Quillen’s plus construction and the D(2) problem

W. H. Mannan

MSC: 57M20, 19D06, 57M05 Keywords: D2 problem, Quillen plus construction

> Given a finite connected 3–complex with cohomological dimension 2, we show
> it may be constructed up to homotopy by applying the Quillen plus
> construction to the Cayley complex of a finite group presentation. This
> reduces the D(2) problem to a question about perfect normal subgroups.

§1 Introduction

Given a finite cell complex one may ask what the minimal dimension of a finite
cell complex in its homotopy type is. If $n\neq 2$ and the cell complex has
cohomological dimension $n$ (with respect to all coefficient bundles), then
the cell complex is in fact homotopy equivalent to a finite $n$–complex (a
cell complex whose cells have dimension at most $n$). Although this has been
known for around forty years (for $n>2$ it is proved in [14] and for $n=1$ it
follows from [12], [13]), it is an open question whether or not this holds
when $n=2$. This question is known as Wall’s D(2) problem:

Let $X$ be a finite 3–complex with $H^{3}(X;\beta)=0$ for all coefficient
bundles $\beta$. Must $X$ be homotopy equivalent to a finite 2–complex?

If $X$ (as above) is not homotopy equivalent to a finite 2–complex, we say it
is a counterexample which solves the D(2) problem.

For connected $X$ with certain fundamental groups, it has shown been shown
that $X$ must be homotopy equivalent to a finite 2–complex (see for example
[8], [5], [10]). However no general method has been forthcoming.

Also, whilst potential candidates for counterexamples have been constructed
(see [2], [3] ), no successful method has yet emerged for verifying that they
are not homotopy equivalent to finite 2–complexes.

In §2 we apply the Quillen plus construction to connected 2–complexes,
resulting in cohomologically 2–dimensional 3–complexes. These are therefore
candidates for counterexamples which solve the D(2) problem. In §3 we show
that in fact all finite connected cohomologically 2–dimensional 3–complexes
arise this way, up to homotopy equivalence.

Finally, in §4 we use these results to reduce the D(2) problem to a question
about perfect normal subgroups. This allows us to generalize existing
approaches to the D(2) problem such as [9, Theorem I] and [7, Theorem 3.5].

Before moving on to the main argument we make a few notational points. All
modules are right modules except where a left action is explicitly stated. The
basepoint of a Cayley complex is always assumed to be its 0–cell.

If $X$ is a connected cell complex with basepoint, we denote its universal
cover $\tilde{X}$. Given two based loops $\gamma_{1},\gamma_{2}\in\pi_{1}(X)$
their product $\gamma_{1}\gamma_{2}$ is the composition whose initial segment
is $\gamma_{2}$ and final segment is $\gamma_{1}$. With this convention, we
have a natural right action of $\pi_{1}(X)$ on the cells of $\tilde{X}$. Let
$G=\pi_{1}(X)$. We can regard the associated chain complex of $\tilde{X}$ as
an algebraic complex of right modules over $\mathbb{Z}[G]$. We follow [9] in
denoting this algebraic complex $C_{*}(X)$. Note that this differs from the
convention in other texts. Thus in particular $C_{*}(X)$ and
$C_{*}(\tilde{X})$ have the same underlying sequence of abelian groups, but
the former is a sequence of modules over $\mathbb{Z}[G]$ whilst the latter is
a sequence of modules over $\mathbb{Z}[\pi_{1}(\tilde{X})]=\mathbb{Z}$.

If $Y$ is a subcomplex of $X$ then $C_{*}(Y)$ is a sequence of right modules
over $\pi_{1}(Y)$. Let $E=\pi_{1}(Y)$. The induced map $E\to G$ yields a left
action of $E$ on $\mathbb{Z}[G]$. Thus we have an algebraic complex
$C_{*}(Y)\otimes_{E}\mathbb{Z}[G]$ over $\mathbb{Z}[G]$. The inclusion
$Y\subset X$ induces a chain map
$C_{*}(Y)\otimes_{E}\mathbb{Z}[G]\longrightarrow C_{*}(X)$. The complex
$C_{*}(X,Y)$ is defined to be the relative chain complex associated to this
chain map.

The basepoint allows us to interchange between coefficient bundles over $X$
and right modules over $\mathbb{Z}[G]$. Thus for a right module $N$ we have:

$H^{n}(X;N)=H^{n}(C_{*}(X);N)$

A left module over $\mathbb{Z}[G]$ may be regarded as a right module over
$\mathbb{Z}[G]$, where right multiplication by a group element is defined to
be left multiplication by its inverse. Hence a left module $M$ may also be
regarded as a coefficient bundle and we have:

$H_{n}(X;M)=H_{n}(C_{*}(X);M),\qquad\qquad H_{n}(X,Y;M)=H_{n}(C_{*}(X,Y);M)$

Given a finitely generated Abelian group $A$ we may regard it as a finitely
generated module over $\mathbb{Z}$. Thus $A\otimes_{\mathbb{Z}}\mathbb{Q}$ is
a finite dimensional vector space over $\mathbb{Q}$. The dimension of this
vector space will be denoted ${\rm rk}_{\mathbb{Z}}(A)$.

Finally given a group $G$ and elements $g,h\in G$, we follow the convention
that $[g,h]$ denotes the element $ghg^{-1}h^{-1}$.

§2 The plus construction applied to a Cayley complex

Let $\varepsilon=\langle g_{1},\cdots,g_{n}\,|\,R_{1},\cdots,R_{m}\rangle$ be
a finite presentation for a group $E$. We say a normal subgroup of $E$ is
finitely closed when it is the normal closure in $E$ of a finitely generated
subgroup. Let $K\lhd E$ be finitely closed and perfect (so $K=[K,K]$). Let
$\mathcal{K}_{\varepsilon}$ denote the Cayley complex associated to
$\varepsilon$.

Theorem 2.1 (Quillen, see [11, Theorem 5.2.2]) There is a 3–complex
$\mathcal{K}_{\varepsilon}^{+}$, containing $\mathcal{K}_{\varepsilon}$ as a
subcomplex, such that the inclusion
$\mathcal{K}_{\varepsilon}\hookrightarrow\mathcal{K}_{\varepsilon}^{+}$
induces the quotient map $E\to E/K$ on fundamental groups and
$H_{*}(\mathcal{K}_{\varepsilon}^{+},\mathcal{K}_{\varepsilon};M)=0$ for all
left modules $M$ over $\mathbb{Z}[E/K]$. Further, given another such 3–complex
$X$, there is a homotopy equivalence $\mathcal{K}_{\varepsilon}^{+}\to X$
extending the identity map of the common subspace $\mathcal{K}_{\varepsilon}$.

In fact we may construct $\mathcal{K}_{\varepsilon}^{+}$ explicitly, using the
fact that $K$ is finitely closed to ensure that we end up with a finite cell
complex. Let $k_{1},\cdots k_{r}\in K$ generate a subgroup of $E$ whose normal
closure (in $E$) is $K$. As $K=[K,K]$, each $k_{i}$ may be expressed as a
product of commutators: $k_{i}=\prod_{j=1}^{m_{i}}[a_{ij},b_{ij}]$ with each
$a_{ij},b_{ij}\in K$. Then each $a_{ij},b_{ij}$ may be represented by words
$A_{ij},B_{ij}$ in the $g_{l}$, $l=1,\cdots,n$. For each $i=1,\cdots,r$ attach
a 2–cell $E_{i}$ to $\mathcal{K}_{\varepsilon}$ whose boundary corresponds to
the word $\prod_{j=1}^{m_{i}}[A_{ij},B_{ij}]$. Denote the resulting chain
complex $\mathcal{K}_{\varepsilon}^{\prime}$.

The chain complex $C_{*}(\mathcal{K}_{\varepsilon})$ may written:

$C_{*}(\mathcal{K}_{\varepsilon})\colon
C_{2}(\mathcal{K}_{\varepsilon})\stackrel{{\scriptstyle\partial_{2}}}{{\to}}C_{1}(\mathcal{K}_{\varepsilon})\stackrel{{\scriptstyle\partial_{1}}}{{\to}}C_{0}(\mathcal{K}_{\varepsilon})$

Note that the boundary map $\partial_{2}$ applied to a 2–cell is the Fox free
differential: $\partial\colon\thinspace F_{\\{g_{1},\cdots,g_{n}\\}}\to
C_{1}(\mathcal{K}_{\varepsilon})$, applied to the word which the 2–cell bounds
(see [9, §48] and [6]). Let $\boldsymbol{e_{i}}$ denote the generator in
$C_{1}(\mathcal{K}_{\varepsilon})$ representing the generator $g_{i}$. The
free Fox differential is then characterized by:

i) $\partial g_{i}=\boldsymbol{e_{i}}$ for all $i=1,\cdots,n$

ii) $\partial(AB)=\partial(A)B+\partial(B)$ for all words $A,B$.

Clearly the inclusion
$\mathcal{K}_{\varepsilon}\hookrightarrow\mathcal{K}_{\varepsilon}^{\prime}$
induces the quotient map $E\to E/K$ on fundamental groups. There is a right
action of $\mathbb{Z}[E/K]$ on itself. Further there is a left action of $E$
on $\mathbb{Z}[E/K]$.

Lemma 2.2 As an algebraic complex of right $\mathbb{Z}[E/K]$ modules
$C_{*}(\mathcal{K}_{\varepsilon}^{\prime})$ may be written:

$C_{*}(\mathcal{K}_{\varepsilon}^{\prime})\colon
C_{2}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\oplus\mathbb{Z}[E/K]^{r}\stackrel{{\scriptstyle\partial_{2}\oplus
0}}{{\to}}C_{1}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\stackrel{{\scriptstyle\partial_{1}}}{{\to}}C_{0}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]$

Proof: The boundary of $E_{i}$ is given by the free Fox differential
$\partial$, applied to the word $\prod_{j=1}^{m_{i}}[A_{ij},B_{ij}]$. However,

$\partial\prod_{j=1}^{m_{i}}[A_{ij},B_{ij}]=\sum_{j=1}^{m_{i}}[\partial
A_{ij}+\partial B_{ij}-\partial A_{ij}-\partial B_{ij}]=0$

as each $A_{ij},B_{ij}$ represents an element of $K$ and hence is trivial in
$\pi_{1}(\mathcal{K}_{\varepsilon}^{\prime})=E/K$. $\Box$

Each $E_{i}$ therefore generates an element of
$H_{2}(\tilde{\mathcal{K}_{\varepsilon}^{\prime}};\mathbb{Z})$. By the
Hurewicz isomorphism Theorem we have isomorphisms
$H_{2}(\tilde{\mathcal{K}_{\varepsilon}^{\prime}};\mathbb{Z})\cong\pi_{2}(\tilde{\mathcal{K}_{\varepsilon}^{\prime}})\cong\pi_{2}(\mathcal{K}_{\varepsilon}^{\prime})$
coming from the Hurewicz homomorphism and the covering map respectively. Let
$\psi_{i}\colon\thinspace S^{2}\to\mathcal{K}_{\varepsilon}^{\prime}$
represent the element of $\pi_{2}(\mathcal{K}_{\varepsilon}^{\prime})$ which
corresponds to $E_{i}$ under these isomorphisms.

For each $i\in 1,\cdots r$ we then attach a 3–cell $B_{i}$ to
$\mathcal{K}_{\varepsilon}^{\prime}$ via the attaching map:
$\psi_{i}\colon\thinspace\partial B_{i}\to\mathcal{K}_{\varepsilon}^{\prime}$.
Let $\mathcal{K}_{\varepsilon}^{\prime\prime}$ denote the resulting 3–complex.
Then we have $C_{*}(\mathcal{K}_{\varepsilon}^{\prime\prime}):$

$\mathbb{Z}[E/K]^{r}\stackrel{{\scriptstyle\partial_{3}}}{{\to}}C_{2}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\oplus\mathbb{Z}[E/K]^{r}\stackrel{{\scriptstyle\partial_{2}\oplus
0}}{{\to}}C_{2}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\stackrel{{\scriptstyle\partial_{1}}}{{\to}}C_{0}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]$

where $\partial_{3}$ is inclusion of the second summand.

Hence we have:

Lemma 2.3
$H_{*}(\mathcal{K}_{\varepsilon}^{\prime\prime},\mathcal{K}_{\varepsilon};M)=0$
for all left modules $M$ over $\mathbb{Z}[E/K]$.

Proof: We have the following relative complex:

$C_{*}(\mathcal{K}_{\varepsilon}^{\prime\prime},\mathcal{K}_{\varepsilon})\colon\quad\mathbb{Z}[E/K]^{r}\tilde{\longrightarrow}\mathbb{Z}[E/K]^{r}\to
0\to 0$

$\Box$

Thus by Theorem 2.1 we may conclude that
$\mathcal{K}_{\varepsilon}^{\prime\prime}$ has the homotopy type of
$\mathcal{K}_{\varepsilon}^{+}$.

Lemma 2.4 The complex $\mathcal{K}_{\varepsilon}^{\prime\prime}$ is
cohomologically 2–dimensional.

Proof: The inclusion
$\iota\colon\thinspace\mathcal{K}_{\varepsilon}\hookrightarrow\mathcal{K}_{\varepsilon}^{\prime\prime}$
induces a chain homotopy equivalence:

$C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\to
C_{*}(\mathcal{K}_{\varepsilon}^{\prime\prime})$

$\Box$

Corollary 2.5 We may choose $\mathcal{K}_{\varepsilon}^{+}$ to be the
cohomologically 2–dimensional finite 3–complex
$\mathcal{K}_{\varepsilon}^{\prime\prime}$.

§3 Cohomologically 2–dimensional 3–complexes

Let $X$ be a finite connected 3–complex with $H^{3}(X;\beta)=0$ for all
coefficient bundles $\beta$. In this section we will show that up to homotopy,
$X$ arises as the Quillen plus construction applied to a finite Cayley
complex.

Let $T$ be a maximal tree in the 1–skeleton of $X$. The quotient map $X\to
X/T$ is a homotopy equivalence. Hence we may assume without loss of generality
that $X$ has one $0$–cell. We take this to be the basepoint of $X$ and any
complexes obtained from $X$ by adding or removing cells. Also we set
$G=\pi_{1}(X)$ with respect to this basepoint.

Let $C_{*}(X)$ be denoted by:

$F_{3}\stackrel{{\scriptstyle\partial_{3}}}{{\to}}F_{2}\stackrel{{\scriptstyle\partial_{2}}}{{\to}}F_{1}\stackrel{{\scriptstyle\partial_{1}}}{{\to}}F_{0}$

where the $F_{i}$, $i=0,1,2,3$, are free modules over $\mathbb{Z}[G]$ and the
$\partial_{i}$ are linear maps over $\mathbb{Z}[G]$.

We have $H^{3}(X;F_{3})=0$ so in particular there exists $\phi$ such the
following diagram commutes:

|  |
---|---|---
$\textstyle{F_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1}$$\scriptstyle{\partial_{3}}$$\textstyle{F_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi}$$\scriptstyle{\partial_{2}}$$\textstyle{F_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\partial_{1}}$$\textstyle{F_{0}}$$\textstyle{F_{3}}$

Hence $\partial_{3}$ is the inclusion of the first summand
$\partial_{3}\colon\thinspace F_{3}\hookrightarrow\partial_{3}(F_{3})\oplus
S=F_{2}$, where $S$ is the kernel of $\phi$. Let $X^{\prime}$ denote the wedge
of $X$ with one disk for each 3–cell in $X$. Then the inclusion of cell
complexes $X\hookrightarrow X^{\prime}$ is a homotopy equivalence and:

$C_{*}(X^{\prime})\colon\thinspace\qquad
F_{3}\stackrel{{\scriptstyle\partial_{3}^{\prime}}}{{\to}}F_{2}\oplus
F_{3}^{\prime}\stackrel{{\scriptstyle\partial_{2}^{\prime}}}{{\to}}F_{1}\oplus
F_{3}^{\prime}\stackrel{{\scriptstyle\partial_{1}^{\prime}}}{{\to}}F_{0}$

Here $F_{3}^{\prime}\cong F_{3}$ and the maps are defined as follows:

$\partial_{1}^{\prime}$ restricts to $\partial_{1}$ on $F_{1}$ and restricts
to $0$ on $F_{3}^{\prime}$,

$\partial_{2}^{\prime}=\left(\begin{array}[]{cc}\partial_{2}&0\\\
0&1\end{array}\right)$

and $\partial_{3}^{\prime}$ is $\partial_{3}\colon\thinspace F_{3}\to F_{2}$
composed with the natural inclusion: $F_{2}\hookrightarrow F_{2}\oplus
F_{3}^{\prime}$. Thus $\partial_{3}^{\prime}$ is the inclusion into the first
summand: $\partial_{3}^{\prime}\colon\thinspace
F_{3}\hookrightarrow\partial_{3}^{\prime}F_{3}\oplus S\oplus F_{3}^{\prime}$.

Let $m$ denote the number of 2–cells in $X$. The submodule $S\oplus
F_{3}^{\prime}\subset(\partial_{3}^{\prime}F_{3}\oplus S)\oplus
F_{3}^{\prime}$ is isomorphic to $S\oplus F_{3}\cong F_{2}$ and hence has a
basis $\boldsymbol{x_{1}},\cdots,\boldsymbol{x_{m}}\in F_{2}\oplus
F_{3}^{\prime}$.

The cell complex $X^{\prime}$ has one 0–cell, so $F_{0}\cong\mathbb{Z}[G]$.
Let $n$ denote the number of 1–cells in $X^{\prime}$. Then each 1–cell
corresponds to a generator $g_{i},\,i\in[1,\cdots,n]$ of $G$. Let
$\\{\boldsymbol{e_{1}},\cdots,\boldsymbol{e_{n}}\\}$ form the corresponding
basis for $F_{1}\oplus F_{3}^{\prime}$.

Let $r$ denote the number of 2–cells in $X^{\prime}$. The attaching map for
each 2–cell maps the boundary of a disk round a word in the $g_{i}$. For each
2–cell let $R_{j},\,j\in[1,\cdots,r]$ denote this word. Let
$\\{\boldsymbol{E_{1}},\cdots,\boldsymbol{E_{r}}\\}$ form the corresponding
basis for $F_{2}\oplus F_{3}^{\prime}$. Thus we have a presentation $G=\langle
g_{1},\cdots,g_{n}\,|\,R_{1},\cdots,R_{r}\rangle$.

We may therefore express each $\boldsymbol{x_{i}}$ as a linear combination of
the $\boldsymbol{E_{j}}$. Thus for some integers $v_{i}$ and sequences
$j_{i1},\cdots,j_{iv_{i}}\in\\{1,\cdots,r\\}$ we have:

$\boldsymbol{x_{i}}=\sum_{l=1}^{v_{i}}\boldsymbol{E_{j_{il}}}\lambda_{il}\sigma_{il}$

with each $\lambda_{il}\in G$ and $\sigma_{il}\in\\{1,-1\\}$. For each
$i\in[1,\cdots,m]$, $l\in[1,\cdots,v_{i}]$ let $w_{il}$ be a word in the
$g_{k},\,k=1,\cdots,n$, representing $\lambda_{il}$. Now for each $i=1,\cdots
m$, let:

$S_{i}=\prod_{l=1}^{v_{i}}w_{il}^{-1}R_{j_{il}}^{\sigma_{il}}w_{il}$

For each $i\in\\{1,\cdots m\\}$, attach a 2–cell $a_{i}$ to $X^{\prime}$ by
mapping the boundary of the disk around the path in the 1–skeleton of
$X^{\prime}$ corresponding to the word $S_{i}$. Let $Z$ denote the resulting
finite cell complex. Note that each word $S_{i}$ corresponds to a trivial
element of $G$, so the inclusion $X^{\prime}\subset Z$ induces an isomorphism
$\pi_{1}(X^{\prime})\cong\pi_{1}(Z)$. Hence we may write $C_{*}(Z)\colon$

$C_{*}(Z)\colon\quad
F_{3}\stackrel{{\scriptstyle\partial_{3}^{\prime\prime}}}{{\to}}(F_{2}\oplus
F_{3}^{\prime})\oplus
F_{2}^{\prime}\stackrel{{\scriptstyle\left(\begin{array}[]{cc}\partial_{2}^{\prime}&\partial_{2}^{\prime\prime}\end{array}\right)}}{{\longrightarrow}}(F_{1}\oplus
F_{3}^{\prime})\stackrel{{\scriptstyle\partial_{1}^{\prime}}}{{\to}}F_{0}$

where $\partial_{3}^{\prime\prime}$ is understood to be
$\partial_{3}^{\prime}\colon\thinspace F_{3}\to(F_{2}\oplus F_{3}^{\prime})$
composed with the natural inclusion $(F_{2}\oplus
F_{3}^{\prime})\hookrightarrow(F_{2}\oplus F_{3}^{\prime})\oplus
F_{2}^{\prime}$.

For $i=1,\cdots,m$ let $\boldsymbol{A_{i}}$ be the basis element of
$F_{2}^{\prime}$ corresponding to the 2–cell $a_{i}$. Recall the Fox free
differential, $\partial$. We have:

$\partial_{2}^{\prime\prime}\boldsymbol{A_{i}}=\partial
S_{i}=\sum_{l=1}^{v_{i}}\partial(w_{il}^{-1}R_{j_{il}}^{\sigma_{il}}w_{il})=\sum_{l=1}^{v_{i}}\partial_{2}^{\prime}\boldsymbol{E_{j_{il}}}\lambda_{il}\sigma_{il}=\partial_{2}^{\prime}\boldsymbol{x_{i}}$

Thus $\boldsymbol{A_{i}}-\boldsymbol{x_{i}}$ represents a class in
$H_{2}(\widetilde{Z^{(2)}};\mathbb{Z})$ which is isomorphic to
$\pi_{2}(Z^{(2)})$ via the Hurewicz isomorphism composed with the map
$\pi_{2}(\widetilde{Z^{(2)}})\to\pi_{2}(Z^{(2)})$ induced by the covering map.
Let $\psi_{i}\colon\thinspace S^{2}\to{Z^{(2)}}$ represent the corresponding
element of $\pi_{2}({Z^{(2)}})$.

Then for each $i=1,\cdots,m$ we may attach a 3–cell $b_{i}$ to $Z$ via the map
$\psi_{i}$. We denote the resulting complex $X^{\prime\prime}$.

Lemma 3.1 The inclusion $\iota\colon\thinspace X^{\prime}\subset
X^{\prime\prime}$ is a homotopy equivalence.

Proof: Starting with $X^{\prime}$, for each $i$ we attached a 2–cell $a_{i}$
with contractable boundary in $X^{\prime}$, and then attached a 3–cell $b_{i}$
with $a_{i}$ as a free face. Thus $X^{\prime\prime}$ is obtained from
$X^{\prime}$ through a series of cell expansions and the inclusion
$X^{\prime}\subset X^{\prime\prime}$ is a simple homotopy equivalence. $\Box$

Let $Y$ denote the subcomplex of $X^{\prime\prime}$ consisting of the
1–skeleton, $X^{\prime\prime(1)}$, together with the $a_{i}$, $i=1,\cdots,m$.
Let $\varepsilon$ denote the group presentation $\langle
g_{1},\cdots,g_{n}\,|\,S_{1},\cdots,S_{m}\rangle$ and let $E$ denote the
underlying group. By construction we have: $Y=\mathcal{K}_{\varepsilon}$.

Let $k_{1},\cdots,k_{r}\in E$ denote the elements represented by the words
$R_{1},\cdots,R_{r}$. Let $K$ denote the normal closure in $E$ of
$k_{1},\cdots,k_{r}$. By construction then, $K$ is finitely closed and we have
a short exact sequence of groups:

$1\to K\to E\to G\to 1$

Lemma 3.2 $K$ is a perfect group.

Proof: Clearly $\mathbb{Z}[G]$ is a right module over itself and there is a
left action of $E$ on $\mathbb{Z}[G]$. The algebraic complex
$C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[G]$ is given by:

$F_{2}^{\prime}\stackrel{{\scriptstyle\partial_{2}^{\prime\prime}}}{{\rightarrow}}(F_{1}\oplus
F_{3}^{\prime})\stackrel{{\scriptstyle\partial_{1}^{\prime}}}{{\to}}F_{0}$

Now consider $C_{*}(X^{\prime})$:

$F_{3}\stackrel{{\scriptstyle\partial_{3}^{\prime}}}{{\to}}F_{2}\oplus
F_{3}^{\prime}\stackrel{{\scriptstyle\partial_{2}^{\prime}}}{{\to}}F_{1}\oplus
F_{3}^{\prime}\stackrel{{\scriptstyle\partial_{1}^{\prime}}}{{\to}}F_{0}$

As $\tilde{X}^{\prime}$ is simply connected, we have ${\rm
ker}(\partial_{1}^{\prime})={\rm Im}(\partial_{2}^{\prime})$.

Recall that $F_{2}\oplus F_{3}^{\prime}=\partial_{3}^{\prime}(F_{3})\oplus
S\oplus F_{3}^{\prime}$ and that $S\oplus F_{3}^{\prime}$ has basis
$\boldsymbol{x_{1}},\cdots,\boldsymbol{x_{m}}$. Clearly
$\partial_{2}^{\prime}$ restricts to $0$ on $\partial_{3}^{\prime}(F_{3})$, so
${\rm ker}(\partial_{1}^{\prime})={\rm Im}(\partial_{2}^{\prime})$ which is
generated by the $\partial_{2}^{\prime}(\boldsymbol{x_{i}})$.

Also recall that
$\partial_{2}^{\prime}\boldsymbol{x_{i}}=\partial_{2}^{\prime\prime}\boldsymbol{A_{i}}$.
Hence ${\rm ker}(\partial_{1}^{\prime})={\rm Im}(\partial_{2}^{\prime\prime})$
and $H_{1}(C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[G])=0$.

However by restricting coefficients $C_{*}(\mathcal{K}_{\varepsilon})$ may be
regarded as an algebraic complex of free modules over $\mathbb{Z}[K]$. Hence
we have:

$K/[K,K]=H_{1}(K;\mathbb{Z})=H_{1}(C_{*}(\mathcal{K}_{\varepsilon})\otimes_{K}\mathbb{Z})=H_{1}(C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[G])=0$.

where $\mathbb{Z}$ is regarded as having a trivial left $K$–action. $\Box$

Lemma 3.3 $X^{\prime\prime}=\mathcal{K}_{\varepsilon}^{+}$ where $+$ is taken
with respect to $K$.

Proof: We may identify $\mathcal{K}_{\varepsilon}$ with the subcomplex
$Y\subset X^{\prime\prime}$. The inclusion
$\ell\colon\thinspace\mathcal{K}_{\varepsilon}\hookrightarrow
X^{\prime\prime}$ then induces the quotient map $E\to E/K$ on fundamental
groups. By Theorem 2.1 it is sufficient to show that
$H_{*}(X^{\prime\prime},Y;M)=0$ for all left coefficient modules $M$.
Let $\ell_{*}\colon\thinspace
C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[G]\to
C_{*}(X^{\prime\prime})$ be the chain map induced by the inclusion
$\ell\colon\thinspace\mathcal{K}_{\varepsilon}\hookrightarrow
X^{\prime\prime}$. We have the following commutative diagram:

$\displaystyle
F_{2}^{\prime}\,\,\,\quad\stackrel{{\scriptstyle\partial_{2}^{\prime\prime}}}{{\longrightarrow}}\quad\,\,\,\,(F_{1}\oplus
F_{3}^{\prime})\stackrel{{\scriptstyle\partial_{1}^{\prime}}}{{\to}}F_{0}\,\,\,\,$
$\displaystyle\downarrow^{\ell_{2}}\qquad\,\,\,\,\qquad\qquad\downarrow^{\ell_{1}}\,\,\,\,\qquad\downarrow^{\ell_{0}}$
$\displaystyle F_{3}\oplus
F_{2}^{\prime\prime}\stackrel{{\scriptstyle\left(\begin{array}[]{cc}\partial_{3}^{\prime\prime}&\partial_{3}^{\prime\prime\prime}\end{array}\right)}}{{\to}}(F_{2}\oplus
F_{3}^{\prime})\oplus
F_{2}^{\prime}\stackrel{{\scriptstyle\left(\begin{array}[]{cc}\partial_{2}^{\prime}&\partial_{2}^{\prime\prime}\end{array}\right)}}{{\longrightarrow}}(F_{1}\oplus
F_{3}^{\prime})\stackrel{{\scriptstyle\partial_{1}^{\prime}}}{{\to}}F_{0}\,\,\,\,$

where $F_{2}^{\prime\prime}$ has a basis
$\boldsymbol{D_{1}},\cdots,\boldsymbol{D_{m}}$ corresponding to the 3–cells
$b_{1},\cdots,b_{m}$, so for $i=1,\cdots,m$ we have
$\partial_{3}^{\prime\prime\prime}(D_{i})=\boldsymbol{A_{i}}-\boldsymbol{x_{i}}$.
Here $\ell_{0}$ and $\ell_{1}$ are the identity maps and $\ell_{2}$ is the
inclusion of the second summand.

We have $(F_{2}\oplus
F_{3}^{\prime})=\partial_{3}^{\prime\prime}F_{3}\oplus(S\oplus
F_{3}^{\prime})$. Hence we have $(F_{2}\oplus F_{3}^{\prime})\oplus
F_{2}^{\prime}=\partial_{3}^{\prime\prime}F_{3}\oplus(S\oplus
F_{3}^{\prime})\oplus F_{2}^{\prime}$.

The submodule $(S\oplus F_{3}^{\prime})$ has basis
$\boldsymbol{x_{1}},\cdots,\boldsymbol{x_{m}}$. The submodule $F_{2}^{\prime}$
has basis $\boldsymbol{A_{1}},\cdots,\boldsymbol{A_{m}}$. Also
$\partial_{3}^{\prime\prime\prime}F_{2}^{\prime\prime}$ has basis
$\boldsymbol{A_{1}}-\boldsymbol{x_{1}},\cdots,\boldsymbol{A_{m}}-\boldsymbol{x_{m}}$.
Hence we have the following equality of submodules: $(S\oplus
F_{3}^{\prime})\oplus
F_{2}^{\prime}=\partial_{3}^{\prime\prime\prime}F_{2}^{\prime\prime}\oplus
F_{2}^{\prime}$

Thus:

$(F_{2}\oplus F_{3}^{\prime})\oplus
F_{2}^{\prime}=\partial_{3}^{\prime\prime}F_{3}\oplus\partial_{3}^{\prime\prime\prime}F_{2}^{\prime\prime}\oplus
F_{2}^{\prime}$

The relative chain complex $C_{*}(X^{\prime\prime},Y)$ is therefore given by:

$F_{3}\oplus
F_{2}^{\prime\prime}{\tilde{\longrightarrow}}\partial_{3}^{\prime\prime}F_{3}\oplus\partial_{3}^{\prime\prime\prime}F_{2}^{\prime\prime}\longrightarrow
0\longrightarrow 0$

and $H_{*}(X^{\prime\prime},Y;M)=0$ for all left coefficient modules $M$ as
required. $\Box$

As $X\sim X^{\prime\prime}$, we have proved the following theorem:

Theorem 3.4 Let $X$ be a finite connected 3–complex with $H^{3}(X;\beta)=0$
for all coefficient bundles $\beta$. Then $X$ has the homotopy type of
$\mathcal{K}_{\varepsilon}^{+}$ for some finite presentation $\varepsilon$ of
a group $E$, where $+$ is taken with respect to some perfect finitely closed
normal subgroup $K\lhd E$.

§4 Implications for the D(2) problem

The D(2) problem asks if every finite cohomologically 2–dimensional 3–complex
must be homotopy equivalent to a finite 2–complex. Clearly a counterexample
must have a connected component which is also cohomologically 2–dimensional
but not homotopy equivalent to a finite 2–complex. By Theorem 3.4 this
component must have the homotopy type of $\mathcal{K}_{\varepsilon}^{+}$ for
some finite presentation $\varepsilon$ of a group $E$, where $+$ is taken with
respect to some perfect finitely closed normal subgroup $K\vartriangleleft E$.

Conversely, by Corollary 2.5, given any finite presentation $\varepsilon$ of a
group $E$ together with some perfect finitely closed normal subgroup
$K\vartriangleleft E$ we have a cohomologically 2–dimensional finite
3–complex, $\mathcal{K}_{\varepsilon}^{+}$. It follows that the D(2) problem
is equivalent to:

Given a finite presentation $\epsilon$ for a group $E$, and a finitely closed
perfect normal subgroup $K\vartriangleleft E$, must
$\mathcal{K}_{\varepsilon}^{+}$ be homotopy equivalent to a finite 2–complex?

Suppose that we have a homotopy equivalence $\mathcal{K}_{\varepsilon}^{+}\sim
Y$ for some finite 2–complex $Y$. Let $T$ be a maximal tree in the
$1$–skeleton of $Y$. The quotient map $Y\to Y/T$ is a homotopy equivalence so
$Y\sim\mathcal{K}_{\mathcal{G}}$ for some finite presentation $\mathcal{G}$ of
$\pi_{1}(Y)=\pi_{1}(\mathcal{K}_{\varepsilon}^{+})=E/K$.

Hence the affirmative answer to the the D(2) problem would be equivalent to:

For all finitely presented groups $E$ and all perfect finitely closed normal
subgroups $K\vartriangleleft E$ and all finite presentations $\varepsilon$ of
$E$, there exists a finite presentation $\mathcal{G}$ of $E/K$ and a homotopy
equivalence $\mathcal{K}_{\varepsilon}^{+}\sim\mathcal{K}_{\mathcal{G}}$
inducing the identity $1\colon E/K\to E/K$ on fundamental groups.

Lemma 4.1 The following are equivalent:

i) There exists a homotopy equivalence:
$\mathcal{K}_{\varepsilon}^{+}\sim\mathcal{K}_{\mathcal{G}}$ inducing the
identity $1\colon E/K\to E/K$ on fundamental groups.

ii) There exists a chain homotopy equivalence:
$C_{*}(\mathcal{K}_{\varepsilon}^{+})\sim C_{*}(\mathcal{K}_{\mathcal{G}})$
over $\mathbb{Z}[E/K]$.

Proof: i) $\Rightarrow$ ii) is immediate. Conversely, from ii) we have a chain
homotopy equivalence between the algebraic complexes associated to a finite
cohomologically 2–dimensional 3–complex and a finite 2–complex (with respect
to an isomorphism of fundamental groups). To show that ii)$\Rightarrow$ i) we
must construct a homotopy equivalence between the spaces, inducing the same
isomorphism on fundamental groups. For finite fundamental groups this is done
in [9, proof of Theorem 59.4]. The same argument holds for all finitely
presented fundamental groups [9, Appendix B, Proof of Weak Realization
Theorem]. $\Box$

From the proof of lemma 2.4,
$C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\sim
C_{*}(\mathcal{K}_{\varepsilon}^{+})$. Hence we have:

Theorem 4.2 The following two statements are equivalent:

i) Let $X$, a finite 3–complex with $H^{3}(X;\beta)=0$ for all coefficient
bundles $\beta$. Then $X$ is homotopy equivalent to a finite 2–complex.

ii) Let $K$ be a perfect finitely closed normal subgroup of a finitely
presented group $E$. For each finite presentation $\varepsilon$ of $E$, there
exists a finite presentation $\mathcal{G}$ of $E/K$, such that we have a chain
homotopy equivalence over $\mathbb{Z}[E/K]$:

$C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[E/K]\to
C_{*}(\mathcal{K}_{\mathcal{G}})$

Suppose we have a short exact sequence:

$1\to L\to F\to G\to 1$

where $G$ is a finitely presented group and $F$ is a free group generated by
elements $g_{1},\cdots,g_{n}$. Let $R_{1},\cdots,R_{m}$ be elements of $L$.

Definition 4.3  $\langle g_{1},\cdots,g_{n}|R_{1},\cdots,R_{m}\rangle$ is
called a finite partial presentation for $G$ when the normal closure
$N_{F}(R_{1},\cdots,R_{m})$ surjects onto $L/[L,L]$ under the quotient map
$L\to L/[L,L]$.

Note that a finite partial presentation $\varepsilon=\langle
g_{1},\cdots,g_{n}|R_{1},\cdots R_{m}\rangle$ as above is an actual finite
presentation of some group $E$, so it has a well defined Cayley complex
$\mathcal{K}_{\varepsilon}$.

Let $K$ denote the kernel of the homomorphism $E\to G$ sending each $g_{i}$ to
the corresponding element in $G$. If $G$ is finitely presented then it is
finitely presented on the generators in $\varepsilon$ [4, Chapter 1,
Proposition 17]. As $K$ is the normal closure in $E$ of the images of this
finite set of relators we have that $K$ is finitely closed.

Further $K$ is perfect as every $k\in K$ may be lifted to an element of $L$
which may be written in the form $ab$ where $a\in[L,L]$ and $b\in
N_{F}(R_{1},\cdots R_{m})$. Thus $k$ is equal to the image of $a$ in $E$, so
$k\in[K,K]$. Thus a finite partial presentation $\varepsilon$ of a finitely
presented group $G$ may be viewed as a presentation satisfying the hypothesis’
of statement ii) in Theorem 4.2.

Conversely, given $\varepsilon$ as in statement ii) Theorem 4.2, we have that
$\varepsilon$ is a finite partial presentation of $E/K$ (as $K=[K,K]$), and
$E/K$ is finitely presented (as K is finitely closed).

Thus statement ii) is equivalent to:

ii)’ Given a finite partial presentation $\varepsilon$ of a finitely presented
group $G$, there exists a finite presentation $\mathcal{G}$ of $G$, such that
we have a chain homotopy equivalence:

$C_{*}(\mathcal{K}_{\varepsilon})\otimes_{E}\mathbb{Z}[G]\to
C_{*}(\mathcal{K}_{\mathcal{G}})$

where $E$ is the group presented by $\varepsilon$ and each $x\in E$ acts on
$\mathbb{Z}[G]$ by left multiplication by its image in $G$.

One approach to the D(2) problem is to use Euler characteristic as an
obstruction. That is, given a finite cohomologically 2–dimensional 3–complex
$X$, if we can show that every finite 2–complex $Y$ with
$\pi_{1}(Y)=\pi_{1}(X)$ satisfies $\chi(X)<\chi(Y)$ then clearly $X$ cannot be
homotopy equivalent to any such $Y$. It has been shown that certain
constructions involving presentations of a group would allow one to construct
such a space [7, Theorem 3.5]. A candidate for such a space is given in [3].
In light of Corollary 2.5 and Theorem 3.4 we are able to generalize this
approach.

The deficiency ${\rm Def}(\mathcal{G})$ of a finite presentation $\mathcal{G}$
is the number of generators minus the number of relators. We say a
presentation of a group is minimal if it has the maximal possible deficiency.
A finitely presented group $G$ always has a minimal presentation, because an
upper bound for the deficiency of a presentation is given by ${\rm
rk}_{\mathbb{Z}}(G/[G,G])$. The deficiency ${\rm Def}(G)$ of a finitely
presented group $G$ is defined to be the deficiency of a minimal presentation.

Again let $K\vartriangleleft E$ be a perfect finitely closed normal subgroup.
Then if $\varepsilon$ is a finite presentation of $E$ and $\mathcal{G}$ is a
finite presentation for $E/K$ we have:

$\chi(\mathcal{K}_{\varepsilon}^{+})=\chi(\mathcal{K}_{\varepsilon})=1-{\rm
Def}(\varepsilon)\qquad\qquad\qquad\chi(\mathcal{K}_{\mathcal{G}})=1-{\rm
Def}(\mathcal{G})$

Lemma 4.4 If ${\rm Def}(E)>{\rm Def}(E/K)$ then given a minimal presentation
$\varepsilon$ of $E$ we have that
$\chi(\mathcal{K}_{\varepsilon}^{+})<\chi(\mathcal{K}_{\mathcal{G}})$ for any
finite presentation $\mathcal{G}$ of $E/K$.

Proof:

$\chi(\mathcal{K}_{\mathcal{G}})=1-{\rm Def}(\mathcal{G})\geq 1-{\rm
Def}(E/K)>1-{\rm Def}(E)=1-{\rm
Def}(\varepsilon)=\chi(\mathcal{K}_{\varepsilon}^{+})$ $\Box$

Suppose we have a short exact sequence of groups:

$1\to K\to E\to G\to 1$

with $E$, $G$ finitely presented. Then given a finite presentation for $E$,
the images in $G$ of the generators will generate $G$. We may present $G$ on
these generators with a finite set of relators [4, Chapter 1, Proposition 17].
Let $k_{1},\cdots,k_{r}$ denote the elements of $K$ represented by these
relators. Then $K$ is the normal closure in $E$ of $k_{1},\cdots,k_{r}$ and so
$K$ is finitely closed in $E$. In particular $K/[K,K]$ is generated by the
$k_{1},\cdots,k_{r}$ as a right module over $\mathbb{Z}[G]$ (where $G$ acts on
$K/[K,K]$ by conjugation). Let ${\rm rk}_{G}(K)$ denote the minimal number of
elements required to generate $K/[K,K]$ over $\mathbb{Z}[G]$.

Theorem 4.5 The following statements are equivalent:

i) There exists a connected finite cohomologically 2–dimensional 3–complex
$X$, such that for all finite connected 2–complexes $Y$ with
$\pi_{1}(Y)=\pi_{1}(X)$ we have $\chi(X)<\chi(Y)$.

ii) There exists a short exact sequence of groups $1\to K\to E\to G\to 1$ with
$E$, $G$ finitely presented and

${\rm rk}_{G}(K)+{\rm Def}(G)<{\rm Def}(E)$

Proof: i)$\Rightarrow$ii): By Theorem 3.4, $X$ is homotopy equivalent to
$\mathcal{K}_{\varepsilon}^{+}$ for some finite presentation $\varepsilon$ of
some group $E$ and some perfect finitely closed normal subgroup $K$. Let
$G=E/K$. We have a short exact sequence:

$1\to K\to E\to G\to 1$

As $K$ is finitely closed, $G$ is finitely presented. As $K$ is perfect we
have ${\rm rk}_{G}(K)=0$. Let $\mathcal{G}$ be some finite presentation of
$G$. We have:

$1-{\rm
Def}(\varepsilon)=\chi(\mathcal{K}_{\varepsilon}^{+})<\chi(\mathcal{K}_{\mathcal{G}})=1-{\rm
Def}(\mathcal{G})$

Thus ${\rm Def}(\mathcal{G})<{\rm Def}(\varepsilon)$. As $\mathcal{G}$ was
chosen arbitrarily, we have ${\rm Def}(G)<{\rm Def}(\varepsilon)\leq{\rm
Def}(E)$. Hence ${0+\rm Def}(G)<{\rm Def}(E)$ as required.

ii)$\Rightarrow$i): We start with the short exact sequence $1\to K\to E\to
G\to 1$. Let $k_{1},\cdots,k_{r}\in K$ generate $K/[K,K]$ over
$\mathbb{Z}[G]$, where $r={\rm rk}_{G}(K)$. Let $K^{\prime}$ denote the normal
closure in $E$ of $k_{1},\cdots,k_{r}$. Then we have a short exact sequence:

$1\to K/K^{\prime}\to E/K^{\prime}\to G\to 1$

Then $K=K^{\prime}[K,K]$ so $K/K^{\prime}$ is perfect. From the discussion
preceding this theorem we know that $K$ is finitely closed in $E$, so
$K/K^{\prime}$ must be finitely closed in $E/K^{\prime}$. Also $E/K^{\prime}$
may be presented by taking a minimal presentation of $E$ and adding $r$
relators (representing to $k_{1},\cdots,k_{r}$). Hence:

${\rm Def}(E/K^{\prime})\geq{\rm Def}(E)-{\rm rk}_{G}(K)>{\rm Def}(G)$

Take a minimal presentation $\varepsilon$ of $E/K^{\prime}$ and let
$X=\mathcal{K}_{\varepsilon}^{+}$, where $+$ is taken with respect to
$K/K^{\prime}$. Any finite connected 2–complex $Y$ with
$\pi_{1}(Y)=\pi_{1}(X)$ is homotopy equivalent to $\mathcal{K}_{\mathcal{G}}$
for some finite presentation $\mathcal{G}$ of $G$. Therefore by lemma 4.4 we
have $\chi(X)<\chi(Y)$ as required. $\Box$

We note that Michael Dyer proved ii)$\Rightarrow$i) in the case where
$H^{3}(G;\mathbb{Z}[G])=0$ and $E$ is a free group whose generators are the
generating set for some minimal presentation of $G$ [7, Theorem 3.5].

## References

  * [1]
  * [2] F. Rudolf Beyl and Nancy Waller ; Examples of exotic free 2–complexes and stably free non-free modules for quaternion groups : Algebraic & Geometric Topology 8 (2008) pp. 1 – 17
  * [3] Martin R. Bridson and Michael Tweedale ; Deficiency and abelianized deficiency of some virtually free groups : Math. Proc. Camb. Phil. Soc. 143 (2007) pp. 257 – 264
  * [4] Daniel E. Cohen, ‘Combinatorial Group Theory: a topological approach’, LMS Student Texts 14, Cambridge University Press (1989)
  * [5] Tim Edwards ; Generalised Swan modules and the D(2) problem : Algebraic & Geometric Topology 6 (2006) pp. 71 – 89
  * [6] R. H. Fox ; Free differential calculus, V. The Alexander matrices re-examined : Ann. of Math. 71 (1960) pp. 408 – 422
  * [7] J. Harlander ; Some aspects of efficiency : Groups-Korea ’98 (Pusan), de Gruyter (2000) pp. 165–180.
  * [8] F.E.A. Johnson ; Explicit homotopy equivalences in dimension two : Math. Proc. Camb. Phil. Soc. 133 (2002) pp. 411 – 430
  * [9] F.E.A. Johnson, ‘Stable Modules and the D(2) Problem’, LMS Lecture Note Series 301, Cambridge University Press (2003)
  * [10] W. H. Mannan ; The D(2) property for D8 : Algebraic & Geometric Topology 7 (2007) pp. 517 – 528
  * [11] Jonathan Rosenberg, ‘Algebraic $K$–Theory and Its Applications’, Graduate Texts in Mathematics 147, Springer (1994)
  * [12] R.G. Swan ; Groups of cohomological dimension one : J. of Algebra 12 (1969) pp. 585 – 610
  * [13] John R. Stallings  ; On torsion free groups with infinitely many ends : Ann. of Math. 88 pp. 312 – 334 (1968)
  * [14] C.T.C. Wall ; Finiteness conditions for CW–complexes : Ann. of Math. 81 (1965) pp. 56 – 69.

Address: Building 54
School of Mathematics
University of Southampton
University Road
Southampton SO17 1BJ
Telephone: +44 (0)2380 592 141
<EMAIL_ADDRESS>


# The Smirnov class for de Branges–Rovnyak spaces

Emmanuel Fricain  Laboratoire Paul Painlevé, Université Lille 1, 59 655
Villeneuve d’Ascq Cédex<EMAIL_ADDRESS>,  Andreas
Hartmann  Institut de Mathématiques de Bordeaux, Université Bordeaux 1, 351
cours de la Libération 33405 Talence Cédex, France
<EMAIL_ADDRESS>,  William T. Ross  Department of
Mathematics and Computer Science, University of Richmond, Richmond, VA 23173,
USA<EMAIL_ADDRESS>and  Dan Timotin  Simion Stoilow Institute of
Mathematics of the Romanian Academy, PO Box 1-764, Bucharest 014700, Romania
<EMAIL_ADDRESS>
###### Abstract.

Using an explicit construction, we show that the de Branges-Rovnyak spaces
$\mathscr{H}(b)$, corresponding to rational $b$, are contained in their
associated Smirnov classes.

###### Key words and phrases:

Smirnov class, multipliers, de Branges Rovnyak spaces

###### 2010 Mathematics Subject Classification:

30J05, 30H10, 46E22

The first author was supported by the Labex CEMPI (ANR-11-LABX-0007-01) and
the project FRONT (ANR-17-CE40-0021).

##  1\. Introduction

Let $H^{2}$ denote the Hardy space of analytic functions $f$ on the open unit
disk $\mathbb{D}$ for which

$\sup_{0<r<1}\int_{\mathbb{T}}|f(r\xi)|^{2}dm(\xi)$

is finite. In the above, $\mathbb{T}=\partial\mathbb{D}$ is the unit circle
and $m$ is normalized Lebesgue measure on $\mathbb{T}$. One can see that
$H^{2}$ contains $H^{\infty}$, the bounded analytic functions on $\mathbb{D}$,
and that any $\varphi\in H^{\infty}$ satisfies $\varphi H^{2}\subseteq H^{2}$.
In fact, these are the only analytic functions on $\mathbb{D}$ which multiply
$H^{2}$ into itself. Furthermore, $\varphi H^{2}$ is dense in $H^{2}$ if and
only if $\varphi$ is a bounded outer function. A well-known theorem of Smirnov
[5, p. 16] says that every function $f\in H^{2}$ can be written as
$f=\varphi/\psi,$ where $\varphi,\psi\in H^{\infty}$ and $\psi$ is outer.

Taking inspiration from $H^{2}$, one can define, for a general reproducing
Hilbert space $\mathcal{H}$ of analytic functions on $\mathbb{D}$, the Smirnov
class of $\mathcal{H}$, denoted by $\mathcal{N}^{+}(\mathcal{H})$, to be the
set of quotients $\varphi/\psi$, where $\varphi$ and $\psi$ are multipliers of
$\mathcal{H}$ ($\varphi\mathcal{H}\subseteq\mathcal{H}$ and
$\psi\mathcal{H}\subseteq\mathcal{H}$) and the denominator $\psi$ is a cyclic
multiplier ($\psi\mathcal{H}$ is dense in $\mathcal{H}$). We say that
$\mathcal{H}$ satisfies the Smirnov property if
$\mathcal{H}\subseteq\mathcal{N}^{+}(\mathcal{H})$. In general, multipliers of
$\mathcal{H}$ belong to $H^{\infty}$ but not every $H^{\infty}$ function is a
multiplier of $\mathcal{H}$. Furthermore, the density of $\psi\mathcal{H}$ in
$\mathcal{H}$ is usually more complicated than the simple criterion that
$\psi$ is a bounded outer function. Indeed, when the polynomials are dense in
$\mathcal{H}$ and the shift operator $S_{\mathcal{H}}$ (the multiplication by
$z$) acts as a bounded operator on $\mathcal{H}$, then $\psi\mathcal{H}$ is
dense in $\mathcal{H}$ if and only if $\psi$ is a cyclic vector for
$S_{\mathcal{H}}$. This problem of cyclicity is known to be notoriously
difficult and is only completely understood for a handful number of cases.

A remarkable result in [1] says that reproducing kernel Hilbert spaces
$\mathcal{H}$ satisfying the complete Nevanlinna–Pick property, along with the
normalizing condition $k_{z_{0}}(z)=1$ for all $z\in\mathbb{D}$ and some
$z_{0}\in\mathbb{D}$, have the property that
$\mathcal{H}\subseteq\mathcal{N}^{+}(\mathcal{H})$. Two related papers [2, 15]
show that the harmonically weighted Dirichlet spaces $D(\mu)$ and a class of
de Branges–Rovnyak spaces $\mathscr{H}(b)$ enjoy the complete Nevanlinna-Pick
property and the above mentioned normalizing condition and thus belong to
their associated Smirnov classes. The purpose of this paper is to give an
explicit construction which proves the Smirnov property for a large class of
$\mathscr{H}(b)$ spaces including the results mentioned above.

To state our main theorem (Theorem 1.1), we make the following definitions.
Let

$H^{\infty}_{1}=\Big{\\{}\varphi\in
H^{\infty}:\|\varphi\|_{\infty}:=\sup_{z\in\mathbb{D}}|\varphi(z)|\leqslant
1\Big{\\}}$

denote the closed unit ball of $H^{\infty}$ and for $b\in H^{\infty}_{1}$ the
de Branges–Rovnyak space $\mathscr{H}(b)$ is the reproducing kernel Hilbert
space of analytic functions whose kernel is

$k^{b}_{\lambda}(z)=\frac{1-b(z)\overline{b(\lambda)}}{1-\overline{\lambda}z},\quad\lambda,z\in\mathbb{D}.$

See [10, 11, 14] for the details concerning $\mathscr{H}(b)$ spaces. Recall
that $f\in H^{\infty}$ is an inner function if the radial boundary function
$f(e^{i\theta})$ satisfies $|f(e^{i\theta})|=1$ almost everywhere with respect
to normalized Lebesgue measure $m$ on the unit circle $\mathbb{T}$. An $f\in
H^{\infty}$ is outer if

$f(z)=\exp\Big{(}\int_{\mathbb{T}}\frac{\xi+z}{\xi-z}\log|f(\xi)|dm(\xi)\Big{)},\quad
z\in\mathbb{D}.$

The main result of our paper is the following.

######  Theorem 1.1.

If $b\in H^{\infty}_{1}$ is rational but not inner, then
$\mathscr{H}(b)\subseteq\mathcal{N}^{+}(\mathscr{H}(b))$.

We discuss an extension of this result when $b$ is outer in Corollary 4.3.
Note that if $b$ is rational and inner, it must be a finite Blaschke product.
In this situation $\mathscr{H}(b)$ becomes a model space $(bH^{2})^{\perp}$
(in fact a finite dimensional one) and the multipliers of any model space are
just the constant functions [7]. Thus, Theorem 1.1 is no longer true. The
multipliers of $\mathscr{H}(b)$ for general $b\in H_{1}^{\infty}$ were studied
in [4, 11, 12, 13] but are far from being completely understood.

Isolated cases of Theorem 1.1 can be gleaned from results in [1, 2, 15]. More
precisely, when $b$ is rational and not inner, there is a unique rational
function $a$ such that $|a|^{2}+|b|^{2}=1$ almost everywhere on $\mathbb{T}$
and $a(0)>0$. First, one can prove that if the zeros of $a$ on $\mathbb{T}$
are all simple, then $\mathscr{H}(b)$ coincides with a $\mathcal{D}(\mu)$
space (with equivalent norms) [3] and so, by [1] and [15],
$\mathscr{H}(b)\subseteq\mathcal{N}^{+}(\mathscr{H}(b))$. Second, if $b(0)=0$
and there exists an analytic function $h$ on $\mathbb{D}$ such that
$h(b(z))=z$ for all $z\in\mathbb{D}$ and the function $(z-b(0))/h(z)$ extends
to be analytic on $\mathbb{D}$ with

$\left|\frac{z-b(0)}{h(z)}\right|\leqslant|1-\overline{b(0)}z|,\quad\mbox{for
$z\in\mathbb{D}$},$

then by [2] $\mathscr{H}(b)$ has the compete Nevanlinna Pick property; since
$k^{b}_{0}\equiv 1$, [1] implies that
$\mathscr{H}(b)\subseteq\mathcal{N}^{+}(\mathscr{H}(b))$. In particular, the
hypothesis of the above results exclude a large class of rational $b$.

One of the main reasons for this paper is to prove Theorem 1.1 directly for
all rational $b$ (not just the cases mentioned in the previous paragraph), and
only using the tools of $\mathscr{H}(b)$ spaces. In addition, our construction
is explicit.

The main idea that makes all of this work is that when $b\in H_{1}^{\infty}$
is rational and not inner, then $\mathscr{H}(b)=\mathscr{M}(\overline{a})$
(with equivalent norms), where
$\mathscr{M}(\overline{a})=T_{\overline{a}}H^{2}$ (equipped with the range
norm), $a\in H^{\infty}_{1}$ satisfies $|a|^{2}+|b|^{2}=1$ almost everywhere
on $\mathbb{T}$, and $T_{\overline{a}}$ is the co-analytic Toeplitz operator
on $H^{2}$. Furthermore, we know the exact structure of
$\mathscr{M}(\overline{a})$ (and hence $\mathscr{H}(b)$), as well as its
multiplier algebra. Along the way to proving Theorem 1.1, we prove the
following curious representation theorem for $H^{2}$ functions.

######  Theorem 1.2.

If $p$ is a polynomial whose roots all lie on $\mathbb{T}$, then any $h\in
H^{2}$ can be written as

(1.3) $h=\frac{u}{pv+1},$

where $u,v\in H^{2}$, $pu,pv\in H^{\infty}$, and $pv+1$ is outer.

Our methods also yield a characterization of the cyclic vectors for the shift
operator $f\mapsto zf$ on $\mathscr{M}(\overline{a})$, and hence
$\mathscr{H}(b)$, for rational $b$ (Proposition 3.8).

##  2\. The Smirnov class for general spaces

Before getting to the Smirnov class for $\mathscr{H}(b)$, let us make a few
remarks about the general case where $\mathcal{H}$ is any reproducing kernel
Hilbert space of analytic functions on $\mathbb{D}$. Define the multiplier
algebra of $\mathcal{H}$ as

$\mathfrak{M}(\mathcal{H})=\\{\varphi\in\operatorname{Hol}(\mathbb{D}):\varphi\mathcal{H}\subseteq\mathcal{H}\\}.$

Standard facts say that $\mathfrak{M}(\mathcal{H})\subseteq H^{\infty}$ and if
the constant function $1$ belongs to $\mathcal{H}$, which will be the case of
$\mathscr{H}(b)$ spaces we consider, then
$\mathfrak{M}(\mathcal{H})\subseteq\mathcal{H}$ and hence
$\mathfrak{M}(\mathcal{H})\subseteq\mathcal{H}\cap H^{\infty}$. Moreover, when
$\varphi\in\mathfrak{M}(\mathcal{H})$, an application of the closed graph
theorem, says that the multiplication operator $M_{\varphi}f=\varphi f$ is a
bounded operator on $\mathcal{H}$. The quantity

$\|M_{\varphi}\|=\sup_{f\in\mathcal{H},\|f\|=1}\|\varphi f\|,$

the operator norm of $M_{\varphi}$, is called the multiplier norm of $\varphi$
and satisfies $\|\varphi\|_{\infty}\leqslant\|M_{\varphi}\|$. For the Hardy
space $H^{2}$ it is a standard fact that $\mathfrak{M}(H^{2})=H^{\infty}$ and
$\|M_{\varphi}\|=\|\varphi\|_{\infty}$. For many other spaces,
$\mathfrak{M}(\mathcal{H})$ is a proper subset of $H^{\infty}$ and it may
happen that $\|\varphi\|_{\infty}<\|M_{\varphi}\|$.

The Smirnov class of $\mathcal{H}$ is

$\mathcal{N}^{+}(\mathcal{H}):=\Big{\\{}\frac{\varphi}{\psi}:\varphi,\psi\in\mathfrak{M}(\mathcal{H}),\overline{\varphi\mathcal{H}}=\mathcal{H}\Big{\\}}.$

In the above $\overline{\varphi\mathcal{H}}$ denotes the closure of
$\varphi\mathcal{H}$ in the norm of $\mathcal{H}$. The following fact is
standard but we recall it here for the reader’s convenience.

######  Lemma 2.1.

$\mathcal{N}^{+}(\mathcal{H})$ is an algebra.

###### Proof.

For a multiplier $\varphi$ we have $\overline{\varphi\mathcal{H}}=\mathcal{H}$
if and only if $\operatorname{Ker}M_{\varphi}^{*}=\\{0\\}$. For
$f_{1},f_{2}\in\mathcal{N}^{+}(\mathcal{H})$, we need to prove that
$f_{1}+f_{2}$ and $f_{1}f_{2}$ belong to $\mathcal{N}^{+}(\mathcal{H})$. Write

$f_{j}=\frac{\varphi_{j}}{\psi_{j}},\quad\varphi_{j},\psi_{j}\in\mathfrak{M}(\mathcal{H}),\quad\operatorname{Ker}M_{\psi_{j}}^{*}=\\{0\\}.$

Since

$f_{1}+f_{2}=\frac{\varphi_{1}\psi_{2}+\varphi_{2}\psi_{1}}{\psi_{1}\psi_{2}}$

and $\mathfrak{M}(\mathcal{H})$ is an algebra, it follows that
$\varphi_{1}\psi_{2}+\varphi_{2}\psi_{1}$ and
$\psi_{1}\psi_{2}\in\mathfrak{M}(\mathcal{H})$. Furthermore, since
$M_{\psi_{1}\psi_{2}}^{*}=M_{\psi_{2}}^{*}M_{\psi_{1}}^{*}$ is injective, we
see that $f_{1}+f_{2}\in\mathcal{N}^{+}(\mathcal{H})$. The proof that
$f_{1}f_{2}\in\mathcal{N}^{+}(\mathcal{H})$ is done in a similar way. ∎

While the following lemma might appear quite obvious, it is important to be
stated here since the complete Nevanlinna-Pick property, which plays quite
some role in connection with the Smirnov property, heavily depends on the norm
of the space whereas the Smirnov property does not.

######  Lemma 2.2.

Suppose $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ are reproducing kernel Hilbert
spaces of analytic functions on $\mathbb{D}$ that are equal as sets. Then
$\mathcal{N}^{+}(\mathcal{H}_{1})=\mathcal{N}^{+}(\mathcal{H}_{2})$.

###### Proof.

The closed graph theorem shows that the norms on $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ are equivalent in that there are positive constants
$c_{1},c_{2}$ such that

$c_{1}\|f\|_{\mathcal{H}_{1}}\leqslant\|f\|_{\mathcal{H}_{2}}\leqslant
c_{2}\|f\|_{\mathcal{H}_{1}},\quad f\in\mathcal{H}_{1}=\mathcal{H}_{2}.$

Thus $\mathfrak{M}(\mathcal{H}_{1})=\mathfrak{M}(\mathcal{H}_{2})$, and, for a
multiplier $\varphi$, it follows that $\varphi\mathcal{H}_{1}$ is dense in
$\mathcal{H}_{1}$ if and only if $\varphi\mathcal{H}_{2}$ is dense in
$\mathcal{H}_{2}$. Thus
$\mathcal{N}^{+}(\mathcal{H}_{1})=\mathcal{N}^{+}(\mathcal{H}_{2})$. ∎

##  3\. Some reminders about $\mathscr{H}(b)$ spaces

Here are some of the basics of $\mathscr{H}(b)$ spaces needed for this paper.
The details can be found in [10, 11, 14]. For $b\in H^{\infty}_{1}$, the de
Branges–Rovnyak space $\mathscr{H}(b)$ is the reproducing kernel Hilbert space
induced by the positive definite kernel

$k^{b}_{\lambda}(z)=\frac{1-b(z)\overline{b(\lambda)}}{1-\overline{\lambda}z},\quad\lambda,z\in\mathbb{D}.$

More properties of the specific class of $\mathscr{H}(b)$ spaces that we will
consider are given in Proposition 3.3.

When $b$ is not an extreme point of $H^{\infty}_{1}$, which is equivalent to
the condition $\log(1-|b|)\in L^{1}(\mathbb{T})$ (which will always be the
case in this paper since $b$ is rational but not inner), then there exists a
unique (up to a unimodular scalar) outer $a\in H_{1}^{\infty}$ such that
$|a|^{2}+|b|^{2}=1$ almost everywhere on $\mathbb{T}$. The pair $(a,b)$ is
call a Pythagorean pair and $a$ is known as the Pythagorean mate to $b$. Below
are some facts from [6] about Pythagorean pairs.

######  Proposition 3.1.

Suppose $b\in H^{\infty}_{1}$ is rational but not inner. Then

  1. (a)

$a$ is also rational;

  2. (b)

$(a,b)$ form a corona pair in that

$\inf_{z\in\mathbb{D}}(|a(z)|+|b(z)|)>0.$

Two important classes of functions in $\mathscr{H}(b)$, where $b$ is not an
extreme point of $H^{\infty}_{1}$, are

$\mathscr{M}(a):=aH^{2}\quad\mbox{and}\quad\mathscr{M}(\overline{a}):=T_{\overline{a}}H^{2}.$

Here $T_{\overline{a}}$ is the co-analytic Toeplitz operator
$T_{\overline{a}}f=P_{+}(\overline{a}f)$ on $H^{2}$. We mention that $a$ is
assumed outer (guaranteeing uniqueness up to a unimodular scalar factor). We
will denote the zeros of $a$ on $\mathbb{T}$ by $\xi_{1},\ldots,\xi_{n}$ and
the corresponding multiplicity by $m_{j}$, and define

(3.2) $a_{1}(z)=\prod_{j=1}^{n}(z-\xi_{j})^{m_{j}},$

Since $a$ cannot vanish on $\mathbb{D}$ and zeros outside the closure of
$\mathbb{D}$ define invertible factors, we have
$\mathscr{M}(\overline{a})=\mathscr{M}(\overline{a_{1}})$ [9, Prop. 2.7]. The
proposition below is from [3, 6].

######  Proposition 3.3.

Suppose $b\in H^{\infty}_{1}$ is rational but not inner. Then

  1. (a)

$\mathscr{M}(a_{1})\subseteq\mathscr{M}(\overline{a_{1}})=\mathscr{H}(b)$;

  2. (b)

If $\xi_{1},\ldots,\xi_{n}$ are the zeros of $a_{1}$ on $\mathbb{T}$ with
corresponding multiplicities $m_{1},m_{2},\ldots,m_{n}$, then

(3.4)
$\mathscr{H}(b)=\mathscr{M}(\overline{a_{1}})=\Big{(}\prod_{j=1}^{n}(z-\xi_{j})^{m_{j}}\Big{)}H^{2}\dotplus\mathscr{P}_{N-1},$

where $N=m_{1}+m_{2}+\cdots+m_{n}$ and $\mathscr{P}_{N-1}$ are the polynomials
whose degree is at most $N-1$, and $\dotplus$ signifies the algebraic direct
sum.

  3. (c)

An equivalent norm on $\mathscr{H}(b)$ is given by

(3.5) $\|a_{1}g+q\|^{2}=\|g\|_{H^{2}}^{2}+\|q\|_{H^{2}}^{2}.$

where $f=a_{1}g+q$ is the decomposition given by (3.4).

We will prove that $\mathscr{H}(b)\subseteq\mathcal{N}^{+}(\mathscr{H}(b))$
(Theorem 1.1) by using the fact that
$\mathscr{H}(b)=\mathscr{M}(\overline{a_{1}})$ (as sets with equivalent norms)
and then employing Lemma 2.2. So we will now focus on proving the following
result.

######  Theorem 3.6.

If $a_{1}$ is a polynomial whose zeros are contained in $\mathbb{T}$, then
$\mathscr{M}(\overline{a_{1}})\subseteq\mathcal{N}^{+}(\mathscr{M}(\overline{a_{1}}))$.

So far, we can readily identify the contents of
$\mathscr{M}(\overline{a_{1}})$. We can also readily identify the multipliers.
Recall that an analytic function $\varphi$ is a multiplier of
$\mathscr{M}(\overline{a_{1}})$ if
$\varphi\mathscr{M}(\overline{a_{1}})\subseteq\mathscr{M}(\overline{a_{1}})$.
The multiplier algebra for $\mathscr{M}(\overline{a_{1}})$ is denoted by
$\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))$. As mentioned previously,
$\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))\subseteq
H^{\infty}\cap\mathscr{M}(\overline{a_{1}})$ but equality does not always
hold. The multiplier algebra for general $\mathscr{H}(b)$ spaces has been
studied in [4, 12, 13] and is often a complicated class of functions that is
related to the invertibility of the product of certain Hankel operators and
their adjoints [12]. Fortunately, in our case, they are particularly simple
[9].

######  Proposition 3.7.

If $a_{1}$ is a polynomial of degree $N$ whose zeros lie in $\mathbb{T}$, then

$\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))=\mathscr{M}(\overline{a_{1}})\cap
H^{\infty}.$

In particular, $\varphi\in\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))$ if and
only if $\varphi=a_{1}\widetilde{\varphi}+r$, where $\widetilde{\varphi}\in
H^{2}$, $r\in\mathscr{P}_{N-1}$, and $a_{1}\widetilde{\varphi}\in H^{\infty}$.

The Smirnov class of $\mathscr{M}(\overline{a_{1}})$ is then

$\mathcal{N}^{+}(\mathscr{M}(\overline{a_{1}}))=\Big{\\{}\frac{\varphi}{\psi}:\varphi,\psi\in\mathfrak{M}(\mathscr{M}(\overline{a_{1}})),\overline{\psi\mathscr{M}(\overline{a_{1}})}=\mathscr{M}(\overline{a_{1}})\Big{\\}}.$

Proposition 3.7 describes $\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))$. To
describe $\mathcal{N}^{+}(\mathscr{M}(\overline{a_{1}}))$, it remains to
characterize the cyclic vectors for the shift operator on
$\mathscr{M}(\overline{a_{1}})$. This is done in the following. We remind the
reader that $a_{1}$ is a polynomial of the form in (3.2).

######  Proposition 3.8.

Let $a_{1}$ be a polynomial of degree $N$ whose zeros lie in $\mathbb{T}$ and
let
$\psi=a_{1}\widetilde{\psi}+r\in\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))$,
where $\widetilde{\psi}\in H^{2}$, with $a_{1}\widetilde{\psi}\in H^{\infty}$,
and $r\in\mathscr{P}_{N-1}$. Then the following are equivalent:

  1. (a)

$\psi\mathscr{M}(\overline{a_{1}})$ is dense in
$\mathscr{M}(\overline{a_{1}})$;

  2. (b)

$\psi$ is outer and $\operatorname{gcd}(a_{1},r)=1$.

###### Proof.

(b) $\Rightarrow$ (a): Assume $\psi$ is outer and
$\operatorname{gcd}(a_{1},r)=1$. In view of (3.5), we use the following inner
product on $\mathscr{M}(\overline{a_{1}})$:

$\langle a_{1}h_{1}+q_{1},a_{1}h_{2}+q_{2}\rangle=\langle
h_{1},h_{2}\rangle_{H^{2}}+\langle q_{1},q_{2}\rangle_{H^{2}}.$

Let $f\in\mathscr{M}(\overline{a_{1}})$ such that
$f\perp\psi\mathscr{M}(\overline{a_{1}})$, that is,

(3.9) $\langle f,\psi g\rangle=0,\quad g\in\mathscr{M}(\overline{a_{1}}).$

Writing $f=a_{1}\widetilde{f}+p$ and applying (3.9) to $g=a_{1}h$ for every
$h\in H^{2}$ we get $0=\langle a_{1}\widetilde{f}+p,a_{1}\psi
h\rangle=\langle\widetilde{f},\psi h\rangle_{H^{2}}.$ Since $\psi$ is outer,
we have $\overline{\psi H^{2}}=H^{2}$ and hence $\widetilde{f}\equiv 0$. It
remains to check that $p\equiv 0$. Since $\operatorname{gcd}(a_{1},r)=1$,
apply the Euclidean algorithm to the ring of polynomials to obtain polynomials
$q$ and $s$ such that $qr=a_{1}s+1.$ Now apply (3.9) to
$g=pq\in\mathscr{M}(\overline{a_{1}})$ (recall that
$\mathscr{M}(\overline{a_{1}})$ contains the polynomials). Then

$0=\langle p,\psi pq\rangle=\langle p,a_{1}pq\widetilde{\psi}+pqr\rangle.$

But note that

$pqr=p(a_{1}s+1)=pa_{1}s+p$

and $p\perp a_{1}pq\widetilde{\psi}+a_{1}ps$. Hence, $0=\langle
p,p\rangle_{H^{2}}$ which yields $p\equiv 0$.

(a) $\Rightarrow$ (b): Assume that $\psi\mathscr{M}(\overline{a_{1}})$ is
dense in $\mathscr{M}(\overline{a_{1}})$. Since
$\mathscr{M}(\overline{a_{1}})$ embeds boundedly into $H^{2}$ and polynomials
belong to $\mathscr{M}(\overline{a_{1}})$ we have

$\mathbb{C}[z]\subseteq\mathscr{M}(\overline{a_{1}})=\overline{\psi\mathscr{M}(\overline{a_{1}})}^{\mathscr{M}(\overline{a_{1}})}\subseteq\overline{\psi\mathscr{M}(\overline{a_{1}})}^{H^{2}}\subseteq\overline{\psi
H^{2}}^{H^{2}}$

and so $\overline{\psi H^{2}}=H^{2}$. By Beurling’s theorem [5], $\psi$ is
outer. Now we need to check that $\operatorname{gcd}(a_{1},r)=1$. It follows
from (3.4) (see [8, Cor. 4.11] for the details) that the radial limit of every
function in $\mathscr{M}(\overline{a_{1}})$ exists at each $\xi_{j}$ (the
zeros of the polynomial $a_{1}$) and

$|f(\xi_{j})|\leqslant C_{j}\|f\|,\quad f\in\mathscr{M}(\overline{a_{1}}).$

Using our assumption again that
$\overline{\psi\mathscr{M}(\overline{a_{1}})}=\mathscr{M}(\overline{a_{1}})$
we produce a sequence $g_{n}\mathscr{M}(\overline{a_{1}})$ such that $\|1-\psi
g_{n}\|\to 0$ as $n\to\infty$. If $\psi(\xi_{j})=0$ then $1\leqslant
C_{j}\|1-\psi p_{n}\|\to 0$ which is an obvious contradiction. Thus
$\psi(\xi_{j})\not=0$ for every $1\leqslant j\leqslant n$. From the
representation $\psi=a_{1}\widetilde{\psi}+r$ we see that $r(\xi_{j})\not=0$
for all $j$. This means that the zeros of $r$ do not meet the zeros of $a_{1}$
and so $\operatorname{gcd}(a_{1},r)=1$. ∎

##  4\. $\mathscr{H}(b)$ is contained in its Smirnov class

In this section, we give the proof of Theorem 1.2 and Theorem 1.1.

### Proof of Theorem 1.2

For $h\in H^{2}$, set

(4.1)
$F(z)=2\,p(z)\int_{-\pi}^{\pi}\frac{e^{it}}{e^{it}-z}\,\frac{|p(e^{it})|}{p(e^{it})}|h(e^{it})|\frac{dt}{2\pi},\quad
z\in\mathbb{D}.$

A partial fraction decomposition with respect to the variable $e^{it}$ yields

(4.2)
$\frac{e^{it}p(z)}{(e^{it}-z)p(e^{it})}=\frac{z}{e^{it}-z}+\sum_{k=1}^{n}\sum_{l=1}^{m_{k}}\frac{p_{k,l}(z)}{(\xi_{k}-e^{it})^{l}},$

where $p_{k,l}(z)$ are polynomials.

Since $2z=e^{it}+z-(e^{it}-z)$, and with the above decomposition in mind, we
get

$\displaystyle F(z)$
$\displaystyle=\int_{-\pi}^{\pi}\frac{e^{it}+z}{e^{it}-z}\,|p(e^{it})h(e^{it})|\frac{dt}{2\pi}-\int_{-\pi}^{\pi}|p(e^{it})h(e^{it})|\frac{dt}{2\pi}$
$\displaystyle\quad+2\sum_{k=1}^{n}\sum_{l=1}^{m_{k}}{p_{k,l}(z)}\,\int_{-\pi}^{\pi}\frac{|p(e^{it})|}{(\xi_{k}-e^{it})^{l}}|h(e^{it})|\frac{dt}{2\pi}.$

Note that $|p(e^{it})|/|\xi_{k}-e^{it}|^{l}$ is bounded, and so all the
integrals appearing in the above double sum are well defined coefficients. As
a result we can write

$F(z)=\underbrace{\int_{-\pi}^{\pi}\frac{e^{it}+z}{e^{it}-z}|p(e^{it})h(e^{it})|\frac{dt}{2\pi}}_{F_{0}(z)}+q(z),$

where $q$ is a polynomial. Hence $F=F_{0}+q.$ Since

$\Re\Big{(}\frac{e^{it}+z}{e^{it}-z}\Big{)}=\frac{1-|z|^{2}}{|e^{it}-z|^{2}}>0,$

$F_{0}$ has positive real part. We also let $Q=1-q$ so that

$F+Q=F_{0}+q+Q=F_{0}+1.$

Setting

$\alpha=\frac{Qh}{F_{0}+1},\quad\beta=\frac{F}{F_{0}+1},$

we get

$\frac{\alpha}{1-\beta}=\frac{Qh/(F_{0}+1)}{1-F/(F_{0}+1)}=\frac{qh}{F_{0}+1-F}=\frac{Qh}{Q}=h.$

It remains to show that

$\frac{\alpha}{1-\beta}$

is of the form (1.3).

Since $\Re F_{0}>0$ it follows that $|F_{0}+1|>1$. Since $Q$ is a polynomial
and $h\in H^{2}$, we have $\alpha\in H^{2}$. For the boundedness of $p\alpha$
it suffices to check that $ph/(F_{0}+1)$ is bounded. From the preceding
discussions it is clear that $\Re F_{0}$ is the Poisson extension of $|ph|$,
and so

$\frac{|ph|}{|F_{0}+1|}\leqslant\frac{|ph|}{\Re
F_{0}+1}=\frac{|ph|}{|ph|+1}\leqslant 1.$

The function $\beta$ is bounded by the following argument

$|\beta|=\frac{|F|}{|F_{0}+1|}=\frac{|F_{0}+p|}{|F_{0}+1|}\leqslant\frac{|F_{0}|}{|F_{0}+1|}+\frac{|p|}{|F_{0}+1|},$

and since $\Re F_{0}>0$ both terms are bounded. From (4.1) the function $F$ is
$p$ times the Cauchy transform of an $L^{2}$ function and thus $F=pg$, where
$g\in H^{2}$. Hence

$\beta=\frac{pg}{F_{0}+1}\in pH^{2}.$

Next we need to verify that

$1-\beta=1-F/(F_{0}+1)=\frac{F_{0}+1-F}{F_{0}+1}=\frac{Q}{F_{0}+1}$

is outer. This is equivalent to saying that $q$ does not vanish on
$\mathbb{D}$. Since $Q=1-q$, and the coefficients in $p$ are homogeneous in
$h$, we can apply a rescaling argument on $h$ to make $p$ small and hence $Q$
is zero-free in $\mathbb{D}$. Thus we have checked that $h=\alpha/(1-\beta)$
satisfies the properties of (1.3).

### Proof of Theorem 1.1

Observe that we have reduced this to proving Theorem 3.6.

Since $\mathscr{P}_{N-1}\subseteq\mathscr{M}(\overline{a_{1}})\cap
H^{\infty}=\mathfrak{M}(\mathscr{M}(\overline{a_{1}}))$ and since
$1\cdot\mathscr{M}(\overline{a_{1}})$ is trivially dense in
$\mathscr{M}(\overline{a_{1}})$, then
$p=p/1\in\mathcal{N}^{+}(\mathscr{M}(\overline{a_{1}}))$ and the polynomial
part of $\mathscr{M}(\overline{a_{1}})$ can be written as required.. Thus by
Lemma 2.1 we just need to show that

$a_{1}H^{2}\subseteq\mathcal{N}^{+}(\mathscr{M}(\overline{a_{1}})).$

By Proposition 3.8 this is equivalent to showing that $a_{1}H^{2}$ is a subset
of the class of functions

$\frac{a_{1}f_{1}+r_{1}}{a_{1}f_{2}+r_{2}},$

where $f_{1},f_{2}\in H^{2}$ with $a_{1}f_{1},a_{1}f_{1}\in H^{\infty}$,
$r_{1},r_{2}\in\mathscr{P}_{N-1}$, $a_{1}f_{2}+r_{2}$ is outer, and
$\operatorname{gcd}(a_{1},r_{2})=1$. But this follows from Theorem 1.2. Note
that we can choose $r_{1}=0$ and $r_{2}=1$.

We end with the following generalization of Theorem 1.1.

######  Corollary 4.3.

If $b\in H^{\infty}_{1}$ is rational and outer and $r>0$ then
$\mathscr{H}(b^{r})\subseteq\mathcal{N}^{+}(\mathscr{H}(b^{r}))$.

###### Proof.

Note that $\mathscr{H}(b^{r})=\mathscr{H}(b)$ [6, Thm. 1.6] and now use
Theorem 1.1 along with Lemma 2.2. ∎

######  Remark 4.4.

Note that Theorem 3.6 is true for $\mathcal{M}(\bar{\psi})$, where
$\psi=a_{1}\varphi$ and $\varphi$ is any $H^{\infty}$ outer function that is
invertible in $H^{\infty}$.

## References

  * [1] Alexandru Aleman, Michael Hartz, John E. McCarthy, and Stefan Richter.  The Smirnov class for spaces with the complete Pick property.  J. Lond. Math. Soc. (2), 96(1):228–242, 2017.
  * [2] Cheng Chu.  Which de Branges-Rovnyak spaces have complete Nevanlinna-Pick property?  J. Funct. Anal., 279(6):108608, 15, 2020.
  * [3] Constantin Costara and Thomas Ransford.  Which de Branges-Rovnyak spaces are Dirichlet spaces (and vice versa)?  J. Funct. Anal., 265(12):3204–3218, 2013.
  * [4] B. Mark Davis and John E. McCarthy.  Multipliers of de Branges spaces.  Michigan Math. J., 38(2):225–240, 1991.
  * [5] P. L. Duren.  Theory of ${H}^{p}$ spaces.  Academic Press, New York, 1970.
  * [6] Emmanuel Fricain, Andreas Hartmann, and William T. Ross.  Concrete examples of $\mathscr{H}(b)$ spaces.  Comput. Methods Funct. Theory, 16(2):287–306, 2016.
  * [7] Emmanuel Fricain, Andreas Hartmann, and William T. Ross.  Multipliers between model spaces.  Studia Math., 240(2):177–191, 2018.
  * [8] Emmanuel Fricain, Andreas Hartmann, and William T. Ross.  Range spaces of co-analytic Toeplitz operators.  Canad. J. Math., 70(6):1261–1283, 2018.
  * [9] Emmanuel Fricain, Andreas Hartmann, and William T. Ross.  Multipliers between range spaces of co-analytic Toeplitz operators.  Acta Sci. Math. (Szeged), 85(1-2):215–230, 2019.
  * [10] Emmanuel Fricain and Javad Mashreghi.  The theory of $\mathcal{H}(b)$ spaces. Vol. 1, volume 20 of New Mathematical Monographs.  Cambridge University Press, Cambridge, 2016.
  * [11] Emmanuel Fricain and Javad Mashreghi.  The theory of $\mathcal{H}(b)$ spaces. Vol. 2, volume 21 of New Mathematical Monographs.  Cambridge University Press, Cambridge, 2016.
  * [12] B. A. Lotto and Donald Sarason.  Multipliers of de Branges-Rovnyak spaces.  Indiana Univ. Math. J., 42(3):907–920, 1993.
  * [13] Benjamin A. Lotto and Donald Sarason.  Multipliers of de Branges-Rovnyak spaces. II.  In Harmonic analysis and hypergroups (Delhi, 1995), Trends Math., pages 51–58. Birkhäuser Boston, Boston, MA, 1998.
  * [14] D. Sarason.  Sub-Hardy Hilbert spaces in the unit disk.  University of Arkansas Lecture Notes in the Mathematical Sciences, 10\. John Wiley & Sons Inc., New York, 1994.  A Wiley-Interscience Publication.
  * [15] Serguei Shimorin.  Complete Nevanlinna-Pick property of Dirichlet-type spaces.  J. Funct. Anal., 191(2):276–296, 2002.