pullbacks REVIEWED

5.17.16: I have the picture labeled wrong compared to how it ends up. Am dealing with operators acting on tensor product spaces. Eventually I wrote all this up within Ch 5.6.7.8. Also non-square matrix stuff. This pullbacks.doc is a version that went before pullbacks v1.doc which has same stuff. Preliminary item: Suppose we have the scenario presented below Rn n ≤ m Rm I am trying to make this match the pullback picture so x'-space will then be t-space, very good. Then in tensor doc notation we have Rab ≡ (∂x'a/∂xb) = Sba ≡ (∂xb/∂x'a) = = Now recall from Chapter 10 v2.doc that x = φ(t) (Dφ)ij ≡ (∂φi(t)/∂tj) = ≡ Rij i = 1,2...m j = 1,2..,n (10.6.3) Thus we see that with this association of spaces and variables, what we REALLY have is (Dφ)ij = = ≡ Sij i = 1,2...m j = 1,2..,n (10.6.3) so Dφ = S, not Dφ = R. Very good! Recall also from (2.1.6) that dxi = Sij dx'j and for a "vector" we would then have vi = Sij v'j or v = S v' If we put label r on the vector, we can write S v'r = vr 10.6 Pullbacks 1. Consider the following differential k-form defined on x'-space Rm αx' = Σ'I fI(x') dxI = Σ'I fI(x') λ'I Σ'I fI(x') λ^I = Σ'I fI(x') (λ'i ^ λ'i....^ λ'i) . In the Dirac notation of Section 2.11 this may be written <αx'| = Σ'I fI(x') ( <e'i | ^ <e'i |....^ <e'i | ) <αx'| ϵ Λk(Rm) (Rm)*k = Σ'I fI(x') AltI ( <e'i | <e'i |.... <e'i | ) The notation (Rm)*k is our familiar dual tensor product space V*k with V = Rm. 2. As a review, when a vector in |v'> in V is "transposed", it becomes the dual vector <v'| in V*. Here we choose represent an arbitrary vector in V as v' for compatibility with what will soon follow. Similarly, consider a vector in the tensor product space Vk , | v'1, v'2 .. v'k> = | v'1> |v'2>.... |v'k> ϵ Vk . When this is transposed, it becomes an element of the dual space V*k, <v'1, v'2 .. v'k| = < v'1| <v'2|.... <v'k| ϵ V*k . If an operator Q is defined on V, we can define Q on Vk in the following manner Q | v'1, v'2 .. v'k> ≡ Q| v'1> Q|v'2>.... Q|v'k> <v'1, v'2 .. v'k| Q = < v'1|Q <v'2|Q.... <v'k|Q // transpose If it happens that Qv'r = vr then we have Q| v'r> = | Qv'r> = | vr> and so Q | v'1, v'2 .. v'k> ≡ | Qv'1> |Qv'2>.... |Qv'k> = | v1> |v2>.... |vk> = | v1, v2 .. vk> <v'1, v'2 .. v'k| Q = < Qv'1| <Qv'2|.... <Qv'k| = < v1| <v2|.... <vk| = <v1, v2 .. vk| where the second line is the transpose of the first line. We could apply Q to <αx'| shown above to get <αx'|Q = Σ'I fI(x') AltI ( <Qe'i | <Qe'i |.... <Qe'i | ) Now let Q be the differential matrix S = (Dφ). Then <αx'|S = Σ'I fI(x') AltI ( <Se'i | <Se'i |.... <Se'i | ) *************************************************** 3. Now let R and S be the differentials of a general transformation x' = F(x) from x-space to x'-space in the sense of Chapter 2 Rn n ≤ m Rm where 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) so Sba ≡ (∂xb/∂x'a). a = 1,2...n b = 1,2...m R = n x m The rule for transformation of vectors is then v's = S vs or (vs')i = Sij (vs)j // for contravariant components Here s is a label, not a component index. In Chapter 2 and in Tensor we had in mind that x-space and x'-space had the same dimension n. Here we generalize that situation by having x'-space = Rm and x-space = Rn with m ≥ n. This means that Rab is a matrix which has m rows (first index a) and n columns (second index b). So: x'-space = Rn x-space = Rm matrix R with components Rab has m rows and n columns From (2.1.2) we can write Graphically equation ** can be vs ϵ Question: I need to take v's = R vs and invert it. What does that mean in non-square context? 4. In covariant notation, Rab ≡ (∂x'a/∂xb) = Sba ≡ (∂xb/∂x'a) = The chain rule says Σb=1m = δac or Σb=1m Rab Sbc = δac or RS = 1m Σc=1n = δac or Σc=1n Sac Rcv = δab or SR = 1n I need to see these equations graphically. Suppose Rab has m rows and n columns with m ≥ n. Then So: R is the right inverse of S relative to 1m R = (S-1)r S is the left inverse of R relative to 1m S = (R-1)l S is the right inverse of R relative to 1n S = (R-1)r R is the left inverse of S relative to 1n R = (S-1)l Go back then to R vs = v's Apply S = (R-1)l to both sides to get vs = Sv's = (R-1)l v's Rab ≡ (∂x'a/∂xb) = ∂bx'a Suppose RT is a linear operator which does RTvi = vi . Then applying the tensor space version of RT to | v1, v2 .. vk> gives RT | v1, v2 .. vk> ≡ RT| v1> RT|v2>.... RT|vk> = | RTv1> |RTv2>.... |RTvk> = | w1> |w2>.... |wk> The transpose of this equation (see ***) is then <v1, v2 .. vk| R = <RTv1| <RTv2|.... <RTvk| = < w1| <w2|.... <wk| Therefore, applying this idea to **, we find <αx| R = Σ'I fI(x) AltI [ <RTei | <RTei |.... <RTei | ] We pause to prove the following fact: RT x( ei) = Σj=1m Rij tej If we were to apply an operator (call it RT) to the first equation, we would get RT | v1, v2 .. vk> = RT| v1> RT|v2>.... RT|vk> Thus we may transpose <αx| above to get |αx> = Σ'I fI(x) ( |xei > ^ |xei >....^ |xei > ) |αx> ϵ Lk(Rm) (Rm)k = Σ'I fI(x) AltI (<ei | <xei |.... <xei | ) where |αx> ϵ Lk(Rm) (Rm)k . Suppose we apply to this vector an operator RT : RT |αx> = Σ'I fI(x) AltI ( | RTxei > |RTxei >... |RTxei> )