archive Section 2_11 on 1_21_16

Archive off Old Section 2.11 PhL 1.21.16 My replacement for this section is infinitely much better in about 20 different ways. I will comment in red below where I had things just hazy and wrong. Improvements: 1. Dirac notation and its simple row column understanding, isomorphism not even worth mentioning. 2. Notion of Transformation A which has non-transforming basis vectors. 3. Explanation of the notation. 2.11 The dual spaces V* , V*V* and V*k We denote dual-space vectors and tensors by Greek or script font letters. The dual space V* is by definition the space of linear functionals over V. If α ϵ V*, we can then write α : V → K α(v) = k ϵ K (2.11.1) where K is any field. Since α is a linear functional, α(v) is a linear function. 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(v) as a "function". (a) The dual space V* : basis λi, general vector α, isomorphism V ~ V* The basis λi Given n vectors {ei} which form a basis for V, one can find another set of n basis vectors {ei} in V such that, ei ej = δij . (2.3.2) (2.11.2) These are the same ei discussed in Section 2.3. Section 2.7 gave an example of computing the ei from the ei. We can then define a basis {λi) for V* as a set of n linear functionals λi such that λi(v) = ei v . λi: V → K (2.11.3) Recall that vectors like v in V have components which lie in field K, usually taken to be the reals. And notice that i is a label, not a component. Functional λi is manifestly linear since λi(kv) = kλi(v) and λi(v + v') = λi(v) + λi(v') k ϵ K . (2.11.4) Here v and v' are vectors in V. As noted earlier, the definition of linearity is often combined into a single statement, λi(k1v + k2v') = k1λi(v) + k2λi(v') k1, k2 ϵ K . (2.11.5) The vectors ei are sometimes called "covectors" or "reciprocal vectors" or "dual vectors" which form a "dual basis" to the ei. From (2.11.3) and (2.11.2) one then has, λi(ej) = ei ej = δij . (2.11.6) Therefore, λi(v) = λi(Σj vj ej) = Σj vjλi(ej) = Σj vjδij = vi so λi(v) = vi λi: V → K . (2.11.7) The linear functional λi applied to vector v generates the e-basis component vi of v. For this reason, the λi(v) is often referred to as "the ith coordinate function". I spent all this time above on λi because I never really understood what it really was : λi = <ei|. The Dirac notation clarifies everything. Scalar and Vector This is a somewhat confusing subject, treated in more detail in Section D.4. This was a total mess. I was getting off into those supposed two kinds of transformations , the component one and the vector one, just a confusing mess. The functional λi a vector in the dual vector space V*, but is not displayed in bold the way ei is. Whereas ei has components (ei)i, the object λi has no components (i is a label). In that sense λi is a scalar quantity. What we mean is that λi(v) is a scalar, since (2.11.3) says λi(v) ϵ K. This idea is further reinforced by λi(v) = ei v since the dot product of two vectors is a scalar, as in (2.2.6). So the object λi(v) is a scalar with respect to a standard (2.1.5) component transformation (v'i)a = Rab (vi)b . *****?? On the other hand, we know that, λi(v'k) =λi( Rkjvj) = Rkjλi(vj) λi(v'i) = Rij λi(vj) where v'i = Rij vj ( a "vector transformation") (2.11.8) This is trivially true because λi is a linear functional. Equation (2.11.8) describes the transformation of a vector object, not a scalar object. Object λi is a rank-1 tensor functional, and λi(v) is a rank-1 tensor function. So we have two different underlying transformations here. The object λi is a scalar with respect to the component transformation of vectors, while at the same time it is a vector with respect to the vector transformation of vectors. See Section D.3 and D.4 for more details. Comments: 1. The basis functionals λi can be defined directly from λi(ej) = δij without use of the dot product. Note that (2.11.7) λi(v) = vi does not depend on the existence of a dot product. The dot product implies that the vector space V is also a Hilbert Space and one need not assume this fact, but in our applications V will always be a Hilbert Space. [ who cares, now that it is all totally clear. I was still mystified by this λi business. ] 2. The covector ei is associated with the basis functional λi but one should not identify λi = ei. For one thing, as just noted above, with respect to component transformations λi is a scalar while ei is a vector. One sometimes sees λi(v) written as ei(v) or as e*i(v), but we use λi to emphasize the distinction between λi and ei. [ other authors?? ] [ all wrong! ] 3. Some authors use notation v*i(v) in place of our λi(v) and vi in place of our ei. General vector α in V* Assuming (for the moment) that the λi form a basis for V*, a general linear functional α in V* can be written as a linear combination of the basis functionals λi, where the coefficients αi ϵ K form a vector α in V (note various distinct meanings for symbol α ) α = Σiαiλi = general vector in V* α: V → K . (2.11.9) The corresponding function α(v) is then given by α(v) = Σiαiλi(v) = Σiαivi = α v . (2.11.10) Evaluating at v = ej, α(ej) = α ej = αj (2.11.11) so α(ei) picks off the coefficient αi appearing in the expansion (2.11.9). Isomorphism between V and V* [ who needs this. We know that for every column vector there is a matching row vector, simple as pie. ] There is one-to-one mapping between the set of vectors α in V and the set of linear functionals α in V*. Consider for example, two functionals α1 and α2, α1(v) = α1 v for all v in V α2(v) = α2 v for all v in V Then, α1 ≠ α2 α1(v) cannot equal α2(v) for all v in V, so α1 ≠ α2 α1 ≠ α2 α1 ≠ α2 , otherwise one would have α1 = α2 (2.11.12) This one-to-one mapping V ↔ V* with α ↔ α exhausts both spaces so is really a bijection which we shall call an isomorphism. For this reason, V and V* have the same dimension n. The basis vectors of V are the n basis vectors ei , while the basis vectors for V* are the n basis functionals λi, as we now show. Fact: The λi form a basis for V* [ totally obvious in the new writeup.] (2.11.13) Proof: Let α be an arbitrary vector in V. Then due to the 1-1 relation just described, we may regard the the functional α defined by α(v) = α v to be an arbitrary linear functional in V*. There are no linear functionals in V* for which one cannot write α(v) = α v with α in V. Then consider, using (2.11.7) that λi(v) = vi , α(v) = α v = Σiαivi = Σiαiλi(v) = [ Σiαiλi](v) . Therefore in V* we can express an arbitrary linear functional α as, α = Σiαiλi and therefore the λi must be a basis for V*. The discussion surrounding (2.11.8) about λi being both a scalar and a vector (oy) applies to any linear functional α. Thus, α is a scalar with respect to a component transformation and it is a vector with respect to a vector transformation. (b) Rank-2 tensors in V*V* We want the object λiλj to be a bilinear functional over the space VxV such that λiλj: VxV → K . λiλj ϵ V*V* (2.11.14) The natural way to accomplish this bilinearity desire is to write (λiλj)(v1,v2) = λi(v1)λj(v2) = scalar * scalar = scalar ϵ K . (2.11.15) Then using (2.11.7) this can be written in several interesting ways (λiλj)(v1,v2) = λi(v1)λj(v2) = (v1)i(v2)j = (v1v2)ij = (ei v1)(ej v2) (2.11.16) 1 2 3 4 From form 4 we see immediately that (λiλj)(v1,v2) is a scalar under a component transformation (V'i)a = Rab (Vi)b, because each dot product is a scalar. This is also clear from the first form since we found earlier that each λi object is a scalar. The fact is less obvious just staring at forms 2 and 3 which certainly don't look much like scalars. However, as emphasized in the discussion leading to (2.6.4), component notation is ambiguous. In the x-space of Fig (2.1.1), we have (v1)i = (v1)(e)i, and in x'-space this object would be (v'1)i = (v'1)(e')i . It is true that (v'1)(e')i = (v1)(e)i because (e'i v'1) = (ei v1). Meanwhile, the tensor transformation rule for (λiλj)(v1,v2) is the following, (λiλj)(v'a,v'b) = Raa' Rbb' (λiλj)(va',vb') v'i = Rij vj (2.11.16a) which result is manifestly true since (λiλj)(v1,v2) is bilinear. As in the rank-1 case, we see from (2.11.16) that (λiλj)(v1,v2) is a scalar with respect to the component transformation (v'i)a = Rab (vi)b, but is a rank-2 tensor function with respect to the vector transformation v'i = Rij vj. The most general bilinear functional T in V*V* can be expanded in the following manner, analogous to (2.10.2), T = Σab Tab λaλb // Tab = [T(λ)]ab in the sense of (2.6.8) (2.11.17) This notation is obsolete, now written as T = Σab Tab λaλb where the coefficients Tab are assumed to transform as a covariant rank-2 tensor. Again, due to the isomorphism between the set of such rank-2 tensors and the set of bilinear functionals T, we refer to the functional T as a rank-2 tensor, or just a 2-tensor, in V* V*. One may evaluate the functional T at (v1,v2) using (2.11.16) to get T(v1,v2) = Σab Tab (λaλb)(v1,v2) = Σab Tab (v1)a(v2)b (2.11.18) where now the function T(v1,v2) is manifestly bilinear in its arguments. Evaluation at (e1,e2) yields T(ei,ej) = Σab Tab λa(ei) λb(ej) = ΣabTab δaiδbj = Tij . (2.11.19) This shows that the coefficients Tij of the bilinear functional T can be obtained from the function T(v1,v2) evaluated at the basis vectors (ei,ej), similar to what happened in (2.11.9). Since the rightmost expression in (2.11.18) is a fully contracted outer product of three tensors (Section 3.1), one sees that T(v1,v2) transforms as a scalar field over V x V, T'(v'1,v'2) = T(v1,v2) . (2.11.20) Once again, although T transforms as a scalar, it is referred to as a 2-tensor because its expansion coefficients form a rank-2 tensor. As a special case, consider T = α β. We find that α β = Σab (α β)ab λaλb = Σabαaβb λaλb (2.11.21) where we now use the outer product (2.8.9) for the two vectors α,β ϵ V*. Evaluation at (v1,v2) along with (2.11.16) gives (α β)(v1,v2) = Σabαaβb λaλb(v1,v2) = Σabαaβb (v1)a(v2)b = [Σaαa (v1)a][Σbαb (v1)b] = α(v1)β(v2) . (2.11.22) Once again, α β is called a 2-tensor even though (α β)(v1,v2) transforms a scalar. Evaluation at (e1,e2) with use of (2.11.9) yields (α β)(ei,ej) = α(ei)β(ej) = αiβi = (α β)ij (2.11.23) analogous to (2.11.19). (c) Rank-k tensor in V*k The equations of this subsection appear rather complicated, being written in a systematic notation. The reader is encouraged to refer back to the equations for k = 2 of the previous subsection if confusion arises. The related equation numbers of are shown in italics. As a shorthand notation we write, Vk ≡ VxVx....xV k factors // Cartesian product Vk ≡ VV....V k factors // tensor product V*k ≡ V*V*....V* k factors // tensor product . (2.11.24) Various equations above can be generalized as follows: (λiλi ... λi): Vk → K . (2.11.14) (2.11.25) (λiλi ... λi)(v1,v2....vk) = λi(v1)λi(v2) ... λi(vk) (2.11.15) (2.11.26) (λiλi ... λi)(v1,v2....vk) = (v1)i(v2)i ... (vk)i . (2.11.16) (2.11.27) The full set (eiei ... ei) for is = 1,2..n forms a basis for Vk The full set (λiλi ... λi) for is = 1,2..n forms a basis for V*k. The most general k-multilinear (rank-k) functional T in V*k can be expanded as T = Σii...i Tii...i (λiλi ... λi) (2.11.17) (2.11.28) with T(v1,v2, ...vk) = Σii...i Tii...i (λiλi ... λi)(v1,v2, ...vk) = Σii...i Tii...i λi(v1)λi(v2) ... λi(vk) = Σii...i Tii...i (v1)i(v2)i ... (vk)i . (2.11.18) (2.11.29) The function T(v1,v2, ....vk) is thus seen to be k-multilinear in its k vector arguments. As a special case, T(ej,ej, ... ej) = Σii...i Tii...i λi(ej)λi(ej) ... λi(ej) = Σii...i Tii...i δjiδji ... δji = Tjj...j (2.11.19) (2.11.30) showing how this evaluation of T(v1,v2, ....vk) projects out the rank-k tensor component Tjj....j . Since (2.11.29) shows T(v1,v2, ....vk) as a full contraction of tensors, it transforms as a scalar, T'(v'1,v'2, ....v'k) = T(v1,v2, ....vk) . (2.11.20) (2.11.31) As a special case, consider T = αj αj ... αj, which is the tensor product of k different rank-1 tensors in V*. We find that : (αj αj ... αj) = Σii...i (αj αj ... αj)ii...i (λiλi ... λi) = Σii...i (αj)i(αj)i ...(αj)i (λiλi ... λi) . (2.11.21) (2.11.32) Evaluation at (v1,v2, ....vk) along with (2.11.27) gives (αj αj ... αj)(v1,v2, ....vk) = Σii...i (αj)i(αj)i ...(αj)i (λiλi ... λi)(v1,v2, ....vk) = Σii...i (αj)i(αj)i ...(αj)i (v1)i(v2)i ... (vk)i = [Σi(αj)i(v1)i] [Σi(αj)i(v2)i] ..... [Σi(αj)i(vk)i] = αj(v1)αj(v2) ... αj(vk) . (2.11.22) (2.11.33) Finally, (αj αj ... αj)(ei,ei, .... ei) = αj(ei)αj(ei) ... αj(ei) = (αj)i(αj)i .... (αj)i = (αj αj ... αj)ii....i (2.11.23) (2.11.34) Once again, due to the isomorphism between functionals T and functions T(v1,v2, ....vk), one loosely refers to the scalar field T(v1,v2, ....vk) as a rank-k tensor, although T is the actual rank-k tensor object whose components Tabc... transform by the usual rule e.g. (2.1.6). One might say that the rank-k tensor T is "represented by" the k-multilinear function T(v1,v2, ....vk) , just as it is represented by Tabc... in the sense of tensor components. Consider these equations from above T(v1,v2, ....vk) = Σii....i Tii....i (v1)i(v2)i ... (vk)i T(ej,ej, .... ej) = Tjj....j The official tensor components in terms of Chapter 2 transformations are Tjj....j. The corresponding function T(v1,v2, ....vk) appears to be a continuation (like an analytic continuation) of the tensor components Tjj....j off the specific argument points (ej,ej, .... ej). There is probably a formal description of this structure perhaps in terms of fiber bundles. In any event, Spivak on page 75 refers to our k-multilinear function T(v1,v2, ....vk) as a k-tensor. (d) Multiindex notation much better red and in line The cluttered equations above can be written in a dense multiindex notation. Forms like those shown below commonly appear in the literature. Ultimately one must refer to the original equation to remove any possibly ambiguity of the notation. Italic equation numbers show the equation being abbreviated. I ≡ i1, i2...ik J ≡ j1, j2...jk // ordinary multiindices λI ≡ λiλi ... λi eJ ≡ ej,ej, .... ej vZ ≡ v1,v2....vk // Z is meant to imply labels 1,2....k (2.11.35) λI(vZ) ≡ (λiλi ... λi)(v1,v2....vk) (2.11.26) vZI ≡ (v1)i(v2)i ... (vk)i a k-multilinear function λI(vZ) = vZI a k-multilinear function (2.11.27) T = ΣI TIλI general k-tensor expansion in V*k (2.11.28) T(vZ) = ΣI TIλI(vZ) = ΣI TI vZI a k-multilinear function (2.11.29) T(eJ) = TJ project out tensor components (2.11.30) T'(v'Z) = T(vZ) T(v1,v2, ....vk) transforms as a scalar (2.11.31) αJ ≡ (αj αj ... αj) tensor product of k vectors of V* (αJ)I ≡ (αj αj ... αj)ii....i = (αj)i(αj)i .... (αj)i outer product αJ(vZ) ≡ αj(v1)αj(v2) ... αj(vk) αJ = ΣI (αJ)I λI expansion of the tensor product of k vectors (2.11.32) αJ(vz) = ΣI (αJ)I λI(vZ) = ΣI (αJ)I vZI = (αJ)(vZ) (2.11.33) αJ(eI) = (αJ)I (2.11.34)