Section 10_6 v3

Section 10.6 PhL 3.6.16 10.6 The pullback of a simple differential form Preliminaries In Chapter 2 we discussed the transformation x' = F(x) from x-space to x'-space using this picture, (2.1.1) The vector transformation and differential (the R-matrix) of the transformation was given by V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a ≡ (F)ab ≡ (DF)ab ≡ (DF)ab (2.1.2) dx'a = Rabdxb dx' = Rdx . (2.1.12) Here V'a = RabVb shows the transformation of a contravariant vector under x' = F(x). In matrix notation one would write V' = RV. Notation: Above we have defined F and DF as alternate names for matrix R because authors like Spivak use this notation. In Tensor (E.4.4) we show that this is in fact a "reverse dyadic notation". Often (DF)ab is written unbolded as just (DF)ab so then R = (DF) with the idea that a matrix like R is normally not bolded. (10.6.1) The differential of the inverse mapping x = F-1(x') is called Sab ≡ (∂xa/∂x'b). In either up-tilt or down-tilt matrix notation one has RS = SR = 1 which follows from the chain rule, see (2.1.3). Axis-Aligned Vectors and Tangent Base Vectors It is shown in Tensor Section 3.2 that e'n are axis-aligned basis vectors [ (e'n)i = δni] in x'-space which inverse map to the "tangent base vectors" en in x-space according to en= Se'n and e'n = Ren . One writes en = ∂x/∂x'n so en is tangent to the "coordinate line" of a constant value x'n drawn in x-space, see Tensor (3.2.8), (1.13) and (3.4.3). Then in Tensor Section 3.5 the inverse transformation is discussed, x = F-1(x'), and for that transformation the un are axis-aligned basis vectors [ (un)i = δni] in x-space which map into the "inverse tangent base vectors" u'n in x'-space according to u'n= Run and un= Su'n as in Tensor (3.5.3) In this case the inverse tangent base vector u'n is tangent to the inverse coordinate line in x'-space which correspond to a constant value of xn in x-space. To summarize: x' = F(x) Rab ≡ (∂x'a/∂xb) = ∂bx'a Sab ≡ (∂xa/∂x'b) = ∂'bxa e'a with (e'a)b = δab axis-aligned basis vectors in x'-space (natural basis) ea ea= Se'a tangent base vector in x-space ua with (ua)b = δab axis-aligned basis vectors in x-space (natural basis) u'a u'a= Rua tangent base vector in x'-space (10.6.2) The properties of the vectors ea and e'a are summarized in Tensor (7.18.1), while those of ua and u'a are summarized in Tensor (7.18.3). For example, (ea)b = Rab and (u'a)b = Rba = Sab. It is shown in Tensor (7.19.10) that the u-basis (e'-basis) is the natural default basis for x-space (x'-space). To be compatible with Sjamaar we now rename our two spaces and our transformation to obtain a new picture, Picture F, and a new set of basis vectors x-space → t-space x'-space → x-space F → φ x' = F(x) → x = φ(t) unprimed vector → vector with t subscript primed vector → vector with x subscript (10.6.3) For this picture, x = φ(t) Rab ≡ (∂xa/∂tb) = ∂(t)bxa Sab ≡ (∂ta/∂xb) = ∂(x)bta (xea) with (xea)b = δab axis-aligned basis vectors in x-space (natural basis) (tea) (tea) = S (xea) tangent base vector in t-space (tua) with (tua)b = δab axis-aligned basis vectors in t-space (natural basis) (xua) (xua) = R(tua) tangent base vector in x-space (10.6.4) For example, xua is a tangent base vector in the new x-space. We put the x subscript to the left of the vector name to keep it out of the way of the basis vector label a. The basis vectors xua for a = 1,2..n will be seen below to span the tangent space TxM at a point x on a manifold M. Thus we have a clear connection between the notion of tangent base vectors and the tangent space TxM. Finally, once again to be compatible with Sjamaar, we reflect the above Picture F in order to get t-space on the left and x-space on the right (a purely cosmetic change), (10.6.5) and here is a more specific drawing of Picture F' for our application, (10.6.6) In Chapter 2 (and in the underlying Tensor document) the two spaces had the same dimension, but now t-space = Rn and x-space = Rm and we have in mind that m ≥ n. The R matrix is then in general no longer square, but is in fact an m x n matrix with m rows (first index a) and n columns (second index b). R is a "tall" matrix when m > n. We have chosen the dimension names m and n to be consistent with Spivak and Sjamaar. Basis vectors in the two spaces From (10.6.4) we select as a basis for t-space the set of of n axis-aligned basis vectors tui, {tui } i = 1,2...n basis for t-space (tui )j = δij components of these basis vectors in t-space (10.6.7) These map into set of n tangent base vectors xui in x-space as shown in (10.6.4), (xui) = R(tui) [ next line is correct by "testing a claim" ] or (xui)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.8) Since there are m basis vectors in x-space, we define the rest of the xui arbitrarily such that the m basis vectors {xui} in Rm are linearly independent, so xui = as needed i = n+1, n+2 .....m . (10.6.9) Note that equation (xui) = R(tui) or (xui)j = Σa=1n Rja(tui)a is a "component transformation". "Raising/lower components and maintaining tilts" as below (2.9.2) we can write a version of (10.6.8) for the dual basis vectors xui and tui, (xui) = R(tui) or (xui)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.10) Since (xui)j = Rji and (xui)j = Rji, we conclude from (10.6.8) and (10.6.10) that Fact: The first n x-space basis vectors xui for i = 1 to n are the columns of R** . The first n x-space basis vectors xui for i = 1 to n are the columns of R** . (10.6.11) That is to say, in (xui)j = Rji if we fix i and examine j = 1,2...m, we describe the column i of R**. If the columns of R** [or R**] are denoted ci [ or ci ] then one can write, R** = [ c1, c2.....cn ] = [xu1, xu2 ....xun ] ci = xui R** = [ c1, c2.....cn ] = [xu1, xu2 ....xun ] ci = xui . (10.6.12) Recall that R is a tall matrix with m rows and n columns, so each ci [ or ci ] has m components. Metric Tensor Comment: We have been quiet about the metric tensors tg and xg in the two spaces. Since we have used covariant notation, this entire section should be valid for any choice of metric tensors. For example (tg)ij = (tui) (tuj) i,j = 1,2..n // (2.4.2) translated from Pic A to Pic F (xg)ij = Σrs=1nRirRjs(tg)rs i,j = 1,2..m // (2.2.3) translated from Pic A to Pic F (10.6.13) One normally selects tg in t-space (often Cartesian meaning (tg)ij = δi,j) and then the last line above gives xg in x-space. The last line just states that the metric tensor transforms as a rank-2 tensor. The object (tg)ij is n x n whereas (xg)ij is m x m since the R matrix has m rows and n columns. ok to here 3.17.16 The pullback operator Now we are going to "pull back" all m of the xui basis vectors in x-space to vectors in t-space which vectors will turn out to be the rows of the R matrix. Here we are not inverse-mapping the first n xui back to their corresponding tui in t-space, we are doing something different. We first define, φ*T(xui) ≡ Σj=1n Rij (tuj) i = 1..m (10.6.14) where φ*T is a pullback operator which maps from x-space to t-space, φ*T : Rm→ Rn. The reason for writing this as the transpose of operator φ* will become clear below. The sum on the right of (10.6.9) is a linear combination of t-space basis vectors, it is not a sum of vector components as in (10.6.4), so φ*T(xui) is a vector in t-space, not in x-space. equation numbers only good to this point! Now take the contravariant s component of both sides of (10.6.9), [φ*T(xui)]s = Σj=1n Rij (tuj)s i = 1,2...m s = 1,2...n = Σj=1n Rij δjs // (10.6.3) = Ris . (10.6.10) Reversing the tilts as done earlier, we can rewrite (10.6.9) and (10.6.10) as φ*T(xui) ≡ Σj=1n Rij (tuj) i = 1..m [φ*T(xui)]s = Ris . (10.6.11) Therefore we conclude from (10.6.10) and (10.6.11) that Fact: The m t-space vectors φ*T(xui) for i =1 to m are the rows of the R** matrix. The m t-space vectors φ*T(xui) for i =1 to m are the rows of the R** matrix. (10.6.12) If the rows of R** [ or R**] are denoted ri [ or ri ] then we can write R** = = and ri = φ*T(xui) . R** = = and ri = φ*T(xui) . (10.6.13) Recall that R is a tall matrix with m rows and n columns, so each ri [ or ri ] has n components. Thus, the pullback of an x-space basis vector xui is nothing more than the corresponding row of the R matrix for the transformation x = φ(t) . Every basis vector xui for i = 1 to m has such a pullback. This drawing represents the three mappings so far described, (10.6.14) Here we show the mappings for the basis vector tu1. In Dirac notation one can write (10.6.9) and (10.6.11) as ket equations, φ*T | xui> = Σj=1n Rij | tuj> i = 1,2..m φ*T | xui> = Σj=1n Rij | tuj> . (10.6.15) Since the transpose of A|b> is <b|AT as in (2.11.d.7), and since the R matrix is real, the corresponding bra equations are <xui | φ* = Σj=1n Rij <tuj | i = 1,2..m <xui | φ* = Σj=1n Rij <tuj | . (10.6.16) The bras in these equations are dual space vectors (rank-1 linear functionals) as discussed in Section 2.11. The ket <xui | is an element of dual space (Rm)* while <tuj | inhabits (Rn)*. Note that Fig (10.6.14) displays only the spaces Rn and Rm and not the corresponding dual spaces. In terms of the (2.11.c.2) basis functionals xλi = <xui| of the dual space (Rm)*, this second line of (10.6.16) may be written, φ*(xλi) = Σj=1n Rij (tλj) = Σj=1n (Dφ)ij (tλj) i = 1,2..m (10.6.17a) or in cosmetic notation of Section 10.1 above, φ*(dxi) = Σj=1n Rij (dtj) = Σj=1n (Dφ)ij (dtj) i = 1,2..m . (10.6.17b) Since xλi = dxi is a very simple example of a differential form, we see in (10.6.17) our first example of "the pull back of a differential form". What is being "pulled back" is the dual vector <xei| = xλi ϵ (Rm)* . The pulled back vector is φ*(xλi) ϵ (Rn)*. The mapping is φ*: (Rm)*→ (Rn)*. The Tangent Space Basis Vectors Assume that as t ranges over some portion of t-space, the mapping x = φ(t) describes a "smooth surface" M embedded in x-space, hopefully a manifold. If we start at some t and move to t + dt in t-space, we move from some point x on M to some nearby point x + dx on M. This dx is tangent to the surface M and thus lies in the tangent space TxM of M at point x, as discussed above in Section 10.2. Applying R to each of the n axis-aligned differentials dti = dti(tui) in t-space (no i sum), we thereby generate a set of n differential vectors dxi = Rdti which are in effect a set of short basis vectors which span the tangent space TxM. Since dxi = dxi (xui), we may take the basis vectors {xei, i=1,2..n} as spanning TxM, in agreement with our arrangement of things in Section 6.2. The vectors {xei, i=n+1,n+2..m} are then all orthogonal to the "surface" M. We know from the fact xui xuj = δij that the set {xui, i=1,2..n} also form a basis for the tangent space TxM. This conclusion can be reached as well by raising all i indices in the previous paragraph. In this case, the set {xui, i=n+1,n+2..m} are then all orthogonal to the "surface" M. According to (10.6.7) we may conclude that Fact: The first n basis vectors xui, which are the columns of R** , span the tangent space TxM. The first n basis vectors xui, which are the columns of R** , span the tangent space TxM. (10.6.18) The dual vectors (functionals) in the sense of xλi = <xei| then span the cotangent space of TxM. ok to here except need to update the picture 10.7 The pullback of a general differential form Recall from (10.6.16) the action of operator φ* on the bra <xui |, <xui | φ* = Σj=1n Rij <tuj | i = 1,2..m . (10.6.16) (10.7.1) An operator P which acts on a vector space V, has a natural extension to being an operator on the tensor product space Vn , P | v1,v2....vk> = P [ | v1> |v2>....|vk> ] = P| v1> P|v2>.... P|vk> . (10.7.2) Similarly, an operator Q which acts on a dual vector space V*, has a natural extension to being an operator on V*n , < v1,v2....vk| Q = [ < v1| <v2|....<vk| ] Q = < v1|Q <v2|Q....<vk|Q . (10.7.3) Setting Q = φ* we then define, < xui, xui....xui| φ* = < xui|φ* <xui|φ*.... <xui|φ* = [Σj=1n Rij <tuj | ] [Σj=1m Rij <tuj | ] ... [Σj=1m Rij <tuj | ] = Σjj...j=1n Rij Rij ... Rij ( <tuj | <tuj | ... <tuj| ) = Σjj...j=1n Rij Rij ... Rij | tui, tui....tui > . (10.7.4) In multiindex notation this says (note that ΣJ is a full symmetric sum), < xuI | φ* = ΣJ RIJ <tuJ | . (10.7.5) A nearly identical result can be obtained for wedge space Λk. Apply AltI to both sides of (10.7.5) to get AltI [ < xuI | φ* ] = AltI [ ΣJ RIJ <tuJ | ] . (10.7.6) According to (8.1.2) the left side is just AltI [ < xuI | φ* ] = AltI [ < xuI | ] φ* = < xu^I | φ* where <xu^I| = <xuj | ^ <xuj | ... ^ <xuj| . (10.7.7) The right side is a little tricky: AltI [ ΣJ RIJ <tuJ | ] = ΣJ [AltI(RIJ)] <tuJ | = ΣJ [(1/k!) ΣP(-1)P RP(I)J ] <tuJ | // (A.2.1) def of AltI = (1/k!) ΣP(-1)P ( ΣJ RP(I)J ] <tuJ | ) // reorder = [(1/k!) ΣP(-1)P ( ΣJ RP(I)P(J) ] <tuP(J) | ) // (A.1.20) that ΣJ fJ = ΣJ fP(J) = [(1/k!) ΣP(-1)P ΣJ RIJ ] <tuP(J) | // (A.8.31) that RP(I)P(J) = RIJ = ΣJ RIJ [(1/k!) ΣP(-1)P <tuP(J) | ] // reorder = ΣJ RIJ AltJ <tuJ | // (A.2.1) def of AltJ = ΣJ RIJ <tu^J | where <tu^I| = <tuj | ^ <tuj | ... ^ <tuj| . (10.7.8) Thus we have shown how φ* acts on dual basis vectors of the space Λk(Rm), < xu^I | φ* = ΣJ RIJ <tu^J | . (10.7.9) This is identical in form to (10.7.5) but the basis vectors are wedge products instead of tensor products. A general element of Λk(Rm) at a point x on our surface M can be written from (10.2.1), αx = Σ'I fI(x) xλ^I or <αx | = Σ'I fI(x) < xu^I | (10.7.10) Why do I use < xu^I | instead of < xe^I | at this point?? In Chapter 2 I never said whether you should use the natural basis or the tangent base vectors basis, I just used there a generic basis e. But here I am making a decision without justification. What does Sjamaar do? He writes α = ΣIfIdyI on page 36 and this goes with the famous page 37 picture. But he never says that dyI = <???I | But in Chapter 7 he writes on page 87 that α = ΣI fI dxI. and on page 83 he has written dxi = <ei| and he shows them on p 83A as axis-aligned basis vectors!!!! So what I am doing above is WRONG, since in Picture F it is the xea which are the axis-aligned vectors!!!!! So I will now save this file off into a new v4 and edit from this point onwards. This was a suspicion I have had for a few days now. Then <αx | φ* = Σ'I fI(x) [ < xu^I | φ* ] = Σ'I fI(x) [ΣJ RIJ <tu^J | ] or φ*(αx) = Σ'I fI(x) ΣJ RIJ tλ^J = Σ'I fI(φ(t)) ΣJ RIJ tλ^J (10.7.11) which is then the pullback of an arbitrary differential k-form. We see from our operations above that: Fact: The pullback of a k-form in x-space is a k-form in t-space, where k ≤ n ≤ m. (10.7.12) After all, xλ^I is the wedge product of k dual basis vectors, and so is tλ^J . There are now two paths leading off from this waypoint. First path Write (10.7.11) as φ*(αx) = ΣJ [ Σ'I fI(φ(t)) RIJ ] tλ^J = ΣJ GJ(t) tλ^J GJ(t) ≡ Σ'I fI(φ(t)) RIJ = Σ'J gJ(t) tλ^J gJ(t) ≡ k! AltJ[GJ(t)] , (10.7.13) where the last line comes from (10.1.1) through (10.1.3). We have stated the result both in terms of symmetric ΣJ and ordered Σ'J summations. The coefficient functions gJ(t) can be re-expressed as, gJ(t) ≡ k!AltJ[GJ(t)] = k!AltJ[ Σ'I fI(φ(t)) RIJ] = k! Σ'I fI(φ(t)) AltJ[RIJ] // Alt is linear = k! Σ'I fI(φ(t)) [ (1/k!) det(RIJ) ] // (A.8.30) = Σ'I fI(φ(t)) det(RIJ) . (10.7.14) The pullback of αx along φ is then φ*(αx) = Σ'J Σ'I fI(φ(t)) det(RIJ(t)) tλ^J = Σ'J Σ'I fI(φ(t)) det[(Dφ)IJ(t)] tλ^J ≡ βt (10.7.15) Being a linear combination of tλ^J , φ*(αx) is seen to be an element of Λk(Rn) which is associated with t-space. That is to say, φ*(αx) is a differential form at a point t in t-space, so we give it an arbitrary name βt. We pause once more to rewrite our last several results in cosmetic notation: αx = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I (10.7.10)C φ*(αx) = Σ'I fI(φ(t)) ΣJ RIJ dt^J R = (Dφ) (10.7.11)C φ*(αx) = ΣJ [ Σ'I fI(φ(t)) RIJ ] dt^J = ΣJ GJ(t) dt^J GJ(t) ≡ Σ'I fI(φ(t)) RIJ = Σ'J gJ(t) dt^J gJ(t) ≡ k! AltJ[GJ(t)] , (10.7.13)C φ*(αx) = Σ'J Σ'I fI(φ(t)) det(RIJ(t)) dt^J = Σ'J Σ'I fI(φ(t)) det[(Dφ)IJ(t)] dt^J ≡ βt (10.7.15)C It is probably useful to write out this last equation in full notation : φ*(αx) = Σ1≤j<j<....<j≤n Σ1≤i<i<....<i≤m fii...i(φ(t)) * det [RIJ] ( dti ^ dti .....^ dti ) (10.7.16) where RIJ is this kxk matrix, Rij Rij ... Rij RIJ = Rij Rij ... Rij ..... Rij Rij ... Rij (10.7.17) The object det(RIJ) is a k x k minor of the full m x n matrix R, so k ≤ n ≤ m in our application. And it is probably time for some kind of picture where we add detail to Fig (10.6.1). Consider: (10.7.18) We have added an n-dimensional open domain region U in t-space which maps via x = φ(t) into an open region V which lies on the manifold M (a "surface"), which is of dimension-n. Our point of interest t lies in U, and x lies in V. The picture is a bit symbolic since it shows t-space = Rn and x-space = Rm , but the differential forms αx and βt are really elements of the dual spaces (Rm)* and (Rn)* . A more practical picture for n = 2 and k = 2: (10.7.19) Here the open region U is a unit square [0,1]2 which maps into a patch on a torus. That is, if m = 3 the object on the right is a torus in R3, but we can imagine it to be a torus embedded in Rm for any m ≥ 3. The space of vectors defined on U R2 is a 2-dimensional dual space (R*)2(U). On this space we can define either 1-forms or 2-forms. The above picture suggests a 2-form since the region U is an area, and since we will later associate dt1 ^ dt2 with the calculus differential dt1dt2 which represents an area (we are not there yet). The picture shows the "forward map" x = φ(t), suggesting that forward means left to right in the picture. Then αx is "pulled back" right to left from dual x-space to dual t-space where it becomes βt. One could imagine a set of 16x6 = 96 mappings like the one shown above which would "cover the torus", using one little patch for each mapping. One would then have an atlas of 96 square maps like that on the left which would serve to cover the surface of Planet Toroid. This is the basic idea of a manifold. In the torus example, one could do the job with only 2 maps. The aspect ratio of the 2-cube on the left is not significant. One could change it to be an arbitrary rectangle in t-space and select a φ to make it map to the same small image patch. Or one could construct a mapping φ which maps the unit 2-cube [0,1]2 to the entire left half of the torus. See Sjamaar. The black arrows on the left are the t-space basis vectors tui (only tu2 is labeled). As shown in (10.6.6), these map according to (xui) = R(tui) into basis vectors which are tangent to M, and these vectors then span the tangent space TxM at point x on M. It is clear that the two xei will vary if the point x on M is varied. As another example consider this situation with n = 1 and k = 1, (10.7.20) Now the domain in t-space is a U = 1-cube [0,1] which maps to a (generally non-planar) red curve which is embedded in Rm . Here αx and βt are 1-forms. The red curve V lies on the manifold M as shown, just as the patch of the previous example lay on the torus. There is only one basis vector tu in t-space (not shown) and it maps the black arrow on the right which is xe and is of course tangent to the curve at x. Second path We return to our waypoint (10.7.11) showing the pullback of a general k-form, φ*(αx) = Σ'I fI(x) ΣJ RIJ tλ^J x = φ(t) (10.7.11) or <αx | φ* = Σ'I fI(x) ΣJ RIJ <tu^J | We make this bra into a tensor function by closing it with a ket | vn,vn...vn> of t-space vectors, | vn,vn...vn> = | vn> | vn> ..... | vn> ϵ Vk V = Rn (10.7.21) to get <αx | φ*| vn,vn...vn> = Σ'I fI(x) ΣJ RIJ <tu^J | vn,vn...vn>. (10.7.22) In more conventional notation this reads [φ*(αx)] (vn,vn...vn) = Σ'I fI(x) ΣJ RIJ tλ^J(vn,vn...vn) . (10.7.23) For the time being, we shall ignore the Σ'I fI(x) part and just consider ΣJ RIJ tλ^J(vn,vn...vn). We now process our expression through a battery of steps to get the desired result: ΣJ RIJ tλ^J(vn,vn...vn) = ΣJ RIJ (tλj ^ tλj ....^ tλj)(vn,vn...vn) = ΣJ RIJ AltJ[(tλj tλj .... tλj)(vn,vn...vn)] // (8.3.8) = ΣJ RIJ AltJ[(tλj(vn)tλj(vn) ...tλj(vn)] // (6.1.3) = ΣJ RIJ AltN[(tλj(vn)tλj(vn) ...tλj(vn)] // (A.8.29) = AltN [ ΣJ RIJ (tλj(vn)tλj(vn) ...tλj(vn) ] // (A.8.10) = AltN [ ΣJ RIJ (vn)j(vn)j ... (vn)j ] // (2.11.c.5) = AltN [ Σjj...j=1m RijRij ....Rij (vn)j(vn)j ... (vn)j ] = AltN { [ΣjRij(vn)j] [ΣjRij(vn)j] ... [ΣjRij(vn)j] } = AltN { (Rvn)i(Rvn)i... (Rvn)i } // (Rv)i = ΣjRijvj = AltN {xλi(Rvn)xλi(Rvn)... xλi(Rvn) } // (2.11.c.5) = AltI {xλi(Rvn) xλi(Rvn)... xλi(Rvn) } // (A.8.29) = AltI {(xλi xλi .... xλj) (Rvn, Rvn ...Rvn)} // (6.1.3) = ( xλi ^ xλi... ^ xλi )(Rvn,Rvn...Rvn) . // (8.3.8) = ( xλ^I )(Rvn,Rvn...Rvn) . To summarize: ΣJ RIJ tλ^J(vn,vn...vn) = ( xλ^I )(Rvn,Rvn...Rvn) We then insert this into (10.7.23) to get [φ*(αx)] (vn,vn...vn) = Σ'I fI(x) [ΣJ RIJ tλ^J(vn,vn...vn)] = Σ'I fI(x) [ ( tλ^I )(Rvn,Rvn...Rvn)] = [Σ'I fI(x) xλ^I ](Rvn,Rvn...Rvn) = αx(Rvn,Rvn...Rvn) We end up then with the "tensor function definition" of the pullback of a differential form αx : [φ*(αx)] (vn,vn...vn) = αx(Rvn,Rvn...Rvn) which in Dirac notation is <αx | φ*| vn,vn...vn> = <αx | Rvn,Rvn...Rvn> . This certainly suggests that φ*| vn,vn...vn> = | Rvn,Rvn...Rvn> I will continue here tomorrow! if this is somehow obvious, the proof becomes a LOT simpler!!!