temp4

Lemma: How do you enumerate a set of permutations in a systematic way by using swaps? Here is what can do with just a single swap 12345 00000 21345 11000 13245 01100 12435 00110 12354 00011 32145 10100 14325 01010 12543 00101 42315 10010 15342 01001 52341 10001 Suppose we just alternate swapping in the last three positions this way: 12345 ε12345 12354 1 12534 1 12543 12453 12435 We get 5 new terms. If we try to continue, we replicate the first term, so this cycle is exhausted. P12f(123) = f(213) P23f(123) = f(132) P13f(123) = f(321) P12P23f(123) = P12 f(132) = f(231) P13P12f(123) = P13 f(213) = f(231) Lemma: Consider some function f(m1, m2 ... mN ). Let P represent some permutation of the arguments. Let p indicate the number of pair swaps needed to get from the original order to that permutation, so p is determined by P. There are N! permutations of f(123...N) including the original term. We could write a simple sum of all N! permutations using this notation f(123...N) + f(21...N ) + ..... = ΣP P f(123...N) It is totally unambigous what this sum is. We could create another sum in which all the terms with an odd number of pair swaps have minus signs. f(123...N) – f(21...N ) + ... = ΣP(-1)p P f(123...N) There are still N! terms. Here is the series for N = 2 and N=3, ΣP(-1)p P f(123) = f(123) - f(213) + f(231) – f(312) + f(321) – f(231) Can we represent this using the ε symbol somehow? εijkf(ijk) = ε123f(123) + ε213f(213) + ... This seems to work, so I think we can say f(123...N) – f(21...N ) + ... = ΣP(-1)p P f(123...N) = Σijk εijk f(ijk) = εijk f(ijk) where each sum is over 1...N. Now suppose you have a different kind of function with only N-1 arguments: g( i1i2.....iN-1) where each in takes a value in 1,2....N. How now, brown cow, do we list the permutations of these N-1 indices? There are only N-1 positions subject to swapping. We could still write g( i1i2.....iN-1) – g( i2i1.....iN-1) + .... = ΣP(-1)p P g( i1i2.....iN-1) where P permutes the N-1 objects. The question now is how you use the ε symbol to indicate the sign! The first thing you might try is this (-1)p = εiii....i But the problem here is that this ε only has N-1 indices, but we want to have N integers involved, so there is a mismatch. Let's just stick to N = 3. We have some g(i1i2) object where each i is 1,2 or 3. Our answer is just this ΣP(-1)p P g(i1i2) = g(i1i2) – g(i2i1) = ε12 g(i1i2) + ε21 g(i2i1) ? Here the ε indices are the subscripts on the in indices. The ε indices are not the values of the in . Since there are only two arguments for g, we use the εij tensor with each index doing 1 or 2. Hmmm. ΣP(-1)p P g( i1i2i3) = εlmn g( il im in) = εabc g(iaibic) Now think of δi,m δi,m..... δi,m = f(m1 m2.... mN-1) Suppose we want to add to this term all the permutations of the m's. We could say series = f(mi1 mi2.... miN-1)εi1i2... It is sort of "indirect addressing" or pointer stuff. Maybe write as series = f(mamb.....mx) εabc...x Then we get δi,m δi,m..... δi,m εabc...x So here is a new statement of the Theorem below Σj εjiii....i εjmmm....m = δi,m δi,m..... δi,m + other terms // line 1 = δi,m δi,m..... δi,m εabc...x // line 2 where εabc...x has N-1 indices each of which takes a value in (1...N). Example Σj εjii εjmm = δi,m δi,mεab = δi,m δi,m ε12 + δi,m δi,m1 ε21 so I think that does it. I think trying to replace the abc with a1a2 is too complicated. Theorem 1: In this theorem we assume all indices are present. The theorem claims that Σj εjiii....i εjmmm....m = δi,m δi,m..... δi,m + other terms // line 1 = δi,s δi,s..... δi,s εasss....s εbmmm....m // line 2 where a is the number in {1,2...N} not present among the in , and b the number not in the mn . (1) Each ε on the LHS has N indices, and each index has must take a value in the set {1,2,3...N}. (2a) In order for the first ε to be non-zero, all N-1 in indices must be different from each other. If this is the case, then only one number in the set {1,2,3...N} does not appear among the in. Call this number a. (2b) In order for the second ε to be non-zero, all N-1 mn indices must be different from each other. If this is the case, then only one number in the set {1,2,3...N} does not appear among the in. Call this number b. (3) In order for the product of the two ε's to be non-zero, the value set taken by the in must omit the same integer that the value set taken by the mn omits. Only then can we set j equal to the omitted value so that both ε's can be non-zero. (4) Assuming this is the case, one possibility is that we have i1= m1, i2= m2 and so on to iN-1 = mN-1. Both sets are omitting the same value, though we don't know what that value is. In the j sum, j will take on that omitted value to generate a non-zero εε product. In this case, for the one contributing term in the j sum, the indices on each ε are identical, so whether than ε is +1 or -1, we know that ε2 = +1. So this is where the first term on line 1 comes from. (5) Another possibility would be i1= m2, i2= m1 and all other indices as they were in (4). In this case the indices on the two ε's don't match. But if we swap indices m1 and m2 on the second ε, they would match. But this creates a minus sign! Thus, another term on line 1 would be – δi,m δi,m..... δi,m and it is contained on the "other terms" of this line. (6) In general, the sign of a term will be minus(plus) if an odd(even) number of mn index swaps is needed to make it look like the original first term. (7) Now look at the RHS of the theorem. One term in the implied s sums has s1 = m1, s2 = m2 and so on. The ε for that term will be then εmmm....m . Since we have added a factor which is this exact same ε, we will have ε2 and then we obtain the term shown be. (8) If we look at other terms in the multiple s sums, the εsss....s factor generates exactly the sign we need as stated in (6) above. (9) So we have now shown that the theorem is true in the case that the in and mn sets meet our conditions (2), (2') and (3) above: the in are all different, the mn are all different, and these two sets exclude the same element. We still need to show that our theorem is true if these rules are not met, in which case we know that the LHS = 0. So we have to show that the RHS is also 0 in these cases. (10a) Suppose two mn indices are the same. The RHS is zero due to εmmm....m. (10b) Suppose two in indices are the same. For example, suppose i1 = i2. We then have symmetric δi,s δi,s contracted against antisymmetric ε with the s's, so we get 0. (10c) Suppose the two sets don't exclude the same value? Consider δi,s δi,s..... δi,s . In order for this product of N-1 δ's to be non-zero, every single δ must be 1. We must have therefore i1= s1 , i2 = s2. ... iN-1 = sN-1 Suppose the in set omits value 3, and the mn set omits value 4. Then the δ product δi,s δi,s..... δi,s can have no factor δ3,3. QED. Let's now write this theorem for N = 3 where it will take on a perhaps familiar form.: Σj εjii εjmm = δi,m δi,m + other terms // line 1 = δi,s δi,s εss εmm // line 2 = δi,m δi,m – δi,m δi,m Suppose 2 is missing from the in set and 3 is missing from the mn set. What makes the RHS be zero? Specifically, suppose we have this Σj εj13 εj12 = δ1,s δ3,s εss * ε12 = δ1,s δ3,s εss = δ1,1 δ3,3 ε13 ??? Consider our tensor application now = εjiii....i εjmmm....m [(e1)i(e1)m][(e2)i(e2)m] ..... [(eN)i(eN)m] εjiii....i εjmmm....m = δi,m δi,m..... δi,m εabc...x So we then have δi,m δi,m..... δi,m εabc...x [(e1)i(e1)m][(e2)i(e2)m] ..... [(eN)i(eN)m] Maybe it is better to just use the P notation. Maybe use this [e1 e1] [e2 e2]..... [eN eN] + other terms = [e1 ea] [e2 eb]..... [eN ex] εabc...x And in our application, k is missing, so there are only N-1 factors, and ε has N-1 indices. Then = g'1ag'2b...g'Nx εabc...x This is (N-1)! terms. Example: suppose N = 3 and k=1 is missing. Then we have g'2bg'3c εbc = g'22g'33ε23 + g'23g'32ε32 = g'22g'33 – g'23g'32 = correct answer But here the indices on εbc only take values 2 and 3 since only those are in the set. g'1a g'2b g'2c...g'Nx εabc...x But we know that k is missing, so why not this g'1a g'2b g'2c...g'Nx εkabc...x where now ε has the full N indices. Maybe we put it in its usual place [ (g'1a g'2b g'2c...g'Nx)/ g'2k] εabc..k..x So try this with N=3 and k=1 missing [ (g'1a g'2b g'3c)/ g'1a] ε1bc = g'2b g'3c ε1bc = g'22 g'33 ε123 etc and it works! So maybe I have (Πx;i≠k ei ) (Πx;i≠k ei) = ([e1 e1] [e2 e2]..... [eN eN])/ [ek ek] + other terms (g'1a g'2b g'2c...g'Nx) εabc..k..x // g'ky is missing from the product. [ (g'1a g'2a ... g'Na)/ g'ka ] εaa...k..a Assuming k is out in the middle somewhere, the first term here is (g'11 g'22 ... g'NN)/ g'ka ε12...k..N Perhaps write instead as [ Πi≠k (g'ia) ] εaa...k..a Example N=3 and k = 1: [ Πi≠1 (g'ia) ] ε1aa = (g'2a)(g'3a) ε1aa = (g'22)(g'33) ε123 + (g'23)(g'32) ε132 = (g'22)(g'33) – (g'23)(g'32) agrees. Example N = 4 and k = 2 [ Πi=1,N; i≠k (g'ia) ] εaa...k..a = [ Πi=1,4; i≠2 (g'ia) ] εa2aa = [(g'1a)(g'3a)(g'4a) ] εa2aa = [(g'11)(g'33)(g'44) ] ε1234 + [(g'11)(g'34)(g'43) ] ε1243 + [(g'13)(g'31)(g'44) ] ε3214 + [(g'13(g'34(g'41) ] ε3241 + [(g'14(g'31)(g'43 ] ε4213 + [(g'14)(g'33)(g'41) ] ε4231 Stop. This looks like a minor of a determinant. I have looked at my matrix notes, so go back to εa2aa[(g'1a)(g'3a)(g'4a) ] = cof(g'22) It works. So here is a much simpler notation [ Πi=1,N; i≠k (g'ia) ] εaa...k..a = cof(g'kk) Theorem: εjiii....i εjmmm....m = δi,m δi,m..... δi,m εabc...x = δi,m δi,m..... δi,m + other terms If nothing were "missing", I think this is the determinant of this matrix | δi,m δi,m δi,m ..... δi,m | | δi,m δi,m δi,m ..... δi,m | | δi,m δi,m δi,m ..... δi,m | | δi,m δi,m δi,m ..... δi,m | This matrix is (N-1)2. The result quoted in my Levi section would have N = 2. Suppose we have our εε object summed against a product of matrix elements this way εjiii....i εjmmm....m gi,m gi,m ...... gi,m Then this is the determinant of a matrix you could construct just like the δ matrix above. We have Σi Σm εjiii....i εjmmm....m gim gim ...... gim = Σi Σm δi,m δi,m..... δi,m εabc...x gim gim ...... gim = Σm εabc...x gmm gmm ...... gmm = Σm [ gmm gmm ...... gmm + other terms ] Is this perhaps N! times det g ? Each [..] object is the determinant of the g matrix with the rows and columns labeled by m1,m2 .... instead of by 1,2,3... I am sure that det of a matrix does not change if you permute the rows AND the columns by some arbitrary permutation, same for both.