In [5] we have determined the flat torsion-free connections on the 2-dimensional Lie algebras which are compatible with a symplectic form and obtained their equiv-alence classes.. We sho
Trang 1Induced geometry onR4n
Nguyen Viet Hai(a)
Abstract The present paper is a continuation of Nguyen Viet Hai's ones [4], [5] In this the author give a method to construct hypersymplectic structures onR4n from affine-symplectic data onR2n
1 Preliminaries
A hypersymplectic structure on a 4n-dimensional manifold M is given by (J, E, δ) whereJ,E are endomorphisms of the tangent bundle ofM such that
J2 = −1, E2 = 1, J E = −EJ,
δis a neutral metric (that is, of signature(2n, 2n)) satisfying
δ(X, Y ) = δ(J X, J Y ) = −δ(EX, EY ) for all vector fieldsX, Y onM and the following associated2-forms are closed
ω1(X, Y ) = δ(J X, Y ), ω2(X, Y ) = δ(EX, Y ), ω3(X, Y ) = δ(J EX, Y )
In [5] we have determined the flat torsion-free connections on the 2-dimensional Lie algebras which are compatible with a symplectic form and obtained their equiv-alence classes We showed all the flat torsion-free connections that preserve a sym-plectic form on the 2-dimensional Lie algebras, namely, onR2and on aff(R) Those importante results used in the 4-dimensional case In [4] we presented a method
to contruct four-dimensional Lie algebras carrying a hypersymplectic structure from two 2-dimensional Lie algebras equipped with compatible flat torsion-free connec-tions and symplectic forms Using this method we obtained the classification, up
to equivalence, of all left-invariant hypersymplectic structures on 4-dimensional Lie groups All those Lie groups are exponential type
The purpose of this paper is to give a procedure to construct hypersymplectic structures on R4nwith complete and not necessarily flat associated neutral metrics
1 NhËn bµi ngµy 24/11/2006 Söa ch÷a xong ngµy 14/12/2006.
Trang 2The idea behind the construction will be to consider the canonical flat hypersymplec-tic structure onR4nand then translate it by using an appropriate group acting simply and transitively on R4n This group will be a double Lie group (R4n, R2n×{0}, {0}×
R2n)constructed from affine data on R2n
The paper is organized as follows In§2 we give to R4na structure of a nilpotent Lie group Starting with a fixed symplectic structureωon R2nwhich is parallel with respect to a pair of affine structures we form the associated double Lie group
(R4n, R2n× {0}, {0} × R2n) and show that it is at most 3-step nilpotent In§3 we consider canonical symplectic structures on R4n, constructed from the givenωon R2n We analyze in§4 the geome-try of the homogeneous metric obtained by using the double Lie group structure given
to R4n to translate the standard inner product of signature(2n, 2n)on R2n⊕ R2n The resulting metric is hypersymplectic, complete and not necessarily flat
2 Three-step nilpotent group structure on R4n
We shall begin by recalling some definitions which will be used throughout this article
An affine structure (or a left symmetric algebra structure) on Rnis given by a connection∇, that is, a bilinear map ∇ : Rn× RRn → Rnsatisfying the following conditions
for allx, y ∈ Rn Ifωis a non-degenerate skew-symmetric bilinear form on Rn, the affine structure∇is compatible withωif
ω(∇xy, z) = ω(∇xz, y), x, y, z ∈ Rn (3)
We notice that affine structures∇on R2ncompatible withωsatisfy a condition stronger than (2), namely,
Trang 3The last equation follows from
ω(∇x∇yz, w) = ω(∇wx, ∇zy) = −ω(∇z∇wx, y) =
= −ω(∇w∇zx, y) = −ω(∇yw, ∇xz) = −ω(∇x∇yz, w) Let∇and ∇0 be two affine connections on R2n compatible with ω and assume furthermore that∇and∇0 satisfy the following compatibility condition
∇x∇0y = ∇y∇0x for allx, y ∈ R2n (5) From (5) and the compatibility of the connections withω, we obtain the following
∇0x∇y = ∇0y∇x, for allx, y ∈ R2n (6)
∇x∇0y = −∇0y∇x, for allx, y ∈ R2n (7) Indeed, (6) follows from
ω(∇0x∇yz, w) = −ω(∇yz, ∇0xw) = −ω(∇y∇0xw, z) =
= −ω(∇x∇0yw, z) = −ω(∇xz, ∇0yw) = ω(∇0y∇xz, w), and (7) follows from
ω(∇x∇0yz, w) = ω(∇y∇0xz, w) = ω(∇wy, ∇0zx) = −ω(∇0z∇wy, x) =
= −ω(∇0w∇zy, x) = −ω(∇0xw, ∇yz) = −ω(∇0x∇yz, w) = −ω(∇0y∇xz, w)
In the next theorem we shall show that two affine structures∇and∇0 on R2n satisfying (5) and (6) give rise to a Lie group structure on the manifold R4nsuch that (R4n, R2n× {0}, {0} × R2n) is a double Lie group We recall that adouble Lie group
is given by a triple(G, G+, G−)of Lie groups such thatG+, G−are Lie subgroups of
G and the product G+ × G− → G, (g+, g−) 7→ g+g− is a diffeomorphism (see [8]) The next result shows that the additional condition (3) of ∇ and∇0 with a fixedω imposes restrictions on the Lie group obtained
Theorem 2.1 Let ∇ and ∇0be two affine structures on R2ncompatible with a symplectic form ω and satisfying also (5) Then R2n
× R2n with the product given by (x, x0) · (y, y0) = (x + α(x0, y), β(x0, y) + y0) (8)
Trang 4where x, x0, y, y0 ∈ R2n and
α(x0, y) = y + ∇0yx0− 1
2∇0y∇yx0, β(x0, y) = x0− ∇x 0y − 1
2∇x 0∇0x0y (9)
is a 3-step nilpotent double Lie group Furthermore, the associated Lie bracket on its Lie algebra R2n⊕ R2n is
[(x, x0), (y, y0)] = (∇0yx0− ∇0
xy0, ∇xy0− ∇yx0) (10) Chøng minh Let us set R2n
+ := R2n× {0}and R2n
− := {0} × R2n We denote α(x0, y) =
αx0(y)and β(x0, y) = βy(x0)for x0
∈ R2n
−, y ∈ R2n
+ The maps α : R2n
− × R2n + −→ R2n
+ and β : R2n
− × R2n + −→ R2n
− satisfy the conditions
α0 = α(0, R2n+) = 1, αx0(0) = 0, αx0 +y 0 = αx0◦ αy0 (11)
β0 = β(0, R2n−) = 1, βy(0) = 0, βx+y = βy ◦ βx (12) for all x0, y0 ∈ R2n
−, x, y ∈ R2n
+, where we denote αx 0(y) := α(x0, y)and βy(x0) := β(x0, y) for x0
∈ R2n
−, y ∈ R2n
+ The above relations show that α is a left action of R2n
−
on R2n
+ and β is a right action of R2n
+ on R2n
− These maps satisfy also the following compatibility conditions
αx0(x + y) = αx0x + αβxx0y, βx(x0+ y0) = βαy0xx0 + βxy0
According to [8], the product given in (8) defines a Lie group structure on R4n such that (R4n, R2n
+, R2n
−) is a double Lie group Note that the neutral element of this group structure is (0, 0) and the inverse of (x, x0
) ∈ R4n is (α(−x0, −x), β(−x0, −x)) Let us determine now the associated Lie algebra Linearizing the above actions, we obtain representations µ : R2n
− −→ End(R2n
+)and ρ : R2n
+ −→ End(R2n
−)given by
µx0(y) = d
dt
0
(dαtx0)0(y), ρy(x0) = d
dt
0
(dβty)0(x0)
For α and β given in (9), we obtain that
(dαx 0)0 = 1 + ∇0x0, (dβy)0 = 1 − ∇y Hence,
µx0(y) = ∇0x0y, ρy(x0) = −∇yx0,
Trang 5showing that the bracket of its Lie algebra is the one given in (10).
If ad(x,x 0 ), x, x0 ∈ R2n stands for the transformation given by (10), then, using that
∇and ∇0 are torsion-free (see (1)) one has
ad(x,x0 ) =
∇0
x
−∇x0 ∇x
From (4) applied to both ∇ and ∇0, we obtain
ad2(x,x0 )=
∇0
x∇x
−∇x0∇0
x 0 ∇x0∇0
x
Finally, using (7),
ad3(x,x0 ) = 0
Hence R2n
⊕ R2nis a 3-step nilpotent Lie algebra (i.e ad3
= 0) or R2n
× R2n is a 3-step nilpotent Lie group, as claimed
We set the notation to be used in what follows Since the construction of the Lie group structure on R4n in Theorem 2.1 depends on the affine structures ∇and∇0,
we will denote this Lie group byH∇∇0 The corresponding Lie algebra will be denoted
h∇∇0 and the abelian Lie subalgebrasR2n ⊕ {0}, {0} ⊕ R2n will be denoted h+, h−, respectively We note that(h∇∇0, h+, h−)is adouble Lie algebra, that is,h+andh−are Lie subalgebras ofh∇∇0 andh∇,∇ 0 = h+⊕ h−as vector spaces
3 Invariant symplectic structures on R4n
In this section we show thath∇∇0 carries three symplectic structures, obtained from the symplectic formω inR2n compatible with∇and∇0 These forms, defined at the Lie algebra level, give rise to left-invariant symplectic forms on the corresponding
Trang 6Lie grouph∇∇0 Hence,R4ninherits symplectic structures which are invariant by this nilpotent group First, we recall that a symplectic structure on a Lie algebrag is a non-degenerate skew-symmetric bilinear formωsatisfying dω = 0, where
d ω(x, y, z) = ω(x, [y, z]) + ω(y, [z, x]) + ω(z, [x, y]) (13) forx, y, z ∈ g A given symplectic formωonR2nallows us to define the following non degenerate skew-symmetric bilinear forms onR2n⊕ R2n:
ω1((x, x0), (y, y0)) := ω(x, y) + ω(x0, y0),
ω2((x, x0), (y, y0)) := −ω(x, y0) + ω(y, x0),
ω3((x, x0), (y, y0)) := ω(x, y) − ω(x0, y0)
(14)
We show below that the above forms are closed with respect to the Lie bracket given in Theorem 2.1 Therefore, they define symplectic structures onh∇∇0
Proposition 3.1 The 2-forms ω1, ω2 and ω3 are closed on h∇
∇ 0 Chøng minh Since R2n⊕ {0}and {0} ⊕ R2n are abelian subalgebras of h∇
∇ 0, it follows that the forms ωi, i = 1, 2, 3, are closed if and only if
(dωi)((x, 0), (y, 0), (0, z0)) = (dωi)((0, x0), (0, y0), (z, 0)) = 0
for all x, y, z, x0, y0, z0 ∈ R2n But
(dωi)((x, 0), (y, 0), (0, z0)) = ωi([(y, 0), (0, z0)], (x, 0)) + ωi([(0, z0), (x, 0)], (y, 0))
= ωi((−∇0yz0, ∇yz0), (x, 0)) + ωi((∇0xz0, −∇xz0), (y, 0)) and
(dωi)((0, x0), (0, y0), (z, 0)) = ωi([(0, y0), (z, 0)], (0, x0)) + ωi([(z, 0), (0, x0)], (0, y0))
= ωi((∇0zy0, −∇zy0), (0, x0)) + ωi((−∇0zx0, ∇zx0), (0, y0)) Using the expressions of ωi, i = 1, 2, 3given in (14), we compute
(dω1)((x, 0), (y, 0), (0, z0)) = (dω3)((x, 0), (y, 0), (0, z0)) =
= −ω(−∇0yz0, x) + ω(∇0xz0, y),
(dω1)((0, x0), (0, y0), (z, 0)) = −(dω3)((0, x0), (0, y0), (z, 0)) =
= −ω(∇zy0, x0) + ω(∇zx0, y0)
Trang 7and (
(dω2)((x, 0), (y, 0), (0, z0)) = ω(x, ∇0yz0) − ω(y, ∇xz0), (dω2)((0, x0), (0, y0), (z, 0)) = −ω(∇0zy0, x0) + ω(∇0zx0, y0)
Since ∇ and ∇0 satisfy (1) and (3), we obtain that dωi = 0, i = 1, 2, 3
It follows from the definitions of the formsωi, i = 1, 2, 3that: The restrictions of
ω1 andω3 toh+andh−are symplectic forms on these subalgebras
Let the form ω on R2n be given by ω = e1 ∧ e2 + e3 ∧ e4 + · · · + e2n−1 ∧ e2n, where {e1, , e2n} is a fixed basis of R2n and {e1, , e2n} denotes the dual ba-sis Let us set ej := (ej, 0) and fj := (0, ej), j = 1, , 2n (see also [4]) Hence {e1, , e2n, f1, , f2n}is a basis ofR2n
⊕ R2n and the formsωi, i = 1, 2, 3, can be written as
ω1 = e1∧ e2+ · · · + e2n−1∧ e2n+ f1∧ f2+ · · · + f2n−1∧ f2n,
ω2 = −e1∧ f2− · · · − e2n−1∧ f2n+ e2∧ f1+ · · · + e2n∧ f2n−1,
ω3 = e1∧ e2+ · · · + e2n−1∧ e2n− f1∧ f2− · · · − f2n−1∧ f2n
Beingh∇∇0 a double Lie algebra, the endomorphismE given byE(x, y) = (x, −y) forx, y ∈ R2nis aproduct structure(see [4]) onh∇∇0, that is,E2 = 1andEis integrable,
in the sense that it satisfies the condition
E[(x, x0), (y, y0)] = [E(x, x0), (y, y0)] + [(x, x0), E(y, y0)] − E[E(x, x0), E(y, y0)] (15) for allx, x0, y, y0 ∈ R2n The symplectic formω2satisfies
ω2(E(x, x0), E(y, y0)) = −ω2((x, x0), (y, y0)) for all x, x0, y, y0 ∈ R2n
4 Induced geometry on R4n
In this section we analyze the properties of the metric on the manifoldR4n ob-tained by left-translating by the Lie grouph∇∇0, the standard inner product of signa-ture(2n, 2n)onR2n⊕ R2n We show that this metric is always complete and it is flat
if and only if the Lie group h∇∇0 is 2-step nilpotent (see Theorem 4.2) Furthermore, this metric onR4nis hypersymplectic with respect to the structuresJ andEdefined previously
Trang 84.1 The bilinear formη onh∇∇0.
Let us define a bilinear formηonh∇∇0 by
η((x, x0), (y, y0)) = −ω(x, y0) + ω(x0, y) (16) for all(x, x0), (y, y0) ∈ h∇∇0 It is clearly symmetric and non degenerate With respect
to the basis{e1, , e2n, f1, , f2n},ηcan be written as
η = 2 −e1· f2− · · · − e2n−1· f2n+ e2· f1+ · · · + e2n· f2n−1 ,
where·denotes the symmetric product of 1-forms Moreover,ηsatisfies the two fol-lowing conditions
η(J (x, x0), J (y, y0)) = η((x, x0), (y, y0)) (17) η(E(x, x0), E(y, y0)) = −η((x, x0), (y, y0)) (18) forx, x0, y, y0 ∈ R2n Indeed,
η(J (x, x0), J (y, y0)) = η((−x0, x), (−y0, y))
= −ω(−x0, y) + ω(x, −y0)
= η((x, x0), (y, y0)) and
η(E(x, x0), E(y, y0)) = η((x, −x0), (y, −y0))
= ω(x, y0) − ω(x0, y)
= −η((x, x0), (y, y0))
Thus,ηis a Hermitian metric onh∇∇0 with respect to both structuresJ andE
We note that the subalgebrash+andh−are both isotropic subspaces ofh∇∇0 with respect toηand this metric has signature(2n, 2n) Moreover, it is easy to verify that the2-formsω1, ω2 andω3 can be recovered fromgand the endomorphismsJ andE Indeed we have
ω1((x, x0), (y, y0)) = η(J (x, x0), (y, y0)),
ω2((x, x0), (y, y0)) = η(E(x, x0), (y, y0)),
ω3((x, x0), (y, y0)) = η(J E(x, x0), (y, y0))
(19)
Trang 9The endomorphismsJandE ofh∇∇0, as well as the2-formsω1, ω2 andω3and the metricηcan be extended to the grouph∇∇0 by left translations Hence,h∇∇0 is equipped with:
1 A complex structureJ and a product structureE such thatJ E = −EJ;
2 A (pseudo) Riemannian metricηsuch that
η(J (x, x0), J (y, y0)) = η((x, x0), (y, y0)), η(E(x, x0), E(y, y0)) = −η((x, x0), (y, y0)) for allx, x0, y, y0 ∈ Γ(T(h∇
∇ 0));
3 Three symplectic formsω1, ω2 andω3which satisfy (19)
To summarize, we have obtained
Theorem 4.1 The nilpotent Lie group h∇
∇ 0 carries a left-invariant hypersymplectic struc-ture given by the 3-tuple {J, E, η}
4.2 The completeness ofη
Since η is left-invariant, the Levi-Civita connection ∇η can be computed on left-invariant vector fields, i.e., on the Lie algebra h∇∇0 After a computation one finds that
∇η(x,x0 )=
∇x+ ∇0x0 0
0 ∇x+ ∇0x0
One can verify, using the above expression of∇η, thatJ andEare parallel with respect to the Levi-Civita connection
We will show next that this connection need not be flat IfRdenotes the curva-ture of∇η, it is easily seen (using (4)) thatR((x, 0), (y, 0)) = R((0, x0), (0, y0)) = 0 Moreover,
R((x, 0), (0, y0)) = ∇η(x,0)∇η(0,y0 )− ∇η(0,y0 )∇η(x,0)− ∇η(−∇0 y 0 ,∇ y 0 )
Trang 10and using (20) together with (1), (5) and (7) one obtains
R((x, 0), (0, y0)) = 4
∇x∇0
y 0
= −4 ad[(x,0),(0,y0 )]
SinceRand the Lie bracket are skew-symmetric, one finally obtains
R((x, x0), (y, y0)) = −4 ad[(x,x0 )(y,y 0 )], thus showing that∇η will be flat if and only ifh∇∇0 is2-step nilpotent Note also that
R = 0 if and only if ∇x∇0y = 0 (21) for allx, y ∈ R2n
We end this section studying the completeness of ∇η It follows from [2, 4] that
∇η will be complete if and only if the differential equation onh∇∇0
admits solutionsx(t) ∈ g defined for all t ∈ R Here adηx means the adjoint of the transformation adx with respect to the metric η It is easy to verify that the right hand side of (22) is given byadηx(t)x(t) = −∇ηx(t)x(t)for all tin the domain ofxand thus we have to solve the equation
The curvex(t)onh∇∇0 can be written asx(t) = (a(t), b(t)), wherea(t), b(t) ∈ R2n
are smooth curves onR2n Hence, using (20), equation (23) translates into the system
(
˙a = −∇aa − ∇0ba,
Let us differentiate the first equation of the system above We have
¨
a = −2∇a˙a − ∇0a˙b − ∇0
b˙a
= 2∇a∇aa + 2∇a∇0ab + ∇0a∇ab + ∇0a∇0bb + ∇0b∇aa + ∇0b∇0ab
= 0,
Trang 11using (4), (5), (6) and (7) In the same fashion, we differentiate the second equation
of (24) and obtain
¨b = −∇
a˙b − ∇b˙a − 2∇0b˙b
= ∇a∇ab + ∇a∇0bb + ∇b∇aa + ∇b∇0ab + 2∇0b∇ab + 2∇0b∇0bb
= 0,
Using again (4), (5), (6) and (7) Thus, there exist constant vectorsA, B, C, D ∈
R2n such that
a(t) = At + B, b(t) = Ct + D
The explicit solution of the system (24) with initial condition x(0) = (a0, b0)is given by
a(t) = (−∇a0a0− ∇0a0b0)t + a0, b(t) = (−∇a0b0− ∇0b0b0)t + b0
Therefore,x(t)is defined for allt ∈ Rand, in consequence,∇η is complete Thus,
we have obtained
Theorem 4.2 Hypersymplectic metrics on h∇
∇ 0 are always complete
tµi liÖu tham kh¶o
[1] A Andrada, S Salamon, Complex product structures on Lie algebras, to appear
in Forum Math
[2] Guediri, M Sur la completude des pseudo-metriques invariantes a gauche sur les groupes de Lie nilpotentes Rend Sem Mat Univ Pol Torino52(1994), 371 376
[3] Nguyen Viet Hai, Quantum co-adjoint orbits ofM D4-groups, Vietnam J Math Vol29, IS 02/2001, pp.131-158
[4] Nguyen Viet Hai, Four-dimensional Lie algebras carrying a hypersymplectic structure, Journal of science-Vinh University,XXXIV, 4A, 2005, 29-39
... Levi-Civita connection ∇η can be computed on left-invariant vector fields, i.e., on the Lie algebra h∇∇0 After a computation one finds that...
One can verify, using the above expression of∇η, thatJ andEare parallel with respect to the Levi-Civita connection
We will show next that this connection need not... claimed
We set the notation to be used in what follows Since the construction of the Lie group structure on R4n in Theorem 2.1 depends on the affine structures ∇and∇0,