#  On stratifications and poset-stratified spaces

Lukas Waas, Jon Woolf and Shoji Yokura  Ruprecht–Karls–Universität Heidelberg,
Mathematisches Institut, Im Neuenheimer Feld 205, 69120 Heidelberg,
Deutschland<EMAIL_ADDRESS>Department of Mathematical
Sciences, University of Liverpool, L69 7ZL United Kingdom
<EMAIL_ADDRESS>Graduate School of Science and Engineering,
Kagoshima University, 1-21-35 Korimoto, Kagoshima, 890-0065, Japan
<EMAIL_ADDRESS>
###### Abstract.

A stratified space is a topological space equipped with a _stratification_ ,
which is a decomposition or partition of the topological space satisfying
certain extra conditions. More recently, the notion of poset-stratified space,
i.e., topological space endowed with a continuous map to a poset with its
Alexandrov topology, has been popularized. Both notions of stratified spaces
are ubiquitous in mathematics, ranging from investigations of singular
structures in algebraic geometry to extensions of the homotopy hypothesis in
higher category theory. In this article we study the precise mathematical
relation between these different approaches to stratified spaces.

_keywords_ : stratification, poset-stratified space, decomposition space,
poset, Alexandrov topology
_Mathematics Subject Classification 2000_ : 54B15, 57N80, 32S60, 06A99

##  1\. Introduction

Stratified spaces are ubiquitous in topology and geometry. For example, they
arise naturally in the presence of group actions, singularities, and when the
space considered is a configuration or moduli space. Classically, a
stratification is a decomposition of the space satisfying certain conditions.
More recently, stratified spaces have been defined as spaces over posets,
again usually satisfying certain conditions. The purpose of this paper is to
explain the relationship between these two perspectives.

We begin by recalling the classical definition, as given for example in [27,
8.2.4 Stratifications and Thom’s $a_{f}$-Regularity]. This defines the basic
structure which underlies any flavour of stratified space. Depending on
context, extra topological or geometric conditions may be imposed on the
strata, and on how they fit together. Common examples of these extra
structures include

  * •

topological stratifications: each stratum is a _topological manifold_ ,

  * •

smooth stratifications: each stratum is a _smooth manifold_ ,

  * •

Whitney stratifications [37]: each pair of smooth strata satisfies _Whitney’s
conditions A and B_ concerning _limits of tangent spaces and tangent lines_ ,

  * •

Thom–Mather stratifications [33, 19, 20]: each stratum is smooth and is
equipped with ‘ _control data_ ’ describing a neighborhood of it.

######  Definition 1.1.

A _decomposition_ of a topological space $X$ is a set $\\{X_{i}\mid i\in I\\}$
of non-empty, disjoint subspaces $X_{i}$, referred to as the _strata_ of the
decomposition, such that $\bigcup_{i\in I}X_{i}=X$.

######  Definition 1.2.

A stratification of a topological space $X$ is a decomposition $\\{X_{i}\mid
i\in I\\}$ satisfying:

Local finiteness:

each $x\in X$ has an open neighborhood $U_{x}$ intersecting only finitely many
strata, i.e., $\\{i\in I\,\,|\,\,U_{x}\cap X_{i}\not=\emptyset\\}$ is finite,

Local closure:

each $X_{i}$ is _locally closed_ , i.e. an intersection of an open and a
closed set,

Frontier condition:

if $X_{i}\cap\overline{X_{j}}\not=\emptyset$ then
$X_{i}\subset\overline{X_{j}}$.

The name of the last condition above refers to the fact that $X_{i}$ is
contained in the _frontier_ $\overline{X_{j}}\setminus X_{j}$ of $X_{j}$ when
$i\neq j$. It is equivalent to the condition that _the closure of each stratum
is a union of strata_ , and this alternative formulation is sometimes used as
the definition, see, for example, [28, Appendix C. Stratified Spaces and
Singularities, Remark C.2., p. 434]. When the frontier condition holds, the
set $I$ of strata is partially-ordered by the relation

$i\leq j\iff X_{i}\subseteq\overline{X_{j}}.$

This feature is sometimes incorporated into the definition of stratification,
for example [41] defines an _$I$ -decomposition_ of a space $X$ to be a
stratification, as in Definition 1.2, for which the set of strata $I$ is
partially-ordered by $i\leq j\iff X_{i}\subset\overline{X_{j}}$. Thus a
stratification with set of strata $I$ is an $I$-decomposition, and vice versa.
Another minor variant of the definition is found in [30]:

  * •

a _partition_ is a decomposition with _locally closed_ strata;

  * •

a _stratification_ is a partition $\\{X_{i}\mid i\in I\\}$ of $X$ together
with a _partial order_ on $I$ such that $\overline{X_{j}}\subset\bigcup_{i\leq
j}X_{i}$ for each $j\in I$;

  * •

a _good stratification_ is a partition _satisfying the frontier condition_ ;

A good stratification is a stratification in the above sense — the frontier
condition implies that the set $I$ of strata is partially-ordered with
$\overline{X_{j}}=\bigcup_{i\leq j}X_{i}$ — and is a stratification in the
sense of Definition 1.2 if, in addition, the associated decomposition is
locally finite.

######  Remark 1.3.

In [20], see also [43], John Mather uses the name “ _prestratification_ ” for
the above decomposition as in Definition 1.2 and the name “stratification” is
used for an existence of prestratifications locally at each point.

More recently, the partial ordering of the set of strata has been elevated to
a central role:

######  Definition 1.4.

A _poset-stratified space_ [18] is a continuous surjection $\pi\colon X\to
I_{\preccurlyeq}$ from a topological space $X$ onto the Alexandrov space
$I_{\preccurlyeq}$ of a poset $(I,\preccurlyeq)$.

This definition has several advantages, not least that it has good categorical
properties, and has been used extensively in studying the homotopy theory of
stratified spaces, see for example [6, 11, 12, 16, 18, 25, 34, 35, 36, 38,
39]. The question arises as to what the precise relationship between
stratified (in the sense of 1.2) and poset-stratified spaces is. In order to
answer this we make the following definitions.

######  Definition 1.5.

Let $\\{X_{i}\mid i\in I\\}$ be a decomposition of $X$. The _decomposition
map_ $\pi\colon X\to I$ is defined by $\pi(x)=i$ for all $x\in X_{i}$. The
_decomposition space_ $I_{\pi}$ is the set $I$ of strata equipped with the
quotient topology, i.e., the finest topology for which $\pi$ is continuous.

In answer to the question, first, and this is well-known, any stratified space
is poset-stratified:

######  Theorem A.

Suppose that $\\{X_{i}\mid i\in I\\}$ is a stratification of $X$. Then

  1. (1)

the set $I$ of strata is partially-ordered by $i\leq j\iff
X_{i}\subseteq\overline{X_{j}}$ and

  2. (2)

the decomposition map $\pi\colon X\to I_{\pi}=I_{\leq}$ is a poset-stratified
space with respect to this partial order.

Second, being poset-stratified is a notion weaker than being stratified: the
strata of a poset-stratified space are locally closed, but the associated
decomposition need not be locally finite, nor satisfy the frontier condition.
The next result gives sufficient conditions, in terms of the Alexandrov space
of the poset and the map to it, for a poset-stratified space to be stratified.

######  Theorem B.

Suppose $\pi\colon X\to I_{\preccurlyeq}$ is a poset-stratified space such
that the Alexandrov space $I_{\preccurlyeq}$ is locally finite and $\pi$ is an
open map. Then the decomposition $\\{\pi^{-1}(i)\mid i\in
I,\pi^{-1}(i)\neq\emptyset\\}$ is a stratification of $X$.

When the conditions of Theorem B are satisfied, the decomposition
$\\{\pi^{-1}(i)\mid i\in I,\pi^{-1}(i)\neq\emptyset\\}$ is a stratification.
Thus $\pi\colon X\to I_{\leq}$ is a poset-stratified space with respect to the
partial order

$i\leq j\iff\pi^{-1}(i)\subseteq\overline{\pi^{-1}(j)}$

by Theorem A. In fact, in this case, the partial order $\leq$ is initial
amongst those partial orders on the set $I$ of strata for which the
decomposition map is a poset-stratified space, i.e. the identity
$(I,\leq)\to(I,\preccurlyeq)$ is monotone111Equivalently, the identity map
$I_{\leq}\to I_{\preccurlyeq}$ is continuous, since for any open set $U$ of
$I_{\preccurlyeq}$ the inverse image $\pi^{-1}(U)$ is an open set by the
continuity of $\pi\colon X\to I_{\preccurlyeq}$, thus $U$ is an open set in
$I_{\pi}=I_{\leq}$ because $I_{\pi}$ has the quotient topology., but need not
be an order isomorphism.

In summary the contents of the paper are as follows. In §2 we recall the
notion of an Alexandrov space and review the well-known adjunction between
preordered sets and topological spaces, which restricts to an equivalence
between preorders and Alexandrov spaces.

In §3 we introduce the notion of an Alexandrov decomposition, that is one
whose decomposition space is an Alexandrov space. This is a slightly weaker
notion than that of a poset-stratified space, but which still turns out to
have good properties, and forms a useful bridge between the worlds of
stratified and poset-stratified spaces. In particular we show, Corollary 3.4,
that every locally finite decomposition is an Alexandrov decomposition. In
fact we derive this as a consequence of a more general result, Proposition
3.3, about decompositions of spaces with a final topology for a suitable
family of maps. We end the section by showing, Proposition 3.9, that an
Alexandrov decomposition is a poset-stratified space when each stratum is open
in the minimal closed union of strata containing it. This is a slightly
stronger condition than asking that each stratum is locally closed.

In §4 we discuss the frontier condition. This is the key property which
ensures a close relationship between the properties of a decomposition and of
its decomposition map and space. For example, although the strata of a poset-
stratified space are always locally closed, the converse is not true in the
absence of the frontier condition. For example, the decomposition

$S^{1}=\left\\{e^{\sqrt{-1}\theta}\mid
0<\theta\leq\pi\right\\}\sqcup\left\\{e^{\sqrt{-1}\theta}\mid\pi<\theta\leq
2\pi\right\\}$

of the unit circle in $\mathbb{C}$ has locally closed strata, but the
decomposition space is the indiscrete space with two points. This is not the
Alexandrov space of (any) partial-order on the set of strata, so the
decomposition map does not exhibit $S^{1}$ as a poset-stratified space. The
main result of this section is Proposition 4.1 which states that an Alexandrov
decomposition satisfies the frontier condition if, and only if, its
decomposition map is open. Both conditions fail in the above example.

Finally in §5 we assemble the results of §3 and 4 to prove Theorem A and
Theorem B above, and give a number of simple, but typical, examples to
illustrate them.

There are two appendices. In Appendix A we prove a generalisation of the
classical result that a locally finite space is Alexandrov, namely we prove
that a locally Alexandrov space is Alexandrov, see Proposition A.3. In
Appendix B we briefly discuss Moore’s notions of upper and lower
semicontinuous decompositions. Lower semicontinuous decompositions are those
for which the decomposition map is open, and so are closely related to
stratifications and the contents of this paper. Upper semicontinuous
decompositions appear in several famous results, including Moore’s and Bing’s
theorems on decompositions of Euclidean spaces into continua, and Freedman’s
proof of the Poincaré Conjecture in dimension $4$. Loosely, upper
semicontinuous decompositions seem to appear in ‘wild topology’ and lower
semicontinuous ones in ‘stratification theory’, which is often an attempt to
tame the topology of singular spaces.

### Acknowledgements

L.W. is supported by the Landesgraduiertenförderung Baden-Württemberg.
S.Y. is supported by JSPS KAKENHI Grant Numbers JP19K03468 and JP23K03117.

##  2\. Preorders and Alexandrov Spaces

A preorder on a set $P$ is a relation $\leq$ which is reflexive and
transitive. A set $(P,\leq)$ equipped with a preorder $\leq$ is called a
_proset_. If a preorder is in addition anti-symmetric, then it is a partial
order; a set with a partial order is called a _poset_. A map $f\colon P\to Q$
between prosets is monotone if $p\leq p^{\prime}\implies f(p)\leq
f(p^{\prime})$.

A preorder $\leq$ on $P$ defines an equivalence relation $p\sim p^{\prime}\iff
p\leq p^{\prime}$ and $p^{\prime}\leq p$. The set $P/\\!\\!\sim$ of
equivalence classes inherits a natural partial order defined by
$[p]\preccurlyeq[p^{\prime}]\iff p\leq p^{\prime}$. Let $\mathsf{Proset}$ be
the category of prosets and monotone maps, and $\mathsf{Poset}$ the full
subcategory of posets. The assignment $P\mapsto P/\\!\\!\\!\sim$ extends to a
functor $\mathsf{Proset}\to\mathsf{Poset}$ which is left adjoint to the
inclusion $\mathsf{Poset}\hookrightarrow\mathsf{Proset}$.

A preorder $\leq$ determines a natural topology on $P$ in which $U\subset P$
is open if, and only if, it is upward-closed, i.e.,

$p\in U\ \text{and}\ p\leq q\implies q\in U.$

Each $p\in P$ has a minimal open neighborhood $U_{p}=\\{q\in P\mid p\leq q\\}$
in this topology, and the set of minimal open neighborhoods $\\{U_{p}\,|\,p\in
P\\}$ is a base. This topology is referred to as the _Alexandrov topology_ on
$P$ because it makes $P$ into an Alexandrov space in the sense of the
following definition. We denote the space $P$ equipped with this topology by
$P_{\leq}$.

######  Definition 2.1 (Alexandrov space).

Let $X$ be a topological space. If the intersection of any family of open sets
is open, equivalently the union of any family of closed sets is closed, then
the topology is called an _Alexandrov topology_ and the space is called an
_Alexandrov space_.

######  Example 2.2.

Any finite topological space, i.e., a topological space with finitely many
points, is Alexandrov.

The following lemma is well-known, but as it plays a crucial role in this
paper we give a proof.

######  Lemma 2.3.

A topological space $X$ is an Alexandrov space if, and only if, each point in
$X$ has a minimal open neighborhood.

###### Proof.

If $X$ is an Alexandrov space, then the intersection $U_{x}$ of all the open
neighborhoods of a point $x\in X$ is a minimal open neighborhood of $x$.
Conversely, suppose that each point $x\in X$ has a minimal open neighborhood
$U_{x}$. Let $U=\bigcap_{j\in J}U_{j}$ be an intersection of open sets. If
$U=\emptyset$, then we are done. If $U\not=\emptyset$, then $U_{x}\subset
U_{j}$ for all $j\in J$ and $x\in U$, so $x\in U_{x}\subset U$. Hence

$U=\bigcap_{j\in J}U_{j}=\bigcup_{x\in U}U_{x}$

is a union of open sets, and so is open. Thus $X$ is an Alexandrov space. ∎

######  Remark 2.4.

  1. (1)

For a proset $(P,\leq)$ the downward-closed set $D_{p}:=\\{q\,\,|\,\,q\leq
p\\}$ is the minimal _closed_ neighborhood of $p$. This is because its
complement

$P-D_{p}=\bigcup_{r\not\leq p}U_{r}$

is the maximal open set not containing $p$.

  2. (2)

The closed sets of an Alexandrov space also form an Alexandrov topology,
because they are closed under arbitrary intersections and unions, equivalently
because the reverse or opposite of a preorder is again a preorder. In this
topology $U_{p}$ is the minimal closed, and $D_{p}$ the minimal open,
neighborhood of $p$. This complementary topology is sometimes referred to as
‘the’ topology on the proset, see for example [3], [5], [21] and [29].
However, when stratification theory or poset-stratified spaces are considered,
as in [9] and [38], it is more convenient to take $U_{p}$ to be open, see also
[18, Definition A.5.1] and [31, Definition 2.1].

######  Proposition 2.5.

There is an adjunction
$\mathsf{T}\colon\mathsf{Proset}\longleftrightarrow\mathsf{Top}\colon\mathsf{P}$
between prosets and topological spaces. The left adjoint $\mathsf{T}$ takes a
proset to its Alexandrov space, and the right adjoint $\mathsf{P}$ takes a
topological space to its set of points equipped with the specialization
preorder

$x\leq y\iff x\in\overline{\\{y\\}}.$

The adjunction restricts to an equivalence
$\mathsf{T}\colon\mathsf{Proset}\longleftrightarrow\mathsf{Alex}\colon\mathsf{P}$
between the category of prosets and the full subcategory of Alexandrov spaces.

###### Sketch proof.

This is also well-known, so we provide only a sketch. The assignment
$P\mapsto\mathsf{T}(P)$ extends to a functor because any monotone map $f\colon
P\to Q$ is continuous with respect to the Alexandrov topologies on $P$ and
$Q$. Similarly, the assignment $X\mapsto\mathsf{P}(X)$ extends to a functor
because a continuous map $f\colon X\to Y$ is monotone as a map between the
specialization preorders on $X$ and $Y$.

The specialization preorder on the Alexandrov space $\mathsf{T}(P)$ of a
proset $P$ is the original preorder because $p\leq q\iff p\in D_{q}$ for both
preorders. Thus the identity defines a natural isomorphism
$P\cong(\mathsf{P}\circ\mathsf{T})(P)$. In the other direction, the Alexandrov
topology of the specialization preorder of a space $X$ is always finer than
the original topology. To see this note that if $D\subseteq X$ is closed,
$y\in D$ and $x\leq y$ then $x\in\overline{\\{y\\}}\subseteq D$, so that
$D\subseteq(\mathsf{T}\circ\mathsf{P})(X)$ is closed too. Thus the identity
defines a natural map

$(\mathsf{T}\circ\mathsf{P})(X)\to X.$

These natural maps are respectively the unit and counit of the adjunction.

When $X$ is an Alexandrov space, then we can write any closed $D\subseteq X$
as the union $D=\bigcup_{x\in D}\overline{\\{x\\}}$ of closed subsets in
$(\mathsf{T}\circ\mathsf{P})(X)$. Therefore $D$ is closed in
$(\mathsf{T}\circ\mathsf{P})(X)$ and the identity
$(\mathsf{T}\circ\mathsf{P})(X)\cong X$ is a homeomorphism. Thus the
adjunction restricts to an equivalence between prosets and Alexandrov spaces.
∎

Finally, we give a characterization of the Alexandrov spaces of posets which
will be useful later.

######  Lemma 2.6.

A proset $P$ is a poset if, and only if, each singleton $\\{p\\}$ locally
closed in the Alexandrov space $\mathsf{T}(P)$.

###### Proof.

Recall that each $p\in P$ has a minimal open neighborhood $U_{p}=\\{q\in P\mid
p\leq q\\}$ and a minimal closed neighborhood $D_{p}=\\{q\in P\mid p\geq q\\}$
in the Alexandrov space $\mathsf{T}(P)$. Moreover, $P$ is a poset precisely
when $\\{p\\}=D_{p}\cap U_{p}$ for all $p\in P$. Therefore, if $P$ is a poset,
each singleton is locally closed. Conversely, if $\\{p\\}$ is locally closed,
then $\\{p\\}=D\cap U$ for some closed $D$ and open $U$. The minimality of
$D_{p}$ and $U_{p}$ then shows

$\\{p\\}=D\cap U\supset D_{p}\cap U_{p}\supset\\{p\\}.$

Hence $\\{p\\}=D_{p}\cap U_{p}$ and each singleton is locally closed. ∎

For more on the Alexandrov topology or Alexandrov spaces see, e.g., [1], [2],
[9, §4.2.1 Alexandrov Topology], [38, Appendix A Pre-orders and spaces].

##  3\. Alexandrov decompositions

A _decomposition_ of a space $X$ is a partition into subspaces, i.e. a set
$\\{X_{i}\mid i\in I\\}$ of disjoint, non-empty subspaces whose union is $X$.
We refer to the $X_{i}$ as the strata of the decomposition, and (by a mild
abuse of notation) to the indexing set $I$ as the set of strata.

There are two equivalent descriptions which we will use. Firstly, a
decomposition can be viewed as the equivalence relation

$x\sim x^{\prime}\iff x,x^{\prime}\in X_{i}\ \text{for some}\ i\in I$

with equivalence classes the strata, and set of equivalence classes
$X/\\!\\!\sim\ \cong I$. Secondly, it can be viewed as the quotient map
$\pi\colon X\to I$ defined by

$\pi(x)=i\iff x\in X_{i}.$

We refer to this as the decomposition map. The _decomposition space_ $I_{\pi}$
is the set $I$ of strata equipped with the quotient topology from $\pi$, i.e.
with the finest topology (most open sets) such that $\pi$ is continuous. The
quotient topology is the finest topology amongst topologies on $I$ such that
$\pi$ is continuous.

The set $I$ of strata has a second interesting topology. Let
$I_{\leq}=\left(\mathsf{T}\circ\mathsf{P}\right)(I_{\pi})$ be the Alexandrov
space of the specialization preorder of $I_{\pi}$. We refer to this as the
_decomposition preorder_ on $I$. It is defined by $i\leq j\iff
i\in\overline{\\{j\\}}$ where the latter denotes the closure in $I_{\pi}$. By
definition

$\overline{\\{j\\}}=\\{i\in I\mid i\leq j\\}=D_{j}$

is the minimal (closed) subset containing $j$ whose preimage in $X$ is closed.
Equivalently, $\pi^{-1}(D_{j})$ is the minimal closed union of strata
containing $X_{j}$.

The above two topologies on $I$ are related: the identity
$I_{\leq}=\left(\mathsf{T}\circ\mathsf{P}\right)(I_{\pi})\to I_{\pi}$ is
continuous, because it is the counit of the adjunction
$\mathsf{T}\dashv\mathsf{P}$, and is a homeomorphism precisely when $I_{\pi}$
is an Alexandrov space.

######  Definition 3.1.

A decomposition $\\{X_{i}\mid i\in I\\}$ of $X$ is _Alexandrov_ if its
decomposition space $I_{\pi}$ is an Alexandrov space.

######  Proposition 3.2.

The following are equivalent:

  1. (1)

the decomposition $\\{X_{i}\mid i\in I\\}$ of $X$ is Alexandrov,

  2. (2)

the identity $I_{\leq}\to I_{\pi}$ is a homeomorphism,

  3. (3)

the map $\pi\colon X\to I_{\leq}$ is continuous.

###### Proof.

(1) $\implies$ (2) because
$I_{\leq}=\left(\mathsf{T}\circ\mathsf{P}\right)(I_{\pi})=I_{\pi}$ when
$I_{\pi}$ is Alexandrov. It is immediate that (2) $\implies$ (1) and (3).
Finally, (3) $\implies$ (2) because the quotient topology is initial amongst
topologies on $I$ for which $\pi$ is continuous, i.e. when $\pi\colon X\to
I_{\leq}$ there is a commutative diagram

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\textstyle{I_{\pi}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{id}}$$\textstyle{I_{\leq}}$

of continuous maps. Since the identity $I_{\leq}\to I_{\pi}$ is also
continuous, it is a homeomorphism. ∎

The pointwise decomposition $\\{\\{x\\}\mid x\in X\\}$ of a space $X$ is
Alexandrov if, and only if, the space $X$ is Alexandrov (because the
decomposition space is just $X$ itself). For example, the pointwise
decomposition of the real line is not Alexandrov. In contrast, every finite
decomposition is Alexandrov, because every finite topological space is
evidently an Alexandrov space. The following result generalizes this example
to give a wide class of Alexandrov decompositions.

######  Proposition 3.3.

Let $\\{X_{i}\mid i\in I\\}$ be a decomposition of $X$. Suppose $X$ has the
final topology with respect to a family of maps $f_{j}:Y_{j}\to X$ for $j\in
J$ such that $\\{i\in I\mid f_{j}^{-1}(X_{i})\neq\emptyset\\}$ is finite for
each $j\in J$. Then the decomposition is Alexandrov.

Recall that the _final topology_ is the finest topology on $X$ such that
$f_{j}:Y_{j}\to X$ is continuous for each $j\in J$. Explicitly, $U\subseteq X$
is open in the final topology if, and only if, $f_{j}^{-1}(U)$ is open in
$Y_{j}$ for all $j\in J$. For example, the final topology for a family
consisting of a single map $f\colon X\to Y$ is the quotient topology. The
final topology has the universal property that $g\colon X\to Z$ is continuous
if, and only if, the composite $g\circ f_{j}\colon Y_{j}\to Z$ is continuous
for each $j\in J$. Before proving Proposition 3.3, we give two corollaries.

######  Corollary 3.4.

Any locally finite decomposition is Alexandrov.

###### Proof.

Choose an open covering $X=\bigcup_{j\in J}U_{j}$ by subsets $U_{j}$
intersecting only finitely many strata. The original topology of $X$ is the
final topology of $X$ for the family of the inclusions $\imath_{i}:U_{i}\to X$
(this is true for any open covering, without requiring the above condition
‘subsets $U_{j}$ intersecting only finitely many strata’). So the
decomposition is Alexandrov by Proposition 3.3. ∎

######  Corollary 3.5.

Any decomposition of a compactly generated space for which each compact
subspace intersects only finitely many strata is Alexandrov.

###### Proof.

Recall that a space is compactly generated222As to this notion, there are
several slightly different definitions, e.g., see [26, 42]. if $U\cap K$ is
open in $K$ for every compact $K\subset X$. In other words, $X$ is compactly
generated if it has the final topology for the family of inclusions of compact
subspaces. Therefore, if each compact subspace intersects only finitely many
strata, the decomposition is Alexandrov by Proposition 3.3. ∎

We now prove Proposition 3.3 as a consequence of the following two lemmas.

######  Lemma 3.6.

Let $f\colon X^{\prime}\to X$ be a continuous map. Let $\\{X_{i}\mid i\in
I\\}$ be a decomposition of $X$, and $\\{f^{-1}(X_{i})\mid i\in I^{\prime}\\}$
the induced decomposition of $X^{\prime}$ with indexing set

$I^{\prime}=\\{i\in I\mid f^{-1}(X_{i})\neq\emptyset\\}.$

Then the inclusion $e\colon I^{\prime}\hookrightarrow I$ is continuous in the
Alexandrov topologies of the decomposition preorders on $I^{\prime}$ and $I$.

###### Proof.

By definition there is a commutative diagram

$\textstyle{X^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\pi^{\prime}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{I^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e}$$\textstyle{I}$

where $\pi^{\prime}$ and $\pi$ are the respective decomposition maps. Thus the
inclusion $e$ is continuous in the quotient topologies on $I^{\prime}$ and $I$
respectively, that is, $e\colon I^{\prime}_{\pi^{\prime}}\hookrightarrow
I_{\pi}$ is continuous. Applying the functor $\mathsf{T}\circ\mathsf{P}$ to
this map gives the result. ∎

######  Lemma 3.7.

Suppose $X$ has the final topology with respect to a family of maps
$f_{j}:Y_{j}\to X$ for $j\in J$. Then a decomposition $\\{X_{i}\mid i\in I\\}$
of $X$ is Alexandrov if, and only if, the induced decomposition
$\\{f_{j}^{-1}(X_{i})\mid i\in I_{j}\\}$ of each $Y_{j}$ is Alexandrov, where
$I_{j}=\\{i\in I\mid f_{j}^{-1}(X_{i})\neq\emptyset\\}$.

###### Proof.

For each $j\in J$ let $\pi_{j}\colon Y_{j}\to I_{j}$ be the decomposition map
of the induced decomposition, and $e_{j}\colon I_{j}\to I$ the inclusion. By
definition the diagram

$\textstyle{Y_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{j}}$$\scriptstyle{\pi_{j}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{I_{j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{e_{j}}$$\textstyle{I}$

commutes. By Lemma 3.6 the inclusion $e_{j}$ is continuous in the Alexandrov
topologies from the respective decomposition preorders on $I_{j}$ and $I$.
Using Proposition 3.2 and the facts that $X$ has the final topology with
respect to the family of maps $f_{j}:Y_{j}\to X$ for $j\in J$ , and that a map
to a subspace is continuous if, and only if, the composite with the inclusion
of the subspace is continuous, we have

$\displaystyle\pi\colon X\to I_{\leq}\ \text{is continuous}$
$\displaystyle\iff\pi\circ f_{j}\colon Y_{j}\to I_{\leq}\ \text{is continuous
for all}\ j\in J$ $\displaystyle\iff e_{j}\circ\pi_{j}\colon Y_{j}\to
I_{\leq}\ \text{is continuous for all}\ j\in J$
$\displaystyle\iff\pi_{j}\colon Y_{j}\to(I_{j})_{\leq}\ \text{is continuous
for all}\ j\in J.$

Hence, by Proposition 3.2, the decomposition of $X$ is Alexandrov if, and only
if, the induced decompositions of each of the $Y_{j}$ are Alexandrov. ∎

###### Proof of Proposition 3.3.

This follows immediately from Lemma 3.7 and the fact that any finite
decomposition is Alexandrov. ∎

Alexandrov decompositions are closely related to poset-stratified spaces.
Indeed, any Alexandrov decomposition $\\{X_{i}\mid i\in I\\}$ has a natural
coarsening which is poset-stratified. Let $(I,\leq)$ be the proset with the
decomposition preorder and let $I/\sim$ be the set of equivalence classes of
$i\sim j\iff i\leq j$ and $j\leq i$, equipped with the induced partial order
$[i]\preccurlyeq[j]\iff i\leq j$ (cf. [40, §5]). The quotient map $I\to
I/\\!\sim\colon i\mapsto[i]$ is monotone, thus the assoicated map
$I_{\leq}\to(I/\sim)_{\preccurlyeq}$ is continuous, so the composite

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{I_{\leq}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}}$$\textstyle{(I/\sim)_{\preccurlyeq}}$

is continuous. This exhibits $X$ as a poset-stratified space: its strata
$X_{[i]}=\bigcup_{j\in[i]}X_{j}$ are unions of strata of the original
decomposition. These coarsened strata have a natural topological description.
Recall that $D_{i}=\\{j\in I\mid j\leq i\\}$ is the closure of $\\{i\\}$ in
$I_{\pi}$ and that $\pi^{-1}(D_{i})$ is the minimal closed union of strata
containing $X_{i}$. Then

$X_{[i]}=X_{[j]}\iff\pi^{-1}(D_{i})=\pi^{-1}(D_{j})$

i.e., the minimal closed unions of strata containing $X_{i}$ and $X_{j}$
agree.

We are interested in when the original stratification is poset-stratified, so
we now consider conditions under which the decomposition preorder is a partial
order.

######  Definition 3.8.

A decomposition $X=\bigsqcup_{i\in I}X_{i}$ is _poset-stratified_ with respect
to a partial order $\preccurlyeq$ on the set $I$ of strata if the
decomposition map $\pi\colon X\to I_{\preccurlyeq}$ is continuous with respect
to the Alexandrov topology defined by $\preccurlyeq$.

######  Proposition 3.9.

Suppose $\\{X_{i}\mid i\in I\\}$ is an Alexandrov decomposition of $X$. Then
the following are equivalent:

  1. (1)

the decomposition is poset-stratified with respect to a partial order
$\preccurlyeq$ on $I$,

  2. (2)

the decomposition is poset-stratified with respect to the decomposition
preorder $\leq$ on $I$,

  3. (3)

each stratum $X_{i}$ is open in $\pi^{-1}(D_{i})$, in particular each stratum
locally closed.

When these equivalent conditions hold, the decomposition preorder $\leq$ is a
partial order on $I$ and the identity $(I,\leq)\to(I,\preccurlyeq)$ is
monotone.

###### Proof.

(1) $\iff$ (2): Suppose $\pi\colon X\to I_{\preccurlyeq}$ is a poset-
stratified space. Since the decomposition is Alexandrov, it follows from
Proposition 3.2 (2) that the identity $I_{\leq}\to I_{\pi}$ is continuous (in
fact, a homeomorphism). If $\pi\colon X\to I_{\preccurlyeq}$ is continuous,
then the identity $I_{\pi}\to I_{\preccurlyeq}$ is continuous, because
$I_{\pi}$ has the quotient topology induced from the decomposition map
$\pi:X\to I$. Composing these two continuous identity maps shows that the
identity $I_{\leq}\to I_{\preccurlyeq}$ is continuous. Equivalently, the
identity $(I,\leq)\to(I,\preccurlyeq)$ is monotone. Thus the decomposition
preorder $\leq$ is a partial order because

$i\leq j\leq i\implies i\preccurlyeq j\preccurlyeq i\implies
i=j\quad\text{(because $\preccurlyeq$ is a partial order)}$

Finally, $\pi\colon X\to I_{\leq}$ is continuous by Proposition 3.2 (3) since
the decomposition is Alexandrov.

The converse is immediate.

(2) $\implies$ (3): When $\pi\colon X\to I_{\leq}$ is a poset-stratified
space, the stratum

$X_{i}=\pi^{-1}(i)=\pi^{-1}(U_{i}\cap
D_{i})=\pi^{-1}(U_{i})\cap\pi^{-1}(D_{i})$

where $U_{i}=\\{j\in I\mid i\leq j\\}$ and $D_{i}=\\{j\in I\mid j\leq i\\}$
are respectively the minimal open and closed neighborhoods of $i$ in the
Alexandrov space $I_{\leq}$. Therefore $X_{i}$ is an open subset of
$\pi^{-1}(D_{i})$.

(3) $\implies$ (2): Since the decomposition is Alexandrov, the map $\pi\colon
X\to I_{\pi}=I_{\leq}$ is continuous. Therefore we only need to show that the
decomposition preorder $\leq$ is a partial order. The subspace topology on
$D_{i}$ from $I_{\pi}$ is the quotient topology induced from the restriction
$\pi^{-1}(D_{i})\to D_{i}$ of $\pi$ to its preimage. (This is a well-known
general fact333For example, see R. Engelking’s book [13, 2.4.15. Proposition].
$D_{i}$ being a closed set is a key, and in fact, this fact also holds in the
case of $\pi^{-1}(B)\to B$ for any open set $B$. However, this fact _does not
necessarily hold_ if $B$ is neither an open set nor a closed set (e.g., see
[13, 2.4.17. Example]). A surjective continuous map $f:X\to Y$ is called
_hereditarily quotient_ if for every subset $B\subset Y$ the restriction
$f_{B}:f^{-1}(B)\to B$ is a quotient map [13, 2.4.F]. In this sense we can say
that _any quotient map $f:X\to Y$ is hereditarily quotient ‘with respect to
open subsets and closed subsets’ of $Y$_.  about the quotient topology.)
Therefore, since $X_{i}=\pi^{-1}(i)$ is open in $\pi^{-1}(D_{i})$ by
assumption, $\\{i\\}$ is open in $D_{i}$. This means that $\\{i\\}$ is locally
closed in $I_{\pi}=I_{\leq}$. Hence the decomposition preorder is a partial
order by Lemma 2.6 as required. ∎

######  Remark 3.10.

We do not claim, and indeed it is not true, that there is always a unique
partial order with respect to which a decomposition is poset-stratified. For
example, the two point discrete space $X=\\{0,1\\}$ with decomposition
$X=\\{0\\}\sqcup\\{1\\}$ is poset-stratified with respect to any of the three
partial orders (distinguished by whether $0\leq 1$ or $0\geq 1$ or neither) on
its set $\\{0,1\\}$ of strata.

However, the final statement of the proposition implies that the decomposition
preorder is initial amongst those partial orders on the set of strata for
which the given decomposition is poset-stratified.

######  Remark 3.11.

The third equivalent condition in the above result is stronger than the
assumption that each stratum locally closed. For example consider the
decomposition

$S^{1}=\left\\{e^{\theta\sqrt{-1}}\mid
0<\theta\leq\pi\right\\}\sqcup\left\\{e^{\theta\sqrt{-1}}\mid\pi<\theta\leq
2\pi\right\\}$

of the unit circle. This is an Alexandrov decomposition, because it is finite,
and both strata are locally closed. Nevertheless, the decomposition space is
the indiscrete space with two points, which is not the Alexandrov space of any
partial order on the set of strata. So the decomposition is not poset-
stratified.

##  4\. The Frontier Condition

A decomposition $\\{X_{i}\mid i\in I\\}$ of $X$ satisfies the _frontier
condition_ if

$X_{i}\cap\overline{X_{j}}\neq\emptyset\implies X_{i}\subset\overline{X_{j}}.$

Equivalently, it satisfies the frontier condition if the closure of each
stratum is a union of strata.

Recall from the previous section that we set $D_{j}=\\{i\in I\mid i\leq
j\\}=\overline{\\{j\\}}$ where the closure is taken in the decomposition space
$I_{\pi}$ and that $\pi^{-1}(D_{j})$ is then the minimal closed union of
strata containing the stratum $X_{j}$.

######  Proposition 4.1.

Let $\\{X_{i}\mid i\in I\\}$ be an Alexandrov decomposition of $X$. Then the
following are equivalent:

  1. (1)

the decomposition satisfies the frontier condition,

  2. (2)

for all $j\in I$, the equality $\overline{X_{j}}=\pi^{-1}(D_{j})$, holds,
i.e., the closure of $X_{j}$ is the minimal closed union of strata containing
$X_{j}$,

  3. (3)

the decomposition preorder is given by $i\leq j\iff
X_{i}\subset\overline{X_{j}}$,

  4. (4)

the decompostion map $\pi\colon X\to I_{\pi}$ is an open map.

###### Proof.

(1) $\implies$ (2): When the frontier condition holds the closure of a stratum
$X_{j}$ is a union of strata, hence is the minimal closed union of strata
containing $X_{j}$. Therefore $\overline{X_{j}}=\pi^{-1}(D_{j})$, since
$D_{j}$ is the minimal closed subset containing $j$.

(2) $\implies$ (3): The decomposition preorder is defined by

$i\leq j\iff i\in D_{j}\iff X_{i}\subseteq\pi^{-1}(D_{j})\iff
X_{i}\subseteq\overline{X_{j}}$

where we use (2) at the last step.

(3) $\implies$ (4): Suppose that $U\subseteq X$ is open, and that $i\in\pi(U)$
and $i\leq j$. Then $U\cap X_{i}\neq\emptyset$ and
$X_{i}\subseteq\overline{X_{j}}$. So $U\cap X_{j}\neq\emptyset$, which implies
$j\in\pi(U)$ too. Thus $\pi(U)$ is open in the Alexandrov topology, and
because the decomposition is assumed to be Alexandrov this is equivalent to
$\pi(U)$ being open in the quotient topology, i.e., open in $I_{\pi}$. Hence
$\pi\colon X\to I_{\pi}$ is an open map as claimed.

(4) $\implies$ (1): Suppose that $\pi\colon X\to I_{\pi}$ is an open map.
Hence $\pi\colon X\to I_{\pi}$ is both continuous and open. Recall that444The
following are well-known: (1) $f:X\to Y$ is continuous if and only if
$\overline{f^{-1}(B)}\subset f^{-1}(\overline{B})$ for any $B\subset Y$ and
(2) $f:X\to Y$ is open if and only if
$f^{-1}(\overline{B})\subset\overline{f^{-1}(B)}$ for any $B\subset Y$. this
implies $\pi^{-1}\left(\overline{J}\right)=\overline{\pi^{-1}(J)}$ for any
$J\subseteq I$. In particular
$\overline{X_{j}}=\overline{\pi^{-1}(j)}=\pi^{-1}(\overline{\\{j\\}})$ and
therefore

$X_{i}\cap\overline{X_{j}}=\pi^{-1}(i)\cap\overline{\pi^{-1}(j)}=\pi^{-1}(i)\cap\pi^{-1}\left(\overline{\\{j\\}}\right)=\pi^{-1}\left(\\{i\\}\cap\overline{\\{j\\}}\right).$

Thus if $X_{i}\cap\overline{X_{j}}\neq\emptyset$, the above implies that
$\\{i\\}\cap\overline{\\{j\\}}\not=\emptyset$, i.e., $i\in\overline{\\{j\\}}$,
which implies that
$\pi^{-1}(i)\subset\pi^{-1}(\overline{\\{j\\}})=\overline{\pi^{-1}(j)}$, i.e.,
$X_{i}\subseteq\overline{X_{j}}$. Hence the frontier condition holds. ∎

######  Remark 4.2.

See [32, Proposition 3.2] for a similar result about poset-stratified spaces:
they prove that a poset-stratified space $\pi:X\to I_{\preccurlyeq}$ is an
open map if, and only if,
$X_{i}\subset\overline{X_{j}}\,\Longleftrightarrow\,i\,\preccurlyeq\,j$.

######  Corollary 4.3.

Let $\\{X_{i}\mid i\in I\\}$ be an Alexandrov decomposition of $X$. The
following are equivalent

  1. (1)

each stratum $X_{i}$ is locally closed and the frontier condition holds;

  2. (2)

$\pi\colon X\to I$ is poset-stratified (for some partial order) and $\pi\colon
X\to I_{\pi}$ is an open map.

Moreover, when these equivalent conditions hold, the decomposition preorder
$\leq$ is a partial order and $\pi\colon X\to I_{\leq}$ is a poset-stratified
space.

###### Proof.

(1) $\implies$ (2): Suppose that each $X_{i}$ is locally closed and the
frontier condition holds. Then Proposition 4.1 implies that $\pi$ is an open
map, and that $\overline{X_{i}}=\pi^{-1}(D_{i})$. Since $X_{i}$ is locally
closed, $X_{i}=U\cap D$ for an open set $U$ and a closed set $D$ in $X$. So,
$X_{i}\subset\overline{X_{i}}\subset D$, since $X_{i}\subset D$ and $D$ is a
closed set. Hence we have

$X_{i}\subset U\cap\overline{X_{i}}\subset U\cap D=X_{i},$

which implies that $X_{i}=U\cap\overline{X_{i}}$, thus $X_{i}$ is open in
$\overline{X_{i}}$ , hence it is open in $\pi^{-1}(D_{i})$. Therefore
$\pi\colon X\to I$ is poset-stratified for some partial order (in fact with
respect to the decomposition preorder) by Proposition 3.9.

(2) $\implies$ (1): Suppose $\pi\colon X\to I$ is poset-stratified and the
decomposition map is open. By Proposition 4.1 the frontier condition holds.
Moreover $X_{i}$ is open in $\pi^{-1}(D_{i})$ by Proposition 3.9, and so in
particular locally closed. ∎

######  Remark 4.4.

The frontier condition plays a key role here. Recall the example in Remark
3.11 of an Alexandrov decomposition with locally closed strata (not satisfying
the frontier condition), which is not poset-stratified.

##  5\. Main results

We assemble the results of the previous sections to prove our two main
theorems. The first states that a stratified space is a poset-stratified
space, in a natural way. See A

###### Proof.

Since $\\{X_{i}\mid i\in I\\}$ is a stratification, it is a locally finite
decomposition, with locally closed strata satisfying the frontier condition.
As it is locally finite, Corollary 3.4 implies that it is an Alexandrov
decomposition. As it satisfies the frontier condition, Proposition 4.1 then
shows that the decomposition preorder on the set $I$ of strata is given by

$i\leq j\iff X_{i}\subseteq\overline{X_{j}}$

and is, in particular, therefore a partial order. Finally, as the strata are
locally closed and satisfy the frontier condition, Corollary 4.3 implies that
$\pi\colon X\to I_{\leq}$ is poset-stratified. ∎

######  Remark 5.1.

This is ‘well-known folklore’ and variants of it can be found in the
literature. For example, see [40, Corollary 4.8] for a version under the extra
assumption that there are finitely many strata.

In contrast, not every poset-stratified space is stratified. Our second main
result gives necessary and sufficient conditions under which a poset-
stratified space is stratified. See B

###### Proof.

Since the decomposition space $I_{\pi}$ has the quotient topology and
$\pi\colon X\to I_{\preccurlyeq}$ is continuous, there is a commutative
diagram

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\textstyle{I_{\pi}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{id}}$$\textstyle{I_{\preccurlyeq}}$

of continuous maps. In particular, any open set in $I_{\preccurlyeq}$ is also
open in $I_{\pi}$. Therefore $I_{\pi}$ is also a locally finite space and
$\pi\colon X\to I_{\pi}$ is also an open map. It follows that the
decomposition $\\{\pi^{-1}(i)\mid i\in I\\}$ is locally finite, and hence
Alexandrov by Corollary 3.4. Then, by Corollary 4.3, each stratum
$\pi^{-1}(i)$ is locally closed and the strata satisfy the frontier condition.
Thus the decomposition is a stratification, as claimed. ∎

######  Remark 5.2.

When a poset-stratified space $\pi\colon X\to I_{\preccurlyeq}$ is stratified
with respect to some partial order $\preccurlyeq$ on the set $I$ of strata,
Theorem B implies the decomposition $\\{\pi^{-1}(i)\mid i\in
I,\pi^{-1}(i)\neq\emptyset\\}$ is a stratification, and then Theorem A implies
that it is poset-stratified with respect to its decomposition preorder.
Moreover, the decomposition preorder is initial amongst those partial orders
on $I$ for which it is poset-stratified by Remark 3.10.

The conditions of Theorem B are sufficient for the poset-stratified space to
be stratified, but the following example shows that they are not necessary.

######  Example 5.3.

Suppose that $\\{X_{i}\mid i\in I\\}$ is a stratification of $X$. Then
$\pi\colon X\to I_{\leq}$ is poset-stratified by Theorem A where $i\leq j\iff
X_{i}\subseteq\overline{X_{j}}$ is the decomposition preorder.

Let $\preccurlyeq$ be any (strict) refinement of $\leq$, i.e. any different
partial order such that $i\leq j\implies i\preccurlyeq j$. Strict refinements
exist unless $\leq$ is a total order. Then the identity $\mathrm{id}\colon
I_{\leq}\to I_{\preccurlyeq}$ is continuous, and therefore $\pi\colon X\to
I_{\preccurlyeq}$ is also poset-stratified. This construction illustrates how
poset-stratified spaces over ‘exotic’ partial orders, different from the
decomposition preorder, arise. Note that, by construction the decomposition
preorder is initial amongst its refinements, hence $I_{\leq}$ is initial
amongst their Alexandrov spaces as claimed in Remark 5.2.

The Alexandrov space $I_{\preccurlyeq}$ of a strict refinement of $\leq$ need
not be locally finite and the map $\pi\colon X\to I_{\preccurlyeq}$ is never
open. (To see the latter, note that there is some $U\subset I$ which is open
in $I_{\leq}$ but not open in $I_{\preccurlyeq}$. Then $\pi^{-1}(U)$ is open
in $X$, but its image $\pi(\pi^{-1}(U))=U$ is not open in $I_{\preccurlyeq}$.)
Of course, the associated decomposition is unchanged, and so is still a
stratification. This shows that the conditions of Theorem B are not necessary.

To give a concrete example, let $X=\mathbb{N}$ with the pointwise
decomposition $\\{\\{n\\}\mid n\in\mathbb{N}\\}$. The decomposition preorder
$\leq$ on $\mathbb{N}$ is trivial, i.e. $m\leq n\iff m=n$. Thus any other
partial order $\preccurlyeq$ is a refinement, and so yields a poset-
stratification $\mathrm{id}\colon\mathbb{N}\to\mathbb{N}_{\preccurlyeq}$ which
is not an open map. If, for instance, we choose $\preccurlyeq$ to be the
standard order, then $\mathbb{N}_{\preccurlyeq}$ is not a locally finite space
either.

A more traditional example is to decompose the positive quadrant
$X=\mathbb{R}_{\geq 0}\times\mathbb{R}_{\geq 0}$ into four strata
$X_{0}=\\{(0,0)\\}$, $X_{1}=\mathbb{R}_{\geq 0}\times\\{0\\}$,
$X_{2}=\\{0\\}\times\mathbb{R}_{\geq 0}$ and
$X_{3}=\mathbb{R}_{>0}\times\mathbb{R}_{>0}$. The decomposition preorder on
$I=\\{0,1,2,3\\}$ is the partial order $0\leq 1,2\leq 3$. If we refine this to
the standard order $0\preccurlyeq 1\preccurlyeq 2\preccurlyeq 3$, then
$\pi\colon X\to I_{\preccurlyeq}$ is poset-stratified, but not an open map,
because $X_{1}\cup X_{3}$ is open, but $\pi(X_{1}\cup X_{3})=\\{1,3\\}$ is
not.

Our final result is that the sufficient conditions of Theorem B under which a
poset-stratified space is stratified are also necessary when the partial order
is the associated decomposition preorder, i.e. the specialization preorder of
the quotient topology on the set of strata.

######  Corollary 5.4.

Suppose $\pi\colon X\to I_{\leq}$ is poset-stratified with respect to the
decomposition preorder

$i\leq j\iff\pi^{-1}(i)\subseteq\overline{\pi^{-1}(j)}$

on $I$. Then the decomposition $\\{\pi^{-1}(i)\mid i\in I\\}$ is a
stratification if, and only if, the Alexandrov space $I_{\leq}$ is locally
finite and the map $\pi$ is open.

###### Proof.

Since the decomposition $\\{\pi^{-1}(i)\mid i\in I\\}$ is poset-stratified, it
is an Alexandrov decomposition. If it is a stratification, then each stratum
is locally closed and the frontier condition is satisfied, so, by Corollary
4.3, the map $\pi\colon X\to I_{\leq}$ is open. Moreover, the decomposition is
locally finite, which implies that the Alexandrov space $I_{\leq}$ is also
locally finite.

The other direction is a special case of Theorem B. ∎

We end with some further examples of poset-stratified spaces which are not
stratified, either because the decomposition space is not locally finite or
the decomposition map is not open.

######  Example 5.5.

The Alexandrov space $P_{\leq}$ of a poset is trivially poset-stratified over
itself via the identity $\mathrm{id}_{P}\colon P_{\leq}\to P_{\leq}$. The
associated decomposition is the pointwise one $\\{\\{p\\}\mid p\in P\\}$ and
the decomposition preorder is the given partial order $\leq$ on $P$. The
identity is evidently an open map, so by Corollary 5.4 the pointwise
decomposition of a poset is a stratification if, and only if, $P_{\leq}$ is
locally finite. This is not always the case, for example it fails for the
standard order on $\mathbb{N}$.

(There is an unfortunate clash of terminology here. A poset $P$ is said to be
locally finite if each bounded interval $[p,q]=\\{r\in P\mid p\leq r\leq q\\}$
is finite. This is not the same as its Alexandrov space $P_{\leq}$ being a
locally finite space, which is equivalent instead to each bounded below
interval $[p,\infty)=\\{q\in P\mid p\leq q\\}$ being finite. For example,
$\mathbb{N}$ with the standard order is a locally finite poset, but its
Alexandrov space $\mathbb{N}_{\leq}$ is not a locally finite space.)

######  Example 5.6.

Consider the locally finite decomposition $\mathbb{R}=X_{0}\cup X_{1}$ into
locally closed strata $X_{0}=(-\infty,0]$ and $X_{1}=(0,\infty)$. The
decomposition preorder is the partial order $0\leq 1$, its Alexandrov space is
locally finite (even finite), and $\pi\colon\mathbb{R}\to\\{0,1\\}$ is a
poset- stratified space. However, the decomposition is not a stratification
because $\pi$ is not an open map, equivalently the frontier condition fails.

##  Appendix A locally finite and locally Alexandrov spaces

A topological space $X$ is _locally finite_ if every point of $X$ is contained
in a finite open set. In [17], see also [24, Corollary 3.5], I. M. James
proved the following result, which we deduce from Corollary 3.4.

######  Corollary A.1.

Any locally finite space is an Alexandrov space.

###### Proof.

The pointwise decomposition of a space $X$ has decomposition space $X$, and
the decomposition map is the identity. When $X$ is a locally finite space, the
pointwise decomposition space $X$ is locally finite. Therefore the
decomposition space $X$ is Alexandrov by Corollary 3.4. ∎

######  Remark A.2.

A locally finite space is an Alexandrov space, but not vice versa. For
example, the Alexandrov space of the natural numbers $\mathbb{N}$ with the
usual order is _not_ locally finite, because the minimal open neighborhood
$U_{m}$ of $m\in\mathbb{N}$ is the infinite set $\\{n\mid m\leq n\\}$.

A topological space $X$ is locally Alexandrov if each point has an open
neighborhood which is an Alexandrov space in the subspace topology. Since a
finite space is an Alexandrov space, the following is a generalization of
Corollary A.1:

######  Proposition A.3.

Any locally Alexandrov space is an Alexandrov space.

###### Proof.

Let $U_{i}\subseteq X$ be open for $i\in I$. To show that $X$ is Alexandrov,
we must show that $\bigcap_{i\in I}U_{i}$ is an open subset of $X$. Since $X$
is locally Alexandrov, there is an open covering $X=\bigcup_{j\in J}V_{j}$
such that each $V_{j}$ is an Alexandrov space in the subspace topology. Then

$\bigcap_{i\in I}U_{i}=X\cap\left(\bigcap_{i\in
I}U_{i}\right)=\left(\bigcup_{j\in J}V_{j}\right)\cap\left(\bigcap_{i\in
I}U_{i}\right)=\bigcup_{j\in J}\left(\bigcap_{i\in I}(V_{j}\cap
U_{i})\right).$

Since each $V_{j}$ is an Alexandrov space, $\bigcap_{i\in I}(V_{j}\cap U_{i})$
is an open subset of $V_{j}$, hence also an open subset of $X$. Thus the right
hand side above is an open subset of $X$, and therefore so is $\bigcap_{i\in
I}U_{i}$. ∎

##  Appendix B Semicontinuous decompositions

We briefly discuss R. L. Moore’s notions of semicontinuity and continuity for
decompositions ([22], cf. [23]) . In this appendix we consider a decomposition
$\mathcal{D}$ of a topological space $X$ to be an equivalence relation, and
therefore denote the decomposition space by $X/\mathcal{D}$. The strata
$X_{i}$ are the equivalence classes, that is the points of $X/\mathcal{D}$.

######  Definition B.1.

A decomposition $\mathcal{D}$ of a space $X$ is

  1. (1)

_upper semicontinuous_ if $\bigcup_{X_{i}\cap U\neq\emptyset}X_{i}$ is open
for any open set $U$ of $X$;

  2. (2)

_lower semicontinuous_ if $\bigcup_{X_{i}\cap F\neq\emptyset}X_{i}$ is closed
for any closed set $F$ of $X$;

  3. (3)

_continuous_ if it is both upper and lower semicontinuous.

These notions can be phrased in terms of the quotient, or decomposition, map
$X\to X/\mathcal{D}$ as follows. The decomposition $\mathcal{D}$ is

  1. (1)

upper semicontinuous if, and only if, $X\to X/\mathcal{D}$ is a closed map;

  2. (2)

lower semicontinuous if, and only if, $X\to X/\mathcal{D}$ is an open map;

  3. (3)

continuous if, and only if, $X\to X/\mathcal{D}$ is an open and closed map.

The theory of decomposing a (metric) space into continua (compact connected
spaces) was developed by R. L. Moore in the 1920s and later by R.H. Bing in
the 1950s (e.g., see [10]). Moore’s famous theorem [22] is that if
$\mathcal{D}$ is an upper semicontinuous decomposition of the 2-dimensional
Euclidean space $\mathbb{R}^{2}$ into continua, none of which separates
$\mathbb{R}^{2}$, then the decomposition space $\mathbb{R}^{2}/\mathcal{D}$ is
homeomorphic to the Euclidean space $\mathbb{R}^{2}$. In contrast, Bing [7]
proved that the analogue in $3$ dimensions is false: there exists an upper
semicontinuous decomposition $\mathcal{D}$ of the 3-dimensional Euclidean
space $\mathbb{R}^{3}$ into continua, none of which separates
$\mathbb{R}^{3}$, such that the decomposition space
$\mathbb{R}^{3}/\mathcal{D}$ is neither homeomorphic to the Euclidean space
$\mathbb{R}^{3}$ nor even a manifold, but nevertheless
$(\mathbb{R}^{3}/\mathcal{D})\times\mathbb{R}^{1}\cong\mathbb{R}^{4}$, see
[8]. This decomposition space $\mathbb{R}^{3}/\mathcal{D}$ is Bing’s famous
_dogbone space_.

A similar ‘wild topology’ is used in Freedman’s proof of the $4$-dimensional
Poincaré Conjecture, see for example [14, 15], where again upper
semicontinuous decompositions play a key role, namely, _Casson’s ‘kinky’
handles_. In contrast, the notion of _lower_ semicontinous decompostion
(together with local-finiteness) seems to play a key role in stratification
theory, singularity theory, and in algebraic and differential geometry.

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# Riesz-type criteria for the Riemann hypothesis

Archit Agarwal  Archit Agarwal
Department of Mathematics
Indian Institute of Technology Indore
Simrol, Indore, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>,  Meghali Garg  Meghali Garg
Department of Mathematics
Indian Institute of Technology Indore
Simrol, Indore, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Bibekananda Maji  Bibekananda Maji
Department of Mathematics
Indian Institute of Technology Indore
Simrol, Indore, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In 1916, Riesz proved that the Riemann hypothesis is equivalent to the bound
$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}\exp\left(-\frac{x}{n^{2}}\right)=O_{\epsilon}\left(x^{-\frac{3}{4}+\epsilon}\right),$
as $x\rightarrow\infty$, for any $\epsilon>0$. Around the same time, Hardy and
Littlewood gave another equivalent criteria for the Riemann hypothesis while
correcting an identity of Ramanujan. In the present paper, we establish a one-
variable generalization of the identity of Hardy and Littlewood and as an
application, we provide Riesz-type criteria for the Riemann hypothesis. In
particular, we obtain the bound given by Riesz as well as the bound of Hardy
and Littlewood.

2010 Mathematics Subject Classification. Primary 11M06; Secondary 11M26 .
Keywords and phrases. Riemann zeta function, Non-trivial zeros, Möbious
function, Riemann hypothesis.

##  1\. introduction

Ramanujan discovered many beautiful identities in his short life span of 32
years, and most of the identities are correct, but a few are incorrect. Over
the years, these erroneous identities also inspired mathematicians to develop
many interesting results. One of such incorrect identities is the following
identity: For any $x>0$,

$\displaystyle\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left({-\frac{x}{n^{2}}}\right)=\sqrt{\frac{\pi}{x}}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left(-\frac{\pi^{2}}{n^{2}x}\right),$
(1.1)

where $\mu(n)$ denotes the Möbius function. Ramanujan, during his stay at
Trinity, showed the above identity to Hardy and Littlewood from his Second
Notebook [15, p. 312]. Unfortunately, all the identities on page 312 are
incorrect. Bruce Berndt111There is a typo in [2, Equation (37.3), p. 470], in
which the expression $\frac{\pi}{p}$ in the right-side sum over $\rho$ should
be replaced by $\frac{\pi}{\sqrt{p}}$.  [2, p. 468-469] gave a very nice
numerical explanation about the wrongness of the identity (1.1). Hardy and
Littlewood [9, p. 156, Section 2.5] established a corrected version of (1.1).
Mainly, they proved that, for any $x>0$,

$\displaystyle\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left({-\frac{x}{n^{2}}}\right)=\sqrt{\frac{\pi}{x}}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left(-\frac{\pi^{2}}{n^{2}x}\right)-\frac{1}{2\sqrt{\pi}}\sum_{\rho}\left(\frac{\pi}{\sqrt{x}}\right)^{\rho}\frac{\Gamma\left(\frac{1-\rho}{2}\right)}{\zeta^{\prime}(\rho)},$
(1.2)

where the right-side sum over $\rho$ runs through the non-trivial zeros of the
Riemann zeta function $\zeta(s)$. The identity (1.2) holds under the
assumption that, all non-trivial zeros of $\zeta(s)$ are simple. Replacing $x$
by $\alpha^{2}$ and $\beta=\pi/\alpha$, one can immediately show that

$\displaystyle\sqrt{\alpha}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left({-\left(\frac{\alpha}{n}\right)^{2}}\right)-\sqrt{\beta}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left({-\left(\frac{\beta}{n}\right)^{2}}\right)=-\frac{1}{2\sqrt{\beta}}\sum_{\rho}\frac{\Gamma\left(\frac{1-\rho}{2}\right)\beta^{\rho}}{\zeta^{\prime}(\rho)}.$
(1.3)

The convergence of the above right-side sum is delicate. Hardy and Littlewood
showed that this infinite sum over the non-trivial zeros of $\zeta(s)$ is
convergent under the assumption of bracketing the terms, that is, the terms
corresponding to the non-trivial zeros $\rho_{1}$ and $\rho_{2}$ will be
inside the same bracket if for some positive constant $A_{0}$, we have

$\displaystyle|\operatorname{Im}(\rho_{1})-\operatorname{Im}(\rho_{2})|<\exp\left(-\frac{A_{0}\operatorname{Im}(\rho_{1})}{\log(\operatorname{Im}(\rho_{1}))}\right)+\exp\left(-\frac{A_{0}\operatorname{Im}(\rho_{2})}{\log(\operatorname{Im}(\rho_{2}))}\right).$
(1.4)

To know more about the identity (1.3), readers can see Berndt [2, p. 470],
Paris and Kaminski [14, p. 143], and Titchmarch [17, p. 219].

Riesz [16], in 1916, proved that the Riemann hypothesis is equivalent to the
fact that

$P_{2}(x):=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}\exp\left(-\frac{x}{n^{2}}\right)=O_{\epsilon}\left(x^{-\frac{3}{4}+\epsilon}\right),\quad{\rm
as}\,\,x\rightarrow\infty,$ (1.5)

for any positive $\epsilon$. Around the same time, as an application of (1.3),
Hardy and Littlewood [9, p. 161] showed that the Riemann hypothesis is
equivalent to the following bound

$\displaystyle
P_{1}(x):=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\exp\left({-\frac{x}{n^{2}}}\right)=O_{\epsilon}\left(x^{-\frac{1}{4}+\epsilon}\right),\quad\mathrm{as}\,\,x\rightarrow\infty,$
(1.6)

for any positive $\epsilon$.

In the current paper, we establish a one-variable generalization of the
identity (1.2) and as an application of our result, we obtain an equivalent
criteria for the Riemann hypothesis. In particular, we are able to recover the
equivalent criteria (1.5) given by Riesz, as well as the bound (1.6) given by
Hardy and Littlewood.

Now we define an important special function. Let $c_{1},\cdots,c_{p}$ and
$d_{1},\cdots,d_{q}$ be $p+q$ complex numbers. The generalized hypergeometric
series [13, p. 404, Equation 16.2.1] is defined by the following series
expansion

$\displaystyle{}_{p}F_{q}\left(c_{1},\cdots,c_{p};\,d_{1},\cdots,d_{q};\,z\right):=\sum_{n=0}^{\infty}\frac{(c_{1})_{n}\cdots(c_{p})_{n}}{(d_{1})_{n}\cdots(d_{q})_{n}}\frac{z^{n}}{n!},$

where $(c)_{n}:=\frac{\Gamma(c+n)}{\Gamma(c)}$. This series converges for all
complex values of $z$ if $p\leq q$. However, if $p=q+1$, then it converges for
$|z|<1$, and in this case, it can be analytically continued to $\mathbb{C}$ if
we consider a branch cut from $1$ to $+\infty$. In particular, when $p=q=1$,
that is, ${}_{1}F_{1}\left(c_{1};\,d_{1};\,z\right)$ is an entire function and
if $c_{1}=d_{1}$, it reduces to the exponential function.

Now we are ready to state the main identity.

######  Theorem 1.1.

Let $k\geq 1$ be a real number. Assume that all non-trivial zeros of
$\zeta(s)$ are simple. Then for any positive $x$, we have

$\displaystyle\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)=\frac{\Gamma(\frac{k}{2})}{x^{\frac{k}{2}}}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}{}_{1}F_{1}\left(\frac{k}{2};\frac{1}{2};-\frac{\pi^{2}}{n^{2}x}\right)+\frac{1}{2}\sum_{\rho}\frac{\Gamma(\frac{k-\rho}{2})}{\zeta^{\prime}(\rho)}x^{-\frac{(k-\rho)}{2}},$
(1.7)

where the sum over $\rho$ runs through all non-trivial zeros of $\zeta(s)$ and
satisfying the condition(1.4).

Substituting $k=1$ in (1.7), and replacing $x$ by $\pi^{2}/x$, one can
immediately obtain (1.2). Letting $x\rightarrow\infty$ in (1.7) and assuming
the absolute convergence of the right side series over the non-trivial zeros
of $\zeta(s)$, one can immediately see that

$\displaystyle
P_{k}(x):=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)=O\bigg{(}x^{-\frac{k}{2}+\frac{1}{4}}\bigg{)},$

under the assumption of the Riemann hypothesis. This observation suggests us
to obtain the following equivalent criteria for the Riemann hypothesis.

######  Theorem 1.2.

The Riemann hypothesis is equivalent to the bound

$P_{k}(x)=O_{\epsilon}\bigg{(}x^{-\frac{k}{2}+\frac{1}{4}+\epsilon}\bigg{)},\quad\mathrm{as}\,\,x\rightarrow\infty,$
(1.8)

for any positive $\epsilon$.

In particular, letting $k=1$ in (1.8) allows us to recover the bound (1.6) of
Hardy and Littlewood, and $k=2$ gives the bound (1.5) of Riesz.

##  2\. Preliminaries

In this section, we state a few well-known results, which will play a vital
role in proving our main identity. We begin with an important summation
formula, namely, Euler’s summation formula.

######  Lemma 2.1.

Let $\\{a_{n}\\}$ be a sequence of complex number and $f(t)$ be a continuously
differentiable function on $[1,x]$. Consider $A(x):=\sum_{1\leq n\leq
x}a_{n}$. Then we have

$\displaystyle\sum_{1\leq n\leq
x}a_{n}f(n)=A(x)f(x)-\int_{1}^{x}A(t)f^{\prime}(t){\rm d}t.$ (2.1)

  * Proof.

Proof of this result can be found in [12, p. 17] ∎

The next result gives an important information about the asymptotic expansion
of the gamma function.

######  Lemma 2.2.

In a vertical strip $\alpha\leq\sigma\leq\beta$, we have

$|\Gamma(\sigma+iT)|=\sqrt{2\pi}|T|^{\sigma-1/2}e^{-\frac{1}{2}\pi|T|}\left(1+O\left(\frac{1}{|T|}\right)\right)\quad{\rm
as}\quad|T|\rightarrow\infty.$ (2.2)

This is popularly known as Stirling’s formula. Now we will state an important
result, which will provide us a bound for $1/\zeta(s)$.

######  Lemma 2.3.

Suppose, for every non-trivial zero $\rho$ of $\zeta(s)$, there exist a
sequence of arbitary larger positive numbers $T$ with
$|T-\operatorname{Im}(\rho)|>\exp\left(-\frac{C_{0}\operatorname{Im}(\rho)}{\log(\operatorname{Im}(\rho))}\right)$,
where $A$ is some positive constant. Then,

$\frac{1}{|\zeta(\sigma+iT)|}<e^{C_{1}T},$

for some constant $0<C_{1}<\pi/4$.

  * Proof.

One can find a proof of this result in [17, p. 219]. An important point to
note that the assumption of the bracketing condition (1.4) guarantee the
existence of such a sequence. ∎

The next result will be one of the crucial ingredients in providing an
equivalence criteria for the Riemann hypothesis.

######  Lemma 2.4.

Let $k\geq 1$. In the region $\frac{1-k}{2}<\operatorname{Re}(s)<1$, except at
$s=0$, we have

$\displaystyle\int_{0}^{\infty}x^{-s-1}P_{k}(x){\rm
d}x=\frac{\Gamma(-s)}{\zeta(2s+k)}.$

  * Proof.

This lemma was proved by Riesz [16] for $k=1$, which was used by Hardy and
Littlewood [9, p. 161, Equn (2.544)] to give an equivalent criteria for the
Riemann hypothesis. Here we shall prove this lemma for $k>1$. Using the
definition of $P_{k}(x)$, one can show that

$\displaystyle
P_{k}(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\sum_{m=0}^{\infty}\frac{(-1)^{m}x^{m}}{m!n^{2m}}=\sum_{m=0}^{\infty}\frac{(-1)^{m}x^{m}}{m!\zeta(k+2m)}.$
(2.3)

Note that, when $k=1$, the last sum will start from $m=1$ since the term
corresponding to $m=0$ will become zero. For
$\operatorname{Re}(s)>\frac{1-k}{2}$, use the series expansion of
$\zeta(2s+k)$ to write

$\displaystyle\zeta(2s+k)\int_{0}^{\infty}x^{-s-1}P_{k}(x){\rm d}x$
$\displaystyle=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\int_{0}^{\infty}\frac{x^{-s-1}}{n^{2s}}P_{k}(x){\rm
d}x$
$\displaystyle=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\int_{0}^{\infty}x^{-s-1}P_{k}\left(\frac{x}{n^{2}}\right){\rm
d}x$
$\displaystyle=\int_{0}^{\infty}x^{-s-1}\sum_{n=1}^{\infty}\frac{1}{n^{k}}P_{k}\left(\frac{x}{n^{2}}\right){\rm
d}x.$ (2.4)

Here in the second last step we replaced $x$ by $x/n^{2}$ and in the final
step interchanging of summation and integration was possible due to
Weierstrass M-test as one can easily show that
$\left|\frac{1}{n^{k}}P_{k}\left(\frac{x}{n^{2}}\right)\right|\leq\frac{\zeta(k)}{n^{k}}$
for $k>1$. Now using (2.3), one can find that

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{k}}P_{k}\left(\frac{x}{n^{2}}\right)=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\sum_{m=0}^{\infty}\frac{(-1)^{m}x^{m}}{m!n^{2m}\zeta(k+2m)}=\sum_{m=0}^{\infty}\frac{(-1)^{m}x^{m}}{m!}=e^{-x}.$
(2.5)

On substituting (2.5) in (Proof.), we get

$\displaystyle\zeta(2s+k)\int_{0}^{\infty}x^{-s-1}P_{k}(x){\rm
d}x=\int_{0}^{\infty}x^{-s-1}e^{-x}{\rm d}x=\Gamma(-s),$

for $\operatorname{Re}(s)<0$. By analytic continuation it can be extended to
the right half plane $\operatorname{Re}(s)<1$ except at $s=0$.

∎

Now we define an important special function, namely, the Meijer $G$-function
[13, p. 415, Definition 16.17], which is the generalization of many well-known
special functions. Let $m,n,p,q$ be non-negative integers with $0\leq m\leq
q$, $0\leq n\leq p$. Let $a_{1},\cdots,a_{p}$ and $b_{1},\cdots,b_{q}$ be
complex numbers with $a_{i}-b_{j}\not\in\mathbb{N}$ for $1\leq i\leq n$ and
$1\leq j\leq m$. Then the Meijer $G$-function is defined by

$\displaystyle G_{p,q}^{\,m,n}\\!\left(\,\begin{matrix}a_{1},\cdots,a_{p}\\\
b_{1},\cdots,b_{q}\end{matrix}\;\Big{|}z\right):=\frac{1}{2\pi
i}\int_{L}\frac{\prod_{j=1}^{m}\Gamma(b_{j}-s)\prod_{j=1}^{n}\Gamma(1-a_{j}+s)z^{s}}{\prod_{j=m+1}^{q}\Gamma(1-b_{j}+s)\prod_{j=n+1}^{p}\Gamma(a_{j}-s)}\mathrm{d}s,$
(2.6)

where the line of integration $L$, from $-i\infty$ to $+i\infty$, separates
the poles of the factors $\Gamma(1-a_{j}+s)$ from those of the factors
$\Gamma(b_{j}-s)$. The above integral converges if $p+q<2(m+n)$ and
$|\arg(z)|<\left(m+n-\frac{p+q}{2}\right)\pi$.

Now we state Slater’s theorem [13, p. 415, Equation 16.17.2], which will allow
us to express Meijer $G$-function in terms of generalized hypergeometric
functions. If $p\leq q$ and $b_{j}-b_{k}\not\in\mathbb{Z}$ for $j\neq k$,
$1\leq j,k\leq m$, then

$\displaystyle G_{p,q}^{\,m,n}\\!\left(\,\begin{matrix}a_{1},\cdots,a_{p}\\\
b_{1},\cdots,b_{q}\end{matrix}\;\Big{|}z\right)$ (2.7)
$\displaystyle\quad=\sum_{k=1}^{m}A_{p,q,k}^{m,n}(z){}_{p}F_{q-1}\left(\begin{matrix}1+b_{k}-a_{1},\cdots,1+b_{k}-a_{p}\\\
1+b_{k}-b_{1},\cdots,*,\cdots,1+b_{k}-b_{q}\end{matrix}\Big{|}(-1)^{p-m-n}z\right),$

where $*$ means that the entry $1+b_{k}-b_{k}$ is omitted and

$\displaystyle A_{p,q,k}^{m,n}(z):=\frac{z^{b_{k}}\prod_{j=1,j\neq
k}^{m}\Gamma(b_{j}-b_{k})\prod_{j=1}^{n}\Gamma(1+b_{k}-a_{j})}{\prod_{j=m+1}^{q}\Gamma(1+b_{k}-b_{j})\prod_{j=n+1}^{p}\Gamma(a_{j}-b_{k})}.$

Now we are ready to provide the proof of the main results.

##  3\. Proof of main results

  * Proof of Theorem 1.1.

It is well-known that the gamma function $\Gamma(s)$ is the Mellin transform
of $e^{-x}$. Therefore, $e^{-x}$ is the inverse Mellin transform of
$\Gamma(s)$. That is,

$\displaystyle e^{-x}=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)x^{-s}{\rm d}s,$

valid for any $c>0$. If we shift the line of integration to the line
$c\in(-1,0)$, then we will have

$\displaystyle e^{-x}-1=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)x^{-s}{\rm d}s.$ (3.1)

First, we recall that the prime number theorem is equivalent to the fact that
$\sum_{n=1}^{\infty}\frac{\mu(n)}{n}=0$. One can easily show that, for any
$k\in\mathbb{N}$, the series
$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)$
is convergent. In fact, for $k\geq 2$, it is absolutely convergent, whereas
for $k=1$, it is conditionally convergent. However, our proof is valid for any
$k\geq 1$. With the help of (3.1), we can write

$\displaystyle\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)$
$\displaystyle=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}-\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\left(1-\exp\left({-\frac{x}{n^{2}}}\right)\right)$
$\displaystyle=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}+\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\left(\frac{x}{n^{2}}\right)^{-s}{\rm
d}s$ $\displaystyle=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}+\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm d}s,$
(3.2)

here in the ultimate step we interchanged the summation and integration since
the series $\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k-2s}}$ is absolutely
absolutely and uniformly convergent in any compact subset of the domain
$\operatorname{Re}(k-2s)>1$. Now we shall try to simplify the line integral
present in (3.2). Let us denote this line integral as

$\displaystyle I_{k}(x):=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm d}s.$
(3.3)

First, we observe that the integrand function has simple poles at non-positive
integers due to the simple poles of $\Gamma(s)$. Again, for any
$m\in\mathbb{N}$, $\frac{k}{2}+m$ is also a simple pole of the integrand
function since $\zeta(k-2s)$ has trivial zeros at $-2m$. Another important
observation is that $\zeta(k-2s)$ has infinitely many non-trivial zeros in the
critical strip $\frac{k}{2}-1<\operatorname{Re}(s)<\frac{k}{2}$. If one
assumes the Riemann hypothesis, then we can say all the non-trivial zeros of
$\zeta(k-2s)$ will lie on the line
$\operatorname{Re}(s)=\frac{k}{2}-\frac{1}{4}$. However, to obtain our main
identity, we do not require the assumption of the Riemann hypothesis.
Nevertheless, as we are interested to collect the contributions of the
residual terms corresponding to the non-trivial zeros of $\zeta(k-2s)$, so we
have to shift the line of integration to the line
$\operatorname{Re}(s)=\lambda$, where
$\lambda\in\left(\frac{k}{2},\frac{k}{2}+1\right)$.

Thus, we consider the contour $\mathcal{C}$ determined by the line segments
$[\lambda-iT,\lambda+iT],[\lambda+iT,c+iT],[c+iT,c-iT]$, and $[c-iT,\lambda-
iT]$, where $-1<c<0$ and $\frac{k}{2}<\lambda<\frac{k}{2}+1$, and $T$ is some
large positive number. Now appealing to Cauchy’s residue theorem, we have

$\displaystyle\frac{1}{2\pi
i}\int_{\mathcal{C}}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm
d}s=R_{0}+\mathcal{R}_{T}(x),$ (3.4)

where $R_{0}$ denotes the residual term due to the simple pole of $\Gamma(s)$
at $s=0$ and $\mathcal{R}_{T}(x)$ denotes the sum of the residual terms
corresponding to the non trivial zeroes $\rho$ of $\zeta(k-2s)$ with
$|\operatorname{Im}(\rho)|<T$. Next, we shall try to show that the horizontal
integrals

$\displaystyle H_{1}(x;T):=\frac{1}{2\pi
i}\int_{\lambda+iT}^{c+iT}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm
d}s,\quad{\rm and}\quad H_{2}(x;T):=\frac{1}{2\pi i}\int_{c-iT}^{\lambda-
iT}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm d}s$

vanish as $T$ approaches to infinity. Here, we employ Starling’s formula for
$\Gamma(s)$ i.e., Lemma 2.2 and along with Lemma 2.3, we find that

$\displaystyle|H_{1}(x;T)|\ll\int_{\lambda}^{c}|T|^{C}\exp\left(2C_{1}T-\frac{\pi}{2}|T|\right){\rm
d}T,$

where $C_{1}$ is some constant, and $0<C_{1}<\pi/4$. This immediately implies
that $H_{1}(x;T)$ vanishes as $T\rightarrow\infty$. In a similar way, one can
show that $H_{2}(x;T)$ goes to zero as $T\rightarrow\infty$. Now letting
$T\rightarrow\infty$ in (3.4), we arrive at

$\displaystyle\frac{1}{2\pi
i}\int_{c-iT}^{c+iT}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm d}s=\frac{1}{2\pi
i}\int_{\lambda-iT}^{\lambda+iT}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm
d}s-R_{0}-\mathcal{R}(x),$ (3.5)

where $\mathcal{R}(x)$ is an infinite series, that is, the sum of the residual
terms corresponding to the non-trivial zeros of $\zeta(k-2s)$. One can easily
calculate the residue at $s=0$,

$\displaystyle R_{0}=\lim_{s\rightarrow
0}\frac{s\Gamma(s)}{\zeta(k-2s)}x^{-s}=\frac{1}{\zeta(k)}.$ (3.6)

If we assume that all the non-trivial zeros of $\zeta(s)$ are simple, then we
get

$\displaystyle\mathcal{R}(x)=\sum_{\rho}\lim_{s\rightarrow\frac{k-\rho}{2}}\frac{\left(s-\frac{k-\rho}{2}\right)\Gamma(s)}{\zeta(k-2s)}x^{-s}=-\frac{1}{2}\sum_{\rho}\frac{\Gamma(\frac{k-\rho}{2})}{\zeta^{\prime}(\rho)}x^{-\frac{k-\rho}{2}},$
(3.7)

where the sum runs through all the non-trivial zeros $\zeta(s)$. More
generally, if we do not assume the simplicity hypothesis, that is, if
$n_{\rho}$ is the multiplicity of $\rho$, then

$\displaystyle\mathcal{R}(x)=\sum_{\rho}\frac{1}{(n_{\rho}-1)!}$
$\displaystyle\lim_{s\rightarrow\frac{k-\rho}{2}}\frac{{\rm
d}^{n_{\rho}-1}}{{\rm
d}s^{n_{\rho}-1}}\Bigg{\\{}\left(s-\frac{k-\rho}{2}\right)^{n_{\rho}}\frac{\Gamma(s)}{\zeta(k-2s)x^{s}}\Bigg{\\}}.$

Substituting (3.6) and (3.7) in (3.5) and combining with (3.2) and (3.3), we
obtain

$\displaystyle\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)=\frac{1}{2\pi
i}\int_{\lambda-i\infty}^{\lambda+i\infty}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm
d}s+\frac{1}{2}\sum_{\rho}\frac{\Gamma(\frac{k-\rho}{2})}{\zeta^{\prime}(\rho)}x^{-\frac{k-\rho}{2}}.$
(3.8)

At this juncture, we shall try to simplify the right vertical integral

$\displaystyle V(x,k):=\frac{1}{2\pi
i}\int_{\lambda-i\infty}^{\lambda+i\infty}\frac{\Gamma(s)}{\zeta(k-2s)}x^{-s}{\rm
d}s,$ (3.9)

where $\frac{k}{2}<\lambda<\frac{k}{2}+1$. Here we make use of the symmetric
form of the functional equation of $\zeta(s)$, that is,

$\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s),$

replace $s$ by $k-2s$ and simplify to see

$\displaystyle\frac{1}{\zeta(k-2s)}=\frac{\pi^{\frac{1}{2}-k+2s}\Gamma\left(\frac{k-2s}{2}\right)}{\Gamma\left(\frac{1-k+2s}{2}\right)\zeta(1-k+2s)}.$
(3.10)

Plugging (3.10) in (3.9), one can see that

$\displaystyle V(x,k):=\frac{\pi^{\frac{1}{2}-k}}{2\pi
i}\int_{\lambda-i\infty}^{\lambda+i\infty}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)\zeta(1-k+2s)}\left(\frac{x}{\pi^{2}}\right)^{-s}{\rm
d}s.$ (3.11)

Here we can check that $1<\operatorname{Re}(1-k+2s)<3$ since
$\frac{k}{2}<\operatorname{Re}(s)=\lambda<\frac{k}{2}+1$. Thus, we can write

$\displaystyle\frac{1}{\zeta(1-k+2s)}=\sum_{n=1}^{\infty}\frac{\mu(n)n^{k-1}}{n^{2s}}.$

Making use of the above series expansion in (3.11) and then interchanging the
summation and integration, (3.11) reduces to

$\displaystyle
V(x,k):=\pi^{\frac{1}{2}-k}\sum_{n=1}^{\infty}\mu(n)n^{k-1}I(X_{n},k),$ (3.12)

where

$\displaystyle I(X_{n},k):=\frac{1}{2\pi
i}\int_{\lambda-i\infty}^{\lambda+i\infty}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)}X_{n}^{-s}{\rm
d}s,$ (3.13)

and $X_{n}=\frac{n^{2}x}{\pi^{2}}$. Now our main goal is the simplify the
integral $I(X_{n},k)$. We shall try to express this integral in terms of the
Meijer $G$-function. We know that $\Gamma(s)$ has poles at $s=0,-1,-2,\cdots$,
and the poles of $\Gamma\left(\frac{k}{2}-s\right)$ are at
$\frac{k}{2},\frac{k}{2}-1,\frac{k}{2}-2,\cdots$. Thus, we can easily observe
that the line of integration $\operatorname{Re}(s)=\lambda$ does not separate
the poles of $\Gamma(s)$ with the poles of $\Gamma\left(\frac{k}{2}-s\right)$
since $\frac{k}{2}<\lambda<\frac{k}{2}+1$. Therefore, we shift the line of
integration $\operatorname{Re}(s)=\lambda$ to a new line
$\operatorname{Re}(s)=\lambda_{1}$, where $0<\lambda_{1}<\frac{k}{2}$, so that
this new line of integration does separate the poles $\Gamma(s)$ from the
poles of $\Gamma\left(\frac{k}{2}-s\right)$. With the help of this new line of
integration, we construct a contour $\mathcal{C}_{1}$ determined by line
segments $[\lambda-
iT,\lambda+iT],[\lambda+iT,\lambda_{1}+iT],[\lambda_{1}+iT,\lambda_{1}-iT]$,
and $[\lambda_{1}-iT,\lambda-iT]$. Again, utilizing the residue theorem, we
have

$\displaystyle\frac{1}{2\pi i}\int_{\lambda-
iT}^{\lambda+iT}+\int_{\lambda+iT}^{\lambda_{1}+iT}+\int_{\lambda_{1}+iT}^{\lambda_{1}-iT}+\int_{\lambda_{1}-iT}^{\lambda-
iT}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)}X_{n}^{-s}{\rm
d}s=\mathop{\rm{Res}}_{s=\frac{k}{2}}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)}X_{n}^{-s}.$
(3.14)

Letting $T\rightarrow\infty$ and using Stirling’s formula (2.2) for
$\Gamma(s)$, one can show that both of the horizontal integrals vanish.
Therefore, (3.14) becomes

$\displaystyle\frac{1}{2\pi
i}\int_{\lambda-i\infty}^{\lambda+i\infty}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)}X_{n}^{-s}{\rm
d}s=\frac{1}{2\pi
i}\int_{\lambda_{1}-i\infty}^{\lambda_{1}-i\infty}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)}X_{n}^{-s}{\rm
d}s-\frac{\Gamma(\frac{k}{2})X_{n}^{-\frac{k}{2}}}{\sqrt{\pi}}.$ (3.15)

We have already seen that the line of integration
$\operatorname{Re}(s)=\lambda_{1}$ does separate the poles of $\Gamma(s)$ from
the poles of $\Gamma\left(\frac{k}{2}-s\right)$. Thus, making use of the
definition (2.6) of the Meijer $G$-function, with $m=n=p=1,q=2$, and
$a_{1}=1,b_{1}=\frac{k}{2},b_{2}=\frac{1+k}{2}$, we can write

$\displaystyle\frac{1}{2\pi
i}\int_{\lambda_{1}-i\infty}^{\lambda_{1}-i\infty}\frac{\Gamma(s)\Gamma\left(\frac{k}{2}-s\right)}{\Gamma\left(\frac{1-k}{2}+s\right)}X_{n}^{-s}{\rm
d}s=G_{1,2}^{1,1}\left(\begin{matrix}1\\\
\frac{k}{2},\frac{k+1}{2}\end{matrix}\Big{|}\frac{1}{X_{n}}\right).$ (3.16)

Here one can easily check all the necessary conditions for the convergence of
this Meijer $G$-function. Now invoking Slater’s theorem (2.7), one can show
that

$\displaystyle G_{1,2}^{1,1}\left(\begin{matrix}1\\\
\frac{k}{2},\frac{k+1}{2}\end{matrix}\Big{|}\frac{1}{X_{n}}\right)=\frac{\Gamma\left(\frac{k}{2}\right)}{X_{n}^{\frac{k}{2}}\sqrt{\pi}}{}_{1}F_{1}\left(\frac{k}{2};\frac{1}{2};-\frac{1}{X_{n}}\right).$
(3.17)

Substituting (3.17) in (3.16) and together with (3.15) and (3.13), we see that
the right vertical integral $V(x,k)$ (3.12) takes the final shape as

$\displaystyle
V(x,k)=\frac{\Gamma\left(\frac{k}{2}\right)}{\pi^{k}}\sum_{n=1}^{\infty}\frac{\mu(n)n^{k-1}}{X_{n}^{\frac{k}{2}}}\left\\{{}_{1}F_{1}\left(\frac{k}{2};\frac{1}{2};-\frac{1}{X_{n}}\right)-1\right\\}.$

Now plugging $X_{n}=\frac{n^{2}x}{\pi^{2}}$, the above expression reduces to

$\displaystyle V(x,k)$
$\displaystyle=\frac{\Gamma\left(\frac{k}{2}\right)}{x^{\frac{k}{2}}}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\left\\{{}_{1}F_{1}\left(\frac{k}{2};\frac{1}{2};-\frac{\pi^{2}}{n^{2}x}\right)-1\right\\}$
(3.18)
$\displaystyle=\frac{\Gamma\left(\frac{k}{2}\right)}{x^{\frac{k}{2}}}\sum_{n=1}^{\infty}\frac{\mu(n)}{n}\left\\{{}_{1}F_{1}\left(\frac{k}{2};\frac{1}{2};-\frac{\pi^{2}}{n^{2}x}\right)\right\\}.$
(3.19)

In the last step, we utilized the fact that
$\sum_{n=1}^{\infty}\frac{\mu(n)}{n}=0$, which is equivalent to the prime
number theorem. Here we note that the expression (3.18) is more effective than
the expression (3.19) to check numerically since the series (3.18) is
absolutely convergent. Finally, in view of (3.19) and (3.9), and together with
(3.8), we complete the proof of Theorem 1.1.

∎

  * Proof of Theorem 1.2.

First, let us assume that the Riemann hypothesis is true. Under the assumption
of the Riemann hypothesis, Littlewood showed that, for any $\epsilon>0$,

$\displaystyle A(x):=\sum_{1\leq n\leq
x}\mu(n)=O_{\epsilon}(x^{\frac{1}{2}+\epsilon}).$ (3.20)

We invoke Euler’s partial summation formula (2.1), with $a_{n}=\mu(n)$ and
$f(t)=t^{-k}$, to see that

$\displaystyle M(\ell;n):=\sum_{m=\ell}^{n}\frac{\mu(m)}{m^{k}}$
$\displaystyle=A(n)f(n)-A(\ell-1)f(\ell-1)-\int_{\ell-1}^{n}A(t)f^{\prime}(t){\rm
d}t.$ (3.21)

Making use of (3.20) in (3.21) and simplifying, one can easily show that

$\displaystyle M(\ell;n)=O_{\epsilon}(\ell^{\frac{1}{2}-k+\epsilon}),$ (3.22)

uniformly in $n$. Our main aim is to find an estimate for the following sum

$\displaystyle
P_{k}(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right).$

We separate the sum into two parts, that is,

$\displaystyle P_{k}(x^{2})$
$\displaystyle=\sum_{n=1}^{\ell-1}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x^{2}}{n^{2}}}\right)+\sum_{n=\ell}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x^{2}}{n^{2}}}\right)$
$\displaystyle:=S_{1}(x^{2})+S_{2}(x^{2}),$ (3.23)

where $\ell=[x^{1-\epsilon}]+1$. First, we shall try to find an estimate for
$S_{2}(x^{2})$. Using the construction (3.21) of $M(l;n)$, for an integer
$N>\ell$, one can check that

$\displaystyle\sum_{n=\ell}^{N}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x^{2}}{n^{2}}}\right)$
$\displaystyle=\frac{\mu(\ell)}{\ell^{k}}\exp\left({-\frac{x^{2}}{\ell^{2}}}\right)+\sum_{\ell<n\leq
N}\left(M(l;n)-M(l;n-1)\right)\exp\left({-\frac{x^{2}}{n^{2}}}\right)$
$\displaystyle=\sum_{n=\ell}^{N-1}M(l;n)\left(e^{-\frac{x^{2}}{n^{2}}}-e^{-\frac{x^{2}}{(n+1)^{2}}}\right)+M(l;N)e^{-\frac{x^{2}}{N^{2}}}.$
(3.24)

Letting $N\rightarrow\infty$ in ( Proof of Theorem 1.2.) and utilizing (3.22),
we arrive at

$\displaystyle
S_{2}(x^{2})=S_{3}(x^{2})+O_{\epsilon}\left(\ell^{\frac{1}{2}-k+\epsilon}\right),$
(3.25)

where

$\displaystyle
S_{3}(x^{2}):=\sum_{n=\ell}^{\infty}M(l;n)\left(e^{-\frac{x^{2}}{n^{2}}}-e^{-\frac{x^{2}}{(n+1)^{2}}}\right).$

To estimate $S_{3}(x^{2})$, we consider a function
$f(y)=e^{-\frac{x^{2}}{y^{2}}}$. Using mean value theorem, one can find
$y_{n}\in(n,n+1)$ such that
$f(n)-f(n+1)=-f^{\prime}(y_{n})=-\frac{2x^{2}}{y_{n}^{3}}e^{-\frac{x^{2}}{y_{n}^{2}}}$.
Utilizing this fact and together with (3.22), we get

$\displaystyle|S_{3}(x^{2})|\ll_{\epsilon}\ell^{\frac{1}{2}-k+\epsilon}\sum_{n=\ell}^{\infty}\frac{2x^{2}}{y_{n}^{3}}e^{-\frac{x^{2}}{y_{n}^{2}}}.$

Further, we divide the sum into two parts in the following way,

$\displaystyle|S_{3}(x^{2})|\ll_{\epsilon}\ell^{\frac{1}{2}-k+\epsilon}\left(\sum_{n=\ell}^{[x]}+\sum_{n=[x]+1}^{\infty}\right)\frac{x^{2}}{y_{n}^{3}}e^{-\frac{x^{2}}{y_{n}^{2}}}.$
(3.26)

The finite sum can be trivially bounded by

$\displaystyle\sum_{n=\ell}^{[x]}\frac{x^{2}}{y_{n}^{3}}e^{-\frac{x^{2}}{y_{n}^{2}}}\ll\sum_{n=\ell}^{\infty}\frac{x^{2}}{n^{3}}\ll\frac{x^{2}}{l^{2}}\ll
x^{2\epsilon},$ (3.27)

as $\ell=[x^{1-\epsilon}]+1$. Now the infinite part can be bounded by

$\displaystyle\sum_{n=[x]+1}^{\infty}\frac{x^{2}}{y_{n}^{3}}e^{-\frac{x^{2}}{y_{n}^{2}}}\ll\sum_{n>x}\frac{x^{2}}{n^{3}}\ll
O(1).$ (3.28)

Here and in the previous inequality, we have used the fact that
$\sum_{n>x}n^{-s}=O(x^{1-s})$ for $\operatorname{Re}(s)>1$. Plugging (3.27)
and (3.28) in (3.26), we obtain

$\displaystyle|S_{3}(x^{2})|\ll_{\epsilon}x^{\frac{1}{2}-k+\epsilon}.$

Finally, substitute this bound of $S_{3}(x^{2})$ in (3.25) to see

$\displaystyle|S_{2}(x^{2})|=O_{\epsilon}\left(x^{\frac{1}{2}-k+\epsilon}\right).$
(3.29)

Now we shall concentrate on $S_{1}(x^{2})$. We write

$\displaystyle|S_{1}(x^{2})|$
$\displaystyle=\left|\sum_{n=1}^{\ell-1}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x^{2}}{n^{2}}}\right)\right|\leq\sum_{n=1}^{\ell-1}\exp\left({-\frac{x^{2}}{l^{2}}}\right).$

This implies that

$\displaystyle S_{1}(x^{2})=O\left(x^{1-\epsilon}\exp(-x^{2\epsilon})\right),$
(3.30)

as $\ell=[x^{1-\epsilon}]+1$. In view of (3.30) and (3.29), one can observe
that the bound for $S_{1}(x^{2})$ goes to zero much faster than the bound for
$S_{2}(x^{2})$ as $x\rightarrow\infty$. Therefore, ( Proof of Theorem 1.2.)
yields that

$\displaystyle
P_{k}(x^{2})=O_{\epsilon}\left(x^{\frac{1}{2}-k+\epsilon}\right).$

Finally, replacing $x$ by $\sqrt{x}$, one can deduce (1.8). Now assuming
(1.8), we shall try to show that the Riemann hypothesis is true. From Lemma
2.4, we know that

$\displaystyle\zeta(2s+k)\int_{0}^{\infty}x^{-s-1}P_{k}(x){\rm d}x=\Gamma(-s)$
(3.31)

valid in the region $\frac{1-k}{2}<\operatorname{Re}(s)<1$, except at $s=0$.
Next, we try to extend the region of validity of this lemma on the left half
plane. For a large positive integer $N$, we separate the above integral in the
following way:

$\displaystyle\zeta(2s+k)\left(\int_{0}^{N}+\int_{N}^{\infty}\right)x^{-s-1}P_{k}(x){\rm
d}x=\Gamma(-s)$

Employing the bound (1.8) for $P_{k}(x)$, one can easily show that the
unbounded part of the above integration is finite if
$\operatorname{Re}(s)>\frac{1}{4}-\frac{k}{2}$. This will allow us to extend
the analyticity of the identity (3.31) in the region
$\frac{1}{4}-\frac{k}{2}<\operatorname{Re}(s)<\frac{1-k}{2}$. Now we observe
that the right side of (3.31) is never zero in the region
$\frac{1}{4}-\frac{k}{2}<\operatorname{Re}(s)<\frac{1-k}{2}$, which implies
that $\zeta(2s+k)$ has no zero in the same region. Equivalently, we can say
that $\zeta(s)$ has no zero in the region $\frac{1}{2}<\operatorname{Re}(s)<1$
and thus the functional equation of $\zeta(s)$ implies that it has no zero
$0<\operatorname{Re}(s)<\frac{1}{2}$. This proves that all the non-trivial
zeros of $\zeta(s)$ must lie on $\operatorname{Re}(s)=\frac{1}{2}$.

∎

##  4\. Final Remarks

In the current paper, we established a one-variable generalization of the
identity (1.2). Hardy and Littlewood [9, p. 161] pointed out that the absolute
convergence of the series in (1.2), over the non-trivial zeros of $\zeta(s)$,
is not so immediate even after the assumption of the Riemann hypothesis. We
remark that the identity (1.2) and our identity (1.7) are both valid under the
assumption of the simplicity hypothesis of the non-trivial zeros of $\zeta(s)$
and we do not require the assumption of the Riemann hypothesis. In [11],
authors explained in more details why the convergence of this kind of series
with $\zeta^{\prime}(\rho)$ in the denominator is more delicate. Another
important observation is that the series
$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{k}}\exp\left({-\frac{x}{n^{2}}}\right)$
is absolutely convergent for any $k>1$, whereas for $k=1$ the convergence is
not absolute.

The identity (1.2) has been studied by many mathematicians in various
directions. A connection with Wiener’s Tauberian theory and the Fourier
reciprocity has been established by Bhaskaran [3]. A character analogue of
(1.3) was given by Dixit [4], and further a one-variable generalization
involving ${}_{1}F_{1}$ hypergeometric function was obtained by Dixit, Roy and
Zaharescu [7]. Again, Dixit et al. [6] established a generalization of (1.3)
for Hecke eigenform. In the same paper, they also obtained a Riesz-type
criteria for the Riemann hypothesis. Recently, an extension for the Dedekind
zeta function obtained by Dixit, Gupta and Vatwani [5]. They also established
an equivalent criteria for the extended Riemann hypothesis. Banerjee and Kumar
[1] derived an analogous identity corresponding to $L$-functions associated to
the primtive cusp forms over $\Gamma_{0}(N)$ and gave an equivalent criteria
for the grand Riemann hypothesis. In a forthcoming paper, we [8] are trying to
find a character analogue of (1.7), which will enable us to find an equivalent
criteria for the generalized Riemann hypothesis for the Dirichlet
$L$-function.

Acknowledgement: The third author wants to thank SERB for the Start-Up
Research Grant SRG/2020/000144.

## References

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