sjamaar 2_3 REVIEWED

Sjamaar equations (2-3) and (2-4) verification A working doc, the conclusions now appears in wedge Section 10.3 , do not edit here. The issue here is how to write an exterior derivative in standard notation. I took a lot of wrong turns below! I have been wondering just how these fit into my picture, and something is wrong with the k = 2 case! α = Σ1≤i<i<...<i≤n Aii...i(x)λ^I = Σ'I AI(x) λ^I dα = Σ1≤i<i<...<i≤n Σj=1n [∂jAii...i(x)] λj ^ λi ^ λi ...^ λi . (10.3.6) = Σ'I Σj=1n [∂jAii...i(x)] λj ^ λ^I Note that this is a mixed sum, so not clear how to make it be a single pure ordered or pure symmetric sum without doing some work (which work Sjamaar does). Now I clam that you can also write α this way α = ΣI TI λ^I where AI = k!Alt(TI) I show in 10.1 that a viable T is this TI = (1/k!) AI Thus I can write α = ΣI { (1/k!) AltI(AI) } λ^I = ΣI { (1/k!) AltI(Aii...i(x)) }λ^I = ΣI { (1/k!)2 ΣP(-1)P (AP(I)(x)) }λ^I Now I claim that you get the same dα if you apply d to this thing, so then dα = ΣI Σj{ (1/k!)2 ΣP(-1)P ∂j(AP(I)(x)) } λj ^λ^I α = ΣI { (1/k)! AltI(dAii...i(x)) }λ^I = ΣI { (1/k)! Σj AltI(∂jAii...i(x) } λj ^ λ^I = ΣI { (1/k)! Σj ∂jAltI(Aii...i(x)) } λj ^ λ^I Now define enlarged index J ≡ i1,i2.....ik,j = j1.....jk+1 . Then the above can be written dα = ΣJ (1/k)! AltJlo(∂jAjj...j(x) } (-λ^J) But now I am stuck because Alt is not for the full set! How might I fix this? Go back to dα = ΣIΣj { (1/k)! ∂jAltI(Aii...i(x) } λj ^ λ^I Maybe think of the product of two tensors, one being T = ∂j I = j S = AltI(Aii...i(x)) I' = ii...i I = j,ii...i (TS)I = ∂jAltI(Aii...i(x)) So what I have is dα = - (1/k!)ΣI (TS)I λ^I = - (1/k!) T^^ S^ = - (1/k!) Alt(T^ S^) ??? What does my product rule say? Alt(TS)(vJ) α = ΣITI(x) λ^I dα = ΣI Σr=1n [∂rTI(x)] (λr ^ λ^I) // since I claim can use Let r = ik+1 and write this as dα = - ΣIΣr=1n [∂rTI(x)] (λ^I ^ λr) = - ΣI [∂iTii...i(x)] λ^I where now I = i1.....ik+1 = - Σ'I {(k+1)! AltI [∂iTii...i(x)]} λ^I (A.8.33) There seem to be no "product form" restrictions on this theorem. OK, this is all wrong, have to start over. The j sum will be a full sum. I am motivated to get a full ΣI as well so they can be combined, but that makes a mess a seen above. Example. If α is a 1-form (so k = 1) then dα = - Σ'I {2! AltI[∂iTi(x)]} λ^I = - Σ'I [ ∂iTi(x) - ∂iTi(x) ] λ^I = Σ'I [ ∂iTi(x) - ∂iTi(x) ] λ^I = Σ1≤i<i≤n [ ∂iTi(x) - ∂iTi(x) ] λi ^ λi = Σ1≤i<j≤n [ ∂iTj(x) - ∂jTi(x) ] dxi ^ dxj // compare Sjamaar (2.2) Example. If α is a 3-form (so k = 2) then dα = - ΣI [[∂iTii(x)] λi ^ λi ^ λi = - Σ'I {3! AltI[∂iTii(x)]} λ^I I = i1,i2,i3 = - Σ'I {∂iTii - ∂iTii + ∂iTii - ∂iTii + ∂iTii - ∂iTii ]} λi ^ λi ^ λi = - Σ'I {∂iTii - ∂iTii + ∂iTii - ∂iTii + ∂iTii - ∂iTii ]} λi ^ λi ^ λi = - Σ1≤i<j<k≤n {∂kTij - ∂kTji + ∂iTjk - ∂iTkj + ∂jTki - ∂jTik ]} λi ^ λj ^ λk Why is Sjamaar missing 3 terms? Or am I doing something wrong! The implication is this: Start Over 1: Here is the ordered sum and the symmetric sum form for α α = Σ'I fI λ^I α = ΣI { k! AltI(fI) } λ^I NOT ok Those are your two choices. Now for each, compute dα dα = Σ'I Σj ∂jfI λj ^λ^I dα = ΣI Σj{ k! ∂j[AltI(fI)] } λj ^λ^I ok The second form is a full symmetric sum with k+1 indices, so seems promising. With the first form you have to do what Sjamaar does and add up all the insertion points. Suppose in the second case I define gI = AltI(fI) ok which is something I can at least write out. Then the second form says dα = ΣI Σj { k! ∂jgI } λj ^λ^I ok Now define J to be the enlarged set with j tacked on in the k+1 position. We then have dα = ΣJ {k! ∂jgjj...j } [-λ^J] ok ok Then write using the ordered form using (10.1.13), dα = Σ'J { - ∂jgjj...j } λ^J ok = Σ'J hJ(x') λ^J where hJ = - ∂jgjj...j - ∂jAltJ'(fJ') where J' is the first k of the set , last one missing. Let's try some examples! For k = 2 we have a function fjj so we get hjjj = - (2/3) ∂j (1/2) [ fjj - fjj] But this does not agree with old Sja p 20 (2-4). So this is the nasty thing that I don't know how to process! There are two Alt operators and they are of different sizes. It look similar to my "pre-antisymmetrization" idea. So maybe you could say -(k!AltJ [∂jfJ'] But in the k = 2 case this is -k!)AltJ [∂jfjj] and then you get 6 distinct terms in the sum, whereas the right answer has only three terms (I think). Just continuing anyway. dα = Σ'J { -k!AltJ [∂jfjj]] } λ^J Now can I just to back to a full sum notation to get dα = ΣJ { -[∂jfjj] } λ^J and then there are a zillion terms , there are n3 terms. End of day! Start Over 2: α = Σ'I fI(x) λ^I = ΣI FI(x) λ^I where FI = Then compute dα = Σ'I Σj=1n [∂jfI(x)] λj ^ λ^I = ΣI Σj=1n [∂jFI(x)] λj ^ λ^I Now define J = j1,j2.....jk,jk+1 = i1,i2.....ik,j Then we can write dα = ΣJ [- ∂jFjj...j(x)] λ^J Try this out for the case k = 2. It then reads dα = Σjjj [- ∂jFjj(x)] λi ^ λi ^ λi = Σj<jj [- ∂jfjj(x)] λi ^ λi ^ λi Then you just have to write out all the terms! But then I have not saved any time really. So write as = Σj<j Σj [- ∂jfjj(x)] λi ^ λi ^ λi = [ Σj<j<j + Σj<j<j + Σj<j<j ] [- ∂jfjj(x)] λi ^ λi ^ λi and I am back to the Sjamaar method. So write out as three separate terms = Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi + Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi + Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi Now make the first two terms look like the third term by renaming indices = Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi 3→1, 1→2,2→3 + Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi 2↔3 + Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi Now reorder the wedge products to get λi ^ λi ^ λi = λi ^ λi ^ λi no sign change since two flips λi ^ λi ^ λi = - λi ^ λi ^ λi one sign flip Then rewrite the above as = Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi - Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi + Σj<j<j [- ∂jfjj(x)] λi ^ λi ^ λi = Σj<j<j [ - ∂jfjj(x) + ∂jfjj(x) - ∂jfjj(x)] λi ^ λi ^ λi This is similar to Sjamaar, but he has two terms + and one term - !! Start Over 3: α = Σ'I fI(x) λ^I dα = Σ'I Σj ∂j fI(x) λj ^ λ^I Now define k+1 multiindex J and k multiindex J' such that J' = j1,j2.....jk J = j1,j2.....jk,jk+1 = i1,i2.....ik,j = J', jk+1 Then we have dα = Σj<j...<j Σj [∂jfI(x)] λj ^ λ^J' Now we know that λj ^ λ^J' = (-1)k λ^J' ^ λj = (-1)k λ^J So we then have dα = (-1)k Σj<j...<j Σj [∂jfI(x)] λ^J Now write Σj<j...<j Σj = Σj<j...<j<j + Σj<j...<j<j + ... + Σjj<j...<j Then we have (-1)kdα = Σj<j...<j<j<j [∂jfjj...j(x)] λj ^ λj .... ^ λj + Σj<j...<j<j<j [∂jfjj...j(x)] λj ^ λj .... ^ λj + Σj<j...<j<j<j [∂jfjj...j(x)] λj ^ λj .... ^ λj + ... + Σjj<j...<j<j [∂jfjj...j(x))] λj ^ λj .... ^ λj Now I need to invent some kind of swap operator which swaps two indices of the following expression S(r,s) Fjj. j...j...j = Fjj. j...j...j Now any dummy swaps are allows for summation indices, so rewrite the above as (-1)kdα = { Σj<j...<j<j<j [∂jfjj...j(x)] (λj ^ λj .... ^ λj) } +S(k,k+1) {Σj<j...<j<j<j [∂jfjj...j(x)] (λj ^ λj .... ^ λj) } +S(k,k+1)S(k-1,k+1) {Σj<j...<j<j<j [∂jfjj...j(x)] (λj ^ λj .... ^ λj) } ... +S(k,k+1)S(k-1,k+1)...S(1,k+1){Σj<j<j...<j<j [∂jfjj...j(x)] (λj ^ λj .... ^ λj) } where nothing has changed in value. The swaps adjust the ordered sum on all subsequent lines so they match that on the first line. The swaps on the basis vectors can be readjusted to the ordering on the first line adding a factor (-1)s where s is the number of pairwise swaps. So we end up with dα = Σj<j...<j<j<j [ Q ] (λj ^ λj .... ^ λj) or dα = (-1)k Σ'J QJ λ^J Where Q = [ 1 - S(k,k+1) +S(k,k+1) S(k-1,k+1) - S(k,k+1)S(k-1,k+1)S(k-2,k+1) ... + (-1)k S(k,k+1)S(k-1,k+1.....S(1,k+1) ... ] ∂jfjj...j(x) Try this for k = 2: Q = [ 1 - S(2,3) + S(2,3)S(1,3) ] ∂jfjj(x) = ∂jfjj(x) - ∂jfjj(x) + S(2,3)∂jfjj(x) = ∂jfjj(x) - ∂jfjj(x) + ∂jfjj(x) and then dα = (-1)k Σ'J QJ λ^J = Σ'J [∂jfjj(x) - ∂jfjj(x) + ∂jfjj(x)] λ^J and finally this agrees with Sjamaar! Try if for k = 1 Q = [ 1 - S(1,2) ] ∂jfj(x) = ∂jfj(x) - ∂jfj(x) Then dα = - Σ'J [ ∂jfj(x) - ∂jfj(x)] λ^J = Σ'J [∂jfj(x) - ∂jfj(x) ] λ^J and this also agrees with Sjamaar. Try if for k = 3 Q = [ 1 - S(3,4) +S(3,4) S(2,4) - S(3,4) S(2,3) S(1,4) ]∂jfjjj(x) term 1 = ∂jfjjj(x) term 2 = - S(3,4) ∂jfjjj(x) = - ∂jfjjj(x) term 3 = +S(3,4) S(2,4)∂jfjjj(x) = S(3,4) ∂jfjjj(x) = ∂jfjjj(x) term 4 = -S(3,4) S(2,3) S(1,4)∂jfjjj(x) = -S(3,4) S(2,3)∂jfjjj(x) = -S(3,4) ∂jfjjj(x) = -∂jfjjj(x) Result is then Q = ∂jfjjj(x) - ∂jfjjj(x) + ∂jfjjj(x) - ∂jfjjj(x)