Extending the R matrix to be m x m REVIEWED

Extending the R matrix to be m x m PhL 3.24.16 Here I show how to extend a "tall R matrix" which is m x n to a square m x m matrix which is invertible and which then rescues all of tensor doc! [ But I don't really like this approach.] I think Bucks did this from time to time, but I did not find a place for this in wedge doc. It is a bit messy. I instead stuck with the non-square R and figured out the implications of non-squareness. // 5.17.16 In the scenario of the tall R matrix, we have these equations x'1 = F1(x1, x2, x3... xn) x = φ(t) x'2 = F2(x1, x2, x3... xn) n = small m = large ... x'm = Fm(x1, x2, x3... xn) . m equations (1.2) F : Rn (x-space) → Rm(x'-space) embedding idea if m > n. The idea is to add m-n extra dimensions to x-space, but to never use these dimensions. Anything used in x-space will lie only in the original Rn subspace of the artificial new Rm in x-space. To do this we write x'1 = F1(x1, x2, x3... xn, xn+1....xm) x = φ(t) x'2 = F2(x1, x2, x3... xn, xn+1....xm) n = small m = large ... x'm = Fm(x1, x2, x3... xn, xn+1....xm) . m equations (1.2) Suppose I write in the extra x-space variables as shown, BUT claim there is no actual dependence on any of these red variables. What then does the R matrix look like? Use DN so Rij = (∂x'i/∂xj) = as before for i = 1..m and j = 1..n. Rij = (∂x'i/∂xj) = 0 for for i = 1..m and j =n+1,,m If we do this simple idea, we end up with an m x m matrix which has the last m-n columns all zero! But this is a problem because then R has rank n and cannot be inverted to obtain S and again tensor doc is hobbled. So let's try a different idea. Write x'1 = F1(x1, x2, x3... xn, xn+1....xm) = F1(x1, x2, x3... xn) x'2 = F2(x1, x2, x3... xn, xn+1....xm) = F2(x1, x2, x3... xn) ... x'n = Fn(x1, x2, x3... xn, xn+1....xm) = Fn(x1, x2, x3... xn) x'n+1 = Fn+1(x1, x2, x3... xn, xn+1....xm) = Fn+1(x1, x2, x3... xn) + xn+1 x'n+2 = Fn+2(x1, x2, x3... xn, xn+1....xm) = Fn+2(x1, x2, x3... xn) + xn+2 ... x'm = Fm(x1, x2, x3... xn, xn+1....xm) = Fm(x1, x2, x3... xn) + xm In the last m-n equations I add arbitrary terms as shown. Since in any application we will always have these last variables being 0, such as xn+1 = 0, nothing is really changed. Now what happens? Rij = (∂x'i/∂xj) = as before for i = 1..m and j = 1..n. the original tall R matrix Rij = (∂x'i/∂xj) = 0 for for i = 1..n and j = n+1..m to the right of the first n rows = 0 = δi,j for for i = n+1..m and j = n+1..m lower right submatrix Here is what this new extended R' matrix looks like Now we have det(R) = det(Rtt) Rtall_top so the new R matrix will be invertible providing Rtt has full rank n. This is the simplest way I can think of to extend the R matrix to an m x m invertible matrix. Then S = R-1 at least exists, and you would think that all of tensor doc would be restored. The new extended x' = F(x) is invertible at any local point provided rank(Rtt) = full value n. Let's compute the g' metric tensors in DN. The first one has always been OK, g' = R g RT I played with this on scratch. You could assume that the upper left n x n of g was your original arbitrary g metric tensor, call it g0. Then you could assume the full g has block diagonal form with g0 as the upper left block, and 1 for the lower right block and zero outside. You can then compute g' in a straightforward manner and it is not too messy. Conclusion: I think this plan does allow you to use tensor doc in full, and everything is well defined. The argument for validity of the method is to say that for any point of interest in the new Rm x-space you always have xn+1 = xn+2 = ... = xm = 0, so the equations x' = F(x) are then exactly as they were before this enlargement was performed. We have just embedded things in a larger x-space. Now that I think this is all possible and well-defined and rescues tensor doc, probably it won't be needed in Section 10.6, but nice to know it is workable. Have I seen this done anywhere before? I thought I saw something like it in Buck, but maybe not. But I think I saw this idea used somewhere. Continue on 3.25.16 Given the above R matrix, what would the inverse matrix S look like? I conjecture this form Example each of the four regions of the right side. 1) top left: the problem is simply to find S1 such that R1S1 = 1. Assuming rank OK, can do. Then we have S1 = R1-1 and so S1 is then known. 2) top right : This region is 0 because the top right of S is 0 and the top right of R is 0, need both. You can loosely say top right = R1*0 + 0*1 = 0. 3) bottom right: Again, get R2*0 + 1*1 = 1. 4) bottom left: Here we need R2*S1 + 1*S2 = 0. But we know R2 and S1 so set S2 = - R2 S1 I think this all works barring a rank problem, so the solution is this S1 = R1-1 S2 = - R2 S1 = - R2 R1-1 It would seem then that you can say Rij ≡ (∂x'i/∂xj) = R Sij ≡ (∂xi/∂x'j) = S and then tensor doc is happy again. Let's try some Maple examples: Here we have a 3+1 case and "it works" in that we have one column of 0's on the right Here is a 3+2 case and it works as well matching my format for S: So I think my matrix work above is OK. What happens to various things in Section 10.6? We are now F: Rm → Rm so what new things are there? 1) there are m-n new unused axis-aligned basis vectors ui in x-space 2) under the new enlarged mapping, we have u'i= Rui i = 1 to n give the tangent space i = n+1 to m give m-n new u'i vectors I guess these extra ones will be orthogonal to the tangent space in order for us to have a basis (wrong). In all my section 10.6 docs to this point I always say compute these u'i "as needed". Well, here is how they get computed. I would say that u'1 = (R11,R21.....Rm1) u'2 = (R12,R22.....Rm2) .... u'n = (R1n,R2n.....Rmn) u'n+1 = (0,0,0... 1,0...0) u'n+2 = (0,0,0... 0,1...0) .... u'm = (0,0,0... 0,0...1) OK, but these last m-n basis vectors do not seem to be orthogonal to the tangent space. So maybe I don't define them that way. Maybe I just define u'i= Rui for i = 1,2...n and then I manually define the other m-n basis vectors "as needed" so they are orthogonal to the tangent space and then all of them form a true basis. How would a Section 10.6 rewrite look in this enlarged elbow-room framework? (1) The kinematics package would be the full tensor doc package, BUT the R and S matrices would be the extended ones shown above. Rij ≡ (∂x'i/∂xj) = the enlarged R matrix shown above Sij ≡ (∂xi/∂x'j) = enlarged S matrix shown above (2) I then write |u'i> = R |ui> i = 1,2,3...n only and where both basis vectors have m components, but |ui> = (0,0...1...0; 0..0) 1 in the ith position where the last m-n components are 0 for all of them i = 1,2..n. I can then have RT and S-1 and all those objects, so then we can write |ui> = S |u'i> = RT |u'i> and this then gives the desired <ui| = <u'i| R i = 1,2...n only and this R operator then "pulls back" the basis vectors from x'-space to x-space. Go back now to |u'i> = R |ui> i = 1,2,3...n only or u'i = R ui or (u'i)i = Σj=1m Rij (ui)j = Σj=1n Rij (ui)j This last sum involves only the R1 portion of the R matrix. Fine. Now look at the inverse equation uk = S u'k k = 1,2..n only or (uk)i = Σj=1m Sij (u'k)j = Σj=1n Sij (u'k)j // last columns of S are 0 ! This last sum involves only the S1 portion of the S matrix. Fine. How do I fit RT into my picture, where I claim that RT = S. Well, they are the same matrix, so nothing to do. I can write |u'i> = R |ui> |ui> = S |u'i> = RT |u'i> Then I can transpose these to get <ui | = <u'i | ST = <u'i | R I see more fancy notation needed. Consider u'i = R* u*i = R ui