Choosing bases  4D
The bases are the vectors e1, e2, e3 and e4, 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 trivector by drawing the
matrix formed by putting the basis vectors sidebyside, 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 
e1y 
e2y 
e3y 
e4y 
e1z 
e2z 
e3z 
e4z 
e1w 
e2w 
e3w 
e4w 

=e1x* 
e2y 
e3y 
e4y 
e2z 
e3z 
e4z 
e2w 
e3w 
e4w 

e2x* 
e1y 
e3y 
e4y 
e1z 
e3z 
e4z 
e1w 
e3w 
e4w 

+e3x* 
e1y 
e2y 
e4y 
e1z 
e2z 
e4z 
e1w 
e2w 
e4w 

e4x* 
e1y 
e2y 
e3y 
e1z 
e2z 
e3z 
e1w 
e2w 
e3w 

To choose trivectors for 4D multivectors we start with the psudoscalar e1^e2^e3^e4 which represents the whole determinant. We then split it up, taking into account the sign, as above:
e1^e2^e3^e4 = e1 * e2^e3^e4  e2 * e1^e3^e4 + e3 * e1^e2^e4  e4 * e1^e2^e3
Where the sign is negative then we invert the order, which gives the basis trivectors as follows:
 e2^e3^e4 is the minor of e1
 e1^e3^e4 is the minor of e2
 e1^e2^e4 is the minor of e3
 e1^e2^e3 is the minor of e4
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:
 e2^e3^e4 is the minor of e1
 e4^e3^e1 is the minor of e2
 e1^e2^e4 is the minor of e3
 e3^e2^e1 is the minor of e4
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?
dual(e1) =  e2^e3^e4
dual(e2) = +e4^e3^e1
dual(e3) =  e1^e2^e4
dual(e4) =  e3^e2^e1
What about the other way round:
dual(e2^e3^e4) =  e1
dual(e4^e3^e1) =  e2
dual(e1^e2^e4) =  e3
dual(e3^e2^e1) = + e4
Almost symmetrical, taking the dual function converts between vectors and trivectors and inverts the sign (except for two cases). This also has the disadvantage that its not a superset of 3D multivectors, for example, it uses e321 instead of e123.
So this is the best I can do to get the order for the trivectors.
What about the bivectors? 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 trivectors as follows:
The minor of e1 
The minor of e2 
The minor of e3 
The minor of e4 
e2^e3^e4 
e4^e3^e1 
e1^e2^e4 
e3^e2^e1 
Now take the minor of e2:
which gives: e3^e4 
Now take the minor of e4:
which gives: e3^e1 
Now take the minor of e1:
which gives: e2^e4 
Now take the minor of e3:
which gives: e2^e1 
Or take the minor of e3:
which gives: e4^e2 
Or take the minor of e3:
which gives: e1^e4 
Or take the minor of e2:
which gives: e4^e1 
Or take the minor of e2:
which gives: e1^e3 
Or take the minor of e4:
which gives: e2^e3 
Or take the minor of e1:
which gives: e4^e3 
Or take the minor of e4:
which gives: e1^e2 
Or take the minor of e1:
which gives: e3^e2 




so to summarise:
minor of e1 followed by e2 gives: 
e3^e4 
minor of e1 followed by e3 gives: 
e4^e2 
minor of e1 followed by e4 gives: 
e2^e3 
minor of e2 followed by e4 gives: 
e3^e1 
minor of e2 followed by e3 gives: 
e1^e4 
minor of e2 followed by e1 gives: 
e4^e3 
minor of e3 followed by e1 gives: 
e2^e4 
minor of e3 followed by e2 gives: 
e4^e1 
minor of e3 followed by e4 gives: 
e1^e2 
minor of e4 followed by e3 gives: 
e2^e1 
minor of e4 followed by e2 gives: 
e1^e3 
minor of e4 followed by e1 gives: 
e3^e2 
replacing minor of x followed by y with x^y
e1^e2 
e3^e4 
dual(e1^e2) = e3^e4 
e1^e3 
e4^e2 
dual(e1^e3) = e4^e2 
e1^e4 
e2^e3 
dual(e1^e4) = e2^e3 
e2^e4 
e3^e1 
dual(e2^e4) = e3^e1 
e2^e3 
e1^e4 
dual(e2^e3) = e1^e4 
e2^e1 
e4^e3 
dual(e2^e1) = e4^e3 
e3^e1 
e2^e4 
dual(e3^e1) = e2^e4 
e3^e2 
e4^e1 
dual(e3^e2) = e4^e1 
e3^e4 
e1^e2 
dual(e3^e4) = e1^e2 
e4^e3 
e2^e1 
dual(e4^e3) = e2^e1 
e4^e2 
e1^e3 
dual(e4^e2) = e1^e3 
e4^e1 
e3^e2 
dual(e4^e1) = e3^e2 
There seem to be two options here, the ones with the green background or reverse the order to get the yellow background. I cant think of anything to choose between them. I therefore chose the green values because the indexes are in order (except one) and the dual function is symmetrical.
 e1^e2
 e1^e3
 e1^e4
 e2^e3
 e4^e2
 e3^e4
So the full multivector and its dual is:
A_{r} 
dual(A_{r}) = A_{r}^{*} = e_{1234} A_{r} 
1 
e_{1234} 
e_{1} 
e_{1234}e_{1}= e_{234} 
e_{2} 
e_{1234}e_{2 }= e_{431} 
e_{3} 
e_{1234}e_{3 }= e_{124} 
e_{4} 
e_{1234}e_{4} = e_{321} 
e_{12} 
e_{1234}e_{12}= e_{34} 
e_{13} 
e_{1234}e_{13}= e_{42} 
e_{14} 
e_{1234}e_{14}= e_{23} 
e_{23} 
e_{1234}e_{23}= e_{14} 
e_{42} 
e_{1234}e_{42}= e_{13} 
e_{34} 
e_{1234}e_{34}=e_{12} 
e_{321} 
e_{1234}e_{321}= e_{4} 
e_{124} 
e_{1234}e_{124}= e_{3} 
e_{431} 
e_{1234}e_{431}= e_{2} 
e_{234} 
e_{1234}e_{234}= e_{1} 
e_{1234} 
e_{1234}e_{1234}= 1 _{} 
This site may have errors. Don't use for critical systems.