wedge sections

4.7 The tensor product of k dual vectors 1 4.8. The wedge product of k dual vectors 3 I ≡ {i1, i2, .....ik} is = 1,2....n (4.5.6) 4.7 The tensor product of k dual vectors In this dual-space reprise of Section 4.5, we skip most of the words and supporting material and provide equation numbers which correspond to those of Section 4.5 (leaving some holes in the numbering sequence). We continue our convention of using Greek or script letters for objects in V* and V*k. Our task is now to generalize the tensor products of Section 4.4 from V*2 to V*k, where V*k ≡ V*V* .... V* // tensor product of k spaces, each one is V* (4.7.1) A generic pure element of Vk is α1 α2 ..... αk . all αi ϵ V* (linear functionals as vectors in V*) (4.7.2) The basis elements of Vk are λi λi ..... λi . total number of basis elements = nk where n = dim(V*) (4.7.3) dim(V*k) = nk. (4.7.4) In the full set of such tensor product basis elements, two or more of the λi 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 Tii....i (λiλi .....λi) (4.7.5) or T = ΣI TI λI. I ≡ {i1, i2, .....ik}, an ordinary multiindex is = 1,2....n λI ≡ λiλi .....λi TI ≡ Tii....i (4.7.6-8) 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 (4.7.9) 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*). V*k is a vector space, with same proof as near (4.5.10). The full dual tensor algebra is given by V*(V) = V*0 V*1 V*2 V*3 + .... = Σk=1∞ V*k . (4.7.13) Here V*0 = the space of scalars, V*1 = V* is the space of vectors (that is, linear functionals on V), V*2 = V*V* = the space rank-2 bilinear functionals on V, and so on. The most general element of the space V*(V) would have the form (s and all coefficients ϵ field K), τ = s + ΣiTi λi + Σij Tij λiλj + Σijk Tijk λiλjλk + ...... (4.7.14) This space V*(V) is itself a vector space, repeating the discussion below (4.5.14). This is also clear since V*(V) is a direct sum of the V*k which are themselves vector spaces, being tensor products of vector spaces V*. For later comparison with the corresponding wedge picture, here we have: Object lin comb is Rank(grade) Space s scalar ϵ K 0 V*0 α vector 1 V*1 αβ rank-2 tensor 2 V*2 αβδ rank-3 tensor 3 V*3 αβδγ rank-4 tensor 4 V*4 αβδγ.... rank-k tensor k V*k ..... arbitrary element of V*(V) multivector mixed V* (4.5.17) The dimensionality of the space V*(V) is as follows, where n = dim(V*), dim[V*(V)] = 1 + n + n2 + n3 + ... = ∞ (4.7.18) The k-multilinear functional T in (4.7.5) is a tensor product of the linear functionals λr . The components like Tii....i transform as a rank-k tensor. The full space of T functionals is isomorphic to the full space of k-multilinear functions according to this connection, T(v1,v2....vk) = Σii....i Tii....i (λiλi .....λi)(v1,v2....vk) = Σii....i Tii....i λi(v1) λi(v2) .....λi(vk) (4.7.19) so one may regard T : VxVx...xV → K. The rank-k tensors T can be represented either by the coefficients Tii....i or by the functions T(v1,v2....vk) . Evaluating this function at the basis vectors gives, similar to (4.2.23), T(e1,e2....ek) = T12....k or T(ei,ei....ei) = Tii....i (4.7.20) providing an interpretation for the expansion coefficients Tii....i . Fact: The vector space V*k is equivalent to the vector space of k-multilinear functions on V. (4.2.21) The full dual tensor algebra shown in (4.7.13) is closed under both + and , just as in the regular tensor algebra (4.5.13). It seems evident how one would add two elements of V*(V) as in (4.7.14), but how would one multiply two elements? Consider two elements of rank k and k' : T = Σii....i Tii....i (λiλi .....λi) rank k (4.7.5) S = Σjj....j Sjj....j (λjλj .....λj') rank k' (4.7.5) Multiplying these together one gets ( see (3.1.9) for example ), TS = Σii....i Σjj....jTii....iSjj....j(λiλi .....λi) (λjλj .....λj') = Σii....i jj....j [Tii....iSjj....j] (λiλi .....λiλjλj .....λj') = Σii....i [Tii....iSii....i] (λiλi . .....λi') In this last step, we renamed the jr summation indices so that j1 = ik+1 , j2 = ik+2 and so on. Clearly the product TS is an element of V 4.8. 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?