Htykuut

In Geometric Algebra, any bivector $BinLambda^2mathbb R^n$ is a sum of blades:
$$B = B_1 + B_2 + cdots$$
$$= vec v_1wedgevec w_1 + vec v_2wedgevec w_2 + cdots$$
Each blade's component vectors $vec v$ and $vec w$, if they're not already orthogonal to each other, can easily be made so by the Gram-Schmidt process:
$$B_1 = vec v_1wedgevec w_1 = vec v_1wedgeleft(vec w_1-Big(frac{vec w_1cdotvec v_1}{vec v_1cdotvec v_1}Big)vec v_1right) = vec v_1wedgevec w_1'$$
$$vec v_1cdotvec w_1' = 0$$
(This can even be generalized to pseudo-Euclidean space where $vec v$ may square to zero: project $vec v$ away from $vec w$ instead of vice-versa, or if they both square to zero, take $vec v'=frac{vec v+vec w}{sqrt2}$ , $vec w'=frac{vec w-vec v}{sqrt2}$. Then $vec vwedgevec w=vec v'wedgevec w'$, and $vec v'cdotvec w'=0$.)

But I don't know how to make each blade orthogonal to the other blades. Orthogonal means that their geometric product is their (grade 4) wedge product; all lower-grade parts are zero.
$$B_1 + B_2 = B_1' + B_2'$$
$$B_1'B_2' = (B_1'cdot B_2')+(B_1'times B_2')+(B_1'wedge B_2') = B_1'wedge B_2'$$
$$B_1'cdot B_2' = 0 = B_1'times B_2'$$

From Wikipedia: In $Lambda^2mathbb R^4$,

"every bivector can be written as the sum of two simple bivectors. It is useful to choose two orthogonal bivectors for this, and this is always possible to do."

Here's a simple example, with $n = 4$:
$$B_1 = e_1wedge e_2 = e_1e_2$$
$$B_2 = (e_1 + e_3)wedge e_4 = e_1e_4 + e_3e_4$$
$$B = B_1 + B_2 = e_1e_2 + e_1e_4 + e_3e_4$$
$$B_1B_2 = -e_2e_4 + e_1e_2e_3e_4$$
$$B_1cdot B_2 = 0 neq B_1times B_2 = -e_2e_4$$

How can I rewrite $B = B_1' + B_2'$ with $B_1'cdot B_2' = 0 = B_1'times B_2'$ ?

EDIT1

After doing some algebra, I arrived at these equations:

$$B_1' = frac{B+Q}{2}$$

$$B_2' = frac{B-Q}{2}$$

$$Q^2 = Bcdot B - Bwedge B$$

$$B^2 = Qcdot Q - Qwedge Q$$

$$Btimes Q = 0$$

$$Bwedge Q = 0$$

We only need to solve for $Q$ in terms of $B$. I was able to take a square root of the third equation (by guessing that $Q = xe_1e_2+ye_3e_4$) but I didn't find the specific root that satisfies the other equations.

EDIT2

After doing some more algebra, I find that, if $Q$ is defined as the reflection of $B$ along some unknown vector $vneq0$,

$$Q=v^{-1}Bv$$

$$B_1=frac{v^{-1}vB+v^{-1}Bv}{2}=v^{-1}(vwedge B)$$

$$B_2=frac{v^{-1}vB-v^{-1}Bv}{2}=v^{-1}(vcdot B)$$

then $B_1$ and $B_2$ are blades, and $B_1cdot B_2=0$ regardless of $v$, and $B_1times B_2=0$ if and only if $vwedgebig((vcdot B)cdot Bbig)=0$. This means that $(vcdot B)cdot B$ must be parallel to $v$; in other words, $v$ is an eigenvector of the operator $(B,cdot)^2$. It follows that $vcdot B=w$ is also an eigenvector with the same eigenvalue, and $vcdot w=0$.

Generalizing, it looks like we want to find an orthogonal set of eigenvectors $v_1,v_2,v_3,cdots$ of $(B,cdot)^2$, so that

$$B_1=v_1^{-1}(v_1cdot B),quad B_2=v_2^{-1}(v_2cdot B),quad B_3=v_3^{-1}(v_3cdot B),quadcdots$$

Of course, all vectors $v$ must also be orthogonal to all $w=vcdot B$.

By using one of Doran's methods involving the exponential function (link, section 2.1.1), I found a formula for bivectors in 4D:

$$B_1 = left(frac{|Bcdot B|+sqrt{|Bcdot B|^2-lVert Bwedge BrVert^2}+Bwedge B}{2sqrt{|Bcdot B|^2-lVert Bwedge BrVert^2}}right)B$$

$$B_2 = left(frac{-|Bcdot B|+sqrt{|Bcdot B|^2-lVert Bwedge BrVert^2}-Bwedge B}{2sqrt{|Bcdot B|^2-lVert Bwedge BrVert^2}}right)B$$

(Obviously, this is undefined if $|Bcdot B|=lVert Bwedge BrVert$. That corresponds to an isoclinic rotation, where the two planes of rotation are not unique.)

Applying this to the example problem,

$$B = e_1e_2+e_1e_4+e_3e_4$$

$$B^2 = -3 + 2e_1e_2e_3e_4;quad Bcdot B = -3,quad Bwedge B = 2e_1e_2e_3e_4$$

$$B_1 = frac{(1+sqrt5)(e_1e_2+e_3e_4)+(3+sqrt5)e_1e_4-2e_2e_3}{2sqrt5}$$

$$B_2 = frac{(-1+sqrt5)(e_1e_2+e_3e_4)+(-3+sqrt5)e_1e_4+2e_2e_3}{2sqrt5}$$

$$B_1wedge B_1 = 0 = B_2wedge B_2$$

(The vanishing wedge product means that they are actually blades, though we don't know what vectors they're made of.)

$$B_1cdot B_2 = 0 = B_1times B_2$$

$$B_1 + B_2 = B$$

This is only a partial answer; it doesn't work when $|Bcdot B| = lVert Bwedge BrVert$, and I still don't know what to do in higher dimensions.