Basis vectors _ Component vs Vector Transformations REVIEWED

Basis Vectors: Component versus Vector Transformations of Vectors PhL 3.4.16 (5.16.16). Several different things are happening here. Some is bad, but the rest is incorporated into wedge doc. I think this is the first time I got the tensor-function definition of the pullback, hence the next comment. But other stuff here is just plain wrong. I think some of the last stuff here may have led to some of the higher numbered theorems in Appendix A. Not doing a detailed review. This document finally "has it right", do not lose!!! Part A Let ei be some basis vectors in Rm. We can talk about x ≡ Rei 1. On the one hand, we can write a component transformation like so: xa = Σb=1m Rab (ei)b . I call the above a "component transformation" of a vector. 2. On the other hand, we expect that Rei must be a linear combination of the ej so we write Rei = Σb=1m aib eb = lin comb of basis vectors and in this equation we show no components of any basis vectors. I call this a "vector transformation" of a vector. My names are not very good, but they are what I have used before. [ the above sum type distinction is now made somewhere in Chapter 10 where I give examples of the two types of transformations involving vectors ] [ The last equation above cannot be true for m x n R because Rei = e'i has m components in Rm while eb has only n components in Rn. For m x m I guess it is true, but not very interesting because e'i lies in x'-space while ei lies in x-space. ] Theorem: aij = Rji Proof: Apply ej to both sides of the above ej Rei = Σb=1m aib ej eb = aij . Now use the "component" breakdown of the left side ej Rei = Σa=1m (ej)a [Rei]a = Σa=1m Σb=1m(ej)a Rab (ei)b . So now we have shown that aij = Σa=1m Σb=1m(ej)a Rab (ei)b Now for ANY "standard e-basis" basis vectors as I discuss in Chapter 2 of wedge doc, we can write (ej)a = δja and (ei)b = δib In this basis we find then that aij = Rji QED aib = Rbi Therefore, our vector expansion may be written Rei = Σb=1m Rbi eb = lin comb of basis vectors and notice the non-abutment of the b summation index. [ 5.16.16: this says e'i = Rbi eb , see comment above how meaningless for non-square R ] Part B Let's try to restate the above in the case of two spaces Rn and Rm. Rn = t-space with n basis vectors tei Rm = s-space with m basis vectors xei , the first n of which span TxM R = tall matrix with m rows and n columns m > n 1. Assume the usual vector rule V' = RV which becomes xv = R tv so we have (xei) ≡ R (tei) tei has n components, a "short vector". (xei) ≡ R (tei) tei has n components, a "short vector". (1) (xv) = R(tv) for any vector The first line is a "component transformation" and we would write (xei)a = Σb=1n Rab (tei)b // conforms a = 1..m We might assume (tei)b = δib which would then say (xei)a = Rai , but I don't think this is important. 2. On the other hand, we expect that R(tei) is a linear combination of the n tei vectors : [ wrong! R(tei) = (xei) is a vector having m components, but the tei have n components. ] R (tei) = Σb=1m aib (teb) Following the exact same argument as above, we could get aib = Rbi so this becomes R (tei) = Σb=1m Rbi (teb) = lin comb of basis vectors in t-space' and this is then the "vector" transformation within t-space. I think I could then claim that R (tei) = Σb=1m Rbi (teb) = lin comb of basis vectors in t-space (2) and then I write (1) and (2) on one line as (xei) ≡ R (tei) = Σj=1m Rji (tej) (3) No component indices are showing in the above equation. Part C Here is a result I want to "derive" or at least somehow explain. It is the k = 1 result from Chapter 10.doc φ*(xλi) ≡ Σj=1m (Dφ)ij tλj = Σj=1m Rij tλj i = 1,2...n . (10.6.2) [ the above now appears as (10.7.20) item 5 // 5.16.16 ] In Dirac notation <φ*(xei)| = Σj=1m Rij <tej| I am pretty sure that the transpose of the above with real Rij would be this |φ*(xei)> = Σj=1m Rij |tej> (4) Notice that this is different from (3) above: equation (3) has Rji while (4) has Rij . So we have |φ*(xei)> = Σj=1m Rij |tej> <φ*(xei)| = Σj=1m Rij <tej| φ*(xλj) = Σj=1m Rij tλj and this last is the desired result with matrix mult abutment. What happens if this is closed onto a t-space vector v? [φ*(xλj)](v) = Σj=1m Rij tλj(v) = Σj=1m Rij tvj Suddenly we have a component transformation on the right side, so we can write the above as Σj=1m Rij tvj = [Rv]i = xλi(Rv) and we then end up with the desired result [φ*(xλj)](v) = xλi(Rv) [ a special case of (10.7.20) item 9 // 5.16.16 ] = (Rv)i = Rijvj = Rij tλj(v) [φ*(xλj)] = Rij tλj which is (10.7.20) # 5 So somewhere in here I have the correct threading. It is 1:30 PM on 3.4.16. Now back up and look again at |φ*(xei)> = Σj=1m Rij |tej> (4) Maybe write this as φ* |xei> = Σj=1m Rij |tej> this is the goal! φ*(xei) = Σj=1m Rij (tej) This equation then tells what the φ* operator does to an x-space basis vector. The result is in a different space, t-space! You cannot write this as φ* |xei> = R |tej> because from (3) we know that R |tej> = Σj=1m Rji |tej> = |xei> no components appear! ok The "tilt" on R is "the wrong way". How could we define a new matrix Q such that φ* |xei> = Q |tei> = Σj=1m Qji |tej> = Σj=1m Rij |tej> lincomb of vectors φ* (xei) = Q (tei) = Σj=1m Qji |tej> = Σj=1m Rij |tej> lincomb of vectors Well, it seems that we would need Qji = Rij Now in tensor doc I claim that Rij = (RT)ji = Qji vertical line swap Then we can replace Q by RT to get φ* |xei> = RT |tei> = | RTtei> This is the most concise way I can write things. The transformation is a vector one, not a component one. So one writes RT |xei> = Σj=1m (RT)ji |tej> = Σj=1m Rij |tej> = φ* |xei> The key here is to NOT confuse the two kinds of transformations. Part D In the general case then we have THREADING 1: |αx> = Σ'I fI(x) ( |xei > |xei >....^ |xei > ) φ*|αx> = Σ'I fI(x) ( φ* |xei > φ* |xei >....^ φ* |xei > ) (a) = Σ'I fI(x) [ Σj=1m Rij |tej >] [Σj=1m Rij |tej >] .... [] ) (b) = Σ'I fI(x) ΣJ Rij Rij .... ( |tej > ^ |tej >....^ |tej > ) (c) = ΣJ [Σ'I fI(x) RIJ ] ( |tej > ^ |tej >....^ |tej > ) (d) = ΣJ GJ(t) ( |tej > ^ |tej >....^ |tej > ) GJ(t) ≡ Σ'I fI(φ(t)) RIJ = Σ'J gJ(t) ( |tej > ^ |tej >....^ |tej > ) gJ(t) ≡ k! AltJ[GJ(t)] But as shown in Chapter 10.doc gJ(t) = Σ'I fI(φ(t)) det(RIJ) So we end up with φ*|αx> = Σ'J [Σ'I fI(φ(t)) det(RIJ) ] ( |tej > ^ |tej >....^ |tej > ) If we transpose this equation, we get an equation which then translates to φ*(αx) = Σ'J [Σ'I fI(φ(t)) det(RIJ) ] λ^J = Sjamaar page 42 A THREADING 2: |αx> = Σ'I fI(x) ( |xei > |xei >....^ |xei > ) = Σ'I fI(x) AltI( |xei > |xei >.... |xei > ) φ*|αx> = Σ'I fI(x) AltI( φ*|xei > φ*|xei >.... φ*|xei > ) (a') = (a) ok = Σ'I fI(x) AltI( [ Σj=1m Rij |tej >] [Σj=1m Rij |tej >] .... [] ) (b') = just sub = Σ'I fI(x) ΣJ AltI( Rij Rij .... |tej > |tej >.... |tej > ) (c') STOP. This (c') does not seem the same as (c) in the first threading, which was this: = Σ'I fI(x) ΣJ Rij Rij .... ( |tej > ^ |tej >....^ |tej > ) (c) = Σ'I fI(x) ΣJ Rij Rij .... AltJ ( |tej > |tej >.... |tej >) I will put this mystery on hold for the moment. ******* Resume this in Appendix A below.// I did so and verified that in fact (c) = (c') due to the symmetric sum ΣJ! So now CONTINUE this threading 2 φ*|αx> = Σ'I fI(x) ΣJ AltI( Rij Rij .... |tej > |tej >.... |tej > ) (c') = (c) ok = Σ'I fI(x) ΣJ AltI( RIJ |tej > |tej >.... |tej > ) = (1/k!) Σ'I fI(x) ΣJ ΣP(-1)S(P) RP(I)J | eJ> = (1/k!) Σ'I fI(x) ΣP(-1)S(P) ΣJ RP(I)J | eJ> Now use (A.1.20) that ΣJ fJ = ΣJ fQ(J) for any permutation Q. Select Q = P so then ΣJ RP(I)J | eJ> = ΣJ RP(I)P(J) | eP(J)> = ΣJ RIJ | eP(J)> Now since RIJ has a simple factored form, it only gets reordered by writing RP(I)P(J) = RIJ, producing the final result above. Making the replacement in φ*|αx> we get φ*|αx> = (1/k!) Σ'I fI(x) ΣP(-1)S(P) ΣJ RIJ | eP(J)> = Σ'I fI(x) ΣJ RIJ (1/k!)ΣP(-1)S(P)| eP(J)> = Σ'I fI(x) ΣJ RIJ AltJ| eJ> = Σ'I fI(x) ΣJ RIJ | e^J> (d') = (d) above. We then just finish this off as in Thread 1. Meanwhile, let's continue on the original issue. So back up to: |φ*αx> = Σ'I fI(x) AltI( φ*|xei > φ*|xei >.... φ*|xei > ) = Σ'I fI(x) AltI( [ Σj=1m Rij |tej >] [Σj=1m Rij |tej >] .... [] ) Transpose to get <φ*αx| = Σ'I fI(x) AltI( [ Σj=1m Rij <tej |] [Σj=1m Rij <tej |] .... [] ) Now close with | v1,v2...vk> to get (vectors are in t-space) <φ*αx| v1,v2...vk> = Σ'I fI(x) AltI( [ Σj=1m Rijvj ] [Σj=1m Rij vj ] .... [] ) = Σ'I fI(x) AltI( [Rv1]i[Rv2]i ... [Rvk]i ) = Σ'I fI(x) AltI( tλi[Rv1]λi[Rv2] ... λi[Rvk] ) = Σ'I fI(x) AltI( λiλi ... λi )(Rv1,Rv2...Rvk) = Σ'I fI(x) ( tλi ^ λi ^ ... ^λi )(Rv1,Rv2...Rvk) = Σ'I fI(φ(t))tλI(Rv1,Rv2...Rvk) = [ Σ'I fI(φ(t))tλI] (Rv1,Rv2...Rvk) = αφ(t)(Rv1,Rv2...Rvk) and thus we get <φ*αx| v1,v2...vk> = [φ*αx] (v1,v2...vk) = αφ(t)(Rv1,Rv2...Rvk) which is the correct result, and the vi are all in t-space. Now back up again to this point |αx> = Σ'I fI(x) AltI( |xei > |xei >.... |xei > ) φ*|αx> = Σ'I fI(x) AltI( φ*|xei > φ*|xei >.... φ*|xei > ) = Σ'I fI(x) AltI( [ Σj=1m Rij |tej >] [Σj=1m Rij |tej >] .... [] ) I should be able to use RT in there. So define RT on the big space by RT|αx> = Σ'I fI(x) AltI( RT|xei > RT|xei >.... RT|xei > ) Now above in Part C I showed that φ*|xei > = RT |xei> = Σj=1m (RT)ji |tej> = Σj=1m Rij |tej> so it is then OK in this general case to write φ*|αx> = RT|αx> where RT is not a matrix, but is an operator defined on the tensor product space as shown above. Appendix A: Show that (c) = (c') above. In other words, show that = Σ'I fI(x) ΣJ AltI( Rij Rij .... |tej > |tej >.... |tej > ) = Σ'I fI(x) ΣJ Rij Rij .... AltJ ( |tej > |tej >.... |tej >) What is the simplest non-trivial case here? First, go to k = 2 and want to show that 1 = Σi<i fii(x) ΣJ AltI( Rij Rij |tej > |tej > ) 2 = Σi<i fii(x) ΣJ Rij Rij AltJ ( |tej > |tej >) Write out both Alt operations 1 = Σi<i fii(x) Σjj [ Rij Rij - Rij Rij ] |tej > |tej > /2 2 = Σi<i fii(x) Σjj Rij Rij [ |tej > |tej > - |tej > |tej > ]/2 Trying to show these are equal. I think this means want to show these are equal 1' = Σjj [ Rij Rij - Rij Rij ] |tej > |tej > 2' = Σjj Rij Rij [ |tej > |tej > - |tej > |tej > ] Write 2' as two terms 2' = Σjj Rij Rij |tej > |tej > - Σjj Rij Rij |tej > |tej > In the second term do j1 ↔ j2 dummies, 2' = Σjj Rij Rij |tej > |tej > - Σjj Rij Rij |tej > |tej > = Σjj [ Rij Rij - Rij Rij ] |tej > |tej > = Σjj [ Rij Rij - RijRij ] |tej > |tej > = 1' So I have verified this for k = 2. How do I generalize that verification? Now more generally, show that these are equal 1 = Σ'I fI(x) ΣJ AltI( Rij Rij .... |tej > |tej >.... |tej > ) 2 = Σ'I fI(x) ΣJ Rij Rij .... AltJ ( |tej > |tej >.... |tej >) Want to show that these are equal 1' = ΣJ AltI( Rij Rij .... |tej > |tej >.... |tej > ) 2' = ΣJ Rij Rij .... AltJ ( |tej > |tej >.... |tej >) Write in permutation notation 1' = (1/k!) ΣJ ΣP(-1)S(P)( Rij Rij .... ) (|tej > |tej >.... |tej > ) 2' = (1/k!) ΣJ Rij Rij .... ΣP(-1)S(P)( |tej > |tej >.... |tej >) Reorder sums and abbreviate and drop common factor (1/k!) , 1' = ΣP(-1)S(P) ΣJ RP(I)J |eJ> 2' = ΣP(-1)S(P) ΣJ RIJ |eP(J)> Now claim that RIJ = RP(I)P(J) since RIJ = Rij Rij .... . Line ** just reorders the factors rendering the same product. Thus we have 1' = ΣP(-1)S(P) ΣJ RP(I)J |eJ> 2' = ΣP(-1)S(P) ΣJ RP(I)P(J) |eP(J)> Now use (A.1.20) which says ΣJ fJ = ΣJ fQ(J) for any permutation Q. Select Q = P-1 so then ΣJ RP(I)P(J) |eP(J)> = ΣJ RP(I)PQ(J) |ePQ(J)> = ΣJ RP(I)J |eJ> Then we have 2' = ΣP(-1)S(P) ΣJ RP(I)P(J) |eP(J)> = ΣP(-1)S(P) ΣJ RP(I)J |eJ> = 1' QED