200 D.V. Millionschikov
Now we are going to consider examples of solvmanifolds that are not nilman-
ifolds.
- LetG 1 be a solvable Lie group of matrices
⎛
⎜
⎜
⎝
ekz 00 x
0e−kz 0 y
001 z
0001⎞
⎟
⎟
⎠, (11.19)
where ek+e−k=n∈N,k=0.
G 1 can be regarded as a semi-direct productG 1 =RR^2 , whereRacts
onR^2 (with coordinatesx, y)via
z→φ(z)=(
ekz 0
0e−kz)
.
A latticeΓ 1 inG 1 is generated by the following matrices:
⎛
⎜
⎜
⎝
ek 000
0e−k 00
0011
0001⎞
⎟
⎟
⎠,
⎛
⎜
⎜
⎝
100 u 1
010 v 1
001 0
000 1⎞
⎟
⎟
⎠,
⎛
⎜
⎜
⎝
100 u 2
010 v 2
001 0
000 1⎞
⎟
⎟
⎠,
where
∣
∣
∣
∣
u 1 v 1
u 2 v 2∣
∣
∣
∣=0.
The corresponding Lie algebrag 1 has the following basis:e 1 =⎛
⎜
⎜
⎝
k 000
0 −k 00
0001
0000⎞
⎟
⎟
⎠,e^2 =⎛
⎜
⎜
⎝
0001
0000
0000
0000
⎞
⎟
⎟
⎠,e^3 =⎛
⎜
⎜
⎝
0000
0001
0000
0000
⎞
⎟
⎟
⎠,
and the following structure relations:
[e 1 ,e 2 ]=ke 2 , [e 1 ,e 3 ]=−ke 3 , [e 2 ,e 3 ]=0.The left-invariant 1-forms
e^1 =dz, e^2 =e−kzdx, e^3 =ekzdy (11.20)are the dual basis toe 1 ,e 2 ,e 3 and
de^1 =0, de^2 =−ke−kzdz∧dx=−ke^1 ∧e^2 , de^3 =ke^1 ∧e^3. (11.21)As the solvable Lie groupGis simply connected the fundamental group
π 1 (G/Γ) is naturally isomorphic to the latticeΓ:π 1 (G/Γ)∼=Γ.
The Lie algebrag 1 ofG 1 considered earlier is an example of completely
solvable Lie algebra. A Lie algebragis called completely solvable if∀X∈g
operator ad(X) has only real eigenvalues.