# On Ehresmann semigroups

Mark V. Lawson  Mark V. Lawson, Department of Mathematics and the Maxwell
Institute for Mathematical Sciences, Heriot-Watt University, Riccarton,
Edinburgh EH14 4AS, UNITED KINGDOM<EMAIL_ADDRESS>This paper is
dedicated to the memory of Peter M. Neumann

###### Abstract.

We formulate an alternative approach to describing Ehresmann semigroups by
means of left and right étale actions of a meet semilattice on a category. We
also characterize the Ehresmann semigroups that arise as the set of all
subsets of a finite category. As applications, we prove that every restriction
semigroup can be nicely embedded into a restriction semigroup constructed from
a category, and we describe when a restriction semigroup can be nicely
embedded into an inverse semigroup.

I would like to thank Prof. V. A. R. Gould of the University of York for her
comments on an earlier draft of this paper. I would also like to thank the
anonymous referee for their careful reading of the submitted paper and helpful
suggestions.

##  1\. Introduction

Ehresmann semigroups were introduced in [14] as generalizations of inverse
semigroups. We recall their definition here. An Ehresmann semigroup is a
semigroup $S$ with a distinguished subset $U\subseteq\mathsf{E}(S)$ of the set
of all idempotents, called the set of projections, equipped with two functions
$\ast,+\colon S\rightarrow U$ satisfying the following four axioms:

(ES1):

$U$ is a commutative subsemigroup; we may therefore, equivalently, view $U$ as
a meet semilattice.

(ES2):

The maps $\ast$ and $+$ are the identity on $U$.

(ES3):

$aa^{\ast}=a$ and $a^{+}a=a$ for all $a\in S$.

(ES4):

$(a^{\ast}b)^{\ast}=(ab)^{\ast}$ and $(ab^{+})^{+}=(ab)^{+}$ for all $a,b\in
S$.

Observe that $a^{\ast}$ is the smallest projection $e$ such that $ae=a$; a
dual result holds for $a^{+}$. Ehresmann semigroups will be denoted by $(S,U)$
to make the set of projections clear or by $S$ alone if the set of projections
is already evident. A morphism of Ehresmann semigroups $(S,U)$ and $(T,V)$ is
a semigroup homomorphism $\theta\colon S\rightarrow T$ such that
$\theta(a)^{\ast}=\theta(a^{\ast})$ and $\theta(a)^{+}=\theta(a^{+})$ for all
$a\in S$. Observe that if $e\in U$ then
$\theta(e)^{\ast}=\theta(e^{\ast})=\theta(e)$. It follows that $\theta$ maps
$U$ to $V$. We say that such a morphism is an isomorphism if $\theta$ is also
an isomorphism of semigroups. Observe that such an isomorphism induces an
isomorphism between $U$ and $V$. Ehresmann semigroups can be equivalently
described as algebras of type $(2,1,1)$; see [1]. The fact that Ehresmann
semigroups thereby form a variety means that free objects exist. We refer the
reader to [12] and [7] for more information; we also recommend the references
to be found in both these papers.

######  Remark 1.1.

Ehresmann semigroups arose within the York School of semigroup theory led by
J. B. Fountain [10]. The paper [6] was particularly influential; in
particular, the conditions below which we refer to below as ‘deterministic’
and ‘codeterministic’ were first introduced in this paper. The theme of this
School became that of determining which properties of (von Neumann) regular
semigroups could be generalized to a non-regular setting. There was a
particular emphasis on the non-regular generalizations of inverse semigroups
with the focus of attention being on the abstract relationship between an
element $a$ and its idempotents $a^{-1}a$ and $aa^{-1}$. Subsequently, these
ideas were then developed not only for semigroups but also for categories [3].
My own paper [14] was written within the York School framework with the goal
being to describe the most general class of non-regular semigroups that could
be regarded as natural generalizations of inverse semigroups. This was
achieved by developing ideas due to Charles Ehresmann and his students. The
key stumbling block became that of order: in an inverse semigroup, the natural
partial order encodes algebraically the order induced by subset-inclusion. For
more general classes of Ehresmann semigroups, this neat relationship between
algebra and order does not hold. Trying to deal with this problem led to the
notions of ‘bideterministic element’ and ‘partial isometry’ which play major
rôles in this paper.

The following are standard results about Ehresmann semigroups but they are
also easy to prove directly.

######  Lemma 1.2.

Let $S$ be an Ehresmann semigroup with set of projections $U$.

  1. (1)

$(ab)^{\ast}\leq b^{\ast}$.

  2. (2)

$(ab)^{+}\leq a^{+}$.

  3. (3)

If $U$ has a zero then $a=0$ if and only if $a^{\ast}=0$.

  4. (4)

If $U$ has a zero then $a=0$ if and only if $a^{+}=0$.

Ehresmann semigroups come equipped with two partial orders which are
algebraically defined:

  * •

$x\leq_{r}y$ if and only if $x=ey$ for some $e\in U$. Observe that $x^{+}\leq
y^{+}$.

  * •

$x\leq_{l}y$ if and only if $x=yf$ for some $f\in U$. Observe that
$x^{\ast}\leq y^{\ast}$.

Define

$\leq\,=\,\leq_{l}\cap\leq_{r}.$

This order will play a special rôle in this paper. It is easy to check that
the following hold:

  * •

$x\leq_{r}y$ if and only if $x=x^{+}y$.

  * •

$x\leq_{l}y$ if and only if $x=yx^{\ast}$.

  * •

$x\leq y$ if and only if $x=x^{+}y=yx^{\ast}$.

Although these orders generalize the natural partial order on an inverse
semigroup (where they are all the same) they do not share such nice
properties. This is an issue we shall have to confront in Section 3. However,
on the set of projections $U$ they agree and we denote their common order by
$\leq$.

Ehresmann semigroups have emerged as an interesting class [1, 2, 4, 16]. In
particular, they are closely allied to categories in two ways. First of all,
underlying every Ehresmann semigroup is a category. The following was proved
as [14, Theorem 3.17].

######  Proposition 1.3.

Let $S$ be an Ehresmann semigroup with set of projections $U$. On the set $S$
define the restricted product $a\cdot b=ab$ when $a^{\ast}=b^{+}$ and
undefined otherwise. Then $(S,\cdot)$ is a category in which
$a^{\ast}=\mathbf{d}(a)$ and $b^{+}=\mathbf{r}(b)$. In addition, for any
$x,y\in S$ we have that $xy=(xe)\cdot(ey)$ where $e=x^{\ast}y^{+}$.

Second of all, examples of Ehresmann semigroups directly arise from
categories; the following is a special case of [13]

######  Example 1.4.

Let $C$ be a small category with set of identities $C_{o}$. We denote the
domain and codomain maps on $C$ by $\mathbf{d}$ and $\mathbf{r}$,
respectively. The product $xy$ is defined in the category if and only if
$\mathbf{d}(x)=\mathbf{r}(y)$; in this case, we shall also write $\exists xy$.
Let $S=\mathsf{P}(C)$ be the set of all subsets of $C$ equipped with the
multiplication of subsets. This is a semigroup. Put $U=\mathsf{P}(C_{o})$, the
set of all subsets of the set of identities. This set consists of idempotents
and is closed under multiplication. For $A\subseteq C$ define
$A^{\ast}=\\{\mathbf{d}(a)\colon a\in C\\}$ and $A^{+}=\\{\mathbf{r}(a)\colon
a\in C\\}$. We now show that with these definitions $\mathsf{P}(C)$ is an
Ehresmann monoid. Let $E,F\subseteq C_{o}$. Then $EF=E\cap F$. It follows that
$U$ is a commutative subsemigroup of the set of all idempotents of $S$ so that
(ES1) holds. It is immediate from the definitions that (ES2) holds. It is
evident that (ES3) holds. It remains to show that (ES4) holds. By symmetry, it
is enough to show that $(A^{\ast}B)^{\ast}=(AB)^{\ast}$ holds. In a category
$ab$ is defined if and only if $\mathbf{d}(a)=\mathbf{r}(b)$. Thus $ab$ is
defined if and only if $\mathbf{d}(a)b$ is defined in which case
$\mathbf{d}(ab)=\mathbf{d}(\mathbf{d}(a)b)$. We have therefore proved that we
are dealing with an Ehresmann monoid. A special case of this construction
shows that the monoid of all binary relations on a set $X$, which we denote by
$\mathscr{B}(X)$, is also an Ehresmann monoid; the category in this special
case is the set $X\times X$ where $\mathbf{d}(x,y)=(y,y)$ and
$\mathbf{r}(x,y)=(x,x)$ and product $(x,y)(y,z)=(x,z)$.

In [14], we showed what additional structure a category needed to be equipped
with in order that it arise from an Ehresmann semigroup. In Section 2 of this
paper, we shall prove Theorem 2.4, which describes a different (though,
obviously, equivalent) way, in which this can be accomplished. It arose from
reading [15, Page 184]. In Section 3 of this paper, we shall characterize the
Ehresmann monoids of the form $\mathsf{P}(C)$, where $C$ is a finite category,
in Theorem 3.13. This result uses ideas first described in [13]. In Section 4,
we prove two theorems, as Theorem 4.2 and Theorem 4.3, about special kinds of
Ehresmann semigroups called restriction semigroups.

##  2\. Ehresmann biactions

The goal of this section is to formulate a different categorical approach to
characterizing Ehresmann semigroups from the one described in [14]. In fact,
our approach can now be seen as a special case of [5] although it in fact
arose from reading [15, Page 184]. Our goal is to prove a kind of converse of
Proposition 1.3. To do this, we shall need a class of actions called étale
actions [8, 17] though we prefer the term ‘supported actions’. Let $S$ be an
inverse semigroup and $X$ a set. Let $p\colon X\rightarrow\mathsf{E}(S)$ be a
function. A supported action is a left action $S\times X\rightarrow X$,
denoted by $(s,x)\mapsto s\cdot x$, such that $p(x)\cdot x=x$ and $p(s\cdot
x)=sp(x)s^{-1}$.

######  Remark 2.1.

A feature of étale actions is that they can be restricted and then
reconstructed from this restriction. Observe that $s\cdot x=s\cdot(p(x)\cdot
x)=(sp(x))\cdot x$ and that $(sp(x))^{-1}sp(x)=p(x)s^{-1}s\leq p(x)$. Thus we
may restrict the action to those pairs $(s,x)$ where $s^{-1}s\leq p(x)$ and
lose nothing. We shall not use this result below but it explains why this new
approach using actions is equivalent to the one adopted in [14] which uses
restrictions and corestrictions.

We shall only be interested in supported actions where the acting inverse
semigroup is a meet semilattice, in which case the above properties simplify
somewhat. We can now define an Ehresmann biaction starting from a category $C$
by the following six axioms:

(E1):

The set of identities $C_{o}$ of $C$ is equipped with the structure of a
commutative, idempotent semigroup.

(E2):

There are two supported actions: there is a left action $C_{o}\times
C\rightarrow C$ denoted by $(e,a)\mapsto e\cdot a$ such that
$\mathbf{r}(a)\cdot a=a$ and $\mathbf{r}(e\cdot a)=e\mathbf{r}(a)$; there is a
right action $C\times C_{o}\rightarrow C$ denoted by $(a,e)\mapsto a\cdot e$
such that $a\cdot\mathbf{d}(a)=a$ and $\mathbf{d}(a\cdot e)=\mathbf{d}(a)e$.
Observe that we do not assume, for example, that $e\cdot(ab)=(e\cdot a)(e\cdot
b)$. See Axiom (E6) below.

(E3):

The biaction property $(e\cdot a)\cdot f=e\cdot(a\cdot f)$ holds.

(E4):

We require that $e\cdot a=ea$ and $a\cdot e=ae$ if $a\in C_{o}$.

(E5):

$\mathbf{d}(e\cdot a)\leq\mathbf{d}(a)$ and $\mathbf{r}(a\cdot
e)\leq\mathbf{r}(a)$.

(E6):

When $\exists xy$ then

$e\cdot(xy)=(e\cdot x)(\mathbf{d}(e\cdot x)\cdot y)\text{ and }(xy)\cdot
e=(x\cdot\mathbf{r}(y\cdot e))(y\cdot e).$

######  Remark 2.2.

The essential difference between the approach to Ehresmann semigroups
described here and the one described in [14] is that the latter requires
orders whereas here we use only actions. Remark 2.1 provides some insight into
the relationship between the two approaches.

We now prove that every Ehresmann semigroup gives rise to an Ehresmann
biaction.

######  Proposition 2.3.

Let $(S,U)$ be an Ehresmann semigroup. Put $C=S$ equipped with the restricted
product. Then $C_{o}=U$, a commutative idempotent semigroup, and $C$ is an
Ehresmann biaction.

###### Proof.

The left action $C_{o}\times C\rightarrow C$ and the right action $C\times
C_{o}\rightarrow C$ are both defined by multiplication. All the axioms are
easy to check from the axioms and properties of an Ehresmann semigroup. ∎

Proposition 2.3 tells us that Ehresmann semigroups give rise to Ehresmann
biactions. We now prove the first theorem of this paper which shows us that we
can construct Ehresmann semigroups from Ehresmann biactions.

######  Theorem 2.4.

Let $C$ be an Ehresmann biaction. Given $x,y\in C$, put
$e=\mathbf{d}(x)\mathbf{r}(y)$ and define $x\bullet y=(x\cdot e)(e\cdot y)$,
called the pseudoproduct. Then $(C,\bullet)$ is an Ehresmann semigroup with
set of projections $C_{o}$.

###### Proof.

The fact that the pseudoproduct is a binary operation follows from axioms (E1)
and (E2). We prove that the pseudoproduct is associative. First, observe that

$(x\bullet y)\bullet z=[(x\cdot f)\cdot\mathbf{r}(f\cdot y\cdot e)](f\cdot
y\cdot e)(e\cdot z)$

where $e=\mathbf{d}(x\bullet y)\mathbf{r}(z)$ and
$f=\mathbf{d}(x)\mathbf{r}(y)$ and using axioms (E6) and (E3). It is easy to
check from the axioms that the following hold:

  * •

$e=\mathbf{d}(f\cdot y)\mathbf{r}(z)$.

  * •

$f\cdot y=\mathbf{d}(x)\cdot y$ by axiom (E2); where we use the properties of
the action and the fact that $\mathbf{r}(y)\cdot y=y$.

  * •

$f\cdot y\cdot e=\mathbf{d}(x)\cdot y\cdot\mathbf{r}(z)$. This follows from
the fact that $(f\cdot y)\cdot e=(f\cdot y)\cdot\mathbf{d}(f\cdot
y)\mathbf{r}(z)$ which is equal to $(f\cdot y)\cdot\mathbf{r}(z)$ using axiom
(E2) and the result now follows by one of the results above.

  * •

$e\cdot z=\mathbf{d}(\mathbf{d}(x)\cdot y)\cdot z$. This uses axiom (E2) and
some of the results above.

  * •

$[(x\cdot f)\cdot\mathbf{r}(f\cdot y\cdot
e)]=x\cdot\mathbf{r}(y\cdot\mathbf{r}(z))$. Observe that $(x\cdot
f)\cdot\mathbf{r}(f\cdot y\cdot e)=x\cdot\mathbf{r}(f\cdot y\cdot e)$ using
axiom (E2). By one of the results above and using axiom (E2) we get the
result.

We have therefore shown that

$(x\bullet y)\bullet
z=(x\cdot\mathbf{r}(y\cdot\mathbf{r}(z)))(\mathbf{d}(x)\cdot
y\cdot\mathbf{r}(z))(\mathbf{d}(\mathbf{d}(x)\cdot y)\cdot z).$

Second, observe that

$x\bullet(y\bullet z)=(x\cdot i)(i\cdot y\cdot j)(\mathbf{d}(i\cdot y\cdot
j)\cdot j\cdot z)$

where $i=\mathbf{d}(x)\mathbf{r}(r\bullet z)$ and
$j=\mathbf{d}(y)\mathbf{r}(z)$ and using axioms (E6) and (E3). It is easy to
check from the axioms that the following hold with similar proofs to the ones
above:

  * •

$x\cdot i=x\cdot\mathbf{r}(x\cdot\mathbf{r}(z))$.

  * •

$i\cdot y\cdot j=\mathbf{d}(x)\cdot y\cdot\mathbf{r}(z)$.

  * •

$\mathbf{d}(i\cdot y\cdot j)\cdot j\cdot z=\mathbf{d}(\mathbf{d}(x)\cdot
y)\cdot z$.

We have therefore shown that

$x\bullet(y\bullet
z)=(x\cdot\mathbf{r}(y\cdot\mathbf{r}(z)))(\mathbf{d}(x)\cdot
y\cdot\mathbf{r}(z))(\mathbf{d}(\mathbf{d}(x)\cdot y)\cdot z).$

It now follows that $\bullet$ is associative.

Observe that if $e\in C_{o}$ then $a\bullet e=a\cdot e$ and that if $e,f\in
C_{o}$ then $ef=e\bullet f$. It follows that $\bullet$ is commutative and
idempotent on $C_{o}$. Define

$a^{\ast}=\mathbf{d}(a)\text{ and }a^{+}=\mathbf{r}(a).$

Then

$a\bullet a^{\ast}=a\bullet\mathbf{d}(a)=a\cdot\mathbf{d}(a)=a$

by (E2). We calculate

$(a^{\ast}\bullet b)^{\ast}=\mathbf{d}(\mathbf{d}(a)\bullet
b)=\mathbf{d}(e\cdot b)$

where $e=\mathbf{d}(a)\mathbf{r}(b)$. Thus $(a^{\ast}\bullet
b)^{\ast}=\mathbf{d}(\mathbf{d}(a)\cdot b)$. On the other hand

$(a\bullet b)^{\ast}=\mathbf{d}(a\bullet b)=\mathbf{d}((a\cdot e)(e\cdot b))$

where $e=\mathbf{d}(a)\mathbf{r}(b)$. Thus $(a\bullet
b)^{\ast}=\mathbf{d}(e\cdot b)=\mathbf{d}(\mathbf{d}(a)\cdot b)$. The dual
results are proved by symmetry. Thus $(C,\bullet)$ is an Ehresmann semigroup
with set of projections $C_{o}$, as claimed. ∎

######  Remark 2.5.

We could continue and show that with suitably defined maps between Ehresmann
biactions our construction is functorial. However, this is straightforward and
will not be carried out here.

##  3\. A class of finite Ehresmann monoids

In this section, we shall characterize the finite Ehresmann monoids which are
isomorphic to those arising from finite categories as in Example 1.4. We have
found [11, Chapter 1] a useful reference in helping us to clarify our
thoughts. We have also used ideas from [13] in an essential way.

We shall need to define two classes of elements within an Ehresmann semigroup.
The first requires only the algebraic structure of an Ehresmann semigroup. We
use the terminology from [3]. An element $a\in S$ of an Ehresmann semigroup is
said to be deterministic if $ea=a(ea)^{\ast}$ for all $e\in U$; it is said to
be codeterministic if $ae=(ae)^{+}a$ for all $e\in U$. An element is said to
be bideterministic if it is both deterministic and codeterministic.

An Ehresmann semigroup is said to be a birestriction semigroup if every
element is bideterministic; in what follows, we shall usually just say
‘restriction semigroup’ rather than ‘birestriction semigroup’. We shall say
more about restriction semigroups in Section 4.

We now characterize the bideterministic elements of Ehresmann monoids of the
form $\mathsf{P}(C)$ where $C$ is a category.

######  Lemma 3.1.

Let $C$ be a category with $\mathsf{P}(C)$ being its associated Ehresmann
monoid.

  1. (1)

A non-empty subset $A\subseteq C$ is deterministic if and only if whenever
$a,b\in A$ and $\mathbf{d}(a)=\mathbf{d}(b)$ then
$\mathbf{r}(a)=\mathbf{r}(b)$.

  2. (2)

A non-empty subset $A\subseteq C$ is codeterministic if and only if whenever
$a,b\in A$ and $\mathbf{r}(a)=\mathbf{r}(b)$ then
$\mathbf{d}(a)=\mathbf{d}(b)$.

###### Proof.

We prove (1) since (2) follows by symmetry. Let $A$ be a non-empty subset of
$C$. Suppose that $a,b\in A$ are such that $\mathbf{d}(a)=\mathbf{d}(b)=e$ but
$\mathbf{r}(a)=i$ is different from $\mathbf{r}(b)=j$. We prove that $A$ is
non-deterministic. The singleton set $\\{i\\}$ is an element of $U$. The
product $\\{i\\}A$ is an element of $\mathsf{P}(S)$ which contains $a$ but
which does not contain $b$. On the other hand, $(\\{i\\}A)^{\ast}$ contains
$e$. It follows that $A(\\{i\\}A)^{\ast}$ contains both $a$ and $b$. Thus $A$
cannot be deterministic since $\\{i\\}A\neq A(\\{i\\}A)^{\ast}$. Suppose now
that $A$ is such that whenever $a,b\in A$ and $\mathbf{d}(a)=\mathbf{d}(b)$
then $\mathbf{r}(a)=\mathbf{r}(b)$. We prove that $A$ is deterministic. Let
$E$ be any projection. Let $x\in A(EA)^{\ast}$. Then $x=ae$ where $a\in A$ and
$e\in(EA)^{\ast}$. Thus $e=\mathbf{d}(e^{\prime}a^{\prime})$ where
$e^{\prime}\in E$ and $a^{\prime}\in A$. It follows that $x,a^{\prime}\in A$
and $\mathbf{d}(x)=\mathbf{d}(a^{\prime})$. By assumption,
$\mathbf{r}(x)=\mathbf{r}(a^{\prime})$. But then $x=e^{\prime}x$ where
$e^{\prime}\in E$. ∎

Ehresmann monoids of the form $\mathsf{P}(C)$ come equipped with subset
inclusion as an order. We shall formalize its properties below. Let $(S,U)$ be
an Ehresmann monoid with set of projections $U$ where $1\in U$. We shall say
that $S$ is Boolean if $S$ is equipped with a partial order, denoted by
$\subseteq$, with respect to which $(S,\subseteq)$ is a Boolean algebra — we
denote the top element of $S$ by $t$ — such that the following properties
hold:

(OE1):

$c(a\cup b)=ca\cup cb$, and dually.

(OE2):

If $e\in U$ and $a\subseteq e$ then $a\in U$.

(OE3):

If $e,f\in U$ then $e\leq f$ if and only if $e\subseteq f$.

(OE4):

If $a=eb$ then $a\subseteq b$ and if $a=be$ then $a\subseteq b$ where $e\in
U$.

(OE5):

$(a\cup b)^{\ast}=a^{\ast}\cup b^{\ast}$, and dually.

In a Boolean Ehresmann monoid, the set of projections is the order-ideal
determined by the identity element of the monoid by axiom (OE2).

The proof of the following lemma is immediate from axioms (OE1) and (OE5).

######  Lemma 3.2.

Let $(S,U)$ be Boolean Ehresmann monoid. Then the following two properties
hold:

  1. (1)

If $a\subseteq b$ and $c\subseteq d$ then $ac\subseteq bd$.

  2. (2)

If $a\subseteq b$ then $a^{\ast}\subseteq b^{\ast}$ and $a^{+}\subseteq
b^{+}$.

The following result assures us that the set of projections of a Boolean
Ehresmann monoid is also a Boolean algebra.

######  Lemma 3.3.

Let $(S,U)$ be a Boolean Ehresmann monoid. Then the following hold:

  1. (1)

$t^{\ast}=1=t^{+}$.

  2. (2)

$U$ is a Boolean algebra in its own right.

###### Proof.

(1) The top element $t$ has the property that $a\subseteq t$ for all $a\in S$.
Thus $a^{\ast}\subseteq t^{\ast}$ by Lemma 3.2. It follows that $t^{\ast}$ is
an upper bound for all projections. Thus $1\subseteq t^{\ast}\subseteq 1$. We
have proved that $t^{\ast}=1$. It follows by symmetry that $t^{+}=1$.

(2) By axiom (OE3), if $e$ and $f$ are projections then $ef=e\wedge f$. In a
Boolean Ehresmann monoid, we are told that $(S,\subseteq)$ is a Boolean
algebra. We prove that $U$ is a Boolean algebra. Let $e,f\in U$. By
assumption, both $e\wedge f$ and $e\vee f$ exist in $S$. Since $e\wedge
f\subseteq e$ we know by axiom (OE2) that $e\wedge f\in U$. The fact that
$e\vee f\in U$ follows by axiom (OE5) and the definition of an Ehresmann
semigroup; alternatively, $e,f\subseteq 1$ and so $e\vee f\subseteq 1$. It now
follows that $e\vee f\in U$ by axiom (OE2). Thus $U$ is closed under binary
greatest lower bounds and least upper bounds. Let $e\in U$. We need to prove
that $U$ has complements. Let $e\in U$. Then $\bar{e}\in S$. Thus $\neg
e=1\wedge\bar{e}\in U$. We calculate $e\wedge\neg e=e\wedge 1\wedge\bar{e}=0$
and $e\vee\neg e=e\vee(1\wedge\bar{e})=1\wedge t=1$. Thus $\neg e$ is the
complement in $U$ of $e$. ∎

The motivation for the above definition comes, of course, from Example 1.4 as
the following lemma shows. The proof is by routine verification.

######  Lemma 3.4.

Let $C$ be a (finite) category. Then $\mathsf{P}(C)$ is a Boolean Ehresmann
monoid.

The following notion first arose in [15] but was then extended in [13]. In
Boolean Ehresmann monoids, we shall need a stronger notion than that of a
bideterministic element. Let $S$ be a Boolean Ehresmann monoid. An element
$a\in S$ is said to be a partial isometry if whenever $b\subseteq a$ then
$b\leq a$. Recall that $b\leq a$ means that $b=ea=af$ for some $e,f\in U$. The
following was proved as [13, Lemma 2.26].

######  Lemma 3.5.

In a Boolean Ehresmann monoid, the set of partial isometries is an order-
ideal.

###### Proof.

Let $a$ be a partial isometry and let $b\subseteq a$. Since $a$ is a partial
isometry, we have that $b\leq a$. Thus $b=b^{+}a=ab^{\ast}$. We prove that $b$
is a partial isometry. Let $c\subseteq b$. Then $c\subseteq a$. It follows
that $c\leq a$. Thus $c=c^{+}a=ac^{\ast}$. By axiom (OE5) applied to
$c\subseteq b$, we deduce that $c^{+}\leq b^{+}$ and $c^{\ast}\leq b^{\ast}$.
Thus $c^{+}b=c^{+}(b^{+}a)=c^{+}a=c$ and
$bc^{\ast}=(ab^{\ast})c^{\ast}=ac^{\ast}=c$. It follows that $c\leq b$, as
required. ∎

The following was proved in [13, Lemma 2.26] but we give a direct proof here.

######  Lemma 3.6.

In a Boolean Ehresmann monoid, every partial isometry is bideterministic.

###### Proof.

Let $a$ be a partial isometry. We prove that it is deterministic; the proof
that it is codeterministic follows by symmetry. Let $e$ be any projection.
Then $ae\subseteq a$ by axiom (OE4). But $a$ is a partial isometry and so
$ae\leq a$. By definition, $ae=a(ae)^{\ast}$. This proves that $a$ is
deterministic. ∎

We next describe the partial isometries in the ordered Ehresmann monoids
$\mathsf{P}(C)$.

######  Lemma 3.7.

Let $C$ be a category. Then a non-empty subset $A\subseteq C$ is a partial
isometry in $\mathsf{P}(C)$ if and only if the following two conditions hold:

  1. (1)

If $a,b\in A$ and $\mathbf{d}(a)=\mathbf{d}(b)$ then $a=b$.

  2. (2)

If $a,b\in A$ and $\mathbf{r}(a)=\mathbf{r}(b)$ then $a=b$.

###### Proof.

Let $A$ be a subset that satisfies condition (1) and suppose that $B\subseteq
A$. Clearly, $B\subseteq AB^{\ast}$. We shall show that $B=AB^{\ast}$. Let
$x\in AB^{\ast}$. Then $x=a\mathbf{d}(b)$ where $a\in A$ and $b\in B$. Thus
$\mathbf{d}(x)=\mathbf{d}(b)$. Now, $x,b\in A$. By the assumption that $A$
satisfies the condition (1), we must have that $x=b$. We have therefore proved
that $B=AB^{\ast}$. The fact that $B=B^{+}A$ if $A$ satisfies condition (2)
follows by symmetry. We now prove the converse. Suppose that $A$ is a partial
isometry. We prove that condition (1) holds; the fact that condition (2) holds
follows by symmetry. Suppose that $a,b\in A$ be distinct elements such that
$\mathbf{d}(a)=\mathbf{d}(b)=e$. Clearly, $\\{a\\},\\{b\\}\subseteq A$, but we
do not have that $\\{a\\}\leq A$. The reason is that $\\{a\\}^{\ast}=\\{e\\}$
and $A\\{a\\}^{\ast}=A\\{e\\}$ contains both $a$ and $b$. Thus this set cannot
be equal to $\\{a\\}$. ∎

######  Remark 3.8.

Subsets of groupoids satisfying both the conditions of Lemma 3.7 are called
local bisections. The set of all local bisections of a groupoid forms an
inverse semigroup. See [15, page 164].

######  Lemma 3.9.

Let $S$ be a Boolean Ehresmann monoid. Then any atom is a partial isometry.

###### Proof.

Let $a$ be an atom. If $b\subseteq a$ then either $b=a$ or $b=0$. In both
cases, $b\leq a$. Thus $a$ is a partial isometry. ∎

The following example shows that the concepts we have introduced are distinct.

######  Example 3.10.

Consider the following category $C$:

$\textstyle{e}$$\textstyle{f\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{b}$

The monoid $\mathsf{P}(C)$ has 16 elements. The element $\\{a\\}$ is a partial
isometry; the elements $\\{a,b\\}$ is bideterministic but not a partial
isometry (since $\\{a\\}\subseteq\\{a,b\\}$ but $\\{a\\}\nleq\\{a,b\\}$); the
element $\\{a,e\\}$ is not bideterministic.

######  Lemma 3.11.

Let $C$ be a category. The product of partial isometries in $\mathsf{P}(C)$ is
also a partial isometry.

###### Proof.

We use our characterization of partial isometries from Lemma 3.7. Let $A$ and
$B$ be non-empty partial isometries. We prove that $AB$ is a partial isometry.
Let $x,y\in AB$ and suppose that $\mathbf{d}(x)=\mathbf{d}(y)$. Then $x=ab$
and $y=a^{\prime}b^{\prime}$ where $\exists ab$ and $\exists
a^{\prime}b^{\prime}$ in the category $C$ and $a,a^{\prime}\in A$ and
$b,b^{\prime}\in B$. We have that $\mathbf{d}(b)=\mathbf{d}(b^{\prime})$. But
$B$ is a partial isometry and so $b=b^{\prime}$. We may similarly show that
$a=a^{\prime}$ and so $x=y$. The other case follows by symmetry. ∎

######  Proposition 3.12.

Let $C$ be a category. Then $\mathsf{P}(C)$ is a Boolean Ehresmann monoid with
the order being subset-inclusion in which the product of partial isometries is
a partial isometry.

###### Proof.

This follows by Lemma 3.4 and Lemma 3.11. ∎

We now prove the second theorem of this paper which characterizes the monoids
arising in Proposition 3.12.

######  Theorem 3.13.

Let $(S,U)$ be a finite Boolean Ehresmann monoid in which the product of
partial isometries is a partial isometry. Then there is a finite category $C$
such that $S$ is isomorphic to $\mathsf{P}(C)$ as an Ehresmann monoid and in
such a way that the order on $S$ is isomorphically mapped to the order on
$\mathsf{P}(C)$.

###### Proof.

Let $S$ be a Boolean Ehresmann monoid having the stated properties. Our proof
is in seven steps. In the first four steps, we construct the category $C$. To
do this, we shall use the atoms of $S$. Recall that in a finite Boolean
algebra every element is a finite join of atoms and, by Lemma 3.9, every atom
is a partial isometry.

(1) If $a$ is an atom then both $a^{+}$ and $a^{\ast}$ are atoms. Suppose that
$a$ is an atom. Let $e\subseteq a^{+}$, where $e$ is a projection by axiom
(OE2). Then $e\leq a^{+}$ by axiom (OE3). By definition, $ea\leq_{r}a$ and so
$ea\subseteq a$ by axiom (OE4). But $a$ is an atom and so either $ea=a$ or
$ea=0$. In the first case, $a^{+}\leq e$ and so $a^{+}\subseteq e$ by axiom
(OE4). We therefore deduce that $e=a^{+}$. In the second case, $ea^{+}=0$ by
the properties of $+$ and Lemma 1.2. Thus $e=0$. We have therefore proved that
$a^{+}$ is also an atom. By symmetry, if $a$ is an atom, then so too is
$a^{\ast}$.

(2) If $a$ and $b$ are atoms then $ab\neq 0$ if and only if $a^{\ast}=b^{+}$.
Suppose first that $a^{\ast}=b^{+}$. If $ab=0$ then $(ab)^{\ast}=0$ and so by
(ES4), we have that $(a^{\ast}b)^{\ast}=0$ and so $b^{\ast}=0$ since
$a^{\ast}=b^{+}$. It follows by Lemma 1.2 that $b=0$ which is a contradiction
since $b$ is an atom. Suppose now that $ab\neq 0$. Then
$ab=(ab^{+})(a^{\ast}b)$ which is a restricted product. Now $ab^{+}\leq a$ and
so $ab^{+}\subseteq a$ by axiom (OE4). But $a$ is an atom. Thus $ab^{+}=0$ or
$ab^{+}=a$. But $ab\neq 0$ and so $ab^{+}=a$. Similarly, $a^{\ast}b=b$. It
follows that $a^{\ast}=b^{+}$ and so the product is a restricted one.

(3) If $a$ and $b$ are atoms and $ab$ is a restricted product then $ab$ is an
atom. In (Step 2), we showed that $ab\neq 0$. Both $a$ and $b$ are atoms and
so by Lemma 3.9 each of $a$ and $b$ is a partial isometry. By assumption,
their product $ab$ is a partial isometry. Let $c\subseteq ab$. Then $c\leq ab$
since $ab$ is a partial isometry. Thus, in particular, $c=abc^{\ast}$. Now,
$bc^{\ast}\leq b$ and so $bc^{\ast}\subseteq b$ by axiom (OE4). But $b$ is an
atom. It follows that $bc^{\ast}=0$ or $bc^{\ast}=b$. Suppose first that
$bc^{\ast}=0$. Then $b^{\ast}c^{\ast}=0$ by axiom (ES4). But from $c\subseteq
ab$ we get that $c^{\ast}\subseteq(ab)^{\ast}\leq b^{\ast}$ where we have used
Lemma 3.2 and Lemma 1.2. Thus $c^{\ast}\leq b^{\ast}$ by axiom (OE3). It
follows that $c^{\ast}=0$ and so $c=0$ by Lemma 1.2. Now suppose that
$bc^{\ast}=b$. Then $c=ab$. We have therefore proved that $ab$ is an atom.

(4) Let $a$ be an atom. Define $\mathbf{d}(a)=a^{\ast}$ and
$\mathbf{r}(a)=a^{+}$. By (Step 1) above these are both atoms. Put $C$ equal
to the set of all atoms of $S$. By (Step 3), the set $C$ is a category under
the restricted product. We have therefore proved that $C$ is a category whose
set of identities is the set of atoms in $U$.

We may accordingly construct the Ehresmann monoid $\mathsf{P}(C)$ whose set of
projections is $\mathsf{P}(C_{o})$. We prove that $S$ is isomorphic to
$\mathsf{P}(C)$ as Ehresmann monoids.

(5) Since we are in a Boolean algebra, every non-zero element is a join of the
atoms below it. If $a$ is an element of $S$ define $\phi(0)=\varnothing$ and
if $a\neq 0$ define $\phi(a)$ to be the set of all atoms below it. The map
$\phi$ determines a bijection from $S$ to $\mathsf{P}(C)$. Observe that by
axiom (OE2), the elements below a projection are all projections. Thus each
projection is a finite join of atoms which are themselves projections. It
follows that $\phi$ also determines a bijection from $U$ to
$\mathsf{P}(C_{o})$. These bijections are actually isomorphisms of Boolean
algebras.

(6) By (Step 3) above, we have that $\phi(a)\phi(b)\subseteq\phi(ab)$. We
prove the reverse inclusion. Let $x$ be an atom such that $x\subseteq ab$. We
are working in a Boolean algebra, and so each of $a$ and $b$ can be written as
unions of atoms. Let $a=\bigcup_{i=1}^{m}a_{i}$ and
$b=\bigcup_{j=1}^{n}b_{j}$, where $a_{i}$ and $b_{j}$ are atoms. Then
$ab=\bigcup_{1\leq i\leq m,1\leq j\leq n}a_{i}b_{j}$ where we omit all
products of atoms which are zero. Thus by (Step 3), we know that each
$a_{i}b_{j}$ is an atom and a restricted product. We now use the
distributivity property of Boolean algebras to deduce that
$x=\bigcup_{i,j}(x\wedge a_{i}b_{j})$. Now $x$ is an atom and so $x\wedge
a_{i}b_{j}=0$ or $x\subseteq a_{i}b_{j}$. Suppose that $x\subseteq a_{i}b_{j}$
for some $i$ and some $j$. Then, since $a_{i}b_{j}$ is an atom, we must have
that $x=a_{i}b_{j}$. We have therefore written $x$ as a product of atoms where
$a_{i}\subseteq a$ and $b_{j}\subseteq b$.

We have proved that $\phi$ is an isomorphism of semigroups between $S$ and
$\mathsf{P}(C)$ and between $U$ and $\mathsf{P}(C_{o})$.

(7) $\phi$ is an isomorphism of Ehresmann semigroups. Let $a\in S$ be a non-
zero element. Then $a$ is the join of all the atoms below it. Thus by axiom
(OE5), we have that $a^{\ast}$ is equal to the join of all the atoms of the
form $e$ where $e\subseteq a^{\ast}$. We now prove that every atom $e$ below
$a^{\ast}$ is of the form $b^{\ast}$ where $b$ is an atom and $b\subseteq a$.
Let $e\leq a^{\ast}$ where $e$ is an atom. Then $ae\leq_{l}a$ and so
$ae\subseteq a$ by axiom (OE4). Observe that $(ae)^{\ast}=e$. If $ae$ is an
atom then we are done. If $ae$ is not an atom then $x\leq ae$ where $x$ is an
atom since we are working in a Boolean algebra. But
$x^{\ast}\subseteq(ae)^{\ast}=e$. But $e$ is an atom. Thus either $x^{\ast}=0$
which implies that $x=0$, which is ruled out since $x$ is an atom, or
$x^{\ast}=e$. We have therefore found an atom $x\leq a$ such that
$x^{\ast}=e$. Consequently, $\theta(a^{\ast})=\theta(a)^{\ast}$. A dual result
holds for $+$. ∎

##  4\. Restriction semigroups

Recall that a restriction semigroup is an Ehresmann semigroup in which each
element is bideterministic. In this section, we shall, in effect, combine
results we proved in the previous two sections. The following results are all
well-known. We include them for the sake of completeness.

######  Lemma 4.1.

Let $(S,U)$ be a restriction semigroup.

  1. (1)

The partial orders $\leq_{l}$ and $\leq_{r}$ are the same and so each is equal
to $\leq$.

  2. (2)

The semigroup $S$ is partially ordered with respect to $\leq$.

  3. (3)

If $a\leq bc$ then there exist $b^{\prime}\leq b$ and $c^{\prime}\leq c$ such
that $b^{\prime}c^{\prime}$ is a restricted product and
$a=b^{\prime}c^{\prime}$.

  4. (4)

The set $U$ is an order-ideal of $S$.

  5. (5)

If $a,b\leq c$ and $a^{\ast}=b^{\ast}$ (respectively, $a^{+}=b^{+})$ then
$a=b$.

  6. (6)

If $a\leq b$ then $a^{\ast}\leq b^{\ast}$ and $a^{+}\leq b^{+}$.

###### Proof.

(1) Suppose that $a\leq_{r}b$. We prove that $a\leq_{l}b$; the proof of the
other direction follows by symmetry. We are given that $a=eb$ where $e$ is a
projection. But $eb=b(eb)^{\ast}$ since $b$ is deterministic. It follows that
$a=b(eb)^{\ast}$ and so $a\leq_{l}b$.

(2) This follows by (1) above because $\leq_{l}$ is left compatible with the
multiplication and $\leq_{r}$ is right compatible.

(3) Suppose that $a\leq bc$. Then $a=(a^{+}b)c$. Put $e=(a^{+}b)^{\ast}c^{+}$.
Then $a=(a^{+}be)(ec)$. Put $b^{\prime}=a^{+}be$ and $c^{\prime}=ec$. Then
$a=b^{\prime}c^{\prime}$, the product $b^{\prime}c^{\prime}$ is a restricted
product, and $b^{\prime}\leq b$ and $c^{\prime}\leq c$.

(4) This is immediate.

(5) The proof of this is straightforward.

(6) This follows by the properties of the partial orders. ∎

Let $C$ be a category. Denote the set of partial isometries in $\mathsf{P}(C)$
by $\mathscr{PI}(C)$. This is a restriction monoid by Lemma 3.4, Lemma 3.11,
and Lemma 3.6.

######  Theorem 4.2.

Let $(S,U)$ be a restriction semigroup. Then there is a category $C$ and an
injective morphism $\alpha\colon S\rightarrow\mathscr{PI}(C)$.

###### Proof.

For the category $C$, we take the set $S$ equipped with the restricted product
according to Proposition 1.3. Then $\mathscr{PI}(C)$ is a restriction monoid.
Define $\alpha\colon S\rightarrow\mathscr{PI}(C)$ by putting $\alpha(a)$ equal
to the set of all elements less than or equal to $a$. This is well-defined by
part (5) of Lemma 4.1 and injective. By part (4) of Lemma 4.1 elements of $U$
are mapped to elements of $\mathsf{P}(C_{o})$. It is a homomorphism by part
(3) of Lemma 4.1. It remains to check that it is a morphism. By symmetry, it
is enough to check the morphism property for $\ast$. If $b\leq a$ then
$b^{\ast}\leq a^{\ast}$ by part (6) of Lemma 4.1. On the other hand, if $e\leq
a^{\ast}$ then $ae\leq a$ and $(ae)^{\ast}=a^{\ast}$. ∎

The following is related to questions discussed in [9], although our approach
is quite different. A natural question is the following: given a restriction
semigroup $S$, under what circumstances can $S$ be embedded into an inverse
semigroup $T$ in such a way that $a^{\ast}=a^{-1}a$ and $a^{+}=aa^{-1}$. We
call such an embedding of a restriction semigroup in an inverse semigroup a
nice embedding. This problem is answered below modulo the problem of embedding
categories into groupoids.

######  Theorem 4.3.

Let $S$ be a restriction semigroup. Denote the set $S$ equipped with the
restricted product by $C$. Then $S$ admits a nice embedding if and only if the
category $C$ can be embedded into a groupoid $G$.

###### Proof.

Suppose first that the category $C$ can be embedded into a groupoid $G$. By
Theorem 4.2, there is an injective morphism $\alpha\colon
S\rightarrow\mathscr{PI}(C)$. Let $C\subseteq G$. Then
$\mathscr{PI}(C)\rightarrow\mathscr{PI}(G)$ is an injective morphism which is
actually an embedding. But the elements of $\mathscr{PI}(G)$ are simply the
local bisections of $G$ and so form an inverse semigroup; see Remark 3.8. It
follows that $S$ admits a nice embedding into the inverse semigroup
$\mathscr{PI}(G)$. To prove the converse, suppose that $S$ can be nicely
embedded into the inverse semigroup $T$. Let $G$ be the set $T$ equipped with
the restricted product. Then $C$ embeds into $G$ as a subcategory. ∎

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# A possible experimental test of the nonlinear phononics interpretation of
light-induced superconductivity

M. Altarelli
Max Planck Institute for Structure and Dynamics of Matter
22761 Hamburg, Germany

(22 February 2019)

###### Abstract

Experimental evidence for a transient enhancement of the superconducting
critical temperature in presence of an intense THz or IR pump pulse was
ascribed to nonlinear phononics effects. Here I introduce a simple
phenomenological Ginzburg-Landau model of this phenomenon, to explore further
consequences and possible experimental tests of this interpretation. Upon
cooling below $T_{c}$ in the absence of THz pumping, both an abrupt softening
of a Raman-active mode frequency and a spontaneous lattice distortion, growing
linearly with $(T_{c}-T)$, are predicted to occur. Numerical estimates for
$YBa_{2}Cu_{3}O_{6+x}$ indicate that the frequency softening should be easily
observable, whereas the lattice distortion may be too small. Comparison with
Raman experiments for $YBa_{2}Cu_{3}O_{6+x}$ is far from conclusive; however
very large (up to 18 %) phonon frequency softening below $T_{c}$, with
behaviour strikingly similar to the predictions of the present model, was
observed over 20 years ago in $HgBa_{2}Ca_{3}Cu_{4}O_{10+x}$; its explanation
has been controversial. Light induced superconductivity was never investigated
in this material and it may be of interest to explore if it is present and
connected, via the mechanism discussed here, to the observed anomalous phonon
behaviour.

##  1 Introduction

The purpose of this note is to explore further possible experimental tests for
the non-linear lattice dynamics explanation [1, 2, 3] of light-enhanced
superconductivity in $YBa_{2}Cu_{3}O_{6+x}$ [4, 5, 6]. According to this
explanation, one (or more) optical phonon $A_{1g}$-symmetric mode coordinate,
$Q$, is pushed off its equilibrium value by excitation of an infrared-active
mode $Q_{IR}$, due to a nonlinear coupling that, according to Mankowsky’s et
al. arguments is of the form $-a_{1,2}QQ_{IR}^{2}$. Indeed, when this term is
added to the harmonic potentials and the kinetic energies of the $Q_{IR},Q$
oscillators, the Barbanis Hamiltonian [7] is obtained, which has been first
investigated in the context of celestial mechanics, then in studies of quantum
ergodicity and molecular physics [8],[9], [10]; and it turns out that an
accurate approximation to the exact eigenfunctions of this Hamiltonian is
given by eigenfunctions of the form [9], [10]:

$\Phi_{n_{1},n_{2}}=N\phi_{n_{1}}(aQ+b)\phi_{n_{2}}(a_{IR}Q_{IR})$ (1)

in terms of harmonic oscillator eigenfunctions for scaled oscillator
coordinates, and, for the coordinate $Q$, with a displaced oscillator
coordinate; the scaling and displacement terms $a,a_{IR}$ and $b$ can be
determined variationally or within a SCF approach for a given value of
$a_{1,2}$. $N$ is a normalization constant. When the $Q$ lattice coordinate is
pushed to an off-equilibrium value by excitation of the infrared mode by a
pump pulse, evidence that could indicate a transient state with a substantial
increase of $T_{c}$ [4, 5, 6] has been reported.

If a given phononic coordinate affects the electronic states at the Fermi
energy and considerably modifies $T_{c}$, as argued in Ref [3], one could
suspect that changes of the electronic states at the Fermi energy, such as the
usual onset of equilibrium superconductivity when cooling below $T_{c}$, with
opening of a gap, should have a corresponding effect on the phononic
coordinate. This paper investigates the possibility that this could lead to
observable effects that validate the nonlinear phononics interpretation of
superconductivity.

To this aim, we adopt a very simple-minded Ginzburg-Landau description of the
superconducting transition [11], including an additional dependence of the
critical temperature $T_{c}$ on the $Q$ lattice coordinate. Reasoning based on
this phenomenological description implies that, even in the absence of the
pump, when the system is cooled below $T_{c}$, a softening of this phonon
frequency and a spontaneous distortion of the lattice should occur. An attempt
to estimate the sizes of the softening and of the distortion is carried out,
at least roughly, to determine whether they could be observable; measured
experimental properties of $YBa_{2}Cu_{3}O_{6+x}$ are used to infer the values
of the parameters entering the Ginzburg-Landau free-energy. It turns out that,
for plausible values of the parameters, the frequency shift should be
observable, whereas the lattice distortion could be harder to detect.

##  2 Ginzburg-Landau functional for the free-energy

In the phenomenological Ginzburg-Landau theory, the free-energy per unit
volume of a superconductor in the vicinity of $T_{c}$ is written as a
functional of the order parameter $\Psi$; in the simplest case of a uniform
superconductor without external fields, it takes the form:

$F(T,\mid\Psi\mid)=a\times(T-T_{c}){\mid\Psi\mid}^{2}+b{\mid\Psi\mid}^{4}$ (2)

where the coefficients $a,b$ are positive. Looking for a minimum of this
functional as a function of $\mid\Psi\mid$ one immediately finds
$\mid\Psi\mid=0$ for $T>T_{c}$, and, for $T<T_{c}$ the value

${\mid\Psi\mid}^{2}=\frac{a(T_{c}-T)}{2b}$ (3)

The free energy value corresponding to this minimum (the condensation energy)
is evaluated plugging Eq. (3) into (2):

$F_{min}(T)=a(T-T_{c})\frac{a(T_{c}-T)}{2b}+b\frac{a^{2}(T_{c}-T)^{2}}{4b^{2}}=\frac{-a^{2}(T_{c}-T)^{2}}{4b}$
(4)

It can be argued that this mean-field type of formulation is oversimplified
for the cuprates, with their short coherence length, and their symmetry lower
than cubic. But for the sake of simplicity we shall stick to this description.
The observation of very similar phenomena in totally different compounds, such
as $K_{3}C_{60}$ [12], supports a rather generic approach.
We further assume that an optical phonon coordinate $Q$, with a vanishing
expectation value in the equilibrium state above $T_{c}$, is affecting the
transition temperature, when it takes a non vanishing amplitude. A simple
description of this state of affairs can be given by modifying the free-energy
functional in the following way:

$F(T,\mid\Psi\mid,Q)=(1/2)\rho
kQ^{2}+a[T-T_{c}-\Theta_{c}(Q)]{\mid\Psi\mid}^{2}+b{\mid\Psi\mid}^{4}$ (5)

Here, the first term on the right-hand side is the elastic energy (per unit
volume) corresponding to the lattice distortion, with $\rho$ being the number
density of deformed bonds and $k=m^{\ast}\omega^{2}$ the product of effective
mass and square frequency of the $Q$ mode. The second term expresses an
increase of the critical temperature to $T_{c}+\Theta_{c}(Q)$, when $Q$ is
nonvanishing. The function $\Theta_{c}(Q)$ reflects the complex physics that
leads to a $T_{c}$ enhancement in a distorted lattice. According to the
arguments in ref. [1, 2], the lattice distortion induces a modification of the
orbital character and density of electronic states at the Fermi energy,
leading to the $T_{c}$ enhancement. Lacking a microscopic theory, the
functional relationship between $Q$ and $T_{c}$ remains out of our reach.
Once again, a word of caution on the questionable aspects of this approach is
necessary. As it was stated above, the Ginzburg-Landau theory is limited to
$T$ near $T_{c}$. Just shifting the critical temperature keeping the same
functional form and $a,b$ parameters is really justified if the shift
satisfies

$\Theta_{c}(Q)<<T_{c}$ (6)

This restriction certainly does not apply to the situation described in ref.
[6], which, taken at face value, implies e.g. a shift of some $250K$ from
$50K$ to room temperature for the critical temperature of
$YBa_{2}Cu_{3}O_{6.5}$ after a THz pulse. So one must understand Eq. (5) as
describing the ”small Q” regime, where Eq. (6) is satisfied, and not the
regime of intense THz pumping that the experiments access. The quotation marks
around ”small Q” are justified by the circumstance that, according to ref.
[3], the $250K$ rise in $T_{c}$ corresponds to a $0.63\%$ elongation of the
intra-Cu bilayer distance, which in other contexts one could call a small
distortion.

In order to make progress we make a linearization assumption for
$\Theta_{c}(Q)$ i.e. assume that in some range of Q

$\Theta_{c}(Q)\simeq cQ$ (7)

(remember that $\Theta_{c}(Q=0)=0$), where c is a coefficient to be
determined. This is very likely the case for a sufficiently small Q range (the
exception being if, for some reason, $(\partial\Theta_{c}/\partial Q)_{Q=0}$
vanishes). Within this assumption Eq. (3) becomes:

$F(T,\mid\Psi\mid,Q)=(1/2)\rho
kQ^{2}+a[T-T_{c}-cQ]{\mid\Psi\mid}^{2}+b{\mid\Psi\mid}^{4}$ (8)

Additional support for this form of the free-energy functional comes from the
discussion of the coupling of the strain tensor to the superconducting order
parameter by Joynt and Rice [13], Ozaki [14] and Millis and Rabe [16] who,
starting from a simple description of the condensation energy in terms of the
band structure and the gap function, suggest coupling terms linear in the
strain and quadratic in the order parameter (for a summary and extensive
information on a generalization of the argument for anisotropic
superconductors, see Ref. [15]). This is exactly the character of the
additional term in Eq. (8), $-cQ|\Psi|^{2}$. Although the strain tensor is
representative of acoustical deformations, and here $Q$ represents an optical
deformation, this should not modify the argument.

If a certain value of $Q\neq 0$ is imposed by the external THz pump, we can
minimize the free-energy as in the previous case and get:

${\mid\Psi\mid}^{2}=\frac{a(T_{c}+cQ-T)}{2b}$ (9)

and

$F_{min}(T,Q)=(1/2)\rho kQ^{2}-\frac{a^{2}(T_{c}+cQ-T)^{2}}{4b}$ (10)

Comparing to the $Q=0$ case, one finds:

$\displaystyle F_{min}(T,Q)-F_{min}(T,Q=0)$ $\displaystyle=$
$\displaystyle(1/2)\rho
kQ^{2}-\frac{a^{2}(T_{c}+cQ-T)^{2}}{4b}+\frac{a^{2}(T_{c}-T)^{2}}{4b}$ (11)
$\displaystyle=$ $\displaystyle(1/2)(\rho
k-\frac{a^{2}c^{2}}{2b})Q^{2}-\frac{a^{2}c(T_{c}-T)}{2b}Q$

or equivalently:

$F_{min}(T,Q)=F_{min}(T,Q=0)+(1/2)(\rho
k-\frac{a^{2}c^{2}}{2b})Q^{2}-\frac{a^{2}c(T_{c}-T)}{2b}Q$ (12)

##  3 Consequences in the absence of external pump pulse

In this Section we shall consider the situation in which no external pump is
present, to examine whether the $F_{min}(T,Q))$, Eq. (12) may lie below
$F_{min}(T,Q=0)$ so that a spontaneous distortion takes place. Before
proceeding in an analytical fashion to this simple task, inspection of Eq.
(12) immediately reveals that the search for the minimum of $F_{min}(T,Q))$
makes sense only when the following inequality holds:

$\rho k>\frac{a^{2}c^{2}}{2b}$ (13)

because in the opposite case both the quadratic and the linear term in the
r.h.s. of Eq. (11) are negative and grow in size without limit with growing
$Q$. Violation of the inequality (13) would signal a lattice instability
triggered by the very strong increase of the critical temperature with the
lattice distortion. Manifestly this does not happen in YBCO, and this is a
first constraint on numerical values of the parameters $a,b,c,\rho$ and $k$,
that we shall discuss in the next Section. As a matter of fact the lattice
stability implies that $c^{2}$ is bound to stay below the value $c_{max}^{2}$:

$c^{2}<c_{max}^{2}\equiv\frac{2b\rho k}{a^{2}}$ (14)

In terms of $c_{max}^{2}$, it is easy to express the softening of the bare
frequency $\omega$ as a function of $c$ in the form:

$\omega_{soft}^{2}=\omega^{2}(1-\frac{c^{2}}{c_{max}^{2}})$ (15)

In the case that $c$ is below this limit, corresponding to a softening of the
phonon mode (reduction of the spring constant), it makes sense to minimize Eq.
(12) as a function of $Q$, a simple exercise that yields the equation:

$(\rho k-\frac{a^{2}c^{2}}{2b})Q_{min}-\frac{a^{2}c(T_{c}-T)}{2b}=0,$ (16)

and therefore

$Q_{min}(T)=\frac{a^{2}c(T_{c}-T)}{2b\rho k-a^{2}c^{2}}.$ (17)

This can also be cast in an appealing form in terms of $c_{max}$:

$\frac{Q_{min}}{T_{c}-T}=\frac{c}{c_{max}^{2}-c^{2}}$ (18)

The corresponding free-energy value is, after some simple algebra:

$\displaystyle F_{min}(T,Q_{min})$ $\displaystyle=$ $\displaystyle
F_{min}(T,0)-\frac{a^{4}c^{2}(T_{c}-T)^{2}}{4b(2b\rho k-a^{2}c^{2})}$ (19)
$\displaystyle=$
$\displaystyle-\frac{a^{2}(T-T_{c})^{2}}{4b}[1+\frac{c^{2}}{c_{max}^{2}-c^{2}}]$

It is therefore easy to see that, provided Eq. (13) holds, two consequences of
the $Q$ dependence of the effective critical temperature arise:

1\. A softening of the phonon frequency (reduction of the ”spring constant” in
Eq. (11)), taking place below $T_{c}$ in agreement with Eq. (15)

2\. A spontaneous distortion from $Q=0$ to $Q_{min}(T)$ given by Eq. (18),
linearly increasing in size as the temperature is lowered.

Questions to address are whether the simple-minded treatment adopted here can
lead to credible predictions, and whether the size of the predicted effects is
possibly leading to observable effects. In order to explore the latter, one
needs to establish numerical values of the model parameters in a realistic
case.

##  4 Numerical estimates for YBCO

According to Ref. [3] the main effect of the THz pumping pulse is to generate
a non-equilibrium elongation of the intra-bilayer distance, via non-linear
phonon couplings. This elongation corresponds (ref. [3], Extended Data Figure
3) to the modes labelled $A_{g}15$ and $A_{g}21$, with wave numbers
$94.52cm^{-1}$ and $125.19cm^{-1}$. This assignment is not unambiguous, as
more recent work by Fechner and Spaldin [17], also based on DFT calculations,
suggests that the relevant $A_{g}$ mode is a different one with a
substantially higher frequency, of order $470cm^{-1}$. Nonetheless, just to
get an order of magnitude estimate, let us take Mankowsky et al.’s estimate
and further simplify this to a single mode; to evaluate the parameter
$k=m^{\ast}\omega^{2}$, let us assume a wavenumber of $\simeq 100cm^{-1}$,
corresponding to $\omega\simeq 2\times 10^{13}s^{-1}$. The effective mass
corresponding to the relative coordinate of two $Cu$ atoms is
$(1/2)m_{Cu}\simeq(1/2)63.5amu$, where $1amu=1.66\times 10^{-24}g$ and $63.5$
is the isotope-averaged mass of a $Cu$ atom. One thus comes to

$k\simeq 2.1\times 10^{4}g/s^{2}.$ (20)

(Note that this purely indicative estimate of $m^{\ast}\omega^{2}$ is
unchanged for a different, O-related phonon mode, if the effective mass is
reduced by a factor 4 (from $Cu$ to $O$) and the frequency is increased by a
factor of 2).

The density of Cu pairs facing each other across a bilayer in the ortho II
structure of YBCO is one pair in a volume corresponding to $3.827\times
3.893\times 11.699\AA^{3}$ [18], that establishes another parameter:

$\rho\simeq 5.73\times 10^{21}cm^{-3}.$ (21)

In order to obtain the value of the parameter $a^{2}/2b$, which determines
(together with the linearization coefficient $c$) the phonon softening, Eq.
(15), and the lattice distortion, Eq. (18), we use a relation involving the
specific heat jump per unit volume at $T_{c}$ (see Eq. (45.8) in ref.[11]):

$(1/V)(C_{s}-C_{n})=\frac{a^{2}T_{c}}{2b}$ (22)

Experimental determinations of the specific heat jump at $T_{c}$ are found in
the literature. For $YBa_{2}Cu_{3}O_{6.5}$ Loram et al. [19] obtain for
$\Delta\gamma=\gamma_{s}-\gamma_{n}$, where $\gamma=C/T$:

$\Delta\gamma(T_{c})\simeq 0.5\frac{mJ}{gram-atomK^{2}}$ (23)

whereas Wühl et al. [20] measure

$\Delta\gamma(T_{c})\simeq 10\frac{mJ}{moleK^{2}}.$ (24)

To compare these two results it is useful to note that $1gram-atom\equiv
1/(12+x)moles$ in $YBa_{2}Cu_{3}O_{6+x}$ [19], so that the value in Eq. (23)
actually corresponds to:

$\Delta\gamma(T_{c})\simeq 6.2\frac{mJ}{moleK^{2}}$ (25)

So an average gives an estimate of $8mJ/(moleK^{2})$. The critical temperature
of $YBa_{2}Cu_{3}O_{6.5}$ is about $50K$, and from Wikipedia [21] we learn
that the density of the material is $6.3g/cm^{3}$ and the molar volume is
$666.2g/mole$, so that we can find the required expression:

$\frac{C_{s}-C_{n}}{VT_{c}}=\frac{a^{2}}{2b}=\frac{8.5\times
10^{-5}J}{cm^{3}K^{2}}$ (26)

Now all the ingredients are there to determine the numerical value of
$c_{max}$. One obtains:

$c_{max}^{2}=\frac{2b\rho k}{a^{2}}=\simeq 1.4\times 10^{23}K^{2}/cm^{2}$ (27)

or $c_{max}\simeq 3.8\times 10^{11}K/cm$. With the help of Eq. (15) and (18),
we can now plot the lattice distortion and the softening of the frequency as a
function of the linearization parameter $c$ (Figures 1 and 2).

##  5 Preliminary comparisons with experiment

The question whether these are observable effects is difficult to answer in
the absence of a reasonable way to estimate the parameter $c$. If we take, on
purely dimensional grounds, the ratio of the inferred increase of the critical
temperature ($250K$) and the measured elongation of the intralayer distance
under strong $THz$ excitation, $2pm$ [3], we obtain a value ( $\simeq
120K/pm$) exceeding $c_{max}$ by a factor of 3. If on a mere guess, we take a
value of one half of $c_{max}$, this would result in a reduction of the phonon
frequency by about $10\%$, which should be detectable in Raman or in other
experiments, especially because of its sudden onset as the temperature is
lowered below $T_{c}$. The lattice distortion, which grows linearly with
$T_{c}-T$, on the other hand, would be limited to $0.2pm$ at $10K$ below
$T_{c}$, which is maybe below the detection limit (I do not feel comfortable
about lowering the temperature further, as this would drive us outside of the
region where Ginzburg-Landau arguments can possibly apply).

Coming to the Raman experiments on YBaCuO in the literature, a thorough study
is reported by Limonov et al. in Ref. [22]. Their attention is focussed on a
mode at $\simeq 340cm^{-1}$, that among all Raman active $A_{1g}$ modes (in
the orthorombic $D_{2h}$ classification) stands out for its larger distinct
temperature dependence below $T_{c}$. In fact, a softening of over 2% is shown
for overdoped ($T_{c}=86K$) and optimally doped ($T_{c}=93K$)) samples, but
much smaller for underdoped ($T_{c}=80K$) ones. Their analysis is based on the
theory of the phonon self-energies in a strong coupling superconductor as put
forward by Zeyher and Zwicknagl [23], modified for the anisotropic d-wave case
by Nicol et al. [24]. All other modes have much smaller frequency shifts
(positive or negative) at and below $T_{c}$. According to the letter of the
analysis of Zeyher and Zwicknagl, the mode is much more affected because in
the tetragonal $D_{2h}$ point group classification this mode would be an
”allowed” $B_{1g}$ mode, while the other considered modes are ”forbidden”
$A_{1g}$ (forbidden in the sense that selection rules would impose vanishing
values of the matrix elements for the electron-phonon interactions affecting
the phonon self-energy). This takes the point of view that the orthorombic
distortions are very small, and disorder in the samples essentially washes
their effects out, so that tetragonal concepts still apply. A further point to
stress is that while the arguments of the present paper predict a sudden drop
of the frequency at $T_{c}$, the observed temperature behaviour is rather
steep but not really sudden. Finally, the mode in question is also not
directly identifiable with the one singled out in Ref [17].

So the situation for YBaCuO is not encouraging, as far as comparison of the
present paper with experiments is concerned. The mode identified by Mankowsky
et al. in Ref. [3] is not sensitive to temperature, and the one displaying
much larger sensitivity may be interpretable by a different mechanism, i.e. in
terms of self-energy effects. This could, among other possibilities,
correspond to the fact that the parameter $c$ describing the electron lattice
coupling in the small $Q$ regime happens to be very small.

Interestingly, experiments on cuprates with more than two layers, in
particular $HgBa_{2}Ca_{3}Cu_{4}O_{10+x}$ [25], a compound in which light-
enhanced superconductivity was, as far as we know, never looked for, are more
in agreement with the predictions of the present analysis. For two modes of
$A_{1g}$ symmetry in this tetragonal structure, with frequencies near $250$
and $390cm^{-1}$ respectively, very large softenings (respectively 6% and 18%)
are observed to occur rather abruptly in a $10-15K$ interval below
$T_{c}\simeq 120K$. The interpretation of these experiments is still
controversial. According to Munzar and Cardona [26], the large phonon
anomalies are related to the appearance, below $T_{c}$, in three- and four
layers cuprates, of a Raman-active transverse Josephson plasma resonance; the
charge fluctuations in the cuprate layers induced by the plasmon interact
strongly with the charge distortions induced by the phonons, leading to a
strong level repulsion and transfer of oscillator strength, with a
corresponding enhancement of the Raman intensity for the softened phonon.
Indeed, similar phonon anomalies are observed not only in the four-layer
$HgBa_{2}Ca_{3}Cu_{4}O_{10+x}$ but also in the three-layer
$Bi_{2}Si_{2}Ca_{2}Cu_{3}O_{10+x}$ system [27] with an $8\%$ softening of a
Raman active phonon at $390cm^{-1}$. However Limonov et al. observed that the
intensity enhancement disappears when the incoming photon energy is increased
from $\simeq 1.9eV$ to a $2.2$ to $2.5eV$ range, suggesting a resonance effect
as the explanation. Following a theoretical suggestion by Venturini et al.
[28] that a Raman active $A_{1g}$ mode arises from collective spin
excitations, Gallais et al. [29] compare Raman spectra for the single layer
$HgBa_{2}CuO_{4+x}$ and the three-layer $HgBa_{2}Ca_{2}Cu_{3}O_{8+x}$, and
conclude that resonance behaviour and scaling with $T_{c}$ of the $A_{1g}$
modes are rather independent of the number of layers, and therefore do not
support the Josephson plasmon explanation.

As far as $Bi_{2}Si_{2}Ca_{2}Cu_{3}O_{10+x}$ is concerned, however, Hu et al.
[30] report the lack of any signatures of light-enhanced critical temperature
in this system. Therefore, whatever the reason for the phonon anomaly reported
by Limonov et al., it is not related to the mechanism discussed here.

##  6 Conclusions

In this paper an attempt was made to explore possible experimentally
verifiable consequences of the non-linear phononics interpretation of the
phenomenon of light induced superconductivity. On the basis of a simple
Ginzburg-Landau free-energy description, an abrupt softening below $T_{c}$ of
one (or more) Raman active phonon(s) was predicted. Inserting reasonable
values for YBCO, it was concluded that the effect should be observable in this
material. Comparison with existing experiments cannot be claimed to support
this prediction, at least for the phonons that are identified as directly
related to the $T_{c}$ enhancement.

Observed phonon softenings in Bi (2223) and Hg (1234) cuprates with $3$ or $4$
$Cu-O$ layers are on the other hand quite similar to those predicted here.
However, a recent search for light induced superconductivity in Bi (2223) gave
a negative result [30], while no search was carried out in Hg(1234).

On the basis of the presently existing data, the nonlinear phononics
explanation cannot be directly and unambiguously supported by the arguments
presented here. It may be of interest to explore the observability of a light
induced enhancement of $T_{c}$ in Hg (1234) and in general to extend the
search for the specific phonon modes driving this spectacular phenomenon.

##  7 Acknowledgments

I am greatly indebted to Andrea Cavalleri for his continuing interest in this
work and for many valuable discussions. I am also grateful to Michael Fechner,
Michael Först, Andrew J. Millis, Angel Rubio and Michael Sentef for useful
discussions.

Figure 1: Lattice distortion in pm divided by the temperature below $T_{c}$ in
K, as a function of the linearization parameter c, in $K/pm$. The vertical
dashed line marks the $c_{max}$ value, $38K/pm$ Figure 2: Relative softening
of the phonon frequency as a function of c

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