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Unexpected output from Cross[ ]?

Posted 3 months ago
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I admit that

Cross[{0,1,0,0},{0,0,1,0},{0,0,0,1}]= - {1,0,0,0}       (1)

applied to four 4-dimensional basis vectors

e1={0,1,0,0}, e2={0,0,1,0}, e3={0,0,0,1}], and e4={1,0,0,0}

can successfully lead us to a calculation of the volume V

V = Cross[v1,v2,v3].v4      (*nner product of Cross[v1,v2,v3] with v4*)

of a parallelohedron made up of four 4-dimensional vectors

vi = {xi,yi,zi,wi} = e1 xi + e2 yi +e3 zi +e4 wi  (i=1,2,3,4) 

represented on the four 4-dimensional orthonormal unit basis vectors

e1={0,1,0,0}, e2={0,0,1,0}, e3={0,0,0,1}, and e4={1,0,0,0}

On the other hand, however, the most natural extension of a well known cross product

Cross[{1,0,0},{0,1,0}]={0,0,1} 

would be

Cross[{0,1,0,0},{0,0,1,0},{0,0,0,1}]= {1,0,0,0},  not  -{1,0,0,0}.     (2)

I will be happy if someone kindly explains me why Cross of 3 unit vector be given by (2).

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I don't know, but maybe because with this definition the four vectors are positively oriented:

In[8]:= v2 = {0, 1, 0, 0};
v3 = {0, 0, 1, 0};
v4 = {0, 0, 0, 1};
cross = Cross[v2, v3, v4];
Det[{v2, v3, v4, cross}]

Out[12]= 1

Thank you for your suggestive note regarding my query on Cross[e2, e3, e4]= -e1. (In this note, previous symbol "v" for unit vectors are changed to "e".)

While looking at your note, I came across a geometrical interpretation of why and how Cross[e2, e3, e4] be given by -e1, It goes as follows. (The following example concerns a 3-dimensional case.) Supppose one is interested in the evaluation of the volume V of a parallelopiped defined by 3 unit vectors e1={1,0,0}, e2={0,1,0}, and e3={0,0,1}.

    V = Det[(e1
e2


e3

)]= \[LeftBracketingBar] {
  {{
    {{
      {{
        {1},
        {0}
       }},
      {0}
     }, {
      {{
        {0},
        {1}
       }},
      {0}
     }}
   }, {
    {{
      {0},
      {0}
     }},
    {1}
   }}
 } \[RightBracketingBar]=1 

Then he likes to change the order of counting these vectors, starting with Overscript[e1, ^] (= e3), Overscripte2, ^, and then Overscript[e3, ^] (=Cross[Overscript[e1, ^],Overscript[e2, ^]] , so that the volume will be

Overscript[V, ^] = Det[(Overscript[e1, ^]
Overscript[e2, ^]


Overscript[e3, ^]

)]

However, since the parallelopipet itself has not changed, except for the order of counting the 3 ridges, its volume V remains unchanged, i.e. Overscript[V, ^]=V. Further, under the right-handed screw rule, Overscript[e3, ^](=Cross[Overscript[e1, ^],Overscript[e2, ^]]= ordinary cross product Overscript[e1, ^]*Overscript[e2, ^]) =-e1, so that we should define Cross[Overscript[e1, ^],Overscript[e2, ^]] as follows:

Cross[Overscript[e1, ^],Overscript[e2, ^]] = -e1.

This is in accord with the following manipulation of V:

V = Det[(e1
e2


e3

)]= \[LeftBracketingBar]1
0


0

    0
1


0



    0
0


1



\[RightBracketingBar]=\[LeftBracketingBar]0
0


-1

    0
1


0



    1
0


0



\[RightBracketingBar]= Det[(Overscript[e1, ^]
Overscript[e2, ^]


Overscript[e3, ^]

)]

It seems to me that the same logic hold in the 4-dimensional case, as verified in your note.

Thanks again.

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