# Geometric algebra

1. Dec 10, 2003

### scariari

can anyone explain how commutators act on tri-vectors (in orthonormal conditions)?

on bi-vectors i know that it ends up to be a bivector again,
but with tri-vectors it vanishes if its lineraly dependent.
what about the case if its not linearly dependent,
does that mean it remains a tri-vector?

how does a vector transform under a transformation generated by exponentiation of a trivector ?

a transformation is a rotation or reflection,
but who can explain the exponentiation?

2. Dec 11, 2003

### arcnets

Hint:
$$\exp{(ix)}=\cos{(x)}+i\sin(x)$$

3. Dec 12, 2003

### scariari

so using exp(ix)=cos(x)+ isin(x), multiplying it by the vector, this will result in a rotation, correct?

i have found examples for bi-vectors, but how does this change with tri-vectors?

for a bivector:

I^2=-1
K^2=1

exp(Ix)=cos(x)+I sin(x)
exp(Kx)=cos(hx)+K sin(hx)

cos (hx)=0.5(exp(x)+exp(-x)
sin(hx) = 0.5(exp(x)-exp(-x)

v'=exp(Kx)vexp(-Kx)

abs(v)=sin(hx)/cos(hx)=tan(hx)

is this a lorentz transformation?

also, i read that complex numbers represent vectors as points with the transformation...?

4. Dec 12, 2003

### arcnets

Yes, R+iI <-> (R,I).

5. Dec 12, 2003

### arcnets

Ah, now I think I understand what you mean.
Of course, you can't usually define the exponentiation of a vector. But you can define the eponentiation of a linear operator (matrix).
Like this: Let A be a matrix, then
$$\exp{(A)}=\sum_{k=0}^\infty \frac{A^k}{k!}.$$
Let's say a transformation can be written in the form
$$x' = \exp{(A)} \cdot x.$$
Now, if A = 1 + Gt with some parameter t, then G is called the generator of this transformation. For rotations, t is the angle.

6. Dec 12, 2003

### scariari

if the commutator of a bi-vector [A,B] is found by AB-BA, is the commutator of a tri-vector then ABC-BCA-CAB?