# Is there a multivector that's a non-trivial cube root of $1$?

This answer (or this site, in case the answer gets deleted) defines a certain 3-dimensional Real algebra by declaring that $j^3={^-}1$, and that $1,j,j^2$ are linearly independent.

(By a simple change of variables $j'={^-}j$, this is isomorphic to an algebra with $j^3={^+}1$, which Reinko seems to have overlooked in his years of study on the subject.)

So I wondered if this system could be embedded in Geometric Algebra, as with Complex numbers, Perplex and Dual numbers, and Quaternions.

Does there exist a multivector $J$ such that $J^3=1$, and $J$ is linearly independent of $J^2$ and $J^3$?

Of course, there is the Complex number $\omega=-\frac12+\frac{\sqrt3}2i$, with $\omega^3=1$, but $\omega^2+\omega+1=0$, so these are linearly dependent.

Yes!

But trivectors are necessary; the multivector $J$ cannot be a sum of only scalars, vectors, and bivectors. Proof by contradiction (you can skip this section):

$$s=\langle s\rangle_0,\quad v=\langle v\rangle_1,\quad B=\langle B\rangle_2$$

$$J = s+v+B$$

Its square is

\begin{align}J^2 &= s^2+2sv+v^2+2sB+(vB+Bv)+(B^2) \\&= s^2+2sv+v^2+2sB+2(v\wedge B)+(B\cdot B+B\wedge B) \\&= \big(s^2+v^2+B\cdot B\big)+\big(2sv\big)+\big(2sB\big)+\big(2v\wedge B\big)+\big(B\wedge B\big) \end{align}

Those are, respectively, the grade 0, 1, 2, 3, 4 components of $J^2$.

And its cube (I'll skip the calculations) is

\begin{align}\langle J^3\rangle_0 &= s^3+3sv^2+3sB\cdot B \\ \langle J^3\rangle_1&= 3s^2v+v^3+v(B\cdot B)+2B\cdot(v\wedge B) \\ \langle J^3\rangle_2&= 3s^2B+2v\cdot(v\wedge B)+v^2B+B(B\cdot B)+B\cdot(B\wedge B) \\ \langle J^3\rangle_3&= 6sv\wedge B+v\cdot(B\wedge B)+2B\times(v\wedge B) \\ \langle J^3\rangle_4&= 3sB\wedge B \\ \langle J^3\rangle_5&=3v\wedge B\wedge B \\ \langle J^3\rangle_6&=B\wedge B\wedge B \end{align}

where $\times$ denotes an intermediate part of the geometric product (in this case, the grade 3 part). We want $J^3=1$ (scalar), so all higher-grade parts must be $0$. This implies $\langle J^3\rangle_4=3sB\wedge B=0$. If $s=0$, then the scalar part of $J^3$ is $0\neq1$, so that doesn't work. Divide by $s$ to get $B\wedge B=0$, which means that $B$ is a blade (only 2-dimensional), and that $B\times(v\wedge B)=0$. What remains of the grade 3 part must be $\langle J^3\rangle_3=6sv\wedge B=0$, which implies that $v$ is in the plane of $B$.

Look back to $J^2$, and the last two terms vanish.

\begin{align}J^2 &= \big(s^2+v^2+B\cdot B\big)+2s\big(v+B\big) \\&= \big({^-}s^2+v^2+B\cdot B\big)+2s\big(s+v+B\big) \\&= \big({^-}s^2+v^2+B\cdot B\big)1+(2s)J \end{align}

This shows that $1,J,J^2$ are linearly dependent, which contradicts our requirements. End of proof.

Here are the solutions in various 3D pseudo-Euclidean spaces. The $\sigma$'s and $\tau$'s are orthonormal basis vectors, $\sigma^2={^+}1$, and $\tau^2={^-}1$. (More solutions can be gotten by rotating and reflecting, but I believe these are unique up to orientation:)

$$\text{3+0D}:\quad J = \frac14+\frac34\sigma_3+\frac{\sqrt3}4\sigma_1\sigma_2-\frac{\sqrt3}4\sigma_1\sigma_2\sigma_3$$

$$\text{2+1D}:\quad J = \frac14+\frac{\sqrt3}4\tau_3+\frac{\sqrt3}4\sigma_1\sigma_2+\frac34\sigma_1\sigma_2\tau_3$$

$$\text{1+2D}:\quad J = \frac14+\frac{\sqrt3}4\tau_3+\frac34\sigma_1\tau_2-\frac{\sqrt3}4\sigma_1\tau_2\tau_3$$

$$\text{0+3D}:\quad J = \frac14+\frac{\sqrt3}4\tau_3+\frac{\sqrt3}4\tau_1\tau_2+\frac34\tau_1\tau_2\tau_3$$

I'll show the verification of the 2+1D case (which is the first one I found); the others are similar. Expanding the square from left to right,

$$J^2 = \Big(\frac14+\frac{\sqrt3}4\tau_3+\frac{\sqrt3}4\sigma_1\sigma_2+\frac34\sigma_1\sigma_2\tau_3\Big)^2$$

$$= \frac1{16}+\frac{\sqrt3}8\tau_3-\frac3{16}+\frac{\sqrt3}8\sigma_1\sigma_2+\frac38\sigma_1\sigma_2\tau_3-\frac3{16}+\frac38\sigma_1\sigma_2\tau_3-\frac{3\sqrt3}8\sigma_1\sigma_2-\frac{3\sqrt3}8\tau_3+\frac9{16}$$

$$= \frac14-\frac{\sqrt3}4\tau_3-\frac{\sqrt3}4\sigma_1\sigma_2+\frac34\sigma_1\sigma_2\tau_3$$

(So $J^2$ is equivalent to $J$ with reversed orientation of $\sigma_1$ and $\tau_3$.)

And finally, the cube,

$$J^3 = JJ^2$$

$$= \bigg(\Big(\frac14+\frac34\sigma_1\sigma_2\tau_3\Big)+\Big(\frac{\sqrt3}4\tau_3+\frac{\sqrt3}4\sigma_1\sigma_2\Big)\bigg)\bigg(\Big(\frac14+\frac34\sigma_1\sigma_2\tau_3\Big)-\Big(\frac{\sqrt3}4\tau_3+\frac{\sqrt3}4\sigma_1\sigma_2\Big)\bigg)$$

(The expression $(a+b)(a-b)=a^2-ab+ba-b^2$ only simplifies to a difference of squares when $ab=ba$. That's true in this case, because the trivector and scalar commute with everything.)

$$= \Big(\frac14+\frac34\sigma_1\sigma_2\tau_3\Big)^2-\Big(\frac{\sqrt3}4\tau_3+\frac{\sqrt3}4\sigma_1\sigma_2\Big)^2$$

$$= \Big(\frac1{16}+\frac38\sigma_1\sigma_2\tau_3+\frac9{16}\Big)-\Big(-\frac3{16}+\frac38\sigma_1\sigma_2\tau_3-\frac3{16}\Big)$$

$$= 1$$

And it's easy to see that they're linearly independent.