# Complex manifolds of Sobolev mappings and a Hartogs-type theorem in loop
spaces

M. Anakkar

###### Abstract.

We recall the complex structure on the generalised loop spaces $W^{k,2}(S,X)$,
where $S$ is a compact real manifold with boundary and $X$ is a complex
manifold, and prove a Hartogs-type extension theorem for holomorphic maps from
certain domains in generalized loop spaces.

###### Key words and phrases:

Hilbert-Hartogs manifold, analytic continuation, Sobolev spaces.

<EMAIL_ADDRESS>
###### Contents

  1. 1 Sobolev maps between smooth manifolds
  2. 2 Complex structure on the space of Sobolev maps
  3. 3 A Hartogs-type theorem

## Introduction

All manifolds in this paper are supposed to be Hausdorff and second countable.
In the first section we recall the notion of the Sobolev class
$W^{k,2}(\Omega)$ for a domain $\Omega\subset\mathbb{R}^{n}$ as well as its
basic properties. We explain that a composition of a smooth map with a map of
class $W^{k,2}$ is still in the class $W^{k,2}$. Namely we prove the following
theorem which goes back to [M].

######  Theorem 1.

Let $k>\frac{n}{2}$, then for every $u\in
W^{k,2}_{\text{loc}}(\Omega,\mathbb{R}^{m})$ and every
$f\in\mathcal{C}^{k}(\mathbb{R}^{m})$ the function $f\circ u$ is in
$W^{k,2}_{\text{loc}}(\Omega)$.

This permits us to define correctly the space of Sobolev maps $W^{k,2}(S,X)$
between real manifolds $S$ and $X$ and provide the natural structure of a
Hilbert manifold on this space. In the second section following Lempert [L] we
discuss the complex Hilbert structure of the Sobolev manifold $W^{k,2}(S,X)$,
where $X$ is now a complex manifold. S everywhere is a compact real manifold
with boundary.

For positive integers $q\geqslant 1$, $n\geqslant 1$ and real $r\in]0,1[$ the
$q$-concave Hartogs figure in $\mathbb{C}^{q+n}$ is defined as

(1)
$H_{q}^{n}(r):=\left(\Delta^{q}\times\Delta^{n}(r)\right)\cup\left(A^{q}_{1-r,1}\times\Delta^{n}\right).$

where $A^{q}_{1-r,1}=\Delta^{q}\backslash\bar{\Delta}^{q}_{1-r}$. Here
$\Delta^{q}_{r}$ stands for the polydisk in $\mathbb{C}^{q}$ centered at zero
of radius $r$. The envelope of holomorphy of $H_{q}^{n}(r)$ is $\Delta^{q+n}$.
We say that a complex manifold $X$ is $q$-Hartogs if every holomorphic mapping
$f:H_{q}^{1}(r)\to X$ extends to a holomorphic mapping
$\tilde{f}:\Delta^{q+1}\to X$. If the same holds for a complex Hilbert
manifold $\mathcal{X}$ we say that $\mathcal{X}$ is $q$-Hilbert-Hartogs. We
proved in [A-Z] that if $\mathcal{X}$ is $q$-Hilbert-Hartogs then every
holomorphic mapping $f:H_{q}^{n}(r)\to\mathcal{X}$ extends to a holomorphic
mapping $\tilde{f}:\Delta^{q+n}\to\mathcal{X}$. For finite dimensional $X$
this was proved in [I]. In the last section of this paper we prove the
following Hartogs-type extension theorem.

######  Theorem 2.

Let $\mathcal{X}$ be a $q$-Hilbert-Hartogs manifold. Then every holomorphic
map
$F:W^{k,2}(S,H_{q}^{n}(r))\to\mathcal{X}$ extends to a holomorphic map
$\tilde{F}:W^{k,2}(S,\Delta^{q}\times\Delta^{n})\to\mathcal{X}$.

This statement gives us an example of pairs of open sets
$\mathcal{U}\subsetneqq\hat{\mathcal{U}}$ in a complex Hilbert manifold such
that holomorphic mappings with values in $q$-Hilbert-Hartogs manifolds extend
from $\mathcal{U}$ to $\hat{\mathcal{U}}$. It shows that
$\hat{\mathcal{U}}{:=}W^{k,2}(S,\Delta^{q}\times\Delta^{n})$ is, in some
sense, the “envelope of holomorphy” of
$\mathcal{U}{:=}W^{k,2}(S,H_{q}^{n}(r))$. For compact $S$ without boundary
this theorem was proved in [A-I].

##  1\. Sobolev maps between smooth manifolds

Our goal in this section is to define the space of Sobolev $W^{k,2}$-maps from
a manifold $S$ to a manifold $X$ and explain that this space possesses a
natural topology. Let $\Omega\subset\mathbb{R}^{n}$ be a domain.

######  Definition 1.1.

The Sobolev class $W^{k,2}(\Omega)$ is defined as follows.

$W^{k,2}(\Omega):=\left\\{u\in L^{2}(\Omega)\ |\
\forall\alpha\in\mathbb{N}^{n},\ |\alpha|\leqslant k,\ D^{\alpha}u\in
L^{2}(\Omega)\right\\},$

where $L^{2}(\Omega)$ is the Hilbert space of equivalence classes of real
valued square integrable functions with respect to the Lebesgue measure in
$\mathbb{R}^{n}$.

The space $W^{k,2}(\Omega)$ endowed with the scalar product

$<u,v>_{W^{k,2}(\Omega)}=\sum_{|\alpha|\leqslant k}{<D^{\alpha}u\ ,\
D^{\alpha}v>_{L^{2}(\Omega)}}$

is a Hilbert space. Recall that the Fourier transform of a function from the
Schwarz space $u\in\mathcal{S}(\mathbb{R}^{n})$ is defined as

$\hat{u}(\xi):=\frac{1}{(\sqrt{2\pi})^{n}}\int_{\mathbb{R}^{n}}{u(x)e^{-i<x,\xi>}d\mu(x)},$

where $\mu$ is the Lebesgue measure in $\mathbb{R}^{n}$. By the density of
$\mathcal{S}(\mathbb{R}^{n})$ in $L^{2}(\mathbb{R}^{n})$ and using the
Placherel identity we can extend the Fourier transform to the isometry of
$L^{2}(\mathbb{R}^{n})$ onto itself, i.e. for $f\in L^{2}(\mathbb{R}^{n})$ one
has that
$||\hat{f}||_{L^{2}(\mathbb{R}^{n})}=\left\|f\right\|_{L^{2}(\mathbb{R}^{n})}$.
One can give the characterisation of Sobolev maps by the Fourier transform.

$u\in
W^{k,2}(\mathbb{R}^{n})\Longleftrightarrow\forall\alpha\in\mathbb{N}^{n},|\alpha|\leqslant
k,\ \ D^{(\alpha)}u\in L^{2}(\mathbb{R}^{n})$

$\Longleftrightarrow\forall\alpha\in\mathbb{N}^{n},|\alpha|\leqslant k,\ \
\xi^{\alpha}\hat{u}\in L^{2}(\mathbb{R}^{n})$

$\Longleftrightarrow\ (1+|\xi|^{2})^{\frac{k}{2}}\hat{u}\in
L^{2}(\mathbb{R}^{n})$

Moreover the corresponding norms are equivalent. More precisely we have the
following.

######  Proposition 1.1.

For $u\in W^{k,2}(\mathbb{R}^{n})$ we have the inequalities:

  * i)

$\left\|u\right\|_{W^{k,2}(\mathbb{R}^{n})}\leqslant\left\|(1+|\xi|^{2})^{\frac{k}{2}}\hat{u}\right\|_{L^{2}(\mathbb{R}^{n})}$;

  * ii)

there exists $a\in\mathbb{R}^{+}$ wich depends on $k$ and $n$ only such that

(2)
$\left\|(1+|\xi|^{2})^{\frac{k}{2}}\hat{u}\right\|_{L^{2}(\mathbb{R}^{n})}\leqslant
a\left\|u\right\|_{W^{k,2}(\mathbb{R}^{n})}.$

###### Proof.

First we compute the following:

$\displaystyle\left\|u\right\|_{W^{k,2}}^{2}$ $\displaystyle=$
$\displaystyle{\sum_{|\alpha|\leqslant
k}\left\|D^{(\alpha)}u\right\|^{2}_{L^{2}}}={\sum_{|\alpha|\leqslant
k}\left\|\widehat{D^{(\alpha)}u}\right\|^{2}_{L^{2}}}={\sum_{|\alpha|\leqslant
k}\left\|\xi_{1}^{\alpha_{1}}\ldots\xi_{n}^{\alpha_{n}}\hat{u}\right\|^{2}_{L^{2}}}$
$\displaystyle=$ $\displaystyle{\sum_{|\alpha|\leqslant
k}\int_{\mathbb{R}^{n}}\xi_{1}^{2\alpha_{1}}\ldots\xi_{n}^{2\alpha_{n}}|\hat{u}(\xi)|^{2}d\xi}={\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leqslant
k}\xi^{2\alpha}|\hat{u}(\xi)|^{2}d\xi}.$

So we can establish the following equality

$\left\|u\right\|_{W^{k,2}}=\sqrt{\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leqslant
k}\xi^{2\alpha}|\hat{u}(\xi)|^{2}d\xi}.$

For $(i)$ we have
$(1+|\xi|^{2})^{k}=(1+\xi_{1}^{2}+\ldots+\xi_{n}^{2})^{k}=\sum_{|\alpha|\leqslant
k}a_{\alpha}\xi^{2\alpha}\geqslant\sum_{|\alpha|\leqslant k}\xi^{2\alpha}$
since $a_{\alpha}\geqslant 1$. And therefore

$\displaystyle\left\|u\right\|_{W^{k,2}}$ $\displaystyle=$
$\displaystyle\sqrt{\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leqslant
k}\xi^{2\alpha}|\hat{u}(\xi)|^{2}d\xi}\leqslant\sqrt{\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{k}|\hat{u}(\xi)|^{2}d\xi}\leqslant$
$\displaystyle\leqslant$
$\displaystyle\left\|(1+|\cdot|^{2})^{\frac{k}{2}}\hat{u}\right\|_{L^{2}}.$

As for $(ii)$, since $(1+|\xi|^{2})^{k}=\sum_{|\alpha|\leqslant
k}a_{\alpha}\xi^{2\alpha}\leqslant a^{2}\sum_{|\alpha|\leqslant
k}\xi^{2\alpha}$ with $a=\max_{|\alpha|\leqslant k}\sqrt{a_{\alpha}}$, we
obtain

$\displaystyle\left\|(1+|\cdot|^{2})^{\frac{k}{2}}\hat{u}\right\|_{L^{2}}$
$\displaystyle=$
$\displaystyle\sqrt{\int_{\mathbb{R}^{n}}(1+|\xi|^{2})^{k}|\hat{u}(\xi)|^{2}d\xi}\leqslant
a\sqrt{\int_{\mathbb{R}^{n}}\sum_{|\alpha|\leqslant
k}\xi^{2\alpha}|\hat{u}(\xi)|^{2}d\xi}$ $\displaystyle\leqslant$
$\displaystyle a\left\|u\right\|_{W^{k,2}}.$

Proposition is proved. ∎

Denote by $W^{k,2}_{\text{loc}}(\Omega)$ the space of distributions
$u\in\mathcal{D}^{\prime}(\Omega)$ which are locally in $W^{k,2}$, i.e. for
every relatively compact $D\Subset\Omega$ the restriction $u|_{D}\in
W^{k,2}(D)$. It is clear that for $u\in W^{k,2}_{\text{loc}}(\Omega)$ and a
mapping $f:B\to\Omega$ from a domain $B\subset\mathbb{R}^{m}$ to $\Omega$ of
class $\mathcal{C}^{k}$ the composition $u\circ f\in W^{k,2}_{\text{loc}}(B)$.

Recall the Sobolev imbedding theorem:  if $f\in W^{k,2}(\Omega)$ and
$k>m+\frac{n}{2}$, then there exists a $m$-times continuously differentiable
function on $\Omega$ that is equal to $f$ almost everywhere. Therefore the
condition $k>\frac{n}{2}$ will be always assumed along this text in order to
insure that all $u\in W^{k,2}$ are at least continuous.

######  Theorem 1.1.

Let $k>\frac{n}{2}$, then for every $u\in W^{k,2}_{\text{loc}}(\Omega)$ and
$f\in\mathcal{C}^{k}(\mathbb{R})$ the function $f\circ u$ is in
$W^{k,2}_{\text{loc}}(\Omega)$.

Proof. The proof will be achieved in three steps.

Step 1: We shall state it in the form of a lemma.

######  Lemma 1.1.

For $\xi,\eta\in\mathbb{R}^{n}$ and $t\geqslant 0$ we have the following
inequality:

(3) $(1+|\xi|^{2})^{t}\leqslant
4^{t}\big{(}(1+|\xi-\eta|^{2})^{t}+(1+|\eta|^{2})^{t}\big{)}.$

###### Proof.

Since
$|\xi-\eta+\eta|^{2}\leqslant\Big{(}|\xi-\eta|+|\eta|\Big{)}^{2}\leqslant
2\Big{(}|\xi-\eta|^{2}+|\eta|^{2}\Big{)}$

$\displaystyle(1+|\xi|^{2})^{t}$ $\displaystyle=$
$\displaystyle(1+|\xi-\eta+\eta|^{2})^{t}\leqslant(1+2|\xi-\eta|^{2}+2|\eta|^{2})^{t}$
$\displaystyle\leqslant$
$\displaystyle(4+2|\xi-\eta|^{2}+2|\eta|^{2})^{t}\leqslant
2^{t}(1+|\xi-\eta|^{2}+1+|\eta|^{2})^{t}$ $\displaystyle\leqslant$
$\displaystyle 2^{t}\Big{(}2\max(1+|\xi-\eta|^{2},1+|\eta|^{2})\Big{)}^{t}$
$\displaystyle\leqslant$ $\displaystyle
4^{t}\Big{(}\max(1+|\xi-\eta|^{2},1+|\eta|^{2})\Big{)}^{t}$
$\displaystyle\leqslant$ $\displaystyle
4^{t}\Big{(}\max(1+|\xi-\eta|^{2},1+|\eta|^{2})\Big{)}^{t}+4^{t}\Big{(}\min(1+|\xi-\eta|^{2},1+|\eta|^{2})\Big{)}^{t}$
$\displaystyle=$ $\displaystyle
4^{t}(1+|\xi-\eta|^{2})^{t}+4^{t}(1+|\eta|^{2})^{t}.$

Lemma is proved ∎

Step 2: Let us state it again as a lemma.

######  Lemma 1.2.

If $k>n/2$ then the product of two Sobolev maps $u,v\in
W^{k,2}(\mathbb{R}^{n})$ is in $W^{k,2}(\mathbb{R}^{n})$ with the estimate

$\left\|uv\right\|_{W^{k,2}(\mathbb{R}^{n})}\leqslant
C\left\|u\right\|_{W^{k,2}(\mathbb{R}^{n})}\left\|v\right\|_{W^{k,2}(\mathbb{R}^{n})}.$

###### Proof.

Since $\widehat{uv}=\hat{u}*\hat{v}$ we have

$\left\|uv\right\|_{W^{k,2}}\leqslant\left\|\widehat{uv}(1+|\cdot|^{2})^{\frac{k}{2}}\right\|_{L^{2}}=\sqrt{\int_{\mathbb{R}^{n}}|(\hat{u}*\hat{v})(\xi)|^{2}(1+|\xi|^{2})^{k}d\xi}=$

$\displaystyle=$
$\displaystyle\sqrt{\int_{\xi\in\mathbb{R}^{n}}\left|\int_{\eta\in\mathbb{R}^{n}}\hat{u}(\xi-\eta)\hat{v}(\eta)(1+|\xi|^{2})^{\frac{k}{2}}\
d\eta\right|^{2}d\xi}$ $\displaystyle\leqslant$
$\displaystyle\sqrt{\int_{\xi\in\mathbb{R}^{n}}\left(\int_{\eta\in\mathbb{R}^{n}}|\hat{u}(\xi-\eta)\hat{v}(\eta)|(1+|\xi|^{2})^{\frac{k}{2}}\
d\eta\right)^{2}d\xi}$ $\displaystyle\leqslant$
$\displaystyle\left\|\int_{\mathbb{R}^{n}}4^{k}|\hat{u}(\xi-\eta)\hat{v}(\eta)|(1+|\xi-\eta|^{2})^{\frac{k}{2}}d\eta\right\|_{L^{2}}+\left\|\int_{\mathbb{R}^{n}}4^{k}|\hat{u}(\xi-\eta)\hat{v}(\eta)|(1+|\eta|^{2})^{\frac{k}{2}}\
d\eta\right\|_{L^{2}}$

The second inequality here follows from lemma 1.1. In the second integral, we
make the change of variable $s=\xi-\eta$ and get

$\left\|\int_{\eta\in\mathbb{R}^{n}}4^{k}|\hat{u}(\xi-\eta)\hat{v}(\eta)|(1+|\eta|^{2})^{\frac{k}{2}}\
d\eta\right\|_{L^{2}}=\hskip 227.62204pt$
$\displaystyle=\sqrt{\int_{\xi\in\mathbb{R}^{n}}\left(4^{k}\int_{\eta\in\mathbb{R}^{n}}|\hat{u}(\xi-\eta)\hat{v}(\eta)|(1+|\eta|^{2})^{\frac{k}{2}}\
d\eta\right)^{2}d\xi}$
$\displaystyle=4^{k}\sqrt{\int_{\mathbb{R}^{n}}\left(\int_{s\in\mathbb{R}^{n}}|\hat{u}(s)\hat{v}(\xi-s)|(1+|\xi-s|^{2})^{\frac{k}{2}}\
ds\right)^{2}d\xi}$ $\displaystyle\leqslant
4^{k}\left\|\int_{s\in\mathbb{R}^{n}}|\hat{u}(s)\hat{v}(\xi-s)|(1+|\xi-s|^{2})^{\frac{k}{2}}\
ds\right\|_{L^{2}}.$

Let recall the Minkowski’s inequality for integrals, see [S] §A.1. For
$p\in]1,+\infty[$ and $F(x,y)$ a measurable function on the product measure
space $X\times Y$ one has

(4)
$\left(\int_{Y}\left|\int_{X}F(x,y)dx\right|^{p}dy\right)^{\frac{1}{p}}\leqslant\int_{X}\left(\int_{Y}|F(x,y)|^{p}dy\right)^{\frac{1}{p}}dx.$

In other words, we have the inequality :

$\left\|\int_{X}F(x,y)dx\right\|_{L^{p}}\leqslant\int_{X}\left\|F(x,y)\right\|_{L^{p}}dx.$

Therefore by Minkowski integral inequality (4) we obtain

$\displaystyle\left\|uv\right\|_{W^{k,2}}\leqslant$ $\displaystyle\leqslant$
$\displaystyle
4^{k}\left\|\int_{\eta\in\mathbb{R}^{n}}|\hat{u}(\xi-\eta)\hat{v}(\eta)|(1+|\xi-\eta|^{2})^{\frac{k}{2}}d\eta\right\|_{L^{2}}+4^{k}\left\|\int_{s\in\mathbb{R}^{n}}|\hat{u}(s)\hat{v}(\xi-s)|(1+|\xi-s|^{2})^{\frac{k}{2}}\
ds\right\|_{L^{2}}$ $\displaystyle\leqslant$ $\displaystyle
4^{k}\int_{\eta\in\mathbb{R}^{n}}\left\|\hat{u}(\xi-\eta)\hat{v}(\eta)(1+|\xi-\eta|^{2})^{\frac{k}{2}}\right\|_{L^{2}}d\eta+4^{k}\int_{s\in\mathbb{R}^{n}}\left\|\hat{u}(s)\hat{v}(\xi-s)(1+|\xi-s|^{2})^{\frac{k}{2}}\right\|_{L^{2}}ds$
$\displaystyle\leqslant$ $\displaystyle
4^{k}\int_{\eta\in\mathbb{R}^{n}}|\hat{v}(\eta)|\left\|\hat{u}(\xi-\eta)(1+|\xi-\eta|^{2})^{\frac{k}{2}}\right\|_{L^{2}}d\eta+4^{k}\int_{\mathbb{R}^{n}}|\hat{u}(s)|\left\|\hat{v}(\xi-s)(1+|\xi-s|^{2})^{\frac{k}{2}}\right\|_{L^{2}}ds$
$\displaystyle\leqslant$ $\displaystyle
4^{k}\int_{\eta\in\mathbb{R}^{n}}|\hat{v}(\eta)|a\left\|u\right\|_{W^{k,2}}d\eta+4^{k}\int_{s\in\mathbb{R}^{n}}|\hat{u}(s)|a\left\|v\right\|_{W^{k,2}}ds.$

In the first integral we transform as follows.

(5) $\displaystyle
4^{k}\int_{\eta\in\mathbb{R}^{n}}|\hat{v}(\eta)|a\left\|u\right\|_{W^{k,2}}d\eta$
$\displaystyle\leqslant$ $\displaystyle
4^{k}a\left\|u\right\|_{W^{k,2}}\int_{\eta\in\mathbb{R}^{n}}|\hat{v}(\eta)|(1+|\eta|^{2})^{\frac{k}{2}}(1+|\eta|^{2})^{-\frac{k}{2}}d\eta\leqslant$
(6) $\displaystyle\leqslant$ $\displaystyle
4^{k}a^{2}\left\|u\right\|_{W^{k,2}}\left\|v\right\|_{W^{k,2}}\left\|(1+|\xi|^{2})^{-\frac{k}{2}}\right\|_{L^{2}}.$

We make the same with the second integral and obtain

$\displaystyle\left\|uv\right\|_{W^{k,2}}$ $\displaystyle\leqslant$
$\displaystyle
4^{k}a^{2}\left\|u\right\|_{W^{k,2}}\left\|v\right\|_{W^{k,2}}\left\|(1+|\xi|^{2})^{-\frac{k}{2}}\right\|_{L^{2}}+4^{k}a^{2}\left\|u\right\|_{W^{k,2}}\left\|v\right\|_{W^{k,2}}\left\|(1+|\xi|^{2})^{-\frac{k}{2}}\right\|_{L^{2}}$
$\displaystyle\leqslant$
$\displaystyle\Big{(}2^{2k+1}a^{2}\left\|(1+|\xi|^{2})^{-\frac{k}{2}}\right\|_{L^{2}}\Big{)}\left\|u\right\|_{W^{k,2}}\left\|v\right\|_{W^{k,2}}.$

Lemma is proved. ∎

This step implies that the product of two $W^{k,2}_{\text{loc}}(\Omega)$ maps
are still in $W^{k,2}_{\text{loc}}(\Omega)$ and consequently this implies that
for any polynomial $P$ and for any $u\in W^{k,2}_{\text{loc}}(\Omega)$ the
function $P(u)\in W^{k,2}_{\text{loc}}(\Omega)$.

Step 3 : For $I$ a closed interval one defines the norm on the space
$\mathcal{C}^{k}(I)$ by

$\left\|f\right\|_{\mathcal{C}^{k}(I)}=\sum_{j=0}^{k}{\sup_{x\in
I}|f^{(j)}(x)}|.$

Remark that:

  * •

The norm $\left\|f\right\|_{\mathcal{C}^{0}(I)}=\sup_{x\in I}|f(x)|$
correspond to the usual sup-norm.

  * •

The space $\mathcal{C}^{k}(I)$ endowed with the norm
$\left\|\cdot\right\|_{\mathcal{C}^{k}(I)}$ is a Banach space.

Recall the approximation theorem of Weirstrass:

Let $f\in\mathcal{C}^{0}(I)$. Then for every $\epsilon>0$, there exists a
polynomial $P$ such that for all $x\in[a,b]$, we have $|f(x)-P(x)|<\epsilon$,
i.e. $\left\|f-P\right\|_{\mathcal{C}^{0}(I)}<\epsilon$.

From that statement, one can deduce an approximation with respect to the norm
$\mathcal{C}^{k}$:

######  Proposition 1.2.

Any function $f:I\to\mathbb{R}$ of class $\mathcal{C}^{k}$ defined on a closed
interval $I$ can be uniformly approximated by polynomials with respect to the
norm $\mathcal{C}^{k}$, i.e.

$\forall\epsilon>0,\ \ \exists\ P\in\mathbb{R}[X],\ \text{such that}\
\left\|f-P\right\|_{\mathcal{C}^{k}(I)}<\epsilon.$

###### Proof.

Let $I=[a,b]$ with $a,b\in\mathbb{R}$. Since $f$ is $\mathcal{C}^{k}$, use the
approximation theorem on $f^{(k)}$. It exists $P_{k}$ in $\mathbb{R}[X]$ such
that $\left\|P_{k}-f^{(k)}\right\|_{\mathcal{C}^{0}(I)}<\epsilon$. Then we
define

$\forall\ l\vskip 3.0pt plus 1.0pt minus 1.0pt\in[\\![1,k]\\!],\
P_{k-l}(x)=\int_{a}^{x}P_{k-l+1}(t)dt+f^{(k-l)}(a).$

We can prove by induction on $l$ that

(7) $\forall\ l\vskip 3.0pt plus 1.0pt minus 1.0pt\in[\\![0,k]\\!],\ \forall
x\in[a,b],\
\left|P_{k-l}(x)-f^{(k-l)}(x)\right|\leqslant\epsilon\frac{(x-a)^{l}}{l!}.$

Indeed for $l=0$ the assertion (7) is true and we suppose that (7) is true for
$l-1$. Then we can write the following.

(8) $\displaystyle|P_{k-l}(x)$ $\displaystyle-$ $\displaystyle
f^{(k-l)}(x)|=\left|\int_{a}^{x}P_{k-l+1}(t)dt+f^{(k-l)}(a)-f^{(k-l)}(x)\right|=$
(9) $\displaystyle=$
$\displaystyle\left|\int_{a}^{x}\Big{(}P_{k-l+1}(t)-f^{(k-l+1)}(t)\Big{)}dt\right|\leqslant\int_{a}^{x}\left|P_{k-l+1}(t)-f^{(k-l+1)}(t)\right|dt$
(10) $\displaystyle\leqslant$
$\displaystyle\int_{a}^{x}\epsilon\frac{(t-a)^{l-1}}{(l-1)!}dt\leqslant\epsilon\frac{(x-a)^{l}}{l!}.$

For all $l\in[\\![0,k]\\!]$, we obtain that
$\left\|P_{0}^{k-l}-f^{(k-l)}\right\|_{\mathcal{C}^{0}(I)}\leqslant\epsilon\frac{(b-a)^{l}}{l!}$
and then

$\left\|P_{0}-f\right\|_{\mathcal{C}^{k}(I)}\leqslant\epsilon\sum_{l=0}^{k}{\frac{(b-a)^{l}}{l!}}.$

Proposition is proved. ∎

Now we can finish the proof of the theorem. Fix a relatively compact subdomain
$D\Subset\Omega$. Since $u$ is continuous there exists a closed interval
$I\supset u(\bar{D})$. Let $(P_{s})_{s\in\mathbb{N}}$ be a sequence of
polynomials that converges to $f$ in $\mathcal{C}^{k}(I)$. Then
$(P_{s})_{s\in\mathbb{N}}$ is a Cauchy sequence : for every $\epsilon>0$ there
exists $N\in\mathbb{N}$ such that for any $l,s>N$ we have
$\left\|P_{l}-P_{s}\right\|_{\mathcal{C}^{k}(I)}<\epsilon$. By Step 2 for any
$s\in\mathbb{N}$ the function $P_{s}(u)$ is in $W^{k,2}(D)$. Let us prove that
$(P_{s}(u))_{s\in\mathbb{N}}$ is a Cauchy sequence in the Hilbert space
$W^{k,2}(D)$. Note $P_{sl}=P_{s}-P_{l}$.

For $\alpha$ with $|\alpha|\leqslant k$ and any polynomial $P$ we have

(11)
$D^{\alpha}P(u)=\sum_{r=1}^{|\alpha|}\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right)P^{(r)}(u).$

For $P(X)=X^{p}$ relation (11) becomes

$\sum_{\underset{\alpha_{1},\cdots,\alpha_{p}\neq
0}{\alpha_{1}+...+\alpha_{p}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{p}}u=\frac{1}{p!}\left(D^{\alpha}(u^{p})-\sum_{r=1}^{p-1}\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right)\frac{p!}{(p-r)!}u^{p-r}\right).$

Let us prove by induction on $p\leqslant|\alpha|$ that

$\sum_{\underset{\alpha_{1},\cdots,\alpha_{p}\neq
0}{\alpha_{1}+...+\alpha_{p}=\alpha}}D^{\alpha_{1}}u\cdots D^{\alpha_{p}}u\in
L^{2}(D).$

For $p=1$ it is true because $\frac{\partial^{|\alpha|}u}{\partial
x_{j}^{\alpha}}\in L^{2}(D)$ for every $j=1,\dots,n$ since $u\in W^{k,2}(D)$.
Let suppose that is true for all $s\leqslant p-1$. Then for
$p\leqslant|\alpha|$ using the previous expression we obtain

$\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{p}\neq
0}{\alpha_{1}+...+\alpha_{p}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{p}}u\right\|_{L^{2}}\leqslant\hskip 284.52756pt$
$\leqslant\frac{1}{p!}\left\|D^{\alpha}(u^{p})\right\|_{L^{2}}+\sum_{r=1}^{p-1}\left\|\frac{1}{(p-r)!}u^{p-r}\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right\|_{L^{2}}.$

By Lemma 1.2 the map $u^{p}\in W^{k,2}(D)$, therefore $D^{\alpha}(u^{p})\in
L^{2}(D)$ for $|\alpha|\leqslant k$. For the second term since $u$ is
continuous we have that $u$ is bounded on $\bar{D}$ by $M$. Then

$\left\|\frac{1}{(p-r)!}u^{p-r}\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right\|_{L^{2}}\leqslant\frac{M^{p-r}}{(p-r)!}\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right\|_{L^{2}},$

and by induction hypothesis

$\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right\|_{L^{2}}<\infty.$

We obtain from (11) that

$\displaystyle\left\|P_{s}(u)-P_{s}(u)\right\|_{W^{k,2}}^{2}$ $\displaystyle=$
$\displaystyle\left\|P_{sl}(u)\right\|_{W^{k,2}}^{2}=\sum_{|\alpha|\leqslant
k}\left\|D^{\alpha}P_{sl}(u)\right\|_{L^{2}}^{2}\leqslant$
$\displaystyle\leqslant$ $\displaystyle\sum_{|\alpha|\leqslant
k}\sum_{r=1}^{|\alpha|}\left\|\sum_{\alpha_{1}+...+\alpha_{r}=\alpha}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right\|_{L^{2}}^{2}|P_{sl}^{(r)}(u)|^{2}$
$\displaystyle\leqslant$ $\displaystyle\sum_{|\alpha|\leqslant
k}\sum_{r=1}^{|\alpha|}\left\|\sum_{\alpha_{1}+...+\alpha_{r}=\alpha}D^{\alpha_{1}}u\cdots
D^{\alpha_{r}}u\right\|_{L^{2}}^{2}\epsilon^{2}.$

Then we can conclude that $(P_{s}(u))_{s\in\mathbb{N}}$ is a Cauchy sequence
on the Hilbert space $W^{k,2}(D)$. Therefore the sequence converge to $f(u)\in
W^{k,2}(D)$. Theorem is proved. ∎

######  Definition 1.2.

Let $\Omega\subset\mathbb{R}^{n}$ be a domain. The Sobolev space
$W^{k,2}(\Omega,\mathbb{R}^{m})$ is defined as

$W^{k,2}(\Omega,\mathbb{R}^{m})=\\{u\in L^{2}(\Omega,\mathbb{R}^{m})\ |\
\forall\alpha\in\mathbb{N}^{n},|\alpha|\leqslant k,D^{\alpha}u\in
L^{2}(\Omega,\mathbb{R}^{m})\\}.$

As in the one dimensional case one can prove an analogous of the Theorem 1.1.
Namely

######  Theorem 1.2.

Let $k>\frac{n}{2}$, then for every $u\in
W^{k,2}_{\text{loc}}(\Omega,\mathbb{R}^{m})$ and
$f\in\mathcal{C}^{k}(\mathbb{R}^{m})$ the function $f\circ u$ is in
$W^{k,2}_{\text{loc}}(\Omega)$.

###### Proof.

The idea of the proof follows the one dimensional case. Fix a relatively
compact subdomain $D\Subset\Omega$.

For any polynomial $P(T_{1},\dots,T_{n})$ and $|\alpha|\leqslant k$ one has

(12) $\displaystyle D^{\alpha}(P(u))$ $\displaystyle=$
$\displaystyle\sum_{r=1}^{|\alpha|}\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}(d^{r}P)_{u}\big{(}D^{\alpha_{1}}u,\dots,D^{\alpha_{r}}u\big{)}$
(13) $\displaystyle=$
$\displaystyle\sum_{r=1}^{|\alpha|}\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}\left(\sum_{i_{1},\cdots,i_{r}=1}^{m}\frac{\partial^{r}P}{\partial
t_{i_{1}}\cdots\partial t_{i_{r}}}(u)D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right)$ (14) $\displaystyle=$
$\displaystyle\sum_{r=1}^{|\alpha|}\sum_{i_{1},\cdots,i_{r}=1}^{m}\frac{\partial^{r}P}{\partial
t_{i_{1}}\cdots\partial
t_{i_{r}}}(u)\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right)$

Fix $\alpha\in\mathbb{N}^{n}$ with $|\alpha|\leqslant k$. We prove by
induction on $p$ with $1\leqslant p\leqslant|\alpha|$ that

$\forall I=(j_{1},\dots,j_{p})\in[\\![1,m]\\!]^{p},\
\sum_{\underset{\alpha_{1},\cdots,\alpha_{{p}}\neq
0}{\alpha_{1}+...+\alpha_{{p}}=\alpha}}D^{\alpha_{1}}u_{j_{1}}\dots
D^{\alpha_{p}}u_{j_{p}}\in L^{2}(D).$

For $p=1$ the property is true because $D^{\alpha}u_{j}\in L^{2}(D)$ for
$j=1,\dots,m$ since $D^{\alpha}u\in L^{2}(D,\mathbb{R}^{m})$. Let suppose that
is true for all $s\leqslant p-1$. For $p\leqslant|\alpha|$ one can define for
every $J=(j_{1},\dots,j_{p})\in[\\![1,m]\\!]^{p}$
$\beta=e_{j_{1}}+\dots+e_{j_{p}}$ with $(e_{j})_{i=1,...,m}$ the standard
basis in $\mathbb{R}^{m}$. Then consider
$P(T)=T^{\beta}=T_{1}^{\beta_{1}}\cdots T_{m}^{\beta_{m}}$ in relation (14)

$\displaystyle D^{\alpha}(u^{\beta})$ $\displaystyle=$
$\displaystyle\sum_{r=1}^{|\alpha|}\sum_{i_{1},\cdots,i_{r}=1}^{m}\frac{\partial^{r}P}{\partial
t_{i_{1}}\cdots\partial
t_{i_{r}}}(u)\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right)$ $\displaystyle=$
$\displaystyle\sum_{r=1}^{p}\sum_{i_{1},\cdots,i_{r}=1}^{m}\frac{\partial^{r}T^{\beta}}{\partial
t_{i_{1}}\cdots\partial
t_{i_{r}}}(u)\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right).$

Note that $\frac{\partial^{p}T^{\beta}}{\partial t_{j_{1}}\cdots\partial
t_{j_{p}}}=\beta!$ and the number of indices $J$ such that
$\beta=e_{j_{1}}+\dots+e_{j_{p}}$ is equal again to $\beta!$. The other term
are equal to zero. Then we obtain

$\displaystyle(\beta!)^{2}$
$\displaystyle\sum_{\underset{\alpha_{1},\cdots,\alpha_{p}\neq
0}{\alpha_{1}+...+\alpha_{p}=\alpha}}D^{\alpha_{1}}u_{j_{1}}\dots
D^{\alpha_{r}}u_{j_{p}}=$
$\displaystyle=D^{\alpha}(u^{\beta})-\sum_{r=1}^{p-1}\sum_{i_{1},\cdots,i_{r}=1}^{m}\frac{\partial^{r}T^{\beta}}{\partial
t_{i_{1}}\cdots\partial
t_{i_{r}}}(u)\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right),$

where $\beta!=\beta_{1}!\cdots\beta_{m}!$.

The term $D^{\alpha}(u^{\beta})=D^{\alpha}(u_{1}^{\beta_{1}}\dots
u_{m}^{\beta_{m}})$ is in $L^{2}(D)$ since $u_{1},\dots u_{m}$ are in
$W^{k,2}(D)$ and Lemma 1.2. For the other term one can remark first that
$\frac{\partial^{r}X^{\beta}}{\partial x_{i_{1}}\cdots\partial x_{i_{r}}}(u)$
is a polynomial evaluate on $u$ and since $u$ is bounded on $D$ this
polynomial is bounded by $M\in\mathbb{R}$. Therefore for every
$(i_{1},\dots,i_{r})\in[\\![1,m]\\!]^{r}$ one has

$\left\|\frac{\partial^{r}X^{\beta}}{\partial x_{i_{1}}\cdots\partial
x_{i_{r}}}(u)\left(\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right)\right\|\leqslant
M\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right\|_{L^{2}(D)}$

and by induction hypothesis this is in $L^{2}(D)$. Then for every
$|\alpha|\leqslant k$

$\forall p\leqslant|\alpha|,\ \forall
I=(j_{1},\dots,j_{p})\in[\\![1,m]\\!]^{p},\
\sum_{\underset{\alpha_{1},\cdots,\alpha_{{p}}\neq
0}{\alpha_{1}+...+\alpha_{{p}}=\alpha}}D^{\alpha_{1}}u_{j_{1}}\dots
D^{\alpha_{p}}u_{j_{p}}\in L^{2}(D).$

Now, since $u$ is continuous there exists a compact set $K\supset u(\bar{D})$.
Let $(P_{s})_{s\in\mathbb{N}}$ be a sequence of polynomials that converges to
$f$ in $\mathcal{C}^{k}(K)$. Then $(P_{s})_{s\in\mathbb{N}}$ is a Cauchy
sequence : for every $\epsilon>0$ there exists $N\in\mathbb{N}$ such that for
any $l,s>N$ we have
$\left\|P_{l}-P_{s}\right\|_{\mathcal{C}^{k}(K)}<\epsilon$. By Lemma 1.2 for
any $s\in\mathbb{N}$ the function $P_{s}(u)$ is in $W^{k,2}(D)$. Let us prove
that $(P_{s}(u))_{s\in\mathbb{N}}$ is a Cauchy sequence in the Hilbert space
$W^{k,2}(D)$. Note $P_{sl}=P_{s}-P_{l}$.

(15) $\displaystyle\left\|D^{\alpha}(P_{sl}(u))\right\|_{L^{2}(D)}^{2}$
$\displaystyle\leqslant$
$\displaystyle\sum_{r=1}^{|\alpha|}\sum_{i_{1},\cdots,i_{r}=1}^{m}\left|\frac{\partial^{r}P_{sl}}{\partial
t_{i_{1}}\cdots\partial
t_{i_{r}}}(u)\right|^{2}\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right\|_{L^{2}(D)}^{2}$ (16) $\displaystyle\leqslant$
$\displaystyle\sum_{r=1}^{|\alpha|}\sum_{i_{1},\cdots,i_{r}=1}^{m}\epsilon^{2}\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right\|_{L^{2}(D)}^{2}$ (17) $\displaystyle\leqslant$
$\displaystyle\epsilon^{2}\sum_{r=1}^{|\alpha|}\sum_{i_{1},\cdots,i_{r}=1}^{m}\left\|\sum_{\underset{\alpha_{1},\cdots,\alpha_{r}\neq
0}{\alpha_{1}+...+\alpha_{r}=\alpha}}D^{\alpha_{1}}u_{i_{1}}\dots
D^{\alpha_{r}}u_{i_{r}}\right\|_{L^{2}(D)}^{2}$

Therefore $(P_{s}(u))_{s}$ is a Cauchy sequence in the Hilbert space
$W^{k,2}(D)$ so the sequence converge to $f(u)\in W^{k,2}(D)$. Theorem is
proved. ∎

From that one can define Sobolev spaces of maps between manifolds.

######  Definition 1.3.

Let $S$ and $X$ be real manifolds of class $\mathcal{C}^{k}$. A map $u:S\to X$
is said to be in $W^{k,2}(S,X)$ if for every $s\in S$, every coordinate chart
$(V,\psi)$ which contains $s$ and every coordinate chart $(U,\phi)$ which
contains $u(s)$ one has that $\phi\circ u\circ\psi^{-1}\in W^{k,2}_{loc}$.

Theorem 1.2 insures that this definition is correct. The space of
$W^{k,2}$-maps from $S$ to $X$ we denote as $W^{k,2}(S,X)$. Notice that
$W^{k,2}(S,X)$ inherits the natural topology from $W^{k,2}_{loc}$. We shall
call this topology the Sobolev topology.

##  2\. Complex structure on the space of Sobolev maps

From now on let $S$ be a compact connected n-dimensional real manifold with
boundary. Let $X$ be a finite dimensional complex manifold. Our goal in this
section to equip the Sobolev space $W^{k,2}(S,X)$ with a natural structure of
a complex Hilbert manifold.

For a set $U\subset S\times X$ and $s\in S$ we write

$U^{s}:=\\{x\in X\ |\ (s,x)\in U\\}\hskip 28.45274pt\text{and}\hskip
28.45274pt\begin{array}[]{ccccc}\epsilon^{s}&:&U^{s}&\to&U\\\
&&x&\mapsto&(s,x).\\\ \end{array}$

######  Lemma 2.1.

Given $g\in W^{k,2}(S,X)$ with $k>\frac{n}{2}$, there exists a
$W^{k,2}$-diffeomorphism $G$ between a neighborhood $U\subset S\times X$ of
the graph $\\{(s,g(s))\ |s\in S\\}$ of $g$ and a neighborhood of the zero
section in $g^{*}TX$ such that

  * i)

$\\{G(s,g(s))\ |\ s\in S\\}$ is the zero section of $g^{*}TX$;

  * ii)

$G^{s}=G\circ\epsilon^{s}$ maps $U^{s}$ biholomorphically on a neighborhood of
$0\in T_{g(s)}X$;

  * iii)

$dG^{s}_{g(s)}$ is the identity map.

###### Proof.

i) We recall the argument from [L] pointing out the smoothness of $G$. Let
$(\Omega_{j},\phi_{j})$ be an atlas of the complex manifold $X$. Then the sets
$S_{j}=g^{-1}(\Omega_{j})\subset S$ form an open covering of $S$. Consider
$U_{j}\subset S_{j}\times\Omega_{j}$ a neighborhood of the graph of
$g|_{S_{j}}$. We can construct locally the diffeomorphism $G_{j}$ by

$\begin{array}[]{cccc}G_{j}:&U_{j}&\longrightarrow&g^{*}TX\\\
&(s,x)&\mapsto&\Big{(}s,(d\phi_{j}^{-1})_{\phi_{j}(g(s))}\big{[}\phi_{j}(x)-\phi_{j}(g(s))\big{]}\Big{)}.\end{array}$

Notice that $G_{j}$ is of class $W^{k,2}$ because such is $g$. It remains to
glue all the $G_{j}$. Take $\\{\eta_{j}\\}$ a $\mathcal{C}^{k}$-partition of
unity subordinated to the covering $\\{S_{j}\\}$ and define
$G(s,x)=\sum_{j}\eta_{j}(s)G_{j}(s,x)$. Choose then the restriction of $G$ to
a suitable $U\subset\bigcup_{j}U_{j}$. Remark that $G(s,g(s))=(s,0)$.

ii) We have that
$G_{j}^{s}(x)=G_{j}\circ\epsilon^{s}(x)=(d\phi_{j}^{-1})_{\phi_{j}(g(s))}\big{[}\phi_{j}(x)-\phi_{j}(g(s))\big{]}$.
The map $G_{j}^{s}:U_{g}^{s}\to g^{*}TX$ is holomorphic for every $s\in S_{j}$
since
$x\mapsto(d\phi_{j}^{-1})_{\phi_{j}(g(s))}\big{[}\phi_{j}(x)-\phi_{j}(g(s))\big{]}$
is holomorphic. Here $U_{g}^{s}=\\{x\in X\ |\ (s,x)\in U_{g}\\}$.
Holomorpicity of $G^{s}(x)=\sum_{j}\eta_{j}(s)G^{s}_{j}(x)$ follows.

iii) We compute the differential of $G^{s}$ and obtain

$dG^{s}_{g(s)}=(d\phi_{j}^{-1})_{\phi_{j}(g(s))}(d\phi_{j})_{g(s)}=\text{Id}$

∎

For $g\in W^{k,2}(S,X)$ choose $U_{g}$ and $G$ as in the previous lemma. Those
$h\in W^{k,2}(S,X)$ whose graph $\Gamma_{h}:=\\{(s,h(s))\ |\ s\in S\\}$ is
contained in $U_{g}$ form a neighborhood $\mathcal{U}_{g}\subset W^{k,2}(S,X)$
of $g$:

$\mathcal{U}_{g}=\left\\{h\in W^{k,2}(S,X)\ |\ \Gamma_{h}\subset
U_{g}\right\\}.$

For $h\in\mathcal{U}_{g}$ define the section $\psi_{g}(h)=G(\cdot,h(\cdot))\in
W^{k,2}(S,g^{*}TX)$. Thus $\psi_{g}$ is a homeomorphism between
$\mathcal{U}_{g}$ and an open neighborhood of zero section in
$W^{k,2}(S,g^{*}TX)$. We can define the chart $(\mathcal{U}_{g},\psi_{g})$
where local coordinates are in a complex Hilbert space $W^{k,2}(S,g^{*}TX)$.
Now we need to verify that transition maps are holomorphic.

Let $h\in W^{k,2}(S,X)$ be such that $\Gamma_{h}\subset U_{g}\cap
U_{g^{\prime}}$, i.e. $h\in\mathcal{U}_{g}\cap\mathcal{U}_{g^{\prime}}$. For
$s\in S$ we have

$\psi_{g^{\prime}}(h)(s)=\left(\psi_{g^{\prime}}\circ\psi_{g}^{-1}\right)[\psi_{g}(h)(s)]=\left[{G^{\prime}}^{s}\circ{G^{s}}^{-1}\right][\psi_{g}(h)(s)]=\left[G^{\prime}(s,\cdot)\circ
G(s,\cdot)^{-1}\right][\psi_{g}(h)(s)]$

Due to item $ii)$ of the Lemma just proved we have that
${G^{\prime}}^{s}\circ{G^{s}}^{-1}$ is a biholomorphism between an appropriate
open subsets of $T_{g(s)}X$ and $T_{g^{\prime}(s)}X$. Therefore the value
$\psi_{g^{\prime}}(h)(s)$ depends holomorphically on $\psi_{g}(h)(s)$.

In more details we have a $W^{k,2}$-regular maps $P=G^{\prime}\circ G^{-1}$
between open sets $V\subset g^{*}TX$ and $V^{\prime}\subset{g^{\prime}}^{*}TX$
such that for every $s\in S$ $P(s,v)$ holomorphically depends on $v\in V\cap
T_{g(s)}X$ where $P(s,v)=(s,P^{s}(v))$. Let $\mathcal{V}$ be the open set of
sections of $g^{*}TX$ which are contained in $V$. The same for
$\mathcal{V}^{\prime}$. We obtain a mapping
$\mathcal{P}:\mathcal{V}\to\mathcal{V}^{\prime}$ defined as
$\mathcal{P}(h)(s)=P(s,P^{s}(h(s)))$, where $h$ is a $W^{k,2}$-section of
$g^{*}TX$ contained in $V$. Then $\mathcal{P}$ is holomorphic. This easily
follows from Gâteaux differentiability and continuity of $\mathcal{P}$. Thus
$W^{k,2}(S,X)$ has the structure of a complex Hilbert manifold.

######  Lemma 2.2.

Let $D$ and $X$ be finite dimensional complex manifolds and let $S$ be an
$n$-dimensional compact real manifold with boundary. A mapping $F:D\times S\to
X$ represents a holomorphic map $F_{*}:D\to W^{k,2}(S,X)$ if and only if the
following holds:

  * i)

for every $s\in S$ the map $F(\cdot,s):D\to X$ is holomorphic;

  * ii)

for every $z\in D$ one has $F(z,\cdot)\in W^{k,2}(S,X)$ and the correspondence
$D\ni z\mapsto F(z,\cdot)\in W^{k,2}(S,X)$ is continuous with respect to the
Sobolev topology on $W^{k,2}(S,X)$ and the standard topology on $D$.

###### Proof.

$\Rightarrow$ Given $F_{*}:D\to W^{k,2}(S,X)$ we construct $F:D\times S\to X$
as follows. For $s\in S$ and $z\in D$ we define $F(z,s)=F_{*}(z)(s)$. By the
assumtion about $F_{*}$ for any $z\in D$ the map $F(z,\cdot)$ is in
$W^{k,2}(S,X)$ and the map $z\mapsto F(z,\cdot)=F_{*}(z)$ is continuous (in
fact it is holomorphic). To prove the holomorphicity of $F(z,s)$ for a fixed
take any $s\in S$ in a neighborhood of some $z_{0}$ take any chart
$(\mathcal{U},\psi_{g})$ which conains the graph of $F_{*}(z_{0})$. We have
$\psi_{g}(F(z,s))(s)=G\big{(}s,F_{*}(z)(s)\big{)}=G(s,F(z,s))$ for $z$ close
to $z_{0}$ by definition of $\psi_{g}$. This is holomorphic by the definition
of the complex structure of $W^{k,2}(S,X)$. Therefore $F:D\times S\to X$ is
holomorphic in $z$.

Conversely given $F:D\times S\to X$ satisfying $i)$ and $ii)$ we can construct
$F_{*}$ as follows. For $z\in D$ we define $F_{*}(z)=F(z,\cdot)$. Mapping
$F_{*}$ is well defined since $F(z,\cdot)\in W^{k,2}(S,X)$ and is continuous
by $ii)$. For the holomorphicity again we take a chart
$(\mathcal{U},\psi_{g})$ and we consider the map $\psi_{g}\circ F_{*}$ defined
for $z\in D$ by $\psi_{g}\circ
F_{*}(z)=G\big{(}\cdot,F(z,\cdot)(\cdot)\big{)}$. By $i)$ for every $s\in S$
the map $s\mapsto G\big{(}s,F(z,s)(s)\big{)}$ is holomorphic by composition of
holomorphic maps $F(\cdot,s)$ and $G^{s}$. Then $\psi_{g}\circ F_{*}$ is
holomorphic. ∎

Remark. If $X$ is a smooth real manifold one can repeat the same contruction
to ensure that $W^{k,2}(S,X)$ has a structure of a smooth Hilbert manifold. In
fact the holomorphy of item $ii)$ in lemma 2.1 should be replaced by
smoothness since coordinate charts $\psi_{g}$ are smooth.

##  3\. A Hartogs-type theorem

In [I] it was proved the following result.

######  Theorem 3.1.

If a complex manifold $X$ is $q$-Hartogs then for any $(p,n)$ with $p\geq
q,n\geq 1$ the map $f:H_{p}^{n}(r)\to X$ extends holomorphically to
$\Delta^{q+n}$.

For Hilbert $\mathcal{X}$ it was proved in [A-Z]. The proof of this result for
Hilbert $\mathcal{X}$ lies on the following two statements proved in [A-Z],
and we shall need them here too. Recall first the definition of a $1$-complete
neighborhood.

######  Definition 3.1.

A $1$-complete neighborhood of a compact $\mathcal{K}\subset\mathcal{X}$ is an
open set $\mathcal{U}\supset\mathcal{K}$ such that

  * i)

$\mathcal{U}$ is contained in a finite union of open coordinate balls centered
at points of $\mathcal{K}$, i.e.

$\mathcal{U}\subset\bigcup_{\alpha=1}^{n}{}\mathcal{B}_{\alpha}\ \text{with}\
\mathcal{B}_{\alpha}=\phi_{\alpha}^{-1}(B^{\infty})\ \text{and}\
\phi_{\alpha}^{-1}(0)=k_{\alpha}\in\mathcal{K},$

  * ii)

$\mathcal{U}$ possesses a strictly plurisubharmonic exhaustion function
$\psi:\mathcal{U}\to[0,t_{0})$, i.e.

    * –

for every $t<t_{0}$ one has that
$\overline{\psi^{-1}\left([0,t)\right)}\subset\mathcal{U}$.

Here by a strictly plurisubhamonic function we mean the following.

######  Definition 3.2.

Let $U$ be an open subset of $\mathcal{X}$. A function
$f\in\mathcal{C}^{2}(U,\mathbb{R})$ is said to be strictly plurisubharmonic on
$U$ if the Levi form $\mathcal{L}_{f,a}$ is positive definite for every $a\in
U$, i.e

$\mathcal{L}_{f,a}(v)>0\hskip 8.53581pt\text{for each}\hskip 8.53581pta\in
U\hskip 8.53581pt\text{and}\hskip 8.53581ptv\in
T_{a}\mathcal{X}\backslash\\{0\\}.$

In [A-Z] we give a strongest definition of a strictly plurisubharmonic
function, see definition 2.1 there. In fact we states that the Levi form
staisfies

${\mathcal{L}_{f,a}(v)}\geqslant c(a)||v||^{2}\quad\text{for each}\hskip
8.53581pta\in U\hskip 8.53581pt\text{and}\hskip 8.53581ptv\in
T_{a}\mathcal{X}\backslash\\{0\\},$

with $c$ a positive function in $\mathcal{C}^{0}(U,\mathbb{R})$. These
definition are not equivalent as it shows the counter-example given by
Lempert:

Take for example the function $f$ defined by
$f(z)=\sum_{j=1}^{\infty}{\frac{|z_{j}|^{2}}{j}}$ for $z\in l^{2}$. The Levi
form on a point $a\in l^{2}$ is

$\forall v\in l^{2},\hskip
8.53581pt\mathcal{L}_{f,a}(v)=\sum_{j=1}^{\infty}\frac{1}{j}|v_{j}|^{2},$

and this cannot be bound from below by $c||v||^{2}$. Nevertheless
plurisubharmonic functions we need in this paper satisfies this strongest
definition.

######  Theorem 3.2.

Let $\phi:\bar{D}\to\mathcal{X}$ be an imbedded analytic $q$-disk in a complex
Hilbert manifold $\mathcal{X}$. Then $\phi(\bar{D})$ has a fundamental system
of $1$-complete neighborhoods.

Here an analytic $q$-disk in a complex Hilbert manifold $\mathcal{X}$ is a
holomorphic imbedding $\phi$ of a neighborhood of a closure of a relatively
compact strongly pseudoconvex domain $D\Subset\mathbb{C}^{q}$ into
$\mathcal{X}$. We shall also need the following lemma from [A-Z], see Lemmma
3.1 there.

######  Lemma 3.1.

Let $\phi_{n}:\bar{D}\to\mathcal{X}$ be a sequence of analytic $q$-disks in a
complex Hilbert manifold $\mathcal{X}$ and let $\Phi_{n}$ be their graphs.
Suppose that there exists an analytic disk $\phi_{0}:\bar{D}\to\mathcal{X}$
with the graph $\Phi_{0}$ such that for any neighborhood
$\mathcal{V}\supset\Phi_{0}$ one has $\Phi_{n}\subset\mathcal{V}$ for $n>>1$.
Then $\phi_{n}$ converges uniformly on $\bar{D}$ to $\phi_{0}$.

Now we return to the proof of the theorem 2 from the introduction.

######  Theorem 3.3.

Let $\mathcal{X}$ be a $q$-Hilbert-Hartogs manifold. Then every holomorphic
map
$F:W^{k,2}(S,H_{q}^{n}(r))\to\mathcal{X}$ extends to a holomorphic map
$\tilde{F}:W^{k,2}(S,\Delta^{q}\times\Delta^{n})\to\mathcal{X}$.

###### Proof.

The proof will be achieved in a number of steps. First we shall construct some
“natural extension” of $F$. After that we shall prove that the extension is
continuous and finally that it is holomorphic.

Step 1. Natural extension. Let
$f=(f^{q},f^{n}):S\to\Delta^{q}\times\Delta^{n}$ be an element of
$W^{k,2}(S,\Delta^{q}\times\Delta^{n})$. Here $f^{q}$ and $f^{n}$ are
componnents of $f$. We want to extend $F$ to $f$. This will be done along an
appropriate analytic disc which passes through $f$. Consider the mapping
$\phi_{f}:\bar{\Delta}^{q}\times S\to\Delta^{q}\times\Delta^{n}$ defined as

(18) $\phi_{f}:(z,s)\to\left(h_{f^{q}(s)}(z),f^{n}(s)\right).$

Here $h_{a}$ is the following automorphism of $\Delta^{q}$ interchanging
$a=(a_{1},...,a_{q})$ and $0$:

$h_{a}(z)=\left(\frac{a_{1}-z_{1}}{1-\bar{a}_{1}z_{1}},...,\frac{a_{q}-z_{q}}{1-\bar{a}_{q}z_{q}}\right).$

Notice that due to the compactness of $f^{q}(S)\subset\Delta^{q}$
authomorphisms $h_{f^{q}(s)}$ are defined for $z$ in a fixed (independant on
$s$) neighborhood of $\bar{\Delta}^{q}$, and therefore such is $\phi_{f}$.
Denote by $\phi_{f*}:\bar{\Delta}^{q}\to W^{k,2}(S,\Delta^{q+n})$ the analytic
$q$-disk in $W^{k,2}(S,\Delta^{q}\times\Delta^{n})$ represented by $\phi_{f}$,
i.e $\phi_{f*}(z)\in W^{k,2}(S,\Delta^{q+1})$ acts as follows

(19)
$\phi_{f*}(z):s\to\left(\frac{f_{1}^{q}(s)-z_{1}}{1-\bar{f}_{1}^{q}(s)z_{1}},...,\frac{f_{q}^{q}(s)-z_{q}}{1-\bar{f}_{q}^{q}(s)z_{q}},f^{n}(s)\right).$

Here $f^{q}(s)=(f_{1}^{q}(s),...,f_{q}^{q}(s))$ is the $q$-component of $f$.
Denote by $\Phi_{f}=\phi_{f*}(\bar{\Delta}^{q})$ the image of $\phi_{f*}$. As
in lemma 3.1 we mark here and everywhere with $*$ analytic disks $D\to
W^{k,2}(S,X)$ represented by maps $D\times S\to X$ and by capital letters we
mark the images of these discs. Our $\phi_{f}$ possesses the following
properties:

  * i)

$\phi_{f*}(0)=f$, i.e. is our loop $f$.

  * ii)

For $z\in\partial\Delta^{q}$ one has that $\phi_{f*}(z)(S)\subset
A^{q}_{1-r,1}\times\Delta^{n}$, therefore

$\partial\Phi_{f}{:=}\phi_{f*}(\partial\Delta^{q})\subset
W^{k,2}(S,H_{q}^{n}(r)).$

Indeed, for $z$ close to ${\partial}\Delta^{q}$ some $z_{j}$ is close to
${\partial}\Delta$ and then the $j$-component of $h_{f^{q}(s)}(z)$ also is
close to ${\partial}\Delta$ for all $s\in S$ as required.

Remark that due to the second item above our map $F$ is defined and
holomorphic near the boundary of the analytic $q$-disc $\Phi_{f}$. In order to
assign to $f$ some “natural” value $\tilde{F}(f)$, which we shall call a
“natural extension” of $F$, we shall extend $F$ holomorphically to the
$q$-disc $\Phi_{f}\ni f$.

Consider the analytic $(q+1)$-disk $\psi_{*}$ in
$W^{k,2}(S,\Delta^{q}\times\Delta^{n})$ represented by

(20)
$\psi^{t}(z,s){:=}\psi(z,t,s){:=}\left(h_{f^{q}(s)}(z),tf^{n}(s)\right),\quad
z\in\bar{\Delta}^{q},\quad|t|\leq 1+\delta,\quad s\in S$

for an appropriate $\delta>0$ small enough. Here we use the compactness of
$f^{n}(S)$. Remark that

  * iii)

$\psi^{0}(\cdot,\cdot)$ takes its values in $\Delta^{q}\times\\{0\\}\subset
H_{q}^{n}(r)$ and therefore for $|t|$ small $\psi^{t}(.,.)$ takes its values
in $H_{q}^{n}(r)$ by continuity.

  * iv)

$\psi^{1}=\phi_{f}$.

  * v)

For all $t\in\bar{\Delta}_{1+\delta}$ one has that
${\partial}\Psi^{t}=\Psi_{*}^{t}(\partial\Delta^{q})\subset
W^{k,2}(S,H_{q}^{n}(r))$ by the second item as before.

Here $\Psi$ and $\Psi^{t}$ stand for the images of $\psi_{*}$ and
$\psi^{t}_{*}$ respectively, i.e. for every t we see $\Psi_{*}^{t}$ as an
analytic $q$-disk, while $\Psi_{*}$ as an analytic $(q+1)$-disk. Therefore for
$\delta>0$ small enough the Hartogs figure

(21) $H_{q}^{1}(\delta){:=}\\{(z,t):||z||<1+\delta,|t|<\delta\text{ or
}1-\delta<||z||<1+\delta,|t|<1+\delta\\}$

is mapped by $\psi_{*}$ to $W^{k,2}(S,H_{q}^{n}(r))$ and consequently the
composition $F\circ\psi_{*}:H_{q}^{1}(\delta)\to\mathcal{X}$ is well defined
and holomorphic. Due to the assumed $q$-Hartogsness of $\mathcal{X}$ this
composition $F\circ\psi_{*}$ holomorphically extends to the associated
polydisc $\Delta^{q+1}_{1+\delta}$. In particular it extends to
$\Delta^{q}\times\\{1\\}$, i.e. $F$ holomorphically extends onto the $q$-disc
$\Phi_{f}\ni f$.

(22) $\text{\it Denote by }\quad\widetilde{F\circ\psi_{*}}\quad\text{\it this
extension and set }\quad\tilde{F}(f){:=}(\widetilde{F\circ\psi_{*}})(0,1).$

Remark that $\tilde{F}$ is indeed an extension of $F$, i.e. that
$\tilde{F}(f)=F(f)$ for $f\in W^{k,2}(S,H_{q}^{n}(r))$ because $\tilde{F}$ is
a holomorphic extension of $F$ from $\Psi\cap W^{k,2}(S,H_{q}^{n}(r))$. We
call $\tilde{F}$ the “natural extension” of $F$ .

Step 2. The natural extension is continuous. Let $f^{\prime}\in
W^{k,2}(S,\Delta^{q}\times\Delta^{n})$ be close to $f$. Construct $q$-disc
$\phi_{f^{\prime}*}$ and $(q+1)$-disc $\psi^{\prime}_{*}$ for $f^{\prime}$ as
we did for $f$. Denote by $\Gamma_{\Psi}$ and $\Gamma_{\Psi^{\prime}}$ the
graphs of $\widetilde{F\circ\psi_{*}}$ and
$\widetilde{F\circ\psi^{\prime}_{*}}$ in
$\Delta^{q+1}_{1+\delta}\times\mathcal{X}$ correspondingly. Due to Lemma 3.1
all we need to prove is that $\Gamma_{\Psi^{\prime}}$ enters to a given
neighborhood of $\Gamma_{\Psi}$ provided $f^{\prime}$ is sufficently close to
$f$. Due to Theorem 3.2 one can choose a 1-complete neighborhood $\mathcal{V}$
of $\Gamma_{\Psi}$. Moreover, for an appropriate $\delta>0$ the graph
$\Gamma_{\Psi^{\prime}}$ over the Hartogs figure $H_{q}^{1}(\delta)$ as in
(21) enters to the neighborhood $\mathcal{V}$ of the graph $\Gamma_{\Psi}$
because $\psi^{\prime}$ is close to $\psi$ and takes it values in
$W^{k,2}(S,H^{n}_{q}(r))$ when restricted to $H_{q}^{1}(\delta)$. A priori
there is no reason that all the graph enters in $\mathcal{V}$. But the maximum
principle applied to the plurisubharmonic exhausting function of $\mathcal{V}$
all the graph $\Gamma_{\Psi^{\prime}}$ enters in the neighborhood
$\mathcal{V}$. The lemma 3.1 permits us to obtain the continuity of the map
and the step is proved.

Step 3. Holomorphicity. All what is left to prove is that our natural
extension $\tilde{F}$ is Gâteaux holomorphic. Fix $f\in
W^{k,2}(S,\Delta^{q}\times\Delta^{n})$, $g\in W^{k,2}(S,\mathbb{C}^{q+n})$ and
consider the complex affine line $L{:=}\\{f+\lambda
g:\lambda\in\mathbb{C}\\}\subset W^{k,2}(S,\mathbb{C}^{q+1+n})$. We need to
prove that $\tilde{F}|_{L}$ is holomorphic in a neighborhood of zero. Fix an
$\epsilon>0$ sufficiently small and consider an analytic $(q+2)$-disc
$\theta_{*}:\bar{\Delta}^{q}\times\bar{\Delta}_{\epsilon}^{2}\to
W^{k,2}(S,\mathbb{C}^{q+n})$ represented by the mapping :

(23) $\theta:(z,\lambda,\mu,s)\to\left(\frac{f_{1}^{q}(s)+\lambda
g_{1}^{q}(s)-z_{1}}{1-(\bar{f}_{1}^{q}(s)+\mu\bar{g}_{1}^{q}(s))z_{1}},...,\frac{f_{q}^{q}(s)+\lambda
g_{q}^{q}(s)-z_{q}}{1-(\bar{f}_{q}^{q}(s)+\mu\bar{g}_{q}^{q}(s))z_{q}},f^{n}(s)+\lambda
g^{n}(s)\right).$

Denote by $\Theta=\text{Im}\theta_{*}$ its image. Notice that for $\epsilon>0$
small enough the $q$-component of (23) defines for every fixed
$|\lambda|,|\mu|\leq\epsilon$ and $s\in S$ a holomorphic imbedding of
$\bar{\Delta}^{q}$ to a neighborhood of $\bar{\Delta}^{q}$ which is uniformly
close to the authomorphism

$z\to\left(\frac{f_{1}^{q}(s)-z_{1}}{1-\bar{f}_{1}^{q}(s)z_{1}},...,\frac{f_{q}^{q}(s)-z_{q}}{1-\bar{f}_{q}^{q}(s)z_{q}}\right)$

of $\Delta^{q}$. Therefore for every $|\lambda|,|\mu|\leq\epsilon$ the disk
$\theta_{*}$ has values in $W^{k,2}(S,\Delta^{q}_{1+r}\times\Delta^{n})$ and
the arguments of Step 1 can be repeated to $\theta(\cdot,\lambda,\mu,\cdot)$
on the place of $\phi_{f}$:

Consider an analytic $(q+3)$-disk $\psi_{*}$ in $W^{k,2}(S,\mathbb{C}^{q+n})$
represented by

(24)
$\psi^{t}(z,\lambda,\mu,s){:=}\psi(z,\lambda,\mu,t,s){:=}\left(\theta^{q}(z,\lambda,\mu,s),t\theta^{n}(z,\lambda,\mu,s)\right),\quad
z\in\bar{\Delta}^{q},\quad|t|\leq 1+\delta,\quad s\in S$

for an appropriate $\delta>0$ small enough. Here $\theta^{q}$ and $\theta^{n}$
are components of $\theta=(\theta^{q},\theta^{n})$. Remark that

  * •

$\psi^{0}$ takes its values in $\Delta^{q}\times\\{0\\}\subset H_{q}^{n}(r)$
and therefore for $|t|$ small $\psi^{t}$ takes its values in $H_{q}^{n}(r)$ by
continuity.

  * •

$\psi^{1}=\theta$.

  * •

For all $t\in\bar{\Delta}_{1+\delta}$ one has that ${\partial}\Psi^{t}\subset
W^{k,2}(S,H_{q}^{n}(r))$.

This gives a holomorphic extension of $F\circ\theta_{*}$ from
${\partial}\Delta^{q}\times\Delta^{2}_{\epsilon}$ to
$\bar{\Delta}^{q}\times\Delta^{2}_{\epsilon}$. We denote this extension as
$\hat{F}$. Notice now the following properties of $\theta$ and $\hat{F}$:

i) $\theta(\cdot,\lambda,\bar{\lambda},\cdot)=\phi_{f+\lambda
g}(\cdot,\cdot)$;

ii) $\theta(0,\lambda,\mu,\cdot)=f(\cdot)+\lambda g(\cdot)$, in particular,
this doesn’t depend on $\mu$.

Properties iv) and v) imply that

$\tilde{F}(f+\lambda g)=\left(\sf{ext}F|_{\phi_{f+\lambda
g}}\right)|_{z=0}=\left(\sf{ext}F|_{\theta(\cdot,\lambda,\bar{\lambda},\cdot)}\right)|_{z=0}=\left(\sf{ext}F|_{\theta(\cdot,\lambda,\mu,\cdot)}\right)|_{\mu=0,z=0}=$
$=\hat{F}|_{\mu=0,z=0}=\hat{F}(0,\lambda,0).$

Here by $\sf{ext}F|_{\phi_{f+\lambda g}}$ we denote the extension of $F$ along
the $q$-disc $\phi_{f+\lambda g}$ (and then taking the value of this extension
at $z=0$). This was the definition of the natural extension. Therefore the
first equality is justified. As for second it is justified by (i) . The third
equality folows from (ii) since at $z=0$ nothing depends on $\mu$ and
$\mu=\bar{\lambda}$ can be replaced by $\mu=0$. But this is the extension
$\hat{F}$ evaluated at $z=0,\lambda,\mu=0$ and the latter holomorphically
depends on $\lambda$. Therefore such is the left hand side
$\tilde{F}(f+\lambda g)$. The holomorphicity of $\tilde{F}$ is proved.

Remark. The theorem just proved was stated in [A-I], see Theorem 3.1 there,
for the case of compact $S$ without boundary. The step 1 of the proof was
presented there as well. As for step 2 and 3 the details were missing. ∎

## References

  * [1]
  * [A-Z] Anakkar M., Zagorodnyuk A.: On Hilbert-Hartogs manifolds.  Complex Variables and Elliptic Equations, doi:10.1080/17476933.2019.1687458 (2019).
  * [A-I] Anakkar M., Ivashkovich S.: Loop Spaces as Hilbert-Hartogs Manifolds.  Arxiv der Mathematik, doi:10.1007/s00013-020-01485-w (2020).
  * [I] Ivashkovich S.: An example concerning extension and separate analyticity properties of meromorphic mappings.  Amer. J. Math. 121, 97-130 (1999).
  * [L] Lempert L.: Holomorphic functions on (generalized) loop spaces.  Math. Proc. R. Ir. Acad. 104A, no: 1, 35-46 (2004).
  * [M] Moser J.: A rapidly convergent iteration method and non-linear differential equations  Annali della Scuola Normale Superiore di Pisa - Classe di Scienze , Ser. 3, 20(3):499-535,(1966)
  * [S] Stein Elias M.: Singular Integrals and Differentiability Properties of Functions (PMS-30)  Princeton University Press (1970)


# On the chromatic index
of generalized truncations

Brian Alspach and Aditya Joshi
School of Mathematical and Physical Sciences
University of Newcastle
Callaghan, NSW 2308, Australia
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Abstract

We examine the chromatic index of generalized truncations of graphs and
multigraphs.

Keywords: generalized truncation, chromatic index

AMS Classification: 05C15

##  1 Introduction

A broad definition of generalized truncations of graphs was introduced in [1].
We give this definition now for completeness, but first a brief word about
terminology is in order. The term multigraph is used if multiple edges are
allowed. Thus, a graph does not have multiple edges. We use $V(X)$ to denote
the set of vertices of a multigraph $X$ and $E(X)$ to denote the set of edges.
The order of $X$ is $|V(X)|$ and the size of $X$ is $|E(X)|$. Finally, the
valency of a vertex $u$, denoted $\mathrm{val}(u)$, is the number of edges
incident with $u$. The term $k$-valent multigraph is used for regular
multigraphs of valency $k$. Throughout the paper, $\Delta(X)$ denotes the
maximum valency of the multigraph $X$ and $\Delta$ often is used when the
graph $X$ involved is apparent.

Given a multigraph $X$, a generalized truncation of $X$ is obtained as follows
via a two-step operation. The first step is the excision step. Let $M$ denote
an auxiliary matching (no two edges have a vertex in common) of size $|E(X)|$.
Let $F:E(X)\rightarrow M$ be a bijective function and for $[u,v]\in E(X)$,
label the ends of the edge $F([u,v])$ with $u$ and $v$. Let $M_{F}$ denote the
vertex-labelled matching thus obtained. So $M_{F}$ represents the edges of $X$
completely disassembled.

The second step is the assemblage step. For each $v\in V(X)$, the set of
vertices of $M_{F}$ labelled with $v$ is called the cluster at $v$ and is
denoted $\mathrm{cl}(v)$. Insert an arbitrary graph on $\mathrm{cl}(v)$. The
inserted graph on $\mathrm{cl}(v)$ is called the constituent graph at $v$ and
is denoted $\mathrm{con}(v)$. The resulting graph

$M_{F}\cup_{v\in V(X)}\mathrm{con}(v)$

is a generalized truncation of $X$. We usually think of the labels on the
vertices of $F(M)$ as being removed following the assemblage stage, but there
are many times when the labels are useful in the exposition. We use
$\mathrm{TR}(X)$ to denote a generalized truncation of the multigraph $X$.

Truncations arise via action involving the edges incident with a vertex.
Consequently, isolated vertices are useless and we make the important
convention that the multigraphs from which we are forming generalized
truncations do not have isolated vertices. This will not be mentioned in the
subsequent material, but is required for the validity of a few statements.

Recall that a proper edge coloring of a multigraph $X$ is a coloring of the
edges so that adjacent edges do not have the same color. The chromatic index
of $X$, denoted $\chi^{\prime}(X)$, is the fewest number of colors for which a
proper edge coloring exists. We now state Vizing’s well-known theorem [4]
which plays a fundamental role in studying chromatic index.

###### Theorem 1.1

If $X$ is a graph, then its chromatic index is either $\Delta(X)$ or
$\Delta(X)+1$.

The preceding theorem leads to a classification of graphs as follows. A graphs
is class I if its chromatic index is equal to its maximum valency $\Delta$ and
is class II otherwise. There is a generalization for multigraphs and to ease
subsequent exposition we shall say a multigraph is class I if its chromatic
index equals its maximum valency.

The following result was proved in [1]. A generalized truncation
$\mathrm{TR}(X)$ is called complete when every constituent is a complete
subgraph.

###### Theorem 1.2

If $X$ is a class I graph, then its complete truncation also is class I. If
$X$ is a class II graph and its maximum valency is even, then its complete
truncation is class I.

The purpose of this paper is to extend the preceding result and explore edge
colorings of generalized truncations in more detail.

##  2 Some Useful Results

In the following material we shall be considering the relationship between
multigraphs and their generalized truncations that are class I. Doing so
requires the use of some well-known edge coloring results which we now give
for completeness.

Let $\sigma$ be a permutation of the vertex set of the complete graph $K_{n}$
of order $n$. If $Y$ is a subgraph of $K_{n}$, then $\sigma(Y)$ denotes the
subgraph of $K_{n}$ whose edge set is $\\{[\sigma(u),\sigma(v)]:[u,v]\in
E(Y)\\}$. Let $n$ be even and $\rho$ be the permutation defined by
$\rho=(u_{0})(u_{1}\;\;u_{2}\;\;u_{3}\;\cdots\;u_{n-1})$. So $\rho$ fixes one
vertex and cyclically rotates the remaining vertices. Let $Y$ be the perfect
matching of $K_{n}$ consisting of the edges

$[u_{0},u_{1}],[u_{2},u_{n-1},[u_{3},u_{n-2}],\ldots,[u_{n/2},u_{(n+2)/2}].$

We then obtain a proper edge coloring of $K_{n}$ by letting the color classes
be $Y,\rho(Y),\rho^{2}(Y),\ldots,\rho^{n-2}(Y)$. We call this proper edge
coloring the canonical edge coloring of $K_{n}$. Figure 1 shows $Y$ for
$K_{8}$.

Figure 1$u_{0}$$u_{1}$$u_{2}$$u_{3}$$u_{4}$$u_{5}$$u_{6}$$u_{7}$

We obtain a proper edge coloring of $K_{n}$ with $n$ colors, $n$ odd, by
starting with the canonical edge coloring of $K_{n+1}$ and then removing the
central vertex $u_{0}$ and the edges incident with it. Note that each vertex
is missing an edge of one color and the set of missing colors has cardinality
$n$. This fact is true for any proper edge coloring of $K_{n}$, $n$ odd.

Because we use the preceding canonical edge coloring for both even and odd
orders, we describe it by referring to the edge of length 1 as the anchor.
When $n$ is even, the anchor is the edge $[u_{n/2},u_{1+n/2}]$. When $n$ is
odd, the anchor is the edge $[u_{(n+1)/2},u_{(n+3)/2}]$. To obtain proper edge
colorings of complete graphs, we cyclically rotate the canonical edge coloring
which, of course, cyclically rotates the anchor. So we may determine a color
class by specifying the anchor.

###### Lemma 2.1

Every proper edge coloring of $K_{n}$ with $n$ colors, $n$ odd, has the
property that there is one color missing on the edges incident with a given
vertex, and the set of missing colors at the vertices has cardinality $n$.

Proof. There are at most $(n-1)/2$ edges of a fixed color in a proper edge
coloring of $K_{n}$ because $n$ is odd. Thus, there is no proper edge coloring
using just $n-1$ colors because $(n-1)^{2}/2<\binom{n}{2}$. There is a proper
edge coloring using $n$ colors by Vizing’s Theorem. Hence, each color class
contains precisely $(n-1)/2$ edges from which the conclusion follows.

We use a result about list chromatic index so that we discuss it briefly here
and present the result. Given a graph $X$ and for each edge $[u,v]$ a list
$L_{[u,v]}$ of colors we may use for the edge $[u,v]$, a proper list edge
coloring is a proper edge coloring of $X$ so that the color on each edge
$[u,v]$ belongs to $L_{[u,v]}$. The list chromatic index of $X$ is the
smallest $N$ such that if every list $L_{[u,v]}$ has cardinality $N$, then $X$
admits a proper list edge coloring. We denote this value by
$\chi^{\prime}_{L}(X)$. The following important result was proved by Häggkvist
and Jansen in [2].

###### Theorem 2.2

The complete graph $K_{n}$ satisfies $\chi^{\prime}_{L}(K_{n})\leq n$.

##  3 Main Result

We are interested in determining which generalized truncations of a given
multigraph $X$ are class I. Theorem 3.1 below provides a general answer and in
subsequent sections we use the theorem to describe class I generalized
truncations of specific types. Some definitions and notation are required
before stating the theorem.

Given a generalized truncation $\mathrm{TR}(X)$, for convenience we call the
subgraph induced by the edges of $\mathrm{con}(v)$ together with the edges of
$M_{F}$ incident with the vertices of $\mathrm{con}(v)$ the sun centered at
$\mathrm{con}(v)$. Given that $\mathrm{con}(v)$ is regular of valency $d-1$,
when is the sun centered at $\mathrm{con}(v)$ class I? If it is class I, then
there is a proper edge coloring using $d$ colors. We use a vector of length
$d$ to describe the number of edges of $M_{F}$ of the various colors in the
sun centered at $\mathrm{con}(v)$.

Given a vertex $v\in X$ of valency $r$, we say that a vector
$(x_{1},x_{2},\ldots,x_{d})$ is admissible if $\sum_{i}x_{i}=r$, the edges of
$M_{F}$ are colored according to the vector, and there is a $(d-1)$-regular
graph on $\mathrm{con}(v)$ such that the sun centered at $\mathrm{con}(v)$ is
class I. We say the vector is totally inadmissible if there is no
$(d-1)$-regular graph on $\mathrm{con}(v)$ which makes the sun class I.

###### Theorem 3.1

If $(x_{1},x_{2},\ldots,x_{d})$ satisfies $x_{1}+x_{2}+\cdots+x_{d}=r\geq d$,
then $(x_{1},x_{2},\ldots,x_{d})$ is admissible if and only if all of
$x_{1},x_{2},\dots,x_{d}\mbox{ and }r$ have the same parity and when the
parity is odd, $d$ also is odd. Otherwise, the vector is totally inadmissible.

Proof. Suppose that $\mathrm{val}(v)=r$ in a multigraph $X$ and that there is
a class I sun centered at $\mathrm{con}(v)$ which is regular of valency $d$.
Let $(x_{1},x_{2},\ldots,x_{d})$ be the vector for the colors of $M_{F}$. For
an arbitrary color $c(i)$, if the number of edges of color $c(i)$ in
$\mathrm{con}(v)$ is $k$, then the number of edges of color $c(i)$ in $M_{F}$
must be $r-2k$. It then follows that $x_{1},x_{2},\ldots,x_{d}\mbox{ and }r$
all have the same parity. Moreover, when $r$ is odd, then the graph
$\mathrm{con}(v)$ is regular of valency $d-1$ and order $r$ which implies that
$d-1$ is even. This completes the necessity.

Now suppose that all coordinates of $(x_{1},x_{2},\ldots,x_{d})$ are odd, $r$
is odd and $x_{1}+x_{2}+\cdots+x_{d}=r$. In this case we also have that $d$ is
odd. Label the $r$ vertices of the constituent
$u_{1},u_{2},u_{3},\ldots,u_{r}$ so that the first $x_{1}$ vertices are
incident with the $x_{1}$ edges of $M_{F}$ of color $c(1)$, the next $x_{2}$
vertices are incident with the $x_{2}$ edges of $M_{F}$ of color $c(2)$ and
continue in this way. Also carry out subscript arithmetic modulo $r$ on the
residues $1,2,3,\ldots,r$.

Consider the $x_{i}$ successive vertices incident with the $x_{i}$ edges of
$M_{F}$ of color $c(i)$. Because $x_{i}$ is odd, there is a central edge of
$M_{F}$ of color $c(i)$ and let it be incident with $u_{a}$. Letting
$\alpha=(x_{i}-1)/2$, the vertices incident with edges of $M_{F}$ of color
$c(i)$ are
$u_{a-\alpha},u_{a-\alpha+1},\ldots,u_{a},u_{a+1},\ldots,u_{a+\alpha}$.

Consider the canonical color class whose anchor is
$[u_{a+(r-1)/2},u_{a+(r+1)/2}]$. Color the edges of this class with color
$c(i)$ starting with the anchor and stopping with the edge
$[u_{a-\alpha-1},u_{a+\alpha+1}]$. This yields edges colored with $c(i)$ so
that each vertex of the constituent is incident with one edge of color $c(i)$.
After doing this for each of the $d$ colors, the resulting sun centered at the
constituent is class I and the constituent is regular of valency $d-1$.

This leaves us with the case that all the coordinates of
$(x_{1},x_{2},\ldots,x_{d})$ are even so that their sum $r$ also is even.
There is no restriction on the parity of $d$ in this case. Also note that
$x_{i}=0$ is possible for various values of $i$. Without loss of generality,
let $x_{1},x_{2},\ldots,x_{a}$ be the non-zero entries of the vector.

We label the vertices of the constituent a little differently using the
canonical edge coloring of Figure 1 as a template. The central vertex is
labelled $u_{0}$ and the others are labelled $u_{1},u_{2},\ldots,u_{r-1}$.

Let the $x_{1}$ edges of $M_{F}$ of color $c(1)$ be incident with
$u_{0},u_{1},\ldots,u_{x_{1}-1}$. Let the remaining edges of $M_{F}$ be
incident with vertices of the constituent as in the preceding case, that is,
edges of the same color are incident with successively labelled vertices.

The edges of color $c(1)$ have a different form than the other colored edges
making the completion of the coloring a little different for them. Vertices
$u_{1},u_{2},\ldots,u_{x_{1}-1}$ are incident with edges of $M_{F}$ of color
$c(1)$. This corresponds to the color class with anchor
$[u_{(x_{1}+r-4)/2},u_{(x_{1}+r-2)/2}]$. So color the edges of that class with
color $c(1)$ starting with the anchor until finishing with
$[u_{r-1},u_{x_{1}}]$. Every vertex of the constituent now is incident with an
edge of color $c(1)$.

For all the other colors $c(j)$, $1<j\leq a$, there are an even number of
successive vertices, say $u_{j},u_{j+1},\ldots,u_{j+t}$, incident with edges
of $M_{F}$ of color $c(j)$. Then the edge $[u_{j+(t-1)/2},u_{j+(t+1)/2}]$ is
the anchor of a color class. So complete the coloring of the edges of this
color class starting with $[u_{j-1},u_{j+t+1}]$ moving away from the anchor.
For the $x_{i}=0$, simply include an entire color class from the canonical
coloring scheme using an unused anchor.

Doing the preceding yields a class I coloring of the sun centered at the
constituent so that the constituent graph is regular of valency $d-1$ and
completes the proof.

We now obtain three corollaries from Theorem 3.1, but first require a couple
of definitions. An edge coloring of a multigraph $X$ is said to be parity-
balanced if for each vertex $v$ of $X$, the parity of the number of edges of
each color incident with $v$ is the same as the parity of $\mathrm{val}(v)$. A
regular truncation is a generalized truncation for which every vertex has the
same valency. A generalized truncation is said to be semiregular if each sun
centered at a constituent $\mathrm{con}(v)$ is class I and the subgraph
$\mathrm{con}(v)$ is regular. The distinction between a semiregular truncation
and a regular truncation is that valencies of regularity for a semiregular
truncation may differ over the constituents. Note that the source multigraph
need not be regular in order to have a regular truncation.

###### Corollary 3.2

Let every entry of the feasible vector $(x_{1},x_{2},\ldots,x_{d})$ be even
and $x_{1}+x_{2}+\cdots+x_{d}=r$. If $d^{\prime}$ is the number of non-zero
entries of the vector, then there are class I suns centered at the appropriate
constituent such that the constituents are regular of every valency from
$d^{\prime}$ through $r-1$.

Proof. This result follows from the proof of Theorem 3.1 rather than the
statement. Because the number of colors on edges of $M_{F}$ is $d^{\prime}$,
the scheme used in the proof of Theorem 3.1 introduces no new colors so that
the valency of regularity is $d^{\prime}$. Because $r$ is even, we may add
canonical coloring classes one at a time until reaching $K_{r}$. This
completes the proof.

###### Corollary 3.3

A multigraph $X$ has a class I semiregular truncation if and only if $X$ has a
parity-balanced edge coloring.

Proof. If $X$ has a parity balanced edge coloring, it follows immediately from
Theorem 3.1 that it has a semiregular truncation. On the other hand, if $X$
has a semiregular truncation, then performing the standard contraction and
retention of the colors on the edges of $M_{F}$ produces a parity-balanced
edge coloring of $X$.

###### Corollary 3.4

A multigraph $X$ has a class I regular truncation of valency $d$ if and only
if it has a parity-balanced coloring and one of the following conditions
holds:

(i) When $d$ is odd, every vertex of odd valency has precisely $d$ colors on
its incident edges, and every vertex of even valency has valency at least
$d+1$ and at most $d$ colors on its incident edges ; or

(ii) When $d$ is even, every vertex of $X$ has even valency at least $d$.

Proof. Let $X$ be a mutigraph and suppose it has a generalized truncation $Y$
which is regular of valency $d$. Consider the case that $d$ is even. This
means that a constituent $\mathrm{con}(v)$ is regular of valency $d-1$ and the
latter is odd. Thus, the constituent has even order. This implies that every
vertex of $X$ has even valency. Clearly every valency is at least $d$.

When $d$ is odd, each constituent must itself be regular of valency $d-1$.
Because $d-1$ is even, there is no restriction on the order $N$ of the
constituent other than it must be at least $d$. If $N$ is odd, then for each
of the $d$ distinct colors, there must be at least one edge of that color
belonging to the edges of $M_{F}$ incident with vertices of the constituent.
So the corresponding vertex of $X$ is incident with edges of precisely $d$
different colors. When $N$ is even there is no restriction on the number of
colors on edges of $M_{F}$ incident with vertices of the constituent other
than it is at most $d$.

For the other direction, when $d$ is even, every constituent has even order
and by Corollary 3.2 it is easy to obtain regular valency of $d-1$ on the
constituent. When $d$ is odd, the result follows from Theorem 3.1.

##  4 Complete Truncations

Let $X$ be a multigraph with maximum valency $\Delta$. If $\Delta$ is odd and
$X$ has an edge coloring with $\Delta$ colors such that the edges incident
with every vertex of valency $\Delta$ have distinct colors, then we say that
$X$ is edge-feasible. The next result characterizes graphs whose complete
truncations are class I. Of course, a complete truncation is a semi-regular
truncation.

###### Theorem 4.1

The complete truncation of a multigraph $X$ is class I if and only if either
the maximum valency $\Delta$ of $X$ is even, or $\Delta$ is odd and $X$ is
edge-feasible.

Proof. This theorem is an improvement on Theorem 1.2 as the latter does not
include multigraphs as part of the hypotheses. Because a complete truncation
is a semiregular truncation, Corollary 3.3 implies that the coloring of the
edges in $M_{F}$ must correspond to a parity-balanced coloring of $X$. When
$\Delta$ is even, coloring all the edges of $M_{F}$ with a single color
corresponds to a parity balanced coloring of $X$. Any constituent of order
$\Delta$ can be properly edge-colored with $\Delta-1$ colors because $\Delta$
is even. Any constituent of order less than $\Delta$ can be properly edge-
colored with at most $\Delta-1$ colors. Hence, the complete truncation of $X$
is class I.

For the remainder of the proof we assume that $\Delta$ is odd. First suppose
that the complete truncation $\mathrm{TR}(X)$ is class I. If $u\in V(X)$ has
valency $\Delta$, then $\mathrm{con}(u)=K_{\Delta}$ the complete graph of
order $\Delta$. We conclude that the edges of $M_{F}$ incident with
$\mathrm{con}(u)$ all have different colors by Lemma 2.1. So if we contract
each constituent to a single vertex, delete all the loops formed and keep the
colors of the edges of $M_{F}$, we obtain an edge coloring of $X$. Clearly,
all the edges incident to any vertex of valency $\Delta$ in $X$ have distinct
colors. Thus, $X$ is edge-feasible.

Now let $X$ be edge-feasible and choose an edge coloring of $X$ which is edge-
feasible. Let $\mathrm{TR}(X)$ be the complete truncation of $X$. Color the
edges of $M_{F}$ with the same color they had in the edge coloring of $X$.
These are the edges between the constituents all of which are complete
subgraphs. At this point we have used $\Delta$ colors. It suffices to show
that we can color the edges of the constituents without introducing any new
colors so that the resulting edge coloring of $\mathrm{TR}(X)$ is proper.

Any constituent $\mathrm{con}(u)$ of order $\Delta$ corresponds to a vertex
$u$ of $X$ of valency $\Delta$. This implies that the edges of $M_{F}$
incident with the vertices of $\mathrm{con}(u)$ all have distinct colors
because the coloring of $X$ was edge-feasible. From Lemma 2.1 it is clear that
we may color the edges of $\mathrm{con}(u)$ with $\Delta$ colors so that the
missing color at each vertex is the color of the edge of $M_{F}$ at the
vertex. This coloring of the edges does not violate the definition of a proper
edge coloring.

Now consider any constituent $\mathrm{con}(u)$ of order $r\leq\Delta-2$.
Because each edge of $\mathrm{con}(u)$ has one edge of $M_{F}$ at each of its
end vertices, the number of possible colors for the edge that do not violate
the proper edge coloring condition is $\Delta-2$. So each edge has a list of
$\Delta-2$ possible colors, and Theorem 2.2 implies that we may color the
edges of $\mathrm{con}(u)$ without violating the proper edge coloring
condition because $r\leq\Delta-2$.

This leaves us with the case that $\mathrm{con}(u)$ has order $\Delta-1$ and
this is the most complicated case. The first observation we make is that the
coloring pattern of the edges of $M_{F}$ incident with the vertices of
$\mathrm{con}(u)$ can vary all the way from having $\Delta-1$ distinct colors
to every color being the same. So we introduce a sequence describing the color
pattern. Let $(s_{1},s_{2},\dots,s_{t})$ satisfy $s_{1}\leq
s_{2}\leq\cdots\leq s_{t}$ and $s_{1}+s_{2}+\cdots+s_{t}=\Delta-1$. The
sequence means there are $t$ distinct colors on the edges of $M_{F}$ incident
with vertices of $\mathrm{con}(u)$ and $s_{i}$ such edges have the same color
$c(i)$ for $i=1,2,\ldots,t$.

We need to show that no matter what the color sequence is we may color the
edges of $\mathrm{con}(u)$ so that the sun centered at $\mathrm{con}(u)$ is
properly edge colored with $\Delta$ colors. As a first step, label the
vertices of $\mathrm{con}(u)$ with $v_{0},v_{1},\ldots,v_{a}$, where
$a=\Delta-2$. Let the vertices $v_{1},v_{2},\ldots,v_{s_{1}}$ be incident with
the $s_{1}$ edges of $M_{F}$ of color $c(1)$. Let the next $s_{2}$ vertices be
incident with the $s_{2}$ edges of $M_{F}$ of color $c(2)$. Continue labelling
the vertices in the obvious manner and let the edge of $M_{F}$ incident with
$v_{0}$ have color $c(t)$.

Carry out an initial coloring of the edges of $\mathrm{con}(u)$ using the
canonical edge coloring described in Section 2 with $v_{0}$ acting as the
fixed vertex and $\rho=(v_{0})(v_{1}\;\;v_{2}\;\;\cdots\;\;v_{a})$. The
strategy now is to choose the colors for the color classes in such a way that
we may re-color some edges to obtain a proper edge coloring for the sun
centered at $\mathrm{con}(u)$.

The first observation we make is that only $\Delta-2$ colors are required for
a canonical edge coloring of $\mathrm{con}(u)$. So we do not use the colors
$c(t-1)$ or $c(t)$ for the canonical edge coloring of $\mathrm{con}(u)$.
Hence, if $t=1$ or $t=2$, we already have an edge coloring of
$\mathrm{con}(u)$ that does not violate the conditions for a proper edge
coloring. Thus, we assume that $t\geq 3$.

The anchor for a color class plays an active role as follows. Suppose we
require two edges of a color class so that the two edges use four successive
vertices under the cyclic labelling of $\\{v_{1},v_{2},\ldots,v_{a}\\}$. If
the anchor is $[v_{i},v_{i+1}]$, then adding the edge $[v_{i-1},v_{i+2}]$ from
the same color class easily does the job. It is now easy to see how we may
obtain $k$ edges from the same color class so that they cover $2k$ successive
vertices. With this in mind, we now describe an iterative process for
determining the colors of certain color classes.

If $s_{1}$ is odd, then we color the color class for which
$[v_{(s_{1}+1)/2},v_{(s_{1}+3)/2}]$ is the anchor with $c(1)$. If $s_{1}$ is
even, then we color the color class for which $[v_{s_{1}/2},v_{(s_{1}+2)/2}]$
is the anchor with $c(1)$. There are two points to observe about the preceding
choices. When $s_{1}$ is odd, the edge from $v_{1}$ to $v_{s_{1}+1}$ has color
$c(1)$ and note that the edge of $M_{F}$ incident with $v_{s_{1}+1}$ has color
$c(2)$. When $s_{1}$ is even, the edge from $v_{1}$ to $v_{s_{1}}$ has color
$c(1)$ and vertex $v_{s_{1}}$ is the last vertex for which the edge of $M_{F}$
incident with it has color $c(1)$.

We continue in the manner suggested by the preceding paragraph, but discuss it
further to make it clearer. When $s_{i}$ is odd and the first vertex incident
with an edge of $M_{F}$ of color $c(i)$ is $v_{d}$, then the anchor is the
edge $[v_{d+(s_{i}-1)/2},v_{d+(s_{i}+1)/2}]$ and we color this color class
with $c(i)$. Note that the edge $[v_{d},v_{d+s_{i}}]$ is in this color class,
but the edge of $M_{F}$ incident with $v_{d+s_{i}}$ has color $c(i+1)$.

When $s_{i}$ is even and the first vertex incident with an edge of $M_{F}$ of
color $c(i)$ is $v_{d}$, then the anchor is
$[v_{d+(s_{i}-2)/2},v_{d+s_{i}/2}]$ and we color this color class with $c(i)$.
In this case, the edge from $v_{d}$ to $v_{d+s_{i}-1}$ is colored $c(i)$ and
$v_{d+s_{i}-1}$ is the last vertex incident with an edge of $M_{F}$ of color
$c(i)$.

The preceding procedure is carried out for the colors $c(1)$ through $c(t-2)$
at which point it stops because colors $c(t-1)$ and $c(t)$ are not used for
the canonical edge coloring of $\mathrm{con}(u)$. The edges of
$\mathrm{con}(u)$ that need to be re-colored are those that are adjacent with
edges of $M_{F}$ having the same color. We have seen that when $s_{i}$ is odd,
there is an edge in $\mathrm{con}(u)$ of color $c(i)$ with one end vertex
incident with an edge of $M_{F}$ of color $c(i)$ and the other end vertex
incident with an edge of $M_{F}$ of color $c(i+1)$. So at the end vertex
incident with an edge of $M_{F}$ of color $c(i+1)$, the procedure gives
another edge of color $c(i+1)$ which also needs to be re-colored.

Thus, the edges that need to be recolored are isolated and possibly paths.
Luckily we have two colors to use for the re-coloring and we arbitrarily re-
color isolated edges with either $c(t-1)$ or $c(t)$, and alternately re-color
the edges of the paths making certain that if one of the paths terminates with
a vertex whose incident edge from $M_{F}$ has color $c(t-1)$, we color the
edge of the path terminating there with $c(t)$. This removes all potential
color conflicts for $\mathrm{con}(u)$ and completes the proof of the theorem.

###### Corollary 4.2

Let $X$ be a class I graph and $\mathrm{TR}(X)$ a generalized truncation. If
$\Delta(X)=\Delta(\mathrm{TR}(X))$, then $\mathrm{TR}(X)$ is class I.

Proof. Let $\mathrm{TR}(X)$ be a generalized truncation of $X$ satisfying
$\Delta(\mathrm{TR}(X))=\Delta(X)$. Then a proper edge coloring of
$\mathrm{TR}(X)$ requires at least $\Delta$ colors. We know the complete
truncation $Y$ of $X$ is class I by Theorem 4.1, that is, it has a proper edge
coloring using $\Delta$ colors. Remove any edges of $Y$ not belonging to
$\mathrm{TR}(X)$ and we are left with a proper edge coloring of
$\mathrm{TR}(X)$ using $\Delta$ colors. So $\mathrm{TR}(X)$ is class I.

It is well known that both the Petersen graph and its complete truncation are
class II graphs. The extension to a regular multigraphs is an immediate
corollary of Theorem 4.1.

###### Corollary 4.3

A regular multigraph $X$ of odd valency is class I if and only if its complete
truncation is class I.

##  5 Cyclic Truncations

Complete truncations have been used by various authors, and there is another
generalized truncation that has been employed frequently. Namely, the
generalized truncation obtained by letting each constituent graph be a cycle.
We shall call these cyclic truncations. Probably the best known example of
this is the cube-connected cycles graph first introduced in [3]. Of course,
the ancient Greeks studied truncations of Platonic and Archimedian solids, and
these resulted in cyclic truncations.

When moving from a multigraph $X$ to a generalized truncation $Y$, we
frequently wish $Y$ to inherit properties of $X$. This can be problem for
cyclic truncations. For example, poorly chosen cyclic truncations can play
havoc with automorphisms.

We are interested in determining conditions for which a cyclic truncation is
class I. Of course, a cyclic truncation is a regular truncation so that we may
use Theorem 3.1. We use the same vector describing the numbers of colors used
on the edges of $M_{F}$ belonging to a sun centered at a constituent. In this
case the vector has length 3 as we are looking at cyclic truncations all of
which are 3-valent. There is a new term not used before, namely, a vector
$(x_{1},x_{2},x_{3})$ is universal if the sun centered at the corresponding
constituent is class I for every cycle on the vertices of the constituent.

###### Corollary 5.1

If $(x_{1},x_{2},x_{3})$ satisfies $x_{1}+x_{2}+x_{3}=d\geq 3$, then
$(x_{1},x_{2},x_{3})$ is admissible if and only $x_{1},x_{2},x_{3}\mbox{ and
}d$ all have the same parity. Otherwise, the vector is totally inadmissible.
Moreover, if just one of $x_{1},x_{2}\mbox{ and }x_{3}$ is non-zero, then
$(x_{1},x_{2},x_{3})$ is universal when $d$ is even.

Proof. The portion of the statement prior to the sentence about universality
is simply Theorem 3.1 restricted to $d=3$. When the vector has the form
$(x_{1},0,0)$ and $x_{1}=d$ is even, this corresponds to all the edges of
$M_{F}$ incident with the vertices of the constituent having the same color.
Clearly, no matter which even length cycle we form on the vertices of the
constituent, the resulting sun is class I.

Corollary 5.1 gives us an easy way to construct a class I cyclic truncation of
a multigraph $X$ based on parity conditions for the colors on the edges of
$M_{F}$. Thus, we need to concentrate on coloring the edges in the source
multigraph.

###### Corollary 5.2

If every vertex of the multigraph $X$ has even valency, then every cyclic
truncation of $X$ is class I.

Proof. This follows from Corollary 5.1 by coloring all edges of $M_{F}$ with a
single color.

###### Corollary 5.3

If $X$ is a regular multigraph of odd valency $d>1$ and is class I, then there
are class I cyclic truncations of $X$.

Proof. Let $X$ be a class I regular multigraph of odd valency $d\geq 3$.
Choose a proper edge coloring of $X$ using $d$ colors. In forming a cyclic
truncation $Y$ of $X$, color the edges of $M_{F}$ according to the colors
$c(1),c(2),\ldots,c(d)$ the edges have in the proper edge coloring of $X$. Now
change the color of any edge of colors $c(4),c(5),\ldots,c(d)$ to $c(3)$. The
vector for each constituent becomes $(1,1,d-2)$ all of which are odd. We now
may find a cycle for each constituent such that the cyclic truncation is class
I by Corollary 5.1.

We now give a sufficient condition for a multigraph to have a class I cyclic
truncation. A definition is required. Let $X$ be an arbitrary multigraph and
$V_{0},V_{1},V_{2}\mbox{ and }V_{3}$ be a partition of $V(X)$, where $V_{i}$
contains the vertices whose valencies are congruent to $i$ modulo 4. Given a
submultigraph $Y$ of $X$, let $X\setminus Y$ be the submultigraph obtained by
removing the edges of $Y$ from $X$. Moreover, let
$V_{0}^{\prime},V_{1}^{\prime},V_{2}^{\prime}\mbox{ and }V_{3}^{\prime}$
denote the vertices of $V(X)$ whose valencies in $X\setminus Y$ are congruent
to $i$ modulo 4. A submultigraph $Y$ of $X$ is called an enabling
submultigraph if satisfies:

  * •

if $v\in V_{0}$, then $v\in V_{0}^{\prime}$;

  * •

if $v\in V_{1}$, then $v\in V_{2}^{\prime}$;

  * •

if $v\in V_{2}$, then $v\in V_{0}^{\prime}$; and

  * •

if $v\in V_{3}$, then $v\in V_{2}^{\prime}$.

The preceding conditions are not as complicated as they may look at first. The
first thing to notice is that the components of $X\setminus Y$ are eulerian.
Second, for many graphs the conditions are simple. For example, if $X$ is
3-regular, then a perfect matching is an enabling subgraph.

###### Corollary 5.4

Let $X$ be a multigraph with an enabling submultigraph $Y$. If every component
of $X\setminus Y$ has even size, then there is a class I cyclic truncation of
$X$.

Proof. Because each component of $X\setminus Y$ has even size, we may
alternately color the edges of an Euler tour of the component with two colors
so that we finish with same number of colors on the edges incident with every
vertex. Doing this for each component yields a 2-coloring so that each vertex
of $X\setminus Y$ has the same number of colors incident with it.

If we now color all the edges of $Y$ with a third color, it is easy to verify
that the conditions of Corollary 5.1 are met for $X$. We check one possibility
for illustrative purposes. If $v\in V_{3}$, it belongs to $V_{2}^{\prime}$ in
$X\setminus Y$. Thus, the number of edges of $Y$ incident with $v$ is
congruent to 1 modulo 4, that is, it is incident to an odd number of edges of
the third color in $X$. Because it is in $V_{2}^{\prime}$, it is incident with
an odd number of edges of each of the first two colors. Thus, the vector for
$v$ has three odd components.

###### Proposition 5.5

If a trivalent graph $X$ has a cut edge, then $X$ is class II.

The proof of the preceding proposition is immediate and leads to examples such
as the following. Let $X$ be the graph of order 10 obtained by taking two
vertex-disjoint copies of $K_{5}$ and joining the two copies with a single
edge from a vertex of one copy to a vertex of the other copy. Lemma 2.1
implies that $X$ is class I. If we take any cyclic truncation $\mathrm{TR}(X)$
of $X$, then $\mathrm{TR}(X)$ is class II by Proposition 5.5 because the edge
of $M_{F}$ connecting the two copies still is a cut edge in $\mathrm{TR}(X)$.

##  6 Other Truncations

We now examine other generalized truncations. An arboreal truncation is a
generalized truncation for which every constituent graph is a forest. The
following result is useful for arboreal truncations.

###### Theorem 6.1

Let $\mathrm{TR}(X)$ be a generalized truncation of a multigraph $X$. If the
maximum valency of $\mathrm{TR}(X)$ is $\Delta$ and every constituent of
$\mathrm{TR}(X)$ having a vertex of valency $\Delta$ is class I, then
$\mathrm{TR}(X)$ is class I.

Proof. Let $\mathrm{con}(v)$ have a vertex of valency $\Delta$ in
$\mathrm{TR}(X)$. Then the maximum valency in the subgraph $\mathrm{con}(v)$
is $\Delta-1$. Its edges may be properly edge-colored with $\Delta-1$ colors
because it is class I. Any constituent not having a vertex of valency $\Delta$
may have its edges properly colored with at most $\Delta-1$ colors because its
maximum valency is $\Delta-2$. Then the edges of $M_{F}$ may be colored with a
new color yielding a proper edge coloring of $\mathrm{TR}(X)$ with $\Delta$
colors. The conclusion now follows.

###### Corollary 6.2

Every arboreal truncation of a multigraph $X$ is class I.

Proof. This follows immediately from Theorem 6.1 because forests are class I.

##  7 Acknowledgement

The second author wishes to thank the University of Newcastle for support from
a 2020 - 2021 Summer Research Scholarship during which time this research was
begun.

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