How to calculate the exponential of a 4D bivector?

Question

I know that rotors are the exponentials of bivectors. If B is simple, then $e^B = cos|B| + \hat{B}sin|B| $. All Bivectors in 3D are simple, but the same is not true in 4D. Is there a matching formula for the exponent of a general 4D bivector?

I wikipedia they write that if B1 and B2 are orthogonal then: $$R=e^{{{\frac {{\mathbf {B}}_{1}+{\mathbf {B}}_{2}}{2}}}}=e^{{{\frac {{\mathbf {B}}_{1}}{2}}}}e^{{{\frac {{\mathbf {B}}_{2}}{2}}}}=e^{{{\frac {{\mathbf {B}}_{2}}{2}}}}e^{{{\frac {{\mathbf {B}}_{1}}{2}}}}$$

And that it is possible to separate any Bivector to a sum of two simple orthogonal bivectors but I didn't understand how to do that. It also seems to me there should a simpler way of doing that.

Answer 1

There is a way to calculate the exponential of a 4D bivector, without splitting it into two orthogonal simple bivectors.

The idea is to split the algebra itself (at least the even part) into two "orthogonal" subalgebras.

Let $e_i$ denote basis vectors, as usual, and $e_{ij}$ basis bivectors, etc. The "left-handed subalgebra" is spanned by

$$1_L=\tfrac12(1-e_{1234}),\quad i_L=\tfrac12(e_{14}+e_{23}),\quad j_L=\tfrac12(e_{24}+e_{31}),\quad k_L=\tfrac12(e_{34}+e_{12}),$$

and the "right-handed subalgebra" is spanned by

$$1_R=\tfrac12(1+e_{1234}),\quad i_R=\tfrac12(e_{14}-e_{23}),\quad j_R=\tfrac12(e_{24}-e_{31}),\quad k_R=\tfrac12(e_{34}-e_{12}).$$

Note that these are both isomorphic to the quaternions. For example, $i_R^2=-1_R$, and $i_Rj_R=k_R$, but $j_Li_L=-i_Lj_L=k_L$ (though we could rename $k_L$ as $-k_L$, to get a proper isomorphism rather than an anti-isomorphism). Also note that the product of a left thing and a right thing is $0$.

So we can write a bivector as $B=B_L+B_R$, with the left part being $B_L=1_LB$ and the right part being $B_R=1_RB$. Then $B_LB_R=B_RB_L=0$, and the exponential is

$$\exp(B)=\exp(B_L)\exp(B_R)=\exp(B_R)\exp(B_L).$$

And we already know how to calculate the exponential of a 3D bivector, which $B_L$ or $B_R$ is isomorphic to. One thing to be careful about is that the multiplicative identity is not the same: $1\neq1_L\neq1_R$; instead $1=1_L+1_R$. Another thing is that the 3D norms are off by a factor of $\sqrt2$ from the 4D norm; we'll use the 3D norms:

$$B_L\!^2=-\lVert B_L\rVert^21_L,\qquad B_R\!^2=-\lVert B_R\rVert^21_R.$$

Thus, the exponentials are

$$\exp(B_L)=(1_L+1_R)+\frac{B_L}{1!}+\frac{B_L\!^2}{2!}+\frac{B_L\!^3}{3!}+\cdots$$ $$=1_R+\left(1_L+\frac{B_L\!^2}{2!}+\frac{B_L\!^4}{4!}+\cdots\right)+\left(\frac{B_L}{1!}+\frac{B_L\!^3}{3!}+\frac{B_L\!^5}{5!}+\cdots\right)$$ $$=1_R+1_L\cos\lVert B_L\rVert+\frac{B_L}{\lVert B_L\rVert}\sin\lVert B_L\rVert$$

and

$$\exp(B_R)=1_L+1_R\cos\lVert B_R\rVert+\frac{B_R}{\lVert B_R\rVert}\sin\lVert B_R\rVert$$

which combine to give us

$$\exp(B)=1_L\cos\lVert B_L\rVert+\frac{B_L}{\lVert B_L\rVert}\sin\lVert B_L\rVert+1_R\cos\lVert B_R\rVert+\frac{B_R}{\lVert B_R\rVert}\sin\lVert B_R\rVert.$$

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See also "isoclinic rotations".

Answer 2

Nicholas refered me to https://arxiv.org/abs/2107.03771. In section 6 it is explained how to to split any bivector into orthogonal blades. Let $B$ be a bivector in a geometric algebra with dimension 4, we aim to find two orthogonal bovectors $b_1, b_2$ such that $B = b_1 + b_2$.

It is shown there that:

$$b_1,_2=\frac {\lambda_i + \frac{1}{2} B \wedge B} {B}$$

where

$$\lambda_1,_2 = \frac{1}{2}B \cdot B \pm \frac{1}{2}\sqrt{(B \cdot B)^2 - (B \wedge B)^2}$$

I've used Wolfram Alpha to make sure that $\Delta = (B \cdot B)^2 - (B \wedge B)^2$ is never < 0 in a 4D euclidian space.

It is said in that paper that the only case the above is not defined, is when $\Delta=0$, because $B$ is not invertible in that case.

However, in this case $B^3 = 2(B \cdot B)B$ and the exponent can be calculated like of a simple bivector with $e^B = cos|B| + \hat{B}sin|B|$.

Once we have $b_1$ and $b_2$ we can calculate the exponent like this (wikipedia): $$ e^{b_1 + b_2} = e^{b_1} e^{b_2}$$