4d Vector Product Derivation

Deriving the 4 Dimensional Cross Product of 2 Vectors  W. F. Esterhuyse October 2009 It is possible to derive the four dimensional cross product in the following way. We first define the operation "dimensional superposition" (symbol = &) as the operation to obtain a n dimensional space from n-k dimensional spaces (n > k > 1). We define the operation to respect both the (+ - + -) structure of the cross product and the signature of the (flat) n dimensional space. To obtain a 3D space from 2D spaces we need to superimpose three 2 dimensional planes at right angles to each other. A 2D plane is defined by three points. Choosing two planes and four vectors in them such that each pair of vectors are orthogonal, we form "&" of the two planes by superimposing two vectors from different planes such that the other two are orthogoal and the origin of both pairs coincide. In this way we can form a right handed or left handed 3D space by the appropriate labelling of the resulting three vectors spannning the 3D space. The three vectors can be normalised to produce an orthonormal basis. We use [i] as the i'th unit vector. Taking the 3D space as right handed we may conceivably choose two vectors u and v in the flat 3D space such that: uxv = & uxv|12 & uxv|13 & uxv|23

(1)

= & uxv|12not3 & uxv|13not2 & uxv|23not1 where the RS is defined as: RS = & [3] | u1 U2 | & [2] | u1 u3 | & [1] | u2 u3 | | v1 v2 | | v1 v3 | | v2 v3 | that is: RS = & [1]x[2] | u1 U2 | & [1]x[3] | u1 u3 | & [2]x[3] | u2 u3 | | v1 v2 | | v1 v3 | | v 2 v3 | obtained by taking the vector product in the 2D subspaces as in RS (1). We need the first "&" just

like we would if "&" was "-". We see that the above construction applies. Superimposing two 2-planes by leaving out the plane of [2] and [3] is not valid (if considering the 3D space spanned by surfaces) since then projection onto the 2-3-plane is impossible. Now, since we know how the 3D cross product is defined, we can work out what the "&" operation reduces to by comparing RS (1) with this definition. The RS must thus reduce to: RS = [3] u1 u2 - [2] u1 u3 + [1] u2 u3 v1 v2 v1 v3 v2 v3 .

(3)

Since (3) can be obained from: [1] [2] [3] u1 u2 u3 v1 v2 v3

(4)

the rule to take "&" to (+ or -) is development of (4) while respecting the + - + order of the determinant. We note that this is a flat 3D space with signature: (+++). Thus the construction (1) is proved valid under the stated "&"- rule. Taking this process further we let u and v be two vectors in a flat 4D space. Then by the above analog we have by constructing a 4D space from 3D spaces: uxv = & uxv|123 & uxv|124 & uxv|134 & uxv|234

(4)

with: RS = & [1] [2] [3] & [1] [2] [4] & [1] [3] [4] & [2] [3] [4] u1 u2 u3 u1 u2 u4 u1 u3 u4 u2 u3 u4 v1 v2 v3 v1 v2 v4 v1 v3 v4 v2 v3 v4

(5)

by the 3D vector product definition (applied in the subspaces in RS (4)). This is equivalent to a superposition of all 3x3 subdeterminants of: [1] [2] [3] [4] u1 u2 u3 u4 v1 v2 v3 v4

(6)

and the "&" operation must reflect development of (6) to obtain (5) with the correct signs. The process to determine the correct transitions from "&" to "+ or -" requires a bit of a leap but the results can be tested for fit into the properties of the cross product. The process

to determine this is by explicitly deleting the i'th column in (6), and shifting the remaining columns left such that all deleted columns are at rightmost while indices still increases to the right. Count the amount of shifts required (say this is s) and change "&" of the term into "+" if s is even and into "-" of s is odd. Therefore (6) reduces as: [1] [2] [3] [4] = & [1] [2] [3] [] & [1] [2] [] [4] & [1] [] [3] [4] & [] [2] [3] [4] (7) u1 u2 u3 u4 u1 u2 u3 [] u1 u2 [] u4 u1 [] u3 u4 [] u2 u3 u4 v1 v2 v3 v4 v1 v2 v3 [] v1 v2 [] v4 v1 [] v3 v4 [] v2 v3 v4 =+

...

-

...

+

...

-

...

where the ellipses are the corresponding 3x3 subdeterminant of the term above it, just after the shifts and with the deleted column left out. This is the correct process to determine what "&" changes into since in 5D we would have two deleted columns (then assigning a "+ or - " by just looking what column got deleted is invalid). This method is valid for evaluating any non-square determinant (by row) having rows longer than columns. This method is equivalent to first developing (6) by row 1 like an ordinary determinant to obtain unit vectors mutiplied by 2x3 determinants and then using the method to evaluate them. We state the change if the signature is not (++++) ( the first row signs is + - + - with this signature). Say the signature is (+++ -) then the first row signs changes to: + - + +, and this is implemented by developing LS (7) by row 1 untill the 2x3 stage (using + - + -) and flipping the sign of the term with [4] in it, or just developing LS (7) by row 1 using the altered pattern. The rule in general is that the determinant sign stays the same on a signature of "+" and flips on a signature of "-". The subspaces in (4) are all assumed right handed. In the case that the terms are a combination of right handed or left handed we would need another type of signature that does not work the same as the metric signature. For a signature of RRRL, (4) would change to: uxv = & uxv|123 & uxv|124 & uxv|134 & ( -uxv|234 )

(4)

(it seems reasonable by looking at term 1 of (7)). In (3) we may say that the superposition due to [1]x[3] is negative since right handedness says [1]x[3] points in the -[2] direction. If we considered

the [1]x[3] generated plane as left handed it would have gone to +[2]. Length Condition The length condition not holding does not disqualify this nD vector product as a candidate for the generalisation of the cross product since: any two dimensional hyperplane is defined by three points therefore by two vectors (if we chose one of the points at the origin of the two vectors), and we may define in 4 dimensions four 2-hyperplanes by having various restrictions on the vector product of u and v and having areas: |[(uxv)|1,2,3]|2 or |[(uxv)|2,3,4 ]|2 or |[(uxv)|1,3,4 ]|2 or |[(uxv)|1,2,4 ]|2 and we may even define a compound 2-plane in 4 dimensions as having area: star |& (uxv)|1,2,3 & (uxv)|2,3,4 & (uxv)|1,3,4 & (uxv)|1,2,4 |2 = star |uxv|2 = (d-2) [(u.u)(v.v) - (u.v)2] (5) where we specify by "&" the operation of "dimensional superposition". We use this definition since it holds in any dimension (d). The operator is not the Hodge star and this fomula is proved in another article of mine. By the formula for the angle between two vectors we have four "angles" between any two vectors in 4D: |[(uxv)|1,2,3]|2 / [(u.u)(v.v)|123] = (sin t)2

(6)

etc. It is clear from (5) and the derivation in 3D that a "compound angle" may exist as: star |uxv|2/[(d-2)(u.u)(v.v)] = (sin t)2 provided u and/or v is not zero. _______________________________ Email me for any comments/equiries: [email protected]