Departments of Mathematics
Montana State University
Fall 2015
Prof. Kevin Wildrick
An introduction to non-smooth analysis and geometry
Lecture 12: The Heisenberg group
1. An introduction to the Heisenberg group
The sub-Riemannian Heisenberg group is an excellent example of a non-trivial metric
measure space that supports a Poincaré inequality and has an easy to understand Cheeger
structure. It is a very interesting space for a variety of reasons, but here is one to get
whet your appetite: the sub-Riemannian Heisenber group has a two-dimensional Cheeger
structure, is homeomorphic to R3 , and is Ahlfors 4-regular. It shares many properties
with Euclidean space, but it does not bi-Lipschitzly embed into Hilbert space.
Let us begin by describing the Heisenberg group itself. The most familiar way to
describe the Heisenberg group is as the group of upper triangular 3 × 3 real matrices with
1’s on the diagonal, with respect to matrix multiplication:

 


1 a + a0 c + c0 + ba0
1 a c
1 a0 c 0
.
 0 1 b   0 1 b0  =  0
1
b + b0
0
0
1
0 0 1
0 0 1
In otherwords, the group has underlying set R3 and multiplication
(a, b, c)(a0 , b0 , c0 ) = (a + a0 , b + b0 , c + c + ba0 ).
This is clearly a non-commutative group operation, but it lacks a symmetry which will
be convenient for us.
To that end, we denote by H the group (R3 , ∗), where
(x, y, t) ∗ (x0 , y 0 , t0 ) = (x + x0 , y + y 0 , t + t0 − 2(xy 0 − yx0 )).
We may also consider the underlying set as C × R. Identifying z = x + iy, the group
operation becomes
(z, t) ∗ (z 0 , t0 ) = (z + z 0 , t + t0 + 2 im(z z̄)).
The factor accounting for the non-commutativity of the group operation is the term
im(z z̄). This term can be interpreted as the signed area of the parallelogram spanned by
z and z 0 , i.e., the determinant of the linear mapping defined by
1 7→ z and i 7→ z 0 .
Note that the identity element is (0, 0, 0) and
(x, y, t)−1 = (−x, −y, −t),
just as in the usual group structure on R3 .
The two groups described above are isomorphic via the isomorphism
x = a, y = b, t = 4c − 2ab.
One of the first things we notice about the group operation in H is that the formulas
describing it are smooth. This fact makes the Heisenberg group an example of a Lie
group. A Lie group is a smooth manifold M endowed with a group operation ∗ so that
the mapping ∗ : M × M → M defined
∗(g, h) = g −1 ∗ h
is smooth. Note that this implies that the mapping g 7→ g −1 is smooth, and for each
g0 ∈ M , the mappings g 7→ g0 ∗ g are smooth.
1
Noticing this, the first thing that comes to mind to try is to differentiate some of these
smooth mappings we have hanging around!
Let’s pick a point g0 = (x0 , y0 , t0 ) ∈ H, and consider the mapping Lg0 : R3 → R3 defined
by
Lg0 (h) = g0 ∗ h.
Its derivative at a point p ∈ H will be a linear mapping from Tp (R3 ) to Tg∗p (R3 ):
D(Lg0 )p : Tp (R3 ) → Tg0 ∗p (R3 ).
Let’s examine this mapping when p = 0. To figure out what it does, we need to evaluate
it on a basis of T0 (R3 ). What are elements of T0 (R3 )? They are equivalence classes of
smooth curves γ : (−, ) → R3 satisfying γ(0) = 0 ∈ R3 , where two such curves γ and β
are equivalent if γ 0 (0) = β 0 (0).
With this perspective, a basis for T0 (R3 ) is given by the equivalence classes of the curves
γx (s) = (s, 0, 0), γy (s) = (0, s, 0), γt (s) = (0, 0, s).
The corresponding vectors in T0 (R3 ) are usually denoted by
∂
= (1, 0, 0) = [γx0 (0)],
∂x
∂
= (0, 1, 0) = [γy0 (0)],
∂y
∂
= (0, 0, 1) = [γt0 (0)].
∂t
To see how D(Lg0 )0 behaves, we push forward these curves with the mapping Lg0 , and
then differentiate.
Lg0 ◦ γx (s) = (x0 , y0 , t0 ) ∗ (s, 0, 0) = (x0 + s, y0 , t0 + 2sy0 ),
Lg0 ◦ γy (s) = (x0 , y0 , t0 ) ∗ (0, s, 0) = (x0 , y0 + s, t0 − 2sx0 ),
Lg0 ◦ γt (s) = (x0 , y0 , t0 ) ∗ (0, 0, s) = (x0 , y0 , t0 + s).
Hence,
∂
∂
+ 2y0 ,
∂x
∂t
∂
∂
− 2x0 ,
(Lg0 ◦ γy )0 (0) = (0, 1, −2x0 ) =
∂y
∂t
∂
(Lg0 ◦ γt )0 (0) = (0, 0, 1) = .
∂t
(Lg0 ◦ γx )0 (0) = (1, 0, 2y0 ) =
Another way to look at this calculation is to consider the full differential of the mapping:
Lg0 (x, y, t) = (x0 + x, y0 + y, t0 + t − 2(x0 y − y0 x)).
Differentiating with respect to (x, y, t) and evaluating

1
0

0
1
D(Lg0 )0 =
2y0 −2x0
at (x, y, t) = 0 yields the matrix

0
0 .
1
Then


1
D(Lg0 )0  0  = (1, 0, 2y0 ),
0
agreeing with the above computation. Notice that the above matrix would have been the
same if we evalutated at any value of (x, y, t). Since it is non-singular, this implies that
Lg0 is a diffeomorphism, and its differential maps bases to bases.
Proposition 1.1. Let g = (x, y, t) ∈ R3 . The vector fields
∂
∂
X(g) :=
+ 2y ,
∂x
∂t
∂
∂
Y (g) :=
− 2x ,
∂x
∂t
∂
T (g) :=
∂t
on R3 are invariant under the action of H, i.e., for any g0 ∈ H
D(Lg0 )g X(g) = X(Lg0 (g)),
D(Lg0 )g Y (g) = Y (Lg0 (g)),
D(Lg0 )g T (g) = T (Lg0 (g)).
Moreover, they form a basis of Tg (R3 ).
Proof. Given a vector V (0) ∈ T0 R3 , we may create an invariant vector field V by defining
V (g) := D(Lg )0 (V (0)).
To see this, note that then, for each g0 ∈ H, the chain rule yields
D(Lg0 )g (V (g)) = D(Lg0 )g D(Lg )0 V (e) = D(Lg0 ◦ Lg )0 V (e) = D(Lg0 ∗g )0 V (e) = V (Lg0 (g)).
This is exactly how we defined X, Y , and T .
The proof shows that a left-invariant vector field is defined by its value at the identity.
Notice that D(Lg0 )g in fact has determinant 1. This implies that Lg0 is a volume
preserving transformation; if Ω is an open set, the
Z
3
L (Lg0 (Ω)) =
det D(Lg0 ) dL3 = L3 (Ω).
Ω
This quickly implies that for any measurable set E ⊆, the Lebesgue measure of E is the
same as the Lebesgue measure of the image of E under any left translation. Thus we have
shown
Proposition 1.2. The Lebesgue measure L3 is a left-invariant measure on H.
One can in fact show that L3 is also right-invariant, and it is, up to scaling, the unique
such measure.
In quantum mechanics, position and momentum correspond non-commutating selfadjoint operators on a complex Hilbert space. The Heisenberg group was first considered
in this context - the vector fields X and Y above provide the simplest possible model of
this behavior (this can be made precise).
Definition 1.3. Given vector fields V and W on R3 , their Lie bracket or commutator or
Lie derivative is defined by
[V, W ] := V W − W V.
The above definition requires some discussion about what exactly vector fields are. If
one takes the view that vector fields are functions R3 → R3 ; the above definition cannot
be properly understood. By V W , we do not mean composition of functions. We can better
understand this by identifying vector fields with derivations or as derivatives of curves,
or better yet as differential equations.
Given a vector field V : R3 → R3 , we can define an ordinary differential equation as
follows. A solution to the differential equation is a path γ : (−, ) → R3 that satisfies
γ 0 (s) = V (γ(s)).
If V is smooth enough (and ours always will be), then solutions to this differential
equation exist and are unique after specification of γ(0).
Let us try understand [V, W ] in this way. Let g ∈ R3 and let s0 > 0. Let γ1 solve
γ10 (s) = V (γ1 (s)), γ1 (0) = g.
Let γ2 solve
γ20 (s) = W (γ2 (s)), γ2 (0) = γ1 (s0 ).
Let γ3 solve
γ30 (s) = −V (γ3 (s)), γ3 (0) = γ2 (s0 ).
Let γ4 solve
γ40 (s) = −W (γ4 (s)), γ4 (0) = γ3 (s0 ).
Now define g(s0 ) to be the end point of the concatenation of γ1 , γ2 , γ3 and γ4 .
We define
1 d2
[V, W ](g) =
g(s) .
2
2 dt
s=0
The idea is the following: we start at g, flow with V for time s, flow with W for time
s, flow backwards in time with V for time s, and then flow backwards in time with W for
time s. Now, we look at how much we now differ from g and take the second derivative.
Let’s compute [X, Y ](e).
First, we flow with X from e for time s0 . The first coordinate of e will grow with
speed 1, the second won’t change, and the third will grow with speed 2 times the second
coordinate. But, the second coordinate is stuck at 0, so
γ1 (s) = (s, 0, 0), and γ1 (s0 ) = (s0 , 0, 0)
Now we flow with Y from (s, 0, 0) for time s. The first coordinate doesn’t change, the
second grows with speed 1, and the third will grow with speed −2 times the first coordinate
(which is constant):
γ2 (s) = (s0 , s, −2s0 s). and γ2 (s0 ) = (s0 , s0 , −2s20 ),
Now reverse
γ3 (s) = (s0 − s, s0 , −2s20 − 2s0 s) and γ3 (s0 ) = (0, s0 , −4s20 ).
Finally
γ4 (s) = (0, s0 − s, −6s20 ), and γ4 (s0 ) = (0, 0, −4s20 ).
Thus
[X, Y ](e) = −4T.
It’s a good exercise to show that if V and W are left-invariant vector fields, then so is
their commutator [V, W ]. It’s also fun to check that if we treat
∂
∂
∂
= (1, 0, 0),
= (0, 1, 0),
= (0, 0, 1)
∂x
∂y
∂t
as differential operators on the space of smooth functions on R3 , e.g.
∂f
∂f
(x0 , y0 , z0 ) + 2y0 (x0 , y0 , z0 ),
X(x0 , y0 , z0 )f =
∂x
∂y
then the commutator can be viewed as iterated differentiation (this is because the tangent
bundle has an alternate description as the space of derivations).
It’s not hard to see that [X, T ] = 0 = [Y, T ] using a similar method of computation.
Thus, we have seen that X and Y do not commute, but instead their commutator is a
vector field that commutes with all others! This is the original motivation for considering the Heisenberg group coming from quantum mechanics and is at the heart of the
Heisenberg uncertainty principle.
Definition 1.4. The Lie algebra of the Heisenberg group is the three-dimensional real
vector space generated by X, Y , and T , and equipped with the operation [·, ·].
Representations of the Heisenberg algebra in the Hilbert space of “states” play an
important role in quantum mechanics.
2. The Carnot-Carathéodory metric on the Heisenberg group
The fact that the vector fields X and Y and their commutator [X, Y ] = −4T generate the entire Heisenberg algebra makes the situation very different from the standard
∂
∂
orthonormal frame for the tangent bundle: the vector fields ∂x
and ∂y
commute, and so
they do not generate (in the sense of Lie algebras) the entire tangent bundle.
To put this in context, a related important result from differential geometry is Frobenius’ theorem. Let M be a smooth manifold, and let B ⊆ T M be a subbundle of the
tangent bundle. Then B arises from a foliation of M if and only B is closed under taking
commutators.
∂
∂
We can see this in action here: the subbundle generated by ∂x
and ∂y
arises from the
3
foliation of R by planes parallel to the x − y plane. However, try as you might, you will
never find a foliation of R3 by smooth surfaces whose tangent planes lie in the subbundle
generated by X and Y .
Definition 2.1. The horizontal tangent bundle HT H of H is the subbundle of R3 generated by the vector fields X and Y .
Of course, since the vector fields X and Y are left-invariant, the horizontal tangent
bundle is also left-invariant:
(DLg )p HTp H = HTg∗p H.
There is a general result, called the Chow-Rashevskii theorem, that is a sort of converse
to Frobenius’ theorem: if a subbundle generates (by taking commutators) the entire
tangent bundle, then any two points of the manifold can be connected by paths whose
tangents lie in the subbundle.
Definition 2.2. Let γ : [0, 1] → R3 be an absolutely continuous path. We say that γ is
(Heisenberg) horizontal if for almost every s ∈ [0, 1], the tangent γ 0 (s) lies in the horizontal
tangent space HT Hγ(s) .
Notice that the notion of a horizontal path is left-invariant under the action of the
Heisenberg group: if g ∈ H and γ : [0, 1] → H horizontal, then for almost every s ∈ [0, 1],
(Lg ◦ γ)0 (s) = (DLg )γ(s) γ 0 (s) ∈ (DLg )γ(s) HTγ(s) H = HTLg ◦γ(s) H.
When we calculated that [X, Y ] = −4T , we did so by creating a path that connected
the origin to a point on the T axis only by flowing with multiples of X and Y . We can
use a similar construction to show that
Theorem 2.3. Any two points of R3 can be connected by a horizontal path.
Proof. It suffices to connect any point to the origin. First, note that for any point
(x0 , y0 , 0) ∈ H, the path
γ(s) = s(x0 , y0 , 0)
is a horizontal path connecting it to the origin. Combining this path with a flow along
X, Y , −X, and −Y (or its negation) as before gives the result.
For such paths, there is a preferred notion of length that is adapted to this structure.
Namely, let γ : [0, 1] → R3 be a horizontal path. Then, for almost every s ∈ [0, 1], we have
the vector equation
γ 0 (s) = (γ10 (s), γ20 (s), γ30 (s)) = aγ (s)X(γ(s)) + bγ (s)Y (γ(s)).
This implies that
(
a(s) = γ10 (s),
b(s) = γ20 (s).
(2.1)
Moreover,
γ30 (s) = 2(γ10 (s)γ2 (s) − γ20 (s)γ1 (s)).
(2.2)
The equality (2.2) is an instance of a contact equation. For an absolutely continuous curve
γ : [0, 1] → R3 , it is equivalent to the horizontality of the curve. It implies that the third
coordinate of γ is determined by the first two! More concretely, if (γ1 , γ2 ) : [0, 1] → R2 is
an absolutely continuous planar path, there is a unique way to lift it to a horizontal path
γ : [0, 1] → R3 , once γ(0) is specified:
Z s0
(2.3)
γ3 (s0 ) = γ3 (0) + 2
γ10 (s)γ2 (s) − γ20 (s)γ1 (s) ds.
0
Now, notice that for g = (x0 , y0 , t0 ),
d
(x0 + γ1 (s), y0 + γ2 (s), t0 + γ3 (s) + 2(x0 γ2 (s) − y0 γ1 (s))
ds
= (γ10 (s), γ20 (s), 2(x0 γ20 (s) − y0 γ10 (s)).
(Lg ◦ γ)0 (s) =
This shows that the standard Euclidean coordinates of γ 0 are not (in general) invariant
∂ ∂
under the action of H, but the X,Y coordinates are invariant - exactly because the ∂x
, ∂y
coordinates are invariant.
Thus, we may define a left-invariant notion of length for a horizontal path.
Definition 2.4. Let γ : [a, b] → R3 be a horizontal path. The Carnot-Carathéodory
length of γ is defined to be
Z b
1/2
length(γ) =
γ10 (s)2 + γ20 (s)2
ds.
cc
a
In otherwords, we have defined {X, Y } to be an orthonormal frame of the horizontal
tangent bundle HT H. It’s interesting to note the following fact: The cc-length of a
horizontal curve is equal to the length of its projection to R2 × {0}. Moreover, the change
in the height of a horizontal curve is given by (2.3):
Z s0
γ3 (s0 ) − γ3 (0) = 2
γ10 (s)γ2 (s) − γ20 (s)γ1 (s) ds.
0
Some of you calc-3 teachers might recall the formula for the signed area enclosed by a
closed planar curve (γ1 , γ2 ) : [0, s0 ] → R2 . To see this, recall that the signed area of an
open set U ⊆ R2 is given by the integral of the volume form:
Z
dx ∧ dy.
U
By Stoke’s theorem, this is equal to the integral over the boundary of U of a differential
one form whose exterior derivative is dx ∧ dy. This one-form is
1
α = (xdy − ydx),
2
since
1
dα = (dx ∧ dy − dy ∧ dx) = dx ∧ dy.
2
Thus
Z
Z
2 dx ∧ dy =
xdy − ydx.
U
∂U
If the boundary ∂U is parameterized by γ, then the change of variables formula gives
Z
Z
xdy − ydx =
γ1 (s)γ20 (s) − γ2 (s)γ10 (s) ds.
∂U
[0,s0 ]
In particular,
Z
−2
Z
dx ∧ dy =
U
s0
Z
ydx − xdy =
γ10 (s)γ2 (s) − γ20 (s)γ1 (s) ds.
0
γ
Since γ is a closed curve, the horizontality condition (2.3) now shows
Z s0
Z
0
0
0 = γ3 (s0 ) − γ3 (0) = 2
γ1 (s)γ2 (s) − γ2 (s)γ1 (s) ds = −4 dx ∧ dy.
0
U
Thus, we have shown:
Theorem 2.5. Let γ be a closed horizontal curve in H. Then the signed area of the
projection of γ to R2 × {0} is 0.
We can also give a second proof that any two points can be connected by a horizontal
curve: Let p = (x0 , y0 , t0 ), and let α(s) = s(x0 , y0 ) be the planar segment connecting the
origin to the projection of p. If β is any absolutely continuous planar path connecting
(x0 , y0 ) to 0, then the concatenation is a closed path and by our argument above, the
signed area enclosed is
Z
Z
1
xdy − ydx.
dx ∧ dy =
2 β◦α
U
However, α contributes noting to this integral, since it is horizontal:
α1 (s)α20 (s) − α2 (s)α10 (s) = 0.
Thus
Z
−2
Z
dx ∧ dy =
U
ydx − xdy.
β
Choose β so that the signed area enclosed by β ◦ α is t0 /4, and lift β to a horizontal curve
γ starting at p = (x0 , y0 , t0 ). We claim that γ ends at the origin. To see this, note that
by definition
Z
Z
γ3 (1) − γ3 (0) = 2 ydx − xdy = −4 dx ∧ dy = −t0 .
β
U
Thus γ3 (1) = 0, as desired.
We are now ready to define a left-invariant metric on H.
Definition 2.6. Let p and q be points of H. Then
dcc (p, q) = inf{length(γ) : γ(0) = p, γ(1) = q, γ horizontal}.
cc
A horizontal curve γ is called a geodesic if
dcc (γ(0), γ(1)) = length γ.
cc
Exericse 2.7. Let p = (x0 , y0 , t0 ) be a point of H. We will try to calculate dc c(0, p) and
find the horizontal curve that realizes it.
• Let γ be any horizontal curve connecting p to 0 ∈ H, and let β = πR2 γ. Let α be
the line segment connecting 0 to p in R2 . Denote by U the interior of β ◦ α. Show
that
Z
t0 = 4 dx ∧ dy.
U
• Show (note?) that
length(γ) = length β.
cc
R2
• Find the shortest curve β in R2 connecting 0 to (x0 , y0 ) so that the concatentation
with α has signed area t0 /4. (hint: it is an arc of a circle).
• Show that for this choice of β, it’s lift γ starting at p is a Heisenberg geodesic
connecting p to 0.
• When is the geodesic connecting p to 0 unique?
Exericse 2.8. The mapping dcc : R3 × R3 → [0, ∞) is continuous.
3. Scaling in the Heisenberg group
In R3 , we have an action of [0, ∞) by scaling that preserves the space and interacts
well with both metric and measure. In the Heisenberg group, this exists but is a bit more
delicate.
What we would like is a one-parameter family of mappings {δτ : H → H}τ ≥0 so that
dcc (δτ (p), δτ (q)) = τ dcc (p, q).
In order to prove such an equality, we would ideally want δτ to preserve the family of
horizontal paths (since they define the metric), and to transform the length of a horizontal
path by a factor of τ . Since the length of a horizontal path is simply the length of the
projection to the plane, it follows that δτ must be the standard scaling by τ on the first
two coordinates:
δτ (x, y, t) = (τ x, τ y, ???).
However, we can deduce how the dilation should behave on the third coordinate if it is
to preserve the class of horizontal curves. Suppose that γ and δτ ◦ γ are both horizontal
curves. Then
(δτ ◦ γ3 )0 (s) = (δτ ◦ γ10 (s))(δτ ◦ γ2 (s)) − (δτ ◦ γ20 (s))(δτ ◦ γ1 (s))
= τ 2 (γ10 (s)γ2 (s) − γ20 (s)γ1 (s))
= τ 2 γ30 (s).
Hence, we see that the correct definition should be
δτ (x, y, t) = (τ x, τ y, τ 2 t).
It is now easy to check the following statement:
Proposition 3.1. For each τ > 0, the mapping δτ : H → H preserves the class of horizontal paths, and satisfies
dcc (δτ (p), δτ (q)) = τ dcc (p, q)
for each p, q ∈ H.
Now we see the strange nature of Heisenberg geometry - the scaling of the metric
behaves differently in different directions!
For example, let γ(s) = (s, 0, 0). Then γ is a horizontal curve, and
γ 0 (s) = 1X(γ(s)) + 0Y (γ(s)).
Thus
length(γ|[0,s0 ] = s0 .
cc
In otherwords, γ|[0,s0 ] is isometric to the line segment [0, s0 ] in R. Accordingly,
dcc (γ(s), 0) = dcc ((s, 0, 0), (0, 0, 0) = sdcc ((1, 0, 0), (0, 0, 0).
However, if α(s) = (0, 0, s), then
dcc (α(s), 0) = dcc (δ√s (0, 0, 1), (0, 0, 0)) =
√
sdcc ((0, 0, 1), (0, 0, 0)).
This means that α|[0,s0 ] is a scaled copy of R equipped with the square root metric!
4. The Koranyi norm
Now that the we know how Heisenberg metric scales, we can easily construct another
left-invariant metric that scales the same way, but is much easier to compute.
For p = (z, t) ∈ H, define the Koranyi norm of p by
||p|| = (|z|4 + |t|2 )1/4 .
It is a computation to show that
dK (p, q) = ||p−1 ∗ q||
is a metric on H; the triangle inequality for this metric is exactly why we needed the
factor two in the definition of the Heisenberg group law (otherwise we would have some
inconvenient constants involved in the norm).
It’s trivial to check that dK is left-invariant:
dK (Lg (p), Lg (q)) = ||(g ∗ p)−1 ∗ (g ∗ q)|| = ||p−1 ∗ g −1 ∗ g ∗ q|| = ||p−1 ∗ q|| = dK (p, q).
Moreover,
dK (δτ (p), δ(τ )(q)) = τ dK (p, q).
We can immediately determine the topology of this metric from its explicit form.
Exericse 4.1. The metric space (H, dK ) is homeomorphic as a topological space to R3 .
The Koranyi norm and distance are much easier to compute than the cc-distance.
Moreover, we see directly that the t-axis is a snowflake of R. However, the Koranyi
distance does not reflect the sub-Riemannian structure of the Heisenberg group. Despite
this, the two metrics are closely related.
Theorem 4.2. The identity mapping (H, dcc ) → (H, dK ) is bi-Lipschitz.
We can use the above result to show the following:
Theorem 4.3. Let E ⊆ R3 be a compact set. Then there is a constant CE ≥ 1 depending
on E such that for any pair of points p, q ∈ E,
||p − q||R3
1/2
≤ dcc (p, q) ≤ CE ||p − q||R3 .
CE
And moreover,
Theorem 4.4. There is a constant C ≥ 1 such that for any ball B in (H, dcc ) of radius
r,
r4
≤ L3 (B) ≤ Cr4 .
C
Proof of Theorem 4.2. The set S = {p ∈ H : ||p||K = 1} is compact, and so the continuous
function
p 7→ dcc (p, 0)
has a minimum m and a maximum M on S. Since 0 ∈
/ S, the minimum m is positive. In
otherwords, for any p ∈ S,
0 < m ≤ dcc (p, 0) ≤ M < ∞.
Now, if p ∈ H is arbitrary, then δ||p||−1 (p) ∈ S. Hence
K
m ≤ ||p||−1
K dcc (p, 0) ≤ M,
as desired. The statement now follows by left-invariance of the metrics.
Proof of Theorem 4.3. By our construction of Heisenberg geodesics, we know that there
is a compact set E 0 ⊇ E such that if p, q ∈ E, then the Heisenberg geodesic γpq connecting
p to q is in E 0 . Now, note that
Z
1/2
0
0
dcc (p, q) =
|(γpq
)1 |2 + |(γpq
)2 |2
while
Z
||p − q||R3 ≤
0
0
0
|(γpq
)1 |2 + |(γpq
)2 |2 + |(γpq
)3 |2
1/2
.
Since γpq is horizontal we may write
0
0
0
0
(γpq
)3 = (γpq
)1 (γpq )2 − (γpq
)2 (γpq
)1 .
Since γpq is contained in E 0 , there is a number M ≥ 1 such that
0
0
0
|(γpq
)3 | ≤ M (|(γpq
)1 | + |(γpq
)2 )|.
This implies that there is some M 0 satisfying
1/2
1/2
0
0
0
0
0
|(γpq
)1 |2 + |(γpq
)2 |2 + |(γpq
)3 |2
≤ M 0 |(γpq
)1 |2 + |(γpq
)2 |2
.
Integrating now show that
||p − q||R3 ≤ M 0 dcc (p, q).
For the upper bound on dcc (p, q), by Theorem 4.2 it suffices to show that
1/2
||p−1 ∗ q||K ≤ CE ||p − q||R3 .
Let p = (zp , tp ) and q = (zq , tq ). Then
||p−1 ∗ q||K ≤ C(|zp − zq | + (|tp − tq | + 2| im(zp zq )|)1/2 ).
Since p and q are in the compact set E, we may find CE so that
|zp − zq | ≤ CE |zp − zq |1/2
and
im(zp zq )1/2 ≤ CE |zp − zq |1/2 .
This completes the proof.
Proof of Theorem 4.4. By translation invariance of the Lebesgue measure, it suffices to
consider balls centered at the origin. Theorem 4.2 implies that it suffices to consider balls
in the Koranyi metric. However, there is a universal constant c ≥ 1 such that
{(x, y, t) : |x| < c−1 r, |y| < c−1 r, |t| < (c−1 r)2 } ⊆ BK (0, r)
and
BK (0, r) ⊆ {(x, y, t) : |x| < cr, |y| < cr, |t| < (cr)2 }.
This implies the result.
5. The Poincaré inequality in the Heisenberg group
Theorem 5.1. The Heisenberg group (H, dcc , L3 ) supports a 1-Poincaré inequality.
Let u : H → R be a Lipschitz function and let g be an upper gradient of u. Fix a ball
B = B(p0 , r) ⊆ H. Since left-translation is an isometry, we may assume without loss of
generality that p0 is the origin.
For q ∈ H, set dq = dcc (0, q), and let γq : [0, dq ] → H be the arclength parameterized
Heisenberg geodesic connecting 0 to q. Then for any p ∈ H, p ∗ γq is the arclength parameterized Heisenberg geodesic connecting p to p ∗ q. Thus the upper gradient inequality
implies that
Z dq
g(p ∗ γq (s)) ds.
|u(p) − u(p ∗ q)| ≤
0
3
Now, the left-invariance of L allows us to change variables as in the Euclidean case, and
apply the above inequality:
Z
Z Z
1
3
|u(p) − uB | dL (p) ≤ 3
|u(p) − u(q)| dL3 (q)dL3 (q)
L
(B)
B
ZB Z B
1
≤ 3
χB (p)χB (p ∗ q)|u(p) − u(p ∗ q)| dL3 (p)dL3 (q)
L (B) H H
Z Z Z dq
1
χB (p)χB (p ∗ q)g(p ∗ γq (s)) ds dL3 (p) dL3 (q)
≤ 3
L (B) H H 0
Z Z dq Z
1
≤ 3
χB (p)χB (p ∗ q)g(p ∗ γq (s)) dL3 (p) ds dL3 (q)
L (B) H 0
H
Now, we change variables again: w = p ∗ γq (s). Using the right-invariance of the measure:
Z
Z
3
χB (p)χB (p ∗ q)g(p ∗ γq (s)) dL (p) =
χB∗γq (s) (w)χB∗q−1 ∗γq (s) (w)g(w) dL3 (w).
H
H
Now, if the integrand above is non-zero, then
w = a ∗ γq (s) = b ∗ q −1 ∗ γq (s)
for some a, b ∈ B. Thus
q = b ∗ a−1 ∈ 2B.
As a result,
Z
3
Z
χB∗γq (s) (w)χB∗q−1 ∗γq (s) (w)g(w) dL (w) ≤ χ2B (q)
H
2B
g(w) dL3 (w).
Plugging this in, we get
Z
|u(p) − uB | dL3 (p) ≤
B
Z
Z Z dq
1
χ2B (q)
g(w) dL3 (w)dsdL3 (q)
L3 (B) H 0
2B
Z
Z
1
≤ 3
dq
g(w)dL3 (w)dL3 (q)
L (B) 2B
2B
Z
g(w)dL3 (w).
≤ Cr
2B
This implies the result.