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\newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{inverse semigroup} \hypertarget{contents}{}\section*{{Contents}}\label{contents} \noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak \noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak \noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak \noindent\hyperlink{idempotents_form_a_subsemigroup}{Idempotents form a subsemigroup}\dotfill \pageref*{idempotents_form_a_subsemigroup} \linebreak \noindent\hyperlink{_is_an_antiinvolution}{$\ast$ is an anti-involution}\dotfill \pageref*{_is_an_antiinvolution} \linebreak \noindent\hyperlink{order_structure}{Order structure}\dotfill \pageref*{order_structure} \linebreak \noindent\hyperlink{connection_with_ordered_groupoids}{Connection with ordered groupoids}\dotfill \pageref*{connection_with_ordered_groupoids} \linebreak \noindent\hyperlink{warning_on_the_definition}{Warning on the definition}\dotfill \pageref*{warning_on_the_definition} \linebreak \noindent\hyperlink{cayleys_theorem_for_inverse_semigroups}{``Cayley's Theorem'' for inverse semigroups}\dotfill \pageref*{cayleys_theorem_for_inverse_semigroups} \linebreak \noindent\hyperlink{links}{Links}\dotfill \pageref*{links} \linebreak \noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak \hypertarget{definition}{}\subsection*{{Definition}}\label{definition} An \textbf{inverse semigroup} is a [[semigroup]] $S$ (a set with an associative binary operation) such that for every element $s\in S$, there exists a \emph{unique} ``inverse'' $s^*\in S$ such that $s s^* s = s$ and $s^* s s^* = s^*$. It is evident from this that $s^{\ast\ast} = s$. \hypertarget{examples}{}\subsection*{{Examples}}\label{examples} Needless to say, a [[group]] is an inverse semigroup. More to the point however: \begin{itemize}% \item The fundamental example is the following: for any set $X$, let $I(X)$ be the set of all partial [[bijections]] on $X$, i.e. bijections between subsets of $X$. The composite of partial bijections is their composite as [[relations]] (or as [[partial functions]]). In fact, any inverse semigroup is isomorphic to a sub-inverse-semigroup of $I(X)$ by the ``Cayley's theorem'' for inverse semigroups (see below). \end{itemize} This inverse semigroup plays a role in the theory similar to that of [[permutation]] groups in the theory of [[groups]]. It is also paradigmatic of the general philosophy that \begin{quote}% Groups describe global symmetries, while inverse semigroups describe local symmetries. \end{quote} Other examples include: \begin{itemize}% \item If $X$ is a topological space, let $\Gamma(X)\subseteq I(X)$ consist of the homeomorphisms between open subsets of $X$. Then $\Gamma(X)$ is a [[pseudogroup of transformations]] on $X$ (a general pseudogroup of transformations is a sub-inverse-semigroup of $\Gamma(X)$). \item If $L$ is a [[meet-semilattice]], then $L$ is an inverse semigroup under the meet operation. \end{itemize} \hypertarget{properties}{}\subsection*{{Properties}}\label{properties} Lots to say here: the meet-[[semilattice]] of [[idempotents]], the connection with [[ordered groupoid]]s, various representation theorems. \hypertarget{idempotents_form_a_subsemigroup}{}\subsubsection*{{Idempotents form a subsemigroup}}\label{idempotents_form_a_subsemigroup} \begin{prop} \label{triv}\hypertarget{triv}{} For any $x$ in an inverse semigroup, $x x^\ast$ and $x^\ast x$ are idempotent. If $e$ is idempotent, then $e^\ast = e$. \end{prop} The proof is trivial. \begin{lemma} \label{comm}\hypertarget{comm}{} In an inverse semigroup, the product of any two idempotents $e, f$ is idempotent, and any two idempotents commute. \end{lemma} \begin{proof} One easily checks that $(e f)^\ast = f(e f)^\ast e$, and that $f(e f)^\ast e$ is an idempotent. So $(e f)^\ast$ is idempotent; as a result, $(e f)^\ast = e f$. Thus $e f$ and similarly $f e$ are idempotent. Next we have \begin{displaymath} e f (f e) e f = e f^2 e^2 f = e f e f = e f, \qquad f e (e f) f e = f e^2 f^2 e = f e f e = f e \end{displaymath} since $e, f, e f, f e$ are all idempotent, and so $f e = (e f)^\ast = e f$, which completes the proof. \end{proof} Thus the idempotents in an inverse semigroup form a subsemigroup which is commutative and idempotent. Such a structure is the same as a meet-semilattice except for the fact that there might not have an empty meet or top element; that is, we define an order $\leq$ on idempotents by $e \leq f$ if and only if $e = e f$, whence multiplication of idempotents becomes the binary meet. \hypertarget{_is_an_antiinvolution}{}\subsubsection*{{$\ast$ is an anti-involution}}\label{_is_an_antiinvolution} \begin{lemma} \label{invol}\hypertarget{invol}{} For any two elements $x, y$ in an inverse semigroup, $(x y)^\ast = y^\ast x^\ast$. \end{lemma} \begin{proof} Since the idempotents $x^\ast x, y y^\ast$ commute, we have \begin{displaymath} x y (y^\ast x^\ast) x y = x (y y^\ast)(x^\ast x) y = x x^\ast x y y^\ast y = x y \end{displaymath} and similarly $y^\ast x^\ast (x y)y^\ast x^\ast = y^\ast y y^\ast x^\ast x x^\ast = y^\ast x^\ast$, which is all we need. \end{proof} \hypertarget{order_structure}{}\subsubsection*{{Order structure}}\label{order_structure} \begin{prop} \label{ord}\hypertarget{ord}{} For elements $x, y$ in an inverse semigroup, the following are equivalent: \begin{enumerate}% \item There exists an idempotent $e$ such that $x = e y$, \item $x = x x^\ast y$, \item There exists an idempotent $f$ such that $x = y f$, \item $x = y x^\ast x$. \end{enumerate} \end{prop} \begin{proof} We show $3. \Rightarrow 2.$; a similar proof shows $1. \Rightarrow 4.$ Clearly then we have $3. \Rightarrow 2. \Rightarrow 1. \Rightarrow 4. \Rightarrow 3.$ Given an idempotent $f$ such that $x = y f$, we have \begin{displaymath} \itexarray{ x & = & y f & \\ & = & y y^\ast y f & \\ & = & y f y^\ast y & since \; idempotents \; commute \\ & = & y f f y^\ast y & \\ & = & y f f^\ast y^\ast y & \text{Proposition 1} \\ & = & y f (y f)^\ast y & \text{Lemma 2} \\ & = & x x^\ast y & } \end{displaymath} which gives $3. \Rightarrow 2.$ \end{proof} A [[preorder]] $\leq$ is defined on an inverse semigroup by saying $x \leq y$ if any of the four conditions of Proposition \ref{ord} is satisfied; transitivity follows by equivalence to 1. and closure of idempotents under multiplication. When restricted to idempotents, this preorder coincides with the meet-semilattice order. \begin{prop} \label{order}\hypertarget{order}{} If $a \leq b$ and $x \leq y$ in an inverse semigroup, then $a x \leq b y$ and $x^\ast \leq y^\ast$. \end{prop} \begin{proof} Writing $a = e b$ for some idempotent $e$, we have $a x = e (b x)$ and so $a x \leq b x$. Similarly $b x \leq b y$, so $a x \leq b y$ by transitivity. This gives $a x \leq b y$. If $x = e y$ for an idempotent $e$, then $x^\ast = (e y)^\ast = y^\ast e^\ast$; this gives $x^\ast \leq y^\ast$, \end{proof} \begin{cor} \label{}\hypertarget{}{} The preorder $\leq$ on an inverse semigroup is a [[partial order]], i.e., if $x \leq y$ and $y \leq x$, then $x = y$. \end{cor} \begin{proof} From $x \leq y$ we derive $x^\ast \leq y^\ast$ and $x x^\ast \leq y y^\ast$, and similarly from $y \leq x$ we derive $y y^\ast \leq x x^\ast$. Thus $x x^\ast = y y^\ast$ since the preorder on idempotents is a meet-semilattice, which is a partial order. Then from $x \leq y$ we derive $x = x x^\ast y = y y^\ast y = y$. \end{proof} Thus an inverse semigroup is naturally regarded as an internal semigroup in the category of posets (equivalently, a finite-product preserving functor from the [[Lawvere theory]] of semigroups to [[Pos]]). \hypertarget{connection_with_ordered_groupoids}{}\subsubsection*{{Connection with ordered groupoids}}\label{connection_with_ordered_groupoids} In this section, an \emph{ordered groupoid} means an [[internal category|internal]] [[groupoid]] in the [[finitely complete category]] of posets [[Pos]]. For any finitely complete category $C$, we observe that the forgetful functor $Gpd(C) \to SemiCat(C)$, taking an internal groupoid in $C$ to the underlying [[semicategory]] (remembering only composition of morphisms, forgetting presence of inverses and identity morphisms), has a right adjoint which takes a semicategory to the [[core]] groupoid of the category of idempotents attached to a semicategory (see \href{/nlab/show/semicategory#idempotents}{here} for details). (This observation is formulated in [[internal logic|finite limit logic]], and thus by a Yoneda lemma argument, its validity reduces to that of the observation in the special case $C = Set$.) In particular, this construction may be applied to an inverse semigroup seen as a semigroup in $Pos$: \begin{defn} \label{}\hypertarget{}{} The groupoid $Ind(S)$ attached to an inverse semigroup $S$ is the core of the category of idempotents $Idem(S)$ of $S$, which as a semigroup in $Pos$ is viewed as a one-object semicategory $B S$ in $Pos$. \end{defn} In more detail: an arrow $e \to e'$ in $Idem(S)$ is a triple $(e, x, e')$ of elements in $S$, where $e, e'$ are idempotent elements and $x$ is an element such that $x e = x = e' x$. Such an arrow is invertible precisely when $e = x^\ast x$ and $e' = x x^\ast$, with inverse $x^\ast$. Thus the core consists of such arrows $x: x^\ast x \to x x^\ast$. A key example to keep in mind is the inverse semigroup of partial bijections $\phi$ on a set, where the arrows of the corresponding groupoid are actual invertible maps $\dom(\phi) \to range(\phi)$ between subsets. In general, the object part of the associated groupoid is not just a poset, but a poset with binary meets. The reason for the notation $Ind(S)$ is that this ordered groupoid is a so-called \emph{inductive groupoid}, defined as follows: \begin{defn} \label{}\hypertarget{}{} An \textbf{inductive groupoid} is an internal groupoid $G$ in $Pos$ with the following additional properties: \begin{itemize}% \item The object part $G_0$ admits binary meets; \item Given $x: e \to f$ in $G_1$ and $e' \leq e$ in $G_0$, there exists a unique $x': e' \to f$ in $G_1$ with $x' \leq x$, called the \emph{restriction} $[x|_\ast e']$. \item Given $x: e \to f$ in $G_1$ and $f \geq f'$ in $G_0$, there exists a unique $x': e \to f'$ in $G_1$ with $x \geq x'$, called the \emph{corestriction} $[f' _\ast| x]$. \end{itemize} \end{defn} In fact conditions 2. and 3. in this definition are equivalent. A \emph{morphism} of inductive groupoids $G \to G'$ is an internal functor from $G$ to $G'$ in $Pos$. For $G$ an inductive groupoid, a tensor product $\otimes: G_1 \times G_1 \to G_1$ may be defined by the rule \begin{displaymath} x \otimes y = [x|_\ast dom(x) \wedge cod(y)] \cdot [dom(x) \wedge cod(y) _\ast|y] \end{displaymath} where $\cdot$ indicates composition in $G$. It may be shown that $(G_1, \otimes)$ is an inverse semigroup $Inv(G)$, and the two notions are equivalent: \begin{theorem} \label{}\hypertarget{}{} \textbf{(Ehresmann-Schein-Nampooripad)} There are canonical isomorphisms $S \to Inv(Ind(S))$ and $G \to Ind(Inv(G))$, providing an equivalent of categories $InvSemiGrp \simeq IndGpd$. \end{theorem} \hypertarget{warning_on_the_definition}{}\subsubsection*{{Warning on the definition}}\label{warning_on_the_definition} With only a subtle change in definition, the result is that one gets only groups: \begin{prop} \label{}\hypertarget{}{} Let $S$ be an [[inhabited set|inhabited]] semigroup with the property that for every $a \in S$ there exists a unique $x \in S$ such that $a x a = a$. Then $S$ is a group. \end{prop} \begin{proof} Since $S$ is inhabited, say by an element $b$, it has an idempotent $e$, for example $b b^\ast$. We will show that $x e = x$ for any $x$; by a similar argument $e x = x$, so that any idempotent $e$ is an identity (\emph{the} identity $1$), whence the idempotents $a a^\ast$ and $a^\ast a$ equal $1$ for any $a$ and $S$ is a group. If $a y a = a$ for unique $y$, then from $(a y a) y a = a y a = a$ it follows $y a y = y$ and hence $S$ is an inverse semigroup. The same observation means it is enough to show $(x e) x^\ast (x e) = x e$, since then also $x^\ast (x e) x^\ast = x^\ast$, which by uniqueness implies $x e = x$. The above results on inverse semigroups apply and we derive \begin{displaymath} \itexarray{ x e x^\ast x e & = & x e e x^\ast x e & \\ & = & x e e^\ast x^\ast x e & Proposition \; 1 \\ & = & x e (x e)^\ast x e & Lemma 2 \\ & = & x e } \end{displaymath} as was to be shown. \end{proof} \hypertarget{cayleys_theorem_for_inverse_semigroups}{}\subsubsection*{{``Cayley's Theorem'' for inverse semigroups}}\label{cayleys_theorem_for_inverse_semigroups} \begin{theorem} \label{}\hypertarget{}{} Every inverse semigroup $S$ can be realized as a semigroup of partial bijections on a set. \end{theorem} \begin{proof} First use the conventional Cayley's theorem to embed $S$ in $\text{End}(S)$ through the map sending $s\in S$ to the map $x\mapsto sx$. We can now consider $S$ a subset of $\text{End}(S)$. For each map $s\in S$, $s^\ast s s^\ast=s^\ast$, so $s|_{s^\ast(S)}:s^\ast(S)\rightarrow s(S)$ has left inverse $s^\ast|_{s(S)}:s(S)\rightarrow s^\ast(S)$. Similarly, as $s s^\ast s=s$, this map is a right inverse, so $s|_{s^\ast(S)}$ is a bijection. Consider the map $S\rightarrow I(S)$ defined by $s\mapsto (s|_{s^\ast(S)}: s^\ast(S)\rightarrow s(S))$. We now check that this map is a homomorphism. For $s,t\in S$, we want to compose the partial bijections $s|_{s^\ast(S)}$ and $t|_{t^\ast(S)}$. To do this, we first compute the overlap between the codomain of the former and the domain of the latter, which is $s(S)\cap t^\ast (S)$. The domain of $t|_{t^\ast(S)}\circ s|_{s^\ast(S)}$ is then $s^\ast (s(S)\cap t^\ast (S))=s^\ast t^\ast(S)=(ts)^\ast (S)$ as required. Finally, we check injectivity. If $s,t\in S$ and $s_{s^\ast(S)}=t_{t^\ast(S)}$, then $s^\ast(S)=t^\ast(S)$ so $ss^\ast=ts^\ast$. But then $s^\ast s s^\ast = s^\ast t s^\ast$, so $s=t$. \end{proof} \hypertarget{links}{}\subsection*{{Links}}\label{links} \begin{itemize}% \item [[homotopy theory of inverse semigroups]] \item [[cohomology of inverse semi-groups]] \end{itemize} \hypertarget{references}{}\subsection*{{References}}\label{references} \begin{itemize}% \item [[Mark V. Lawson]], tutorial lectures on semigroups in Ottawa, notes available \href{http://www.ma.hw.ac.uk/~markl/ottawa.html}{here}. \item [[Mark V. Lawson]], \emph{Constructing ordered groupoids}, Cahiers de Topologie et G\'e{}om\'e{}trie Diff\'e{}rentielle Cat\'e{}goriques, 46 no. 2 (2005), p. 123--138, \href{http://www.numdam.org/item?id=CTGDC_2005__46_2_123_0}{URL stable} \item [[Mark V. Lawson]], Inverse semigroups: the theory of partial symmetries, World Scientific, 1998. \item [[Mark V. Lawson]], G. Kurdyavtseva, \emph{The classifying space of an inverse semigroup}, Period. Math. Hungar., to appear, \href{http://arxiv.org/pdf/1210.4421.pdf}{as preprint: ArXiv 1210.4421} \item Alan L. T. Paterson, \emph{Groupoids, inverse semigroups, and their operator algebras}, Progress in Mathematics \textbf{170}, Birkh\"a{}user 1999, \href{http://www.ams.org/mathscinet-getitem?mr=1724106}{MR 1724106} \item Alcides Buss, Ruy Exel, [[Ralf Meyer]], \emph{Inverse semigroup actions as groupoid actions}, Semigroup Forum \textbf{85} (2012), 227--243, \href{http://arxiv.org/abs/1104.0811}{arxiv/1104.0811} \item Ruy Exel, \emph{Inverse semigroups and combinatorial $C^\ast$-algebras}, Bull. Braz. Math. Soc. (N.S.) 39 (2008), no. 2, 191--313, \href{http://dx.doi.org/10.1007/s00574-008-0080-7}{doi} MR 2419901 \item Alcides Buss, Ruy Exel, \emph{Fell bundles over inverse semigroups and twisted \'e{}tale groupoids}, J. Oper. Theory \textbf{67}, No. 1, 153-205 (2012) \href{http://www.ams.org/mathscinet-getitem?mr=2821242}{MR2821242} \href{http://zbmath.org/?q=an:1249.46053}{Zbl 1249.46053} \href{http://arxiv.org/abs/0903.3388}{arxiv/0903.3388}\href{http://www.mathjournals.org/jot/2012-067-001/0000-000-000-000.html}{journal}; \emph{Twisted actions and regular Fell bundles over inverse semigroups}, \href{http://arxiv.org/abs/1003.0613}{arxiv/1003.0613} \item [[Pedro Resende]], \emph{Lectures on \'e{}tale groupoids, inverse semigroups and quantales}, Lecture Notes for the GAMAP IP Meeting, Antwerp, 4-18 Sep 2006, 115 pp. \href{http://www.math.ist.utl.pt/%7Epmr/poci55958/gncg51gamap-version2.pdf}{pdf}; \emph{\'E{}tale groupoids and their quantales}, Adv. Math. 208 (2007) 147-209; also published electronically: \href{http://dx.doi.org/10.1016/j.aim.2006.02.004}{doi} \href{http://arxiv.org/abs/math/0412478}{math/0412478}; \emph{A note on infinitely distributive inverse semigroups}, Semigroup Forum 73 (2006) 156-158; \href{http://dx.doi.org/10.1007/s00233-005-0547-4}{doi} \href{http://arxiv.org/abs/math.RA/0506454}{math/0506454} \item [[B. Steinberg]], \emph{Strong Morita equivalence of inverse semigroups}, Houston J. Math. 37 (2011) 895-927 \item Darien DeWolf and Dorette Pronk, \emph{The Ehresmann-Schein-Nampooripad Theorem for Inverse Categories}, \href{http://arxiv.org/pdf/1507.08615v1.pdf}{arXiv pdf link}. \end{itemize} [[!redirects inverse semigroups]] \end{document}