Can the Bianchi algebra $VI_0$ be represented as a 5D bivector algebra inducing a positive-definite metric on vectors?

by mr_e_man   Last Updated September 12, 2019 23:20 PM - source

See Bianchi algebras.

Find a vector $e$ and a trio of bivectors $X,Y,Z$, in a Clifford algebra over a 5-dimensional space, such that $$X\times Y=Y,\quad X\times Z=-Z,\quad Y\times Z=0,$$ and $\{e\times X,\;e\times Y,\;e\times Z\}$ span a 3-dimensional positive-definite subspace.

Bivector products can be defined in terms of Clifford's geometric product by

$$A\bullet B=\langle AB\rangle_0,\;A\times B=\langle AB\rangle_2,\;A\wedge B=\langle AB\rangle_4,$$

or in terms of vector products by

$$\begin{align} &(a\wedge b)\bullet(c\wedge d)=(b\cdot c)(a\cdot d)-(a\cdot c)(b\cdot d), \\ &(a\wedge b)\times(c\wedge d)=(b\cdot c)(a\wedge d)-(a\cdot c)(b\wedge d)-(b\cdot d)(a\wedge c)+(a\cdot d)(b\wedge c), \\ &(a\wedge b)\wedge(c\wedge d)=a\wedge b\wedge c\wedge d, \end{align}$$

$$a\bullet(b\wedge c)=a\times(b\wedge c)=(a\cdot b)c-(a\cdot c)b.$$

From this it follows that the bivector "triple product" is "associative":

$$(A\times B)\bullet C=A\bullet(B\times C).$$

Applying this to the Bianchi algebra equations, we get

$$X\bullet Y=X\bullet Z=Y\bullet Y=Y\bullet Z=Z\bullet Z=0,$$

and nothing for $X\bullet X$. Over a positive-definite vector space, this would imply $Y=Z=0$, which is not allowed; so the space must have indefinite signature.

If "5D" is replaced with "6D", there is a solution in signature $(4,2)$, that is, $++++--$ :





$$e\times X=e_2+e_4$$

$$e\times Y=e_1$$

$$e\times Z=-e_3.$$

This is related to a model of Thurston's solv geometry; see my post at higherspace. There's a hyperbolic plane (hyperboloid model) in the $e_1e_2e_5$ space, and another in the $e_3e_4e_6$ space.

I want to know whether there's a model in 5D. I've ruled out all signatures (including degenerate ones) except $+++--$ . I've also ruled out simple bivectors; $X\wedge X\neq0$, $Y\wedge Y\neq0$, $Z\wedge Z\neq0$.