scraps2

Footnote: In proper up/down tensor notation, one can define both λi and λi functionals, λi(ej) = (qi)Tej = δij = δi,j ei = Σjgijej qi = Σjgijqj λi(ej) = (qi)Tej = δij = δi,j gij = (contravariant) metric tensor for V and then the tensor F can be written four different ways, as was done with T at the end of Section 1. As noted earlier, we shall not in this document stress this notational aspect of our direct product analysis. We do note that in order to have a metric tensor gij, the space V must, in addition to being a vector space, also at least be a normed vector space with some notion of length of vectors so ||dx||2 = (ds)2 = Σijgijdxidxj ds = "length of vector dx" Normally one deals with vector spaces which, besides being normed vector spaces, are also metric spaces with metric d(a,b) = ||a-b||, a convention known as taking the natural metric (ie, metric defined in terms of the norm). In this type of space one has a notion of "distance" between two vectors a and b, d2(a,b) = ||a-b||2 = Σijgijaibj ≡ a b a,b = vectors In the last step the inner (dot) product is defines as shown, and the space is then a Hilbert Space. Footnote: In proper covariant tensor notation, if the vector space V is also a metric space with metric tensor gij. all vector objects exist in both contravariant and covariant form. The dual basis vectors discussed above should really be labeled qk so that qk ei = δki and, as noted above, qk is usually written as ek so then ek ei = δki . One then has λi(ej) = (qi)T ej = (ei)T ej = ei For example qk = Σigkiqi which is the same as qk = Σigkiqi one can define both λi and λi functionals, λi(ej) = (ei)Tej = δij = δi,j ei = Σjgijej λi = Σjgijλj λi(ej) = (ei)Tej = δij = δi,j gij = (contravariant) metric tensor for V and then the tensor F can be written four different ways, as was done with T at the end of Section 1. In order to have a metric tensor gij, the space V must, in addition to being a vector space, also at least be a normed vector space with some notion of length of vectors so ||dx||2 = (ds)2 = Σijgijdxidxj ds = "length of vector dx" Normally one deals with vector spaces which, besides being normed vector spaces, are also metric spaces with metric d(a,b) = ||a-b||, a convention known as taking the natural metric (ie, metric defined in terms of the norm). In this type of space one has a notion of "distance" between two vectors a and b, d2(a,b) = ||a-b||2 = Σijgijaibj ≡ a b a,b = vectors In the last step the inner (dot) product is defines as shown, and the space is then a Hilbert Space. ************************** ei ^ ej = - ej ^ ei and ei ^ ei = 0 . ei ^ ej ≡ eiej - ejei Notice therefore that ei ^ ej is an element of V x V, since it is a linear combination of elements of V x V. It is "antisymmetrized" under i ↔ j. Since not all elements of V x V can be written this way, the set of elements ei ^ ej spans a subspace of V x V (as one can verify from the definition of "subspace" ). The above definition implies that ei ^ ej = - ej ^ ei and ei ^ ei = 0 . Above we stated certain scalar and distributive properties of the operator. These properties are transferred onto the wedge ^ operator by the above definition. For example (αei) ^ ej = (αei)ej - ej(αei) = α [ eiej - ejei ] = α (ei ^ ej) α = scalar (ei+ek) ^ ej = (ei+ek)ej - ej (ei+ek) = eiej + ekej - ejei - ejek = [ eiej - ejei ] + [ ekej - ejek ] = (ei ^ ej) + (ek ^ ej) distributive ei ^ (ej + ek) = (ei ^ ej) + (ei ^ ek) by similar argument distributive ******************************** The n2 basis elements of V2 are eiej. We can linearly combine these basis vectors of V2 into two sets of basis vectors in this manner: ei^ ej = [eiej - ejei] n(n-1)/2 independent elements in this set ei * ej ≡ [eiej+ ejei] n (n/2) independent elements in this set As expected, the total number of linearly independent basis vectors is n(n-1)/2+ n (n/2) = n2. A general element of V2 can then be expressed as F = Σij [ Fij ei^ ej + Gij ei * ej] We claim that a general element of W2 ********************** If a, b ϵ V, then the previous rules for allow us to write a = Σiaiei b = Σjbjej a b = ( Σiaiei) ( Σjbjej) = Σijaibj(eiej) Similarly, we find that a ^ b = (Σiaiei) ^ ( Σjbjej) = Σijaibj (ei^ej) // rules for ^ above = Σi≠jaibj (ei^ej) // since ei ^ ei = 0 = Σi<jaibj (ei^ej) + Σi>j aibj (ei^ej) // break into two parts = Σi<jaibj (ei^ej) + Σi<j ajbi (ej^ei) // i↔j in the second sum = Σi<jaibj (ei^ej) – Σi<j ajbi (ei^ej) //(ej^ei) = - (ei^ej) = Σi<j [ aibj - ajbi] (ei^ej) = Σi<j det (ei^ej) . From the last expression one sees that (b ^ a ) = - (a ^ b) // similar to b x a = - a x b for cross product in R3 (a ^ a) = 0. // similar to a x a = 0 for cross product in R3 ********************************* To answer this question, we need to make use of the following two Lemmas. Examples: Fii = Tii - Tii Fiii = Tiii - Tiii + Tiii - Tiii + Tiii - Tiii ____________________________________________________________________________________ Lemma 1: Consider an n x k matrix M having column vectors m1,m2, ... mk so we can write M = (m1,m2,m3...mk). Each column vector has n components where n ≥ k, so one can think of M as a "tall" matrix. Now from the set of n integers {1,2...n} select a subset of k unequal integers {j1,j2 ,..jk}. Let these integers label k rows of the matrix M. We can then define a square matrix N to be N ≡ rows {j1,j2...jk} of M Since matrix N is square, k x k, it has a determinant which can be written in this standard manner, det(N) = Σii...i εii...i N1,i N2,i ...... Nk,i Given the connection between N and M, we then have det(N) = Σii...i εii...i Mj,i Mj,i ...... Mj,i This determinant is the minor of M obtained by selecting only rows {j1,j2 ,..jk}. So rewrite as det [ M(jj...j)] = Σii...i εii...i Mj,i Mj,i ...... Mj,i ____________________________________________________________________________________ such that the following is true P(ej ^ ej ^ .... ^ ej) = ( eP(j) ^ eP(j) ^ .... ^ eP(j)) = (ei ^ ei ^ .... ^ ei) where i1 < i2 < ..... < ik If P involves S pairwise swaps of indices, then we know that P(ej ^ ej ^ .... ^ ej) = (-1)S (ei ^ ei ^ .... ^ ei) where i1 < i2 < ..... < ik (ej ^ ej ^ .... ^ ej) Since V is spanned by the n ei, the above represents nk objects. Of these nk, the number of non-vanishing objects is n*(n-1)*...*(n-k+1) = n!/(n-k)! For k = 3 this smaller set would include e1^e2^e3 , e3^e2^e1, and four other combinations for a total of 3! objects. All objects in this smaller set are either ± e1^e2^e3 so we can regard only e1^e2^e3 as a basis vector of L3 representing that group of 3! elements. If we agree to represent each such group by a combination Applying ** to a set of basis vectors ei of V we find that (ej ^ ej ^ .... ^ ej) = εjj....j (e1^ e2^ .....^ ek) For each basis vector on the left one has n choices from the n ej which span V. Thus, nk basis vectors Any basis ei of V has n basis elements. The left side of the above equation represents nk possible wedge products, of which only The left side of this equation represents nn different "basis" elements of which only k! are non-zero, and all those k! elements are either + or - e1^ e2^ .....^ ek Thus, the space Lk has exactly one linearly independent basis element which we can take to be the one with increasing index order (e1^ e2^ .....^ ek). The most general element of Lk is then T = T123..k (e1^ e2^ .....^ ek) ej ^ ej ^ .... ^ ej = εjj....j Σii....i εii....i (ei ei ..... ei) = εjj....j ΣP (-1)S(P) ( eP(1) eP(2) ..... eP(k)) The wedge space Lk .We define the space Lk to contain all linear combinations of the basis vectors, T = Σii....i Tii....i ( ei ^ ei ^ .... ^ ei ) For example, if Tii....i = (v1)i(v2)i .....(vk)i we obtain T = Σii....i (v1)i(v2)i .....(vk)i ( ei ^ ei ^ .... ^ ei ) = (Σi(v1)iei) ^ (Σi(v2)iei) ^ .... ^ (Σi(vk)iei) = v1^ v2^ .....^ vk and so Lk contains all wedge products of k vectors of V. The ordered basis for Lk. It is clear from the "change sign under swap" rule that the kk "basis elements" of the form ei ^ ei ^ .... ^ ei are not all linearly independent, so in fact they are not really a basis for Lk. One way to select a linearly independent subset of these elements is to form the following set: ei ^ ei ^ .... ^ ei with 1 ≤ i1 < i2 < ..... < ik ≤ n where the {ei} form a basis for V. Since there are at most n linearly independent ei for V, we know that the index i on ei ranges from 1 to n, giving the left and right limits shown above. Size of Lk compared to Vk (b) How big is the space Lk compared to the space Vk? As noted earlier, the most general element of Vk has the form F = Σii....i Fii....i (eiei .....ei) // each sum is 1 to n so each "tensor" F with n indices defines an element of Vk. Again, if the reals had only N values instead of ∞, we would say that the number of elements of Vk as *************************** 4. Summary to this point Everything was built upon the notion of a direct product space V x V with elements of the form ab. By selecting a basis of the form eiej in V x V we were able to construct the notion of a rank-2 "tensor" T with the expansion T ≡ ΣijTij eiej. The inhabitants of the direct product space V x V = V2 are precisely such rank-2 tensors. Over in the dual world of V* x V* we constructed similar rank-2 tensors of the form F = Σij Fij λiλj as expansions on basis elements λiλj where λi are linear functionals. These rank-2 tensors F are the inhabitants of V* x V* = V*2. By evaluating these bilinear functionals (like F) at points in V x V = V2, we produced bilinear functions of the form F(v1,v2). One can think of the rank-2 tensor either in terms of these functions, or in terms of the functionals F with no evaluation arguments. The vectors in V* and in V* x V* are really functionals, not functions, but one can always talk about the associated functions like F(v1,v2) or λi(v1). Sometimes the space of these functionals (or functions) is called T2(V), where T is for Tensor, and the superscript 2 matches V* x V* = V*2. We then introduced the wedge product in the V x V world. The wedge products like a ^ b occupy only a subspace of the full V x V direct product space. For example, since a ^ a = 0, elements of the form aa which of course exist in the direct product space V x V do not play a role in the discussion of wedge products. This fact is by construction, since a ^ a = aa - aa. The wedge products a ^ b are antisymmetric in the sense that a^b = -b^a and so only antisymmetric linear combinations of xy elements can appear in the space inside V2 which is spanned by the wedge products. We did not give this subspace of V2 a specific name. One can represent this space formally as a quotient space V2/I where I is an "ideal" consisting of these aa elements which are in effect removed from (divided out from) V2 and set to 0. We make no attempt to prove this fact, but is seems reasonable. One thing we noted with the a ^ b wedge product was a connection to the geometry of R2. The shape of the parallelogram spanned by two vectors a and b of a ^ b in R2 is suggestive of the name this object is given when we later discuss the wedge product of two vectors in the context of the Grassmann or Clifford Algebras. It is called a "2-blade". We then moved back into the dual V* x V* world. There we defined and then computed the wedge product α ^ β of two arbitrary linear functionals in V* (linear functionals are the "vectors" of V*). We found that such functionals inhabit a subspace of V* x V* = V*2 to which we gave the name Λ2(V).Thus we can say that Λ2(V) T2(V). When our general wedge product of two V* vectors was evaluated at a point in V x V, the function (α ^ β)(v1,v2) was found to be a bilinear antisymmetric function. The notion that the wedge product is antisymmetric in any of the above senses -- b ^ a = - a ^ b or β ^ α = - α ^ β or (β ^ α)(v1,v2) = - (α ^ β)(v1,v2) or (α ^ β)(v1,v2) = - (α ^ β)(v2,v1) -- is intentionally injected into the definition of the wedge products within V2 and within V*2. One might ask: why this is done? Since one is certainly allowed to define things in any way one wants, perhaps a better question is: how does doing this help me in my work? Up to this point, we have developed the wedge product as a stand-alone concept having nothing to do with things like Clifford Algebras or Differential Forms. Here is one possible motivating factor for introducing the antisymmetry: in the theory of Differential Forms, when one integrates over the area of a flat object in R2, one can replace the usual differential dxdy with a wedge product dx ^ dy, similar to our a ^ b in V2 discussed above. The wedge product dx ^ dy can be thought of as a little square with a direction which is normally thought of as [dx ] x [dy ] = dxdy , this being a piece of differential area with a direction (one of the normals to the area). The "other side" of this patch of area is said to have the area - dxdy which is associated with dy ^ dx = - dx ^ dy . In this manner, the fact that an area has one of two "orientations" is represented by the wedge products dx ^ dy and dy ^ dx. In a similar vague vein, one knows that [dx ] x [dx ] = 0 and one can associate this with the fact that dx ^ dx = 0. A line segment also has two orientations, as does a little cube dx ^ dy ^ dz in that the cube can be right-handed or left-handed. This dual handedness or orientation exists for geometric primitives in all dimensions. Again, this is just meant as some dim motivation. The antisymmetry is inserted into the wedge product theory because it turns out to be useful in various applications of the wedge product. The reader will find with a quick search that the wedge product is the basis of "exterior algebra", as opposed to "interior algebra", and there are an endless number of connections to the constructs of abstract algebra, concepts in the bailiwick of the pure mathematician, perhaps an Algebraic Topologist. Our intention is not to bring out these higher level connections, but instead to poke around in the engine room and see how wedge products actually work. Our next tour stop will be the study the wedge products of objects which are fancier than vectors, and wedge products which have more that two factors. 5. Wedge products of three vectors of V Without justification, we shall require by fiat that the wedge product of three vectors in V shall be antisymmetric under the interchange of any two of the three vectors involved. Thus we want to have a ^ b ^ c = - b ^ a ^ c a ^ b ^ c = - a ^ c ^ b a ^ b ^ c = - c ^ b ^ a The fiat rule then implies that a ^ b ^ c vanishes if any two vectors are the same. This is in some sense the logical extension of the idea that a ^ b = - b ^ a for the wedge product of two vectors in V. As a vague motivation from the differential forms world, if we associate dx ^ dy ^ dz with an integration volume element spanned by three small vectors dx = dx , dy = dy , and dz = dz , then we might expect to find that dx ^ dx ^ dz = 0. A simple solution is the following : a ^ b ^ c = abc - acb + cab - cba + bca - bac = abc + five signed permutations Here a,b,c are all vectors in V. This triple wedge product is fully antisymmetrized. To make things easier notationally, we rename the three vectors to be v1, v2 and v3 so that v1 ^ v2 ^ v3 = v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3 = v1v2v3 + five signed permutations of the subscripts This sum can be expressed in a fairly standard notation, v1 ^ v2 ^ v3 = ΣP (-1)S vP(1)vP(2)vP(3) where the sum is over all 6 permutations P of the integers {1,2,3} including the identity permutation. The integer S is the number of swaps it takes to link two permutations. For example, for P{1,2,3} = {2,1,3) one would have S = 1. For this particular permutation, one has P(1) = 2, P(2) = 1, P(3) = 3 so that the corresponding term is (-1)1 vP(1)vP(2)vP(3) = - v2v1v3 . An alternative notation is this v1 ^ v2 ^ v3 = Σijk εijk vivjvk where εijk is the usual permutation tensor with ε123 = 1, ε changes sign whenever two indices are swapped, and ε is zero when any two indices are the same. ε is often called "the totally antisymmetric tensor". Notice that vi is the name of a vector, not the ith component of the vector v. We are building these triple wedge products of vectors in V on top of the direct product space V x V x V = V3. It is all quite mechanical. One might have approached the triple wedge product by grouping things and trying to use our earlier wedge products of two objects. For example (a ^ b) ^ c = ? = ( ab - ba) c - c ( ab - ba) // wrong This approach fails to achieve full antisymmetrization, it only produces a↔b antisymmetrization. This is evident in that the result has only 4 terms, but 6 are required. The problem here is that when we do the second wedge product (a ^ b) ^ c, the object (a ^ b) is an antisymmetric rank-2 tensor and not a vector (a vector is a rank-1 tensor), and we have never discussed how to take the wedge product of any objects other than vectors in V. In particular, it must be that d ^ c ≠ dc - cd when c is vector and d a rank-2 tensor. Eventually we will learn how to do arbitrary wedge products, and it will then be true that one can group things any way one wants, and we will find that (a ^ b) ^ c = a ^ (b ^ c) = a ^ b ^ c . We should note that, starting with the basis eiejek for V3 , we can express an arbitrary element of V3 in this manner T ≡ ΣijkTijk eiejek. Here T is a rank-3 tensor and we show its official contravariant components Tijk in the correct tensor notation. So the elements of V3 are rank-3 tensors, but only a subset of this space V3 is spanned by "vectors" of the form a ^ b ^ c, where by "vector" we mean an element of the vector space VxVxV = V3. Only totally antisymmetrized combinations of xyz are allowed in this smaller space (which again we give no name). Let us compute a ^ b ^ c for three vectors in V in terms of their components and the basis vectors. We first impose rules similar to those earlier for the double wedge product, α(a ^ b ^ c) = (αa) ^ b ^ c = a ^ (αb) ^ c = a ^ b ^ (αc) α scalar a ^ b ^ (c1 + c2) = a ^ b ^ c1 + a ^ b ^ c2 and so on so a ^ b ^ c is a "trilinear" object -- it is separately linear in each of its factors. Then given a = Σiaiei b = Σjbjej c = Σkckek We find that a ^ b ^ c = ( Σiaiei) ^ (Σjbjej) ^ (Σkckek) = Σijk aibjck (ei^ej^ek) = Σi≠j≠k aibjck (ei^ej^ek) . We now break up the summation into a somewhat tedious set of regions, = (Σi<j<k + Σi<k<j + Σj<i<k + Σj<k<i + Σk<i<j + Σk<j<i) aibjck (ei^ej^ek) . We could write out each of these 6 terms and change dummy summation index names and use the rules ** for the (ei^ej^ek) to get this triple wedge product always in the same order (ei^ej^ek) and we would end up with something of the form = Σi<j<k f(a,b,c) (ei^ej^ek) Rather than do that brute force calculation, we instead use some obscure permutation summation methods. These methods work for wedging together any number of vectors. To make these methods fly, we need a few Lemmas. The first will be familiar to the reader, the second maybe less so. ____________________________________________________________________________________ Lemma 1: Suppose a 3x3 matrix M has column vectors m1,m2,m3 so we can write M = (m1,m2,m3). The determinant of this matrix can be written in several equivalent ways, using notations shown earlier, det(M) = (m1)1(m2)2(m3)3 + 5 signed permutations of the subscripts = ΣP (-1)S (m1)P(1)(m2)P(2)(m3)P(3) = Σijk εijk (m1)i(m2)j(m3)k = Σijk εijk Mi1Mj2Mk3 where the final form is probably familiar to the reader. For the case M = (a,b,c) then we find from the second line above, det(M) = det(a,b,c) = ΣP (-1)S aP(1)bP(2)cP(3) . For an n x n matrix M with column vectors a,b,c.... the same argument concludes that det(M) = det(a,b,c,.....) = ΣP (-1)S aP(1)bP(2)cP(3) ..... where there are n column vectors, there are n factors on the right, and the P are permutations of n objects. ____________________________________________________________________________________ Lemma 2: We do n = 2, then n = 3, then n = n and state the Lemma claim at the end. Consider the sum Q = [Σi<j + Σi>j] fij = Σi<jfij + Σj<ifij . Swapping the dummy summation indices in the second term gives Q = Σi<jfij + Σi<jfji = Σi<j [ fij + fji] This can be written as Q = Σi<j [ΣP fP(i)P(j)] where ΣP is over all permutations of two objects, the first being the identity permutation. By the same argument, we can write Q = (Σi<j<k + Σi<k<j + Σj<i<k + Σj<k<i + Σk<i<j + Σk<j<i) fijk = Σi<j<k [ΣP fP(i)P(j)P(k)] where ΣP is now a sum over the permutations of three objects, such as P{1,2,3} = {2,1,3}. In the first line above, the sum of sums can itself be written as (Σi<j<k + Σi<k<j + Σj<i<k + Σj<k<i + Σk<i<j + Σk<j<i) = ΣP [ΣP(i)<P(j)<P(k)] So for n = 3 our Lemma can be stated as ΣP [ΣP(i)<P(j)<P(k)] fijk = Σi<j<k [ΣP fP(i)P(j)P(k)] where the P are permutations of three objects. The simple idea is that one can move the permutation operators P() from the summation Σ to the summand without changing the sum. For n = n we then have ΣP [ΣP(i)<P(j)<P(k)<.... ] fijk.... = Σi<j<k<.... [ΣP fP(i)P(j)P(k)....] where now P are the permutations of n objects. ____________________________________________________________________________________ With these Lemmas stated, we now continue our interrupted development. a ^ b ^ c = (Σi<j<k + Σi<k<j + Σj<i<k + Σj<k<i + Σk<i<j + Σk<j<i) aibjck (ei^ej^ek) = ΣP [ΣP(i)<P(j)<P(k)] aibjck (ei^ej^ek) = Σi<j<k [ΣP aP(i)bP(j)cP(k) (eP(i)^eP(j)^eP(k)) ] . // Lemma 2 Since swapping any two terms of (eP(i)^eP(j)^eP(k)) creates a minus sign, one may write (eP(i)^eP(j)^eP(k)) = (-1)S (ei^ej^ek) where S is the number of swaps required. Then we end up with a ^ b ^ c = Σi<j<k [ΣP aP(i)bP(j)cP(k) (-1)S (ei^ej^ek) ] = Σi<j<k [ ΣP (-1)S aP(i)bP(j)cP(k) ] (ei^ej^ek) = Σi<j<k det(a,b,c) (ei^ej^ek) // Lemma 1 We can rewrite this result in fancier notation, changing the names of the wedged vectors, v1 ^ v2 ^ v3 = Σi<i<i det(v1,v2,v3) (ei^ ei^ ei) A key idea here is that a determinant magically appears when the summation regions are reordered so that only a single inequality region is left, and that one region has increasing index values. We can similarly rewrite our earlier result for the wedge product of two arbitrary vectors a ^ b = Σi<j det (ei^ej) as v1 ^ v2 = Σi<i det(v1,v2) (ei^ ei) 6. Wedge products of n vectors of V Extending the ideas of the previous section (and with the two Lemmas in place), we can consider the wedge product of n vectors of V. Using the Lemmas stated in their most general forms, we have also shown that a ^ b ^ c ^....... = Σi<j<k<...... det(a,b,c, .....) (ei^ej^ek^ ..... ) For example, a ^ b ^ c ^ d = Σi<j<k<l det(a,b,c.d) (ei^ej^ek^el ) Using more proper notation (subscripts on subscripts), we can write our most general result this way v1 ^ v2 ^ ......^ vn = Σi<i<....<i det(v1,v2,...vn) (ei^ei .....^ei) To reduce notational clutter, one can make these definitions I ≡ {i1, i2.....in} where i1< i2<....<.in I = an ascending multi-index eI ≡ (ei^ei .....^ei) eI = a basis vector in Vn with ascending multi-index Then the above wedge product can be compactly written v1 ^ v2 ^ ......^ vn = ΣI det(v1,v2,...vn) eI Regardless of notation, the key idea is that a determinant magically appears in a wedge product of vectors when the summation index chain is put into ascending order . One can use an ordinary multi-index (that is, not one that ascends) to obtain compact forms of various statements involving the Vn direct product space. For example, generalizing our earlier result, we can write out the expansion of a rank-n tensor in this manner T = Σijk...Tijk... (eiejek....) . In fancier notation, T = Σii...i Tii...i (ei^ei .....^ei) // rank-n tensor in Vn . We can then define I ≡ {i1, i2.....in} I = a multi-index eI ≡ (ei^ei .....^ei) eI = a basis vector in Vn so the tensor expansion then takes this very compact form T = ΣI TI eI I = multi-index of order n.