Comments on A Neighborhood of Infinity: Automata and the A-D-E classification.

I would use the word indecomposable for what you c...

2009-07-01T20:03:03.347-07:00

I would use the word indecomposable for what you call "irreducible."

I'd call a VA irreducible if it has no proper nonzero sub-VAs, whose meaning is hopefully clear.

The two notions are not the same: consider a NFA with one state A and one transition f from A to A.

Define an associated VA with A two-dimensional and f(1, 0) = (1, 1); f(0, 1) = (0, 1).

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.

migmit, in other words, there is always a represen...

2009-06-29T12:17:19.488-07:00

migmit, in other words, there is always a representation (is that word acceptable here? ;) ) where one of case 1,2 or 3 holds.

As always, great post. Very well explained. Finally, I can say I wish you had written at a slightly higher level!

migmit, To continue...if f(1,0)=u and f(0,1)=au, ...

2009-06-29T06:58:37.694-07:00

migmit,

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.

@migmit >u and v are both non-zero and multiple...

2009-06-28T23:10:44.783-07:00

@migmit
>u and v are both non-zero and multiples of each other
As I understand it's the case (2)

Anonymous, There's something described as a p...

2009-06-28T13:53:10.243-07:00

Anonymous,

There's something described as a proof here: http://www.math.neu.edu/~king_chris/webster.pdf

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.

Sorry for the word conjecture, i understand after ...

2009-06-28T12:53:55.657-07:00

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.

2009-06-28T06:57:23.591-07:00

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Minor correction: > (1) u and v are distinct, ...

2009-06-28T01:27:34.462-07:00

Minor correction:

> (1) u and v are distinct, non-zero, and not multiples of each other.
> (2) u is non-zero but v is zero
> (3) both u and v are zero

What about the case when u and v are both non-zero and multiples of each other?

I'm a bit curious, is there any place for Bn, Cn, F4 and G2 in this picture?

Anonymous, Somewhat surprisingly, you don't n...

2009-06-27T21:33:33.730-07:00

Anonymous,

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.

BTW It's a theorem (proved by someone called Gabriel I presume), not a conjecture.

just wanted to say yours is one of the coolest blo...

2009-06-27T19:09:10.036-07:00

just wanted to say yours is one of the coolest blogs ever and Thanks!

You have to explain. You edges are arrows when you...

2009-06-27T18:19:23.884-07:00

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 ?