fredholm determinant R1

The det(1+A) determinant expansion and the Fredholm Determinant PhL 3.23.09 These are "Round 1" notes written when I was still confused about many things. The whole subject is now better presented in my "Round 2" notes in a separate document. Still, there are some things in these notes that are worth keeping, so I won't delete this document. A reader interested in learning about this subject should 0. Overview of This Discussion. 1 1. Preliminary Facts concerning the Determinants of Certain Submatrices of a Matrix. 5 2. Application: computation of det(1+A) 10 3. Application: The Fredholm Determinant 13 4. Preliminary Facts concerning the Determinants of Certain Other Submatrices of a Matrix. 18 5. Figuring out the B0 and B1 terms of the Fredholm First Minor expansion 20 6. Show that these B0 and B1 terms agree with our full Fredholm First Minor Result 22 7. Consider the B2 terms, but then stop this approach for now. 23 9. Consider the B2 terms one more time 25 0. Overview of This Discussion. I wanted to learn a little about this "Fredholm stuff" because it was omitted from Stakgold's integral equations chapter. When the Fredholm integral equations are put on a lattice, they are reduced to N x N matrix problems. One of these equations is "the eigenvalue problem" and the eigenvalues can be determined from a certain determinant (The Fredholm Determinant), and the general inhomogeneous equation can then be solved by inverting a certain matrix which as usual can be done using cofactors which invoke minors (Fredholm's First Minor). One can do this for any N one wants. In the case that the kernel K of the integral equation set is "small" so it is written λK, it is possible to find expansions in powers of λ for both the Fredholm Determinant and the Fredholm First Minor, and it is really these expansions that have the official Fredholm names. If one then takes the limit N→∞ to move back off the lattice, these expansions are still well defined, but each has an infinite number of terms. Again, if λ is small, one usually deals with just the first few terms of these N=∞ expansions. Various buzzwords occur in this development that I will mention here. First of all, the inhomogeneous integral equation in the "Neumann form" can be written φ = f + λKφ where I show the smallness parameter λ. The formal solution is φ = (1-λK)-1f . So it is the matrix 1-λK that we want to "invert" to solve our problem, and this brings up the Fredholm determinant det(1-λK) and this also brings up the minors since we know that (1 - λK)-1 = [ cof (1-λK) ]T / det(1-λK). Now, every high school student knows how to write (1-λK)-1 as a series 1 + Σm=1λmKm where we break out the first term which is 1. The so-called resolvent Γ is defined by 1 + Γ = (1-λK)-1 so we can equate the resolvent with the formal series solution Γ = Σm=1λmKm. One can then write various integral equations which involve the resolvent, such as Γ = λK + λKΓ = λK + λΓK . And our solution can then be written φ = f + Γf, so finding the resolvent solves the problem. I will now outline each of the sections below, by number. 1. In the first section below, I define certain determinants such as det456 which are "diagonal" subdeterminants of a larger matrix. I use the word diagonal for this reason: if you assemble the 3x3 array of numbers of which det456 is the determinant, you will find that all the diagonal elements of your array came from the diagonal of the original full matrix. So these subdeterminants like det456 are a certain small subset of all possible subdeterminants of a matrix. They are those that you get by crossing out the same set of rows and columns (in this case 1,2 and 3). It turns out that these are the only subdeterminants we have to worry about to derive the expansion formula for the Fredholm determinant. A general diagonal subdeterminant of a matrix can be written in the following somewhat obscure manner: detI1 I2 I3...In = εi1 i2 i3....in AI1 i1 AI2 i2 ..... AIn in // n factors of A where i1 i2 i3....in {I1,I2,I3...In} , εi1 i2 i3....in anti-symmetric, εI1 I2 I3....In = +1 where the matrix A is NxN and 2 ≤ n ≤ N, and where all the indices i1 through in are implicitly summed. 2. Using this general formula, and setting A = 1+B, I am able to show that: det(1+B) = 1 + tr(B) + (1/2!) Σij detij + (1/3!) Σijk detijk + (1/4!) Σijkl detijkl + ... + det(B) which is basically the Fredholm determinant expansion. For finite N, starting with the third term the dimensionality of the subdeterminants increases by 1 with each term until it reaches the full N. In the case N=5, all terms are displayed above and we see there are 6 terms, so in general there are N+1 terms in the expansion, where the first is 1 and the last is det(B). Each summation index has full range 1 to N, and of course any detij...object is zero if any two subscripts are the same, as follows from the ε expression above. The factorials appearing in the expansion are overcounting corrections which allow free range of the summation indices. An alternate formula can be written where there is no overcounting: det(1+B) = 1 + tr(B) + Σi<j detij + Σi<j<k detijk + Σi<j<k<l detijkl + ... + det(B) 3. For application with Fredholm integral equations, as was just shown above, we want B = -λK. With this replacement, the Fredholm determinant expansion takes on a more familiar form, det(1-λK) = 1– λtr(K) + (λ2/2) Σij detij - (λ3/3!) Σijk detijk+ (λ4/4!) Σijkl detijkl + ... + (-λ)ndet(K) where the various detijk.. objects are subdeterminants of the matrix K. Now we see that the expansion is in fact a power series in λ, which is the main idea since λ is a smallness parameter. Also, we pick up alternating signs. The above is often presented in a more graphical but equivalent notation det(1-λK) = 1 - λi Kii + (λ2/2)ij - (λ3/3!)ijk + ... where now you "see" the various subdeterminants involved. To take the limit N→∞, we make these changes: (1) replace things like Kxy by k(x,y). (3) replace summation indices with integrations. When this is done, the Fredholm Determinant takes on its traditional appearance which is this: det(1-λK) = 1 - λ ∫dx k(x,x) + (λ2/2) ∫dx dy - (λ3/3!) ∫dxdydz + .... (goes on forever) Although the series never ends, for small λ we might do well keeping only the first few terms in the series if we are doing approximation work. 4. I derived a formula for Fredholm's First Minor in a separate document. Although I believe this proof (it is based on 1970's notes I had, probably from the Mikhlin book), I don't much like the proof, and I wanted to find a derivation similar to the one above for Fredholm's Determinant. However, in this case I need general formulas for a different class of subdeterminants: those obtained from a full matrix by crossing out row p and column q. Unless p = q, these subdeterminants are not "diagonal ones" as described above, so a new formula was needed. It turns out that the formula for a general subdeterminant can be written in a form very similar to that shown above for a diagonal one. Here is the general result: detI1 I2 I3...In; a1 a2 a3...an; = εi1 i2 i3....in AI1 i1 AI2 i2 ..... AIn in // n factors of A where i1 i2 i3....in {a1 a2 a3...an} , εi1 i2 i3....in anti-symmetric, εI1 I2 I3....In = +1 The new feature is that we now have two groups of integer labels on our det object. The first controls the row indices of the A factors, while the second group affects the column indices of the A factors. For a diagonal subdeterminant, these two sets of integer labels are the same, so det456 = det456;456 in our fancier more general notation. In our new more general case, the first index set denotes the labels of the rows of the original matrix which are contributing to our submatrix, and the second set denotes the labels of the columns which are contributing to our submatrix. Indices not shown in these sets represent rows and columns that are "crossed out" of the full matrix to expose the elements of the subdeterminant. As a special case of the above general formula, we can set n = N-1 and think about the subdeterminant obtained. Unfortunately, I have not been able to find an elegant way to express the det objects for this special case, other than the above formula. For example, if N = 4 and we cross out row p=1 and column q=3, we get det234;124 = A21A32A44 + signed permutations = det ; where "signed permutations" means other terms obtained by permuting the second (column) indices on the A factors, such that an odd number of index swaps implies a minus sign, otherwise a plus sign. For example det234;124 = A21A32A44 – A21A34A42 + other signed permutations = det ; where I show a permutation obtained by swapping one pair of indices. This is just the ε object doing its work, but I am suppressing the messy ε factor by just saying " + signed permutations". 5. I then consider the case N = 5, set B = 1+A as with the determinant calculation, so no λ floating around here. I consider that we cross out row p and column q, so we are faced with a 4x4 subdeterminant. I write this first as detI1,I2,I3,I4 ; a1,a2,a3,a4 = (δI1,a1 + BI1,a1) (δI2,a2 + BI2,a2) (δI3,a3 + BI3,a3) (δI4,a4 + BI4,a4) + signed perms I then analyze those terms with no B factors, and then those terms with one B factor. These cases were plenty hard to do, and I did not go on to do the more general cases of three or more B factors. I found these results for these first two cases (after much work) : no B factors: δp,q one B factor: tr(B) δp,q – (-1)p+q Bq,p 6. I then defined a new matrix N = det(1+B)Γ where Γ is the resolvent discussed earlier. Using our usual method of simply "inverting a matrix", one finds that Npq = – det(1+B)δp,q + (-1)p+q minor [(1+B)qp] and one sees here the need for our special subdeterminant with a row and column crossed out, since that is what the word "minor" means. The indices are swapped since the matrix inversion formula involves the transpose of the cofactor matrix, as outlined above. If we insert into this formula our Fredholm Determinant results for det(1+B) through first order in B, we get Npq = – [ 1 + tr(B) ] δp,q + (-1)p+q minor [(1+B)qp] // through order B1 If we then insert our results above which say (no B and single B terms only) minor [(1+B)qp] = δp,q + tr(B) δp,q – (-1)p+q Bp,q we then find that Npq = – Bpq The good news is that this is in fact correct for the B0 and B1 terms of Npq. If we set B = -K, the correct formula for N is given by (this is the Fredholm First Minor expansio∂n) Nij= Kij – Σk + (1/2!) Σkm – (1/3!) Σkmn +. Although I did what seemed "a lot of work", all I got was the first term showing in this expansion! 7. I took a quick look at that B2 terms, and then "gave up" on this approach, at least in the way I am doing it right now. I feel there is some trick that I am missing, but I do feel that my approach using direct formulas for the "minor" type subdeterminants would completely validate the above formula, if I had enough time and energy to follow through. ______________________________________________________________________ 1. Preliminary Facts concerning the Determinants of Certain Submatrices of a Matrix. (0) The determinant of a 3x3 matrix can be written in this way det = εijk A1iA2jA3k = A11A22A33 + 5 other terms We know there are 6 terms in all because εijk has 3*2*1 = 6 non-zero values such as ε123 = 1. Remember that ε is a totally antisymmetric tensor and so changes sign when any index pair is swapped and of course therefore must be 0 if any two indices are the same. The other terms can be generated "one at a time" by just swapping alternating pairs of second indices on the previous term, and adding a minus sign which that swap creates due to the ε factor. Thus = A11A22A33 - A12A21A33 + A12A23A31 - A13A22A31 + A13A21A32 - A11A23A32 where the underlined indices show the swap we are about to do to make the term to the right. A 4x4 determinant of course has 4*3*2*1 = 24 terms and the terms could be "generated" in a similar (though not identical) fashion. Basically you just set the set of second indices to all possible permutations of 123...n , and if it took an odd number of swaps to reach a certain permutation, that term gets a minus sign. Now, let A be a 6x6 matrix. Later we shall generalize to an N x N matrix. (1) Consider: det456 = ε123ijk A4iA5jA6k = A44A55A66 + other terms This is the determinant of the 3x3 matrix obtained by crossing our rows and columns 1,2,3 of the matrix A. This fact should be clear based on our item (0) above -- the only difference is that the indices now take values4,5,6 instead of 1,2,3. Notice that the indices 456 on det456 are in the "natural order" corresponding to the natural order in which the rows and columns of A are numbered, 1,2,3,4,5,6. That is to say, 4,5,6 are in increasing order. We now want to have det546 be the determinant of this same 3x3 matrix but which has rows and columns 4 and 5 swapped. We know that if we swap a pair of rows AND a pair of columns, the determinant does not change, so we know that det546 = det456 . The question is how to write det546 using the ε symbol. Here is the answer det546 = ε123ijk A5jA4iA6k = A55A44A66 + other terms The rule is that we start with det456 and make the change A4iA5j → A5jA4i, as I shall prove immediately below. But we see that this is really "no change" so det546 = det456 as expected. Proof: Here is my proof that this is the right result for swapping both rows and columns 4 and 5. This proof is indirect, there is probably a faster way to do it, but at least this works. Start with det456 and make the change A4iA5j → A5iA4j . This accomplishes the swapping of rows 4 and 5 (row index = first index on A). We then have this intermediate result, having started with det456, det456 = ε123ijk A4iA5jA6k → ε123ijk A5iA4jA6k We then rewrite this intermediate result using the fact that the determinant of a matrix is the same as that of its transpose, so we get ε123ijk A5iA4jA6k = ε123ijk Ai5Aj4Ak6 Now we swap columns 5 and 4 (second indices on A) so make the change Ai5Aj4 → Ai4Aj5 . So now after doing both the row and column swaps we have ε123ijk Ai5Aj4Ak6 → ε123ijk Ai4Aj5Ak6 Finally, we use our transpose rule a second time to write this last as ε123ijk Ai4Aj5Ak6 = ε123ijk A4iA5jA6k So far, then, we have det456 = ε123ijk A4iA5jA6k → (swap row and cols 4,5) → ε123ijk A4iA5jA6k Notice that we never mess with the ε indices in all the steps. Rewrite one more time: det546 = ε123ijk A4iA5jA6k = ε123ijk A5jA4iA6k = as claimed above, QED. (2) In general, any permutation of the indices on det456 yields the same result. We can think of a general permutation being accomplished by a sequence of row/column swaps of the type shown in (1). Thus we know that d456 = det465 = det546 = det564= det645 = det654 (3) We now seek a way to write a general formula for dIJK . Let's start going back to d456: det456 = ε123ijk A4iA5jA6k = A44A55A66 + other terms This full notation is a little clumsy, because we have to put the ε indices in the right place, for example, det136 = εi2j45k A4iA5jA6k = A11A33A66 + other terms So let's rewrite det456 in a slightly different notation: det456 = εijk A4iA5jA6k where ijk {4,5,6}, εijk = anti-symmetric, and ε456 = +1. Notice that εijk is basically "the usual" rank 3 totally antisymmetric tensor, but its indices take values 4,5,6 instead of 1,2,3 AND we are going to set a certain specific element of this tensor to +1 and this element may not be the one you think it should be. All other elements can be obtained from the one we set by doing permutations with minus signs as appropriate. Here then is our claimed generalized formula: detIJK = εijk AIiAJjAKk where ijk {I,J,K}, εijk = anti-symmetric, and εIJK = +1. For example, we get (in our first example, our set value is ε546 = 1 ) det546 = εijk A5iA4jA6k = ε546 A55A44A66 + .... = A55A44A66 + ... det215 = εijk A2iA1jA5k = ε215 A22A11A55 + .... = A22A11A55 + ... There are 6 terms in one of these 3x3 determinants, and it is our goal to avoid having to write them out! That is why we are using this slightly strange compressed ε notation. It is clumsy to write out the 6 terms in a linear fashion, and it is also clumsy to write it out as a 3x3 matrix surrounded by vertical lines, but we could of course write things out in either of these forms it we really wanted to, for example detijk = = AiiAjjAkk + other terms (4) Let's go back to our general formula above, det456 = ε123ijk A4iA5jA6k where ε here is the true antisymmetric tensor in our 6D space. If the matrix A has the same numbers in rows 4 and 5, the RHS above = 0 because ε is antisymmetric on ij while A4iA5j = A4iA4j is symmetric on ij. Let's define this as the meaning of det446 . This is still the determinant of a 3x3 matrix, but two rows are the same, so we know det446= 0. Similarly, we will define detIJK = 0 if any of the two indices are the same. We can just define this to be true with no justification needed, or we can think of it in terms of two rows having the same numbers. And of course it would also be zero if all three indices are the same. (5) Because all the permutations are the same, we can write det456 = 1/6 ΣIJK{4,5,6} detIJK and 1/6 = 1/3! We have in mind of course that we are summing over all possible sets IJK where IJK are different integers taken from the set {4,5,6} . But we can also think of this as including all choices IJK {4,5,6} since we know that detIJK = 0 when any two indices are the same. (6) Suppose we were to encounter the following sum: d123 + d124 + d125 + d126 + d134+ d135 + d136 + d145 + d146 + d156 + d234 + d235 + d236 + d245+ d246 + d256 + d345 + d346 + d356 + d456 Here we include all sets of indices which are in natural order. We could write this sum as Σi<j<k dijk We know that there are 3! = 6 permutation of each term in the sum. Thus, we could write this as Σi<j<k dijk = 1/3! Σijk dijk where we can regard the Σijk as being over all sets of ijk that are different integers, or as being over all sets of ijk whether or not two indices are the same, since we know that dijk = 0 when two indices are the same. (7) In sections 1-6 above, we focused on the determinants of 3x3 submatrices of a matrix A. In similar fashion, we could think about 2x2 submatrices. Here are all the above results cast into this form: [1] det45 = ε123ij6 A4iA5j = A44A55 - A45A54 det54 = ε123ij6 A5jA4i = A55A44 - A54A45 [2] det45 = det54 // only 2 = 2! terms [3] detIJ = εij AIiAJj where ij {I,J}, εij = anti-symmetric, and εIJ = +1. = εIJ AIIAJJ + ... = AIIAJJ + ... // = AIIAJJ - AIJAJI in this simple case det13 = εij A1iA3j = A11A33 - A13A31 detij = [4] det44 = 0 and in general detII = 0, two rows equal numbers [5] det45 = 1/2 ΣIJ{4,5} detIJ and 1/2 = 1/2! [6] Σi<j dij = 1/2! Σij dij (8) Now let's do it all again for 4x4 submatrices: [1] det3456 = ε12ijkl A3iA4jA5kA6l = A33A44A55A66 + 23 other terms = det of 4 x 4 submatrix obtained by crossing out row/columns 1 and 2 det4356 = ε12ijkl A4jA3iA5kA6l = A44A33A55A66 + 23 other terms [2] det3456 = 23 other permutations, all are equal, total permutations = 4*3*2*1 = 24 = 4! [3] detIJKL = εijkl AIiAJjAKkALl where ijkl {I,J,K.L}, εijkl = anti-symmetric, and εIJKL = +1. = εIJKL AIIAJJAKK ALL + .... = AIIAJJAKK ALL + .... det1256 = εijkl A1iA2jA5k A6l = ε1256 A11A22A55 A66 + ... = A11A22A55 A66 + ... [4] det1156 = 0, etc for any two indices the same, two rows of 4x4 have same numbers [5] det3456 = 1/24 ΣIJKL{3,4,5,6} detIJKL and 1/24 = 1/4! [6] Σi<j<k<l dijkl = 1/4! Σijkl dijkl (9) Now we shall try to express our main results for general n x n submatrix determinants, now that we have studied the cases n = 3,2 and 4. We do this using the above notation which conceals the fact that we started with A = 6 x 6 matrix. These results apply to A = N x N matrix, as long as n ≤ N. [2] detI1 I2 I3...In has n! permutations, all of which are equal and which can be interpreted in terms of sets of row/column swaps. This thing is the determinant of an n x n submatrix of A where we have crossed out the rows and columns corresponding to the integers which do not appear in I1 I2 I3...In. [3] detI1 I2 I3...In = εi1 i2 i3....in AI1 i1 AI2 i2 ..... AIn in // n factors of A where i1 i2 i3....in {I1,I2,I3...In} , εi1 i2 i3....in anti-symmetric, εI1 I2 I3....In = +1 detI1 I2 I3...In = AI1,I1 AI2,I2 ..... AIn,In + other terms found from above ε notation [4] detI1 I2 I3...In = 0 if any two (or more) indices are the same [6] Σi1<i2<i3 ....<in deti1 i2 i3...in = (1/n!) Σi1,i2,i3... in deti1 i2 i3...in 2. Application: computation of det(1+A) (1) Consider the general determinant formula applied to a matrix 1+A in the case n = 4: det(1+A) = (1+A)1i(1+A)2j(1+A)3k(1+A)4l εijkl // implied sum on all letter indices I give some proofs of this general formula for a determinant in my notes, so let's accept it for now [ We have been using the formula all along. ] Write out the above as: (δ1i + A1i) (δ2j + A2j) (δ3k + A3k) (δ4l + A4l ) εijkl (2) If we take the left side of each factor we get just ε1234 = 1, so det(1+A) = 1 + other stuff. (3) If we take the first three δ's and the last A4l we get A4l ε123l = A44. There are three similar terms and when we add them we get tr(A). Thus, we know that det(1+A) = 1 + tr(A) + other stuff. (4) Now let's select all terms which have two factors of A. There are (4,2) = 6 ways to do this. Each of these 6 ways has the first indices of the A factors in the "natural" increasing order. For example, if we take the last two A factors, the term is δ1i δ2j A3k A4l εijkl = A3k A4l ε12kl = A33A44 - A34A43 This is exactly the object det34 discussed in section (7) of our Preliminaries above. In the det(1+A) above we are obtaining all 6 terms detij where i,j are in natural increasing order. Thus, we can say the two-A terms = Σi<j detij = (1/2!) Σij detij where on the right we use our result [6] to rewrite as a general sum including i = j values, where 2! is our overcount correction factor since each term has 2! permutations. (5) Now let's select all terms which have three factors of A. There are (4,3) = 4 ways to do this. Each of these 4 ways has the first indices of the A factors in the "natural" increasing order. For example, if we take the last three A factors, the term is δ1i A2j A3k A4l εijkl = A2j A3k A4l ε1jkl = A22A33A44 + other terms This is exactly the object det234 discussed in our Preliminaries above. We get all the natural order terms, so we have: the three-A terms = Σi<j<k detijk = (1/3!) Σijk detijk (6) We can see now the pattern. The next group will be the four-A terms = Σi<j<k<l detijkl = (1/4!) Σijkl detijkl If we started with A = 4x4 matrix, then we can simplify this "last term" because there is only one 4x4 submatrix of A, so all 24 of the detijkl are exactly the same and are just detA. Another way to say this is that Σi<j<k<l detijkl contains just one term which is det1234 = det(A). (7) Here then is the general result for an NxN matrix A: det(1+A) = 1 + tr(A) + (1/2!) Σij detij + (1/3!) Σijk detijk+ (1/4!) Σijkl detijkl + ... + det(A) where detI1 I2 I3...In = εi1 i2 i3....in AI1 i1 AI2 i2 ..... AIn in so that in particular detIJ = εij AIiAJj where ij {I,J}, εij = anti-symmetric, and εIJ = +1. detIJK = εijk AIiAJjAKk where ijk {I,J,K}, εijk = anti-symmetric, and εIJK = +1. detIJKL = εijkl AIiAJjAKk ALl where ijkl {I,J,K.L}, εijkl = anti-symmetric, and εIJKL = +1. etc Notice what is appearing in the above expression: tr(A) is the sum of all diagonal elements, or the sum of all 1x1 diagonal subdeterminants Σij detij is the sum of all 2x2 diagonal subdeterminants Σijk detijk is the sum of all 3x3 diagonal subdeterminants and so on The following more graphical formula states the same result: det(1+A) = 1 + i Aii + ij + ijk + ... + det(A) Another way to write our expansion would be this: det(1+A) = 1 + tr(A) + Σi<j detij + Σi<j<k detijk + Σi<j<k<l detijkl + ... + det(A) where now we don't do the overcounting. (8) If A has elements all >> 1, then the last term is the dominant term and det(1+A) ≈ det(A). (9) If A is a "small matrix" we might write it as A→ λA. Then our theorem is: det(1+λA) = 1 + λtr(A) + (λ2/2) Σij detij + (λ3/3!) Σijk detijk+ (λ4/4!) Σijkl detijkl + ... + λndet(A) where we now have a power series expansion for det(1+λA) and we might keep just the first few terms in the expansion if we are doing some kind of approximation work. If we keep only the trace term, we are then making "the trace approximation". (10) Suppose Aij = aibj, a matrix which is "factorizable". Then detI1 I2 I3...In = εi1 i2 i3....in AI1 i1 AI2 i2 ..... AIn in = εi1 i2 i3....in aI1 bi1 aI2 bi2 ..... aIn bin = aI1 aI2.... aIn * εi1 i2 i3....in bi1 bi2.... bin For n ≥ 2, we have a contraction between an antisymmetric and a symmetric tensor, so det = 0. Thus, our entire result boils down to this: det(1+λA) = 1 + λtr(A) where we can say tr(A) = ΣiAii = Σi aibi . So in the factorizable case, the trace approximation is exact. (11) There is yet another way to write our expansion. We know this general matrix theorem: det(M) = etr[ ln(M) ] which is really just det(eA) = etr(A) with A = ln(M) Now set M = 1+A. We then have det(1+A) = exp [ tr( { ln(1+A)} ] But we know that ln(1+A) = A - A2/2 + A3/3 - ... = - Σn=1∞ (-1)nAn/n Then we can say tr( { ln(1+A)} = - tr { Σn=1∞ (-1)nAn/n } = - Σn=1∞ (-1)n/n * tr(An) Then we have shown that det(1+A) = exp [- Σn=1∞ (-1)n/n * tr(An)] or det(1-A) = exp [- Σn=1∞ tr(An)/n] or det(1-λA) = exp [- Σn=1∞ λn tr(An)/n] which agrees with page 16 of my Mikhlin notes. 3. Application: The Fredholm Determinant (1) Consider the eigenvalue problem: Au = μu or λAu = u where λ = 1/u. We can write this last as: (λA-1)u = 0 or (1-λA)u = 0. This eigenvalue equation will have solutions u only if det(1-λA) = 0. But we have an expansion for this determinant from above det(1-λA) = 1– λtr(A) + (λ2/2) Σij detij - (λ3/3!) Σijk detijk+ (λ4/4!) Σijkl detijkl + ... + (-λ)ndet(A) where now we have a minus sign so we just take λ → -λ in our previous λ result. (2) If A is a finite dimensional matrix of degree N, then we have written det(1-λA) as a polynomial of degree N in λ. If A is symmetric, this is going to have N real roots, and these are going to be the eigenvalues of our eigenvalue equation. (3) In general you would never write this polynomial using the formula above unless perhaps N was somewhat large and λ was small, so you would keep only the first several terms and in this way you would find the "dominant" eigenvalues. (4) If we are dealing with an integral equation so A = K is an integral operator with kernel k(x,y), we can approximate the equation on an N x N lattice and then the secular equation det(1-λK) = 0 would give us an approximation to the some of the eigenvalues. I suspect these might be the smallest eigenvalues in λ and the largest in μ, but I am not sure, have to reread my Stak notes on this. (5) We can also consider the above expansion in the limit that n→∞. The first two terms would be: det(1-λK) = det(1 - λkx,y ) = 1 – λ tr(k) = 1 - λ ∫dx k(x,x) Notice that for λ = 0, you would say det(1) = 1 = the product of an infinite number of 1's on the diagonal of an ∞ x ∞ matrix. (6) What would the next term be? detIJ = εij AIiAJj where ij {I,J}, εij = anti-symmetric, and εIJ = +1. = AIIAJJ - AIJAJI +(λ2/2) Σij detij = (λ2/2) ∫dx dy [k(x,x) k(y,y) - k(x,y) k(y,x) ] (*) = (λ2/2) ∫dx dy and I think the higher terms generalize similarly. Of course in this case the series goes on forever, but for small λ or perhaps small λ ||K|| we could keep the first few terms as an approximation, but this is a different kind of approximation from the lattice one described above. Again, we set det(1-λK)= 0 to find some eigenvalues λ. (7) Expressing Fredholm terms in terms of traces. Let's rewrite the above again for n = N: (λ2/2) ΣIJ detIJ = (λ2/2) ΣIJ { εij AIiAJj } = (λ2/2) ΣIJ { AIIAJJ - AIJAJI} = (λ2/2) { ΣIAII ΣJAJJ - ΣIJ AIJAJI } = (λ2/2) { [ tr(A)]2 - tr(A2) } = (λ2/2) { T12- T2 } so this replicates our n=∞ result (*) above, where we define Tn = tr(An). Let's try the next term in terms of traces: - (λ3/3!) ΣIJK detIJK = - (λ3/3!) ΣIJK{ εijk AIiAJjAKk } We have no choice at this point but to write out all the terms: εijk AIiAJjAKk = AIIAJJAKK – AIJAJIAKK + AIJAJKAKI - AIKAJJAKI + AIKAJIAKJ - AIIAJKAKJ where I exchange one pair for each term. Now we can look at our sum ΣIJK { AIIAJJAKK – AIJAJIAKK + AIJAJKAKI - AIKAJJAKI + AIKAJIAKJ - AIIAJKAKJ} = [ tr(A)]3 – tr(A) tr(A2) + tr(A3) - tr(A)tr(A2) + tr(A3) - tr(A) tr(A2) = [ tr(A)]3 - 3 tr(A) tr(A2) + 2 tr(A3) So our term is then - (λ3/3!) { [ tr(A)]3 - 3 tr(A) tr(A2) + 2 tr(A3)} = (λ3/3!) { (∫dx k(x,x))3 - 3 ∫dx k(x,x) ∫dy k2(y,y) + 2 ∫dx k3(x,x) } = (λ3/3!) { (∫dxdydz k(x,x) k(y,y) k(z,z) - 3 ∫dxdydz k(x,x) k(x,y) k(y,z) + 2 ∫dxdydz k(x,y) k(y,z) k(z,x) = (λ3/3!) ∫dxdydz { k(x,x) k(y,y) k(z,z) - 3 k(x,x) k(x,y) k(y,z) + 2 k(x,y) k(y,z) k(z,x)} and of course this is going to have the expected form in terms of |...|. That is to say, start over and say - (λ3/3!) Σijk and then replace Σi with ∫dx and so on, so we get - (λ3/3!) ∫dxdydz But ultimately, the simplest way to write this thing is: - (λ3/3!) { [ tr(A)]3 - 3 tr(A) tr(A2) + 2 tr(A3)} - (λ3/3!) { T13 - 3 T1T2 + 2T3} where Tn = tr(Tn) = tr(TT..T) = ∫dx kn(x,x). (8) It is in this n→∞ situation that we refer to det(1-λK) as The Fredholm Determinant. (9) Compare results of (7) to the trace expansion (11) below. Let's now examine our alternate form given in (11) above, det(1-λA) = exp [- Σn=1∞ λn tr(An)/n] For an NxN matrix A we can expand e-x = 1 - x + x2/2! - x3/3! + .. = Σi=0∞ (-x)i/i ! Thus we get det(1-λA) = exp [- Σn=1∞ λn tr(An)/n] x = Σn=1∞ λn tr(An)/n = Σi=0∞ (-{Σn=1∞ λn tr(An)/n } )i/i! = Σi=0∞ (-1)i/ i! * { Σn=1∞ λn tr(An)/n}i I can only keep the first few terms in the n sum to see what is going on here, = Σi=0∞ (-1)i/ i! * {λtr(A) + λ2tr(A2)/2 + λ3tr(A3)/3 + ...}i Let's define these traces such that T3 = tr(A3) etc, so we then have = Σi=0∞ (-1)i/ i! * {λT1 + λ2T2/2 + λ3T3/3 + ...}i = 1 - {λT1 + λ2T2/2 + λ3T3/3 + ...} + {λT1 + λ2T2/2 + λ3T3/3 + ...}2/2! - {λT1 + λ2T2/2 + λ3T3/3 + ...}3/3! + ... We can now gather up terms of a given power of λ, where λ0 term is just the 1 and the λ1 term is -λT1. Then λ2: - λ2T2/2 + λ2T12/2 = (λ2/2) ( T12 – T2) // agrees with above λ3: - λ3T3/3 + λ3(T1T2)/2 - λ3T13/3! = -(λ3/3!) { T13 - 3T1T2 + 2 T3 } which also agrees with our result above. So we have shown directly through order λ3 that our two expansions give the same result. (10) So we can now summarize the above several results: det(1-λK) = 1 - λT1 + (λ2/2) { T12- T2 } – (λ3/3!) { T13 - 3T1T2 + 2 T3 } + O(λ4) Tn = tr(Kn) = tr(KKK..K) = ∫dx kn(x,x) = ∫dx1dx2dx3....dxn . k(x1,x2) k(x2,x3) k(x3,x4).... k(xn-1,xn) det(1-λK) = 1 - λ ∫dx k(x,x) + (λ2/2) ∫dx dy - (λ3/3!) ∫dxdydz + .... (goes on forever) det(1-λA) = 1 - i Aii + ij - ijk + ... + (-λ)n det(A) = 1 - λT1 + (λ2/2) { T12- T2 } – (λ3/3!) { T13 - 3T1T2 + 2 T3 } + O(λ4) where Tn = tr(An) det(1-λA) = exp [- Σn=1∞ λn tr(An)/n] = etr[ ln(1-λA)] = Σi=0∞ (-1)i/ i! * { Σn=1∞ λn tr(An)/n}i (11) It is not easy to find these results on the web, but I have found them in a google scanned book which is this one: http://books.google.com/books?id=WHjO9K6xEm4C and the result is presented in this way: where my result is the first object above so that det(1-λK) = D(λ). The other object will be commented on soon. 4. Preliminary Facts concerning the Determinants of Certain Other Submatrices of a Matrix. (1) In our above preliminary notes, we learned how to write formulas for submatrices of a matrix which were centered on the diagonal of the main matrix, for example det456 = ε123ijk A4iA5jA6k = A44A55A66 + other terms det456 = εijk A4iA5jA6k where ijk {4,5,6}, εijk = anti-symmetric, and ε456 = +1. These were the only "objects" we needed to develop the results for the Fredholm Determinant. (2) But how would we write a formula for an arbitrary submatrix of a matrix? Theorem: Start with some NxN matrix A. We are interested in the determinant of an arbitrary nxn submatrix where n ≤ M. Let the set of n increasing integers IJK.... label the rows of the full matrix which are part of the submatrix. And let the set of n increasing numbers abc... label the columns of the full matrix which are part of the submatrix. Then the subdeterminant is given by: detIJK.. ; abc... = εijk.. AIiAJjAKk ... where ijk... {abc...} and εabc.. = 1. Using the transpose determinant formula, we could also write this as detIJK.. ; abc... = εijk.. AiaAjbAkc ... where ijk... {IJK...} and εIJK.. = 1. Proof: The only real issue here is whether we have the sign right. We know that we are correctly accessing the right rows and columns. We imagine that we "assemble" the n2 numbers of interest into a contiguous n x n matrix without changing ordering of columns or rows, and it is the determinant of that contiguous "ancillary" matrix that we are interested in. By requiring that our two sets of indices be in "natural order", we make sure that no such shuffles of columns or rows occur. The first term in our result is always going to be this, with a plus sign: + AIaAJbAKc ... so I think our theorem is basically proved. (3) Let's try a fancier way to write the first result above: detI1,I2,I3..In ; a1,a2,a3...an = εi1 i2..in AI1,i1AI2,i2AI3,i3 ... AIn,in where i1,i2,...in... { a1,a2,a3...an} and εa1 a2 a3..an = 1. Thus we can say this: detI1,I2,I3..In ; a1,a2,a3...an = AI1,a1AI2,a2AI3,a3 ... AIn,an + signed permutations of pairs of second indices and we could write AI1,i1AI2,i2AI3,i3 ... AIn,in = !Syntax Error, I AIs,is The full NxN determinant would be written: |A| = det1,2,3..N ; 1,2,3..N = Σi1,i2..iN εi1 i2..iN A1,i1A2,i2A3,i3 ... AN,iN = Σi1,i2..iN {1,2,3...N } εi1 i2..iN !Syntax Error, IAs,is Here there are N indices on the ε tensor, and there are N factors of A. (4) Let's try to make hay with this non-ε version of our formula: detI1,I2,I3..In ; a1,a2,a3...an = AI1,a1AI2,a2AI3,a3 ... AIn,an + signed permutations of pairs of second indices and we could write AI1,a1AI2,a2AI3,a3 ... AIn,an = !Syntax Error, I AIs,as Then we have for our arbitrary nxn subdeterminant of an NxN matrix A: detI1,I2,I3..In ; a1,a2,a3...an = !Syntax Error, I AIs,as + signed permutations Now let's knock out row p and column q and have n = N-1 as a special case. Let's just start with an example where N = 4 and we knock out p = 1 and q = 3. Then the above reads det234;124 = A21A32A44 + permutations = det ; The full 4x4 determinant would be this det1234;1234 = A11A22A33A44 + permutations So here is my basic problem: I don't know a good algebraic notational way to express the RHS of this equation: det ; = A21A32A44 + perms How do I write the RHS in terms of the numbers "1" and "3" ? 5. Figuring out the B0 and B1 terms of the Fredholm First Minor expansion Let's go with the case N=5. Let's take as our general starting point this: detI1,I2,I3..In ; a1,a2,a3...an = AI1,a1AI2,a2AI3,a3 ... AIn,an + perms where we are dealing with an n x n submatrix. I am still interested in the subdeterminant you get when you cross out row,column = p,q. So setting N = 5, that will be a 4x4 subdeterminant. Write detI1,I2,I3,I4 ; a1,a2,a3,a4 = AI1,a1AI2,a2AI3,a3 AI4,a4 + perms where I1,I2,I3,I4 is the set of integers 1,2,3,4,5 with one integer "p" knocked out. where a1,a2,a3,a4 is the set of integers 1,2,3,4,5 with one integer "q" knocked out. Now right away, let's go to our A = 1+B form. We avoid the extra baggage of λ and minus signs for the moment, later we can replace K → -λK. So the above becomes detI1,I2,I3,I4 ; a1,a2,a3,a4 = (δI1,a1 + BI1,a1) (δI2,a2 + BI2,a2) (δI3,a3 + BI3,a3) (δI4,a4 + BI4,a4) + perms Some comments in passing: (1) this has the same number of factors as my N=4 fredholm det situation, that is why I chose N=5 to start with; (2) using the +perms avoids the problem of figuring out what the "sign" is. (3) Remember that in the first term shown above, both index sets are in "natural order". (4) the notation + perms means that you permute the ai index set and add signs as appropriate. But in all of these terms the Ii indices will remain in natural order. Let's study the B0 terms. We write: (δI1,a1) (δI2,a2) (δI3,a3) (δI4,a4) + perms ai If p ≠ q, then the two integer sets Ii and ai are different. The first term shown here must be 0 because there is no way that we can make every δ = 1. And this is true for all the perms as well. If you have two sets of integers like boys = 1234 and girls = 2345 there is no way you can pair the boys and girls into four couples such that all four couples have the same number labels. You could do 22,33,44 but then you are left with 15. That is the best you could do in this little example, and of course δ15 = 0 so the product of the four deltas = 0. If p = q, then the two sets are the same, and there will be one arrangement of couples that gives each δ = 1. For example, if boys = 2345 and girls = 2345 then the pairs would be 22,33,44,55 = 1. Notice that this is in fact the term shown above, not one of the permutations, since 2345 are in natural order for both boys and girls, so we don't have to worry about any permutation sign here. Therefore we find this very nice and simple result (δI1,a1) (δI2,a2) (δI3,a3) (δI4,a4) + perms ai = δp,q I know this is going to be the 1 matrix of 1 + Γ, somehow. Let's study the B1 terms. We start again with this: (δI1,a1 + BI1,a1) (δI2,a2 + BI2,a2) (δI3,a3 + BI3,a3) (δI4,a4 + BI4,a4) + perms There are four terms (ignoring perms). The one of these terms with the last B is this: (δI1,a1) (δI2,a2) (δI3,a3) (BI4,a4) + perms ai but let's think of all four terms plus their permutations in the following discussion. If p ≠ q, reconsider (p=5) boys = 1234 and (q=1) girls = 2345 . In all our terms and their permutations, there is only one surviving term here, the one where the three δ's are 22,33,44. In this example, the B would be B1,5 because that is the left-over pairing. Remember that the first index has to be a boy and the second a girl. That is why it is B1,5 and not the other way around. This term would not be in the term shown above, but in the term where the B is all the way to the left. It seems pretty clear that the general result here is this: if p ≠ q, only one term survives and it is Bq,p . But now we have a permutation sign to think about. Our example is really B1,5 δ2,2 δ3,3 δ4,4. We put the B first because I1 = 1 and that is where it goes. The Ii indices are in natural order 1234, but the ai are not! How many pair swaps does it take to get from the order 5234 to order 2345? Three. But how would we express this sign in terms of integers p and q? That is the big problem! We can restate this problem in terms of (p=5) boys = 1234 and (q=1) girls = 2345. The question is: how many swaps to move the 1 of the boys group to the position of the 5 in the girls group. At this point in my writeup, we need to insert a "subroutine". I will make a section below to solve this problem for general p and q, and then I will report back that result here. I do this because that subroutine is probably going to be large. Result: The sign change produced by the pairwise swaps required to move the q of the boys group to the p position in the girls group = (-1)p+q+1 . If p = q, reconsider boys = 2345 and girls = 2345 with p = q = 1. Now there are four different terms contributing, not just one. First, we get 22,33,44 and B55. Second we get 22,33,B44,55. Of course permutations like 22,33,45,B54 = 0, so lots of the perms are 0. The answer in this case is pretty clear. We get B22 + B33 + B44+ B55 which we could write as Σi≠p Bii = tr(B) - Bpp. In all four terms, both index sets are in normal order, so there is no "permutation sign" to worry about. So here is our result for the B1 terms: B1 terms = {Bq,p signp,q}(1 - δp,q) + { tr(B) - Bpp}δp,q B1 terms = {Bq,p (-1)p+q+1}(1 - δp,q) + { tr(B) - Bpp}δp,q = Bq,p (-1)p+q+1 + δp,q { tr(B) - Bpp - Bp,p (-1)p+p+1} = Bq,p (-1)p+q+1 + δp,q { tr(B) - Bpp - Bp,p (-1)} = Bq,p (-1)p+q+1 + δp,q { tr(B) } = tr(B) δp,q – (-1)p+q Bq,p 6. Show that these B0 and B1 terms agree with our full Fredholm First Minor Result From our separate document with λ = 1 and B = -K, we got this result: Γ = N/ |1-K| Nij = Kij – Σk + (1/2!) Σkm – (1/3!) Σkmn But we also know this fact: (where I set λ = 1) 1 + Γ = (1-K)-1 = cofT(1-K)/det(1-K) // just from the usual A-1 formula We could write this in finite matrix notation as follows: δij + Γij = cof [(1-K)ji] / det(1-K) If we recall these definitions from matrix theory cof(apq) = (-1)p+q minor(apq). and minor(apq) = det of the N-1 x N-1 matrix you get by crossing out row p and column q. we could write that δij + Γij = (-1)j+i minor [(1-K)ji] / det(1-K) Let's now set K = -B and ij = pq to get δp,q + Γpq = (-1)p+q minor [(1+B)qp] / det(1+B) which rewrite again as det(1+B)δp,q + det(1+B)Γpq = (-1)p+q minor [(1+B)qp] (*) But from above we have Γ = N/ |1-K| = N/ |1+B| => det(1+B) Γ = N => det(1+B) Γpq = Npq Then (*) becomes det(1+B)δp,q + Npq = (-1)p+q minor [(1+B)qp] => Npq = – det(1+B)δp,q + (-1)p+q minor [(1+B)qp] Now we are in a position to check our first "minor" determinant result obtained above. If we keep only terms through B1 , we can replace det(1+B) = 1 + tr(B) . Then we have Npq = – [ 1 + tr(B) ] δp,q + (-1)p+q minor [(1+B)qp] // through order B1 In our work (so far) above, we found that: minor [(1+B)pq ] = B0 piece + B1 piece = δp,q + tr(B) δp,q – (-1)p+q Bq,p // through order B1 This then suggests that Npq = – [ 1 + tr(B) ] δp,q + (-1)p+q minor [(1+B)qp] = – [ 1 + tr(B) ] δp,q + (-1)p+q { δp,q + tr(B) δp,q – (-1)p+q Bp,q } = – Bp,q + δp,q [ -1 - tr(B) +1 + tr(B) ] – Bp,q This agrees (!) with our formula above which says Nij = Kij + higher orders. So yes, we are on the right track! (But it is a long track). 7. Consider the B2 terms, but then stop this approach for now. Let's study the B2 terms. We start again with this: (δI1,a1 + BI1,a1) (δI2,a2 + BI2,a2) (δI3,a3 + BI3,a3) (δI4,a4 + BI4,a4) + perms and there are (4,2) = 12 different contributing terms, each with its permutations. A typical term looks like this: (δI1,a1 ) (δI2,a2 ) (BI3,a3) (BI4,a4) If p ≠ q, reconsider (p=5) boys = 1234 and (q=1) girls = 2345. We have various contributors now: B14δ22δ33B45 B15δ22δ33B44 B12B25δ33δ44 B15B22δ33δ44 Once again, we have the painful question of getting the sign of each term right. I think the first of the four shown here needs 2 swaps to get the 4 of B14 to the right of the two δ's. Maybe rule is (-1)p+q . I suspect this is the rule for even number of δ's, and the previous rule I got is for odd number of deltas, but maybe some work to show that to be true. In any event, you can see the 2x2 determinants of B starting to form here, which show up in the N formula, but beyond that the details are pretty hairy. OK, I agree to stop now on this line of development. This is really not a very good method of proof because you always have det(1+B) as part of the result, and things are all mushed together. I think my original proof in the Fredholm minor paper is the better path, but at least here I have shown that it can in theory be done using my formulas for subdeterminants with one row and column crossed out 8.Dealing with the "permutation sign problem" at the B1 level. This is the subroutine referred to from above. "(p=5) boys = 1234 and (q=1) girls = 2345. The question is: how many swaps to move the 1 of the boys group to the position of the 5 in the girls group." Let's start looking at lots of cases: boys girls swaps p=1 q=2 2345 1345 move the 2 of boys to 1 position of girls 0 p=1 q=3 2345 1234 move the 3 of boys to 1 position of girls 1 left p=1 q=4 2345 1235 move the 4 of boys to 1 position of girls 2 left p=1 q=5 2345 1234 move the 5 of boys to 1 position of girls 3 left Let's try a situation with larger N p = 3 q = 6 12456789 12345789 move the 6 of boys to 3 position of girls 2 left p = 3 q = 7 12456789 12345689 move the 7 of boys to 3 position of girls 3 left It seems pretty clear now that the number of required swaps is related to the distance q-p-1. A formula that works for this last set of examples would be # swaps = q-p-1 so sign = (-1)q-p-1. This also works for the first set of examples. If I repeated all these examples with p↔q, we just switch the boys and girls groups, for example p = 6 q = 3 12345789 12456789move the 3 of boys to 6 position of girls 2 right p = 7 q = 3 12345689 12456789move the 3 of boys to 7 position of girls 3 right The only change is in the direction of the shifts. Here we shift p-q-1 to the right, sign (-1)p-q-1. But the original formula still works : (-1)q-p-1 = (-1)q-p-1+2p-2q = (-1)p-q-1 . We can also express this as (-1)p-q-1 = (-1)p-q-1+2q = - (-1)p+q which I like, anticipating the "cofactor" work we are about to do. Theorem: The sign change produced by the pairwise swaps required to move the q of the boys group to the p position in the girls group = - (-1)p+q . 9. Consider the B2 terms one more time First, let's look again at our general formula for a subdeterminant (from the Overview above) "It turns out that the formula for a general subdeterminant can be written in a form very similar to that shown above for a diagonal one. Here is the general result: " detI1 I2 I3...In; a1 a2 a3...an; = εi1 i2 i3....in AI1 i1 AI2 i2 ..... AIn in // n factors of A where i1 i2 i3....in {a1 a2 a3...an} , εi1 i2 i3....in anti-symmetric, εI1 I2 I3....In = +1 For our case with N=5 we write our minor-sized general subdeterminant, detI1,I2,I3,I4; a1,a2,a3,a4; = Σi ={a1,a2,a3,a4} εi1 i2 i3 i4 AI1,i1 AI2,i2 AI1,i1 AI2,i2 = Σi1 ={a1,a2,a3,a4} Σi2 ={a1,a2,a3,a4} Σi3 ={a1,a2,a3,a4} Σi4 ={a1,a2,a3,a4} εi1 i2 i3 i4 (δI1,i1 + BI1,i1) (δI2,i2 + BI2,i2) (δI3,i3 + BI3,i3) (δI4,i4 + BI4,i4) = Σi1 Σi2 Σi3 Σi4 εi1 i2 i3 i4 BI1,i1 BI2,i2 δI3,i3 δI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 i2 i3 i4 BI1,i1 δI2,i2 BI3,i3 δI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 i2 i3 i4 BI1,i1 δI2,i2 δI3,i3 BI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 i2 i3 i4 δI1,i1 BI2,i2 BI3,i3 δI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 i2 i3 i4 δI1,i1 BI2,i2 δI3,i3 BI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 i2 i3 i4 δI1,i1 δI2,i2 BI3,i3 BI4,i4 I cannot actually DO any of the sums because we don't know whether or not we get hits with the deltas. But I could relabel the δ associated ε indices replacing i with I as follows: = Σi1 Σi2 Σi3 Σi4 εi1 i2 I3 I4 BI1,i1 BI2,i2 δI3,i3 δI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 I2 i3 I4 BI1,i1 δI2,i2 BI3,i3 δI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 I2 I3 i4 BI1,i1 δI2,i2 δI3,i3 BI4,i4 + Σi1 Σi2 Σi3 Σi4 εI1 i2 I3 i4 δI1,i1 BI2,i2 BI3,i3 δI4,i4 + Σi1 Σi2 Σi3 Σi4 εI1 i2 i3 I4 δI1,i1 BI2,i2 δI3,i3 BI4,i4 + Σi1 Σi2 Σi3 Σi4 εi1 i2 I3 I4 δI1,i1 δI2,i2 BI3,i3 BI4,i4 I could then move certain summations to their deltas: = Σi1 Σi2 εi1 i2 I3 I4 BI1,i1 BI2,i2 (Σi3δI3,i3) (Σi4δI4,i4) + Σi1 Σi3 εi1 I2 i3 I4 BI1,i1 (Σi2 δI2,i2) BI3,i3 (Σi4δI4,i4) + Σi1 Σi4 εi1 I2 I3 i4 BI1,i1 (Σi2δI2,i2) (Σi3δI3,i3) BI4,i4 + Σi2 Σi3 εI1 i2 I3 i4 (Σi1δI1,i1) BI2,i2 BI3,i3 (Σi4δI4,i4 ) + Σi2 Σi4 εI1 i2 i3 I4 (Σi1δI1,i1) BI2,i2 (Σi3δI3,i3) BI4,i4 + Σi3 Σi4 εi1 i2 I3 I4 (Σi1δI1,i1) (Σi2δI2,i2) BI3,i3 BI4,i4 where I am trying to maintain the I1 I2 I3 I4 order of things. Idea: let's make a simple name for each index set: R = I1,I2,I3,I4 C = a1,a2,a3,a4; Then (Σi1δI1,i1) = (I1C) OK, enough for this little shot, I am not succeeding very well.