tensor and wedge old

1.3 Outer Products 1 1.4 Tensor products of the form VV...V : The Tensor Algebra 5 1.5 Kronecker Products 8 2. The Wedge Product built on the Tensor Product 13 2.1 The tensor product of 2 vectors 13 2.2 The tensor product of 2 dual vectors. 17 2.3 The wedge product of 2 vectors 21 2.4 The wedge product of 2 dual vectors 27 2.5 The tensor product of k vectors 30 2.6 The wedge product of k vectors 32 2.7 The tensor product of k dual vectors 41 2.8. The wedge product of k dual vectors 43 Put in red stuff that is already in Chapter 2 3.1 Outer Products By either development above, we have a definition of the tensor product VW with pure elements of the form vw, and with general elements of the form F = Σij Fij(eie'j), along with the bilinear rules noted earlier, (v1+v2) w = (v1w) + (v2w) for all v1,v2 ϵ V and all w ϵ W v (w1+w2) = (vw1) + (vw2) for all v ϵ V and all w1, w2 ϵ W α(vw) = (αv)w = v(αw) for all v ϵ V and all w ϵ W and all α ϵ K . (1.3.1) The essence of the tensor product is this bilinearity, and there is no requirement to describe the objects vw in more detail. In the formal sense we are done and fini. However, for "engineering purposes", it is useful to add more structure to the tensor product by defining "tensor components". We have already written a vector v ϵ V as v = Σiviei where the coefficient vi may be regarded as the "component of the vector v in the ei basis. So now we write, [vw]ij ≡ viwj (1.3.2) where vi and wj are components of vector v in V and w in W. These vi and wj can be the elements of any field K and the juxtaposition of viwj implies multiplication in that field. but we have in mind that K = R, the real numbers. The key point: because the function viwj is manifestly bilinear, this extra specification does not conflict with any of the earlier tensor product "rules". For example we can evaluate, [(v1+v2) w]ij = (v1+v2)iwj (v1w)ij + (v2w)ij = (v1)i wj + (v2)i wj The first rules says the left sides of these two equations must be equal, but we can see that the right sides are also equal, so things are consistent. When this component level structure is glommed onto the raw tensor product, we end up with something called an outer product, though it is indicated by the same notations VW and vw. Thus an outer product is a tensor product which has been "extended" so we can talk about components of the various tensors. Most readers probably think of tensors a priori as having "components", but here we have made a thin distinction between tensors "in the large" and tensors with components which can be manipulated. Consider then the following most-general expansion of a tensor T of VW, T = Σij Fij eie'j . Suppose the basis vectors ei are simple "unit vectors" so (ei)k = δi,k, and similarly for the e'i of W, (e'i)k = δi,k ( but remember V has n axes and W has n' axes). Then we can take the "ab-component" of the above tensor expansion as follows : Tab = [ Σij Fij eie'j]ab = Σij Fij [eie'j]ab = Σij Fij (ei)a(e'j)b = Σij Fij δi,a δj,b = Fab . In the special case that Fij = viwj, we know that T = ab and so Tij = [ab]ij = viwj Whereas [v]i = vi in any basis {ei}, we find that Tij = Fij Thus the expansion coefficients Fij in this simple basis are exactly the tensor components Tij. The same thing happens in the V world: v = Σiaiei vk = Σiai(ei)k = Σi aiδi,j = ak . _________________________________________________________________________________ Exercise: Show how Fij and Tij are related for a general basis eie'j. It will be shown in Section ** that the bases ei and e'i have dual bases qi and q'i such that, in the matrix/vector notation of ***, qiT ej = δi,j q'iT e'j = δi,j which can also be written eiT qj = δi,j e'iT q'j = δi,j . For the simple basis (ei)k = δi,k one has qi = ei, but for other bases, the qi are linear combinations of the ei. Let is reconsider then in a completely general basis eie'j, T = Σij Fij (eie'j) Tab = Σij Fij (eie'j)ab = Σij Fij(ei)a(e'j)b . Multiply both sides by (qmq'n)ab = (qm)a(q'n)b and sum on a and b, Σab (qm)a(q'n)bTab = Σab Σij Fij (qm)a(q'n)b(ei)a(e'j)b = Σij Fij [Σa(qm)a(ei)a] [Σb(q'n)b(e'j)b] = Σij Fij δm,i δn,j = Fmn so Fij = Σab (qi)a(q'j)bTab ** Meanwhile, a basis is complete with respect to its dual basis (as we show just below), so we also know that Σi(qi)a(ei)b = δa,b Σi(q'i)a(e'i)b = δa,b . Then multiply both sides of ** by (ei)m(e'j)n and sum on i and j to get Σij(ei)m(e'j)nFij = Σij(ei)m(e'j)n Σab (qi)a(q'j)bTab = Σab Tab [Σi (qi)a(ei)m ][Σj(q'j)b(e'j)n] = Σab Tab δa,m δb,n = Tmn Our conclusions just obtained can be written as Fij = Σab (qi)a(q'j)bTab // sum is on component indices Tij = Σab(ea)i(e'b)jFab // sum is on basis vector labels Thus the coefficients Fij are linear combinations of the Tij, and vice versa, with coefficients as shown. We can write the first equation the matrix/vector notation of *** as Fij = (qi)T T (q'j) For comparison, we found earlier that Tij = (ei)T T (e'j) . In each case, one can think of the tensor T as a basis-independent operator which we sandwich between different sets of vectors to create the Fij and Tij. In the case W = V, one would write the above equations in quantum mechanics notation as Fij = <qi | T | qj > = matrix elements of operator T in the {qi} basis which is dual to general {ei} Tij = <ei | T | ej > = matrix elements of operator T in the simple {ei} basis Completeness relation. Write v = Σiviei. Apply qjT from the left to get qjTv = qjT(Σiviei) = ΣiviqjTei = Σiviδj,i = vj Then, v = Σiviei = Σi[qiTv]ei so va = Σi[qiTv](ei)a or va = Σi[Σb(qi)bvb](ei)a = Σb vb [ Σi(qi)b(ei)a] . For this equation to be valid for any component of any vector v, we must have [ Σi(qi)b(ei)a] = δb,a . Some authors write this completeness relation as ΣiqieiT = 1 with the meaning shown above. _____________________________________________________________________________ By adding information to the tensor product to create an outer product, we create machinery which is useful in component manipulations of tensors. This is especially useful when we have V = Rn and W = Rn', and even more useful when V = W = Rn. Usually the term "rank-2 tensor" assumes that V = W. The idea of an outer product can easily be extended to the tensor product of more than two spaces. Here are upgrades of some equations above for the case k = 3 where we have VWX : [vwx]ijk ≡ viwjxk // manifestly trilinear! T = Σijk Fijk (eie'je"j) . // most general element of VWX (e"j are a basis for X). Fijk = Σabc (qi)a(q'j)b(q"k)cTabc . The generalization to an outer product of k vector spaces is then the following: [vwx....]ijk ≡ viwjxk ..... // manifestly k-multilinear T = Σijk... Fijk... (eie'je"j...) . // most general element of VWX .... Fijk... = Σabc... (qi)a(q'j)b(q"k)c....Tabc... Below we shall extend the idea of an outer product to apply to tensors other than vectors. 1.4 Tensor products of the form VV...V : The Tensor Algebra We have tried to keep things general up this point by using VW...Z where all the vector spaces could be different, but in this section we assume they are all the same, and this is our main interest. There is then only one set of basis functions {ei} to worry about, the basis for V. We now introduce the compact notation: Vk ≡ VV....V // k copies, fancy notation Πi=1k V In Section 1.3 we defined the outer produce as a specialization of the tensor product where [abc....]ijk... = aibjck...... which is a manifestly k-multilinear function. Here a,b,c... are all vectors in V. We would like to generalize the notation so it can act on objects other than vectors. This is fairly easy to do. We start by defining: Aij ≡ [ab]ij = aibj rank-2 tensor We refer to Aij as the components of a rank-2 tensor A = ab, whereas ai and bi are components of a rank-1 tensors a and b. The word "tensor" has a weak and a strong meaning as discussed in ***. In the weak meaning, a rank-2 tensor is something that has components with two indices like Aij. In the strong meaning, a rank-2 tensor is a set of components Aij which transform in a certain manner with respect to some transformation. This is the meaning one works with when the ei are the tangent base vectors of a transformation, and the ei are the reciprocal base vectors, as already discussed. In either sense of the word tensor, the statement Aij = aibj defines a rank-2 tensor as the outer product of two rank-1 tensors, and this is a profitable way to construct higher rank tensors from lower rank ones. The outer product form also specifies exactly how a rank-2 tensor should "transform" in terms of how rank-1 vectors "transform" under a transformation. For example under a rotation a vector goes as v'a = ΣiRaivi and then the transformation of the rank-2 tensor is A'ab = ΣijRaiRbjAij . A prime here indicates the vector or tensor that has been rotated and thus has different components. The term "rank-2 tensor" also indicates the total object A which has components Aij, just as a "vector" is a total object v which has components vi. Some authors replace the word "rank" with "order" or "degree" so then A is a tensor of order 2, or perhaps A is a "2-tensor" where the words rank, order and degree are avoided altogether. We shall use the word "rank" even though it has other unrelated mathematical meanings (e.g. rank of a matrix). It should be understood that not all rank-2 tensors can be written as outer products of vectors. For example, in the general expansion T = Σij Tij eiej in the simple basis noted above we have T being the total rank-2 tensor, and Tij being its components. But in general, we don't have Tij = aibj. The key point is that Tij "transforms" and "behaves" the same way as aibj. We could construct an explicit Tij = aibj if we wanted by saying a b = (Σiaiei) (Σjbjej) = Σij (aibj) eiej. So, let's first define two outer product rank-2 tensors A = ab Aij = [ab]ij = aibj B = cd Bij = [cd]ij = cidj We can then define the tensor product of A and B in a fairly obvious manner, AB ≡ (ab)(cd) = abcd . // associative Then we take components in the following way, [AB]ijkl = [ abcd]ijkl = aibjckdl = AijBkl The above lines shows that AB is in fact a rank-4 tensor constructed by taking the "outer product" of two rank-2 tensors (or the outer product of four rank-1 tensors). Later we shall have use for an object defined in this strange manner [AB]ik,jl ≡ [AB]ijkl = AijBkl and we just mention it here in passing. Note that the indices are shuffled. Using the same method as above, we could construct a rank-6 tensor from the tensor product of three rank-2 tensors, [ABC]abcdef = AabBcdCef or from the tensor product of two rank-3 tensors [AB]abcdef = AabcBdef . At a slightly lower level, we could construct a rank-3 tensor from a rank-1 tensor and a rank-2 tensor, another form of the outer product, [v B]abc = va Bbc . In general, one can take the tensor product of any set of tensors to create a new tensor whose rank is the sum of the ranks of the tensors that were combined by the symbol. If α,β,γ... are arbitrary tensors, having multiindices I,J,K (for example I = {i1,i2,i3} if α is rank-3), we could write a general formula for the components of the tensor product of any number of pure tensor objects in this manner, [ α β γ ....]IJK... = αI βJ γK...... (*) The tensor here is α β γ .... and the equation specifies its components. The Direct Sum of Vector Spaces The direct sum of two or more vector spaces is a very simple concept. Consider G ≡ V2V3. The vectors in this space have 5 components and can be visualized by just stacking vectors into tall column vectors. For example, v = ϵ V2 w = ϵ V3 = ϵ V2V3 . One could write the above as vw ϵ V2V3. Since α = , one has α(vw) = (αv)(αw). And to the extent that a vector's position within the larger vector is immaterial, one has vw = wv . Regardless, one sees that each piece of the direct sum vector resides in its own private region within the tall column vector. Whereas dim(AB) = dimA * dimB, one has dim(AB) = dimA + dimB. The Tensor Algebra Suppose we define a very large vector space as the following direct sum of vector spaces, T = V0 V V2 V3 ....... on forever // = Σi=0∞ Vi The space V0 represents the space of scalars. The elements of this space T are then tensors of any rank. A vector would fit into the V part, a rank-2 tensor would fit into the V2 part, and so on. One could in fact have a linear combination of a vector and a rank-2 tensor in this space T. Since T contains arbitrary linear combinations of tensors of different ranks, T is called a "graded algebra" where the ranks of the pieces are the grades. The "algebra" part is that we have a set elements with rules + and for adding and multiplying elements. The set of elements of T is closed under each of these operations. This last fact should be clear to the reader. For example, one can think of the product of any number of tensors as [**...][**...][**...] ..... = ***** .... ϵ Vk (if k factors) ϵ T Then the most general elements of the space T are linear combinations (with coefficients in field K) of the tensors shown in (*). Thus huge vector space with all its many elements is known as the tensor algebra over field K", and again, we usually use K = R = reals. Of course vectors in the space T defined above have an infinite number of components. Usually only a finite number of these components are non-zero. In the context of this large space T, we have two meanings for "mixed" tensors which we don't want to confuse together. A pure rank-2 tensor has the form ab. A mixed rank-2 tensor is ΣijFijei ej . But a "mixed tensor" in the space T could be any linear combination of tensors of different rank, such as 5.3 + 2v + ΣijFijei ej - π abc Suppose t is such a mixed tensor composed of a linear combination of tensors whose ranks range from s to r. And suppose t' has ranks ranging from s' to r'. Then tt' can contain tensors whose ranks range from |s-s'| to r+r' 1.5 Kronecker Products The subject here is the tensor product of two linear operators, but as will be seen, it boils down combining two rank-2 tensors to make a rank-4 tensor. The reader can regard this section as an exercise in using the tensor product machinery, and the Kronecker product just arises along the way. Let V and X be vector spaces of dimension n and m. Basis(V) = ei Basis(X) = ei Let W and Y be vector spaces of dimension n' and m'. Basis(W) = e'i Basis(Y) = e'i Consider linear operators S and T such that, x = Sv = a vector in X S: V→X xi = Σa=1n Siava i = 1,2..m y = Tw = a vector in Y T:W→Y yj = Σb=1n'Tjbwb j = 1,2..m' The linear operator S is represented by matrix Sia which has m rows and n columns (m x n). The linear operator T is represented by matrix Tjb which has m' rows and n' columns (m' x n'). We want to create a meaning for ST which is the tensor product of these two operators S and T. A candidate definition for this meaning is the following, (ST)(vw) = (xy) = (Sv)(Tw) ST : VW → XY . Consider the following processing steps, (ST)([αv1 + βv2]w) = (S[αv1 + βv2])(Tw) // definition of action of (ST) = (α Sv1+ βSv2) (Tw) // S:V→X is linear = α (Sv1)(Tw) + β(Sv2)(Tw) // using the first rule for vectors = α (ST)(v1w) + β (ST)(v2w) . // definition of action of (ST) This shows that (ST)(vw) is linear in v. A similar argument shows it is also linear in w. Thus, the operator (ST) as defined above is a bilinear operator on VW, and we confirm the essential characteristic of the tensor product, which is its bilinearity. We accept the candidate. ____________________________________________________________________ Exercise: Compute the action of (ST) on a general element of VW . Applying (ST) to a general element of VW we get (ST)[ ΣijFij eie'j] = ΣijFij (ST)( eie'j) = ΣijFij (Sei)(Te'j) . The action of S on a vector v (and T on w) can be written as (Sv) = Σa[Sv]aea = Σa(ΣbSabvb)ea = Σab(Sabvb)ea (Tw) = Σc[Tw]ce'c = Σc(ΣdTcdwd)e'c = Σcd(Tcdwd)e'c . Therefore (Sei) = ΣabSab(ei)b ea (Te'j) = ΣcdTcd(e'j)d e'c . Then (Sei)(Te'j) = [ ΣabSab(ei)b ea] [ ΣcdTcd(e'j)d e'c] = Σabcd Sab(ei)bTcd(e'j)d (eae'c) and so (ST)[ ΣijFij eie'j] = ΣijFij (Sei)(Te'j) = Σijabcd FijSab(ei)bTcd(e'j)d (eae'c) = Σac { Σijbd FijSab(ei)bTcd(e'j)d } (eae'c) = Σac Gac (eae'c) where Gac = Σijbd FijSab(ei)bTcd(e'j)d . In the special case that ei and e'j are the standard unit vector bases for V and W, the result simplifies, Gac = Σijbd FijSabδi,bTcd δj,d = Σij FijSaiTcj = Σij SaiFijTTjc = (SFTT)ac or G = SFTT . // G(m x m') = S(m x n) F(n x n') TT (n' x m'), so matrices "conform" This shows explicitly how bilinear operator (ST) acts on a general element of VW to produce an element of the space XY which has basis eie'j. _______________________________________________________________ It is useful now to consider the component analysis of the action of ST on a pure element of VW in the sense of outer products. Then (xy) = (ST)(vw) = (Sv)(Tw) so (xy)ii' = [(ST)(vw)]ii' = [(Sv)(Tw)]ii' . (**) The right side of (**) is easily processed as above, [(Sv)(Tw)]ii' = (Sv)i(Tw)i' = (Σj Sijvj)(Σj'Ti'j'wj') (*) = Σjj' SijTi'j' vjwj' = Σjj' SijTi'j' (vw)jj' . (***) It is helpful to visualize the object in the middle of (**) in this manner, [(ST)(vw)]ii' = Σjj' (ST)ii',jj' (vw)jj' (****) so then (**) becomes (xy)ii' = Σjj' (ST)ii',jj' (vw)jj' as if we were multiplying a vector (vw) by a matrix (ST), but the usual summation index is replaced by two summation indices j and j'. In a multiindex notation one might write the above as (xy)I = [(ST)(vw)]I = ΣJ (ST)I,J (vw)J I = {i,i'} J = {j,j'} . Comparing (***) and (****) we find that, (ST)ii',jj' = SijTi'j' Note carefully how the indices are arranged: S gets the firsts, T gets the seconds. This same equation appeared earlier in Section 1.4. The object (ST)ii',jj' is a rank-4 tensor, since it is the outer product of two rank-2 tensors Sij and Ti'j', and as such it has four indices. As shown earlier, normally one would write (ST)iji'j' with no comma and with the indices in the same order as those in SijTi'j'. The alternate comma notation allows the quasi-matrix multiplication point of view shown above. Is there some way to write ST as a standard matrix with two indices instead of four? Go back to our equation (xy)ii' = Σjj' (ST)ii',jj' (vw)jj' or (xiyi') = Σjj' (ST)ii',jj' (vjwj') (ST)ii',jj' = (SijTi'j') . We want to write this somehow in a form q'r = Σs Mrs qs . For illustration purposes, assume n = 2 and n' = 3. Then write the components (vjwj') as a single column vector in this obvious manner, where the w component index moves fastest, = = q with components qs where s = 1,2....n*n' . If vjwj' → qs, one can compute s from j,j' as follows: ( here 3 = n' = dim(W) for this special case ) s = (j-1)3 + j' (s-1) = (j-1)3 + (j'-1) = (j-1) + int() = j-1 and rem () = j'-1 . Thus for general n' we can compute j and j' from s in this way (integer part and remainder) j = 1+int( ) j' = 1+rem( ) . s = 1,2....n*n' One can similarly consider xiyi'→ q'r where the column vector q' has m*m' components. The rules here are i = 1+int( ) i' = 1+rem( ) . r = 1,2...m*m' Therefore, the desired Mrs is given by Mrs = (ST)ii',jj' = SijTi'j' where i = 1+int( ) j = 1+int( ) s = 1,2....n*n' i' = 1+rem( ) j' = 1+rem( ) r = 1,2...m*m' . It is a bit tedious to compute and display one of these M matrices by hand, so we let Maple do it for us. For this example we use S = m x n = 2 x 3 rows = m*m' = 6 T = m' x n' = 3 x 4 cols = n*n' = 12 Symbolically we can write M = (ST), with the meaning shown above. Matrices M of this general type are known as Kronecker products. Staring at the above matrix, one can see that the T submatrix is repeated many times, and one can write this matrix in a shorthand notation as M = where T = . This provides an easy way to manually construct such matrices. This construction is explained if we look back at the M matrix definition, Mrs = (ST)ii',jj' = SijTi'j' where i = 1+int( ) j = 1+int( ) s = 1,2....n*n' i' = 1+rem( ) j' = 1+rem( ) r = 1,2...m*m' . The indices i,j on S select a rectangular subregion of the M matrix due to their integer part definitions. Then within each subregion the i'j' indices run through their full ranges so a copy of matrix T appears in that subregion, multiplied by the Sij for that subregion. One is commonly interested in the case where S: V→V S = n x n matrix T: W→W T = n' x n' matrix With n = m = 2 and n' = m' = 2 the above code generates this matrix M, which can be compared with a result quoted on the wiki tensor product page. 2. The Wedge Product built on the Tensor Product We now back up a bit and reconsider the space VW and its elements vw. The goal of the next two subsections is to establish the parallelism between the vector space VW and the "dual" vector space V*W*. Some repetition is used to reinforce earlier stated facts. Then the following two subsections introduce the wedge product developed in a similar parallel fashion. 2.1 The tensor product of 2 vectors Consider two vector spaces V and W of dimension n and n'. Let {ei} = basis of V dim(V) = n v = Σi=1n vi ei = general vector in V {ei'} = basis of W dim(W) = n' w = Σj=1n'wj ej' = general vector in W {eie'j} = basis for the tensor product space VW dim(VW) = n*n' vw = a pure "vector" in the tensor product space VW : VxW → VW : (v,w) ↦ vw The last line shows as a mapping → between two sets, while ↦ shows how set elements map. The basis vectors of V and W. The basis vectors of V are denoted ei. Expanding a vector v on these basis elements, one writes v = Σi viei. The coefficients vi take values in the field K (normally reals). These expansion coefficients vi are the "components of vector v in the ei basis". Since ek is itself a vector in V, one can write ek = Σi (ek)iei . Notice that k is a label on ek, whereas the i on (ek)i is a component index. So the components of the vector ek in the ei basis are (ek)i. Given any two vectors v1 and v2 in V, consider the sum f(v1,v2) ≡ Σk=1n(v1)k(v2)k which clearly is an element of the field K. This function f: VxV → K is manifestly bilinear in its two vector arguments, see *** . This sum can of course be applied to any two basis vectors, so that fij ≡ f(ei,ej) = Σk=1n(ei)k(ej)k We now make two definitions: The {ei} are orthogonal if fij = giδi,j : Σk=1n(ei)k(ej)k = giδi,j The {ei} are orthonormal if fij = δi,j : Σk=1n(ei)k(ej)k = δi,j In our work below, we sometimes use a simple basis ej where (ej)k = δj,k. Each such basis vector has all components zero except for the component whose index matches the label of the basis vector, and that component has magnitude 1. If V = Rn one would think of the en as en = which are axis-aligned unit vectors. This simple basis is orthonormal according to the above definition, since Σk=1n(ei)k(ej)k = Σk=1nδi,kδj,k = δi,j . In the discussion below where ei appear, we do NOT assume the ei are orthogonal or orthonormal. We just assume they are linearly independent, as must be true for any basis. The exact same comments apply to the basis vectors e'i of W. Comment: Notice in the above paragraphs there is no mention of metrics, metric tensors, metric spaces, norms, normed linear spaces, inner (scalar or dot) products, Hilbert spaces, Cartesian space, or any of the topics that arise when a vector space (= linear space) "grows up" and applies itself to some useful purpose. Normally one would add the following structure to the vector space V, <v1,v2> ≡ f(v1,v2) = Σk=1n(v1)k(v2)k // inner product in Cartesian space ||v1||2 = <v1,v1> // implied norm d2(v1,v2) = ||v1-v2||2 = <v1-v2,v1-v2> // implied metric Then along with the technical requirement of completeness, V becomes a Hilbert Space. We are trying to avoid the need for all this extra structure and keep the discussion at the level of the primordial vector space. Outer Product Revisited. The notion of an outer product was discussed in Section 1.3 above. We had for example (where ai and bj are the components of vectors a and b), (a b)ij = aibj // outer product of two vectors (here a,b are vectors) (A B)abcd = AabBcd // outer product of two rank-2 tensors (here a,b,c,d are indices) The outer product of two vectors a and b may be written in vector/matrix notation as follows, a b = abT = ( b1. b2....bn) = (a b)ij = (abT)ij and we have seen in ** how the outer product of AB can be written as a Kronecker product. The same vector/matrix notation used above can also be used to express the "inner product" function f(a,b) appearing in ***, f(a,b) = aTb = ( a1. a2....an) = Σk=1n akbk = Σk=1n bkak = bTa = f(b,a) Tensor Product Revisited. By convention one represents an element of a tensor product space using the symbol. It is a certain kind of "product" between a vector in one vector space and a vector in another vector space. On can treat as an operator : VxW → (VW) in the sense that (v,w) = (v) (w) = (vw) = element of tensor product space (VW). Certain rules were declared in Section 1 which make the tensor product be a vector space, and which in an intuitive sense just seem "reasonable", (kv) w = v (kw) = k (vw) // k = scalar (ϵ K) v (w1+ w2) = vw1 + vw2 // left distributive property (v1 + v2) w = v1w + v2w // right distributive property In the last two equations, the inside + represents addition in either W or V, whereas the + on the right side represents addition in VW. These lines say that multiplication "distributes" over addition +. The scalar rule can be combined with the distributive rules to obtain this equivalent rules restatement: v (k1w1+ k2w2) = k1(vw) + k2(vw2) // k1,k2 = scalar (ϵ K) (k1v1 + k2v2) w = k1(v1w)+ k2(v2w) . // k1,k2 = scalar (ϵ K) The above rules in effect say that defines a "bilinear" operation -- it is linear separately in each of its operands. Notice that the following two rules are incorrect: v w = w v // wrong! (unless V = W and v = w) (kv) (kw) = k (vw) // wrong! (unless k = 1) As noted earlier last rules apply to the direct sum , not the tensor product . Using the correct "rules" above, one may write v w = ( Σiviei)( Σjwjej') = Σijviwj (eie'j) showing how this pure tensor product vector can be expressed in terms of the basis functions. A general "vector" (tensor) in W x V can be written as a linear combination of the basis vectors, T ≡ ΣijFij eie'j T ϵ V x W . Now if it happens that the two spaces being tensor-multiplied are the same, so W = V, we refer to the space V x W = V x V as V2, and the most general "vector" of this tensor product space may be then written as in terms of an arbitrary basis as, T ≡ ΣijFij eiej . T ϵ V x V = V2 Although we have said T is a "vector" in the abstract sense that a vector space (even a tensor product vector space) has "vectors" as elements, the usual terminology is to say that T is a "rank-2 tensor" in the space V2. The word "tensor" has a weak and a strong meaning as noted earlier. In the weak meaning, a rank-2 tensor is something that has components with two indices like Fij. In the strong meaning, a rank-2 tensor is a set of components Fij which transform in a certain manner with respect to some transformation. As noted in ***, the components Fij are linear combinations of the simple-basis components Tij and so both component sets transform the same way under a transformation. For example under rotation R the rotated components T'ij are related to the unrotated components by T'ij = Σab RiaRjbTab F'ij = Σab RiaRjbFab . For a general non-linear transformation, Ria is the "differential" of the transformation. It is shown in tensor doc Appendix Q that if the {ei} are the "tangent base vectors" of that transformation, then the Fij are the "contravariant components" of the tensor T, which in standard notation are usually written Tij . On the other hand, if the {ei} are taken to be the "reciprocal base vectors" {ei), then the coefficients Fij are the "covariant components" of the tensor T, written as Tij. Written in proper up/down tensor notation, one then has T ≡ ΣijTij eiej ei = tangent base vectors Tij = contravariant components T ≡ ΣijTij eiej ei = reciprocal base vectors Tij = covariant components The other two combinations result in coefficients which are "mixed" tensors having an upper contravariant index and a lower covariant index, T ≡ ΣijTij eiej T ≡ ΣijTij eiej The reciprocal base vectors ei are just the dual vectors qi discussed earlier and in the next section. Notice again that it is the object T which is the real "tensor", and it can be expanded in various ways on different choices of basis elements resulting in components of various natures. 2.2 The tensor product of 2 dual vectors. Let V* and W* be the "dual spaces" of V and W. In our treatment of the dual space side of things, we shall attempt to use Greek or script-font letters for all dual space vectors and tensors encountered. Vectors in V or W or VW will continue to be represented by lower-case Latin letters. Assume that {λi} = basis of V* {λi'} = basis of W* Like all elements of V*, the object λi ϵ V* is a "linear functional over V". This is a linear function which, when evaluated at a point in V, produces a scalar: λi : V → K. As usual, we think of K as the reals R, but we try to stay more general by having K be an arbitrary field. In normal calculus, if f: V → R, one refers to f as a function, and f(v) as that function evaluated at some point in V, though loosely speaking f(v) is also called a function. To emphasize the distinction, we shall refer to f as a "functional" and f(x) as a "function". In particular, if the n vectors {ei} are a basis for V (basis vectors are linearly independent), one can find another set of n vectors {qi} in V such that, qiTej = δi,j. or ( (ej)1, (ej)2...) = or Σi=1n (qi)n(ej)n = δi,j A method for finding these unique qi from the ei is given in *** . The action of the linear functional λi can then be taken as λi(v) = qiTv which is manifestly linear since λi(αv) = αλi(v) and λi(v + v') = λ(v) + λ(v') α = scalar in field K Here v and v' are vectors in V. As noted earlier, the definition of linearity is often combined into a single statement, λi(αv + βv') = αλ(v) + βλ(v'), α,β = scalars in field K The vectors qi are sometimes called "covectors" or "reciprocal vectors" or "dual vectors" which form a "dual basis". From ** one then has λi(ej) = qiTej = δi,j . General linear functionals in V* and W* can be written as linear combinations of the basis functionals, α = Σiαiλi = general vector in V* α(v) = Σiαiλi(v) α: V → K β = Σjβjλj' = general vector in W* β(v) = Σjβjλj'(v) β: W → K where on the right we show the corresponding functions α(v) and β(v). Using **, one sees that α(ei) = αi β(e'j) = βj There is a dual tensor product space V*W* which has the basis λiλ'j . We want this object λiλ'j to be a functional over the space V*W* such that λiλ'j: V*xW* → K . The natural way to accomplish this desire is to write (λiλ'j)(v,w) = λi(v)λj'(w) = scalar * scalar = scalar ϵ K This function is manifestly "bilinear" in that it is separately linear in each of the arguments v and w. For example, (λiλ'j)(v1+v2,w) = λi(v1+v2)λj'(w) = [λ(v1) + λ(v2)]λj'(w) = λ(v1)λj'(w) + λ(v2)λj'(w) = (λiλ'j)(v1,w) + (λiλ'j)(v2,w) . The "rules" for the operator in the space V*W* are the same as those for in the space VW, (kα) β = α (kβ) = k (α β) k = scalar α (β1+ β2) = α β1 + α β2 // distributive property (α1 + α2) β = α1β + α2β . // same idea as above A general element of the dual tensor product space V*W* can be written T = Σij Fij λiλ'j where the Fij are coefficients in the field K. Evaluating at a point (v,w) in VxW one gets, T(v,w) = Σij Fij (λiλ'j)(v,w) = Σij Fij λi(v) λj'(w) ϵ K so one may regard T : VxW → K. Setting v = ei and w = e'j and using ***, one finds that T(ei,e'j) = Fij ϵ K which provides an interpretation of Fij as the function T(v,w) evaluated at two basis vectors. This may be compared with α(ei) = αi. More specifically, consider α ϵ V*and β ϵ W* as shown above. Then α β = (Σiαiλi)(Σjβjλ'j) = Σij αiβj λiλ'j ϵ V*W* (α β)(v,w) = Σij αiβj (λiλ'j)(v,w) = Σij αiβj λi(v)λj'(w) ϵ K which then is just a particular example of the more general T(v,w) above. Another way to write the above line is (α β)(v,w) = Σij αiβj λi(v)λj'(w) = (Σiαi λi(v))(Σjβj λ'j(v)) = α(v) β(v) where now the bilinear function (α β)(v,w) is equal to the product of the two functions α(v) and β(w). In particular (α β)(ei,e'j) = α(ei) β(e'j) = αiβj . If it happens that W = V, then W* = V* and we write V*W* = V*V* = V*2. Then the most general element T of V*2 can be expressed as T = Σij Fij (λiλj) T(v1,v2) = Σij Fij (λiλj)(v1,v2) = ΣijFij λi(v1)λj(v2) . T(ei,ej) = Fij In analogy with Section 2.1 above, one says that the linear functional T in the dual space V*2 is a "tensor" of rank 2, or order 2, or it is a 2-tensor. Whereas T earlier was a rank-2 tensor in V2, this object T is a rank-2 tensor in the dual space V*2. In the abstract, it is T which is the real tensor and it is shown expanded on a particular basis {λi} which in turn is associated with a particular basis {ei} of V as discussed above. What are the "components" of tensor T ? In a sense they are the coefficients Fij of the expansion shown above onto the basis (λiλj), in analogy with the non-dual space. But in the dual space this is not the sense of "component" we are so interested in. In the dual world, the function vector arguments play the role that vector component indices play in the outer products of the non-dual world. For example, compare these equations: (a b)ij = aibj // VW (α β)(v,w) = α(v)β(w) // V*W* T = Σij Fij eie'j // VW Tab = Σij Fij (eie'j)ab = Σij Fij (ei)a(e'j)b T = Σij Fij λiλ'j // V*W* T(v,w) = Σij Fij λiλ'j(v,w) = Σij Fij λi(v)λ'j(w) . So it is really the T(v,w) which are the "tensor components" of the tensor T, whereas the Fij are the coefficients in the above expansion, In the non-dual space the "tensor components" were Tij . One can then regard T(v,w) as the tensor of interest. For example, Spivak on page 75 refers (in effect) to our function T(v1,v2) as a being a 2-tensor. The space of bilinear functionals on V2 = VV (which includes any T above) is just V*2 = V*V*. As just noted, one can regard V*2 also as the space of all bilinear functions T(v1,v2) . Fact: The vector space V*2 is equivalent to the vector space of bilinear functions on V. Comment: It is possible to emphasize the parallelism between the dual-world and the non-dual world by considering, in analogy with the functions λi(v), alternate functions vi(λ). Either world is then the dual of the other world. This is the approach taken on page 2 of Benn and Tucker. 2.3 The wedge product of 2 vectors (a) Definition of the wedge product and the space L2 Momentarily jumping ahead, we shall find that the wedge product of v ϵ V and w ϵ W is going to be v ^ w = vw - wv . If V and W are different vector spaces, this does not make any sense since the second term wv implies that w lies in the left space V and v lies in the right space W. So in our discussion of wedge products, we require that W = V. This being the case, instead of using letters v and w as representative vectors, we shall a and b. Then ei are the basis vectors for both component spaces in the tensor product space VV. So, we start off by defining the following "wedge product" of two vectors a and b of V, a ^ b ≡ ab - ba . dim(V) = n Notice therefore that a ^ b is an element of VV = V2, since it is a linear combination of elements of VV. It is "antisymmetrized" under a ↔ b. Since not all elements of VV can be written this way, the set of elements a ^ b exist in a subset of VV which we shall call L2, so L2 V2. Some authors write L2 as Λ^Λ (projmec ref), but this notation is uncommon. The above definition implies that a ^ b = - b ^ a a,b ϵ V so a ^ a = 0 a ϵ V . In Section *.* we stated certain scalar and distributive properties of the operator. These properties are transferred onto the wedge ^ operator by the above definition. For example, (ka) ^ b = (ka)b - b(ka) = k [ ab - ba ] = k (a ^ b) k = scalar (a+c) ^ b = (a+c)b - b (a+c) = ab + cb - ba - bc = [ ab - ba ] + [ cb - bc ] = (a ^ b) + (c ^ b) distributive and similarly for a ^ (kb) and a ^ (b + c). To summarize: (ka) ^ b = k (a ^ b) (a+c) ^ b = (a ^ b) + (c ^ b) a ^ (kb) = k (a ^ b) a ^ (b + c) = (a ^ b) + (a ^ c) The operator ^ is then seen to be "bilinear" over elements of V: it is separately linear in each operand. All the above equations are valid for the special case where a,b,c are basis vectors of V, a = ei b = ej c = ek. why is this here? To more precisely define the space L2, we claim that the most general element of the space L2 can be written this way, T = Σij Fij ei ^ ej . For example, if Fij = aibj this would be T = Σij aibj ei ^ ej = (Σiaiei) ^ ( Σjbjej) = a ^ b and then a ^ b is included in L2 for any vectors a and b in V. One could rearrange the n2 basis vectors of VV into these two groups: (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 for a total of n(n-1)/2+ n(n)/2 = n2 basis vectors. One would say then that L2 is spanned by just the first set of basis vectors. It was noted above that L2 is a subset of V2. A stronger statement is that L2 is a subspace of V2. First of all, L2 is obviously closed under addition of vectors since Σij Fij ei ^ ej + Σij F'ij ei ^ ej = Σij (Fij+F'ij) ei ^ ej . And if (a ^ b) is an element of L2 then so is k(a ^ b) = (kα) ^ b ϵ L2 . Finally, since a ^ a = 0, L2 includes the 0 element. So L2 then is a vector space which is a subspace of V2. (b) How big is the space L2 compared to the space V2? Consider this most general element of L2: T = Σij Fij (ei ^ ej) = Σi≠ j Fij (ei ^ ej) // (ei ^ ei) = 0 = Σi<j Fij (ei ^ ej) + Σi>j Fij (ei ^ ej) = Σi<j Fij (ei ^ ej) + Σj>i Fji (ej ^ ei) // i↔j in second sum = Σi<j Fij (ei ^ ej) - Σi<j Fji (ei ^ ej) // (ej ^ ei) = - (ei ^ ej) = Σi<j (Fij- Fji) (ei ^ ej) = Σi<j Aij (ei ^ ej) Aij ≡ (Fij- Fji) Aij = - Aji Thus, the number of elements in L2 is equal to the number of antisymmetric n x n matrices A one can construct which contain elements of field K. An n x n antisymmetric matrix has only n(n-1)/2 places to insert independent values since the diagonal is all zeros and one triangle is the negative of the other. If the scalar space K contains N elements ( N = ∞ for the reals), one could then construct exactly Nn(n-1)/2 antisymmetric matrices A. Meanwhile, the most general element of V2 can be written T = Σij Fij (ei ej) . Now each matrix Fij defines an element of V2. Using the same counting method as above, the total number of elements of V2 is Nn2. We conclude that = = (1/2) = (1/2) (1 - ) The conclusion is that L2 contains less than half the number of elements in V2. This ratio is of course the same as the count ratio of L2 basis vectors to V2 basis vectors: [n(n-1)/2] / [n2] = (n-1)/2n. (c) Wedge products and determinants: the geometry connection. For the case discussed above where Fij = aibj we found a ^ b = Σij aibj (ei ^ ej) = Σi<j (aibj- ajbi) (ei ^ ej) = Σi<j det (ei^ej) . The determinants which appear here are minors of a matrix having n rows and 2 columns. The two columns are the vectors a and b, each of which has n components. That matrix is shown on the left, and some of the minors are shown in gray on the right: If V = R2 (so n=2), the above matrix is square and there is then only one minor which is the determinant of the entire matrix, a ^ b = det e1^e2 = det(a,b) e1^e2 = [ a1b2 - a2b1] e1^e2 . It is easy to show (draw a picture) that det(a,b) is the area of the parallelogram (2-piped) spanned by vectors a and b. There is then some connection between the wedge product of two vectors in R2 and the geometry of R2. Later we will show that for V = R3 the triple wedge product of three vectors is given by, a ^ b ^ c = det(a,b,c) (e1^ e2^ e3) *** make sure this is shown somewhere ! and here det(a,b,c) is the volume of the 3-piped spanned by the vectors a,b,c. However, for R3 the wedge product of two vectors is more complicated. Using the above expression, we find a ^ b = det e1^e2 + e1^e3 + e2^e3 = [a1b2 - a2b1] e1^e2 + [a1b3 - a3b1] e1^e3 + [a2b3 - a3b2] e2^e3 . The coefficients are (apart from a sign) those which appear in the normal "cross product" of two vectors, a x b = [a1b2 - a2b1] e3 - [a1b3 - a3b1] e2 + [a2b3 - a3b2] e1 . We do not wish, however, to identify for example e1^e2 with e3. After all, e3 is a basis vector in V, whereas e1^e2 is a vector in the tensor product space VV. One can, on the other hand, define a correspondence of sorts where one says (each line in cyclic order, and ↔ means "corresponds to") e1^e2 ↔ e3 e2^e3 ↔ e1 e3^e1 ↔ e2 // = - e1^e3 in which case one can say a ^ b = [a1b2 - a2b1] e1^e2 + [a1b3 - a3b1] e1^e3 + [a2b3 - a3b2] e2^e3 ↔ a x b = [a1b2 - a2b1] e3 + [a1b3 - a3b1] e2 + [a2b3 - a3b2] e1 so there is then a correspondence between the wedge product and the cross product in R3. This correspondence was described by Scottish mathematician William Hodge (1903-1975) around 1941 and the relationship ↔ is formalized by the "Hodge dual operator * " where for example *(e1^e2) = e3 in R3. For Rn with n > 3 there is no cross product of two vectors, but there is a wedge product. For V = R4 for example, using the result stated above, a ^ b = det e1^e2 + dete1^e3 + dete1^e4 + dete2^e3 +dete2^e4 + dete3^e4 . There are enthusiastic workers (e.g. Denker) who recommend deep-sixing the cross product altogether and replacing it with the wedge product for the study of topics like angular momentum (Ref...). (d) Components Here we study the outer product components of various equations stated above. (eie'j)rs = (ei)r(e'j)s // VW (eiej)rs = (ei)r(ej)s // VV = V2 (ei^ ej)rs = (ei)r(ej)s - (ej)r(ei)s = det // L2 = V^V (ei^ ej)rs = - (ei^ ej)sr = - (ej^ ei)rs . // two forms of antisymmetry Using the regular double sum form, one has (a ^ b)rs = Σij aibj (ei ^ ej)rs = Σij aibj [ (ei)r(ej)s - (ej)r(ei)s] = Σij aibj det . For the simple basis (ei)k = δi,k this becomes (a ^ b)rs = Σij aibj [δi,rδj,s - δj,rδi,s] = arbs - asbr // simple basis Using the ordered double sum, we find instead (a ^ b)rs = Σi<j (aibj- ajbi) (ei ^ ej)rs = Σi<j det (ei ^ ej)rs = Σi<j det [(ei)r(ej)s - (ej)r(ei)s] = Σi<j det det . For the simple basis (ei)k = δi,k, evaluation of the ordered sum requires a bit more work : (a ^ b)rs = Σ1≤i<j≤n det [δi,rδj,s - δj,rδi,s] . For r = s, one clearly has (a ^ b)rs = 0. Otherwise we may write, using θ(Boolean) = 1 if true else 0, (a ^ b)rs = det θ(r<s) - det θ(s<r) = det θ(r<s) + det θ(s<r) // swap rows 2nd term = det [ θ(r<s) + θ(s<r)] = det = arbs - asbr Combining the results for r=s and r≠s we get (a ^ b)rs = arbs - asbr // simple basis producing the same result as ** . For components of a general element of L2 we find, using the regular double sum, Trs = Σij Fij (ei ^ ej)rs = Σij Fij [(ei)r(ej)s - (ej)r(ei)s] = Σij Fij det . For the simple basis (ei)k = δi,k this becomes Trs = Σij Fij[δi,rδj,s - δj,rδi,s] = Frs - Fsr = Ars Using the ordered double sum, Trs = Σi<j Aij (ei ^ ej)rs = Σi<j Aij [δi,rδj,s - δj,rδi,s] For r = s one has [..] = 0 so Trs= 0. Otherwise, Trs = Ars θ(r<s) - Asrθ(s<r) = Ars θ(r<s) + Arsθ(s<r) = Ars[ θ(r<s) + θ(s<r)] = Ars with the conclusion that Trs = Ars for all r,s ϵ (1,n) 2.4 The wedge product of 2 dual vectors We mimic the approach of Section 2.3 for the tensor product space VV, but now we have V*V*. (a) Definition of the wedge product and the space Λ2 We start off by defining the following "wedge product" of two vectors α and β of V* α ^ β ≡ α β - β α . Notice therefore that α ^ β is an element of V* x V* = V*2, since it is a linear combination of elements of V*V*. It is "antisymmetrized" under α ↔ β. Since not all elements of V*V* = V*2 can be written this way, the set of elements α ^ β exist in a subspace of V*2 which we shall call Λ2, so Λ2 V*2. The proof that Λ2 is a subspace and not just a subset of V*2 is the same as in the previous section, and some authors write Λ2 = V* ^ V*. The above definition implies that α ^ β = - β ^ α α, β ϵ V* so α ^ α = 0 α ϵ V* . The "rules" for the ^ operator in V*2 are found as they were for V2, to wit : (kα) ^ β = k (α ^ β) (α+γ) ^ β = (α ^ β) + (γ ^ β) α ^ (kβ) = k (α ^ β) α ^ (β + γ) = (α ^ β) + (α ^ γ) where α,β,γ are vectors in V* and k is a scalar in K. All the above equations are valid for the special case where α,β,γ are basis vectors of V* : α = λi β = λj γ = λk. To more precisely define the space Λ2, we claim that the most general element of the space Λ2 can be written this way (that is, Λ2 is the space spanned by the λi ^ λj basis vectors) T = Σij Fij (λi ^ λj) . For example, if Fij = αiβj this would be T = Σij αiβj (λi ^ λj) = (Σiαiλi) ^ ( Σjβjλj) = α ^ β and then α ^ β is included in Λ2 for any vectors a and b in V. (b) How big is the space Λ2 compared to the space V2? Just as in Section 3, we can show that T = Σij Fij (λi ^ λj) = Σi<j Aij (λi ^ λj) Aij ≡ (Fij- Fji) Aij = - Aji where Aij is an antisymmetric n x n matrix. Using the same argument of Section 2.3, we find = = (1/2) = (1/2) (1 - ) (c) Wedge products and determinants For the case discussed above where Fij = aibj we found α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- αjβi) (λi ^ λj) = Σi<j det (λi ^ λj) For V* = R2 (so n=2) α ^ β = det λ1^ λ2 = det(α,β) λ1^ λ2 = [ α1β2 - α2β1] λ1^ λ2 . For V* = R2 (so n=2) α ^ β ^ γ = det(α,β,γ) (λ1^ λ2^ λ3) (d) Components (functions) (λiλ'j)(v,w) = λi(v)λj'(w) // V*W* (λiλj)(vr,vs) = λi(vr)λj(vs) // V*V* (λi^ λj)(vr,vs) = (λiλj)(vr,vs) - (λjλi)(vr,vs) = λi(vr)λj(vs) - λj(vr)λi(vs) = det // Λ2 = V* ^ V* (λi^ λj)(vr,vs) = - (λi^ λj)(vr,vs) = - (λj^ λi)(vr,vs) From ** we see that the function (λi^ λj)(vr,vs) is antisymmetric in its arguments. From ** we see in addition that (λi^ λj)(vr,vs) is a bilinear function because λk(vk) is a linear function, see **. Using the regular double some form one has (α ^ β)(vr,vs) = Σij αiβj (λi ^ λj)(vr,vs) = Σij αiβj [ λi(vr)λj(vs) - λj(vr)λi(vs)] = Σij αiβj det . Evaluated at vr = er and vs = es this becomes (α ^ β)(er,es) = Σij αiβj [ λi(er)λj(es) - λj(er)λi(es)] = Σij αiβj [δi,rδj,s - δj,rδi,s] = αrβs - αsβr Using the ordered double some form one has, using Fij = αiβj (α ^ β)(vr,vs) = Σi<j (αiβj- αjβi) (λi ^ λj)(vr,vs) = Σi<j det (λi ^ λj)(vr,vs) = Σi<j det [ λi(vr)λj(vs) - λj(vr)λi(vs)] = Σi<j det det . Evaluation at vr = er and vs = es may be done as in the previous section producing a repeat of the result **. For components of a general element of Λ2 we find, using the regular double sum, T(vr,vs) = Σij Fij (λi ^ λj)(vr,vs) = Σij Fij [ λi(vr)λj(vs) - λj(vr)λi(vs)] = Σij Fij det Moreover, the function T(vr,vs) is bilinear in its arguments since (λi ^ λj)(vr,vs) is bilinear, being equal to [ λi(vr)λj(vs) - λj(vr)λi(vs)] and since λk(v) is linear in its argument. Evaluated ** at vr = er and vs= es this becomes T(er,es) = Σij Fij [ λi(er)λj(es) - λj(er)λi(es)] = Σij Fij [δi,rδj,s - δj,rδi,s] = Frs - Fsr = Ars Using the ordered double sum T(vr,vs) = Σi<j Aij (λi ^ λj)rs = Σi<j Aij [λi(er)λj(es) - λj(er)λi(es)] = Σi<j Aij det . Evaluated at vr = er and vs = es this becomes T(er,es) = Σi<j Aij [λi(er)λj(es) - λj(er)λi(es)] = Σi<j Aij [ δi,rδj,s - δj,rδi,s] For r = s it is clear that T(vr,vs) = 0, otherwise T(er,es) = Ars θ(r<s) - Asr θ(s<r) = Ars θ(r<s) + Ars θ(s<r) = Ars [θ(r<s) + θ(s<r)] = Ars and we find that T(er,es) = Ars for all r,s ϵ (1,n) It was noted above that (λi^ λj)(vr,vs) is an antisymmetric bilinear functions of its arguments. This fact passes through to (α ^ β)(vr,vs) and more generally to T(vr,vs), as is evident from ** or **. We conclude that Fact: The vector space Λ2(V) is equivalent to the vector space of antisymmetric bilinear functions on V. This may be compared to our earlier statement: Fact: The vector space V*2 is equivalent to the vector space of bilinear functions on V. 2.5 The tensor product of k vectors Our task is now to generalize the tensor product from V2 to Vk, where Vk ≡ VV .... V // k component spaces, each one is V We are setting up for a parallel treatment in the next section where becomes ^, so certain rather obvious statements will be made here to allow for comparison later with the wedge product. A generic pure ("decomposable") element of Vk is v1 v2 ..... vk . all vi ϵ V The basis elements of Vk are ei ei ..... ei . If n = dim(V), the total number of such basis elements is nk, so dim(Vk) = nk. In the full set of such tensor product basis elements, two or more of the ek might be the same. This will always be the case if k > n where n ≡ dim(V). A rank-k tensor T in Vk has this general expansion T = Σii....i Fii....i (ei ei ..... ei) . Using the notion of a multiindex I (an ordinary multiindex), I ≡ {i1, i2, .....ik} is = 1,2....n and a shorthand notation for the basis vectors eI ≡ ei ei ..... ei the general rank-k tensor in Vk can be expanded in the following compact notation, T = ΣI FI eI . The tensor product of k vectors is "k-multilinear" meaning it is linear in each of its k factors. For example v1(v2 + v2')v3.....vk = v1v2v3 .....vk + v1v2'v3 .....vk v1(sv2)v3 ..... vk = s(v1v2v3 .....vk) s = scalar Here we show linearity in the 2nd factor. All the other factors have similar equations. We impose this k-multilinearity by fiat with the result that Vk is then a vector space, and this can be verified by mimicking our earlier work with V2. The above equations are meaningful for any integer k, regardless of the value n = dim(V). In Section 1.4 we introduced the so-called tensor algebra. Here we repeat that discussion with a bit more detail. Normally one does not add apples and oranges, so one does not add items of the form ab ϵ V2 to those of the form abc ϵ V3. However, as one writer notes, fruit salad is great, and so we could define a very large vector space of the form T(V) = V0 V V2 V3 ....... Here V0 = the space of scalars, V1 = V the space of vectors, V2 = VV = the space of "bivectors", and so on. The most general element t of the space T(V) would have the form t = s + ΣiFi ei + Σij Fij eiej + Σijk Fijk eiejek + ...... This large space T(V) is in fact itself a vector space. It is closed under addition and has the right scalar rule. For example, if k1 and s are scalars, k1 + a + bc + fgh = sum of 4 elements of T(V) = an element of T(V) s(k1 + a + bc + fgh) = (sk1) + (sb) c + f(sg)h = element of T(V) The space is also closed under the multiplication operation . For example (bc)(fgh) = bcfgh = ϵ V5 = ϵ T(V) . One could make the following definitions with regard to the space T(V): k1 scalar 0-blade 0 a vector 1-blade 1 ab bivector 2-blade 2 abc trivector 3-blade 3 abcd quadvector 4-blade 4 ..... arbitrary element of T(V) multivector linear combination of any the above However, these terms are reserved for objects within the wedge space we shall define below. Since T(V) is closed under the operations + and , it is "an algebra" (the space Vk alone is not an algebra because it is not closed under ). The T(V) algebra is different from that of the reals, due its definition as a direct sum of vector spaces. The elements of T(V) have different "grades" as shown in the right column above, and is known therefore as a "graded algebra". Sometimes T(V) is called "the tensor algebra" over V, see for example Benn and Tucker page 3. The dimensionality of the space T(V) is as follows, where n = dim(V), dim(T) = 1 + n + n2 + n3 + ... = ∞ 2.6 The wedge product of k vectors Turning now to wedge products, we want to define the wedge product of k vectors in V, v1^ v2^ .....^ vk . We impose the requirement that this wedge product changes sign when any two vectors are swapped. This leads to the following candidate wedge product definition which we write in three equivalent forms : v1^ v2^ .....^ vk = Σii....i εii....i (vi vi ..... vi) = ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) = v1 v2 ..... vk + all signed permutations (***) In the first line, ε is the usual permutation tensor (here with k indices) and the sum over each index ranges from 1 to k. If the first term in this sum is taken to be that term which has ir= r, one obtains the third line since ε123..k = 1. On the second line the sum is over all permutations P of the set of integers {1,2,3....k}, and S(P) is the number of pair swaps involved in the permutation P. For example P{1,2,3...k} = {2,1,3...k} has S(P) = 1. If the first term in this sum involves the identity permutation P = I, then again one obtains the third line. The third line is really just a vague symbolic way of writing out either of the first two lines, where we expose the first term. The sum contains k! terms, The number of permutations P of k objects is k!, so the second line says there are k! terms. In the first line, although there are kk terms in the sum, only those terms for which ε has all different indices contribute, so again there are k! terms. We shall verify the "changes sign" rule momentarily, thus vetting our candidate expressions above. Meanwhile, as encouragement, here is the first expansion above for k = 2 and k = 3: v1^ v2 = Σa,b =12 εab va vb // 2! = 2 terms = v1 v2 - v2 v1 // agrees with *** v1 ^ v2 ^ v3 = Σa,b,c =13 εabc va vb vc // 3! = 6 terms = v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3 . ok to here The wedge product is k-multilinear. Since the wedge product is a linear combination of tensor products, the fact that the operator is k-multilinear passes through to the wedge ^ operator (see ***). Thus, for example, the rules given above for become v1^(v2 + v2')^v3^.....^vk = v1^v2^v3^ .....^vk + v1^v2'^v3^ .....^vk v1^(sv2)^v3^ .....^ vk = s(v1^v2^v3^ .....^vk) s,s1,s2 = scalar or v1^(s1v2 + s2v2')^v3^.....^vk = s1(v1^v2^v3^ .....^vk) + s2(v1^v2'^v3^ .....^vk) and these rules apply independently to every vector in the wedge product. Verification of the "changes sign under pair swap" rule. Our candidate equivalent forms from above are v1^ v2^ .....^ vk = Σii....i εii....i (vi vi ..... vi) = ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) . For the first line, consider a swap of 1 and 2: v2^ v1^ .....^ vk = Σii....i εii....i (vi vi ..... vi) = Σii....i εii....i (vi vi ..... vi) // swap dummy indices i1↔i2 = Σii....i [-εii....i] (vi vi ..... vi) // swap ε indices i1↔i2 = - v1^ v2^ .....^ vk . For the second line we start off with v2^ v1^ .....^ vk = ΣP (-1)S(P) ( vP(2) vP(1) ..... vP(k)) Define the permutation P' by P'{1,2...k} = {2,1...k} which has S(P') = 1. Then the above sum my be written this way, since for example 2 = P'(1) and then P(2) = P(P'(1)) = PP'(1), = ΣP (-1)S(P) ( vPP'(1) vPP'(2) ..... vPP'(k)) . The swap count S(PP') = S(P) + S(P') = S(P) + 1 so that (-1)S(PP') = - (-1)S(P). Then = - ΣP (-1)S(PP') ( vPP'(1) vPP'(2) ..... vPP'(k)) . Finally we can use the "rearrangement theorem" of group theory applied to the permutation group. This theorem says that ΣP f(PQ) = ΣP f(P) where Q is any group element. The first sum is just a reordering of the second sum and so equals the second sum. Using Q = P' we continue to get, = - ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) = - v1^ v2^ .....^ vk . Wedge product vanishes if any two vectors are the same. Given a sign change for any pair swap of vectors in the wedge product, we know that v1^ v2^ .....^ vk = 0 if any two (or more) vectors are the same. For example. a ≡ v2^ v1^ .....^ vk = - v1^ v2^ .....^ vk = -a if 1 = 2, so a = 0. We can then write vj ^ vj ^ .... ^ vj = εjj....j ( v1^ v2^ .....^ vk ) . If the subscripts jr are a permutation of {1,2...k}, the above expression generates the correct sign for the wedge product relative to v1^ v2^ .....^ vk . If one or more subscripts jr are the same, the ε forces the result to be zero. Our two expressions for the wedge product stated above then become vj ^ vj ^ .... ^ vj = εjj....j Σii....i εii....i (vi vi ..... vi) = εjj....j ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) . Wedge product vanishes if vectors are linearly dependent. We have just shown that the wedge product vanishes if any two vectors are the same. It is also true that the wedge product v1^ v2^ .....^ vk vanishes if the vectors vi are linearly dependent. Linear dependence means one can write at least one vector in the set as a linear combination of the others, so perhaps v2 = ( Σi≠2 aivi). Then v1^ v2^ .....^ vk = v1^ ( Σi≠2 aivi)^ .....^ vk = Σi≠2 ai (v1^ vi ^ .....^ vk) // since ^ is k-multilinear, see above But (v1^ vi ^ .....^ vk) = 0 for any i ≠2 since then two vectors are the same. QED. Wedge product vanishes if k > n . If dim(V) = n, we know there can be at most n linearly independent vectors in V. If k > n, any set of k vectors must be linearly dependent. Thus, the wedge product of any set of k vectors must vanish if k > n. Therefore for a given vectors space V of dimension n, the only wedge products of interest are those for k = 1,2,3....n. Basis elements for Lk. Consider the following objects in Lk obtained by wedging together k basis elements of V, where each ei is selected from the set of n available for V which has dimension n, (ej ^ ej ^ .... ^ ej) . Of these putative nk objects, only n*(n-1)*...*(n-k+1) = n!/(n-k)! are non-zero because all the others have at least two vectors the same, so we can assume that all the subscripts jr are different. Now there exists a unique permutation P of the all-different subscripts {jr} such that { j1, j2....jk} = P{ i1, i2....ik} where i1 < i2 < ..... < ik If this permutation involves S pairwise swaps of indices, then we know that (ej ^ ej ^ .... ^ ej) = (-1)S (ei ^ ei ^ .... ^ ei) where i1 < i2 < ..... < ik because each pairwise swap of vectors in a wedge product creates a minus sign. Since there are k! possible permutations P, there are k! equations like the above which relate different objects to the same object (ei ^ ei ^ .... ^ ei) which has i1 < i2 < ..... < ik .Thus, if we want to count our number of independent basis elements of Lk, we have to divide our earlier count of n!/(n-k)! non-vanishing objects by k!. The conclusion is that there are (n,k) independent basis elements for Lk and they all have this form (ei ^ ei ^ .... ^ ei) where i1 < i2 < ..... < ik basis elements Example: for k = 3 and n = 6 we have e1^e3^e5 = (-1)0 e1^e3^e5 = +e1^e3^e5 135 e1^e5^e3 = (-1)1 e1^e3^e5 = - e1^e3^e5 153→135 e3^e1^e5 = (-1)1 e1^e3^e5 = - e1^e3^e5 315 →135 e3^e5^e1 = (-1)2 e1^e3^e5 = +e1^e3^e5 351→315→135 e5^e1^e3 = (-1)2 e1^e3^e5 = +e1^e3^e5 513→153→135 e5^e3^e1 = (-1)3 e1^e3^e5 = - e1^e3^e5 531→513→153→135 Each of this group of 3! objects is equal to + or - the same object. We now define Lk to be the space whose objects can be written in this form T = Σi<i<....<i Tii...i (ei ^ ei ^ .... ^ ei) because the set (ei ^ ei ^ .... ^ ei) with i1 < i2 < ..... < ik forms a complete basis for Lk. The coefficients of the (n,k) basis elements of the sum are the Tii...i where i1 < i2 < ..... < ik. On the other hand, we know we can also expand this same Q as T = Σii...i Fii...i (ei ^ ei ^ .... ^ ei) What then is the connection between the Tii...i and the Fii...i coefficients? Let's start with this last form: T = Σii...i Fii...i (ei ^ ei ^ .... ^ ei) = Σi≠i≠...≠i Fii...i (ei ^ ei ^ .... ^ ei) We then partition the summation space as follow: Σi≠i≠...≠i = [ Σi<i<...<i + Σi<i<...<i + many similar reorderings ] The total sum can be written in this manner, using the permutation sum notation, Σi≠i≠...≠i = ΣP ΣP(i)<P(i)<...<P(i) where P are the k! permutations of the k integers {1,2,...k}. Thus, we can write T = ΣP ΣP(i)<P(i)<...<P(i) Fii...i (ei ^ ei ^ .... ^ ei) . Just below we shall prove the following Lemma, which in the meantime we hope seems at least plausible to the reader, ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . Within the ΣP permutation sum, the index permutation operators have moved from the summation indices to the summand indices. Accepting this Lemma, we then have T = Σi<i<...<i ΣP [ FP(i)P(i)...P(i) (eP(i) ^ eP(i) ^ .... ^e P(i)) ] . But we know that ( eP(i) ^ eP(i) ^ .... ^e P(i)) = (-1)S(P) (ei ^ ei ^ .... ^ ei) where S(P) is the number of swaps associated with permutation P. Thus. T = Σi<i<...<i [ΣP (-1)S(P) FP(i)P(i)...P(i)] (ei ^ ei ^ .... ^ ei) which we can compare with our other sum T = Σi<i<....<i Tii...i (ei ^ ei ^ .... ^ ei) . Thus, the relation between the T and F coefficients is given by Tii...i = ΣP (-1)S(P) FP(i)P(i)...P(i) i1 < i2 < ..... < ik = Fii...i - Fii...i + other signed permutations = a total of k! terms Examples: Tab = Fab - Fba k = 2 a < b Tabc = Fabc - Facb + Fcab - Fcba + Fbca - Fbac k = 3 a < b < c Exercise: Show that an arbitrary wedge product of k vectors lies in the space Lk Consider the expansion T = Σii...i Fii...i (ei ^ ei ^ .... ^ ei) where Fii...i = ai bi ... Then get T = Σii...i ai bi ...(ei ^ ei ^ .... ^ ei) = a ^ b ^ ... ^ q To obtain this same result using the other expansion, one would set Tii...i = ΣP (-1)P aP(i)bP(i)...qP(i) = aibi...qi - aibi...qi + other signed terms Number of elements in Lk compared with Vk. From above we have found that dim(Vk) = nk // number of basis elements of Vk dim(Lk) = // number of basis elements of Lk If the number of elements of field K is N, then ratio = = = = / nk . For a given n, this is a strongly decreasing function of k. For example, for n = 10 we can plot the log of the ratio for k = 0 to 10, Multi-index notations. This is done in two different ways. First, using the "redundant" expansion, T = Σii...i Fii...i (ei ^ ei ^ .... ^ ei) T = ΣI FI eI where eI ≡ ei ^ ei ^ .... ^ ei and I ≡ 1 ≤ i1, i2,.... ik ≤ n = ordinary multiindex n = dim(V) The more significant notation involves the other expansion which has only one term for each linearly independent basis element T = Σi<i<....<i Tii...i (ei ^ ei ^ .... ^ ei) T = ΣI TI eI where eI ≡ ei ^ ei ^ .... ^ ei and I ≡ 1 ≤ i1 < i2 <....< ik ≤ n = increasing multiindex n = dim(V) The full wedge tensor algebra. We now repeat the step taken above for tensor products of vectors for wedge products of vectors, cribbing the text and editing in changes. Define a very large vector space of the form, L(V) = L0 L1 L2 L3 + .... Here L0 = the space of scalars, L1 = V the space of vectors, L2 = V ^ V = the space of (wedge) "bivectors", and so on. The most general element of the space L(V) would have the form T = s + ΣiFi ei + Σij Fij ei^ej + Σijk Fijk ei^ej^ek + ..... or T = s + ΣiTi ei + Σi<j Tij ei^ej + Σi<j<k Tijk ei^ej^ek + ..... This large space L(V) is in fact itself a vector space. It should be clear to the reader that it is closed under addition and has the right scalar rule. For example, if k1 and s are scalars, k1 + a + b^c + f^g^h = sum of 4 elements of L(V) = an element of L(V) s(k1 + a + b^c + f^g^h) = (sk1) + (sb) ^c + f^(sg)^h = element of L(V) The space is also closed under the multiplication operation ^. For example (b^c)^(f^g^h) = b^c^f^g^h = ϵ L5 = ϵ L . One then makes the following definitions with regard to the space L: k1 scalar 0-blade 0 a vector 1-blade 1 a^b bivector 2-blade 2 k = 2 a^b^c trivector 3-blade 3 k = 3 a^b^c^d quadvector 4-blade 4 k = 4 ..... a^b^c^d^.... k-vector k-blade 4 k = k .... a^b^c^d^.... n-vector n-blade n k = n arbitrary element of L(V) multivector linear combination of any of the above Since L(V) is closed under the operations + and ^, it is "an algebra" (the space Lk alone is not an algebra because it is not closed under ^). The L algebra is different from that of the reals, due its definition as a sum of vector spaces. The elements of L have different "grades" as shown in the right column above, and is known therefore as a "graded algebra". Sometimes L(V) is called "the wedge tensor algebra" over V. Unlike in the tensor product world, in the wedge product world the above list is finite for a given n = dim(V). For k = n, where n = dim(V), there is exactly one linearly independent basis vector which is the wedge product of all the basis vectors of v. For k > n, all wedge products vanish since the vectors in the wedge product are linearly dependent. The dimensionality of the space L is as follows, dim[L(V)] = 1 + n + + + ... + = Σk=0n = 2n = a finite number Lemma: (as promised above). Show that : ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . Rather than present a formal proof, we look at the two simplest cases. First, consider Q ≡ Σi≠j fij = [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. Consider next the triple sum case where there are six sums. For each of the last five sums, we rename the summation indices to cause that sum to have the same form as the first sum. Q ≡ Σi≠j≠k fijk = (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 the triple sum our Lemma can be stated as ΣP [ΣP(i)<P(j)<P(k)] fijk = Σi<j<k [ΣP fP(i)P(j)P(k)] The argument for a k-fold sum proceeds in the same manner, and we end up with ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . 2.7 The tensor product of k dual vectors In this dual-space reprise of Section 2.5, we skip most of the words and supporting material of Section 2.5 and provide equation numbers which correspond to those of Section 2.5. Latin letters are roughly converted to Greek or script-font ones to maintain our past convention. Our task is now to generalize the tensor products from V*2 to V*k, where V*k ≡ V* x V* x .... x V* // k component spaces, each one is V* A generic element of Vk is α1 α2 ..... αk . all αi ϵ V* (linear functionals on V) The basis elements of Vk are λi λ ..... λi . total number of basis elements = nk where n = dim(V*) In the full set of such tensor product basis elements, two or more of the λk might be the same. This will always be the case if k > n where n ≡ dim(V*). A rank-k tensor T in V*k has this general expansion T = Σii....i Fii....i (λiλi .....λi) or T = ΣI FI λI. I ≡ {i1, i2, .....ik}, an ordinary multiindex is = 1,2....n λI ≡ λiλi .....λi FI ≡ Fii....i The tensor product of k vectors is "k-multilinear" meaning it is linear in each of its k factors. For example α1(α2 +α2')α3.....αk = α1α2α3 .....αk + α1α2'α3 .....αk α1(sα2)α3 ..... αk = s(α1α2α3 .....αk) s = scalar The above equations are meaningful for any integer k, regardless of the value n = dim(V*). The tensor T is a tensor product of the linear functionals λr . The "components" of T are obtained by evaluating the functional T at a point (v1,v2....vk) in VxV...xV, and this leads to the following set of k-multilinear functions T(v1,v2....vk) = Σii....i Fii....i (λiλi .....λi)(v1,v2....vk) = Σii....i Fii....i λi(v1) λi(v2) .....λi(vk) so one may regard T : VxVx...xV → K. Define: V*(V) = V*0 V*1 V*2 V*3 + .... Here V*0 = the space of scalars, V*1 = V* the space of vectors (that is, linear functionals on V), V*2 = V* x V* = the space of "bivectors" (bilinear functionals on V), and so on. The most general element of the space V*(V) would have the form τ = s + ΣiFi λi + Σij Fij λiλj + Σijk Fijk λiλjλk + ...... This large space V*(V) is in fact itself a vector space. One could make the following definitions with regard to the space V*(V) k1 scalar 0-blade 0 α vector 1-blade 1 αβ bivector 2-blade 2 αβκ trivector 3-blade 3 αβκδ quadvector 4-blade 4 ..... arbitrary element of V*(V) multivector linear combination of any the above However, these terms are really reserved for objects within the wedge space we shall define below. The dimensionality of the space V*(V) is as follows, where n = dim(V*), dim[V*(V)] = 1 + n + n2 + n3 + ... = ∞ Need something about functions!!! 2.8. The wedge product of k dual vectors In this dual-space reprise of Section 2.6, we skip most of the words and supporting material of Section 2.6 and provide equation numbers which correspond to those of Section 2.6. Latin letters are roughly converted to Greek or script font ones to maintain our past convention. Turning now to wedge products, we want to define the wedge product of k vectors in V*, α1^ α2^ .....^ αk . αi = linear functional on V We impose the requirement that this wedge product changes sign when any two vectors are swapped. This leads to the following wedge product definition which we write in three equivalent forms : α1^ α2^ .....^ αk = Σii....i εii....i (αi αi ..... αi) = ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) = α1 α2 ..... αk + all signed permutations See Section 6 for an explanation of these expansions and why the "changes sign" rule works. The sum ΣP contains k! terms. The wedge product is k-multilinear. Wedge product vanishes if any two vectors are the same. αj ^ αj ^ .... ^ αj = εjj....j α1^ α2^ .....^ αk . αj ^ αj ^ .... ^ αj = εjj....j Σii....i εii....i (αi αi ..... αi) = εjj....j ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) . Wedge product vanishes if vectors are linearly dependent. Wedge product vanishes if k > n . Basis elements for Λk. (λi ^ λi ^ .... ^ λi) where i1 < i2 < ..... < ik basis elements T = Σii...i Fii...i (λi ^ λi ^ .... ^ λi) T = Σi<i<....<i Tii...i (λi ^ λi ^ .... ^ λi) Fii...i = ΣP (-1)P TP(i)P(i)...P(i) i1 < i2 < ..... < ik Examples: Tab = Fab - Fba k = 2 a < b Tabc = Fabc - Facb + Fcab - Fcba + Fbca - Fbac k = 3 a < b < c Exercise: Show that an arbitrary wedge product of k vectors lies in the space Λk T = Σii...i Fii...i (λi ^ λi ^ .... ^ λi) Fii...i = αi βi ... T = Σii...i αi βi ...(λi ^ λi ^ .... ^ λi) = α ^ β ^ ... ^ q Number of elements in Λk compared with V*k. From above we have found that ratio = = = = / nk . Multiindex notations. T = Σii...i Fii...i (λi ^ λi ^ .... ^ λi) T = ΣI FI λI where λI ≡ λi ^ λi ^ .... ^ λi FI ≡ Fii...i and I ≡ 1 ≤ i1, i2,.... ik ≤ n = ordinary multiindex n = dim(V*) __________________________________________________________________________ T = Σi<i<....<i Tii...i (λi ^ λi ^ .... ^ λi) T = ΣI TI eI where λI ≡ λi ^ λi ^ .... ^ λi TI ≡ Tii...i and I ≡ 1 ≤ i1 < i2 <....< ik ≤ n = increasing multiindex n = dim(V*) __________________________________________________________________________ The full wedge tensor algebra. Λ(V) = Λ0 Λ1 Λ2 Λ3 .... τ = s + ΣiFi λi + Σij Fij λi^λj + Σijk Fijk λi^λj^λk + ...... or τ = s + ΣiTi λi + Σi<j Tij λi^λej + Σi<j<k Tijk λi^λj^λk + ...... This large space Λ(V) is in fact itself a vector space. One then makes the following definitions with regard to the space Λ: k1 scalar 0-blade 0 α vector 1-blade 1 α^β bivector 2-blade 2 k = 2 α^β^κ trivector 3-blade 3 k = 3 α^β^κ^δ quadvector 4-blade 4 k = 4 ..... α^β^κ^δ^.... k-vector k-blade 4 k = k .... α^β^κ^δ^.... n-vector n-blade n k = n arbitrary element of Λ(V) multivector linear combination of any of the above dim[Λ(V)] = 1 + n + + + ... + = Σk=0n = 2n = a finite number What comes next? 1. Differential forms comment? 2. Clifford algebra comment?