# How to decompose a bivector into a sum of _orthogonal_ blades?

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 $$w=v\cdot B$$.

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.