dealing with an n-dimensional TxM volume element, so let's now restrict to k = n. In that case, ∫φ αx = (!Syntax Error, I!Syntax Error, I ... !Syntax Error, I)Σ'J Σ'I fI(φ(t)) det(RIJ)dtjdtj ... dtjn where det(RIJ) is a full-width minor of R which is an m x n matrix. I think there is only one term in the J' sum so we rewrite the above as ∫φ αx = (!Syntax Error, I!Syntax Error, I ... !Syntax Error, I) Σ'I fI(φ(t)) det(RIZ) dt1dt2 ... dtn Now det(RIZ) = minorI and we get ∫φ αx = (!Syntax Error, I!Syntax Error, I ... !Syntax Error, I) Σ'I fI(φ(t)) [minorI dt1dt2 ... dtn] The theorem says Σ'I [minorI]2 = det(RTR) = (1/n!) ΣI [minorI]2 Look at the general-case 2-form formulas, ∫φ αx = !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 G(t) = Σ'I fI(φ(t)) Now think of this as the dot product of two vectors each having (m,2) components, so G(t) = f(φ(t)) = f(φ(t)) det(RIZ) = f(φ(t)) minor = Σ'I fI(φ(t)) minorI = ( f(φ(t)) minor-hat ) | minor | And then we would say | minor |2 = Σ'I [minorI]2 = det(RTR) and THAT is now it gets into the works. Suppose you had f(φ(t)) = | f(φ(t)) | minor-hat . Then in the case you would have G(t) = | f(φ(t)) | | minor | = | f(φ(t)) | and finally ∫φ αx = !Syntax Error, I!Syntax Error, I | f(φ(t)) | dt1dt2 // this idea now in F.7 Asynchronous Claim: Volume of tangent space is dV = dt1dt2...dtn Is this true? TBC. Continuing 5.5.16