Timeline for an inverse semigroup (and perhaps a $C^*\!$-algebra) associated with a directed graph
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2018 at 20:40 | comment | added | Benjamin Steinberg | @DavidHillman, I still owe you a longer response but have been busy this week. Sorry. | |
| Apr 11, 2018 at 19:54 | vote | accept | David Hillman | ||
| Mar 26, 2018 at 21:03 | comment | added | Benjamin Steinberg | You are essentially looking at Kellendonks tiling semigroup associated to the two-sided shift of finite type and then going to the tight algebra I think. | |
| Mar 26, 2018 at 21:02 | comment | added | Benjamin Steinberg | The usual graph C*-algebras are about 1-sided shifts. This may be the difference | |
| Mar 26, 2018 at 20:37 | comment | added | Benjamin Steinberg | I am not sure. I don't think the 0 relation would be automatic for sinks from the tight algebra. You might have to impose that if v is a sink then the empty path is 0 in the inverse semigroup as part of the relations for the inverse semigroup. | |
| Mar 26, 2018 at 20:15 | comment | added | David Hillman | I've thought for other reasons that there is some sort of linear independence condition relevant to this, but haven't quite figured out what that is. But I see now that this applies to the source sink question. What I want to happen there is: any edge that hits a source or sink becomes zero. In (3) what we get is something like $aa^*\!+bb^*\!=0$, and I'd want that to imply that $a=b=0$. Doesn't something like this hold in $C^*\!$-algebras? | |
| Mar 26, 2018 at 19:57 | comment | added | Benjamin Steinberg | McAlister monoid consists of those Munn trees which are a straight-line. | |
| Mar 26, 2018 at 19:56 | comment | added | Benjamin Steinberg | 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. | |
| Mar 26, 2018 at 19:55 | comment | added | Benjamin Steinberg | 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. | |
| Mar 26, 2018 at 18:26 | comment | added | David Hillman | 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). | |
| Mar 26, 2018 at 18:25 | comment | added | David Hillman | 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. | |
| Mar 26, 2018 at 17:17 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Mar 26, 2018 at 17:08 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Mar 26, 2018 at 16:59 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |