scraps save

THESE ARE NOW JUST SCRATCH NOTES, BUT SAVE FOR POSSIBLE FUTURE USE done when developing the curvilinear meta notes 11.20.09 or so. (e) Assignment: How can all of the above be extended from N=3 to N ≥ 4 dimensions? Let's try a more systematic notation for our two cases so far (ei)a =(1/2!) εi,n1,n2 εa,m1,m2 (en1)m1(en2)m2 /v N = 3 (ei)a =(1/3!) εi,n1,n2,n3 εa,m1,m2,m3 (en1)m1(en2)m2(en3)m3 /v N = 4 Our general case will be (ei)a =(1/(N-1)!) εi,n1,n2,n3..n(N-1) εa,m1,m2,m3..n(N-1) [ Πk=1N-1 (enk)mk ]/v N = N where n(N-1) means nN-1 . We might compact down the ε labels a bit in this way, where like {n} is the set n1, n2.....nN-1 and similarly for {m}. Then we get (ei)a = (1/(N-1)!) εi,{n} εa,{m} [ Πk=1N-1 (enk)mk ]/v N = N Things are a little clearer if we rotate the indices on both ε factors to the left one place (ei)j = (1/(N-1)!) ε{n},i ε{m},j [ Πk=1N-1 (enk)mk ]/v N = N Now we put this notation to its first use. Consider ea ei =(ea)j (ei)j = (ea)j { (1/(N-1)!) ε{n},i ε{m},j [ Πk=1N-1 (enk)mk ]/v } In order to make the leading factor (ea)j look like the missing last term in the product, we set j = mN = a summation index a = nN = a fixed value Then enN ei = (ea)mN (ei) mN = (enN)mN { (1/(N-1)!) ε{n},i ε{m},mN [ Πk=1N-1 (enk)mk ]/v } = (1/(N-1)!) ε{n},i ε{m},mN [ Πk=1N (enk)mk ]/v In any given non-zero term in the sum, the sets {n} and {m} contain all the integers between 1 and N-1, in some order. For example, a term for N= 5 might be (WRONG!) ε1,3,2,4,i ε4,5,1,2,nM The only way this factor cannot vanish is it i = nM = 5. Applying this to all the terms in the sum, we may conclude that enN ei= 0 unless nN = i . In the case nN = i , we have ei ei = (1/(N-1)!) ε{n},i ε{m},i [ Πk=1N (enk)mk ]/v // no sum on i From our Lemma below we know that v = det(eab) = (1/N!) ε{n},nN ε{m},mN [ Πk=1N (enk)mk ] Now recall from above the expression for v as a determinant v = εijklm... (e1)i (e2)j (e3)k(e4)l(e5)m .... = ε{m},mN (e1)m1 (e2)m2 (e3)m3(e4)m4(e5)m5 .... (eN)mN This result is unchanged if we replace the set of lower indices with some other partition of the integers 1 through N, as long as we compensate with a sign reflecting the number of "column swaps" we did. But that sign is going to be This is of course true for any rearrangement of the set 1,2,3,4... Comments on determinants The formula for the determinant of a matrix can be written det(A) = εm1,m2,m3..mN A1,m1 A2,m2 A3,m3 ...... AN,mN If the fixed set of integers ( n1,n2,n3..nN) is a rearrangement of (1,2,3..N), then if we replace the first indices of the A factors with this rearranged set, the determinant is the same except for a sign which is determined by the number of row swaps implied by the rearrangement. That sign is precisely εn1,n2,n3...nN, so we can rewrite the above as ( sums on the mi, but no sums on the ni) det(A) = εn1,n2,n3..nN εm1,m2,m3...mN An1,m1 An2,m2 An3,m3 ...... AnN,mN ni not summed = εn1,n2,n3..nN εm1,m2,m3...mN [ Πk=1N Ank,mk ] Suppose now we pick a specific value for nN, but we sum over all the remaining ni indices. Then we get (N-1)! replications of det(A) in the sum, so we could say det(A) = (1/(N-1)!) εn1,n2,n3..nN εm1,m2,m3...mN [ Πk=1N Ank,mk ] // no sum on nN There are N! possible fixed rearrangements of the integers1 to N, so if we now allow summation over all the ni indices as well as the mi ones, we have det(A) = (1/N!) εn1,n2,n3..nN εm1,m2,m3...mN An1,m1 An2,m2 An3,m3 ...... AnN,mN If we adopt our notation above and replace A with e and put the second index up, we get det(eab) = (1/N!) ε{n},nN ε{m},mN [ Πk=1N (enk)mk ] det(A) = εn1,n2,n3..nN εm1,m2,m3...mN An1,m1 An2,m2 An3,m3 ...... AnN,mN ni not summed = Σm1,m2...mN εm1,m2,m3...mN { [ Πk=1N Ank,mk ] εn1,n2,n3..nN } Now in each term in the multiple sum, suppose we choose nN = mN. Then we can write det(A) = Σm1,m2...mN εm1,m2,m3...mN { [ Πk=1N Ank,mk ] εn1,n2,n3..mN } Now if we sum over all possible rearrangements of n1,n2,,,,,n(N-1), we will generate (N-1)! copies of det(A), so we can then say det(A) = (1/(N-1)!) Σm1,m2...mN Σn1,n2...n(N-1) εm1,m2,m3...mN { [ Πk=1N Ank,mk ] εn1,n2,n3..mN } *******************************************************8 Appendix on Matrix Forms Let's make sure things are well-defined: (Tup)ab ≡ Tab [(Tup)T]ab = (Tup)ba = Tba [(Tup)-1]ab = ??? Does up commute with inverse??? Consider this equation (Tup) (Tup)-1 = 1 which nobody can deny. Then (Tup)ab (Tup)-1bc = δac Tab (Tup)-1bc = δac we know that (Tup)-1bc = Tcb from orthonormality Question: does the notation T-1 all by itself mean anything? Well, does T mean anything? T could mean either of two different matrices, so it is ambiguous, therefore T-1 is also ambiguous. But certainly the quantity (Tup)-1 is not ambiguous. Maybe we could make this definition: (T-1)up ≡ (Tup)-1 [(T-1)up] ab = [(Tup)-1]ab (T-1)ab = [(Tup)-1]ab = Tba Then we have Tab (T-1)bc = δac So, T and T-1 solo are ambiguous, but Tup and (T-1)up are not ambiguous.