section 5 6 7 8

This has been copied out on 9.26.15, do not edit 5. The tensor product of k vectors in Vk 1 6. The wedge product of k vectors in Vk 3 7. The tensor product of k vectors in V*k 12 8. The wedge product of k vectors in V*k 13 2.5 The tensor product of k vectors in Vk 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 in Vk 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 vectors in V*k 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*). 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 + ... = ∞ 2.8. The wedge product of k vectors in V*k 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