Maths - choosing bases - 5D

The bases are the vectors e1, e2, e3, e4 and e5, however the higher level products such as: e1^e2, e3^e1, e2^e3, e1^e4, e4^e2 and e3^e4 is also as a basis for the bivectors.

The question is what order do we choose for this (which is equivalent to saying what sign do we use since e1^e2=-e2^e1).

The methodology used for choosing the order of these indexes is explained here.

We calculate the psuedoscalar by drawing the matrix formed by putting the basis vectors side-by-side, then taking its minor by removing the row associated with its own coordinate type and removing the column of the basis vector not associated with.

We have to be very careful with signs as the sign alternates with terms as follows:

e1x e2x e3x e4x e5x
e1y e2y e3y e4y e5y
e1z e2z e3z e4z e5z
e1w e2w e3w e4w e5w
e1v e2v e3v e4v e5v
=e1x*
e2y e3y e4y e5y
e2z e3z e4z e5z
e2w e3w e4w e5w
e2v e3v e4v e5v
-e2x*
e1y e3y e4y e5y
e1z e3z e4z e5z
e1w e3w e4w e5w
e1v e3v e4v e5v
+e3x*
e1y e2y e4y e5y
e1z e2z e4z e5z
e1w e2w e4w e5w
e1v e2v e4v e5v
-e4x*
e1y e2y e3y e5y
e1z e2z e3z e5z
e1w e2w e3w e5w
e1v e2v e3v e5v
+e5x*
e1y e2y e3y e4y
e1z e2z e3z e4z
e1w e2w e3w e4w
e1v e2v e3v e4v

To choose tri-vectors for 5D multivectors we start with the psudoscalar e1^e2^e3^e4^e5 which represents the whole determinant. We then split it up, taking into account the sign, as above:

e1^e2^e3^e4^e5 = e1 * e2^e3^e4^e5 - e2 * e1^e3^e4^e5 + e3 * e1^e2^e4^e5 - e4 * e1^e2^e3^e5 + e5 * e1^e2^e3^e4

Where the sign is negative then we invert the order, which gives the basis tri-vectors as follows:

In order to remove the minus terms from the 2nd and 4th terms we can just swap any two of the terms, but which two do we swap?

how about:

I'm not an expert at suduko, so I cant get one number in each row, but the above seems to have a patern. Can anyone tell me a better way to choose the order?

What are the properties of these under the dual function?

So this is the best I can do to get the order for the quad-vectors.

What about the bivectors and tri-vectors? we can reduce the deteminant in two stages, but how does that correspond to the clifford algebra functions?

I think I will try finding the minors of the above tri-vectors as follows:

The minor of e1

The minor of e2

The minor of e3

The minor of e4 The minor of e5
e2^e3^e4^e5 e1^e3^e4^e5 e1^e2^e4^e5 e1^e2^e3^e5 e1^e2^e3^e4
 

Now take the minor of e1:

which gives: e3^e4^e5

Now take the minor of e1:

which gives: -e2^e4^e5

Now take the minor of e1:

which gives: e2^e3^e5

Now take the minor of e1:

which gives: -e2^e3^e4

Now take the minor of e2:

which gives: e3^e4^e5

 

Or take the minor of e2:

which gives: e1^e4^e5

Or take the minor of e2:

which gives: -e1^e3^e5

Or take the minor of e2:

which gives: e1^e3^e4

Or take the minor of e3:

which gives: -e2^e4^e5

Or take the minor of e3:

which gives: e1^e4^e5

 

Or take the minor of e3:

which gives: e1^e2^e5

Or take the minor of e3:

which gives: -e1^e2^e4

Or take the minor of e4:

which gives: e2^e3^e5

Or take the minor of e4:

which gives: -e1^e3^e5

Or take the minor of e4:

which gives: e1^e2^e5

 

Or take the minor of e4:

which gives: e1^e2^e3

Or take the minor of e5:

which gives: -e2^e3^e4

Or take the minor of e5:

which gives: e1^e3^e4

Or take the minor of e4:

which gives: -e1^e2^e4

Or take the minor of e5:

which gives: e1^e2^e3

 

so to summarise:

minor of e1 followed by e2 gives: e1^e2 e3^e4^e5 (e1^e2)(e3^e4^e5)
minor of e1 followed by e3 gives: e1^e3 -e2^e4^e5 -(e1^e3)(e2^e4^e5)
minor of e1 followed by e4 gives: e1^e4 e2^e3^e5 (e1^e4)(e2^e3^e5)
minor of e1 followed by e5 gives: e1^e5 -e2^e3^e4 -(e1^e5)(e2^e3^e4)
minor of e2 followed by e1 gives: e2^e1 -e3^e4^e5 -(e2^e1)(e3^e4^e5)
minor of e2 followed by e3 gives: e2^e3 e1^e4^e5 (e2^e3)(e1^e4^e5)
minor of e2 followed by e4 gives: e2^e4 -e1^e3^e5 -(e2^e4)(e1^e3^e5)
minor of e2 followed by e5 gives: e2^e5 e1^e3^e4 (e2^e5)(e1^e3^e4)
minor of e3 followed by e1 gives: e3^e1 e2^e4^e5 (e3^e1)(e2^e4^e5)
minor of e3 followed by e2 gives: e3^e2 -e1^e4^e5 -(e3^e2)(e1^e4^e5)
minor of e3 followed by e4 gives: e3^e4 e1^e2^e5 (e3^e4)(e1^e2^e5)
minor of e3 followed by e5 gives: e3^e5 -e1^e2^e4 -(e3^e5)(e1^e2^e4)
minor of e4 followed by e1 gives: e4^e1 -e2^e3^e5 -(e4^e1)(e2^e3^e5)
minor of e4 followed by e2 gives: e4^e2 e1^e3^e5 (e4^e2)(e1^e3^e5)
minor of e4 followed by e3 gives: e4^e3 -e1^e2^e5 -(e4^e3)(e1^e2^e5)
minor of e4 followed by e5 gives: e4^e5 e1^e2^e3 (e4^e5)(e1^e2^e3)
minor of e5 followed by e1 gives: e5^e1 e2^e3^e4 (e5^e1)(e2^e3^e4)
minor of e5 followed by e2 gives: e5^e2 -e1^e3^e4 -(e5^e2)(e1^e3^e4)
minor of e5 followed by e3 gives: e5^e3 e1^e2^e4 (e5^e3)(e1^e2^e4)
minor of e5 followed by e4 gives: e5^e4 -e1^e2^e3 -(e5^e4)(e1^e2^e3)

So the full multivector and its dual is:

Ar
dual(Ar) = Ar* = e12345 Ar
1 e12345
e1 e12345e1= e2345
e2 e12345e2 = e3145
e3 e12345e3 = e1245
e4 e12345e4 = e1253
e5 e12345e5 = e1234
e12 e12345e12= -e345
e13 e12345e13= -e425
e14 e12345e14= -e235
e15 e12345e15= -e324
e23 e12345e23= -e145
e24 e12345e24= -e315
e25 e12345e25= -e134
e34 e12345e34= -e125
e35 e12345e35= -e214
e45 e12345e45=-e123
e123 e12345e123= -e45
e214 e12345e214= -e35
e125 e12345e125= -e34
e134 e12345e134= -e25
e315 e12345e315= -e24
e145 e12345e145= -e23
e324 e12345e324= -e15
e235 e12345e235= -e14
e425 e12345e425= -e13
e345 e12345e345=-e12
e1234 e12345e1234= e5
e1253 e12345e1253= e4
e1245 e12345e1245= e3
e3145 e12345e3145= e2
e2345 e12345e2345= e1
e12345 e12345e12345= 1

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