Activities in the Dual Direct Product Space

Activities in the Dual Direct Product Space PhL 9.9.15 1. Two vector spaces V, W and their direct product space V x W 1 2. Two vector spaces V*, W* and their direct product space V* x W* 2 3. The wedge product in the direct product space V x V 5 4. The wedge product in the direct product space V* x V* 9 1. Two vector spaces V, W and their tensor product space VW Consider two vector spaces V and W of dimension n and n'. Let {ei} = basis of V v = Σi=1n vi ei = general vector in V {ei'} = basis of W w = Σj=1n'wj ej' = general vector in W {eie'j} = basis for the tensor product space VW vw = a "vector" in the tensor product space VW The dimension of the tensor product space VW is then n * n'. Terminology and Comparison of the Tensor Product with other kinds of Products (1) If V and W are simply sets, then VxW with elements (v,w) is called a Cartesian product. (2) There is a 1-to-1 correspondence between (v,w) in the Cartesian product space VxW and vw in the tensor product space VW. We shall sometimes blur these two notions together. (3) The tensor product of two vector spaces is sometimes called a tensor direct product, or simply a direct product. In quantum mechanics in particular, if V and W are vector spaces representing the states of two particles, the combined two-particle states |v> |w> are elements of the tensor product of V and W, but this tensor product is often called a direct product. This is confusing to mathematicians because for them the direct product of a finite number of vector spaces is the same as the direct sum of those spaces, a completely different animal from the tensor product. A direct sum VW has dimension n + n', while the tensor product VW has dimension n * n'. (4) In an unrelated usage, if G and H are groups, then GxH is called a direct product of groups. This is essentially a Cartesian product where GxH has elements (g,h) and (g,h)*(g',h') = (gg',hh'). (5) As shown below, the tensor product space (VW) is itself a vector space, so one can refer to the elements of this space like vw as "vectors" in this vector space. This is true as well if W = V. In this case, however, we shall more commonly refer to v1v2 in VV as a "2-tensor" with respect to V, so we have a potential confusion that v1v2 is a vector in one sense, and a 2-tensor in another sense. About the Basis Vectors ei. If the vector space V is also a metric space, it has some covariant metric tensor ij which defines differential distance according to, (ds)2 = Σij ijdxidxj . This same metric tensor is then used to construct an "inner product" a b on V according to a b = Σijijaibj = aT b = bT a = b a (ds)2 = dx dx Once this inner product is introduced, one can define the "norm" of a vector by ||x||2 = x x and then a "metric" by d2(x,y) = ||x-y||2 = (x-y)(x-y). The space V is then a Hilbert Space, though perhaps an unofficial one if ij is such that x x < 0 for certain vectors x. For Euclidean spaces, one has ij = δi,j so a b = Σijδi,jaibj = Σiaibi = aTb (ds)2 = Σi (dxi)2 . In this context, one has for the basis vectors of V an inner product ei ej and a norm ||ei||2 = ei ei. If ei ej = 0 for all i ≠ j, the basis is orthogonal. If ei ej = δi,j the basis is orthonormal. However, when considering the {ej}, we do not assume the basis vectors are orthogonal or orthonormal. They are, however, linearly independent as must be the case for any "basis". The Kronecker Product. We mention this only to fill out our plate of product types and to make sure this product is not confused with the tensor product. The Kronecker product uses the same symbol as the tensor product, but is a completely different object. Here are some examples: (a b)ij = aibj // Kronecker product of two vectors, also known as an outer product (A B)ij,ab = AiaBjb // Kronecker product of two 2-tensors (A B)ij,ab,αβ = AiaαBjbβ // Kronecker product of two 3-tensors Notice that the outer product of two vectors a and b may be written a b = abT = ( b1. b2....bn) = which may be compared to the inner product of two vectors with a Euclidean metric tensor a b = aTb = ( a1. a2....an) = Σi=1n aibi In our current document, we do not talk about Kronecker products and the symbol does not have the meaning shown here. We now resume our discussion of the tensor product VW. 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). For the notation to be self-consistent, one must assume these properties of : (kv) w = v (kw) = k (vw) // k = scalar 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 the tensor product space. These lines say that multiplication "distributes" over addition +. 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 will be seen below, these last rules apply to the direct sum , not the direct product . Using the correct "rules" above, one may write (v,w) = v w = ( Σiviei)( Σjwjej') = Σijviwj (eie'j) showing how this tensor product vector can be expressed in terms of the basis functions. A general vector in W x V can be written as a linear combination of the basis vectors, T ≡ ΣijTij 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, T ≡ ΣijTij 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. Some authors replace the word "rank" with "order" so then T is a tensor of order 2, or perhaps T is a "2-tensor" and the words rank or order are avoided altogether. The coefficients Tij are then the components of this rank-2 tensor in the {ei} basis. Since Tij are elements of a square matrix T of size n x n, the Tij are sometimes called "the matrix elements of T". 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 Tij. In the strong meaning, a tensor is a set of components Tij which transform in a certain manner with respect to some transformation. It is shown in tensor doc Appendix Q that if the {ei} are the "tangent base vectors" of that transformation, then the Tij 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 Tij are the "covariant components" of the tensor T. 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 vectors qi discussed in the next section. They are vectors in V. Notice in the above 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. ok to here 2. Two vector spaces V*, W* and their direct product space V* x W* 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 letters for all vectors and tensors encountered. Vectors in V or W or V x W will continue to be represented by Latin letters. Assume that {λi} = basis of V* {λi'} = basis of W* 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 real number: λi : V → R. 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, in "matrix notation", qiTej = δi,j. or ( e1, e2...) = diag (1,1,....) = identity matrix or Σi=1n (qi)n(ej)n = δi,j If it happened that V was not just a vector space but was also a Hilbert Space with an inner product , one could express the above as qi ej = δi,j which is perhaps more familiar to the reader, but the Hilbert space assumption is not needed. The action of the linear functional λi can then be written as λi(v) = qiTv which is manifestly linear in that λi(αv) = αλi(v) and λi(v + v') = λ(v) + λ(v') α = scalar The definition of linearity is often combined into a single statement, λi(αv + βv') = αλ(v) + βλ(v'), α,β = scalars where α and β are scalars which here we take to lie in R, the real numbers (more generally, they might lie in some field K). The vectors qi are sometimes called "covectors" or "reciprocal vectors" or "dual vectors" which form a "dual basis". They are uniquely determined by the {ei} as shown in tensor doc ***. From ** one then has λi(ej) = qiTej = δi,j Footnote: In covariant notation, the vectors qi are written as ei ( i is a label, not a component index), and the above dot product is written as ei ej = δij // δij is a covariant way of writing 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 → R β = Σjβjλj' general vector in W* β(v) = Σiβiλi'(v) β: W → R where on the right we show the corresponding functions α(v) and β(v). There is a dual product space V* x W* in which one has λiλ'j = basis for the dual direct product space V* x W* . The object λiλ'j is a "vector" in the vector space which is the dual direct product space V* x W*. We want this object λiλ'j to be a functional over the space V* x W* such that λiλ'j: V* x W* → R The natural way to accomplish this desire is to write (λiλ'j)(v,w) = λi(v)λj'(w) = real times real = real This function is manifestly "bilinear" in that it is separately linear in each of the arguments v and w. The "rules" for the operator in the space V* x W* are the same as those for in the space V x W,' (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 direct product space V*x W* can be written T = Σij Tij λiλ'j where the Tij are real coefficients and so T: V* x W* → R . Thus, evaluating at a point in V x W one gets, T(v,w) = Σij Tij (λiλ'j)(v,w) = Σij Tij λi(v) λj'(w) = real More specifically, consider α ϵ V*and β ϵ W* as shown above. Then α β = (Σiαi λi)(Σjβj λ'j) = Σij αiβj λiλ'j (α β)(v,w) = Σij αiβj (λiλ'j)(v,w) = Σij αiβj λi(v)λj'(w) = real number which then is just a particular example of the more general Φ(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). If it happens that W = V, then W* = V* and we write V* x W* = V* x V* = V*2. Then the most general element Φ of V*2 can be expressed as T = Σij Tij (λiλj) T(v1,v2) = Σij Tij (λiλj)(v1,v2) = ΣijTij λi(v1)λj(v2) . In analogy with Section 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. The Tij are then tensor components in the λiλj basis. Although the bilinear functional T above is the real tensor, one can regard the tensor alternatively as being the bilinear function T(v1,v2) which is just T evaluated at some point in V2. 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 V*2 = V* x V* (which includes any T above) is just V*2 = V* x 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 of the direct products of two linear functionals of V* over V is equivalent to the vector space of bilinear functions on V2. 3. The wedge product in the direct product space V x V (a) Definition of the wedge product and the space L2 Momentarily jumping ahead, we shall find that the wedge product of two 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 direct product space V x V. So, we start off by defining the following "wedge product" of two vectors a and b of V a ^ b ≡ ab - ba . Notice therefore that a ^ b is an element of V x V = V2, since it is a linear combination of elements of V x V. It is "antisymmetrized" under a ↔ b. Since not all elements of V x V can be written this way, the set of elements a ^ b exist in a subset of V x V which we shall call L2, so L2 V2. The above definition implies that a ^ b = - b ^ a a,b ϵ V and a ^ a = 0 a ϵ V . In Section 1 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: a = ei b = ej c = ek. 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 V x V 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 One would say then that L2 is spanned by the first set of basis vectors and none of the second set. 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 closed under addition of vectors since (a ^ b) + (a' ^ b') = Σij aibj (ei ^ ej) + Σij a'ib'j (ei ^ ej) = Σij [aibj+ a'ib'j] (ei ^ ej) = Σij Cij (ei ^ ej) ϵ L2 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 real numbers. 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. Suppose the number of reals were some large number N instead of ∞. One could then construct exactly Nn(n-1)/2 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. (c) How does the wedge product of two vectors involve determinants? For the case discussed above where Fij = aibj we wrote 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 the 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 and one notes 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 shall 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 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 direct product space V x V. 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 (zealots?) who recommend deep-sixing the cross product altogether and replacing it with the wedge product for the study of topics like angular momentum (Ref...). 4. The wedge product in the direct product space V* x V* We mimic the approach of Section 3 for the direct product space V x V, but now we have V* x V*. So, 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* x V*. It is "antisymmetrized" under α ↔ β. Since not all elements of V* x 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. The above definition implies that α ^ β = - β ^ α α, β ϵ V* and 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. 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 Φij (λi ^ λj) . For example, if Φij = 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. Just as in Section 3, we can show that T = Σij Φij (λi ^ λj) = Σi<j Tij (λi ^ λj) Tij ≡ (Φij- Φji) Tij = - Tji where Tij is an antisymmetric n x n matrix. We then reach the same conclusion which we restate in the current context, = = (1/2) = (1/2) (1 - ) Recall from Section 2 in the V* x W* discussion that. λiλ'j: V* x W* → R (λiλ'j)(v,w) = λi(v)λj'(w) . Now with W* = V* we have instead λi λj: V*2 → R (λiλj)(v1,v2) = λi(v1)λj(v2). and then λi^ λj: Λ2 → R (λi^ λj) = (λiλj) - (λjλi) // special case of ** above If we evaluate the V*2 functional (λi^ λj) at the V2 location (v1,v2) we find, (λi^ λj)(v1,v2) = (λiλj)(v1,v2) - (λjλi)(v1,v2) = λi(v1)λj(v2) - λj(v1)λi(v2) = det Notice that : (1) the basis function (λi^ λj)(v1,v2) is linear on both v1 and v2, so it is a "bilinear" function. This follows from the fact that the λk(v) functions are linear as was shown in ***. (2) the basis function (λi^ λj)(v1,v2) is antisymmetric under v1↔ v2, (λi^ λj)(v2,v1) = - (λi^ λj)(v1,v2) This is obvious from the determinant form since we switch two rows. As in Section 3, the wedge product of two vectors in V* can be expressed in terms of certain determinants which are minors of a "tall matrix" whose columns are α and β, α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- ajβi) (λi ^ λj) = Σi<j det (λi ^ λj) . Because we are now in the dual space of linear functionals, we claim no particular geometric significance of the above wedge product when n = 2. One could of course blithely set V*2 = R2 and repeat the previous geometric discussion, but usually one does not discuss "geometry" in the dual space context. Instead, one notes that, if the linear functional α ^ β is evaluated at (v1,v2) in V2, one gets a statement about functions, (α ^ β)(v1,v2) = Σij αiβj (λi ^ λj)(v1,v2) = Σi<j αiβj (λi ^ λj)(v1,v2) = Σi<j (αiβj- ajβi) (λi ^ λj)(v1,v2) = Σi<j det (λi ^ λj)(v1,v2) = Σi<j det [ λi(v1)λj(v2) - λj(v1)λi(v2) ] = Σi<j det det The function (α ^ β)(v1,v2) is bilinear and antisymmetric because (λi ^ λj)(v1,v2) is bilinear and antisymmetric. As noted above, the most general functional in Λ2 may be written T = Σi<j Tij (λi ^ λj) Aij = - Aji so the most general function is then T(v1,v2) = Σi<j Tij (λi ^ λj)(v1,v2) = Σi<j Tij [ λi(v1)λj(v2) - λj(v1)λi(v2) ] Since this function is manifestly bilinear and antisymmetric, we conclude with this claim: Fact: The vector space Λ2(V) of the wedge products of two linear functionals of V* over V is equivalent to the vector space of antisymmetric bilinear functions on V2. This may be compared to our earlier statement: Fact: The vector space V*2 of the direct products of two linear functionals of V* over V is equivalent to the vector space of bilinear functions on V2. ***************************************************