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Reciprocal Frame Vectors Reciprocal Frame Vectors Peeter Joot March 29, 2008 1 Approach without Geometric Algebra. Without employing geometric algebra, one can use the projection operation expressed as a dot product and calculate the a vector orthogonal to a set of other vectors, in the direction of a reference vector. Such a calculation also yields RN results in terms of determinants, and as a side effect produces equations for parallelogram area, parallelopiped volume and higher dimensional analogues as a side effect (without having to employ change of basis diagonalization arguments that don’t work well for higher di- mensional subspaces). 1.1 Orthogonal to one vector The simplest case is the vector perpendicular to another. In anything but R2 there are a whole set of such vectors, so to express this as a non-set result a reference vector is required. Calculation of the coordinate vector for this case follows directly from the dot product. Borrowing the GA term, we subtract the projection to calculate the rejection. Rejuˆ (v) = v − v · uˆ uˆ 1 = (vu2 − v · uu) u2 1 = v e u u − v u u e u2 ∑ i i j j j j i i 1 vi vj = ∑ ujei u2 ui uj 1 ui uj = (u e − u e ) 2 ∑ i j j i v v u i<j i j Thus we can write the rejection of v from uˆ as: 1 1 ui uj ui uj Rej (v) = (1) uˆ 2 ∑ v v e e u i<j i j i j Or introducing some shorthand: uv ui uj Dij = vi vj ue ui uj Dij = ei ej equation 1 can be expressed in a form that will be slightly more convient for larger sets of vectors: 1 ( ) = uv ue Rejuˆ v 2 ∑ Dij Dij (2) u i<j Note that although the GA axiom u2 = u · u has been used in equations 1 and 2 above and the derivation, that was not necessary to prove them. This can, for now, be thought of as a notational convenience, to avoid having to write u · u, or kuk2. This result can be used to express the RN area of a parallelogram since we just have to multiply the length of Rejuˆ (v): 1 2 k ( )k2 = ( ) · = uv Rejuˆ v Rejuˆ v v 2 ∑ Dij u i<j with the length of the base kuk. [FIXME: insert figure.] Thus the area (squared) is: 2 2 uv Au,v = ∑ Dij (3) i<j For the special case of a vector in R2 this is uv ui uj Au,v = |D12 | = abs (4) vi vj 1.2 Vector orthogonal to two vectors in direction of a third. The same procedure can be followed for three vectors, but the algebra gets messier. Given three vectors u, v, and w we can calculate the component w0 of w perpendicular to u and v. That is: 2 v0 = v − v · uˆ uˆ =⇒ w0 = w − w · uˆ uˆ − w · vˆ0vˆ0 After expanding this out, a number of the terms magically cancel out and one is left with w00 = w0(u2v2 − (u · v)2) = u −u · wv2 + (u · v)(v · w) + v −u2(v · w) − (u · v)(u · w) + w u2v2 − (u · v)2 And this in turn can be expanded in terms of coordinates and the results collected yielding 00 vj vk uj uk uj uk w = ∑ eiujvk ui − vi wi wj wk wj wk vj vk ui uj uk = e u v v v v ∑ i j k i j k wi wj wk ui uj uk uj uk = ∑ ei vi vj vk vj vk i,j<k wi wj wk ! ui uj uk uj uk = ∑ + ∑ + ∑ ei vi vj vk . vj vk i<j<k j<i<k j<k<i wi wj wk Expanding the sum of the denominator in terms of coordinates: 2 ui uj u2v2 − (u · v)2 = ∑ v v i<j i j and using a change of summation indexes, our final result for the vector perpendicular to two others in the direction of a third is: ui uj uk ui uj uk ∑ v v v v v v i<j<k i j k i j k wi wj wk ei ej ek Rej (w) = (5) uˆ ,ˆv 2 ui uj ∑i<j vi vj 3 ui uj As a small aside, it is notable here to observe that span is the ei ej ui uj uk null space for the vector u, and the set span v v v is the null space i j k ei ej ek for the two vectors u and v respectively. Since the rejection is a normal to the set of vectors it must necessarily in- clude these cross product like determinant terms. uvw As in equation 2, use of a Dijk notation allows for a more compact result: !−1 2 uv uvw uve Rejuˆ vˆ (w) = ∑ Dij ∑ Dijk Dijk (6) i<j i<j<k And, as before this yields the Volume of the parallelopiped by multiplying perpendicular height: !−1 2 2 uv uvw kRejuˆ vˆ (w)k = Rejuˆ vˆ (w) · w = ∑ Dij ∑ Dijk i<j i<j<k by the base area. Thus the squared volume of a parallelopiped spanned by the three vectors is: 2 2 uvw Vu,v,w = ∑ Dijk . (7) i<j<k The simplest case is for R3 where we have only one summand: u1 u2 u3 uvw Vu,v,w = |D | = abs v v2 v3 . (8) ijk 1 w1 w2 w3 1.3 Generalization. Inductive Hypothesis. There are two things to prove 1. hypervolume of parallelopiped spanned by vectors u1, u2,..., uk ui ui ···ui 2 V2 = D 1 2 k (9) u1,u2,··· ,uk ∑ i1i2···ik i1<i2<···<ik 2. Orthogonal rejection of a set of vectors in direction of another. u ···u u ···u e i1 ik i1 ik−1 ∑i <···<i Di ···i Di ···i Rej (u ) = 1 k 1 k 1 k (10) uˆ 1···uˆ k−1 k u ···u 2 i1 ik−1 ∑ <···< D i1 ik−1 i1···ik−1 4 I cannot recall if I ever did the inductive proof for this. Proving for the initial case is done (since it’s proved for both the two and three vector cases). For the limiting case where k = n it can be observed that this is normal to all the others, so the only thing to prove for that case is if the scaling provided by hypervolume equation 9 is correct. 1.4 Scaling required for reciprocal frame vector. Presuming an inductive proof of the general result of 10 is possible, this rejec- tion has the property Rej (u ) · u ∝ δ uˆ 1···uˆ k−1 k i ki With the scaling factor picked so that this equals δki, the resulting “recipro- cal frame vector” is u ···u u ···u e i1 ik i1 ik−1 ∑i <···<i Di ···i Di ···i uk = 1 k 1 k 1 k (11) ui ···ui 2 ∑ D 1 k i1<···<ik i1···ik The superscript notation is borrowed from Doran/Lasenby, and denotes not a vector raised to a power, but this this special vector satisfying the follow- ing orthogonality and scaling criteria: k u · ui = δki. (12) Note that for k = n − 1, equation 11 reduces to u1···un−1e D1···(n−1) un = u1···un . (13) D1···n This or some other scaled version of this is likely as close as we can come to generalizing the cross product as an operation that takes vectors to vectors. 1.5 Example. R3 case. Perpendicular to two vectors. 3 0 3 Observe that for R , writing u = u1, v = u2, w = u3, and w = u3 this is: u u u 1 2 3 v v v 1 2 3 0 e1 e2 e3 u × v w = = (14) u u u (u × v) · w 1 2 3 v v v 1 2 3 w1 w2 w3 This is the cross product scaled by the (signed) volume for the parallelop- iped spanned by the three vectors. 5 2 Derivation with GA. Regression with respect to a set of vectors can be expressed directly. For vectors ui write B = u1 ∧ u2 ··· uk. Then for any vector we have: 1 x = xB B 1 = xB B 1 1 = (x · B + x ∧ B) B 1 All the grade three and grade five terms are selected out by the grade one operation, leaving just 1 1 x = (x · B) · + (x ∧ B) · . (15) B B This last term is the rejective component. 1 (x ∧ B) · B† Rej (x) = (x ∧ B) · = (16) B B BB† Here we see in the denominator the squared sum of determinants in the denominator of equation 10: u ···u 2 BB† = D i1 ik ∑ i1···ik i1<···<ik In the numerator we have the dot product of two wedge products, each expressible as sums of determinants: ui ···ui B† = (−1)k(k−1)/2 D 1 k e e ··· e ∑ i1···ik i1 i2 ik i1<···<ik And xu ···u i1 ik x ∧ B = D ei ei ··· ei ∑ i1···ik+1 1 2 k+1 i1<···<ik+1 Dotting these is all the grade one components of the product. Performing that calculation would likely provide an explicit confirmation of the inductive hypothesis of equation 10. This can be observed directly for the k + 1 = n case. That product produces a Laplace expansion sum. † xu1···un−1 u1···un−1 u1···un−1 u1···un−1 (x ∧ B) · B = D12···n e1D234···n − e2D134···n + e3D124···n xu ···u eu ···u 1 D 1 n−1 D 1 n−1 (x ∧ B) · = 12···n 12···n (17) B ui ···ui 2 ∑ D 1 k i1<···<ik i1···ik 6 Thus equation 10 for the k = n − 1 case is proved without induction.
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