It's more convenient here to use a non-orthogonal basis:
ei • ej = 0 for all i,j;
fi • fj = 0 for all i,j;
fi • ej = 1 for i=j, and 0 for i≠j.
(For an orthogonal basis, take ei + ½fi and ei - ½fi.)
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.