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tation of the algebra g • g with the bracket:

[(X, U ), (Y, V )] = ([X, Y ]G , [X, V ]G + [U, Y ]G + [V, U ]N ) (4.3.14)



Proof. The equivalence of (a), (b), (c) has been explained. Suppose now that (4.3.13)
is true. It is then straightforward to show that the space of all elements

a(X,Y ) = adG ’ adN
X Y

forms a Lie algebra with the linear commutator as bracket. Moreover, we have:

[a(X,U ) , a(Y,V ) ] = a[(X,U ),(Y,V )]

This shows that AdG AdN is a group and the property of its algebra. In order to ¬nish
the proof remark that a(X,X) = aX , de¬ned at (4.3.12).

Corollary 4.17 The atlas A gives to G a C 1 N (D, G) manifold structure if and only
if (4.3.13) is true and for any µ > 0, X, Y ∈ g
’1 ’1
δµ [X, δµ Y ]N ’ δµ [X, δµ Y ]G = [X, Y ]N ’ [X, Y ]G (4.3.15)
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4 SUB-RIEMANNIAN LIE GROUPS


Proof. The atlas A gives to G a C 1 N (D, G) manifold structure if and only if
n
J (G, D) ‚ HL(g, ·) and any of its elements commute with dilatations δµ . This
is equivalent with (4.3.13) and (4.3.15).

Remark 4.18 A Lie group is a manifold endowed with a smooth operation. In what
sense is then G a (sub-Riemannian) Lie group? We already have problems to assign an
atlas with smooth transition functions to G. The real meaning of the corollary 4.17 is
that even if we succeed to give to G an atlas with smooth transition functions then the
left translations and the exponential map will not be smooth. This is a problem. If we
renounce to give G a manifold structure, at least the operation is smooth in the sense
of proposition 4.12.
Returning to the problem of the atlas, if N (G, D) is a Heisenberg group, or equiv-
alently the distribution has codimension one, then the commentary of Bella¨che [3],
±
page 73, shows that such an atlas exists, as a consequence of Darboux theorem on nor-
mal forms for contact di¬erential forms. But such an atlas is not compatible with the
operation.

We shall see now how it can be possible that the operation is commutative smooth
but the right translations are not.
We shall look to the double G(2) . This group is isomorphic with G2 = G — G with
componentwise multiplication by the isomorphism

F : G(2) ’ G2 , F (x, u) = (x, xu)

The Lie algebra of G(2) is easily then g(2) = g — g, with Lie bracket

[(x, u), (y, v)](2) = ([x, u], [x + u, y + v] ’ [x, y])

Take in g(2) the distribution D (2) = D — D. This distribution generates g(2) , and the
dilatations are
(2)
δµ (x, u) = (δµ x, δµ u)
We translate the distribution at left all over G(2) .
We know that the operation is commutative smooth. We shall check if the classi-
cal derivative of the operation transports the distribution D (2) in the distribution D.
Straightforward computation shows that

D op(e, e)D (2) = D

where D here means classical derivative. Indeed, D op(e, e)(X, Y ) = X + Y and D is
a vector space, therefore (X, Y ) ∈ D (2) implies X + Y ∈ D.
For general (x, u) ∈ G(2) this is no longer true. Indeed, we have:

D op(x, u)(Y, V ) = DLG (0) Y + V + (AdG ’ AdG )Y
xu xu u

therefore a su¬cient condition for D op(x, u)D (2) = D is that x = e.
In conclusion the fact that the operation is commutative smooth does not imply
that so are the right translations.
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We can hope that by a slight modi¬cation in the de¬nition of smoothness we shall
be able to give to G a manifold structure and simultaneously to have smooth right
translations. For the manifold structure we have to ”smooth” the group generated by
the elements with the form
’1
AdG AdN
X X

To solve the problem of the right translations we need that the group AdG be ”smoothed”.
Before doing this we shall see that the pure metric point of view does not feel these
problems but in such a precise way.

4.4 Metric tangent cone
We shall prove the result of Mitchell [23] theorem 1, that G admits in any point a
metric tangent cone, which is isomorphic with the nilpotentisation N (G, D). Mitchell
theorem is true for regular sub-Riemannian manifolds. The proof that we give here is
based on the lemma 4.13 and Gromov [13] section 1.2.
Because left translations are isometries, it is su¬cient to prove that G admits a
metric tangent cone in identity and that the tangent cone is isometric with N (G, D).
For this we transport all in the algebra g, endowed with two brackets [·, ·]G , [·, ·]N
g n
and with two operations · and ·. Denote by dG , dN the Carnot-Carath´odory distances
e
g n
corresponding to the ·, respectively · left invariant distributions on (a neighbourhood of
0 in) g. lG , lN are the corresponding length functionals. We shall denote by BG (x, R),
BN (x, R) the balls centered in x with radius R with respect to dG , dN .
We can re¬ne lemma 4.13 in order to obtain the Ball-Box theorem in this more
general situation.

Theorem 4.19 (Ball-Box Theorem) Denote by

Box1 (µ) = x ∈ G : ∆1 (x) < µ
G G

Box1 (µ) = x ∈ N : ∆1 (x) < µ
N N

For small µ > 0 there is a constant C > 1 such that

expG Box1 (µ) ‚ Box1 (Cµ) ‚ expG Box1 (C 2 µ)
N G N




Proof. Reconsider the proof of lemma 4.13. This time, instead of the trick of replacing
commutators expG [X, Y ](t) with four letters words

expG X(t1 ) expG Y (t2 ) expG X(t3 ) expG Y (t4 )

we shall use a smarter replacement (which, important! , works equally for the nilpo-
tentisation N (G, D)).
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4 SUB-RIEMANNIAN LIE GROUPS


We start with a basis {Y1 , ..., Yn } of the algebra g, constructed from multibrackets of
elements {X1 , ..., Xp } which form a basis for the distribution D. Introduce an Euclidean
distance by declaring the basis of g orthonormal. Set
g g g
[X(t), Y (t)]—¦ = (tX) · (tY ) · (’tX) · (’tY )

It is known that for any set of vectors Z1 , ..., Zq ∈ g, if we denote by ±—¦ (Z1 (t), ..., Zq (t))
a [·, ·]—¦ multibracket and by ±(Z1 , ..., Zq ) the same constructed [·, ·] multibracket, then
we have
±—¦ (Z1 (t), ..., Zq (t)) = tq ±(Z1 , ..., Zq ) + o(tq )
with respect to the Euclidean norm.
Remark as previously that the function
n
(t1 , ..., tn ) ’ (ti Yi ) (4.4.16)
i=1

is invertible in a neighbourhood of 0 ∈ Rn . Each Xi from the basis of g can be written
as a multibracket
Xi = ±i (Xi1 , ..., Xj i )
which has the length li = j i ’ 1. If li is odd then replace (ti Xi ) by

±—¦ (Xi1 (t1/li ), ..., Xj i (t1/li ))

If li is even then the multibracket ±i can be rewritten as

±i (Xi1 , ..., Xj i ) = βi (Xi1 , ..., [Xik , Xik+1 ], ..., Xj i )

Replace then (ti Xi ) by

βi (Xi1 (| t |1/li ), ..., [Xik (| t |1/li ), Xik+1 (| t |1/li )]—¦ , ..., Xj i (| t |1/li ))
—¦ if t ≥ 0
βi (Xi1 (| t |1/li ), ..., [Xik+1 (| t |1/li ), Xik (| t |1/li )]—¦ , ..., Xj i (| t |1/li ))
—¦ if t ¤ 0

After this replacements in the expression (4.4.16) one obtains a function EG which
is still invertible in a neighbourhood of 0. We obtain a function EN with the same
algebraic expression as EG , but with [·, ·]—¦ brackets instead of [·, ·]—¦ ones. Use these
N G
functions to (obviously) end the proof of the theorem.

Theorem 4.20 The Gromov-Hausdor¬ limit of pointed metric spaces (g, 0, »dG ) as
» ’ ∞ exists and equals (g, dN ).

Proof. We shall use the proposition 1.24. For this we shall construct µ isometries
’1
between Box1 (1) and Box1 (Cµ). These are provided by the function EG —¦ EN —¦ δµ .
N G
The trick consists in the de¬nition of the nets. We shall exemplify the construction
for the case of a 3 dimensional algebra g. The basis of g is X1 , X2 , X3 = [X1 , X2 ]G .
Divide the interval [0, X1 ] into P equal parts, same for the interval [0, X2 ]. The interval
[0, X3 ] though will be divided into P 2 intervals. The net so obtained, seen in N ≡ g,
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4 SUB-RIEMANNIAN LIE GROUPS

’1
turn EG —¦ EN —¦ δµ into a µ isometry between Box1 (1) and Box1 (Cµ). To check this
N G
is mostly a matter of smart (but still heavy) notations.
The proof of this theorem is basically a re¬nement of Proposition 3.15, Gromov [13],
pages 85“86, mentioned in the introduction of these notes. Close examination shows
that the theorem is a consequence of the following facts:

(a) the identity map id : (g, dG ) ’ (g, dN ) has ¬nite dilatation in 0 (equivalently
expG : N (D, G) ’ G has ¬nite dilatation in 0),

(b) the change from the G-invariant distribution induced by D to the N invariant
distribution induced by the same D is classically smooth.

It is therefore natural that the result does not feel the non-derivability of id map in
x = 0.

4.5 Noncommutative smoothness in Carnot groups
Let N be a Carnot group and

N = V1 + ... + Vm (4.5.17)

the graduation of its algebra. Let End(N ) be the group of endomorphisms of N .

De¬nition 4.21 The group of vertical endomorphisms of N , noted by V L(N ), is the
subgroup of End(N ) of all N endomorphisms A which admit the form A = (Ai,j ) with
respect to the direct sum decomposition of N (4.5.17), such that

Aii = idVi , Aij = 0 ∀i < j ∈ {1, ..., m}



Proposition 4.22 (a) The elements of the linear group HL(N ) have diagonal form
with respect to the direct sum decomposition (4.5.17).

(b) Any A ∈ End(N ) admits the unique decomposition

A = Av Ah , Av ∈ V L(N ) , Ah ∈ HL(N )

(c) Ad(N ) is a normal subgroup of V L(N ). The quotient group is Carnot.

Proof. (b) Proof by induction. Just write what the algebra morphism condition for
A ∈ End(N ) means and get the inferior diagonal form of A. Such a matrix (with ele-
ments linear transformations) decomposes uniquely in a product of a vertical morphism
Av and a diagonal morphism Ah . Check that the latter commutes with dilatations.
(a) We know from (b) that any element A ∈ HL(N ) has inferior diagonal form.
Commutation with dilatations forces A to have diagonal form.
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4 SUB-RIEMANNIAN LIE GROUPS


(c) Ad(N ) is normal in End(N ) and it is a subgroup of V L(N ). It is therefore
normal in V L(N ). The quotient group is nilpotent, because V L(N ) is nilpotent. It is
Carnot because it has dilatations. Consider the map

δµ : End(N ) ’ End(N ) , δµ A = δµ Aδµ
’1


From (b) we get that δµ is an endomorphism of V L(N ). From the fact that δµ ∈ End(N )
we get that
δµ Ad(N ) = Ad(N )
Therefore δµ factorizes to an endomorphism of the quotient V L(N )/Ad(N ). The rest
is trivial.
We shall call the quotient

R(N ) = V L(N )/Ad(N )

the rest of N .
The de¬nition of noncommutative derivative follows. For any function f : N ’ N
and x ∈ N set
fx : N ’ N , fx (y) = f (x)’1 f (xy)
f is Pansu derivable in x if and only if fx is Pansu derivable in 0; moreover the Pansu
derivative of f in x equals the Pansu derivative of fx in 0.
From the proof of Pansu™s Rademacher theorem one can extract the following in-
formation:

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