Timeline for General linear inverse monoid
Current License: CC BY-SA 4.0
14 events
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| Apr 30, 2020 at 14:06 | comment | added | YCor | Viewing it as groupoid looks a bit clumsy: it is the disjoint union over all cardinals $\alpha\le\dim(V)$ of the groupoid of isomorphisms between $\alpha$-dimensional subspaces of $V$. In particular this groupoid has far more many automorphisms than $\mathrm{GLI}(V)$. A more possibly more interesting forgetful operation would be the "semigroupoid", aka small category (but I don't want to think of it as category), allowing composition $gf$ exactly when $codom(f)\subset dom(g)$ (rather than asking for equality). At least it keeps track of embeddings of small subspaces into larger ones. | |
| Apr 30, 2020 at 13:51 | comment | added | YCor | @RyanBudney You don't have to remove zero. $M=\mathrm{GLI}(V)$ is canonically the set of arrows of some groupoid, whose set of objects is the set of subspaces of $V$, and whose set of arrows are the isomorphisms between those subspaces, defining $gf$ only if $codom(f)=dom(g)$. So viewing it as a groupoid is just losing some structure (as one restricts the composition): this groupoid is purely defined from the monoid structure: $codom(f)=dom(g)$ indeed means that for every idempotent $e$, we have $(eg=g)\Leftrightarrow (fe=f)$. Removing zero just artificially removes an isolated point. | |
| Apr 30, 2020 at 13:34 | history | edited | YCor | CC BY-SA 4.0 |
added tags, minor formatting
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| Oct 1, 2010 at 21:25 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Oct 1, 2010 at 21:25 | history | bounty ended | CommunityBot | ||
| Oct 1, 2010 at 19:21 | comment | added | user6976 | @Ryan: If you take an inverse monoid, and remove the 0, you get a groupoid. So the difference is cosmetic. | |
| Oct 1, 2010 at 19:14 | comment | added | Ryan Budney | I'm a little confused as to why you call it a monoid and then a groupoid. Do some people have more flexible notions of monoids -- that the binary operation need not be defined on the entire product? | |
| Oct 1, 2010 at 18:34 | answer | added | Ryan Budney | timeline score: 1 | |
| Oct 1, 2010 at 18:03 | history | edited | user6976 | CC BY-SA 2.5 |
added 673 characters in body
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| Oct 1, 2010 at 13:31 | comment | added | user6976 | I want to know standard information about it (say, what is the co-homology ring, singularities, etc.) If somebody considered such varieties before, I would like to see a reference. From the topological point of view, I would like to see if there exists a boundary similar to the boundary of Lie groups. | |
| Oct 1, 2010 at 11:32 | comment | added | Vivek Shende | I think you answered your own question -- it's a disjoint union of principal bundles over products of Grassmannians. What more do you want? | |
| Oct 1, 2010 at 8:35 | history | bounty started | CommunityBot | ||
| Sep 25, 2010 at 22:17 | answer | added | Carl Futia | timeline score: 0 | |
| Sep 25, 2010 at 22:08 | history | asked | user6976 | CC BY-SA 2.5 |