an inverse semigroup (and perhaps a $C^*\!$-algebra) associated with a directed graph The following inverse semigroup associated to a directed graph came up in my research. I've read that from an inverse semigroup one may derive a $C^*\!$-algebra whose generators are partial isometries (but I don't yet understand how to do it). I am aware of the existence of the well-known $C^*\!$-algebras derived from directed graphs, called graph $C^*\!$-algebras...but these seem to be different. My questions are:


*

*does anyone recognize this (as an inverse semigroup or as a $C^*\!$-algebra)?

*any hints as to how to derive the $C^*\!$-algebra from the inverse semigroup? (for instance, what is the norm?)


Here is the definition. I use $*$ to denote the inverse. The inverse semigroup has 0, and also a partially defined sum: if $x$ and $y$ are orthogonal ($xy^*\!=x^*\!y=0$) then we define $x+y$. Multiplication uses the distributive law, and $*$ distributes over sums. The generators are the edges of the graph. Suppose that $a_m\colon w_m\to v$ and $b_n\colon v\to x_n$ are the incoming and outgoing edges at vertex $v$. Then we have these relations:


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*if $i\ne j$ then $a_i a_j^*=0$

*if $i\ne j$ then $b_i^* b_j=0$

*$\sum_{i=1}^m a_i^* a_i=\sum_{j=1}^n b_j b_j^*$


That's all the relations if you think of this as an inverse category rather than an inverse semigroup. For each edge $e\colon v\to w$ add an edge $e^*\colon w\to v$ to the directed graph. The allowed multiplications are those that follow directed paths. If you want an inverse semigroup, make all the nonallowed multiplications equal to zero.
 A: Your inverse monoid without the partial sum relations seems like a variation of McAlister's monoid here but for paths in a graph instead of words over a set.
The main difference is that the McAlister monoid is the case the graph is a bouquet of loops at a vertex and satisfies just 1,2.  I suspect if you take the tight algebra in the sense of Exel of the McAlister monoid it would impose your relations (3).  One should check that though.
If you forget about property (3), which I think will come from going to the tight algebra, non-zero elements of you inverse semigroup can be represented by directed paths in your graph together with a distinguished in vertex of the path and out vertex of the path (they don't have to be the first or last vertex and they can be the same).  You can multiply if you can line up the out vertex of the first path with the in vertex of the second and take the union and get a valid birooted path.  
I believe that the empty path at a vertex gets identified with your two sums in the with tight algebra although maybe if there are sources or sinks one has to be a little careful. 
