Maths - 2D Rotations and Translations using multivectors


You may want to review vectors:

Representing 2D transforms

To represent isometry transforms in 2D we only need one scalar value to represent rotation and we need 2 scalar values (a 2D vector) to represent translation. A multivector based on 2D vectors therefore has enough values to represent a 2D isometries , for instance e12 might hold the rotation value and (e1,e2) hold the translation value, but does the algebra work out? Can we do useful operations on the multivector using such a notation? I have tried below but, as you can see, I cant find a way to do it. Also, on this page, I have tried to find a transformation based on 2D that is reversable - again, it does not work out.

We want to find which multivectors are reversible, that is we need to always find the inverse, so:

if ,

multivector a translates b into c

then we need to be able to find the inverse a-1 which translates c into b

one condition which meets this requirement is:

a a†=scalar

as explained here.

So using the multiplication table here and multiplying out the terms of a a† gives:


e = scalar = e * e + e1 * e1 + e2 * e2 + e12 * e12
e1 = 0 = e1 * e + e * e1 - e12 * e2 - e2 * e12
e2 = 0 = e2 * e + e12 * e1 + e * e2 + e1 * e12
e12 = 0 = e12 * e + e2 * e1 - e1 * e2 - e * e12

scalar = e2 + e12 + e22 + e122
0 = 2*e*e1 - 2*e2*e12
0 = 2*e*e2 + 2*e1*e12
0 = 0

rearrange equation 2 to give

e12 = e*e1/e2

substitute in equation 3:

e*e2 =-e1*e12

e*e2 =-e1*e*e1/e2

e2*e2 =-e1*e1

e1 = √(-e22)

e1 = i e2

so perhaps these equations don't have a scalar solution? or have I made an error?

Applying a multivector to a point

To calculate the rotated point from the original position of the point.

Pout=a * Pin * a -1


In some sources I have seen this transform shown as:
A(b) =Ab(A#)-1


a# = Main involution sumi(-1)ia<i>

Such transformations satisfy A(b) |A(c) = b |c and are known as Lorentz transforms, or as isometries of inner product- |
If A is even then A(b) = AbA-1 and we have A(bc)) = A(b)A(c) .
If A is a k-versor then A# = (-1)kA so that A(bc) = (-1)kA(b)A(c).



since we cant always invert a we will use a† instead of a -1 which gives:

Pout=a * Pin * a†

expanding out the terms gives:


simplifying the terms and combineing the 'in' terms gives:

out.e= ( a.e*a.e +a.e1*a.e1 +a.e2*a.e2 +a.e12*a.e12)*in.e +( 2*a.e*a.e1 -2*a.e2*a.e12)*in.e1 +( 2*a.e*a.e2 +2*a.e1*a.e12)*in.e2
out.e1= ( 2*a.e*a.e1 +2*a.e2*a.e12)*in.e +( a.e*a.e +a.e1*a.e1 -a.e2*a.e2 -a.e12*a.e12)*in.e1 +( 2*a.e*a.e12 +2*a.e1*a.e2)*in.e2
out.e2= ( 2*a.e*a.e2 -2*a.e1*a.e12)*in.e +( -2*a.e*a.e12 +2*a.e1*a.e2)*in.e1 +( a.e*a.e -a.e1*a.e1 +a.e2*a.e2 -a.e12*a.e12)*in.e2
out.e12= ( a.e*a.e -a.e1*a.e1 -a.e2*a.e2 +a.e12*a.e12)*in.e12

We want to represent a transform of a vector, in other words we want to transform (in.e1,in.e2) into (out.e1,out.e2)

So if we let in.e=0 and in.e12=0 we get:

out.e= ( 2*a.e*a.e1 -2*a.e2*a.e12)*in.e1 +( 2*a.e*a.e2 +2*a.e1*a.e12)*in.e2
out.e1= ( a.e*a.e +a.e1*a.e1 -a.e2*a.e2 -a.e12*a.e12)*in.e1 +( 2*a.e*a.e12 +2*a.e1*a.e2)*in.e2
out.e2= ( -2*a.e*a.e12 +2*a.e1*a.e2)*in.e1+( a.e*a.e -a.e1*a.e1 +a.e2*a.e2 -a.e12*a.e12)*in.e2

In order to represent a rotation in 2D we would need to get the equations into this form:

cos theta -sin theta
sin theta cos theta

so top left = bottom right,

a.e*a.e +a.e1*a.e1 -a.e2*a.e2 -a.e12*a.e12 = a.e*a.e -a.e1*a.e1 +a.e2*a.e2 -a.e12*a.e12

a.e1*a.e1 -a.e2*a.e2 = -a.e1*a.e1 +a.e2*a.e2

a.e1*a.e1 = a.e2*a.e2

a.e1 = a.e2 or -a.e2

also top right = -bottom left,

2*a.e*a.e12 +2*a.e1*a.e2 = -( -2*a.e*a.e12 +2*a.e1*a.e2)

a.e*a.e12 +a.e1*a.e2 = a.e*a.e12 - a.e1*a.e2

a.e1*a.e2 = - a.e1*a.e2

a.e1*a.e2 = 0

So this does not seem to work, as this is only true when a.e1 = a.e2 = 0

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