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In Geometric Algebra, any bivector $B\in\Lambda^2\mathbb R^n$ is a sum of blades: $$B = B_1 + B_2 + \cdots$$ $$= \vec v_1\wedge\vec w_1 + \vec v_2\wedge\vec w_2 + \cdots$$ 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: $$B_1 = \vec v_1\wedge\vec w_1 = \vec v_1\wedge\left(\vec w_1-\Big(\frac{\vec w_1\cdot\vec v_1}{\vec v_1\cdot\vec v_1}\Big)\vec v_1\right) = \vec v_1\wedge\vec w_1'$$ $$\vec v_1\cdot\vec w_1' = 0$$ (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 $\vec v\wedge\vec w=\vec v'\wedge\vec w'$, and $\vec v'\cdot\vec w'=0$.)

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 $\Lambda^2\mathbb R^4$,

"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 $n = 4$: $$B_1 = e_1\wedge e_2 = e_1e_2$$ $$B_2 = (e_1 + e_3)\wedge e_4 = e_1e_4 + e_3e_4$$ $$B = B_1 + B_2 = e_1e_2 + e_1e_4 + e_3e_4$$ $$B_1B_2 = -e_2e_4 + e_1e_2e_3e_4$$ $$B_1\cdot B_2 = 0 \neq B_1\times B_2 = -e_2e_4$$

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:

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

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

$$Q^2 = B\cdot B - B\wedge B$$

$$B^2 = Q\cdot Q - Q\wedge Q$$

$$B\times Q = 0$$

$$B\wedge Q = 0$$

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 $v\neq0$,

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

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

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

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

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

$$B_1=v_1^{-1}(v_1\cdot B),\quad B_2=v_2^{-1}(v_2\cdot B),\quad B_3=v_3^{-1}(v_3\cdot B),\quad\cdots$$

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


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

$$B_1 = \left(\frac{|B\cdot B|+\sqrt{|B\cdot B|^2-\lVert B\wedge B\rVert^2}+B\wedge B}{2\sqrt{|B\cdot B|^2-\lVert B\wedge B\rVert^2}}\right)B$$

$$B_2 = \left(\frac{-|B\cdot B|+\sqrt{|B\cdot B|^2-\lVert B\wedge B\rVert^2}-B\wedge B}{2\sqrt{|B\cdot B|^2-\lVert B\wedge B\rVert^2}}\right)B$$

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


Applying this to the example problem,

$$B = e_1e_2+e_1e_4+e_3e_4$$

$$B^2 = -3 + 2e_1e_2e_3e_4;\quad B\cdot B = -3,\quad B\wedge B = 2e_1e_2e_3e_4$$

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

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

$$B_1\wedge B_1 = 0 = B_2\wedge B_2$$

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

$$B_1\cdot B_2 = 0 = B_1\times B_2$$

$$B_1 + B_2 = B$$


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


...In fact the result is false for pseudo-Euclidean spaces in general. Take an orthogonal basis $\{\sigma_1,\sigma_2,\tau_1,\tau_2\}$ with $\sigma_1\!^2=\sigma_2\!^2={^+}1,\;\tau_1\!^2=\tau_2\!^2={^-}1$, and consider the bivector

$$J=\sigma_1\frac{\sigma_2+\tau_2}{2}+\tau_1\frac{\sigma_2-\tau_2}{2}\quad=\frac{\sigma_1+\tau_1}{2}\sigma_2+\frac{\sigma_1-\tau_1}{2}\tau_2$$

(The $1/2$ is only to normalize $J^4=1$.)

$$(\sigma_1\cdot J)\cdot J=\frac{-\tau_1}{2},\quad(\tau_1\cdot J)\cdot J=\frac{\sigma_1}{2}$$

$$(\sigma_2\cdot J)\cdot J=\frac{-\tau_2}{2},\quad(\tau_2\cdot J)\cdot J=\frac{\sigma_2}{2}$$

It's easy to see that $(J\,\cdot\,)^2$ has no eigenvectors, so $J$ is not orthogonally decomposable.

(If it were orthogonally decomposable as $J=v_1\wedge w_1+v_2\wedge w_2$, then $v_1$ would be an eigenvector, with eigenvalue $(v_1\wedge w_1)^2$.)

I suspect that the result is still true for Euclidean and Lorentzian spaces.

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