An inverse semigroup is a semigroup $S$ (a set with an associative binary operation) such that for every element $s\in S$, there exists a unique “inverse” $s^*\in S$ such that $s s^* s = s$ and $s^* s s^* = s^*$. It is evident from this that $s^{\ast\ast} = s$.
Needless to say, a group is an inverse semigroup. More to the point however:
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
Groups describe global symmetries, while inverse semigroups describe local symmetries.
Other examples include:
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 $\Gamma(X)$).
If $L$ is a meet-semilattice, then $L$ is an inverse semigroup under the meet operation.
Lots to say here: the meet-semilattice of idempotents, the connection with ordered groupoid?s, various representation theorems.
For any $x$ in an inverse semigroup, $x x^\ast$ and $x^\ast x$ are idempotent. If $e$ is idempotent, then $e^\ast = e$.
The proof is trivial.
In an inverse semigroup, the product of any two idempotents $e, f$ is idempotent, and any two idempotents commute.
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. We then have
since $e, f, e f, f e$ are all idempotent, and so $f e = (e f)^\ast = e f$, completing the 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.
For any two elements $x, y$ in an inverse semigroup, $(x y)^\ast = y^\ast x^\ast$.
Since the idempotents $x^\ast x, y y^\ast$ commute, we have
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.
For elements $x, y$ in an inverse semigroup, the following are equivalent:
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
which gives $3. \Rightarrow 2.$
A partial order $\leq$ is defined on an inverse semigroup by saying $x \leq y$ if any of the four conditions of Proposition 2 is satisfied. When restricted to idempotents, this order coincides with the meet-semilattice order.
If $a \leq b$ and $x \leq y$ in an inverse semigroup, then $a x \leq b y$ and $x^\ast \leq y^\ast$.
We observe from Proposition 2 that for any elements $c, x$ we have $c x^\ast x = x x^\ast c$. From the hypotheses $a = a a^\ast b$ and $x = y x^\ast x$ we have $a x = a a^\ast b y x^\ast x = (x x^\ast) a a^\ast b y$ by our observation. But $e = (x x^\ast)(a a^\ast)$ is idempotent by Lemma 1. This gives $a x \leq b y$. If $x = e y$ for an idempotent $e$, then $x^\ast = (e y)^\ast = y^\ast e^\ast = y^\ast e$; this gives $x^\ast \leq y^\ast$,
cohomology of inverse semi-groups?