# Moduli of Legendrian foliations and quadratic differentials in the
Heisenberg group

Robin Timsit

###### Abstract

The aim of the paper is to prove the following result concerning moduli of
curve families in the Heisenberg group. Let $\Omega$ be a domain in the
Heisenberg group foliated by a family $\Gamma$ of legendrian curves. Assume
that there is a quadratic differential $q$ on $\Omega$ in the kernel of an
operator defined in [Tim2] and every curve in $\Gamma$ is a horizontal
trajectory for $q$. Let $l_{\Gamma}:\Omega\rightarrow]0,+\infty[$ be the
function that associates to a point $p\in\Omega$, the $q$-length of the leaf
containing $p$. Then, the modulus of $\Gamma$ is

$M_{4}(\Gamma)=\int_{\Omega}\frac{|q|^{2}}{(l_{\Gamma})^{4}}\mathrm{d}L^{3}.$

## Introduction and statement of results

Holomorphic quadratic differentials on Riemann surfaces are well-studied
objects that play a central role in Teichmüller theory; in particular in
Teichmüller’s theorem (see [Ahl, Hub, IT]) and because they are in one-to-one
correspondance with measured foliation [HM]. It leads us to curve families and
especially to the modulus (or its reciprocal, the extremal length) of a curve
family which is a classical conformal invariant studied since the work of
Grötzsch and his famous length-area argument. The interplay between
holomorphic quadratic differentials and moduli (or extremal lengths) of curve
families was then naturally considered (see [Jen, Str, Iva, Ker]). The purpose
of this work is to study the interplay between quadratic differentials and
moduli of curve families in the setting of $3$-dimensional spherical CR
geometry.

A holomorphic quadratic differential on a Riemann surface $X$ is a holomorphic
section of the line bundle $K(X)\otimes K(X)$ where $K(X)$ is the canonical
bundle of $X$. In a local holomorphic coordinate $z$, a quadratic differential
writes as $q(z)\mathrm{d}z^{2}$ for a holomorphic function $q$ of $z$. A
holomorphic quadratic differential naturally gives rise to a flat singular
metric on $X$ locally given by $\sqrt{|q(z)|}|\mathrm{d}z|$. Thus, the
following were defined.

  * •

The $q$-length of a curve $\gamma$ is

$l_{q}(\gamma)=\int_{\gamma}\sqrt{|q|}.$

  * •

The $q$-area of $X$ is

$\mathrm{Area}_{q}(X)=\int_{X}|q|.$

We emphase that the holomorphicity of $q$ is not mandatory to define these
notions.

Associated to a holomorphic quadratic differential are two (measured) singular
foliations said horizontal and vertical: a curve $\gamma$ on $X$ is called a
horizontal trajectory of $q$ if $q(\gamma^{\prime})>0$ and a vertical
trajectory of $q$ if $q(\gamma^{\prime})<0$. What lies behind the terminology
is that $q$, away from a zero, defines a local holomorphic coordinate (called
natural) $w=x+iy$ such that $q=\mathrm{d}w^{2}$ and $\gamma$ is a horizontal
(resp. vertical) trajectory of $q$ if and only if $y(\gamma)$ (resp.
$x(\gamma)$) is constant. The computation of the extremal lengths of
foliations has been done in a slightly different form, see for instance the
remark [Ker, p. 34] that extends a work a Jenkins [Jen] and Strebel [Str] (see
also [Ker, Theorem 3.1]). For a precise statement of Kerckhoff’s remark, see
[PT, Theorem 2.21] and for a detailed proof see [Iva, section 1.2]. The
precise result we will be interested in states as follow (refer to section 1.2
for the definition of the modulus of a curve family):

######  Theorem 1 (Theorem 2.1.1).

Let $D$ be a domain in $\mathbb{C}$ foliated by a family $\Gamma$ of
horizontal trajectories of a holomorphic quadratic differential $q$ on $D$.
Let $l_{\Gamma}:D\rightarrow]0,+\infty[$ be the function that associates to a
point $z\in D$, the $q$-length of the leaf containing $z$. Then the modulus of
$\Gamma$ is

$M_{2}(\Gamma)=\int_{D}\frac{|q|}{(l_{\Gamma})^{2}}\mathrm{d}L^{2}.$

Let us point out that the computation of extremal length ([Ker]) is usually
done using natural coordinates and that we don’t have access to those in the
setting of spherical CR geometry. Thus, we need to find another strategy to
prove our result. This is why an elementary proof (because of regularity
assumptions) of Theorem 1 will be given in section 2.1. It relies on a
decomposition of the $q$-area (see Lemma 2.1.2). As stated, the purpose of
this paper is to prove an analog result in the setting of ($3$-dimensional)
spherical CR geometry.

#### CR manifolds

For completeness, we give (spherical/strictly pseudoconvex) CR manifolds a
pretty long introduction. The interested reader can learn more about CR
manifolds for instance in [Jac, Bog]. The conventions used in those books are
not the same as here. In [Jac], the CR structure $V$ is used as our
$\overline{V}$ (that is its sections serve as Cauchy-Riemann operator whereas
here we understand sections of $\overline{V}$ as Cauchy-Riemann operators). In
[Bog], CR manifolds are defined without specifying the rank of the CR
structure and thus, what we call CR manifolds here are called CR manifolds of
hypersurface type there.

A CR manifold is a manifold $M^{2n+1}$ endowed with (complex) rank $n$
subbundle $V$ of the complexified tangent bundle $\mathbb{C}TM$ satisfying
$V\cap\overline{V}=\\{0\\}$ and $[V,V]\subset V$. $V$ is called the CR
structure on $M$. Equivalently (though we won’t use this point of view here),
a CR manifold is a $2n+1$ manifold $M$ endowed with a (real) rank $2n$
distribution P and a complex structure $J$ on P.

######  Remark 0.0.1.

  * •

Notice that for a $3$-dimensional manifold, the condition $[V,V]\subset V$ is
always fulfilled.

  * •

Some authors omit the rank $n$ dimension of the subbundle $V$ in the
definition of a CR manifold and call a manifold endowed with a rank $n$ CR
structure a CR manifold of hypersurface type.

  * •

Every hypersurface $M$ of $\mathbb{C}^{n+1}$ inherits a CR structure from the
complex structure on $\mathbb{C}^{n+1}$ simply by taking $V=\mathbb{C}TM\cap
T^{1,0}\mathbb{C}^{n+1}$. In particular, the unit sphere $S^{2n+1}$ in
$\mathbb{C}^{n+1}$ has such a CR structure said standard.

When a new structure is introduced, we usually like to understand the maps
preserving this structure. A map $f:(M,V)\rightarrow(M^{\prime},V^{\prime})$
between CR manifold is called a CR map if $f_{\ast}(V)\subset V^{\prime}$. A
function $f:(M,V)\rightarrow\mathbb{C}^{k}$ is called a CR function if
$f_{\ast}(V)\subset T^{1,0}\mathbb{C}^{k}$. A CR diffeomorphism between CR
manifolds is a CR map which is also a diffeomorphism. Two CR manifolds are
called equivalent (or CR equivalent) if there is a CR diffeomorphism from one
to the other.

A CR manifold $(M,V)$ is called strictly pseudoconvex if $V\oplus\overline{V}$
is a (complex) contact structure on $M$ (that is, locally
$V\oplus\overline{V}=\ker(\omega)$ for a $1$-form $\omega$ such that
$\omega\wedge(\mathrm{d}\omega)^{n}$ is nowhere zero). Using Rumin complex
[Rum], one can define quadratic differentials on strictly pseudoconvex CR
$3$-manifolds [Tim2] and recover the notions of trajectories, $q$-length and
$q$-volume. For completeness, we give a short overview about them in section
1.1.

#### Spherical CR manifolds and the Heisenberg group

A ($3$-dimensional) spherical CR manifold [BS] is a manifold locally modeled
on the unit sphere $S^{3}\subset\mathbb{C}^{2}$ endowed with its standard CR
structure. Equivalently (because of a Liouville type theorem: a CR
diffeomorphism between open subsets of $S^{3}$ is the restriction of the
action of an element of $\mathrm{PU}(2,1)$ on $S^{3}$; see [BS, p. 224] for
instance), it is $(G,X)$-manifold for $X=S^{3}$ and $G=\mathrm{PU}(2,1)$.
Another local model for a spherical CR manifold is the Heisenberg group
${\mathcal{H}}$ which is $\mathbb{C}\times\mathbb{R}$ endowed with the group
law

$(z,t)(z^{\prime},t^{\prime})=(z+z^{\prime},t+t^{\prime}+2\Im(z\overline{z}^{\prime})).$

As for its CR structure, it is given by $V_{\mathcal{H}}=span(Z)$ where $Z$ is
the complex vector field

$Z=\frac{\partial}{\partial z}+i\overline{z}\frac{\partial}{\partial t}.$

It is easy to see that
$V_{\mathcal{H}}\oplus\overline{V}_{\mathcal{H}}=\ker(\omega)$ with
$\omega=\mathrm{d}t-i\overline{z}\mathrm{d}z+iz\mathrm{d}\overline{z}$. Since
$\omega$ is a contact form, $({\mathcal{H}},V_{\mathcal{H}})$ is a strictly
pseudoconvex CR manifold and it is known to be equivalent to the unit sphere
minus a point endowed with its standard CR structure. Thus, one can see a
spherical CR manifold as locally modeled on $({\mathcal{H}},V_{\mathcal{H}})$,
that is a manifold endowed with an atlas of charts (called a spherical CR
atlas) $(U_{i},\varphi_{i}:U_{i}\rightarrow
U_{i}^{\prime}\subset{\mathcal{H}})$ such that every
$\varphi_{j}\circ\varphi_{i}^{-1}$ is a CR diffeomorphism.

Moreover, a function $f:U\subset{\mathcal{H}}\rightarrow\mathbb{C}$ is CR if
and only if $\overline{Z}f=0$ and it is easy to see that a map
$g=(g_{1},g_{2}):U\rightarrow V$ between open subsets of ${\mathcal{H}}$ is CR
if and only if both $g_{1}$ and $g_{2}+i|g_{1}|^{2}$ are CR functions. With
this angle several differential operators can be defined on quadratic
differentials on spherical CR manifolds and especially one: $B_{2}$ that
remained quite mysterious in [Tim2]. The main result of this paper gives a
geometric interpretation of quadratic differentials in the kernel of this
$B_{2}$, this is the result announced in the abstract (refer to section 1.2
for the modulus of a family of a legendrian curves in the Heisenberg group).

######  Theorem 2 (Theorem 2.2.1).

Let $\Omega$ be a domain in the Heisenberg group foliated by a family $\Gamma$
of legendrian (everywhere tangent to the contact distribution) curves. Assume
that there is a quadratic differential $q$ on $\Omega$ such that $B_{2}q=0$
and every curve in $\Gamma$ is a horizontal trajectory for $q$. Let
$l_{\Gamma}:\Omega\rightarrow]0,+\infty[$ be the function that associates to a
point $p\in\Omega$, the $q$-length of the leaf containing $p$.Then, the
modulus of $\Gamma$ is

$M_{4}(\Gamma)=\int_{\Omega}\frac{|q|^{2}}{(l_{\Gamma})^{4}}\mathrm{d}L^{3}.$

If in addition we ask the leaves to have the same $q$-length $C>0$, then the
modulus expresses with the $q$-volume (denoted $\mathrm{Vol}_{q}$, see
Definition 1.1.2) and $C$ (see Corollary 2.2.4):

$M_{4}(\Gamma)=\frac{1}{C^{4}}\mathrm{Vol}_{q}(\Omega).$

As for the proof given here of the result in the complex plane, the proof of
this theorem relies on a decomposition of the $q$-volume. It is for this
decomposition that we need the quadratic differential to be in the kernel of
$B_{2}$.

The paper is organized as follow. The first section begins with a short
overview about quadratic differentials in spherical CR geometry and continues
with moduli of curve families in the complex plane and in the Heisenberg
group. In the second section, we prove theorems 1 and 2 and we give some
examples and counter examples) of application of theorem 2.

##  1 Quadratic differentials and moduli of curve families

###  1.1 Quadratic differentials on strictly pseudoconvex CR manifolds

We first give a short overview about quadratic differentials on a CR manifold.
Most of this section can be found in [Tim2].

Let $(M,V)$ be a $3$-dimensional strictly pseudoconvex CR manifold. For a
canonical bundle over $M$ we want to consider $V^{\ast}$. In order to have
natural operators on it, we would want to see $V^{\ast}$ as a subbundle of
$\mathbb{C}T^{\ast}M$. However, this implies a choice of a complement of
$V\oplus\overline{V}$. In order to avoid that choice, we define a line bundle
$\wedge^{1,0}$ as follow.

Let $B^{1,0}$ be the subbundle of $\mathbb{C}T^{\ast}M$ of forms vanishing on
$\overline{V}$ and let $A$ be the subbundle of forms vanishing on
${\text{{P}}}=V\oplus\overline{V}$. Then, we define

$\wedge^{1,0}=\faktor{B^{1,0}}{A}$

which is a line bundle naturally isomorphic to $V^{\ast}$.

######  Definition 1.1.1 (Quadratic differential).

A quadratic differential on a strictly pseudoconvex CR manifold is a smooth
section of the line bundle $\wedge^{1,0}\otimes\wedge^{1,0}$.

The first geometric data induced by a quadratic differential on a strictly
pseudoconvex manifold can be defined by analogy with the Riemann surface case.
However, since quadratic differentials are defined up to a contact form, we
shall restrict to legendrian curves (that is curves everywhere tangent to the
contact distribution).

######  Definition 1.1.2 (Trajectories, length, volume).

Let $q$ be a quadratic differential on a $3$-dimensional strictly pseudoconvex
CR manifold $(M,V)$. Let also $\gamma:I\rightarrow M$ be a legendrian
parametrized curve on $M$

  1. 1.

$\gamma$ is called a

     * •

horizontal trajectory of $q$ if $q(\gamma^{\prime}(s))>0$ for all $s\in I$;

     * •

vertical trajectory of $q$ if $q(\gamma^{\prime}(s))<0$ for all $s\in I$.

  2. 2.

$\sqrt{|q|}$ is a length element that can be integrated along any legendrian
curve. Thus, we call the $q$-length of the curve $\gamma$, the number

$l_{q}(\gamma)=\int_{\gamma}\sqrt{|q|}.$

  3. 3.

The $q$-volume of $M$ is

$\mathrm{Vol}_{q}(M)=\int_{M}|q|^{2}.$

Three differential operators on quadratic differentials on spherical CR
manifolds that allowed to obtain analogue of half-translation structures were
defined in [Tim2]. First, let us emphase that on a spherical CR manifold, one
can understand quadratic differentials in terms of cocycle relations. Namely,
assume that the manifold $M$ is now endowed with a spherical CR atlas
$(U_{i},\varphi_{i}:U_{i}\rightarrow U_{i}^{\prime}\subset{\mathcal{H}})_{i\in
I}$. On every open subset of ${\mathcal{H}}$, a quadratic differential is the
class modulo the contact form of a form of type $f\mathrm{d}z^{2}$. We write
such classes $[f\mathrm{d}z^{2}]_{\omega}$. Then, if $q$ is a quadratic
differential on $M$, in every chart $(U_{i},\varphi_{i})$ we have

$q=\varphi_{i}^{\ast}[q_{i}\mathrm{d}z^{2}]_{\omega}$

with $q_{i}\in C^{\infty}(U_{i}^{\prime},\mathbb{C})$. Since $q$ is globally
defined on $M$, if $U_{i}\cap U_{j}\neq\emptyset$ we must have

$[q_{i}\mathrm{d}z^{2}]_{\omega}=g_{j,i}^{\ast}[q_{j}\mathrm{d}z^{2}]_{\omega}\text{
on $\varphi_{i}(U_{i}\cap U_{j})$ where
$g_{j,i}=\varphi_{j}\circ\varphi_{i}^{-1}$.}$

That is

$[q_{i}\mathrm{d}z^{2}]_{\omega}=[(q_{j}\circ
g_{j,i})\left(Zg_{j,i}^{1}\right)^{2}\mathrm{d}z^{2}]_{\omega}\text{ on
$\varphi_{i}(U_{i}\cap U_{j})$}$

where $g_{j,i}=\varphi_{j}\circ\varphi_{i}^{-1}$ and
$g_{j,i}=(g_{j,i}^{1},g_{j,i}^{2})$. Which legitimates the following
definition.

######  Definition 1.1.3 (Quadratic differential on a spherical CR manifold).

A quadratic differential on a spherical CR manifold $M$ with spherical CR
atlas $(U_{i},\varphi_{i})_{i\in I}$ is a collection of smooth complex valued
functions $q_{i}\in C^{\infty}(\varphi_{i}(U_{i}))$ satisfying for every
$i,j\in I$

$q_{i}=(q_{j}\circ g_{j,i})\left(Zg_{j,i}^{1}\right)^{2}\text{ on
$\varphi_{i}(U_{i}\cap U_{j})$ where
$g_{j,i}=\varphi_{j}\circ\varphi_{i}^{-1}$.}$

We can now define operators on quadratic differentials on ${\mathcal{H}}$ and
"lift" them using charts on the manifold. On ${\mathcal{H}}$, the operators
are the following: let $q$ be a quadratic differential defined in an open
subset $U$ of ${\mathcal{H}}$

$\mathrm{D}^{\prime}_{2,{\mathcal{H}}}[q\mathrm{d}z^{2}]_{\omega}=\left(2qZ\overline{Z}q-Zq\overline{Z}q-4iqTq\right)[\mathrm{d}z^{3}]_{\omega}\otimes\omega\wedge\mathrm{d}z\text{,}$
$\mathrm{D}^{\prime\prime}_{2,{\mathcal{H}}}[q\mathrm{d}z^{2}]_{\omega}=\left(2q\overline{Z}^{2}q-\left(\overline{Z}q\right)^{2}\right)[\mathrm{d}z^{3}]_{\omega}\otimes\omega\wedge\mathrm{d}\overline{z}\text{
and}$
$B_{2,{\mathcal{H}}}[q\mathrm{d}z^{2}]_{\omega}=\left(\overline{Z}\left(|q|^{2}\right)+\overline{q}\overline{Z}q\right)\omega\wedge\mathrm{d}\overline{z}\otimes\omega.$

######  Remark 1.1.4.

  * •

The two first operators are extensions to quadratic differentials of a
splitting by Garfield and Lee [GL] of Rumin operator [Rum] on $(1,0)$-forms. A
quadratic differential $q$ annihilating these two operators is locally, away
from a zero, the square of the differential of a complex valued CR function.
To be precise, if $q$ is a quadratic differential in the kernels of both
$\mathrm{D}_{2,{\mathcal{H}}}^{\prime}$ and
$\mathrm{D}_{2,{\mathcal{H}}}^{\prime\prime}$, then, locally, away from a
zero, $q=[(Zf)^{2}\mathrm{d}z^{2}]_{\omega}$ for a complex valued CR function
$f$ (see [Tim2, Proposition 2.3.4]).

  * •

The operator $B_{2,{\mathcal{H}}}$ appeared when the following question was
asked: for a complex valued CR function $f$, how can we ensure that (locally)
there is a real valued function $h$ such that $(f,h)$ is a CR map ? That is,
under what condition does the equation $\overline{Z}(h+i|f|^{2})=0$ have a
real valued solution $h$ ? It turns out that, assuming $f$ CR,
$[(Zf)^{2}\mathrm{d}z^{2}]_{\omega}$ being in the kernel of $B_{2}$ is a
necessary and sufficient condition for the equation
$\overline{Z}(h+i|f|^{2})=0$ to have a real valued solution. For details,
refer to [Tim2] right before Definition 2.3.5.

Thus, if a quadratic differential $q$ annihilates
$\mathrm{D}_{2,{\mathcal{H}}}^{\prime}$,
$\mathrm{D}_{2,{\mathcal{H}}}^{\prime\prime}$ and $B_{2,{\mathcal{H}}}$ then
$q$ is locally, away from a zero, of the form
$g^{\ast}[\mathrm{d}z^{2}]_{\omega}$ for a CR map $g$ (see [Tim2] and
Proposition 1.1.7 below).

In order to define the new line bundles involved here, first denote
$\wedge^{2}$ the subbundle of $2$-forms vanishing on ${\text{{P}}}^{2}$. Now,
define $\wedge^{1,1}$ the subbundle of forms in $\wedge^{2}$ with vanishing
interior product with every vector of $V$ and $\wedge^{2,0}$ the subbundle of
forms in $\wedge^{2}$ with vanishing interior product with every vector of
$\overline{V}$. Then,

######  Definition 1.1.5.

Let $M$ be a spherical CR manifold $M$ with atlas $(U_{i},\varphi_{i})_{i}$.
Then the following operators are well-defined (where $\Gamma(E)$ refers to the
space of smooth sections of any line bundle $E$ over $M$):

$\begin{array}[]{cccc}\mathrm{D}_{2}^{\prime}:&\Gamma\left(\left(\wedge^{1,0}\right)^{\otimes
2}\right)&\longrightarrow&\Gamma\left(\left(\wedge^{1,0}\right)^{\otimes
3}\otimes\wedge^{2,0}\right)\\\
&q&\longmapsto&\varphi_{i}^{\ast}\mathrm{D}^{\prime}_{2,{\mathcal{H}}}[q_{i}\mathrm{d}z^{2}]_{\omega}\text{
on $U_{i}$ where
$q=\varphi_{i}^{\ast}[q_{i}\mathrm{d}z^{2}]_{\omega}$,}\end{array}$
$\begin{array}[]{cccc}\mathrm{D}_{2}^{\prime\prime}:&\Gamma\left(\left(\wedge^{1,0}\right)^{\otimes
2}\right)&\longrightarrow&\Gamma\left(\left(\wedge^{1,0}\right)^{\otimes
3}\otimes\wedge^{1,1}\right)\\\
&q&\longmapsto&\varphi_{i}^{\ast}\mathrm{D}^{\prime\prime}_{2,{\mathcal{H}}}[q_{i}\mathrm{d}z^{2}]_{\omega}\text{
on $U_{i}$ where $q=\varphi_{i}^{\ast}[q_{i}\mathrm{d}z^{2}]_{\omega}$
and}\end{array}$
$\begin{array}[]{cccc}B_{2}:&\Gamma\left(\left(\wedge^{1,0}\right)^{\otimes
2}\right)&\longrightarrow&\Gamma\left(\wedge^{1,1}\otimes A\right)\\\
&q&\longmapsto&\varphi_{i}^{\ast}B_{2,{\mathcal{H}}}[q_{i}\mathrm{d}z^{2}]_{\omega}\text{
on $U_{i}$ where
$q=\varphi_{i}^{\ast}[q_{i}\mathrm{d}z^{2}]_{\omega}$.}\end{array}$

######  Example 1.1.6.

  * •

For an example of a quadratic differential annihilating
$\mathrm{D}_{2}^{\prime}$ and $\mathrm{D}_{2}^{\prime\prime}$, as we said just
take the square of the differential of a CR function. For instance, we will
discuss later about the trajectories of the quadratic differential
$\left[\frac{\overline{z}^{2}}{(t+i|z|^{2})^{2}}\mathrm{d}z^{2}\right]_{\omega}$
on ${\mathcal{H}}$.

  * •

$\left[\frac{\overline{z}^{2}(t^{2}+|z|^{4})^{\frac{2}{3}}}{|z|^{\frac{8}{3}}(t+i|z|^{2})^{2}}\mathrm{d}z^{2}\right]_{\omega}$
is an example of a quadratic differential on
${\mathcal{H}}\backslash\\{0\\}\times\mathbb{R}$ in the kernel of $B_{2}$. We
will be interested in it again in section 2

  * •

Finally,
$\left[\frac{(t-i|z|^{2})^{2}}{(t+i|z|^{2})^{4}}\mathrm{d}z^{2}\right]_{\omega}$
is a quadratic differential on ${\mathcal{H}}\backslash\\{0\\}$ annihilating
$\mathrm{D}_{2}^{\prime}$, $\mathrm{D}_{2}^{\prime\prime}$ and $B_{2}$

Those operators were defined in order to obtain the following result, giving a
structure on a spherical CR manifold analog to a half-translation structure.

######  Proposition 1.1.7 (Proposition 2.3.1 in [Tim2]).

Let $q$ be a non-zero quadratic differential on a spherical CR manifold $M$
such that $\mathrm{D}^{\prime}_{2}q=0$, $\mathrm{D}^{\prime\prime}_{2}q=0$ and
$B_{2}q=0$. Then, around every point $p\in M$ where $q(p)\neq 0$, there is a
chart $(U,\varphi)$ such that $q=\varphi^{\ast}[\mathrm{d}z^{2}]_{\omega}$ on
$U$.

Moreover, two such charts differ only by translation and sign, that is by
$(z,t)\mapsto\left(\pm z+z_{0},t+t_{0}+2\Im(\pm z\overline{z}_{0})\right)$
which is an analogue situation as half-translation structures.

###  1.2 Moduli of curve families

First, we recall what is the modulus of a family of curves in the complex
plane (see [Vas] for instance). Let $U$ be an open subset of $\mathbb{C}$ and
$\Gamma$ be a family of $C^{1}$ curves lying in $U$. Denote
$\operatorname*{adm}(\Gamma)$ the set of measurable Borel functions
$\rho:U\rightarrow[0,\infty]$ such that

$\int_{\gamma}\rho\mathrm{d}l=\int_{a}^{b}\rho(\gamma(s))|\dot{\gamma}_{1}(s)|\mathrm{d}s\geq
1$

for every curve $\gamma:]a,b[\longrightarrow\Omega$ in $\Gamma$. Elements of
$\operatorname*{adm}(\Gamma)$ are called densities. Define the $2$-modulus of
$\Gamma$ by

$M_{2}(\Gamma)=\underset{\rho\in\operatorname*{adm}(\Gamma)}{\inf}\int_{U}\rho^{2}\mathrm{d}L^{2}$

where $\mathrm{d}L^{2}$ is the Lebesgue measure on $\mathbb{R}^{2}$. A density
$\rho_{0}$ is said extremal for $\Gamma$ if

$M_{2}(\Gamma)=\int_{U}\rho_{0}^{2}\mathrm{d}L^{2}.$

######  Remark 1.2.1.

  * •

Sometimes, authors give another (though equivalent) definition of the modulus
of a curve family. For a non negative measurable Borel function
$\rho:U\rightarrow[0,+\infty]$, denote
$\mathrm{L}_{\rho}(\Gamma)=\underset{\gamma\in\Gamma}{\inf}\int_{\gamma}\rho\mathrm{d}l$
and $\mathrm{A}_{\rho}(U)=\int_{U}\rho^{2}\mathrm{d}L^{2}$. Then, it is clear
that

$M_{2}(\Gamma)=\underset{\rho}{\inf}\frac{\mathrm{A}_{\rho}(U)}{(\mathrm{L}_{\rho}(\Gamma))^{2}}$

where the infimum is taken over all non negative measurable Borel functions
$\rho$ in $U$. We chose the other definition here because with it, the
essential uniqueness of an extremal density is more clear (see Theorem 2.1.2
in [Vas] for instance).

  * •

Using the same notations, the extremal length of $\Gamma$ is

$E(\Gamma)=\underset{\rho}{\sup}\frac{(\mathrm{L}_{\rho}(\Gamma))^{2}}{\mathrm{A}_{\rho}(U)}.$

For curves in the Heisenberg group, things are quite the same. One needs to
consider legendrian curves because these are the locally rectifiable curves
for the usual left invariant metric
$d_{\mathcal{H}}(p,q)=\|p^{-1}q\|_{\mathcal{H}}$ with
$\|(z,t)\|_{\mathcal{H}}=\left(t^{2}+|z|^{4}\right)^{\frac{1}{4}}$. Let
$\gamma=(\gamma_{1},\gamma_{2}):I\longrightarrow{\mathcal{H}}$ be a $C^{1}$
curve. $\gamma$ is legendrian if and only if

$\dot{\gamma}_{2}(s)=-2\Im(\overline{\gamma}_{1}(s)\dot{\gamma}_{1}(s))\text{
for every $s\in I$.}$

Then, let $\Gamma$ be a family of legendrian curves in an open subset $\Omega$
of ${\mathcal{H}}$. Denote $\operatorname*{adm}(\Gamma)$ the set of measurable
Borel functions $\rho:\Omega\longrightarrow[0,\infty]$ such that

$\int_{\gamma}\rho\mathrm{d}l=\int_{a}^{b}\rho(\gamma(s))|\dot{\gamma}_{1}(s)|\mathrm{d}s\geq
1$

for every curve $\gamma:]a,b[\longrightarrow\Omega$ in $\Gamma$. Elements of
$\operatorname*{adm}(\Gamma)$ are called densities. Define the $4$-modulus of
$\Gamma$ by

$M_{4}(\Gamma)=\underset{\rho\in\operatorname*{adm}(\Gamma)}{\inf}\int_{\Omega}\rho^{4}\mathrm{d}L^{3}$

where $\mathrm{d}L^{3}$ is the Lebesgue measure on $\mathbb{R}^{3}$. A density
$\rho_{0}$ is said extremal for $\Gamma$ if

$M_{4}(\Gamma)=\int_{\Omega}\rho_{0}^{4}\mathrm{d}L^{3}.$

As in the complex case, an extremal density is essentially unique (see [BFP2,
Proposition 3.4] for instance). The modulus of a family of curves is a
conformal invariant [KR2] and a classical tool in the theory of quasiconformal
maps in the Heisenberg group [KR, KR2], with various purposes such as finding
extremal quasiconformal maps (see [BFP, BFP2, Tim, Tan]) or for proving non-
existence of quasiconformal maps between certain CR manifolds (see [Kim,
Min]). We will be interested in computing the modulus of a family of curves
when this family is a foliation by horizontal trajectories of a quadratic
differential.

##  2 Modulus of a foliation by trajectories of a quadratic differential

###  2.1 The result in the complex plane

As stated in the introduction, Theorem 1 is a local formulation of a well
known ([PT, Theorem 2.21]) result dealing with the computation of the modulus
(or extremal length) of a foliation in the complex plane. We give it here an
elementary proof that doesn’t use all the potential of conformal geometry
because these are tools we don’t have to our disposal in the setting of
families of legendrian curves in the Heisenberg group. To be precise, we give
a proof that uses only the annihilation of the Cauchy-Riemann operator by the
quadratic differential.

######  Theorem 2.1.1.

Let $I,J$ be intervals and $\Phi:I\times J\longrightarrow U\subset\mathbb{C}$
be a $C^{2}$-diffeomorphism. Denote

$\Gamma=\\{\Phi(.,p)\ |\ p\in J\\}.$

Let $q$ be a holomorphic quadratic differential on $U$ such that every curve
in $\Gamma$ is a horizontal trajectory of $q$. Then,

$M_{2}(\Gamma)=\int_{U}\frac{|q|}{(l_{\Gamma})^{2}}\mathrm{d}L^{2}$

where, for every $z\in U$, $l_{\Gamma}(z)=l_{q}(\Phi(.,\Phi^{-1}_{2}(z)))$.

The proof is a consequence of the following lemma:

######  Lemma 2.1.2.

Let $q:U\rightarrow\mathbb{C}$ be a holomorphic quadratic differential on a
domain $U$ of $\mathbb{C}$. Let $I,J$ be intervals and $\Phi:I\times
J\rightarrow U$ be a diffeomorphism such that every curve $\Phi(.,p)$ for
$p\in J$ is a horizontal trajectory for $q$. Then, there is a nowhere
vanishing function $\lambda:J~\rightarrow\mathbb{R}$ such that

$\displaystyle(|q|\circ\Phi)\mathrm{J}_{\Phi}=\sqrt{(|q|\circ\Phi)}|\partial_{s}\Phi|\lambda$
(1)

where $\mathrm{J}_{\Phi}$ is the Jacobian determinant of $\Phi$.

######  Remark 2.1.3.

  * •

What the lemma says is that, on $I\times J$, the $q$-area decomposes as the
product of the $q$-length and a measure on $J$.

  * •

We will prove and use the lemma with the conclusion written in another form,
easier for computations. We use here the same notations and hypotheses than in
the lemma. First, we write

$\mu=\sqrt{(|q|\circ\Phi)}|\partial_{s}\Phi|=\sqrt{(q\circ\Phi)(\partial_{s}\Phi)^{2}}.$

We then multiply both sides of equation 1 by $|\partial_{s}\Phi|^{2}$ (which
is nowhere zero since $\Phi$ is a diffeomorphism) to obtain

$(|q|\circ\Phi)|\partial_{s}\Phi|^{2}\mathrm{J}_{\Phi}=\mu|\partial_{s}\Phi|^{2}\lambda.$

Then, notice that $(|q|\circ\Phi)|\partial_{s}\Phi|^{2}=\mu^{2}$ and divide
both sides by $\mu$ (which is nowhere zero) and we finally have the form of
the conclusion we will prove:

$\displaystyle\sqrt{(q\circ\Phi)(\partial_{s}\Phi)^{2}}\mathrm{J}_{\phi}=\lambda|\partial_{s}\Phi|^{2}.$
(2)

###### Proof of Theorem 2.1.1.

Assuming the lemma for now, let $\rho\in\mathrm{adm}(\Gamma)$ be an admissible
density for $\Gamma$. Then,

$\int_{U}\rho^{2}\mathrm{d}L^{2}=\int_{J}\int_{I}\rho^{2}(\Phi)|\mathrm{J}_{\Phi}|\mathrm{d}s\mathrm{d}p.$

However, for every $p$ in $J$, Cauchy-Schwarz inequality gives

$\left(\int_{I}\left(\rho(\Phi)|\mathrm{J}_{\Phi}|^{\frac{1}{2}}\right)^{2}\mathrm{d}s\right)\left(\int_{I}\left(\frac{|\partial_{s}\Phi|}{|\mathrm{J}_{\Phi}|^{\frac{1}{2}}}\right)^{2}\mathrm{d}s\right)\geq\left(\int_{I}\rho(\Phi)|\partial_{s}\Phi|\mathrm{d}s\right)^{2}\underset{\text{$\rho\in\mathrm{adm}(\Gamma)$}}{\geq}1.$

Using equation 2, we have for every $p\in J$

$\int_{I}\left(\frac{|\partial_{s}\Phi|}{|\mathrm{J}_{\Phi}|^{\frac{1}{2}}}\right)^{2}\mathrm{d}s=\int_{I}\frac{\sqrt{(q\circ\Phi)\left(\partial_{s}\Phi\right)^{2}}}{|\lambda|}\mathrm{d}s=\frac{l_{q}(\Phi(.,p))}{|\lambda(p)|}.$

Thus,

$\int_{U}\rho^{2}\mathrm{d}L^{2}\geq\int_{J}\frac{|\lambda(p)|}{l_{q}(\Phi(.,p))}\mathrm{d}p.$

Since it is true for every $\rho\in\mathrm{adm}(\Gamma)$,

$M_{2}(\Gamma)\geq\int_{J}\frac{|\lambda(p)|}{l_{q}(\Phi(.,p))}\mathrm{d}p.$

Now, let $\rho_{0}$ be defined for every $z$ by
$\rho_{0}(z)=\frac{\sqrt{|q(z)|}}{l_{q}(\Phi(.,\Phi^{-1}_{2}(z)))}$. Then, if
$\Phi(.,p_{0})$ is a curve in $\Gamma$, one has

$\int_{\Phi(.,p_{0})}\rho_{0}\mathrm{d}l=\frac{1}{l_{q}(\Phi(.,p_{0}))}\int_{I}\sqrt{|q(\Phi(s,p_{0}))|}|\partial_{s}\Phi(s,p_{0})|\mathrm{d}s=1$

and so $\rho_{0}$ is admissible for $\Gamma$. Moreover,

$\int_{U}\rho_{0}^{2}\mathrm{d}L^{2}=~\int_{J}\int_{I}\frac{|q(\Phi)||\mathrm{J}_{\Phi}|}{(l_{q}(\Phi(.,p)))^{2}}\mathrm{d}s\mathrm{d}p.$

Using Lemma 2.1.2, we obtain

$\int_{U}\rho_{0}^{2}\mathrm{d}L^{2}=\int_{J}\int_{I}\frac{\sqrt{|q(\Phi)|}|\partial_{s}\Phi||\lambda|}{(l_{q}(\Phi(.,p)))^{2}}\mathrm{d}s\mathrm{d}p=\int_{J}\frac{|\lambda(p)|}{l_{q}(\Phi(.,p))}\mathrm{d}p.$

Thus,

$M_{2}(\Gamma)\leq\int_{J}\frac{|\lambda(p)|}{l_{q}(\Phi(.,p))}\mathrm{d}p.$

∎

###### Proof of Lemma 2.1.2.

As in remark 2.1.3, we write

$\mu=\sqrt{(q\circ\Phi)\left(\partial_{s}\Phi\right)^{2}}$

which is a positive valued function. Moreover,

$~\mathrm{J}_{\Phi}=\Im\left(\overline{\partial_{s}\Phi}\partial_{p}\Phi\right).$

Consequently, since $\Phi$ is a diffeomorphism, $\partial_{s}\Phi$ is nowhere
zero,

$\mu\mathrm{J}_{\Phi}=\Im\left(\mu\overline{\partial_{s}\Phi}\partial_{p}\Phi\right)=|\partial_{s}\Phi|^{2}\Im\left(\frac{\mu\partial_{p}\Phi}{\partial_{s}\Phi}\right).$

Thus, to prove the lemma (in the form given by equation 2), we need to show
that
$\partial_{s}\left(\Im\left(\frac{\mu\partial_{p}\Phi}{\partial_{s}\Phi}\right)\right)=\Im\left(\partial_{s}\left(\frac{\mu\partial_{p}\Phi}{\partial_{s}\Phi}\right)\right)=0$.
For that, we first write

$q=\left(\frac{\mu(\Phi^{-1}_{1})}{\partial_{s}\Phi(\Phi^{-1})}\right)^{2}.$

From $\partial_{\overline{w}}q=0$, we obtain

$\partial_{s}\Phi\left(\partial_{s}\mu\partial_{\overline{w}}\Phi^{-1}_{1}(\Phi)+\partial_{p}\mu\partial_{\overline{w}}~\Phi^{-1}_{2}(\Phi)\right)=\mu\left(\partial_{s,s}\Phi\partial_{\overline{w}}\Phi^{-1}_{1}(\Phi)+\partial_{s,p}\Phi\partial_{\overline{w}}\Phi^{-1}_{2}(\Phi)\right).$

Usual formulas for inverse functions give

$\partial_{\overline{w}}\Phi^{-1}_{1}=-i\frac{\partial_{p}\Phi}{2\mathrm{J}_{\Phi}}\text{
and
}\partial_{\overline{w}}\Phi^{-1}_{2}=i\frac{\partial_{s}\Phi}{2\mathrm{J}_{\Phi}}.$

Leading to

$\partial_{s}\Phi\left(\partial_{p}\mu\partial_{s}\Phi-\partial_{s}\mu\partial_{p}\Phi\right)=\mu\left(\partial_{s,p}~\Phi\partial_{s}\Phi-\partial_{s,s}\Phi\partial_{p}\Phi\right)$

and so

$\displaystyle\partial_{s}\Phi\left(\partial_{s}\mu\partial_{p}\Phi+\mu\partial_{s,p}\Phi\right)-\mu\partial_{s,s}\Phi\partial_{p}\Phi$
$\displaystyle=$
$\displaystyle\partial_{p}\mu\left(\partial_{s}\Phi\right)^{2}.$ (3)

Now, we finally compute:

$\displaystyle\partial_{s}\left(\frac{\mu\partial_{p}\Phi}{\partial_{s}\Phi}\right)$
$\displaystyle=$
$\displaystyle\frac{\partial_{s}\Phi\left(\partial_{s}\mu\partial_{p}\Phi+\mu\partial_{s,p}\Phi\right)-\mu\partial_{p}\Phi\partial_{s,s}\Phi}{\left(\partial_{s}\Phi\right)^{2}}$
$\displaystyle\underset{\text{using equation \ref{eq1.1}}}{=}$
$\displaystyle\frac{\partial_{p}\mu\left(\partial_{s}\Phi\right)^{2}}{\left(\partial_{s}\Phi\right)^{2}}$
$\displaystyle=$ $\displaystyle\partial_{p}\mu$

and $\partial_{p}\mu$ is a real valued function. ∎

If in addition to the hypotheses of the theorem we ask the curves $\Phi(.,p)$
to all have the same $q$-length $C>0$, then $l_{\Gamma}$ becomes constant
equal to $C$ and we have the following corollary.

######  Corollary 2.1.4.

Let $D$ be a domain in $\mathbb{C}$ foliated by a family $\Gamma$ of
horizontal trajectories of a holomorphic quadratic differential $q$ on $D$.
Assume that each curve in $\Gamma$ has $q$-length $C>0$. Then the modulus of
$\Gamma$ is

$M_{2}(\Gamma)=\frac{1}{C^{2}}\mathrm{Area}_{q}(D).$

###  2.2 The result in the Heisenberg group

We now show Theorem 2. It’s proof follows the same steps as the one of Theorem
2.1.1 with a lemma similar to Lemma 2.1.2.

######  Theorem 2.2.1.

Let $I$ be an interval, $\Lambda$ an open subset of $\mathbb{R}^{2}$ and
$\Phi:I\times\Lambda\longrightarrow\Omega\subset{\mathcal{H}}$ be a
$C^{2}$-diffeomorphism such that for every $p\in\Lambda$, $\Phi(.,p)$ is a
legendrian curve. Denote

$\Gamma=\\{\Phi(.,p)\ |\ p\in\Lambda\\}.$

Let $q$ be quadratic differential on $\Omega$ satisfying $B_{2}q=0$ and such
that every curve in $\Gamma$ is a horizontal trajectory of $q$. Then,

$M_{4}(\Gamma)=\int_{\Omega}\frac{|q|^{2}}{(l_{\Gamma})^{4}}\mathrm{d}L^{3}$

where, for every $(z,t)\in\Omega$,
$l_{\Gamma}(z,t)=l_{q}(\Phi(.,\Phi^{-1}_{2}(z,t)))$.

The main interest of that result is that the quadratic differential only needs
to be in the kernel of $B_{2}$ and so doesn’t have to be the square of the
differential of (the complex part of) a CR map. The proof is a consequence of
the following lemma:

######  Lemma 2.2.2.

Let $q:\Omega\rightarrow\mathbb{C}$ be a quadratic differential on an open
subset $\Omega$ of ${\mathcal{H}}$ such that $B_{2}q=0$. Let $I$ be an
interval, $\Lambda$ be an open subset of $\mathbb{R}^{2}$ and
$\Phi:I\times\Lambda\longrightarrow\Omega\subset{\mathcal{H}}$ be a
diffeomorphism such that for every $p\in\Lambda$, $\Phi(.,p)$ is a horizontal
trajectory for $q$. Then, there is a nowhere vanishing real valued function
$\lambda$ on $\Lambda$ such that

$\displaystyle(|q|^{2}\circ\Phi)\mathrm{J}_{\Phi}$ $\displaystyle=$
$\displaystyle\sqrt{(|q|\circ\Phi)}|\partial_{s}\Phi_{1}|\lambda$ (4)

where $\mathrm{J}_{\Phi}$ is the Jacobian determinant of $\Phi$ and
$\Phi=(\Phi_{1},\Phi_{2})$.

######  Remark 2.2.3.

  * •

As for the complex case, we can read the lemma as: on $I\times\Lambda$, the
$q$-volume decomposes as the product of the $q$-length and a measure on
$\Lambda$.

  * •

We will prove and use the lemma with the conclusion written in another form,
easier for computations. We use here the same notations and hypotheses than in
the lemma. First, we write

$\mu=\sqrt{(|q|\circ\Phi)}|\partial_{s}\Phi_{1}|=\sqrt{(q\circ\Phi)(\partial_{s}\Phi_{1})^{2}}.$

We then multiply both sides of equation 4 by $|\partial_{s}\Phi_{1}|^{4}$
(which is nowhere zero since $s\mapsto\Phi(s,p)$ is legendrian for every $p$
and $\Phi$ is a diffeomorphism; see the beginning of the proof of Lemma 2.2.2)
to obtain

$(|q|^{2}\circ\Phi)|\partial_{s}\Phi_{1}|^{4}\mathrm{J}_{\Phi}=\mu|\partial_{s}\Phi_{1}|^{4}\lambda.$

Then, notice that $(|q|^{2}\circ\Phi)|\partial_{s}\Phi_{1}|^{4}=\mu^{4}$ and
divide both sides by $\mu$ (which is nowhere zero) and we finally have the
form of the conclusion we will mostly use:

$\displaystyle\left(\sqrt{(q\circ\Phi)(\partial_{s}\Phi_{1})}\right)^{3}\mathrm{J}_{\phi}=\lambda|\partial_{s}\Phi_{1}|^{4}.$
(5)

###### Proof of Theorem 2.2.1.

Assuming the lemma for now, let $\rho\in\mathrm{adm}(\Gamma)$ be an admissible
density for $\Gamma$. Then,

$\int_{\Omega}\rho^{4}\mathrm{d}L^{3}=\int_{\Lambda}\int_{I}\rho^{4}(\Phi)|\mathrm{J}_{\Phi}|\mathrm{d}s\mathrm{d}L^{2}(p).$

However, for every $p$ in $\Lambda$, Hölder inequality gives

$\displaystyle\left(\int_{I}\left(\rho(\Phi)|\mathrm{J}_{\Phi}|^{\frac{1}{4}}\right)^{4}\mathrm{d}s\right)\left(\int_{I}\left(\frac{|\partial_{s}\Phi_{1}|}{|\mathrm{J}_{\Phi}|^{\frac{1}{4}}}\right)^{\frac{4}{3}}\mathrm{d}s\right)^{3}$
$\displaystyle\geq$
$\displaystyle\left(\int_{I}\rho(\Phi)|\partial_{s}\Phi_{1}|\mathrm{d}s\right)^{4}$
$\displaystyle\geq$ $\displaystyle 1\text{ (since
$\rho\in\mathrm{adm}(\Gamma)$).}$

Using equation 5, we have for every $p\in\Lambda$

$\left(\int_{I}\left(\frac{|\partial_{s}\Phi_{1}|}{|\mathrm{J}_{\Phi}|^{\frac{1}{4}}}\right)^{\frac{4}{3}}\mathrm{d}s\right)^{3}=\left(\int_{I}\frac{\sqrt{(q\circ\Phi)\left(\partial_{s}\Phi_{1}\right)^{2}}}{|\lambda|^{\frac{1}{3}}}\mathrm{d}s\right)^{3}=\frac{(l_{q}(\Phi(.,p)))^{3}}{|\lambda|}$

for a nowhere vanishing function $\lambda:\Lambda\rightarrow\mathbb{R}$. Thus,

$\int_{\Omega}\rho^{4}\mathrm{d}L^{3}\geq\int_{\Lambda}\frac{|\lambda(p)|}{(l_{q}(\Phi(.,p)))^{3}}\mathrm{d}L^{2}(p).$

Since it is true for every $\rho\in\mathrm{adm}(\Gamma)$,

$M_{4}(\Gamma)\geq\int_{\Lambda}\frac{|\lambda(p)|}{(l_{q}(\Phi(.,p)))^{3}}\mathrm{d}L^{2}(p).$

Now, let $\rho_{0}$ being defined for every $(z,t)$ by
$\rho_{0}(z,t)=\frac{\sqrt{|q(z,t)|}}{l_{q}(\Phi(.,\Phi^{-1}_{2}(z,t)))}$.
Then, if $\Phi(.,p_{0})$ is a curve in $\Gamma$, one has

$\int_{\Phi(.,p_{0})}\rho_{0}\mathrm{d}l=\frac{1}{l_{q}(\Phi(.,p_{0}))}\int_{I}\sqrt{|q(\Phi(s,p_{0}))|}|\partial_{s}\Phi_{1}(s,p_{0})|\mathrm{d}s=1$

and so $\rho_{0}$ is admissible for $\Gamma$. Moreover,

$\int_{\Omega}\rho_{0}^{4}\mathrm{d}L^{3}=~\int_{\Lambda}\int_{I}\frac{|q(\Phi)|^{2}|\mathrm{J}_{\Phi}|}{(l_{q}(\Phi(.,p)))^{4}}\mathrm{d}s\mathrm{d}L^{2}(p).$

Using Lemma 2.2.2, we obtain

$\displaystyle\int_{\Omega}\rho_{0}^{4}\mathrm{d}L^{3}$ $\displaystyle=$
$\displaystyle\int_{\Lambda}\frac{|\lambda(p)|}{(l_{q}(\Phi(.,p)))^{4}}\left(\int_{I}\sqrt{|q(\Phi)|}|\partial_{s}\Phi_{1}|\mathrm{d}s\right)\mathrm{d}L^{2}(p)$
$\displaystyle=$
$\displaystyle\int_{\Lambda}\frac{|\lambda(p)|}{(l_{q}(\Phi(.,p)))^{3}}\mathrm{d}L^{2}(p).$

Thus,

$M_{4}(\Gamma)\leq\int_{\Lambda}\frac{|\lambda(p)|}{(l_{q}(\Phi(.,p)))^{3}}\mathrm{d}L^{2}(p)$

which ends the proof. ∎

###### Proof of Lemma 2.2.2.

As in remark 2.2.3, we write

$\mu=\sqrt{(q\circ\Phi)\left(\partial_{s}\Phi_{1}\right)^{2}}$

which is a positive valued function. Moreover, using the fact that the curves
$\Phi(.,p)$ are legendrian, we easily compute

$\mathrm{J}_{\Phi}=-\Im\left(\overline{\partial_{s}\Phi_{1}}A\right)$

with

$A=\partial_{p_{1}}\Phi_{2}\partial_{p_{2}}\Phi_{1}-\partial_{p_{2}}\Phi_{2}\partial_{p_{1}}\Phi_{1}+2\Phi_{1}\Im\left(\partial_{p_{1}}~\Phi_{1}\overline{\partial_{p_{2}}\Phi_{1}}\right).$

Since $\Phi$ is a diffeomorphism, $\partial_{s}\Phi_{1}$ is nowhere zero.
Thus,

$\mu^{3}\frac{\mathrm{J}_{\Phi}}{|\partial_{s}\Phi_{1}|^{4}}=-\Im(B)$

with

$B=\frac{\mu^{3}}{\partial_{s}\Phi_{1}|\partial_{s}\Phi_{1}|^{2}}A.$

Consequently, we only have to prove that $\Im\left(\partial_{s}B\right)=0$.

For that, we compute

$\partial_{s}B=\frac{3\mu^{2}\partial_{s}\mu|\partial_{s}\Phi_{1}|^{2}-\mu^{3}\left(2\overline{\partial_{s}\Phi_{1}}\partial_{s,s}\Phi_{1}+\partial_{s}\Phi_{1}\overline{\partial_{s,s}\Phi_{1}}\right)}{\partial_{s}\Phi_{1}|\partial_{s}\Phi_{1}|^{4}}A+\frac{\mu^{3}|\partial_{s}\Phi_{1}|^{2}\partial_{s}A}{\partial_{s}\Phi_{1}|\partial_{s}\Phi_{1}|^{4}}$

which leads to

$\partial_{s}B=\frac{\mu^{2}}{|\partial_{s}\Phi_{1}|^{4}}\left(\left(3\overline{\partial_{s}\Phi_{1}}\partial_{s}\mu-2\mu\frac{\overline{\partial_{s}\Phi_{1}}}{\partial_{s}\Phi_{1}}\partial_{s,s}\Phi_{1}-\mu\overline{\partial_{s,s}\Phi_{1}}\right)A+\mu\overline{\partial_{s}\Phi_{1}}\partial_{s}A\right).$

Denoting

$\displaystyle D$ $\displaystyle=$
$\displaystyle\left(3\overline{\partial_{s}\Phi_{1}}\partial_{s}\mu-2\mu\frac{\overline{\partial_{s}\Phi_{1}}}{\partial_{s}\Phi_{1}}\partial_{s,s}\Phi_{1}-\mu\overline{\partial_{s,s}\Phi_{1}}\right)A+\mu\overline{\partial_{s}\Phi_{1}}\partial_{s}A,$
(6)

we have to prove that $\Im(D)=0$.

For that, we write

$q=\left(\frac{\mu(\Phi^{-1}_{1})}{\partial_{s}\Phi_{1}(\Phi^{-1})}\right)^{2}.$

From $B_{2}q=0$, we obtain

$3|\partial_{s}\Phi_{1}|^{2}\left(\partial_{s}\mu\overline{Z}\Phi^{-1}_{1}(\Phi)+\partial_{p_{1}}\mu\overline{Z}\Phi^{-1}_{2}(\Phi)+\partial_{p_{2}}\mu\overline{Z}\Phi^{-1}_{3}(\Phi)\right)\\\
=\overline{Z}\Phi^{-1}_{1}(\Phi)\left(2\overline{\partial_{s}\Phi_{1}}\partial_{s,s}\Phi_{1}+\partial_{s}\Phi_{1}\overline{\partial_{s,s}\Phi_{1}}\right)\\\
+\overline{Z}\Phi^{-1}_{2}(\Phi)\left(2\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{1}}\Phi_{1}+\partial_{s}\Phi_{1}\overline{\partial_{s,p_{1}}\Phi_{1}}\right)\\\
+\overline{Z}\Phi^{-1}_{3}(\Phi)\left(2\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{2}}\Phi_{1}+\partial_{s}\Phi_{1}\overline{\partial_{s,p_{2}}\Phi_{1}}\right).$
(7)

Usual formulas for inverse functions give

$\displaystyle\overline{Z}\Phi^{-1}_{1}(\Phi)$ $\displaystyle=$
$\displaystyle\frac{i}{2\mathrm{J}_{\Phi}}A\ \ (\neq 0);$ (8)
$\displaystyle\overline{Z}\Phi^{-1}_{2}(\Phi)$ $\displaystyle=$
$\displaystyle\frac{i\partial_{s}\Phi_{1}}{2\mathrm{J}_{\Phi}}\left(\partial_{p_{2}}\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{2}}~\Phi_{1}\right)\right)$
(9)

and

$\displaystyle\overline{Z}\Phi^{-1}_{3}(\Phi)$ $\displaystyle=$
$\displaystyle\frac{i\partial_{s}\Phi_{1}}{2\mathrm{J}_{\Phi}}\left(-\partial_{p_{1}}\Phi_{2}-2\Im\left(\overline{\Phi_{1}}\partial_{p_{1}}~\Phi_{1}\right)\right).$
(10)

Thus, using equations 8, 9 and 10, equation 7 becomes

$3\overline{\partial_{s}\Phi_{1}}\partial_{s}\mu
A-2\mu\frac{\overline{\partial_{s}\Phi_{1}}}{\partial_{s}\Phi_{1}}\partial_{s,s}\Phi_{1}A-\mu\overline{\partial_{s,s}\Phi_{1}}A\\\
=-3\partial_{p_{1}}~\mu|\partial_{s}\Phi_{1}|^{2}\left(\partial_{p_{2}}~\Phi_{2}+2\Im(\overline{\Phi_{1}}\partial_{p_{2}}~\Phi_{1})\right)\text{
($\in\mathbb{R}$)}\\\
+3\partial_{p_{2}}~\mu|\partial_{s}\Phi_{1}|^{2}\left(\partial_{p_{1}}~\Phi_{2}+2\Im(\overline{\Phi_{1}}\partial_{p_{1}}~\Phi_{1})\right)\text{
($\in\mathbb{R}$)}\\\
+\mu\left(\partial_{p_{2}}\Phi_{2}+2\Im(\overline{\Phi_{1}}~\partial_{p_{2}})\right)\left(2\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{1}}\Phi_{1}+\partial_{s}\Phi_{1}+\overline{\partial_{s,p_{1}}\Phi_{1}}\right)\\\
-\mu\left(\partial_{p_{1}}\Phi_{2}+2\Im(\overline{\Phi_{1}}~\partial_{p_{1}})\right)\left(2\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{2}}\Phi_{1}+\partial_{s}\Phi_{1}+\overline{\partial_{s,p_{2}}\Phi_{1}}\right).$
(11)

We can now use equation 11 in equation 6 and take advantage of terms that are
real valued to find:

$\frac{\Im(D)}{\mu}=\left(\partial_{p_{2}}\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{2}}~\Phi_{1}\right)\right)\Im(\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{1}}\Phi_{1})\\\
-\left(\partial_{p_{1}}\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{1}}~\Phi_{1}\right)\right)\Im(\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{2}}\Phi_{1})+\Im(\overline{\partial_{s}\Phi_{1}}\partial_{s}A)$

Finally, we need to show that

$\Im\left(\overline{\partial_{s}\Phi_{1}}\partial_{s}A\right)=-\left(\partial_{p_{2}}\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{2}}~\Phi_{1}\right)\right)\Im\left(\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{1}}\Phi_{1}\right)\\\
+\left(\partial_{p_{1}}\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{1}}~\Phi_{1}\right)\right)\Im\left(\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{2}}\Phi_{1}\right).$

For that, we compute $\partial_{s}A$:

$\partial_{s}A=\partial_{s,p_{1}}\Phi_{2}\partial_{p_{2}}\Phi_{1}+\partial_{p_{1}}\Phi\partial_{s,p_{2}}\Phi_{1}-\partial_{s,p_{2}}\Phi_{2}\partial_{p_{1}}\Phi_{1}-\partial_{p_{2}}\partial_{s,p_{1}}\Phi_{1}\\\
+2\partial_{s}\Phi_{1}\Im\left(\partial_{p_{1}}\Phi_{1}\overline{\partial_{p_{2}}\Phi_{1}}\right)+2\Phi_{1}\Im\left(\partial_{s,p_{1}}~\Phi\overline{\partial_{p_{2}}~\Phi_{1}}+\partial_{p_{1}}~\overline{\partial_{s,p_{2}}~\Phi_{1}}\right).$

Since
$\partial_{s}\Phi_{2}=-2\Im\left(\overline{\Phi_{1}}\partial_{s}\Phi_{1}\right)$,
we find

$\partial_{s}A=\partial_{s,p_{1}}\Phi_{1}\left(-\partial_{p_{2}}~\Phi_{2}-2\Im\left(\overline{\Phi_{1}}\partial_{p_{2}}~\Phi_{1}\right)\right)\\\
+\partial_{s,p_{2}}\Phi_{1}\left(\partial_{p_{1}}~\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{1}}~\Phi_{1}\right)\right)\\\
+4\partial_{s}\Phi_{1}\Im\left(\partial_{p_{1}}\Phi_{1}\overline{\partial_{p_{2}}\Phi_{1}}\right).$

Consequently,

$\Im\left(\overline{\partial_{s}\Phi_{1}}\partial_{s}A\right)=\left(-\partial_{p_{2}}~\Phi_{2}-2\Im\left(\overline{\Phi_{1}}\partial_{p_{2}}\Phi_{1}\right)\right)\Im\left(\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{1}}\Phi_{1}\right)\\\
+\left(\partial_{p_{1}}\Phi_{2}+2\Im\left(\overline{\Phi_{1}}\partial_{p_{1}}\Phi_{1}\right)\right)\Im\left(\overline{\partial_{s}\Phi_{1}}\partial_{s,p_{2}}\Phi_{1}\right)$

∎

As for the complex case, if in addition to the hypotheses of the theorem we
ask the curves $\Phi(.,p)$ to all have the same $q$-length $C>0$, then
$l_{\Gamma}$ becomes constant equal to $C$ and we have the following
corollary.

######  Corollary 2.2.4.

Let $\Omega$ be a domain in the Heisenberg group foliated by a family $\Gamma$
of legendrian curves. Assume that there is a quadratic differential $q$ on
$\Omega$ such that $B_{2}q=0$ and every curve in $\Gamma$ is a horizontal
trajectory for $q$ with $q$-length $C>0$. Then, the modulus of $\Gamma$ is

$M_{4}(\Gamma)=\frac{1}{C^{4}}\mathrm{Vol_{q}}(\Omega).$

###  2.3 Examples

#### Explicit computation of moduli

As an explicit example of application of theorem 2.2.1, we want to discuss
about the quadratic differential
$q_{0}=\left[\frac{\overline{z}^{2}(t^{2}+|z|^{4})^{\frac{2}{3}}}{|z|^{\frac{8}{3}}(t+i|z|^{2})^{2}}\mathrm{d}z^{2}\right]_{\omega}$
in the annulus $A_{r}=\\{(z,t)\in{\mathcal{H}}\ |\
1<\|(z,t)\|_{\mathcal{H}}<r\\}$ for $r>1$. The horizontal trajectories of
$q_{0}$ are the spherical arcs:

$\gamma_{x,\alpha}(s)=\left(\sqrt{e^{x}\sin(s)}\alpha
e^{\frac{is}{2}},e^{x}\cos(s)\right)\text{ for $|\alpha|=1$ and
$x\in]0,2\ln(r)[$}$

with $s\in]0,\pi[$. Denote $\Gamma=\\{\gamma_{x,\alpha}\ |\ 0<x<2\ln(r)\ \&\
|\alpha|=1\\}$. Then $\Gamma$ is the foliation of
$A_{r}\backslash(\\{0\\}\times\mathbb{R})$ by horizontal trajectories for
$q_{0}$ which satisfies $B_{2}q_{0}=0$. For every $(s,x,\alpha)$, we easily
compute

$q_{0}(\gamma_{x,\alpha}^{\prime}(s))=\frac{1}{4\sin^{\frac{4}{3}}(s)}$

and thus, for every $(x,\alpha)$, we have

$l_{q_{0}}(\gamma_{x,\alpha})=\frac{1}{2}\int_{0}^{\pi}\sin^{-\frac{2}{3}}(s)\mathrm{d}s.$

Notice that all curves in $\Gamma$ have the same $q_{0}$-length which we
denote $C=\frac{1}{2}\int_{0}^{\pi}\sin^{-\frac{2}{3}}(s)\mathrm{d}s$.

Now, the $q_{0}$-volume of $A_{r}\backslash(\\{0\\}\times\mathbb{R})$ is easy
to compute (by substitution) in terms of $l_{q_{0}}(\gamma_{x,\alpha})$, we
find

$\displaystyle\mathrm{Vol}_{q_{0}}(A_{r}\backslash(\\{0\\}\times\mathbb{R}))$
$\displaystyle=.$
$\displaystyle\int_{A_{r}\backslash(\\{0\\}\times\mathbb{R})}\frac{\mathrm{d}L^{3}}{|z|^{\frac{4}{3}}(t^{2}+|z|^{4})^{\frac{2}{3}}}$
$\displaystyle=$ $\displaystyle
2\pi\int_{0}^{2\ln(r)}\int_{0}^{\pi}\frac{e^{2x}}{2e^{\frac{2x}{3}}\sin^{\frac{2}{3}}(s)e^{\frac{4x}{3}}}\mathrm{d}s\mathrm{d}x$
$\displaystyle=$ $\displaystyle 4\pi C\ln(r).$

Then, Corollary 2.2.4 gives the modulus of $\Gamma$:

$M_{4}(\Gamma)=\frac{4\pi\ln(r)}{C^{3}}.$

The same thing can be done for the vertical trajectories of $q_{0}$. The
theorem still applies to vertical trajectories since vertical trajectories of
a quadratic differential $q$ are horizontal trajectories of $-q$. The vertical
trajectories are the radii

$\delta_{y,\beta}(s)=\left(\sqrt{e^{s}\sin(y)}\beta
e^{-\frac{is}{2}\cot(y)},e^{s}\cos(y)\right)\text{ for $|\beta|=1$ and
$y\in]0,\pi[$}$

with $s\in]0,2\ln(r)[$. Denote $\Delta=\\{\delta_{y,\beta}\ |\ 0<y<\pi\ \&\
|\beta|=1\\}$. $\Delta$ is the foliation of
$A_{r}\backslash(\\{0\\}\times\mathbb{R})$ by vertical trajectories for
$q_{0}$. For every $(s,y,\beta)$, we easily compute

$q_{0}(\delta_{y,\beta}^{\prime}(s))=-\frac{1}{4\sin^{\frac{4}{3}}(y)}$

and thus, for every $(y,\beta)$,
$l_{q_{0}}(\delta_{y,\beta})=\frac{\ln(r)}{\sin^{\frac{2}{3}}(y)}$ which leads
to

$l_{\Delta}(z,t)=\frac{(t^{2}+|z|^{4})^{\frac{1}{3}}\ln(r)}{|z|^{\frac{4}{3}}}.$

Finally, we use Theorem 2.2.1 to compute (by substitution) the modulus of
$\Delta$:

$\displaystyle M_{4}(\Delta)$ $\displaystyle=$
$\displaystyle\int_{A_{r}}\frac{|q_{0}|^{2}}{(l_{\Delta})^{4}}$
$\displaystyle=$
$\displaystyle\frac{1}{\ln(r)^{4}}\int_{A_{r}}\frac{|z|^{4}}{(t^{2}+|z|^{4})^{2}}\mathrm{d}L^{3}$
$\displaystyle=$
$\displaystyle\frac{2\pi}{\ln(r)^{3}}\int_{0}^{\pi}\sin^{2}(y)\mathrm{d}y$
$\displaystyle=$ $\displaystyle\frac{\pi^{2}}{\ln(r)^{3}}.$

#### General strategy

As an end, let us give a general strategy to compute the modulus of a
foliation by Legendrian curves, applying theorem 2.2.1.
Let $\Phi:I\times\Lambda\rightarrow\Omega$ be a diffeomorphism. Then, the
family of quadratic differentials for which the $\Phi(.,p)$ are horizontal
trajectories is the family

$\left\\{q_{f}=\left[\frac{f}{(\partial_{s}\Phi_{1})^{2}\circ\Phi^{-1}}\mathrm{d}z^{2}\right]_{\omega}\
|\ f:\Omega\rightarrow]0,+\infty[\right\\}.$

Thus, to apply theorem 2.2.1 one needs to find a positive valued function $f$
such that $B_{2}q_{f}=0$.

######  Remark 2.3.1.

A quadratic differential with prescribed trajectories in the kernel of $B_{2}$
is (if it exists) unique up to a positive constant. Indeed, assume that $q$ is
a quadratic differential and $q^{\prime}$ are quadratic differentials with the
same horizontal trajectories. We just saw that there must be a positive valued
function $f$ such that $q^{\prime}=fq$. If we assume moreover that both $q$
and $fq$ annihilate $B_{2}$. Then,

$0=B_{2}(fq)=f^{2}B_{2}q+\left(|q|^{2}\left(\overline{Z}(f^{2})+f\overline{Z}f\right)\right)\omega\wedge\mathrm{d}\overline{z}\otimes\omega=\left(3f|q|^{2}\overline{Z}f\right)\omega\wedge\mathrm{d}\overline{z}\otimes\omega.$

Thus, $f$ is a positive CR function, hence constant.

## References

E-mail address<EMAIL_ADDRESS>


|\xi|=q\end{smallmatrix}}\Delta(\lambda_{\xi})\Delta(x)^{\xi}+\sum_{\begin{smallmatrix}\nu\subset
I\\\
|\nu|=q-1\end{smallmatrix}}\Delta(\mu_{\nu})\Delta(x)^{\nu},\sum_{\begin{smallmatrix}\nu\subset
I\\\ |\nu|=q-1\end{smallmatrix}}\Delta(\mu_{\nu})\Delta(x)^{\nu}\big{)},$

and this element is by assumption in $J_{\langle 2^{n}\rangle}^{q+1}$. Let us
analyse the two components separately, starting from the second one. As
observed above Lemma 3.8, these components must in fact belong to $J_{\langle
1\rangle}^{q+1}$, and therefore

$\sum_{\begin{smallmatrix}\nu\subset I\\\
|\nu|=q-1\end{smallmatrix}}\Delta(\mu_{\nu})\Delta(x)^{\nu}\in J_{\langle
1\rangle}^{q+1}.$

By Lemma 3.7, $V(\mu_{\nu}(dx)^{\nu})$ vanishes in the de Rham Witt complex
modulo $2$, and therefore so does

$dV^{n}(b)=dV^{n-1}(\sum_{\begin{smallmatrix}\nu\subset I\\\
|\nu|=q-1\end{smallmatrix}}V(\mu_{\nu}(dx)^{\nu})).$

Our original element $c=V^{n}(a)+dV^{n}(b)$ is then equal to $V^{n}(a)$ in the
de Rham-Witt complex modulo $2$, and the map $u_{n}$ sends this element to

$u_{n}(c)=V^{n}(a)=\big{(}\sum_{\begin{smallmatrix}\xi\subset I\\\
|\xi|=q\end{smallmatrix}}\Delta(\lambda_{\xi})\Delta(x)^{\xi},0\big{)}.$

Moreover, this element is by assumption in $J_{\langle 2^{n}\rangle}^{q+1}$.
By applying Lemma 3.8, the first component
$\sum_{\begin{smallmatrix}\xi\subset I\\\
|\xi|=q\end{smallmatrix}}\Delta(\lambda_{\xi})\Delta(x)^{\xi}$ in fact belongs
to $J_{\langle 1\rangle}^{q+2}$. Again by Lemma 3.7, we find that

$c=V^{n}(a)=V^{n-1}V(\sum_{\begin{smallmatrix}\xi\subset I\\\
|\xi|=q\end{smallmatrix}}\lambda_{\xi}(dx)^{\xi})=0$

in $(\operatorname{W}_{\langle 2^{n}\rangle}\Omega^{q}_{k})/2$, proving that
$u_{n}$ is injective. ∎

###  3.3 The Milnor conjecture and TCR

Let us recall that the $2$-typical topological cyclic homology spectrum
$\operatorname{TC}(k;2)$ of $k$ can be defined as the equaliser

$\textstyle{\operatorname{TC}(k;2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\operatorname{TR}(k;2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{id}}$$\scriptstyle{F}$$\textstyle{\operatorname{TR}(k;2)}$

of the identity and the Frobenius map of $\operatorname{TR}(k;2)$. Let us
denote $\operatorname{W}_{\langle 2^{\infty}\rangle}\Omega^{\ast}_{k}$ the
limit of $\operatorname{W}_{\langle 2^{n}\rangle}\Omega^{\ast}_{k}$ over the
map $R$, and define $\nu_{\ast}^{dRW/2}(k;2)$ and
$\epsilon_{\ast}^{dRW/2}(k;2)$ respectively as the equaliser and coequaliser
of the parallel group homomorphisms

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0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
126.98589pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
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0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
211.73033pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\epsilon^{\ast}_{dRW/2}(k;2)}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

Then, since the parallel arrows are ring homomorphisms,
$\nu^{\ast}_{dRW/2}(k;2)$ is a graded ring, and $\epsilon^{\ast}_{dRW/2}(k;2)$
is a graded $\nu^{\ast}_{dRW/2}(k;2)$-module (where the action on the latter
is defined via either of the maps $\operatorname{id}$ or $F$). We now prove a
version of the Milnor conjecture for $\nu^{\ast}_{dRW/2}(k;2)$ and
$\epsilon^{\ast}_{dRW/2}(k;2)$, which describes these in terms of the graded
ring associated to the kernel of the restriction map

$K:=\ker\big{(}\operatorname{res}^{\operatorname{\mathbb{Z}}/2}_{e}\colon\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}\longrightarrow(\pi_{0}\operatorname{TC}(k;2))^{\operatorname{\mathbb{Z}}/2}/Im(1+w)\big{)}.$

The restriction map is defined as we did at the beginning of §2.3. Let us also
define the
$\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}$-module

$T_{-1}:=\pi_{-1}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2},$

so that the quotients $K^{\ast}T_{-1}/K^{\ast+1}T_{-1}$ form a graded
$K^{\ast}/K^{\ast+1}$-module.

######  Theorem 3.9.

Let $k$ be a field of characteristic $2$. There is an isomorphism of graded
rings

$\nu_{dRW/2}^{\ast}(k;2)\cong K^{\ast}/K^{\ast+1},$

and an isomorphism of graded $K^{\ast}/K^{\ast+1}$-modules

$\epsilon^{\ast}_{dRW/2}(k;2)\cong K^{\ast}T_{-1}/K^{\ast+1}T_{-1}.$

###### Proof.

Let $\nu^{\ast}(k)$ and be $\epsilon^{\ast}(k)$ be respectively the equaliser
and coequaliser of the projection map and the inverse Cartier operator

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3.0pt}}}}}}\ignorespaces{\hbox{\kern
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70.89212pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\Omega^{\ast}_{k}/d(\Omega^{\ast-1}_{k})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
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145.80807pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{\epsilon^{\ast}}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

The map $R$ of the de Rham-Witt complex induces multiplicative maps
$\nu_{dRW/2}^{\ast}(k;2)\to\nu^{\ast}(k)$ and
$\epsilon_{dRW/2}^{\ast}(k;2)\to\epsilon^{\ast}(k)$, which are isomorphisms by
the proof of [CMM21, Proposition 2.26]. Moreover, by [Kat82, Theorem (2)], the
graded ring associated to the fundamental ideal $I$ of the symmetric Witt
group is isomorphic to $\nu^{\ast}(k)$, and the graded module
$I^{\ast}\operatorname{W}^{q}(k)/I^{\ast+1}\operatorname{W}^{q}(k)$ to
$\epsilon^{\ast}(k)$. Thus, since [Kat82, Theorem (1)] and Corollary 2.4
identify $\operatorname{W}^{s}(k)$ and
$\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}$, as rings,
with the equaliser of

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and $\operatorname{W}^{q}(k)$ and $T_{-1}$, as modules, with their
coequaliser, it suffices to show that $I$ and $K$ correspond to the same ideal
under these identifications.

For the symmetric Witt group, the isomorphism with the equaliser is given by
the unique additive map that sends the rank $1$ form $\langle a\rangle$, with
$a\in k^{\times}$, to $a^{-1}\otimes a$. For
$\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}$, it is
induced by the map

$\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}\stackrel{{\scriptstyle
R}}{{\longrightarrow}}\pi_{0}\operatorname{TRR}^{2}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}\stackrel{{\scriptstyle
c}}{{\longrightarrow}}\pi_{0}(\operatorname{THR}(k)^{\phi\operatorname{\mathbb{Z}}/2})^{C_{2}},$

followed by the identification of the target with $(k\otimes_{S}k)^{C_{2}}$
from Proposition 2.2. Let us note that, after including the fixed points into
$k\otimes_{S}k$, this is the map

$\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}\stackrel{{\scriptstyle
R}}{{\longrightarrow}}\pi_{0}\operatorname{THR}(k)^{\phi\operatorname{\mathbb{Z}}/2}\cong
k\otimes_{S}k,$

where the isomorphism is from [DMPR21, Theorem 5.1]. Thus, in order to compare
$I$ and $K$, it suffices to show that, under the isomorphism of Corollary 2.4,
the restriction map from
$\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}$ to
$(\pi_{0}\operatorname{TC}(k;2))^{\operatorname{\mathbb{Z}}/2}/Im(1+w)$ sends
$a^{-1}\otimes a$ to $1$. As we do not have a good handle of
$\pi_{0}\operatorname{TC}(k;2)$ for a general field $k$, we found ourselves
unable to prove this by direct calculation.

Instead, we can employ the existence of a trace map of
$\operatorname{\mathbb{Z}}/2$-equivariant spectra
$\operatorname{tr}\colon\operatorname{KR}(k)\to\operatorname{TCR}(k;2)$ which
lifts the trace map $\operatorname{K}(k)\to\operatorname{TC}(k;2)$ from
[BHM93]. This trace is constructed in [HNS21], or can be produced by showing
that the trace map of [DO19], which lands in $\operatorname{THR}(k)$, lifts to
the $C_{2^{n}}$ fixed points of $\operatorname{THR}(k)$, in a way which
commutes with $F$ and $R$. After identifying the components of the geometric
fixed points of $\operatorname{KR}(k)$ with $\operatorname{W}^{s}(k)$, and the
restriction map with the rank (see [CDH+20a] and [CDH+20b]), we obtain a
commutative diagram

---
$\textstyle{\operatorname{W}^{s}(k)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{rk}$$\scriptstyle{\operatorname{tr}}$$\textstyle{\pi_{0}\operatorname{TCR}(k;2)^{\phi\operatorname{\mathbb{Z}}/2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{res}^{\operatorname{\mathbb{Z}}/2}_{e}}$$\scriptstyle{R}$$\textstyle{\pi_{0}\operatorname{THR}(k)^{\phi\operatorname{\mathbb{Z}}/2}\cong
k\otimes_{S}k}$$\textstyle{\operatorname{\mathbb{Z}}/2\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{tr}}$$\textstyle{(\pi_{0}\operatorname{TC}(k;2))^{\operatorname{\mathbb{Z}}/2}/Im(1+w)}$

where the map $R$ on the top row is injective. The bottom horizontal trace map
is a ring homomorphism, and it therefore must be injective. Thus, the
respective vertical kernels $I$ and $K$ are isomorphic, provided the top
horizontal trace map is an isomorphism. This follows from the discussion
above, if we can show that the composite of the top row sends the rank $1$
form $\langle a\rangle$ to $a^{-1}\otimes a$. This composite is the effect, on
geometric fixed-points, of a trace map of
$\operatorname{\mathbb{Z}}/2$-spectra
$\operatorname{KR}(k)\to\operatorname{THR}(k)$. This map is characterised in a
way analogous to the usual Bökstedt trace map for algebraic K-theory, which we
now recall (see [DO19, §3.2]). The map is obtained from maps

$B^{\sigma}GL_{n}(k)\stackrel{{\scriptstyle s}}{{\to}}B^{di}GL_{n}(k)\to
B^{di}M_{n}(k)\to\Omega^{\infty}\operatorname{THR}(M_{n}(k))\stackrel{{\scriptstyle
m}}{{\cong}}\Omega^{\infty}\operatorname{THR}(k),$

by taking the disjoint union over $n\geq 0$, and group-completing the source
with respect to direct sums. Here $B^{\sigma}GL_{n}(k)$ is the classifying
space of $GL_{n}(k)$ with the involution of [BF84, Proposition 1.1.3], so that
its $\operatorname{\mathbb{Z}}/2$-fixed-points space are the bar construction
of $GL_{n}(k)$ acting on the symmetric matrices
$GL_{n}(k)^{\operatorname{\mathbb{Z}}/2}$ by conjugation (see also [DO19,
§2.1]). The second space is the dihedral nerve $B^{di}GL_{n}(k)$, which is the
cyclic nerve of $GL_{n}(k)$ with the involution analogous to the one of
$\operatorname{THR}(k)$ from §1. Its $\operatorname{\mathbb{Z}}/2$-fixed-
points space is the two-sided bar construction of the same action. The map $s$
is the canonical section, which is defined on an $n$-simplex
$(g_{1},\dots,g_{n})$ by

$s(g_{1},\dots,g_{n})=((g_{1}\dots g_{n})^{-1},g_{1},\dots,g_{n}).$

The second map is the inclusion of invertible matrices into all matrices, and
the third map is adjoint to the canonical map

$\Sigma^{\infty}_{+}B^{di}M_{n}(k)\cong
B^{di}\Sigma^{\infty}_{+}M_{n}(k)\longrightarrow
B^{di}HM_{n}(k)=\operatorname{THR}(M_{n}(k)).$

Finally, the last map is a certain Morita invariance map $m$, which is the
identity for $n=1$. After passing to $\operatorname{\mathbb{Z}}/2$-fixed-
points spaces, the map $s$ corresponds to the map

$B(GL_{n}(k);GL_{n}(k)^{\operatorname{\mathbb{Z}}/2})\stackrel{{\scriptstyle
s}}{{\longrightarrow}}B(GL_{n}(k)^{\operatorname{\mathbb{Z}}/2};GL_{n}(k);GL_{n}(k)^{\operatorname{\mathbb{Z}}/2})$

which, in simplicial degree $n$, sends $(g_{1},\dots,g_{n},x)$, with $g_{i}\in
GL_{n}(k)$ and $x\in GL_{n}(k)^{\operatorname{\mathbb{Z}}/2}$, to

$s(g_{1},\dots,g_{n},x)=((g_{1}\dots g_{n}xg_{n}^{T}\dots
g^{T}_{1})^{-1},g_{1},\dots,g_{n},x),$

where $(-)^{T}$ is matrix transposition. After applying $\pi_{0}$ and
identifying the components of $\operatorname{THR}$ using [DMPR21, Theorem
5.1], this becomes a map

$(GL_{n}(k)^{\operatorname{\mathbb{Z}}/2})/_{conj}\stackrel{{\scriptstyle
s}}{{\to}}GL_{n}(k)^{\operatorname{\mathbb{Z}}/2}\times_{GL_{n}(k)}GL_{n}(k)^{\operatorname{\mathbb{Z}}/2}\to
M_{n}(k)^{\operatorname{\mathbb{Z}}/2}\otimes_{N}M_{n}(k)^{\operatorname{\mathbb{Z}}/2}\stackrel{{\scriptstyle
m}}{{\cong}}k\otimes_{S}k,$

where $N$ is the subgroup of the tensor product generated by the elements
$gxg^{T}\otimes y-x\otimes gyg^{T}$, for $g\in GL_{n}(k)$ and $x,y\in
GL_{n}(k)^{\operatorname{\mathbb{Z}}/2}$. By construction, this map sends the
isomorphism class of a form $x$ of rank $n$ to $m(x^{-1}\otimes x)$. In
particular, since $m$ is the identity for $n=1$, this map sends the
isomorphism class of an invertible $1\times 1$-matrix $a$ to $a^{-1}\otimes
a$. Then, on the Witt group, this induces the map that sends $\langle
a\rangle$ to the same element. ∎

######  Remark 3.10.

In the proof of Theorem 3.9, we are using the Milnor conjecture twice: once in
order to identify $\nu^{\ast}(k)$ with the graded ring of $I$, and then in
order to identify $I$ with the kernel of $\pi-\phi$. It seems plausible that
one could find a proof of the theorem which does not use Kato’s Theorems.
Define $W$ and $T$ respectively as the equaliser and coequaliser of

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32.41667pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{k\otimes_{S}k\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
70.2199pt\raise 6.65971pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{\pi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 87.21498pt\raise
2.15277pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
70.12956pt\raise-8.26387pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{\phi}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern
87.21498pt\raise-2.15277pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
87.21498pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{(k\otimes_{S}k)/Im(1+w)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
205.62431pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
205.62431pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{T}$}}}}}}}\ignorespaces}}}}\ignorespaces,$

and $K$ as the kernel of the multiplication map $\mu\colon
W\to\operatorname{\mathbb{Z}}/2$. One can then try to directly show that the
induced sequence

$0\to K^{n}/K^{n+1}\to
J^{n}/J^{n+1}\xrightarrow{\pi-\phi}J^{n}/((J^{n+1}+B)\cap J^{n})\to
K^{n}T/K^{n+1}T\to 0$

remains exact, where $B$ is the subgroup of $k\otimes_{S}k$ generated by the
elements $a\otimes b+b\otimes a$. This would then prove Theorem 3.9 because,
under the isomorphism $\Omega^{\ast}_{k}\cong J^{\ast}/J^{\ast+1}$, the kernel
of the middle map corresponds to $\nu^{\ast}(k)$, and the cokernel to
$\epsilon^{\ast}(k)$ (see [Ara20, Fact 6]).

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