Section 2.11 retired on Oct 31

Section 2.11 retired on Oct 31, 2015 PhL 10.31.15 2.11 The dual space V* In our treatment of the dual space side of things, we shall attempt to use Greek or script letters for all dual space vectors and tensors encountered. Vectors in V will continue to be represented by Latin letters. Just as we used V as our prototype vector in space V, here we use V as our prototype vector in space V* . The dual space V* is by definition the space of linear functionals over V. If V ϵ V*, we can then write V : V → K V(v) = k ϵ K (2.11.1) where K is any field such as the reals. Since V is a linear functional, V(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(x) as a "function". The basis λi Given n vectors {ei} which are 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 linear functionals λi such that λi(v) = ei v . // = (ei)a va (2.11.3) 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". From (2.11.3) and (2.3.2) one then has λi(ej) = ei ej = δij . (2.11.6) Therefore, λi(v) = λi(Σj vj ej) = Σj vjλi(ej) = Σj vjδij = vi . (2.11.7) The linear functional λi applied to vector v thus gives the e-basis component vi of v. A component of λi is given by THIS MAKES NO SENSE (λi)j = (ei(e))j = δij (2.11.8) where we use (2.6.4) col 4 row 4. The functional basis λi of V* is analogous to the basis ui of V discussed in Section 2.4. In particular, (2.4.1) says (ui)j = δij . General vector in V* A general linear functional V in V* can be written as a linear combinations of the basis functionals λi, V = ΣiViλi = general vector in V* V(v) = ΣiViλi(v) V: V → K (2.11.9) where on the right we show the corresponding function V(v). Evaluating at v = ej, V(ej) = ΣiViλi(ej) = ΣiViδij = Vj . // using (2.11.6) (2.11.10) Consider now, V(v) = ΣiViλi(v) = ΣiVi vi = V v = ΣiVivi . (2.11.11) Imagine a version of Picture A where we transform from x-space with basis vectors ei to some x'-space with basis vectors e'i. Under such a transformation, a dot product like V v transforms as a scalar. Whereas v and V transform as vectors, the function V(v) transforms as a scalar field, just as in (2.2.6). That is to say, V(v) = V v = V' v' = V'(v') (2.11.12) It is not unexpected that the contraction of a contravariant linear functional with a covariant vector results in a scalar. Clearly there is a 1-to-1 isomorphism between the set of linear functionals V in V* and the set of linear functions V(v) on V. One can therefore regard V* as the being the space of such functions V(v) . Since the functional V transforms as a rank-1 tensor, we loosely speak of V(v) as being a rank-1 tensor, despite the fact that it transforms as a scalar field as shown just above. Rank-2 tensor in V*V* We want the object λiλj to be a linear functional over the space V*V* such that λiλj: V*xV* → K . (2.11.13) The natural way to accomplish this desire is to write (λiλj)(v1,v2) = λi(v1)λj(v2) = scalar * scalar = scalar ϵ K (2.11.14) Object (λiλj) is bilinear in its arguments because λi(v1) is linear in its argument. A rank-2 tensor M in V*V* has the following expansion, analogous to (2.10.2) M = Σab [M(λ)]ab λaλb (2.11.15) and we do the verification similar to (2.10.3), [M(λ)]ij = Σab [M(λ)]ab (λaλb)ij = Σab [M(λ)]ab (λa)i (λb)j (2.11.16) = Σab [M(λ)]ab δaiδbj = [M(λ)]ij . The components [M(λ)]ab transform as a covariant rank-2 tensor, [M'(λ')]ij = Σab [M(λ)]ab [λ'aλ'b]ij = Σab [M(λ)]ab (λ'a)i(λ'b)j = Σab [M(λ)]ab [Ric (λa)c] [Rjd (λa)d] = Σab [M(λ)]ab [Ric δac] [Rjd δbd ] = Σab[M(λ)]ab Ria Rjb = ΣabRiaRjb [M(λ)]ab (2.11.17) We evaluate the tensor M at (v1,v2) to get M(v1,v2) = Σab [M(λ)]ab (λaλb)(v1,v2) = Σab [M(λ)]ab λa(v1) λb(v2) (2.11.18) M(ei,ej) = Σab [M(λ)]ab λa(ei) λb(ej) = Σab [M(λ)]ab δaiδbj = [M(λ)]ij (2.11.19) This shows that the tensor components [M(λ)]ij are obtained from the function M(v1,v2) evaluated at the basis vectors (ei,ej). One can rewrite M(v1,v2) as M(v1,v2) = Σab [M(λ)]ab λa(v1) λb(v2) = Σab [M(λ)]ab (v1)a(v2)b (2.11.20) Since this is a fully contracted outer product of three tensors (Section 3.1), we see that M(v1,v2) transforms as a scalar field over V* x V*, M'(v'1,v'2) = M(v1,v2) (2.11.21) Notice that the function M(v1,v2) is bilinear in its vector arguments because λa(v1) and λb(v2) are each linear. There is an isomorphism between the space of rank-2 tensor functionals M in V*V* and the set of bilinear functions on V. Although the functional M is really the rank-2 tensor, one loosely speaks of M(v1,v2) as being a rank-2 tensor even though it transforms as a scalar field of two variables as shown above. As a special case, consider M = α β . Then we get a result similar to (3.1.2), (α β)ij = Σab αaβb (λaλb)ij = Σab αaβb λaiλbj = Σab αaβb δaiδbj = αi βj (2.11.22) and (α β)(v1,v2) = Σabαaβb (v1)a(v2)b = [Σaαa (v1)a][Σbαb (v1)b] = α(v1)β(v2) (2.11.23) Again, α β is called a 2-tensor even though (α β)(v1,v2) is a scalar. Notice from (2.11.23) and (2.11.10) that (α β)(ei,ej) = α(ei)β(ej) = αiβi = (α β)ij (2.11.24) Rank-k tensor in V*V* ..... A rank-k tensor functional M in V*k ≡ V*V*.... can be expanded as in (2.10.14), M = Σabc... [M(λ)]abc... (λaλbλc ...) (2.11.25) with M(v1,v2, ....vk) = Σabc... [M(λ)]abc... λa(v1) λb(v2)λc(v3).... (2.11.26) The function M(v1,v2, ....vk) is thus seen to be k-multilinear in its k vector arguments. M(v1,v2, ....vk) is a scalar field over V*k , M'(v'1,v'2, ....v'k) = M(v1,v2, ....vk) . (2.11.27) For the special case where M = α βδ..... , (α βδ.....)ijk.... = αi βjδk....... (2.11.28) (α βδ.....)(v1,v2, ...vk) = α(v1)β(v2)δ(v3)...... (2.11.29) Once again, due to the isomorphism between functionals M and functions M(v1,v2, ....vk) , one loosely refers to the scalar field M(v1,v2, ....vk) as a rank-k tensor, although M is the actual tensor object whose components [M(λ)]abc... transform by the usual rule. One might say that the rank-k tensor M is "represented by" the k-multilinear function M(v1,v2, ....vk) , just as it is represented by [M(λ)]abc... in the sense of tensor components.