Saturday, June 20, 2009

Automata and the A-D-E classification.


The A-D-E classification is a strange ubiquitous pattern that appears in many branches of mathematics. Typically it appears when you try to classify certain types of mathematical construction. If the A-D-E classification applies then you end up with two infinite sequences of cryptically named objects (A1,A2,A3,...) and (D1,D2,D3,...) as well as three leftover objects called E6, E7 and E8. Unfortunately, most of these objects and their classifications are tricky to define using only elementary mathematics. However, there is one type of object that is classified in this way that can be given a relatively straightforward computational description involving a little linear algebra and assuming you know a tiny bit about automata.

But first: why care about the A-D-E classification? Well I tried to say a little bit about how symmetries relate to nature a while back. Certain types of possible symmetry of particle physics can be classified the A-D-E way. The symmetry group corresponding to E8 is the now famous exceptional group E8. I won't be able to get to an explanation of how groups are involved. But at least I'll be able to give a hint about the bigger picture that E8 is part of.

Non-deterministic Finite State Automata

Here's a diagram representing a very simple non-deterministic finite state automaton (NDA):

It can be in one of two states. When in state A it can transition to state B and in state B it can only transition back to state B, but it can do so in two different ways.

Vector Automata

Now I'll introduce a more general kind of automaton: a vector automaton (VA). (I made that term up, it's not meant to correspond with anyone else's terminology.) Every vector automaton is built from an NDA. But each state corresponds to a finite dimensional vector space and each transition corresponds to a linear function mapping from the vector space of the source state to the vector space of the destination state. We could turn the above example into a VA by assigning a 1D vector space VA to A, a 2D vector space to VB and defining linear functions:

f : (x) -> (x,0)
g : (x,y) -> (-y,x)
h : (x,y) -> (y,-x)

A VA is just like an NDA in that it transitions from state to state according to the given transitions. But additionally it keeps track of a vector in the vector space corresponding to the current state. Each time it makes a transition the linear function corresponding to that function is applied to the vector. So in the example above, the NDA might start in state A with a scalar value x (ie. a 1D vector). When it makes its first transition its vector becomes the 2D vector (x,0) and after that each transition rotates the vector through 90 degrees clockwise or anticlockwise.

There's a lot of freedom in defining a VA given its underlying NDA. For each node you can pick any vector space you like of any finite dimension, and for each transition you can pick any linear function you like mapping between the source and target vector spaces.

Let's make this a little more formal. An NDA is a finite set of states combined with a finite set of transitions. Each transition has a source and destination, each of which is a state. That's it. You're allowed any finite number of transitions between states and a transition can have the same state as source and destination.

A VA is an NDA combined with a finite dimensional vector space attached to each state and a linear function for each transition such that the function maps from the vector space of the of the source to the vector space of the target.

Now consider this really simple NDA:

When we build a VA from this NDA, for convenience I'll call the vector spaces corresponding to A and B, A and B. And I'll call the linear function from A to B, f. How many VAs can be made from this VA? Clearly an infinite number. But a lot of them are very similar.

Suppose we assume A and B are 2-dimensional and write their elements as pairs (x,y). Suppose f(1,0)=u and f(0,1)=v and that u is not a multiple of v and both are non-zero. Then we can use u and v as a basis for B. If we write f as f' in this new basis we get f'(1,0)=(1,0) and f'(0,1)=(0,1). So by relabelling the basis of B we have actually revealed that an infinite number of choices for f reduce to the same thing apart from a change of basis in B.

Up to change of basis in A and B we find there are only three possibilities:
(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

We started with an infinite number of possibilities for our particular choice of dimensions and ended up with just 3. We'll say that two VAs are equivalent if we can get one from the other by changing basis like this.

On the other hand, consider this NDA:

Let's choose A, B and C to be 1-dimensional vector spaces and define f, g and h as:

f(x) = x
g(x) = x
h(x) = λx

where λ is any real number. Then h(g(f(x)))=λx. We can choose any λ we like so we have an infinity of possibilities. No amount of basis change is going to change this fact. This is different from the case above because now we're comparing x and h(g(f(x))) which lie in the same vector space. So when our NDAs have loops in them, the space of possible VAs, for most choices of dimension, is infinite.

The sum of two VAs

If we have two VAs corresponding to the same NDA we can combine them together to make a single machine. The state is given by a state in the shared underlying NDA, but we now have a pair of vectors. Each time there is a transition we apply the pair of transforms to transform the two vector simultaneously. But we can encode a pair of vectors as a single vector simply by concatenating together the vector components in some basis. So this machine is just another VA. The dimension of the vector space for each state is simply the sum of the dimensions of the vector spaces in the original VAs. So, given two VAs we can sum them to get a third.

Given a VA for an NDA it may or may not be equivalent to the sum of two simpler VAs. If it isn't, it's said to be irreducible. If it is, then we can ask the same question about the simpler VAs. In this way, every VA is the sum of irreducible VAs.

Main Theorem

Now comes the main result I want to give:

If the graph underlying the NDA is one of the following list then (and only then) there is a finite number of inequivalent irreducible VAs for the NDA. All other VAs for that NDA are simply finite sums of machines from this finite set. Here's the list:

That's it! Weird huh? (Note that diagram Xn has n nodes.)

Strangely, those same diagrams (and a few more) appear (in a quite different way) when classifying the possible symmetries of fundamental particles. The symmetries are given the same names as these diagrams. And in just the same way we get those 'sporadic' symmetries leading up to E8. Those same diagrams also arise in Catastrophe Theory.

I'd like to sketch the proof, at least in one direction, but I seem to have run out of time. Another day perhaps. But note that none of these diagrams have loops for the reasons I gave above, and I've already shown that A2 gives only a finite number of choices for a certain choice of dimension.

Meanwhile, I should give the proper mathematician names for the things above. The diagrams listed above are examples of Dynkin diagrams. A non-deterministic automaton as described above is known as a quiver. A vector automaton is normally known as a representation of a quiver. And the theorem above is half of Gabriel's theorem.


Anonymous Anonymous said...

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 ?

Saturday, 27 June, 2009  
Anonymous talisman said...

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

Saturday, 27 June, 2009  
Blogger sigfpe said...


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.

Saturday, 27 June, 2009  
Anonymous Anonymous said...

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?

Sunday, 28 June, 2009  
Blogger sigfpe said...

This comment has been removed by the author.

Sunday, 28 June, 2009  
Anonymous Anonymous said...

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.

Sunday, 28 June, 2009  
Blogger sigfpe said...


There's something described as a proof here:

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.

Sunday, 28 June, 2009  
Anonymous Anonymous said...

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

Sunday, 28 June, 2009  
Blogger sigfpe said...


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.

Monday, 29 June, 2009  
Blogger Cory said...

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!

Monday, 29 June, 2009  
Anonymous Jason McCarty said...

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.

Wednesday, 01 July, 2009  

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