Chapter 9 wedge product as quotient

The Quotient Approach to the Wedge Product Installed on 1.26.16, do not edit here. 9. The Wedge Product as a Quotient Space 1 9.1. Development of Lk as Vk/S. 1 9.2. Development of L as T/I. 4 9. The Wedge Product as a Quotient Space We conclude our document with a wedge product "theory section" which really should be part of Chapter 1, but we wanted to have the reader first immersed in the nuts and bolts approach to the wedge product presented in Chapters 4, 7 and 8. As is the case for Chapter 1, this chapter makes no mention of the components of vectors or tensors. 9.1. Development of Lk as Vk/S. The presentation below is based on the paragraph titled Definition 3.1 on page 5 of Conrad. Consider the vector space V defined over some field K (the scalars, normally reals). If V used coefficients in a ring R instead of a field K, V would be called an R-module. Since any field K is also a ring, we can regard our usual V as an R-module (any vector space is also an R-module). Statements about R-modules are more general that statements about vector spaces, so for that reason one sees the R-module moniker in discussions of our current topic. We shall use the bare term module. Thus, the vector space Vk = VV...V can be regarded as a module since its vectors are defined over the field K which is also a ring. The pure elements of Vk have the form v1v2v3.... vk (k factors). Consider the subset S of Vk whose elements have a repeat of one of the vectors. That is, suppose we have vi = vj for some i ≠ j in v1v2v3.... vk. There could be other vectors which are also equal to vi, so at least two vectors are the same. For example, if k = 4 one would say axbx and xxbx were in the subset S. Adding elements of this subset produces another element of the subset, so this subset is itself a module. Thus we are talking about elements of a submodule S of the module Vk. Notice that 0 is an element of S, which can be represented by any element of Vk having one or more vectors being the 0 vector of V, as in (1.1.9). For k = 4, consider this element of Vk, A' = 3 abcd + 5abca - 2 aacd + 3 aaca . (9.1.1) If we were to throw out elements of the set S, we would get A = 3 abcd . (9.1.2) The set of elements of Vk that is generated by adding all elements of set S to A is called the coset of A, usually written [A]. Thus, the coset of A is A + s where s ϵ S. The elements of Vk can be partitioned into an array in this manner, where each row (coset) involves all the si ϵ S : row name coset → [0] 0 0 + s1 0 + s2 ..... [A] A A + s1 A + s2 .... [B] B B + s1 B + s2 .... (9.1.3) ... For example our Vk element A' lies somewhere in the row of this chart labeled on the left by [A]. It turns out that the rows themselves (the cosets) form a module called Vk/S . The elements of this module can be regarded as being those in the first column of the cosets. So A is an element of Vk/S , but A' is not. Strictly speaking, there is an isomorphism between A and [A], but we ignore such details. Fact: To enumerate the elements of the module Vk/S we write down all the elements of Vk and just set to 0 all terms in which a vector is repeated, such as the last three terms of A' above. We thus filter out such terms, they are "modded out", which is why Vk/S is sometimes called Vk mod S. (9.1.4) Define Lk to be Lk ≡ Vk/S . (9.1.5) The fact that Vk elements lying in S (those that have repeated vectors) are "thrown out" (modded out, set equal to 0) is reminiscent of the construction (1.1.4) that F(VxW)/N = VW and certain elements of the full set F(VxW) were similarly modded out (set to 0, such as (v2, w1+w2) – (v2,w1) – (v2,w2)). Elements of Vk are written v1v2v3.... vk. This product is "associative" in that parentheses can be placed any way one wants, such as v1(v2v3).... vk, with no change in value. (9.1.6) Elements of Lk ≡ Vk/S are written v1^v2^v3.... ^vk . This product is declared to be "associative" in that parentheses can be placed any way one wants, such as v1^(v2^v3).... ^vk, with no change in value. (9.1.7) Using this definition of the wedge product of k vectors, we can derive some of its properties. Fact 1: v1^v2^v3.... ^vk = 0 if two (or more) vectors are the same. (9.1.8) Proof: This follows from the definition of Lk ≡ Vk/S and the Fact (9.1.4) stated above. Fact 2: v1^v2 = - v2^v1 (9.1.9) Proof: We know that (v1+v2) ^ (v1+v2) = 0 since this has the form v3 ^ v3 which is 0 by Fact 1. Expanding, 0 = (v1+v2) ^ (v1+v2) = v1^v1 + v1^v2 + v2^v1 + v2^v2 = v1^v2 + v2^v1 so 0 = v1^v2 + v2^v1 and v1^v2 = - v2^v1 QED Fact 3: Swapping any pair of vectors in v1^v2^v3.... ^vk creates a minus sign. (9.1.10) Proof by example: (swap v1 and v3 by making use of Fact 2 three times) : v3^v2^v1.... ^vk = + v3^(v2^v1).... ^vk = - v3^(v1^v2).... ^vk = - (v3^v1)^v2.... ^vk = + (v1^v3)^v2.... ^vk = + v1^(v3^v2).... ^vk = - v1^(v2^v3).... ^vk = - v1^v2^v3.... ^vk QED Fact 4: vj ^ vj ^ .... ^ vj = εjj....j ( v1^ v2^ .....^ vk ) (9.1.11) Proof: Fact 4 is the combination of Fact 3 and Fact 1. Fact 4 appears as (4.6.12). Fact 5: vj ^ vj ^ .... ^ vj = 0 if the vectors are linearly dependent. (9.1.12) Proof: See (4.6.14). In this manner, we can derive all the properties of the wedge product stated in Section 4.6 without having to lean on the construction of the wedge product as a linear combination of tensor products. However, we know that the elements of Lk are linear combinations of the elements of Vk. We have written in (4.6.2) that v1^ v2^ .....^ vk = (1/k!) Σii....i εii....i (vi vi ..... vi) = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) . (4.6.2) Since in Section 4.6 this linear combination generates all the Facts listed above, and does not contradict any of them, we conclude that this must be the linear combination of Vk elements that equals v1^ v2^ .....^ vk (apart from a possible normalization factor). Alternate Language. Looking at A and A' above, we could say that A and A' are in the same equivalence class so that A ~ A'. Two elements of Vk are in the same equivalence class if they differ by an element of S, so we have A' - A = s ϵ S. The elements of the equivalence class of A are then just the coset [A]. The submodule of Vk called Vk/S is called a quotient module. Using category diagrams, one can consolidate this notion with that of quotient rings and quotient groups. 9.2. Development of L as T/I. Start with the tensor algebra (vector space) shown in (5.4.1), T(V) = V0 V V2 V3 ....... (9.2.1) The elements of the vector space T(V) form a ring with operations and . It is easy to show that T(V) is closed under addition and multiplication and has the other required ring properties. Ideal Example 1: Consider the set S of elements of T(V) which are linear combinations of elements of the form ABC (with coefficients in field K) where B is some fixed element of T(V) and A,C ϵ T(V) are allowed to vary. This set is closed under addition. For example, ABC + A'BC' ϵ S. Since coefficients in K can be absorbed into A, one could just say that the elements of S are sums of elements of the form ABC. One could take any 0BC as the "0" element, and -ABC is the additive inverse. The set S is commutative and associative under addition. Therefore S forms an additive subgroup of the ring T(V). Moreover if we left or right multiply (using ) any element of this set by any element of T(V), the result clearly lies in T(V). Q(ABC) ϵ T(V) (ABC)Q ϵ T(V) . (9.2.2) Therefore this set S is a two-sided ideal of the ring T(V). Ideal Example 2: S = sums of elements of the form ABCDE where elements B and D are fixed and A,C are E varied, all letters being ϵ T(V) . Ideal Example 3: S = sums of elements of the form AxCxE where vector x is fixed and A,C,E ϵ T(V) are varied. This set is the set of all sums of elements of T(V) in which the vector x appears at least twice. Let's call this particular ideal by the name S = I, because this is our ideal of interest. Now suppose we declare the following equivalence relation AxCxE ~ 0 x,A,C,E ϵ T(V) . (9.2.3) Sums of such elements form the ideal I discussed above, and we are in effect setting all elements of this ideal equal to 0. There then exists a subset of T(V) which we shall call T(V)/I, or T(V) "mod" I. This is a standard algebraic structure where one takes the quotient of a ring R divided by a two-sided ideal I of that ring. The upshot is that the elements of the new quotient set T(V)/I consist of all sums of T(V) elements except that any term which matches the form (9.2.3) is filtered out ("modded out") by setting it equal to 0. Example: t' = k1 abcd + k2 ab + k3 bcc + k4 abca = element of T(V) t = k1 abcd + k2 ab = element of T(V)/I (9.2.4) In algebra terminology, adding all elements of the form AxCxE to t generates a coset associated with t called [t], and T(V)/I is in effect the set of all such cosets. Element t' is one element of the t coset. The elements of T(V)/I themselves form a new ring called the quotient ring or factor ring. The ring/ideal situation is quite similar to that discussed above for the module/submodule situation Vk/S. Recall from (7.8.1) that the full wedge (exterior) tensor algebra is given by the direct sum space L(V) = L0 L1 L2 L3 + .... . (9.2.5) The claim then is that L(V) = T(V)/I where I = the ideal of Example 3 above. (9.2.6) This is then the space of all T(V) elements where all terms in which a vector is repeated are set to 0 and thus are not part of L(V). Notice that the quotient of Section 9.1 has a finer granularity. It deals with individual Lk Vk spaces, whereas Section 5.2 deals with the entire L(V) T(V). Many texts refer to Lk as Λk(V) and L(V) as Λ(V). We have reserved the Λ names for the dual spaces. The category theory approaches to Lk and L(V) are similar to the discussion of Section 1.2 with the main point being that Lk and L(V) are "universal" and therefore uniquely defined up to isomorphism. The role played by k-multilinear functions is played by antisymmetric k-multilinear functions.