ÿþ<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD><TITLE>paper0</TITLE> <META http-equiv=Content-Type content="text/html; charset=unicode"> <META content="MSHTML 6.00.2800.1479" name=GENERATOR></HEAD> <BODY> <H4 align=center>Chaotic manifolds</H4><BR><BR> <P> <P><BR><BR><BR>E-mail: paper0@byandbygones.com &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; IE6x: Text optimized at: up to ~ 700px <P> <P><BR><BR><BR><U>I</U>: <U>Introduction:</U><BR>&nbsp; &nbsp; Tensor analysis assures us of the existence of its rank two tensor of dimensionality n, n an integer, n<FONT face=symbol>³</FONT>2. Therefore, by<BR>its process/operation of tensor inner product we are guaranteed of the existence of rank one tensors of dimensionality n.<BR>Rank one tensors are identical to ordinary vectors (in euclidean space/cartesian frame). Therefore there exists ordinary<BR>vectors of dimensionality n (and in fact the n-dimensional vector is posited as a priori fact in tensor analysis).<BR>&nbsp;&nbsp; Tensor analysis further assures us that the inner product of these posited or derived n-dimensional rank one tensors is<BR>defined, meaningful, and valid: that is, the inner/dot product of vectors of dimensionality n is an indispensable fact of n-<BR>vector/tensor analysis. In fact to smother any implication whatsoever that the existence of the rank one tensor/vector of<BR>dimensionality n is in any way dependent upon the existence of any higher or lower rank tensor whatsoever of whatever<BR>dimensionality (and vice versa) tensor analysis a priori postulates the independent existence of rank one tensors/vectors<BR>of dimensionality n<FONT face=symbol>³</FONT>2.<BR>&nbsp;&nbsp; The linear orthogonal transformation in E<SUP>n</SUP>of an n-dimensional (virtually always cartesian) reference frame of n linearly<BR>independent direction axes thus defines/posits the arbitrary n-dimensional tensor of arbitrary rank. The linear orthogonal<BR>transformation is a linear rotation of the reference frame/tensor in n dimensions - n-space. Thus when the n-dimensional<BR>unit basis set vector, say, <B>i</B>=<B>e</B><SUB>1</SUB>, is rotated in n-space, E<SUP>n</SUP>, the new vector <B>e</B>'<SUB>1</SUB>, as a consequence of the linear orthogonal<BR>transformation, is some linear combination of the orthonormal basis set <B>e</B><SUB>i</SUB>, <B>e</B><SUB>i</SUB>.<B>e</B><SUB>j</SUB>=1, j=i, and <B>e</B><SUB>i</SUB>.<B>e</B><SUB>j</SUB>=0, j<FONT face=symbol>¹</FONT>i. <BR>&nbsp;&nbsp; Therefore the linear orthogonal transformation/rotation of a vector/tensor, and the resulting vector/tensor of the linear<BR>transformation is each totally independent of dimensionality.<BR>&nbsp;&nbsp; Like the linear rotation operation, the differential operation is a rigorously defined, valid, linear difference operation<BR>totally independent of dimensionality: what is generated/derived by the linear operation in m-space, m&gt;1, is preserved<BR>in n<FONT face=symbol>¹</FONT>m - nothing qualitatively new or novel created, nothing substantive wiped out. Thus there is no substantive/analytic<BR>structure difference at all in the instance, <DIV align=center><BR><BR>d<B>R</B> = (dR<SUB>1</SUB>)<B>e</B><SUB>1</SUB> + (dR<SUB>2</SUB>)<B>e</B><SUB>2</SUB> + (dR<SUB>3</SUB>)<B>e</B><SUB>3</SUB>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<BR> <P>d<B>R</B> = (<FONT face=symbol>¶</FONT><SUB>x</SUB><B>R</B>)dx + (<FONT face=symbol>¶</FONT><SUB>y</SUB><B>R</B>)dy + (<FONT face=symbol>¶</FONT><SUB>z</SUB><B>R</B>)dz, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>R</B>=<B>R</B>(x,y,z) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <P>d<B>R</B> = <FONT face=symbol size=5>S</FONT><SUP><SUP>n</SUP></SUP><SUB>i=<FONT size=2>1</FONT></SUB>(<FONT face=symbol>¶</FONT><SUB>x<SUB>i</SUB></SUB><B>R</B>)dx<SUB>i</SUB>, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>R</B>=<B>R</B>(x<SUB>1</SUB>,...,x<SUB>i</SUB>,...,x<SUB>n</SUB>) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</P></DIV><BR>or even <P> <DIV align=center>d<B>R</B> = (<FONT face=symbol>¶</FONT><SUB>t</SUB><B>R</B>)dt + (<FONT face=symbol>¶</FONT><SUB>x</SUB><B>R</B>)dx + (<FONT face=symbol>¶</FONT><SUB>y</SUB><B>R</B>)dy + (<FONT face=symbol>¶</FONT><SUB>z</SUB><B>R</B>)dz, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>R</B>=<B>R</B>(x,y,z,t) <P>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;d<B>R</B> = (<FONT face=symbol>¶</FONT><SUB>t</SUB><B>R</B>)dt + (<FONT face=symbol>¶</FONT><SUB>x</SUB><B>R</B>)dx + (<FONT face=symbol>¶</FONT><SUB>y</SUB><B>R</B>)dy + (<FONT face=symbol>¶</FONT><SUB>z</SUB><B>R</B>)dz, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>R</B>=<B>R</B>( x(t),y(t),z(t) ) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2).</P></DIV><BR> <P> <P><BR><BR><BR><U>II</U>: <U>Differentials of <B>R</B></U>: <BR>&nbsp; &nbsp; The premise that an observer stationed at a point fixed relative to an n-space n-dimensional cartesian (x<SUB>1</SUB>,...,x<SUB>i</SUB>,...,x<SUB>n</SUB>)<BR>coordinate system with origin O rotating wrt an (x'<SUB>1</SUB>,...,x'<SUB>i</SUB>,...,x'<SUB>n</SUB>)-coordinate system taken as fixed in n-space and having<BR>origin also at O, is logically and analytically valid. Indeed, it is the raison d'etre of the basis of the equations of the linear<BR>orthogonal transformation of orthogonal reference frames and the Klein definition of the vector. <BR>&nbsp;&nbsp; The linear combination of two successive linear operations is a linear operation. Thus to the fixed observer, due to its<BR>linear rotation, the orthonormal basis/direction vectors <B>e</B><SUB>1</SUB>,...,<B>e</B><SUB>i</SUB>,...,<B>e</B><SUB>n</SUB> in the moving (x<SUB>1</SUB>,...,x<SUB>i</SUB>,...,x<SUB>n</SUB>)-frame change during<BR>the duration of the rotation (infinitesimal or discrete). Therefore, to take the 4-dimensional case for algebraic simplicity,<BR>the observer computes the differential of <B>R</B>=<FONT face=symbol size=5>S</FONT><SUP><SUP>n</SUP></SUP><SUB>i=<FONT size=2>1</FONT></SUB>R<SUB>i</SUB><B>e</B><SUB>i</SUB> as <DIV align=center><BR>d<B>R</B> = <FONT face=symbol size=5>S</FONT><SUP><SUP>n</SUP></SUP><SUB>i=<FONT size=2>1</FONT></SUB>[(dR<SUB>i</SUB>)<B>e</B><SUB>i</SUB> + R<SUB>i</SUB>d<B>e</B><SUB>i</SUB>] &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (3) </DIV><BR>To determine the d<B>e</B><SUB>i</SUB> in terms of the linearly independent <B>e</B><SUB>i</SUB> we note that <B>e</B><SUB>i</SUB>.<B>e</B><SUB>i</SUB>=1 implies <B>e</B><SUB>i</SUB>.d<B>e</B><SUB>i</SUB>=0, which in turn implies<BR>that <B>e</B><SUB>i</SUB> and d<B>e</B><SUB>i</SUB> are perpendicular (real euclidean) n-vectors. Thus d<B>e</B><SUB>i</SUB>, simplified, must be an n-vector restricted to the<BR>i<FONT face=symbol>¹</FONT>j (<B>e</B><SUB>1</SUB>,...,<B>e</B><SUB>i</SUB>,...,<B>e</B><SUB>n</SUB>)-surface of E<SUP>n</SUP>. Thus R<SUB>i</SUB>d<B>e</B><SUB>i</SUB> is (in array format): <div align=center><BR><IMG height=169 src="http://www.byandbygones.com///untitle.GIF" width=636></div> <P align=left>or</FONT> <DIV align=center><BR>[<FONT face=symbol size=5>S</FONT><SUP><SUP>n</SUP></SUP><SUB>i=<FONT size=2>1</FONT></SUB>R<SUB>i</SUB><B>e</B><SUB>i</SUB>] = [1<SUB>n</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][d<B>e</B><SUB>i</SUB>] = [1<SUB>n</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][<B>e</B><SUB>i</SUB>] &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (4b) </DIV><BR>in which [1<SUB>n</SUB>], utilized for the convenience of summing the array entries to ordinary vector format, is an nx1 row array in<BR>which every entry is 1, [R<SUB>ii</SUB>] of <B>R</B>=[1<SUB>n</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][<B>e</B><SUB>i</SUB>], is the nxn version of the nx1 [R<SUB>i</SUB>] array for <B>R</B>, R<SUB>ij</SUB>=0, i<FONT face=symbol>¹</FONT>j, and <FONT face=symbol>s</FONT><SUB>ij</SUB>=0, i=j,<BR>i,j=1,2,...,n. <BR>&nbsp;&nbsp; Evidently the <FONT face=symbol>s</FONT><SUB>ii</SUB>=0 are the <B>e</B><SUB>i</SUB>.d<B>e</B><SUB>i</SUB>=0 from <B>e</B><SUB>i</SUB>.<B>e</B><SUB>i</SUB>=1, and evidently too the <FONT face=symbol>s</FONT><SUB>ij</SUB> tab the <B>e</B><SUB>i</SUB>.d<B>e</B><SUB>j</SUB>. That is, from <B>e</B><SUB>i</SUB>.<B>e</B><SUB>j</SUB>=0, j<FONT face=symbol>¹</FONT>i,<BR>we have [<B>e</B><SUB>i</SUB>.d<B>e</B><SUB>j</SUB>+<B>e</B><SUB>j</SUB>.d<B>e</B><SUB>i</SUB>]=0. Since we define <B>e</B><SUB>j</SUB>.d<B>e</B><SUB>i</SUB>=<FONT face=symbol>s</FONT><SUB>ij</SUB> and <B>e</B><SUB>i</SUB>.d<B>e</B><SUB>j</SUB>=<FONT face=symbol>s</FONT><SUB>ji</SUB> we see that <FONT face=symbol>s</FONT><SUB>ij</SUB>=-<FONT face=symbol>s</FONT><SUB>ji</SUB>. Thus, the [<FONT face=symbol>s</FONT><SUB>ij</SUB>] array is an<BR>antisymmetric array for all n<FONT face=symbol>³</FONT>2, but we add, the array is a computing and calculating picture, not quite a formal matrix. <BR>&nbsp;&nbsp; For the instance n=4 the first order differential d<B>R</B> in (3), fully written out is: <DIV align=center><BR>d<B>R</B> = <FONT color=black size=6>[</FONT><FONT size=4>{</FONT>(dR<SUB>1</SUB>)<B>e</B><SUB>1</SUB> + (dR<SUB>2</SUB>)<B>e</B><SUB>2</SUB> + (dR<SUB>3</SUB>)<B>e</B><SUB>3</SUB> + (dR<SUB>4</SUB>)<B>e</B><SUB>4</SUB> + ... + (dR<SUB>n</SUB>)<B>e</B><SUB>n</SUB><FONT size=4>}</FONT><BR><BR>+ <FONT size=4>{</FONT>R<SUB>1</SUB>d<B>e</B><SUB>1</SUB> + R<SUB>2</SUB>d<B>e</B><SUB>2</SUB> + R<SUB>3</SUB>d<B>e</B><SUB>3</SUB> + R<SUB>4</SUB>d<B>e</B><SUB>4</SUB> + ... + R<SUB>n</SUB>d<B>e</B><SUB>n</SUB><FONT size=4>}</FONT><FONT color=black size=6>]</FONT> <BR><BR>= <FONT color=black size=6>[</FONT><FONT size=4>{</FONT>(dR<SUB>1</SUB>)<B>e</B><SUB>1</SUB> + (dR<SUB>2</SUB>)<B>e</B><SUB>2</SUB> + (dR<SUB>3</SUB>)<B>e</B><SUB>3</SUB> + (dR<SUB>4</SUB>)<B>e</B><SUB>4</SUB> + ... + (dR<SUB>n</SUB>)<B>e</B><SUB>n</SUB><FONT size=4>}</FONT><BR><BR>- <FONT color=blue size=4>{</FONT>(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>R<SUB>n</SUB>)<B>e</B><SUB>1</SUB><BR><BR>+ (<FONT face=symbol>s</FONT><SUB>21</SUB>R<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>R<SUB>n</SUB>)<B>e</B><SUB>2</SUB><BR><BR>+ (<FONT face=symbol>s</FONT><SUB>31</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>32</SUB>R<SUB>2</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>34</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>3n</SUB>R<SUB>n</SUB>)<B>e</B><SUB>3</SUB><BR><BR>+ (<FONT face=symbol>s</FONT><SUB>41</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>42</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>43</SUB>R<SUB>3</SUB> + 0 + ... + <FONT face=symbol>s</FONT><SUB>4n</SUB>R<SUB>n</SUB>)<B>e</B><SUB>4</SUB><BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;+ ... ... ... ... + (<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB> + ... + <FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB> + 0)<B>e</B><SUB>n</SUB><FONT color=blue size=4>}</FONT><FONT color=black size=6>]</FONT>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (5) </DIV><BR>and the second order differential d<SUP>2</SUP><B>R</B> of (3) for n=4 is: <DIV align=center><BR>d<SUP>2</SUP><B>R</B> = <FONT color=black size=6>[</FONT><FONT size=4>{</FONT>(d<SUP>2</SUP>R<SUB>1</SUB>)<B>e</B><SUB>1</SUB> + (d<SUP>2</SUP>R<SUB>2</SUB>)<B>e</B><SUB>2</SUB> + (d<SUP>2</SUP>R<SUB>3</SUB>)<B>e</B><SUB>3</SUB> + (d<SUP>2</SUP>R<SUB>4</SUB>)<B>e</B><SUB>4</SUB> + ... + (d<SUP>2</SUP>R<SUB>n</SUB>)<B>e</B><SUB>n</SUB><FONT size=4>}</FONT><BR><BR>- <FONT color=blue size=4>{</FONT>(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>dR<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>dR<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>dR<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>dR<SUB>n</SUB>)<B>e</B><SUB>1</SUB><BR><BR>+ (<FONT face=symbol>s</FONT><SUB>21</SUB>dR<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>dR<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>dR<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>dR<SUB>n</SUB>)<B>e</B><SUB>2</SUB><BR><BR>+ (<FONT face=symbol>s</FONT><SUB>31</SUB>dR<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>32</SUB>dR<SUB>2</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>34</SUB>dR<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>3n</SUB>dR<SUB>n</SUB>)<B>e</B><SUB>3</SUB><BR><BR>+ (<FONT face=symbol>s</FONT><SUB>41</SUB>dR<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>42</SUB>dR<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>43</SUB>dR<SUB>3</SUB> + 0 +... + <FONT face=symbol>s</FONT><SUB>4n</SUB>dR<SUB>n</SUB>)<B>e</B><SUB>4</SUB><BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;+ ... ... ... ... + (<FONT face=symbol>s</FONT><SUB>n1</SUB>dR<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>n2</SUB>dR<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>n3</SUB>dR<SUB>3</SUB> + ... + <FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>dR<SUB>n-1</SUB> + 0)<B>e</B><SUB>n</SUB><FONT color=blue size=4>}</FONT>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<BR><BR>- <FONT color=orange size=4>{</FONT>[d(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>R<SUB>n</SUB>)]<B>e</B><SUB>1</SUB><BR><BR>+ [d(<FONT face=symbol>s</FONT><SUB>21</SUB>R<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>R<SUB>n</SUB>)]<B>e</B><SUB>2</SUB><BR><BR>+ [d(<FONT face=symbol>s</FONT><SUB>31</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>32</SUB>R<SUB>2</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>34</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>3n</SUB>R<SUB>n</SUB>)]<B>e</B><SUB>3</SUB><BR><BR>+ [d(<FONT face=symbol>s</FONT><SUB>41</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>42</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>43</SUB>R<SUB>3</SUB> + 0 + ... + <FONT face=symbol>s</FONT><SUB>4n</SUB>R<SUB>n</SUB>)]<B>e</B><SUB>4</SUB><BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;+ ... ... ... ... + [d(<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB> + ... + <FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB> + 0)]<B>e</B><SUB>n</SUB><FONT color=orange size=4>}</FONT>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<BR><BR>- <FONT color=red size=4>{</FONT><FONT color=red><B>[</B></FONT><FONT>(..)<FONT face=symbol>s</FONT><SUB></SUB><SUB>11</SUB></FONT> + (<FONT face=symbol>s</FONT><SUB>21</SUB>R<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB><U><FONT color=blue>21</FONT></U></SUB> + (<FONT face=symbol>s</FONT><SUB>31</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>32</SUB>R<SUB>2</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>34</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>3n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>31</SUB> +<BR><BR>(<FONT face=symbol>s</FONT><SUB>41</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>42</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>43</SUB>R<SUB>3</SUB> + 0 + ... + <FONT face=symbol>s</FONT><SUB>4n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>41</SUB> + ... ... ... + (<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB> + ... + <FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB> + 0)<FONT face=symbol>s</FONT><SUB>n1</SUB><FONT color=red><B>]</B></FONT><B>e</B><SUB>1</SUB><BR><BR>+ <FONT color=red><B>[</B></FONT>(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>12</SUB> + <FONT>(..)<FONT face=symbol>s</FONT><SUB></SUB><SUB>22</SUB></FONT> + (<FONT face=symbol>s</FONT><SUB>31</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>32</SUB>R<SUB>2</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>34</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>3n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>32</SUB> +<BR><BR>(<FONT face=symbol>s</FONT><SUB>41</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>42</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>43</SUB>R<SUB>3</SUB> + 0 + ... + <FONT face=symbol>s</FONT><SUB>4n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>42</SUB> + ... ... ... + (<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB> + ... + <FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB> + 0)<FONT face=symbol>s</FONT><SUB>n2</SUB><FONT color=red><B>]</B></FONT><B>e</B><SUB>2</SUB><BR><BR>+ <FONT color=red><B>[</B></FONT>(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>13</SUB> + (<FONT face=symbol>s</FONT><SUB>21</SUB>R<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB><U><FONT color=blue>23</FONT></U></SUB> + <FONT>(..)<FONT face=symbol>s</FONT><SUB></SUB><SUB>33</SUB></FONT> +<BR><BR>(<FONT face=symbol>s</FONT><SUB>41</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>42</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>43</SUB>R<SUB>3</SUB> + 0 + ... + <FONT face=symbol>s</FONT><SUB>4n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>43</SUB> + ... ... ... + (<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB> + ... + <FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB> + 0)<FONT face=symbol>s</FONT><SUB>n3</SUB><FONT color=red><B>]</B></FONT><B>e</B><SUB>3</SUB><BR><BR>+ <FONT color=red><B>[</B></FONT>(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>14</SUB> + (<FONT face=symbol>s</FONT><SUB>21</SUB>R<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB><U><FONT color=blue>24</FONT></U></SUB> +<BR><BR>(<FONT face=symbol>s</FONT><SUB>31</SUB>R<SUB>1</SUB> + <FONT face=symbol>s</FONT><SUB>32</SUB>R<SUB>2</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>34</SUB>R<SUB>4</SUB> +...+ <FONT face=symbol>s</FONT><SUB>3n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>34</SUB> + <FONT>(..)<FONT face=symbol>s</FONT><SUB></SUB><SUB>44</SUB></FONT> + ... ... . + (<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB>+<FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB>+<FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB>+...+<FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB>+0)<FONT face=symbol>s</FONT><SUB>n4</SUB><FONT color=red><B>]</B></FONT><B>e</B><SUB>4</SUB><BR><BR>+ .... ... .... + <FONT color=red><B>[</B></FONT>(0 + <FONT face=symbol>s</FONT><SUB>12</SUB>R<SUB>2</SUB> + <FONT face=symbol>s</FONT><SUB>13</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>14</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>1n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>1n</SUB> + (<FONT face=symbol>s</FONT><SUB>21</SUB>R<SUB>1</SUB> + 0 + <FONT face=symbol>s</FONT><SUB>23</SUB>R<SUB>3</SUB> + <FONT face=symbol>s</FONT><SUB>24</SUB>R<SUB>4</SUB> + ... + <FONT face=symbol>s</FONT><SUB>2n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB><U><FONT color=blue>2n</FONT></U></SUB> +<BR><BR>(<FONT face=symbol>s</FONT><SUB>31</SUB>R<SUB>1</SUB>+<FONT face=symbol>s</FONT><SUB>32</SUB>R<SUB>2</SUB>+0+<FONT face=symbol>s</FONT><SUB>34</SUB>R<SUB>4</SUB>+...+ <FONT face=symbol>s</FONT><SUB>3n</SUB>R<SUB>n</SUB>)<FONT face=symbol>s</FONT><SUB>3n</SUB>+ ... ... +(<FONT face=symbol>s</FONT><SUB>n1</SUB>R<SUB>1</SUB>+<FONT face=symbol>s</FONT><SUB>n2</SUB>R<SUB>2</SUB>+<FONT face=symbol>s</FONT><SUB>n3</SUB>R<SUB>3</SUB>+...+<FONT face=symbol>s</FONT><SUB>n<FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT></SUB>R<SUB>n-1</SUB>+0)<FONT face=symbol>s</FONT><SUB><FONT size=1><B>(</B></FONT>n-1<FONT size=1><B>)</B></FONT>n</SUB>+<FONT>(..)<FONT face=symbol>s</FONT><SUB></SUB><SUB>nn</SUB></FONT>)<FONT color=red><B>]</B></FONT><B>e</B><SUB>n</SUB><FONT color=red size=5>}</FONT><FONT color=black size=5>]</FONT> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (6) </DIV><BR>the generalization of which to all n is quite obvious - making it clear that rather than being some derivation of some sort,<BR>(4) is literally a one to one description in array format of what's already in (3). Clearly, exactly as for n=3 (and even for<BR>n=2) which is obvious from (3,4,5,6) the <FONT face=symbol>s</FONT><SUB>ij</SUB>R<SUB>j</SUB> terms of (5) result from a non-scalar/non-inner product of vectors since<BR>a rotation and differential of <B>R</B> and <B>e</B><SUB>i</SUB> cannot create non-vector objects and structures. From actually working through <BR>the algebra for n=2,3,4 we <I>calculate</I> from (4,5) <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B><B>.R</B>=0, where <B><FONT face=symbol>d</FONT></B>, ( akin <B><FONT face=symbol>w</FONT></B>, n=3), is meant to represent the angular<BR>motion n-vector constituted of the <FONT face=symbol>s</FONT><SUB>ij</SUB> of (5), and the <SUB><FONT size=4>*</FONT></SUB> is the generalized product for all n<FONT face=symbol>³</FONT>2. From blue {..} of (5) and<BR>red {..} of (6) we further <I>calculate</I> that (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.</B>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)=-<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B> - and which is true for in particular n=3. It is clear<BR>from extending the pattern in (5,6) to all n that these results hold too for all n<FONT face=symbol>³</FONT>4. [The blue subscripts of (6) may facilitate<BR>computing a typical sum term in <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B>] <P> <P><BR><BR><BR><U>III</U>: <U>Arrays in d<B>R</B> and d<SUP>2</SUP><B>R</B></U>: <BR>&nbsp; &nbsp; Recasting/literally copying the calculated product in blue {..} of (5) into array format we have for n=4 - extended to<BR>arbitrary n: <P align=center><IMG height=169 src="http://www.byandbygones.com///untitle2.GIF" width=636> <P align=left>or</FONT> <DIV align=center><BR><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> = [1<SUB>n</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][<B>e</B><SUB>i</SUB>] = ([<FONT face=symbol>s</FONT><SUB>ij</SUB>][R<SUB>i</SUB>])<SUP>T</SUP>[<B>e</B><SUB>i</SUB>] &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (7b); </DIV><BR>in which R<SUB>ij</SUB>=R<SUB>i</SUB>, j=i, and R<SUB>ij</SUB>=0, j<FONT face=symbol>¹</FONT>i; and similarly for the actual calculated product in red {..} of (6) for n=4 - again<BR>extended to arbitrary n: <P align=center><IMG height=169 src="http://www.byandbygones.com///untitle3.GIF" width=636> <P align=left>or</FONT> <DIV align=center><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>) = [1<SUB>n</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][<B>e</B><SUB>i</SUB>] = ([<FONT face=symbol>s</FONT><SUB>ij</SUB>][R<SUB>i</SUB>])<SUP>T</SUP>[<FONT face=symbol>s</FONT><SUB>ij</SUB>][<B>e</B><SUB>i</SUB>] &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (8b); </DIV><BR>in all of which we tucked in the generalization to all n since the generalization of (5,6) from n=4 to n&gt;4, and hence (7,8),<BR>is so instantaneously and transparently evident; for (7,8) the <FONT face=symbol>s</FONT><SUB>ii</SUB>=0, <FONT face=symbol>s</FONT><SUB>ij</SUB>=-<FONT face=symbol>s</FONT><SUB>ji</SUB>, i,j=1,2,...,n, and T is the 'transpose'. <BR>&nbsp;&nbsp; The <B><FONT face=symbol>s</FONT></B>=[<FONT face=symbol>s</FONT><SUB>ij</SUB>] arrays are antisymmetric for all n, including for n=3, the basis for which can be seen in the four arrays for<BR><B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> - which is <B>A</B>x<B>B</B> for n=3 - in the fully written out first principles form converted to array format for arbitrary three-<BR>dimensional vectors <B>A</B>, <B>B</B> - and extended to arbitrary n: <META content="Internet Assistant for Word Version 3.0" name=Generator> <P align=center><IMG height=260 src="http://www.byandbygones.com///untitle4.GIF" width=627> <P align=left>(9a,b) showing (<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>)<SUP>T</SUP>=-<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B> if [<B>e</B><SUB>i</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>j</SUB>] is antisymmetric. <BR>&nbsp;&nbsp; Evidently there are other array permutations that equate the expanded expression for <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>; or <B>A</B>x<B>B</B> for n=3, and n=2.<BR>And evidently too we can also use the very same arrays to represent the dot product instead of the <SUB><FONT size=4>*</FONT></SUB>-product. Thus the<BR>presence of the antisymmetric array in an analysis of vectors appears to be somehow a result of the perspective in the<BR>analysis rather than the logical preclusion of the non-antisymmetric array, and vice versa. In fact as written, since (7) is<BR>not the standard array for <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> in standard matrix analysis - i.e., since the array [<FONT face=symbol>s</FONT><SUB>ij</SUB>] is <U>not</U> the array form of the vector<BR><B><FONT face=symbol>d</FONT></B>, as is [R<SUB>ii</SUB>] of <B>R</B>, (7) is evidently not reflected in (9), nor can (7), without specifying whether [<FONT face=symbol>s</FONT><SUB>ij</SUB>] results from <B><FONT face=symbol>d</FONT></B>, or<BR>from <SUB><FONT size=4>*</FONT></SUB> or from both, handle <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> becoming redundancies like <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B><FONT face=symbol>d</FONT></B> or <B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>, though it evidently will logically handle, as<BR>a matrix under matrix properties, self-repeat symmetry like <B>R.</B><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> and (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.</B>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>), and analogously (8) will handle<BR>the likes of <B>R.</B><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>), <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R.</B><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>), and <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.<FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>). But since we cannot extricate <B><FONT face=symbol>d</FONT></B> from <B>R</B> in the basic<BR><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> root/nucleus as it were to get [<B><FONT face=symbol>d</FONT></B>]<SUP>T</SUP>[Q<SUB>ij</SUB>][<B>R</B>] in conformity, as at (9), with [<B>y</B>]<SUP>T</SUP>[Q<SUB>ij</SUB>][<B>x</B>] or [<B>x</B>]<SUP>T</SUP>[Q<SUB>ij</SUB>][<B>x</B>] of standard<BR>matrix analysis, neither (7) nor (8) can accommodate the likes of <B><FONT face=symbol>d</FONT>.</B>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>), <B><FONT face=symbol>d</FONT>.</B><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>) or even known array equivalent<BR>of structures of (5) and (6) and further m<SUP>th</SUP>-order differentials like, unsymmetric with (7) and (8), (d<SUP>m</SUP><B><FONT face=symbol>d</FONT></B>)<SUB><FONT size=4>*</FONT></SUB><B>R</B>, in which the<BR>(d<SUP>m</SUP><B><FONT face=symbol>d</FONT></B>)={[1]<SUP>T</SUP>(d<SUP>m</SUP>[<FONT face=symbol>s</FONT><SUB>ij</SUB>])[<B>e</B><SUB>i</SUB>]} and which is obvious in (6) for m=1. Actually from actual calculation all <B>R.</B>(d<SUP>m</SUP><B><FONT face=symbol>d</FONT></B>)<SUB><FONT size=4>*</FONT></SUB><B>R</B>=0, and<BR>so too all <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(d<SUP>m</SUP><B>R</B>)<B>.</B>(d<SUP>m</SUP><B>R</B>)=0, with (d<SUP>m</SUP><B>R</B>)={(d<SUP>m</SUP>[R<SUB>i</SUB>])<SUP>T</SUP>[<B>e</B><SUB>i</SUB>]} - that is, the <B>e</B><SUB>i</SUB> are not included in the differential operation<BR>in these terms.<BR>&nbsp;&nbsp; Returning to (5,6) again it is clear by actual calculation that from {..} of (5) that dotting <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> with itself is exactly equal<BR>to dotting <B>R</B> with <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>) from red {..} of (6) for all n. Thus, <DIV align=center><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.</B>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>) = - <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (10), </DIV><BR>and the layout of relevant terms in (5,6) and (7,8) makes it evident that (10) is true for all n. <BR>&nbsp;&nbsp; In terms of the representational arrays (7) then becomes, <DIV align=center><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.</B>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>) = [<B>e</B><SUB>i</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][R<SUB>ii</SUB>][<B>e</B><SUB>i</SUB>]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(11), </DIV><BR>again with diagonal [R<SUB>ii</SUB>] such that R<SUB>ii</SUB>=R<SUB>i</SUB>, R<SUB>ij</SUB>=0, j=i; and dotting <B>R</B>=[<B>e</B><SUB>i</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>] into (8) means (8) now becomes, <DIV align=center><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <B>R.</B><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>) = [<B>e</B><SUB>i</SUB>]<SUP>T</SUP>[R<SUB>ii</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][<FONT face=symbol>s</FONT><SUB>ij</SUB>][R<SUB>ii</SUB>][<B>e</B><SUB>i</SUB>] &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (12), </DIV><BR>and here because of the evident symmetry in the arrays making use of the transpose properties of [R<SUB>ii</SUB>] and [<FONT face=symbol>s</FONT><SUB>ij</SUB>] in (11)<BR>the arrays of (12) and (11) make (10) evident for all, E<SUP>n</SUP>, orthogonal frames, n&gt;1 - though without necessarily making<BR>transparent what the <SUB><FONT size=4>*</FONT></SUB>-product incontestably necessarily is. <BR>&nbsp;&nbsp; Since the linear rotation (orthogonal transformation) of an orthogonal frame is per se evidently independent, as for<BR>n=3, of dimensionality (and since even if we linearly rotate an (n-m)-dimensional surface of the n-dimensional frame/<BR>manifold the linear rotation is still by definition a linear infinitesimal rotation <FONT face=symbol>dq</FONT>), analytically, the inherent properties of<BR>the <SUB><FONT size=4>*</FONT></SUB>-product are, for n=3, evidently identical of the intrinsic properties of the x-product. <BR>&nbsp;&nbsp; Thus, if we linearly rotate an n-dimensional frame in one sense (counterclockwise?), then the vector <B>e</B><SUB>i</SUB> changes from<BR>an initial time zero vector <B>e</B><SUB>i</SUB> to a time t final frame position vector <B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB>: that is, the change is [(<B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB>)-<B>e</B><SUB>i</SUB>]=<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB>. Now,<BR>since there is no preferred/absolute (reference) plane, and, in general, no preferred/absolute (n-m)-surface, in/of an n-<BR>dimensional space or frame and in an arbitrary plane/surface no preferred/absolute direction and since every plane in<BR>any/every E<SUP>n</SUP> space can a priori likewise be orthogonalized as the (<B>e</B><SUB>1</SUB>-<B>e</B><SUB>2</SUB>)-basis of E<SUP>2</SUP>, then the nature of 'angle' in any<BR>arbitrary plane cross-section is the same in all spaces. Thus, linear (infinitesimal) rotation is linearly reversible. Thus, if<BR>now the <B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB> vector/frame is now linearly rotated back to its original time zero orientation then <B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB> must return to<BR>a final t' equal time zero state. Thus {[(<B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB>)-<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB>] + [<B>e</B><SUB>i</SUB>-(<B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB>)]}=<B>0</B> is the vector sum of the changes <B>e</B><SUB>i</SUB> to <B>e</B><SUB>i</SUB>+<FONT face=symbol>d</FONT><B>e</B><SUB>i</SUB> to<BR><B>e</B><SUB>i</SUB> in every space n&gt;1. <BR>&nbsp;&nbsp; Thus when an arbitrary linear rotation compounds arbitrary n-dimesional vectors <B>A</B> and <B>B</B>, n&gt;1, to yield the resultant<BR>n-dimensional vector <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> then an arbitrary counter linear rotation that compounds <B>B</B> and <B>A</B> to yield the n-vector <B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B><BR>means [<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>+<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>]=<B>0</B>. Thus <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>=-<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B> for all n<FONT face=symbol>³</FONT>2 since <FONT face=symbol>dq</FONT><FONT face=symbol>¹</FONT><FONT face=symbol>dq</FONT>(n). <BR>&nbsp;&nbsp; For <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>=-<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B> setting <B>B</B>=<B>A</B> implies <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>=-<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>. Thus <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>=<B>0</B>. Thus since <B>A</B> and <B>A</B> are parallel, <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>=-<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>=<B>0</B><BR>when/implies <B>B</B> is parallel to <B>A</B>. Thus for arbitrary <B>A</B> and <B>B</B> if there is a component of <B>B</B> (that can be resolved) parallel<BR>to <B>A</B> then the vector <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> cannot have any component in any way parallel to <B>A</B>. Similarly for any component of <B>A</B> (that<BR>can be resolved) parallel to <B>B</B>. Thus the n-vector <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> is perpendicular to (the plane of) both <B>A</B> and <B>B</B>. <BR>&nbsp;&nbsp; A more analytical derivation of the (same) properties of the <SUB><FONT size=4>*</FONT></SUB>-product is perhaps more concrete. To begin with we<BR>are assured of the existence as analytical mathematical fact of the n-vector <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> by (5) and (6) and all d<SUP>m</SUP><B>R</B> beyond (6).<BR>The n-vectors <B>A</B>, <B>B</B> exist a priori, and the vector operation we characterize <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>, or more precisely, <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> does yield -<BR>in fact, are - the vectors of (5), (6). They are hard facts of valid analysis of a logically and mathematically non-optional<BR>fundamental a priori of vector analysis and the very definition of the vector. The n<FONT face=symbol>³</FONT>2 n-dimensional vector <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> is not<BR>a priori decree, not notional fancy, not axiom, not dictum, not surmise not premise, not hypothesis, nor even airy-fairy<BR>abracadabra wand-waving; it is an irremovable fact of the linear rotation and the linear differential operation. <BR>&nbsp;&nbsp; The n-vector <B>R</B> is arbitrary and since rotation is susceptible of arbitrary rate, of arbitrary increment/infinitesimal <FONT face=symbol>dq</FONT>, of<BR>arbitrary inclination, of arbitrary direction, of arbitrary angle displacement then evidently <B><FONT face=symbol>d</FONT></B> is arbitrary. In fact not only is<BR><B><FONT face=symbol>d</FONT></B> an arbitrary n-vector, since it comes from the ordinary sum of ordinary scalar, R<SUB>i</SUB>, multiples of ordinary n-dimensional<BR>d<B>e</B><SUB>i</SUB>, it must necessarily be an ordinary n-dimensional vector as evidenced at (4) and (3). Granted, a <B><FONT face=symbol>d</FONT></B> that would seem<BR>transparently independent of and totally separable from the R<SUB>i</SUB> of <B>R</B>, as for n=3, would be ideal, but its independence of<BR>and separability from the R<SUB>i</SUB> transparent or not evidently the scalar R<SUB>i</SUB> cannot affect the vector nature of the d<B>e</B><SUB>i</SUB> by simple<BR>scalar multiplication R<SUB>i</SUB>d<B>e</B><SUB>i</SUB>. Thus, even a <B><FONT face=symbol>d</FONT></B>=<B><FONT face=symbol>d</FONT></B>(R<SUB>i</SUB>,<FONT face=symbol>s</FONT><SUB>ij</SUB>) even with scalars, R<SUB>i</SUB>, of <B>R</B>, must be an arbitrary vector. Thus given<BR>that the n-vector <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> is a non-optional, analytic, fact, and <B><FONT face=symbol>d</FONT></B> arbitrary and ordinary, then the vector properties of all/any<BR>arbitrary <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>, regardless of origin context of <B>A</B>, <B>B</B> are the properties of <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>. <BR>&nbsp;&nbsp; Now, since under linear rotation (regardless of sense) the length/magnitude of an n-vector (by definition of the vector<BR>under linear orthogonal transformation/infinitesimal rotation) is preserved (a scalar) then <B>A</B>/|<B>A</B>| and <B>B</B>/|<B>B</B>| are (even with<BR>components variable) unit n-vectors, and since the absolute angular distance relation of <B>A</B> to <B>B</B> is the same as that of <B>B</B><BR>to <B>A</B> then these unit n-vectors must satisfy the scalar relation |(<B>A</B>/|<B>A</B>|)<SUB><FONT size=4>*</FONT></SUB>(<B>B</B>/|<B>B</B>|)|=|(<B>B</B>/|<B>B</B>|)<SUB><FONT size=4>*</FONT></SUB>(<B>A</B>/|<B>A</B>|)|<FONT face=symbol>£</FONT>1 unless <SUB><FONT size=4>*</FONT></SUB> is a left-right<BR>untisymmetric boost function. But evidently since by definition sense of rotation can change only their direction it cannot<BR>affect |d<B>e</B><SUB>i</SUB>|. Thus by (5,6) <SUB><FONT size=4>*</FONT></SUB> has no such properties. Thus: <DIV align=center><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>| = |<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>| &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (13) </DIV><BR>since |<B>A</B>|, |<B>B</B>| are scalars. <BR>&nbsp;&nbsp; Now, since <B><FONT face=symbol>d</FONT></B>, <B>R</B> arbirtrary and by (5), (6) <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B> is an ordinary n-vector whose unit n-vector evidently exists and can<BR>be calculated from its actual components in (5) then <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>/|<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>| exists and is a unit n-vector <B>u</B><SUB>o</SUB> parallel to <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>. Thus,<BR>from <B>u</B><SUB>o</SUB>.<B>u</B><SUB>o</SUB>=1 and (13), <DIV align=center><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B><SUP>2</SUP> = |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|<SUP>2</SUP> = |<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>|<SUP>2</SUP> = <B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B><SUP>2</SUP> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (14) </DIV><BR>&nbsp;&nbsp; From (<B>A</B>/|<B>A</B>|)<SUB><FONT size=4>*</FONT></SUB>(<B>B</B>/|<B>B</B>|)=<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B> with unit n-vectors <B>u</B>'=(<B>A</B>/|<B>A</B>|) and <B>u</B>=(<B>B</B>/|<B>B</B>|) and (<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>/|<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|)=<B>u</B><SUB>o</SUB> it is evident that <B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><BR>and <B>u</B><SUB>o</SUB> are parallel since |<B>A</B>||<B>B</B>| is a scalar multiplier of <B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B> and |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>| is a scalar multiplier of <B>u</B><SUB>o</SUB>. The dot product of the<BR>two <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> yields <DIV align=center><BR><B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B><SUP>2</SUP> = <B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><B>.u</B><SUB>o</SUB>|<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>||<B>A</B>||<B>B</B>| = |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|<SUP>2</SUP> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (15a), </DIV><BR>and thus <DIV align=center><BR>|<B>A</B>||<B>B</B>||<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>| = |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>| = |<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>|= |<B>A</B>||<B>B</B>||<B>u</B><SUB><FONT size=4>*</FONT></SUB><B>u</B>'| &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (15b) </DIV><BR>in which |<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>| is the angle or some function of the angle between <B>u</B>', </SUB><B>u</B> or <B>A</B> , <B>B</B>. <BR>&nbsp;&nbsp; Clearly by (14), (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<SUP>2</SUP>=-<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B>, which is (10), by symmetry, must imply <DIV align=center><BR>(<B>R</B><SUB><FONT size=4>*</FONT></SUB><B><FONT face=symbol>d</FONT></B>)<SUP>2</SUP> = - <B><FONT face=symbol>d</FONT>.</B><B>R</B><SUB><FONT size=4>*</FONT></SUB>(<B>R</B><SUB><FONT size=4>*</FONT></SUB><B><FONT face=symbol>d</FONT></B>) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (16a), </DIV><BR>and thus <DIV align=center><BR><B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B> = <B>R</B><SUB><FONT size=4>*</FONT></SUB>(<B>R</B><SUB><FONT size=4>*</FONT></SUB><B><FONT face=symbol>d</FONT></B>)<B>.<FONT face=symbol>d</FONT></B> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (16b) </DIV><BR><BR>for all n-vectors. That is, evidently <SUB><FONT size=4>*</FONT></SUB> and <B>.</B> are interchangeable for all n-vectors in the manner of n=3. In fact that, the<BR>interchangeability of <SUB><FONT size=4>*</FONT></SUB> and <B>.</B>, and additionally the relation <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>=-<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>, is what (10) always stated - and it taken straight<BR>out of (5) and (6) evidently requiring virtually nothing more than mere observation. <BR>&nbsp;&nbsp; Thus, the results, by actual calculation, (d<SUP>m</SUP><B><FONT face=symbol>d</FONT></B>)<SUB><FONT size=4>*</FONT></SUB><B>R</B><B>.R</B>=0, etc., and, in particular, (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B>=0, also by actual calculation<BR>from (5) or (7) now, and noting that by actual calculation too, (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.Q</B><FONT face=symbol>¹</FONT>0, does imply (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B>=<B><FONT face=symbol>d</FONT>.</B>(<B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)=0, which in<BR>turn, since <B><FONT face=symbol>d</FONT></B> is arbitrary, implies <B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>=<B>0</B>. Evidently too, by interchange of <B>.</B> and <SUB><FONT size=4>*</FONT></SUB>, <B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.</B>(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)=0 for all n-vectors, <BR>a result by actual calculation at (5) and (6). However, although the analysis at (5,6) seems to know, or to define, what <BR> <B><FONT face=symbol>d</FONT></B> is, (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.<FONT face=symbol>d</FONT></B>=(<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B><FONT face=symbol>d</FONT></B>)<B>.R</B>=0 is not computed from (5) as is its parallel (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B>R</B>)<B>.R</B>=0, since, for n&gt;3, how to construct the<BR>n-dimensional <B><FONT face=symbol>d</FONT></B> n-vector from it's organic (1/2)n(n-1) <FONT face=symbol>s</FONT><SUB>ij</SUB> 'components' is unclear. Nevertheless, <B><FONT face=symbol>d</FONT></B> explicit or implicit,<BR>as a consequence of the interchange of <SUB><FONT size=4>*</FONT></SUB> and <B>.</B>, (<B><FONT face=symbol>d</FONT></B><SUB><FONT size=4>*</FONT></SUB><B><FONT face=symbol>d</FONT></B>)<B>.R</B> must be trivial since non-trivial <B>A.</B>(<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>C</B>)=<B>B.</B>(<B>C</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>)=<B>C.</B>(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>)<BR>becomes, when <B>B</B>=<B>A</B>, say, <B>A.</B>(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>C</B>)=(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>)<B>.C</B> and <B>A.</B>(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>C</B>)=<B>B.</B>(<B>C</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>)=<B>A.</B>(<B>C</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>)=-<B>A.</B>(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>C</B>)=-(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>)<B>.C</B>, thus forcing<BR>a regulation 0 since +(<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>)<B>.C</B> should logically be retrieved if that is not necessarily analytically the case. <BR>&nbsp;&nbsp; Now since <B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>=<B>0</B> now means the <SUB><FONT size=4>*</FONT></SUB>-product of parallel (or anti-parallel) n-vectors is 0 it implies that for arbitrary <B>A</B>,<BR><B>B</B> the n-vector <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> is perpendicular to <B>A</B>, <B>B</B> since <B>A.</B><B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>=0 and <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B><B>.B</B>=0. <BR>&nbsp;&nbsp; From <B>R</B><B>.R</B>=<B>R</B><SUP>2</SUP> we have <B>R.</B>d<B>R</B><FONT face=symbol>¹</FONT>0 in general. Therefore [<B>R</B><SUB><FONT size=4>*</FONT></SUB>d<B>R</B>+d<B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>]=<B>0</B> implies <B>R</B><SUB><FONT size=4>*</FONT></SUB>d<B>R</B>=-d<B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>. For the case <B>R</B><BR>parallel d<B>R</B> with <B>R</B><SUB><FONT size=4>*</FONT></SUB>d<B>R</B>=<B>0</B> there is the observation that |<B>R</B><SUB><FONT size=4>*</FONT></SUB>d<B>R</B>|=RdR|<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>| can yield 0 for |<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>|=sin<FONT face=symbol>q</FONT> but not cos<FONT face=symbol>q</FONT>; and<BR>for <B>R</B> perpendicular d<B>R</B> sin(<FONT face=symbol>±</FONT><FONT face=symbol>p</FONT>/2) satisfies <B>R</B><SUB><FONT size=4>*</FONT></SUB>d<B>R</B>=-d<B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>. In fact since it's a determination of |<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>| that is required, in<BR>effect <B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B> is the situation <B>e</B><SUB>i</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>f</SUB> which implies |<B>e</B><SUB>i</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>f</SUB>|=|<B>e</B><SUB>f</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>|=|f(<FONT face=symbol>±</FONT><FONT face=symbol>p</FONT>/2)|. Now, since <B>e</B><SUB>i</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>f</SUB>=-<B>e</B><SUB>f</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>=(-<B>e</B><SUB>f</SUB>)<SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>=<B>e</B><SUB>f</SUB><SUB><FONT size=4>*</FONT></SUB>(-<B>e</B><SUB>i</SUB>) we<BR>then have |<B>e</B><SUB>i</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>f</SUB>|=|-<B>e</B><SUB>f</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>|=|(-<B>e</B><SUB>f</SUB>)<SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>| which implies f<SUP>2</SUP>(<FONT face=symbol>p</FONT>/2)=f<SUP>2</SUP>(-<FONT face=symbol>p</FONT>/2)=f<SUP>2</SUP>(-(<FONT face=symbol>p</FONT>/2)-<FONT face=symbol>p</FONT>). Thus, along with |<B>e</B><SUB>i</SUB><SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>|<SUP>2</SUP>=f<SUP>2</SUP>(0)=0 and<BR>|(-<B>e</B><SUB>i</SUB>)<SUB><FONT size=4>*</FONT></SUB><B>e</B><SUB>i</SUB>|<SUP>2</SUP>=f<SUP>2</SUP>(<FONT face=symbol>p</FONT>)=0 these imply f is evidently a periodic function in the (<B>e</B><SUB>i</SUB>-<B>e</B><SUB>j</SUB>)-plane such that f(<FONT face=symbol>±</FONT><FONT face=symbol>p</FONT>)=0 - evidently the sine<BR>function. <BR>&nbsp;&nbsp; For arbitrary <FONT face=symbol>q</FONT>, we take <B>A</B>, <B>B</B> separated by angle <FONT face=symbol>q</FONT>. Since we know only the perpendicular component <B>B</B><SUB>n</SUB> of <B>B</B> to <B>A</B><BR>contributes to <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> we find |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|=|<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B><SUB>n</SUB>|=|<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>|Bsin<FONT face=symbol>q</FONT>=|<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>|ABsin<FONT face=symbol>q</FONT> and similarly |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|=|<B>A</B><SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>B</B>|=|<B>u</B>'<SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>u</B>|ABsin<FONT face=symbol>q</FONT> for<BR>which <B>u</B>', <B>u</B>'<SUB>n</SUB> are the unit n-vectors of <B>A</B>, <B>A</B><SUB>n</SUB> and likewise <B>u</B>, <B>u</B><SUB>n</SUB> are those of <B>B</B>, <B>B</B><SUB>n</SUB>. Since |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|=|<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>|AB we find for<BR>|<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>|<SUP>2</SUP>=f<SUP>2</SUP>(<FONT face=symbol>q</FONT>) the relations |<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>|=|<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>|sin<FONT face=symbol>q</FONT>=|<B>u</B>'<SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>u</B>|sin<FONT face=symbol>q</FONT> yield <DIV align=center><BR>f<SUP>2</SUP>(<FONT face=symbol>q</FONT>) = f<SUP>2</SUP>(<FONT face=symbol>±</FONT><FONT face=symbol>p</FONT>/2)sin<SUP>2</SUP><FONT face=symbol>q</FONT> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (17).</DIV><BR>&nbsp;&nbsp; Now, (<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>)<SUP>2</SUP>=<B>u</B>'<SUB><FONT size=4>*</FONT></SUB>(<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>)<B>.u</B><SUB>n</SUB>=<B>u</B><SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB>(<B>u</B><SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>u</B>')<B>.u</B>' and similarly (<B>u</B>'<SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>u</B>)<SUP>2</SUP>=<B>u</B>'<SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB>(<B>u</B>'<SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>u</B>)<B>.u</B>=<B>u</B><SUB><FONT size=4>*</FONT></SUB>(<B>u</B><SUB><FONT size=4>*</FONT></SUB><B>u</B>'<SUB>n</SUB>)<B>.u</B>'<SUB>n</SUB>. The n-vector <B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB><BR>is perpendicular to <B>u</B>', <B>u</B><SUB>n</SUB>, thus the triple product n-vector <B>u</B>'<SUB><FONT size=4>*</FONT></SUB>(<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>) is parallel, or antiparallel, to <B>u</B><SUB>n</SUB>. Similarly for each<BR>of the other three such analogous n-vectors, and they are all equal; the dot product angle in all must be the same 0 or <FONT face=symbol>p</FONT><BR>since cos<FONT face=symbol>q</FONT><FONT face=symbol>¹</FONT>cos<FONT face=symbol>p</FONT>. Thus, for each of the n-vectors, equating their magnitudes to the magnitude (<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>)<SUP>2</SUP>, that is, from the<BR>fact that (<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B>)<SUP>2</SUP>=(<B>u</B>'<SUB><FONT size=4>*</FONT></SUB><B>u</B><SUB>n</SUB>)<SUP>2</SUP>sin<SUP>2</SUP><FONT face=symbol>q</FONT>=(<B>u</B>'<SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>u</B>)<SUP>2</SUP>sin<SUP>2</SUP><FONT face=symbol>q</FONT> we now find f<SUP>2</SUP>(<FONT face=symbol>q</FONT>)=f<SUP>2</SUP>(<FONT face=symbol>p</FONT>/2)sin<SUP>2</SUP><FONT face=symbol>q</FONT>=f<SUP>2</SUP>(-<FONT face=symbol>p</FONT>/2)sin<SUP>2</SUP><FONT face=symbol>q</FONT>, or |f(<FONT face=symbol>p</FONT>/2)|=|f(-<FONT face=symbol>p</FONT>/2)| as well<BR>as |f(<FONT face=symbol>p</FONT>/2)|=|f((<FONT face=symbol>p</FONT>/2)+<FONT face=symbol>p</FONT>), |f(-<FONT face=symbol>p</FONT>/2)|=|f(-(<FONT face=symbol>p</FONT>/2)-<FONT face=symbol>p</FONT>)|. <BR>&nbsp;&nbsp; Now, for n-vectors <B>A</B>, <B>B</B> we can always find <B>A</B>=(<B>B</B>-<B>C</B>) with coplanar <B>A</B>, <B>B</B>, <B>C</B> comprising the law of vector addition<BR>triangle and parallelogram. For the parallelogram so formed the angle between <B>A</B> and <B>B</B> is <FONT face=symbol>f</FONT>, that between <B>B</B> and <B>C</B> <FONT face=symbol>a</FONT>,<BR>that between <B>A</B> and <B>C</B> then will be <FONT face=symbol>a</FONT>+<FONT face=symbol>f</FONT> such that the angle between the image n-vector of <B>A</B> at the opposite side of the<BR>parallelogram and <B>C</B> is (<FONT face=symbol>p</FONT>-<FONT face=symbol>a</FONT>-<FONT face=symbol>f</FONT>)=<FONT face=symbol>b</FONT>. Thus in their n-vector triangle <B>A</B> is opposite <FONT face=symbol>a</FONT>, the angle opposite <B>B</B> is <FONT face=symbol>b</FONT>, and that<BR>opposite <B>C</B>, <FONT face=symbol>f</FONT>. With |<B>A</B>|=A and |<B>B</B>|=B we have then |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|<SUP>2</SUP>=A<SUP>2</SUP>B<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>f</FONT>); when <B>A</B>=(<B>B</B>-<B>C</B>), |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|<SUP>2</SUP>=|-<B>C</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|<SUP>2</SUP>=B<SUP>2</SUP>C<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>a</FONT>),<BR>while for <B>B</B>=(<B>A</B>+<B>C</B>), |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|<SUP>2</SUP>=|<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>C</B>|<SUP>2</SUP>=A<SUP>2</SUP>C<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>b</FONT>)=A<SUP>2</SUP>C<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>p</FONT>-<FONT face=symbol>a</FONT>-<FONT face=symbol>f</FONT>)= A<SUP>2</SUP>C<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>a</FONT>+<FONT face=symbol>f</FONT>). Equating terms therefore now yields<BR>[A<SUP>2</SUP>/f<SUP>2</SUP>(<FONT face=symbol>a</FONT>)]=[B<SUP>2</SUP>/f<SUP>2</SUP>(<FONT face=symbol>b</FONT>)]=[C<SUP>2</SUP>/f<SUP>2</SUP>(<FONT face=symbol>f</FONT>)]. <BR>&nbsp;&nbsp; In the n-space triangle the n-vector normal to <B>C</B> from the vertex at <FONT face=symbol>f</FONT> is such that (Bsin<FONT face=symbol>a</FONT>)=(Asin<FONT face=symbol>b</FONT>). Similarly for C.<BR>Thus [A/sin<FONT face=symbol>a</FONT>]=[B/sin<FONT face=symbol>b</FONT>]=[C/sin<FONT face=symbol>f</FONT>] with, since the n-vector triangle is a plane triangle, the clear inference is f(<FONT face=symbol>a</FONT>)=sin<FONT face=symbol>a</FONT>,<BR>f(<FONT face=symbol>b</FONT>)=sin<FONT face=symbol>b</FONT>, f(<FONT face=symbol>f</FONT>)=sin<FONT face=symbol>f</FONT>. <BR>&nbsp;&nbsp; However, noting from <B>R</B><SUB><FONT size=4>*</FONT></SUB><B>R</B>=<B>0</B> that <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B> is reduced to <B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B><SUB>n</SUB> or <B>A</B><SUB>n</SUB><SUB><FONT size=4>*</FONT></SUB><B>B</B> and that in the <B>A</B><B>B</B><B>C</B>-parallelogram/triangle<BR>these <SUB><FONT size=4>*</FONT></SUB>-products are actually the area of the parallelogram or its two triangles then we see |<B>A</B><SUB><FONT size=4>*</FONT></SUB><B>B</B>|=|<B>B</B><SUB><FONT size=4>*</FONT></SUB><B>C</B>|=|<B>C</B><SUB><FONT size=4>*</FONT></SUB><B>A</B>|, which<BR>implies A<SUP>2</SUP>B<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>f</FONT>)=B<SUP>2</SUP>C<SUP>2</SUP>f<SUP>2</SUP>(<FONT face=symbol>a</FONT>)=A<SUP>2</SUP>C<SUP>2</