tag:blogger.com,1999:blog-11295132.post3505988620847364705..comments2014-04-16T10:57:46.206-07:00Comments on A Neighborhood of Infinity: Automata and the A-D-E classification.Dan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-11295132.post-29474320890280617842009-07-01T20:03:03.347-07:002009-07-01T20:03:03.347-07:00I would use the word indecomposable for what you c...I would use the word indecomposable for what you call "irreducible."<br /><br />I'd call a VA irreducible if it has no proper nonzero sub-VAs, whose meaning is hopefully clear.<br /><br />The two notions are not the same: consider a NFA with one state A and one transition f from A to A.<br /><br />Define an associated VA with A two-dimensional and f(1, 0) = (1, 1); f(0, 1) = (0, 1).<br /><br />This VA is indecomposable (think Jordan normal form), but it has a sub-VA consisting of span{(0, 1)} and the identity function, so it is not irreducible.Jason McCartynoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-34453320538916780222009-06-29T12:17:19.488-07:002009-06-29T12:17:19.488-07:00migmit, in other words, there is always a represen...migmit, in other words, there is always a representation (is that word acceptable here? ;) ) where one of case 1,2 or 3 holds.<br /><br />As always, great post. Very well explained. Finally, I can say I wish you had written at a slightly higher level!Coryhttp://www.blogger.com/profile/07752328226179627747noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-53121549906958554332009-06-29T06:58:37.694-07:002009-06-29T06:58:37.694-07:00migmit,
To continue...if f(1,0)=u and f(0,1)=au, ...migmit,<br /><br />To continue...if f(1,0)=u and f(0,1)=au, then f(a,-1)=0 and we can use (a,-1) and (0,1) as a new basis for A.sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-87203011005335364982009-06-28T23:10:44.783-07:002009-06-28T23:10:44.783-07:00@migmit
>u and v are both non-zero and multiple...@migmit<br />>u and v are both non-zero and multiples of each other<br />As I understand it's the case (2)mirror2imagehttp://mirror2image.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-20948412339306966882009-06-28T13:53:10.243-07:002009-06-28T13:53:10.243-07:00Anonymous,
There's something described as a p...Anonymous,<br /><br />There's something described as a proof here: http://www.math.neu.edu/~king_chris/webster.pdf<br /><br />I'd call that more of a sketch of a proof myself and I may write it up in more detail in the near future.sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-79067839482370484622009-06-28T12:53:55.657-07:002009-06-28T12:53:55.657-07:00Sorry for the word conjecture, i understand after ...Sorry for the word conjecture, i understand after my post was sent. But when will you give the proof ? Or could you give a link to find it ? It's an interesting result.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-28447074210533935212009-06-28T06:57:23.591-07:002009-06-28T06:57:23.591-07:00This comment has been removed by the author.sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-65412930605297432702009-06-28T01:27:34.462-07:002009-06-28T01:27:34.462-07:00Minor correction:
> (1) u and v are distinct, ...Minor correction:<br /><br />> (1) u and v are distinct, non-zero, and not multiples of each other.<br />> (2) u is non-zero but v is zero<br />> (3) both u and v are zero<br /><br />What about the case when u and v are both non-zero and multiples of each other?<br /><br />I'm a bit curious, is there any place for Bn, Cn, F4 and G2 in this picture?migmithttp://migmit.vox.com/noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-13118546748006675302009-06-27T21:33:33.730-07:002009-06-27T21:33:33.730-07:00Anonymous,
Somewhat surprisingly, you don't n...Anonymous,<br /><br />Somewhat surprisingly, you don't need to know the direction of the arrows for the theorem. All that is needed is the underlying graph in order to tell if there's a finite number of irreducibles.<br /><br />BTW It's a theorem (proved by someone called Gabriel I presume), not a conjecture.sigfpehttp://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-26632393930020347582009-06-27T19:09:10.036-07:002009-06-27T19:09:10.036-07:00just wanted to say yours is one of the coolest blo...just wanted to say yours is one of the coolest blogs ever and Thanks!talismannoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-24442692446544393272009-06-27T18:19:23.884-07:002009-06-27T18:19:23.884-07:00You have to explain. You edges are arrows when you...You have to explain. You edges are arrows when you introduce VA and irreducibility, but in your theorem they are just links, not arrows. Is it meaningless for you conjecture ?Anonymousnoreply@blogger.com