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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:

  1. does anyone recognize this (as an inverse semigroup or as a $C^*\!$-algebra)?
  2. 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:

  1. if $i\ne j$ then $a_i a_j^*=0$
  2. if $i\ne j$ then $b_i^* b_j=0$
  3. $\sum_{i=1}^m a_i^* a_i=\sum_{j=1}^n b_j b_j^*$

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.

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  • $\begingroup$ How do you want to interpret (3) if there are sources or sinks? $\endgroup$
    – Benjamin Steinberg
    Mar 26, 2018 at 17:09
  • $\begingroup$ Do you allow empty paths? $\endgroup$
    – Benjamin Steinberg
    Mar 26, 2018 at 17:50
  • $\begingroup$ (Note: I just edited my question to add that last paragraph about partially defined multiplication.) The sum of zero items is taken to be zero (in answer to your question about sources and sinks). $\endgroup$
    – David Hillman
    Mar 26, 2018 at 17:52
  • $\begingroup$ I take it you mean length-0 paths? Yes: these correspond to the quantities given in (3): you could call that idempotent $1_v$, the length-0 path beginning and ending at $v$. $\endgroup$
    – David Hillman
    Mar 26, 2018 at 17:57
  • $\begingroup$ A weird thing about this is: the above construction is very useful for understanding flow equivalence maps of SFTs from an algebraic point of view. And yet, the $C^*\!$-algebras that have been used to obtain flow equivalence invariants are, as far as I know, the graph $C^*\!$-algebras, which, as you mentioned, are different! $\endgroup$
    – David Hillman
    Mar 26, 2018 at 18:03

1 Answer 1

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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.

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  • $\begingroup$ Your third paragraph I recognize as describing the Munn tree associated with a free inverse semigroup. Which is relevant here too. Here the Munn tree is a line. So we can describe an element like this: a[bcd]ef where the letters are (unstarred) edges and the [ and ] are the incoming and outgoing distinguished vertices. The inverse of that would be a]bcd[ef. $\endgroup$
    – David Hillman
    Mar 26, 2018 at 18:25
  • $\begingroup$ What the third relation does is: it lets you think of an algebraic element as representing the set of all bi-infinite strings-with-two-pointers having the given restriction. For instance, if the edges that can follow the vertex v are a and b, then 1_v (the set of bi-infinite strings containing the vertex v, with both pointers pointing at v) is exactly the same as []a + []b (the set of bi-infinite strings with both pointers pointing at v and then the edge a following it, disjoint union with the set of bi-infinite strings with both pointers pointing at v and then the edge b following it). $\endgroup$
    – David Hillman
    Mar 26, 2018 at 18:26
  • $\begingroup$ You are using the Munn tree representation for the free inverse category on a graph. Your construction without (3) is just the analogue of McAlister monoids for free inverse categories. The relations (3) are a consequence of performing Exel's tight C*-algebra construction. However, I think equation 3 has some issues if you have sources and sinks because then one of the summands can be empty. I think you need to modify it so that you don't apply the relations at all for isolated vertices and for sources you only put in one of the sums is the empty path and for sinks the other ones. $\endgroup$
    – Benjamin Steinberg
    Mar 26, 2018 at 19:55
  • $\begingroup$ Your explanation for (3) is essentially the proof that the tight algebra gives (3). However that is only true for two finite graphs with no sources or sinks. Otherwise you need to modify slightly. $\endgroup$
    – Benjamin Steinberg
    Mar 26, 2018 at 19:56
  • $\begingroup$ McAlister monoid consists of those Munn trees which are a straight-line. $\endgroup$
    – Benjamin Steinberg
    Mar 26, 2018 at 19:57

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