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.



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.