Hyper Rogue

[mr_e_man]

Yes.


It's more convenient here to use a non-orthogonal basis:

e_i • e_j = 0 for all i,j;
f_i • f_j = 0 for all i,j;
f_i • e_j = 1 for i=j, and 0 for i≠j.

(For an orthogonal basis, take e_i + ½f_i and e_i - ½f_i.)

Equivalently, the dot product is defined by either of these matrices, depending on whether the basis vectors are ordered as e₁, e₂, ..., e₆, f₁, f₂, ..., f₆, or as e₁, f₁, e₂, f₂, ..., e₆, f₆:

Code: Select all

⌈. . . . . . 1 . . . . .⌉
|. . . . . . . 1 . . . .|
|. . . . . . . . 1 . . .|
|. . . . . . . . . 1 . .|
|. . . . . . . . . . 1 .|
|. . . . . . . . . . . 1|
|1 . . . . . . . . . . .|
|. 1 . . . . . . . . . .|
|. . 1 . . . . . . . . .|
|. . . 1 . . . . . . . .|
|. . . . 1 . . . . . . .|
⌊. . . . . 1 . . . . . .⌋

or

⌈0 1 . . . . . . . . . .⌉
|1 0 . . . . . . . . . .|
|. . 0 1 . . . . . . . .|
|. . 1 0 . . . . . . . .|
|. . . . 0 1 . . . . . .|
|. . . . 1 0 . . . . . .|
|. . . . . . 0 1 . . . .|
|. . . . . . 1 0 . . . .|
|. . . . . . . . 0 1 . .|
|. . . . . . . . 1 0 . .|
|. . . . . . . . . . 0 1|
⌊. . . . . . . . . . 1 0⌋


Nil geometry corresponds to the Heisenberg algebra.
We cannot use the 3×3 matrices given there; we want to represent the group by isometries, not general linear transformations. Instead we can use these bivectors:

X = e₁∧f₆ + 2e₆∧f₂ + e₅∧f₄,
Y = e₂∧f₄ + 2e₄∧f₃ + e₆∧f₅,
Z = e₁∧f₅ + 2e₅∧f₃ + e₆∧f₄,

or equivalently these matrices:

Code: Select all

    ⌈. . . . . 1  . . . . . .⌉        ⌈. . . . . .  . . . . . .⌉        ⌈. . . . 1 .  . . . . . .⌉
    |. . . . . .  . . . . . .|        |. . . 1 . .  . . . . . .|        |. . . . . .  . . . . . .|
    |. . . . . .  . . . . . .|        |. . . . . .  . . . . . .|        |. . . . . .  . . . . . .|
    |. . . . . .  . . . . . .|        |. . 2 . . .  . . . . . .|        |. . . . . .  . . . . . .|
    |. . . 1 . .  . . . . . .|        |. . . . . .  . . . . . .|        |. . 2 . . .  . . . . . .|
X = |. 2 . . . .  . . . . . .| ,  Y = |. . . . 1 .  . . . . . .| ,  Z = |. . . 1 . .  . . . . . .|
    |. . . . . .  . . . . . .|        |. . . . . .  . . . . . .|        |. . . . . .  . . . . . .|
    |. . . . . .  . . . . .-2|        |. . . . . .  . . . . . .|        |. . . . . .  . . . . . .|
    |. . . . . .  . . . . . .|        |. . . . . .  . . .-2 . .|        |. . . . . .  . . . .-2 .|
    |. . . . . .  . . . .-1 .|        |. . . . . .  .-1 . . . .|        |. . . . . .  . . . . .-1|
    |. . . . . .  . . . . . .|        |. . . . . .  . . . . .-1|        |. . . . . . -1 . . . . .|
    ⌊. . . . . . -1 . . . . .⌋        ⌊. . . . . .  . . . . . .⌋        ⌊. . . . . .  . . . . . .⌋


The bivectors satisfy the defining relations of the Heisenberg algebra (where × is the commutator):

X×Y = -Y×X = Z, X×Z = 0, Y×Z = 0.

(To convert a bivector to a matrix: multiply it with each of the basis vectors, and put the resulting 12 vectors as the columns of the matrix.)

Our representation of the Nil geometry is exp(B/2) v exp(-B/2) where the bivector B varies in the algebra (so B is any linear combination of X,Y,Z), or equivalently exp(M) v where the matrix M varies in the algebra, and where v is a fixed vector. For concreteness, take v = ½(e₂ + e₃ - f₁ - f₂), that is,

Code: Select all

    ⌈  0  ⌉
    | 1/2 |
    | 1/2 |
    |  0  |
    |  0  |
v = |  0  |
    |-1/2 |
    |-1/2 |
    |  0  |
    |  0  |
    |  0  |
    ⌊  0  ⌋


The tangent space at B=0 is spanned by the three vectors

X×v = e₆ + ½f₆,
Y×v = e₄ + ½f₄,
Z×v = e₅ + ½f₅;

these are unit vectors, orthogonal to each other, so the geometry has a positive-definite metric.


I don't know if this is possible in less than 12 dimensions. It's certainly not possible in less than 6 dimensions (according to my calculations), if we want a positive-definite metric.

[mr_e_man]

In fact even this may be insufficient; I haven't incorporated the circle group O(ℝ²) mentioned here.