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(b1) Area. (general N) In N dimensions for N≠3, the notation q = A x B has no meaning so we must replace it with something that does have meaning. Consider the N=3 case again where the cross product is written this way (repeated indices are summed unless stated otherwise) q = A x B qk = εkab Aa Bb The generalization to N dimensions is this q = A x B x C x ... x X qk = εkabc..x Aa BbCc ... Xx There must be N-1 vectors crossed together, so there will then be N-2 cross's. If N were 5, we would use X = D, the 4th letter of the alphabet , but since we don't know N, we just use X. A more proper notation would be to write qk = εkiii....i A(1)i A(2)i .... A(N-1)i but this requires a lot more symbols which cause much clutter as we continue, so we will try to stick with the simpler though less precise notation. ε is the totally antisymmetric tensor in N dimensions (it has N indices), meaning that if any two indices are swapped, it changes sign. The cross product notation suggests that q might be perpendicular to all the vectors from which it is composed, and this is in fact the case. For example, consider q B = qkBk = εkabc..x Aa BbCc ... Xx Bk = Σcd..[ΣabBbBk εkabc..x] AaCc ... Xx But [ΣabBbBk εkabc..x] = 0 since we have a symmetric object BbBk contracted on both indices against an antisymmetric object. Therefore, q B = 0 and similarly q is perpendicular to the other vectors from which q is built. We are then led to this obvious generalization of the formula for differential area in N dimensions, dS1 = ds2 x ds3 x ds4 ........ x dsN dS2 = ds1 x ds3 x ds4 ......... x dsN .... dSN = ds1 x ds2 x ds3 x ds4 ... dsN-1 We can summarize these N equations by one equation dSk = Πx;i≠k dsi (1) where the first subscript x of Π indicates "cross products", and the second tells us to omit dsk. [ Please don't confuse this x with the x used above as an ε subscript.] If we omit the first subscript on a Π symbol, we imply regular multiplication of the factors. The N vectors dsi scaffold an N dimensional parallelogram in N-space, and each of these vectors can be written in terms of its tangent base vector ei dsi = ei dqi i = 1...N (2) so we can write the area element this way dSk = Πx;i≠k(x) dsi = (Πx;i≠k ei) (Πi≠k dqi) (3) For example, in N=5 dimensions we would have dS3 = e1 x e2 x e4 x e5 dq1dq2dq4dq5 We are interested in the magnitude dSk so we have to compute the magnitude2 of (Πx;i≠k ei ): (Πx;i≠k ei ) (Πx;i≠k ei) = (Πx;i≠k ei )j (Πx;i≠k ei )j Now using our cross product ε formula from above we can write (Πx;i≠k ei )j = εjiii...i (e1)i (e2)i ..... (eN)i // ik and (ek)i missing where ε is missing index ik and the product is missing (ek)i Then we have (Πx;i≠k ei )j (Πx;i≠k ei )j = (εjiii....i (e1)i (e2)i .... (eN)i )(εjmmm....m (e1)m (e2)m .... (eN)m) = εjiii....i εjmmm....m [(e1)i(e1)m][(e2)i(e2)m] ..... [(eN)i(eN)m] (4) // ik and (ek)i missing AND mk and (em)i missing Now we have to deal with this εε product. For the case where nothing is "missing", we show elsewhere that if we contract the product of two ε symbols only on the first index, we get εjiii....i εjmmm....m = ΣP p P2(δi,mδi,m δi,m... δi,m) p = parity of P (5) By way of explanation, P2 represents an arbitrary permutation P acting on the 2nd indices of the δ's. We might write such a permutation this way P2(δi,mδi,m δi,m... δi,m) = (δi,m"δi,m" δi,m"... δi,m") where P(m1m2m3....mN) = m1"m2"m3"....mN" with parity p The parity of a permutation is p = (-1)S where S is the number of pairwise swaps it takes to get from the original order to the permuted order. Although there might be many pairwise swap sequences to do this, they will all have S even or all have S odd so the parity is well-defined for a given permutation P. Notice that the first term in the sum is obtained with P = 1 and p = 1 so we can write εjiii....i εjmmm....m = δi,mδi,m δi,m... δi,m + other signed terms Hopefully the reader is convinced of the validity of (5) . Since we know that ΣP p P2(M11M22M33...MNN) = det(M) we can "draw" εjiii....i εjmmm....m as the determinant of a certain matrix, but it takes a lot of space to draw such things, and the permutation notation seems more compact. Here is an example of such a determinant with simpler subscript notation Now in our case, we have certain items "missing" , but the permutation sum formula "still works" and we can write εjiii....i εjmmm....m // ik and (ek)i missing AND mk and (em)i missing = ΣP p P2(δi,mδi,m δi,m... δi,m) p = parity of P (6) where now there are only N-1 δ's in the product and δi,m is missing. The P operator is now acting on the second indices of the δ's, but there are only N-1 such indices, so P permutes N-1 variables, and as before the parity p is determined by P. We now apply (6) to (4) to get (Πx;i≠k ei )j (Πx;i≠k ei )j = (εjiii....i (e1)i (e2)i .... (eN)i )(εjmmm....m (e1)m (e2)m .... (eN)m) = εjiii....i εjmmm....m [(e1)i(e1)m][(e2)i(e2)m] ..... [(eN)i(eN)m] (4) // ik and (ek)i missing AND mk and (em)i missing = ΣP p P2(δi,mδi,m δi,m... δi,m) [(e1)i(e1)m][(e2)i(e2)m] ..... [(eN)i(eN)m] = ΣP p P2[(e1e1) (e2e2) (e3e3).... (eNeN)] = ΣP p P2[g'11g'22 g'22....g'nn] // g'kk is missing This is the determinant of the metric tensor submatrix obtained by crossing out the kth row and the kth column! The leading term is g'11g'22 g'22....g'nn where g'kk is missing. Therefore in this very long-winded discussion we have shown that (Πx;i≠k ei )j (Πx;i≠k ei )j = min(g'kk) Each of the "other terms" has the same form as the first term, but the set of mi labels has been permuted (and the ii labels stay fixed). Such a term has a sign ± depending on whether this permutation is obtained by an even or odd number of index swaps from the original order. Including the first term, the total number of permutations is (N-1)!, so the portion "other terms" contains (N-1)! - 1 terms. For N=3, there is only one "other term". Notice that no mi in any term (including the first term) ever takes the value k needs repair Therefore, we have shown that (Πx;i≠k ei ) (Πx;i≠k ei) = [e1 e1] [e2 e2]..... [eN eN] + other terms where again [ek ek] is missing. In those other terms, the second set of e's is permuted. We recognize that dot products as metric tensor elements, so we have (Πx;i≠k ei ) (Πx;i≠k ei) = g'11 g'22..... g'NN + other terms Now that we have a relatively simple product, we can account for the signs due to the swaps mentioned above with and ε symbol having N-1 indices. Thus (Πx;i≠k ei ) (Πx;i≠k ei) = g'1m g'2m..... g'Nm εmmm....m // mk missing For example, if N=4 and k=2 then each index mi in the implied sums only take the values mi = 1,3,4 because we can never have In the implied sums here over the mi, we are listing off the permutations of {mi...} = { 1,2....N} where k is missing from the list. Therefore the sums are all of the form Σm≠m . We can add this reminder to the notation by writing (Πx;i≠k ei ) (Πx;i≠k ei) = Σm≠m g'1m g'2m..... g'Nm εmmm....m // mk missing So there are two issues here. (1) ε is missing index mk and g'km is missing from the product of g' objects. (2) each sum excludes mk We have therefore shown that dSk = Πx;i≠k(x) dsi = (Πx;i≠k ei) dSk = [Σm≠m g'1m g'2m..... g'Nm εjmmm....m]1/2(Πi≠k dqi) // mk missing As a check on this result, for N=2 it says dS1 = [Σm≠1g'2m g'3m εmm]1/2 dq2 dq3 = [g'2m g'3m εmm]1/2 dq2 dq3 g'2m g'3m εmm