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‚x 2 ‚ x ¯
¯
‚f 1 ‚f
’ ω(f, )=1 (3.2.6)
‚x¯ 2 ‚x
¯
The cancellation condition (3.2.3) and relation (3.2.6) give
¯
‚f 1 ‚f 1
y) ’ ω(x, y)
y = ω(f, (3.2.7)
‚x 2 ‚x 2
These conditions describe completely the class of volume preserving di¬eomor-
phisms of H(n). Conditions (3.2.6) and (3.2.7) are in fact di¬erential equations for
¯
the function f when f is given. However, there is a compatibility condition in terms
¯
of f which has to be ful¬lled for (3.2.7) to have a solution f . Let us look closer to
(3.2.7). We can see the symplectic form ω as a closed 2-form. Let » be a 1-form such
that d» = ω. If we take the (regular) di¬erential with respect to x in (3.2.7) we quickly
obtain the compatibility condition
‚f
∈ Sp(n) (3.2.8)
‚x
and (3.2.7) takes the form:
¯
df = f — » ’ » (3.2.9)
(all functions seen as functions of x only).
Conditions (3.2.8) and (3.2.5) imply: there is a scalar function µ = µ(x, x) such
¯
that
‚f
= µ Jx
‚x
¯
Let us see what we have until now:
‚f
∈ Sp(n) (3.2.10)
‚x

T
¯
‚f 1 ‚f
Jf ’ Jx
= (3.2.11)
‚x 2 ‚x
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3 THE HEISENBERG GROUP


¯
‚f 1 ‚f
= 1 + ω(f, ) (3.2.12)
‚x
¯ 2 ‚x
¯
‚f
= µ Jx (3.2.13)
‚x
¯
Now, di¬erentiate (3.2.11) with respect to x and use (3.2.13). In the same time di¬er-
¯
entiate (3.2.12) with respect to x. From the equality
¯ ¯
‚2f ‚2f
=
‚x‚ x
¯ ‚ x‚x
¯
we shall obtain by straightforward computation µ = 0.

3.3 Hamilton™s equations
˜
For any element φ ∈ Sympl2 (R2n ) we de¬ne the lift φ to be the image of (φ, 0) by the
isomorphism described in the theorem 3.3.
Let A ‚ R2n be a set. Sympl2 (A)c is the group of symplectomorphisms with
regularity C 2 , with compact support in A.

De¬nition 3.4 For any ¬‚ow t ’ φt ∈ Sympl2 (A)c denote by φh (·, x)) the horizontal
˜
¬‚ow in H(n) obtained by the lift of all curves t ’ φ(t, x) and by φ(·, t) the ¬‚ow obtained
by the lift of all φt . The vertical ¬‚ow is de¬ned by the expression
˜
φv = φ’1 —¦ φh (3.3.14)



Relation (3.3.14) can be seen as Hamilton equation.

Proposition 3.5 Let t ∈ [0, 1] ’ φv be a curve of di¬eomorphisms of H(n) satisfying
t
the equation:
dv
φt (x, x) = (0, H(t, x)) , φv = idH(n)
¯ (3.3.15)
0
dt
Then the ¬‚ow t ’ φt which satis¬es (3.3.14) and φ0 = idR2n is the Hamiltonian ¬‚ow
generated by H.
Conversely, for any Hamiltonian ¬‚ow t ’ φt , generated by H, the vertical ¬‚ow
t ’ φv satis¬es the equation (3.3.15).
t


˜ ˜ ™
Proof. Write the lifts φt and φh , compute then the di¬erential of the quantity φt ’ φh
t t
and show that it equals the di¬erential of H.
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3 THE HEISENBERG GROUP


Flows of volume preserving di¬eomorphisms We want to know if there is any
nontrivial smooth (according to Pansu di¬erentiability) ¬‚ow of volume preserving dif-
feomorphisms.
˜
Proposition 3.6 Suppose that t ’ φt ∈ Dif f 2 (H(n), vol) is a ¬‚ow such that
- is C 2 in the classical sense with respect to (x, t),
˜
- is horizontal, that is t ’ φt (x) is a horizontal curve for any x.
Then the ¬‚ow is constant.

˜
Proof. By direct computation, involving second order derivatives. Indeed, let φt (x, x) =
¯
˜t ∈ Dif f 2 (H(n), vol) we obtain
(φt (x), x + Ft (x)). From the condition φ
¯
‚Ft 1 ‚φt 1
(x)y) ’ ω(x, y)
y = ω(φt (x),
‚x 2 ‚x 2
˜
and from the hypothesis that t ’ φt (x) is a horizontal curve for any x we get
dFt 1 ™
(x) = ω(φt (x), φt (x))
dt 2
Equal now the derivative of the RHS of the ¬rst relation with respect to t with the
derivative of the RHS of the second relation with respect to x. We get the equality, for
any y ∈ R2n :
1 ‚φt ™
0 = ω( (x)y, φt (x))
2 ‚x

therefore φt (x) = 0.
One should expect such a result to be true, based on two remarks. The ¬rst,
general remark: take a ¬‚ow of left translations in a Carnot group, that is a ¬‚ow t ’
φt (x) = xt x. We can see directly that each φt is smooth, because the distribution is left
invariant. But the ¬‚ow is not horizontal, because the distribution is not right invariant.
The second, particular remark: any ¬‚ow which satis¬es the hypothesis of proposition
3.6 corresponds to a Hamiltonian ¬‚ow with null Hamiltonian function, hence the ¬‚ow
is constant.
At a ¬rst glance it is disappointing to see that the group of volume preserving
di¬eomorphisms contains no smooth paths according to the intrinsic calculus on Carnot
groups. But this makes the richness of such groups of homeomorphisms, as we shall
see.

3.4 Volume preserving bi-Lipschitz maps
We shall work with the following groups of homeomorphisms.

De¬nition 3.7 The group Hom(H(n), vol, Lip) is formed by all locally bi-Lipschitz,
volume preserving homeomorphisms of H(n), which have the form:
˜¯
φ(x, x) = (φ(x), x + F (x))
¯
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3 THE HEISENBERG GROUP


The group Sympl(R2n , Lip) of locally bi-Lipschitz symplectomorphisms of R2n , in
the sense that for a.e. x ∈ R2n the derivative Dφ(x) (which exists by classical Rademacher
theorem) is symplectic.
˜
Given A ‚ R2n , we denote by Hom(H(n), vol, Lip)(A) the group of maps φ which
belong to Hom(H(n), vol, Lip) such that φ has compact support in A (i.e. it di¬ers
from identity on a compact set relative to A).
The group Sympl(R2n , Lip)(A) is de¬ned in an analogous way.

˜
Remark that any element φ ∈ Hom(H(n), vol, Lip) preserves the ”vertical” left
invariant distribution
(x¯) ’ (x, x)Z(H(n))
x ¯
for a.e. (x, x) ∈ H(n).
¯
We shall prove that any locally bi-Lipschitz volume preserving homeomorphism of
H(n) belongs to Hom(H(n), vol, Lip).
˜
Theorem 3.8 Take any φ locally bi-Lipschitz volume preserving homeomorphism of
˜
H(n). Then φ has the form:
˜¯
φ(x, x) = (φ(x), x + F (x))
¯

Moreover
φ ∈ Sympl(R2n , Lip)
F : R2n ’ R is Lipschitz and for almost any point (x, x) ∈ H(n) we have:
¯
1 1
ω(φ(x), Dφ(x)y) ’ ω(x, y)
DF (x)y =
2 2


˜x x ¯x
Proof. Set φ(˜) = (φ(˜), φ(˜)). Also denote by

| (y, y ) |2 = | y |2 + | y |
¯ ¯
˜
By theorem 2.21 φ is almost everywhere derivable and the derivative can be written
in the particular form:
A0
01
with A = A(x, x) ∈ Sp(n, R).
¯
˜
For a.e. x ∈ R2n the function φ is derivable. The derivative has the form:
˜¯
Dφ(x, x)(y, y ) = (Ay, y )
¯ ¯

where by de¬nition

˜¯ ’1 ˜
Dφ(x, x)(y, y ) = lim δµ φ(x, x)’1 φ((x, x)δµ (y, y ))
¯ ¯ ¯ ¯
µ’0

Let us write down what Pansu derivability means.
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3 THE HEISENBERG GROUP


For the R2n component we have: the limit
1
lim | [φ((x, x)δµ (y, y )) ’ φ(x, x)] ’ A | = 0
¯ ¯ ¯ (3.4.16)
µ’0 µ

is uniform with respect to y ∈ B(0, 1). We shall rescale all for the ball B(0, R) ∈ H(n).
˜
Relation (3.4.16) means: for any » > 0 there exists µ0 (») > 0 such that for any R > 0
and any µ > µ0 /R and y ∈ H(n) such that | y |< R we have
˜ ˜
1
| (φ(˜δµ y ) ’ φ(˜)) ’ Ay | < »R
x˜ x (3.4.17)
µ
Suppose that x = (x, x) satis¬es | x |¤ R with R > 1. Fix now » > 0 and consider
˜ ¯
2n , | y |< 1 and µ > 0. Set
arbitrary y ∈ R
1
yµ =
¯ ω(y, x)

There exists a constant C > 0 such that
C
| (y, yµ |2 ¤ 1 + (1 + R)2
¯
µ
For µ < µ0 (») choose
1
C 2 (1 + R)4 + 4µ C(1 + R)2
µ=µ=
4
1/2
Then the inequality (3.4.17) is true for R replaced by 1 + C (1 + R)2 , y = (y, yµ ),

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