# Indecomposable algebraic integers

D. M. H. van Gent  Mathematisch Instituut, Universiteit Leiden, The
Netherlands<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
  2. 2 Inner products and Hilbert spaces
  3. 3 Hilbert lattices
  4. 4 Decompositions
  5. 5 Hilbert lattice decompositions
  6. 6 Voronoi cells
  7. 7 The lattice of algebraic integers
  8. 8 Indecomposable algebraic integers
  9. 9 Decompositions of the lattice of algebraic integers
  10. 10 Universal gradings
  11. 11 Enumeration of indecomposable algebraic integers of degree 2
  12. 12 Geometry of numbers
  13. 13 Szegő capacity theory
  14. 14 Computing decompositions of algebraic integers of large square-norm
  15. 15 Bounds on indecomposable algebraic integers
  16. 16 Fekete capacity theory
  17. 17 Reduction to exponentially bounded polynomials
  18. 18 Volume computation
  19. 19 Proof of the main theorem
  20. 20 Remarks on the proof of the main theorem
  21. 21 Computing decompositions of algebraic integers of small square-norm
  22. 22 Working out an example
  23. 23 Enumeration of indecomposable algebraic integers of degree 3

##  1\. Introduction

In number theory, in particular the theory of the geometry of numbers, one
equips an order $R$, and by extension its field of fractions, with an inner
product, turning $R$ into a lattice in $\mathbb{R}\otimes_{\mathbb{Z}}R$.
After normalizing this inner product, we may define it on an algebraic closure
$\overline{\mathbb{Q}}$ of $\mathbb{Q}$ as

$\langle\alpha,\beta\rangle=\frac{1}{[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]}\sum_{\sigma:\mathbb{Q}(\alpha,\beta)\to\mathbb{C}}\sigma(\alpha)\cdot\overline{\sigma(\beta)},$

where the sum ranges over all field embeddings of $\mathbb{Q}(\alpha,\beta)$
in $\mathbb{C}$. We write $\overline{\mathbb{Z}}$ for the ring of integers of
$\overline{}\mathbb{Q}$ and we call its elements the algebraic integers.
Although $\overline{\mathbb{Z}}$ is not of finite rank, we may still
meaningfully call it a lattice in a more general sense: We show it is a
discrete subgroup of a Hilbert space (Theorem 7.9). In this document we prove
some basic facts about such ‘Hilbert lattices’, but we are mainly concerned
with $\overline{\mathbb{Z}}$.

A decomposition of an element $\alpha$ of a Hilbert lattice $\Lambda$ is a
pair $(\beta,\gamma)\in\Lambda^{2}$ such that $\alpha=\beta+\gamma$ and
$\langle\beta,\gamma\rangle\geq 0$. If $\alpha$ is non-zero and has only the
trivial decompositions $(\alpha,0)$ and $(0,\alpha)$, then $\alpha$ is called
indecomposable, or Voronoi-relevant. Indecomposable elements are of interest
because they determine the Voronoi polyhedron (Theorem 6.11), and can be used
to detect whether $\Lambda$ is indecomposable (Theorem 5.4), meaning $\Lambda$
is non-zero and cannot be written as a non-trivial orthogonal sum of two
sublattices. Moreover, indecomposable elements can be used in the context of
gradings of orders [12]. We prove the following theorem in Section 9.

######  Theorem 1.1.

The Hilbert lattice of algebraic integers is indecomposable.

Our main goal is to determine the indecomposable algebraic integers. Formally
we ask the question: Does there exist an algorithm that, given a non-zero
algebraic integer, decides whether it is indecomposable and if not produces a
non-trivial decomposition? We give a partial answer. The following is a result
derived from classical capacity theory by T. Chinburg, for which we give a
proof in Section 13.

######  Theorem 1.2.

If $\alpha\in\overline{\mathbb{Z}}$ satisfies $4<\langle\alpha,\alpha\rangle$,
then $\alpha$ has infinitely many decompositions.

The proof of this theorem is constructive and we are able to derive an
algorithm to compute arbitrarily many non-trivial decompositions in this case.
A consequence of Theorem 1.2 is that for every $n\in\mathbb{Z}_{>0}$ there
exist only finitely many indecomposable algebraic integers of degree $n$. We
have effective upper and lower bounds, which are proven in Section 15.

######  Theorem 1.3.

There are least $\exp(\frac{1}{4}(\log 2)n^{2}+O(n\log n))$ and at most
$\exp(\frac{1}{2}(1+\log 2)n^{2}+O(n\log n))$ indecomposable algebraic
integers of degree up to $n$.

Our main result is the following, which we prove in Section 19.

######  Theorem 1.4.

If $\alpha\in\overline{\mathbb{Z}}$ satisfies
$\langle\alpha,\alpha\rangle<2\sqrt{\textup{e}}\approx 3.297$, then $\alpha$
has only finitely many decompositions.

The proof is inspired by capacity theory. Although the proof is not
constructive, we are able to construct an algorithm to compute all
decompositions which is provably correct for
$\langle\alpha,\alpha\rangle<2\sqrt{\textup{e}}$. We use it in Section 23 to
compute decompositions for some algebraic integers of degree 3. We derive the
following numerical results.

######  Theorem 1.5.

There are exactly $2$ indecomposable algebraic integers of degree $1$, there
are exactly $14$ of degree $2$, and there are at least $354$ and at most $588$
of degree $3$.

Theorem 1.2 and Theorem 1.4 combined still leave a gap between
$2\sqrt{\textup{e}}$ and $4$. It is still an open problem what happens in this
case. We would also like to know what the supremum of
$\langle\alpha,\alpha\rangle$ is, where $\alpha$ ranges over the
indecomposable algebraic integers. Theorem 1.2 gives an upper bound of $4$. We
prove the following lower bound in Section 8.

######  Theorem 1.6.

It holds that
$2\sqrt{2}\leq\sup\\{\langle\alpha,\alpha\rangle\,|\,\textup{$\alpha\in\overline{\mathbb{Z}}$
is indecomposable}\\}$.

A grading of a ring $R$ is a pair $(G,\\{R_{g}\\}_{g\in G})$ such that $G$ is
an abelian group, $R_{g}$ is a subgroup of $R$ for all $g\in G$ such that the
natural map $\bigoplus_{g\in G}R_{g}\to R$ is an isomorphism of groups, and
such that for all $g,g^{\prime}\in G$ we have $R_{g}R_{g^{\prime}}\subseteq
R_{gg^{\prime}}$. We say a grading $(G,\\{R_{g}\\}_{g})$ of a ring $R$ is
universal if for each grading $(H,\\{S_{h}\\}_{h})$ of $R$ there exists a
unique group homomorphism $f:G\to H$ such that $S_{h}=\sum_{g\in
f^{-1}h}R_{g}$ for all $h\in H$. Using the theory of Hilbert lattices we
generalize two results of Lenstra and Silverberg [12] in Section 10.

######  Theorem 1.7.

Every subring of $\overline{\mathbb{Z}}$ has a universal grading.

The universal grading of $\overline{\mathbb{Z}}$ is the trivial grading, which
is a consequence of Theorem 1.1.

######  Theorem 1.8.

Every integrally closed subring of $\overline{\mathbb{Z}}$ has a universal
grading with a subgroup of $\mathbb{Q}/\mathbb{Z}$.

We also show that every subgroup of $\mathbb{Q}/\mathbb{Z}$ occurs for some
integrally closed subring of $\overline{\mathbb{Z}}$.

The structure of this document is as follows. In sections 2 through 6.13 we
develop general theory on Hilbert lattices. We specialize to the Hilbert
lattice $\overline{\mathbb{Z}}$ in sections 7 through 8. In sections 9 through
10.10 we treat some applications of the theory. In section 11 and 23 we do
some explicit computations on indecomposable integers of degree 2 respectively
3. Sections 12 through 15 we dedicate to the proof of Theorem 1.2 and its
algorithmic counterpart. In sections 16 through 22 we prove Theorem 1.4 and
its algorithmic counterpart.

##  2\. Inner products and Hilbert spaces

######  Definition 2.1.

Let $R\subseteq\mathbb{C}$ be a subring. An $R$-norm on an $R$-module $M$ is a
map $\|\cdot\|:M\to\mathbb{R}_{\geq 0}$ that satisfies:

  * (Absolute homogeneity)

For all $x\in M$ and $a\in R$ we have $\|ax\|=|a|\cdot\|x\|$;

  * (Triangle inequality)

For all $x,y\in M$ we have $\|x+y\|\leq\|x\|+\|y\|$;

  * (Positive-definiteness)

For all non-zero $x\in M$ we have $\|x\|\in\mathbb{R}_{>0}$.

A normed $R$-module is an $R$-module $M$ together with an $R$-norm on $M$. For
normed $R$-modules $M$ and $N$ an $R$-module homomorphism $f:M\to N$ is called
an isometric map if $\|x\|=\|f(x)\|$ for all $x\in M$. The isometric maps are
the morphisms in the category of normed $R$-modules.

Note that an isometric map is injective, but not necessarily surjective.

######  Definition 2.2.

Let $R\subseteq\mathbb{C}$ be a subring and $M$ be an $R$-module. An $R$-inner
product on $M$ is a map $\langle\,\cdot\,,\,\cdot\,\rangle:M^{2}\to\mathbb{C}$
that satisfies:

  * (Conjugate symmetry)

For all $x,y\in M$ we have $\langle x,y\rangle=\overline{\langle y,x\rangle}$;

  * (Left linearity)

For all $x,y,z\in M$ and $a\in R$ we have $\langle x+ay,z\rangle=\langle
x,z\rangle+a\langle y,z\rangle$;

  * (Positive-definiteness)

For all non-zero $x\in M$ we have $\langle x,x\rangle\in\mathbb{R}_{>0}$.

We say it is a real inner product if $\langle M,M\rangle\subseteq\mathbb{R}$,
which implies $R\subseteq\mathbb{R}$ when $M\neq 0$. An $R$-inner product
space is an $R$-module together with an $R$-inner product.

######  Remark 2.3.

An $R$-inner product space $M$ comes with an $R$-norm given by
$\|x\|=\sqrt{\langle x,x\rangle}$, which in turn induces a metric
$d(x,y)=\|x-y\|$ and a topology. One can then speak about the completeness of
$M$ with respect to this metric.

######  Lemma 2.4.

Suppose $R\subseteq\mathbb{C}$ is a subring and $M$ is a real $R$-inner
product space. Then the induced norm satisfies the parallelogram law: For all
$x,y\in M$ we have

$\displaystyle\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}.\hfill\qed$

The following is an exercise in many standard texts.

######  Theorem 2.5 (Jordan–von Neumann [8]).

Let $R\subseteq\mathbb{Q}$ be a subring, $M$ an $R$-module and suppose a map
$\|\cdot\|:M\to\mathbb{R}_{\geq 0}$ satisfies positive-definiteness and the
parallelogram law. Then $\|\cdot\|$ is an $R$-norm on $M$ induced by a real
$R$-inner product $\langle\,\cdot\,,\,\cdot\,\rangle:M^{2}\to\mathbb{R}$ given
by

$\langle x,y\rangle=\tfrac{1}{2}(\|x+y\|^{2}-\|x\|^{2}-\|y\|^{2}).$

###### Proof.

Note that taking $x=y=0$ in the parallelogram law shows
$2\|0\|^{2}=4\|0\|^{2}$, hence $\|0\|=0$. For all $x\in M$ we have
$\|x+x\|^{2}=2\|x\|^{2}+2\|x\|^{2}-\|x-x\|^{2}=4\|x\|^{2}$, hence $\langle
x,x\rangle=\tfrac{1}{2}(\|2x\|^{2}-2\|x\|^{2})=\|x\|^{2}$. It now suffices to
show that $\langle\,\cdot\,,\,\cdot\,\rangle$ is an inner product, as
$\|\cdot\|$ is then the associated norm as in Remark 2.3. Clearly
$\langle\,\cdot\,,\,\cdot\,\rangle$ satisfies conjugate symmetry and positive
definiteness, so it remains to prove left linearity. It suffices to show for
all $x\in M$ that $x\mapsto\langle x,z\rangle$ is $\mathbb{Z}$-linear: Since
$R$ is in the field of fractions of $\mathbb{Z}$, any $\mathbb{Z}$-linear map
to $\mathbb{R}$ is also $R$-linear. Let $x,y,z\in M$ and note that $\langle
x,y\rangle=\tfrac{1}{4}(\|x+y\|^{2}-\|x-y\|^{2})$. By the parallelogram law we
have

$\displaystyle
2\|y+z\|^{2}+2\|x\|^{2}-\|-x+y+z\|^{2}=\|x+y+z\|^{2}=2\|x+z\|^{2}+2\|y\|^{2}-\|x-y+z\|^{2}.$

so

(1) $\displaystyle
2\|x+y+z\|^{2}=2\|x+z\|^{2}+2\|y+z\|^{2}+2\|x\|^{2}+2\|y\|^{2}-\|-x+y+z\|^{2}-\|x-y+z\|^{2}.$

Applying (1) also with $z$ replaced by $-z$, we obtain

$\displaystyle 8\langle x+y,z\rangle$
$\displaystyle=2\|x+y+z\|^{2}-2\|x+y-z\|^{2}$
$\displaystyle=2\|x+z\|^{2}+2\|y+z\|^{2}-2\|x-z\|^{2}-2\|y-z\|^{2}$
$\displaystyle=8\langle x,z\rangle+8\langle y,z\rangle,$

as was to be shown. We conclude that $\langle\,\cdot\,,\,\cdot\,\rangle$ is in
fact an $R$-inner product. ∎

Inner product spaces over $\mathbb{Z}$ or $\mathbb{Q}$ can be extended to
$\mathbb{R}$ in a ‘canonical’ way. This can best be expressed in a categorical
sense in terms of universal morphisms. We proceed as in Chapter III of [13].

######  Definition 2.6.

Let $\mathcal{C}$ be a category. An object $U$ of $\mathcal{C}$ is called
universal if for each object $X$ of $\mathcal{C}$ there exists a unique
morphism $U\to X$ in $\mathcal{C}$.

######  Definition 2.7.

Let $\mathcal{C}$ and $\mathcal{D}$ be categories. Let
$F:\mathcal{C}\to\mathcal{D}$ be a functor and $Z$ an object of $\mathcal{D}$.
For $X$ an object of $\mathcal{C}$ we say
$\eta\in\operatorname{Hom}_{\mathcal{D}}(Z,F(X))$ is a universal morphism from
$Z$ to $F$ if for all objects $Y$ of $\mathcal{C}$ and every
$g\in\operatorname{Hom}_{\mathcal{D}}(Z,F(Y))$ there exists a unique
$f\in\operatorname{Hom}_{\mathcal{D}}(X,Y)$ such that $F(f)\circ\eta=g$.
Equivalently, for all objects $Y$ of $\mathcal{C}$ and morphisms $g:Z\to F(Y)$
we have the following diagram:

${Z}$${F(X)}$${X}$${F(Y)}$${Y}$$\scriptstyle{\eta}$$\scriptstyle{g}$$\scriptstyle{F(f)}$$\scriptstyle{!f}$

For a reader familiar with category theory we remark that, if for a functor
$F:\mathcal{C}\to\mathcal{D}$ every object $Z$ of $\mathcal{D}$ has a
universal morphism, then $F$ is a right adjoint functor. For a reader less
familiar with category theory we leave it as an exercise to construct, given a
functor $F:\mathcal{C}\to\mathcal{D}$ and an object $Z$ of $\mathcal{D}$, a
category for which the universal objects naturally correspond to the universal
morphisms from $Z$ to $F$, justifying the name universal morphism.

######  Example 2.8.

We will give a concrete example of a universal morphism.

(i) Consider the forgetful functor $F$ from the category of
$\mathbb{Q}$-vector spaces to the category of abelian groups, i.e. the functor
that sends a $\mathbb{Q}$-vector space to its underlying abelian group. Now
let $A$ be an abelian group. We take $V=\mathbb{Q}\otimes_{\mathbb{Z}}A$,
which is a $\mathbb{Q}$-vector space, and $\eta:A\to F(V)$ the map $z\mapsto
1\otimes z$. Because $F$ is a forgetful functor, as will always be the case in
our applications, we may omit it in the notation for simplicity and state that
$\eta$ is a morphism $A\to V$ of abelian groups. One can verify that $\eta$ is
a universal morphism.

Since $\eta$ is universal the vector space $V$ corresponding to $A$ is
‘uniquely unique’, meaning that any other other vector space $V^{\prime}$ with
a universal morphism $\eta^{\prime}:A\to V^{\prime}$ gives rise to a unique
isomorphism $\varphi:V\to V^{\prime}$ such that
$\varphi\circ\eta=\eta^{\prime}$.

Note that $\eta$ need not be injective, as this is only the case when $A$ is
torsion-free. Then $\eta$ can be thought of as a canonical embedding.

(ii) Similarly, we can consider a forgetful functor $F$ from the category of
$\mathbb{Q}$-inner product spaces to the category of $\mathbb{Z}$-inner
product spaces. The underlying $\mathbb{Q}$-vector space $V$ and map $\eta$
are the same as before, and we equip $V$ with the inner product we extend
$\mathbb{Q}$-linearly from $A$. To show that this inner product is positive
definite we use that $A$ is torsion-free, being a $\mathbb{Z}$-inner product
space.

######  Definition 2.9.

A Hilbert space is an $\mathbb{R}$-module $\mathcal{H}$ equipped with a real
$\mathbb{R}$-inner product such that $\mathcal{H}$ is complete with respect to
the induced metric. The morphisms of Hilbert spaces are the isometric maps.

######  Theorem 2.10 (Theorem 3.2-3 in [10]).

Let $F$ be the forgetful functor from the category of Hilbert spaces to the
category of $\mathbb{Q}$-inner product spaces. Then every $\mathbb{Q}$-inner
product space $V$ has an injective universal morphism to $F$, and a morphism
$f:V\to\mathcal{H}$ for some Hilbert space $\mathcal{H}$ is universal
precisely when the image of $f$ is dense in $\mathcal{H}$. ∎

The Hilbert space constructed for $V$ in Theorem 2.10 can be obtained as the
topological completion of $V$, similar to how $\mathbb{R}$ is constructed from
$\mathbb{Q}$ using Cauchy sequences.

######  Definition 2.11.

For a set $B$ and $p\in\mathbb{R}_{>0}$ we define the $\mathbb{R}$-vector
space

$\ell^{p}(B)=\Big{\\{}(x_{b})_{b}\in\mathbb{R}^{B}\,\Big{|}\,x_{b}=0\text{ for
all but countably many $b\in B$ and }\sum_{b\in
B}|x_{b}|^{p}<\infty\Big{\\}}\vspace{-.7em}$

and $\|x\|_{p}=(\sum_{b\in B}|x_{b}|^{p})^{1/p}$ for all
$x=(x_{b})_{b}\in\ell^{p}(B)$.

######  Theorem 2.12 (Minkowski’s inequality, Theorem 1.2-3 in [10]).

For any set $B$ and $p\in\mathbb{R}_{\geq 1}$ the map $\|\cdot\|_{p}$ is an
$\mathbb{R}$-norm on $\ell^{p}(B)$. ∎

######  Lemma 2.13 (Example 3.1-6 in [10]).

For any set $B$ the space $\ell^{2}(B)$ is a Hilbert space with inner product
given by $\langle x,y\rangle=\sum_{b\in B}x_{b}\cdot y_{b}$ for
$x=(x_{b})_{b},y=(y_{b})_{b}\in\ell^{2}(B)$, such that $\langle
x,x\rangle=\|x\|_{2}^{2}$.∎

######  Lemma 2.14.

Let $n\in\mathbb{Z}_{\geq 1}$, $x\in\mathbb{R}^{n}$ and let $0<p\leq q$ be
real. Then we have

$\|x\|_{q}\leq\|x\|_{p}\quad\text{and}\quad n^{-1/p}\cdot\|x\|_{p}\leq
n^{-1/q}\cdot\|x\|_{q}.$

###### Proof.

Clearly we may assume $x\neq 0$. For the first inequality, consider
$y=x/\|x\|_{p}$. Then $|y_{i}|\leq 1$ for all $i$, from which
$|y_{i}|^{q}\leq|y_{i}|^{p}$ follows. Now

$\|y\|_{q}^{q}=\sum_{i=1}^{n}|y_{i}|^{q}\leq\sum_{i=1}^{n}|y_{i}|^{p}=\|y\|_{p}^{p}=1.$

Hence $\|x\|_{q}/\|x\|_{p}=\|y\|_{q}\leq 1$, as was to be shown. For the
second inequality, note that $x\mapsto x^{q/p}$ is a convex function on
$\mathbb{R}_{\geq 0}$. We have by Jensen’s inequality (Theorem 7.3 in [3])
that

$\|x\|_{q}^{q}=\sum_{i=1}^{n}|x_{i}|^{q}=\sum_{i=1}^{n}|x_{i}^{p}|^{q/p}\geq
n\left(\frac{1}{n}\sum_{i=1}^{n}|x_{i}|^{p}\right)^{q/p}=n^{1-q/p}\|x\|_{p}^{q},$

so $n^{-1/p}\cdot\|x\|_{p}\leq n^{-1/q}\cdot\|x\|_{q}$. ∎

######  Definition 2.15.

Let $\mathcal{H}$ be a Hilbert space. A subset $S\subseteq\mathcal{H}$ is
called orthogonal if $0\not\in S$ and $\langle x,y\rangle=0$ for all distinct
$x,y\in S$. The orthogonal dimension of $\mathcal{H}$, written
$\operatorname{orth\,dim}\mathcal{H}$, is the cardinality of a maximal
orthogonal subset of $\mathcal{H}$.

That the orthogonal dimension is well-defined, i.e. that maximal orthogonal
subsets of a given Hilbert space have the same cardinality follows from
Proposition 4.14 in [2].

######  Theorem 2.16 (Theorem 5.4 in [2]).

Let $\mathcal{H}$ be a Hilbert space and $B$ a set. Then the Hilbert spaces
$\mathcal{H}$ and $\ell^{2}(B)$ are isomorphic if and only if the cardinality
of $B$ equals $\operatorname{orth\,dim}\mathcal{H}$.

##  3\. Hilbert lattices

######  Definition 3.1.

A Hilbert lattice is an abelian group $\Lambda$ together with a map
$q:\Lambda\to\mathbb{R}$, which we then call the square-norm of $\Lambda$,
that satisfies:

  * (Parallelogram law)

For all $x,y\in\Lambda$ we have $q(x+y)+q(x-y)=2q(x)+2q(y)$;

  * (Positive packing radius)

There exists a $r\in\mathbb{R}_{>0}$ such that $q(x)\geq r$ for all non-zero
$x\in\Lambda$.

We write $\textup{P}(\Lambda)=\inf\\{q(x)\,|\,x\in\Lambda\setminus\\{0\\}\\}$.

The following lemma gives an equivalent definition of a Hilbert lattice.

######  Lemma 3.2.

A Hilbert lattice $\Lambda$ with square-norm $q$ is a discrete
$\mathbb{Z}$-inner product space with inner product given by
$(x,y)\mapsto\frac{1}{2}(q(x+y)-q(x)-q(y))$. Conversely, every discrete
$\mathbb{Z}$-inner product space $M$ is a Hilbert lattice with square norm
given by $x\mapsto\langle x,x\rangle$.

###### Proof.

The first statement is Theorem 2.5 with the observation that the positive
packing radius implies non-degeneracy and discreteness. The second statement
is Lemma 2.4 with the observation that discreteness implies a positive packing
radius. ∎

######  Example 3.3.

Consider for some $n\in\mathbb{Z}_{\geq 0}$ the vector space $\mathbb{R}^{n}$
with the standard inner product. If $\Lambda\subseteq\mathbb{R}^{n}$ is a
discrete subgroup, then $\Lambda$ is a Hilbert lattice when $q$ is given by
$x\mapsto\|x\|^{2}$.

######  Example 3.4.

Let $B$ be a set. Then
$\mathbb{Z}^{(B)}=\\{(x_{b})_{b}\in\mathbb{Z}^{B}\,|\,x_{b}=0\text{ for all
but finitely many $b$}\\}$ is a Hilbert lattice in $\ell^{2}(B)$ when $q$ is
given by $x\mapsto\|x\|^{2}$. In fact, any discrete subgroup of $\ell^{2}(B)$
is a Hilbert lattice.

######  Example 3.5.

The infimum defining $\textup{P}(\Lambda)$ of a Hilbert lattice $\Lambda$ need
not be attained. Certainly if $\Lambda=0$ we have that
$\textup{P}(\Lambda)=\infty$ is not attained. For an example of a non-
degenerate $\Lambda$ consider the following. For a set $I$ and a map
$f:I\to\mathbb{R}_{\geq 0}$ we write $\smash{\Lambda^{f}}$ for the group
$\mathbb{Z}^{(I)}$ together with the map $q((x_{i})_{i})=\sum_{i\in
I}f(i)^{2}x_{i}^{2}$. Note that
$\inf\\{q(x)\,|\,x\in\smash{\Lambda^{f}}\setminus\\{0\\}\\}=\inf\\{f(i)^{2}\,|\,i\in
I\\}$, so $\smash{\Lambda^{f}}$ is a Hilbert lattice if and only if
$\inf\\{f(i)\,|\,i\in I\\}>0$. We now simply take
$f:\mathbb{Z}_{>0}\to\mathbb{R}_{\geq 0}$ given by $n\mapsto 1+1/n$.

######  Lemma 3.6.

Let $\Lambda$ be a Hilbert lattice with square-norm $q$. Then any subgroup
$\Lambda^{\prime}\subseteq\Lambda$ is a Hilbert lattice when equipped with the
square-norm $q|_{\Lambda^{\prime}}$. ∎

######  Theorem 3.7.

Let $F$ be the forgetful functor from the category of Hilbert spaces to the
category of $\mathbb{Z}$-inner product spaces. Then every $\mathbb{Z}$-inner
product space $L$ has an injective universal morphism $\eta$ to $F$. For every
$\mathbb{Z}$-inner product space $L$, Hilbert space $\mathcal{H}$ and
injective morphism $f:L\to\mathcal{H}$ we have that $f$ is universal if and
only if $\mathbb{Q}\cdot f(L)$ is dense in $\mathcal{H}$, and $L$ is a Hilbert
lattice if and only if $f(L)$ is discrete in $\mathcal{H}$.

It follows from this theorem that the Hilbert lattices are, up to isomorphism,
precisely the discrete subgroups of Hilbert spaces. Hence Theorem 3.7 allows
us to assume without loss of generality that a Hilbert lattice is a discrete
subgroup of a Hilbert space.

###### Proof.

The first and second statement are just a combination of Example 2.8.ii and
Theorem 2.10, while the third is trivial when taking the equivalent definition
of Lemma 3.2. ∎

######  Remark 3.8.

Let $\Lambda$ be a Hilbert lattice in a Hilbert space $\mathcal{H}$ and
suppose that $\Lambda$ is finitely generated. Then $\mathbb{R}\Lambda$ is a
finite dimensional $\mathbb{R}$-inner product space and thus complete. It
follows that $\Lambda\to\mathbb{R}\Lambda$ is a universal morphism because
$\mathbb{Q}\Lambda$ is dense in $\mathbb{R}\Lambda$. Since $\mathbb{R}\Lambda$
is finite dimensional, $\Lambda$ is a lattice in the classical sense: a
discrete subgroup of a Euclidean vector space.

######  Lemma 3.9.

Let $\Lambda$ be a Hilbert lattice in a Hilbert space $\mathcal{H}$. Then the
natural map $\mathbb{R}\otimes_{\mathbb{Z}}\Lambda\to\mathcal{H}$ is
injective.

###### Proof.

To show $\mathbb{R}\otimes_{\mathbb{Z}}\Lambda\to\mathcal{H}$ is injective we
may assume by Lemma 3.6 without loss of generality that $\Lambda$ is finitely
generated, as any element in the kernel is also in
$\mathbb{R}\otimes_{\mathbb{Z}}\Lambda^{\prime}$ for some finitely generated
sublattice $\Lambda^{\prime}\subseteq\Lambda$. Write
$V=\mathbb{R}\Lambda\subseteq\mathcal{H}$. We may choose an $\mathbb{R}$-basis
for $V$ in $\Lambda$, and let $\Lambda^{\prime}$ be the group generated by
this basis. Then $\mathbb{R}\otimes_{\mathbb{Z}}\Lambda^{\prime}\to V$ is an
isomorphism. As $\Lambda$ is discrete in $V$, also $\Lambda/\Lambda^{\prime}$
is discrete in $V/\Lambda^{\prime}$. Now $V/\Lambda^{\prime}$ is compact, so
the quotient $\Lambda/\Lambda^{\prime}$ is finite. Then
$\Lambda\subseteq\tfrac{1}{n}\Lambda^{\prime}$, where
$n=\\#(\Lambda/\Lambda^{\prime})$. Now the natural map
$\mathbb{R}\otimes_{\mathbb{Z}}\Lambda\to\mathcal{H}$ is injective because it
is the composition of the map
$\mathbb{R}\otimes_{\mathbb{Z}}\Lambda\to\mathbb{R}\otimes_{\mathbb{Z}}(\tfrac{1}{n}\Lambda^{\prime})=\mathbb{R}\otimes_{\mathbb{Z}}\Lambda^{\prime}$,
which is injective since $\mathbb{R}$ is flat over $\mathbb{Z}$, and the map
$\mathbb{R}\otimes_{\mathbb{Z}}\Lambda^{\prime}\to V$, which is injective by
construction. ∎

######  Proposition 3.10.

Let $\Lambda$ be a Hilbert lattice in a Hilbert space $\mathcal{H}$ and
suppose $\Lambda$ is finitely generated as $\mathbb{Z}$-module. Then $\Lambda$
has a $\mathbb{Z}$-basis and any $\mathbb{Z}$-basis is $\mathbb{R}$-linearly
independent.

###### Proof.

Since $\Lambda$ is finitely generated and torsion free, it is clear that
$\Lambda$ is free. By Lemma 3.9, any $\mathbb{Z}$-linearly independent subset
of $\Lambda$ is $\mathbb{R}$-linearly independent. ∎

######  Proposition 3.11.

Suppose $\Lambda$ is a Hilbert lattice in a Hilbert space $\mathcal{H}$ and
let $\Lambda^{\prime}\subseteq\Lambda$ be a finitely generated subgroup. Let
$\pi:\mathcal{H}\to\mathcal{H}$ be the orthogonal projection onto the
orthogonal complement of $\Lambda^{\prime}$. Then for each $0\leq
t<\tfrac{1}{4}\textup{P}(\Lambda)$ there are only finitely many
$z\in\pi\Lambda$ such that $q(z)\leq t$, and $\pi\Lambda$ is a Hilbert
lattice.

###### Proof.

Suppose that $\pi\Lambda$ contains infinitely many points $z$ with $q(z)\leq
t$, or equivalently there exists some infinite set $S\subseteq\Lambda$ such
that $\pi|_{S}$ is injective and $q(\pi(x))\leq t$ for all $x\in S$. Consider
the map $\tau:\mathcal{H}\to\mathbb{R}\Lambda^{\prime}$, the complementary
projection to $\pi$. As $(\mathbb{R}\Lambda^{\prime})/\Lambda^{\prime}$ is
compact, there must exist distinct $x,y\in S$ such that
$q(\tau(x)-\tau(y)+w)<\textup{P}(\Lambda)-4t$ for some $w\in\Lambda^{\prime}$.
Then

$0<q(x-y+w)=q(\pi(x-y))+q(\tau(x-y)+w)<2\big{(}q(\pi(x))+q(\pi(y))\big{)}+\textup{P}(\Lambda)-4t\leq\textup{P}(\Lambda),$

a contradiction. Hence there are only finitely many $z\in\pi\Lambda$ such that
$q(z)\leq t$. To verify that $\pi\Lambda$ is a Hilbert lattice it suffices to
show that it is discrete in $\mathcal{H}$, which follows from the previous by
taking any non-zero value for $t$. ∎

######  Lemma 3.12.

Let $\Lambda$ be a Hilbert lattice which is finitely generated as
$\mathbb{Z}$-module and let $S\subseteq\Lambda$ be a set of vectors that forms
a basis for $\Lambda\cap(\mathbb{R}S)$. Then there exists a basis $B\supseteq
S$ of $\Lambda$.

###### Proof.

Let $\pi$ be the projection onto the orthogonal complement of $S$. Then
$\pi\Lambda$ is a Hilbert lattice by Proposition 3.11 with a basis $B_{\pi}$
by Proposition 3.10. Now choose for every $b_{\pi}\in B_{\pi}$ a lift
$b\in\Lambda$ and let $T$ be the set of those elements. It is easy to show
that $B=S\cup T$ is a basis of $\Lambda$. ∎

We say an abelian group is almost free if all its countable subgroups are
free.

######  Theorem 3.13.

If $\Lambda$ is a Hilbert lattice, then $\Lambda$ is almost free as an abelian
group.

###### Proof.

By Lemma 3.6 suffices to show that if $\Lambda$ is countable then $\Lambda$ is
free. We may write $\Lambda=\\{x_{1},x_{2},\dotsc\\}$ and let
$V_{i}=\sum_{j=1}^{i}\mathbb{R}x_{j}$ and $\Lambda_{i}=V_{i}\cap\Lambda$. We
claim that there exist bases $B_{i}$ for $\Lambda_{i}$ such that
$B_{i}\subseteq B_{j}$ for all $i\leq j$. Indeed, take $B_{0}=\emptyset$ and
inductively for $\Lambda_{i+1}$ note that $B_{i}$ is a basis for
$\Lambda_{i}=\Lambda_{i+1}\cap V_{i}$, so that by Lemma 3.12 there exists some
basis $B_{i+1}$ for $\Lambda_{i+1}$ containing $B_{i}$. Then
$B=\bigcup_{i=0}^{\infty}B_{i}$ is a basis for $\Lambda$, so $\Lambda$ is
free. ∎

######  Question 3.14.

We have by Example 3.4 and Theorem 3.13 two inclusions

$\\{\textup{free abelian groups}\\}\subseteq\\{\textup{underlying groups of
Hilbert lattices}\\}\subseteq\\{\textup{almost free abelian groups}\\}.$

Is one of these inclusions an equality, and if so, which?

######  Example 3.15.

There are abelian groups which are almost free but not free. Let $X$ be a
countably infinite set and consider the Baer–Specker group $B=\mathbb{Z}^{X}$.
Then by Theorem 21 in [9], we have that $B$ is not free. Since $B$ is a
torsion-free $\mathbb{Z}$-module, so is any countable subgroup, which is then
free by Theorem 16 in [9], i.e. $B$ is almost free.

######  Definition 3.16.

For a Hilbert lattice $\Lambda$ we define its rank as
$\operatorname{rk}\Lambda=\dim_{\mathbb{Q}}(\Lambda\otimes_{\mathbb{Z}}\mathbb{Q})$.
We will say a Hilbert lattice $\Lambda$ is of full rank in an ambient Hilbert
space $\mathcal{H}$ if $\mathbb{Q}\Lambda$ is dense in $\mathcal{H}$.

For free Hilbert lattices $\Lambda$ we have
$\Lambda\cong\mathbb{Z}^{(\operatorname{rk}\Lambda)}$ as abelian group. By
Theorem 3.7 every Hilbert lattice has a uniquely unique Hilbert space in which
it is contained and of full rank.

######  Lemma 3.17.

Let $\mathcal{H}$ be a Hilbert space and let $S,T\subseteq\mathcal{H}$ be
subsets such that $S$ is infinite, the $\mathbb{Q}$-vector space generated by
$S$ is dense in $\mathcal{H}$ and $\inf\\{\|x-y\|\,|\,x,y\in T,\,x\neq
y\\}>0$. Then $\\#S\geq\\#T$.

###### Proof.

Because $S$ is infinite, the set $S$ and the $\mathbb{Q}$-vector space $V$
generated by $S$ have the same cardinality. Let
$\rho=\inf\\{\|x-y\|\,|\,x,y\in T,\,x\neq y\\}$. Since $V$ is dense in
$\mathcal{H}$ we may for each $x\in T$ choose $f(x)\in V$ such that
$\|x-f(x)\|<\rho/2$. If $f(x)=f(y)$, then
$\|x-y\|=\|(x-f(x))-(y-f(y))\|\leq\|x-f(x)\|+\|y-f(y)\|<\rho$, so $x=y$. Hence
$f$ is injective and we have $\\#S=\\#V\geq\\#T$. ∎

The following theorem was independently proven by O. Berrevoets and
independently by B. Kadets.

######  Theorem 3.18.

If $\Lambda$ is a Hilbert lattice in a Hilbert space $\mathcal{H}$, then
$\operatorname{rk}\Lambda\leq\operatorname{orth\,dim}\mathcal{H}$ with
equality if $\Lambda$ is of full rank in $\mathcal{H}$.

###### Proof.

First suppose $\operatorname{rk}\Lambda$ is finite. It follows from Lemma 3.9
that
$\operatorname{rk}\Lambda=\dim_{\mathbb{R}}(\mathbb{R}\otimes_{\mathbb{Z}}\Lambda)=\dim_{\mathbb{R}}(\mathbb{R}\Lambda)\leq\dim_{\mathbb{R}}\mathcal{H}$.
If $\Lambda$ is of full rank, then $\mathbb{R}\Lambda$ is dense in
$\mathcal{H}$, but $\mathbb{R}\Lambda$ is complete as it is finite-
dimensional, so $\mathbb{R}\Lambda=\mathcal{H}$ and
$\operatorname{rk}\Lambda=\dim_{\mathbb{R}}\mathcal{H}$. Lastly, it follows
from Theorem 2.16 that
$\dim_{\mathbb{R}}\mathcal{H}=\operatorname{orth\,dim}\mathcal{H}$ when
$\dim_{\mathbb{R}}\mathcal{H}$ is finite.

Now suppose $\operatorname{rk}\Lambda$ is infinite and thus
$\\#\Lambda=\operatorname{rk}\Lambda$. By Theorem 2.16 we may assume without
loss of generality that $\mathcal{H}=\ell^{2}(B)$ for some set $B$ of
cardinality $\operatorname{orth\,dim}\mathcal{H}$, which must be infinite.
Observe that the $\mathbb{Q}$-vector space generated by $B$ is dense in
$\mathcal{H}$. We may apply Lemma 3.17 by discreteness of $\Lambda$ to obtain
$\operatorname{orth\,dim}\mathcal{H}=\\#B\geq\\#\Lambda=\operatorname{rk}\Lambda$,
as was to be shown. If $\mathbb{Q}\Lambda$ is dense in $\mathcal{H}$, then we
may apply Lemma 3.17 since $\|b-c\|^{2}=\|b\|^{2}+\|c\|^{2}=2$ for all
distinct $b,c\in B$ to conclude that
$\operatorname{rk}\Lambda=\\#\Lambda\geq\\#B=\operatorname{orth\,dim}\mathcal{H}$,
and thus we have equality. ∎

##  4\. Decompositions

######  Definition 4.1.

Let $\Lambda$ be a Hilbert lattice. A decomposition of an element
$z\in\Lambda$ is a pair $(x,y)\in\Lambda^{2}$ such that $z=x+y$ and $\langle
x,y\rangle\geq 0$. A decomposition $(x,y)$ of $z\in\Lambda$ is trivial if
$x=0$ or $y=0$. We say $z\in\Lambda$ is indecomposable if it has exactly two
decompositions, i.e. $z\neq 0$ and the only decompositions of $z$ are trivial.
Write $\text{dec}(z)$ for the set of decompositions of $z\in\Lambda$ and
$\text{indec}(\Lambda)$ for the set of indecomposable elements of $\Lambda$.

Indecomposable elements are in the compute science literature often called
Voronoi-relevant vectors, for example in [7]. This name is clearly inspired by
Theorem 6.11.

######  Example 4.2.

Let $f:I\to\mathbb{R}_{\geq 0}$ be such that $\smash{\Lambda^{f}}$ as in
Example 3.5 is a Hilbert lattice. We will compute the indecomposables of
$\Lambda^{f}$. Let $x=(x_{i})_{i}\in\textup{indec}(\smash{\Lambda^{f}})$ and
write $e_{i}$ for the $i$-th standard basis vector. Note that $x$ must be
primitive, i.e. not be of the form $ny$ for any $y\in\smash{\Lambda^{f}}$ and
$n\in\mathbb{Z}_{>1}$, because otherwise $\langle y,(n-1)y\rangle=(n-1)\langle
y,y\rangle>0$ shows $(y,(n-1)y)$ is a non-trivial decomposition. If $x_{i}$
and $x_{j}$ are non-zero for distinct $i,j\in I$, then
$q(x-e_{i}x_{i})+q(e_{i}x_{i})=q(x)$ and we have a non-trivial decomposition
of $x$. Hence $x=\pm e_{i}$ for some $i\in I$. Note that $e_{i}$ is indeed
indecomposable for all $i\in I$: Any decomposition
$(x,y)\in\textup{dec}(e_{i})$ must have $|x_{i}|+|y_{i}|=1$ and
$x_{j}=y_{j}=0$ for $i\neq j$, so $x=0$ or $y=0$. As $z\mapsto-z$ is an
isometry of $\smash{\Lambda^{f}}$, we have that $-e_{i}$ is indecomposable as
well. Hence $\textup{indec}(\smash{\Lambda^{f}})=\\{\pm e_{i}\,|\,i\in I\\}$.

######  Lemma 4.3.

Let $\Lambda$ be a Hilbert lattice and let $x,y,z\in\Lambda$. Then the
following are equivalent:

  1. (i)

The pair $(x,y)$ is a decomposition of $z$.

  2. (ii)

We have $x+y=z$ and $q(x)+q(y)\leq q(z)$.

  3. (iii)

We have $x+y=z$ and $q(x-z/2)\leq q(z/2)$.

  4. (iv)

We have $x+y=z$ and $q(z-2y)\leq q(z)$.

For a visual aid to this lemma see Figure 1.

###### Proof.

(i $\Leftrightarrow$ ii) By bilinearity we have

$q(z)=\langle x+y,x+y\rangle=q(x)+q(y)+2\langle x,y\rangle.$

(ii $\Leftrightarrow$ iii) By the parallelogram law we have

$q(x)+q(y)=2q\Big{(}\frac{x+y}{2}\Big{)}+2q\Big{(}\frac{x-y}{2}\Big{)}=2q\Big{(}\frac{z}{2}\Big{)}+2q\Big{(}x-\frac{z}{2}\Big{)},$

so $q(x)+q(y)-q(z)=2\cdot[q(x-z/2)-q(z/2)]$. The claim then follows trivially.

(iii $\Leftrightarrow$ iv) Note that $z-2y=x-y=2(x/2-y/2)=2(x-z/2)$. By the
parallelogram law we have $q(2w)=4q(w)$ for all $w\in\mathbb{Q}\Lambda$, from
which this equivalence trivially follows. ∎

By Lemma 4.3, finding decompositions of $z\in\Lambda$ amounts to finding
$x\in\Lambda$ sufficiently close to $z/2$.

######  Lemma 4.4.

Let $z\in\Lambda$ such that $0<q(z)\leq 2\textup{P}(\Lambda)$. Suppose that
the latter inequality is strict or $\textup{P}(\Lambda)$ is not attained by
any vector in $\Lambda$. Then $z\in\textup{indec}(\Lambda)$.

###### Proof.

If $(x,y)\in\textup{dec}(z)$ is non-trivial, then by Lemma 4.3 we have
$2\textup{P}(\Lambda)\geq q(z)\geq
q(x)+q(y)\geq\textup{P}(\Lambda)+\textup{P}(\Lambda)$ with either the first or
last inequality strict, which is a contradiction. Since $z\neq 0$ it follows
that $z$ is indecomposable. ∎

$0$$x$$z$$y$$z/2$$z-2y$ Figure 1. A decomposition $z=x+y$

######  Proposition 4.5.

If $\Lambda$ is a non-zero Hilbert lattice, then every $z\in\Lambda$ can be
written as a sum of at most $q(z)/\textup{P}(\Lambda)$ indecomposables
$z_{1},\dotsc,z_{n}\in\Lambda$ such that $\sum_{i}q(z_{i})\leq q(z)$.

###### Proof.

By scaling $q$ we may assume without loss of generality that
$\textup{P}(\Lambda)=1$. We apply induction to $\lfloor q(z)\rfloor$. When
this equals $0$ we have $q(z)<\textup{P}(\Lambda)$, so $z=0$, which we can
write as a sum of zero indecomposables. Now suppose $\lfloor q(z)\rfloor\geq
1$ and thus $z\neq 0$. If $z$ is indecomposable then indeed it is the sum of
$1\leq\lfloor q(z)\rfloor$ indecomposable, so suppose there is a non-trivial
$(x,y)\in\text{dec}(z)$. Then $q(x)+q(y)\leq q(z)$ by Lemma 4.3 and since
$y\neq 0$ also $q(y)\geq\textup{P}(\Lambda)=1$. Hence $\lfloor
q(x)\rfloor\leq\lfloor q(z)-q(y)\rfloor<\lfloor q(z)\rfloor$, so by the
induction hypothesis we may write $x=\sum_{i}x_{i}$ with
$x_{1},\dotsc,x_{a}\in\text{indec}(\Lambda)$ and $a\leq q(x)$ such that
$\sum_{i}q(x_{i})\leq q(x)$. By symmetry we may similarly write $y$ as a sum
of at most $q(y)$ indecomposables $y_{1},\dotsc,y_{b}$. Hence we can write
$z=\sum_{i}x_{i}+\sum_{i}y_{i}$ as a sum of $a+b\leq q(x)+q(y)\leq q(z)$
indecomposables such that $\sum_{i}q(x_{i})+\sum_{i}q(y_{i})\leq q(x)+q(y)\leq
q(z)$. The proposition follows by induction. ∎

######  Lemma 4.6.

Suppose $\Lambda$ is a Hilbert lattice and $z\in\Lambda$ is the sum of some
non-zero $z_{1},\dotsc,z_{n}\in\Lambda$ and $n\in\mathbb{Z}_{\geq 2}$. If
$\sum_{i}q(z_{i})\leq q(z)$, then $z$ has a non-trivial decomposition.

###### Proof.

We have

$\sum_{i=1}^{n}\langle z_{i},z-z_{i}\rangle=\sum_{i=1}^{n}\langle
z_{i},z\rangle-\sum_{i=1}^{n}\langle
z_{i},z_{i}\rangle=q(z)-\sum_{i=1}^{n}q(z_{i})\geq 0,$

so $\langle z_{i},z-z_{i}\rangle\geq 0$ for some $i$. As neither $z_{i}$ nor
$z-z_{i}$ are $0$, we conclude that $(z_{i},z-z_{i})$ is a non-trivial
decomposition of $z$. ∎

######  Proposition 4.7.

Let $\Lambda$ be a Hilbert lattice. The group $\\{\pm 1\\}$ acts on
$\textup{indec}(\Lambda)$ by multiplication, and the natural map
$\textup{indec}(\Lambda)/\\{\pm 1\\}\to\Lambda/2\Lambda$ is injective.

###### Proof.

Note that $\\{\pm 1\\}$ acts on $\Lambda$ and thus also on
$\textup{indec}(\Lambda)$. By (i $\Leftrightarrow$ iv) of Lemma 4.3 we have

$\text{dec}(z)=\\{(\tfrac{z-a}{2},\tfrac{z+a}{2})\,|\,a\in
z+2\Lambda,\,q(a)\leq q(z)\\}.$

Let $z\in\textup{indec}(\Lambda)$. Then $\\{(0,z),(z,0)\\}=\text{dec}(z)$, so
the only $a\in z+2\Lambda$ such that $q(a)\leq q(z)$ are $a=z$ and $a=-z$.
Thus $z$ is a $q$-minimal element of its coset in $\Lambda/2\Lambda$ and this
minimal element is unique up to sign. Consequently, the map
$\textup{indec}(\Lambda)/\\{\pm 1\\}\to\Lambda/2\Lambda$ is injective. ∎

######  Corollary 4.8.

Let $\Lambda$ be a Hilbert lattice. Then $\textup{indec}(\Lambda)$ is finite
if and only if $\operatorname{rk}\Lambda<\infty$.

###### Proof.

Recall that $\textup{indec}(\Lambda)$ generates $\Lambda$ by Proposition 4.5.
Hence if $\textup{indec}(\Lambda)$ is finite, then
$\operatorname{rk}\Lambda<\infty$. If $\operatorname{rk}\Lambda<\infty$, then
Proposition 4.7 implies $\\#\textup{indec}(\Lambda)\leq
2\cdot\\#(\Lambda/2\Lambda)=2^{1+\operatorname{rk}\Lambda}<\infty$. ∎

The zero coset of $\Lambda/2\Lambda$ is never in the image of the map of
Proposition 4.7, as any non-zero element of the form $2x$ with $x\in\Lambda$
has a non-trivial decomposition $(x,x)$. A non-zero coset $C$ of
$\Lambda/2\Lambda$ can fail to be in the image for two reasons: Either $C$ has
no minimal element or a minimal element exists but is not unique up to sign.
In the latter case, with $z\in C$ minimal, there exists a
$(x,y)\in\text{dec}(z)$ with $x,y\neq 0$ and $q(x)+q(y)=q(z)$ and thus
$\langle x,y\rangle=0$, i.e. $z$ has an orthogonal decomposition. This is
exhibited, for example, by the lattice $\mathbb{Z}^{2}\subseteq\mathbb{R}^{2}$
with the standard inner product and $z=(1,1)$, where $(-1,1)\in
z+2\mathbb{Z}^{2}$ gives rise to the orthogonal decomposition $(1,0)+(0,1)=z$.
In the former case, $\operatorname{rk}\Lambda$ has to be infinite: If
$\operatorname{rk}\Lambda$ is finite, then for any $x\in\Lambda$ there are
only finitely many $y\in\Lambda$ with $q(y)\leq q(x)$, so $q$ assumes a
minimum on any non-empty subset of $\Lambda$, in particular $C$. An example is
the following.

######  Example 4.9.

We will exhibit a Hilbert lattice $\Lambda$ and a coset of $2\Lambda$ on which
$q$ does not attain a minimum. Let $f:I\to\mathbb{R}_{\geq 0}$ be such that
$\smash{\Lambda^{f}}$ as in Example 3.5 is a Hilbert lattice. We define the
$\smash{\Lambda_{2}^{f}}$ to be the sublattice

$\Lambda_{2}^{f}=\ker(\Lambda^{f}\xrightarrow{\Sigma}(\mathbb{Z}/2\mathbb{Z}))=\Big{\\{}(x_{i})_{i}\in\mathbb{Z}^{(I)}\,\Big{|}\,\sum_{i\in
I}x_{i}\equiv 0\mod 2\Big{\\}}.$

Consider $f:\mathbb{Z}_{\geq 0}\to\mathbb{R}_{\geq 0}$ strictly decreasing,
write $f(\infty)$ for its limit, assume $f(\infty)>0$, and let
$\Lambda=\smash{\Lambda_{2}^{f}}$. Let $z=2e_{k}\in\Lambda$ for any
$k\in\mathbb{Z}_{\geq 0}$. Then for all $y=(y_{i})_{i}\in\Lambda$ we have

$q(z-2y)=4\Big{(}(1-y_{k})^{2}f(k)^{2}+\sum_{i\neq
k}y_{i}^{2}f(i)^{2}\Big{)}>4f(\infty)^{2},$

since either $y_{i}^{2}\geq 1$ for some $i\neq k$ or $y_{k}$ is even. However,
we also have for $y=e_{k}+e_{i}$ that $q(z-2y)=q(2e_{i})=4f(i)^{2}\to
4f(\infty)^{2}$ as $i\to\infty$. Hence $\\{q(z-2y)\,|\,y\in\Lambda\\}$ does
not contain a minimum.

If we eliminate the zero coset in the proof of Corollary 4.8 the upper bound
on the number of indecomposables becomes $2(2^{\operatorname{rk}(\Lambda)}-1)$
and this bound is tight. If one took the effort to define a sensible
probability measure on the space of all Hilbert lattices of given finite rank,
then this upper bound will in fact be an equality with probability $1$.

##  5\. Hilbert lattice decompositions

The main motivation for considering indecomposable elements is found in the
study of decompositions of lattices. In this section we generalize a result of
Eichler [5] on the existence of a universal decomposition in lattices to
Hilbert lattices.

######  Definition 5.1.

Let $\Lambda$ be a Hilbert lattice. We call a family $\\{\Lambda_{i}\\}_{i\in
I}$ of sub-Hilbert lattices $\Lambda_{i}\subseteq\Lambda$ for some index set
$I$ a decomposition of $\Lambda$, and if so write $\mathop{\mathchoice{\vbox{
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{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptscriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}}\displaylimits_{i\in
I}\Lambda_{i}=\Lambda$, if $\langle\Lambda_{i},\Lambda_{j}\rangle=0$ for all
$i\neq j$ and the natural map of abelian groups $\bigoplus_{i\in
I}\Lambda_{i}\to\Lambda$ is an isomorphism. We make the set of decompositions
of $\Lambda$ into a category, where the morphisms from
$\\{\Lambda_{i}\\}_{i\in I}$ to $\\{\textup{M}_{j}\\}_{j\in J}$ are the maps
$f:I\to J$ such that $\textup{M}_{j}=\mathop{\mathchoice{\vbox{
\hbox{$\displaystyle\raisebox{2.4pt}{ \leavevmode\hbox to13.8pt{\vbox
to13.8pt{\pgfpicture\makeatletter\hbox{\hskip 6.90227pt\lower-6.90227pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{6.57227pt}{0.0pt}\pgfsys@curveto{6.57227pt}{3.6298pt}{3.6298pt}{6.57227pt}{0.0pt}{6.57227pt}\pgfsys@curveto{-3.6298pt}{6.57227pt}{-6.57227pt}{3.6298pt}{-6.57227pt}{0.0pt}\pgfsys@curveto{-6.57227pt}{-3.6298pt}{-3.6298pt}{-6.57227pt}{0.0pt}{-6.57227pt}\pgfsys@curveto{3.6298pt}{-6.57227pt}{6.57227pt}{-3.6298pt}{6.57227pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{6.57227pt}\pgfsys@lineto{0.0pt}{-2.78864pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-5.9748pt}{-2.78864pt}\pgfsys@lineto{5.9748pt}{-2.78864pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\textstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptscriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}}\displaylimits_{i\in
f^{-1}\\{j\\}}\Lambda_{i}$ for all $j\in J$. We say $\Lambda$ is
indecomposable if $\Lambda\neq 0$ and for all
$\Lambda_{1},\Lambda_{2}\subseteq\Lambda$ such that
$\Lambda_{1}\mathop{\mathchoice{\displaystyle\raisebox{2.4pt}{
\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys<EMAIL_ADDRESS>\leavevmode\hbox to6.75pt{\vbox to6.75pt{\pgfpicture\makeatletter\hbox{\hskip
3.3742pt\lower-3.3742pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{3.23921pt}{0.0pt}\pgfsys@curveto{3.23921pt}{1.78899pt}{1.78899pt}{3.23921pt}{0.0pt}{3.23921pt}\pgfsys@curveto{-1.78899pt}{3.23921pt}{-3.23921pt}{1.78899pt}{-3.23921pt}{0.0pt}\pgfsys@curveto{-3.23921pt}{-1.78899pt}{-1.78899pt}{-3.23921pt}{0.0pt}{-3.23921pt}\pgfsys@curveto{1.78899pt}{-3.23921pt}{3.23921pt}{-1.78899pt}{3.23921pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{3.23921pt}\pgfsys@lineto{0.0pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.27pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-2.94473pt}{-1.3744pt}\pgfsys@lineto{2.94473pt}{-1.3744pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}}}\Lambda_{2}=\Lambda$
we have $\Lambda_{1}=0$ or $\Lambda_{2}=0$.

######  Definition 5.2.

A graph is a pair $(V,E)$ where $V$ is a (potentially infinite) set and $E$ is
a set of size-2 subsets of $V$. For a graph $(V,E)$ we call the elements of
$V$ its vertices and those of $E$ its edges. A connected component of $(V,E)$
is a non-empty set $S\subseteq V$ such that there is no $\\{u,v\\}\in E$ such
that $u\in S$ and $v\not\in S$ and which is minimal with respect to inclusion
given these properties. We say a graph is connected if it has precisely one
connected component.

######  Lemma 5.3.

Let $G=(V,E)$ be a graph. Then the connected components of $G$ are pairwise
disjoint, and if for $S\subseteq V$ there exist no $\\{u,v\\}\in E$ such that
$u\in S$ and $v\not\in S$, then $S$ is a union of connected components.

###### Proof.

Let $\mathcal{C}$ be the set of $S\subseteq V$ such that there are no
$\\{u,v\\}\in E$ such that $u\in S$ and $v\not\in S$, so that the connected
components of $G$ become the minimal non-empty elements of $\mathcal{C}$ with
respect to inclusion. Note that $\mathcal{C}$ is closed under taking
complements, arbitrary unions and arbitrary intersections, i.e. $\mathcal{C}$
is a clopen topology on $V$. Suppose $S,T\in\mathcal{C}$ are connected
components that intersection non-trivially, then $S\cap T\in\mathcal{C}$ is
non-empty, so by minimality $S=T$. Hence the connected components are pairwise
disjoint.

Now let $S\in\mathcal{C}$ and for all $s\in S$ let
$A_{s}=\\{T\in\mathcal{C}\,|\,s\in T\\}$ and $C_{s}=\bigcap_{T\in A_{s}}T$,
which is an element of $A_{s}$. For all $T\in\mathcal{C}$ we either have $T\in
A_{s}$ or $V\setminus T\in A_{s}$, and thus either $C_{s}\subseteq T$ or
$C_{s}\cap T=\emptyset$. It follows that no non-empty $T\in\mathcal{C}$ is
strictly contained in $C_{s}$, i.e. $C_{s}$ is a connected component of $G$.
As $S\in A_{s}$ we have $s\in C_{s}\subseteq S$ and thus $S=\bigcup_{s\in
S}C_{s}$ is a union of connected components. ∎

The following is a generalization of a theorem due to Eichler [5], although
the proof more closely resembles that of Theorem 6.4 in [15].

######  Theorem 5.4 (Eichler).

Let $\Lambda$ be a Hilbert lattice and $V\subseteq\textup{indec}(\Lambda)$
such that $V$ generates $\Lambda$ as a group. Let $G$ be the graph with vertex
set $V$ and with an edge between $x$ and $y$ if and only if $\langle
x,y\rangle\neq 0$. Let $U$ be the set of connected components of $G$ and for
$u\in U$ let $\textup{Y}_{u}\subseteq\Lambda$ be the subgroup generated by the
elements in $u$. Then $\\{\textup{Y}_{u}\\}_{u\in U}$ is a universal
decomposition of $\Lambda$.

As corollary to this theorem we have that $\Lambda$ is indecomposable if and
only if $G$ is connected. Note that $V=\textup{indec}(\Lambda)$ generates
$\Lambda$ as a group by Proposition 4.5. It then follows from the theorem that
every Hilbert lattice has a universal decomposition. This universal
decomposition is unique and thus independent of choice of $V$.

###### Proof.

We have $V\subseteq\bigcup_{u\in U}\textup{Y}_{u}$ by Lemma 5.3, so
$\sum_{u\in U}\textup{Y}_{u}=\Lambda$ by assumption on $V$. For $u,v\in U$
distinct we have $\langle u,v\rangle=\\{0\\}$ by definition of $G$, so
$\langle\textup{Y}_{u},\textup{Y}_{v}\rangle=\\{0\\}$. We conclude that
$\Lambda=\mathop{\mathchoice{\vbox{ \hbox{$\displaystyle\raisebox{2.4pt}{
\leavevmode\hbox to13.8pt{\vbox to13.8pt{\pgfpicture\makeatletter\hbox{\hskip
6.90227pt\lower-6.90227pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{6.57227pt}{0.0pt}\pgfsys@curveto{6.57227pt}{3.6298pt}{3.6298pt}{6.57227pt}{0.0pt}{6.57227pt}\pgfsys@curveto{-3.6298pt}{6.57227pt}{-6.57227pt}{3.6298pt}{-6.57227pt}{0.0pt}\pgfsys@curveto{-6.57227pt}{-3.6298pt}{-3.6298pt}{-6.57227pt}{0.0pt}{-6.57227pt}\pgfsys@curveto{3.6298pt}{-6.57227pt}{6.57227pt}{-3.6298pt}{6.57227pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{6.57227pt}\pgfsys@lineto{0.0pt}{-2.78864pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-5.9748pt}{-2.78864pt}\pgfsys@lineto{5.9748pt}{-2.78864pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\textstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptscriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}}\displaylimits_{u\in
U}\textup{Y}_{u}$ is a decomposition.

To show it is universal, let $\\{\Lambda_{i}\\}_{i\in I}$ be a family of
sublattices of $\Lambda$ such that $\mathop{\mathchoice{\vbox{
\hbox{$\displaystyle\raisebox{2.4pt}{ \leavevmode\hbox to13.8pt{\vbox
to13.8pt{\pgfpicture\makeatletter\hbox{\hskip 6.90227pt\lower-6.90227pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{6.57227pt}{0.0pt}\pgfsys@curveto{6.57227pt}{3.6298pt}{3.6298pt}{6.57227pt}{0.0pt}{6.57227pt}\pgfsys@curveto{-3.6298pt}{6.57227pt}{-6.57227pt}{3.6298pt}{-6.57227pt}{0.0pt}\pgfsys@curveto{-6.57227pt}{-3.6298pt}{-3.6298pt}{-6.57227pt}{0.0pt}{-6.57227pt}\pgfsys@curveto{3.6298pt}{-6.57227pt}{6.57227pt}{-3.6298pt}{6.57227pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{6.57227pt}\pgfsys@lineto{0.0pt}{-2.78864pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-5.9748pt}{-2.78864pt}\pgfsys@lineto{5.9748pt}{-2.78864pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\textstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}{\vbox{
\hbox{$\scriptscriptstyle\raisebox{2.4pt}{ \leavevmode\hbox to10.05pt{\vbox
to10.05pt{\pgfpicture\makeatletter\hbox{\hskip 5.0245pt\lower-5.0245pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}{{}}{}{{}}{}{{{}}{}{}{}{}{}{}{}{}}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@moveto{4.6945pt}{0.0pt}\pgfsys@curveto{4.6945pt}{2.59273pt}{2.59273pt}{4.6945pt}{0.0pt}{4.6945pt}\pgfsys@curveto{-2.59273pt}{4.6945pt}{-4.6945pt}{2.59273pt}{-4.6945pt}{0.0pt}\pgfsys@curveto{-4.6945pt}{-2.59273pt}{-2.59273pt}{-4.6945pt}{0.0pt}{-4.6945pt}\pgfsys@curveto{2.59273pt}{-4.6945pt}{4.6945pt}{-2.59273pt}{4.6945pt}{0.0pt}\pgfsys@closepath\pgfsys@moveto{0.0pt}{0.0pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{0.0pt}{4.6945pt}\pgfsys@lineto{0.0pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{}{{}}{}
{}{}\pgfsys@beginscope\pgfsys@invoke{
}\pgfsys@setlinewidth{0.66pt}\pgfsys@invoke{
}{}\pgfsys@moveto{-4.26773pt}{-1.9919pt}\pgfsys@lineto{4.26773pt}{-1.9919pt}\pgfsys@stroke\pgfsys@invoke{
} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}$}}}}\displaylimits_{i\in
I}\Lambda_{i}=\Lambda$. Let $x\in\textup{indec}(\Lambda)$ and write
$x=\sum_{i\in I}\lambda_{i}$ with $\lambda_{i}\in\Lambda_{i}$ for all $i\in
I$. If $j\in I$ is such that $\lambda_{j}\neq 0$, then
$\langle\lambda_{j},\sum_{i\neq j}\lambda_{i}\rangle=0$ and thus
$\lambda_{j}=x$, because otherwise we obtain a non-trivial decomposition of
$x$. Therefore every indecomposable of $\Lambda$ is in precisely one of the
$\Lambda_{i}$. We conclude that the $S_{i}=\Lambda_{i}\cap V$ for $i\in I$ are
pairwise disjoint and have $V$ as their union. Then by Lemma 5.3 every
connected component $u\in U$ is contained in precisely one of the $S_{i}$, say
in $S_{f(u)}$. By definition of the map $f:U\to I$ and the $\textup{Y}_{u}$ we
have $\mathop{\mathchoice{\vbox{ \hbox{$\displaystyle\raisebox{2.4pt}{
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f^{-1}\\{i\\}}\textup{Y}_{u}\subseteq\Lambda_{i}$ for all $i$, and since both
the $\textup{Y}_{u}$ and the $\Lambda_{i}$ sum to $\Lambda$ we must have
equality for all $i$. Hence a map $f$ as in the theorem exists and it follows
trivially from the construction that it is unique. We conclude that
$\\{\textup{Y}_{u}\\}_{u\in U}$ is a universal decomposition of $\Lambda$. ∎

We will use this theorem in Section 10 to generalize some theorems from [12].

##  6\. Voronoi cells

We will generalize the Voronoi cell as defined for classical lattices to
Hilbert lattices and extend some known definitions and properties. Some of
these definitions relate to the ambient Hilbert space of the Hilbert lattice,
which exists and is uniquely unique by Theorem 3.7 when we require the
$\mathbb{Q}$-vector space generated by the lattice to lie dense in the Hilbert
space. In this section we will write $\mathcal{H}_{\Lambda}$ for this ambient
Hilbert space of a Hilbert lattice $\Lambda$.

######  Definition 6.1.

Let $\Lambda$ be a Hilbert lattice. The packing radius of $\Lambda$ is

$\rho(\Lambda)=\inf\\{\tfrac{1}{2}\|x-y\|\,|\,x,y\in\Lambda,\,x\neq
y\\}=\tfrac{1}{2}\sqrt{\textup{P}(\Lambda)},$

the covering radius of $\Lambda$ is

$r(\Lambda)=\inf\\{b\in\mathbb{R}_{>0}\,|\,(\forall
z\in\mathcal{H}_{\Lambda})\,(\exists\,x\in\Lambda)\,\|z-x\|\leq b\\}$

and the Voronoi cell of $\Lambda$ in $\mathcal{H}$ is the set

$\operatorname{Vor}(\Lambda)=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall
x\in\Lambda\setminus\\{0\\})\ \|z\|<\|z-x\|\\}.$

We call $\rho(\Lambda)$ the packing radius because it is the radius of the
largest sphere $B\subseteq\mathcal{H}_{\Lambda}$ such that the spheres $x+B$
for $x\in\Lambda$ are pairwise disjoint. Similarly $r(\Lambda)$ is the radius
of the smallest sphere $B\subseteq\mathcal{H}_{\Lambda}$ for which
$\bigcup_{x\in\Lambda}(x+B)=\mathcal{H}_{\Lambda}$. Note that $r(\Lambda)=0$
only for $\Lambda=0$ by discreteness.

######  Example 6.2.

The covering radius of a Hilbert lattice need not be finite. Take
$\Lambda^{f}$ as in Example 3.5 but with $f:\mathbb{Z}_{\geq
0}\to\mathbb{R}_{\geq 0}$ diverging to infinity. The lattice point closest to
$\frac{1}{3}e_{i}$ is $0$ for all $i\in\mathbb{Z}_{\geq 0}$, so it has
distance $\frac{1}{3}f(i)$ to the lattice. Hence
$r(\smash{\Lambda^{f}})\geq\sup\\{\frac{1}{3}f(i)\,|\,i\in\mathbb{Z}_{\geq
0}\\}=\infty$.

######  Example 6.3.

The Voronoi cell does not need to be an open set. Consider the lattice
$\Lambda=\Lambda^{f}_{2}$ as in Example 4.9 with $f:\mathbb{Z}_{\geq
0}\to\mathbb{R}_{>0}$ strictly decreasing. Let $i\in\mathbb{Z}_{\geq 0}$ and
$\textup{A}=(1+f(\infty)^{2}f(i)^{-2})/2$ and write $e_{i}$ for the $i$-th
standard basis vector. We claim that $\alpha
e_{i}\in\operatorname{Vor}(\Lambda)$ for $\alpha\in\mathbb{R}$ precisely when
$|\alpha|\leq\textup{A}$, which proves the Voronoi cell is not open.

Let $x=\sum_{j}x_{j}e_{j}\in\Lambda$ such that $x_{i}\neq 0$. Then $|x_{i}|=1$
or $q(x)/|x_{i}|\geq|x_{i}|f(i)^{2}>f(i)^{2}+f(i+1)^{2}$. It follows that

$\inf\Big{\\{}\frac{q(x)}{2|x_{i}|f(i)^{2}}\,\Big{|}\,x\in\Lambda,\,x_{i}\neq
0\Big{\\}}=\frac{f(i)^{2}+f(\infty)^{2}}{2f(i)^{2}}=\textup{A}$

and that the infimum is not attained. By definition $\alpha
e_{i}\in\operatorname{Vor}(\Lambda)$ if and only if $g(x):=q(\alpha
e_{i}-x)-q(\alpha e_{i})>0$ for all $x\in\Lambda\setminus\\{0\\}$. Note that
$g(x)=q(x)-2\alpha x_{i}f(i)^{2}$.

Suppose $|\alpha|\leq\textup{A}$ and let $x\in\Lambda\setminus\\{0\\}$. If
$\alpha x_{i}\leq 0$, then $g(x)\geq q(x)>0$, so suppose $\alpha x_{i}>0$.
Then

$\frac{g(x)}{2|x_{i}|f(i)^{2}}=\frac{q(x)}{2|x_{i}|f(i)^{2}}-|\alpha|>\textup{A}-|\alpha|\geq
0,$

so $g(x)>0$. We conclude that $\alpha e_{i}\in\operatorname{Vor}(\Lambda)$.
Conversely, if $|\alpha|>\textup{A}$, then

$\inf\Big{\\{}\frac{g(x)}{2|x_{i}|f(i)^{2}}\,\Big{|}\,x\in\Lambda,\,x_{i}\neq
0\Big{\\}}=\textup{A}-|\alpha|<0,$

hence there exists some $x\in\Lambda$ such that $g(x)<0$ and thus $\alpha
e_{i}\not\in\operatorname{Vor}(\Lambda)$.

######  Definition 6.4.

Let $\mathcal{H}$ be a Hilbert space and let $S\subseteq\mathcal{H}$ be a
subset. We say $S$ is symmetric if for all $x\in S$ also $-x\in S$. We say $S$
is convex if for all $x,y\in S$ and $t\in[0,1]$ also $(1-t)x+ty\in S$.

######  Lemma 6.5.

Let $\mathcal{H}$ be a Hilbert space. For $X\subseteq\mathcal{H}$ write
$\overline{X}$ for the topological closure of $X$. Then

  1. (i)

The intersection $\bigcap_{i}S_{i}$ of convex sets $(S_{i})_{i\in I}$ in
$\mathcal{H}$ is convex;

  2. (ii)

The topological closure $\overline{S}$ of a convex set $S$ in $\mathcal{H}$ is
convex;

  3. (iii)

For all $S$ in $\mathcal{H}$ open convex, $x\in S$, $y\in\overline{S}$ and
$t\in[0,1)$ we have $(1-t)x+ty\in S$.

  4. (iv)

For convex open sets $(S_{i})_{i\in I}$ in $\mathcal{H}$ with non-empty
intersection we have
$\overline{\bigcap_{i}S_{i}}=\bigcap_{i}\overline{S_{i}}$.

###### Proof.

(i) Trivial. (ii) Let $x,y\in\overline{S}$ and let $(x_{n})_{n}$ and
$(y_{n})_{n}$ be sequences in $S$ with limit $x$ respectively $y$. For all
$t\in[0,1]$ we have $(1-t)x_{n}+ty_{n}\in S$, and since addition and scalar
multiplication are continuous also
$(1-t)x+ty=\lim_{n\to\infty}[(1-t)x_{n}+ty_{n}]\in\overline{S}$. (iii) By
translating $S$ we may assume without loss of generality that $(1-t)x+ty=0$.
Since $x\in S$ and $S$ is open there exists some $r_{x}>0$ such that the open
ball $B_{x}$ of radius $r_{x}$ around $x$ is contained in $S$. For $r_{y}>0$
sufficiently small (in fact $r_{y}=r_{x}\cdot(1-t)/t$ suffices, see Figure 2)
it holds that for any $z$ in the open ball $B_{y}$ of radius $r_{y}$ around
$y$ the line through $0$ and $z$ intersects $B_{x}$. Taking $z\in B_{y}\cap
S$, which exists because $y$ is in the closure of $S$, there exists some $w\in
B_{x}\subseteq S$ such that $0$ lies on the line segment between $w$ and $z$.
By convexity $0\in S$ follows, as was to be shown. (iv) Since
$\bigcap_{i}\overline{S_{i}}$ is closed and contains $\bigcap_{i}S_{i}$,
clearly $\overline{\bigcap_{i}S_{i}}\subseteq\bigcap_{i}\overline{S_{i}}$. By
(i) the set $\bigcap_{i}S_{i}$ is convex and by assumption it contains some
$x$. For $t\in[0,1)$ and $y\in\bigcap_{i}\overline{S_{i}}$ we have
$z_{t}=(1-t)x+ty\in S_{i}$ by (iii). Thus $z_{t}\in\bigcap_{i}S_{i}$ for all
$t\in[0,1)$, so $y=\lim_{t\to 1}z_{t}\in\overline{\bigcap_{i}S_{i}}$, proving
the reverse inclusion. ∎

$0$$x$$y$$(1-t)\|x-y\|\phantom{abc}$$\phantom{ab}t\|x-y\|$ Figure 2.
Computation of $r_{y}$ from $r_{x}$ via similar triangles.

######  Lemma 6.6.

Let $\mathcal{H}$ be a Hilbert space. Then for $x,y\in\mathcal{H}$ we have
$\|y\|\leq\|y-x\|$ if and only if $2\langle x,y\rangle\leq\langle x,x\rangle$,
and similarly with $\leq$ replaced by $<$.

###### Proof.

We have $\|y-x\|^{2}-\|y\|^{2}=\langle x,x\rangle-2\langle x,y\rangle$, from
which the lemma trivially follows. ∎

######  Proposition 6.7.

For all Hilbert lattices $\Lambda$ the set $\operatorname{Vor}(\Lambda)$ is
symmetric, convex and has topological closure
$\overline{\operatorname{Vor}}(\Lambda)\vcentcolon=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall
x\in\Lambda)\ \|z\|\leq\|z-x\|\\}$.

###### Proof.

It follows readily from the definition that $\operatorname{Vor}(\Lambda)$ is
symmetric. Now for $x\in\Lambda$ consider
$H_{x}=\\{z\in\mathcal{H}_{\Lambda}\,|\,2\langle x,z\rangle<\langle
x,x\rangle\\}$. It is easy to show for all $x\in\Lambda$ that $H_{x}$ is
convex: For $a,b\in H_{x}$ and $t\in[0,1]$ we have

$2\langle(1-t)a+tb,x\rangle=(1-t)\cdot 2\langle a,x\rangle+t\cdot 2\langle
b,x\rangle<(1-t)\langle x,x\rangle+t\langle x,x\rangle=\langle x,x\rangle,$

so $(1-t)a+tb\in H_{x}$. As $\operatorname{Vor}(\Lambda)$ is the intersection
of all $H_{x}$ with $x\in\Lambda$ non-zero by Lemma 6.6, it follows from Lemma
6.5.i that $\operatorname{Vor}(\Lambda)$ is convex. The $H_{x}$ are all open,
and for $x$ non-zero we have $0\in H_{x}$. Hence the topological closure of
$\operatorname{Vor}(\Lambda)$ equals
$\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall x\in\Lambda\setminus\\{0\\})\
\|z\|\leq\|z-x\|\\}$ by Lemma 6.5.iv, from which the proposition follows. ∎

######  Example 6.8.

We do not have in general that
$\mathcal{H}_{\Lambda}=\Lambda+\overline{\operatorname{Vor}}(\Lambda)$ for all
Hilbert lattices $\Lambda$, as in the finite-dimensional case. Note that
$z\in\mathcal{H}_{\Lambda}$ is in
$\Lambda+\overline{\operatorname{Vor}}(\Lambda)$ if and only if the infimum
$\inf\\{\|z-x\|\,|\,x\in\Lambda\\}$ is attained for some $x\in\Lambda$.
Consider Example 4.9, where we exhibit a lattice $\Lambda$ and a coset
$z+2\Lambda$ of $\Lambda/2\Lambda$ where $\inf\\{q(z+2x)\,|\,x\in\Lambda\\}$
is not attained. Equivalently, $\inf\\{\|\frac{1}{2}z-x\|\,|\,x\in\Lambda\\}$
is not attained, so
$\frac{1}{2}z\not\in\Lambda+\overline{\operatorname{Vor}}(\Lambda)$.

######  Proposition 6.9.

Let $\Lambda$ be a Hilbert lattice. Then for all $\epsilon>0$ we have
$\Lambda+(1+\epsilon)\operatorname{Vor}(\Lambda)=\mathcal{H}_{\Lambda}$.

###### Proof.

Without loss of generality we assume $\Lambda\neq 0$ and
$\textup{P}(\Lambda)=2$. Let $z\in\mathcal{H}_{\Lambda}$. Choose $y\in\Lambda$
such that $q(z-y)\leq\epsilon+\inf\\{q(z-w)\,|\,w\in\Lambda\\}$. Suppose
$x\in\Lambda\setminus\\{0\\}$. Then

$\displaystyle(1+\epsilon)\langle x,x\rangle-2\langle x,z-y\rangle=\epsilon
q(x)+(q(z-y-x)-q(z-y))\geq\epsilon
q(x)-\epsilon\geq\epsilon(\textup{P}(\Lambda)-1)=\epsilon>0.$

It follows that $2\langle x,(z-y)/(1+\epsilon)\rangle<\langle x,x\rangle$ for
all $x\in\Lambda\setminus\\{0\\}$, so
$(z-y)/(1+\epsilon)\in\operatorname{Vor}(\Lambda)$ by Lemma 6.6. Thus
$z\in\Lambda+(1+\epsilon)\operatorname{Vor}(\Lambda)$, as was to be shown. ∎

######  Proposition 6.10.

For all Hilbert lattices $\Lambda$ the set $\operatorname{Vor}(\Lambda)$
contains the open sphere of radius $\rho(\Lambda)$ around $0\in\Lambda$ and
$\overline{\operatorname{Vor}}(\Lambda)$ is contained in the closed sphere of
radius $r(\Lambda)$ around $0$.

###### Proof.

Let $z\in\mathcal{H}_{\Lambda}$ be such that $\|z\|<\rho(\Lambda)$ and let
$x\in\Lambda\setminus\\{0\\}$. By Cauchy–Schwarz we have $\langle
x,z\rangle\leq\|x\|\cdot\|z\|<\|x\|\cdot\frac{1}{2}\|x\|=\frac{1}{2}\langle
x,x\rangle$, so $z\in\operatorname{Vor}(\Lambda)$ by Lemma 6.6.

Let $z\in\overline{\operatorname{Vor}}(\Lambda)$. For each $r>r(\Lambda)$
there exists $x\in\Lambda$ such that $\|z-x\|\leq r$ by definition of
$r(\Lambda)$. Then by Proposition 6.7 we have $\|z\|\leq\|z-x\|\leq r$. Taking
the limit of $r$ down to $r(\Lambda)$ proves the second inclusion. ∎

######  Theorem 6.11.

Let $\Lambda$ be a Hilbert lattice. Write $C$ for the collection of subsets
$S\subseteq\Lambda$ for which
$\operatorname{Vor}(\Lambda)=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall x\in
S)\ \|z\|<\|z-x\|\\}$. Then $\textup{indec}(\Lambda)\in C$, and it is the
unique minimum with respect to inclusion.

###### Proof.

For $S\subseteq\Lambda$ write $V(S)=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall
x\in S)\ \|z\|<\|z-x\|\\}$.

First suppose $V(S)=\operatorname{Vor}(\Lambda)$ for some $S\subseteq\Lambda$.
Let $z\in\text{indec}(\Lambda)$ and note that
$\frac{1}{2}z\not\in\operatorname{Vor}(\Lambda)$ since
$\|\frac{1}{2}z\|\geq\|\frac{1}{2}z-z\|$. As
$V(S)=\operatorname{Vor}(\Lambda)$ there must be some $x\in S$ such that
$\|\frac{1}{2}z\|\geq\|\frac{1}{2}z-x\|$. Hence $(x,z-x)$ is a decomposition
of $z$ by Lemma 4.3, so $0=x$ or $z=x$ since $z$ is indecomposable. If $0=x\in
S$, then $V(S)=\emptyset\neq\operatorname{Vor}(\Lambda)$, hence $z=x\in S$. We
conclude that $\text{indec}(\Lambda)\subseteq S$.

It remains to show that
$V(\text{indec}(\Lambda))=\operatorname{Vor}(\Lambda)$. We clearly have that
$\operatorname{Vor}(\Lambda)\subseteq V(\text{indec}(\Lambda))$. Suppose $z\in
V(\text{indec}(\Lambda))$ and let $x\in\Lambda\setminus\\{0\\}$. By
Proposition 4.5 we may write $x=\sum_{i=1}^{n}x_{i}$ for some
$n\in\mathbb{Z}_{\geq 1}$ and $x_{i}\in\textup{indec}(\Lambda)$ such that
$\sum_{i=1}^{n}\langle x_{i},x_{i}\rangle\leq\langle x,x\rangle$. Then by
Lemma 6.6 we have

$2\langle x,z\rangle=\sum_{i=1}^{n}2\langle
x_{i},z\rangle<\sum_{i=1}^{n}\langle x_{i},x_{i}\rangle\leq\langle x,x\rangle$

and thus $z\in\operatorname{Vor}(\Lambda)$. We conclude that
$\operatorname{Vor}(\Lambda)=V(\text{indec}(\Lambda))$. ∎

######  Corollary 6.12.

Let $\Lambda$ be a Hilbert lattice. Then
$\overline{\operatorname{Vor}}(\Lambda)=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall
x\in\textup{indec}(\Lambda))\ \|z\|\leq\|z-x\|\\}$.

###### Proof.

By Lemma 6.5.iv we have for $S\subseteq\Lambda$ not containing $0$ that
$\overline{V}(S)=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall x\in S)\
\|z\|\leq\|z-x\|\\}$ is the topological closure of $V(S)$ as defined in the
proof of Theorem 6.11. The corollary then follows from Proposition 6.7 and
Theorem 6.11. ∎

######  Example 6.13.

For a Hilbert lattice $\Lambda$ the set $\text{indec}(\Lambda)$ can fail to be
the minimum among all sets $S\subseteq\Lambda$ such that
$\overline{V}(S)=\\{z\in\mathcal{H}_{\Lambda}\,|\,(\forall x\in S)\
\|z\|\leq\|z-x\|\\}$ equals $\overline{\operatorname{Vor}}(\Lambda)$. We will
give a counterexample.

Let $I=\mathbb{Z}_{\geq 0}\cup\\{\infty\\}$ and let $f:I\to\mathbb{R}_{\geq
0}$ such that $f|_{\mathbb{Z}_{\geq 0}}$ is strictly decreasing with limit
$f(\infty)>0$. Consider the lattice $\Lambda=\smash{\Lambda_{2}^{f}}$ as in
Example 4.9. Note that $\textup{P}(\Lambda)=2f(\infty)^{2}$ and that
$\textup{P}(\Lambda)$ is not attained by any vector. Hence
$2e_{\infty}\in\Lambda$ is indecomposable by Lemma 4.4. Now let
$S=\textup{indec}(\Lambda)\setminus\\{\pm 2e_{\infty}\\}$. We claim that
$\overline{V}(S)=\overline{V}(\text{indec}(\Lambda))$, the latter being equal
to $\overline{\operatorname{Vor}}(\Lambda)$ by Corollary 6.12. It remains to
show for all $z=(z_{i})_{i}\in\overline{V}(S)$ that
$\|z\|\leq\|z-2e_{\infty}\|$ by symmetry.

For all $i$ let $s_{i}\in\\{\pm 1\\}$ such that $s_{i}z_{i}=|z_{i}|$. Since
$f$ is strictly decreasing there exists some $N\in\mathbb{Z}_{\geq 0}$ such
that $f(n)<\sqrt{3}f(\infty)$ for all integers $n\geq N$. For all integers
$n\geq N$ we have $s_{n}e_{n}+e_{\infty}\in S$ by Lemma 4.4 as
$q(s_{n}e_{n}+e_{\infty})=f(n)^{2}+f(\infty)^{2}<4f(\infty)^{2}=2\textup{P}(\Lambda)$.
Then

$0\leq\|z-(s_{n}e_{n}+e_{\infty})\|^{2}-\|z\|^{2}=(1-2|z_{n}|)f(n)^{2}+(1-2z_{\infty})f(\infty)^{2}.$

As $z\in\mathcal{H}_{\Lambda}$ we must have $\lim_{n\to\infty}|z_{n}|=0$, so
taking the limit over the above inequality we get $0\leq
2(1-z_{\infty})f(\infty)^{2}$ and thus $z_{\infty}\leq 1$. But then
$\|z-2e_{\infty}\|^{2}-\|z\|^{2}=4f(\infty)^{2}(1-z_{\infty})\geq 0$ and we
are done.

##  7\. The lattice of algebraic integers

We will write $\overline{\mathbb{Q}}$ for an algebraic closure of
$\mathbb{Q}$. An algebraic integer is an element
$\alpha\in\overline{\mathbb{Q}}$ for which there exists a monic
$f\in\mathbb{Z}[X]$ such that $f(\alpha)=0$. The algebraic integers form a
subring of $\overline{\mathbb{Q}}$, which we denote $\overline{\mathbb{Z}}$.
In this section we will prove that $\overline{\mathbb{Z}}$ together with a
natural choice of square-norm is a Hilbert lattice.

######  Definition 7.1.

For a ring $K$ we define the fundamental set to be the set
$\operatorname{X}(K)$ of ring homomorphisms from $K$ to $\mathbb{C}$. For a
ring $L$ with subring $K$ and $\sigma\in\operatorname{X}(K)$ we define
$\operatorname{X}_{\sigma}(L)=\\{\rho\in\operatorname{X}(L)\,|\,\rho|_{K}=\sigma\\}$.

######  Lemma 7.2.

Let $\alpha\in\overline{\mathbb{Q}}$ and $\mathbb{Q}(\alpha)\subseteq
L\subseteq\overline{\mathbb{Q}}$ subfields with $[L:\mathbb{Q}]<\infty$. Then
the quantities

$\prod_{\sigma\in\operatorname{X}(L)}|\sigma(\alpha)|^{1/[L:\mathbb{Q}]}\quad\text{and}\quad\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}|\sigma(\alpha)|^{2}$

are in $\mathbb{R}_{\geq 0}$, equal to zero if and only if $\alpha=0$, and do
not depend on the choice of $L$. ∎

######  Definition 7.3.

We define the maps $N,q:\overline{\mathbb{Q}}\to\mathbb{R}_{\geq 0}$ by

$N(\alpha)=\prod_{\sigma\in\operatorname{X}(\mathbb{Q}(\alpha))}|\sigma(\alpha)|^{1/[\mathbb{Q}(\alpha):\mathbb{Q}]}\quad\text{and}\quad
q(\alpha)=\frac{1}{[\mathbb{Q}(\alpha):\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(\mathbb{Q}(\alpha))}|\sigma(\alpha)|^{2}.$

######  Lemma 7.4.

For $\alpha,\beta\in\overline{\mathbb{Q}}$ we have
$q(\alpha+\beta)+q(\alpha-\beta)=2q(\alpha)+2q(\beta)$.

###### Proof.

By Lemma 7.2 the restriction of $q$ to $L=\mathbb{Q}(\alpha,\beta)$ is given
by
$q(\gamma)=\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}|\sigma(\gamma)|^{2}$.
The norm $|\cdot|$ on $\mathbb{C}$ satisfies the parallelogram law and we may
apply this term-wise to the sum defining $q$ to obtain the lemma. ∎

######  Lemma 7.5 (AM-GM inequality, Theorem 5.1 in [3]).

Let $n\in\mathbb{Z}_{\geq 1}$ and $x_{1},\dotsc,x_{n}\in\mathbb{R}_{\geq 0}$.
Then

$\sqrt[n]{x_{1}\dotsm x_{n}}\leq\frac{x_{1}+\dotsm+x_{n}}{n},$

with equality if and only if $x_{1}=x_{2}=\dotsm=x_{n}$. ∎

######  Definition 7.6.

An element $\delta\in\overline{\mathbb{Q}}$ is called uniform if
$|\sigma(\delta)|=|\tau(\delta)|$ for all
$\sigma,\tau\in\operatorname{X}(\overline{\mathbb{Q}})$.

######  Lemma 7.7.

For all $\alpha\in\overline{\mathbb{Q}}$ we have $N(\alpha)^{2}\leq q(\alpha)$
with equality if and only if $\alpha$ is uniform.

###### Proof.

This lemma follows from a straightforward application of Lemma 7.5:

$q(\alpha)=\frac{1}{[\mathbb{Q}(\alpha):\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(\mathbb{Q}(\alpha))}|\sigma(\alpha)|^{2}\geq\Big{(}\prod_{\sigma\in\operatorname{X}(\mathbb{Q}(\alpha))}|\sigma(\alpha)|^{2}\Big{)}^{1/[\mathbb{Q}(\alpha):\mathbb{Q}]}=N(\alpha)^{2},$

with equality if and only if $|\sigma(\alpha)|^{2}=|\rho(\alpha)|^{2}$ for all
$\sigma,\rho\in\operatorname{X}(\mathbb{Q}(\alpha))$. ∎

######  Proposition 7.8.

If $\alpha\in\overline{\mathbb{Z}}$, then $q(\alpha)\leq 1$ if and only if
$\alpha=0$ or $\alpha$ is a root of unity. If $\alpha$ is a root of unity,
then $q(\alpha)=1$.

###### Proof.

Let $\alpha\in\overline{\mathbb{Z}}$. The ‘if’ part of the implication follows
directly from the definition. For the ‘only if’ part, suppose $\alpha$ is non-
zero. If $q(\alpha)\leq 1$, then $N(\alpha)^{2}\leq 1$ by Lemma 7.7. Then
$N(\alpha)^{[\mathbb{Q}(\alpha):\mathbb{Q}]}=|N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)|\in\mathbb{Z}_{\geq
1}$, so $N(\alpha)^{2}=q(\alpha)=1$. By Lemma 7.7 we have $|\sigma(\alpha)|=1$
for all $\sigma\in\operatorname{X}(\mathbb{Q}(\alpha))$, so $\alpha$ is a root
of unity by Kronecker’s theorem [4]. ∎

Recall for a Hilbert lattice $\Lambda$ with square-norm $q$ the definition
$\textup{P}(\Lambda)=\\{q(x)\,|\,x\in\Lambda\setminus\\{0\\}\\}$ from
Definition 3.1.

######  Theorem 7.9.

The group $\overline{\mathbb{Z}}$ with square-norm $q$ is a Hilbert lattice
and $\textup{P}(\overline{\mathbb{Z}})=1$.

###### Proof.

By Lemma 7.4 and Proposition 7.8 respectively the group
$\overline{}\mathbb{Z}$ together with $q$ satisfies the parallelogram law and
has
$\inf\\{q(\alpha)\,|\,\alpha\in\overline{\mathbb{Z}}\setminus\\{0\\}\\}=1$. ∎

######  Remark 7.10.

Since $\overline{\mathbb{Z}}$ is a Hilbert lattice, it is a discrete subgroup
of a Hilbert space by Theorem 3.7, which has an inner product. For
$\alpha,\beta\in\overline{\mathbb{Q}}$ and $\mathbb{Q}(\alpha,\beta)\subseteq
L\subseteq\overline{\mathbb{Q}}$ with $[L:\mathbb{Q}]<\infty$ it is given by

$\langle\alpha,\beta\rangle=\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}\sigma(\alpha)\overline{\sigma(\beta)}.$

######  Lemma 7.11.

Suppose $\alpha,\delta\in\overline{\mathbb{Q}}$ and $\delta$ is uniform. Then
$q(\alpha\delta)=q(\alpha)q(\delta)$. If also
$\alpha,\delta\in\overline{\mathbb{Z}}$ and $\alpha\delta$ is indecomposable,
then $\alpha$ is indecomposable.

###### Proof.

Let $L\supseteq\mathbb{Q}(\alpha,\delta)$. Then for all
$\sigma\in\operatorname{X}(L)$ we have $q(\delta)=|\sigma(\delta)|^{2}$.
Moreover,

$q(\alpha\delta)=\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}|\sigma(\alpha\delta)|^{2}=\frac{1}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(L)}|\sigma(\alpha)|^{2}\cdot
q(\delta)=q(\alpha)q(\delta).$

Now suppose $\alpha,\delta\in\overline{\mathbb{Z}}$ and let
$(\beta,\gamma)\in\textup{dec}(\alpha)$. Then
$\alpha\delta=\beta\delta+\gamma\delta$ and

$q(\alpha\delta)=q(\alpha)q(\delta)\geq(q(\beta)+q(\gamma))q(\delta)=q(\beta\delta)+q(\gamma\delta),$

so $(\beta\delta,\gamma\delta)\in\textup{dec}(\alpha\delta)$. If
$\alpha\delta$ is indecomposable, then $\delta\neq 0$ and
$0\in\\{\beta\delta,\gamma\delta\\}$, so $0\in\\{\beta,\gamma\\}$ and
$(\beta,\gamma)$ must be a trivial decomposition. Hence $\alpha$ is
indecomposable. ∎

######  Definition 7.12.

Write $\mu_{\infty}$ for the group of roots of unity in
$\overline{\mathbb{Z}}$ and $\textup{Gal}(\overline{\mathbb{Q}})$ for the
group of ring automorphisms of $\overline{\mathbb{Q}}$. Note that
$\textup{Gal}(\overline{\mathbb{Q}})$ naturally acts on $\mu_{\infty}$, and
write $\mu_{\infty}\rtimes\textup{Gal}(\overline{\mathbb{Q}})$ for their semi-
direct product with respect to this action.

######  Lemma 7.13.

The group $\mu_{\infty}\rtimes\textup{Gal}(\overline{\mathbb{Q}})$ acts
faithfully on the Hilbert lattice $\overline{\mathbb{Z}}$, where
$\mu_{\infty}$ acts by multiplication and
$\textup{Gal}(\overline{\mathbb{Q}})$ by application.

###### Proof.

Let $\alpha\in\overline{\mathbb{Z}}$, $\zeta\in\mu_{\infty}$ and
$\rho\in\textup{Gal}(\overline{\mathbb{Q}})$. Let $K$ be the normal closure of
$\mathbb{Q}(\zeta,\alpha)$ and $n=[K:\mathbb{Q}]$.

First we show that the individual group actions on $\overline{\mathbb{Z}}$ are
well-defined. Clearly $\zeta\alpha\in\overline{}\mathbb{Z}$ and note that
$\zeta$ is uniform with $q(\zeta)=1$. Hence multiplication by $\zeta$ map is
an isometry, i.e. preserves length, by Lemma 7.11. Recall that automorphisms
preserve integrality and thus $\rho(\alpha)\in\overline{\mathbb{Z}}$. Since
$K$ is normal over $\mathbb{Q}$ we have $\rho K=K$ and thus
$\operatorname{X}(K)\circ\rho=\operatorname{X}(K)$. Hence applying $\rho$ to
$\alpha$ simply results in a reordering of the terms in the sum defining $q$
with respect to $K$, and thus $\rho$ is an isometry.

Note that for
$(\chi,\sigma),(\xi,\tau)\in\mu_{\infty}\rtimes\textup{Gal}(\overline{\mathbb{Q}})$
we have

$(\chi,\sigma)\big{(}(\xi,\tau)\,\alpha\big{)}=\chi\cdot\sigma(\xi\cdot\tau(\alpha))=(\chi\sigma(\xi))((\sigma\tau)(\alpha))=\big{(}(\chi,\sigma)\cdot(\xi,\tau)\big{)}\,\alpha,$

so the semi-direct product acts on $\overline{\mathbb{Z}}$ as well. Finally,
suppose $(\zeta,\rho)$ acts as the identity. Note that
$\textup{Gal}(\overline{\mathbb{Q}})$ fixes $1$, so letting $(\zeta,\rho)$ act
on $1$ shows that $\zeta=1$, and thus $\rho=\textup{id}$. Hence the action is
faithful. ∎

######  Question 7.14.

Is $\mu_{\infty}\rtimes\textup{Gal}(\overline{\mathbb{Q}})$ the entire
isometry group of $\overline{\mathbb{Z}}$?

######  Proposition 7.15.

Let $\alpha\in\overline{\mathbb{Z}}$, $r\in\mathbb{Z}_{\geq 0}$ and
$s\in\mathbb{Z}_{>0}$ such that $r/s\leq 1$. Then any root $\beta$ of
$X^{s}-\alpha^{r}$ satisfies $q(\beta)\leq q(\alpha)^{r/s}$.

###### Proof.

Let $\beta$ be a root of $X^{s}-\alpha^{r}$, let $K=\mathbb{Q}(\alpha,\beta)$
and $n=[K:\mathbb{Q}]$. The case $r=0$ follows from Proposition 7.8, so
suppose $r>0$. Then

$\displaystyle q(\beta)$
$\displaystyle=\frac{1}{n}\sum_{\sigma\in\operatorname{X}(K)}|\sigma(\beta)|^{2}=\frac{1}{n}\sum_{\sigma\in\operatorname{X}(K)}|\sigma(\beta^{s})|^{2/s}=\frac{1}{n}\sum_{\sigma\in\operatorname{X}(K)}|\sigma(\alpha^{r})|^{2/s}$
$\displaystyle=\frac{1}{n}\sum_{\sigma\in\operatorname{X}(K)}\big{(}|\sigma(\alpha)|^{2}\big{)}^{r/s}\leq\Big{(}\frac{1}{n}\sum_{\sigma\in\operatorname{X}(K)}|\sigma(\alpha)|^{2}\Big{)}^{r/s}=q(\alpha)^{r/s},$

where the inequality is the inequality $n^{-1/r}\|x\|_{r}\leq
n^{-1/s}\|x\|_{s}$ from Lemma 2.14 applied to the vector
$x=(|\sigma(\alpha)|^{2/s})_{\sigma\in\operatorname{X}(K)}$, using that
$0<r\leq s$. ∎

##  8\. Indecomposable algebraic integers

We will now focus on the indecomposables of the lattice
$\overline{\mathbb{Z}}$.

$\alpha$01i$\beta$$\gamma$ Figure 3. Integral
$\alpha=\frac{1}{2}(1+\textup{i}\sqrt{7})$ with $q(\alpha)=2$.

######  Proposition 8.1.

Let $\alpha\in\overline{\mathbb{Z}}$. If $0<q(\alpha)<2$, then $\alpha$ is
indecomposable. If $q(\alpha)=2$, then $\alpha$ is decomposable if and only if
it is the sum of two roots of unity. Such roots of unity are necessarily
orthogonal, unique up to reordering, and of degree at most $2$ over
$\mathbb{Q}(\alpha)$.

###### Proof.

If $0<q(\alpha)<2$, then $\alpha$ is indecomposable by combining Theorem 7.9
and Lemma 4.4. Suppose $q(\alpha)=2$. If $\alpha=\zeta+\xi$ for roots of unity
$\zeta,\xi\in\overline{\mathbb{Q}}$, then $q(\alpha)=2=q(\zeta)+q(\xi)$, so
$(\zeta,\xi)\in\textup{dec}(\alpha)$ is non-trivial. Conversely, suppose
$(\beta,\gamma)\in\text{dec}(\alpha)$ is non-trivial. By Theorem 7.9 we have
$q(\beta),q(\gamma)\geq 1$. Then $0\leq
q(\alpha)-q(\beta)-q(\gamma)=2-q(\beta)-q(\gamma)\leq 0$, so we must have
$q(\beta)=q(\gamma)=1$. It follows that $\beta$ and $\gamma$ are orthogonal,
and by Proposition 7.8 they are roots of unity.

Suppose $(\beta,\gamma)\in\text{dec}(\alpha)$ is non-trivial. For any
$\sigma\in\operatorname{X}(\mathbb{Q}(\alpha))$ and
$\rho\in\operatorname{X}_{\sigma}(\mathbb{Q}(\alpha,\beta))$ the points $0$,
$\rho(\alpha)$, $\rho(\beta)$ and $\rho(\gamma)$ form the vertices of a
rhombus with unit length sides, as can be seen in Figure 3. It follows that
$\\{\rho(\beta),\rho(\gamma)\\}$ is uniquely determined by
$\rho(\alpha)=\sigma(\alpha)$. As $\rho$ is uniquely determined by
$\rho(\beta)$ there are at most two elements in
$\operatorname{X}_{\sigma}(\mathbb{Q}(\alpha,\beta))$, in other words
$[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]\leq 2$. ∎

######  Remark 8.2.

Proposition 8.1 gives us a way to decide whether an
$\alpha\in\overline{\mathbb{Z}}$ with $q(\alpha)=2$ is indecomposable, as it
puts an upper bound on the degree of the roots of unity, leaving only finitely
many to check. Knowledge of $\mathbb{Q}(\alpha)$ can further reduce this
number.

######  Example 8.3.

There exist $\alpha\in\overline{\mathbb{Z}}$ with $q(\alpha)=2$ which are
indecomposable. Consider $f=X^{2}-X+2$ with root
$\alpha\in\overline{\mathbb{Z}}$, as in Figure 3. Note that the roots of $f$
in $\mathbb{C}$ are $\frac{1}{2}(1\pm\textup{i}\sqrt{7})$ with absolute value
$\frac{1}{2}\sqrt{1^{2}+7}=\sqrt{2}$, so $q(\alpha)=2$. Suppose
$(\beta,\gamma)\in\textup{dec}(\alpha)$ is a non-trivial decomposition.
Proposition 8.1 shows that $\beta$ and $\gamma$ are roots of unity. Note that
$|\alpha^{2}|=2$ under all embeddings of $\alpha$ in $\mathbb{C}$, and
$\alpha^{2}=|\alpha|^{2}\cdot\beta\gamma$. Hence
$\alpha^{2}/2=\frac{1}{2}\alpha-1$ is a root of unity. Either one notes that
$\alpha^{2}/2$ is not even integral, or that $\alpha^{2}/2=\pm 1$, as those
are the only roots of unity in $\mathbb{Q}(\alpha)$, which is clearly absurd.
Hence we have a contradiction and $\alpha$ is indecomposable.

######  Lemma 8.4.

Let $\mathbb{Q}\subseteq K\subseteq L\subseteq\overline{\mathbb{Q}}$ be fields
with $[L:\mathbb{Q}]<\infty$. Then for all $\alpha\in K$ and $\beta\in L$ we
have

$[L:K]\cdot\langle\alpha,\beta\rangle=\langle\alpha,\operatorname{Tr}_{L/K}(\beta)\rangle.$

###### Proof.

Recall for $\sigma\in\operatorname{X}(K)$ the definition
$\operatorname{X}_{\sigma}(L)=\\{\rho\in\operatorname{X}(L)\,|\,\rho|_{K}=\sigma\\}$
from Definition 7.1. For all $\sigma\in\operatorname{X}(K)$ and $\beta\in L$
we have
$\sigma(\operatorname{Tr}_{L/K}(\beta))=\sum_{\rho\in\operatorname{X}_{\sigma}(L)}\rho(\beta)$.
Then with $\alpha\in K$ and $\beta\in L$ we have

$\displaystyle[L:K]\cdot\langle\alpha,\beta\rangle$
$\displaystyle=\frac{[L:K]}{[L:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(K)}\sum_{\rho\in\operatorname{X}_{\sigma}(L)}\rho(\alpha)\overline{\rho(\beta)}=\frac{1}{[K:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(K)}\sigma(\alpha)\overline{\sum_{\rho\in\operatorname{X}_{\sigma}(L)}\rho(\beta)}$
$\displaystyle=\frac{1}{[K:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(K)}\sigma(\alpha)\overline{\sigma(\operatorname{Tr}_{L/K}(\beta))}=\langle\alpha,\operatorname{Tr}_{L/K}(\beta)\rangle,$

as was to be shown. ∎

######  Proposition 8.5.

Roots of unity $\zeta,\xi\in\overline{\mathbb{Z}}$ are orthogonal, i.e.
$\langle\zeta,\xi\rangle=0$, if and only if $\zeta^{-1}\xi$ does not have
square-free order.

###### Proof.

Let $K=\mathbb{Q}(\zeta^{-1}\xi)$. We have
$[K:\mathbb{Q}]\cdot\langle\zeta,\xi\rangle=[K:\mathbb{Q}]\cdot\langle
1,\zeta^{-1}\xi\rangle=\operatorname{Tr}_{K/\mathbb{Q}}(\zeta^{-1}\xi)$ by
Lemma 7.13 and Lemma 8.4. Recall that the trace of an $n$-th root of unity
equals $\mu(n)$, the Möbius function, which is zero precisely when $n$ has a
square divisor in $\mathbb{Z}_{>1}$. ∎

For $\alpha,\beta\in\overline{\mathbb{Z}}$ we say $\beta$ divides $\alpha$,
and write $\beta\mid\alpha$, if there exists some
$\gamma\in\overline{\mathbb{Z}}$ such that $\alpha=\beta\gamma$. We write
$\beta\nmid\alpha$ if $\beta$ does not divide $\alpha$. Recall from Definition
7.6 that for $\delta\in\overline{\mathbb{Z}}$ we say $\delta$ is uniform if
$|\sigma(\delta)|=|\tau(\delta)|$ for all
$\sigma,\tau\in\operatorname{X}(\overline{\mathbb{Q}})$

######  Proposition 8.6.

If $\alpha\in\overline{\mathbb{Z}}$ is such that $\sqrt{2}\mid\alpha$ or
$\sqrt{3}\mid\alpha$, then
$\alpha\not\in\textup{indec}(\overline{\mathbb{Z}})$.

###### Proof.

Let $\zeta\in\overline{\mathbb{Q}}$ be a primitive $8$-th root of unity, which
we may choose such that $\zeta+\zeta^{-1}=\sqrt{2}$. Thus
$(\zeta,\zeta^{-1})\in\textup{dec}(\sqrt{2})$, because
$\langle\zeta,\zeta^{-1}\rangle=0$ by Proposition 8.5. Moreover, $\sqrt{2}$,
$\zeta$ and $\zeta^{-1}$ are all uniform. For any
$\beta\in\overline{\mathbb{Z}}$ we get from Lemma 7.11 that

$q(\zeta\beta)+q(\zeta^{-1}\beta)=(q(\zeta)+q(\zeta^{-1}))\cdot
q(\beta)=q(\sqrt{2})\cdot q(\beta)=q(\sqrt{2}\beta),$

so $\sqrt{2}\beta$ has a non-trivial decomposition.

With $\xi$ a twelfth root of unity we have $\xi+\xi^{-1}=\sqrt{3}$ with $\xi$,
$\xi^{-1}$ and $\sqrt{3}$ uniform. We have a decomposition because
$\langle\xi,\xi^{-1}\rangle=\langle 1,\xi^{-2}\rangle=\frac{1}{2}\geq 0$, so
the argument from before applies. ∎

We will now begin the proof of Theorem 1.6.

######  Lemma 8.7.

If $\alpha\in\overline{\mathbb{Z}}$ is such that $\sqrt{2}\nmid\alpha\mid 2$
and $\alpha$ is uniform, then
$\alpha\in\textup{indec}(\overline{\mathbb{Z}})$.

###### Proof.

By assumption we may write $2=\alpha\gamma$ for some non-zero
$\gamma\in\overline{\mathbb{Z}}$. Note that $\gamma$ is not a unit, since
otherwise $\sqrt{2}\mid 2\mid\alpha$. Now let
$(\beta,\alpha-\beta)\in\textup{dec}(\alpha)$. Then by Lemma 4.3 and Lemma
7.11 we have $q(\alpha)\geq
q(\alpha-2\beta)=q(\alpha-\alpha\beta\gamma)=q(\alpha)\cdot q(1-\beta\gamma)$,
so $q(1-\beta\gamma)\leq 1$. As $\gamma$ is not a unit we have
$\beta\gamma\neq 1$, so $\beta\gamma=1-\zeta$ for some root of unity $\zeta$
of order say $n$ by Proposition 7.8. Suppose $n$ is not a power of $2$. Then
$1-\zeta$ and $2$ are coprime. As $2\mid 2\beta=\alpha(1-\zeta)$ we have that
$2\mid\alpha$, which contradicts $\sqrt{2}\nmid\alpha$. Hence $n$ is a power
of $2$. If $n>2$, then $1-\zeta\mid\sqrt{2}$ so $\sqrt{2}\mid\alpha$, which is
again a contradiction. Therefore $n=1$ or $n=2$, which correspond to the
trivial decompositions with $\beta=0$ and $\beta=\alpha$ respectively. We
conclude that $\alpha$ is indecomposable. ∎

Theorem 1.6. It holds that
$2\sqrt{2}\leq\sup\\{\langle\alpha,\alpha\rangle\,|\,\textup{$\alpha\in\overline{\mathbb{Z}}$
is indecomposable}\\}$.

###### Proof.

We will prove that for each $r\in\mathbb{Q}\cap[1,3/2)$ there exists
$\alpha\in\textup{indec}(\overline{\mathbb{Z}})$ such that $q(\alpha)=2^{r}$.

Consider $\beta=\frac{1+\sqrt{-7}}{2}$ as in Example 8.3 and write
$\overline{\beta}=1-\beta$ for its conjugate. Write $r=\frac{a}{b}$ with
integers $a\geq b>0$ and let $\gamma\in\overline{\mathbb{Z}}$ be a zero of
$X^{b}-\beta$. Now take $\alpha=\overline{\beta}\cdot\gamma^{a-b}$. We will
show $\alpha$ satisfies the conditions to Lemma 8.7. Because
$|\sigma(\alpha)|=|\sigma(\beta)|^{r}=2^{r/2}$ for all
$\sigma\in\operatorname{X}(\overline{\mathbb{Q}})$, and hence $\alpha$ is
uniform, we then have that $\alpha$ is indecomposable and $q(\alpha)=2^{r}$.

Note that $\alpha\cdot\gamma^{2b-a}=\overline{\beta}\cdot\beta=2$, so
$\alpha\mid 2$. Let $v:K\to\mathbb{R}\cup\\{\infty\\}$ be a valuation over $2$
for some number field $K$ which is Galois over $\mathbb{Q}$ containing the
relevant elements. Because
$0=v(1)=v(\overline{\beta}+\beta)\geq\min\\{v(\overline{\beta}),v(\beta)\\}$,
we have $v(\beta)=0$ or $v(\overline{\beta})=0$. By potentially composing $v$
with an automorphism swapping $\beta$ and $\overline{\beta}$ we obtain a
valuation $v^{\prime}$ such that $v^{\prime}(\overline{\beta})=0$. We have
$1=v^{\prime}(2)=v^{\prime}(\beta\cdot\overline{\beta})=v^{\prime}(\beta)$ and
thus $v^{\prime}(\gamma)=1/b$. Then
$v^{\prime}(\alpha)=r-1<1/2=v^{\prime}(\sqrt{2})$, from which we conclude that
$\sqrt{2}\nmid\alpha$. Thus $\alpha$ satisfies the conditions to Lemma 8.7, as
was to be shown. ∎

##  9\. Decompositions of the lattice of algebraic integers

In this section we will show that $\overline{\mathbb{Z}}$ is indecomposable as
a Hilbert lattice. The following is a standard result from linear algebra.

######  Lemma 9.1.

Let $V$ be a vector space over an infinite field and let $S$ be a finite set
of subspaces of $V$. If $\bigcup_{U\in S}U=V$, then $V\in S$. ∎

######  Proposition 9.2.

Let $S\subseteq\overline{\mathbb{Z}}$ with $S$ finite and $0\not\in S$. Then
there exist $\alpha\in\textup{indec}(\overline{\mathbb{Z}})$ such that
$\langle\alpha,\beta\rangle\neq 0$ for all $\beta\in S$.

###### Proof.

Let $K$ be the field generated by $S$ and fix $1<r<\sqrt{2}$.

We will construct an element $u\in\mathcal{O}_{K}$ such that $0\not\in\langle
u,S\rangle$ and $|\sigma(u)|>r$ for all $\sigma\in\operatorname{X}(K)$. For
$x\in K$ write $x^{\bot}=\\{y\in K\,|\,\langle x,y\rangle=0\\}$, which is a
proper $\mathbb{Q}$-vector subspace of $K$ when $x\neq 0$ because $x\not\in
x^{\bot}$. Hence $\bigcup_{x\in S}x^{\bot}\neq K$ by Lemma 9.1, so there
exists some non-zero $u\in K$ such that $0\not\in\langle u,S\rangle$. By
scaling $u$ by some non-zero integer we may assume $u\in\overline{\mathbb{Z}}$
as well. By further scaling $u$ with integers we may assume $|\sigma(u)|>r$
for all $\sigma\in\operatorname{X}(K)$, as was to be shown.

As $|\sigma(u)|>r$ for all $\sigma\in\operatorname{X}(K)$ we have $N(u)>r>1$,
where $N$ is as in Definition 7.3, so $u$ is not a unit. Let
$\mathfrak{p}\subseteq\mathcal{O}_{K}$ be a prime containing $u$ and let
$v\in\mathfrak{p}\setminus\mathfrak{p}^{2}$. Let
$f_{n}=X^{n}-uX^{n-1}-v\in\mathcal{O}_{K}[X]$ for $n\geq 2$ and note that it
is Eisenstein at $\mathfrak{p}$ and therefore irreducible. Let
$\alpha_{n}\in\overline{\mathbb{Z}}$ be a root of $f_{n}$. It suffices to show
that for $n$ sufficiently large $\alpha_{n}$ is indecomposable and satisfies
$0\not\in\langle\alpha_{n},S\rangle$. By Lemma 8.4 and by construction of $u$
it holds for any $n\geq 2$ that

$\langle\alpha_{n},S\rangle=\frac{\langle\operatorname{Tr}_{K(\alpha_{n})/K}(\alpha_{n}),S\rangle}{[K(\alpha_{n}):K]}=\frac{\langle
u,S\rangle}{[K(\alpha):K]}\not\ni 0,$

so it remains to be shown that $\alpha_{n}$ is indecomposable for $n$
sufficiently large.

Let $D\subseteq\mathbb{C}$ be the closed disk of radius $r$ around $0$. Let
$n$ be sufficiently large such that we have $|\sigma(v)|\cdot
r^{1-n}<|\sigma(u)|-r$ for all $\sigma\in\operatorname{X}(K)$. Fix
$\sigma\in\operatorname{X}(K)$. For all $x$ on the boundary of $D$ we have

$|x^{n}-\sigma(v)|\leq r^{n}+|\sigma(v)|=r^{n-1}(r+|\sigma(v)|\cdot
r^{1-n})<|\sigma(u)|\cdot r^{n-1}=|\sigma(u)\cdot x^{n-1}|.$

Hence by Rouché’s Theorem $\sigma(u)X^{n-1}$ and
$\sigma(f_{n})=(X^{n}-\sigma(v))-\sigma(u)X^{n-1}$ have the same number of
zeros in $D$, counting multiplicities, which for $\sigma(u)X^{n-1}$ clearly is
$n-1$. For the remaining zero $x_{\sigma,n}\in\mathbb{C}$ of $\sigma(f_{n})$
with $|x_{\sigma,n}|>r$ we have
$x_{\sigma,n}^{n-1}(x_{\sigma,n}-\sigma(u))=\sigma(v)$ and thus

$|x_{\sigma,n}-\sigma(u)|=|\sigma(v)|\cdot|x_{\sigma,n}|^{1-n}<|\sigma(v)|\cdot
r^{1-n}\to 0\quad(\text{as }n\to\infty),$

i.e. $\lim_{n\to\infty}x_{\sigma,n}=\sigma(u)$. Now summing over all
$\sigma\in\operatorname{X}(K)$ we get

$\displaystyle q(\alpha_{n})$
$\displaystyle=\frac{1}{n\cdot[K:\mathbb{Q}]}\sum_{\sigma\in
X(K)}\sum_{\rho\in\operatorname{X}_{\sigma}(K(\alpha_{n}))}|\rho(\alpha_{n})|^{2}\leq\frac{1}{n\cdot[K:\mathbb{Q}]}\sum_{\sigma\in
X(K)}\big{(}(n-1)r^{2}+|x_{\sigma,n}|^{2}\big{)}$ $\displaystyle\leq
r^{2}+\frac{1}{n\cdot[K:\mathbb{Q}]}\sum_{\sigma\in\operatorname{X}(K)}|x_{\sigma,n}|^{2}\to
r^{2}\quad(\text{as }n\to\infty).$

Because $r^{2}<2$ we have for sufficiently large $n$ that $q(\alpha_{n})<2$.
From Proposition 8.1 we may then conclude that $\alpha_{n}$ is indecomposable,
as was to be shown. ∎

Theorem 1.1. The Hilbert lattice $\overline{\mathbb{Z}}$ is indecomposable.

###### Proof.

Let $\beta,\gamma\in\textup{indec}(\overline{}\mathbb{Z})$. Then there exists
some $\alpha\in\textup{indec}(\overline{\mathbb{Z}})$ such that
$\langle\alpha,\beta\rangle\neq 0\neq\langle\alpha,\gamma\rangle$ by
Proposition 9.2. Hence $\alpha$, $\beta$ and $\gamma$ must be in the same
connected component of the graph of Theorem 5.4. As this holds for all $\beta$
and $\gamma$ the graph is connected and hence $\overline{\mathbb{Z}}$ is
indecomposable. ∎

##  10\. Universal gradings

######  Definition 10.1.

Let $R$ be a commutative ring. A grading of $R$ is a pair $(G,\mathcal{R})$
where $G$ is an abelian group and $\mathcal{R}$ is a $G$-indexed collection
$\\{R_{g}\\}_{g}$ of additive subgroups of $R$ such that the natural map
$\bigoplus_{g\in G}R_{g}\to R$ is an isomorphism and for all $g,h\in\Gamma$ we
have $R_{g}\cdot R_{h}\subseteq R_{gh}$. We equip the set of gradings of $R$
with a category structure, where the morphisms $(G,\\{R_{g}\\}_{g\in G})$ to
$(H,\\{R_{h}^{\prime}\\}_{h\in H})$ are the group homomorphisms $f:G\to H$
such that $R_{h}^{\prime}=\bigoplus_{g\in f^{-1}\\{h\\}}R_{g}$ for all $h\in
H$. A universal grading of $R$ is a universal object in the category of
gradings of $R$.

Note that a universal grading, when it exists, is uniquely unique, i.e. for
any two universal gradings of a given commutative ring there exists a unique
isomorphism between them.

An order is a commutative ring for which the additive group is isomorphic to
$\mathbb{Z}^{r}$ for some $r\in\mathbb{Z}_{\geq 0}$. Lenstra and Silverberg
[12] showed using a classical version of Theorem 5.4, that every grading of a
reduced order, in particular every order in $\overline{\mathbb{Q}}$, has a
universal grading. Theorem 5.4 is powerful enough to prove Theorem 1.7, which
states that every subring of $\overline{\mathbb{Z}}$ has a universal grading.
We quickly go through the proof, which generalizes that of Lenstra and
Silverberg.

######  Lemma 10.2.

Suppose $R$ is a commutative ring with a grading $(G,\\{R_{g}\\}_{g})$. Then
$R_{1}\subseteq R$ is a subring. If $y\in R_{h}$ for some $h\in G$ is
invertible in $R$, then $y^{-1}\in R_{h^{-1}}$ and $yR_{1}=R_{h}$.

###### Proof.

It follows directly from the definition that $R_{1}$ is an additive group and
closed under multiplication. Write $1=\sum_{g\in G}x_{g}$ with $x_{g}\in
R_{g}$ for all $g$. Then $x_{h}=1\cdot x_{h}=\sum_{g\in G}x_{g}x_{h}$ with
$x_{g}x_{h}\in R_{gh}$ for all $h\in G$. Since $\\{R_{g}\\}_{g}$ is a
decomposition of $R$ we have $x_{1}x_{h}=x_{h}$ and thus $x_{1}\cdot 1=1$.
Hence $1=x_{1}\in R_{1}$, and $R_{1}$ is a subring of $R$.

As before write $y^{-1}=\sum_{g\in G}z_{g}$ with $z_{g}\in R_{g}$ for all $g$.
As $R_{1}\ni 1=yy^{-1}=\sum_{g\in G}yz_{g}$ it follows that $yz_{h^{-1}}=1$
and thus $y^{-1}=z_{h^{-1}}\in R_{h^{-1}}$. Finally $yR_{1}\subseteq
R_{h}=yy^{-1}R_{h}\subseteq yR_{1}$, so that $yR_{1}=R_{h}$. ∎

######  Lemma 10.3.

Suppose $R\subseteq\overline{\mathbb{Z}}$ is a subring and
$(G,\\{R_{g}\\}_{g})$ is a grading of $R$. Then $H=\\{g\in G\,|\,R_{g}\neq
0\\}$ is a countable torsion subgroup of $G$. Moreover, if
$(G,\\{R_{g}\\}_{g})$ is universal, then $G=H$.

###### Proof.

Since $0$ is the only zero-divisor in $\overline{\mathbb{Z}}$, we have for
$g,h\in H$ that $0\subsetneq R_{g}R_{h}\subseteq R_{gh}$, so $gh\in H$. For
$g\in H$ and $x\in R_{g}$ non-zero we have $x^{n}=\sum_{i=0}^{n-1}a_{i}x^{i}$
for some $n\in\mathbb{Z}_{\geq 1}$ and $a_{i}\in\mathbb{Z}$, so $0\neq
x^{n}\in R_{g^{n}}\cap\sum_{i=0}^{n-1}R_{g^{i}}$. Hence $g^{n}=g^{i}$ for some
$0\leq i<n$, so the order of $g$ is finite, and $H$ is a torsion group. That
it is countable follows from the countability of $R$.

Suppose now that $(G,\\{R_{g}\\}_{g})$ is universal. By what we have shown