Chapter 2 rewrite 4_2_16 v3

Rewrite of Chapter 2, started 4/2/16 Rewrite of Chapter 2, started 4/2/16 1 2. A Brief Review of Tensors in Covariant Notation 1 2.1 R, S and how tensors transform : Picture A 2 2.2 The metric tensors g and g' and the dot product 7 2.3 The basis vectors en and en 8 2.4 The basis vectors un and un 10 2.5 The basis vectors e'n and u'n and a summary 11 2.6 How to compute a viable x' = F(x) from a set of constant basis vectors en 13 2.7 Expansions of vectors onto basis vectors 16 2.8 The Outer Product of Tensors and Use of 18 2.9 The Inner Product (Contraction) of Tensors 22 Dot products in spaces VV, VW, VVV and VWX 23 2.10 Tensor Expansions 24 (a) Rank-2 Tensor Expansion and Projection 24 (b) Rank-k Tensor Expansions and Projections 26 2.11 Dual Spaces and Tensor Functions 27 (a) The Dual Space V* in Matrix and Dirac Notation 28 (b) Functional notation 29 (c) Basis vectors for the dual space V* 29 (d) Rank-2 functionals and tensor functions 33 (e) Rank-k functionals and tensor functions 36 (f) The Covariant Transpose 39 (g) Linear Dirac Space Operators 39 2. A Brief Review of Tensors in Covariant Notation In Chapter 1 we generally avoided mentioning components of vectors and tensor products. But in many ways, "components" is what tensors are all about. Anyone who wants to use tensor analysis to actually do something practical is going to use tensor components. The whole notion of what it means to be a tensor of some rank requires components and component indices. Later when we deal with tensor functions, the components will in morph into the vector arguments of multilinear functions. Tensor analysis provides some very heavy-duty machinery to handle manipulations of tensors and tensor components. A key idea is that a true tensor is something that transforms in a certain manner relative to some defined underlying transformation which below is called x' = F(x). In the following notes, we review this machinery, The review is based on our document Tensor which follows the unusual path of developing tensor analysis in a "developmental notation" where all indices are down and covariant objects have overbars, then later this notation is converted to "standard notation" with the usual up and down indices. It is a very large and complicated world, and below we report out only those facts which are useful for our efforts here. Equation numbers referring to Tensor are followed by a prime ' . Bolding Vectors. For the time being we shall display all vectors in bold font because we feel it helps the reader when dealing with covariant dot products and is compatible with Tensor. However, vector components are not bolded. Thus vector V will have components Va and Va. The exception is when vectors have extra labels, such as for the basis vectors en. It's components are written (en)a and (en)a. Eventually in Section 3.1 where we finally tie back to Sections 1.1 and 1.2 we shall quietly stop bolding vectors and will then be compatible with those earlier sections. Higher rank tensors are never bolded. 2.1 R, S and how tensors transform : Picture A Tensor is in large part based on the following "picture", (1.11)' (2.1.1) Below we shall be thinking of x-space as a vector space V having a set of basis vectors {ei} or {ui}. Then x'-space is a vector space V'. Below we shall use V as a prototype vector in space V, so V ϵ V . The two vector spaces V and V' have the same dimension N. In what follows, repeated indices are implicitly summed (Einstein convention) so for example RabVb means Σb=1N RabVb. Hanging indices like a in RabVb can take any value in the range a = 1,2...N. The implied summation convention reduces symbol clutter especially when there are many summed indices in an equation. Sometimes however we will display sums for emphasis. Figure (2.1.1) summarizes a generally non-linear transformation x' = F(x) between two spaces called x-space on the right (metric tensor g) and x'-space on the left (metric tensor g'). The coordinates of x-space are called x, and those of x'-space are called x'. Quantities in x'-space always have a prime, while those in x-space have no prime. A vector V in x-space has contravariant components Va and covariant components Va. The corresponding components V'a and V'a of V' in x'-space are these, V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a (7.5.3)' (7.5.2)' V'a = SbaVb Sba ≡ (∂xb/∂x'a) = ∂'axb (7.5.4)' (2.1.2) As noted, primed equation numbers refer to Tensor. The matrices R and S in (2.1.2) and Fig (2.1.1) are linearizations of the generally non-linear transformation x' = F(x) and x = F-1(x') in the close neighborhood of a selected point x in x-space and x' = F(x) in x'-space. Therefore, R and S are in general functions of x, though we suppress this dependence. R is sometimes call the differential of the transformation x' = F(x). We call it the R matrix. Many texts don't make up a symbol like R for the differential, and so tensor equations such as (2.1.8) below are strewn with partial derivatives of the form Rab = . This is useful in doing chain rules, but otherwise obscures how the indices work. We settled on symbols R and S after rejecting various reasonable alternatives. One should understand that Rab is in general not a simple rotation matrix despite the letter R. From the chain rule, one can see that the matrices R and S are inverses of each other, Sab Rbc = δac . // SR = 1 (7.6.1)' (2.1.3) If desired, the matrix S can be eliminated from the discussion by the fact that (reflect indices in a vertical line between the indices) Sab = Rba Sab = Rba . (7.5.13)' (2.1.4) Then (2.1.2) can be then be written with only R's , V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a // V' = RV V'a = RabVb Rab ≡ (∂xb/∂x'a) = ∂'axb . (2.1.5) Here is a table summarizing different forms of the differentials R and S : Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b . (7.5.16)' (2.1.6) Notice how an upper index in a derivative denominator acts as a lower index and vice versa. The fact that each item can be represented by two partial derivatives follows from (2.1.4), (2.1.2) and the raising and lowering operations described in Section 2.2 below. Vectors which transform under (according to) transformation F as in (2.1.5) are called rank-1 tensors. Here is how the four forms of a rank-2 tensor M transform under F, M'ab = Raa' Rbb' Ma'b' // pure contravariant M'ab = Raa' Rbb' Ma'b' // mixed M'ab = Raa' Rbb' Ma'b' // mixed M'ab = Raa' Rbb' Ma'b' . // pure covariant (7.5.8)' (2.1.7) We sometimes refer to Raa' as the "down-tilt" R matrix, and Raa' as the "up-tilt" R matrix. One sees that a down-tilt R transforms each contravariant (up) index, while an up-tilt R transforms each covariant (down) index. From the above, one can intuit the way an arbitrary rank-n tensor transforms under F. For example T 'abcde = Raa' Rbb' Rcc' Rdd' Ree' Ta'b'c'd'e' . (7.10.1)' (2.1.8) The various forms of the R matrix have these four orthogonality rules, 1: Rba Rbc = δac 2: Rba Rbc = δac // sum is on 1st index 3: Rab Rcb = δac 4: Rab Rcb = δac . // sum is on 2nd index (7.6.4)' (2.1.9) These are just renditions of (2.1.3) with (2.1.4) ( that is to say, SR = RS = 1 ). Using these rules, one can show (proof below) that the inverses of the vector transforms shown above in (2.1.5) are these Va = RbaV'b (7.6.7)' Va = RbaV'b (7.6.8)' (2.1.10) where the summation index b on R is not abutted against the following vector. The inversion of any tensor equation can be obtained instantly using the following simple rule: Inversion Rule: For each R, reflect the indices in a vertical line between the indices. (2.1.11) Examples: V'n = Rnm Vm , inversion is: Vn = Rmn V'm Rn| m → Rmn V'n = Rnm Vm , inversion is: Vn = Rmn V'm Rn| m → Rmn Proof : (2.1.8)#1 V'n = Rnm Vm RniV'n = RniRnmVm = δimVm = Vi Vi = RniV'n Vn = RmnV'm . The proof of the 2nd example follows from application of the tilt reversal rule (2.9.1) to the first example General Proof: Recall from (2.1.4) that Sab = Rba. The vertical line reflection Ra| b → Rba = Sab just changes R into S, and S is the inverse of R as in (2.1.3). So V' = RV gives V = SV' and similarly for higher tensor cases. Exercise: Invert equation (2.1.7) which says M'ab = Raa' Rbb' Ma'b' : Result: Mab = Ra'a Rb'b M'a'b' . The canonical vectors are the differential distances dxi in x-space and dx'i in x'-space (near some point x and corresponding x'). From (2.1.5) we then have, dx'a = Rabdxb // dx' = Rdx dx'a = Rabdxb . (2.1.12) From rule (2.1.11) the inverses of (2.1.12) are, dxa = Rbadx'b dxa = Rbadx'b . (2.1.13) The derivative operator ∂a ≡ ∂/∂xa transforms like any other covariant vector. It is in fact the canonical covariant vector for the transformation F. Thus, from (2.1.5), ∂'a = Rab∂b Rab ≡ (∂x'a/∂xb) = ∂bx'a ∂a = ∂/∂xa ∂'a = ∂/∂x'a ∂'a = Rab∂b Rab ≡ (∂xb/∂x'a) = ∂'axb ∂a = ∂/∂xa ∂'a = ∂/∂x'a (2.1.14) For example, if φ(x) is a scalar field (rank-0 tensor), ∂aφ(x) transforms as a covariant vector under F, and ∂aφ(x) transforms as a contravariant vector. Derivatives of tensors above rank-0 are more complicated, see Comment 1 below. The R matrix is really four matrices, and we have seen two of its forms above. The R object is not a tensor because, as Rab ≡ (∂x'a/∂xb) suggests, R has one foot in x'-space and one foot in x-space. In fact, the first index of R is raised or lowered by g', while the second index is raised or lowered by g: Rab = (∂x'a/∂xb) Rab = Rab' gb'b = (∂x'a/∂xb) // g pulls up the second index of Rab Rab = g'aa'Ra'b = (∂x'a/∂xb) // g' pulls down the first index of Rab Rab = g'aa'Ra'b' gb'b = (∂x'a/∂xb) . // both actions at once (7.5.9)' (2.1.15) Tensor fields We have suppressed the fact that in general everything above is a function of x (or equivalently x'). For example, when we compute Rab ≡ (∂x'a/∂xb) we generally obtain Rab(x). The transformation of a vector from x-space to x'-space was given in (2.1.5) as V'a = RabVb. For general x' = F(x) this really a statement about the transformation of vector fields: V'a(x') = Rab(x)Vb(x). The rank-2 tensor transformation in (2.1.6) really says M'ab(x') = Raa'(x) Rbb'(x) Ma'b'(x) and we are transforming a rank-2 tensor field. In special relativity it happens that x' = F(x) is linear so x'a = Fabxb (usually written with non-bold 4-vectors and Greek indices like x'μ = Λμνxν). In this situation Rab does not depend on x, and one can then have vectors which are not fields like p'μ = Λμνpν (momentum of a point particle) and vectors that are fields like A'μ(x') = ΛμνAν(x) (electromagnetic vector potential). Notice on the x'-space side of the equation that the vector field A'μ is a function of the x'-space coordinate x', while on the x-space side Aμ has argument x. In continuum mechanics and general relativity, Rab is a function of x so everything is a field. In the following examples, matrix R is the linearization of x' = F(x). In the neighborhood of the point x one has dx' = R(x)dx as a "linear fit" to the generally non-linear x' = F(x). If F(x) is a linear transformation, then x' = Fx so dx' = Fdx and then R = F = independent of x. (a) T' a'b'c'd' = Ra'aRb'bRc'cRd'd Tabcd tensor (rank 4, mixed) (b) T' a'b'c'd'(x') = Ra'a(x)Rb'b(x)Rc'c(x)Rd'd(x)Tabcd tensor which is not a field in x-space (c) T' a'b'c'd'(x') = Ra'aRb'bRc'cRd'd Tabcd(x) tensor field, linear F(x) (d) T' a'b'c'd'(x') = Ra'a(x)Rb'b(x)Rc'c(x)Rd'd(x) Tabcd(x) general tensor field (2.1.16) Item (a) is the generic form we use for a transformation of a rank-4 tensor. If F(x) is linear (as in special relativity), then Rab does not depend on position, and it is possible for T'abcd and Tabcd not to be fields. Item (b) is for a non-linear x' = F(x) where Tabcd is not a field. Obviously T'a'b'c'd' must depend therefore on x, and hence x', and so it is a field. In this unusual situation, the entire dependence of T' on x' is induced by the non-linearity of the transformation. Item (c) is more standard, where the transformation is linear and a tensor field is being transformed. This is the case in special relativity. Item (d) is the same thing, but the transformation is non-linear so one has Rij(x). Comments: 1. For situations where Rab is a function of x, it is easy to see why there is trouble with derivatives. One need only consider : V'a(x') = Rab(x)Vb(x) and ∂'b = Rbc(x)∂c so V'a,b(x') ≡ ∂'bV'a(x') // Va,b is just a new notation for the derivative = (Rbc(x)∂c)[Rad(x)Vd(x)] = Rbc(x) Rad(x) (∂cVd(x)) + Rbc(x)(∂cRad(x)) Vd(x) . (2.1.17) It is this second term that causes ∂cVd not to transform as a rank-2 tensor. It only transforms as a tensor if it happens that x' = F(x) is linear so Rab is constant (as in special relativity). This problem is remedied by introducing the covariant derivative Vd;c as discussed in Tensor Appendix F, see for example (F.9.5). This new object then properly transforms as a rank-2 tensor, V'b;a(x') = Rbc(x) Rad(x)Vd;c(x) . (2.1.18) 2. We have chosen to write (∂x'a/∂xb) as Rab as a space-saving notation. This object is often called "the differential" of the transformation x' = F(x) at point x. Tensor deals only with transformations where x and x' have the same number of components N, but the idea (∂x'a/∂xb) as Rab generalizes beyond this restriction. Of course then the matrix Rab is no longer square and the associated linear algebra is more complicated. This situation arises in Chapter 10 below. 3 Although we have used the letter R in Rab, one should not think that R is a rotation. It could be a rotation, but in general it is more complicated, involving both rotation and stretching. It could be a rotation which, although being a rotation, is a different rotation at every point in space, like Ry(θ(x)). 2.2 The metric tensors g and g' and the dot product Within each space (x-space and x'-space in the (2.1.1) Picture A), the metric tensor lowers or raises vector indices, Va = gabVb V'a = g'abV'b Va = gab Vb V'a = g'ab V'b (7.4.4)' (2.2.1) In the same way, the metric tensor lowers or raises any index on any tensor. The contravariant and covariant metric tensors are inverses of each other, gabgbc = gac = δac = δa,c // note that δij = δij = δi,j . (2.2.2) Here the gab lowers the first index on gbc to make gac which is δac = δa,c so gdngup = 1. The objects gab, gab, gba and gab are true tensor objects whereas δac and δa,c are not. It just happens that the value of gac is δac. In writing covariant equations, one should replace δac by gac before attempting to raise index a or lower index c. The metric tensor is a rank-2 tensor like any other rank-2 tensor, and so, looking at the first and last lines of (2.1.6), g'ab = Raa' Rbb' ga'b' g'ab = Raa' Rbb' ga'b' . (7.5.7)' (2.2.3) Any metric tensor is symmetric, gab = gba g'ab = g'ba gab = gba g'ab = g'ba . (5.4.3)' (2.2.4) The metric tensor defines a dot product in each space a b = gijaibj = gijaibj = aibi = aibi x-space a ' b' = g'ija'ib'j = g'ija'ib'j = a'ib'i = a'ib'i x'-space . (2.2.5) The dot product is a scalar (rank-0 tensor) so it must be the same in either space a ' b' = a b . (2.2.6) An exception to this rule is noted for fluid flow, see Tensor Section 5.2. When applied to the canonical differential vector dxi we find dx dx = gijdxidxj = || dx ||2 ≡ (ds)2 x-space dx' dx' = g'ijdx'idx'j = || dx' ||2 ≡ (ds')2 x'-space (2.2.7) Thus from (2.2.6) ds = ds' (invariant distance). In special relativity, ds is called the proper time dτ. The metric tensor gets its name from these last equations. The distance between two points x and y in a metric space is determined by a function called "the metric" d(x,y). A commonly used metric is d(x,y) = ||x-y|| where the metric is defined by the norm. The distance between two close points x and x+dx is then given by d(x,x+dx) = ||x+dx-x|| = || dx || = ds. Squaring, d2(x,x+dx) = || dx ||2 = gijxixj which shows how the metric tensor gij describes the squared metric d(x,x+dx) in the metric space of interest. One might recall that in a raw vector space, there is no distance concept d(v1,v2). A vector space with a metric and an inner product, such as that shown above as is then a Hilbert Space. It happens that the same metric tensor gij has two roles to play: it determines differential distance, and it lowers an index. See Chapter 5 of Tensor for more details. As shown in (2.1.16) (d), for a general transformation everything (including g) is a function of x. For example, Va(x) = gab(x)Vb(x) V'a(x') = g'ab(x')V'b(x') (2.2.1) g'ab(x') = Raa'(x) Rbb'(x) ga'b'(x) . (2.2.3) (2.2.8) 2.3 The basis vectors en and en There are two sets of basis vectors called en and en which exist in x-space (vector space V). The integer n is a label, not a component index. These basis vectors are defined as en = ∂x/∂x'n = ∂'nx tangent base vectors en = ∂x/∂x'n = ∂'nx . reciprocal base vectors (7.13.5)' (2.3.1) where x = F-1(x'). The tangent base vectors en are tangent to the "coordinate lines" in x-space. If in x'-space a particular x'n is allowed to vary while all other x'i are held fixed, the trace in x'-space is a line parallel to the n axis, while the mapping of that line in x-space is the (often curved) coordinate line associated with x'n. Then en = ∂x/∂x'n evaluated at a point x on that coordinate line is a vector in x-space tangent to that coordinate line. For coordinate line examples, see e.g. Tensor (3.2.8)', (3.4.3)' or (3.4.7)' . Meanwhile, the reciprocal vectors en are "dual to" the en in that en em = δnm . In fact we have, en em = g'nm g'in en = ei en em = δnm = g'nm en em = g'nm g'in en = ei (7.18.1)' (2.3.2) The equations on the left imply those on the right which show how to raise and lower basis vector labels. Although the en and en are vectors in x-space, it is the metric tensor of x'-space which raises and lowers. For "duality" see text above Tensor (6.2.8)'. It is shown there that a unique dual basis bn always exists for any given basis bn. For a general transformation F, the tangent base vectors are functions of location and should be written en(x), and of course the same is true for en(x). Looking at (2.3.2), we see that en(x) em(x) = δnm manages to be valid at every point in x-space. On the other hand, g'nm(x) = en(x) em(x) shows that the metric tensor is also a function of x. Example: In polar coordinates (r,θ) one has er = and eθ = r , both of which obviously depend on location in space. The metric tensor is gab = and also depends on spatial location through r. The coordinate lines for θ are circles whose tangents are eθ , while those for r are rays whose tangents are er. These basis vectors are like any other vectors in x-space, and so they have contravariant and covariant components such as (en)i and (en)i . Going back to our definition (2.3.1) we see that en ≡ ∂x/∂x'n (en)i = ∂xi/∂x'n = Rni . // from (2.1.5) Looking at (en)i = Rni we make these observations: (1) the n label of (en)i goes up and down with g' as shown in (2.3.2). (2) the n index of Rni also goes up and down with g' as shown in (2.1.13). (3) the i index of (en)i goes up and down with g as shown in (2.2.1). (4) the i index of Rni also goes up and down with g as shown in (2.1.13). Therefore the equation (en)i = Rni is "covariant", even though it is not a tensor equation (since R is not a tensor), so we can raise and lower indices at will on both sides. Therefore (en)i = Rni (en)i = Rni (en)i = Rni (en)i = Rni . (2.3.4) The en and en also satisfy a "completeness relation", (en)a(en)b = δab (2.3.5) where the implied sum is over the label n. From (2.3.4) this says Rna Rnb = δab which in fact is just orthogonality rule #2 in (2.1.8). This completeness relation is different from the "orthogonality relation" en em = δnm of (2.3.2) which, with x-space components, is written (en)i (em)i = δnm . Here the implied sum is over the component index i. Recall from above that en em = g'nm . (2.3.2) There are two cases that are often of interest: en em = fn δn,m the {en} form an orthogonal basis for V en em = δn,m the {en} form an orthonormal basis for V (2.3.6) Remember that the vectors en exist in x-space, despite the fact that the metric tensor g'nm is for x'-space in our transformation Picture A. 2.4 The basis vectors un and un One can also define a set of "axis-aligned" basis vectors in x-space as follows (un)i = δni = gni (un)i = δni = gni (7.18.3)' (2.4.1) which can be compared with (2.3.4) for en and en. Relations involving the u basis vectors are un um = gnm gin un = ui un um = δnm = gnm un um = gnm gin un = ui . (7.18.3)' (2.4.2) Notice the similarity to the relations for the en and en. The un are dual to the un. Whereas the en and en involve the x'-space metric tensor g', the un and un involve the x-space metric tensor g. We can easily calculate from (2.3.4) and (2.4.1) that en um = (en)i(um)i = Rni δmi = Rnm . (2.4.3) According to (2.3.2), g' raises and lowers the label n on en. According to (2.1.13), g' raises and lowers the first index n on Rnm. Similarly, the label on um and the second index of Rnm are raised and lowered by g. Thus our equation (2.4.3) is "covariant" (even though it is not a true tensor equation), so we can at once write out all four forms of the dot products between the e and u basis vectors on the left below, en um = Rnm en = Rnm um = Rnm um en um = Rnm en = Rnm um = Rnm um en um = Rnm un = Rmn em = Rmn em en um = Rnm un = Rmn em = Rmn em (2.4.4) The left column implies the right column (implied sum on m). For example, for the third line un = Σm Rmn em ek un = Σm Rmn ek em = Σm Rmn δkm = Rkn . (2.4.5) The equations on the right of (2.4.4) show that the R matrix is the "basis change matrix" relating the two different basis of x-space. The basis ui has a special place among possible bases for x-space. Above we call it an "axis-aligned" basis since (un)i = δni so for example in R2 we would have u1* = (1,0) and u2* = (0,1). Here the little asterisk is a notation to show we are talking about contravariant components. To say that (un)i = δni does not say that x-space is Cartesian, since x-space could have any metric tensor gij. What is really being said by the statement (un)i = δni is that the components of vectors (and higher tensors) are being defined in a specific way. If v is a vector, then vi has the following meaning: vi = ui v. Another way to say this is that the vi are the components of v when v is expanded in the u basis: v = Σi viui. In particular, since un is itself a vector, we have un = Σi (un)iui = Σi δniui = un. So the components of vectors are always defined relative to the u basis when u is selected as the "axis aligned basis" with (un)i = δni. The component indices on all tensors of Chapter 2 are referred to the u basis, for example the i and j of the metric tensor gij and of any tensor Mij or Mijk. We shall see below that one could expand vector v on the en basis, but then one gets v = Σi v'iei where v'i = ei v . The coefficient v'i here is of course different from vi since the bases are different. In fact v'i = Rijvj. 2.5 The basis vectors e'n and u'n and a summary Basis vectors e'n and u'n are just mappings of en and un from x-space to x'-space, u'n = R un u'n = R un e'n = R en e'n = R en . (2.5.1) Each of these mappings is like any vector mapping V' = RV. The primed basis vectors exist in x'-space, whereas the unprimed ones exist in x-space. The components of these vectors are easily computed, (u'n)i = Rij (un)j = Rij δnj = Rin // (2.4.1) (e'n)i = Rij (en)j = Rij Rnj = δin = g'in // (2.3.4) and (2.1.9) #3 . (2.5.2) Whereas the un were the axis-aligned basis vectors in x-space, we see that the e'n are the axis-aligned basis vectors in x'-space. Consider : (2.5.3) The figure shows just one base vector of each type. Here red e1 is the tangent base vector for coordinate x'1, whereas blue u'1 is the tangent base vector of the inverse transformation x = F-1(x') for coordinate x1. In each case a light black curve represents a piece of a coordinate line (curve) whose tangent is the tangent base vector. The associated dot products are obtained from (2.4.2) and (2.3.2), u'n u'm = gnm // = un um gin u'n = u'i u'n u'm = gnm u'n u'm = gnm gin u'n = u'i (2.5.4) e'n e'm = g'nm // = en em g'in e'n = e'i e'n e'm = g'nm e'n e'm = g'nm g'in u'n = u'i (2.5.5) and the basis vectors can therefore be raised and lowered as shown on the right by an appropriate metric tensor. So far we have computed these basis vector components, (un)i = gni = δni (2.4.1) (en)i = Rni (2.3.4) (u'n)i = Rin (2.5.2) (e'n)i = g'in = δin (2.5.2) (2.5.6) By raising n, lowering i, or doing both, one arrives at 12 more equations to get this full set of 16, 1 (un)i = gni = δni (en)i = Rni 2 (u'n)i = Rin (e'n)i = g'in = δin 3 (un)i = gni (en)i = Rni 4 (u'n)i = Rin (e'n)i = g'in 5 (un)i = gni (en)i = Rni 6 (u'n)i = Rin (e'n)i = g'in 7 (un)i = gni (en)i = Rni 8 (u'n)i = Rin (e'n)i = g'in (2.5.7) The reason one is allowed to do this follows from the label raising and lowering relations shown on the right side of (2.3.2), (2.4.2),(2.5.4), (2.5.5), and finally from (2.1.15) concerning indices on Rij . Since (u'n)i = Rin , the contravariant u'n vectors are the columns of R**. Since (en)i = Rni , the covariant en vectors are the rows of R** : R** = [u'1*,u'2*, ...u'N* ] = (2.5.8) Fact: One can treat the en as an arbitrary set of basis vectors (2.5.9) Above we started off by assuming some arbitrary transformation x' = F(x) and then the en(x) are the tangent base vectors for this transformation F. From a different viewpoint, one can assume some arbitrary expressions for the en(x) and try to find a corresponding x' = F(x) for which those en(x) are the tangent base vectors. Given the functions en(x), one would know the matrix of functions Rni(x) from (2.5.7) item 1. One could then attempt to integrate (2.1.10) which says dx'a = Rab(x)dxb to find x' = F(x). Let's assume this is all doable so x' = F(x) can always be found. From this point of view, one can regard the above equations concerning the en(x) to apply to an arbitrary set of basis functions en(x). Of course they have to be linearly independent at each value of x. The next section provides a very simple example. 2.6 How to compute a viable x' = F(x) from a set of constant basis vectors en As a first step, select g = 1 so that x-space is the usual Cartesian space and un = un are the usual orthonormal axis-aligned unit vectors of Cartesian space. Suppose we are handed a set of constant-in-space basis vectors en specified by their components relative to the x-space axes. From (2.5.7) item 1, (em)n = Rmn (2.6.1) so we know the matrix Rab (the em are the rows of the R matrix). What is the simplest way to fit this scenario into the tensor environment of Picture A in (2.1.1)? We try a linear transformation of the form x' = F(x) = Fx where F is a constant matrix (independent of x). Since x'a = Fabxb we find Rab ≡ (∂x'a/∂xb) = Fab, and then of course Rab = Fab. So we have found a linear transformation F that works: Fab = Rab. Then the en are the tangent base vectors for this transformation F, and en are the reciprocal base vectors (the dual vectors of en). The metric tensor g'nm can be computed from the dot product (2.3.2), g'nm = en em = (en)i (em)i = (en)i (em)i = RniRmi . (2.6.2) Since g = 1, the second index on R is allowed to move up and down "for free". This g'nm can then be inverted to determine g'nm. The reciprocal base vectors are then given by (2.3.2), en = g'nm em (2.6.3) with components (en)i = g'nm (em)i = g'nm Rmi . (2.6.4) We then rewrite the above equations as (em)n = Rmn // the em are the rows of matrix R, (2.6.1) lower n g'nm = RniRmi = RniRTim = (RRT)nm g' = RRT // (2.6.2) lower i en = g'nm em = hnm em // where we define hnm ≡ gnm // (2.3.2) (en)i = hnmRmi = (hR)ni. // the en are the rows of matrix (hR), (2.6.4) (2.6.5) Exercise: You are handed these three constant vectors en in a 3-dimensional Cartesian x-space, e1 = (2,-1,3) // for example, (e1)2 =( e1)2 = - 1 e2 = (-1,2,4) e3 = (1,3,2) // e3 = 1 u1 + 3 u2 + 2 u3 so R** = // = F; "the em are the rows of matrix R" from (2.6.5) (2.6.6) Use Maple to compute g, h and then (hR). Here we are just implementing the equations in (2.6.5). First enter the three en vectors and construct matrix R**, FIX CODE!!!! ****** Then compute the covariant metric tensor g' = RRT , From this compute the contravariant metric tensor h = g'-1 and then matrix (hR) The rows of (hR) are the vectors en (called En in the code), As a check, we verify that en em = δnm : 2.7 Expansions of vectors onto basis vectors Given vector V in x-space and the corresponding vector V' in x'-space, one may write the following expansions ( Tensor (7.13.10,11) ), 1 V = Σn Vn un where Vn = un V axis-aligned basis 2 V = Σn Vn un where Vn = un V axis-aligned basis 3 V = Σn V'n en where V'n = en V tangent base vector basis 4 V = Σn V'n en where V'n = en V tangent base vector basis 5 V' = Σn V'n e'n where V'n = e'n V' axis-aligned basis 6 V' = Σn V'n e'n where V'n = e'n V' axis-aligned basis 7 V' = Σn Vn u'n where Vn = u'n V' tangent base vector basis 8 V' = Σn Vn u'n where Vn = u'n V' tangent base vector basis (2.7.1) Notice that expansions 7,8 are obtained from 1,2 by applying the R matrix, since V' = RV and u'n = Run. Similarly, expansions 5,6 are obtained from 3,4 Any expansion can be directly verified by dotting the left column into an appropriate basis vector. For example, for expansion 6, using e'm e'n = δmn , V' = Σn V'n e'n e'm V' = e'm (Σn V'n e'n ) = Σn V'n e'm e'n = Σn V'n δmn = V'm . (2.7.2) Notice that each set of coefficients appears twice on the right in (2.7.1), once for V and once for V'. This duplication arise because a b = a' b' for any pair of vectors in either space. For example, Vn = un V = u'n V' appears in lines 1 and 7 (2.7.3) Looking a bit ahead, we shall be extending the notion of a vector expansion to that of tensors of any rank, and the meaning of component indices is still governed by Fact (2.7.5). First, for expansions on the axis-aligned un we write, V = Σn Vn un rank-1 tensor, (2.7.1) line 1 M = Σnm Mnm unum rank-2 tensor T = Σii....i Tii....i (ui ui ..... ui) . rank-k tensor (2.7.8) As discussed in (2.5.9) we may regard the basis vectors en as being an arbitrary basis. The primed tensor components are then the components of the x'-space version of the tensor under the transformation x' = F(x) generated by those arbitrary en. The corresponding expansions are then, V = Σn V'n en rank-1 tensor, (2.7.1) line 3 M = Σnm M'nm enem rank-2 tensor T = Σii....i T' ii....i (ei ei ..... ei) . rank-k tensor (2.7.9) In the above, any pair of tilted matching indices can be tilted the other way. The meaning of the symbol is discussed below. For an expansion on a mixed basis like enum , we will use the following notation M = Σnm [M(e,u)]nm enum . (2.7.10) Using this same notation we could write, M = Σnm M'nm enem = Σnm [M(e,e)]nm enem ≡ Σnm [M(e)]nm enem . (2.7.11) Exercise: Consider these two expansions shown above of the rank-2 tensor M, M = Σab Mab uaub // (2.7.8) (2.7.12) M = Σab M'ab eaeb // (2.7.9) (2.7.13) Verify that the coefficients M'ab and Mab are related as expected. Use the result in (2.4.4) line 2 that em = Rmiui to get for (2.7.12), M = Mab uaub // all implied sums = Mab (Ria ei) (Rjb ej) = (RiaRjbMab) ei ej so comparing to (2.7.13) one concludes that M'ab = RiaRjbMab which is the correct statement that M transforms as a rank-2 tensor as shown in (2.1.7). Confusion about vectors and scalars The following issue is a subtle one that is worth nailing down early on because it can lead to confusions and seeming paradoxes. Consider this fact, un V = u'n V' (2.7.14) Being the dot product of two vectors, this object transforms as a scalar under x' = F(x) as shown in (2.2.6). We might express this fact by writing s(n) = un V s'(n) = u'n V' s(n) = s'(n) n = 1,2....N . (2.7.15) What we have here is a set of N scalars, s(n) for n = 1,2..N, where n is just a label. One would never claim that these scalars s(n) transform as a vector, which would require that s'(n) = Rnm s(m). We don't have such a relation; what we have is s'(n) = s(n). Now it happens that s(n) = s'(n) = Vn where Vn is the component of a vector. Notice that : V'n = RnmVm true : the Vn transform as a vector V'n = Vn false s(n) = s'(n) true : each s(n) = un V transforms as a scalar s'(n) = Rnms(m) false (2.7.16) We can summarize this discussion as follows: Fact: Just because the scalars s(n)= un V take the values Vn does not mean that the scalars s(n) transform as vectors, nor does it mean that the vector components Vn transform as a scalar. (2.7.17) 2.8 The Outer Product of Tensors and Use of Consider two vectors which transform in the usual rank-1 tensor manner relative to some underlying transformation x' = F(x) (for which R is the linearization), a'i = Σj Rij aj b'k = Σm Rkm bm from (2.1.5) (2.8.1) where we temporarily show the summation symbols. Multiplying these equations together gives (a'i)(b'k) = (ΣjRijaj)(ΣmRkmbm) = Σjm RijRkm(ajbm) and then hiding the sums again, (a'ib'k) = RijRkm(ajbm) . (2.8.2) Looking at the first line of (2.1.7), we see that this object is transforming as a rank-2 tensor, therefore it is a rank-2 tensor, and we can write it as M'ik = RijRkmMjm where Mij ≡ aibj . (2.8.3) The rank-2 tensor Mij = aibj is said to be the outer product of two rank-1 tensors (vectors). This idea can be generalized ad infinitum. For example, if K is a rank-2 tensor and v is a vector, then Mijk = Kijvk (2.8.4) is a rank-3 tensor because it transforms as one, using the same argument shown above. Next, consider, Mabcde = KabKcd ve . (2.8.5) If K is a rank-2 tensor and v is a rank-1 tensor, then M is a rank-5 tensor. Of course since this is a "true tensor equation" (a covariant one), indices may be shuffled any way one wants, such as Mabcde = KabKcd ve . (2.8.6) Just imagine applying g** several times to both sides of (2.8.5) to get (2.8.6). There are so many possibilities for creating outer product tensors that one sometimes forgets that not all tensors can be "factored" into products of lower rank tensors. The Symbol Appears subtitle: "the rabbit goes into the hat" In Sections 1.1 and 1.2 we had vw being an element of a "tensor product space" VW and we described two approaches to the development of the meaning of the symbol : quotient space and category theory. Here we provide a third approach to the meaning of which is equivalent to that of the first two approaches. This third approach is geared to dealing with tensor components so there are lots of indices floating around, whereas in Sections 1.1 and 1.2 components were not even mentioned. Recall our previous two equations Mabcde = KabKcd ve . (2.8.5) Mabcde = KabKcd ve . (2.8.6) In order to display the outer product as a unified entity, we had to make up a new symbol M to represent the outer product tensor. We can avoid having to do this by writing M = KKv, so that the symbol in our "third approach" is just a way to name an outer product tensor. The above equations are then (KKv)abcde = KabKcd ve . (2.8.7) (KKv)abcde = KabKcd ve . (2.8.8) Here K and v are tensors, they are not spaces, so this is more like vw than VW. In fact, as a special case we can use this idea to name the outer product of two vectors to be rank-2 tensor ab, (ab)ij = aibj a,b ϵ V . (2.8.9) Notice that is a non-commuting operator: ab ≠ ba . In our "third approach" the symbol exists only within the context of VV, since vectors a and b both belong to the x-space of Picture A (2.1.1) which we identify with vector space V. However, one can extend this meaning of to apply more generally as the outer product of vectors in different vector spaces, (vw)ij = viwj v ϵ V w ϵ W . (2.8.10) If we try to fit this into our notion of tensor transformations, we would need two copies of Picture A, one for U→V and the other for X→W with vector transformations v(V)i = R(V)ijv(U)j R(V)ij = linearization of some transformation x' = F(V)(x) w(W)i = R(W)ijw(X)j . R(W)ij = linearization of some transformation y' = F(W)(y) (2.8.11) Then the transformation of the outer product "tensor" would be written as, (v(V)iw(W)i) = R(V)iaR(W)jb (v(U)aw(X)b) or [(vw)(V,W)] ij = R(V)iaR(W)jb [(vw)(U,X)]ab . UX → VW (2.8.12) One might refer to (vw)(V,W) as a "cross-space rank-2 tensor" (cross tensor). Normally the word "tensor" is used when W = V. Then the above reads, [(vw)(V,V)] ij = R(V)iaR(V)jb [(vw)(U,U)]ab UU → VV or (vw)' ij = RiaRjb (vw)ab . // Picture A (2.1.1) (2.8.13) The outer product thus has the same form in x'-space and in x-space, being a rank-2 tensor, (ab)ij = aibj a,b ϵ V . (ab)'ij = a'ib'j a',b' ϵ V' . (2.8.14) As a final outer product example, consider the outer product of three vectors, Mijk = aibjck a,b,c ϵ V , (2.8.15) Using our naming method for outer products, this becomes (abc)ijk = aibjck (2.8.16) with this obvious extension to the outer product of any number of vectors (abc....)ijk.... = aibjck.... (2.8.17) Associativity of The outer product operator as defined here is an associative operator, because multiplication of real numbers is associative. Consider for example, (ABv)abcde = AabBcdve (multiplication of reals is associative) [(AB)v]abcde = [(AB)abcd] ve = [ AabBcd] ve = AabBcdve . (2.8.19) Adding the parentheses on the second line in (AB)v does not alter the value of the components. This is true for the tensor product of any number of tensors, (T1T2T3....TN)III...I = T1IT2IT3I .... TNI (2.8.20) where each Ii represents a set of indices to go with Ti. For example, (T1(T2T3)....TN)III...I = T1I(T2T3)II ... TNI = T1I[ T2IT3I] ... TNI = T1IT2IT3I .... TNI (2.8.21) Therefore we have, Fact: The operator is associative for any tensor product, so parentheses can be added anywhere in a tensor product. (2.8.22) The associativity of the product of a set of real numbers along with the outer product definition of is what causes the operator to be associative. With the abstract definitions of Chapter 1, associativity of is added by fiat as an axiom. 2.9 The Inner Product (Contraction) of Tensors When any tensor structure contains a pair of implicitly summed indices which are "tilted", one says that those indices are contracted. It is easy to show that, due to the orthogonality rules (2.1.9), such internal index contractions behave as a scalar, which is to say, behave as if they weren't there at all with respect to a transformation. A proof appears in Tensor (7.12.2). Such contractions in a tensor structure reduce the rank of the tensor by two, resulting in an inner product. The contracting sum must occur only on a "tilted pair" of indices. Tilt Reversal Rule: Any such tilted index pair can have its tilt reversed "for free". (2.9.1) Proof: Using (2.2.1) and (2.2.2), [-----a---------a----] = gab gac [-----b---------c----] = gba gac [-----b---------c----] = δbc [-----b---------c----] = [-----b---------b----] = [-----a---------a----] where dashes indicate up or down tensor indices we don't care about. This "tilt reversal rule" applies to any contracted index within a tensor expression. It applies as well in other cases where g raises and lowers things so the above proof still works. The classic example involves expansions of the form (2.5.1) V = V'n en = V'n en . (2.9.2) The tilt can be reversed even though the n on en is a label and not a tensor index. The reason is that en = g'ni ei en = g'ni ei (2.3.2) V'n = g'nbV'b V'n = g'nb V'b . (2.2.1) The standard first example of an inner product is the inner product of two vectors. Consider, Mij = aibj = a rank-2 tensor, which we now contract to form, s = Mii = aibi = aibi = a rank-0 tensor (a scalar) (2.9.3) Using our notation (2.2.5) this is written s = a b // <a | b> in Dirac notation (2.9.4) which is an "inner product" of two vectors. This is of course the inner product / scalar product / dot product which makes our vector space V be a Hilbert space. In this example, creating an "inner product" of the two vectors ai and bj which has rank-0 goes in the opposite direction of the "outer product" that creates Mij = Mij = aibj of rank-2. The term "contraction" is more often applied to reducing the rank of tensors than is "inner product", and perhaps it is best to reserve the term "inner product" for the above dot product of two vectors. Here are other examples of rank reduction by contraction. Define Mabcd ≡ KabQcd = rank-4 tensor (2.9.5) Tac ≡ Mabcb = KabQcb = rank-2 tensor . (2.9.6) In this last example, contraction on the b index happens to occur between the two rank-2 tensors from which M was constructed as an outer product. One more step, S ≡ Taa = KabQab = rank-0 tensor (scalar) . (2.9.7) Using the notation introduced in the previous section, we can write the inner product s = a b as a contraction of the outer product ab s = (ab)ii = (ab)ii and ||a||2 ≡ aiai = (aa)ii . (2.9.8) Similarly (2.9.5,6,7) can be written (KQ)abcd = KabQcd = rank-4 tensor (2.9.9) Tac = (KQ)abcb = rank-2 tensor (2.9.10) S = Taa = (KQ)abab = rank-0 tensor (scalar) . (2.9.11) Dot products in spaces VV, VW, VVV and VWX Recall that (2.2.5) defines the (covariant) dot product of two vectors in V a b = gijaibj = gijaibj = aibi = aibi . x-space = V (2.2.5) It is possible to define an inner product operator for use between two elements of VV : (ab) (cd) ≡ Σij(ab)ij(cd)ij (ab), (cd) ϵ VV (2.9.12) = Σijaibjcidj . With this definition, we have a tiny theorem: Theorem: (ab) (cd) = (ac)(bd) a,b,c,d ϵ V (2.9.13) Proof: (ac)(bd) = (Σi aici)( Σj bjdj) = Σij aicibjdj = Σij aibj cidj = Σij(ab)ij(cd)ij = (ab) (cd) . Suppose dim(V) = n and dim(W) = n'. Then we can extend the above theorem to VW in this way. First define the dot product as, (ab') (cd') ≡ Σi=1nΣj=1n'(ab')ij(cd')ij (vw),(v'w') ϵ VW (2.9.14) = Σijaib'jcid'j . The corresponding Theorem is then Theorem: (ab') (cd') = (a c)(b' d') a,c ϵ V b',d' ϵ W (2.9.15) Proof: (ac)(b'd') = (Σi=1n aici)( Σj=1n b'jd'j) = Σijaicib'jd'j = Σijaib'jcid'j = Σij (ab')ij (cd')ij = (ab') (cd') . In a similar fashion one can show using (2.8.17) that with the following definition, (abc) (def) ≡ Σijk (abc)ijk (def)ijk VVV (2.9.16) one obtains Theorem: (abc) (def) = (a d)(b e)(c f) all vectors ϵ V (2.9.17) with a similar extension to VWX, Theorem: (ab'c") (de'f") = (a d)(b' e')(c" f") a,d ϵ V; b',e' ϵ W; c",f"' ϵ X (2.9.18) 2.10 Tensor Expansions Having a name for the outer product of two vectors allows us to write expansions of tensors of rank greater than 1 in a compact notation. The template is the vector expansion from (2.7.1) line 3, V = Σa V'a ea . (2.10.1) (a) Rank-2 Tensor Expansion and Projection As shown in (2.7.9) and (2,7,13), one can expand a rank-2 tensor on the tangent base vectors as follows, M = Σab M'ab eaeb . (2.10.2) To verify that this is the correct expansion, we take the components of both sides, [M]'ij = [ Σab M'ab eaeb]'ij = Σab M'ab (eaeb)'ij = Σab M'ab (e'a)i (e'b)j // (2.8.14), outer product in x'-space = Σab M'ab δaiδbj // (2.5.7) line 2 used twice = M'ij (2.10.3a) so the expansion is correct. Similarly, M = Σab Mab uaub [M]ij = [ Σab Mab uaub]ij = Σab Mab (uaub)ij = Σab Mab (ua)i (ub)j // (2.8.14), outer product in x-space = Σab M'ab δaiδbj // (2.5.7) line 1 used twice = Mij . (2.10.3b) A convenient notational method for projecting out the coefficients of any tensor expansion is the use of tensor-product-space dot products defined in Section 2.9. To demonstrate, we use a tensor expansion in VW where the basis vectors are en and e'n for V and W, using notation of (2.7.9), M = Σab [M(e,e')]ab eae'b . M ϵ VW . (2.10.4) The appropriate projector is (eie'j), which is just the expansion's basis eae'b with up/down toggled on the indices, and dummy labels like i,j selected. Using this projector one finds, (eie'j) M = Σab [M(e,e')]ab (eie'j) (eae'b) = Σab [M(e,e')]ab (ei ea)(e'j e'b) // theorem (2.9.15) = Σab [M(e,e')]ab δia δjb // dual pairs as in (2.3.2) = [M(e,e')]ij (2.10.5) and indeed, the coefficient is duly projected out of M. Here is more complicated example where M is now a rank-3 tensor, and where we use a perverse mixed basis, M = Σabc [M(e,u',e")]abc ea u'b e"c M ϵ VWX . (2.10.6) The projector is (ei u'j e"k) and we use it to project out the coefficient in (2.10.6) : (ei u'j e"k) M = (ei u'j e"k) Σab [M(e,u',e")]abc ea u'b e"c = Σab [M(e,u',e")]abc (ei u'j e"k) (ea u'b e"c) = Σab [M(e,u',e")]abc (ei ea)(u'j u'b)(e"k e"c) // theorem (2.9.18) = Σab [M(e,u',e")]abc δia δjb δkc // each pair is dual as in (2.3.2) = [M(e,u',e")]ijk . (2.10.7) (b) Rank-k Tensor Expansions and Projections A rank-k tensor T in Vk has this expansion on the er basis, T = Σii....i T'ii....i (ei ei ..... ei) . (2.10.8) To verify, we take components of both sides, [T]jj...j = Σii....i T'ii....i (ei ei ..... ei)jj...j = Σii....i T'ii....i (ei)j (ei)j .....(ei)j // (2.8.17) = Σii....i T'ii....i Rij Rij .....Rij // (2.5.7) line 1 = Σii....i [ Rij Rij .....Rij T'ii....i ] = Tjj...j . (2.10.9) To get the last step, we use the inversion rule (2.1.11) applied to the known tensor transformation T'jj...j = Rji Rji .....Rji Tii....i . (2.10.10) The coefficients T'ii....i can be projected out from T as in (2.10.5), (eiei... ei) T = T'ii...i (2.10.11) with an appropriate generalization of the dot product to the space Vk = VV...V , (v1v2...vk) (w1w2...wk) ≡ Σii....i (v1v2...vk)ii....i (w1w2...wk)ii....i = Σii....i (v1)i(v2)i... (vk)i (w1)i(w2)i... (wk)i // outer products = (v1 w1) (v2 w2) .... (vk wk) . (2.10.12) Using the notion of a multiindex I (an ordinary multiindex), I ≡ i1, i2, .....ik // each ir ranges 1,2....n n = dim(V) (2.10.13) and a shorthand notation for the basis vectors eI ≡ ei ei ..... ei eI ≡ ei ei ..... ei (2.10.14) the expansion (2.10.8) can be stated in the following compact form, T = ΣI T'I eI (2.10.8) (2.10.15) and the coefficients T'I can be projected out according to (2.10.11), eI T = T'I . (2.10.11) (2.10.16) With no comments, we now repeat the above set of steps for the expansion of T on the ur basis: T = Σii....i Tii....i (ui ui ..... ui) (2.10.17) [T]jj...j = Σii....i Tii....i (ui ui ..... ui)jj...j = Σii....i Tii....i (ui)j (ui)j .....(ui)j // (2.8.17) = Σii....i Tii....i δij δij .....δij // (2.5.7) line 1 = Tjj...j (2.10.18) (uiui... ui) T = Tii...i (2.10.19) uI ≡ ui ui ..... ui uI ≡ ui ui ..... ui (2.10.20) T = ΣI TI uI (2.10.21) uI T = TI . (2.10.22) 2.11 Dual Spaces and Tensor Functions 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 (but we always use the reals). 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". Comments: Much of the rest of this section will be repeated in later Chapters. We have found that the notations involved can be a major stumbling block, and feel it is important to exercise the notation in many ways to make the reader (and author) feel comfortable with it. As with most endeavors, it is a matter of practice. We also try to explain why certain notations are used. (a) The Dual Space V* in Matrix and Dirac Notation For every column vector v in V, there exists a row vector vT such that (vT)i = vi. For example, for N=2, v = = |v> vT = (a, b) = <v| (2.11.a1) Here we have snuck in the Dirac bra-ket notation where the ket |v> is a column vector and the bra <v| is the corresponding row vector. The notation vT means that the row vector is the Transpose of the column vector. We now have multiple ways to write the dot (inner, scalar) products of Section 2.9 : v v' = vTv' = (a, b) = aa'+bb' = <v | v' > . (2.11.a2) Because our vectors have real components, the above can also be written v' v = v'Tv = (a', b') = aa'+bb' = <v' | v > . (2.11.a3) We mention real components only because in the Dirac notation one has <a|b> = <b|a>* where * means complex conjugation, so if this scalar product is real, then <a|b> = <b|a> . We regard v or |v> as being a vector in the vector space V, while vT or <v| (the row vector) is a vector in the dual space V*. This is really a simple concept. Sometimes the dual-space vector vT = <v| is referred to as the covector of v = | v > . Suppose αT = <α | is a vector in the dual space V* . One can regard this dual-space vector <α | as being a functional which acts on vectors in the space V. Then, α = <α | = functional α(v) = <α | v> = α v = function = a scalar number (2.11.a4) α : V → K . // K = any real field, such as the real numbers Just as the space of column vectors V is a linear space (a vector space), so also the dual space of row vectors V* is a linear space, so we know that the functional <α | is a linear functional. That in turn implies that the function α(v) is a linear function, which we now show directly: α(sv) = α (sv) = s ( α v ) = s α(v) α(v+v') = α (v+v') = α v + α v' = α(v) + α(v') . (2.11.a5) (b) Functional notation We have now a slight notational conundrum. We like to write a scalar-valued function F(v) in non-bold font, whereas a vector-valued function would be F(v). Thus we have written α(v) above with a non-bold α, since α(v) is a scalar-valued function. On the other hand, α(v) is really a function of the vector α, so it seems misleading to refer to it as α(v), and we ought to call it α(v) so then α(v) = <α | v> has everything bolded on both sides. But then the functional would have to be called α = <α | . But this contradicts our notation earlier that α is a vector, αT is the transpose, and we should write αT = <α |. If we use α = <α | then we avoid that contradiction. This is what authors end up doing, writing a functional as a scalar entity which for us means an unbolded entity. A possible solution would be to say, fα = <α | = functional fα(v) = <α | v> = α v = function (2.11.b1) where f is non-bold, and the subscript label α is bold, but then we have introduced a new symbol f which seems superfluous. So the conclusion is this: α = <α| = functional, α(v) = <α | v> = function, and one must understand that α(v) is a function of the vector quantity α. Obviously there is a unique functional α(v) for each vector α in V (and thus for each αT in V*). The spaces V and V* have the same dimension n and are isomorphic to each other in the sense just noted. (c) Basis vectors for the dual space V* Now recall that our axis-aligned x-space basis vectors ui have dual basis vectors ui where ui uj = δij which is the idea of orthogonality in the covariant world (which might be non-Cartesian). In our notations above, ui uj = (ui)T uj = <ui | uj> = δij = ui uj = (ui)T uj = <ui | uj> = δij = δi,j . (2.11.c1) Since ui and ui are in general different column vectors in V, (ui)T and (ui)T are different row vectors in the dual space V*. Just as the column vectors |ui> and |ui> form two distinct bases for V, the row vectors <ui | and <ui | form two distinct bases for V*. Certainly dim(V) = dim(V*). Definition of λi Above we discussed α = <α| as a vector functional, and α(v) = <α|v> as the corresponding scalar function. Whereas <α| is some general vector in V*, we now consider in its place a basis vector <ui| in V*. With what notation shall we represent this functional? In analogy with α and α(v) we could use ui and ui(v) where the ui is unbolded to indicate a scalar function. Or we could use fu = <ui| and fu(v) = <ui| v> . The first notation is not uncommon (see wiki dual space where ui = ei), while the latter notation is unpleasant. Other common notations are v*i(v) or e*i(v) which for us would be u*i(v). We shall use the following notation, λi ≡ <ui| basis functional in V* // λi = (ui)T so λi(v) = <ui|v> basis function in V*f // λi(v) = (ui)Tv . (2.11.c2) The λ is unbolded, consistent with α(v). λ is a Greek letter consistent with our plan to use Greek or script letters for dual space objects. The index on λi is up, matching the index on ui in <ui|. Notice that λi(uj) = <ui|uj> = δij . (2.11.c3) We think of λi ≡ <ui| as being in the dual space V* while λi(v) = <ui|v> lies in a directly corresponding space of functions which we call V*f. Comment: For the dual space basis functionals Sjamaar used symbol λi in his 2006 notes (p 84), but changed to βi in his 2015 update (p 91). Spivak uses φi (p 76). Wiki (dual basis) uses basis vectors vi instead of ei so their λi is called vi. Wiki (dual space) uses ei while Lang Algebra uses fi (p 143). There seems to be no standard notation as in physics where F = ma is universally recognized as Newton's Second Law which would be hard to identify if written G = nb. Probably ui or ui (unbolded) is the most logical choice if the V basis vectors are ui, but it is so easy to confuse functional ui with the vector ui (especially when we drop our bolding of vectors starting in Chapter 3) that we shall stick with λi. Eq. (2.7.1) line 3 gives the expansion of a vector v onto the ui v = Σi vi ui where vi = ui v or |v> = Σi vi |ui> where vi = <ui| v > = ui v . (2.11.c4) Notice therefore that λi(v) = <ui|v> = ui v = vi . (2.11.c5) The function λi(v) is sometimes called "the ith coordinate function" since it projects out the ith component the vector v. As summarized in (2.7.17), since each dot product ui v is a scalar, the functions λi(v) i = 1..N transform as scalars despite the fact that the values of these scalars are the components of the vector vi. Transposing (2.11.c4) produces a vector functional expansion in V*, vT = Σi vi (ui)T or <v| = Σi vi <ui| = Σi vi <ui| . (2.11.c6) Using Greek letters for dual space objects we write this as α = <α| = Σn αi <ui| = Σn αiλi . (2.11.c7) Then, α = Σiαiλi functional (2.11.c8) α(v) = Σiαiλi(v) = Σiαivi = α v function (2.11.c9) or in bra-ket notation, <α| = Σiαi<ui| functional α(v) = Σiαi<ui|v> =Σiαivi α v = <α |v> function (2.11.c10) and we replicate the result (2.11.b1). Definition of λ'i We have defined λi ≡ <ui| as a notation for a certain basis functional in dual x-space. We would like to somehow define an object λ'i which is a basis functional in dual x'-space. How should this be done? One might intuitively feel that one should set λ'i ≡ <u'i| . Or one might think that once λi is defined as above, then the meaning of λ'i is forced upon us by some equation like λ'i = Rijλj . Both these notions are not what we want to do. We are not forced to say λ'i ≡ <u'i| just because λi ≡ <ui| since we are making two separate definitions. And λ'i = Rijλj is complete nonsense for the following reason. The N functionals λi for i = 1..N are each vectors in V*, so {λi} is a set of vectors, not a set of numbers, whereas when one tries to write λ'i = Rijλj one is implying that λj is a set of numbers which form a vector. Recall that the ui are the "axis-aligned" basis vectors in x-space since (ui)j = (ui)j = δi,j. Recall that the e'i are the "axis-aligned" basis vectors in x'-space since (e'i)j = (e'i)j = δi,j. This suggests that the proper definition of λ'i is the following: λ'i ≡ <e'i| (2.11.c11) One then finds that, for v' a vector in x'-space, λ'i(v') ≡ <e'i| v'> = v'i (2.11.c12) which is then analogous to λi(v) ≡ <ui| v'> = vi (2.11.c5) In both cases then λi and λ'i are the "ith coordinate functions", projecting out the ith coordinate from a vector. Since v'i = Rijvj we can certainly write λ'i(v') = Rij λj(v) (2.11.c13) as a statement relating two vectors of scalars. Notice this does not say λ'i = Rijλj which we already noted above does not even make sense. If we display the fact that Rij in general is Rij(x) then λ'i(v') = Rij(x) λj(v) (2.11.c14) Since this does not fit into any of the molds shown in (2.1.16), one cannot quite claim that λj(v) transforms as a vector field, but the transformation is similar. We can study (2.11.c13) in Dirac notation as follows (see below for Dirac notation details). λ'i(v') = <e'i|v'> = <ei|v> = <ei| 1 | v> = <ei|uj >< uj| v> = ei uj λj(v) = Rijλj(v) (2.11.c15) where the last step comes from (2.4.3). Once we have a functional λ'i, we can define a general rank-1 functional α' in dual x'-space as follows: α' = Σiα'iλ'i functional in V'* (2.11.c16) α'(v') = Σiα'iλ'i(v') = Σiα'iv'i = α' v' function in V'*f (2.11.c17) It then follows that α'(v') = α' v' = α v = α(v) (2.11.c18) and in some sense one could say that α(v) transforms as a scalar field, where v plays the role normally occupied by the position vector x. On the other hand, the vector |v> and the dual vector (functional) α = <α| transform as vectors and so α is a vector functional. We now define α(v) to be a "rank-1 tensor function". Spivak would call it a "1-tensor". We have this seeming contradiction that α(v) is a rank-1 tensor function, yet that function transforms as a rank-0 scalar. The rank-1 description really applies to the functional α = <α| which is in fact a vector and transforms as a vector. When this is closed with the ket |v> one obtains the scalar object α(v) = <α | v>. (2.11.c19) Vector space names: V, V*, V*f and V', V'*, V'*f (2.11.c20) Here we have associated vector space names V and V* with x-space in Picture A (2.1.1), while V' and V'* are associated with x'-space. All these spaces have the same dimension N and all are isomorphic. There is a 1-to-1 relationship between V and dual space V* as noted above, and there is a 1-to-1 relationship between V and V' since for every vector v in x-space there is a unique corresponding vector v' = Rv in x'-space. We refer to V* as dual x-space and V'* as dual x'-space. Associated with the dual space V* of functionals is the space V*f of corresponding functions, and similarly for V'* and V'*f. (d) Rank-2 functionals and tensor functions A rank-2 tensor may be represented as T = Σab Tab ua ub (2.11.d1) V = Σa Va ua where on the second line for comparison we show a general rank-1 tensor (vector). In Dirac notation, we write | ua, ub> ≡ |ua> |ub> ↔ ua ub (2.11.d2) which represents any of the n2 basis vectors of the tensor product space V2 = VV. We could write this as | ua, ub>2 ≡ |ua>1 |ub>1 to distinguish the fact that some kets are in V1 and others in V2, but the contents of the ket usually make it obvious to which vector space a ket belongs. In Dirac notation, the tensor T is written |T> = ΣabTab |ua> |ub> = ΣabTab | ua, ub> (2.11.d3) and this is a general element of the space V2. The corresponding rank-2 linear functional in the dual space V*2 is given by <T| = ΣabTab <ua | <ub| = ΣabTab < ua, ub| . (2.11.d4) This is done in analogy with the vector case |V> = Σa Va |ua> <V| = Σa Va<ua | (2.11.d5) where we are careful to have the index "tilt" have the form of a contraction, even though we are not really contracting indices on a tensor. The rank-2 functional <T| is linear in both V* spaces of V*V*, so it is called a bilinear functional. If we let (subscript 1 and 2 are labels of two vectors, not components of v ) | v1, v2> ≡ |v1> |v2> (2.11.d6) represent an arbitrary (but pure) element of V2 = VV, then we may construct T = <T| rank-2 tensor functional T(v1,v2) = <T | v1, v2> rank-2 tensor function (a Spivak "2-tensor") . (2.11.d7) It follows that T(v1,v2) = <T | v1, v2> = ΣabTab < ua, ub| v1, v2> = ΣabTab <ua| v1> <ub| v2> = ΣabTab (v1)a (v2)b (2.11.d8) where we have used the fact that the scalar product for elements of V*2 with elements of V2 is the product of two V*-with-V scalar products, as seen for example in (2.9.13). In the last line above we see that the tensor function T(v1,v2) is the contraction of a rank-2 tensor with two rank-1 tensors, and so is a scalar. Thus, T'(v'1,v'2) = T(v1,v2) (2.11.d9) and a rank-2 tensor function transforms as a "scalar field of two arguments". The "rank-2" description applies to the tensor functional <T| , and when this is closed with an element of V2 the result is a scalar. Note from (2.11.d8) and (2.4.1) that (ui)a = δia , T(ui,uj) = ΣabTab (ui)a (uj)b = ΣabTab δia δjb = Tij (2.11.d10) so the tensor function evaluated at the basis vectors gives a corresponding element of the tensor. Using λi = < ui| as defined above in (2.11.c2), we can rewrite (2.11.d4) <T| = ΣabTab <ua | <ub| as rank-2 tensor functional T = ΣabTab λa λb (2.11.d11) which is in analogy to <α | = Σa αa <ua| rank-1 tensor functional α = Σa αa λa . (2.11.d12) We continue to use script or Greek fonts to represent functionals, such as α and T. Taking the special case of a rank-2 functional which is just λa λb we construct the following rank-2 tensor function, (λa λb)(v1,v2) = < ua, ub| v1, v2> = <ua| v1> <ub| v2> = (v1)a (v2)b = λa(v1) λb(v2) . (2.11.d13) This function is manifestly linear in both arguments, since λa(v1) is linear, so it is a bilinear function. For example, (λa λb)(v1+v1',v2) = (v1+v1')a (v2)b = (v1)a (v2)b + (v'1)a (v2)b = (λa λb)(v1,v2) + (λa λb)(v'1,v2) . (2.11.d14) Whereas <T| shown above is a general rank-2 tensor functional, we can consider the special case of a pure rank-2 functional formed from two vector functionals α = <α| and β = <β| . In that case one finds, (α β) = <α| <β| = <α, β | rank-2 functional (α β)(v1,v2) = <α, β | v1, v2> = <α | v1> <β | v2> = α(v1)β(v2) rank-2 tensor function (α β)(ui,uj) = α(ui)β(uj) = αiβj = (α β)ij rank-2 tensor (2.11.d15) where the very last item is αiβj expressed in the tensor product notation of (2.8.9). Once again, evaluation of a tensor function at two basis vectors creates an element of the tensor. Comment on vertical bars in the Dirac Notation Let |a> be a vector in V, and <b| a vector in the dual space V*. Notice that <b| |a> = <b||a> = <b|a> . (2.11.d16) The official notation for the scalar product is <b | a> not <b || a> so one replaces the || with | . The same replacement is made for example doing a scalar product between elements of V*2 and V2 <a| <b| |c> |d> = <a,b||c,d> = <a,b | c,d> or <a| <b| |c> |d> = ( <a| |c> ) ( <b| |d> ) = <a|c><b|d> . (2.11.d17) (e) Rank-k functionals and tensor functions It is a simple matter to generalize from k = 2 to k = k, so the vector space is Vk and the dual space is V*k, Vk ≡ VxVx....xV k factors // Cartesian product of k spaces Vk ≡ VV....V k factors // tensor product of k vector spaces V*k ≡ V*V*....V* k factors // tensor product of k dual spaces . (2.11.e1) We then have as a most general element of Vk (a rank-k tensor), T = Σii....i Tii....i (ui ui ..... ui ) . T = ΣITI uI (2.11.e2) with (ui ui ..... ui) = |ui> |ui >..... |ui> = | ui, ui ....., ui >k (2.11.e3) = | ui, ui ....., ui > . uI ≡ ui ui ..... ui On the right in red we show our equations expressed in the multi-index notation introduced in (2.10.17-22). The letter Z which appears below is used to represent the set of integers 1,2...k. Then the rank-k tensor T in Vk is represented in Dirac notation as |T> = Σii....i Tii....i | ui, ui ....., ui > . |T> = ΣITI |uI> (2.11.e4) The rank-k tensor functional <T| of V*k is then <T | = Σii....i Tii....i < ui, ui ....., ui | <T| = ΣITI <uI| or (2.11.e5) T = Σii....i Tii....i λi λi ..... λi . T = ΣITI λI A general pure element of Vk is specified by |v1, v2, ...vk> = |v1> |v2> .... |vk> . |vZ> = |v1> |v2> .... |vk> (2.11.e6) The corresponding rank-k tensor function is given by T(v1, v2, ...vk) = <T | v1, v2, ...vk> = Σii....i Tii....i < ui, ui, ..., ui | v1, v2, ...vk> = Σii....i Tii....i < ui|v1>< ui|v2> .... < ui|vk> (2.11.e7) = Σii....i Tii....i (v1)i (v2)i....(vk)i T(vZ) = ΣITI (vZ)I This shows that the rank-k tensor function is a linear combination of the products of the argument components weighted by the components of the corresponding rank-k tensor. Since this is the contraction of a rank-k tensor with k rank-1 tensors, the result transforms as a scalar, so then T'(v'1, v'2, ...v'k) = T(v1, v2, ...vk) . T'(v'Z) = T(vZ) (2.11.e8) That is to say, the rank-k tensor function transforms as a scalar field, where the term "rank-k" is associated with the functional T = <T| which is an element of the dual space V*k . Finally we see that T(uj,uj, .... uj) = <T | uj,uj, .... uj > = Σii....i Tii....i (uj)i (uj)i....(uj)i = Tjj....j . T(uJ) = TJ (2.11.e9) From (2.11.e7) one sees that the tensor function T(v1, v2, ...vk) is manifestly k-multilinear, which is the generalization of linear for k = 1 and bilinear for k = 2. Once can construct a rank-k tensor functional purely from the dual basis vectors, (λiλi ... λi) = <ui| <ui| ... <ui| rank-k tensor functional λI = <uI| (2.11.e10) (λiλi ... λi)(v1,v2....vk) = λi(v1)λi(v2)....λi(vk) = (v1)i(v2)i ... (vk)i rank-k tensor function λI(vZ) = (vZ)I (2.11.e11) (λiλi ... λi)(uj,uj, .... uj) = λi(uj)λi(uj)....λi(uj) = (uj)i(uj)i ... (uj)i = δjiδji ... δji evaluated at basis vectors . λI(uJ) = δIJ (2.11.e12) As an alternative to the most general rank-k tensor functional T and the all-basis-vector rank-k tensor functional (λiλi ... λi), one can consider a "pure" rank-k tensor functional constructed from k dual vectors which we shall call <αi| . In this case we find, <α1, α2....αk| = <α1| <α2| .... <αk| pure rank-k tensor functional = α1 α2... αk = (α1α2...αk) (2.11.e13) (α1α2...αk)(v1, v2, ...vk) = α1(v1)α2(v2) ....αk(vk) = (α1v1)(α2v2) ....(αkvk) pure rank-k tensor function (2.11.e14) (α1α2...αk)(uj,uj, .... uj) = α1(uj)α2(uj) ....αk(uj) = (α1 uj) (α2 uj) ... (αk uj) evaluated at ur = (α1)j(α2)j...(αk)j = (α1α2...αk)jj... j outer product notation (2.11.e15) Hopefully after this long slog, the following paragraph makes some sense to the reader: A rank-k tensor function is the bra-ket closure (inner product) of a rank-k dual tensor functional <T| of V*k with a pure rank-k non-dual tensor |v1,v2...vk> of Vk such that T(v1,v2,...vk) = <T|v1,v2...vk>. The tensor function is k-multilinear in its arguments, and transforms as a scalar field with k vector arguments. When the rank-k tensor function is evaluated at the basis vectors ur, it replicates the non-dual rank-k tensor with which is it associated, which is to say, T(uj,uj, .... uj) = Tjj....j . Spivak on page 75 refers to a rank-k tensor function as a "k-tensor". (2.11.e16) As we shall see later, the motivation for using tensor functions is their crashingly simple description of the tensor product of an arbitrary rank-k tensor with an arbitrary rank-k' tensor to produce a rank-(k+k') tensor : k<T | v1,v2...vk>k k'< S| vk+1,vk+2...vk+k'>k' = [ k<T k'<S| ] [|v1,v2...vk>k |vk+1,vk+2...vk+k'>k'] = k+k'<TS | v1,v2...vk+k'>k+k' (2.11.e17) or T(v1,v2,...vk) S(vk+1,vk+2,...vk+k') = (TS)(v1,v2 .... vk+k') . (2.11.e18) This equation appears below as (6.6.13) and also appears in Spivak page 75. As noted by Benn and Tucker page 2, the relationship between the vector space Vk and the dual vector space V*k is a reciprocal one. One could, as they say, perversely regard V*k as the starting vector space and then Vk would be the dual space of V*k. This amounts to swapping bra ↔ ket in the Dirac notation outlined above. Instead of having a functional α(v) = <α|v>, one would have a functional v(α) = <v|α>. We find that things are hard enough to understand without doing this "perverse" swapping of things right off the bat as they do. They refer to a rank-k tensor as a tensor of degree k, while other authors refer to rank as the order of a tensor. We us the term rank and promise not to confuse it with the different notion of the rank of a matrix which is the number of linearly independent rows or columns, or with various other meanings of the word "rank" in mathematics. (f) The Covariant Transpose Whereas the matrix transpose of a matrix Mab would be (MT)ab = Mba (swap the rows and columns), it is the covariant transpose (MT)ab = Mba that is significant in covariant notation. We quote from Tensor where M is a general rank-2 tensor while R and S are the "differentials" of (2.1.2), (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . (7.9.3)' (2.11.f1) Equations in any column can be obtained by lowering one or both indices in the top equation, so that the covariant transform MT is a rank-2 tensor if M is a rank-2 tensor. For all-up or all-down indices, the two kinds of transposes are the same: (MT)ab = (MT)ab = Mba. The transpose always has the indices reflected in a vertical line between the indices. This subject is discussed in Tensor Section 7.9 where all claims are proved. We quote some of the conclusions: det(M) = det(MT) = det(MT) (7.9.7)' (2.11.f2) RRT = RTR = 1 SST = STS = 1 RS = SR = 1 RT = R-1 = S ST = S-1 = R . (7.9.8)' (2.11.f3) (g) Linear Dirac Space Operators Consider these three ways of writing the same real number, where M is a matrix sandwiched between vector b on the right and transpose vector a on the left, aT (Mb) M acts to the right ( * * *) [ ] (aTM)b M acts to the left, and note that (aTM) = (MTa)T [ (* * *) ] aTM b can think of M acting either to the right or to the left. (2.11.g1) In writing these equations, one normally thinks of M as being a matrix Mij = (M[u])ij or M = M[u] . By default, the matrix elements are taken in the axis-aligned ui basis on both left and right (and this applies to all indices as discussed at the end of Section 2.4) so that (ui)TM (uj) = Σa,b (ui)a Mab (uj)b = Σa,b δia Mab δjb = Mij // = (M[u])ij. (2.11.g2) One could, however, do this in some other basis, for example, (ei) TM (ej) = Σa,b (ei)a Mab (ej)b = Σa,b Ria Mab Rjb = (RMRT)ij // = (M[e])ij (2.11.g3) and the result is a completely different matrix. In this case the matrices are related by a covariant similarity transformation by R M[e] = R M[u]RT . (2.11.g4) It is useful to think of the object M as being a basis-independent abstract linear operator which, when sandwiched between certain basis vectors, has certain matrix elements. Different types of basis vectors yield different matrices. We could even have mixed basis elements, (ui) TM (ej) = Σa,b (ui)aMab(ej)b = Σa,b δiaMabRjb = (MRT)ij // = (M[u,e])ij (2.11.g5) so in this case we get M[u,e] = MRT . (2.11.g6) The abstract operator M only becomes a matrix when it is properly sandwiched. This notion of thinking of the object M as a basis-independent linear operator becomes more pronounced in the Dirac notation. We restate the above equations as follows, all of which evaluate to the same real number, <a| ( M |b>) M acts to the right = <a | Mb > (<a|M ) |b> M acts to the left, and note that <a|M = <MTa | = <MTa | b > <a| M |b> can think of M acting either to the right or to the left. (2.11.g7) The space between the vertical bars is inhabited by abstract linear operators like M. The matrix elements shown above are then <ui | M | uj> = (M[u])ij = Mij <ei | M | ej> = (M[e])ij = (RMRT)ij <ui | M | ej> = (M[u,e])ij = (MRT)ij (2.11.g8) To emphasize this notion of abstract operator, we shall write the operator in a different font, so M is a matrix and M is a Dirac-space operator, and then <a| M |b> = <a| ( M |b>) = <a |M b> = a scalar product of two vectors <a| M |b> = (<a|M ) |b> = <MTa | b > = a scalar product of two vectors <ui | M | uj> = (M[u])ij = Mij etc . (2.11.g9) Here then is a review of the matrix and Dirac notations, a' = (Ma) = (M)a (b')T = (Mb)T = bT MT matrix notation |a'> = |Ma> = M|a> <b'| = <Mb| = <b|MT . Dirac notation (2.11.g10) Then consider the following claim Fact: <a | M | b> = <b| MT | a> (2.11.g11) where both M and MT are the names of abstract linear operators. Proof: <a | M | b> ≡ <a | M b> = a (Mb) = ai(Mb)i = ai[ Mijbj] = ai Mij bj = bj Mij ai = bj (MT)ji ai = bj [MTa]j = b (MTa) = <b| MTa> = <b| MT |a> . Operator M is defined by its action on an arbitrary ket vector M | b> = | M b> Operator MT is defined by its action on an arbitrary ket vector MT | b> = | MT b> Notice in the proof that the covariant transpose MT is the correct transpose to use since Mij = (MT)ji. Exercise: Show that wv is a scalar under any transformation x' = F(x) : w'v' = <w' | v'> = <Rw|Rv> = <w|RTR|v> = <w| 1 |v> = <w|v> = wv . (2.11.g12) In this example R is a matrix, whereas R is the corresponding Dirac space operator. The statement RTR = 1 (2.11.g13) is the operator version of our (2.11.f3) matrix statement RTR = 1 (2.11.g14) which we verify as follows, (RTR)ac = (RT)abRbc = Rba Rbc = δac // (2.1.9) #1 (2.11.g15) and which is valid for any transformation differential matrix Rij. One may represent a Dirac operator M in various ways M = Σij | ui> Mij <uj| = Σij | ei> [M[e]] ij <ej| = Σij | ui> [M[u,e]] ij <ej| (2.11.g16) as can be verified by closing with the appropriate basis vectors. For example, for the last line above, <ua | M | eb> = <ua | { Σij | ui> [M[u,e]] ij <ej|} | eb> = Σij <ua | ui> [M[u,e]] ij <ej| eb> = Σij δai [M[u,e]] ij δjb = [M[u,e]] ab . (2.11.g17) When M = 1 we find 1 = Σij | ui> δij <uj| = Σi | ui><ui| (2.11.g18) which is just a statement that the | ui> basis is complete. Finally, we make a comparison between the abstract Dirac operator M and the abstract rank-2 "vector" M, M = Σij | ui> Mij <uj| // Dirac operator M = Σij Mij ui uj // (2.8.10), "vector" in vector space V2 or |M> = Σij Mij | ui> | uj> . (2.11.g19) The first object M is an operator in the Dirac Hilbert Space V. The second object M or |M> is a vector in the tensor product space V V. M and M are completely different objects, though they both involve the same matrix elements Mij. In each case, we can project out those matrix elements in an appropriate fashion: <ua | M | ub> = <ua | { Σij | ui> Mij <uj|} | ub> = Mab [< ua| < ub| ] | M > = [< ua| < ub| ] Σij Mij | ui> | uj> = Mab . (2.11.g20) The above discussion is presented implicitly for a square matrix M, but only small adjustments are needed for it to apply to a non-square matrix. In this case, in aTM b one thinks of vectors a and b as having different dimensions. Perhaps b lies in x-space which is Rn while a lies in x'-space which is Rm with m > n, and then Mij is an m x n matrix. The x'-space V' has n basis vectors |u'i> while the x-space V has m basis vectors |ui>. Then one would have, for example, <ui | M | u'j> = Mij M = Σi=1m Σj=1n | ui> Mij <u'j| |M> = Σi=1m Σj=1n Mij | ui> | u'j> 1' = Σi | u'i><u'i| completeness in V' 1 = Σi | ui><ui| completeness in V (2.11.g21) This is exactly the situation we shall encounter in Chapter 9 where the matrix R is an m x n matrix. We shall not use a script font to represent Dirac space operators, but one should keep in mind that any non-scalar object which is seen to be operating to the left on a bra, or to the right on a ket, or which is found between the two vertical bars in < | | > is a Dirac space operator, regardless of the font used to represent it. Such operator objects are not matrices, but their matrix elements form matrices.